Day: October 28, 2022

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202210280936 Problem 1.24

Expand f(t)=\sin^2t\cos^3t in Fourier series.

Extracted from Hwei Piao Hsu. (1984). HBJ College Outline of Applied Fourier Analysis.

Roughwork.

One can make use of the identities below:

\begin{aligned} e^{\pm\mathrm{i}n\theta} & = \cos n\theta\pm\mathrm{i}\sin n\theta \\ \cos n\theta & = \frac{e^{\mathrm{i}n\theta}+e^{-\mathrm{i}n\theta}}{2} \\ \sin n\theta & = \frac{e^{\mathrm{i}n\theta}-e^{-\mathrm{i}n\theta}}{2\mathrm{i}} \\ \end{aligned}

and thus

\begin{aligned} & \begin{cases} \sin^2t = \displaystyle{\bigg(\frac{e^{\mathrm{i}t}-e^{-\mathrm{i}t}}{2\mathrm{i}}\bigg)^2 }\\ \cos^3t = \displaystyle{ \bigg(\frac{e^{\mathrm{i}t}+e^{-\mathrm{i}t}}{2}\bigg)^3 }\\ \end{cases} \\ \Longrightarrow & \begin{cases} \sin^2t = \displaystyle{\frac{e^{2\mathrm{i}t}-2e^{\mathrm{i}t}e^{-\mathrm{i}t}+e^{-2\mathrm{i}t}}{-4}} \\ \cos^3t = \displaystyle{\frac{e^{3\mathrm{i}t}+3e^{\mathrm{i}t}+3e^{-\mathrm{i}t}+e^{-3\mathrm{i}t}}{8}} \\ \end{cases} \\ \end{aligned}

Or one can make use of the identity

\sin^2\theta + \cos^2\theta = 1.

and hence

\begin{aligned} f(t) & = (1-\cos^2t)\cos^3t \\ & = \cos^3t-\cos^5t \\ \end{aligned}

where \cos^5t is as expanded

\begin{aligned} \cos^5t & = \bigg(\frac{e^{\mathrm{i}t}+e^{-\mathrm{i}t}}{2}\bigg)^5 \\ & = \frac{e^{5\mathrm{i}t}+5e^{3\mathrm{i}t} +10e^{\mathrm{i}t} +10e^{-\mathrm{i}t}+5e^{-3\mathrm{i}t}+ e^{-5\mathrm{i}t}}{32}\\ \end{aligned}

so that

\begin{aligned} f(t) & = -\frac{e^{5\mathrm{i}t}}{32}+\bigg(\frac{1}{8}-\frac{5}{32}\bigg)e^{3\mathrm{i}t} + \bigg(\frac{3}{8}-\frac{10}{32}\bigg) e^{\mathrm{i}t} \\ & \qquad + \bigg(\frac{3}{8}-\frac{10}{32}\bigg) e^{-\mathrm{i}t} + \bigg(\frac{1}{8}-\frac{5}{32}\bigg) e^{-3\mathrm{i}t} - \frac{e^{-5\mathrm{i}t}}{32}\\ & = -\frac{e^{5\mathrm{i}t}}{32}-\frac{e^{3\mathrm{i}t}}{32} + \frac{e^{\mathrm{i}t}}{16} + \frac{e^{-\mathrm{i}t}}{16} - \frac{e^{-3\mathrm{i}t}}{32} -\frac{e^{-5\mathrm{i}t}}{32} \\ & = \frac{1}{16}\Bigg( 2\bigg(\frac{e^{\mathrm{i}t}+e^{-\mathrm{i}t}}{2}\bigg) - \frac{e^{3\mathrm{i}t}+e^{-3\mathrm{i}t}}{2} - \frac{e^{5\mathrm{i}t}+e^{-5\mathrm{i}t}}{2} \Bigg)\\ & = \cdots \\ \end{aligned}

Answer. \displaystyle{\frac{1}{16}(2\cos t-\cos 3t-\cos 5t)}.

No slightest disrespect to de Moivre’s and Euler’s formulae but in due respect of Fourier’s analysis, this problem had need be treated again.

Background.

A function f(t) satisfying f(t+T)=f(t) is called a periodic function, and the smallest T its period. A periodic function f(t) can be represented by the trigonometric Fourier series

\displaystyle{f(t)=\frac{1}{2}a_0+\sum_{n=1}^{\infty}(a_n\cos n\omega_0t+b_n\sin n\omega_0t)}

or

\displaystyle{f(t)=C_0+\sum_{n=1}^{\infty}C_n\cos (n\omega_0t-\theta_n)}

where \omega_0=2\pi /T. The coefficients of the Fourier series are found by using the orthogonality properties of sine and cosine functions over a period:

\begin{aligned} \begin{Bmatrix} a_n \\ b_n \end{Bmatrix} & = \frac{2}{T}\int_{-T/2}^{T/2}f(t)\begin{Bmatrix}\cos n\omega_0t \\ \sin n\omega_0t \end{Bmatrix}\,\mathrm{d}t \\ C_0 & = \frac{1}{2}a_0 \\ C_n & = \sqrt{a_n^2+b_n^2} \\ \theta_n & = \tan^{-1}\bigg(\frac{b_n}{a_n}\bigg) \\ \end{aligned}

Text on pg. 11

For f(t)=\sin^2t\cos^3t,

\begin{aligned} T & = 2\pi \\ \omega_0 & = 1 \\ \begin{Bmatrix} a_n \\ b_n \end{Bmatrix} & = \frac{1}{\pi}\int_{-\pi}^{\pi}\sin^2t\cos^3t\begin{Bmatrix}\cos nt \\ \sin nt\end{Bmatrix}\,\mathrm{d}t\\ \end{aligned}

When n=0:

\begin{aligned} a_0 & = \frac{1}{\pi}\int_{-\pi}^{\pi}\sin^2t\cos^3t\,\mathrm{d}t \\ \cdots\enspace\textrm{Let }& u=\sin t\textrm{ s.t. }\mathrm{d}t=\frac{\mathrm{d}u}{\cos t}\enspace\cdots \\ & = \frac{1}{\pi}\int_{t=-\pi}^{t=\pi}u^2\cos^2t\,\mathrm{d}u \\ & = \frac{1}{\pi}\int_{t=-\pi}^{t=\pi}u^2(1-u^2)\,\mathrm{d}u \\ & = \frac{1}{\pi}\int_{t=-\pi}^{t=\pi}(u^2-u^4)\,\mathrm{d}u \\ & = \frac{1}{\pi}\bigg[ \frac{u^3}{3}-\frac{u^5}{5} \bigg]\bigg|_{t=-\pi}^{t=\pi} \\ & = \frac{1}{\pi}\bigg[ \frac{\sin^3t}{3}-\frac{\sin^5t}{5} \bigg]\bigg|_{t=-\pi}^{t=\pi} \\ & = 0 \\ \end{aligned}

When n=1:

\begin{aligned} a_1 & = \frac{1}{\pi}\int_{-\pi}^{\pi}\sin^2t\cos^4t\,\mathrm{d}t \\ \dots\enspace &\textrm{By WolframAlpha}\enspace\dots \\ & = \frac{1}{\pi}\bigg[\frac{1}{192}(12t+3\sin (2t)-3\sin (4t)-\sin (6t))\bigg]\bigg|_{-\pi}^{\pi} \\ & = \frac{1}{\pi}\bigg(\frac{24\pi}{192}\bigg) \\ & = \frac{1}{8} \\ \end{aligned}

a bit tiring an exercise.

(to be continued)