So the statement sin(n+1)pi/2 = cos(n)pi/2 is true by this property.

In summary, the Fourier series for f(x) = |cos(x)| in the interval (-π, π) is zero for the sine term and can be written as an = 2/pi * (1 - 2/n^2) * sin(n pi/2) * cos(n x), where n is an odd integer. To find the series, the intervals must be broken at pi/2 and the complementary relation between sin and cos must be used to simplify the final expression.
  • #1
SVXX
35
0
Question - Find the Fourier series for f(x) = |cos(x)| in the interval (-π, π).
Right, I attempted the question and the integration that followed. I'm having trouble in the integration itself..
Firstly I found that f(x) is an even function, so the sine term(bn) of the Fourier expansion would be zero and that I can use a property of integration to write

gif.latex?a_0%20=%20\frac{1}{\pi}%20\int_{-\pi}^{\pi}f(x)dx.gif
as
gif.latex?a_0%20=%20\frac{2}{\pi}%20\int_{0}^{\pi}f(x)dx.gif
and similarly so for an.

%20=%20\frac{1}{\pi}%20\int_{0}^{\pi}[cos(n-1)x%20\hspace{1mm}%20+%20\hspace{1mm}%20cos(n+1)x]dx.gif


frac{1}{\pi}[\frac{sin(n-1)x}{n-1}]_0%20^\pi%20+%20\frac{1}{\pi}[\frac{sin(n+1)x}{n+1}]_0%20^\pi.gif


I computed a0 to be zero, but the trouble is, I'm getting an to be zero too, and if that is so, there will be no Fourier series! Could someone please point out my mistake?
 
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  • #2
Hints :

1. One period of |cos(x)| extends from -pi/2 to pi/2. Check your limits.

2. Beware divide by zero.
 
  • #3
Thanks for the hints...I figured out what I was doing wrong in the integration. I had to break the intervals at pi/2 as |cos(x)| = cos(x) for 0 < x < pi/2 and = - cos(x) for pi/2 < x < pi.
I referred a popular book from the college library today which proved me correct. But what I don't understand is the last part of the solution that was given in that book...it stated that sin(n+1)pi/2 = cos(n)pi/2...how is this so?
 
  • #4
SVXX said:
Thanks for the hints...I figured out what I was doing wrong in the integration. I had to break the intervals at pi/2 as |cos(x)| = cos(x) for 0 < x < pi/2 and = - cos(x) for pi/2 < x < pi.
I referred a popular book from the college library today which proved me correct. But what I don't understand is the last part of the solution that was given in that book...it stated that sin(n+1)pi/2 = cos(n)pi/2...how is this so?

The complementary relation between sin and cos tells us that sin x = cos (pi/2 - x) and visa versa.

So sin (n+1)pi/2 = cos (pi/2 -(n+1)pi/2) = cos (-n pi/2).

Now since cos is an even function then cos (-n pi/2) = cos (n pi/2).
 

What is a Fourier Series for |cos(x)|?

A Fourier Series is a mathematical representation of a periodic function, such as |cos(x)|, as a sum of sine and cosine functions. It allows us to break down a complex function into simpler components and understand its behavior more easily.

Why is the Fourier Series important?

The Fourier Series has many practical applications, such as in signal processing, image compression, and solving differential equations. It also helps us understand the periodic nature of functions and how they can be represented as a combination of simpler functions.

What are the key components of a Fourier Series?

The key components of a Fourier Series are the coefficients, which represent the amplitude and phase of each sine and cosine function, and the fundamental frequency, which determines the period of the function.

How is the Fourier Series for |cos(x)| calculated?

The Fourier Series for |cos(x)| is calculated by finding the coefficients using integration and the fundamental frequency using the period of the function. The final series is then the sum of these coefficients multiplied by the corresponding sine and cosine functions.

What are the limitations of the Fourier Series?

The Fourier Series has some limitations, such as only being applicable to periodic functions and not being able to represent discontinuous functions. It also requires an infinite number of terms to accurately represent a function, which can be computationally expensive.

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