Dated: 30-10-2024
Examples
Sum of Fourier Series at the point of Discontinuity
Let us say we have a point of discontinuty at \(x = x_1\) such that there is a jump in value of \(f(x)\) from \(y_1\) to \(y_2\).
To denote this, we say
For Left Side
\[f(x) = f(x_1 - 0)\]
For Right Side
\[f(x) = f(x_1 + 0)\]
When we find the fourier series, it turns out that the series converges to the average of \(y_1\) and \(y_2\).
Example
Consider the function
\[f(x) = 0 \quad - \pi < x < 0\]
\[f(x) = a \quad 0 < x < \pi\]
\[f(x) = f(x + 2 \pi)\]
Solution
\(a_0\)
\[a_{0}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx\]
\[=\frac{1}{\pi}\int_{0}^{\pi}a~dx\]
\[=\frac{1}{\pi}[ax]_{0}^{\pi}=a\]
\(a_n\)
\[a_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx\]
\[=\frac{1}{\pi}\int_{0}^{\pi}a \cdot \cos(nx)dx\]
\[=\frac{a}{\pi}\left[\frac{\sin~nx}{n}\right]_{0}^{\pi}\]
\[= 0\]
\(b_n\)
\[b_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx\]
\[=\frac{1}{\pi}\int_{0}^{\pi}a \cdot \sin(nx)dx\]
\[=\frac{a}{\pi}\left[\frac{-\cos(nx)}{n}\right]_{0}^{\pi}\]
\[=\frac{a}{n\pi}(1-\cos(n\pi))\]
If \(n\) is even
\[b_n = 0\]
If \(n\) is Odd
\[b_n = \frac {2a}{n \pi}\]
\[\because f(x)=\frac{1}{2}a_{0}+\sum_{n=1}^{\infty}b_{n} \cdot \sin(nx)\]
\[f(x)=\frac{a}{2}+\frac{2a}{\pi}\left(\sin(x)+\frac{1}{3}\sin(3x)+\frac{1}{5}\sin(5x)+\ldots\right)\]
The point of discontinuity in this function is \(x = 0\).
If we put that in \(f(x)\), all the \(\sin(x)\) terms vanish and we are left with \(\frac a 2\) which is the average between \(0\) and \(a\).
Half Range Series
Sometimes, a function is defined over only \(0\) to \(\pi\) range, not \(- \pi\) to \(\pi\) or \(0\) to \(2 \pi\) range.
Example
Determine a half range sin series to represent the following function
\[f(x) = 1 + x \quad 0 < x < \pi\]
\[f(x) = f(x + 2 \pi)\]
Solution
\[\because a_0 = 0 \text{ and } a_n = 0\]
\[b_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(nx)dx\]
\[b_{n}=\frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(nx)dx\]
\[b_{n}=\frac{2}{\pi}\int_{0}^{\pi}(1+x)\sin(nx)dx\]
\[=\frac{2}{\pi}\left(\left[(1+x)\frac{-\cos(nx)}{n}\right]_{0}^{\pi}+\frac{1}{n}\int_{0}^{\pi}\cos(nx)dx\right)\]
\[=\frac{2}{\pi}\left(-\frac{1+\pi}{n}\cos(n\pi)+\frac{1}{n}+\frac{1}{n}\left[\frac{\sin(nx)}{n}\right]_{0}^{\pi}\right)\]
\[=\frac{2}{\pi}\left(\frac{1}{n}-\frac{1+\pi}{n}\cos(n\pi)\right)\]
\[=\frac{2}{\pi n}(1-(1+\pi)\cos(n\pi))\]
If \(n\) is even
\[\because \cos(n\pi) = 1\]
\[\therefore b_n = - \frac 2 n\]
If \(n\) is Odd
\[\because \cos(n\pi) = -1\]
\[\therefore b_n = \frac{4 + 2 \pi}{\pi n}\]
Substituting into the general equation, we have
\[f(x) = \left(\frac 4 \pi + 2\right)\left(\sin (x) + \frac 1 3 \sin (3x) + \frac 1 5 \sin (5x)\right) - 2 \left(\frac 1 2 \sin(2x) + \frac 1 4 \sin(4x)\right)\]
References
Read more about notations and symbols.