LAPLACE TRANSFORM, FOURIER TRANSFORM AND DIFFERENTIAL EQUATIONS

XU WANG

These notes for TMA4135 (first seven weeks) are based on Erwin Kreyszig’s book [2], Dag Wessel-
Berg’s video: http://video.adm.ntnu.no/serier/4fe2d4d3dbe03, and references [1, 3, 4] (among them
[1] is very short and readable).

C ONTENTS
1. Laplace transform 1
1.1. Basic facts 1
1.2. Laplace transform of derivatives, ODEs 2
1.3. More Laplace transforms 3
2. Fourier analysis 9
2.1. Complex and real Fourier series 9
2.2. Fourier Sine and Cosine series 13
2.3. More Fourier series 14
2.4. Fourier transform 17
2.5. Fourier inversion formula 18
2.6. More Fourier transforms 20
3. Partial differential equations 21
3.1. Functions of several variables 21
3.2. Solve wave equation by Fourier series 23
3.3. Solve heat equation by Fourier series 25
3.4. Solve heat equation by Fourier transform 26
4. Appendix: Definition of e, π and Euler’s formula 27
4.1. Where does e come from ? 27
4.2. Definition of the exponential function 28
4.3. Definition of π and trigonometric functions 29
References 30

1. L APLACE TRANSFORM
1.1. Basic facts.
Definition 1.1. Let f (t), t ≥ 0 be a given function. We call
Z ∞
F (s) := e−st f (t) dt,
0
the Laplace transform of f (t) and write
F = L(f ), f = L−1 F.
Date: September 25, 2018.
1
2 XU WANG

Remark: One can prove that the Laplace transform L is injective (see page 9 in [1]), that is the
reason why L−1 is well defined (for a precise formula of L−1 , see page 10 in [1]). To compute Laplace
transforms, we need:
Z b
(1) d(f g) = f dg + gdf, df = f (b) − f (a),
a
where df := f 0 (t)dt.
Example 1:
1 1
(2) L(1) = , L−1 ( ) = 1, s > 0.
s s
Example 2:
1 1
(3) L(ekt ) = , L−1 ( ) = ekt , s > k.
s−k s−k
2
Example 3: L(et ) does not exist,
Z ∞
2
e−st et dt = ∞,
0
for all real number s.
Remark: Laplace transform is linear: By linearity, we mean for all real numbers a, b,
(4) L(af + bg) = aL(f ) + bL(g).

Application 1:
5(s − 3)
L(3 + 2e5t ) = 3L(1) + 2L(e5t ) = , s > 5.
s−5
Application 2: Since
1 1 1 1
= = − .
s2 − 3s + 2 (s − 1)(s − 2) s−2 s−1
linearity gives
1 1 1
L−1 ( ) = L−1 ( ) − L−1 ( ) = e2t − et .
s2 − 3s + 2 s−2 s−1
n!
Proposition 1.2. L(tn ) = sn+1
, for n = 1, 2, · · · , and s > 0.
1.2. Laplace transform of derivatives, ODEs.
Theorem 1.3 (Laplace transform of the derivative).
L(f 0 ) = sL(f ) − f (0).
Example: Solve:
y 0 = y, y(0) = 1.
The answer is
y(t) = et .
Remark: Apply the theorem to f 0 , we get
L(f 00 ) = sL(f 0 ) − f 0 (0) = s(sL(f ) − f (0)) − f 0 (0) = s2 L(f ) − sf (0) − f 0 (0).
 LAPLACE TRANSFORM AND FOURIER TRANSFORM 3

Remark: Apply the Laplace transform to a differential equation

y 00 + ay 0 + by = c(t), a, b ∈ R,
then we get
s2 Y − sy(0) − y 0 (0) + a(sY − y(0)) + bY = C,
i.e
(s2 + as + b)Y = (s + a)y(0) + y 0 (0) + C,
thus the inverse transform gives the solution
0
 
−1 −1 (s + a)y(0) + y (0) + C
y = L (Y ) = L .
s2 + as + b
Example: Consider
y 00 + 4y 0 + 4y = 0, y(0) = 0, y 0 (0) = 1.
then the above formula gives
   
−1 1 −1 1
y=L =L .
s2 + 4s + 4 (s + 2)2
How to compute the inverse Laplace transform of 1
(s+2)2
? Is it related to L−1 ( s12 ) = t ? We will
answer them in the next section.
1.3. More Laplace transforms.
1.3.1. s-Shifting. The following formula
Z ∞
e−st eat f (t) dt = F (s − a), F (s),
0
gives
Theorem 1.4 (s-Shifting theorem).
L(eat f (t)) = F (s − a).
Example:  
−1 1
L = e−2t t.
(s + 2)2
Example:
1
L(ekt ) = .
s−k
Take k = iw,
1 s + iw
L(eiwt ) = = 2 .
s − iw s + w2
Euler’s formula (see the appendix) gives
s w
(5) L(cos wt) = , L(sin wt) = 2 .
s2 +w 2 s + w2
Example:
s+2 s 2
L−1 ( 2
) = L−1 ( 2 ) + L−1 ( 2 ) = cos 2t + sin 2t.
s +4 s +4 s +4
Exercises:
4 XU WANG

s
1. Find the inverse Laplace transform of s2 +2s+2
. The answer is
s
L−1 ( ) = e−t (cos t − sin t).
s2 + 2s + 2

2. Solve y 00 − y = t, y(0) = y 0 (0) = 1. The answer is

1
y = et + (et − e−t ) − t.
2
Definition 1.5 (sinh t and cosh t).
et − e−t et + e−t
sinh t := , cosh t := .
2 2
Exercises:
3. Compute Laplace transform of sinh t and cosh t. The answer is
1 s
L(sinh t) = , L(cosh t) = 2 .
s2 −1 s −1
4. Compute Laplace transform of
f (t) = 1 if 3 < t < 4; f (t) = 0 otherwise.
The answer is
e−3s − e−4s
L(f ) = .
s
1.3.2. Laplace transform of integrals. Put
Z t
g(t) = f (τ ) dτ,
0
then
g 0 = f, g(0) = 0.
Thus
F = L(f ) = L(g 0 ) = sG − g(0) = sG
F
gives G = s, i.e.
Z t 
L(f )
L f (τ ) dτ = .
0 s

Exercises:
1. Show that  
−1 1
L 2
= 1 − cos t.
s(s + 1)

2. Show that  
−1 1 1
L · = et − 1 − t.
s s(s − 1)
 LAPLACE TRANSFORM AND FOURIER TRANSFORM 5

1.3.3. Using step functions.

Definition 1.6 (Step function). Let a ≥ 0, the step function u(t − a) is defined as follows
u(t − a) = 0, for 0 ≤ t < a; u(t − a) = 1, for t ≥ a.
In case a = 0 we call u(t) the Heaviside function.
Exercise:
1. Draw the graphs of
f (t) = u(t − 1) − u(t − 3),

f (t) = (u(t) − u(t − π)) sin t,

f (t) = u(t) + u(t − 1) + · · · + u(t − n) + · · · .

2. Show that
e−as
L(u(t − a)) = .
s
3. Compare the graphs of u(t − a)f (t − a) with that of f (t).
Theorem 1.7 (t-Shifting theorem).
L(u(t − a)f (t − a)) = e−as L(f ).
Example: Since
1
= L(e2t ),
s−2
we have
1
L−1 (e−s ) = u(t − 1)e2(t−1) .
s−2
Exercise:
1. Compute Laplace transform of
∞
X
f (t) = u(t − n).
n=0
P∞ n 1
(Hint: n=0 r = 1−r for |r| < 1) Answer
∞
!
X 1
L u(t − n) = .
s(1 − e−s )
n=0

RC-Circuit equation (see page 29 section 1.5 and page 93 section 2.9 of Kreyszig’s book): R, C
postive constants, i(t), e(t) functions:
1 t
Z
Ri(t) + i(τ ) dτ = e(t).
C 0
Apply the Laplace transform, we get
1 I(s)
RI(s) + · = E(s),
C s
6 XU WANG

i.e.
E(s) s E(s)
I(s) = 1 = 1 .
R + Cs s + RC R

Exercise:
2. Compute i(t) when
e(t) = u(t − 1) − u(t − 2), R = C = 1.
Answer
i(s) = u(t − 1)e−(t−1) − u(t − 2)e−(t−2) .

1.3.4. Dirac delta function. The delta function δ(t − a) is defined by

Z ∞
f (t)δ(t − a) dt = f (a).
0

In case f (t) = e−st , we have

Z ∞
e−st δ(t − a) dt = e−as .
0
Thus
L(δ(t − a)) = e−as .

Exercise: solve
y 00 + y = δ(t − 1), y(0) = y 0 (0) = 0.
Answer
y = L−1 (e−s L(sin t)) = u(t − 1) sin(t − 1).

1.3.5. Convolution. Let f (t), g(t) be two functions for t ≥ 0.

Definition 1.8 (Convolution of f and g).
Z t
(f ? g)(t) := f (τ )g(t − τ ) dτ, t ≥ 0.
0

Examples:

t2
1?t= ,
2

et ? et = tet ,

Z t
f (t) ? 1 = f (τ ) dτ.
0

Theorem 1.9 (Laplace transform of convolution).

L(f ? g) = L(f ) · L(g).
 LAPLACE TRANSFORM AND FOURIER TRANSFORM 7

Exercises:
1. Compute tm ? tn . Hint: use tm ? tn = L−1 L(tm ? tn ), answer
m!n!
tm ? tn = tm+n+1 , m, n = 0, 1, · · ·
(m + n + 1)!

2. Compute L−1 ( (s2 +1)

1
2 ). Hint: use L
−1 ( 1
(s2 +1)2
) = L−1 (L(sin t) · L(sin t)). Answer

1 sin t − t cos t
L−1 ( )= .
(s2 + 1) 2 2

3. Solve y 00 + y = sin t, y(0) = 0, y 0 (0) = 1. Hint: use Exercise 2. Answer

sin t − t cos t 3 sin t − t cos t
y = sin t + = .
2 2
Rt Rt
4. Solve: y − 0 (t − τ )y(τ ) dτ = 1. Hint: use 0 (t − τ )y(τ ) dτ = y ? t. Answer

et + e−t
y= = cosh t.
2

1.3.6. Non-homogeneous linear ODEs. Consider

y 00 + by 0 + cy = r(t),
given y(0) and y 0 (0), we have

s2 Y − sy(0) − y 0 (0) + b(sY − y(0)) + cY = R(s).

Thus
1 sy(0) + y 0 (0) + by(0)
Y = · R(s) + := K(s) · R(s) + G(s),
s2 + bs + c s2 + bs + c
we get
y = k ? r + g.

Example: Consider
y 00 + y = r(t), y(0) = y 0 (0) = 0.
Apply the Laplcae transform, we have
s2 Y + Y = L(r).
Thus
1
Y = · L(r),
s2 +1
which gives
y(t) = sin t ? r.
8 XU WANG

1.3.7. Derivative of the Laplace transform. Apply differential to

Z ∞
F (s) = e−st f (t) dt,
0
we get
∞ ∞
d(e−st )
Z Z
F 0 (s) = f (t) dt = e−st · (−tf (t)) dt = L(−tf (t)).
0 ds 0
Example 1:
1 2s
L(t sin t) = −( )0 = 2 .
s2 +1 (s + 1)2

Example 2: Let F (s) = ln(1 + s−2 ). Then

2s 2
F 0 = (ln(1 + s2 ) − ln(s2 ))0 = 2
− .
1+s s
Thus
L−1 (F 0 ) = 2 cos t − 2.
By the above theorem, we have
L−1 (F 0 ) = −tf (t).
thus
2 − 2 cos t
f (t) = L−1 (ln(1 + s−2 )) = .
t
1.3.8. System of differential equations. Look at this example:
y10 = −y1 + y2 ; y20 = −y1 − y2 + f (t), y1 (0) = y2 (0) = 0.
Apply the Laplace transform, we get
sY1 = −Y1 + Y2 ; sY2 = −Y1 − Y2 + F (s).
Thus
(s + 1)Y1 − Y2 = 0;

Y1 + (s + 1)Y2 = F (s).
The first equation gives Y2 = (s + 1)Y1 , together with the second, we have
Y1 = F (s)(1 + (s + 1)2 )−1 .
Use the first equation again,
Y2 = F (s)(s + 1)(1 + (s + 1)2 )−1 .
Thus
y1 = f (t) ? (e−t sin t), y2 = f (t) ? (e−t cos t).
when f (t) = e−t , we get
Z t
y1 = e−(t−τ ) e−τ sin τ dτ = e−t (1 − cos t), y2 = e−t sin t.
0
 LAPLACE TRANSFORM AND FOURIER TRANSFORM 9

1.3.9. Homework for Laplace transform. Please compute Laplace transform of

1. f (t) = t, if 0 ≤ t ≤ a, f (t) = 0, if t > a. Answer
1 e−as e−as
F (s) = − − a .
s2 s2 s
−πs
2. f (t) = u(t − π) sin t. Answer F = − se2 +1 .
3. Solve the following equation
Z t
0
i (t) + 2 i(t) + i(τ ) dτ = δ(t − 1), i(0) = 0.
0
Answer
i(t) = u(t − 1)(e−(t−1) − e−(t−1) (t − 1)).

2. F OURIER ANALYSIS
2.1. Complex and real Fourier series.
2.1.1. Complex Fourier series. Fix p > 0, if
f (x + p) = f (x), ∀ x ∈ R,
then call f a periodic function with period p.
Example: periodic function:
1. A polynomial is periodic if and only if it is a constant;
2. eλx has period 2π if and only if λ = in, n ∈ Z.
The main theorem in Fourier analysis is the following:
Theorem 2.1 (Fourier 1807). If f has period 2π and is smooth enough then we have
X
f (x) = cn einx , ∀ x ∈ R.
n∈Z
——The proof (see Page 63 in [3]) is not assumed in this course.
What does "smooth enough" mean? It means that f is piecewise smooth and
f (x0 +) + f (x0 −)
f (x0 ) = ,
2
if f is not smooth at x0 .
How to compute cn ? We shall prove that
Z π
1
cn = f (x)e−inx dx.
2π −π
In fact , by the above theorem, we have
Z π X cm Z π
1
f (x)e−inx dx = eimx e−inx dx.
2π −π 2π −π
m∈Z
If m = n then Z π Z π
imx −inx
e e dx = 1 dx = 2π.
−π −π
10 XU WANG

If m 6= n then
π π
ei(m−n)x
Z Z
i(m−n)x
e dx = d( ) = 0.
−π −π i(m − n)
Thus
X cm Z π
eimx e−inx dx = cn .
2π −π
m∈Z

Definition 2.2. We call

X
f (x) = cn einx
n∈Z

the complex Fourier series of f and

Z π
1
cn := f (x)e−inx dx, n ∈ Z,
2π −π

the complex Fourier coefficients of f .

Example: Consider

f (x) = 1, 0 < x < π; f (x) = −1, −π < x < 0,

and
f (0) = f (π) = f (−π) = 0.
Then we know f is smooth enough and
Z π Z π Z 0
−inx −inx
2πcn = f (x)e dx = e dx − e−inx dx.
−π 0 −π

Since
π π
e−inx (−1)n − 1
Z Z
−inx
e dx = d( )= ,
0 0 −in −in
and
0 0
e−inx 1 − (−1)n
Z Z
−inx
e dx = d( )= ,
−π −π −in −in
we have
2(1 − (−1)n )
2πcn = ,
in
i.e.
2
cn =
, n odd; cn = 0, n even.
inπ
Thus the complex fourier series of f is
X 2
f (x) = ei(2m+1)x .
i(2m + 1)π
m∈Z
 LAPLACE TRANSFORM AND FOURIER TRANSFORM 11

2.1.2. (Real) Fourier series. In the previous example, we have

e3ix 2 e−ix e−3ix
   
2 ix
f (x) = e + + ··· + + + ··· ,
iπ 3 iπ −1 −3
thus
e3ix − e−3ix
 
2 ix −ix
f (x) = (e − e ) + + ··· .
iπ 3
Euler’s formula gives
einx − e−inx = 2i sin nx,
thus  
4 sin 3x
f (x) = sin x + + ··· ,
π 3
In particular, it gives  
π 4 1 1 1
1 = f( ) = 1 − + − + ··· .
2 π 3 5 7
Thus
1 1 1 π
1−
+ − + ··· = ,
3 5 7 4
which is a famous formula obtained by Leibniz in 1673 from geometric considerations.
For a general function f , by the Euler formula, we have
X X
f (x) = cn einx = cn (cos nx + i sin nx),
which gives
∞
X ∞
X
f (x) = c0 + cn (cos nx + i sin nx) + c−n (cos nx − i sin nx).
n=1 n=1
Thus we have
∞
X
f (x) = c0 + ((cn + c−n ) cos nx + i(cn − c−n ) sin nx) ,
n=1
Recall that
Z π Z π
1 1
(cn + c−n ) = f (x)(e−inx + einx ) dx = f (x) cos nx dx,
2π −π π −π
and Z π Z π
i −inx inx 1
i(cn − c−n ) = f (x)(e −e ) dx = f (x) sin nx dx,
2π −π π −π
thus we get
Theorem 2.3. If f has period 2π and is smooth enough then it has the following Fourier series
expansion
X∞
f (x) = a0 + (an cos nx + bn sin nx),
n=1
where a0 , an , bn are the Fourier coefficients of f such that
Z π
1
a0 = c0 = f (x) dx,
2π −π
12 XU WANG

and for n = 1, 2 · · · , we have

Z π
1
an = f (x) cos nx dx,
π −π
and Z π
1
bn = f (x) sin nx dx.
π −π

Example: Consider
f (x) = 0, −π < x < 0; f (x) = x, 0 ≤ x < π.
Then
π π
π2
Z Z
2πa0 = f (x) dx = x dx = ,
−π 0 2
and
Z π Z π Z π Z π
sin nx sin nx
πan = f (x) cos nx dx = x cos nx dx = x d( )=− dx.
−π 0 0 n 0 n
Since
π π
(−1)n − 1
Z Z
sin nx cos nx
− dx = d( ) = ,
0 n 0 n2 n2
we get
π −2
a0 = , a2m = 0, a2m−1 = , m = 1, 2 · · · .
4 (2m − 1)2 π
Moreover, we have
Z π π π π
− cos nx π(−1)n+1
Z Z Z
cos nx
πbn = f (x) sin nx dx = x sin nx dx = x d( )= + dx,
−π 0 0 n n 0 n
Notice that Z π Z π
cos nx sin nx
dx = d( ) = 0,
0 n 0 n2
thus
(−1)n+1
bn = .
n
Thus    
π 2 cos 3x sin 2x sin 3x
f (x) = − cos x + + · · · + sin x − + + ··· .
4 π 32 2 3
Take x = 0 then we get
π 2 1
0= − (1 + 2 + · · · ),
4 π 3
i.e.
1 1 π2
1+ + + · · · = .
32 52 8
1 1 π2
Exercise: Use 1 + 32
+ 52
+ ··· = 8 to prove that
1 1 π2
1+ + + · · · = .
22 32 6
 LAPLACE TRANSFORM AND FOURIER TRANSFORM 13

Proposition 2.4. Put δmn = 1 if m = n and δmn = 0 if m 6= n then

Z π Z π
cos nx cos mx dx = sin nx sin mx dx = πδmn , m, n = 1, 2, · · · ,
−π −π
and Z π Z π
cos nx dx = 2πδn0 , cos nx sin mx = 0, m, n = 0, 1, 2, · · · .
−π −π

Proof. Follows from the Euler formula

einx + e−inx einx − e−inx
cos nx = , sin nx = ,
2 2i
and Z π
einx e−imx dx = 2πδmn .
−π
Try to give the details yourself. 
Exercise: Try to use the above proposition to prove the formulas for an , bn in Theorem 2.3.

2.2. Fourier Sine and Cosine series.

Definition 2.5. We say that f is odd if f (−x) = −f (x); f is even if f (−x) = f (x).
Example: For every positive integer n, we know that cos nx is even and sin nx is odd.
Application: If f is even then
Z π Z π
f (x) dx = 2 f (x) dx.
−π 0
If f is odd then Z π
f (x) dx = 0.
−π
In particular, if f is odd then all
Z π
1
an = f (x) cos nx dx = 0;
π −π
if f is even then all Z π
1
bn = f (x) sin nx dx = 0.
π −π
Thus we get:
Theorem 2.6. Assume that f has period 2π and is smooth enough. If f is odd then it can be written
as a Fourier sine series
∞
2 π
X Z
f (x) = bn sin nx, bn = f (x) sin nx dx.
π 0
n=1
If f is even then it can be written as a Fourier cosine series
∞
1 π 2 π
Z X Z
f (x) = f (x) dx + an cos nx, an = f (x) cos nx dx.
π 0 π 0
n=1
14 XU WANG

Odd or Even extension: Let f be a function in (0, π). Then we can extend f to an odd function,
say fo such that
fo (−x) = −f (x), x ∈ (0, π);
we can also extend f to an even function, say fe such that
fe (−x) = f (x), x ∈ (0, π).
Exercise:
1. Draw the graph of the odd extension fo and the even extension fe of the following function
π π π
f (x) = x, 0 < x < ; f (x) = , < x < π.
2 2 2
2. Find the Fourier cosine series of fe . Answer
 
3π 2 2 cos 2x cos 3x cos 5x
fe (x) = + − cos x − − − − ··· .
8 π 22 32 52
3. Find the Fourier sine series of fo . Answer
     
2 1 −2 1
fo (x) = + 1 sin x + 0 − sin 2x + + sin 3x
π 2 32 π 3
   
1 2 1
+ 0− sin 4x + 2
+ sin 5x + · · · .
4 5 π 5
2.3. More Fourier series.

2.3.1. Parseval’s identity. Let f be a smooth enough function with period 2π. Consider the complex
Fourier series expansion of f
X
f= cn einx .
n∈Z
We have Z π Z π X
2
|f (x)| dx = cn cm einx−imx dx.
−π −π n,m∈Z

Now use that Z π

einx−imx dx = 0,
−π
if m 6= n and Z π
einx−imx dx = 2π,
−π
if m = n. We get the following Parseval identity
Z π X
|f (x)|2 dx = 2π |cn |2 .
−π n∈Z

Example: Consider the last example in section 2.1.1:

f (x) = 1, 0 < x < π; f (x) = −1, −π < x < 0,
and
f (0) = f (π) = f (−π) = 0.
 LAPLACE TRANSFORM AND FOURIER TRANSFORM 15

We know that f has the following complex Fourier series expansion:

X 2
f (x) = ei(2m+1)x .
i(2m + 1)π
m∈Z

Thus the Parseval identity gives

Z π  
1 2
X 4 8 1 1
|f (x)| dx = 1 = = 2 1 + 2 + 2 + ··· ,
2π −π (2m + 1)2 π 2 π 3 5
m∈Z

which gives another proof of

1 1 π2
1+ + + · · · = .
32 52 8

2.3.2. Resonance phenomenon. Let us consider

(?) : y 00 (t) + ω 2 y(t) = sin αt,

where α 6= ω are positive constants. We call

y(t) = A cos αt + B sin αt,

a steady-state solution of (?). A direct computation gives

1
A = 0, B = .
ω2 − α2
The "resonance phenomenon" can be explained by the following fact: the amplitude |B| of the steady-
state solution is very big if α is very close to ω (the frequency of the system). In general, a steady
solution of
y 00 (t) + ω 2 y(t) = an cos nt + bn sin nt,
is defined as a solution of the following form

yn (t) := An cos nt + Bn sin nt.

Moreover, consider the following equation

(??) y 00 (t) + ω 2 y(t) = γ(t),

Assume that
∞
X
γ(t) = an cos nt + bn sin nt,
n=0

then we call
y(t) := y0 + y1 + · · · + yn + · · · ,
a steay state solution of (??). When ω = N is a natural number, avoid resonance means that the input
γ(t) satisfies aN = bN = 0.
16 XU WANG

2.3.3. Best L2 –approximation by inx .

P
|n|≤N cn e Let f be a smooth enough function with period
2π. We hope to find an N -series, X
FN := Cn einx ,
|n|≤N
such that Z π
|FN − f |2 dx
−π
is minimal. The main idea is to use orthogonal decomposition.
Orthogonal Decomposition in vector space: Let S be a subspace of a vector space V (think of
S = R2 × {0} and V is R3 ), then we can write a vector, say v, in V as
v = vS + vS ⊥ ,
where vS lies in S and vS ⊥ is orthoginal to S.
Definition 2.7. We call vS the orthogonal projection of v to S.
Then it is very clear from the picture that vS is the unique solution of the following extremal
problem:
||vS − v|| = min{||u − v|| : u ∈ S}.
For a real proof it is enough to use
||u − v||2 = ||u − vS ||2 + ||vS ⊥ ||2 ,
which implies that u = vS is the unique solution.
Orthogonal Decomposition in the space of functions: In our case, we consider V as the space of
complex functions on [−π, π] with the following inner product structure:
Z π
(f, g) := f (x)g(x) dx, ||f ||2 := (f, f ).
−π

Now S is the subspace of V spanned by einx , |n| ≤ N . Let f be a smooth enough function with
period 2π. Then we know that
||fS − f || = min{||u − f || : u ∈ S}
Thus FN = fS (recall that fS means the orthogonal projection of f to S) solves our extremal problem.
What is fS ? The simplest way is to use the complex Fourier series expansion
X
f (x) = cn einx .
n∈Z
Put X
fN = cn einx ,
|n|≤N

since {einx }n∈Z is an orthogonal basis of V we have

(fN , f − fN ) = 0,
which implies that fN is the orthogonal projection of f to S. Now we have
fN = fS ,
which gives:
 LAPLACE TRANSFORM AND FOURIER TRANSFORM 17

Theorem 2.8. Complex Fourier series expansion solves the best L2 –approximation by inx
P
|n|≤N cn e
problem problem.
Bessel’s inequality and Parseval’s identity: Notice that (try!)
||f ||2 = ||fN ||2 + ||f − fN ||2 ,
Thus we get the Bessel inequality
||f ||2 ≥ ||fN ||2 ,
i.e. Z π X
|f (x)|2 dx ≥ 2π · |cn |2 ,
−π |n|≤N
and the Parseval identity Z π X
|f (x)|2 dx = 2π · |cn |2 .
−π n∈Z

2.4. Fourier transform.

Definition 2.9. We call Z ∞
1
ˆ
f (w) := √ f (x)e−iwx dx,
2π −∞
the Fourier transform of f and write fˆ = F(f ).
Example: F1: Fourier transform of f (x) = 1 if |x| < 1 and f (x) = 0 otherwise:
Z ∞ Z 1
1 1
fˆ(w) := √ f (x)e−iwx dx = √ e−iwx dx
2π −∞ 2π −1
if w 6= 0 then
1 1
e−iwx e−iw eiw
Z Z
−iwx 2 sin w
e dx = d( )= − = .
−1 −1 −iw −iw −iw w
Notice that Z 1
2 sin w
lim =2= dx = fˆ(0).
w→0 w −1
Thus we can write   r
1 2 sin w 2 sin w
fˆ(w) = √ = .
2π w π w

Example: F2: Fourier transform of f (x) = e−x if x > 0 and f (x) = 0 otherwise:
Z ∞ Z ∞
1 −iwx 1 1
ˆ
f (w) := √ f (x)e dx = √ e−x e−iwx dx = √ L(e−iwt )(1).
2π −∞ 2π 0 2π
Recall that
1
L(e−iwt )(s) = ,
s + iw
thus
1 1
fˆ(w) = √ · .
2π 1 + iw
18 XU WANG

2.4.1. From Complex Fourier series to inverse Fourier transform. : Assume that f is smooth enough
in −N < x < N and f = 0 when |x| > N . For each L > N , let us define a periodic function fL
such that
fL (x) = f (x), |x| < L; fL (x + 2L) = fL (x).
Then we know that  
Lx
gL (x) = fL ,
π
has period 2π and is smooth enough. Thus
Z π
X 1
gL (x) = cn einx , cn = gL (x)e−inx dx.
2π −π
Thus  πx  X πx
fL (x) = gL = cn ein L .
L
Consider v = Lxπ , we can write
Z L Z ∞ √
1 −in πv πv 1 −in πv 2π ˆ nπ
cn = fL (v)e L d( )= f (v)e L dv = f ( ).
2π −L L 2L −∞ 2L L
which gives
in πx
π X fˆ( nπ
r
L )·e
L
f (x) = .
2 L
n∈Z
Put
π
∆w = ,
L
then we have
1 Xˆ
f (x) = √ f (n∆w) · eix·n∆w ∆w.
2π n∈Z
Assume that fˆ(w)eixw is integrable in −∞ < x < ∞. Let L goes to infty, the above formula gives
the following Fourier inversion formula:
Z ∞
1
f (x) = √ fˆ(w)eixw dw.
2π −∞
We say that f (x) is the inverse Fourier transform of fˆ(w) and write f = F −1 (fˆ).
2.5. Fourier inversion formula. When do we have the Fourier inversion formula ? It is known that
(see Page 141 Theorem 1.9 in [3]) the Fourier inversion formula is true if f is smooth and rapidly
decreaing, in the sense that
sup |x|k |f (l) (x)| < ∞, for every k, l ≥ 0,
x∈R

where f (l) denotes the l-th derivative of f .

x2
Example: Let f (x) = e− . We shall use Fourier inversion formula to prove
2
Z ∞
1 x2 w2
(6) ˆ
f (w) := √ e− 2 e−iwx dx = e− 2 = f (w).
2π −∞
Step 1: Look at the derivative of fˆ(w):
Z ∞
0 1 x2
ˆ
(f ) (w) = √ e− 2 (−ix)e−iwx dx = F(−ixf (x)).
2π −∞
 LAPLACE TRANSFORM AND FOURIER TRANSFORM 19

x2 x2
Notice that (e− 2 )0 = e−
(−x), thus
2

Z ∞ Z ∞
0 i −iwx − x2
2 −i x2
ˆ
f (w) = √ e d(e )= √ e− 2 d(e−iwx ) = −wfˆ(w).
2π −∞ 2π −∞
Now we have
 0  
w2 w2
fˆ(w)e 2 = (−w + w) fˆ(w)e 2 = 0.

w 2
Thus fˆ(w)e 2 is a constant, i.e.
w 2
fˆ(w)e 2 ≡ fˆ(0)e0 = fˆ(0).

Now we have
−w 2
fˆ(w) = fˆ(0)e 2 = fˆ(0)f (w)
Step 2: The Fourier inversion formula implies (notce that f is smooth and rapidly decreasing)

f (x) = F −1 (fˆ) = fˆ(0)F −1 (f ) = fˆ(0)fˆ(−x) = (fˆ(0))2 f (x).

Thus fˆ(0) = ±1. Since

Z ∞
1 x2
fˆ(0) = √ e− 2 dx > 0,
2π −∞

we get
fˆ(0) = 1, fˆ = f.

2.5.1. Normal distribution. fˆ(0) = 1 gives

Z ∞
1 x2
(7) √ e− 2 dx = 1,
2π −∞

(one may also use integration on R2 to compute the following integral directly, see page 138 formula
(6) in [3]), consider
√
u = tx + µ, t > 0, µ ∈ R,
then (7) becomes the following classical formula in Gauss’s normal distribution theory
Z ∞
1 (u−µ)2
(8) √ e− 2t du = 1,
−∞ 2πt
where
1 (u−µ)2
(9) f (u | µ, t) := √ e− 2t ,
2πt
is the probability density of the normal distribution with expectation µ and variance t.
20 XU WANG

2.5.2. Inverse Laplace transform. Recall the definition of the Laplace transform
Z ∞
F (s) = e−st f (t) dt
0
Extend f to a function on R such that
f (x) = 0, ∀ x < 0.
Thus we have Z ∞
F (s) = e−st f (t) dt,
−∞
which gives √
F (iw) = 2π fˆ(w).
Thus the Fourier inversion formula gives
Z ∞ Z ∞
1 ˆ itw 1
Laplace inversion formula : f (t) = √ f (w)e dw = F (iw)eitw dw.
2π −∞ 2π −∞
2.6. More Fourier transforms.
2.6.1. Derivative and convolution formulas in Fourier transform. In this section, we only consider
functions that are smooth and rapidly decreasing. The main result is the following:
R∞
Theorem 2.10. Let F(f )(w) = √12π −∞ f (x)e−iwx dx be the Fourier transform of f . Then
1) F(af + bg) = aF(f ) + bF(g);
2) F(f 0 ) = iwF(f );
3) (F(f ))0 = −iF(xf (x)).
Proof. We only prove 2). Notice that
Z ∞ Z ∞ Z ∞ Z ∞
0 −iwx −iwx −iwx
f (x)e dx = e df = d(e f) − f (x) d(e−iwx ).
−∞ −∞ −∞ −∞
Since f is smooth and repadly decreasing, we have
Z ∞
d(e−iwx f ) = lim e−iwx f (x) − lim e−iwx f (x) = 0.
−∞ x→∞ x→−∞

Thus Z ∞ Z ∞ Z ∞
0 −iwx −iwx
f (x)e dx = − f (x) d(e ) = iw f (x)e−iwx dx,
−∞ −∞ −∞
which implies 2). 
2 2
− x2 − w2
Example: F(xe ) = −iwe : By 3), we have
2 x2 w2 w2
− x2
F(xe ) = i(F(e− 2 ))0 = i(e− 2 )0 = −iwe− 2 .

Convolution: Let us first recall the definition of convolution for functions f, g defined on [0, ∞):
Z t
(f ? g)(t) := f (τ )g(t − τ ) dτ.
0
Notice that if we extend f, g to functions on R such that
f = g = 0, when x ≤ 0.
 LAPLACE TRANSFORM AND FOURIER TRANSFORM 21

Then we can write Z ∞

(f ? g)(x) := f (u)g(x − u) du.
−∞

Definition 2.11. The convolution of two functions on R is defined by

Z ∞
(f ? g)(x) := f (u)g(x − u) du.
−∞

Similar as the Laplace transform, we have

√
(10) F(f ? g) = 2π F(f ) · F(g).

Proof. [Need integration on R2 ]. We have

√ Z ∞ ∞ ∞
Z Z 
−iwx
2πF(f ? g) = (f ? g)(x)e dx = f (u)g(x − u) du e−iwx dx.
−∞ −∞ −∞

Change the order of integration, we get

√ Z ∞ √ Z
Z ∞  ∞
−iwx
2πF(f ? g) = f (u) g(x − u)e dx du = 2π f (u)F(g)(w)e−iwu du,
−∞ −∞ −∞
√
which gives 2πF(f ? g) = 2πF(f ) · F(g). 

2.6.2. From Parseval identity to Parseval formula (*TBA).

2.6.3. Fourier transform of distributions (∗T BA).

3. PARTIAL DIFFERENTIAL EQUATIONS

3.1. Functions of several variables. In Laplace transform and Fourier transform, the functions
e−st , e−iwx depend on two variables. We can look at e−st as a map, say
f : (s, t) 7→ f (s, t) := e−st ,
from R2 to R. We say that the map f defines a function on R2 . It is clear that
f (x, y) = x2 + y 2 ,
is a function on R2 ;
f (x, y, z) = x + y + z,
is a function on R3 and
1 x2
f (t, x) = √ e− 2t ,
2πt
is a function on (0, ∞) × R.

3.1.1. Graph. If f is a function on U ⊂ Rn then we call the following set

Gf,U := {(x, f (x)) ∈ Rn+1 : x ∈ U },
in Rn+1 the graph of f over U .
Exercise: Draw the graph of f (x, y) = x2 + y 2 over the unit disc.
22 XU WANG

3.1.2. Partial derivatives. x-partial derivative of f (x, y) means derivative of f with y fixed:
∂f f (x + h, y) − f (x, y)
(x, y) := lim ,
∂x h→0 h
we write
∂f ∂fx
fx := , fxy := .
∂x ∂y

Example: If f (x, y) = x2 + y 2 + xy then

fx = 2x + y, fy = 2y + x, fxy = 1 = fyx , fxx = 2, fyy = 2.

Example: If
1 x2
f (t, x) = √ e− 2t ,
2πt
then
1 −1 −3 − x2 1 x2 −x2 −1 x2 − t
ft = √ · · t 2 · e 2t + √ e− 2t · · 2 = f,
2π 2 2πt 2 t 2t2
and
x 1 x2 x2 − t
fx = − f, fxx = − f + 2 f = f.
t t t t2
Thus we get that
1
(11) ft = fxx .
2
3.1.3. Directional derivative and gradient. Let n in the unit sphere be a given direction, then we call
f (p + hn) − f (p)
fn (p) := lim ,
h→0 h
the derivative of f along direction n at p. If we write
n = (a, b, c), p = (x, y, z),
then
f (x + ah, y + bh, z + ch) − f (x, y, z)
fn (p) = lim .
h→0 h
Example: If
f (x, y, z) = c0 + c1 x + c2 y + c3 z,
then
f (hn) = f (ah, bh, ch) = c0 + (c1 a + c2 b + c3 c)h,
which gives
(12) fn (0) = c1 a + c2 b + c3 c = (fx (0), fy (0), fz (0)) · n.
Definition 3.1. We call
∇f (p) := (fx1 (p), · · · , fxm (p)),
the gradient of f (x1 , · · · , xm ) at p.
 LAPLACE TRANSFORM AND FOURIER TRANSFORM 23

For a general smooth function, we have the following generalization of (12):

(13) fn (p) = ∇f (p) · n.
Since |n| = 1, the above formula gives
fn (p) = ∇f (p) cos θ,
where θ denotes the angle from ∇f (p) to n.
3.1.4. PDEs. We shall study the following partial differential equations
1. Wave equation
utt = uxx .
2. Heat equation
1
ut = uxx ,
2
Homework: Find the background of the above two equations in [2] or wikipedia.
Example: By (11), we know that
1 x2
f (t, x) := √ e− 2t ,
2πt
satisfies the above heat equation. We call f (t, x) the heat kernel.
3.2. Solve wave equation by Fourier series. Let us solve the wave equation
utt = uxx ,
with boundary conditions
u(t, 0) = u(t, π) = 0, ∀ t ≥ 0;
and initial conditions
u(0, x) = f (x), ut (0, x) = g(x), ∀ 0 ≤ x ≤ π.
Step 1: Separating variables: Find solutions of the form
u(t, x) = G(t)F (x),
Since
utt = G00 F, uxx = GF 00 ,
our equation becomes
G00 F = GF 00 ,
thus
G00 F 00
= ≡ k,
G F
where k is constant (notice that k does not depend on t and x).
Step 2: Fit boundary conditions: Notice that the boundary conditions
G(t)F (0) = G(t)F (π) = 0,
is equivalent to
F (0) = F (π) = 0.
In case k = 0, then F 00
≡ 0, i.e F (x) = ax + b. The boundary conditions give F ≡ 0.
In case k = µ2 > 0, then the general solution for
F 00 = µ2 F,
24 XU WANG

is F = Aeµx + Be−µx , then the boundary conditions give

A + B = 0, Aeµπ + Be−µπ = 0,
thus A = B = 0.
Thus the only possible case is k = −p2 < 0, then the general solution for
F 00 = −p2 F,
is F = A cos px + B sin px, F (0) = 0 gives A = 0. Thus F = B sin px, B 6= 0, but F (π) = 0 gives
sin pπ = 0, i.e.
p = n, n = 1, 2, · · · ,
(notice that sin −px = − sin px, thus up to a constant they give the same solution).
Summary: The boundary condition implies that p = n2 , n = 1, 2, · · · , and
F = Fn (x) = sin nx.
Now let us solve
G00 = −n2 G,
the general solution is
Gn (t) = Bn cos nt + Cn sin nt.
Now we know that each
un (t, x) = Gn (t)Fn (t) = (Bn cos nt + Cn sin nt) sin nx,
satisfies the wave equation and the boundary conditions, so does
X∞
u(t, x) = (Bn cos nt + Cn sin nt) sin nx.
n=1

Step 3: Fit the initial conditions: Choose Bn and Cn such that

u(0, x) = f (x), ut (0, x) = g(x).
i.e.
∞
X
Bn sin nx = f (x);
n=1
and
∞
X
nCn sin nx = g(x);
n=1
Consider odd extension fo , go of f, g, by Theorem 2.6, if fo and go is smooth enough, then it is enough
to choose
2 π
Z
Bn = f (x) sin nx dx,
π 0
and
2 π
Z
nCn = g(x) sin nx dx.
π 0
Now we know that if f, g have smooth enough odd extension then
X∞
u(t, x) = (Bn cos nt + Cn sin nt) sin nx,
n=1
solves the wave equation (of course we have it check that it converges).
 LAPLACE TRANSFORM AND FOURIER TRANSFORM 25

Example: When g = 0 we have

∞
X
u(t, x) = Bn cos nt sin nx.
n=1
Thus we can write
∞
1X 1
u(t, x) = (Bn sin n(x − t) + Bn sin n(x + t)) = (fo (x − t) + fo (x + t)),
2 2
n=1
i.e. u(t, x) is the superposition of two travelings of the initial wave.
Exercise: In case
π π
f (x) = x, 0 ≤ x ≤ , f (x) = π − x, < x ≤ π,
2 2
try to draw the graph of u(t, x) for t = 0, π8 , π4 , π2 , π.
3.3. Solve heat equation by Fourier series. Consider the heat equation
1
ut = uxx ,
2
with boundary conditions
u(t, 0) = u(t, π) = 0, ∀ t ≥ 0;
and initial conditions
u(0, x) = f (x), ∀ 0 ≤ x ≤ π.
Step 1: Separating variables: Find solutions of the form
u(t, x) = G(t)F (x),
Since
ut = G0 F, uxx = GF 00 ,
our equation becomes
1
G0 F = GF 00 ,
2
thus
2G0 F 00
= ≡ k,
G F
Step 2: Fit the boundary conditions: Same as the wave equation, we have
k = −n2 , n = 1, 2 · · · , n
and
Fn (x) = sin nx.
Then
n2
G0 = − G,
2
gives
n2
Gn (t) = Bn e− 2
t
,
Thus the general solution is
∞
X n2
u(t, x) = Bn e− 2
t
sin nx.
n=1
26 XU WANG

Step 3: Fit the initial conditions: We hope

∞
X
u(0, x) = Bn sin nx = f (x).
n=1
By Theorem 2.6, we get Z π
2
Bn = f (x) sin nx dx,
π 0

Example: f (x) = 2 sin x then

B1 = 2, B2 = · · · = Bn = 0.
Thus
t2
u(t, x) = 2e− 2 sin x,
Notice that u goes to zero as t goes to infinity.
Exercise: Find u with
π π
f (x) = x, 0 ≤ x ≤ , f (x) = π − x, ≤ x ≤ π.
2 2
3.4. Solve heat equation by Fourier transform. Still the heat equation
1
ut = uxx .
2
But this time we consider the initial condition for all x in R (thus no boundary conditions):
u(0, x) = f (x), ∀ − ∞ < x < ∞.
Step 1: Reduce to ODE by Fourier transform: Consider Fourier transform of ut with respect to the
x variable Z ∞
1
F(ut ) = ubt (w) = √ ut (t, x)e−ixw dx.
2π −∞
Then we have Z ∞
1 1 1
F(ut ) = F( uxx ) = √ uxx (t, x)e−ixw dx.
2 2π −∞ 2
Recall that if ux is smooth and rapidly decreasing then
Z ∞ Z ∞ Z ∞ Z ∞
−ixw −ixw −ixw
uxx e dx = e d(ux ) = − ux d(e ) = iw ux e−ixw dx,
−∞ −∞ −∞ −∞
the same computation for u gives
Z ∞ Z ∞
−ixw
ux e dx = iw ue−ixw dx,
−∞ −∞
thus we have
∞
−w2
Z
1
F(ut ) = F(u), F(u) := √ u(t, x)e−ixw dx
2 2π −∞
Notice that we also have
F(ut ) = (F(u))t .
Thus F(u) satisfies the following ODE.
−w2
(F(u))t = F(u).
2
 LAPLACE TRANSFORM AND FOURIER TRANSFORM 27

Step 2: Solve the ODE and fit the initial condition: The general solution is
−w2
t
F(u)(t, w) = c(w)e 2 .
Notice that our initial condition implies
Z ∞
1
F(u)(0, w) = √ f (x)e−ixw dx = F(f ).
2π −∞
Thus
c(w) = F(f ).
Now we have
−w2
F(u)(t, w) = F(f ) · e 2 t .
Step 3: use Fourier convolution formula: Recall that (see (6))
Z ∞
−u2 1 −y 2
e 2 =√ e 2 e−iyu dy.
2π −∞
Take √ x
u = w t, y = √ ,
t
we get Z ∞
−w2 1 −x2 1 −x2
e 2
t
= √ e 2t e−ixw dx = √ F(e 2t ).
2πt −∞ t
Thus  
1 −x2
F(u)(t, w) = F(f ) · √ F(e 2t )
t
Now the Fourier convolution formula gives
Z ∞
1 −x2 1 −(x−p)2
u(t, x) = √ (f ? e 2t ) = f (p) √ e 2t dp,
2πt −∞ 2πt
Summary: the solution u(t, x) is given by convolution of the initial temperature distribution with
the heat kernel.

4. A PPENDIX : D EFINITION OF e, π AND E ULER ’ S FORMULA

4.1. Where does e come from ? Recall that: Let A : Cn → Cn be a linear map (here linear map
means A(au + bv) = aA(u) + bA(v) for all a, b in C and all u, v in Cn ). We call u 6= 0 in Cn an
eigenvector of A if
(14) Au = λu,
where λ is a constant in C.
What is an eigenvector of the derivative ?
By (14), we want to find function u : R → C such that
u0 = λu.
Power series method: Assume that
u(x) = a0 + a1 x + a2 x2 + · · · + an xn + · · · .
The following lemma gives:
u0 (x) = a1 + 2a2 x + · · · + nan xn−1 + (n + 1)an+1 xn + · · · .
28 XU WANG

Lemma 4.1. (xn )0 = nxn−1 , n = 1, 2, · · · .

Proof. If n = 1 then
(x + 4x) − x
x0 (x) = lim = 1.
4x→0 4x
Assume the Lemma for n = 1, · · · , N − 1. Then (f g)0 = f 0 g + f g 0 gives
(xN )0 = (xN −1 )0 · x + xN −1 · x0 = (N − 1)xN −2 · x + xN −1 = N xN −1 .
The proof is complete. 
Exercise: Why we have (f g)0 = f 0 g + f g 0 ?
Now
u0 = λu ⇔ λan = (n + 1)an+1 , n = 0, 1, · · · .
Thus
λan λ2 an−1 λn+1 a0 λn+1 a0
an+1 = = = ··· = = ,
(n + 1) (n + 1)n (n + 1)n · · · 1 (n + 1)!
where we define
n! = 1 · 2 · · · n.
Then we have
(λx)n
u(x) = u0 · (1 + λx + · · · + + · · · ).
n!
Put
xn
E(x) := 1 + x + · · · + + ··· .
n!
Since for every C > 0,
Cn
lim
= 0,
n→∞ n!
we know that E(x) converges for all x in C.
Theorem 4.2. E(λx) is a unique solution of the eigenvalue equation
u0 = λu,
with initail condition u(0) = 1.
Definition 4.3. We shall define
1 1
e := E(1) = 1 + 1 + + ··· + + ··· .
2 n!
4.2. Definition of the exponential function. Let us write
e2 = e · e, e3 = e2 · e,
and define em inductively by
en+1 = en · e.
m
Since e is positive, we can take the q − th root of em , we write it as e q . Thus for every x ∈ Q, ex is
well defined. The following lemma tells us that E(x) is an extension of ex from Q to C.
Lemma 4.4. For every x ∈ Q, we have ex = E(x).
 LAPLACE TRANSFORM AND FOURIER TRANSFORM 29

Proof. Since E(1) = e, it suffices to prove

(15) E(λ1 )E(λ2 ) = E(λ1 + λ2 ),
for every λ1 , λ2 in C. Notice that
(E(λ1 x)E(λ2 x))0 = E(λ1 x)0 E(λ2 x) + E(λ2 x)0 E(λ1 x).
Put
G(x) = E(λ1 x)E(λ2 x).
Apply E(λx)0 = λE(λx), we get
G0 = (λ1 + λ2 )G.
Notice that G(0) = 1. Thus Theorem 4.2 implies that
G(x) = E((λ1 + λ2 )x).
Take x = 1, we get E(λ1 )E(λ2 ) = E(λ1 + λ2 ). 
Exercise: Find a direct proof of E(λ1 )E(λ2 ) = E(λ1 + λ2 ) without using Theorem 4.2.
Definition 4.5. We shall use the same symbol ex to denote E(x) for all x in C and call ex the
exponential function. If x > 0 then we define ln x as the unique real solution of eln x = x.
By Theorem 4.2, we know that ex is fully determined by
(ex )0 = ex , e0 = 1.
4.3. Definition of π and trigonometric functions. : Fix P0 = (1, 0) in the unit circle
S 1 := {(x, y) ∈ R2 : x2 + y 2 = 1}.
A counterclockwise rotation of P0 gives a arc P0 P . The length, say θ(P ), of the arc P0 P is a function
of P . It is clear that the circumference diameter ratio is equal to θ(−1, 0).
Definition 4.6 (Definition of π). We shall write the circumference diameter ratio as π.
Denote by
F : θ(P ) 7→ P,
the inverse function of 0 ≤ θ(P ) ≤ 2π.
Definition 4.7. We shall write F (θ) = (cos θ, sin θ).
Notice that
F (0) = (1, 0) = F (2π), F (π) = (−1, 0), |F (θ)| ≡ 1.
In particular, it gives
sin(0) = sin(2π) = 0, cos(0) = cos(2π) = 1.
By definition of θ, we have
Z θ̂
|F 0 (θ)| dθ = θ̂, 0 ≤ θ̂ ≤ 2π,
0
which gives
|F 0 (θ)| ≡ 1.
Now F (θ) · F (θ) ≡ 1 implies
F 0 · F + F · F 0 = 2F · F 0 ≡ 0.
30 XU WANG

Hence F 0 ⊥F , thus we know that

F 0 (θ) = (− sin θ, cos θ), or F 0 (θ) = (sin θ, − cos θ).
But notice that F 0 (0) = (0, 1), thus we must have
F 0 (θ) = (− sin θ, cos θ),
which is equivalent to
(cos θ + i sin θ)0 = i(cos θ + i sin θ).
Notice that cos 0 + i sin 0 = 1, thus Theorem 4.2 gives
Theorem 4.8 (Euler’s formula). eiθ = cos θ + i sin θ.
Take θ = π, we get the following Euler’s identity
eiπ = −1.
Moreover, apply (15), we get
eiθ1 eiθ2 = ei(θ1 +θ2 ) ,
thus by Euler’s formula, we have
(cos θ1 + i sin θ1 )(cos θ2 + i sin θ2 ) = cos(θ1 + θ2 ) + i sin(θ1 + θ2 ),
i.e.
(16) cos(θ1 + θ2 ) = cos θ1 cos θ2 − sin θ1 sin θ2 ,
and
(17) sin(θ1 + θ2 ) = sin θ1 cos θ2 + cos θ1 sin θ2 .

R EFERENCES
[1] B. Berndtsson, Fourier and Laplace transforms, can be found in:
www.math.chalmers.se/Math/Grundutb/CTH/mve025/1718/Dokument/F-analysny.pdf.
[2] E. Kreyszig, Advanced Engineering Mathematics, 10 th edition, internation student version.
[3] E. Stein and R. Shakarchi, Fourier analysis, Princeton lectures in analysis.
[4] A. Vretblad, Fourier Analysis and Its Applications, GTM 233.

D EPARTMENT OF M ATHEMATICAL S CIENCES , N ORWEGIAN U NIVERSITY OF S CIENCE AND T ECHNOLOGY, NO-

7491 T RONDHEIM , N ORWAY
E-mail address: xu.wang@ntnu.no