Why Does the Fourier Coefficient Formula Fail for r=0?

In summary, the expressions for the coefficients of a Fourier series are valid for all integers [0;n]. However, when evaluating the Fourier series of an even function composed only of cosines, the expression for the r'th coefficient may have r in the demoninator, which leads to problems when trying to find the first coefficient, a0. This is because the integral for a0 involves division by zero, which is not allowed. Therefore, even if an expression for the r'th coefficient can be derived, it cannot be trusted to give the correct value for a0. This is demonstrated through examples of integrating cos(nx) and sin(nx) where plugging in r=0 leads to nonsensical results.
  • #1
zezima1
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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.
 
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  • #2
zezima1 said:
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.

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

THE CAKE IS A LIE

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?

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
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
aaaa202 said:
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 :)

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
 

FAQ: Why Does the Fourier Coefficient Formula Fail for r=0?

What are Fourier series coefficients?

Fourier series coefficients are numerical values that represent the amplitude and phase of each frequency component in a periodic signal. They are used in Fourier series, which is a mathematical tool for representing a periodic function as a sum of sinusoidal functions.

How are Fourier series coefficients calculated?

The coefficients are calculated using an integral formula involving the periodic function and the sine and cosine functions. The integral is evaluated over one period of the function, and the resulting values represent the amplitudes and phases of the different frequency components.

What is the significance of Fourier series coefficients?

Fourier series coefficients allow us to break down a complicated periodic function into simpler sinusoidal components, making it easier to analyze and understand. They also have many applications in signal processing, such as in filtering and compression.

Can Fourier series coefficients be negative?

Yes, Fourier series coefficients can be negative. The sign of a coefficient indicates the phase shift of the corresponding frequency component. A positive coefficient represents a cosine function, while a negative coefficient represents a sine function with a 90-degree phase shift.

How do changes in a function affect its Fourier series coefficients?

If a function is changed in any way (e.g. translated, scaled, or reflected), its Fourier series coefficients will also change. These changes can be predicted by using properties of the Fourier transform, such as linearity and time shifting.

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