# Fourier Series Coefficient Symmetries

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1. Mar 28, 2017

### Marcus95

1. The problem statement, all variables and given/known data
Let $f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx)$
What can be said about the coefficients $a_n$ and $b_n$ in the following cases?
a) f(x) = f(-x)
b) f(x) = - f(-x)
c) f(x) = f(π/2+x)
d) f(x) = f(π/2-x)
e) f(x) = f(2x)
f) f(x) = f(-x) = f(π/2-x)

2. Relevant equations
Sine is odd and cosine is even. Even functions can only contain cosine terms and odd functions only sine terms in their Fourier Series.

3. The attempt at a solution
Here is how far I have gotten:
a) All b=0 because even function
b) All a=0 because odd function
c) Funtion has period π/2 (?)
d) Funtion has period π/2-2x (?)
e) Function has period x (not neccessarily the shortest period)
f) All b=0 because even function.

But here I am stuck, because I have no idea how to relate the coefficients to the period in c)-f), and in c)-e) I don't know if the funtion is odd or even.
Many thanks for any help! :)

2. Mar 28, 2017

### eys_physics

Use the identities for sin and cos:
$sin(x+y)=sin(x)cos(y)+sin(y)cos(x)$
$cos(x+y)=cos(x)cos(y)-sin(x)sin(y)$

for the different for terms in the sum. What is happening with $a_n$ and $b_n$ for odd and even values of $n$?
By putting different values for $y$ you can study the different cases in c), d), e) and f).

3. Mar 29, 2017

### Marcus95

Alright, I tried using the formulas in case c):
$\sin(\pi/2 +x) = \cos x$
$\cos(pi/2 +x) = -\sin x$
which is as expected. However, I am confused about how this should be applied to the problem. We have a general function $f(x)$ and I can't really see how the above properties reveal anything about the coefficients.

4. Mar 29, 2017

### eys_physics

Yes, but this is only valid for $n=1$. What do you get for $cos n(\pi/2+x)$ and $sin n(\pi/2+x)$, for $n$ odd and even? You can then write the Fourier serie for $f(\pi/2+x)$ and simplify the sum. An advice is to split into odd and even $n$. I hope this helps.

5. Mar 29, 2017

### Marcus95

Alrigth, I feel like the solution is close but I am still not really there. I have found the following:
$\sin n(\pi/2 +x) = \pm \cos nx$ for odd n
$\sin n(pi/2 +x) = \pm \sin nx$ for even n
$\cos n(\pi/2 +x) = \pm \cos nx$ for even n
$\cos n(pi/2 +x) = \pm \sin nx$ for odd n

But to me these are still "just" some trigonometric identities which I don't know how to apply to the function.
$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos nx + b_n \sin nx)$

6. Mar 29, 2017

### eys_physics

'
You can write the above relations as
$\sin n(\pi/2+x)=(-1)^{(n-1)/2}\cos nx$
$\sin n(pi/2 +x) = (-1)^{n/2} \sin nx$ for even n
$\cos n(\pi/2 +x) = (-1)^{(n-1)/2} \cos nx$ for even n
$\cos n(pi/2 +x) =-(-1)^{n/2} \sin nx$ for odd n

Furthermore, we can write
$f(\pi/2+x)=\frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos n(\pi/2+x) + b_n \sin n(\pi/2+x))$,

Can you now see how to use the above trignometric identies?

It is also convenient to split the sum into odd and even values of $n$, i.e. something like $\sum_{n=1}^{\infty}=\sum_{odd\: n}+\sum_{even\: n}$. It is then convenient to introduce a variable $k$ through $n=2k+1$ (for the sum over odd $n$) and similarly $n=2k$ (for even n).

7. Mar 29, 2017

### Marcus95

Hmm, I think there migth be an algebraic miss in the exponents in your previous post. shouldn't it be:
$\sin n(\pi/2+x)=(-1)^{n/2}\cos nx$ for odd n
$\sin n(\pi/2 +x) = (-1)^{(n-1)/2} \sin nx$ for even n
$\cos n(\pi/2 +x) = (-1)^{n/2} \cos nx$ for even n
$\cos n(\pi/2 +x) =-(-1)^{(n-1)/2} \sin nx$ for odd n?

Nevertheless,
$f(x) = f(\pi/2+x)=\frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n' \cos n(\pi/2+x) + b_n' \sin n(\pi/2+x))$,
Supposed we have odd n, that would imply: $a_n' \cos n(\pi/2 +x) = a_n'( -(-1)^{(n-1)/2} \sin nx)$ and $b_n' \sin n(\pi/2+x)= b_n'(-1)^{n/2}\cos nx$. But exactly what is this telling us? Just that:
$a_n = b_n'(-1)^{n/2}$ and $b_n = -a_n'(-1)^{(n-1)/2)}$? (I use prime to note that it is the coefficients for $f(\pi/2+x)$, and unprimed for $f(x)$)

Last edited: Mar 29, 2017
8. Mar 29, 2017

### eys_physics

Yes, I did a typo while writing my previous post. Thanks for pointing this out. First, first I don't think you should not introduce other coefficients for $f(\pi/2+x)$ since the function $f$ is the same on both sides. By comparing term by term the series for $f(\pi/2+x)$ and $f(x)$ you get as said
$-(-1)^{(n-1)/2}a_n=b_n$ and $(-1)^{(n-1)/2}b_n=a_n$ if $n$ is odd. But this can only be satisfied if $a_n=b_n=0$. Now, for even $n$ one has $(-1)^{n/2}a_n=a_n$ and $(-1)^{n/2}b_n=b_n$. So, we must have $(-1)^{n/2}=1$ or $a_n=b_n=0$. Therefore, we have only non-zero $a_n$ and $b_n$ for $n$ being a multiple of 4. We have thus
$f(x)=a'_0/2+\sum_{n=1}^{\infty}(a'_n\cos(4nx)+b'_n\sin(4nx))$, where
$a'_n=a_{4n}$ and $b'_n=b_{4n}$. As is discussed on the page http://mathworld.wolfram.com/FourierSeries.html this correspond to a function with period $\pi/2$ which you already concluded from the beginning. This was maybe a long way to see this. But, I hope you are now able to proceed with the other parts of the exercise.

9. Apr 5, 2017

### Marcus95

Thank you very much for the help! I managed to solve all the problems except for the case when $f(x) = f(2x)$, because here the trigonometric relations are much more complicated with squares and mixed terms. Do you have any last advice for this case?

Last edited: Apr 5, 2017