17.6. THE DISCREET FOURIER TRANSFORM 415

=

(2π

1− cos(kπ)

k

)sinkx

The terms when k is even are all 0. Therefore, the above reduces to

4π

n

∑k=1

(1

2k−1

)sin(2k−1)x

In the case where n = 4, the graph of the function f being approximated along with theabove function which is approximating it are as shown in the following picture on [−π,π].This sum which

-4 -2 0 2 4

-1

0

1

delivers the closest point in U will be denoted by Sn f . Note how the approximate func-tion, closest in the mean square norm, is not equal to the given function at very many pointsbut is trying to be close to it across the entire interval [−π,π], except for a small intervalcentered at 0. You might try doing similar graphs on a calculator or computer in whichyou take larger and larger values of n. What will happen is that there will be a little bumpnear the point of discontinuity which won’t go away, but this little bump will get thinnerand thinner. The reason this must happen is roughly because the functions in the sum arecontinuous and the function being approximated is not. Therefore, convergence cannottake place uniformly. This is all I will say about these considerations because this is notan analysis book. See Problem 20 below for a discussion of where the Fourier series doesconverge at jumps.

17.6 The Discreet Fourier TransformEverything done above for the Fourier series on [−π,π] could have been done just as easilyon [0,2π] because all the functions are periodic of period 2π . Thus, for f a function definedon [0,2π] , you could consider the partial sums of the Fourier series on [0,2π]

n

∑k=−n

akeikx

where

ak =1

2π

∫ 2π

0f (y)e−ikydy

and all the results of the last section continue to apply except now it is on the new interval.This is done to make the presentation of what follows easier to write.

The idea is that maybe you don’t know what the function f is at all points, only atcertain points

x j = j2π

N, j = 0,1, · · · ,N−1