Jone133 2 years ago Fourier series help! http://i91.photobucket.com/albums/k285/jay1ty0/P2130029_zps640806c1.jpg I don't know anything about series fourier. Is this the right way? How to continue? Thanks in advance ;)

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1. UnkleRhaukus

that function is not a odd function, so a_n wont be zero ( nor will a_0) to continue with b_n, use integration by parts

2. Jone133

Isn't 2x an odd function?

3. ankolix

i've got: $\frac{ 2 }{ \pi }\int\limits_{0}^{\frac{ \pi }{ 4}}2x \sin (2nx)dx$.. are you sure your process?

4. Jone133

5. mathmate

What you have done seems perfectly ok. Here's the result after summing 500 terms.

6. mathmate

@Jone133 To proceed, start from the last line of your work, $$\large b_n = \frac{\pi}{2}\int_0^\frac{\pi}{4}2x sin(nx) dx$$ Integrate to get an algebraic expression of bn. You will need to integrate by parts. Evaluate between the limits 0 and $$\pi/4$$ to get the following: $\large b_n =\frac{4\,\mathrm{sin}\left( \frac{\pi \,n}{4}\right) -\pi \,n\,\mathrm{cos}\left( \frac{\pi \,n}{4}\right) }{\pi \,{n}^{2}}$ which is basically your answer. The required function is then $f(x)=\sum_1^\infty b_n sin(nx)$ The previous post shows an image after summing the first 500 terms:

7. Jone133

A medal for you, kind sir :) I was thinking, do I have to evaluate between $0 \to \pi/4$ Or $0 \to \pi$ ? I did it for 0 to pi 1st, going to 0 to pi/4 now

8. mathmate

From 0 to pi/4 would be on the right track, otherwise it would not result in the right shape.

9. Jone133

I can't thank you enough :D Now I have the question: "Indicate the value of the series found on the previous question, on the points: -pi/4 and 9pi/5" Do I substitute n or x? :o

10. mathmate

The answer should be 0, because the series is supposed to have a cycle of pi. But it is not zero. The series is -pi/2 at t=-pi/4, since we only had the sine components. I believe we need to work out both the sine and cosine components, because the given pattern is neither even nor odd, as pointed out previously by UncleRhaukus. Also, the standard period of Fourier series is 2pi. Think there is a transformation to be done here as well. So while your solution gives the right shape for the interval [0,pi], it is not the correct periodic function, thanks to the second part. Can you see if you can do the following: 1. map the period [0,pi] to [0,2pi] so that we can have the correct periodic function. 2. include a0, an and bn in the calculations. See what you get. If I get some results, I will post here as well. I won't be able to work on it until later today. So you can get started in the mean time.