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Fourier series expansion and Fourier transform are completely different. In this article, I’ll write about Fourier series expansion.

Fourier series expansion method is to express the periodic function \(f(x)\) by using the sum of \(sin\) and \(cos\). Using the method, the periodic function \(f(x)\) whose period is \(L\) can be expressed as follows.

$$a_0=\frac{2}{L}\int_{-L/2}^{L/2} f(x) dx$$

$$a_n=\frac{2}{L}\int_{-L/2}^{L/2} f(x)cos(\frac{2πnx}{L}) dx$$

$$b_n=\frac{2}{L}\int_{-L/2}^{L/2} f(x)sin(\frac{2πnx}{L}) dx$$

We sometimes find a book in which \(a_0/2\) is written as \(a_0\), so it’s pretty confusing. This difference is only meaning that \(\frac{1}{2}\) is a constant, so it is better to put it together with the constant \(a_ 0\) to improve the appearance of the formula. According to a book with \(a_0\) as the first term after the expansion of f (x), the formula for \(a_0\) should be as follows.

$$a_0=\frac{1}{L}\int_{-L/2}^{L/2} f(x) dx$$

If you remember the formulas with \(\frac{a_0}{2}\), it is easy to remember the formulas like “All formulas of Fourier series expansion method about \(a_0\), \(a_n\) and \(b_n\) have \(\frac{2}{L}\).” Therefore, I recommend to memorize the formula about \(\frac{a_0}{2}\) as the top expression.

## Practice

Here, we will solve a basic practice that you will definitely touch in the Fourier analysis lesson at your university.

Practice: Expand the Fourier series of the periodic function \(f(x)\) with period \(2π\).

$$\begin{eqnarray} f(x)= \begin{cases} -1 & (-π<x<0) \\ 1 & (0<x<π) \end{cases} \end{eqnarray}$$

First, substitute \(L=2π\) along the above formula and find \(a_0\).

\begin{eqnarray}a_0&=&\frac{2}{2π}\int_{-π}^{π}f(x) dx\\&=&\frac{1}{π} \left[ \int_{-π}^{0}(-1) dx+\int_{0}^{π}(+1) dx \right]\\&=&\frac{1}{π}(-π+π)\\&=&0\end{eqnarray}

Next, I’d like to calculate of \ (a_n \) … but there is a trick to make the calculation easy. \(f(x)\) is an odd function because it satisfies \(f(x)=-f(-x)\). Therefore, \(f(x)\) does not include even functions. Then the formula of \(a_n\) includes \(cos\) function which is an even function. This shows \(a_0=0\). If you are worried, you can find \(a_n=0\) by substituting \(2π\) for \(L\) in the \(a_n\) formula.

Finally, find \(b_n\). As with \(a_0\), substitute \(L=2π\).

\begin{eqnarray}b_n&=&\frac{2}{2π}\int_{-π}^{π}f(x)sin(\frac{2πnx}{2π}) dx\\&=&\frac{1}{π} \left[ \int_{-π}^{0}(-1)sin(nx) dx+\int_{0}^{π}(+1)sin(nx) dx \right]\\&=&\frac{1}{π} \left[ \left[ \frac{1}{n}cos(nx) \right]_{-π}^{0}+\left[ -\frac{1}{n}cos(nx) \right]_{0}^{π} \right]\\&=&\frac{1}{π}\left[ \frac{1}{n}(1-cos(-nπ))+\frac{1}{n}(1-cos(nπ)) \right]\\&=&\frac{2}{nπ}(1-cos(nπ))\end{eqnarray}

Here, note that \(cos(-nπ)=cos(nπ)\) if \(n\) is an integer for the transformation of lines 4 to 5. Also note that \(cos(nπ)=(-1)^n\). Using them \(b_n\) is

$$\begin{eqnarray} b_n=\frac{2}{nπ}(1-(-1)^n) =\begin{cases} \frac{4}{nπ} & (n:odd) \\ 0 & (n:even) \end{cases} \end{eqnarray}$$

From the above, replacing \(n\) of \(b_n=\frac{4}{nπ}\) with odd \(2m-1\) (m: natural number) gives the following answer.

$$f(x)=\sum_{n=1}^{∞}{b_{n}sin(\frac{2πnx}{L})}=\sum_{m=1}^{∞}{\frac{4}{(2m-1)π}sin((2m-1)x)}$$

## Why consider Fourier series expansion

When I touched this Fourier series expansion for the first time, I was confused by the notation of \ (\frac{a_0}{2}\) and \(a_0\), the period of \(f(x)\), etc. Why do we have to use Fourier series expansion to study physics?

From the Fourier series expansion formula, all waves denoted by \(f(x)\) can be written by superimposing sinusoidal waves that can be written with sin and cos. I forcibly give a physical example of this.

There is a television with a white screen. What kind of light does this white light from this screen contain? Of course, since the television shows white using the three primary colors of light, so this white light is made up of a superposition of red, blue and green light. The color is determined by the wavelength of light. Therefore, if you can describe the white light as the formula and the formula is periodic, you can obtain the formulas expressing wave of red,blue and green by Fourier series expanding. By the way, if you adapt the Fourier transform to white light, you will find three peaks that indicate red, blue and green light.