# Discontinous functions - fourier series and taylor series [Solved!]

**Gaurav** 17 Dec 2015, 04:03

### My question

can a discontinous function can be develop in fourier series and taylar series?

### Relevant page

2. Full Range Fourier Series

### What I've done so far

Just wondering because I tried to do a few

X

can a discontinous function can be develop in fourier series and taylar series?

Relevant page
<a href="/fourier-series/2-full-range-fourier-series.php">2. Full Range Fourier Series</a>
What I've done so far
Just wondering because I tried to do a few

## Re: Discontinous functions - fourier series and taylor series

**Murray** 18 Dec 2015, 04:46

Hi Gaurav

A function has to be continuous before you can expand it using Fourier or Taylors, and for Fourier the function needs to be periodic.

With Taylors though, you can limit the domain to the continuous part and then expand for that. For example, on this page:

1. Taylor Series

The example on `f(x)=log x` shows that even though it is discontinuous at `x = 0` and not defined for `x < 0`, we can still do a Taylor's series expansion.

Hope that makes sense

X

Hi Gaurav
A function has to be continuous before you can expand it using Fourier or Taylors, and for Fourier the function needs to be periodic.
With Taylors though, you can limit the domain to the continuous part and then expand for that. For example, on this page:
<a href="/series-expansion/1-taylor-series.php">1. Taylor Series</a>
The example on `f(x)=log x` shows that even though it is discontinuous at `x = 0` and not defined for `x < 0`, we can still do a Taylor's series expansion.
Hope that makes sense

## Re: Discontinous functions - fourier series and taylor series

**Gaurav** 19 Dec 2015, 03:28

Yes it does. thanks

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