One of the equations you would come across during higher studies would be the heat equation. Its another partial differential equation which we can solve using something called the Fourier series… Thanks to Joseph Fourier who introduced this series for the same purpose.

Using the Fourier Series, we simply decompose a particular real-valued function into terms containing the sine & cosine functions.

This function, lets say f(x) is taken to be periodic with a period 2pi. We could further extent the definition of the Fourier Series for functions with arbitrary periods.

The Fourier Series of a function f(x) could we written down as…

$\150dpi \large f(x)=\frac{a_{0}}{2}+ \sum_{n=1}^{\infty }(a_{n}cosnx+b_{n}sinnx)$

In the above decomposition, $\120dpi \inline \large a_{0},a_{n},b_{n}$ are called Fourier Coefficients or Euler’s Coefficients.

The former because they appear in the Fourier Series & the latter because we obtain their values using the Euler’s Formula.

Now the Euler’s Formula gives the the values of the Euler’s coefficients & for a function with period 2pi, we have…

$\150dpi \inline \large a_{0}=\frac{1}{\Pi }\int_{c}^{c+2\Pi }f(x)dx$

$\150dpi \inline \large a_{n}=\frac{1}{\Pi }\int_{c}^{c+2\Pi }f(x)cosnxdx$

$\150dpi \inline \large b_{n}=\frac{1}{\Pi }\int_{c}^{c+2\Pi }f(x)sinnxdx$

Now, once you know the values of the coefficients(which would come out to be functions of n) you could find out the complete sum & hence, the complete Fourier series of the function f(x).

The value of ‘c’ here would depend on the interval specified with the function f(x).

For example, Lets say we have a function,

$\150dpi \inline \large f(x)=1+x$

on the interval…

$\150dpi \inline \large [-\pi , \pi ]$

We have here a simple saw tooth function which we’ve made periodic with a period 2pi.

To start with, we have the interval correpsonding to c to c+2pi which gives c=-pi.

Hence, the values of the coefficients would now be given by…

$\150dpi \inline \large a_{0}=\frac{1}{\Pi }\int_{-\Pi }^{\Pi }(1+x)dx$

$\150dpi \inline \large a_{n}=\frac{1}{\Pi }\int_{-\Pi }^{\Pi }(1+x)cosnxdx$

$\150dpi \inline \large b_{n}=\frac{1}{\Pi }\int_{-\Pi }^{\Pi }(1+x)sinnxdx$

Hence, on integration we obtain…

$\150dpi \inline \large a_{0}=2, a_{n}=0, b_{n}=\frac{2(-1)^{n+1}}{n}$

This would generate the following Fourier series for our function f(x)…

$\150dpi \inline \large f(x)=1+\sum_{n=1}^{\infty }\frac{2(-1)^{n+1}}{n}sinnx$

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