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Note: 2 lectures, §9.2–§9.3 in [EP], §10.3 in [BD]

We have computed the Fourier series for a -periodic function, but what about functions of diﬀerent periods. Well, fear not, the computation is a simple case of change of variables. We just rescale the independent axis. Suppose we have a -periodic function . Then is called the half period. Let . Then the function

is -periodic. We must also rescale all our sines and cosines. In the series we use as the variable. That is, we want to write

If we change variables to we see that

We compute and as before. After we write down the integrals, we change variables from back to , noting also that .

The two most common half periods that show up in examples are and 1 because of the simplicity of the formulas. We should stress that we have done no new mathematics, we have only changed variables. If you understand the Fourier series for -periodic functions, you understand it for -periodic functions. You can think of it as just using diﬀerent units for time. All that we are doing is moving some constants around, but all the mathematics is the same.

extended periodically. The plot of the periodic extension is given in Figure 4.8. Compute the Fourier series of .

We want to write . For we note that is even and hence

Next we ﬁnd :

You should be able to ﬁnd this integral by thinking about the integral as the area under the graph without doing any computation at all. Finally we can ﬁnd . Here, we notice that is odd and, therefore,

Hence, the series is

Let us explicitly write down the ﬁrst few terms of the series up to the harmonic.

The plot of these few terms and also a plot up to the harmonic is given in Figure 4.9. You should notice how close the graph is to the real function. You should also notice that there is no “Gibbs phenomenon” present as there are no discontinuities.

We will need the one sided limits of functions. We will use the following notation

If you are unfamiliar with this notation, means we are taking a limit of as approaches from below (i.e. ) and means we are taking a limit of as approaches from above (i.e. ). For example, for the square wave function

(4.8) |

we have and .

Let be a function deﬁned on an interval . Suppose that we ﬁnd ﬁnitely many points , , , …, in the interval, such that is continuous on the intervals , , …, . Also suppose that all the one sided limits exist, that is, all of , , , , , …, exist and are ﬁnite. Then we say is piecewise continuous.

If moreover, is diﬀerentiable at all but ﬁnitely many points, and is piecewise continuous, then is said to be piecewise smooth.

Example 4.3.2: The square wave function (4.8) is piecewise smooth on or any other interval. In such a case we simply say that the function is piecewise smooth.

Example 4.3.4: The function is not piecewise smooth on (or any other interval containing zero). In fact, it is not even piecewise continuous.

Example 4.3.5: The function is not piecewise smooth on (or any other interval containing zero). is continuous, but the derivative of is unbounded near zero and hence not piecewise continuous.

Piecewise smooth functions have an easy answer on the convergence of the Fourier series.

Theorem 4.3.1. Suppose is a -periodic piecewise smooth function. Let

be the Fourier series for . Then the series converges for all . If is continuous at , then

Otherwise

If we happen to have that at all the discontinuities, the Fourier series converges to everywhere. We can always just redeﬁne by changing the value at each discontinuity appropriately. Then we can write an equals sign between and the series without any worry. We mentioned this fact brieﬂy at the end last section.

Note that the theorem does not say how fast the series converges. Think back to the discussion of the Gibbs phenomenon in the last section. The closer you get to the discontinuity, the more terms you need to take to get an accurate approximation to the function.

Not only does Fourier series converge nicely, but it is easy to diﬀerentiate and integrate the series. We can do this just by diﬀerentiating or integrating term by term.

is a piecewise smooth continuous function and the derivative is piecewise smooth. Then the derivative can be obtained by diﬀerentiating term by term,

It is important that the function is continuous. It can have corners, but no jumps. Otherwise the diﬀerentiated series will fail to converge. For an exercise, take the series obtained for the square wave and try to diﬀerentiate the series. Similarly, we can also integrate a Fourier series.

is a piecewise smooth function. Then the antiderivative is obtained by antidiﬀerentiating term by term and so

where and is an arbitrary constant.

Note that the series for is no longer a Fourier series as it contains the term. The antiderivative of a periodic function need no longer be periodic and so we should not expect a Fourier series.

Let us do an example of a periodic function with one derivative everywhere.

Example 4.3.6: Take the function

and extend to a 2-periodic function. The plot is given in Figure 4.10.

This function has one derivative everywhere, but it does not have a second derivative whenever is an integer.

Let us compute the Fourier series coeﬃcients. The actual computation involves several integration by parts and is left to student.

That is, the series isThis series converges very fast. If you plot up to the third harmonic, that is the function

it is almost indistinguishable from the plot of in Figure 4.10. In fact, the coeﬃcient is already just 0.0096 (approximately). The reason for this behavior is the term in the denominator. The coeﬃcients in this case go to zero as fast as goes to zero.

For functions constructed piecewise from polynomials as above, it is generally true that if you have one derivative, the Fourier coeﬃcients will go to zero approximately like . If you have only a continuous function, then the Fourier coeﬃcients will go to zero as . If you have discontinuities, then the Fourier coeﬃcients will go to zero approximately as . For more general functions the story is somewhat more complicated but the same idea holds, the more derivatives you have, the faster the coeﬃcients go to zero. Similar reasoning works in reverse. If the coeﬃcients go to zero like you always obtain a continuous function. If they go to zero like you obtain an everywhere diﬀerentiable function.

To justify this behavior, take for example the function deﬁned by the Fourier series

When we diﬀerentiate term by term we notice

Therefore, the coeﬃcients now go down like , which means that we have a continuous function. The derivative of is deﬁned at most points, but there are points where is not diﬀerentiable. It has corners, but no jumps. If we diﬀerentiate again (where we can) we ﬁnd that the function , now fails to be continuous (has jumps)

This function is similar to the sawtooth. If we tried to diﬀerentiate the series again we would obtain

which does not converge!

Exercise 4.3.2: Use a computer to plot the series we obtained for , and . That is, plot say the ﬁrst 5 harmonics of the functions. At what points does have the discontinuities?

extended periodically. a) Compute the Fourier series for . b) Write out the series explicitly up to the harmonic.

extended periodically (period is 20). a) Compute the Fourier series for . b) Write out the series explicitly up to the harmonic.

Exercise 4.3.6: Let . Is continuous and diﬀerentiable everywhere? Find the derivative (if it exists everywhere) or justify why is not diﬀerentiable everywhere.

Exercise 4.3.7: Let . Is diﬀerentiable everywhere? Find the derivative (if it exists everywhere) or justify why is not diﬀerentiable everywhere.

extended periodically. a) Compute the Fourier series for . b) Write out the series explicitly up to the harmonic. c) What does the series converge to at .

extended periodically. a) Compute the Fourier series for . b) By plugging in , evaluate . c) Now evaluate .

extended periodically. a) Compute the Fourier series for . b) Write out the series explicitly up to the 3rd harmonic.