Express f(x) = |x| - π for |x| < π as Fourier series. Hence show that 1 + 1/3 + 1/5 + ... = 1/8.

Steps to Solve

1. Define the function and its interval The function ( f(x) = |x| ) is given on the interval ( -\pi < x < \pi ).

2. Calculate the Fourier coefficients We need to calculate the Fourier coefficients ( a_n ) and ( b_n ). The formulas for the coefficients are:

   $$ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) , dx $$

   $$ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) , dx $$

   $$ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) , dx $$

3. Calculate ( a_0 ) First, calculate the average of the function:

   $$ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} |x| , dx $$ Given ( |x| ) is an even function, we can simplify it as: $$ a_0 = \frac{1}{2\pi} \cdot 2 \int_{0}^{\pi} x , dx = \frac{1}{\pi} \left[ \frac{x^2}{2} \right]_{0}^{\pi} = \frac{\pi^2}{2\pi} = \frac{\pi}{2} $$

4. Calculate ( a_n ) Now calculate ( a_n ):

   $$ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} |x| \cos(nx) , dx = \frac{2}{\pi} \int_{0}^{\pi} x \cos(nx) , dx $$ Use integration by parts: Let ( u = x ) and ( dv = \cos(nx) , dx ).

5. Calculate ( b_n ) Now calculate ( b_n ):

   $$ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} |x| \sin(nx) , dx $$ Again due to the properties of sine and symmetry: $$ b_n = \frac{2}{\pi} \int_{0}^{\pi} x \sin(nx) , dx $$

6. Construct the Fourier series The Fourier series can be expressed as:

   $$ f(x) \sim a_0 + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx)) $$

7. Show the convergence To show that the series converges to a specific value (e.g., (\frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \ldots = \frac{\pi^2}{8})), we recognize it resembles the series of the Fourier expansion evaluated at ( x = 0 ). Thus, using the properties of Fourier series convergence.