Answer to Question #86240 in Calculus for Ajay

Question

Obtain the Fourier cosine series for the following function:

{1for 0≤ x<1
0 for 1≤ x <4 }

Answer 1

Solution:

If

$$f(x)=\begin{cases} 1 &\text{ if } 0≤ x<1 \\ 0 &\text{if 1≤ x <4 } \end{cases}$$

then the Fourier cosine series for this function is:

$$f(x)=\frac {a_0}{2}+{\displaystyle\sum_{n=1}^\infin}a_ncos\frac{n{\pi}x}{4},$$

$$0≤ x<4,$$

where

$$a_0=\frac{1}{2}{\smallint_0}^1 1dx+\frac{1}{2}{\smallint_1}^4 0dx=\frac{1}{2}$$

and

$$a_n=\frac{1}{2}{\smallint_0}^1 1cos\frac{n{\pi}x}{4}dx+\frac{1}{2}{\smallint_1}^4 0cos\frac{n{\pi}x}{4}dx=\frac{2}{n{\pi}}sin\frac{n{\pi}}{4}.$$

or

$$f(x)=\frac {1}{4}+2{\displaystyle\sum_{n=1}^\infin}\frac{1}{n{\pi}}sin\frac{n{\pi}}{4}cos\frac{n{\pi}x}{4},$$ $$0≤ x<4.$$

Answer:

$$f(x)=\frac {1}{4}+2{\displaystyle\sum_{n=1}^\infin}\frac{1}{n{\pi}}sin\frac{n{\pi}}{4}cos\frac{n{\pi}x}{4},$$

$$0≤ x<4.$$