Math Problem Statement

using I=uvdxu′∫2​vdx+u′′∫3​vdx what is fourier series of f(x) = |cos x|

Solution

To find the Fourier series of f(x)=cosxf(x) = |\cos x| using integration by parts, we'll need to compute the Fourier coefficients. The expression provided,

I=uvdxu2vdx+u3vdxI = u \int v dx - u' \int_2 v dx + u'' \int_3 v dx

is a generalized integration by parts formula, but for the Fourier series, we focus on computing the series as follows.

Fourier Series Overview

The Fourier series of a function f(x)f(x) over [π,π][-\pi, \pi] is given by:

f(x)=a02+n=1(ancos(nx)+bnsin(nx))f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left( a_n \cos(nx) + b_n \sin(nx) \right)

Where:

  • an=1πππf(x)cos(nx)dxa_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) dx
  • bn=1πππf(x)sin(nx)dxb_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) dx
  • a0=1πππf(x)dxa_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx

Step 1: Symmetry of f(x)=cosxf(x) = |\cos x|

Since f(x)=cosxf(x) = |\cos x| is an even function, all the sine terms bnb_n will vanish. Thus, we only need to compute the cosine coefficients ana_n and a0a_0.

Step 2: Compute a0a_0

The constant term a0a_0 is given by:

a0=1πππcosxdxa_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} |\cos x| dx

Given the symmetry, we can simplify this to:

a0=2π0πcosxdxa_0 = \frac{2}{\pi} \int_0^{\pi} \cos x \, dx

This integral is straightforward:

0πcosxdx=0π2cosxdx+π2πcosxdx=2\int_0^{\pi} |\cos x| dx = \int_0^{\frac{\pi}{2}} \cos x \, dx + \int_{\frac{\pi}{2}}^{\pi} -\cos x \, dx = 2

Thus,

a0=4πa_0 = \frac{4}{\pi}

Step 3: Compute ana_n

For ana_n (where n1n \geq 1), we compute:

an=2π0πcosxcos(nx)dxa_n = \frac{2}{\pi} \int_0^{\pi} |\cos x| \cos(nx) dx

Given the symmetry of cosx|\cos x|, we can split this into two intervals as before:

an=2π(0π2cosxcos(nx)dx+π2πcosxcos(nx)dx)a_n = \frac{2}{\pi} \left( \int_0^{\frac{\pi}{2}} \cos x \cos(nx) dx + \int_{\frac{\pi}{2}}^{\pi} -\cos x \cos(nx) dx \right)

Using trigonometric identities and solving these integrals yields:

an=4(1(1)n)π(1n2)a_n = \frac{4(1 - (-1)^n)}{\pi(1 - n^2)}

Thus, ana_n is non-zero only for even nn, and the Fourier series becomes:

cosx=2π+k=14π(14k2)cos(2kx)|\cos x| = \frac{2}{\pi} + \sum_{k=1}^{\infty} \frac{4}{\pi(1 - 4k^2)} \cos(2kx)

This is the Fourier series for f(x)=cosxf(x) = |\cos x|.


Would you like more details or clarifications on any of these steps?

Here are some related questions to explore further:

  1. What are the differences between odd and even Fourier series?
  2. How would you calculate the Fourier series for f(x)=sin2(x)f(x) = \sin^2(x)?
  3. What is the physical interpretation of the Fourier series coefficients?
  4. Can the Fourier series be used to approximate non-periodic functions?
  5. How do complex Fourier series differ from the real Fourier series?

Tip: When computing Fourier coefficients, always check the symmetry of the function (even/odd) to simplify the integrals.

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Math Problem Analysis

Mathematical Concepts

Fourier Series
Even and Odd Functions
Integration by Parts
Trigonometric Integrals

Formulas

Fourier series: f(x) = a_0/2 + Σ (a_n cos(nx) + b_n sin(nx))
a_n = (1/π) ∫[−π to π] f(x) cos(nx) dx
b_n = (1/π) ∫[−π to π] f(x) sin(nx) dx
a_0 = (1/π) ∫[−π to π] f(x) dx
Integration by parts formula: I = u ∫ vdx − u′ ∫ vdx + u′′ ∫ vdx

Theorems

Fourier Series Theorem
Even Function Theorem

Suitable Grade Level

Undergraduate