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Problem 405

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

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

## 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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