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Problem 405

Find the Fourier series of the function \(\mathrm{f}(\mathrm{x})=|\mathrm{x}|,-\pi<\mathrm{x} \leq \pi\).

Short Answer

Expert verified
The Fourier series of the function \(f(x) = |x|\), within the interval \(-\pi < x \leq \pi\), is given by: \[f(x) = \frac{\pi}{4} + \sum_{n=1}^{\infty} \frac{2}{n\pi} (\frac{-1}{n^2}[(-1)^n - 1])\cos(nx)\]
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Step 1: Determine the function's periodicity

The given function f(x) = |x| is periodic within the interval -π < x ≤ π. The periodicity of the given function is 2π since the original interval range (-π to π) is 2π.

Step 2: Determine if the function is even or odd

The given function f(x) = |x| is an even function because f(-x) = |-x| = |x| = f(x). For even functions, we only have to calculate the cosine (a) coefficients because the sine (b) coefficients will be zero for any n.

Step 3: Calculate the a0 coefficient

Firstly, we compute the a0 coefficient by using the formula: \[a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \, dx\] Substituting f(x) with |x|: \[a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} |x| \, dx\] Since the function is an even function, we can rewrite the integral expression: \[a_0 = \frac{2}{2\pi} \int_{0}^{\pi} x \, dx\] Calculate the integral: \[a_0 = \frac{1}{\pi} \left[ \frac{1}{2}x^2 \right]_0^{\pi}\] \[a_0 = \frac{\pi}{2}\]

Step 4: Calculate the an coefficients

Next, we compute the a_n coefficients by using the formula: \[a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx\] Substituting f(x) with |x|: \[a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} |x| \cdot \cos(nx) \, dx\] Since the function is an even function, we can rewrite the integral expression: \[a_n = \frac{2}{\pi} \int_{0}^{\pi} x \cdot \cos(nx) \, dx\] Calculate the integral using integration by parts: Let u = x and dv = cos(nx) dx. Then, du = dx and v = (1/n)sin(nx). Now apply integration by parts formula, ∫udv = uv - ∫vdu: \[a_n = \frac{2}{\pi} \left[(x(\frac{1}{n}\sin(nx)) - \int_0^{\pi} (\frac{1}{n}\sin(nx)) \, dx) \right]\] \[a_n = \frac{2}{n\pi} [(x\sin(nx))\bigg|_0^{\pi} - \int_0^{\pi} \sin(nx) \, dx)\] Further, we evaluate the integral within the square bracket: \[a_n = \frac{2}{n\pi} [(0 - \frac{-1}{n^2}\cos(nx))\bigg|_0^{\pi}]\] \[a_n = \frac{2}{n\pi} [(\frac{-1}{n^2}\cos(n\pi) - \frac{-1}{n^2}\cos(0))]\] \[a_n = \frac{2}{n\pi} (\frac{-1}{n^2}[(-1)^n - 1])\]

Step 5: Write out the Fourier series

Finally, we write the Fourier series using the a0 and an coefficients: \[f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} a_n \cos(nx)\] Substituting with the calculated coefficients, we get: \[f(x) = \frac{\pi}{4} + \sum_{n=1}^{\infty} \frac{2}{n\pi} (\frac{-1}{n^2}[(-1)^n - 1]) \cos(nx)\] This equation represents the Fourier series of the given function f(x) = |x| in the specified interval -π < x ≤ π.

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Most popular questions from this chapter

Chapter 14

The Fourier series for $\mathrm{f}(\mathrm{x})=|\mathrm{x}| \quad-\pi<\mathrm{x} \leq \pi$ and $$ \begin{aligned} &\mathrm{f}(\mathrm{x}+2 \pi)=\mathrm{f}(\mathrm{x}) \text { is } \\ &\qquad(\pi / 2)-(4 / \pi)^{\infty} \sum_{\mathrm{n}=1}\left[\\{\cos (2 \mathrm{n}-1) \mathrm{x}\\} /\left\\{(2 \mathrm{n}-1)^{2}\right\\}\right] \end{aligned} $$ Without computing any Fourier coefficients, find the Fourier series for $$ \begin{array}{ll} g(x)=-1 & -\pi

Chapter 14

State the most general Pointwise Convergence Theorem for Fourier series (i.e. the one with the weakest premises). Discuss its meaning.

Chapter 14

Consider the infinite trigonometric series $\left(a_{0} / 2\right)+{ }^{\infty} \Sigma_{n=1}\left(a_{n} \cos n x+b_{n} \sin n x\right)$ and assume that it converges uniformly for all \(\mathrm{x} \in(-\pi, \pi)\). It can then be considered as a function \(f\) of \(x\) with period \(2 \pi\), i.e. $\mathrm{f}(\mathrm{x})=\left(\mathrm{a}_{0} / 2\right)+{ }^{\infty} \sum_{\mathrm{n}=1} \mathrm{a}_{\mathrm{n}} \cos \mathrm{n} \mathrm{x}+\mathrm{b}_{\mathrm{n}} \sin \mathrm{n} \mathrm{x}$ Determine the values of \(a_{n}, b_{n}\) in terms of \(f(x)\)

Chapter 14

Determine the Fourier series of the function given by $$ \begin{aligned} &f(x)=x^{2}, x \in(-\pi, \pi) \\ &f(x+2 \pi)=f(x), a 11 x \end{aligned} $$

Chapter 14

A piecewise continuous function \(\mathrm{f}(\mathrm{x})\) is to be approximated in the interval \((-\pi, \pi)\) by a trigonometric polynomial of the form $$ g_{n}(x)=\left(A_{0} / 2\right)+{ }^{n} \sum_{k=1} A_{k} \cos k x+B_{k} \sin k x $$ where \(\mathrm{A}_{\mathrm{k}}, \mathrm{B}_{\mathrm{k}}, \mathrm{A}_{\circ}\) are undetermined. Prove that the total square deviation $$ \mathrm{D}_{\mathrm{n}}=\pi \int_{-\pi}\left[\mathrm{f}(\mathrm{x})-\mathrm{g}_{\mathrm{n}}(\mathrm{x})\right]^{2} \mathrm{~d} \mathrm{x} $$ is minimized by choosing \(A_{0}, A_{k}, B_{k}\) to be the Fourier coefficients of $\mathrm{f}, \mathrm{a}_{\mathrm{k}}, \mathrm{b}_{\mathrm{k}}, \mathrm{a}_{0}$.

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