$x^1 = a \cos u^1 \\ x^2 = a \sin u^1 \cos u^2 \\ x^3 = a \sin u^1 \sin u^2 \cos u^3 \\ \vdots \\ x^{N-1} = a \sin u^1 \sin u^2 \sin u^3 \cdots \sin u^{N-2} \cos u^{N-1} \\ \displaystyle x^N = a \prod_1^{N-1} \sin u^i $

My pattern recognition skills are decent enough that I can tell this is a hypersphere. But, why?

Specifically if I call the pattern c, sc, ssc, sssc, ..., ssssssssssssssssc, sssssssssssssssss, then a few aspects of the sequence seem strange:

  • First of all why does s appear so much more often than c in the formula? (edit: it's clear now these are interchangeable)
  • Doesn't a product of $\sin$es often end up smaller than a single $\cos$ine? Since $|\sin \theta| \leq 1$ So where's the symmetry of the sphere there?
  • Apparently I never thought hard enough when I was learning azimuthal angles in the first place. Why do some of the terms get just one trig function whilst others get products of trig functions?
  • Lastly it would seem more "balanced" to have a pattern like scscscscsc than sssssssssc.
  • $\begingroup$ Should $x^3$ be $sin sin cos$ instead of $sin cos cos$? $\endgroup$
    – David H
    Sep 17 '13 at 5:56
  • $\begingroup$ Is it me or the third line should be $x^3 = a \sin u^1 \sin u^2 \cos u^3$? $\endgroup$
    – Pipicito
    Sep 17 '13 at 5:56
  • 1
    $\begingroup$ David H(ilbert?), we are connected :) $\endgroup$
    – Pipicito
    Sep 17 '13 at 5:57
  • $\begingroup$ Yes, thanks @DavidH, that was a typo. $\endgroup$ Sep 17 '13 at 6:09
  • $\begingroup$ Thanks @Pipicito your correction was right. $\endgroup$ Sep 17 '13 at 6:10

I'll describe the idea that allows a rigorous proof by recurrence.

Let's take this sum (I'll put indices in subscripts for clarity):

$$\frac{1}{a^2}\sum_{i=1}^N x_i^2 $$ and study it's two last terms: $$\cos^2 u_{N-1} \prod_1^{N-2} \sin u^2_i + \sin u^2_{N-1}\prod_1^{N-2} \sin u^2_i$$ $$=\prod_1^{N-2} \sin u^2_i.$$

Now we combine this term with the $(N-2)$-th element of the sum to further reduce the number of factors. After repeating this step $N-2$ times we obtain that the total sum is equal to $1$.

  • $\begingroup$ Great, thanks. That makes perfect sense. $\endgroup$ Sep 17 '13 at 6:43

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