This Fourier series calculator is designed to find the Fourier series expansion for given functions with a complete solution. In this way, you can get any variable function's Fourier coefficients.
“Fourier series is a term that is used to indicate the expansion of periodic functions as an infinite sum of simple sine and cosine functions”.
Fourier series allows any arbitrary periodic signal with the combination of sine and cosine.
To synthesize the harmonic content of your signals or solve new differential equations
with our incredible Fourier coefficients calculator. It operates well by having the following values:
Select the function variable and enter its value
Also, insert the lower and upper limit
Fourier Series: Step-by-step Fourier series for the given functions with formula
Coefficient Insights: Obtain the Fourier coefficients a(0), a(n), and b(n)
Harmonic Components: Get the individual harmonic components contributing to the Fourier series expansion.
For periodic functions, the Fourier series is concluded from the Laurent expansions. It defines the orthogonality relation between the sin and cosine functions. The formula for this series of the function f(x) in the interval [-L, L], i.e. -L ≤ x ≤ L is given by:
$$ f(x)=\frac{1}{2}a_{0}+\sum_{n=1}^{\infty}a_{n}cos\;nx+\sum_{n=1}^{\infty}b_{n}sin\;nx $$
Where:
$$ a_0 = \frac{1}{\pi} \int_{- \pi}^{\pi} f(x) dx $$
$$ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x)cos\;nx\;dx $$
$$ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi}f(x)sin\;nx\;dx $$
$$ n = 1, 2, 3 $$
By determining a function as a sum of sines and cosines many problems become easy to analyze. Therefore, a Fourier series calculator is designed to help you to calculate the complex Fourier series of a given function.
A Fourier series is a way to deal with functions in terms of trigonometric functions. It is prominent in partial differential equations. It is calculated by using the simple Fourier series calculator.
So look at the example below that will help you to understand their calculation process in a better way.
Calculate the Fourier series of a function f(x) = 2 - x² over an interval [-1, 1].
Given Data:
To Find: $$ 2 - x^{2} $$
On Interval: $$ \left[- \pi,\pi\right] $$
As we know the formula is considered valuable by the Fourier expansion calculator.
$$ f(x)=\frac{1}{2}a_{0}+\sum_{n=1}^{\infty}a_{n}cos\;nx+\sum_{n=1}^{\infty}b_{n}sin\;nx $$
Find the values:
$$ A_0,A_n \text{ and } B_n $$
$$ A_0=\frac{1}{2\times \pi}\cdot \int _{- \pi}^\pi\left(2 - x^{2}\right)dx = 4 - \frac{2 \pi^{2}}{3} $$
$$ A_n=\frac{1}{\pi}\cdot \int _{- \pi}^\pi\left(2 - x^{2}\right)\cos \left(\frac{n\pi x}{\pi}\right)dx = - \frac{4 \left(-1\right)^{n}}{n^{2}} $$
$$ B_n=\frac{1}{\pi}\cdot \int _{- \pi}^\pi\left(2 - x^{2}\right)\sin \left(\frac{n\pi x}{\pi}\right)dx = 0 $$
Put the values in the formula:
$$ f\left(x\right)=4 - \frac{2 \pi^{2}}{3} +\sum _{n=1}^{\infty \:}- \frac{4 \left(-1\right)^{n}}{n^{2}}\cdot \cos \left(\frac{n\pi x}{L}\right)+\sum _{n=1}^{\infty \:}0\cdot \sin \left(\frac{n\pi x}{L}\right) $$
After Simplifying:
$$ \sum_{n=1}^{\infty} - \frac{4 \left(-1\right)^{n} \cos{\left(n x \right)}}{n^{2}} - \frac{\pi^{2}}{3} + 2 $$
Wikipedia: Fourier series, Common forms of the Fourier series, Table of common Fourier series, Table of basic properties.
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