Expert: Paul Klarreich - 1/1/2007

Question
What is meant by the even extention of a function and by the odd extention of a function?Could not find this in detail in the books i have.

Siddharth
(enginnearing aspirant)


Answer
Questioner:  Siddharth
Category:  Calculus
 
Subject:  even and odd extension
Question:  What is meant by the even extension of a function and by the odd extension of a function?  Could not find this in detail in the books i have.

Siddharth
(engineering aspirant)
...............................
Hi, Siddarth,

I have no idea what these mean.  However, that does not stop me from giving you an answer, so treat my answer with the disrespect it probably deserves.

Given a function  f(x) defined on the interval [0,1], you (probably) want to expand it as a Fourier series. (You said you were doing engineering, right?)   So you assume the function is periodic, and  repeats over the intervals [0,1], [1,2], [2,3] etc.  and you expand it, as a series:

f(x) = SUM (ak cos (2pi kx) + bk sin (2pi kx))

The 2pi is there so that the usual 'trig' interval of [0,2pi] matches the given interval [0,1].

But what if you want to expand it JUST using cosines?  Cosine is an even function.  So you make the assumption that the function has a period of 2, not 1, and has a basic interval of [-1,1], twice as large.  Then you define a function as follows:


EVEN(x) =  f(x), on  [0,1]
          f(-x), on [-1,0]

(note that in [-1,0], x is negative, so -x is positive.)  You can easily confirm that  EVEN(-x) = EVEN(x), making it really an even function.

Now, when you do your Fourier series, and get your coefficients, the ones for the sine terms will vanish because of the symmetric interval.  Your rules for finding a coefficient require you to get this integral:

{1
| EVEN(x) sin(2 pi x) dx
}-1

Since f(x) will be replaced by EVEN(x) on this interval, the integrand is an odd function, and the integral of an odd function on a symmetric interval, like [-1,1], is always zero.

So the function EVEN(x) is your even extension.  It will not be hard for you to work out the details for the function:

ODD(x) =   f(x), on  [0,1]
         -f(-x), on [-1,0]

which will have a F.S. with only sine terms.