Simplifying the Fourier Series Function: Tips & Tricks

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SUMMARY

The discussion focuses on simplifying the Fourier series representation of the piecewise function defined as f(x) = 1 for 0 < t < 1 and f(x) = -1 for 1 < t < 2. The solution involves recognizing that the Fourier series can be expressed as a sum over odd integers, leading to the formulation of the series as &frac{4}{\pi}∑_{n=0}^∞ &frac{sin((2n+1)πt)}{2n+1}. Participants explore methods to determine if a function can be rewritten in a different form, emphasizing the importance of identifying the nature of the function's periodicity and symmetry.

PREREQUISITES
  • Understanding of Fourier series and their applications
  • Knowledge of piecewise functions and their properties
  • Familiarity with trigonometric identities and summation techniques
  • Basic calculus, particularly integration and differentiation of functions
NEXT STEPS
  • Study the properties of Fourier series convergence
  • Learn about the Dirichlet conditions for Fourier series
  • Explore the concept of odd and even functions in Fourier analysis
  • Investigate the use of Fourier series in signal processing applications
USEFUL FOR

Mathematics students, educators, and professionals in engineering or physics who are working with Fourier series and seeking to simplify or analyze piecewise functions.

darkmagic
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Homework Statement



f(x) = 1 0<t<1
= -1 1<t<2

How can I simplify this given that function(on the attachment).

What I mean is that how can I write the function in any other way?

In addition, How can I know if the function can be written in other form?
How can I write the function in other form?



Homework Equations





The Attempt at a Solution

 

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If n is even, 1- (-1)n= 1- 1= 0! If n is odd, 1- (-1)n= 1- (-1)= 2.

So
[tex]\frac{2}{\pi}\sum_{n=1}^\infty \frac{[1- (-1)^n] sin(n\pi t)}{n}[/tex]
is just
[tex]\frac{2}{\pi}\sum \frac{2 sin(n\pi t)}{n}[/tex]
where now the sum runs only over odd n. One way to show that is to use 2n+1 rather than n in the body of the sum. That way, as n goes over all non-negative integers, 2n+1 goes over all positive odd integers:
[tex]\frac{4}{\pi}\sum_{n=0}^\infty \frac{sin((2n+1)\pi t)}{2n+1}[/tex]
 
Can 2n+1 be 2n-1 provided that n=1 to infinity?
How can I know if the function can be converted in some form?
 

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