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Fourier Series Involving Hyperbolic Functions

aNxello

New member
Feb 20, 2013
13
[SOLVED] Fourier Series Involving Hyperbolic Functions

Hello everyone!
Sorry if this isn't the appropriate board, but I couldn't think of which board would be more appropriate. I was running through some problems I have to do as practice for a test and I got stuck on one I'm 99% sure they'll ask in the test, in some way or another, and I can't figure it out. This isn't homework, and I'm not asking for you guys to do it for me, but I would love some help to get through it!
Anyways, the problem is simple, I have to expand the function f(x) = cosh(ax) where a is a real number, in a fourier series over the interval (-pi, pi)

I've started to calculate the coefficients, but they don't seem to be able simplify. I found this online but I can't get to the that answer.

Any help/tips? I'd really appreciate it! (Bow)
 
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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Ok , I am not home so I will give you a hint , it is better to use
\(\displaystyle \cosh(ax) =\frac{e^{ax}+e^{-ax}}{2}\) then use integration by parts.
 

aNxello

New member
Feb 20, 2013
13
Ahh! Thank you, I had thought of using this, but I didn't know if it worked when x was next to some constant.
Do you think it would be better to first integrate to obtain 1/a Sinh(ax)? I feel like replacing first will make it harder to find some of the coefficients. Anyways I'll try it out and update!
Thanks!
 

aNxello

New member
Feb 20, 2013
13
So for the first coefficient I got this
\(\displaystyle a_{0} = \frac{e^{a\pi } - e^{-a\pi }}{\pi a}\)
but for the second one I keep looping while trying to do integration by parts (I keep getting back to the same integral over and over)
Any clue on how to deal with this:
\(\displaystyle a_{n} =\frac{1}{\pi } \int Cosh(ax)Cos(nx)\)
or
\(\displaystyle b_{n} =\frac{1}{\pi } \int Cosh(ax)Sin(nx)\)

Maybe some hint or identity I could use?
Thanks! :)
 
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zzephod

Well-known member
Feb 3, 2013
134
So for the first coefficient I got this
\(\displaystyle a_{0} = \frac{e^{a\pi } - e^{-a\pi }}{\pi a}\)
but for the second one I keep looping while trying to do integration by parts (I keep getting back to the same integral over and over)
Any clue on how to deal with this:
\(\displaystyle a_{n} =\frac{1}{\pi } \int Cosh(ax)Cos(nx)\)
or
\(\displaystyle b_{n} =\frac{1}{\pi } \int Cosh(ax)Sin(nx)\)

Maybe some hint or identity I could use?
Thanks! :)
\(\displaystyle \cosh(x)=\frac{e^x+e^{-x}}{2}\)

\(\displaystyle \sin(x)=\frac{e^{ix}-e^{-ix}}{2i}\)

\(\displaystyle \cos(x)=\frac{e^{ix}+e^{-ix}}{2}\)
 

chisigma

Well-known member
Feb 13, 2012
1,704
So for the first coefficient I got this
\(\displaystyle a_{0} = \frac{e^{a\pi } - e^{-a\pi }}{\pi a}\)
but for the second one I keep looping while trying to do integration by parts (I keep getting back to the same integral over and over)
Any clue on how to deal with this:
\(\displaystyle a_{n} =\frac{1}{\pi } \int Cosh(ax)Cos(nx)\)
or
\(\displaystyle b_{n} =\frac{1}{\pi } \int Cosh(ax)Sin(nx)\)

Maybe some hint or identity I could use?
Thanks! :)
Successive integrations by part is a tedious job but it permits to solve the integral...


$\displaystyle \int \cosh (a x)\ \cos (nx)\ dx = \frac{1}{a}\ \sinh (ax)\ cos (nx) + \frac{n}{a}\ \int \sinh (a x)\ \sin (nx)\ dx = $

$\displaystyle = \frac{1}{a}\ \sinh (ax)\ cos (nx) + \frac{n}{a^{2}}\ \cosh (a x)\ \sin (n x) - \frac{n^{2}}{a^{2}}\ \int \cosh (a x)\ \cos (n x)\ dx$

... so that is...

$\displaystyle \int \cosh (a x)\ \cos (nx)\ dx = \frac{a\ \sinh (ax)\ cos (nx) + n\ \cosh (a x)\ \sin (n x)} {a^{2} + n^{2}}$

The other indefinite integral $\displaystyle \int \sinh (a x)\ \sin (n x)\ dx$ can be solved in similar way...


Kind regards


$\chi$ $\sigma$
 

aNxello

New member
Feb 20, 2013
13
Thank you very much, let me try to finish this, and I'll let you guys know how it goes! (Dance)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Here is the solution :

\(\displaystyle \cosh(ax) = A_0 +\sum^{\infty}_{k=1} A_k \cos(kx) +\sum^{\infty}_{k=1} B_k \sin(kx)\)

By definition we know that :

\(\displaystyle A_0 = \frac{1}{2\pi} \int^{\pi}_{-\pi} f(x) \, dx\)

\(\displaystyle A_k = \frac{1}{\pi} \int^{\pi}_{-\pi} f(x) \cos(kx)\, dx\,\, k\geq 1\)

\(\displaystyle B_k = \frac{1}{\pi} \int^{\pi}_{-\pi} f(x) \sin(kx)\, dx\,\, k\geq 1\)


\(\displaystyle A_0 = \frac{1}{2\pi} \int^{\pi}_{-\pi}\,\cosh(ax) dx=\frac{1}{\pi} \int^{\pi}_{0}\,\cosh(ax) dx =\frac{\,\sinh(a\pi)}{a\pi}\)

Here I will use the result by chisigma (no need to expand the hyperbolic func!)


\(\displaystyle A_k = \frac{1}{\pi} \int^{\pi}_{-\pi} \cosh(ax) \cos(kx)\, dx = \frac{2}{\pi}\left(\frac{a\ \sinh (ax)\ cos (kx) + k\ \cosh (a x)\ \sin (k x)} {a^{2} + k^{2}}\right)^{\pi}_0\)

\(\displaystyle A_k=\frac{2}{\pi} \left(\frac{a\ \sinh (a\pi)\cos (k\pi)} {a^{2} + k^{2}}\right)= \frac{2a\ (-1)^k \sinh (a\pi)} {\pi (a^{2} + k^{2})}\)


\(\displaystyle B_k = \frac{1}{\pi} \int^{\pi}_{-\pi} \cosh(ax) \sin(kx)\, dx=0 \) odd fucntion

\(\displaystyle \cosh(ax) = \frac{\sinh(a\pi)}{a\pi} + \frac{2a\,\sinh (a\pi)}{\pi}\,\sum^{\infty}_{k=1} \frac{\ (-1)^k \cos(kx)} {a^{2} + k^{2}}\)
 
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aNxello

New member
Feb 20, 2013
13
Woah! Thank you!
I don't know what to say, you guys have helped me so much!
I changed the title of the thread so it shows that it's solved (that's the right thing to do right?)
I don't know how to thank you, you guys are awesome! (Bow)
I had not realized Cosh(ax)sin(kx) was odd, that helped a lot!
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Here is the solution :

\(\displaystyle B_k = \frac{2}{\pi} \int^{\pi}_{0} \cosh(ax) \sin(kx)\, dx=0 \) odd function
I had not realized Cosh(ax)sin(kx) was odd, that helped a lot!
I think there might be a mistake here. Odd functions do integrate to zero... if the interval is symmetric. You should write your integral with the symmetric intervals with which you're working.

[EDIT] Error fixed.
 
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aNxello

New member
Feb 20, 2013
13
I think there might be a mistake here. Odd functions do integrate to zero... if the interval is symmetric. You should write your integral with the symmetric intervals with which you're working.
So do you you mean like this?
\(\displaystyle B_{n} = \frac{1}{\pi }\int_{-\pi }^{\pi} Cosh(ax)sin(nx)\)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
Yes, or maybe it's

$$B_{n}= \frac{2}{ \pi} \int_{- \pi}^{ \pi} \cosh(ax) \sin(nx) \, dx?$$

By the way, use backslashes in front of standard mathematical functions like cosh and sin:

Code:
 \cosh(ax) \sin(nx)
is the right way to go. No capitalization necessary (unless you're working in Mathematica).
 

aNxello

New member
Feb 20, 2013
13
Yes, or maybe it's

$$B_{n}= \frac{2}{ \pi} \int_{- \pi}^{ \pi} \cosh(ax) \sin(nx) \, dx?$$

By the way, use backslashes in front of standard mathematical functions like cosh and sin:

Code:
 \cosh(ax) \sin(nx)
is the right way to go. No capitalization necessary (unless you're working in Mathematica).
I think that 2 on top of the Pi was there because he had changed the range of the integral, so if we put it back to the original range the 2 shouldn't be there...I might be wrong tho

And thanks! I'm new with latex and I was kinda learning as I was going along, I appreciate the tips!
 
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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
I think there might be a mistake here. Odd functions do integrate to zero... if the interval is symmetric. You should write your integral with the symmetric intervals with which you're working.
Yes , copy-paste mistake , I edited it :cool: