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Summaries of all lecture notes in APM346, you will be good to go if you are able to understand everything shown in the notes
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Summaries of all lecture notes in APM346, you will be good to go if you are able to understand everything shown in the notes

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  • December 12, 2022
  • 14
  • 2019/2020
  • Class notes
  • Justin ko
  • All classes

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  • partial differential equation

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July 5, 2020 APM346 – Week 8 Justin Ko


1 Fourier Series
The Fourier series of f is of the form
∞
a0 X   nπx   nπx 
f (x) = + an cos + bn sin . (1)
2 n=1
L L

There are 3 main types of coefficients:
1. Full Fourier Series: The coefficients of the (full) Fourier series of f : [−L, L] → R is given by
Z L Z L
1  nπx  1  nπx 
an = f (x) cos dx and bn = f (x) sin dx. (2)
L −L L L −L L

This Fourier series satisfies periodic boundary conditions, f (−L) = f (L) and f 0 (−L) = f 0 (L).
2. Fourier Cosine Series: The coefficients of the Fourier cosine series of f : [0, L] → R is given by
the coefficients of the full Fourier series of the even extension of f :
L
2 L
Z Z
1  nπx   nπx 
an = feven (x) cos dx = f (x) cos dx (3)
L −L L L 0 L
1 L
Z  nπx 
bn = feven (x) sin dx = 0. (4)
L −L L

To compute bn , we used the fact that the product of an even and odd function is odd. This
Fourier series satisfies Neumann boundary conditions f 0 (0) = f 0 (L) = 0.
3. Fourier Sine Series: The coefficients of the Fourier sine series of f : [0, L] → R is given by the
coefficients of the full Fourier series of the odd extension of f :
Z L
1  nπx 
an =
fodd (x) cos dx = 0 (5)
−L L L
1 L 2 L
Z  nπx  Z  nπx 
bn = fodd (x) sin dx = f (x) sin dx. (6)
L −L L L 0 L

To compute an , we used the fact that the product of an even and odd function is odd. This
Fourier series satisfies Dirichlet boundary conditions f (0) = f (L) = 0..

1.1 Derivation of the Fourier Series Using Orthogonality
Suppose (Xn )n≥1 is an orthogonal sequence of functions on [a, b]. The general Fourier series of f is a
series of the form
X∞
f (x) = cn Xn (x).
n=1

To solve for the Fourier coefficient ck , we can take the inner product of both sides by Xk
∞

X
∞
X hf, Xk i
hf, Xk i = cn Xn , Xk = cn hXn , Xk i = ck hXk , Xk i =⇒ ck = . (7)
n=1 n=1
hXk , Xk i

since orthogonality implies that hXn , Xk i = 0 unless n = k.




Page 1 of 14

, July 5, 2020 APM346 – Week 8 Justin Ko


We can use this formula to derive the Fourier coefficients of the full Fourier series. By the product
sum identities, it is easy to check that Xn = cos( nπx nπx
L ) and Yn = sin( L ) for n ≥ 1 and X0 = 2 for
1

n = 0 are orthogonal. The full Fourier series is of f : [−L, L] → R is of the form
∞ ∞ 
X  a0 X  nπx   nπx 
f (x) = a0 X0 (x) + an Xn (x) + bn Yn (x) = + an cos + bn sin .
n=1
2 n=1
L L

Its coefficients for n ≥ 1 are
RL RL
hf, Xn i f (x)Xn (x) dx f (x) cos( nπx
L ) dx 1 L
Z  nπx 
−L
an = = RL = −L
RL = f (x) cos dx
hXn , Xn i X 2 (x) dx cos2 ( nπx L −L L
−L n −L L ) dx
RL RL
hf, Yn i f (x)Yn (x) dx f (x) sin( nπx
L ) dx 1 L
Z  nπx 
−L
bn = = RL = −L
RL = f (x) sin dx
hYn , Yn i Y 2 (x) dx sin2 ( nπx L −L L
−L n −L L ) dx

and the coefficient for a0 is given by
RL RL
hf, X0 i f (x)X0 (x) dx f (x) 21 dx L
Z
−L −L 1
an = = RL = RL 1 = f (x) dx
hX0 , X0 i X02 (x) dx dx L −L
−L −L 22

which is what we get if we naively extend the formula for an to a0 .
Remark 1. The coefficients of the Fourier cosine and sine series can be derived in a similar way. We
can also use symmetry and the uniqueness of the Fourier coefficients to recover these coefficients from
the full Fourier series.
Remark 2. It turns out that the Fourier series form a orthogonal basis on the space of square
integrable functions L2 ([−L, L]). This means that every square integrable function can be written as
a linear combination of its orthogonal basis elements. The Fourier coefficients are an explicit formula
for the scalars in this linear combination.

1.2 Derivation of the Fourier Series Using Least-Square Approximation
Suppose (Xn )n≥1 is an orthogonal sequence of functions on [a, b]. The general Fourier series of f is a
series of the form
X∞
f (x) = cn Xn (x).
n=1

For fixed N ≥ 1, we want to find coefficients (cn ) that minimizes the mean-squared error
N
X 2 Z L N
X 2
EN = EN (c1 , . . . , cN ) = f − cn Xn = f (x) − cn Xn (x) dx.
n=1 L2 −L n=1

We used the notation k · k2L2 = h·, ·i to denote the inner product norm. If we expand the square terms,
then
Z L N
X Z L N
X Z L
EN = |f (x)|2 dx − 2 cn f (x)Xn (x) dx + cn cm Xn (x)Xm (x) dx.
−L n=1 −L n,m=1 −L

N
X N
X
= hf, f i − 2 cn hf, Xn i + c2n hXn , Xn i
n=1 n=1




Page 2 of 14

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