tag:blogger.com,1999:blog-3826774396449730052018-03-06T09:14:38.612+01:00Physics intuitionsPhysics, math, quantum ... at hands reach.
<br>
<i>"All we do is draw little arrows..."</i> Richard P. Feynman.Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.comBlogger101125tag:blogger.com,1999:blog-382677439644973005.post-18519083603363182592016-11-27T10:29:00.001+01:002016-12-04T11:44:54.544+01:00Law of sines, chords and similar trianglesRecently, I picked up my pencil and notebook and started to draw circles and triangles again. Swiftly, by drawing similar triangles circumscribed by circles, I got interested in their proportionalities, leading me to the <a href="https://en.wikipedia.org/wiki/Law_of_sines" target="_blank">law of sines</a>.<br />For any triangle <i>ABC</i>, where <i>a</i> is the length of the side opposite to angle <i>A</i>, <i>b</i> the length of the side opposite to angle <i>B</i> and <i>c</i> the length of the side opposite to angle <i>C</i>, the law of sines states that:<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-Yinhk27FBWE/WDsRw7bNsyI/AAAAAAAAHtU/XMYgsA6oVl0RY1rRzKOCv6OPtR0ZdWJoQCEw/s1600/LawOfSinesEquation.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://2.bp.blogspot.com/-Yinhk27FBWE/WDsRw7bNsyI/AAAAAAAAHtU/XMYgsA6oVl0RY1rRzKOCv6OPtR0ZdWJoQCEw/s1600/LawOfSinesEquation.png" /></a></div><br />where <i>d</i> is the diameter of the <a href="https://en.wikipedia.org/wiki/Circumscribed_circle" target="_blank">circle circumscribing</a> <i>ABC</i>, as traced in Figure 1.<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-oIZF79ZqTow/WEPQU9C9hQI/AAAAAAAAHvc/53hvRa8JSu4rAzwF4gTiztx3RAj3A5iDwCLcB/s1600/LawOfSinesCircumscribedCircle.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="400" src="https://2.bp.blogspot.com/-oIZF79ZqTow/WEPQU9C9hQI/AAAAAAAAHvc/53hvRa8JSu4rAzwF4gTiztx3RAj3A5iDwCLcB/s400/LawOfSinesCircumscribedCircle.png" width="400" /></a></div></td></tr><tr><td class="tr-caption" style="text-align: center;">Figure 1: Arbitrary triangle <i>ABC</i> circumscribed in its circle with diameter <i>d</i></td></tr></tbody></table><br />My search in literature and web left me somewhat frustrated, for different reasons:<br /><ul><li>one often omits to mention the diameter <i>d</i>, in its statement and even in its proof,</li><li>one rarely develops this very elegant statement to closely related geometrical principles, like the <a href="https://en.wikipedia.org/wiki/Intercept_theorem" target="_blank">intercept theorem</a> or <a href="https://en.wikipedia.org/wiki/Similarity_(geometry)" target="_blank">similarity transformations</a></li><li>the historical background of the law of sines couldn't be checked easily from its original sources.</li></ul><h4>Why we should not leave out the diameter <i>d</i> of the circumscribed circle</h4>The law is often stated in the reciprocal form and leaving out the diameter.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://wikimedia.org/api/rest_v1/media/math/render/svg/39dce98d307d91bf9a659846fe7c059e20582520" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img alt="{\displaystyle {\frac {\sin A}{a}}\,=\,{\frac {\sin B}{b}}\,=\,{\frac {\sin C}{c}}}" border="0" class="mwe-math-fallback-image-inline" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/39dce98d307d91bf9a659846fe7c059e20582520" style="vertical-align: -2.005ex;" /></a></div>while in this case, we should really write:<br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="https://wikimedia.org/api/rest_v1/media/math/render/svg/9bc6593a87d4bd57d505905909e9c7b1ba9a0c29" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img alt="{\displaystyle {\frac {\sin A}{a}}\,=\,{\frac {\sin B}{b}}\,=\,{\frac {\sin C}{c}}\,=\,{\frac {\sin 90^{\circ }}{d}}}" border="0" class="mwe-math-fallback-image-inline" src="https://wikimedia.org/api/rest_v1/media/math/render/svg/9bc6593a87d4bd57d505905909e9c7b1ba9a0c29" style="height: 5.509ex; vertical-align: -2.005ex; width: 36.802ex;" /></a></div><br />We should have in mind that the sine of an angle is the length of the chord of the same angle inscribed in a circle of unit diameter, as <a href="http://commonsensequantum.blogspot.fr/2010/04/teaching-sine-function-with-spaghetti.html" target="_blank">you can teach children with spaghetti</a>.<br /><br />The law of sines can be understood as a statement relative to proportions between:<br />- chords traced in a circle with unit diameter,<br />- and similar chords traced in a circle scaled up by a factor <i>d</i>.<br /><br />For example, we could trace a circle with unit diameter tangent at A inside the original circle, see Figure 2. With the notation used in this figure, the scale factor <i>d</i> can then be read out easily as different ratios: <i>d = a/a<span style="font-size: x-small;"><sub>0</sub></span> = b/b<span style="font-size: x-small;"><sub>0</sub></span> = c/c<span style="font-size: x-small;"><sub>0</sub></span></i>.<br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody><tr><td style="text-align: center;"><a href="https://2.bp.blogspot.com/-QaW6y0FOBoY/WEKeCqReqtI/AAAAAAAAHu4/7Ob91ukflq4alo-yoAsYEHn0D39W4vM6gCLcB/s1600/LawOfSinesWithUnit.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="400" src="https://2.bp.blogspot.com/-QaW6y0FOBoY/WEKeCqReqtI/AAAAAAAAHu4/7Ob91ukflq4alo-yoAsYEHn0D39W4vM6gCLcB/s400/LawOfSinesWithUnit.png" width="400" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Figure 2: Triangle <i>ABC</i> and tangent unit circle in A</td></tr></tbody></table>This is one of the intuitions behind the law of sines. and we could view it as a natural <a href="https://en.wikipedia.org/wiki/Similarity_(geometry)" target="_blank">law of similarity</a>: "Corresponding lines in similar figures are in proportion, and corresponding angles in similar figures have the same measure."<br /><div><br /></div><div><div>Of course, if we want to set up a formal proof, we can deduce it from other laws :</div><div><br /></div><div>1. <u>The inscribed angle</u>:</div><div>When <a href="https://girlsangle.wordpress.com/2014/01/10/do-you-believe-this/" target="_blank">inscribed in the same circle</a>, all angles, subtending arcs of the same measure, are equal.</div><div>In Figure 2, <i>BC</i> being of the same length as <i>DE</i>, the angles <i><bac i=""> and <i><dae i=""> are equal.</dae></i></bac></i></div><div>Therefore sin <i>A</i> = "opposite side over hypotenuse" = <i>a/d</i>.</div><div>Q.E.D.</div><div>This is also visually explained at "<a href="https://betterexplained.com/articles/law-of-sines/" target="_blank">Better explained</a>" or for those who read French "<a href="https://blogdemaths.wordpress.com/2011/06/18/loi-des-sinus-une-demonstration-simple/" target="_blank">Blog de maths"</a>.</div><div><br /></div><div>2. <u>Central and inscribed angle, with Pythagoras</u>:</div><div>This proof involves an additional notion: the central angle.<br />One can find it on other good sites:<br /><a href="http://pballew.blogspot.fr/2009/02/dont-write-law-of-sines-upside-down.html" target="_blank">Pat Ballew's blog</a><br /><a href="https://mathlesstraveled.com/2008/12/06/sine-of-an-inscribed-angle/" target="_blank">Math less travelled</a><br /><br />3. <u>Using the height of the triangle</u><br />This proof comes in different variations, either through expressing the area respective to the different heights (as given on <a href="https://en.wikipedia.org/wiki/Law_of_sines#Proof" target="_blank">wikipedia</a>), either through expressing one height as ratios with two different sides (this is the academic proof, example <a href="https://opencurriculum.org/5486/the-law-of-sines/" target="_blank">here</a>). Not my favorite one, as it doesn't give any insight in the scale factor <i>d</i>. If you don't need to pass exams, but doing math for fun, please forget this one!<br /><h4>Homothetic transformation with scale factor <i>d</i></h4>A <a href="https://en.wikipedia.org/wiki/Homothetic_transformation" target="_blank">homothety</a> is a transformation where a geometric entity is transformed a similar version with a scale factor. In Figure 2, we represented the homothety from a circle of unit diameter towards a circle with diameter <i>d</i>. The inscribed lines, triangles and other polygons undergo the same scaling. And thus, we can complement the law of sines with a list of other ratios that also equal <i>d</i>.<br />In Figure 3, I draw the unit circle at an arbitrary place in space. Then joining similar points. The intersection of the lines is the homothetic center O.<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="https://3.bp.blogspot.com/-7NskG68HfTM/WEPoIu-_K3I/AAAAAAAAHvs/jXvl3bdF-HknUbwAvF4LSkWO5DRhQoztQCLcB/s1600/LawOfSinesHomothety.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="444" src="https://3.bp.blogspot.com/-7NskG68HfTM/WEPoIu-_K3I/AAAAAAAAHvs/jXvl3bdF-HknUbwAvF4LSkWO5DRhQoztQCLcB/s640/LawOfSinesHomothety.png" width="640" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Figure 3: Triangle <i style="font-size: 12.8px;">ABC</i><span style="font-size: 12.8px;"> as a homothety from <i>A<sub>0</sub>B<sub>0</sub>C<sub>0</sub></i> centered in <i>O</i></span></td></tr></tbody></table>Now, any line passing through <i>O</i> and intersecting with the small circle, will also intersect with the large circle at similar points (example <i>D</i> and <i>D<span style="font-size: x-small;"><sub>0</sub></span></i>, <i>C</i> and <i>C<span style="font-size: x-small;"><sub>0</sub></span></i>, <i>B</i> and <i>B<span style="font-size: x-small;"><sub>0</sub></span></i>, etc.) The ratios of various line segments that are created if we trace pairs of parallels from these points will be the same, for example:<br /><i>CD/C</i><i><span style="font-size: x-small;"><sub>0</sub></span></i><i>D<span style="font-size: x-small;"><sub>0</sub></span></i> = <i>BD/</i><i>B<span style="font-size: x-small;"><sub>0</sub></span></i><i>D<span style="font-size: x-small;"><sub>0</sub></span></i> = <i>AD/</i><i>A<span style="font-size: x-small;"><sub>0</sub></span></i><i>D<span style="font-size: x-small;"><sub>0</sub></span></i> = <i>OD/O</i><i>D<span style="font-size: x-small;"><sub>0</sub></span></i> = <i>d</i><br />This is the <a href="https://en.wikipedia.org/wiki/Intercept_theorem" target="_blank">intercept theorem</a>.<br /><br />Curiously, when searching on the web, both laws, the law of sines and the intercept theorem aren't often associated, while they are, in my opinion, stemming from the same basic principle of conservation of proportions.<br /><br />I refer to two interesting posts that are related:<br />At Math is fun: <a href="https://www.mathsisfun.com/geometry/triangles-similar-theorems.html" target="_blank">Theorems about Similar Triangles</a><br />At Girls' Angle: <a href="https://girlsangle.wordpress.com/2014/01/10/do-you-believe-this/" target="_blank">Do you believe this?</a><br /><h4>Historical background</h4>And in the history of geometry?<br /><br />I've looked up sources about Apollonius of Perga, Ptolemy, Regiomontanus, Viete, Coignet (<a href="http://logica.ugent.be/albrecht/math/bosmans/R007.pdf">http://logica.ugent.be/albrecht/math/bosmans/R007.pdf</a>), Simson (<a href="https://archive.org/details/elementsofconics00sims" target="_blank">Elements of the conic sections</a>), Jakob Steiner, but couldn't always find the original sources. I would be interested to have access to them.<br /><br />The same for a paper by Richard Brandon Kershner. "The Law of Sines and Law of Cosines for Polygons." <i>Mathematics Magazine</i>, vol. 44, no. 3, 1971, pp. 150–153. <a href="http://www.jstor.org/stable/2688227">www.jstor.org/stable/2688227</a>.</div></div>Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com0tag:blogger.com,1999:blog-382677439644973005.post-30607517409894627212013-03-30T17:54:00.001+01:002013-03-30T17:55:17.228+01:00It From Bit or Bit From It? FQXi essay<a href="http://fqxi.org/community/forum/topic/1589" target="_blank">2013 FQXi essay contest is announced</a>. Topic "It From Bit or Bit From It?" This topic is somehow connected with the 2011 topic "Is Reality Digital or Analog?", not my favorite one, as I have grown my conviction that the it from physics is what really what underlies reality. Talking about "Bits" then just talking about data, information, sensations perceptions, formulations that originate from the "It"<br /><br />I've been writing FQXi essays three times in row:<br /><ul><li>"<a href="http://www.fqxi.org/community/forum/topic/545" target="_blank">Ordinary Analogues for Quantum Mechanics</a>" in 2009, for the "What's Ultimately Possible in Physics?" essay.</li><li>"<a href="http://www.fqxi.org/community/forum/topic/919" target="_blank">Reality Will Ultimately Be Analog and Digital</a>" in 2011, for the "Is Reality Digital or Analog?" essay.</li><li>"<a href="http://www.fqxi.org/community/forum/topic/1501" target="_blank">Dreaming in Geneva</a>" in 2012, for the "Which of Our Basic Physical Assumptions Are Wrong?" essay.</li></ul><div>Never in the prizes, but enjoying the writing. This time, I think I'll pass my turn as I need to finish my PhD thesis before this summer, which enables me to apply some ideas of my 2009 essay to some unsolved experimental issues in semiconductor nanophysics. A question of focus.</div><div><br /></div><div>For those who'll compete, enjoy and good luck!</div>Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com2tag:blogger.com,1999:blog-382677439644973005.post-4404572601604561612012-09-04T22:27:00.001+02:002012-09-05T09:31:15.546+02:00Dreaming in Geneva - FQXi essayThe theme for this year's FQXi contest topic is "Questioning the Foundations: Which of Our Basic Physical Assumptions are Wrong?". I had some difficulty to start with this topic (I didn't seem to be the only one, see Ajit Jadhav's blog <a href="http://ajitjadhav.wordpress.com/2012/07/21/so-you-think-physicists-got-it-wrong/" target="_blank">here</a> and <a href="http://ajitjadhav.wordpress.com/2012/08/31/i-am-not-participating-in-this-fqxi-competition/" target="_blank">there</a>). I had a lot of things to say about what has gone wrong with physics, which assumptions had to be reconsidered. So, since the opening of the contest, I regularly put some ideas in a draft, being confident that I would be able to arrange them into a coherent thesis for the essay. However by the 20th of August (ten days before closing), I still didn't know how I could write them together into an essay without being suspected of "trotting out my pet theory" (see warning in the <a href="http://fqxi.org/community/essay" target="_blank">Evaluation Criteria</a>).<br /><br />My "pet theory" is simple: the fundamental entity in physics is "THE quantum particle" which you can represent as an arrow (a vector, a ket). From the mechanical interactions between such rod-like particles, you may deduce all of physics, provided that you assume some complementary parameters (such as the velocity at which two particles fly one from another = c, the length of the rod = Bohr diameter of hydrogen). No mass, no force, no charge, etc. Just paths of rotating arrows that interact with each other through contact (collision). This is the way I reason about photons, electrons, quarks, fields, waves, etc. But I can't reasonably write it that way in an essay. I would need to recall a lot of history of science. So I chose to bring up some ideas that have emerged in history of science that we could reconsider, not necessarily in the same way, but gaining insight with hindsight.<br /><br />Also I prefer to avoid abstract mathematics when talking physics. Mathematics is just a language, very convenient though, but really just a language that can hinder us in our intuitive understanding. Instead of math, scientists could as well use words, fantasy, dreams, pictures, poems maybe. It is an art and sometimes it is necessary to change the expression of this art. I hope you'll enjoy my <a href="http://fqxi.org/community/forum/topic/1501" target="_blank">dreaming in Geneva</a>.Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com0tag:blogger.com,1999:blog-382677439644973005.post-30004992276938187992012-08-23T22:57:00.000+02:002013-05-08T11:09:02.583+02:00Auguste Bravais - 201st birthday anniversaryOn August 23, 1811, during a relatively calm period of Napoleon's reign, Aurélie-Adelaïde Thomé, spouse of physician François-Victor Bravais, gave birth to <a href="http://en.wikipedia.org/wiki/Auguste_Bravais" target="_blank">Auguste</a> in <a href="http://en.wikipedia.org/wiki/Annonay" target="_blank">Annonay</a>. Annonay is located just south east of the Pilat massif, in the French mild climate department of the Ardèche.<br /><br />Some 30 years earlier, the people of Annonay witnessed the first public hot-air balloon flights, as it was the hometown of the <a href="http://en.wikipedia.org/wiki/Montgolfier_brothers" target="_blank">Montgolfier brothers</a>. Buth both brothers died before Auguste was given the privilege to nest in Annonay. As last one, Joseph-Michel died almost one year earlier. Auguste surely benefited from the scientific entrepreneurial spirit of that town.<br /><br /><table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; margin-left: 1em; text-align: right;"><tbody><tr><td style="text-align: center;"><a href="http://3.bp.blogspot.com/-hgGzK4XnHYM/UDaYjVFuN0I/AAAAAAAABuM/w3P05gN4DAA/s1600/2012-08-23+17.09.32.jpg" imageanchor="1" style="clear: right; margin-bottom: 1em; margin-left: auto; margin-right: auto;"><img border="0" height="320" src="http://3.bp.blogspot.com/-hgGzK4XnHYM/UDaYjVFuN0I/AAAAAAAABuM/w3P05gN4DAA/s320/2012-08-23+17.09.32.jpg" width="240" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Stanislas College caption</td></tr></tbody></table>He was sent to Paris for his studies, first at the <a href="http://en.wikipedia.org/wiki/Coll%C3%A8ge_Stanislas_de_Paris" target="_blank">college Stanislas</a>. And consequently was admitted to <a href="http://en.wikipedia.org/wiki/%C3%89cole_Polytechnique" target="_blank">Polytechnique</a>.<br /><br />Auguste Bravais is best known for pointing out that there are in total 14 types of crystallographic lattices. His <a href="http://en.wikipedia.org/wiki/Bravais_lattice" target="_blank">ordering and denomination of lattices</a> is still in use today.<br /><br />In his young years, his main interest was in meteorological observations. At age 10, he climbed alone the Pilat mountain hoping to better understand cloud formation. Later in his life, together with two other scientists, he participated in the first scientific mission at the top of the Mont-Blanc, as well as in numerous observations on the <a href="http://en.wikipedia.org/wiki/Faulhorn" target="_blank">Faulhorn</a> with his brother Louis.<br /><br />With Louis, he also shared a passion for botany, which was given to them by their father. Together, they investigated the arrangement of leaves on the stem of plants, which shows <a href="http://goldenratiomyth.weebly.com/phyllotaxis-the-fibonacci-sequence-in-nature.html" target="_blank">Fibonacci series</a> in their spiraling. They came to the conclusion, that the leaves were never really growing vertically of each other. There was a prevalent tendency that two successive leaves were following each other on a spiral at 137.5 degrees (or, which is the same, at 222.5 degrees counter-wise). This result they published in 1835.<br /><br />In 1868, Wilhelm Hofmeister gave an explanation for that angle, now known as <a href="http://www.math.smith.edu/phyllo/About/math.html" target="_blank">Hofmeister's rule</a>: as the plant grows, each new leave originates at the least crowded spot. A very natural law...<br /><br />Sadly, for the last ten yours of his life, he lost his intellectual capacities, being aware that he could not fulfill the redaction of all his scientific work. He was said to start work at 4 o'clock in the morning with a lot of caffeine The lack of sleep surely didn't arrange things. He died March 30, 1863 near Versailles.Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com0tag:blogger.com,1999:blog-382677439644973005.post-59822306850550622842012-08-15T23:39:00.000+02:002012-08-25T20:48:45.689+02:00Louis de Broglie - 120th birthday anniversaryExactly 120 years ago, on August 15th, 1892, <a href="http://en.wikipedia.org/wiki/Louis_de_Broglie" target="_blank">Louis de Broglie</a> was born in Dieppe, a little town on the coast of Normandy. De Broglie is one of my favorite physicists because he has tried to conciliate quantum theory with intuition. He entered the physics stage after the first World War, where he had served as radiographer on the Eiffel tower. That stimulated his interest in electromagnetic radiation questions. At that time, it became clear that electromagnetic radiation could be explained as well by wave mechanics (constructive and destructive interference as evidenced by <a href="http://en.wikipedia.org/wiki/Young%27s_interference_experiment" target="_blank">Thomas Young in 1803</a>), as by a collection of particles (<a href="http://en.wikipedia.org/wiki/Photoelectric_effect" target="_blank">photoelectric effect explained by Albert Einstein in 1905</a>). Louis de Broglie made an important following step: if light had dual wave-particle behavior, matter also should have that duality.<br /><br />De Broglie tried to interpret this duality as phase matching between a particle embedded in a wave, the pilot wave. There should be phase matching between both: "<em>les photons incidents possèdent une fréquence d’oscillation interne égale à celle de l’onde (</em>my translation<em>: the incident photons have an internal oscillation frequency equal to that of the wave)"</em>. He saw photons, as well as electrons, as little clock-watches embedded in their wave. I am sure this intuition will lead to new physics in the future, because this aspect of duality has hardly been investigated, see <a href="http://phys.org/news78650511.html" target="_blank">Couder's bouncing droplets in pilot wave</a>. Personally I am working with this pilot wave idea in order to explain some properties of quantum dots.<br /><br />As Louis de Broglie lived his last years in a little town, Louveciennes, that is close to where I live, I had a walk there today. Maybe I could find some place related to him. Unfortunately, I didn't find the exact location of his residence (please drop a comment if you know). But surely the scenery of the pictures below near to the royal residence of the Manoir du Coeur Volant must have been very familiar to him.<br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-jVUzks1mpSQ/UCwLpx0BAeI/AAAAAAAABeo/WcTf8rA9e_8/s1600/2012-08-15+18.47.24.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="300" src="http://4.bp.blogspot.com/-jVUzks1mpSQ/UCwLpx0BAeI/AAAAAAAABeo/WcTf8rA9e_8/s400/2012-08-15+18.47.24.jpg" width="400" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><a href="http://louveciennestribune.typepad.com/media/2010/08/louveciennes-dans-lhistoire-contemporaine-4-henri-dorl%C3%A9ans-comte-de-paris.html" target="_blank">Manoir du Coeur Volant</a></td></tr></tbody></table><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://2.bp.blogspot.com/-UsspnwOals8/UCwLp-jkSYI/AAAAAAAABeo/z6qN_QXHWbU/s1600/2012-08-15+18.34.04.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="300" src="http://2.bp.blogspot.com/-UsspnwOals8/UCwLp-jkSYI/AAAAAAAABeo/z6qN_QXHWbU/s400/2012-08-15+18.34.04.jpg" width="400" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><a href="http://fr.topic-topos.com/abreuvoir-marly-le-roi" target="_blank">Abreuvoir of Marly-le-Roi</a></td></tr></tbody></table><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://1.bp.blogspot.com/-RlZRUQEF2Qo/UCwLpxW6SyI/AAAAAAAABeo/5uWli-jZ13o/s1600/2012-08-15+18.24.08.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="300" src="http://1.bp.blogspot.com/-RlZRUQEF2Qo/UCwLpxW6SyI/AAAAAAAABeo/5uWli-jZ13o/s400/2012-08-15+18.24.08.jpg" width="400" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><a href="http://fr.wikipedia.org/wiki/Domaine_national_de_Marly-le-Roi" target="_blank">Royal Domain of Marly-le-Roi</a></td></tr></tbody></table><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://4.bp.blogspot.com/-_aEMhvZNOhU/UCwLpzQ6aGI/AAAAAAAABeo/L3ocFXOnRUY/s1600/2012-08-15+18.47.03.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="300" src="http://4.bp.blogspot.com/-_aEMhvZNOhU/UCwLpzQ6aGI/AAAAAAAABeo/L3ocFXOnRUY/s400/2012-08-15+18.47.03.jpg" width="400" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;"><a href="http://fr.topic-topos.com/manoir-du-coeur-volant-louveciennes" target="_blank">Commemoration plaque of the Manoir du Coeur-Volant</a></td></tr></tbody></table><br />Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com0Côte du Cœur Volant, 78160 Marly-le-Roi, France48.864263057433874 2.103538513183593848.859040057433873 2.0936680131835939 48.869486057433875 2.1134090131835936tag:blogger.com,1999:blog-382677439644973005.post-18769449204767176302012-03-15T08:16:00.000+01:002012-03-15T08:16:00.071+01:00Electronic vibrations in ski poles<a href="http://2.bp.blogspot.com/-MI2m8O99Yrs/T2BrdyoSB8I/AAAAAAAABFc/IXA2L8p9nss/s1600/2012-02-29+18.14.27.jpg" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" height="240" src="http://2.bp.blogspot.com/-MI2m8O99Yrs/T2BrdyoSB8I/AAAAAAAABFc/IXA2L8p9nss/s320/2012-02-29+18.14.27.jpg" width="320" /></a>Two weeks ago, I was skiing in the <a href="http://en.wikipedia.org/wiki/Vosges_Mountains">Vosges mountains</a>. The ski resort Lac Blanc is crossed by 400 kV high tension transmission lines which run over the ski trails at a height about 8 - 10 m. While waiting under them, I had the surprise to "feel" electronic vibrations in the ski pole with the tip of my fingers, as if bunches of electrons were running back and forth on the surface of the pole. Experimenting a bit with them, I noticed that the ski pole had to be planted in the ground (or the snow in this case) to set up these vibrations. For ski poles where the tip was isolated with a plastic material, there were no such vibrations. Also, this worked whatever the orientation of the pole, parallel or perpendicular to the transmission lines, which I found quite surprising. Whatever, I tried not to dwell too long under those lines, not sure to which extent these transmission lines affected my neuronal electrons ;-)<br /><br /><br />Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com1tag:blogger.com,1999:blog-382677439644973005.post-70895295503230282312012-02-04T19:16:00.001+01:002012-02-06T10:28:23.291+01:00Spinning dancers around poles<div class="separator" style="clear: both; text-align: left;">Some ideas for a spinning dancers choreography, representing spinning electrons with spin up and down, inspired by Pauli exclusion principle: <a href="http://commonsensequantum.blogspot.com/2010/10/explaining-electron-spin-and-pauli.html">http://commonsensequantum.blogspot.com/2010/10/explaining-electron-spin-and-pauli.html</a></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">All electrons spinning at same speed:</div><div class="separator" style="clear: both; text-align: left;"><a href="http://1.bp.blogspot.com/-lrYY6djGFo8/Ty11QZh9vgI/AAAAAAAABFQ/D6kMG7P3HAw/s1600/DanseDesElectronsCarre.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em; text-align: center;"><img border="0" src="http://1.bp.blogspot.com/-lrYY6djGFo8/Ty11QZh9vgI/AAAAAAAABFQ/D6kMG7P3HAw/s1600/DanseDesElectronsCarre.gif" /></a></div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">With an excited electron, spinning twice the speed of other electrons. At that speed, it doesn't disturb the dance:</div><div class="separator" style="clear: both; text-align: left;"><a href="http://3.bp.blogspot.com/-fLZ2Rr9S_SU/Ty10Sx1huaI/AAAAAAAABFI/-_5xJt5waGU/s1600/DanseDesElectronsCarreExcite.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-fLZ2Rr9S_SU/Ty10Sx1huaI/AAAAAAAABFI/-_5xJt5waGU/s1600/DanseDesElectronsCarreExcite.gif" /></a></div><br />To be continued with perturbing dancers representing laser light.Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com0tag:blogger.com,1999:blog-382677439644973005.post-23103120037298806142011-04-03T23:19:00.000+02:002016-11-19T12:29:34.524+01:00Forwards multiplying, backwards dividing<a href="http://commonsensequantum.blogspot.com/2011/03/morley-triangle-derived-from-tripling.html">My last post on the Morley triangle theorem</a> got encouraging feedback, namely that it showed the living and breathing side of geometry. Although there are a lot of results discovered in previous, sometimes ancient times, the future of geometry is alike the future of life. You can construct its future in many directions, using different languages, without being constrained by impossibilities which show up on some paths. In case of dead ends, it’s up to us to step back, reexamine the fundamentals, and take another path that has not yet been explored.<br /><br />One of those <a href="http://en.wikipedia.org/wiki/Compass_and_straightedge_constructions#Impossible_constructions">geometrical impossibilities</a> which many people know of, is the division of an arbitrary angle by 3, with only a compass and an unmarked ruler. My advice is: don't try it using the geometry you learned at school, you'll be caught in a dead end. Of course, you might try it for some time to get experience with it, experiencing by yourself the hopes and frustrations that generations of mathematical inquirers have felt, but don't expect to break through in this way. In order to bypass the impossibility, you need to step back and try it differently. That's how inquiring minds discovered <a href="http://www.jimloy.com/geometry/trisect.htm#tools">physical tools</a> or <a href="http://en.wikipedia.org/wiki/Angle_trisection#Origami">paper folding</a> manners that allow to trisect an angle.<br /><br />Another way to explore trisection is to reformulate the problem. Dividing an angle is the inverse operation of multiplying an angle. So, if we want to solve problems involving the trisection of an angle, we might first focus on solutions to the problem of tripling an angle. This may sound trivial, so trivial that hardly anyone emphasizes this point. Before learning the operation of division, we should first learn how to multiply. Through multiplication, we advance constructively from a unit towards a <a href="http://en.wikipedia.org/wiki/Product_(mathematics)">product of factors</a>. Once we know how we got that product through multiplication, we can divide backwards the product through <a href="http://en.wikipedia.org/wiki/Factorization">factorization</a>.<br /><br />Multiplying geometrically means generating one length (or surface or volume) from another. If we do this recursively, we get a series of successive powers. For example, with straight lines, using the <a href="http://en.wikipedia.org/wiki/Intercept_theorem">Thales intercept theorem</a>, we can mark successive powers of a number on lines, see Figure 1. If OA'/OA = <i>x</i> and OA is our unit length and OA' = OB, than you can verify that OB = <i>x</i>, OC = <i>x</i><sup>2</sup>, OD = <i>x</i><sup>3</sup>, OE = <i>x</i><sup>4</sup>, etc.<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-KuMuhcJ4jmM/TZikVrYkoLI/AAAAAAAAAcs/EVfJx0z9mYo/s1600/Figure1.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://3.bp.blogspot.com/-KuMuhcJ4jmM/TZikVrYkoLI/AAAAAAAAAcs/EVfJx0z9mYo/s1600/Figure1.jpg" /></a></div>So forwards, we multiply each time by <i>x</i>. If we go backwards in the series we divide each time by <i>x</i> and we needn't stop at our initial unit length, but we may divide indefinitely and recursively by <i>x</i>. So the geometric representation of multiplication gives us insight on how to divide an initial length OA by <i>x</i>.<br /><br />We can do something similar with circular arc lengths. For example, for the recursive doubling of an arc, we can proceed as illustrated in Figure 2.<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-KiMoErPlxDY/TZjH37CQ2PI/AAAAAAAAAc0/lgiQ2S2BSVk/s1600/Figure2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://1.bp.blogspot.com/-KiMoErPlxDY/TZjH37CQ2PI/AAAAAAAAAc0/lgiQ2S2BSVk/s1600/Figure2.jpg" /></a></div><br /><ol><li>Define a unit arc from an origin O on a circle of center P. Draw also a little circle of diameter OP.</li><li>If from the endpoint 1 of the unit arc, we draw a radius towards P, it intersects the little circle at the (red) point illustrated on the figure. If we double the (red) chord from O to that (red) point, it will end up at length 2 on the large circle.</li><li>If from the (red) endpoint 2 of the arc, we draw the (red) radius towards P, it intersects the little circle at the (blue) point illustrated on the figure. If we double the (blue) chord from O to that (blue) point, it will end up at at length 2*2=4 on the large circle.</li><li>Proceeding further (blue, green, violet...) in the same manner will mark successive powers of 2 on the large circle (as well on the small circle).</li></ol><br /><div>If we reverse the direction of this recurrence, we can halve the length of the arc indefinitely. The proof of this recurrence can be given using the isosceles triangles, two sides of which are successive chords.<br /><br /><u>Update 19 November 2016</u>: <i>the conjectured process below is incorrect. This is not the correct way to multiply angles. It only works for doubling and quadrupling angles (may other powers of 2, I didn't check).</i><br /><br />By adapting the geometric configuration of the two circles, we can do the same for the multiplication of an arc by any integer number, forwards. There is no impossibility to multiply an angle by an integer. Reversing the direction of the recurrence, it becomes therefore possible to access to the division of an arc by any integer number, and henceforth any rational number. As food for thought, the beginning of the conjectured process is illustrated in Figure 3.<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-V9Kd3PVuhcc/TZjdtgdnZiI/AAAAAAAAAc4/4wpWa2hNWoM/s1600/Figure3.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://4.bp.blogspot.com/-V9Kd3PVuhcc/TZjdtgdnZiI/AAAAAAAAAc4/4wpWa2hNWoM/s1600/Figure3.jpg" /></a></div><br /><ol><li>Draw a circle of diameter QP.</li><li>A circle of center P and with radius smaller than QP defines the origin of the arc at one of their intersections O.</li><li>Define a unit arc from origin O on the circle of center P.</li><li>If from endpoint 1 of the unit arc, we draw a radius towards P, it intersects the other circle at the (red) point illustrated on the figure. If we prolongate the (red) chord from Q to the (red) point, it will end up at length <i>x</i> on the arc.</li><li>If from (red) endpoint <i>x</i> of the arc, we draw the (red) radius towards P, it intersects the other circle at the (blue) point illustrated on the figure. If we prolongate the (blue) chord from Q to that (blue) point, it will end up at length <i>x</i>*<i>x</i>=<i>x</i><sup>2</sup> on the arc.</li><li>Proceeding further (blue, green...) in the same manner will mark successive powers of <i>x</i> on the circle.</li></ol></div>Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com0tag:blogger.com,1999:blog-382677439644973005.post-74384107361457567802011-03-27T17:08:00.006+02:002011-03-27T22:20:08.612+02:00Morley triangle derived from the tripling of an angle<table cellpadding="0" cellspacing="0" class="tr-caption-container" style="float: right; text-align: right;"><tbody><tr><td style="text-align: center;"><a href="http://upload.wikimedia.org/wikipedia/commons/6/68/Morley_triangle.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" height="150" src="http://upload.wikimedia.org/wikipedia/commons/6/68/Morley_triangle.png" width="200" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Morley equilateral triangle at the intersection of the trisectors<br />(from <a href="http://en.wikipedia.org/wiki/File:Morley_triangle.png">Wikimedia Commons</a>)</td></tr></tbody></table>In a <a href="http://commonsensequantum.blogspot.com/2011/03/playing-with-angles.html">previous post</a>, I mentioned <a href="http://www.cut-the-knot.org/triangle/Morley/index.shtml">Morley's miracle</a>: an equilateral triangle appearing at the points of intersection of the angle trisectors of any triangle.<br /><div class="MsoNormal"><br />Frank Morley was an excellent teacher and chess player (see some more about him at <a href="http://pballew.blogspot.com/2011/03/trisecting-angle.html">Pat'sBlog</a>). If he had lived today, I like to think of him having his own blog communicating about his passion for math and chess, submitting his recreational problems. If he hadn't discovered the equilateral triangle at the intersection of trisectors, it's probable that this theorem would still be unknown. Trisection of angles is a controversial subject for research, especially among academia, where it is associated with suspicion of crankiness (see <a href="http://books.google.com/books?id=tp3kHvbMjqUC">Underwood Dudley's book about trisectors</a>). A pity... because trying to understand how to divide angles is a "natural" question, of which we shouldn't be ashamed, provided that we modestly take into account what has been found by other people.<br /><br />All the proofs of Morley's theorem that I know of start with a <i>result</i>: an equilateral triangle or a triangle already drawn with its trisectors (see the different proofs referenced on the very complete French site <a href="http://www-cabri.imag.fr/abracadabri/GeoPlane/Classiques/Morley/Morley1.htm">Abracadabri</a> or at the end of <a href="http://www.cut-the-knot.org/triangle/Morley/index.shtml">Alexander Bogolmony's Cut the Knot site</a>). In this sense, they are backward proofs, which keeps some mystery about the physical origin of the equilateral triangle. I tried something different, a bit similar in spirit to my <a href="http://commonsensequantum.blogspot.com/2011/03/archimedes-angle-trisection-or-tripling.html">Archimedes tripling circle</a>: How can we multiply an angle by 3, from which would result a Morley triangle embedded in a triangle of any shape? So here follows an alternative forward proof for Morley's theorem.<br /><br /><u>Step 1</u>: I start with a circle in which I inscribe an angle <span class="Apple-style-span" style="color: #333333; font-family: Georgia, serif; font-size: 13px; line-height: 20px;">α</span> inferior to 60°. The lines are intercepting an arc on the circle.<br /><br /><u>Step 2</u>: I duplicate the angle by tracing a little circle centered at one of the endpoints of the arc and with radius the chord of that arc. I repeat the same operation to triplicate the angle (see Figure 1).<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-HuBLR0tTTLM/TY8Dr3-EqZI/AAAAAAAAAcE/0l5tlzIulSs/s1600/TriplicationAngle.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-HuBLR0tTTLM/TY8Dr3-EqZI/AAAAAAAAAcE/0l5tlzIulSs/s1600/TriplicationAngle.gif" /></a></div><u>Step 3</u>: When two circles of equal size and common radius intersect, two <a href="http://mathschallenge.net/library/constructions/polygonal_constructions">equilateral triangles appear</a>, so I draw both of them: Figure 2.<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/-8SBDCQtRo0k/TY8MzhvpSMI/AAAAAAAAAcI/a7OieUuD20g/s1600/Figure2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/-8SBDCQtRo0k/TY8MzhvpSMI/AAAAAAAAAcI/a7OieUuD20g/s1600/Figure2.jpg" /></a></div><u>Step 4</u>: I draw supplementary equilateral triangles at both sides, with the help of the intersections of the circles and the outer lines of the triplicate angle, see Figure 3.<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/-fe5JkDq6GBE/TY8PXzSaVhI/AAAAAAAAAcM/heWwqegnYng/s1600/Figure3.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/-fe5JkDq6GBE/TY8PXzSaVhI/AAAAAAAAAcM/heWwqegnYng/s1600/Figure3.jpg" /></a></div><u>Step 5</u>: The magenta colored equilateral triangle is the Morley triangle of the arbitrary triangle we are looking for. The Morley triangle is a pivotal triangle, so the other vertices of the arbitrary triangle can be found through symmetric construction of the initial inscribing circle centered towards the other sides of the Morley triangle, see the red circles at Figure 4.<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/-TK9Cmb-569k/TY8aqiL5olI/AAAAAAAAAcQ/GQmFYFyjnq8/s1600/Figure4.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/-TK9Cmb-569k/TY8aqiL5olI/AAAAAAAAAcQ/GQmFYFyjnq8/s1600/Figure4.jpg" /></a></div><u>Step 6</u>: I complete the figure with the triplicated angles <span class="Apple-style-span" style="font-family: arial, sans-serif; font-size: x-small; line-height: 15px;">β</span> and <span class="Apple-style-span" style="-webkit-border-horizontal-spacing: 2px; -webkit-border-vertical-spacing: 2px; font-family: serif; font-size: 14px; line-height: 16px;"><b>γ </b></span>in the red inscribing circles. The sides of the searched triangle are the outer lines of the triplicated angles, see Figure 5.<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/-r6OyTE7WzoA/TY8kof0d7bI/AAAAAAAAAcU/qBXUKaOmwTI/s1600/Figure5.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/-r6OyTE7WzoA/TY8kof0d7bI/AAAAAAAAAcU/qBXUKaOmwTI/s1600/Figure5.jpg" /></a></div>With this tripling angle construction, we always obtain an equilateral triangle at the intersection of the trisectors of a triangle. When the initial vertex is moved on the initial inscribing circle, all triangle shapes can be generated for any initial angle <span class="Apple-style-span" style="color: #333333; font-family: Georgia, serif; font-size: 13px; line-height: 20px;">α</span> between 0 and 60°, which proves the Morley theorem for any triangle.</div>Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com0tag:blogger.com,1999:blog-382677439644973005.post-42352674264869448972011-03-14T23:08:00.000+01:002011-03-14T23:08:33.441+01:00Archimedes angle trisection or tripling?Pi-day today, so a nice occasion to write something about circles. π is the ratio between the circles' perimeter (περίμετρος in Greek) and diameter.<br /><br />In the <a href="http://commonsensequantum.blogspot.com/2011/03/playing-with-angles.html">previous post</a>, I explored some angular tricks using circles. The circle is a key geometrical object when dealing with angles. For example, <a href="http://en.wikipedia.org/wiki/Angle_trisection#With_a_marked_ruler">Archimedes' construction to trisect an angle</a> uses a circle and a straight line with a marked unit. But for clarity's sake, I prefer to draw a set of circles which show how a straight line can mark odd multiples of a unit angle α on intersecting circles, see Figure 1. In that way, it is easier to see how Archimedes' trisection circle relates to the multiplication and division of angles by odd integers, see alternative Figure 2.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://lh3.googleusercontent.com/-FRTb_HS5LnU/TX6RK5QB85I/AAAAAAAAAbo/6KeyRqE87Lw/s1600/ArchimedesMultiplication.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://lh3.googleusercontent.com/-FRTb_HS5LnU/TX6RK5QB85I/AAAAAAAAAbo/6KeyRqE87Lw/s1600/ArchimedesMultiplication.jpg" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://lh5.googleusercontent.com/-R4kgS6LciP0/TX6ReBN6v7I/AAAAAAAAAbs/CiXB3SJ8mHI/s1600/ArchimedesTripling.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="https://lh5.googleusercontent.com/-R4kgS6LciP0/TX6ReBN6v7I/AAAAAAAAAbs/CiXB3SJ8mHI/s1600/ArchimedesTripling.jpg" /></a></div><br />This makes me think that Archimedes' angle trisection is in fact the inverse of "Archimedes' angle tripling".Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com3tag:blogger.com,1999:blog-382677439644973005.post-68292886383586831312011-03-09T20:52:00.000+01:002011-03-09T21:13:17.561+01:00Playing with angles<div class="MsoNormal"><span lang="EN-US">A few months ago, I prepared this post but didn't feel ready to publish it. It raised more questions than it answered, so I wanted to have a little more background about it. Hopeless wish! </span>There's so much about angles that I don't fully understand. <span lang="EN-US">With pi-day approaching, I'm also looking for some inspiration about circles and angles, so I thought publishing might get me in the direction...</span></div><div class="MsoNormal"><span lang="EN-US"><br /></span></div><div class="MsoNormal"><span lang="EN-US">I really became aware of the difficulty to divide an angle into three equal angles (a.k.a. angle trisection) some three years ago. It's bloody hard to find general procedures to divide precisely a circular arc into three equal arcs. There are some </span><span lang="EN-US"><a href="http://www.jimloy.com/geometry/trisect.htm">tricks that are well documented on the web</a></span><span lang="EN-US">, but you won’t find a simple geometric procedure using only a compass and an unmarked ruler. It has even been "proven" impossible. We only know of some "cheating" tricks like using a marked ruler (for </span><span lang="EN-US"><a href="http://en.wikipedia.org/wiki/Angle_trisection#With_a_marked_ruler">Archimedes' trisection</a></span><span lang="EN-US">), </span><span lang="EN-US"><a href="https://www.math.lsu.edu/~verrill/origami/trisect/">origami</a></span><span lang="EN-US"> or </span><span lang="EN-US"><a href="http://www.iri.upc.edu/research/webprojects/cuikweb/Trisector/trisector.html">mechanical tools</a></span><span lang="EN-US"> that will allow you to trisect precisely an angle. And there is an equilateral triangle that appears "</span><span lang="EN-US"><a href="http://www.cut-the-knot.org/triangle/Morley/">by miracle</a></span><span lang="EN-US">" when you trisect all angles of a triangle, a result found by </span><span lang="EN-US"><a href="http://en.wikipedia.org/wiki/Frank_Morley">mathematician Frank Morley</a></span><span lang="EN-US"> (because he himself tried to better understand or crack that impossibility?). So the angle trisection really became to intrigue me and I progressively dug into it.</span></div><a href="http://3.bp.blogspot.com/_0Zi-ITihiW4/TJe7bsMLaPI/AAAAAAAAAV0/UDpULFJL3DI/s1600/MeasureOfAngle.jpg" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5519085952758933746" src="http://3.bp.blogspot.com/_0Zi-ITihiW4/TJe7bsMLaPI/AAAAAAAAAV0/UDpULFJL3DI/s1600/MeasureOfAngle.jpg" style="float: right; margin-bottom: 10px; margin-left: 10px; margin-right: 0px; margin-top: 0px;" /></a> <br /><div class="MsoNormal"><span lang="EN-US">One of the questions that has obsessed me is : How do we characterize an “Angle”? I’ve already written </span><span lang="EN-US"><a href="http://commonsensequantum.blogspot.com/2010/04/teaching-sine-function-with-spaghetti.html">a post on that question</a></span><span lang="EN-US">, pointing out the utility of the chord subtending the circular arc of a circle of unit diameter. The measure of this chord is what we call the sine of the subtended angle. It characterizes the angle and it greatly facilitates our way to count angles, for example by </span><span lang="EN-US"><a href="http://lh4.ggpht.com/_0Zi-ITihiW4/S8ItOtOPiZI/AAAAAAAAASE/JQD2ZUrboRg/s800/Figure3.gif">progressing stepwise with chords in a circle</a></span><span lang="EN-US">.<o:p></o:p></span></div><div class="MsoNormal"><span lang="EN-US">Roughly speaking there are two ways to manipulate angles with circles. Either you use central angles, meaning that you fix the vertex and you radiate towards the circle in order to subtend an arc, or you use inscribed angles, meaning that you put the vertex of the angle on the circle. Figure 1 shows the equivalence between both methods. The central angle of a subtended arc in a circle of radius 1 is equal to the same angle when it is inscribed in a circle of diameter 1. This equivalence between central and inscribed angles is another way of stating that the </span><span lang="EN-US"><a href="http://en.wikipedia.org/wiki/Inscribed_angle#Property">measure of a central angle is exactly twice the measure of the inscribed angle</a></span><span lang="EN-US">, the factor 2 expressing the ratio between diameter and radius.</span></div><div class="MsoNormal"><span lang="EN-US">Figure 2 and 3 illustrate some advantages of inscribed angles. You can put the vertex anywhere on the circle: if the subtended circular arc (or its chord) remains constant in length, the angle remains the same. For the case when the vertex is located between the two endpoints of the subtended arc, you have to add the parts at each side of the vertex.<o:p></o:p></span></div><table style="width: auto;"><tbody><tr><td><a href="http://picasaweb.google.com/lh/photo/gVEFGKx-d1gvt5SHb_r1cV5N3PUMRydn1SbBTOwMW3Y?feat=embedwebsite"><img src="http://lh3.ggpht.com/_0Zi-ITihiW4/TJfBEmdtBoI/AAAAAAAAAWI/6dxtwVckHFA/s800/Figure2.gif" /></a></td></tr><tr><td style="font-family: arial,sans-serif; font-size: 11px; text-align: right;"></td></tr></tbody></table><a href="http://picasaweb.google.com/lh/photo/0XzrusRZmZR31c_JE8pqb15N3PUMRydn1SbBTOwMW3Y?feat=embedwebsite"><img src="http://lh5.ggpht.com/_0Zi-ITihiW4/TJfJP3UGFaI/AAAAAAAAAWQ/GSQz0wWDgnM/s1600/Figure3.gif" /></a><br /><div class="MsoNormal"><span lang="EN-US">This flexibility of inscribed angles opens possibilities that remain hidden for central angles. If you combine different circles at intersections with straight lines, you’re simply calculating with angles. Figure 4 illustrates an alternative way to count angles. Starting with an angle α inscribed in a circle of unit diameter, if we draw a unit circle symmetric to the chord of the subtended angle, we can read off an angle (<o:p></o:p></span>α+α) at the other side of the circle. Adding another circle results in an angle <span lang="EN-US">(2<o:p></o:p></span>α+α) at the far side, etc. When the span of the angle exceeds the height of the unit circle, we can use extra circles in order to take over the continuation of the subtended arc (see the angles > 6α on Figure 4. So this simple procedure allows us to read off integer multiples of any angle.</div><a href="http://2.bp.blogspot.com/_0Zi-ITihiW4/TJfSyZ6z8KI/AAAAAAAAAWU/d5KexxW6pTw/s1600/Figure4.gif" onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5519111631758684322" src="http://2.bp.blogspot.com/_0Zi-ITihiW4/TJfSyZ6z8KI/AAAAAAAAAWU/d5KexxW6pTw/s1600/Figure4.gif" /></a><br /><div class="MsoNormal">The interesting thing is that wherever we shift a unit circle in this setting, the opposite subtended arcs differ always by an angle amount α, as shown in Figure 5 (in fact the circle needn't be aligned on the axis). So this allows us to add or subtract any angle to another angle.</div><div class="MsoNormal"><a href="https://picasaweb.google.com/lh/photo/PCSa5Kx8_et4HKWlxJEhBV5N3PUMRydn1SbBTOwMW3Y?feat=embedwebsite"><img height="362" src="https://lh3.googleusercontent.com/_0Zi-ITihiW4/TXfdlkpd0sI/AAAAAAAAAbk/HtQLTB-7vxc/s640/Figure5.gif" width="640" /></a></div>Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com0tag:blogger.com,1999:blog-382677439644973005.post-65552786712286746772011-02-28T22:00:00.000+01:002011-02-28T22:00:23.767+01:00FQXi essay contest "Is reality digital or analog?"The third edition of FQXi essay contest is already a success, because there is a growth of 140% in the number of competing essays. There are <a href="http://www.fqxi.org/community/forum/category/31417">161 essays dealing with the question "Is reality digital or analog?"</a> that are waiting for your votes and comments. I have less time to participate in it as for the previous edition on "<a href="http://commonsensequantum.blogspot.com/2009/10/whats-ultimately-possible-in-physics.html">What's ultimately possible in Physics</a>", but I found it an important question. Important enough to take my pen on vacation and advocate the fact that physical "Reality" must ultimately be accessible to everyone on earth. Reality can be defined in the way we reach universal agreement about it, whether it be analog or digital (I think it will ultimately always be expressed with both concepts of discreteness and continuity: digital and analog).<br /><div><br /></div><div>Some thoughts you'll find in <a href="http://www.fqxi.org/community/forum/topic/919">my essay</a>:</div><div><br /></div><div><i>"I postulate Reality to be that on which all people can agree"</i></div><div><i>"Physical reality must be potentially and reproducibly accessible through experiment to any human inhabitant of earth. If it is not, the corresponding statement about reality is biased."</i></div><div><i>"Some aspects of physics haven’t reached the stadium of universal agreement yet. This is the case for processes at the cosmological scale or at the atomic and nuclear scale, because the rules governing their behavior are not directly part of our everyday experience. We haven’t yet managed to describe all their reality with concepts or words on which everybody can agree. <b>It is the physicist’s job to enlarge the scope of universal agreement about physical reality.</b>"</i><br /><i>"We will describe anything happening in the submicroscopic world </i><i>with concepts derived from our macroscopic experience, analogically and digitally. Only then </i><i>can we come to universal agreement about its reality."</i><br /><br />There are 160 other essays. Some are from essay writers whom I already know a little from last year. Unfortunately I won't be able to read more than a few, but I love it to catch up with your ideas, even if I don't always agree with them, probably because I haven't yet got all the information needed to understand your reality. Universal agreement about reality can only be reached if we discuss our ideas openly.</div>Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com0tag:blogger.com,1999:blog-382677439644973005.post-68140825593986268112011-01-10T22:17:00.000+01:002011-01-10T22:20:16.498+01:002011 and prime number sievesI'm a slow blogger. It's already been more than a week that we entered the 2011th year of the Christian era. Interestingly 2011 is a <a href="http://en.wikipedia.org/wiki/Prime_number">prime number</a>. Many bloggers have mentioned the fact that the number 2011 is also the sum of 11 (a prime number) consecutive primes, see for example at <a href="http://republicofmath.wordpress.com/2011/01/01/happy-mathematical-new-year-2011-is-the-sum-of-11-consecutive-prime-numbers/">Republic of Math</a>, at <a href="http://pballew.blogspot.com/2011/01/happy-new-prime-year.html">Pat'sBlog</a>, at <a href="http://www.mathlesstraveled.com/?p=831">The Math less Traveled</a>, at <a href="http://luckytoilet.wordpress.com/2011/01/01/is-2011-a-special-number/">Lucky's Notes</a>, at the <a href="http://patternizer.wordpress.com/2010/12/30/2011-primer/">Pattern Connection</a> ... and I apologize to those whom I forget, there must be many more.<br /><br />I would state the obvious if I said that primes are special numbers. But what makes primality so special? <a href="http://en.wikipedia.org/wiki/Don_Zagier">Don Zagier</a>, an eminent specialist, said: <i>Upon looking at these numbers, one has the feeling of being in the presence of one of the inexplicable secrets of creation.</i> That's awesome, isn't it? In antiquity, the Greeks already tried to understand the patterns underpinning the prime numbers. A way to visualize those patterns is through <a href="http://pballew.blogspot.com/2010/03/different-prime-sieve.html">prime number sieves</a>. <a href="http://en.wikipedia.org/wiki/Eratosthenes">Eratosthenes</a> is said to have proposed the first sieve, the <a href="http://en.wikipedia.org/wiki/Sieve_of_Eratosthenes">Eratosthenes sieve</a>. (Hey, have you noticed that Eratosthenes was born 2287 years ago? So we are in year 2287 of the Eratosthenian era. Also a prime number). But there are <a href="http://en.wikipedia.org/wiki/Generating_primes#Prime_sieves">other sieves</a>. More of that later.<br /><br />Prime numbers occur in nature in cyclic processes. Imagine that, somewhere in the universe, there's a star with a huge number of planets, all of them having the same orbiting speed. They are numbered from 1 to N, planet n°1 being the inner planet and the diameters of their circular orbits are as following:<br /><ul><li>The diameter of planet n°2's orbit is twice the diameter of planet n°1's orbit.</li><li>The diameter of planet n°3's orbit is three times the diameter of planet n°1's orbit.</li><li>The diameter of planet n°4's orbit is four times the diameter of planet n°1's orbit.</li><li>And so on, the unit length being the diameter of planet n°1's orbit, the nth planet has an orbit of diameter n, n being an integer between 1 and N.</li></ul><div><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/_0Zi-ITihiW4/TSoiY_1IL7I/AAAAAAAAAac/6kpGgOvTFHc/s1600/TangentCircles.jpg" imageanchor="1" style="clear: right; float: right; margin-bottom: 1em; margin-left: 1em;"><img border="0" src="http://3.bp.blogspot.com/_0Zi-ITihiW4/TSoiY_1IL7I/AAAAAAAAAac/6kpGgOvTFHc/s1600/TangentCircles.jpg" /></a></div>Imagine also that at some timestamp zero, all the planets are aligned at the same side of the star (yes I know, this is very improbable, but it's just a thought experiment). We define the orbital period of the first planet as unit time. "Prime times" occur when two and only two planets are aligned on the initial direction with the star: the outer planet than has a prime distance to the star (the closest planet being always planet n°1). I should make an animation to visualize this effect. As making an animation is very time-consuming, I looked for other ways to illustrate this principle and ended up with another sieve: a circle sieve. Instead of drawing concentric circles for the orbits, I shift each circle such that all are tangent at the origin (for example the location of the center of the star, see Figure 1). For each period of a planet, I successively copy its circle and place it tangentially to the right (see Figure 2 for the first periods of planet 1 and 2).<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://4.bp.blogspot.com/_0Zi-ITihiW4/TSopU3ZYGRI/AAAAAAAAAaQ/IYgJdYxGLBw/s1600/TangentCircles.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://4.bp.blogspot.com/_0Zi-ITihiW4/TSopU3ZYGRI/AAAAAAAAAaQ/IYgJdYxGLBw/s1600/TangentCircles.png" /></a></div>Now, as each planet is orbiting at the same speed, the first alignment along the initial direction will occur at timestamp 2. Planet 1 will have circled twice around the star and planet 2 will have circled once. Circles 1 and 2 touch at abscissa 2. Number 2 is a prime number. The next alignment on the axis will occur for planet 3. Circles 1 and 3 touch at abscissa 3 without circle 2 (see Figure 3). Number 3 is a prime number.<br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/_0Zi-ITihiW4/TSotDUYJuFI/AAAAAAAAAaI/2wNhIg6WZxo/s1600/TangentCircles3.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/_0Zi-ITihiW4/TSotDUYJuFI/AAAAAAAAAaI/2wNhIg6WZxo/s1600/TangentCircles3.png" /></a></div>The next alignment will occur for planets 1, 2 and 4. Number 4 is not a prime. But number 5 is prime. We can see that easily if we draw only circles 2 and 3 (leaving out the unit circles), see Figure 4. Prime numbers can only occur at vacant places along the axis. The symmetries that show up along this axis help to design primality tests. Any prime number greater than 3 must be of the form 3*2<i>n</i>±1. This formula is a condensed form of the intersection of odd numbers (noted 2<i>n</i> +1) and non-multiples of 3 (noted 3n+{1,2}) and generates all primes between 3 and 5².<br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/_0Zi-ITihiW4/TStiyR9lgsI/AAAAAAAAAZ0/0xndglosOJg/s1600/TangentCirclesFig4.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/_0Zi-ITihiW4/TStiyR9lgsI/AAAAAAAAAZ0/0xndglosOJg/s1600/TangentCirclesFig4.png" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div>The succession of circles n°5 together with n°2 and 3 sieves out all primes between 5 and 7², see Figure 5.<br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="http://2.bp.blogspot.com/_0Zi-ITihiW4/TSttJvyET7I/AAAAAAAAAZw/dqSvnxc9sJA/s1600/TangentCirclesFig5.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://2.bp.blogspot.com/_0Zi-ITihiW4/TSttJvyET7I/AAAAAAAAAZw/dqSvnxc9sJA/s1600/TangentCirclesFig5.png" /></a></div>I let you try it further with circles n°7. You will see patterns with concentric circles become more apparent. I like this geometric variant of Eratosthenes' sieve because symmetries can be caught with the eye and suggest prime generating formulas. For example, I came across two simple prime generating polynomials (which must have been noticed by other people because they are less powerful than <a href="http://en.wikipedia.org/wiki/Formula_for_primes#Prime_formulas_and_polynomial_functions">Euler's P(n) = <i>n</i>² − <i>n</i> + 41</a>):<br /><blockquote><i>n</i>²+9<i>n</i>+1</blockquote><blockquote><i>n</i>²+21<i>n</i>+1</blockquote>This way of sieving has been described earlier and more extensively by physicist Imre Mikoss. Check his work at "<i><a href="http://www.ma.utexas.edu/mp_arc/c/06/06-314.pdf">The Prime Numbers Hidden Symmetric Structure and its Relation to the Twin Prime Infinitude and an Improved Prime Number Theorem</a></i>".<br /><br />Happy prime year!</div>Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com4tag:blogger.com,1999:blog-382677439644973005.post-76178578817924704482010-12-23T00:23:00.000+01:002010-12-23T00:23:02.372+01:00Math and Physics: Creative Arts?My favorite approach towards math is through drawing geometric figures like circles, triangles, squares... They link the mathematical abstraction to the physical reality, whatever that may mean. They help to visualize patterns that may remain hidden when described with symbols. Doodling geometric figures allow digressive exploration of otherwise marked out paths and thus favor creative and curiosity-driven approaches towards math. Although I am not a mathematician, I practice math for my job and as a hobby. I subscribe to views expressed by G.H. Hardy in his <i><a href="http://en.wikiquote.org/wiki/G._H._Hardy#A_Mathematician.27s_Apology_.281941.29">Mathematician's Apology</a></i><br /><blockquote><i>I am interested in mathematics only as a creative art.</i></blockquote>or by Paul Halmos in his <i><a href="http://www.google.com/#hl=en&q=halmos+creative+art">Mathematics as a Creative Art</a></i><br /><blockquote><i>Mathematics is far closer to an art than it is to the business of equation-solving.</i></blockquote>or by Paul Lockhart in his <i><a href="http://www.maa.org/devlin/LockhartsLament.pdf">Mathematician's Lament</a></i><br /><blockquote><i>I’m just playing. That’s what math is— wondering, playing, amusing yourself with your imagination.</i></blockquote>Interestingly, in these quotes, <i>math</i> could be replaced by <i>physics</i>. Concerning me, it would likewise express my interest in physics. What if all of math and physics could be expressed through art? Well, it should.Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com2tag:blogger.com,1999:blog-382677439644973005.post-83440634943117406532010-11-28T20:31:00.000+01:002010-11-28T22:38:22.430+01:00Geometric representations of higher degree binomialsA <a href="http://en.wikipedia.org/wiki/Binomial">binomial</a> is a mathematical expression with two terms, say <i>x</i> and <i>y</i>. The simplest binomial is the sum of those two terms <i>x+y</i>. Powers of this binomial (<i>x+y</i>)<i><sup>n</sup></i> are ruled by the so-called <a href="http://en.wikipedia.org/wiki/Binomial_theorem">binomial theorem</a>, which states that the coefficients of the expansion terms <i>x<sup>p</sup>y<sup>q</sup></i> are given by the <a href="http://en.wikipedia.org/wiki/Pascal's_triangle">Pascal triangle</a> <span class="Apple-style-span" style="color: #990000;">entries</span>. So we have:<br />(<i>x+y</i>)<i><sup>0</sup></i> = <span class="Apple-style-span" style="color: #990000;">1</span><br />(<i>x+y</i>)<i><sup>1</sup></i> = <span class="Apple-style-span" style="color: #990000;">1</span>.<i>x</i> + <span class="Apple-style-span" style="color: #990000;">1</span>.<i>y</i><br /><div style="margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px;">(<i>x+y</i>)<i><sup>2</sup></i> = <span class="Apple-style-span" style="color: #990000;">1</span>.<i>x</i><i><sup>2</sup></i>+ <span class="Apple-style-span" style="color: #990000;">2</span>.<i>xy</i> + <span class="Apple-style-span" style="color: #990000;">1</span>.<i>y</i><i><sup>2</sup></i></div><div style="margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px;"><i></i></div><div><div style="margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px;">(<i>x+y</i>)<i><sup>3</sup></i> = <span class="Apple-style-span" style="color: #990000;">1</span>.<i>x</i><i><sup>3</sup></i>+ <span class="Apple-style-span" style="color: #990000;">3</span>.<i>x</i><i><sup>2</sup></i><i>y</i> + <span class="Apple-style-span" style="color: #990000;">3</span>.<i>x</i><i>y</i><i><sup>2</sup></i> + <span class="Apple-style-span" style="color: #990000;">1</span>.<i>y</i><i><sup>3</sup></i></div><div style="margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px;"><i></i></div><div style="margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px;"><i></i></div></div>(<i>x+y</i>)<i><sup>4</sup></i> = <span class="Apple-style-span" style="color: #990000;">1</span>.<i>x</i><i><sup>4</sup></i>+ <span class="Apple-style-span" style="color: #990000;">4</span>.<i>x</i><i><span class="Apple-style-span" style="font-style: normal;"><i><sup>3</sup></i></span>y</i> + <span class="Apple-style-span" style="color: #990000;">6</span>.<i>x</i><i><sup>2</sup></i><i>y</i><i><sup>2</sup></i> + <span class="Apple-style-span" style="color: #990000;">4</span>.<i>x</i><i>y</i><i><sup>3</sup></i> + <span class="Apple-style-span" style="color: #990000;">1</span>.<i>y</i><i><sup>4</sup></i><br /><div><div style="margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px;"><i></i></div><div style="margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px;"><i></i></div><div style="margin-bottom: 0px; margin-left: 0px; margin-right: 0px; margin-top: 0px;"><i></i></div></div>and so forth ...<br /><br />But where do these coefficients come from? You can just be satisfied by the result algebraic calculation, or you could look for geometric representations. For instance, a <a href="http://www.mathaware.org/mam/00/master/essays/B3D/2/binomial.html">nice geometric explanation</a> is given at the <a href="http://www.mathaware.org/">www.mathaware.org</a> site.<br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://www.mathaware.org/mam/00/master/essays/B3D/2/JPG/figure12.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://www.mathaware.org/mam/00/master/essays/B3D/2/JPG/figure12.jpg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Geometric proof of square binomial (original at <a href="http://www.mathaware.org/mam/00/master/essays/B3D/2/binomial.html">The Geometry of the Binomial Theorem</a>, Math Awareness Month site)</td></tr></tbody></table><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="http://www.mathaware.org/mam/00/master/essays/B3D/2/JPG/figure13.jpg" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" src="http://www.mathaware.org/mam/00/master/essays/B3D/2/JPG/figure13.jpg" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">Geometric proof of cubic binomial (original at <a href="http://www.mathaware.org/mam/00/master/essays/B3D/2/binomial.html">The Geometry of the Binomial Theorem</a>, Math Awareness Month site) </td></tr></tbody></table>For the second degree binomial, the Pascal triangle entries emerge naturally as the number of squares or rectangles with area given by the expansion terms <i>x<sup>p</sup>y<sup>q</sup></i>. For third degree binomials, they emerge as the number of cubes or parallelepipeds with volume given by the expansion terms. And for higher degree binomials, they emerge as the number of <a href="http://en.wikipedia.org/wiki/Hypercube">hypercubes</a> and hyperparallelepipeds given by the expansion terms, so an Euclidean geometric representation for those higher degree binomials seems impossible.<br /><br />This suggested impossibility prompted me to have a closer look at geometric representations of binomial expansions. As a matter of fact a number to the fourth power <i>x<sup>4</sup></i> is just a number, but it is also a square <i>x<sup>4</sup></i> = (<i>x<sup>2</sup></i>)<i><sup>2</sup></i>. And there is no impossibility in representing numbers and squares geometrically, so the suggested impossibility is only an impossibility along the common line of thought, which sees cubic expansions as volumes. Along another line of thought, which seems to have been unnoticed, cubic expansions can be seen as areas, and then geometric representations of higher degree binomials become possible. The following figures illustrate this fact for the special case where <i>x+y</i> is normalized to 1 (for arbitrary <i>x+y</i>, one just has to rescale the figure each time, the pattern remains the same).<br /><br />For the binomial (<i>x+y</i>)<i><sup>1</sup></i> = <span class="Apple-style-span" style="color: #990000;">1</span>.<i>x</i> + <span class="Apple-style-span" style="color: #990000;">1</span>.<i>y</i>, we can divide a unit square into two rectangles, <span class="Apple-style-span" style="color: #990000;">one</span> rectangle of area <i>x</i> and <span class="Apple-style-span" style="color: #990000;">one</span> rectangle of area <i>y</i>, see Figure 1.<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/_0Zi-ITihiW4/TPLBwAwIUgI/AAAAAAAAAYg/y7Kvfsmi_wg/s1600/Binomial_1.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/_0Zi-ITihiW4/TPLBwAwIUgI/AAAAAAAAAYg/y7Kvfsmi_wg/s1600/Binomial_1.jpg" /></a></div>For the binomial (<i>x+y</i>)<i><sup>2</sup></i> = <span class="Apple-style-span" style="color: #990000;">1</span>.<i>x</i><i><sup>2</sup></i>+ <span class="Apple-style-span" style="color: #990000;">2</span>.<i>xy</i> + <span class="Apple-style-span" style="color: #990000;">1</span>.<i>y</i><i><sup>2</sup></i>, we can divide the unit square into two squares, <span class="Apple-style-span" style="color: #990000;">one</span> of area <i>x</i><i><sup>2</sup></i> and <span class="Apple-style-span" style="color: #990000;">one</span> of area <i>y</i><i><sup>2</sup></i>, plus <span class="Apple-style-span" style="color: #990000;">two</span> rectangles of area <i>xy</i>, see Figure 2.<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/_0Zi-ITihiW4/TPLBwu2OcTI/AAAAAAAAAYk/p203Os4dEew/s1600/Binomial_2.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/_0Zi-ITihiW4/TPLBwu2OcTI/AAAAAAAAAYk/p203Os4dEew/s1600/Binomial_2.jpg" /></a></div>For the binomial (<i>x+y</i>)<i><sup>3</sup></i> = <span class="Apple-style-span" style="color: #990000;">1</span>.<i>x</i><i><sup>3</sup></i>+ <span class="Apple-style-span" style="color: #990000;">3</span>.<i>x</i><i><sup>2</sup></i><i>y</i> + <span class="Apple-style-span" style="color: #990000;">3</span>.<i>x</i><i>y</i><i><sup>2</sup></i> + <span class="Apple-style-span" style="color: #990000;">1</span>.<i>y</i><i><sup>3</sup></i>, we can divide each of the previous squares and rectangles into proportions <i>x</i> and <i>y</i>, giving <span class="Apple-style-span" style="color: #990000;">eight</span> rectangles, <span class="Apple-style-span" style="color: #990000;">one</span> of area <i>x</i><i><sup>3</sup></i>, <span class="Apple-style-span" style="color: #990000;">one</span> of area <i>y</i><i><sup>3</sup></i>, plus <span class="Apple-style-span" style="color: #990000;">three</span> rectangles of area <i>x</i><i><sup>2</sup></i><i>y</i> and <span class="Apple-style-span" style="color: #990000;">three</span> rectangles of area <i>x</i><i>y</i><i><sup>2</sup></i>, see Figure 3.<br /><i></i><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://1.bp.blogspot.com/_0Zi-ITihiW4/TPLBxBHLneI/AAAAAAAAAYo/krtRgZDcT3Q/s1600/Binomial_3.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://1.bp.blogspot.com/_0Zi-ITihiW4/TPLBxBHLneI/AAAAAAAAAYo/krtRgZDcT3Q/s1600/Binomial_3.jpg" /></a></div>For the binomial (<i>x+y</i>)<i><sup>4</sup></i> = <span class="Apple-style-span" style="color: #990000;">1</span>.<i>x</i><i><sup>4</sup></i>+ <span class="Apple-style-span" style="color: #990000;">4</span>.<i>x</i><i><span class="Apple-style-span" style="font-style: normal;"><i><sup>3</sup></i></span>y</i> + <span class="Apple-style-span" style="color: #990000;">6</span>.<i>x</i><i><sup>2</sup></i><i>y</i><i><sup>2</sup></i> + <span class="Apple-style-span" style="color: #990000;">4</span>.<i>x</i><i>y</i><i><sup>3</sup></i> + <span class="Apple-style-span" style="color: #990000;">1</span>.<i>y</i><i><sup>4</sup></i>, we can again divide each of the previous squares and rectangles into proportions <i>x</i> and <i>y</i>, giving <span class="Apple-style-span" style="color: #990000;">one</span> square of area <i>x</i><i><sup>4</sup></i>, <span class="Apple-style-span" style="color: #990000;">one</span> of area <i>y</i><i><sup>4</sup></i>, four squares and two rectangles (giving <span class="Apple-style-span" style="color: #990000;">six</span>) of area <i>x</i><i><sup>2</sup></i><i>y</i><i><sup>2</sup></i>, <span class="Apple-style-span" style="color: #990000;">four</span> rectangles of area <i>x</i><i><sup>3</sup></i><i>y</i> and <span class="Apple-style-span" style="color: #990000;">four</span> rectangles of area <i>x</i><i>y</i><i><sup>3</sup></i>, see Figure 4.<br /><i></i><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://3.bp.blogspot.com/_0Zi-ITihiW4/TPLHG1L1CUI/AAAAAAAAAYw/Xw77T-49t3Y/s1600/Binomial_4.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://3.bp.blogspot.com/_0Zi-ITihiW4/TPLHG1L1CUI/AAAAAAAAAYw/Xw77T-49t3Y/s1600/Binomial_4.jpg" /></a></div>And we could go further indefinitely, doodling areas of incrementing powers, just in 2D, without any reference to unintuitive hyperspaces.Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com10tag:blogger.com,1999:blog-382677439644973005.post-6273876900218736912010-10-24T20:12:00.000+02:002010-10-24T22:24:50.610+02:00Explaining electron spin and Pauli exclusion principle to children<div>Fundamental particles are the building blocks of nature, of which the photon and the electron have the most visible impact on our everyday life. Photons are all pervasive. If they have the right energy, they can stimulate your eyes' photoreceptor cells. At other energies they will warm you up because they radiate from a warm object. Electrons are more energetic than the photons. They can either be free in space, or bound in atoms. Through their motion, they transmit motion to photons, which in turn can excite other electrons at distant places. This phenomenon, known as electromagnetism, is used in all wireless transmissions. Photons are the electromagnetic force carriers and electrons are the electromagnetic force sources.</div><div><br /></div><div>In order to understand the behavior of photons and electrons, it is important to have analogies that help us keeping track of them. In previous posts, I mentioned some helpful analogies for photons (for example at <a href="http://commonsensequantum.blogspot.com/2008/06/first-video-sequence-of-common-sense.html">this post on polarization</a>). Although electrons also show wave behavior, they act a bit differently from photons. You can not stack electrons near to one another, except if they have compatible spinning motions. For spinning motions to be compatible means that the electrons must:</div><div><ul><li>either spin at rates whose proportions are expressed with integers: for example one electron spins twice as fast as the other electron,</li><li>or spin in different directions, if they spin with the same velocity.</li></ul><div><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_0Zi-ITihiW4/TMQUTLAo1ZI/AAAAAAAAAYU/ufiZlaF68c8/s1600/IMG_2994.JPG"><img style="float:right; margin:0 0 10px 10px;cursor:pointer; cursor:hand;width: 200px; height: 150px;" src="http://3.bp.blogspot.com/_0Zi-ITihiW4/TMQUTLAo1ZI/AAAAAAAAAYU/ufiZlaF68c8/s200/IMG_2994.JPG" border="0" alt="" id="BLOGGER_PHOTO_ID_5531568561923347858" /></a>I sometimes come across situations that remind me of electrons. If you're standing in the bus or in the metro, you grip a pole to keep equilibrium. In the metro-train in my Paris suburb, the poles occur in pairs, like in the picture aside. When my children were younger, one of their favorite games was to spin around those poles. For parents, if you let your kids spin around the poles disorderedly, this game can be quite stressful, ending with fighting or crying. I used to explain to them that they had to spin like electrons in atoms. If one kid spins in one direction, the other kid needs to spin in opposite direction, in order to avoid hard clashes. I recently asked them if they could do it again so that I could put it on movie and post it to illustrate this electron analogy. But they've grown up and are now ashamed to play such games:-) So I decided to create the following simple animations that illustrate the electron spin and the Pauli exclusion principle.</div></div><div><a href="http://picasaweb.google.com/lh/photo/zEBUOi7NaOUu4n6RA2FyIF5N3PUMRydn1SbBTOwMW3Y?feat=embedwebsite"><img src="http://lh6.ggpht.com/_0Zi-ITihiW4/TMMLiwY_REI/AAAAAAAAAX4/87jUwXJfFR8/s800/ElectroSpinDown.gif" height="567" width="756" /></a><br /></div><div><a title="By Right_hand_rule_simple.png:Schorschi2 at de.wikipedia derivative work: Wizard191 (Right_hand_rule_simple.png) [public domain], from Wikimedia Commons" href="http://commons.wikimedia.org/wiki/File:Right-hand_grip_rule.svg"><img style="float:right" width="120px" alt="Right-hand grip rule" src="http://upload.wikimedia.org/wikipedia/commons/thumb/3/34/Right-hand_grip_rule.svg/240px-Right-hand_grip_rule.svg.png" border="0" /></a>A kid spinning around the pole is alike an electron spinning around a proton in its state of minimum energy, see Figure 1. Physicists designate the spinning direction with the help of the <a href="http://en.wikipedia.org/wiki/Right-hand_rule">right hand rule</a>. The kid of Figure 1 therefore has its spin down.</div><br />If your second kid spins in the same direction around the other pole, you can be sure that this game won't last for long. Their motions are incompatible and it ends up with a clash, see Figure 2.<br /><a href="http://picasaweb.google.com/lh/photo/0M36bPU5rG4e5Lk3PZ94Pl5N3PUMRydn1SbBTOwMW3Y?feat=embedwebsite"><img src="http://lh5.ggpht.com/_0Zi-ITihiW4/TMMXoSAgKHI/AAAAAAAAAX8/NlT6F84GhME/s800/ElectronSpinPairDown.gif" height="567" width="756" /></a><br />If you want them to play peacefully, you need to instruct them to follow a natural rule: the <a href="http://en.wikipedia.org/wiki/Pauli_exclusion_principle">Pauli exclusion principle</a>, illustrated in Figure 3. Electrons with same spinning velocity and sharing the same space can only occur if their spins are opposite. Very useful rule to keep harmony in the family!<br /><a href="http://picasaweb.google.com/lh/photo/e86HHT7f3dB4SJkovbLyd15N3PUMRydn1SbBTOwMW3Y?feat=embedwebsite"><img src="http://lh6.ggpht.com/_0Zi-ITihiW4/TMQHleccNcI/AAAAAAAAAYQ/BeVaUcjbpdg/s800/ElectronSpinUpDown.gif" height="567" width="756" /></a>Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com7tag:blogger.com,1999:blog-382677439644973005.post-62440394388426016902010-10-21T20:26:00.000+02:002010-10-22T07:36:58.161+02:00Volume of a bead vs. volume of a sphereImagine you drill a hole through the center of a sphere. The remaining object (a bead, if we think of a little pearl with a hole) has an interesting property as pointed out by Pat Ballew on his blog at <a href="http://pballew.blogspot.com/2010/10/surprisingly-constant.html">the end of this post</a>. Whatever the radius of the initial sphere, the volume of the bead depends only on its height h (see following figure taken from Pat's blog). More precisely, its volume is the same as the volume of a sphere with diameter h. A proof of this property has been given by Pat in a <a href="http://pballew.blogspot.com/2010/10/non-calculus-explanation-for-volume-of.html">later post</a>.<div><br /><div><img src="http://2.bp.blogspot.com/_PQZIFRiZz38/TL_o3jFmhdI/AAAAAAAACXs/Nzo42pYHLD4/s320/sphere+vol1.jpg" /></div><div><br /></div><div>As I was thinking about it, I thought about another way to prove it, inspired by <a href="http://www.mamikon.com/">Mamikon</a>'s visual calculus method.</div><div><img src="http://lh6.ggpht.com/_0Zi-ITihiW4/TMEiXJIMsXI/AAAAAAAAAXg/2uCZYNjs53k/s800/VolumeBead.gif" height="567" width="567" /><br /></div><div>If we think of a plane parallel to the axis of the bead and tangent to its inner cylindrical surface, the intersection of the bead and the plane is a disk of diameter h. If we now rotate the plane around the axis for a whole turn such that it remains tangent to the inner surface, the disk will also rotate a whole turn, as if it sweeps the volume of a sphere of diameter h.</div><div><br /></div><div><br /><img src="http://lh4.ggpht.com/_0Zi-ITihiW4/TMEiWw2P2KI/AAAAAAAAAXc/_8lBOJLOTCQ/s800/VolumeSphere.gif" height="567" width="567" /><br />Post scriptum. Afterwards I found a drawing of such a bead in <a href="http://www.its.caltech.edu/~mamikon/Article.html">Mamikon's original notes</a> (see second page of his drawings).</div><div><br /></div><div><br /></div></div>Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com3tag:blogger.com,1999:blog-382677439644973005.post-36403859639011571612010-10-17T09:56:00.000+02:002010-11-21T18:11:23.639+01:00In memoriam: Georges Charpak and Maurice AllaisTime goes by and people die. And so do great scientists. There is <a href="http://en.wikipedia.org/wiki/Beno%C3%AEt_Mandelbrot">Benoit Mandelbrot</a> (November 20, 1924 – October 14, 2010) who developed the study of fractals, because he "<i>decided to go into fields where mathematicians would never go because problems were badly stated</i>". Two others, less known in the English speaking world and who are an example to me have also left us recently. I name: Georges Charpak and Maurice Allais.<br /><br /><a href="http://picasaweb.google.com/lh/photo/pLZ1GEUQFRkGb3Ac_9mwjl5N3PUMRydn1SbBTOwMW3Y?feat=embedwebsite"><img src="http://lh5.ggpht.com/_0Zi-ITihiW4/TLrPhDY2y_I/AAAAAAAAAW0/QBCs8xINh8g/s288/CHARPAK_Georges-24x50-2005%5B1%5D.jpg" align="right" height="133" width="288" /></a><a href="http://en.wikipedia.org/wiki/Georges_Charpak">Georges Charpak</a> (March 8, 1924 – September 29, 2010) was born in the little town <a href="http://uk.wikipedia.org/wiki/%D0%94%D1%83%D0%B1%D1%80%D0%BE%D0%B2%D0%B8%D1%86%D1%8F">Dubrovytsia</a> located in a region where the political and social situation was very complicated at that time. The region was essentially populated with Ukrainian and Yiddish speaking people. It had suffered the post-WW1 Polish-Russian war and belonged to Poland at the time of his birth. Charpak's family had the opportunity to flee to France, which saved him from later WW2 exterminations of Jews in his natal country. In France, the situation was much better, Georges calling it even "paradise". During the 1920-30s, there was a tolerant spirit in France, allowing him to make friends with people of all origin. The Nazi occupation of France during WW2 brought new dangers for him. He had to change his name to George Charpentier, entered the <a href="http://en.wikipedia.org/wiki/French_Resistance">French Resistance</a>, he was imprisoned, participated to mutiny in the prison, escaped the punishment fusillade for the mutineers (he heard the ball flying around his ears). He was deported to the concentration camp of Dachau and was saved from extermination again because the Nazis could use his young guy's force in Dachau instead of sending him to more severe camps. His career as an experimental physicist started after the war with a thesis on particle detectors in <a href="http://en.wikipedia.org/wiki/Fr%C3%A9d%C3%A9ric_Joliot-Curie">Frédéric Joliot-Curie</a>'s group. He excelled in building simple detectors. His <a href="http://en.wikipedia.org/wiki/Wire_chamber">wire-detectors</a> slightly replaced the historical bubble and ionization-chambers. He further worked at <a href="http://cerncourier.com/cws/article/cern/37861">CERN</a> and one of his detectors, the multi wire proportional chamber ("not the most elegant" in his words), <a href="http://nobelprize.org/nobel_prizes/physics/laureates/1992/">earned him the Nobel in 1992</a>. He also lectured at the <a href="http://www.espci.fr/actualites/editos/georges-charpak-1924-2010">ESPCI</a>, where I'm currently PhD student. Apart from this exceptional course, after his Nobel, he had the nobility of mind to start a hands-on program for elementary school students "<a href="http://xen-lamap.inrp.fr/lamap/">La Main à la Pâte</a>" (literally <a href="http://www.recess.ufl.edu/transcripts/2005/0217.shtml">Hand in the dough</a>). I am fond of such initiatives because it brings experimental physics nearer to us. It is always preferable to first discover by ourselves how Nature works before learning how to formulate its laws through math. Too often, we learn the formula of a physical law before having experimented it personally.<div>"<a href="http://en.wikiquote.org/wiki/Georges_Charpak"><i>If there's one thing to do, it's to engage in education.</i></a>" ~ Georges Charpak.<div><br /><a href="http://picasaweb.google.com/lh/photo/0PcSeUD30SMUzQeYH4GSol5N3PUMRydn1SbBTOwMW3Y?feat=embedwebsite"><img src="http://lh6.ggpht.com/_0Zi-ITihiW4/TLrigMkvaoI/AAAAAAAAAW4/Ds505qZZIr0/s288/ALLAIS_PN_Maurice-24x30-2001%5B1%5D.jpg" height="167" width="288" align="right" /></a><a href="http://en.wikipedia.org/wiki/Maurice_Allais">Maurice Allais</a> (May 31, 1911 – October 9, 2010) was born earlier, before WW1. His father died in a German prison during WW1. Early loss of his father left a profound mark on the rest of his life. He devoted his life to the comprehension of all things he encountered. His passions were history, science, economics, physics. He excelled in all disciplines during his education. He had the opportunity to visit the United States in 1933 and was so impressed by the Great Depression and the inability to solve the crisis, that he studied by himself the principles that would secure economic wealth. The life-long product of this work earned him the <a href="http://nobelprize.org/nobel_prizes/economics/laureates/1988/allais.html">Nobel Economics in 1988</a>. I'm not a specialist in Economics, but as far as I understand, one of his findings (before other economists) is the <a href="http://en.wikipedia.org/wiki/Golden_Rule_savings_rate">Golden rule of savings rate</a>, which states that the rate of interest a banker applies should be equal to the rate of economic growth: an equilibrium law applied to economics. At the beginning of his professional career, Maurice Allais taught economics at the <a href="http://www.gemtech.fr/98011902/1/fiche___pagelibre/">Ecole des Mines</a>. I suppose Georges Charpak, student at that same school, must have followed some of his lectures (*footnote). While Georges Charpak engaged in "normal" physics, Maurice Allais pondered over the foundations of physics. He wasn't satisfied about the interpretation of relativity and quantum theories. As a physicist, he needed to find it out for himself. In the 19th century tradition, against mainstream, he began to conduct experiments on a <a href="http://en.wikipedia.org/wiki/Paraconical_pendulum">pendulum of his invention</a> in order to investigate periodical fluctuations in gravity and electromagnetism and their influence by planetary motion. The interesting thing is that he found unexplained effects, among which the most famous is the "<a href="http://en.wikipedia.org/wiki/Allais_effect">Allais effect</a>", a deviation of the oscillatory plane of the pendulum during solar eclipses. Maurice Allais published some books in French where he details the results of his investigations. These effects remain unexplained today, likewise the <a href="http://en.wikipedia.org/wiki/Pioneer_anomaly">Pioneer anomaly</a>. I have no settled idea about these effects. I think that such effects suffer from capricious cosmological (photon, graviton, muon or whatever other particles) weather. One can find some seasonal regularities though. Further investigation is left to us, curious experimenters, satisfied only by what Nature teaches us.</div></div><div>"<i><a href="http://en.wikiquote.org/wiki/Maurice_Allais">Submission to the experimental data is the golden rule that dominates any scientific discipline.</a></i>" ~ Maurice Allais.</div><div><br /></div><div>*Update November 20, 2010: This was confirmed to me by close relatives to Maurice Allais and Georges Charpak.</div><br /><span class="Apple-style-span" style="font-size: x-small;">Credit of the portraits of both Nobel Prize winners by </span><a href="http://www.studio-harcourt.eu/00.php?lang=en"><span class="Apple-style-span" style="font-size: x-small;">Studio Harcourt Paris</span></a><span class="Apple-style-span" style="font-size: x-small;">.</span>Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com2tag:blogger.com,1999:blog-382677439644973005.post-26979311326222043682010-10-13T21:35:00.001+02:002010-10-13T22:03:01.373+02:00Follow-up of my FQXi essay: Ordinary analogues for Quantum MechanicsToday I had the good surprise to discover the article "<a href="http://www.pnas.org/content/107/41/17455">Quantum mechanics writ large</a>" written by John W. M. Bush, Professor of Applied Mathematics at MIT, promoting the work of Couder, Fort et al. on the bouncing droplets. John Bush writes: "At the time that pilot wave theory was developed and then overtaken by the Copenhagen interpretation as the standard view of quantum mechanics, there was no macroscopic pilot wave analog to draw upon. Now there is."<br /><br />I'm totally in line with this opinion. Quantum mechanics has macroscopic analogues which have so far nearly never been discussed and from which we would learn a lot. There has already been <a href="http://www.fqxi.org/community/forum/topic/545">some discussion along with my 2009 FQXi essay</a>. In the abstract, I wrote something similar to John Bush: "Classical physics was not sufficiently advanced to deal with macroscopic particle-wave systems at the birth of quantum mechanics. Physicists therefore lacked references to compare quantum with analogous macroscopic behaviour. After consideration of some recent experiments with droplets steered by waves, we examine possibilities to give some intuitive meaning to the rules governing the quantum world."<br /><br />So, I hope this new article will gain much attention and foster discussion about macroscopic analogues for quantum behavior.Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com0tag:blogger.com,1999:blog-382677439644973005.post-50623005152558080372010-08-28T07:40:00.000+02:002010-08-28T10:29:47.125+02:00Alternative Pythagorean quadruples and other extensions to Pythagoras theoremThe vertices of an arbitrary triangle can be disposed onto two concentric circles such that the base is the diameter of the first circle (which I call the base circle) and the opposite vertex is on the second circle (which I call the leg circle). As there are three bases, there are generally three ways to arrange this setting. I mentioned in <a href="http://commonsensequantum.blogspot.com/2010/08/pythagorean-relation-for-any-triangle.html">my preceding post</a> that we can apply the Pythagorean-like relation <i>a</i>²+<i>b</i>²=<i>c</i>²±2<i>t</i>² to this triangle, where <i>c</i> is its base, <i>a</i> and <i>b</i> are the legs and <i>t</i> is the tangent to the inner circle emanating from the outer circle. For clarity, I reproduce an illustrative figure from my preceding post, for the case where the inner circle is the base circle.<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_0Zi-ITihiW4/TGg2Th_sbsI/AAAAAAAAAVI/2t3aJ1zglyo/s1600/PythagorasGeneralizedFig5a.gif"><img style="cursor:pointer; cursor:hand;width: 484px; height: 363px;" src="http://1.bp.blogspot.com/_0Zi-ITihiW4/TGg2Th_sbsI/AAAAAAAAAVI/2t3aJ1zglyo/s1600/PythagorasGeneralizedFig5a.gif" border="0" alt="" /></a><br />Now, like for the classical Pythagoras relation, there are many features that can be investigated regarding this alternative relation. For example, I would like to better understand the relation of this 2D formula with the Cartesian 3D variant of Pythagoras: x²+y²+z²=d², which is analogous to the case where the leg circle is smaller than the base circle (see <a href="http://1.bp.blogspot.com/_0Zi-ITihiW4/TGg2UBQaWuI/AAAAAAAAAVY/Ehd2Yiwmqwg/s1600/PythagorasGeneralizedFig5c.gif">Figure 5c</a> of preceding post). Another interesting aspect is the fact that the square of the tangent <i>t²</i> (multiplied by <i>π</i>) determines the area of the annular ring delimited by the base and the leg circles, <a href="http://demonstrations.wolfram.com/MamikonsProofOfThePythagoreanTheorem/">as stated by Mamikon Mnatsakanian</a>. This fact offers possibilities for areal representations of the squares or circles related to the sides of the arbitrary triangle, like we are acquainted to do with the <a href="http://commons.wikimedia.org/wiki/Category:Pythagorean_theorem">squares related to the sides of a right triangle</a>. Also, what would be interesting to develop is its relation with the law of cosines (as <a href="http://pballew.blogspot.com/2010/08/another-almost-pythagorean-relationship.html">noticed by Pat Ballew</a>) and the angle subtended by sides <i>a</i> and <i>b</i>. I guess this will leave enough stuff for future posts or conversations, or even papers in specialized journals.<div><br /><div>As a conclusion, let me mention Pythagorean numerological features which provide stuff for entertaining puzzles. Maybe you know the <a href="http://en.wikipedia.org/wiki/Pythagorean_triple">Pythagorean triples</a>, those sets of integer numbers (<i>n</i>,<i>m</i>,<i>l</i>) that verify <i>n</i>²+<i>m</i>²=<i>l</i>² and of which (<i>3</i>,<i>4</i>,<i>5</i>) is the simplest instance. The 3D Pythagoras relation allows an extension to <a href="http://en.wikipedia.org/wiki/Pythagorean_quadruple">Pythagorean quadruples</a>, sets of integer numbers (<i>n</i>,<i>m</i>,<i>l</i>,<i>k</i>) that satisfy <i>n</i>²+<i>m</i>²+<i>l</i>²=<i>k</i>² and of which (<i>1</i>,<i>2</i>,<i>2</i>,<i>3</i>) is the simplest instance. I couldn't resist to look for some integer quadruples that satisfy <i>a</i>²+<i>b</i>²=<i>c</i>²+2<i>t</i>². Among them I found two nice quadruples with successive <i>a</i>,<i>b</i>,<i>c</i> :</div><div><br /></div><div>(<i>7</i>,<i>8</i>,<i>9</i>,<i>4</i>) for which 7² + 8² = 9² + 2 × 4² = 113</div><div><br /></div><div>(<i>35</i>,<i>36</i>,<i>37</i>,<i>24</i>) for which 35² + 36² = 37² + 2 × 24² = 2521</div></div>Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com0tag:blogger.com,1999:blog-382677439644973005.post-79673274934315336782010-08-15T20:26:00.000+02:002010-08-15T21:54:13.084+02:00A Pythagorean relation for any triangle?Physics makes extensive use of the <a href="http://en.wikipedia.org/wiki/Pythagorean_theorem">Pythagorean law</a> relating the squares of the sides of a right triangle. The well-known <i>a</i>² + <i>b</i>² = <i>c</i>² relation is of special interest for the determination of distances and lengths of vectors, but also for energy conservation laws and Lorentz transformations. There are various relations that resemble the Pythagoras law, but none of them seems to have the usefulness of Pythagoras’ original one, as well as the “beauty” originating from its sole quadratic terms. When the vertex opposite to the hypotenuse runs on the circle determined by the right triangle, we have the Pythagoras relation illustrated in Figure 1.<div><div><div><a href="http://picasaweb.google.com/lh/photo/tzsmSB0zKXGHsla2uTbJpV5N3PUMRydn1SbBTOwMW3Y?feat=embedwebsite"><img src="http://lh3.ggpht.com/_0Zi-ITihiW4/TGg0_KLe1gI/AAAAAAAAAUs/ZFK_u5utoAk/s800/DynamicPythagore.gif" /></a></div><div>Some time ago, I was made aware of another interesting Pythagorean law, discovered by Nguyen Tan Tai and which I mentioned in a <a href="http://commonsensequantum.blogspot.com/2010/01/pythagoras-extended.html">previous post</a>. For any triangle with sides <i>a</i>, <i>b</i> and <i>c</i>, if the vertex opposite to the base (say <i>c</i>) runs on any fixed circle centered at the center of <i>c</i>, we have the relation:</div><div><br /><span class="Apple-tab-span" style="white-space:pre"> </span><i>a</i>² + <i>b</i>² = <i>c</i>² + constant</div><div><br />In order to have a better understanding of this quadratic relation, I tried to re-derive it in my own mental representation. I will call the fixed circle with the running vertex C the “leg circle”, because it is determined by the vertex that is common to legs AC and BC of the triangle (see Figure 2).</div><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_0Zi-ITihiW4/TGg1shzuHfI/AAAAAAAAAUw/0pyVnFZQo9c/s1600/PythagorasGeneralizedFig2.gif"><img style="cursor:pointer; cursor:hand;width: 400px; height: 300px;" src="http://3.bp.blogspot.com/_0Zi-ITihiW4/TGg1shzuHfI/AAAAAAAAAUw/0pyVnFZQo9c/s400/PythagorasGeneralizedFig2.gif" border="0" alt="" id="BLOGGER_PHOTO_ID_5505709583566249458" /></a><br /><div><br />When the leg circle is larger than the base circle, the legs AC and BC intersect the base circle and we can decompose the arbitrary triangle into two right triangles, for example as illustrated in Figure 3, the right triangles ABD and BCD.</div><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_0Zi-ITihiW4/TGg2TCu_QxI/AAAAAAAAAU4/oSEqJmNM9PU/s1600/PythagorasGeneralizedFig3.gif"><img style="cursor:pointer; cursor:hand;width: 400px; height: 300px;" src="http://2.bp.blogspot.com/_0Zi-ITihiW4/TGg2TCu_QxI/AAAAAAAAAU4/oSEqJmNM9PU/s400/PythagorasGeneralizedFig3.gif" border="0" alt="" id="BLOGGER_PHOTO_ID_5505710245239800594" /></a><br /><div><br />With the Pythagoras relation we than have:</div><div><br /><span class="Apple-tab-span" style="white-space:pre"> </span>AC² + BC² <span class="Apple-tab-span" style="white-space:pre"> </span>=<span class="Apple-tab-span" style="white-space:pre"> </span>(AD + CD)² + BC²<br /><span class="Apple-tab-span" style="white-space:pre"> </span>=<span class="Apple-tab-span" style="white-space:pre"> </span>(AD² + CD² + 2.AD.CD) + (BD² + CD²)<br /><span class="Apple-tab-span" style="white-space:pre"> </span>=<span class="Apple-tab-span" style="white-space:pre"> </span>AD² + BD² + 2.CD² + 2.AD.CD<br /><span class="Apple-tab-span" style="white-space:pre"> </span>=<span class="Apple-tab-span" style="white-space:pre"> </span>(AD² + BD²) + 2.(AD + CD).CD<br /><span class="Apple-tab-span" style="white-space:pre"> </span>=<span class="Apple-tab-span" style="white-space:pre"> </span>AB² + 2.AC.CD</div><div><br />But the product AC.CD is the <a href="http://en.wikipedia.org/wiki/Power_of_a_point">power of point C</a> with respect to the base circle, i.e. for every point C on the leg circle, AC.CD is constant and equal to the square of the tangent ray CT (see Figure 4).<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_0Zi-ITihiW4/TGg2TTX5oyI/AAAAAAAAAVA/tKSqXacx5Zw/s1600/PythagorasGeneralizedFig4.gif"><img style="cursor:pointer; cursor:hand;width: 400px; height: 300px;" src="http://1.bp.blogspot.com/_0Zi-ITihiW4/TGg2TTX5oyI/AAAAAAAAAVA/tKSqXacx5Zw/s400/PythagorasGeneralizedFig4.gif" border="0" alt="" id="BLOGGER_PHOTO_ID_5505710249706365730" /></a><br /><br />If we use another notation, AC = <i>a</i>, BC = <i>b</i>, AB = <i>c</i> and CT = <i>t</i>, the relation becomes a nice equation with only quadratic terms:</div><div><br /><span class="Apple-tab-span" style="white-space:pre"> </span><i>a</i>² + <i>b</i>² = <i>c</i>² + <i>t</i>² + <i>t</i>²<br /><br />One can verify that if the leg circle has same size as the base circle, we retrieve the original Pythagoras relation :</div><div><br /><span class="Apple-tab-span" style="white-space:pre"> </span><i>a</i>² + <i>b</i>² = <i>c</i>²</div><div><br />And if the leg circle is smaller than the base circle, we have:</div><div><br /><span class="Apple-tab-span" style="white-space:pre"> </span><i>a</i>² + <i>b</i>² + <i>t</i>² + <i>t</i>² = <i>c</i>²</div><div><br />where <i>t</i> is now the tangent ray to the leg circle emanating from any point of the base circle.<br /><br />We thus have a Pythagoras-like relation for any triangle given its base and leg circles, as illustrated by Figures 5a, 5b and 5c.<br /></div></div></div><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_0Zi-ITihiW4/TGg2Th_sbsI/AAAAAAAAAVI/2t3aJ1zglyo/s1600/PythagorasGeneralizedFig5a.gif"><img style="cursor:pointer; cursor:hand;width: 400px; height: 300px;" src="http://1.bp.blogspot.com/_0Zi-ITihiW4/TGg2Th_sbsI/AAAAAAAAAVI/2t3aJ1zglyo/s400/PythagorasGeneralizedFig5a.gif" border="0" alt="" id="BLOGGER_PHOTO_ID_5505710253631368898" /></a><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_0Zi-ITihiW4/TGg2T1eL_AI/AAAAAAAAAVQ/KZaze6H9aHs/s1600/PythagorasGeneralizedFig5b.gif"><img style="cursor:pointer; cursor:hand;width: 400px; height: 300px;" src="http://1.bp.blogspot.com/_0Zi-ITihiW4/TGg2T1eL_AI/AAAAAAAAAVQ/KZaze6H9aHs/s400/PythagorasGeneralizedFig5b.gif" border="0" alt="" id="BLOGGER_PHOTO_ID_5505710258859539458" /></a><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_0Zi-ITihiW4/TGg2UBQaWuI/AAAAAAAAAVY/Ehd2Yiwmqwg/s1600/PythagorasGeneralizedFig5c.gif"><img style="cursor:pointer; cursor:hand;width: 400px; height: 300px;" src="http://1.bp.blogspot.com/_0Zi-ITihiW4/TGg2UBQaWuI/AAAAAAAAAVY/Ehd2Yiwmqwg/s400/PythagorasGeneralizedFig5c.gif" border="0" alt="" id="BLOGGER_PHOTO_ID_5505710262022986466" /></a>Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com2tag:blogger.com,1999:blog-382677439644973005.post-85468243264122335112010-06-06T10:55:00.000+02:002010-08-15T21:49:49.015+02:00Lost theorem about angular proportionsLast week, I came across a so called <a href="http://groups.google.com/group/geometry.research/browse_frm/thread/5242cefe2554263a/643ce9010fe75d57#643ce9010fe75d57"><i>missing theorem</i></a> about angular proportions in a triangle, discovered (or rediscovered) by <a href="http://www.kafou.com/book/news.html">Leon Romain</a>. This triangle construction is presented by <a href="http://www.youtube.com/watch?v=SaMMxifaAl4">user Linelites on youtube</a>.<div><br /></div><div>For such a "Romain triangle", the missing theorem states that, if one of the inner angles is twice another inner angle, one has the property <i>a</i>² = <i>bc</i> +<i>c</i>², where <i>a</i>, <i>b</i> and <i>c</i> are outlined in Figure 1.</div><div><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_0Zi-ITihiW4/TAtuCcaGuLI/AAAAAAAAAUI/rnLpJB1kb-c/s1600/RomainTriangleFig1.jpg"><img style="cursor:pointer; cursor:hand;width: 400px; height: 300px;" src="http://1.bp.blogspot.com/_0Zi-ITihiW4/TAtuCcaGuLI/AAAAAAAAAUI/rnLpJB1kb-c/s400/RomainTriangleFig1.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5479594359890819250" /></a><br /></div><div><br /></div><div>The sine chord pattern presented in the <a href="http://commonsensequantum.blogspot.com/2010/04/teaching-sine-function-with-spaghetti.html"><i>Teaching sine function with spaghetti</i></a> provides helpful insights for the missing theorem. In this pattern, all angles at the intersection of chords are integer multiples of a chosen unit angle or its complement, modulo 90°. All segment lengths in this pattern can therefore be written as sums or differences of cosine and sine products and ratios of that angle. Figure 2 pictures some measures of sine chords in a circle of unit diameter, for an arbitrary angle <i>θ</i>.</div><div><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_0Zi-ITihiW4/TAtt0bWXBDI/AAAAAAAAAUA/TJED4rt_TuE/s1600/RomainTriangleFig2.jpg"><img style="cursor:pointer; cursor:hand;width: 400px; height: 300px;" src="http://2.bp.blogspot.com/_0Zi-ITihiW4/TAtt0bWXBDI/AAAAAAAAAUA/TJED4rt_TuE/s400/RomainTriangleFig2.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5479594119088505906" /></a><br /></div><div><br /></div><div>Figure 3 shows a Romain triangle for angle α in this pattern. There are numerous other Romain triangles in this pattern. Can you figure them out? With the help of the measures pictured in Figure 2, one can follow visually the elements of a proof for the missing theorem.</div><div><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_0Zi-ITihiW4/TAtt0LQRPlI/AAAAAAAAAT4/ODPKJzYfVLQ/s1600/RomainTriangleFig3.jpg"><img style="cursor:pointer; cursor:hand;width: 400px; height: 300px;" src="http://3.bp.blogspot.com/_0Zi-ITihiW4/TAtt0LQRPlI/AAAAAAAAAT4/ODPKJzYfVLQ/s400/RomainTriangleFig3.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5479594114767994450" /></a><br /></div><div><br /></div><div>Side <i>b</i> is the sine of the chosen angle<i> θ</i> (Figure 4):</div><div><i><br /></i></div><div><i><span class="Apple-tab-span" style="white-space:pre"> </span>b</i> = sin<i>θ</i></div><div><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_0Zi-ITihiW4/TAttz0F-S6I/AAAAAAAAATw/GMQtZqnUv7g/s1600/RomainTriangleFig4.jpg"><img style="cursor:pointer; cursor:hand;width: 400px; height: 300px;" src="http://3.bp.blogspot.com/_0Zi-ITihiW4/TAttz0F-S6I/AAAAAAAAATw/GMQtZqnUv7g/s400/RomainTriangleFig4.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5479594108550794146" /></a><br /></div><div><br /></div><div>Side <i>c</i> times sin<i>3θ</i> equals <i>b</i> times sin<i>θ</i> (Figure 5):</div><div><br /></div><div><i><span class="Apple-tab-span" style="white-space:pre"> </span>c</i> = sin²<i>θ</i>/sin<i>3θ</i></div><div><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_0Zi-ITihiW4/TAttzZ4KWoI/AAAAAAAAATo/gw8NcFj5oIg/s1600/RomainTriangleFig5.jpg"><img style="cursor:pointer; cursor:hand;width: 400px; height: 300px;" src="http://1.bp.blogspot.com/_0Zi-ITihiW4/TAttzZ4KWoI/AAAAAAAAATo/gw8NcFj5oIg/s400/RomainTriangleFig5.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5479594101513542274" /></a><br /></div><div><br /></div><div>Side <i>a</i> equals <i>b </i>cos<i>θ</i> minus <i>c </i>cos<i>3θ</i> (Figure 6)</div><div><br /></div><div><i><span class="Apple-tab-span" style="white-space:pre"> </span>a</i> = sin<i>θ</i> cos<i>θ</i> - sin²<i>θ</i> cos<i>3θ</i> /sin<i>3θ</i></div><div><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_0Zi-ITihiW4/TAyMuTBI5SI/AAAAAAAAAUQ/AO7oZq7l4-s/s1600/RomainTriangleFig6.jpg"><img style="cursor:pointer; cursor:hand;width: 400px; height: 300px;" src="http://4.bp.blogspot.com/_0Zi-ITihiW4/TAyMuTBI5SI/AAAAAAAAAUQ/AO7oZq7l4-s/s400/RomainTriangleFig6.jpg" border="0" alt=""id="BLOGGER_PHOTO_ID_5479909573609973026" /></a><br /></div><div><br /></div><div>Working out <i>a</i>² and <i>bc</i> +<i>c</i>², one finds that they are both equal to <i>2</i> sin²<i>θ</i> cos<i>θ</i> / sin3<i>θ</i>.</div><div><br /></div><div>The interesting thing is that the sine chord pattern hosts plenty of "missing theorems" about angular proportions, which are only waiting to be (re)discovered.</div><div><br /></div>______________________<br />Update (June 7, 2010): I corrected Figure 6, which held a wrong term.Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com3tag:blogger.com,1999:blog-382677439644973005.post-67154802011299447182010-04-11T10:36:00.000+02:002010-08-15T21:50:56.974+02:00Teaching sine function with spaghettiSine functions appear everywhere in physics and mathematics. This seems to be related to the circular symmetry of space and to the periodic behavior of dynamical processes. The usual definition is that <i>the sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse</i> in a right triangle (<a href="http://en.wikipedia.org/wiki/Trigonometric_functions#Sine.2C_cosine_and_tangent">definition at wikipedia</a>). The sine is also often represented as the ordinate of a point running on a unit circle centered at the origin. Any right triangle can be put in that setting. The length of the blue segment on Figure 1 illustrates this definition for an arbitrary angle α, in a circle of unit radius.<div> <div><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_0Zi-ITihiW4/S8HmwfkNs4I/AAAAAAAAARc/azxWzThv01s/s1600/Figure1.jpg"><img style="cursor:pointer; cursor:hand;width: 355px; height: 400px;" src="http://2.bp.blogspot.com/_0Zi-ITihiW4/S8HmwfkNs4I/AAAAAAAAARc/azxWzThv01s/s400/Figure1.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5458897944131253122" /></a></div><div><br /></div><div><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_0Zi-ITihiW4/S8HmwfkNs4I/AAAAAAAAARc/azxWzThv01s/s1600/Figure1.jpg"></a>It is of interest to have in mind other geometric representations of the sine function. This helps to associate figures with sine and cosine operations or identities. The red segment in Figure 1 shows such an alternative representation, because the sine is also a chord of a circle of unit diameter. When we increment the angle by steps α, one endpoint of the chord advances on the little circle, while the other endpoint remains fixed at the origin, as illustrated by the animation in Figure 2.</div><br /><a href="http://picasaweb.google.com/lh/photo/RiFEr8-t4IEKjhdu6YnUy15N3PUMRydn1SbBTOwMW3Y?feat=embedwebsite"><img src="http://lh4.ggpht.com/_0Zi-ITihiW4/S8HpajCQybI/AAAAAAAAARk/PbJLsOY-Cds/s800/Figure2.gif" /></a></div><div><a href="http://picasaweb.google.com/lh/photo/RiFEr8-t4IEKjhdu6YnUy15N3PUMRydn1SbBTOwMW3Y?feat=embedwebsite"></a><br /><div>When it represents the sine of angle α, the chord in the little circle spans an angle 2α. It is therefore a direct way to the angle multiplication and division by 2. Moreover, the endpoints of this chord can be placed two by two in any other direction on this circle. This enables one to find other variations on the recurrence pattern of the angle in the circle. During one of my circle drawing sessions, I was surprised by the stepwise zigzagging pattern of Figure 3.</div><div><br /></div><div><a href="http://picasaweb.google.com/lh/photo/bzm0QpKKz1ICr-K1KxYfJF5N3PUMRydn1SbBTOwMW3Y?feat=embedwebsite"><img src="http://lh4.ggpht.com/_0Zi-ITihiW4/S8ItOtOPiZI/AAAAAAAAASE/JQD2ZUrboRg/s800/Figure3.gif" /></a><br /></div><div><br /></div><div>Teaching the sine function with this zigzag pattern is particularly suited for daily life situations. Children like it, especially when you explain it at the table.</div><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_0Zi-ITihiW4/S8IwyV1UKbI/AAAAAAAAASI/S9ZEiTLl3Hc/s1600/IMG_2506.JPG"><img style="cursor:pointer; cursor:hand;width: 300px; height: 400px;" src="http://2.bp.blogspot.com/_0Zi-ITihiW4/S8IwyV1UKbI/AAAAAAAAASI/S9ZEiTLl3Hc/s400/IMG_2506.JPG" border="0" alt="" id="BLOGGER_PHOTO_ID_5458979339738818994" /></a></div>Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com2tag:blogger.com,1999:blog-382677439644973005.post-29140512081989531972010-04-05T09:40:00.000+02:002010-08-15T21:51:18.222+02:00Keeping track of the circle for integral representations of πThe fundamental constant π is characterized in many ways. Historically, it all began from tentative measurements of the circle's perimeter or area and gradually shifted into more <i>advanced</i> mathematics, in such a way that the link between the circle and modern characterizations of π faded away, see for example the formulas presented at <a href="http://mathworld.wolfram.com/PiFormulas.html">Wolfram MathWorld</a> or <a href="http://en.wikipedia.org/wiki/Pi_approximation">Wikipedia</a>.<div><br /></div>In the <a href="http://commonsensequantum.blogspot.com/2010/03/exhaustion-of-nested-squares-and-wallis.html">preceding posts</a>, I mentioned infinite products as approximations for π. These may be seen geometrically as exhaustion methods, where the area of a polygon approaches the circular area alternately from above, from below, from above, from below, etc.<div>There are also <a href="http://functions.wolfram.com/Constants/Pi/07/01/01/">integral representations of pi</a>. In such integral representations, π appears in the quantitative value of the <a href="http://en.wikipedia.org/wiki/Integral">integral</a> of a <a href="http://http//en.wikipedia.org/wiki/Function_(mathematics)">mathematical function</a>. Visually, this is often represented as <a href="http://en.wikipedia.org/wiki/File:Integral_example.svg">the area delimited by the bounds of the function</a>. However, the relation with the circle is lost, when viewed under <a href="http://en.wikipedia.org/wiki/Cartesian_coordinate_system">Cartesian coordinates</a>. For example, the graph of the simplest instance of the <a href="http://mathworld.wolfram.com/CauchyDistribution.html">Cauchy-Lorentz distribution</a>, <i>f</i>(<i>x</i>)=1/(1+<i>x</i>²), "<i>has nothing at all to do with circles or geometry in any obvious way</i>" as quoted from <a href="http://scienceblogs.com/builtonfacts/2010/03/sunday_function_65.php">last Pi-day Sunday function from Matt Springer's Built on Facts blog</a>.</div><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_0Zi-ITihiW4/S7exTq2bv4I/AAAAAAAAAPY/E0M8s_G0aVY/s1600/Figure1a.jpg"><br /><img style="float:left; margin:0 10px 10px 0;cursor:pointer; cursor:hand;width: 360px; height: 400px;" src="http://3.bp.blogspot.com/_0Zi-ITihiW4/S7exTq2bv4I/AAAAAAAAAPY/E0M8s_G0aVY/s400/Figure1a.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5456024425060876162" /></a><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_0Zi-ITihiW4/S7exSsBBcAI/AAAAAAAAAPQ/OwfKnApwZWo/s1600/Figure1b.jpg"><img src="http://1.bp.blogspot.com/_0Zi-ITihiW4/S7exSsBBcAI/AAAAAAAAAPQ/OwfKnApwZWo/s400/Figure1b.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5456024408193855490" /></a><div><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_0Zi-ITihiW4/S7exSsBBcAI/AAAAAAAAAPQ/OwfKnApwZWo/s1600/Figure1b.jpg"></a><br /><div>In order to view the role of the circle in integral representations of π, we need to switch to alternative ways to visualize math functions. As an example, let's take the constant function <i>y</i>=<i>f</i>(<i>x</i>)=2. The function <i>f</i> maps an element <i>x</i> from a domain to the element <i>y</i> of the target. In this case, for every <i>x</i>, the target <i>y</i> has the constant value 2. With Cartesian coordinates, we are used to represent this function as a horizontal straight line, like in Figure 1a (click on the figure to view it enlarged). If however we write it as <i>R</i>=<i>f</i>(<i>r</i>)=2, where the function <i>f</i> maps any circle of radius <i>r</i> of the domain to a target circle of radius <i>R</i>=2, the same function can be viewed as a circle of constant radius, like in Figure 1b. So the same function <i>f</i> can be equally well viewed as a straight line or as a circle (<i>x</i>, <i>y</i>, <i>r</i> or <i>R</i> are only dummy variables).</div><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_0Zi-ITihiW4/S7eweWItqqI/AAAAAAAAAPI/qQAffH4oPWk/s1600/Figure1c.jpg"><img src="http://4.bp.blogspot.com/_0Zi-ITihiW4/S7eweWItqqI/AAAAAAAAAPI/qQAffH4oPWk/s400/Figure1c.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5456023508967336610" /></a><div>Now if we take another example, the linear function, <i>y</i>=<i>f</i>(<i>x</i>)=2<i>x</i>, we are often used to view it in Cartesian coordinates as a straight line with slope 2, like in Figure 1c. In the circular representation <i>R</i>=<i>f</i>(<i>r</i>)=2<i>r</i>, this works however differently. Because we are relating circles of the input domain to other circles of the target, for each circle of radius <i>r</i>, we need to draw the target circle of radius 2<i>r</i>. A single line won't do. For one value of <i>r</i>, we need to draw two circles. If we use <span class="Apple-style-span" style="color:#3333FF;">blue circles for elements of the input domain</span> and <span class="Apple-style-span" style="color:#FF0000;">red circles for elements of the target</span>, we could visualize it for successive values of <i>r</i> as an animation like in Figure 1d. In that way, we view the progression of the <span class="Apple-style-span" style="color:#FF0000;">target circle</span> as the <span class="Apple-style-span" style="color:#3333FF;">input circle</span> becomes larger.</div><br /><a href="http://picasaweb.google.com/lh/photo/kCt9ms6gn0VgBb7AA56n_w?authkey=Gv1sRgCL3s4JSJk_zoTg&feat=embedwebsite"><img src="http://lh3.ggpht.com/_0Zi-ITihiW4/S7hNZ4VT6MI/AAAAAAAAAQA/xFQgH6IFzEk/s800/Figure1d.gif" style=";cursor:pointer; cursor:hand;width: 360px; height: 400px;" /></a><br />Unlike the Cartesian representation which shows the progression of a function in a static graph, this circular representation needs a dynamic or recurrent process to get grip of the progression of the function. Therefore it isn't very adapted for illustrations in print media. On the other hand, it has the advantage of keeping track of the geometrical form of the circle. And that's exactly what we need in order to perceive the circular nature when π shows up in mathematical functions. The relation of the integral of the Cauchy-Lorentz distribution <i>f</i>(<i>r</i>)=1/(1+<i>r</i>²) with the circle can then be seen with the help of the geometric counterparts of arithmetic operations like addition, squaring and dividing. A convenient procedure is illustrated in the successive steps of Figure 2.<br /><br /><div><b>Step 1.</b> Draw the input circle of radius <i>r</i> and the reference circle of radius unity.</div><div><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_0Zi-ITihiW4/S7mdxhgCzCI/AAAAAAAAAQo/w1wgSbqO6II/s1600/Figure2Step1.jpg"><img style="cursor:pointer; cursor:hand;width: 400px; height: 276px;" src="http://4.bp.blogspot.com/_0Zi-ITihiW4/S7mdxhgCzCI/AAAAAAAAAQo/w1wgSbqO6II/s400/Figure2Step1.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5456565897667988514" /></a><br /></div><div><br /></div><div><b>Step 2</b>. Determine <i>r</i>².</div><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_0Zi-ITihiW4/S7mdxQ06WcI/AAAAAAAAAQg/V52Js0DZ3dQ/s1600/Figure2Step2.jpg"><img style="cursor:pointer; cursor:hand;width: 400px; height: 276px;" src="http://3.bp.blogspot.com/_0Zi-ITihiW4/S7mdxQ06WcI/AAAAAAAAAQg/V52Js0DZ3dQ/s400/Figure2Step2.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5456565893192112578" /></a><div><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_0Zi-ITihiW4/S7mdxQ06WcI/AAAAAAAAAQg/V52Js0DZ3dQ/s1600/Figure2Step2.jpg"></a><br /><div><b>Step 3.</b> Add 1 to <i>r</i>².</div><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_0Zi-ITihiW4/S7mdw6DTg4I/AAAAAAAAAQY/VyCuEaVLW5c/s1600/Figure2Step3.jpg"><img style="cursor:pointer; cursor:hand;width: 400px; height: 276px;" src="http://2.bp.blogspot.com/_0Zi-ITihiW4/S7mdw6DTg4I/AAAAAAAAAQY/VyCuEaVLW5c/s400/Figure2Step3.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5456565887078466434" /></a></div><div><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_0Zi-ITihiW4/S7mdw6DTg4I/AAAAAAAAAQY/VyCuEaVLW5c/s1600/Figure2Step3.jpg"></a><br /><div><b>Step 4.</b> Invert (1+<i>r</i>²). We now have the target circle of radius <i>R</i>=1/(1+<i>r</i>²).</div><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_0Zi-ITihiW4/S7mdwY5-UVI/AAAAAAAAAQQ/ivgZW1xWSsQ/s1600/Figure2Step4.jpg"><img style="cursor:pointer; cursor:hand;width: 400px; height: 276px;" src="http://3.bp.blogspot.com/_0Zi-ITihiW4/S7mdwY5-UVI/AAAAAAAAAQQ/ivgZW1xWSsQ/s400/Figure2Step4.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5456565878180958546" /></a></div><div><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_0Zi-ITihiW4/S7mdwY5-UVI/AAAAAAAAAQQ/ivgZW1xWSsQ/s1600/Figure2Step4.jpg"></a><br /><div><b>Step 5.</b> Find the target ring related to the input ring ranging over [<i>r</i>, <i>r</i> + d<i>r</i>]. This yields a ring of width d<i>r</i>/(1+<i>r</i>²). The location of this ring depends on the relative progression rates of <i>r</i> and <i>r</i>² (I've not yet found a straightforward explanation for this determination).</div><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_0Zi-ITihiW4/S7mdwF0sjhI/AAAAAAAAAQI/6aMCqJDssZM/s1600/Figure2Step5.jpg"><img style="cursor:pointer; cursor:hand;width: 400px; height: 276px;" src="http://4.bp.blogspot.com/_0Zi-ITihiW4/S7mdwF0sjhI/AAAAAAAAAQI/6aMCqJDssZM/s400/Figure2Step5.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5456565873058549266" /></a></div><div><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_0Zi-ITihiW4/S7mdwF0sjhI/AAAAAAAAAQI/6aMCqJDssZM/s1600/Figure2Step5.jpg"></a><br /><div><b>Step 6.</b> Integrate d<i>r</i>/(1+<i>r</i>²) for <i>r</i> running over all space. For <i>r</i> becoming larger and larger, the summed area tends towards the area of a circle of radius 1. For the positive half plane, this corresponds to the <a href="http://functions.wolfram.com/Constants/Pi/07/01/01/0001/">π/2 value found analytically</a>. </div><div><div><a href="http://picasaweb.google.com/lh/photo/wko6Yic1eirAZkpT8yF-ZA?authkey=Gv1sRgCL3s4JSJk_zoTg&feat=embedwebsite"><img style="cursor:pointer; cursor:hand;width: 400px; height: 276px;" src="http://lh4.ggpht.com/_0Zi-ITihiW4/S7msJLd6xTI/AAAAAAAAARI/VBspl_LGgm0/s800/Figure_2_6.gif" /></a><br /></div><div><br /></div><div>The tricky step seems to be the way how to relate the progression between <i>r</i> and 1/(1+<i>r</i>²) in steps 5 and 6. One can verify for example the value of the integral at intermediate steps. For the integral from <i>r</i>=0 to 1, the value in the positive half plane must be π/4, which can be verified on the figure below.</div><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_0Zi-ITihiW4/S7nIly8MvyI/AAAAAAAAARU/AgFOjIwlWj0/s1600/Figure_2_6_IntermediateValue.jpg"><img style="cursor:pointer; cursor:hand;width: 400px; height: 276px;" src="http://3.bp.blogspot.com/_0Zi-ITihiW4/S7nIly8MvyI/AAAAAAAAARU/AgFOjIwlWj0/s400/Figure_2_6_IntermediateValue.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5456612975191047970" /></a></div><div>In order to gain more insight on π, it could be of interest to develop skills for this circular representation.</div></div></div>Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com0tag:blogger.com,1999:blog-382677439644973005.post-43218040680842895712010-03-28T21:14:00.000+02:002010-08-15T21:51:36.875+02:00Wallis product for nested equilateral triangles<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_0Zi-ITihiW4/S6-nrvnGdpI/AAAAAAAAAOM/j45nuxoFGHo/s1600/EquilateralTriangleArea.jpg"><img style="float:right; margin:0 0 10px 10px;cursor:pointer; cursor:hand;width: 378px; height: 400px;" src="http://2.bp.blogspot.com/_0Zi-ITihiW4/S6-nrvnGdpI/AAAAAAAAAOM/j45nuxoFGHo/s400/EquilateralTriangleArea.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5453762043725641362" /></a><br />I'm <a href="http://commonsensequantum.blogspot.com/2010/03/exhaustion-of-nested-squares-and-wallis.html">discovering the Wallis product</a> thanks to an approach involving geometry of areas. In order to relate that to what has already been written on the subject, here are some works which I would like to study in the coming time:<br /><ul><li>the appropriate parts of <a href="http://en.wikipedia.org/wiki/John_Wallis">John Wallis</a>' <a href="http://books.google.com/books?id=Z5w_AAAAcAAJ"><span style="font-style:italic;">Arithmetica Infinitorum</span></a> (is there a translation online ?)</li><li><span style="font-style:italic;"><a href="http://www.informaworld.com/smpp/content~content=a746869068&db=all">Some generalizations of the Wallis product</a><span class="Apple-style-span" style="font-style: normal;"> (1992)</span></span>, by <a href="http://www.napier.ac.uk/business-school/OurStaff/BusinessSchoolStaff/Pages/LesShort.aspx">Les Short</a>, containing a general formula for different starting ratios</li><li><span style="font-style:italic;"><a href="http://www.ep.liu.se/ea/lsm/2005/002/?__printable=1">An Elementary Proof of the Wallis Product Formula for pi</a><span class="Apple-style-span" style="font-style: normal;"> </span></span><span>(2005)</span>, by <a href="http://www.math.chalmers.se/~wastlund/">Johan Wästlund</a></li></ul><div>Maybe you have other suggestions on this subject.</div><div><br /></div><div>Meanwhile, I tried to draw equilateral triangles in the exhaustion scheme of nested circles. We start from the concentric rings of same area as described in Figure 4 of the <span style="font-style:italic;"><a href="http://commonsensequantum.blogspot.com/2010/03/variations-on-dividing-circular-area.html">Variations of circular area division into equal parts post</a></span>. Remember the inner circle has radius 1 and area π. Each subsequent Nth circle has radius square root of N (noted √N) and area Nπ. Then we can draw an equilateral triangle whose sides are tangent to the first circle. The neat thing about this procedure is that we can read out the pertaining numerical values right on the figure. Figure 1 shows us that the area of the equilateral triangle is 3√3. So the ratio between the outscribed equilateral triangle area and the circle area is equal to 3√3/π.</div><div><br /></div><div>By drawing equilateral triangles tangent to circles of incrementing area Nπ, we discover the nested equal area division for equilateral triangles: the area between each equilateral triangle is equal to 3√3 (see Figure 2, you can click on the figure to view it enlarged). Among other interesting features, this figure contains the table of 4. For each triangle outscribing a circle of radius √N, its vertices are located on the circle of order 4×N.</div><div><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_0Zi-ITihiW4/S68_AW7fK-I/AAAAAAAAANU/q9N-O6XL-Zo/s1600/NestedTriangles.jpg"><img style="cursor:pointer; cursor:hand;width: 387px; height: 400px;" src="http://2.bp.blogspot.com/_0Zi-ITihiW4/S68_AW7fK-I/AAAAAAAAANU/q9N-O6XL-Zo/s400/NestedTriangles.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5453646949156596706" /></a><br /></div><div><br /></div><div>If we outscribe alternately a circle, a triangle, a circle, a triangle, a circle, etc., the circles (and the vertices of the triangles) go through the subsequent powers of 4, see Figure 3. By changing the form of the triangle, we have access to other tables and powers. It seems there are plenty of properties that are hidden in this structure.</div><div><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_0Zi-ITihiW4/S69D4K3k9cI/AAAAAAAAANc/tFVn65PT69E/s1600/PowersOf4.jpg"><img style="cursor:pointer; cursor:hand;width: 387px; height: 400px;" src="http://2.bp.blogspot.com/_0Zi-ITihiW4/S69D4K3k9cI/AAAAAAAAANc/tFVn65PT69E/s400/PowersOf4.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5453652306038158786" /></a><br /></div><div>Now, we can try the Wallis exhaustion scheme on the nested circle and triangle structure. The procedure apparently goes something like:</div><div><b>Step 1</b>. Draw the initial circle and triangle with area ratio 3√3/4π.</div><div><a href="http://3.bp.blogspot.com/_0Zi-ITihiW4/S6-hNpoLlwI/AAAAAAAAANk/dLctyVGpRD4/s1600/Figure4Step1.jpg"><img src="http://3.bp.blogspot.com/_0Zi-ITihiW4/S6-hNpoLlwI/AAAAAAAAANk/dLctyVGpRD4/s400/Figure4Step1.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5453754929653716738" style="cursor: pointer; width: 329px; height: 400px; " /></a></div><div><b>Step 2</b>. A better fit would be to inscribe both figures in higher orders of the nested structure. But the area of the triangle is too small with respect to the circle. So we must enlarge the area of the triangle more than the circle. We enlarge the area of the triangle by 3 while leaving unchanged the area of the circle. This is illustrated in the figure below.</div><div><a href="http://3.bp.blogspot.com/_0Zi-ITihiW4/S6-hNw0qdjI/AAAAAAAAANs/5nHvSN5d2tI/s1600/Figure4Step2.jpg"><img src="http://3.bp.blogspot.com/_0Zi-ITihiW4/S6-hNw0qdjI/AAAAAAAAANs/5nHvSN5d2tI/s400/Figure4Step2.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5453754931585119794" style="cursor: pointer; width: 329px; height: 400px; " /></a></div><div><b>Step 3</b>. But now the area of the triangle is too large with respect to the circular area. So for this fit, we must enlarge the area of the triangle a little less than the area of the circle. We therefore enlarge the area of the triangle by 3, while enlarging the area of the circle by 4, as illustrated below.</div><div><a href="http://2.bp.blogspot.com/_0Zi-ITihiW4/S6-hOMTw5LI/AAAAAAAAAN0/MYpXDKmI884/s1600/Figure4Step3.jpg"><img src="http://2.bp.blogspot.com/_0Zi-ITihiW4/S6-hOMTw5LI/AAAAAAAAAN0/MYpXDKmI884/s400/Figure4Step3.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5453754938963322034" style="cursor: pointer; width: 329px; height: 400px; " /></a></div><div><b>Step 4</b>. But now the area of the triangle is too small with respect to the circular area. So for the next fit, we must enlarge the area of the triangle a little more than the area of the circle. We therefore enlarge the area of the circle by 6, while enlarging the area of the circle by 5, as illustrated below.</div><div><a href="http://4.bp.blogspot.com/_0Zi-ITihiW4/S6-hOXRNuqI/AAAAAAAAAN8/Vp11A-EDpb0/s1600/Figure4Step4.jpg"><img src="http://4.bp.blogspot.com/_0Zi-ITihiW4/S6-hOXRNuqI/AAAAAAAAAN8/Vp11A-EDpb0/s400/Figure4Step4.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5453754941905418914" style="cursor: pointer; width: 329px; height: 400px; " /></a></div><div>This exhaustion scheme that makes the area of equilateral triangles alternately larger, smaller, larger, smaller, etc... than the circular area can be written as:</div><div><a href="http://1.bp.blogspot.com/_0Zi-ITihiW4/S6-mqwd0xMI/AAAAAAAAAOE/mEmf4Bb33-4/s1600/3Sqrt3Formula.jpg"><img src="http://1.bp.blogspot.com/_0Zi-ITihiW4/S6-mqwd0xMI/AAAAAAAAAOE/mEmf4Bb33-4/s400/3Sqrt3Formula.jpg" border="0" alt="" id="BLOGGER_PHOTO_ID_5453760927263671490" style="cursor: pointer; width: 400px; height: 54px; " /></a></div><div>This formula corresponds to an infinite product approximating π, which I've found in a French book <a href="http://www.eveandersson.com/pi/le-fascinant-nombre-pi">Le fascinant nombre π</a> by <a href="http://fr.wikipedia.org/wiki/Jean-Paul_Delahaye">Jean-Paul Delahaye</a> but I couldn't find it mentioned elsewhere. As approximation method, it is about twice as fast as Wallis' original product. This equilateral triangle exhaustion product is probably a special case in a large family of Wallis products, each of which pertaining to a particular geometrical configuration from which one tries to approximate π. The starting and recurrence conditions depend on the form and the initial area ratio between the polygon and the circle. The rules governing this exhaustion scheme in the general case still elude me.</div>Arjen Dijksmanhttp://www.blogger.com/profile/09450431291713605237noreply@blogger.com0