0
$\begingroup$

I am new to fourier series and i don't really know if i have understood correctly the whole concept.

I have to solve: calculate the fourier series of $$f(x)= x(\pi^2-x^2)$$ Then $ii)$ calculate the fourier series of f'in two ways i)Develop to fourier series f'(derivative) ii)by differentiating each term of the fourier of f.

So far i writed it as $(x\pi^2-x^3)$ used the odd function property and from that i concluded there are not $a_0$ coefficient and $a_n$.

Is so far correct and what about the $ii)$.

$\endgroup$
2
  • $\begingroup$ You will not avoid the computation of coefficients $b_n$, you probably know how to calculate their integral expressions ? $\endgroup$
    – M. Boyet
    Apr 5, 2017 at 16:22
  • $\begingroup$ Well i didn't said that i will avoid the computaiton of sin but the others $\endgroup$
    – Agaeus
    Apr 5, 2017 at 18:18

1 Answer 1

2
$\begingroup$

$$ f(x) = x\left(\pi^{2} - x^{2} \right) \qquad \Rightarrow \qquad f'(x) = \pi^{2}-3x^{2} $$


A: Fourier series of derivative $\ f'(x) = \pi^{2}-3x^{2}$

The derivative has even parity and will be a sum of cosines.

The Fourier amplitudes are computed via $$ \begin{align} a_{0} &= \frac{1}{\pi} \int_{-\pi }^{\pi } f'(x) dx \\ % a_{k} &= \frac{1}{\pi} \int_{-\pi }^{\pi } f'(x) \cos (k x) dx \\ % b_{k} &= \frac{1}{\pi} \int_{-\pi }^{\pi } f'(x) \sin (k x) dx \end{align} $$ There are two basic integrals: $$ \begin{align} % \int_{-\pi }^{\pi } \cos (k x) dx &= 0 \\ % \int_{-\pi }^{\pi } x^{2} \cos (k x) dx &= \left( -1 \right)^{k+1}\frac{4\pi}{k^{2}} \\ % \end{align} $$ The Fourier amplitudes for the derivative are $$ \begin{align} % a_{0} &= 0 \\ % a_{k} &= \left( -1 \right)^{k+1} \frac{12}{k^{2}} \\ % b_{k} &= 0 \\ % \end{align} $$ The approximation of the derivative is $$ f'(x) = \pi^{2}-3x^{2} = \sum_{k=1}^{\infty} a_{k} \cos (kx) = \color{blue}{\sum_{k=1}^{\infty} \left( -1 \right)^{k+1} \frac{12}{k^{2}} \cos (kx)} $$ The plot sequence below provides a quality check on the computation. The parameter $m$ represents the highest value of $k$.

approximation

B: Derivative of Fourier series for $\ f(x) = x\left(\pi^{2} - x^{2} \right)$

The derivative has odd parity and will be a sum of sines.

Construct the Fourier series using these building blocks: $$ \begin{align} % \int_{-\pi }^{\pi } x \sin (k x) \, dx &= \left( -1 \right)^{k+1} \frac{2\pi}{k}\\ % \int_{-\pi }^{\pi } x^{3} \sin (k x) dx &= \left( -1 \right)^{k+1} \frac{2\pi \left(k^{2}\pi^{2} - 6 \right)}{k^{3}} % \end{align} $$ The Fourier amplitudes are $$ \begin{align} % a_{0} &= 0 \\ % a_{k} &= 0 \\ % b_{k} &= (-1)^{k+1}\frac{12 }{k^3} \\ % \end{align} $$ The approximation of the function is $$ f(x) = x\left(\pi^{2} - x^{2} \right) = \sum_{k=1}^{\infty} b_{k} \sin (kx) = \sum_{k=1}^{\infty} \left( -1 \right)^{k+1} \frac{12}{k^{3}} \sin (kx) $$ The derivative of the Fourier series is $$ f'(x) = \frac{d}{dx} \left( \sum_{k=1}^{\infty} b_{k} \sin (kx) \right) = \sum_{k=1}^{\infty} \frac{d}{dx} \left( b_{k} \sin (kx) \right) = \color{blue}{\sum_{k=1}^{\infty} \left( -1 \right)^{k+1} \frac{12}{k^{2}} \cos (kx)} $$ which matches the series for the derivative term-by-term.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .