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Fourier Coefficients

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In the previous posts, I used \(\cdot\) to refer to the inner product. Given that this is quite confusing as \(\cdot\) is also used for multiplication, I'm now going to use \(\iprod{f,g}\) to refer to the inner product of \(f,g\).

Determining Coefficients of Fourier Series

Suppose I had a function \(f\) that can be written as a fourier series as in equation 1 (we still haven't figured out what conditions need to be imposed on \(f\) to ensure this, but we'll take care of that in due time). Then by equation 2, $$\iprod{f,1}=\iprod{c,1}=2\pi c$$$$c=\frac1{2\pi}\iprod{f,1}\tag{10}$$Similarly, by theorem 1, $$a_k=\frac1\pi\iprod{f,\sin kx}\tag{11}$$$$b_k=\frac1\pi\iprod{f,\cos kx}\tag{12}$$

There is a slight lack of symmetry between equation 10 and equations 11 and 12 because of the extra \(\frac12\) term. Thus, from now on, our standard form for a Fourier Series will be $$f(x)=\frac c2+\sum\limits_{k=1}^\infty(a_k\sin kx+b_k\cos kx)\tag{13}$$
In the derivations of equations 10, 11, and 12, we implictly used the fact $$\Iprod{\frac c2+\sum\limits_{k=1}^\infty(a_k\sin kx+b_k\cos kx),g}=cg+\sum\limits_{k=1}^\infty\iprod{a_k\sin kx+b_k\cos kx,g}$$for some function \(g\). In other words, we need to show that the inner product is linear over infinite sums as well. This requires Fubini's Theorem and Tonelli's Theorem. Note that for \(f\) defined in equation 13$$\Iprod{f,g}=\frac{|c|}2\intp|g|dx+\intp\sum\limits_{k=1}^\infty(a_k\sin kx\cdot g+b_k\cos kx\cdot g)dx$$By Fubini's theorem, if $$\intp\sum\limits_{k=1}^\infty\Card{a_k\sin kx\cdot g+b_k\cos kx\cdot g}dx<\infty$$then we can move the summation outside the integral. By Tonelli's theorem, this is equivalent to $$\sum\limits_{k=1}^\infty\intp\Card{a_k\sin kx\cdot g+b_k\cos kx\cdot g}dx$$Assuming \(g\) is orthogonal to all but one of the terms of \(f\), this summation is finite, and thus Fubini's theorem implies that the procedure was well-defined.
Consider the function \(f(x)=x\) on \([-\pi,\pi)\). $$\frac1\pi\iprod{x,1}=0$$$$\frac1\pi\iprod{x,\cos kx}=0$$$$\frac1\pi\iprod{x,\sin kx}=(-1)^{k+1}\cdot\frac2k$$Thus, the Fourier Series is $$2\sin x-\sin 2x+\frac{2\sin 3x}3+\ldots$$So how well has the Fourier Series done on modeling \(f(x)=x\)? Well, because of the \(2\pi\) periodicity, we know that our Fourier Series will be poor outside of \([-\pi,\pi)\). Here is the first three terms of the series

The first 25 terms

And the first 100 terms

Outside of the funkiness at \(x=k\pi\), the series looks like it is converging well!

Convergence of twice continuously differentiable functions

How do we evaluate the inner products of \(f\) with our trigonometric functions? If we assume that \(f\in C^2\) (the set of functions which have continuous second derivative), then we can use integration by parts to rewrite equations 11-12 as follows:$$b_k=\frac1\pi\iprod{f,\cos kx}=\left[\frac1{k\pi}f(x)\sin kx\right]\Bigg\vert_{-\pi}^\pi-\frac1{k\pi}\iprod{f^\prime,\sin kx}$$$$=\left[\frac1{k\pi}f(x)\sin kx\right]\Bigg\vert_{-\pi}^\pi+\left[\frac1{k^2\pi}f^\prime(x)\cos kx\right]\Bigg\vert_{-\pi}^\pi-\frac1{k^2\pi}\iprod{f^{\prime\prime},\cos kx}$$Note that the non-integral terms are zero, leaving us with $$b_k=\frac1{k^2\pi}\iprod{f^{\prime\prime},\cos kx}\tag{14}$$If \(\sup|f^{\prime\prime}|=M\), then we have $$b_k\leq\frac1{k^2\pi}\iprod{M,1}=\frac{2M}{k^2}$$A parallel argument shows that $$a_k\leq\frac{2M}{k^2}$$Thus, $$\sum\limits_{k=1}^\infty\left(|a_k\sin kx|+|b_k\cos kx|\right)\leq \sum\limits_{k=1}^\infty(|a_k|+|b_k|)\leq\frac{2M\pi^2}3$$with the last inequality due to Tonelli. By the Preliminaries, this implies that the fourier series converges everywhere.

Note: This does not mean that it necessarily converges to the function that determined the coefficients of the fourier series. We'll prove that in the next post for a special category of function.

Given a continuous function \(f\) which has a continuous second derivative everywhere except finitely many points, prove that the fourier series generated by \(f\) converges.


Hint: Partition the interval into segments with endpoints being points where the function isn't twice differentiable continuously. Apply integration by parts to each partition and see what happens.

Riemann's Theorem

One of the key issues with the fourier series so far is that we have no guarantee about how well partial sums approximate the total summation, and how well the total summation approximates \(f\). The latter question is a bit tricky and will be dealt with later.

However, we can show that the partial summations of the fourier series converge well to the total summation for arbitrary functions in \(L_2([-\pi,\pi))\). To see why, we have to invoke the fact that this space is a Hilbert space.

Let \(H\) be a hilbert space, and \(E=\{e_n\}_{n\in\bbN}\) a countable orthonormal set (i.e., pairwise orthogonal with all elements with norm \(1\)). Then for all \(h\in H\)$$\sum\limits_{k=1}^\infty\iprod{h,e_n}^2\leq\Norm{h}^2$$


For some \(n\in\bbN\), we have $$\Norm{h-\sum\limits_{k=1}^n\iprod{h,e_k}e_k}=\Iprod{h-\sum\limits_{k=1}^n\iprod{h,e_k}e_k,h-\sum\limits_{k=1}^n\iprod{h,e_k}e_k}$$By linearity$$=\iprod{h,h}-2\Iprod{h,\sum\limits_{k=1}^n\iprod{h,e_k}e_k}+\Iprod{\sum\limits_{k=1}^n\iprod{h,e_k}e_k,\sum\limits_{k=1}^n\iprod{h,e_k}e_k}$$By orthogonality and linearity of inner product$$\|h\|^2-2\sum\limits_{k=1}^n\Iprod{h,\iprod{h,e_k}e_k}+\sum\limits_{k=1}^n\iprod{h,e_k}^2\cdot\|e_k\|$$Since \(\|e_k\|=1\) for all \(k\), and \(\iprod{h,\iprod{h,e_k}e_k}=\iprod{h,e_k}^2\)$$\|h\|^2-\sum\limits_{k=1}^n\iprod{h,e_k}^2$$This implies that $$\|h\|^2\geq\sum\limits_{k=1}^n\iprod{h,e_k}^2$$for all \(k\), and thus $$\|h\|^2\geq\sum\limits_{k=1}^\infty\iprod{h,e_k}^2$$

With Bessel's inequality, we can show the following.

Given a function \(f\in L_2([-\pi,\pi))\) with Fourier series as given here, $$\lim\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}b_n=0$$


Let $$E=\Set{\frac{\cos kx}\pi\mid k\in\bbN}\cup\Set{\frac{\sin kx}\pi\mid k\in\bbN}$$ Thus, \(E\) is a countably infinite orthonormal set, and \(a_k=\Iprod{f,\frac{\sin kx}\pi}\), \(b_k=\Iprod{f,\frac{\cos kx}\pi}\). By Bessel's inequality$$\sum\limits_{k=1}^\infty a_k^2+b_k^2<\infty$$And thus, the desired result holds.
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