Question

What is the fourier series of dirichlet function?

Answer 1

Given the Dirichlet function defined as:

f(x)={01x∈Qx∈R∖Qf(x)={0x∈Q1x∈R∖Q

Find the corresponding Fourier series.

Before starting, I believe that f(x)f(x) is periodic for any real period greater than zero. Therefore I feel free to choose 2π2π as period. Hence the Fourier series will have the following form:

F(x)=a0+∑n=1∞ancosnx+bnsinnxF(x)=a0+∑n=1∞ancos⁡nx+bnsin⁡nx

With the following coefficients:

a0=12π∫π−πf(x)dxan=1π∫π−πf(x)cosnxdxbn=1π∫π−πf(x)sinnxdxa0=12π∫−ππf(x)dxan=1π∫−ππf(x)cos⁡nxdxbn=1π∫−ππf(x)sin⁡nxdx

My function is not Riemann-integrable, however is Lebesgue-integrable. Knowing the following theorem...

I=I1∪I2g(x)={abx∈I1x∈I2∫Ig(x)dx=∫I1g(x)dx+∫I2g(x)dxI=I1∪I2g(x)={ax∈I1bx∈I2∫Ig(x)dx=∫I1g(x)dx+∫I2g(x)dx

... I can easily calculate the coefficients a0=1,an=bn=0a0=1,an=bn=0. Therefore the Fourier series corresponding to the Dirichlet function is:

F(x)=1F(x)=1

Which does not converge to the function.

Answer 2

the Fourier series corresponding to the Dirichlet function is: F(x)=1