4D-Volume of a 4D-Sphere
 

I saw this sequence problem on a "high IQ" test on the net:

1, 2r, pi r[sup]2[/sup], 4pi r[sup]3[/sup]/3, ___

So what IS the next term?

:unsure::question:

If Wikipedia is to be believed (and if I'm interpreting it correctly), [tex]\begin{matrix} \frac{1}{2} \end{matrix} \pi^2 r^4[/tex] is the hypervolume of a 3-sphere (aka 4D-Sphere).
[URL]http://en.wikipedia.org/wiki/3-sphere[/URL]

[spoiler]Using multiple integration I found the same result that is shown at [url]http://en.wikipedia.org/wiki/Sphere[/url] about 20 years ago.[/spoiler]

One can easily generalize the n-dimensional integration to derive a simple pair of formulae for all the odd and even-dimensional sphere volumes, respectively.

[url]http://mathworld.wolfram.com/Hypersphere.html[/url]

I did this ~30 years ago, and am scratching my head to
remember the jist of it.
V(n,r^2) means nVolume of ndimensional sphere radius r
Something like:
dV(n,r^2)=2*pi*r*V(n-2,R^2-r^2)dr
Turning the handle relates V(N,R) to V(N-2,R)

I sound more like Mally every day:smile:

30 years ago !
 

[QUOTE=davieddy;109515]I did this ~30 years ago, and am scratching my head to
remember the jist of it.
~
I sound more like Mally every day:smile:[/QUOTE]

:wink: If scratching your head wont help you remember, try something else :lol:

Why should x_1^2 +x_2^2 + x_3^2+ x_4^2 =R^2?.

I say that the geometry will change!

If the formula is true for n=3 it does not mean it will be true for n=4,5,6..... Its like saying x^2 + y^2 =z^2 is true but that does not mean that x^3 +y^3 = z^3 for integers.

Mally :coffee:

The hypersphere of n dimensions is expressed by:

(x[sub]1[/sub]-X[sub]1[/sub])[sup]2[/sup] + (x[sub]2[/sub]-X[sub]2[/sub])[sup]2[/sup] + ... + (x[sub]n[/sub]-X[sub]n[/sub])[sup]2[/sup] = R[sup]2[/sup]

This is because it is defined as the set of points whose distance to the point (X[sub]1[/sub], X[sub]2[/sub], ..., X[sub]n[/sub]) is R.

For example, suppose that n=4 and the central point is (0, 0, 0, 0) so we get:

x[sub]1[/sub][sup]2[/sup] + x[sub]2[/sub][sup]2[/sup] + x[sub]3[/sub][sup]2[/sup] + x[sub]4[/sub][sup]2[/sup] = R[sup]2[/sup]

If x[sub]4[/sub] = 0 you should have the projection of the hypersphere on the standard 3-dimensional space, i.e. a sphere.

Replacing you get the formula for the standard sphere:

x[sub]1[/sub][sup]2[/sup] + x[sub]2[/sub][sup]2[/sup] + x[sub]3[/sub][sup]2[/sup] = R[sup]2[/sup]

If you use other exponents as you talked about in another thread this would not work.

[quote=mfgoode;109523]:wink: If scratching your head wont help you remember, try something else :lol:

Why should x_1^2 +x_2^2 + x_3^2+ x_4^2 =R^2?.

Mally :coffee:[/quote]

Because of your friend Pythagoras?

multigrades!
 

[QUOTE=davieddy;109550]Because of your friend Pythagoras?[/QUOTE]

:smile:
Perhaps: Py. only dealt with triplets and these were not for any integers but certain ones.

Here we are dealing with multigrades (that's what J.S. Madachy calls them) like 3^3 +4^3 +5^3 = 6^3 or
1^3 + 6^3 + 8^3 =9^3 but for the exponent 2 which are exceptions to the general rule.

However as long as we have the right angled triangle the formula given holds.
I have to concede that!

Mally :coffee:

[QUOTE=alpertron;109527]A hypersphere of n dimensions … is defined as the set of points … [/QUOTE] equidistant from the center.

This is the point that some are missing. When expressed in Cartesian coordinates, the second power is strictly a result of the distance metric. Anything else is a "red herring".