# Fourier series coefficients

1. Mar 4, 2012

### zezima1

The expressions for the coefficients of a fourier series are valid for all integers [0;n].
Though sometimes when I evaluate the fourier series of an even function (composed only of cosines) I get an expression for the r'th coefficient, which has r in the demoninator. It could be for instance:

ar = (-1)r/(rπ)

Since division by zero is not allowed this expression doesn't hold for the r=0, i.e. the first coefficient in the sum of cosines. What is that, that is not reversible, in the proces of solving the integral for the r'th coefficient? Because the integral expression clearly works for all r.

2. Mar 4, 2012

### LCKurtz

I'm guessing you are thinking of an example where you are given something like $f(x) = 3$ on $(0,\pi)$ and you are trying to write a FS for its even extension. The $b_n$ are all zero for an even function. You have $$a_n=\frac 2 {\pi} \int_0^{\pi} f(x)\cos(nx)\, dx = \frac 2 {\pi} \int_0^{\pi} 3\cos(nx)\, dx =\frac 6 {\pi}\left . \frac{\sin(nx)}{n} \right | _0^{\pi} = 0$$This gives the incorrect value for $a_0$.The reason for this is that for $n=0$ you would be saying that$$\int \cos (0x) = \frac {\sin (0x)}{0} = 0$$That formula doesn't make any sense and doesn't apply when $n=0$ because there would be no cosine in the integral. The integral for $a_0$ is$$a_0=\frac 1 \pi \int_0^\pi f(x)\, dx = \frac 1 \pi \int_0^\pi 3\, dx=3$$

3. Mar 4, 2012

### aaaa202

hmm yes, I see. So in general, even if you get an expression for the r'th coefficient, where putting r=0 would make sense, you can't trust that to get you the right value for a0 right?

4. Mar 4, 2012

### LCKurtz

You just have to be careful that you don't get senseless expressions. For example, when you have an integral like$$\int_0^\pi \sin(nx)\sin(mx)\,dx$$ where $m$ and $n$ are integers, you would use the product formulas to change the product of sines to a sum:$$\sin(mx)\sin(nx)=\frac 1 2 (\cos((m-n)x)-\cos((m+n)x)$$If you integrate that you will get$$\left. \frac 1 2 \left(\frac {\sin(m-n)x}{m-n}-\frac{\sin(m+n)x}{m+n}\right)\right |_0^\pi=0$$If you don't notice that $m-n$ in the denominator, you might say the integral is zero for all $m$ and $n$, which is similar to the problem you are mentioning. But the calculations don't make sense when $m=n$, which is why the answer can't be trusted. But you wouldn't want to actually do the integral that way anyway if $m=n$ because you would use the double angle angle formula:$$\sin^2(nx)=\frac {1-\cos(2nx)}{2}$$not a product formula.

5. Mar 4, 2012

### Office_Shredder

Staff Emeritus
But.... but.... the double angle formula IS the product formula!!!!

THE CAKE IS A LIE

For example, if your function is a constant function all your higher coefficients are going to be 0. Plug in r=0 and you get 0, which is probably wrong

6. Mar 4, 2012

### aaaa202

but like you say if r=0 then integrating cos(nx) is just nonsense. So how would you be able to trust the expression even if it made sense to plug in r=0? Give an example please :)

7. Mar 4, 2012

### Office_Shredder

Staff Emeritus
You can't trust it. See https://www.physicsforums.com/showpost.php?p=3798255&postcount=2 posted above. You just have to do the integral separately