Finding Fourier Expansion of f(t): Part b & c

Click For Summary
SUMMARY

The Fourier expansion of the function f(t) = 1 + |t| for |t| < 1 is established as 1 + (2/π) Σ (from n=0 to ∞) [(-1)^(n+1)/n] sin(nπt). The discussion focuses on solving parts b and c, where part b requires finding the sum of the series Σ (from n=0 to ∞) [(-1)^n/(2n+1)] using the Fourier expansion. In part c, the task is to compute F(1) + F(-5) - 3F(0) based on the Fourier series representation of f.

PREREQUISITES
  • Understanding of Fourier series and expansions
  • Familiarity with trigonometric identities and series summation
  • Knowledge of convergence criteria for infinite series
  • Basic calculus, particularly evaluating functions at specific points
NEXT STEPS
  • Study the derivation of Fourier series coefficients a0, an, and bn
  • Learn about the convergence of Fourier series and their applications
  • Explore techniques for summing infinite series, particularly alternating series
  • Investigate the properties of odd and even functions in Fourier expansions
USEFUL FOR

Mathematicians, physics students, and engineers interested in signal processing, particularly those working with Fourier analysis and series summation techniques.

zyphriss2
Messages
18
Reaction score
0
Find the Fourier expansion of one period of f(t)=1+t absolute value of t<1

I found this to be 1+2/pi Sigma(0 to infinity) ((-1^(n+1))/n)sinnpit by just the standard methods of the a0 an and bn formuals, which I know is correct

Now the parts I am having problems with is part b and c which our teacher has not covered much at all and I cannot find any help online.

Part b. Use the Fourier expansion of f to find the sum of the Series
Sigma(0toinfinity)(-1^n)/(2n+1)

Part C. If f denotes the function defined on (-infinity to infinity) by the Fourier series of f, find F(1)+F(-5)-3F(0)
 
Physics news on Phys.org
Just to make things a bit more clear on the notation side, I take it your expansion is:

1+\frac{2}{\pi}\sum_{n=0}^{\infty}\frac{(-1)^{n+1}}{n} \sin{n\pi t}

As for question b:

Notice that if you set t equal to t_0 \equiv \frac{1}{2}, then

\sin{n\pi t_0} = \sin{\frac{n\pi}{2}}

which is zero whenever n is even and (-1)^{(n-1)/2 whenever n is odd... Do you see where I'm going with this?

part C:
I take it by F(0) you just mean f(0)? Have you tried filling in these numbers into the Fourier series?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K