Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$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.
share|improve this question
    
Should $x^3$ be $sin sin cos$ instead of $sin cos cos$? –  David H Sep 17 '13 at 5:56
    
Is it me or the third line should be $x^3 = a \sin u^1 \sin u^2 \cos u^3$? –  Pipicito Sep 17 '13 at 5:56
1  
David H(ilbert?), we are connected :) –  Pipicito Sep 17 '13 at 5:57
    
Yes, thanks @DavidH, that was a typo. –  isomorphismes Sep 17 '13 at 6:09
    
Thanks @Pipicito your correction was right. –  isomorphismes Sep 17 '13 at 6:10

1 Answer 1

up vote 1 down vote accepted

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 $1$.

share|improve this answer
    
Great, thanks. That makes perfect sense. –  isomorphismes Sep 17 '13 at 6:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.