4D-Volume of a 4D-Sphere - Page 2

davieddy · 2007-07-05, 08:27

Quote:

Originally Posted by Wacky

...
This is the point that some are missing.

Not me Guvnor

davieddy · 2007-07-05, 09:00

I could clarify my "head scratching" post
(which apart from lazy notation was
essentially correct) but I think Mally should
first observe that in 3D, a point (x,y,z) distance R
from the origin satifies R^2= x^2+y^2+z^2.
This follows from successive applications of
Pythagoras' Theorem to two right-angled triangles.

David

mfgoode · 2007-07-05, 15:13

Quote:

Originally Posted by davieddy

I could clarify my "head scratching" post
( but I think Mally should
first observe that in 3D, a point (x,y,z) distance R
from the origin satifies R^2= x^2+y^2+z^2.
This follows from successive applications of
Pythagoras' Theorem to two right-angled triangles.

David

Thank you David but I already meant that.

Quote:

Originally Posted by Mally

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

Mally

davieddy · 2007-07-05, 15:53

Quote:

Originally Posted by mfgoode

Thank you David but I already meant that.
Mally

I noticed that, and you went up in my
estimation my dear old friend

David

davar55 · 2007-07-05, 20:53

The references given in previous posts give the formulas
for the sequence as n increases, as well as a derivation.

The OEIS has several entries related to this sequence.

Also: if the dimension is considered as a REAL (as opposed
to integral) variable, then these coefficients of r^n have
a maximum value in dimension n = 5.25694640486057678013..
(according to OEIS entry A074455).

The related surface area series also has a maximum value,
in dimension n = 7.256946... (according to OEIS entry A074457).

[I suspect there's an OEIS calculational error. The two values
given for these dimensions differ by almost exactly 2, but NOT
exactly. How could they agree in their decimal expansions for 64
digits and then diverge, as the two OEIS entrys would indicate?
I think that's unlikely.]

alpertron · 2007-07-05, 22:16

As you suggest above their difference is exactly 2.

For an sphere of radius R = 1 we get:

$S(n) = 2\frac{\pi^{n/2}}{\Gamma(n/2)}$
$V(n) = \frac{S(n)}{n}$

The maxima for these functions are the maxima for the logarithm of these functions so we will compute them:

$\log S(n) = \log 2+(n/2)\log \pi\,-\,\log \Gamma(n/2)$

$\log V(n) = \log 2+n*(1/2)\log\pi\,-\,\log \Gamma(n/2)-\log n$

$\log V(n) = \log 2+n*(1/2)\log\pi\,-\,\log \Gamma(n/2)-\log (n/2) - \log 2$

$\log V(n) = n*(1/2)\log\pi\,-\,\log \Gamma(n/2+1)$

Let $q=(1/2)\log\pi$

$\log S(n) = \log 2 + nq - \log \Gamma(n/2)$
$\log V(n) = nq - \log \Gamma(n/2+1)$

Given that $\frac{d\log \Gamma(n)}{dn} = \psi(n)$ (the digamma function) we get:

$\frac {d\log S(n)}{dn} = q - (1/2)\psi(n/2)$
$\frac {d\log V(n)}{dn} = q - (1/2)\psi(n/2+1)$
$\frac {d\log V(n)}{dn} = q - (1/2)\psi((n+2)/2)$
$\frac {d\log V(n-2)}{dn} = q - (1/2)\psi((n/2)$
$\frac {d\log V(n-2)}{dn} = \frac {d\log S(n)}{dn}$

So we don't know the value of n, but it is clear that the difference between the zeros of the derivatives should be 2.

2007-07-05, 22:16	#17
alpertron Aug 2002 Buenos Aires, Argentina 1366₁₀ Posts	As you suggest above their difference is exactly 2. For an sphere of radius R = 1 we get: $S(n) = 2\frac{\pi^{n/2}}{\Gamma(n/2)}$ $V(n) = \frac{S(n)}{n}$ The maxima for these functions are the maxima for the logarithm of these functions so we will compute them: $\log S(n) = \log 2+(n/2)\log \pi\,-\,\log \Gamma(n/2)$ $\log V(n) = \log 2+n(1/2)\log\pi\,-\,\log \Gamma(n/2)-\log n$ $\log V(n) = \log 2+n(1/2)\log\pi\,-\,\log \Gamma(n/2)-\log (n/2) - \log 2$ $\log V(n) = n(1/2)\log\pi\,-\,\log \Gamma(n/2+1)$ Let $q=(1/2)\log\pi$ $\log S(n) = \log 2 + nq - \log \Gamma(n/2)$ $\log V(n) = nq - \log \Gamma(n/2+1)$ Given that $\frac{d\log \Gamma(n)}{dn} = \psi(n)$ (the digamma function) we get: $\frac {d\log S(n)}{dn} = q - (1/2)\psi(n/2)$ $\frac {d\log V(n)}{dn} = q - (1/2)\psi(n/2+1)$ $\frac {d\log V(n)}{dn} = q - (1/2)\psi((n+2)/2)$ $\frac {d\log V(n-2)}{dn} = q - (1/2)\psi((n/2)$ $\frac {d\log V(n-2)}{dn} = \frac {d\log S(n)}{dn}$ So we don't know the value of n, but it is clear that the difference between the zeros of the derivatives should be 2. Last fiddled with by alpertron on 2007-07-05 at 22:31*

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2007-07-05, 09:00	#13
davieddy "Lucan" Dec 2006 England 2·3·13·83 Posts	I could clarify my "head scratching" post (which apart from lazy notation was essentially correct) but I think Mally should first observe that in 3D, a point (x,y,z) distance R from the origin satifies R^2= x^2+y^2+z^2. This follows from successive applications of Pythagoras' Theorem to two right-angled triangles. David

2007-07-05, 20:53	#16
davar55 May 2004 New York City 10212₈ Posts	The references given in previous posts give the formulas for the sequence as n increases, as well as a derivation. The OEIS has several entries related to this sequence. Also: if the dimension is considered as a REAL (as opposed to integral) variable, then these coefficients of r^n have a maximum value in dimension n = 5.25694640486057678013.. (according to OEIS entry A074455). The related surface area series also has a maximum value, in dimension n = 7.256946... (according to OEIS entry A074457). [I suspect there's an OEIS calculational error. The two values given for these dimensions differ by almost exactly 2, but NOT exactly. How could they agree in their decimal expansions for 64 digits and then diverge, as the two OEIS entrys would indicate? I think that's unlikely.]