Fourier Series (Part 2)
Introduction and Derivation of Fourier Series
Previously, you looked at Fourier series and derived what an even Fourier series would look like for a function. Now we are going to determine what the Fourier series would be like for an odd function and a general function.
Fourier Series for Odd Functions
Remember that an odd function is a function that has the property \(f(-x)=-f(x)\). The odd Fourier series will have the form
Similar to before, we want:
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Using the definition of an odd function and looking at \(F_{0}(x)\), explain why there cannot be a constant term \(b_{0}\)
\[ \begin{align} F_{0}(x) &= b_0 + b_{1}\sin(x)+b_{2}\sin(2x)+b_{3}\sin(3x)+\cdots\\ F_{0}(-x) &= b_0 + b_{1}\sin(-x)+b_{2}\sin(-2x)+b_{3}\sin(-3x)+\cdots\\ &= b_0 - b_{1}\sin(x)-b_{2}\sin(2x)-b_{3}\sin(3x)-\cdots\\ &= b_0 - F_{0}(x)\\\\ \therefore b_0 &= 0 \text{ given that } F_{0}(x) = -F_{0}(-x) \quad \blacksquare \end{align} \] -
Simplify \(\cos(mx+nx)-\cos(mx-nx)\) and use that to simplify \(\int_{-\pi}^{\pi}\sin(nx)\sin(mx) dx\) when \(m \ne n\) and when \(m=n\). (Assume that \(m\), \(n\) are integers).
\[ \begin{align} &\cos(mx+nx)-\cos(mx-nx)\\ &= -2\sin\left(\frac{(mx+nx)+(mx-nx)}{2}\right)\sin\left(\frac{(mx+nx)-(mx-nx)}{2}\right)\\ &= -2\sin(mx)\sin(nx)\\\\ \end{align} \]When \(n=m\):
\[ \begin{align} &\int_{-\pi}^{\pi}\sin^2(nx) dx\\ &= \int_{-\pi}^{\pi}\frac{1-\cos(2nx)}{2} dx\\ &= \frac{1}{2}\left[x\right]_{-\pi}^{\pi} - \frac{1}{2}\int_{-\pi}^{\pi}\cos(2nx) dx\\ &= \frac{1}{2}\cdot 2\pi\ - \frac{1}{2}\left[\frac{\sin(2nx)}{2n}\right]_{-\pi}^{\pi}\\ &= \pi - \frac{1}{2}\left[\frac{\sin(0)}{2n}-\frac{\sin(0)}{2n}\right]\\ &= \pi \end{align} \]When \(n \ne m\):
\[ \begin{align} &\int_{-\pi}^{\pi}\sin(nx)\sin(mx)dx\\ &= \frac{1}{2}\int_{-\pi}^{\pi}[\cos((m-n)x) + \cos((m+n)x)]dx\\ &= \frac{1}{2}\left[\frac{\sin((m-n)x)}{m-n} + \frac{\sin((m+n)x)}{m+n}\right]_{-\pi}^{\pi}\\ &= \frac{1}{2}\left[\frac{\sin(0)}{m-n} + \frac{\sin(0)}{m+n} - \frac{\sin(0)}{m-n} - \frac{\sin(0)}{m+n}\right]\\ &= 0 \end{align} \] -
Use the previous result to find \(b_{1}\) in terms of \(\int_{-\pi}^{\pi}\sin(x)f(x)dx\).
\[ \begin{align} \int_{-\pi}^{\pi}\sin(x)f(x)dx &= \int_{-\pi}^{\pi}b_{1}\sin^2(x) + b_{2}\sin(x)\sin(2x) + \cdots dx\\ &= b_{1}\int_{-\pi}^{\pi}\sin^2(x)dx\\ &= b_{1}\pi\\\\ \therefore b_{1} &= \frac{1}{\pi}\int_{-\pi}^{\pi}\sin(x)f(x)dx \quad \blacksquare \end{align} \] -
Do the same for \(b_{2}\) and \(b_{3}\) and generalize for \(b_{k}\) over the interval \([-\pi,\pi]\)
\[ \begin{align} \int_{-\pi}^{\pi}\sin(2x)f(x)dx &= \int_{-\pi}^{\pi}b_{1}\sin(x)\sin(2x) + b_{2}\sin^2(2x) + \cdots dx\\ &= b_{2}\int_{-\pi}^{\pi}\sin^2(2x)dx\\ &= b_{2}\pi\\\\ \therefore b_{2} &= \frac{1}{\pi}\int_{-\pi}^{\pi}\sin(2x)f(x)dx \quad \blacksquare \end{align} \]\[ \begin{align} \int_{-\pi}^{\pi}\sin(3x)f(x)dx &= \int_{-\pi}^{\pi} \cdots + b_{2}\sin(2x)\sin(3x) + b_{3}\sin^2(3x) + \cdots dx\\ &= b_{3}\int_{-\pi}^{\pi}\sin^2(3x)dx\\ &= b_{3}\pi\\\\ \therefore b_{3} &= \frac{1}{\pi}\int_{-\pi}^{\pi}\sin(3x)f(x)dx \quad \blacksquare \end{align} \]\[ \begin{align} \int_{-\pi}^{\pi}\sin(kx)f(x)dx &= \int_{-\pi}^{\pi} \cdots + b_{k-1}\sin((k-1)x)\sin(kx) + b_{k}\sin^2(kx) + \cdots dx\\ &= b_{k}\int_{-\pi}^{\pi}\sin^2(kx)dx\\ &= b_{k}\pi\\\\ \therefore b_{k} &= \frac{1}{\pi}\int_{-\pi}^{\pi}\sin(kx)f(x)dx \quad \blacksquare \end{align} \] -
Now let \(f(x)=\sin^{3}x\). Find the coefficients for \(b_{1}\), \(b_{2}\), \(...\), \(b_{5}\) and graph both the Fourier series and the original function over the interval \([-\pi,\pi]\).
\[ \begin{aligned} b_1 &= \frac{1}{\pi}\int_{-\pi}^{\pi}\sin(x)\sin^{3}x dx &= 0.75\\ b_2 &= \frac{1}{\pi}\int_{-\pi}^{\pi}\sin(2x)\sin^{3}x dx &= 0\\ b_3 &= \frac{1}{\pi}\int_{-\pi}^{\pi}\sin(3x)\sin^{3}x dx &= -0.25\\ b_4 &= \frac{1}{\pi}\int_{-\pi}^{\pi}\sin(4x)\sin^{3}x dx &= 0\\ b_5 &= \frac{1}{\pi}\int_{-\pi}^{\pi}\sin(5x)\sin^{3}x dx &= 0\\ \end{aligned} \] -
Show, using identities, that the Fourier series that you generated is EXACTLY the same as \(f(x)\).
General Fourier Series
Now we are going to consider the Fourier series for a general function.
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What is the value of \(\int_{-\pi}^{\pi}\sin(nx)\cos(mx)dx\) for all values of \(n=m\). and \(n\ne m\)? (Hint: Remember that one function is odd, one function is even. What kind of function must their product be? What does that mean about the integral?)
When \(n=m\):
\[ \begin{align} &\int_{-\pi}^{\pi}\sin(nx)\cos(nx)dx\\ &= \frac{1}{2}\int_{-\pi}^{\pi}\sin(nx + nx) + \sin(nx - nx)dx\\ &= \frac{1}{2}\int_{-\pi}^{\pi}\sin(2nx)dx\\ &= 0 \end{align} \]When \(n \ne m\):
\[ \begin{align} &\int_{-\pi}^{\pi}\sin(nx)\cos(mx)dx\\ &= \frac{1}{2}\int_{-\pi}^{\pi}[\sin(nx + mx) + \sin(nx - mx)]dx\\ &= \frac{1}{2}\int_{-\pi}^{\pi}[\sin((n+m)x) + \sin((n-m)x)]dx\\ &= 0 \end{align} \] -
Since we are doing general Fourier series, now we are considering the function
\[ F(x)=a_{0}+a_{1}\cos(x)+b_{1}\sin(x)+a_{2}\cos(2x)+b_{2}\sin(2x)+\cdots \]Use \(\int_{-\pi}^{\pi}f(x)=\int_{-\pi}^{\pi}F(x)\) to find \(a_0\). (Hint: This should look familiar)
\[ \begin{align} \int_{-\pi}^{\pi}f(x) &= \int_{-\pi}^{\pi}a_{0}+a_{1}\cos(x)+b_{1}\sin(x)+a_{2}\cos(2x)+b_{2}\sin(2x)+\cdots dx\\ &= 2\pi a_{0} + \int_{-\pi}^{\pi}a_{1}\cos(x)+b_{1}\sin(x)+\cdots dx\\ \end{align} \]The area under the curve of \(\cos(nx)\) and \(\sin(nx)\) (given \(n\) is an integer where \(n \neq 0\)) over the interval \([-\pi,\pi]\) is zero, since \(n\) describes the number of complete waves within a period of \(2\pi\). In other words:
\[ \int_{-\pi}^{\pi}\cos(nx) dx = 0 \text{ and } \int_{-\pi}^{\pi}\sin(nx) dx = 0 \]\[ \begin{align} \int_{-\pi}^{\pi}f(x) &= 2\pi a_{0} + \int_{-\pi}^{\pi}a_{1}\cos(x)+b_{1}\sin(x)+\cdots dx\\ &= 2\pi a_{0}\\\\ \therefore a_0 &= \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)dx \quad \blacksquare \end{align} \] -
So now think of finding \(a_{1}\) by multiplying \(f(x)\) and \(F(x)\) by \(\cos(x)\). What two integrals will be set equal? Use these integrals to find \(a_{1}\) in terms of \(\int_{-\pi}^{\pi}f(x)\). (Hint: This should look familiar)
\[ \begin{align} \int_{-\pi}^{\pi}f(x)\cos(x) &= \int_{-\pi}^{\pi}a_{0}\cos(x)+a_{1}\cos^2(x)+b_{1}\sin(x)\cos(x)+\cdots dx\\ &= a_{1}\int_{-\pi}^{\pi}\cos^2(x)dx\\ &= a_{1}\pi\\\\ \therefore a_{1} &= \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(x)dx \quad \blacksquare \end{align} \] -
Now multiply \(f(x)\) and \(F(x)\) by \(\cos(nx)\) (\(n=\) integer). Use this to find the pattern for \(a_n\) (Hint: Again, should look familiar). (Hint 2: You should be using information from previous questions)
\[ \begin{align} \int_{-\pi}^{\pi}f(x)\cos(nx) &= \int_{-\pi}^{\pi}\cdots+a_{1}\cos(x)\cos(nx)+b_{1}\sin(x)\cos(nx)+\cdots dx\\ &= a_{n}\int_{-\pi}^{\pi}\cos^2(nx)dx\\ &= a_{n}\pi\\\\ \therefore a_{n} &= \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos(nx)dx \quad \blacksquare \end{align} \] -
Now do similar things as the previous question but multiplying by \(\sin(mx)\) to find the pattern for \(b_{n}\).
\[ \begin{align} \int_{-\pi}^{\pi}f(x)\sin(mx) &= \int_{-\pi}^{\pi}\cdots+a_{1}\cos(x)\sin(mx)+b_{1}\sin(x)\sin(mx)+\cdots dx\\ &= b_{n}\int_{-\pi}^{\pi}\sin^2(mx)dx\\ &= b_{n}\pi\\\\ \therefore b_{n} &= \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin(mx)dx \quad \blacksquare \end{align} \] -
Let \(f(x)=\frac{1}{1+e^{x}}\). Find the coefficients \(a_{0}\), \(a_{1}\), \(...\), \(a_{5}\) and \(b_{1}\), \(b_{2}\), \(...\), b_{5}.
\[ \begin{aligned} a_0 &= \frac{1}{2\pi}\int_{-\pi}^{\pi}\frac{1}{1+e^{x}}dx &= 0.5\\ a_1 &= \frac{1}{\pi}\int_{-\pi}^{\pi}\frac{1}{1+e^{x}}\cos(x)dx &= 0\\ b_1 &= \frac{1}{\pi}\int_{-\pi}^{\pi}\frac{1}{1+e^{x}}\sin(x)dx &\approx -.3913\\ a_2 &= \frac{1}{\pi}\int_{-\pi}^{\pi}\frac{1}{1+e^{x}}\cos(2x)dx &= 0\\ b_2 &= \frac{1}{\pi}\int_{-\pi}^{\pi}\frac{1}{1+e^{x}}\sin(2x)dx &\approx 0.1447\\ a_3 &= \frac{1}{\pi}\int_{-\pi}^{\pi}\frac{1}{1+e^{x}}\cos(3x)dx &= 0\\ b_3 &= \frac{1}{\pi}\int_{-\pi}^{\pi}\frac{1}{1+e^{x}}\sin(3x)dx &\approx -0.0982\\ a_4 &= \frac{1}{\pi}\int_{-\pi}^{\pi}\frac{1}{1+e^{x}}\cos(4x)dx &= 0\\ b_4 &= \frac{1}{\pi}\int_{-\pi}^{\pi}\frac{1}{1+e^{x}}\sin(4x)dx &\approx 0.0733\\ a_5 &= \frac{1}{\pi}\int_{-\pi}^{\pi}\frac{1}{1+e^{x}}\cos(5x)dx &= 0\\ b_5 &= \frac{1}{\pi}\int_{-\pi}^{\pi}\frac{1}{1+e^{x}}\sin(5x)dx &\approx -0.0585\\ \end{aligned} \] -
Graph \(f(x)\) and your 5th order Fourier approximation and compare on the interval \([-\pi,\pi]\)