Interesting pattern in Fourier coefficients

Question

I was playing around with Fourier coefficients and I observed something very interesting

The above is for the function $3x^2+2x^3$ for $100$ terms($-\pi$ to $\pi$ periodic).

The alternate coefficients seem to form a nice decay pattern. Initially, I guessed that this decay was exponential, but to my surprise, it was a polynomial decay, as I observed it with a curve fit.

While the other set of $b_k$s and the two other $a_k$ sets form a very similar decay pattern, they do not exactly coincide to such a great extent.

So my question is why these coefficients form such a nice alternating sequence and if it is possible to find an approximate curve fit function.

I tried to do find a relation between $a_{k+2}$ and $a_{k}$:

$$a_{k+2}-a_{k}=-2\int_{-\pi}^{\pi}f(x)\sin((k+1)x)\sin(x)dx$$

but this doesn't seem to lead anywhere.

Answer 1

A succinct explanation can be given with Bernoulli_polynomials.

I will use interval $[-1,1]$ instead of $[-\pi,\pi]$ for my explanations.

Indeed,

1. As you can see in the Wikipedia reference : the complex Fourier series of $B_n(x)$ is :

$$B_n(x) = -\frac{n!}{(2\pi i)^n}\sum_{\matrix{k=-\infty\\ k\ne 0}}^{+\infty}\frac{1}{k^n}e^{2\pi ikx}\tag{1}$$

or :

$$B_n(x)=2\, \Re\left(-\frac{n!}{(2\pi i)^n}\sum\limits_{k=1}^{+\infty} \frac{1}{k^n}\, e^{2\pi i k x}\right)$$

(Thanks to @Steven Clark who has corrected an error of mine)

Remark : Identity (1), strictly speaking, is valid for $x \in [0,1]$ ; or, instead, for all $x$ under the condition of replacement, in the LHS, of $B_n(x)$ by a periodised version.

Remark : the real coefficients of the Fourier series of the $B_p(x)$ are (with classical notations)

• only with $a_k$ coefficients when $p$ is even.

• only with $b_k$ coefficients when $p$ is odd.

2. Polynomials $B_p(x), \ p \le n$ constitute a basis of the $(n+1)$ dimensional vector space of polynomials with degree $\le n$.

Therefore, any polynomial $P(x)$ with $\text{degree}(P)=n$ can be written :

$$P(x)=d_nB_n(x)+d_{n-1}B_{n-1}(x)+\cdots+d_0\underbrace{B_0(x)}_1$$

with coefficient $d_n \ne 0$ due to the necessary presence of a monomial in $x^n$.

This $d_n$ coefficient will have a long-term decisive influence.

Answer 2

This answer is intended to supplement the original answer posted by Jean Marie.

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The $2 \pi$-periodic Fourier series for $f(x)=2 x^3+3 x^2$ is

$$f(x)=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{k=-K}^K \left(\left\{\begin{array}{cc} \pi ^2 & k=0 \\ \frac{\left(2\, (-1)^k\, \left(k\, \left(3+i\, \pi^2\, k\right)-6\, i\right)\right)}{k^3} & k\ne 0 \\ \end{array}\right.\right)\, e^{i kx}\right),\ -\pi<x<\pi\tag{1}.$$

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Consider the following $2 \pi$-periodic Fourier series for the Bernoulli polynomials $B_0(x)$ to $B_3(x)$

$$B_0(x)=1\tag{2}$$

$$B_1(x)=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{k=-K}^K \left(\left\{\begin{array}{cc} -\frac{1}{2} & k=0 \\ \frac{i (-1)^k}{k} & k\ne 0 \\ \end{array}\right.\right) e^{i k x}\right),\ -\pi<x<\pi\tag{3}$$

$$B_2(x)=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{k=-K}^K \left(\left\{\begin{array}{cc} \frac{1}{6} \left(1+2 \pi ^2\right) & k=0 \\ \frac{(-1)^k (2-i k)}{k^2} & k\ne 0 \\ \end{array}\right.\right) e^{i k x}\right),\ -\pi<x<\pi\tag{4}$$

$$B_3(x)=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{k=-K}^K \left(\left\{\begin{array}{cc} -\frac{\pi ^2}{2} & k=0 \\ \frac{i (-1)^k \left(k \left(6 i+k+2 k \pi ^2\right)-12\right)}{2 k^3} & k\ne 0 \\ \end{array}\right.\right) e^{i k x}\right),\ -\pi<x<\pi\tag{5}$$

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Noting that

$$f(x)=2 x^3+3 x^2=\frac{3}{2} B_0(x)+5 B_1(x)+6 B_2(x)+2 B_3(x)\tag{6}$$

formulas (1) to (5) above lead to the equivalency

$$\left(\left\{\begin{array}{cc} \pi ^2 & k=0 \\ \frac{2 (-1)^k \left(k \left(3+i \pi ^2 k\right)-6 i\right)}{k^3} & k\ne 0 \\ \end{array}\right.\right)=\frac{3}{2} \delta(k)+5 \left(\left\{\begin{array}{cc} -\frac{1}{2} & k=0 \\ \frac{i (-1)^k}{k} & k\ne 0 \\ \end{array}\right.\right)+6 \left(\left\{\begin{array}{cc} \frac{1}{6} \left(1+2 \pi ^2\right) & k=0 \\ \frac{(-1)^k (2-i k)}{k^2} & k\ne 0 \\ \end{array}\right.\right)+2 \left(\left\{\begin{array}{cc} -\frac{\pi ^2}{2} & k=0 \\ \frac{i (-1)^k \left(k \left(6 i+k+2 k \pi ^2\right)-12\right)}{2 k^3} & k\ne 0 \\\end{array}\right.\right)\tag{7}$$

where $\delta(k)$ is the Kronecker delta function.

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Likewise, the 1-periodic Fourier series for $f(x)=2 x^3+3 x^2$ and $B_n(x)$ illustrated in formulas (8) and (9) below and the relationship illustrated in formula (6) above lead to the equivalency illustrated in formula (10) below.

$$f(x)=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{k=-K}^K \left(\left\{\begin{array}{cc} \frac{3}{2} & k=0 \\ \frac{\pi k (6+5 i \pi k)-3 i}{2 \pi ^3 k^3} & k\ne 0 \\ \end{array}\right.\right) e^{i 2 \pi k x}\right),\ 0<x<1\tag{8}$$

$$B_n(x)=\left(\left\{\begin{array}{cc} 1 & n=0 \\ \underset{K\to\infty}{\text{lim}}\left(\sum\limits_{k=-K}^K \left(\left\{\begin{array}{cc} 0 & k=0 \\ -\frac{n!}{(2 \pi i k)^n} & k\ne 0 \\ \end{array}\right.\right) e^{2 \pi i k x}\right) & n>0 \\ \end{array}\right.\right),\ 0<x<1\tag{9}$$

$$\left(\left\{\begin{array}{cc} \frac{3}{2} & k=0 \\ \frac{\pi k (6+5 i \pi k)-3 i}{2 \pi ^3 k^3} & k\ne 0 \\ \end{array}\right.\right) =\frac{3}{2} \delta_k +5 \left(\left\{\begin{array}{cc} 0 & k=0 \\ -\frac{1!}{(2 \pi i k)^1} & k\ne 0 \\ \end{array}\right.\right) +6 \left(\left\{\begin{array}{cc} 0 & k=0 \\ -\frac{2!}{(2 \pi i k)^2} & k\ne 0 \\ \end{array}\right.\right) +2 \left(\left\{\begin{array}{cc} 0 & k=0 \\ -\frac{3!}{(2 \pi i k)^3} & k\ne 0 \\ \end{array}\right.\right)\tag{10}$$