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Home , Fourier series

Math 241: : details and convergence

D. DeTurck

University of Pennsylvania

October 15, 2012

D. DeTurck Math 241 002 2012C: Fourier series 1 / 22 Fourier series

We’ve been using Fourier series to solve the heat and Laplace equations, but we haven’t paid much attention to the following questions: • Do Fourier series converge? • What do they converge to? Do they converge to the functions we expect them to? • Can we differentiate Fourier series and obtain the expected answers (so that they really do represent solutions to PDEs)?

D. DeTurck Math 241 002 2012C: Fourier series 2 / 22 Fourier series ”flavors”

We’ve seen several types of Fourier series: • ”Full” Fourier series (in solutions of Laplace equation on the disk): ∞ X 2nπx  2nπx  f (x) = a + a cos + b sin 0 n L n L n=1 • Fourier series (zero boundary conditions on both ends) ∞ X nπx  f (x) = b sin n L n=1 • Fourier cosine series (zero on both ends, i.e., insulated ends) ∞ X nπx  f (x) = a cos n L n=1

D. DeTurck Math 241 002 2012C: Fourier series 3 / 22 More flavors

We’ve also seen ”mixed” flavors, with f = 0 at one end and f 0 = 0 at the other: ∞ X (2n + 1)πx  • f (x) = a cos has f 0(0) = 0, f (L) = 0 n 2L n=0 ∞ X (2n + 1)πx  • f (x) = b sin has f (0) = 0, f 0(L) = 0 n 2L n=0 One thing we know is that whatever these Fourier series converge to will be a periodic , with period L in the full Fourier case, period 2L in the sine series and cosine series cases, and period 4L in the mixed case.

D. DeTurck Math 241 002 2012C: Fourier series 4 / 22 Even and odd

We know a little more than that: since sin x is an odd function and cos x is an even function, the Fourier sine series will produce an odd function and the Fourier cosine series an even one. So, if you start out with the function f (x) = x on the interval 0 ≤ x ≤ 2:

D. DeTurck Math 241 002 2012C: Fourier series 5 / 22 Full Fourier series

The ”full” Fourier series for f (x) = x on 0 ≤ x ≤ 2 is

∞ X −2 nπx  1 + ∗ sin nπ 2 n=1 which gives the periodic extension of f :

D. DeTurck Math 241 002 2012C: Fourier series 6 / 22 Sine series

The Fourier sine series for f (x) = x on 0 ≤ x ≤ 2 is

∞ X (−1)n+14 nπx  sin nπ 2 n=1 which gives the odd periodic extension of f :

D. DeTurck Math 241 002 2012C: Fourier series 7 / 22 Cosine series

Likewise, the Fourier cosine series for f (x) = x on 0 ≤ x ≤ 2 is

∞ X −8 nπx  1 + cos (2n + 1)2π2 2 n=0 which gives the even periodic extension of f :

D. DeTurck Math 241 002 2012C: Fourier series 8 / 22 Observations

• The sum of the cosine series is continuous (but its derivative is not), and its coefficients go to zero like 1/n2, whereas the full Fourier series and the sine series are not continuous and their coefficients go to zero like 1/n (slower) — this is not a coincidence. • At x = ±2, ±4,... the sine series converges to 0. This is not a coincidence either. • To what do the mixed Fourier series converge?

D. DeTurck Math 241 002 2012C: Fourier series 9 / 22 Inner product spaces

To understand more about the convergence of Fourier series, we will discuss inner-product spaces in general. These are vector spaces (we know what they are) that are equipped with an inner product — that is, a ”positive-definite symmetric bilinear function”. • Biliniear means: hαv + βw , xi = αhv , xi + βhw , xi and hv , αw + βxi = αhv , wi + βhv , xi for all vectors v, w, x and scalars α, β. • Symmetric means: hv , wi = hw , vi • Positive definite means: hv , vi ≥ 0 for all v, and hv , vi = 0 if and only if v is the zero vector.

D. DeTurck Math 241 002 2012C: Fourier series 10 / 22 Geometry

3 The usual dot product of vectors in R isthe basic example of an inner product. And just as in that case, we use the inner product to do geometry: • We define the length of a vector to be |v| = phv , vi (and the distance between two vectors, if we want to consider them as points, is |v − w|. • Then we have the Schwarz inequality |hv , wi| ≤ |v||w| and the triangle inequality |v ± w| ≤ |v| + |w|. • We can define the angle between two vectors via hv , wi cos θ = . This makes sense because of the Schwarz |v||w| inequality.

D. DeTurck Math 241 002 2012C: Fourier series 11 / 22 Functions as vectors

We will think of functions as vectors. Typically, we’ll take the set of functions defined on a specific interval (like 0 ≤ x ≤ L) and satisfying some analytic condition (i.e, continuous, differentiable, integrable, piecewise smooth). We can add such functions together and multiply them by scalars, so we have a . For our inner product, we’ll define:

L hf , gi = f (x)g(x) dx. ˆ0 It’s easy to see that this is bilinear and symmetric. It’s also positive definite if we don’t allow the functions to be too weird (or else we have to get into measure theory).

D. DeTurck Math 241 002 2012C: Fourier series 12 / 22 A specific example — periodic functions with period 2π

To keep all those pesky L’s out of the way, let’s consider the space of (continuous, differentiable) functions that are periodic with period 2π. As an exercise, show that all the functions in the set {1, cos x, sin x, cos 2x, sin 2x,...} are perpendicular to one another. √ √ √ Also show that |1| = 2π, | sin nx| = π and | cos nx| = π. (This shows that this space of functions is infinite-dimensional.) You can also show that for finite N (and even for N = ∞ if the series converge), we have

2 N N X X 2 2 a0 + (an cos nx + bn sin nx) = 2πa0 + π(an + bn). n=1 n=1

D. DeTurck Math 241 002 2012C: Fourier series 13 / 22 Projections

You probably remember the formula for the projection of the vector v onto the vector w: hv , wi proj (v) = w. w hw , wi

(You learned this in Math 114 – and it makes sense: it’s a vector of length |v| cos θ in the direction of w). You can generalize this as

follows: If w1, w2,..., wN are orthogonal vectors (i.e., each is orthogonal to all of the others), then the projection of v onto the subspace W spanned by the wn’s is

N X hv , wni proj (v) = w . W hw , w i n n=1 n n

D. DeTurck Math 241 002 2012C: Fourier series 14 / 22 Fourier coefficients

You should observe the correspondence between the projection formula N X hv , wni projW (v) = wn. hwn , wni i=n and the Fourier series formula

π N π f (x) dx X f (x) cos nx dx f (x) = −π + −π cos nx ´ 2π ´ π n=1 π f (x) sin nx dx + −π sin nx ´ π N hf (x) , 1i X hf (x) , cos nxi hf (x) , sin nxi = 1 + cos nx + sin nx h1 , 1i hcos nx , cos nxi hsin nx , sin nxi n=1

D. DeTurck Math 241 002 2012C: Fourier series 15 / 22 Pythagoras

The projection of v onto the subspace W is the closest vector in W to v. You should check that the difference v − projW (v) is perpendicular to W , and in particular it is perpendicular to projW (v). Therefore the three vectors projW (v), v − projW (v) and v are the sides of a right triangle, having v as the hypotenuse. Thus, 2 2 2 | projW (v)| + |v − projW (v)| = |v| by the Pythagorean theorem. Because squares are not negative, this results in Bessel’s inequality:

2 2 | projW (v)| ≤ |v| .

D. DeTurck Math 241 002 2012C: Fourier series 16 / 22 Bessel’s inequality, Riemann-Lebesgue lemma

For Fourier series, we can rewrite Bessel’s inequality as

N X π 2πa2 + π(a2 + b2) ≤ |f |2 = (f (x))2 dx 0 n n ˆ n=1 −π Since this is true for all N, we can let N → ∞ and obtain:

∞ X π 2πa2 + π(a2 + b2) ≤ |f |2 = (f (x))2 dx 0 n n ˆ n=1 −π Therefore, the series (of positive terms) on the left converges, and so an → 0 and bn → 0 as n → ∞. This is the content of the Riemann-Lebesgue lemma:

π π lim f (x) sin nx dx = lim f (x) cos nx dx = 0. n→∞ ˆ−π n→∞ ˆ−π

D. DeTurck Math 241 002 2012C: Fourier series 17 / 22 A change in perspective

Now, let’s substitute the formula for the Fourier coefficients back into the Nth partial sum Sn(x) of the Fourier series:

N 1 π 1 X  π  S (x) = f (y) dy + f (y) cos(ny) dy cos(nx) n 2π ˆ π ˆ −π n=1 −π  π  + f (y) sin(ny) dy sin(nx) ˆ−π N ! 1 π 1 X = f (y) + cos(ny) cos(nx) + sin(ny) sin(nx) dy π ˆ 2 −π n=1 N ! 1 π 1 X = f (y) + cos(nx − ny) dy π ˆ 2 −π n=1 1 π = f (y)DN (x − y) dy π ˆ−π

D. DeTurck Math 241 002 2012C: Fourier series 18 / 22 The

1 PN The function DN (y) = 2 + n=1 cos(ny) from the last slide is called the Dirichlet kernel of order N.

Some properties of DN (y) that are easy to prove are

• DN is an even with period 2π 1 π • DN (y) dy = 1 π ˆ−π 1 π Using these and the equation Sn(x) = π −π f (y)DN (x − y) dy, we get (assuming f (x) has been extended´ as a periodic function of period 2π): 1 π • SN (x) = f (x + y)DN (y) dy. π ˆ−π 1 π • SN (x) = f (x − y)DN (y) dy. π ˆ−π 1 π • f (x) = f (x)DN (y) dy. π ˆ−π

D. DeTurck Math 241 002 2012C: Fourier series 19 / 22 A surprising sum

We can put the last three equations together to get a formula for how different the Nth partial sum of the Fourier series is from f (x):

1 π SN (x) − f (x) = [f (x + y) + f (x − y) − 2f (x)]DN (y) dy. 2π ˆ−π To make use of this, we need the following surprising fact: We can evaluate the sum which defines DN (y)!

N 1 1 X sin(N + 2 )y DN (y) = + cos ny = . 2 2 sin 1 y n=1 2 (We’ll prove this in class using geometric sums and Euler’s formula — you’ll do another proof for homework that just uses trig identities)

D. DeTurck Math 241 002 2012C: Fourier series 20 / 22 End of the proof (almost)

Knowing the closed form of DN (y) allows us to write 1 π SN (x) − f (x) = [f (x + y) + f (x − y) − 2f (x)]DN (y) dy 2π ˆ−π 1 π sin(N + 1 )y = [f (x + y) + f (x − y) − 2f (x)] 2 dy 2π ˆ 1 −π 2 sin 2 y 1 π f (x + y) + f (x − y) − 2f (x) = [sin(N + 1 )y] dy 2π ˆ 1 2 −π 2 sin 2 y So we’ll let f (x + y) + f (x − y) − 2f (x) Q(y) = 1 2 sin 2 y and our job is to show that π 1 lim Q(y) sin(N + 2 )y dy = 0. N→∞ ˆ−π

D. DeTurck Math 241 002 2012C: Fourier series 21 / 22 The necessary lemma

Lemma π 2 If Q(y) is a function such that −π(Q(y)) dy is finite, then ´ π 1 lim Q(y) sin(N + 2 )y dy = 0. N→∞ ˆ−π To prove this, note that 1 1 1 sin((N + 2 )y) = sin Ny cos 2 y + cos Ny sin 2 y. So we can break the in the lemma into two:

π π 1 1 Q(y) sin(N + 2 )y dy = (Q(y) sin 2 y) cos Ny dy ˆ−π ˆ−π π 1 + (Q(y) cos 2 y) sin Ny dy ˆ−π and use the Riemann-Lebesgue lemma on each.

D. DeTurck Math 241 002 2012C: Fourier series 22 / 22 The last piece

To finish up, we need that

f (x + y) + f (x − y) − 2f (x) Q(y) = 1 2 sin 2 y is square-integrable. This is no problem for reasonable f except where y = 0. But this is no problem where f is continuous and differentiable, nor is it a problem even at a point where f has a jump discontinuity, provided the from the left and right exist (such an f is called piecewise continuous). At such points, the Fourier series will converge to the average of the left and right limits of f .

D. DeTurck Math 241 002 2012C: Fourier series 23 / 22