Complex power series to calculate Fourier series

In summary, the conversation is about trying to solve problem #21 from the book Visual Complex Analysis. The problem involves finding the Fourier series for the function cos(sin θ)e^cos θ. The author suggests substituting z = re^(iθ) into the power series for e^z and then isolating the real and imaginary parts. The person seeking help is unsure about how this substitution helps and is asking for a nudge in the right direction.
  • #1
terhorst
11
0
I never took complex analysis in undergrad and always regretted it, so I'm working through the book Visual Complex Analysis on my own. Really enjoying it so far.

Homework Statement



Actually you can view the problem http://books.google.de/books?id=ogz5FjmiqlQC&printsec=frontcover&hl=en#PPA117,M1", it's #21 at the top of the page.

Show that the Fourier series for [tex][\cos(\sin \theta)]e^{\cos \theta}[/tex] is [tex]\sum^{\infty}_{n=0} \frac{\cos n \theta}{n!}[/tex].

Homework Equations



The author suggests substituting [tex]z=re^{i \theta}[/tex] into the power series for [tex]e^z[/tex] and then isolating the real and imaginary parts.

The Attempt at a Solution



I don't really understand why the substitution. So then you end up with [tex]\sum \frac{r^n [\cos(n \theta) + i \sin (n \theta)]}{n !}[/tex]. But I don't see how this gets me any closer to equating [tex][\cos(\sin \theta)]e^{\cos \theta}[/tex] and [tex]\sum \frac{r^n \cos(n \theta)}{n!}[/tex].

A nudge in the right direction would be appreciated. Thanks for looking.
 
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  • #2
terhorst said:
Show that the Fourier series for [tex][\cos(\sin \theta)]e^{\cos \theta}[/tex] is [tex]\sum^{\infty}_{n=0} \frac{\cos n \theta}{n!}[/tex].

The author suggests substituting [tex]z=re^{i \theta}[/tex] into the power series for [tex]e^z[/tex] and then isolating the real and imaginary parts.

Hi terhorst! :smile:

(have a theta: θ :wink:)

Hints:

i] use r = 1

ii] what is cos(eisinθ) ? :smile:
 

1. What is a complex power series?

A complex power series is a mathematical series that involves complex numbers. It is typically written in the form of n=0∞ cnzn, where z is a complex number, cn is a sequence of complex coefficients, and n is the power to which z is raised.

2. How is a complex power series used to calculate Fourier series?

A complex power series can be used to calculate Fourier series by representing a periodic function as a sum of complex exponentials. This is known as the exponential form of the Fourier series and can be expressed as f(x) = ∑n=0∞ cneinx. The coefficients cn can be found by using the complex power series of the function.

3. What is the relationship between complex power series and Fourier series?

The relationship between complex power series and Fourier series is that a complex power series can be used to calculate the Fourier series of a periodic function. The coefficients of the complex power series correspond to the coefficients in the exponential form of the Fourier series.

4. How do you determine the convergence of a complex power series?

The convergence of a complex power series can be determined by using the ratio test. If the limit of |cn+1|/|cn| is less than 1, then the series converges absolutely. If the limit is greater than 1, then the series diverges. If the limit is equal to 1, then the ratio test is inconclusive and other tests, such as the root test, may be used to determine convergence.

5. Can a complex power series be used to calculate the Fourier series of any function?

No, a complex power series can only be used to calculate the Fourier series of a function if the function is periodic and can be represented by a continuous power series. Additionally, the complex power series must have a radius of convergence greater than or equal to the period of the function in order to accurately represent the Fourier series.

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