Fourier Series

• Polynomials are not the only functions that can be used to approximate the known function.
• Another means for representing known functions are approximations that use sines and cosines, called Fourier series.
  • Any function can be represented by an infinite sum of sine and cosine terms with the proper coefficients, (possibly with an infinite number of terms).
• Any function, $f(x)$, is periodic of period $P$ if it has the same value for any two x-values, that differ by $P$, or
  
  $$f(x) = f(x + P) = f(x + 2P) = \ldots =f(x - P)=f(x - 2P)=\ldots$$
  
  

  Figure 1: Plot of a periodic function of period P.

  
  Figure 1 shows such a periodic function. Observe that the period can be started at any point on the $x$-axis.
  • $Sin(x)$ and $cos(x)$ are periodic of period $2\pi$
  • $Sin(2x)$ and $cos(2x)$ are periodic of period $\pi$
  • $Sin(nx)$ and $cos(nx)$ are periodic of period $2\pi/n$
• We now discuss how to find the $A$s and $B$s in a Fourier series of the form
  

  $$f(x)\approx \frac{A_0}{2}+ \sum_{n=1}^{\infty} [A_ncos(nx)+B_nsin(nx)]$$ (1)

  
  
  The determination of the coefficients of a Fourier series (when a given function,$f(x)$, can be so represented) is based on the property of orthogonality for sines and cosines. For integer values of $n, m$:
  

  $$\int_{-\pi}^{\pi}sin(nx)dx=0$$ (2)

  
  
  
  

  $$\int_{-\pi}^{\pi}cos(nx)dx=\left\{ \begin{array}{ll} 0, & n\neq 0 \\ 2\pi, & n=0\\ \end{array}\right\}$$ (3)

  
  
  
  

  $$\int_{-\pi}^{\pi}sin(nx)cos(mx)dx=0$$ (4)

  
  
  
  

  $$\int_{-\pi}^{\pi}sin(nx)sin(mx)dx=\left\{ \begin{array}{ll} 0, & n\neq m \\ \pi, & n=m \\ \end{array}\right\}$$ (5)

  
  
  
  

  $$\int_{-\pi}^{\pi}cos(nx)cos(mx)dx=\left\{ \begin{array}{ll} 0, & n\neq m \\ \pi, & n=m \\ \end{array}\right\}$$ (6)

  
  
  It is related to the same term used for orthogonal (perpendicular) vectors whose dot product is zero. Many functions, besides sines and cosines, are orthogonal (such as the Chebyshev polynomials).
• To begin, we assume that $f(x)$ is periodic of period $2\pi$ and can be represented as in Eq. 1. We find the values of $A_n$ and $B_n$ in Eq. 1 in the following way;
  • For $A_0$; multiply both sides of Eq. 1 by $cos(0x)=1$, and integrate term by term between the limits of $-\pi$ and $\pi$.
    
    $$\int_{-\pi}^{\pi} f(x)dx=\int_{-\pi}^{\pi}\frac{A_0}{2}dx+ ... ..._ncos(nx)dx+ \sum_{n=1}^{\infty} \int_{-\pi}^{\pi}B_nsin(nx)dx$$
    
    
    Because of Eqs. 2 and 3, every term on the right vanishes except the first, giving
    
    $$\int_{-\pi}^{\pi} f(x)dx=\frac{A_0}{2}(2\pi),~or~A_0=\frac{1}{\pi}\int_{-\pi}^{\pi} f(x)dx$$
    
    
    Hence, $A_0$ is found and it is equal to twice the average value of $f(x)$ over one period.
  • For $A_n$; multiply both sides of Eq. 1 by $cos(mx)$, where $m$ is any positive integer, and integrate:
    
    $$\int_{-\pi}^{\pi} cos(mx)f(x)dx=\int_{-\pi}^{\pi} \frac{A_0}{2}cos(mx)dx+$$
    
    
    
    
    $$\sum_{n=1}^{\infty} \int_{-\pi}^{\pi} A_ncos(mx)cos(nx)dx+\sum_{n=1}^{\infty} \int_{-\pi}^{\pi} B_ncos(mx)sin(nx)dx$$
    
    
    Because of Eqs. 3,4 and 6 the only nonzero term on the right is when $m=n$ in the first summation, so we get a formula for the $A$s;
    
    $$A_n=\frac{1}{\pi}\int_{-\pi}^{\pi} f(x)cos(nx)dx,~n=1,2,3,\ldots$$
    
    

  • For $B_n$; multiply both sides of Eq. 1 by $sin(mx)$, where $m$ is any positive integer, and integrate:
    
    $$\int_{-\pi}^{\pi} sin(mx)f(x)dx=\int_{-\pi}^{\pi} \frac{A_0}{2}sin(mx)dx+$$
    
    
    
    
    $$\sum_{n=1}^{\infty} \int_{-\pi}^{\pi} A_nsin(mx)cos(nx)dx+\sum_{n=1}^{\infty} \int_{-\pi}^{\pi} B_nsin(mx)sin(nx)dx$$
    
    
    Because of Eqs. 2, 4 and 5, the only nonzero term on the right is when $m=n$ in the second summation, so we get a formula for the $B$s;
    
    $$B_n=\frac{1}{\pi}\int_{-\pi}^{\pi} f(x)sin(nx)dx,~n=1,2,3,\ldots$$
    
    
    It is obvious that getting the coefficients of Fourier series involves many integrations.