Fourier transform of the hyperbolic secant function

In summary, The Fourier transform of the hyperbolic secant function can be derived using the geometric series, resulting in the formula 4\sum\ ((-1)^n*(1+2n))/(ω^2*(2n+1)^2). This formula can be obtained by applying the term-by-term method and taking into account the even symmetry of the function. The factor (-1)^n comes from the expansion of 1/(1+a) and alternates sign.
  • #1
Marie_Curie
3
0

Homework Statement



Hi there!

I'm just trying to figure out the Fourier transform of the hyperbolic secant function... I already know the outcome:

4[itex]\sum\ ((-1)^n*(1+2n))/(ω^2*(2n+1)^2) [/itex]

But sadly, I cannot figure out how to work round to it! :( maybe one of you could help me...

Homework Equations



I was thinking of using the geometric series 1/(1+q) = [itex]\sum (-q)^n [/itex] for q = e^(-2*x), as the hyperbolic secant is 2*e^(-x)/(1+e^(-2*x)) .
And then you need to multiply the hyperbolic secant with e^(iωt) and integrate from -∞
up to ∞.
But... frankly, that was it. I have never managed to come to the result
4[itex]\sum\ ((-1)^n*(1+2n))/(ω^2*(2n+1)^2) [/itex]

Maybe someone could give me a hint to to solve this problem...? :)

Lots of greetings,
Marie
 
Last edited:
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  • #2
Marie_Curie said:

Homework Statement



Hi there!

I'm just trying to figure out the Fourier transform of the hyperbolic secant function... I already know the outcome:

4[itex]\sum\ ((-1)^n*(1+2n))/(ω^2*(2n+1)^2) [/itex]

But sadly, I cannot figure out how to work round to it! :( maybe one of you could help me...

Homework Equations



I was thinking of using the geometric series 1/(1+q) = [itex]\sum (-q)^n [/itex] for q = e^(-2*x), as the hyperbolic secant is 2*e^(-x)/(1+e^(-2*x)) .
And then you need to multiply the hyperbolic secant with e^(iωt) and integrate from -∞
up to ∞.
But... frankly, that was it. I have never managed to come to the result
4[itex]\sum\ ((-1)^n*(1+2n))/(ω^2*(2n+1)^2) [/itex]

Maybe someone could give me a hint to to solve this problem...? :)

Lots of greetings,
Marie

f(x) = sech(x) is an even function, so the FT of f is (g + conjugate g), where ##g = \int_0^{\infty} f(x) e^{i \omega x} \, dx.## Apply this term-by-term to
[tex] \text{sech}(x) = 2 \sum (-1)^n e^{-(2n+1)x}.[/tex]
 
  • #3
Ray Vickson said:
f(x) = sech(x) is an even function, so the FT of f is (g + conjugate g), where ##g = \int_0^{\infty} f(x) e^{i \omega x} \, dx.## Apply this term-by-term to
[tex] \text{sech}(x) = 2 \sum (-1)^n e^{-(2n+1)x}.[/tex]


Aaahh, wonderful! Thank you!

Just one more question... where does the Term [tex](-1)^n[/tex] come from?

Thanks a lot!
MARIE
 
  • #4
Marie_Curie said:
Aaahh, wonderful! Thank you!

Just one more question... where does the Term [tex](-1)^n[/tex] come from?

Thanks a lot!
MARIE

It comes from the expansion of 1/(1+a)=1-a+a^2-a^3+... It's the factor that alternates sign.
 
Last edited:
  • #5
Dick said:
It comes from the expansion of 1/(1-a)=1-a+a^2-a^3+... It's the factor that alternates sign.

Oh yes, sure! Now I got it :)

Thank you very much! :) And have a nice evening!

_MARIE_
 
  • #6
Marie_Curie said:
Oh yes, sure! Now I got it :)

Thank you very much! :) And have a nice evening!

_MARIE_

You're welcome. Nice evening to you too. Note I goofed on the post. I meant 1/(1+a) not 1/(1-a).
 

FAQ: Fourier transform of the hyperbolic secant function

1. What is the Fourier transform of the hyperbolic secant function?

The Fourier transform of the hyperbolic secant function (sech) is given by: F(ω) = √(π/2) * sech(πω/2), where ω is the frequency variable.

2. How is the hyperbolic secant function defined?

The hyperbolic secant function is defined as sech(x) = 1/cosh(x) = 2/(e^x + e^(-x)), where x is the input variable.

3. What is the significance of the hyperbolic secant function in mathematics?

The hyperbolic secant function is often used in statistics and probability theory, particularly in the study of random variables. It is also used in signal processing and in the study of wave propagation.

4. Are there any applications of the Fourier transform of the hyperbolic secant function?

Yes, the Fourier transform of the hyperbolic secant function has applications in signal processing, as it can help analyze signals in the frequency domain. It is also used in the study of quantum field theory in physics.

5. How is the Fourier transform of the hyperbolic secant function related to other mathematical functions?

The Fourier transform of the hyperbolic secant function is closely related to the Gaussian function and the inverse tangent function. It can also be expressed in terms of the error function and the Jacobi theta function.

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