MHB Odd & Even Extension of a Function: Explained

[evinda]

Hello! (Wave)

According to my notes:

For the solution of problems with differential equations it is often useful to expand to a Fourier series with period $2L$ a function $f$ that is initially only defined on the interval $[0,L]$. We could

• define a function $g$ with period $2L$ such that $g(x)=\left\{\begin{matrix}
  f(x) & ,0 \leq x \leq L,\\
  f(-x) & ,-L<x<0
  \end{matrix}\right.$.
  
  So the function $g$ is the even extension of $f$.
  
• We could also define a function $h$ with period $2L$ such that $h(x)=\left\{\begin{matrix}
  f(x) & ,0 < x <L,\\
  0 & , x=0,L \\
  -f(-x) & ,-L<x<0
  \end{matrix}\right.$.
  
  So the function $h$ is the odd extension of $f$.
  

I was wondering why we define the extensions so that the functions are defined for $x=L$ but not for $x=-L$.

Suppose that we have the function $f(x)=2-x, 0<x<2$.

Then we get the following even and odd extension respectively:$g(x)=\left\{\begin{matrix}
2-x & , 0 \leq x \leq 2\\
2+x &, -2 <x<0\\
\end{matrix}\right.$

and

$h(x)=\left\{\begin{matrix}
2-x &,0<x<2 \\
0 & , x=0,2\\
-2+x & ,-2<x<0
\end{matrix}\right.$

Right?The graphs of the even and odd extension of $f$ will then be the following, respectively:View attachment 6577

View attachment 6576

Am I right?

 

[Opalg]

The even extension is correct. The odd one is not, because the line segment from $(-2,-4)$ to $(0,-2)$ should actually go from $(-2,0)$ to $(0,-2)$. (The graph of an odd function should look the same when it is rotated through $180^\circ$ around the origin.)

 

[evinda]

The odd one is not, because the line segment from $(-2,-4)$ to $(0,-2)$ should actually go from $(-2,0)$ to $(0,-2)$. (The graph of an odd function should look the same when it is rotated through $180^\circ$ around the origin.)

Oh right... It should be as follows, right?

View attachment 6578

 

[Opalg]

Oh right... It should be as follows, right?

Yes. (Yes)

I was wondering why we define the extensions so that the functions are defined for $x=L$ but not for $x=-L$.

The reason that the function is not defined at $x=-L$ is that it does not need to be. The function has period $2L$, so its value at $x=-L$ must automatically be the same as the value at $x=L$.

 

[evinda]

The reason that the function is not defined at $x=-L$ is that it does not need to be. The function has period $2L$, so its value at $x=-L$ must automatically be the same as the value at $x=L$.

Ah I see... Thanks a lot! (Happy)