Section 11.8 Fourier series
Note: 3–4 lectures
Fourier series is perhaps the most important (and the most difficult) of the series that we cover in this book. We saw a few examples already, but let us start at the beginning.
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Named after the French mathematician Jean-Baptiste Joseph Fourier (1768–1830).
Subsection 11.8.1 Trigonometric polynomials
A trigonometric polynomial is an expression of the form
or equivalently, thanks to Euler’s formula ( ):
So a trigonometric polynomial is really a rational function of the complex variable (we are allowing negative powers) evaluated on the unit circle. There is a wonderful connection between power series (actually Laurent series because of the negative powers) and Fourier series because of this observation, but we will not investigate this further.
Another reason why Fourier series is important and comes up in so many applications is that the functions are eigenfunctions of various differential operators. For example,
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Eigenfunction is like an eigenvector for a matrix, but for a linear operator on a vector space of functions.
That is, they are the functions whose derivative is a scalar (the eigenvalue) times itself. Just as eigenvalues and eigenvectors are important in studying matrices, eigenvalues and eigenfunctions are important when studying linear differential equations.
The functions and are -periodic and hence trigonometric polynomials are also -periodic. We could rescale to make the period different, but the theory is the same, so we stick with the period The antiderivative of is and so
Consider
and for compute
We just found a way of computing the coefficients using an integral of If the integral is 0, so we might as well have included enough zero coefficients to make
Proof.
The complex conjugate goes inside the integral because the integral is done on real and imaginary parts separately.
The functions are also linearly independent.
Proposition 11.8.2.
Proof.
The result follows immediately from the integral formula for
Subsection 11.8.2 Fourier series
We now take limits. The series
is called the Fourier series and the numbers the Fourier coefficients. Using Euler’s formula we could also develop everything with sines and cosines, that is, as the series It is equivalent, but slightly more messy.
Several questions arise. What functions are expressible as Fourier series? Obviously, they have to be -periodic, but not every periodic function is expressible with the series. Furthermore, if we do have a Fourier series, where does it converge (where and if at all)? Does it converge absolutely? Uniformly? Also note that the series has two limits. When talking about Fourier series convergence, we often talk about the following limit:
There are other ways we can sum the series to get convergence in more situations, but we refrain from discussing those. In light of this, define the symmetric partial sums
Conversely, for an integrable function call the numbers
its Fourier coefficients. To emphasize the function the coefficients belong to, we write We then formally write down a Fourier series:
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The notation should seem similar to Fourier transform to those readers that have seen it. The similarity is not just coincidental, we are taking a type of Fourier transform here.
As you might imagine such a series might not even converge. The doesn’t imply anything about the two sides being equal in any way. It is simply that we created a formal series using the formula for the coefficients. We will see that when the functions are “nice enough,” we do get convergence.
Example 11.8.3.
Consider the step function so that on and on extended periodically to a -periodic function. With a little bit of calculus, we compute the coefficients:
A little bit of simplification leads to
For a second example, consider the function on and then extended to a -periodic function. Computing the coefficients, we find
A little simplification yields
See the right hand graph in Figure 11.11.
Note that for both and the even coefficients (except ) happen to vanish, but that is not really important. What is important is convergence. First, at the discontinuity at we find for all so converges to a different number from (at a nice enough jump discontinuity, the limit is the average of the two-sided limits, see the exercises). That should not be surprising; the coefficients are computed by an integral, and integration does not notice if the value of a function changes at a single point. We should remark, however, that we are not guaranteed that in general the Fourier series converges to the function even at a point where the function is continuous. We will prove convergence if the function is at least Lipschitz.
What is really important is how fast the coefficients go to zero. For the discontinuous the coefficients go to zero approximately like On the other hand, for the continuous the coefficients go to zero approximately like The Fourier coefficients “see” the discontinuity in some sense.
Do note that continuity in this setting is the continuity of the periodic extension, that is, we include the endpoints So the function defined on and extended periodically would be discontinuous at the endpoints
In general, the relationship between regularity of the function and the rate of decay of the coefficients is somewhat more complicated than the example above might make it seem, but there are some quick conclusions we can make. We forget about finding a series for a function for a moment, and we consider simply the limit of some given series. A few sections ago, we proved that the Fourier series
converges uniformly and hence converges to a continuous function. This example and its proof can be extended to a more general criterion.
Proposition 11.8.4.
The proof is to apply the Weierstrass -test (Theorem 11.2.4) and the -series test to find that the series converges uniformly and hence to a continuous function (Corollary 11.2.8). We can also take derivatives.
Proposition 11.8.5.
The proof is to note that the series converges to a continuous function by the previous proposition. In particular, it converges at some point. Then differentiate the partial sums
and notice that for all nonzero
The differentiated series converges uniformly by the -test again. Since the differentiated series converges uniformly, we find that the original series converges to a continuously differentiable function, whose derivative is the differentiated series (see Theorem 11.2.14).
Subsection 11.8.3 Orthonormal systems
Let us abstract away the exponentials, and study a more general series for a function. One fundamental property of the exponentials that makes Fourier series work is that the exponentials are a so-called orthonormal system. Fix an interval We define an inner product for the space of functions. We restrict our attention to Riemann integrable functions as we do not have the Lebesgue integral, which would be the natural choice. Let and be complex-valued Riemann integrable functions on and define the inner product
If you have seen Hermitian inner products in linear algebra, this is precisely such a product. We must include the conjugate as we are working with complex numbers. We then have the “size” of that is, the norm by (defining the square)
Remark 11.8.6.
In what follows, we will assume all functions are Riemann integrable.
Definition 11.8.7.
We noticed above that
is an orthonormal system on The factor out in front is to make the norm be 1.
Having an orthonormal system on and an integrable function on we can write a Fourier series relative to Let
and write
In other words, the series is
Notice the similarity to the expression for the orthogonal projection of a vector onto a subspace from linear algebra. We are in fact doing just that, but in a space of functions.
Theorem 11.8.8.
In other words, the partial sums of the Fourier series are the best approximation with respect to the norm.
Proof.
When we do plug in then
and so for all
Note that
by the calculation above. We take a limit to obtain the so-called Bessel’s inequality.
Theorem 11.8.9. Bessel’s inequality.
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Named after the German astronomer, mathematician, physicist, and geodesist Friedrich Wilhelm Bessel (1784–1846).
Then
Subsection 11.8.4 The Dirichlet kernel and approximate delta functions
We return to the trigonometric Fourier series. The system is orthogonal, but not orthonormal if we simply integrate over We can rescale the integral and hence the inner product to make orthonormal. That is, if we replace
(we are just rescaling the really), then everything works and we obtain that the system is orthonormal with respect to the inner product
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Mathematicians in this field sometimes simplify matters with a tongue-in-cheek definition that
Recall the notation for the symmetric partial sums, The inequality leading up to Bessel now reads:
Let the Dirichlet kernel be
We claim that
for such that The left-hand side is continuous on and hence the right-hand side extends continuously to all of To show the claim, we use a familiar trick:
Multiply by
The claim follows.
Expand the definition of
The central peak gets taller and taller as gets larger, and the side peaks stay small. We are convolving (again) with approximate delta functions, although these functions have all these oscillations away from zero. The oscillations on the side do not go away but they are eventually so fast that we expect the integral to just sort of cancel itself out there. Overall, we expect that goes to Things are not always simple, but under some conditions on such a conclusion holds. For this reason people write
where is the “delta function” (not really a function), which is an object that will give something like “ ” We can think of converging in some sense to However, we have not defined (and will not define) what the delta function is, nor what does it mean for it to be a limit of or have a Fourier series.
Subsection 11.8.5 Localization
If satisfies a Lipschitz condition at a point, then the Fourier series converges at that point.
Theorem 11.8.10.
In particular, if is continuously differentiable at then we obtain convergence at (exercise). A function is continuous piecewise smooth if it is continuous and there exist points such that for every restricted to is continuously differentiable (up to the endpoints).
Corollary 11.8.11.
The proof of the corollary is left as an exercise. Let us prove the theorem.
Proof.
Write
By the hypotheses, for small nonzero
As where as we notice that is continuous at the origin. Hence, as a function of is bounded near the origin. As is the only place on where the denominator vanishes, it is the only place where there could be a problem. So, the function is bounded near and clearly Riemann integrable on any interval not including and thus it is Riemann integrable on We use the trigonometric identity
to compute
As functions of and are bounded Riemann integrable functions and so their Fourier coefficients go to zero by Theorem 11.8.9. So the two integrals on the right-hand side, which compute the Fourier coefficients for the real version of the Fourier series go to 0 as goes to infinity. This is because and are also orthonormal systems with respect to the same inner product. Hence goes to 0, that is, goes to
The theorem also says that convergence depends only on local behavior. That is, to understand convergence of we only need to know in some neighborhood of
Corollary 11.8.12.
Suppose is a -periodic function, Riemann integrable on If is an open interval and for all then for all
In particular, if and are -periodic functions, Riemann integrable on an open interval, and for all then for all the sequence converges if and only if converges.
The first claim follows by taking in the theorem. The “In particular” follows by considering which is zero on and So convergence at depends only on the values of the function near However, we saw that the rate of convergence, that is, how fast does converge to depends on global behavior of
Note a subtle difference between the results above and what Stone–Weierstrass theorem gives. Any continuous function on can be uniformly approximated by trigonometric polynomials, but these trigonometric polynomials may not be the partial sums
Subsection 11.8.6 Parseval’s theorem
Finally, convergence always happens in the sense and operations on the (infinite) vectors of Fourier coefficients are the same as the operations using the integral inner product.
Theorem 11.8.13. Parseval.
Proof.
Via Stone–Weierstrass, approximate with a trigonometric polynomial uniformly. That is, there is a trigonometric polynomial such that for all Hence
Hence, the first claim follows.
Next,
We need the Schwarz (or Cauchy–Schwarz or Cauchy–Bunyakovsky–Schwarz) inequality for that is,
Its proof is left as an exercise; it is not much different from the finite-dimensional version. So
The right-hand side goes to 0 as goes to infinity by the first claim of the theorem. That is, as goes to infinity, goes to and the second claim is proved. The last claim in the theorem follows by using
Exercises 11.8.7 Exercises
11.8.1.
Consider the Fourier series
Show that the series converges uniformly and absolutely to a continuous function. Remark: This is another example of a nowhere differentiable function (you do not have to prove that). See Figure 11.13.
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See G. H. Hardy, Weierstrass’s Non-Differentiable Function, Transactions of the American Mathematical Society, 17, No. 3 (Jul., 1916), pp. 301–325. A thing to notice here is the th Fourier coefficient is if and zero otherwise, so the coefficients go to zero like
11.8.2.
Suppose that a -periodic function that is Riemann integrable on and such that is continuously differentiable on some open interval Prove that for every we have
11.8.3.
Prove Corollary 11.8.11, that is, suppose a -periodic function is continuous piecewise smooth near a point then Hint: See the previous exercise.
11.8.4.
Given a -periodic function Riemann integrable on and show that there exists a continuous -periodic function such that
11.8.5.
11.8.6.
11.8.7.
- Show that
is continuous. - Formally (that is, suppose you can differentiate under the sum) find a solution (formal solution, that is, do not yet worry about convergence) to the differential equationof the form
- Then show that this solution
is twice continuously differentiable, and in fact solves the equation.
11.8.8.
11.8.9.
Suppose that for all and converges. Let be the unit disc, and be the closed unit disc. Show that there exists a continuous function that is analytic on and such that on the boundary of we have
Hint: If then
Hint: If
11.8.10.
11.8.11.
11.8.12.
- Let
be the -periodic function defined by if and if letting and be arbitrary. Show that - Let
be a -periodic function Riemann integrable on and there are continuously differentiable and where for all and where for all Then or in other words,
11.8.13.
Let be such that Show that there is a continuous -periodic function whose Fourier coefficients satisfy that for each there is a where
Remark: The exercise says that if is only continuous, there is no “minimum rate of decay” of the coefficients. Compare with Exercise 11.8.11.
Hint: Look at Exercise 11.8.1 for inspiration.
Remark: The exercise says that if
Hint: Look at Exercise 11.8.1 for inspiration.