# I am stuck in one step of a Fourier sine series

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• Jul 20th 2013, 12:48 PM
juanma101285
I am stuck in one step of a Fourier sine series
Hi, I had solved this problem in a very complex and lengthy way and got the same result as what our lecturer says we should get. However, his answer (which skips several steps) is much simpler than what I did, but I don't know how he does the last step.

Here is the problem:

Find the Fourier sine series of $\displaystyle f(t)=2+t$ on $\displaystyle (0, \pi)$.

Solution:
$\displaystyle b_k=\frac{2}{\pi}\int_{0}^{\pi}f(t)sin(kt) dt=\frac{2}{\pi}\int_{0}^{\pi}e^{t}sin(kt) dt$

We know that $\displaystyle e^{ikt}=cos(kt)+i sin(kt)$, so we will only use the imaginary part of $\displaystyle e^{ikt}$

$\displaystyle b_k=\frac{2}{\pi}\Im\int_{0}^{\pi}e^{t} e^{ikt} dt = \frac{2}{\pi}\Im\int_{0}^{\pi}e^{(1+ik)t} dt$

$\displaystyle b_k=\frac{2}{\pi}\Im[\frac{e^{(1+ik)t}}{1+ik} ]_{0}^{\pi}$

$\displaystyle b_k=\Im (\frac{2}{\pi(1+ik)}\{e^{(1+ik)\pi}-1\})$

And what I don't understand is how it goes from the result above to the final result below:

$\displaystyle =\frac{2k}{\pi(1+k^2)}\{1-(-1)^ke^{\pi}\}$

I would really appreciate it if someone could tell me how this is done! I have no clue where the $\displaystyle k$ in "$\displaystyle 2k$" and "$\displaystyle k^2$" come from... Thanks a lot!
• Jul 20th 2013, 01:47 PM
HallsofIvy
Re: I am stuck in one step of a Fourier sine series
Quote:

Originally Posted by juanma101285
Hi, I had solved this problem in a very complex and lengthy way and got the same result as what our lecturer says we should get. However, his answer (which skips several steps) is much simpler than what I did, but I don't know how he does the last step.

Here is the problem:

Find the Fourier sine series of $\displaystyle f(t)=2+t$ on $\displaystyle (0, \pi)$.

Solution:
$\displaystyle b_k=\frac{2}{\pi}\int_{0}^{\pi}f(t)sin(kt) dt=\frac{2}{\pi}\int_{0}^{\pi}e^{t}sin(kt) dt$

What? You said one line above this that "f(t)= 2+ t". How did it become $\displaystyle e^t$???
Quote:

We know that $\displaystyle e^{ikt}=cos(kt)+i sin(kt)$, so we will only use the imaginary part of $\displaystyle e^{ikt}$

$\displaystyle b_k=\frac{2}{\pi}\Im\int_{0}^{\pi}e^{t} e^{ikt} dt = \frac{2}{\pi}\Im\int_{0}^{\pi}e^{(1+ik)t} dt$

$\displaystyle b_k=\frac{2}{\pi}\Im[\frac{e^{(1+ik)t}}{1+ik} ]_{0}^{\pi}$

$\displaystyle b_k=\Im (\frac{2}{\pi(1+ik)}\{e^{(1+ik)\pi}-1\})$

And what I don't understand is how it goes from the result above to the final result below:

$\displaystyle =\frac{2k}{\pi(1+k^2)}\{1-(-1)^ke^{\pi}\}$

I would really appreciate it if someone could tell me how this is done! I have no clue where the $\displaystyle k$ in "$\displaystyle 2k$" and "$\displaystyle k^2$" come from... Thanks a lot!
• Jul 20th 2013, 02:36 PM
juanma101285
Re: I am stuck in one step of a Fourier sine series
Thanks for your reply! Sorry, I wrote it wrong... The problem is:

Find the Fourier sine series of $\displaystyle f(t) = e^t$ on $\displaystyle (0, \pi)$.
• Jul 20th 2013, 09:59 PM
zzephod
Re: I am stuck in one step of a Fourier sine series
Quote:

Originally Posted by juanma101285
..

$\displaystyle b_k=\Im (\frac{2}{\pi(1+ik)}\{e^{(1+ik)\pi}-1\})$

And what I don't understand is how it goes from the result above to the final result below:

$\displaystyle =\frac{2k}{\pi(1+k^2)}\{1-(-1)^ke^{\pi}\}$

I would really appreciate it if someone could tell me how this is done! I have no clue where the $\displaystyle k$ in "$\displaystyle 2k$" and "$\displaystyle k^2$" come from... Thanks a lot!

First look at what is inside the braces:

$\displaystyle \{e^{(1+ik)\pi}-1\} = e^{\pi}e^{ik\pi}-1$

Now observe that $\displaystyle e^{ik\pi}=\cos(k\pi)+i\sin(k\pi)$ and $\displaystyle k\in \mathbb{N}$, so $\displaystyle \cos(k\pi)=(-1)^{k}$ and $\displaystyle \sin(k\pi)=0$.

Hence: $\displaystyle e^{ik\pi}=\cos(k\pi)+i\sin(k\pi) = (-1)^k$ which is real.

So now we need the imaginary part of what is outside the braces, and you get that by first multiplying top and bottom by $\displaystyle (1-ik)$ ...

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