COMPLEX FORM OF FOURIER SERIES

7. COMPLEX FORM OF FOURIER SERIES

Recall that real sines and cosines can be expressed in terms of complex exponentials by the formulas [Chapter 2, (11.3)]

e inx +e −inx (7.1)

e inx −e −inx

sin nx =

cos nx =

2 If we substitute equations (7.1) into a Fourier series like (5.12), we get a series of

2i

terms of the forms e inx and e −inx . This is the complex form of a Fourier series. We can also find the complex form directly; this is often easier than finding the sine- cosine form. We can then, if we like, work back the other way and [using Euler’s formula, Chapter 2, (9.3)] get the sine-cosine form from the exponential form.

We want to see how to find the coefficients in the complex form directly. We assume a series

f (x) = c 0 +c 1 e ix +c −1 e −ix +c 2 e 2ix +c −2 e −2ix +···

n=+∞

c n e inx

n=−∞

Section 7 Complex Form of Fourier Series 359 and try to find the c n ’s. From (5.4) we know that the average value of e ikx on (−π, π)

is zero when k is an integer not equal to zero. To find c 0 , we find the average values of the terms in (7.2):

1 π 1 π average values of terms of the (7.3)

form e −π ikx −π

To find c n , we multiply (7.2) by e −inx and again find the average value of each term. Note the minus sign in the exponent. In finding a n , the coefficient of cos nx in equation (5.1), we multiplied by cos nx; but here in finding the coefficient c n of

e inx , we multiply by the complex conjugate e −inx .

dx = c 0 e −inx dx + c 1 e −inx e ix dx 2π −π

f (x)e −inx

e −inx e −ix dx + · · · .

The terms on the right are the average values of exponentials e ikx , where the k values are integers. Therefore all these terms are zero except the one where k = 0; this is the term containing c n . We then have

dx = c n · dx = c n , 2π −π

f (x)e −inx dx = c

e −inx e inx

f (x)e −inx dx.

Note that this formula contains the one for c 0 (no 1 2 to worry about here!). Also, since (7.6) is valid for negative as well as positive n, you have only one formula to memorize here! You can easily show that for real f (x), c −n =¯ c n (Problem 12).

Example. Let us expand the same f (x) we did before, namely (5.11). We have from (7.6)

e −inx · 0 · dx +

e −inx · 1 · dx

(7.7) = 1 = πin (e , n odd, −inπ − 1) =

1 e −inx π

2π −in 0 −2πin

dx = .

360 Fourier Series and Transforms Chapter 7

Then

e 5ix (7.8)

e −3ix

e −5ix

It is interesting to verify that this is the same as the sine-cosine series we had before. We could use Euler’s formula for each exponential, but it is easier to collect terms like this:

which is the same as (5.12).

PROBLEMS, SECTION 7

1 to 11. Expand the same functions as in Problems 5.1 to 5.11 in Fourier series of complex exponentials e inx on the interval (−π, π) and verify in each case that the answer is equivalent to the one found in Section 5.

12. Show that if a real f (x) is expanded in a complex exponential Fourier series P ∞

−∞ c n e inx , then c −n =¯ c n , where ¯ c n means the complex conjugate of c n .

−∞ c n e , use Euler’s formula to find a n and b n in terms of c n and c −n , and to find c n and c −n in terms of a n