Expanding Function f(x)= x^2+xcosx in Fourier Series

[rayman123]

1. Homework Statement [/b
expand the function in Fourier series$$f(x)= x^2+xcosx$$
I divided the function in separate part and try to expand it in Fourier series separately
i have started with
$$x^2$$
$$a_{0}= \frac{1}{2\pi}\int_{-\infty}^{\infty}x^2dx=\frac{\pi^2}{3}$$

$$a_{n}= \frac{2}{\pi}\int_{0}^{\pi}x^2cosnxdx=\frac{2}{\pi}[\frac{x^2}{n}sinnx-2\int_{0}^{\pi}x\frac{sinx}{n}]dx=\frac{2}{\pi}[\frac{2\pi(-1)^n}{n^2}]=4\cdot\frac{(-1)^n}{n^2}\Rightarrow x^2= 4 \sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}cosnx$$
$$x^2$$ if even function then$$b_{n}=0$$

function f(x)=x is an odd function $$a_{n}= 0$$
$$b_{n}= \frac{2}{\pi}\int_{0}^{\pi}xsinnxdx=\frac{2}{\pi}[-\frac{x}{n}cosnx+\frac{1}{n}\int_{0}^{\pi}cosnxdx]=\frac{2}{\pi}[-\frac{\pi}{n}(-1)^n]=2 \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}sinnx$$
is this correct so far?
can someone help me and show me how to calculate the rest? I mean how to connect everything together with cosx?

Homework Equations

The Attempt at a Solution

 

[Cyosis]

It would be handy if you tell us what the period of that function needs to be. From the looks of it you want it to have a period of $2\pi$.

You're integrating the wrong function in every integral. You're supposed to integrate $f(x)=x^2+x\cos x$. Secondly some of your integrands are odd, therefore you can't just say that evaluating the integral between -pi and pi is the same as 2 times the integral between 0 and pi.

 

[rayman123]

for $$|x|<\pi$$
my teacher did it in almost the same way but at the end he did something else with the xcosx part...

 

[Cyosis]

It doesn't matter what your teacher did. What does matter is that you're doing it wrong. I suspect that your teacher makes good use of his knowledge how even and odd functions behave when integrated over a symmetrical interval. It is not smart to try and copy a method that you don't understand.

For example you get the correct answer for $a_0$ by pure coincidence. In fact your integral for $a_0$ doesn't even converge.

This is the integral you should have calculated:

$$a_0=\frac{1}{2 \pi}\int_{-\pi}^\pi f(x)dx$$

 

[rayman123]

well i calculated this integral once again and i get $$\frac{\pi^2}{3}$$

 

[Cyosis]

That is correct, but you calculated a different integral in your OP, which had the same answer by coincidence.

This is why it gave the same answer.

$$\begin{align*} a_0 & =\frac{1}{2 \pi}\int_{-\pi}^\pi x^2+x \cos x dx \\ & = \underbrace{\frac{1}{2 \pi} \int_{-\pi}^\pi x^2 dx}_\text{even integrand}+\underbrace{\frac{1}{2 \pi}\int_{-\pi}^\pi x \cos x dx}_\text{odd integrand} \end{align*}$$

The second integral goes to zero.

 

[rayman123]

okej, i see it now.
but how to calculate coefficients $$a_{n} , b_{n}$$ then?
the problem is that i am not able to say whether the function is odd or even...$$x^2$$ is even but $$xcosx$$ is odd...but the whole function?

 

[Cyosis]

You can calculate the other coefficients in a similar way. Read post #6 again, because I just fixed the latex.

the problem is that i am not able to say whether the function is odd or even...LaTeX Code: x^2 is even but LaTeX Code: xcosx is odd...but the whole function?

If a function is even then $f(-x)=f(x)$. If a function is odd then $f(-x)=-f(x)$. You can use this to check that f(x) is neither even nor odd.

 

[rayman123]

okej
so i started with $$a_{n}=\frac{1}{\pi}\int_{-\pi}^{\pi}x^2cos(nx)dx+\frac{1}{\pi}\int_{-\pi}^{\pi}xcosxcos(nx)dx$$

the first integral gave me
$$\frac{2}{n^2}[xcos(nx)]_{-\pi}^{\pi}]=\frac{4\pi}{n^2}(-1)^n$$ hope its correct
how would you calculate the second one?

 

[Cyosis]

The $\pi$ shouldn't be there. As for the second integral. Is the integral even or odd?

 

[rayman123]

i would say the second integral is odd, should it give 0 then?
is the final substitution wrong in the first integral? whe $$\pi$$ should not be there?

 

[Cyosis]

Correct. The integrand is odd, and the integration interval is symmetric therefore the integral is 0. You have a $\pi$ in there because you forgot the factor $1/ \pi$ in front of the integral.

Notice how you got the same answer again, this time however without luck.

 

[rayman123]

oh...again, right

$$\frac{2}{n^2}[xcos(nx)]_{-\pi}^{\pi}]=\frac{4\pi}{n^2}(-1)^n$$

$$a_{n}=\frac{4}{n^2}(-1)^ncosnx$$

$$b_{n}= \frac{1}{\pi}\int_{-\pi}^{\pi}x^2sin(nx)dx+ \frac{1}{\pi}\int_{-\pi}^{\pi}xsin(nx)cos(nx)= \frac{1}{\pi}[-\frac{x^2}{n}cosnx+\frac{2x}{n^2}sinnx+\frac{2}{n^3}cosnx]_{-\pi}^{\pi}$$
is this correct so far?
i get from the first integral 0...the second integrand is even so it will give 0

 

[Cyosis]

Why did you suddenly put a cosine in your expression for $a_n$? It shouldn't be there.

The first integral is zero yes, why? The second integral is even yes, but that does not mean the integral is zero.

 

[rayman123]

okej, I just got quite confused...I must admit that I am a real beginner when it comes to Fourier analysis...all this is quite difficult;)
i just calculated the value for the first integral
$$b_{n}= \frac{1}{\pi}[-\frac{x^2}{n}cosnx+\frac{2x}{n^2}sinnx+\frac{2}{n^ 3}cosnx]_{-\pi}^{\pi}=\frac{1}{\pi}[-\frac{\pi^2}{n}(-1)^n+\frac{2}{n^3}(-1)^n-(-\frac{\pi^2}{n}(-1)^n+\frac{2}{n^3}(-1)^n)]=}[-\frac{\pi^2}{n}(-1)^n+\frac{2}{n^3}(-1)^n+\frac{\pi^2}{n}(-1)^n-\frac{2}{n^3}(-1)^n=0$$

you are right with the second integral, i mixed it up with an odd function and I assumpted it should give 0...
however i do not have any clue how do solve the 2nd integral...

 

[Cyosis]

The first integral is odd so you could have set it to zero saving you a lot of trouble.

For the second integral use the well known double angle formula $\sin 2x=2\sin x \cos x$.

 

[rayman123]

okej, i used it and i got
$$\frac{1}{2\pi}\int_{-\pi}^{\pi}xsin(2nx)dx=\frac{1}{2\pi}[-\frac{x}{2n}cos(2nx)+\frac{1}{4n^2}sin(2nx)]_{-\pi}^{\pi}=\frac{1}{2\pi}[-\frac{\pi}{2n}(1)^n-\frac{\pi}{2n}(-1)^n]$$

 

[Cyosis]

You're almost there.

$$\frac{1}{2\pi}\int_{-\pi}^{\pi}xsin(2nx)dx = \underbrace{\frac{1}{2\pi}[-\frac{x}{2n}cos(2nx)+\frac{1}{4n^2}sin(2nx)]_{-\pi}^{\pi}}_\text{correct} =\underbrace{\frac{1}{2\pi}[-\frac{\pi}{2n}(1)^n}_\text{missing a factor}-\underbrace{\frac{\pi}{2n}(-1)^n}_\text{wrong}]$$

Also $1^n=...$?

 

[rayman123]

should it be $$2n$$
hm...the term with sin2nx will give 0 right?
for the rest i get :
$${\frac{1}{2\pi}[-\frac{\pi}{2n}(1)^{2n}+\frac{\pi}{2n}(-1)^{2n}]$$
but according to what you have written it is still wrong,

 

[Cyosis]

You found the correct primitive. Now all you have to do is plug in the limits of integration. You're nearly there, but don't get sloppy with simple arithmetic.

Yes the $\sin 2 n \pi$ term is zero for every $n \in \mathbb{Z}$. Don't forget to answer my question in post #18.

 

[rayman123]

well i am not being sloppy, i am just very tired, have been studying all day long and i have still much left.
Right now i really do not se any other solution, maybe my head is just too tired. Will try to recalculate it again and find a correct answer.
your question in post #18 $$1^{n}=?$$ the value depends on the factor n i suppose

 

[Cyosis]

Plug in some numbers for n and see what you get.

I suggest you take a little break after you finish this problem!

 

[rayman123]

in fact i am not going to calculate any problems tonight, will continue tomorrow because i start making really stupid mistakes. Thank you very much for you help anyway, you have been very helpful.
I will continue with the problem tomorrow.

 

[rayman123]

$$={\frac{1}{2\pi} \left[-\frac{\pi}{2n}+\frac{-\pi}{2n}\right] = \frac{-1}{2n}$$

made correction here

$$\frac{\pi^2}{3}+\sum_{n=1}^{\infty}[\frac{4}{n^2}(-1)^ncos(nx)+\frac{-1}{2n}sin(nx)]$$

 

[Cyosis]

It is almost correct.

$$={\frac{1}{2\pi} \left[-\frac{\pi}{2n}+\frac{-\pi}{2n}\right] = \frac{-1}{2n}$$

The part between brackets is zero, this should not be the case.

The $x^2/3$ in front of the sum isn't correct either, it should be $a_0$ instead.

 

[rayman123]

yes you are right it was a stupid mistake $$a_{0}$$ should be$$a_{0}=\frac{\pi^2}{3}$$
$$\frac{\pi^2}{3}+\sum_{n=1}^{\infty}[\frac{4}{n^2}(-1)^ncos(nx)]$$
is it now correct?

are you sure that the part between the brackets is zero?

 

[Cyosis]

No sorry about that. It is not zero and your result in post #24 is correct if you replace $x^2/3$ with $a_0$. Don't reply when you just got out of bed!

To clear it up:

$b_n=-\frac{1}{2n}$

 

[rayman123]

great:) Thank you very much. You relly helped me to understand the idea of this problem