Question

Find the Fourier series for the function f (x) = |Cosx| in (-π,π)

Answer 1

Given f(x) = |sinx|

f(x) = |sinx| = sinx

f(-x) = |sin(-x)| = sinx

Hence f(x) = f(-x)

∴∴ |sinx| is even function

The Fourier series of an even function contains only cosine terms and is known as Fourier Series and is given by

f(x)=a02+∑n=1∞ancosnxf(x)=a02+∑n=1∞ancosnx

a0=1π∫−ππf(x)dx=2π∫0πf(x)cosnxdxa0=1π∫−ππf(x)dx=2π∫0πf(x)cosnxdx

∴∴ Let us first find

a0=2π∫0πsinxdxa0=2π∫0πsinxdx

∴a0=2π∫0πsinxdx∴a0=2π∫0πsinxdx

∴a0=2π[−cosx]π0=2π=4π∴a0=2π[−cosx]0π=2π=4π

∴a0=4π∴a0=4π

Now

an=2π∫0πf(x)cosnxdxan=2π∫0πf(x)cosnxdx

∴an=2π∫0πsinxcosnxdx∴an=2π∫0πsinxcosnxdx

∴an=2π∫0πsin(1+n)x+sin(1−n)xdx∴an=2π∫0πsin(1+n)x+sin(1−n)xdx

∵sinAsinB=12[sin(A+B)+sin(A−B)]∵sinAsinB=12[sin(A+B)+sin(A−B)]

∴an=1π∫0πsin(1+n)x+sin(1−n)xdx∴an=1π∫0πsin(1+n)x+sin(1−n)xdx

∴an=1π[−cos(1+n)x1+n−cos(1−n)x1−n]π0∴an=1π[−cos(1+n)x1+n−cos(1−n)x1−n]0π

∴an=1π[−cos(1+n)π1+n−cos(1−n)π1−n+11−n+11+n]∴an=1π[−cos(1+n)π1+n−cos(1−n)π1−n+11−n+11+n]

∴an=1π[−cos(π+nπ)1+n−cos(π−nπ)1−n+21−n2]∴an=1π[−cos(π+nπ)1+n−cos(π−nπ)1−n+21−n2]

∴an=1π[−(−cosnπ)1+n−(−cosnπ)1−n+21−n2]∴an=1π[−(−cosnπ)1+n−(−cosnπ)1−n+21−n2]

∴an=1π[2cosnπ1−n2+21−n2]∴an=1π[2cosnπ1−n2+21−n2]

an=2π(1−n2)(cosnπ+1)an=2π(1−n2)(cosnπ+1)

an=2π(1−n2)(2)an=2π(1−n2)(2)

If n is even

an=4π(1−n2)an=4π(1−n2)

an=0an=0

If n is odd

∴f(x)=a02+∑n=1∞ancosnx∴f(x)=a02+∑n=1∞ancosnx

∴f(x)=12∗4π+∑n=1∞ancos2nx∴f(x)=12∗4π+∑n=1∞ancos2nx

(∵an∵an exist only if n is even i.e. put n = 2n)

∴f(x)=2π+∑n=1∞4π(1−(2n)2)cos2nx∴f(x)=2π+∑n=1∞4π(1−(2n)2)cos2nx

(i.e. put n = 2n)

∴f(x)=2π4π∑n=1∞cos2nx(1−4n2)∴f(x)=2π4π∑n=1∞cos2nx(1−4n2)

By re-arranging,

f(x)=2π−4π∑n=1∞cos2nx(4n2−1)