Easy one

There are several different ways to interpret this, so I'm curious to see what people come up with:

What is the average distance from a point in or on a sphere of radius r to its center?

Given any radial line in this sphere (or would that be ball?), the average - or mean - of the set of distances from the center along that raduis is clearly r/2.

Is this a trick question? 😕

-Ray.

Originally posted by royalchicken
There are several different ways to interpret this, so I'm curious to see what people come up with:

What is the average distance from a point in or on a sphere of radius r to its center?

Hmm, If the point is selected to give an equal chance of the point being at any position within the sphere, then:

The area of a shell of radius s within the sphere is 4.pi.s^2

and the contribution of this shell of pts to the total sphere volume is
(4.pi.s^2)/(4/3.pi.r^3) = 3.s^2/r^3

to get the mean distance s we need to sum each possible s, weighted by the chance of getting that s
i.e we need to integrate 3.s^3/r^3 over all s between 0 and r
the answer is then 3/4(r^4/r^3)

= 3r/4

Well done, Iamatiger, your answer was the one I had in mind. Now, notice that for circles, the average distance is 2r/3, for lines, r/2, and for points obviousl 0. So can someone prove/disprove that for an n-dimensional sphere of radius r, and a Pythagorean interpretation of distance, the average distance from center to a point in the n-sphere is nr/(n+1) ?

Originally posted by royalchicken
Well done, Iamatiger, your answer was the one I had in mind. Now, notice that for circles, the average distance is 2r/3, for lines, r/2, and for points obviousl 0. So can someone prove/disprove that for an n-dimensional sphere of rad ...[text shortened]... age distance from center to a point in the n-sphere is nr/(n+1) ?

That would be quite easy! 🙂
Perhaps you mean dist = (n-1)r/n (assuming a point is a 1d sphere)

Originally posted by iamatiger
That would be quite easy! 🙂
Perhaps you mean dist = (n-1)r/n (assuming a point is a 1d sphere)

A point is a 0d sphere, so I think RC had it right when he said nr/(n+1).

Originally posted by richjohnson
A point is a 0d sphere, so I think RC had it right when he said nr/(n+1).

Come to think of it, a point is an n-dimensional sphere with radius 0, for all n in N 😲

Originally posted by richjohnson
A point is a 0d sphere, so I think RC had it right when he said nr/(n+1).

Oops - my mistake - I was a D out.

Originally posted by royalchicken
Come to think of it, a point is an n-dimensional sphere with radius 0, for all n in N 😲

n-dimensional sphere with r=0 dose not exist.

n-dimensional sphere with r=1/infinity is a point, i think.😕

I beg to differ. The sphere of radius r centered on the origin in n-space is the set of points (x1,x2,x3,...x4) such that:

x1^2 + x2^2 + ... +xn^2 = r^2

There is one point, namely (0,0,0...,0) that satisfies this equality when r is 0. Thus a point is a zero-radius n-sphere.

Doesn't the proof follow immedeately from Iamatiger's post?

The only thing we need for that is a formula for the surface and volume of a n-dimensional sphere. Research shows:

Volume V(r): r^n * pi^(n/2) * Gamma (n/2 +1)

Taking the derivative to r:
Surface S(r): n* r^(n-1) * pi^(n/2) * Gamma (n/2 +1)

Where the Gamma function is
- Gamma(z) = integral for 0 to infinity t^(z-1) * e^(-t) dt
- Gamma(z+1) = zGamma(z)
- Gamma(1)=1 and so Gamma(n)=n! for all natural n
- Gamma(1/2) = sqrt(pi)

A shell of radius s has surface S(s), and it contributes S(s)/V(r) to the total sphere.

S(s)/V(r) = n* s^(n-1) / r^n

Just as in Iamatiger's post, we need to integrate this times s from 0 to r over s. Wich gives as result:

nr / n+1

Strangely i founf many formulaes of the surface of a hypersphere, but none of them would give this result, so i resorted to taking the derivative of the volume. Try getting the same result with
S(r) = r^(n-1) * pi^(n/2) / Gamma(n/2) and you'll find yourself concluding the formula 2r/n(n+1)...

Ton

Originally posted by TheMaster37
Doesn't the proof follow immedeately from Iamatiger's post?

The only thing we need for that is a formula for the surface and volume of a n-dimensional sphere. Research shows:

Volume V(r): r^n * pi^(n/2) * Gamma (n/2 +1)

Taking the derivative to r:
Surface S(r): n* r^(n-1) * pi^(n/2) * Gamma (n/2 +1)

Where the Gamma function is
- Gamma(z) ...[text shortened]... -1) * pi^(n/2) / Gamma(n/2) and you'll find yourself concluding the formula 2r/n(n+1)...

Ton

You are correct!

Can someone please explain the title of this thread?

Steffin

"The Great Circle That Went Straight" in infinite inverse steps?

RC...

Can one use the sum of vectors tangent to the circle at all points x on the arc?

I don't really know how to state this mathmatically, so I'll just do it with words.

Given an arc on a circle, A of length R (radius) can there exist n points,
each of which can be "assigned" a vector which has a direction which is tangent to the circle and a "size" which is 1 /n.

If all vectors are then mapped to the points x on the line are linked, and retain the value of the preceeding point, then if the zeroth vector is rotated to the center of the circle, ninety degrees, all vectors rotate in that same direction and their size remains constant. The beginning of a cyclonic spin. Then the Zeroth + 1 n is rotated through ninety degrees. The spin increases inversely to the number of vectors who are realigned from tangent to be aimed at the center. Then repeat until N. In my imagination, there exists a whirlpool, unraveling into the center point, comprised of "infinite" vectors... almost. Kind of neat to envision, but I can't put it into math.