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.

I was browsing this thread when I came across this answer. I can neither make heads nor tail of it. Can someone help me understand it?

This I understand: $$\int_0^{2\pi}e^{ikx}\,dx=\left\{\begin{array}{cl}0&k\ne0\\ 2\pi&k=0\end{array}\right.,$$

But how does he get this? Where does the summation come from?

$$\int_0^{2\pi}\sin^{100}x\,dx=\frac1{2^{100}}\sum_{k=0}^{100}\binom{100}{k}\int_0^{2\pi}e^{ikx}(-1)^{100-k}e^{-i(100-k)x}\,dx=\frac{\binom{100}{50}}{2^{100}}2\pi.$$

EDIT: The answer is so obvious...I can't believe I didn't notice it! I kept expanding $\sin^{100}(x)$ as $\frac{1}{2i} (e^{100ix} + e^{-100ix})$ which got me nowhere.

share|improve this question
2  
It might help to add a link to the answer. –  Git Gud May 30 at 22:57
    
Take care! $\displaystyle\frac1{2i}\left(e^{100ix}-e^{-100ix}\right)\ =\ \sin(100x)\ \ne\ \sin^{100}(x)$ . –  Berci May 30 at 23:14

3 Answers 3

Hint: Using the binomial theorem

$$\sin x=\frac{e^{ix}-e^{-ix}}{2i}\implies \sin^{100}x=\frac1{2^{100}}\sum_{k=0}^{100}\binom{100}ke^{kix}e^{-ix(100-k)}\;\ldots$$

share|improve this answer

We have

$$\sin x=\frac1{2i}(e^{ix}-e^{-ix})$$

so by the binomial formula we get $$\frac1{2^{100}}\sum_{k=0}^{100}(-1)^{100-k}\binom{100}ke^{kix}e^{-ix(100-k)}$$ so $$\int_0^{2\pi}\sin^{100}xdx=\frac1{2^{100}}\sum_{k=0}^{100}(-1)^{k}\binom{100}k\underbrace{\int_0^{2\pi}e^{(2k-100)ix}dx}_{=2\pi\;\text{only for}\; k=50}=\frac{2\pi}{2^{100}}\binom{100}{50}$$

share|improve this answer

Write $$ \sin x = \frac{e^{ix} - e^{-ix}}{2i} $$ Then $$ \sin^{100} x = \frac{(a + b)^{100}}{2^{100}} $$ where $a = e^{ix}$ and $b = -e^{-ix}$.

Now use the binomial theorem: $$ (a + b)^{100} = \sum_{k=0}^{100}\left( \begin{array}{c} 100 \\ k \end{array} \right) a^k b^{100-k} $$ and the first summation step in the formula you were puzzled about you give falls out. And then in that integral, only the term where the factor in front of $x$ in the exponent gives a non-zero integral. So only the term with $k - (100-k) = 0$, or $k=50$, is non-zero. That is where the $\left( \begin{array}{c} 100 \\ 50 \end{array} \right)$ comes from.

share|improve this answer

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.