Calculating Fourier Series Expansions for Piecewise Functions

But i'm not sure it would use a reduction formula.In summary, the Fourier Series Expansion for functions (a) and (b) can be found by using the equations F(x)=\frac{a_0}{2}+\Sigma_{n=1}^{\infty}[a_ncos(nx)+b_nsin(nx)], a_0=\frac{1}{\pi} \int_{- \pi}^{\pi}f(x)dx, a_n= \frac{1}{\pi} \int_{- \pi}^{\pi}f(x)cos(nx)dx, and b_n= \frac{1}{\pi} \int_{- \pi}^{\pi}
  • #1
kreil
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Homework Statement


Find the Fourier Series Expansion for:

(a) f(x) = [pi-2x, 0 < x < pi | pi+2x, -pi < x < 0]

(b) f(x) = [0, -pi < x < 0 | sin(x), 0 < x < pi]

Homework Equations



[tex]F(x)=\frac{a_0}{2}+\Sigma_{n=1}^{\infty}[a_ncos(nx)+b_nsin(nx)][/tex]

[tex]a_0=\frac{1}{\pi} \int_{- \pi}^{\pi}f(x)dx[/tex]

[tex]a_n= \frac{1}{\pi} \int_{- \pi}^{\pi}f(x)cos(nx)dx[/tex]

[tex]b_n= \frac{1}{\pi} \int_{- \pi}^{\pi}f(x)sin(nx)dx[/tex]

The Attempt at a Solution



For (a) my final answer was:

[tex]f(x)=\frac{8}{\pi}(cos(x)+\frac{cos(3x)}{9}+...+\frac{cos(nx)}{n^2})[/tex]

and i think this is correct, but for (b) i got kind of a funny answer imo;

[tex]f(x)=\frac{1}{\pi}+\frac{2}{\pi}(\frac{cos(3x)}{8}+\frac{cos(5x)}{24}+...+\frac{cos(nx)}{n^2-1})[/tex]

if someone could work out b and see if they get the same answer i would appreciate it.

Josh
 
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  • #2
Those are supposed to be piecewise functions, right?

Hmm, in my denominator (for the second one) I got something different. Here's what I got
[tex] f(x) = \sum_{n=0}^\infty \frac{cosnx}{\pi(1+1/n^2)}[/tex]

Anybody else get something similar?
 
  • #3
well here's my work for (b) so it can be checked:

f(x)=[0, -pi < x < 0 | sin(x), 0 < x < pi]

[tex]a_0= \frac{1}{\pi} \int_{- \pi}^{\pi}f(x)dx= \frac{1}{\pi} \int_{0}^{\pi}sin(x)dx= \frac{1}{\pi}(-cos(x))|_{0}^{\pi}=\frac{2}{\pi} \implies \frac{a_0}{2}=\frac{1}{\pi}[/tex]

[tex]a_n=\frac{1}{\pi} \int_{- \pi}^{\pi}f(x)cos(nx)dx= \frac{1}{\pi} \int_0^{\pi}sin(x)cos(nx)dx=\frac{1}{\pi} \frac{cos(x)cos(nx)+nsin(x)sin(nx)}{n^2-1}|_0^{\pi}=\frac{1}{\pi}(\frac{1}{n^2-1}-\frac{cos(n \pi)}{n^2-1})=\frac{1}{\pi (n^2-1)}(1-(-1)^n)[/tex]
[tex]\implies a_n=\frac{1}{\pi (n^2-1)}((-1)^{n+1}+1)[/tex]

[tex] b_n=0[/tex]

[tex]f(x)=\frac{1}{\pi}+\frac{2}{\pi}(\frac{cos(3x)}{8} +\frac{cos(5x)}{24}+...+\frac{cos(nx)}{n^2-1})[/tex]
 
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  • #4
How did you go from the integral of sin(x) cos(nx) to what you have? A reduction formula?

The b_n's are definitely zero. I forgot to put a_0 in my solution too.
 
  • #5

FAQ: Calculating Fourier Series Expansions for Piecewise Functions

What is a Fourier Series Expansion?

A Fourier Series Expansion is a mathematical technique used to express a periodic function as a sum of sinusoidal functions. It is based on the principle that any periodic function can be represented as a sum of sine and cosine functions with different frequencies and amplitudes.

What is the purpose of a Fourier Series Expansion?

The purpose of a Fourier Series Expansion is to simplify the representation of a periodic function and make it easier to analyze and manipulate. It allows for the decomposition of complex functions into simpler components, making it a powerful tool in signal processing, engineering, and physics.

What is the difference between a Fourier Series Expansion and a Fourier Transform?

A Fourier Series Expansion is used for periodic functions, while a Fourier Transform is used for non-periodic functions. The Fourier Series Expansion decomposes a periodic function into a sum of sinusoidal functions, while the Fourier Transform decomposes a non-periodic function into a continuous spectrum of frequencies.

What are the applications of Fourier Series Expansions?

Fourier Series Expansions have many practical applications in various fields such as engineering, physics, and signal processing. Some examples include analyzing and designing electronic circuits, understanding the behavior of vibrating structures, and creating sound and image compression algorithms.

What are the limitations of Fourier Series Expansions?

One limitation of Fourier Series Expansions is that it can only be used for periodic functions. It also assumes that the function being expanded is continuous and has a finite number of discontinuities. Additionally, the accuracy of the expansion depends on the number of terms used, and for some functions, an infinite number of terms may be required for an exact representation.

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