"The enlargement of the periodic characteristic in phrases of infinite sums of sines and cosines is referred to as Fourier series."
check the given method that indicates the periodic feature f(x) in the c programming language \(-L\le \:x\le \:L\:\)
$$ f\left(x\right)=a_0+\sum _{n=1}^{\infty \:}a_n\cdot \cos \left(\frac{n\pi x}{L}\right)+\sum _{n=1}^{\infty \:}b_n\cdot \sin \left(\frac{n\pi x}{L}\right) $$
Where ;
$$ a_0=\frac{1}{2L}\cdot \int _{-L}^Lf\left(x\right)dx $$
$$ a_n=\frac{1}{L}\cdot \int _{-L}^Lf\left(x\right)\cos \left(\frac{n\pi x}{L}\right)dx,\:\quad \:n>0 $$
$$ b_n=\frac{1}{L}\cdot \int _{-L}^Lf\left(x\right)\sin \left(\frac{n\pi x}{L}\right)dx,\:\quad \:n>0 $$
With the assist of the Fourier coefficients calculator, you could without problems locate values towards those coefficients.
Calculate the Fourier collection of the characteristic given underneath:
$$ f(x) = x^2 \text{ for } -\pi \leq x \leq \pi $$
Solution:
The function is:
$$ f(x) = x^2 $$
First, we determine the coefficients \(a_0\), \(a_n\), and \(b_n\):
The formula for \(a_0\) is:
$$ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx $$
Substitute \(f(x) = x^2\):
$$ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \, dx $$
Since \(x^2\) is even, we can simplify:
$$ a_0 = \frac{2}{\pi} \int_{0}^{\pi} x^2 \, dx $$
$$ a_0 = \frac{2}{\pi} \cdot \left[ \frac{x^3}{3} \right]_0^\pi $$
$$ a_0 = \frac{2}{\pi} \cdot \frac{\pi^3}{3} = \frac{2\pi^2}{3} $$
The formula for \(a_n\) is:
$$ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx $$
Substitute \(f(x) = x^2\):
$$ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} x^2 \cos(nx) \, dx $$
Since \(x^2 \cos(nx)\) is even:
$$ a_n = \frac{2}{\pi} \int_{0}^{\pi} x^2 \cos(nx) \, dx $$
Using integration by parts (or an online integral calculator):
$$ a_n = \frac{4(-1)^n}{n^2} $$
The formula for \(b_n\) is:
$$ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx $$
Since \(x^2 \sin(nx)\) is odd:
$$ b_n = 0 $$
Combining the results, the Fourier series is:
$$ f(x) = \frac{2\pi^2}{3} + \sum_{n=1}^\infty \frac{4(-1)^n}{n^2} \cos(nx) $$
Even here, a Fourier series calculator can help simplify such calculations effectively.
Every time you come across complicated capabilities, our unfastened online fourier series calculator is here that will help you out in determining accurate outcomes. you may get a proper situation of the calculations by means of the usage of our calculator.
Input:
Output: The Fourier growth calculator calculates: