Have we discovered or invented maths?

[metrictensor]

wow! cool! you are way ahead of me. i found the formulas for dimension 4,5 with ratios 32/pi^2, 60/pi^2.

this is neat!

Yeah, I wrote you I realized that every two terms the \pi in the result goes up by a power. This is obvious when we look at the general solution given by the formula. BTW, how did you get your results for n=4 and n=5?

yeah, this is cool. I wonder if the sum of general equation I gave converges as N -> infinity. I demonstrated that the demononator is always bigger than the numerator so at least we know it doesn't blow up. Let's see where all this goes.

 

[mathwonk]

(i looked on the web for some formulas) what if we look at you ration a different way?

i.e. instead of V1/V2 = A1/A2, look at it as V1/A1 = V2/A2. i.e. look at the ratio of the volume/area for various figures. it seems you get something simpler then.

i.e. in dimension n you seem to get V/A = r/n. do you get this?


I did not see offhand the limit of your formulas, but they should be interesting.

 

[Berislav]

i.e. in dimension n you seem to get V/A = r/n. do you get this?

If I may add...
It seems to me that that would simply follow from V/V'=A/A'. That is it would follow if we define a n-dimensional cube to be a n-dimensional space bounded by 2n n-1 dimensional boarders (i.e, two boarders for each dimension, which bind it in every other possible direction, hence dimension n-1).
The volume of the n-cube would be simply
V_n=a^n

and the surface area should be:

A_n=2na^{n-1}

and a=2r.

 

[metrictensor]

(i looked on the web for some formulas) what if we look at you ration a different way?

i.e. instead of V1/V2 = A1/A2, look at it as V1/A1 = V2/A2. i.e. look at the ratio of the volume/area for various figures. it seems you get something simpler then.

i.e. in dimension n you seem to get V/A = r/n. do you get this?


I did not see offhand the limit of your formulas, but they should be interesting.

I did get r/n. The formulas I used for the four volumes/surface areas are (I looked these up on the internet)

<br /> V_s=\frac{R^N \pi^{N/2}}{(N/2)!}

A_s=\frac{NR^{N-1}\pi^{N/2}}{(N/2)!}

V_c=2^NR^N

A_c=2^NNR^{N-1}

The one thing I am not 100% certain of is the condition that L=2R for all N. I think this is true but can not visualize or determine how I would prove this for N>3.

 

[mathwonk]

well lengths should be measurable by slicing down to lower dimensions, i.e. for a cube we can slice and get a square and still measure the length of the square. so the length should be the same as in lower dimensions i.e. L = 2R.

I.e. R is the distance from the center to the side, and this should be measurable on a slice? I also agree with Berislav's computation for a cube. of course the point of metrictensor's result is that the same ratio holds for other figures as well, such as spheres, and possibly other shapes. (I checked it for a triangle, and am pretty confident the shape does not affect the ratio in general.)

I.e. the simpler ration r/n still leaves one obliged to calculate r for a complicated shape. it is only easy for cubes and spheres.

for instance, in a tetrahedron, r is the distance from the center of the tetrahedron to the center of one face, i.e. its the radius of an inscribed sphere as metrictensor originally said.

this same slicing principle seems to apply also to spheres. for example one obtains the area of a unit circle by integral calculus, from the length of a slice.

i.e. for a half circle, one integrates, for x going from 0 to 1, the length of a slice lying a distance x away from the center of the circle. by pythagoras that length is 2sqrt(1 -x^2).

the antiderivative is then (after some cazlculation)

xsqrt(1-x^2) + arcsin(x), which evaluated from 0 to 1 gives pi/2, or the area of 1/2 a semicircle.


one gets the volume of a sphere the same way, integrating the area formual for a circle, i.e. of a slice of a unit sphere, where again the radius of this circular slice is sqrt(1-x^2), so the area of this slice is pi(1-x^2).

this is far easier to integrate, the antiderivative being pi(x - x^3/3), and hence the integral from x = 0 to x=1 equals pi(2/3), the volume of the unit hemisphere.

to do the volume of a 4 dimensional sphere one then considers half of it again, the radius of the spherical splice at x being again sqrt(1-x^2), but now the volume of this slice (itself a three dimensional sphere) is (4pi/3)(1-x^2)^(3/2), i.e. 4pi/3 R^3 where R = sqrt(1-x^2).

to integrate this i needed to find the antiderivative of cos^4, which i was unable to quite do in my head in the car on the way home. let me see if i can get it now with pencil and paper... ok, i got ther ight answer (pi)^2/2, but it was a pain to do the antiderivative of cos^4.

i got the antiderivative of (1-x^2)^(3/2) to be something like:


xsqrt(1-x^2)(1-2x^2)/8 + xsqrt(1-x^2)/2 + 3arcsin(x)/8. yaak! but it gave the right answer.

[i hope i remember not to assign this to a class, or perhaps I SHOULD assign the sort of unscripted thing one actually runs into in "real life"?]

I did this because it feels somehow unfulfilling to look these up on the internet as I did yesterday, and frankly I don't fully trust formulas (even my own) given without proof.

 

[mathwonk]

well anyway, i did notice that the area of a circle is r/2 times the circumference formula, but I do not recall noticing before that this also holds for a square, nor that the area formula for a sphere is 3/r times the volume formula, much less that this also holds for the cube, and conjecturally every other regular solid.

say since there are only a few (5?) regular polyhedra in three dimensions, i wonder how many there are in 4 dimensions? [tetrahedron, cube, dodecahedron, icosahedron, octahedron, in 3 dimensions. by duality there should be an even number, but the tetrahedron is "self dual". octahedron - cube, dodecahedron - icosahedron, tetrahedron - tetrahedron, where faces are interchanged with vertices.]

and what about non regular solids? what is the ratio there?

say this holds aalso for any triangle, not just a regular one. and it reminds me of a famous theorem i never learned in school, due to Heron?

well Heron's theorem is a little different but seems to follow from this formula plus Pythagoras.

i.e. this principle shows that the area of any triangle equals the radius of an inscribed circle times the semi perimeter, i.e. half the perimeter times the radius...well anyway.

the key to the ratio principle seems to be not strict regularity but that there exist an inscribed sphere, i.e. a sphere inside which is tangent to every face, not like arectangle for example, but any triangle works. ?

 

[mathwonk]

here is a "proof" of the basic phenomenon found by metrictensor, and shows it is a very fundamental relation, essentially the most fundamental fact underlying most computations of area and volume, namely the rate of change of these quantities, as codified in calculus.

i.e. an n - dimensional volume formula for a figure with a center, has form V = Cr^n, where C is some constant.

if we assume that the distance from the center to each face is the same, [as is true whenever there exists a truly inscribed sphere], then the method of derivatives, shows that the area formula for this figure is the drivative of the volume formula wrt the radius, i.e. A = nCr^(n-1). hence the ratio V/A = r/n.

since this holds for all such symmetrical shapes, we also have V/A = V'/A' for sphere and cube, and hence also V/V' = A/A', as metrictensor originally observed, and the same holds in all dimensions.

metrictensor, i think this is one of the most basic facts about areas and volumes anyone has ever shown me. i.e. it holds over a very wide variety of cases, it is simple, and it lies at the foundation of the subject. also it has computational uses to render actual calculations easier.

congratulations! that's really nice! in many years of teaching, very few people have ever shown me something simultaneously this simple and interesting, and which held up in so many new cases stretching so far from its beginnings.

I think you have a gift.

best wishes

 

[metrictensor]

My next problem is to see if there is a relationship between an ellipse in a rectangle.

 

[mathwonk]

good idea!

if you find a simple relationship here it will be even more interesting, as there is no simple formula for the arclength of an ellipse.

i.e. in calculus books you will notice that the area formulas for an ellipse are given but not the length formulas, although they may refrain from saying why not.

 

[metrictensor]

good idea!

if you find a simple relationship here it will be even more interesting, as there is no simple formula for the arclength of an ellipse.

i.e. in calculus books you will notice that the area formulas for an ellipse are given but not the length formulas, although they may refrain from saying why not.

Yeah, I have been looking into this and found out that you can only approximate the arc length of an ellipse. There is an infinite series but I assume it does not converge. The strange thing is that if you look at an ellipse it certainly has an arc length so why can't we find a closed solution?

 

[selfAdjoint]

Yeah, I have been looking into this and found out that you can only approximate the arc length of an ellipse. There is an infinite series but I assume it does not converge. The strange thing is that if you look at an ellipse it certainly has an arc length so why can't we find a closed solution?

The form of the integral you get when you set up the problem of the arc length of an ellipse was not one the 17th century mathematicians could solve, and it remained that way for a century. In the early 19th century Abel and Jacobi inverted the integral and discovered the elliptic functions. so called from their relation to this very problem. They are new functions, different from the exponentials and circular and hyperbolic functions known up to that time. The theory of the elliptic functions was very big in the 19th and 20th century, and they still come into a lot of research. So you can express the arc length of an ellipse in terms of these functions but not in terms of more traditional functions.

 

[mathwonk]

there is a fascinating exposition of this in chapter one, volume one, of the book "topics in complex analysis" by C. L. Siegel. An Italian count named Fagnano showed that although he could not calculate the length of arc of an ellipse from a beginning point to a later point, he could do something interesting: given one later point, although he could not calculate the arclength that far, he COULD find another point whose arclength was twice as great as the first (unknown) arclength.

so he could not calculate arclengths, but he could double arclengths.

this led eventually to the theory of elliptic functions and to the group law on an elliptic curve,

i.e. euler showed how to generalize this and find, given two points, another point such that the arc length to that third point was the sum of the arclengths of the first two, points, although he could not calculate any of the lengths explicitly!

this is an incredible story. and i thought with your talent for noticing nice things you just might see something.

 

[selfAdjoint]

Yes I worked through Fagnano's original paper once upon a time, it's very short. A really neat result. After Euler, Legendre spent years discovering all kinds of relationships in the elliptic integrals and he wrote a treatise on them, in two dense volumes. It was these properties that Legendre had explicated, which Abel and Jacobi used to develop the identities of their elliptic functions. It was remarked that Legendre could have discovered them but didn't. The point is the analogy between the arcsine integral
u = \int_0^x \frac{dt}{\sqrt{1 - t^2}}
which defines u as the inverse sine of x, and x = sin u, and the elliptic integral
u = \int_0^x\frac{dt}{\sqrt{1 - t^2}\sqrt{1 - k^2t^2}}
which defines u as the inverse of an elliptic function: x = sn u depending on a parameter k.

Lengendre was so focussed on the integrals that he couldn't see to invert them. Commenting on this later, Jacobi suggested "Man muss immer umkehren", or as I clumsily translate it, "Ever evert".

 

[mathwonk]

have you ever read galois works? apparently in the same letter in which galois wrote out the theory of groups and applications to solving polynomials equations, he also wrote out a theory of abelian integrals. for some reason this is essentially never mentioned in the literature but seems to predate riemanns work on the topic by some 25-30 years. on the other hand i do not recall galois using the idea of topology, i.e. connectivity of surfaces, to study these integrals as riemann did.

thus galois did know there was a number p asociated to an abelian integral or algebraic curve, i.e. a number p which counted the number of holomorphic differentials on the curve, but i think he did not link it with the number 2p of loop cuts needed to dissect the surface as riemann did. any comments? (i do not have access to galois here now.)

so, if this is right, the genus p precedes riemann, but riemann first linked the genus to the topology of the surface.

 

[selfAdjoint]

I can't help but adding two facts about elliptic function. One, they have two periods (the circular functions like sine and cosine of course have one, from 1 to 2pi). In the complex plane where they are most fully realized these are two complex numbers, think of them as vectors of different length in different directions. Then they can be regarded as two sides of a parallelogram, and this basic cell will be replicated - tiled - infinitely many time across the complex plane with the same values being taken at corresponding points in each iteration. Or if you prefer you can define them as single valued on a torus (their Riemann surface), and remarkably, their properties can be derived just from the fact that they can be so defined. Well maybe not so remarkable, given that the properties of the sine and cosine can be derived from their relation to the circle.

Two, they have LOTS of applications in engineering and physics.

 

[mathwonk]

this brings me back to an interesting fact about the torus i.e.; the formulas for its area and volume. i.e. take a look at the volume formula for a torus and also for its area, and notice how the area is related to the volume as a partial derivative.

there are also similar relations for the area and volume of a cylinder. maybe there is something like that for an ellipse, (but probably not).

in the interest of generalizing elliptic funtions, there are some interesting functions of several variables called theta functions, that have amazing properties discovered by riemann.