How can I find the Fourier Series of the function f(x)=(1-x)(1+x) on the interval [-1/2, 1/2].
Asked
Active
Viewed 1,863 times
-1
-
what did you try ? where did you stop ? – G Cab Nov 13 '16 at 10:55
-
I found out Ao which is coming out to be 11/12. I tried finding An which came out to be -cos(pi n x)/(pi*n)^2 Am i doing it right? Is there a dfference in formulae for Fourier Coefficients in finding a series from -1/2 to 1/2 from when finding it from -1 to 1?? – Muhammad Anas Athar Nov 13 '16 at 10:58
-
$A_n$ cannot depend on $x$. – Julián Aguirre Nov 13 '16 at 11:46
-
Sorry, thats a typo. x isnt there – Muhammad Anas Athar Nov 13 '16 at 12:22
-
Have I computed Ao correctly? – Muhammad Anas Athar Nov 13 '16 at 12:39
2 Answers
0
The Fourier coefficients on an interval $[-L/2,L/2]$ are given by $$ A_n=\frac{2}{L}\int_{-L/2}^{L/2}f(x)\cos\frac{2\,\pi\,n\,x}{L}\,dx,\quad B_n=\frac{2}{L}\int_{-L/2}^{L/2}f(x)\sin\frac{2\,\pi\,n\,x}{L}\,dx. $$ In your case $L=1$ and, integrating by parts, $$ A_n=2\int_{-1/2}^{1/2}(1+x)(1-x)\,\cos(2\,\pi\,n\,x)\,dx=2\int_{-1/2}^{1/2}(1-x^2)\,\cos(2\,\pi\,n\,x)\,dx=\frac{(-1)^{n+1}}{\pi^2\,n^2}. $$ Since $f$ is even, $B_n=0$.
Julián Aguirre
- 76,354
-
-
-
the final answer would be f(x)=Ao+Sum(An cos (n pi x) Is it so sir?? – Muhammad Anas Athar Nov 13 '16 at 12:37
-
-
-
-
as you said my L in this case is 1. Why should cos(2n pi x) come in the final result and why not cos(n pi x).. Please clear – Muhammad Anas Athar Nov 13 '16 at 13:48
-
Because $f$ is of period $1$. Look at the formula I wrote for the coefficients. The period and the frequency are related to the length of the interval. – Julián Aguirre Nov 13 '16 at 14:01
-
-
Just plug $n=0$. If you need to ask all these questions, I suggest that you go back to your textbook, where I am sure all of this is explained. – Julián Aguirre Nov 13 '16 at 14:18
0
In your reply to my comment, you were asking about the difference in changing
the period from $[-1,1]$ to $[-1/2,1/2]$.
Let us fix this point first of all.
If
$$
f(x/T)\quad \to \quad a_{\,n} \cos \left( {\frac{{2\pi n}}
{T}x} \right)
$$
then
$$
g(x/T') = f(x/T)\quad \to \quad a_{\,n} \cos \left( {\frac{{2\pi n}}
{{T'}}x} \right)
$$
That is:
if you change the $x$ scale and consequently the period,
keeping the ordinate scale, then the Fourier coefficients will be the same, but will apply to frequencies which are in the inverse relation with the periods.
In your case: $$ x^{\,2} = \left( {\frac{x} {1}} \right)^{\,2} \quad \to \quad a_{\,n} \cos \left( {\frac{{2\pi n}} {1}x} \right)\quad \Rightarrow \quad x^{\,2} = \left( {\frac{1} {2}} \right)^{\,2} \left( {\frac{x} {{1/2}}} \right)^{\,2} \quad \to \quad \frac{{a_{\,n} }} {4}\cos \left( {\frac{{2\pi n}} {{1/2}}x} \right) $$ Which means, that the coefficients of the series for: $$ x^{\,2} \quad \left| {\;x \in \left[ { - 1/2,\;1/2} \right]} \right. $$ will be $1/4$ of those corresponding to: $$ x^{\,2} \quad \left| {\;x \in \left[ { - 1,\;1} \right]} \right. $$ and that of course they will apply to harmonics with double frequencies.
That's because you have a scaling either in $x$ and in $y$.
G Cab
- 35,272