Archive | April 2016

Complex Fourier series calculations

Consider a function f:\mathbb{R}\rightarrow\mathbb{C} . Recall  f is periodic (with period  1 ) iff  \forall x f(x+1)=f(x) . For example,  e_{k}=e_{k}(x)=e^{2\pi ikx} is periodic:

e_{k}\left(x+1\right)=e^{2\pi ikx}e^{2\pi ik}=e^{2\pi ikx}\cdot1=e_{k}\left(x\right)

recalling that e^{iy}=\cos y+i\sin y for y\in\mathbb{R} . Assume  f is periodic and continuous. Define, for  k\in\mathbb{Z} ,

{\displaystyle c_{k}=c_{k}(f)=\int_{0}^{1}f(x)\overline{e_{k}}dx=\int_{0}^{1}f(x)e^{-2\pi ikx}dx}

The c_{k} are called the Fourier coefficients of  f , while the Fourier series is

{\displaystyle \sum_{k=-\infty}^{\infty}c_{k}e_{k}}

Let’s try an example. Take f\left(x\right)=x . Of course, this isn’t periodic. However implicity it is meant that \left.f\right|_{\left[0,1\right]}\left(x\right)=x , and f is the periodic extension of this mapping, i.e. \left.f\right|_{[1,2]}=x-1 \left.f\right|_{[-1,0]}=x+1 , etc. For k\neq0 ,

\begin{array}{ccc} c_{k} & = & {\displaystyle \int_{0}^{1}xe^{-2\pi ikx}}\\ \\ & = & {\displaystyle \int_{0}^{1}x\left(\cos\left(2\pi kx\right)-i\sin\left(2\pi kx\right)\right)}\\ \\ & = & {\displaystyle \int_{0}^{1}x\cos\left(2\pi kx\right)-i\int_{0}^{1}x\sin\left(2\pi kx\right)}\end{array}

Evaluating these (real) integrals individually, it is found that

{\displaystyle c_{k}=0-i\left(-\frac{1}{2\pi k}\right)=\frac{i}{2\pi k}}

Also c_{0}=\int_{0}^{1}x=1/2 . Hence the Fourier series of f is

{\displaystyle \sum_{k=-\infty}^{\infty}\frac{i}{2\pi k}e^{2\pi ikx}}

implicitly having the c_{0} -th term to be 1/2 , not i/\left(2\pi\cdot0\right) . However, immediately this a bit difficult to deal with, say if pointwise convergence were in question!

To deal with this, we derive an equivalent representation of the Fourier series. Notice that in general,

{\displaystyle \sum_{k=-n}^{n}c_{k}e_{k}=c_{0}+\sum_{k=1}^{n}(c_{-k}e_{-k}+c_{k}e_{k})}

Continuing, for k\ne0 , \overline{e_{k}}=e^{-2\pi kx}=e_{-k} . Similarly,

\begin{array}{ccc}c_{-k} & = & {\displaystyle \int_{0}^{1}f(x)e^{-2\pi i(-k)x}}\\ \\ & = & {\displaystyle \int_{0}^{1}f(x)e_{k}}\\ \\ & = & {\displaystyle \int_{0}^{1}f(x)\cos\left(2\pi kx\right)+i\int_{0}^{1}f(x)\sin\left(2\pi kx\right)}\end{array}

The same work shows

{\displaystyle c_{k}=\int_{0}^{1}f(x)\cos\left(2\pi kx\right)-i\int_{0}^{1}f(x)\sin\left(2\pi kx\right)}

It is then immediate that  c_{-k}=\overline{c_{k}} . Hence


recalling that  \forall z\in\mathbb{C} , \overline{z}+z=2\mathrm{Re}(z) . Consequently the equivalent expression

{\displaystyle c_{0}+\sum_{k=-n}^{n}2\mathrm{Re}(c_{k}e_{k})}

for the partials sums is obtained, thereby obtaining the equivalent expression

{\displaystyle c_{0}+\sum_{k=1}^{\infty}2\mathrm{Re}(c_{k}e_{k})}

for the Fourier series of f . This is much easier to work with for calculations. For example the recently calculated Fourier series can now be written as

{\displaystyle \sum_{k=1}^{\infty}2\mathrm{Re}\left(\frac{1}{2\pi k}\left(i\cos\left(2\pi kx\right)+i^{2}\sin\left(2\pi kx\right)\right)\right)=\sum_{k=1}^{\infty}-\frac{1}{\pi k}\sin\left(2\pi kx\right)}

Since 1/k\rightarrow0 and are decreasing, for pointwise convergence it is sufficient to show that

{\displaystyle \sum_{k=1}^{n}\sin\left(2\pi kx\right)}

are bounded. But this is true, and a common proof is due to Dirichlet (e.g. as explained in this post–easier to show \sum_{k=1}^{n}\sin\left(ky\right) is bounded and substitute y=2\pi x ). But more than pointwise convergence is true: Uniform convergence of the Fourier series of f\left(x\right)=x is guaranteed, and in fact to f . This by a theorem since f is continuous, periodic, and piecewise continuously differentiable.