Fourier series equation derivation

  • Thread starter merlyn
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15
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Hi all. Could someone work out for me how equation 21 in attachment left side becomes right side. Please show in detail if you could.
It's for exponential Fourier series.

Drforbin

Screenshot_2019-03-06_02-42-49.png

thank you
 

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DrClaude

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Hi all. Could someone work out for me how equation 21 in attachment left side becomes right side. Please show in detail if you could.
It's for exponential Fourier series.
This is not how we work here. We will gladly help you find the solution, but we won't do you work for you.

So, starting from the left-hand side, how far can you go?
 
15
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Look, it's not my work. I'm 48 years old and teaching myself Fourier Transforms.
Don't make it any more difficult than it already is.
If you could show me the derivation I would appreciate it.
But in answer to your question, i've tried converting it to Eulers form than applying trig identities with no luck.

Thank you in advance for your time.
 

f95toli

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I don't see a "trick" of any kind ; it is just matter of solving the integral.

Do you know how to integrate an exponential function?
 

sophiecentaur

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@merlyn Have you tried any other sources? It's been a long time but is your problem with collapsing the product of two exponentials into one (with n-m in it) or doing the (trivial?) definite integral (∫eax dx) after that? Or is it the 1/(n-m) term on the RHS, when n=m?
The argument goes onto the next page of your book. What happens there?
 
15
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@merlyn Have you tried any other sources? It's been a long time but is your problem with collapsing the product of two exponentials into one (with n-m in it) or doing the (trivial?) definite integral (∫eax dx) after that? Or is it the 1/(n-m) term on the RHS, when n=m?
The argument goes onto the next page of your book. What happens there?
Thank you so much in advance for your time.

It's mainly collapsing the the two 'e' into one. The definite integral I think I can handle.
If you could show me a worked out example I would REALLY appreciate it.
I'm sure I am just missing a step here. It's probably my stupid brain.
Sorry for all the trouble.

By the way, where do you get the 1/(n-m) from?

You asked what is continued on next page, please see attached.
Screenshot_2019-03-07_04-10-28.png
 

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jtbell

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It's mainly collapsing the the two 'e' into one.
General rule for multiplying exponentials: ##e^a e^b = e^{a+b}##.
 
15
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General rule for multiplying exponentials: ##e^a e^b = e^{a+b}##.
Right..But that really does not help in this case.
All you get is ##e^(i2pinx-i2pimx)/l##
right?
and

##e^(i2pix(n-m))/l##
 

jtbell

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OK, so now you have the integral $$\int_0^L e^{i 2\pi (n-m) x/L} \, dx$$ Do you know how to integrate exponentials? The exponential looks messy, but it's just a big messy constant times x. Simplify it for a moment by collapsing the big messy constant into a new one, ##a = i 2\pi (n-m) /L##. Can you do this one? $$\int_0^L e^{ax} \, dx$$
 
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OK, so now you have the integral $$\int_0^L e^{i 2\pi (n-m) x/L} \, dx$$ Do you know how to integrate exponentials? The exponential looks messy, but it's just a big messy constant times x. Simplify it for a moment by collapsing the big messy constant into a new one, ##a = i 2\pi (n-m) /L##. Can you do this one? $$\int_0^L e^{ax} \, dx$$
I'll look up my integral tables tonight and see.
Thank you so far.
 

anorlunda

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Ah, sweet memories of the CRC tables and slide rules.

But for those of us who didn't keep their old CRC copy, there are tools like wolframalpha.com where we can just type in "integrate e^(i*a*t) dt from 0 to L" and get an answer.
 

jtbell

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And there are several web sites with tables of integrals similar to the ones in the CRC book, including one on Wikipedia. Just Google for "table of integrals".
 

sophiecentaur

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Right..But that really does not help in this case.
All you get is ##e^(i2pinx-i2pimx)/l##
right?
and

##e^(i2pix(n-m))/l##
Your presentation is very confusing. There is a button ∑ on the menu bar that gives you a whole selection of symbols - including π :smile:
 
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Your presentation is very confusing. There is a button ∑ on the menu bar that gives you a whole selection of symbols - including π :smile:
Sorry...I will retype later tonight.
Didn't see menu bar.
 

jtbell

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Enclose the whole exponent in curly braces.

e^{(i2π(n-m)x)/L} produces ##e^{(i2π(n-m)x)/L}##.

I think it looks better without one pair of parentheses. Also, in LaTeX, you can use \pi instead of the forum's π.

e^{i2 \pi (n-m)x/L} produces ##e^{i2 \pi (n-m)x/L}##.

For more information, see our LaTeX primer.
 

DrClaude

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This made me dig in my closet to see if I still have my old book of math tables. Printed in 1986.
This is off topic, but I use the exact same edition.
 

sophiecentaur

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I'll look up my integral tables tonight and see.
Thank you so far.
I can sympathise that you are out of touch with Calculus. If you really feel you need to get to grips with Fourier then it will be a long hard slog, I think. I can't recommend any particular learning resource but you will need more than just a list of integrals. The way many people look on Fourier is over simplified and they often come to wrong conclusions about what it really involves.
 
15
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I can sympathise that you are out of touch with Calculus. If you really feel you need to get to grips with Fourier then it will be a long hard slog, I think. I can't recommend any particular learning resource but you will need more than just a list of integrals. The way many people look on Fourier is over simplified and they often come to wrong conclusions about what it really involves.

Ok.

Here is worked out example.

##\int_0^L e^{i2 \pi nx/L}e^{-i2 \pi mx/L} \, dx \rightarrow \int_0^L e^{\frac {i2 \pi nx + -i2 \pi mx} L} \, dx \rightarrow\int_0^L e^{\frac {i2 \pi(n-m)x} L } \, dx##

##a= \frac {i2 \pi(n-m)} L##

##\int e^{ax} \, dx = \frac 1 a e^{ax} \rightarrow##

##\int \frac 1 { \frac {i2 \pi(n-m)} L } e^{\frac {i2 \pi(n-m)} L x} \, dx = \frac L{i2 \pi(n-m)} e^{\frac {i2 \pi(n-m)} Lx} ##

Thank you all!
 

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