How to write sin(n*pi/2) as an expression (-1)^f(n)?

[zenterix]

Homework Statement

While calculating the Fourier series of a function I reached the following expression (for one class of Fourier coefficients):

Relevant Equations

$$\frac{1}{n\pi}\sin{\left ( \frac{n\pi}{2} \right )}$$

I could summarize this as

$\frac{1}{n\pi}$ for $n=1, 5, 9, \ldots$

$\frac{-1}{n\pi}$ for $n=3, 7, 11, \ldots$

and $0$ for all other $n$.

How would I go about writing this in a single expression, with $(-1)^{f(n)}$ where $f(n)$ summarizes both cases above?

 

[Orodruin]

How would I go about writing this in a single expression, with $(-1)^{f(n)}$ where $f(n)$ summarizes both cases above?

You wouldn’t. $(-1)^z \neq 0$ for any $z$.

 

[Orodruin]

What you could use is the complex expansion of the sine to write
$$
\sin(n\pi/2) = \frac{e^{in\pi/2} - e^{-in\pi/2}}{2i}
= \frac{i^n - (-i)^n}{2i} = \frac 12 (i^{n-1} + (-i)^{n-1})
$$

 

[zenterix]

You wouldn’t. $(-1)^z \neq 0$ for any $z$.

Perhaps I phrased the question in a way that wasn't clear.

What I ultimately want to do is rewrite

$$\sum\limits_{n=1}^\infty \left (\frac{1}{n\pi} \sin{\left ( \frac{n\pi}{2} \right )}\right ) \cos{nx}$$

in a simpler way.

I was wondering if I could replace the $\sin{\left ( \frac{n\pi}{2} \right )}$ with a simpler expression.

The complex expansion does work, though I still wonder if there is not a simpler way.

I don't need to have $(-1)^{f(n)}=0$ for any $n$. As long as $f(n)$ doesn't take on the values $0,2,4,6,\ldots$ then the summation would just skip the terms where $\sin{\left ( \frac{n\pi}{2} \right )}=0$.

The problem is how to make $(-1)^{f(n)}$ alternate between $1$ and $-1$ as $f(n)$ traverses all odd numbers.

 

[PeroK]

The problem is how to make $(-1)^{f(n)}$ alternate between $1$ and $-1$ as $f(n)$ traverses all odd numbers.

Normally, you use the sequence $2n+1$ or $2n - 1$ in the summand.

 

[zenterix]

For example, can we write $f(n)=1+(2n+1\ \text{mod}\ 4)\ \text{mod}\ 3$?

Then

$$(-1)^{1+(2n+1\ \text{mod}\ 4) \text{mod}\ 3}$$

For $n=0,1,2,\ldots$, we have

$2n+1=1,3,5,7,9,\ldots$

$2n+1\ \text{mod}\ 4= 1,3,1,3,1,3,\ldots$

$(2n+1\ \text{mod}\ 4)\ \text{mod}\ 3=1,0,1,0,1,0,\ldots$.

$1+(2n+1\ \text{mod}\ 4)\ \text{mod}\ 3=2,1,2,1,2,1,\ldots$.

I guess this isn't really simplifying anything, but it seems to work.

 

[pasmith]

$\sin (n\pi/2)$ vanishes for even $n$. Thus you need only consider $$\sum_{k=0}^\infty \frac{1}{(2k+1)\pi}\sin\left( \frac{(2k + 1)\pi}2 \right) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)\pi}.$$

 

[Orodruin]

If you sum only over odd numbers, note that this is equivalent to summing over $k \in\mathbb N$ with $n = 2k + 1$. Consider how your expression depends on $k$.

 

[zenterix]

$\sin (n\pi/2)$ vanishes for even $n$. Thus you need only consider $$\sum_{k=0}^\infty \frac{1}{(2k+1)\pi}\sin\left( \frac{(2k + 1)\pi}2 \right) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)\pi}.$$

This is indeed the solution I was looking for. It's actually simple I just missed it.