*x*and

*y*. The simplest binomial is the sum of those two terms

*x+y*. Powers of this binomial (

*x+y*)

*are ruled by the so-called binomial theorem, which states that the coefficients of the expansion terms*

^{n}*x*are given by the Pascal triangle entries. So we have:

^{p}y^{q}(

*x+y*)

*= 1*

^{0}(

*x+y*)

*= 1.*

^{1}*x*+ 1.

*y*

(

*x+y*)*= 1.*^{2}*x**+ 2.*^{2}*xy*+ 1.*y*^{2}(

*x+y*)*= 1.*^{3}*x**+ 3.*^{3}*x*^{2}*y*+ 3.*x**y**+ 1.*^{2}*y*^{3}*x+y*)

*= 1.*

^{4}*x*

*+ 4.*

^{4}*x*

*+ 6.*

*y*^{3}*x*

^{2}*y*

*+ 4.*

^{2}*x*

*y*

*+ 1.*

^{3}*y*

^{4}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 nice geometric explanation is given at the www.mathaware.org site.

Geometric proof of square binomial (original at The Geometry of the Binomial Theorem, Math Awareness Month site) |

Geometric proof of cubic binomial (original at The Geometry of the Binomial Theorem, Math Awareness Month site) |

*x*. 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 hypercubes and hyperparallelepipeds given by the expansion terms, so an Euclidean geometric representation for those higher degree binomials seems impossible.

^{p}y^{q}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

*x*is just a number, but it is also a square

^{4}*x*= (

^{4}*x*)

^{2}*. 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*

^{2}*x+y*is normalized to 1 (for arbitrary

*x+y*, one just has to rescale the figure each time, the pattern remains the same).

For the binomial (

*x+y*)

*= 1.*

^{1}*x*+ 1.

*y*, we can divide a unit square into two rectangles, one rectangle of area

*x*and one rectangle of area

*y*, see Figure 1.

For the binomial (

*x+y*)

*= 1.*

^{2}*x*

*+ 2.*

^{2}*xy*+ 1.

*y*

*, we can divide the unit square into two squares, one of area*

^{2}*x*

*and one of area*

^{2}*y*

*, plus two rectangles of area*

^{2}*xy*, see Figure 2.

For the binomial (

*x+y*)

*= 1.*

^{3}*x*

*+ 3.*

^{3}*x*

^{2}*y*+ 3.

*x*

*y*

*+ 1.*

^{2}*y*

*, we can divide each of the previous squares and rectangles into proportions*

^{3}*x*and

*y*, giving eight rectangles, one of area

*x*

*, one of area*

^{3}*y*

*, plus three rectangles of area*

^{3}*x*

^{2}*y*and three rectangles of area

*x*

*y*

*, see Figure 3.*

^{2}For the binomial (

*x+y*)

*= 1.*

^{4}*x*

*+ 4.*

^{4}*x*

*+ 6.*

*y*^{3}*x*

^{2}*y*

*+ 4.*

^{2}*x*

*y*

*+ 1.*

^{3}*y*

*, we can again divide each of the previous squares and rectangles into proportions*

^{4}*x*and

*y*, giving one square of area

*x*

*, one of area*

^{4}*y*

*, four squares and two rectangles (giving six) of area*

^{4}*x*

^{2}*y*

*, four rectangles of area*

^{2}*x*

^{3}*y*and four rectangles of area

*x*

*y*

*, see Figure 4.*

^{3}And we could go further indefinitely, doodling areas of incrementing powers, just in 2D, without any reference to unintuitive hyperspaces.

It seems that you've projectivized the volumes to create a sort of level curve graph in R^2. It's very nicely done and a very keen insight. There are so many visual subtleties in math.

ReplyDeleteI think you lost me:-) I don't exactly see how I have reduced the cubes and parallelepipeds to areas. I rather thought of x^3 as (x reduced by x reduced by x) for x<1, or (x increased by x increased by x) for x>1.

ReplyDeleteThere are those who say it's difficult for us to visualize dimensions n>4. I as driving home last night and had this same insight and my search for similar thought 'out there' brought me here. Actually, this representation of higher dimensions that result from higher degree binomials can be used to symbolically represent the hyperspaces and for any system where variables relate via addition/subtraction multiplication/division. Take the colorscale and make solutions < 0 one range; solutions > 0 another. I foresee applications for understanding the complexity of solution space if this is done algorithmically; undefined solutions painted black. moreover, a set a rules could be made such that solution graphs could be related to each and mathematical manipulations of solution spaces could be projected into and executed in this graph space; and then the mathematical solution could be read from the graph. Wolfram might be interested. -lifebiomedguru

ReplyDelete"each" above should read "each other".

ReplyDeleteYes, if someone could develop this way of viewing higher dimensions, it would but great. Maybe Wolfram indeed?

ReplyDeleteYes, viewing would be cool, and the application I suggest goes further; Also can imagine a video of problems being solved by interacting Dijksman graphs. The problems stated as equations have equivalent graphical representations; their relationships in Dijksman graph space can be studied, and their manipulations in Dijksman graph space would be very interesting to watch. Maybe some simple well-known elementary polynomial problems being solved would be nice to see (and they would suggest a set of consistent operations in this space). Some intractable problems could be handily dealt with and here is a conjecture: With a set of consistent rules for abstraction of operations in place in graph space, a solution rendered in Dijksman graph space is equivalent to its proof. NP-Completeness is bunk; electrical charges "find" the shortest path because they are at once an abstraction of the problem, and the solution. You're onto to something profound here, I think. We use graphs to SEE unsolvable problems, why not use your to SOLVE them. - lifebiomedguru

ReplyDelete"electrical charges "find" the shortest path " should read "electrical charges "find" the shortest path through a submerged maze in a salt solution" - lifebiomedguru

ReplyDeleteDijksman graphs? My parents would love it ;-)

ReplyDeletecan we draw geometry of (x+y)^n ; n>=4 ; as drawn for (x+y)^3 in fig-"Geometric proof of cubic binomial (original at The Geometry of the Binomial Theorem, Math Awareness Month site) " above????

ReplyDelete@Shantanu. Yes of course we can draw 3D volumes for (x+y)^n, n=4, 5, 6,... like we can draw 2D areas for those binomial expansions. We just have to divide the volumes into x/(x+y) and y/(x+y).

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