How to simplify $\sin(n\frac{\pi}{2})$ as we do with $\cos(n\pi)$

Question

When calculating Fourier series we almost always get some expression that look like: \begin{equation} ...\left[\cos(nx) \right]_0^\pi. \end{equation} Then we use $\cos(n\pi) = (-1)^n$.

However, lately I have come across expressions like this: \begin{equation} ...\left[\sin(nx) \right]_0^{\pi/2}. \end{equation} Which is equal to: \begin{array}{|c|c|c|c|} \hline n& 1 & 2 & 3 & 4 \\ \hline \sin(nx)&\sin(\frac{\pi}{2}) & \sin(\pi) & \sin(\frac{3\pi}{2})& \sin(2\pi)\\ \hline val &1 &0 &-1 &0\\ \hline \end{array} How do we simplify this so that it looks like the $\cos(n\pi) = (-1)^n$.

Answer 1

You have

$$\sin\frac{n\pi}{2} = \frac{1-(-1)^n}{2}(-1)^{\lfloor n/2\rfloor}.$$

So that's one way. But often, if there are no $\cos$'s messing things up, you can re-index your sum. All the terms for even $n$ are zero, you can rewrite

$$\sum_{n=1}^{\infty} \sin\frac{n\pi}{2}\mbox{crud} = \sum_{k=1}^{\infty} \sin\frac{(2k-1)\pi}{2}\mbox{crud}= \sum_{k=1}^{\infty}(-1)^k \mbox{crud}.$$

Whatever the crud is, replace all occurrences of $n$ with $2k-1.$

Answer 2

Note that $e^{\pm in\pi / 2} = \cos n\pi/2 \pm i\sin n\pi/2$, so $$\sin n\pi/2 = \tfrac{1}{2i}(e^{in\pi / 2} - e^{-in\pi / 2})$$ $$= \tfrac{1}{2i}(i^n - i^{-n})$$ $$=\frac{i^{2n}-1}{2i^{n+1}}$$ $$= \frac{(-1)^n - 1}{2i^{n+1}}$$ $$= \boxed{\frac{1 - (-1)^n}{2i^{n-1}}}\tag{mult top/bottom by $i^{-2} = -1$}$$

This expression vanishes for even $n$, and for odd $n = 2k+1$ it becomes $$\frac{1 - (-1)^{2k+1}}{2i^{2k+1-1}} = \frac{2}{2(-1)^k} = (-1)^k$$ as desired.

Answer 3

You have a sequence with period $4$ (specifically $0,1,0,-1,0,1,\cdots$), but $(-1)^n$ has period $2$, so it can't be used in this form.

However, $(-1)^{\binom n2}=(-1)^{n(n-1)/2}$ does have period $4$:

$$(-1)^0=1,\quad(-1)^1=-1,\quad(-1)^3=-1,\quad(-1)^6=1,$$ $$(-1)^{10}=1,\quad(-1)^{15}=-1,\quad(-1)^{21}=-1,\quad(-1)^{28}=1,$$ $$\cdots$$

To prove periodicity:

$$(-1)^{(n+4)(n+3)/2}=(-1)^{(n^2+7n+12)/2}=(-1)^{4n+6+(n^2-n)/2}\\=((-1)^2)^{2n+3}(-1)^{n(n-1)/2}=(-1)^{n(n-1)/2}$$

Now let's try shifting and combining these:

$$\begin{matrix}n&=&(0,&1,&2,&3,&\cdots) \\ (-1)^n&=&(1,&-1,&1,&-1,&\cdots) \\ (-1)^{n(n-1)/2}&=&(1,&1,&-1,&-1,&\cdots) \\ (-1)^{n(n+1)/2}&=&(1,&-1,&-1,&1,&\cdots) \\ (-1)^{n(n-1)/2}-(-1)^{n(n+1)/2}&=&(0,&2,&0,&-2,&\cdots)\end{matrix}$$

And there's your answer:

$$\sin(n\pi/2)=\frac{(-1)^{n(n-1)/2}-(-1)^{n(n+1)/2}}{2}$$

In fact any $4$-periodic sequence can be expressed in this way:

$$f_n=\frac{1+(-1)^n+(-1)^{n(n-1)/2}+(-1)^{n(n+1)/2}}{4}=(1,0,0,0,1,0,0,0,1,\cdots)$$ $$(a_0,a_1,a_2,a_3,a_0,a_1,a_2,a_3,a_0,\cdots)=a_0f_n+a_1f_{n-1}+a_2f_{n-2}+a_3f_{n-3}$$ $$=a_0\frac{1+(-1)^n+(-1)^{n(n-1)/2}+(-1)^{n(n+1)/2}}{4} \\ +a_1\frac{1-(-1)^n+(-1)^{n(n-1)/2}-(-1)^{n(n+1)/2}}{4} \\ +a_2\frac{1+(-1)^n-(-1)^{n(n-1)/2}-(-1)^{n(n+1)/2}}{4} \\ +a_3\frac{1-(-1)^n-(-1)^{n(n-1)/2}+(-1)^{n(n+1)/2}}{4}$$

See also What is a mathematical expression for the sequence $\{1,1,-1,-1,1,1,-1,-1,\dots\}$? and the linked posts.

Answer 4

If you are multiplying by that sin term, even terms fall out $$ \sum_{n=1}^{2N} f(n)\sin \left(\frac{n\pi}{2}\right) = \sum_{k=1}^N (-1)^{k-1} f(2k-1) $$