Convergence of Fouier Series and Evaluating Sums Using Fouier's Law

  • Thread starter dikmikkel
  • Start date
  • Tags
    Series Sum
In summary, the conversation discusses finding the real Fourier series for the 2\pi-periodic function f(t) = t t in [-Pi;Pi], which is given by f(t) ~ \sum\limits_{n=1}^{\infty}\frac{2}{n}(-1)^{n+1}\sin nt. Using this series, the value of \sum\limits_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{2n-1} is evaluated by substituting t = \pi/2 and applying Fourier's law. The final answer is \frac{\pi}{4}. There is some confusion about whether t or f(t)
  • #1
dikmikkel
168
0

Homework Statement


Consider the 2[itex]\pi[/itex]-periodic function f(t) = t t in [-Pi;Pi]
a) show that the real fouier series for f(t) is:
[itex]f(t) ~ \sum\limits_{n=1}^{\infty}\frac{2}{n}(-1)^{n+1}\sin nt[/itex]
b)
Use the answer to evaluate the following : [itex]\sum\limits_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{2n-1}[/itex]
Hint: Use Fouier's law with t = [itex]\pi[/itex]/2

Homework Equations


Fouiers Law? I'm danish, and therefore I'm not really sure what it's called.

The Attempt at a Solution


Part a i have done by finding the coefficients.
Part b) I can't see where the problem in part b and the answer to a relates. I've tried with Maple 15 to calculate the value and I'm getting Pi/4, but i keep getting something different for the series from a)
Please Help me, i would really like to understand this as I'm studying physics.
 
Last edited:
Physics news on Phys.org
  • #2
The answer should be pi/4 ...


from Fourier series we got ...

t = [itex]\sum\frac{2}{n}(-1)n+1sinnt[/itex]

for t = [itex]\frac{\pi}{2}[/itex] ,

[itex]\frac{\pi}{2}[/itex] = 2 - 2/3 + 2/5 - 2/9 + ...

which followed ,

[itex]\frac{\pi}{2}[/itex] = 2 * [itex]\sum\frac{(-1)n+1}{2n - 1}[/itex]

hence, [itex]\sum\frac{(-1)n+1}{2n - 1}[/itex] = [itex]\frac{\pi}{4}[/itex]
 
  • #3
Hmm, isn't that a backwards proof? Can't i use something like the even terms of sin(n*Pi/2) = 0 and the odd ones alternates between -1 and 1

And how is t = ... isn't it f(t)?
 

What is a Fourier series?

A Fourier series is a mathematical tool used to represent a periodic function as a sum of sinusoidal functions. It is named after French mathematician Joseph Fourier.

How do you calculate the sum of a Fourier series?

The sum of a Fourier series can be calculated by using the Fourier coefficients, which are obtained by integrating the function over a period and then multiplying it by certain trigonometric functions. The sum is then computed by adding all the terms in the series.

What is the purpose of calculating the sum of a Fourier series?

The purpose of calculating the sum of a Fourier series is to approximate a periodic function with a simpler function made up of sinusoidal components. This allows for easier analysis and manipulation of the function.

What are some applications of Fourier series?

Fourier series have many applications in mathematics, physics, and engineering. They are commonly used in signal processing, image and sound compression, and solving differential equations.

Is it possible to calculate the sum of a Fourier series for any function?

No, it is not possible to calculate the sum of a Fourier series for any function. The function must be periodic and have certain conditions, such as being continuous and having a finite number of discontinuities, for the series to converge and accurately represent the function.

Similar threads

  • Calculus and Beyond Homework Help
Replies
1
Views
175
  • Calculus and Beyond Homework Help
Replies
3
Views
395
  • Calculus and Beyond Homework Help
Replies
1
Views
211
  • Calculus and Beyond Homework Help
Replies
1
Views
320
  • Calculus and Beyond Homework Help
Replies
4
Views
256
  • Calculus and Beyond Homework Help
Replies
6
Views
203
  • Calculus and Beyond Homework Help
Replies
2
Views
360
  • Calculus and Beyond Homework Help
Replies
2
Views
155
  • Calculus and Beyond Homework Help
Replies
16
Views
542
  • Calculus and Beyond Homework Help
Replies
1
Views
519
Back
Top