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Question:

Find the value $$f^{(2)}_{n}(x)=\sin^2{x}+\sin^2{(2x)}+\sin^2{(3x)}+\cdots+\sin^2{(nx)}$$

My solution:

since

$$\sin^2{x}=\dfrac{1}{2}(1-\cos{(2x)})$$ so $$f_{n}(x)=\dfrac{n}{2}-\dfrac{1}{2}(\cos{(2x)}+\cos{(4x)}+\cdots+\cos{(2nx)})$$ since $$2\sin{x}\cos{y}=\sin{(x+y)}-\sin{(y-x)}$$ so $$2\sin{x}\cdot[\cos{(2x)}+\cos{(4x)}+\cdots+\cos{(2nx)}]=\sin{(2n+1)x}-\sin{x}$$

Have other methods? Thank you

Question 2: Find this sum closed form $$f^{(3)}_{n}(x)=\sin^3{x}+\sin^3{(2x)}+\cdots+\sin^3{(nx)}$$ $$\cdots$$ $$f^{(m)}_{n}(x)=\sin^m{x}+\sin^m{(2x)}+\cdots+\sin^m{(nx)}=?$$

math110
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  • You have a superscripit (2) missing in the third formula. It does not impact the quality of your post and problem. For m=2, your way is fine and elegant. Unfortunately, I am not been able to find for the other. But I am sure that you have to go to sums of cosines for arguments in arithmetic progressions. Cheers. – Claude Leibovici Dec 25 '13 at 10:00
  • FYI : http://www.wolframalpha.com/input/?i=sum_%28k%3D1%29%5E%28n%29%28sin%28kx%29%29%5E3 – mathlove Dec 25 '13 at 10:35

1 Answers1

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HINT: Using Euler's formula, $\displaystyle 2i\sin y=e^{iy}-e^{-iy}$

$$\sin^m(rx)=\left(\frac{e^{irx}-e^{-irx}}{2i}\right)^m$$

Expand and use Euler's formula to find $\sin^m(rx)$ as summation of multiples angles of $rx$ sine & cosine ratios

For example, $$\sin^3(rx)=\left(\frac{e^{irx}-e^{-irx}}{2i}\right)^3=\frac{e^{i3rx}-e^{-i3rx}+3\left(e^{irx}-e^{-irx}\right)}{-8i}=\frac{2i\sin3rx+3(2i\sin rx)}{-8i}$$

lab bhattacharjee
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  • @chinamath, For $m=3,$ putting $r=1,2,\cdots,n$ we get two sums of sines each has arguments in arithmetic progressions, right? – lab bhattacharjee Dec 25 '13 at 10:48
  • Go one step further and carry out the summation as geometric sums in $e^{ikx}$, $k=-3,-1,1,3$. After that, return to real expressions using sine and cosine. – Lutz Lehmann Dec 25 '13 at 22:27