Question 9.2: Find the Fourier series of the function f (x) = 4x, 0 ≤ x ≤...
Find the Fourier series of the function
f (x) = 4x, 0 ≤ x ≤ 10, with period 2l = 10, l = 5.
The Fourier coefficients are
a_{0} = \frac{1}{5} \int\limits_{0}^{10}{4x dx} = \frac{2}{5} x^{2}|^{10}_{0}= 40,a_{n} = \frac{1}{5} \int\limits_{0}^{10}{4x cos\frac{nπx}{5} dx} = \frac{4x}{nπ} cos\frac{nπx}{5}|^{10}_{0} − \frac{4}{nπ} \int\limits_{0}^{10}{sin \frac{nπx}{5}dx}
= 0+ \frac{20}{n^{2}π^{2}} cos\frac{nπx}{5}|^{10}_{0}= 0,
b_{n} = \frac{4}{5} \int\limits_{0}^{10}{x sin \frac{nπx}{5} dx} =−\frac{4x}{nπ} cos\frac{nπx}{5}|^{10}_{0}+\frac{4}{nπ} \int\limits_{0}^{10}{cos\frac{nπx}{5}dx}
=−\frac{40}{nπ} + \frac{20}{n^{2}π^{2}} sin \frac{nπx}{5}|^{10}_{0}=−\frac{40}{nπ}.
Hence, the Fourier series reads
f (x) = 20−\frac{40}{π} \sum\limits_{n=1}^{∞}{\frac{1}{n} sin \frac{nπx}{5}}.The first partial sums S_{n} of this series are drawn in Fig. 9.2. A comparison of this series with the starting curve f (x) illustrates the convergence of this Fourier series.
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