Derivatves of curves of hyper-sphere volumes and areas See wikipedia "N-sphere".
I need this differentiated with respect to "n", not "r".
This is so I can find when the slope of the curve with respect to n(treated as x-axis), the number of dimensions, equals zero.
Surface Area:
S(n-1)(R) = (R^(n-1)) * (n*(π^(n/2))) / Γ((n/2)+1)
V(n)(R) = (R^n) * (π^(n/2) / Γ((n/2)+1)
You could use the special case where R=1.
I need:
S'(n-1)
and
V'(n)
I need the derivatives of the terms involving the Gamma function to be continuous solutions, not integer solutions, because I'm trying to find the fractional dimension maxima for surface area and volume.
I know surface area maxes between n=7 and n=8, but not exactly on either of them, and I know Volume maxes somewhere around n=5.
Truth be told, the only help I really need is a continous derivative of the Gamma function:
F(n) = Γ((n/2)+1)
So I need:
F'(n) = Γ'((n/2)+1)
It can't be integers only, because the real solution is not an integer value.
Put in browser:
blazelabs.com/f-u-hds.asp
Here it is shown that the Surface Area Maxes at a fractional dimension just above 7 dimensions, which would occur when the derivative is zero, but he didn't show how he proved that, since it requires a continuous derivation of the Gamma function, which I have no idea how to obtain.
I searched for it and none of the solutions on here are continuous either.
This is ridiculously complex, so if anyone answers Thanks ahead of time.
 A: How?
What happened to the Gamma Function from V(N) if Sn = V'n?
By the quotient rule, the derivative of the Γ((n/2)+1) function should end up in the numerator of the solution, and the denominator should have (Γ((n/2)+1))^2.
by inspection, It should be something like:
V'(n) = {{(n*R^(n-1)π^(n/2) + (R^n)(1/2)*π^(n/2)) * Γ((n/2)+1))} - {((R^n) * (π^(n/2)* Γ'((n/2)+1)}} / (Γ((n/2)+1))^2)
The derivative of the entire top is obtained via the Product Rule, and the Chain Rule. This must then be plugged back into the Quotient Rule, which leaves a copy of "Gamma", Γ((n/2)+1) , on the left side of the numerator, and leaves a copy of "Gamma Prime", Γ'((n/2)+1), then multiplied by the original numerator. That completes the numerator, except for the solving Γ'((n/2)+1) part.
Then you have "Gamma Squared," Γ((n/2)+1))^2, for the new denominator.
How does all that simplify back to Sn as a general rule?
S(n) = 4πR^2
V(n) = (4/3)πR^3
Does the derivative of Gamma equal Gamma?!
The specific instance for a 3-sphere works, obviously, I know that, but the general formula does not appear to produce that result.
I feel like I'm going to cry or something.
Oh well, thanks for the help
