math 241: fourier series: details and convergencedeturck/m241/fourier_series.pdffourier series...

Math 241: Fourier series: 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 Laplaceequations, but we haven’t paid much attention to the followingquestions:

• Do Fourier series converge?

• What do they converge to? Do they converge to the functionswe expect them to?

• Can we differentiate Fourier series and obtain the expectedanswers (so that they really do represent solutions to PDEs)?


Fourier series ”flavors”

We’ve seen several types of Fourier series:

• ”Full” Fourier series (in solutions of Laplace equation on thedisk):

f (x) = a0 +∞∑n=1

an cos

(2nπx

L

)+ bn sin

(2nπx

L

)• Fourier sine series (zero boundary conditions on both ends)

f (x) =∞∑n=1

bn sin(nπx

L

)• Fourier cosine series (zero derivative on both ends, i.e.,

insulated ends)

f (x) =∞∑n=1

an cos(nπx

L

)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 ′ = 0at the other:

• f (x) =∞∑n=0

an cos

((2n + 1)πx

2L

)has f ′(0) = 0, f (L) = 0

• f (x) =∞∑n=0

bn sin

((2n + 1)πx

2L

)has f (0) = 0, f ′(L) = 0

One thing we know is that whatever these Fourier series convergeto will be a periodic function, with period L in the full Fouriercase, period 2L in the sine series and cosine series cases, andperiod 4L in the mixed case.


Even and odd

We know a little more than that: since sin x is an odd function andcos x is an even function, the Fourier sine series will produce anodd function and the Fourier cosine series an even one.

So, if you start out with the function f (x) = x on the interval0 ≤ x ≤ 2:


Full Fourier series

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

1 +∞∑n=1

−2

nπ∗ sin

(nπx

2

)which gives the periodic extension of f :


Sine series

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

∞∑n=1

(−1)n+14

nπsin(nπx

2

)which gives the odd periodic extension of f :


Cosine series

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

1 +∞∑n=0

−8

(2n + 1)2π2cos(nπx

2

)which gives the even periodic extension of f :


Observations

• The sum of the cosine series is continuous (but its derivativeis not), and its coefficients go to zero like 1/n2, whereas thefull Fourier series and the sine series are not continuous andtheir coefficients go to zero like 1/n (slower) — this is not acoincidence.

• At x = ±2,±4, . . . the sine series converges to 0. This is nota coincidence either.

• To what do the mixed Fourier series converge?


Inner product spaces

To understand more about the convergence of Fourier series, wewill discuss inner-product spaces in general. These are vectorspaces (we know what they are) that are equipped with an innerproduct — that is, a ”positive-definite symmetric bilinearfunction”.

• Biliniear means: 〈αv + βw , x〉 = α〈v , x〉+ β〈w , x〉 and〈v , αw + βx〉 = α〈v , w〉+ β〈v , x〉 for all vectors v , w , xand scalars α, β.

• Symmetric means: 〈v , w〉 = 〈w , v〉• Positive definite means: 〈v , v〉 ≥ 0 for all v , and 〈v , v〉 = 0

if and only if v is the zero vector.


Geometry

The usual dot product of vectors in R3 isthe basic example of aninner product. And just as in that case, we use the inner productto do geometry:

• We define the length of a vector to be |v | =√〈v , v〉 (and

the distance between two vectors, if we want to consider themas points, is |v − w |.

• Then we have the Schwarz inequality |〈v , w〉| ≤ |v ||w | andthe triangle inequality |v ± w | ≤ |v |+ |w |.

• We can define the angle between two vectors via

cos θ =〈v , w〉|v ||w |

. This makes sense because of the Schwarz

inequality.


Functions as vectors

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

〈f , g〉 =

ˆ L

0f (x)g(x) dx .

It’s easy to see that this is bilinear and symmetric. It’s also positivedefinite if we don’t allow the functions to be too weird (or else wehave to get into measure theory).


A specific example — periodic functions with period 2π

To keep all those pesky L’s out of the way, let’s consider the spaceof (continuous, differentiable) functions that are periodic withperiod 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 theseries converge), we have∣∣∣∣∣a0 +

N∑n=1

(an cos nx + bn sin nx)

∣∣∣∣∣2

= 2πa0 +N∑

n=1

π(a2n + b2n).


Projections

You probably remember the formula for the projection of thevector v onto the vector w :

projw (v) =〈v , w〉〈w , w〉

w .

(You learned this in Math 114 – and it makes sense: it’s a vectorof 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 isorthogonal to all of the others), then the projection of v onto thesubspace W spanned by the wn’s is

projW (v) =N∑

n=1

〈v , wn〉〈wn , wn〉

wn.


Fourier coefficients

You should observe the correspondence between the projectionformula

projW (v) =N∑i=n

〈v , wn〉〈wn , wn〉

wn.

and the Fourier series formula

f (x) =

´ π−π f (x) dx

2π+

N∑n=1

´ π−π f (x) cos nx dx

πcos nx

+

´ π−π f (x) sin nx dx

πsin nx

=〈f (x) , 1〉〈1 , 1〉

1 +N∑

n=1

〈f (x) , cos nx〉〈cos nx , cos nx〉

cos nx +〈f (x) , sin nx〉〈sin nx , sin nx〉

sin nx


Pythagoras

The projection of v onto the subspace W is the closest vector inW to v . You should check that the difference v − projW (v) isperpendicular to W , and in particular it is perpendicular toprojW (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,

|projW (v)|2 + |v − projW (v)|2 = |v |2

by the Pythagorean theorem.Because squares are not negative, this results in Bessel’s inequality:

|projW (v)|2 ≤ |v |2.


Bessel’s inequality, Riemann-Lebesgue lemma

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

2πa20 +N∑

n=1

π(a2n + b2n) ≤ |f |2 =

ˆ π

−π(f (x))2 dx

Since this is true for all N, we can let N →∞ and obtain:

2πa20 +∞∑n=1

π(a2n + b2n) ≤ |f |2 =

ˆ π

−π(f (x))2 dx

Therefore, the series (of positive terms) on the left converges, andso an → 0 and bn → 0 as n→∞. This is the content of theRiemann-Lebesgue lemma:

limn→∞

ˆ π

−πf (x) sin nx dx = lim

n→∞

ˆ π

−πf (x) cos nx dx = 0.


A change in perspective

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

Sn(x) =1

2π

ˆ π

−πf (y) dy +

1

π

N∑n=1

(ˆ π

−πf (y) cos(ny) dy

)cos(nx)

+

(ˆ π

−πf (y) sin(ny) dy

)sin(nx)

=1

π

ˆ π

−πf (y)

(1

2+

N∑n=1

cos(ny) cos(nx) + sin(ny) sin(nx) dy

)

=1

π

ˆ π

−πf (y)

(1

2+

N∑n=1

cos(nx − ny)

)dy

=1

π

ˆ π

−πf (y)DN(x − y) dy


The Dirichlet kernel

The function DN(y) = 12 +

∑Nn=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 periodic function with period 2π

•1

π

ˆ π

−πDN(y) dy = 1

Using these and the equation Sn(x) = 1π

´ π−π f (y)DN(x − y) dy ,

we get (assuming f (x) has been extended as a periodic function ofperiod 2π):

• SN(x) =1

π

ˆ π

−πf (x + y)DN(y) dy .

• SN(x) =1

π

ˆ π

−πf (x − y)DN(y) dy .

• f (x) =1

π

ˆ π

−πf (x)DN(y) dy .


A surprising sum

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

SN(x)− f (x) =1

2π

ˆ π

−π[f (x + y) + f (x − y)− 2f (x)]DN(y) dy .

To make use of this, we need the following surprising fact: We canevaluate the sum which defines DN(y)!

DN(y) =1

2+

N∑n=1

cos ny =sin(N + 1

2)y

2 sin 12y

.

(We’ll prove this in class using geometric sums and Euler’s formula— you’ll do another proof for homework that just uses trigidentities)


End of the proof (almost)

Knowing the closed form of DN(y) allows us to write

SN(x)− f (x) =1

2π

ˆ π

−π[f (x + y) + f (x − y)− 2f (x)]DN(y) dy

=1

2π

ˆ π

−π[f (x + y) + f (x − y)− 2f (x)]

sin(N + 12)y

2 sin 12y

dy

=1

2π

ˆ π

−π

f (x + y) + f (x − y)− 2f (x)

2 sin 12y

[sin(N + 12)y ] dy

So we’ll let

Q(y) =f (x + y) + f (x − y)− 2f (x)

2 sin 12y

and our job is to show that

limN→∞

ˆ π

−πQ(y) sin(N + 1

2)y dy = 0.


The necessary lemma

Lemma

If Q(y) is a function such that´ π−π(Q(y))2 dy is finite, then

limN→∞

ˆ π

−πQ(y) sin(N + 1

2)y dy = 0.

To prove this, note thatsin((N + 1

2)y) = sin Ny cos 12y + cos Ny sin 1

2y . So we can breakthe integral in the lemma into two:

ˆ π

−πQ(y) sin(N + 1

2)y dy =

ˆ π

−π(Q(y) sin 1

2y) cos Ny dy

+

ˆ π

−π(Q(y) cos 1

2y) sin Ny dy

and use the Riemann-Lebesgue lemma on each.


The last piece

To finish up, we need that

Q(y) =f (x + y) + f (x − y)− 2f (x)

2 sin 12y

is square-integrable. This is no problem for reasonable f exceptwhere y = 0. But this is no problem where f is continuous anddifferentiable, nor is it a problem even at a point where f has ajump discontinuity, provided the derivatives from the left and rightexist (such an f is called piecewise continuous).At such points, the Fourier series will converge to the average ofthe left and right limits of f .


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