Find the fourier sine series of cosine.

In summary, the problem is that the text does not define f(t) in the interval -pi<t<0. If its Fourier series contains only sine terms, that suggests that the function is odd, f(t) = -cos(t) if -pi<t<0. The answer is listed as 0.
  • #1
pondzo
169
0

Homework Statement


Hi, so I am doing some past exam papers and there was this question;
fourier sine series.PNG


Homework Equations

The Attempt at a Solution


a0 and an both are equal to zero, this leaves only bn.

Since you can only use the sine series for an odd function, and cos(t) is even, does this mean i have to find the odd half expansion of cos(t) with L=π ? This would be f(t) = -cos(t) for [-π,0] and cos(t) for [0,π]. Or do I use L= π/2 and do the integral from 0 to π for bn?

Letting n=1 since it is the first partial sum we are talking about (can i do this?)then using the first method I get 16/(3π) and if i try the second method I get 8/(3π). But the answer is listed as 0.

Any help would be great thanks.
 
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  • #2
You should write the relevant equations.
The problem text does not define f(t) in the interval -pi<t<0. If its Fourier series contains only sine terms, that suggests that the function is odd, f(t) = -cos(t) if -pi<t<0.
What is the formula for the Fourier coefficients of an odd function?
 
  • #3
Hi ehild, sorry i wasn't quite sure how to typeset them.

I did write that if I perform the odd half expansion of cos(t) I get
f(t) = cos(t) for [0,π] and f(t) = -cos(t) for [-π,0]
Using this (and assuming I can let n=1 since it is this first partial sum, but I am not sure if i can do this) then ##b_1=\frac{1}{\pi}[\int_{-\pi}^0-\cos{t}\sin{2t}dt + \int_{0}^{\pi}\cos{t}\sin{2t}dt] = \frac{8}{3\pi}##

Where am I going wrong?
 
  • #4
Oops, I found where i went wrong.
Instead of ##b_1=\frac{1}{\pi}[\int_{-\pi}^0-\cos{t}\sin{2t}dt + \int_{0}^{\pi}\cos{t}\sin{2t}dt] ##
It should be ##b_1=\frac{1}{\pi}[\int_{-\pi}^0-\cos{t}\sin{t}dt + \int_{0}^{\pi}\cos{t}\sin{t}dt] = 0 ##

I would still like to know if it is correct to let n=1 in the bn integral to find b1. Or should you only sub in n=1 after you find the general form of bn?
 
  • #5
You can set n=1 before doing the integration. You might as well since all you were looking for was ##b_1##.
 

1. What is a Fourier sine series?

A Fourier sine series is a mathematical representation of a periodic function as a sum of sine functions with different amplitudes and frequencies. It is used in the field of Fourier analysis to decompose a function into simpler components.

2. How is the Fourier sine series of cosine calculated?

The Fourier sine series of cosine is calculated by using the formula:
f(x) = a0 + ∑n=1 (ansin(nπx/L) + bncos(nπx/L))
where L is the period of the function, an and bn are the Fourier coefficients, and n is the number of terms in the series.

3. What is the difference between a Fourier sine series and a Fourier cosine series?

A Fourier sine series only contains sine functions, while a Fourier cosine series only contains cosine functions. The main difference is in the coefficients used to calculate the series. In a Fourier sine series, the coefficients are an = 0 and bn ≠ 0, while in a Fourier cosine series, the coefficients are bn = 0 and an ≠ 0.

4. Why is the Fourier sine series important?

The Fourier sine series is important because it allows us to represent a function as an infinite sum of simpler components, making it easier to analyze and understand. It also has many practical applications in fields such as signal processing, image reconstruction, and data compression.

5. Can any function be represented by a Fourier sine series?

No, not every function can be represented by a Fourier sine series. The function must be periodic, have a finite number of discontinuities, and have a finite number of maxima and minima within each period. Also, the function must satisfy the Dirichlet conditions for the Fourier series to converge.

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