Hey NowIsForever.
Use the fact that cos(a+b) = cos(a)cos(b) - sin(a)sin(b) where a = 2nx and b = 2x. This gives cos(2nx)cos(2x) - sin(2nx)sin(2x) for which you get some cancellation and simplification for the cos and sin of 2x.
I found this problem recently on Yahoo Answers! It is gone now because it went unanswered after four days, and it is the policy of YA! to delete such questions.
Given:
Prove:
, for all natural numbers . (Since "natural number" is ambiguous, let's say for non negative—or possibly positive integers.)
I couldn't get passed the singularities at and . WolframAlpha couldn't provide an answer either, even for (e.g.), .
I have my doubts about this being true; however, if it is I would like to see a proof.
Hey NowIsForever.
Use the fact that cos(a+b) = cos(a)cos(b) - sin(a)sin(b) where a = 2nx and b = 2x. This gives cos(2nx)cos(2x) - sin(2nx)sin(2x) for which you get some cancellation and simplification for the cos and sin of 2x.
Ah, yes, that does it; however, you have (n + 2) in the last four integrals where it should be (n + 1).
You side stepped the singularity by clever use of trigonometric identities, and the identity .
I thought it would require something like the Cauchy principal value calculated with limits, but you've made that unnecessary, thanks.
Nice—and elegant.
Hi !
We saw that I(1)=pi/2 which leads to I(n)=((-1)^((n-1)/2))*(Pi/2) in case of odd n.
In case of even n, we have to compute I(0) which requires to follow the computation rules for the principal value of a Cauchy integral. The result is I(0)=0.
As a consequence I(n)=0 in case of even n
.
Thanks you for bringing the typo to attention. Of course, it doesn't change the result.