Chapter 4 - Periodic Functions & Fourier Series

Question 1

Fourier Series

Overview

Answer 1

• considering trigonometric sine and cosine functions as a basis for the linear space of periodic functions on the line
• any reasonable (to be defined) periodic function can be written as an expansion in sine and cosine functions known as Fourier series

Question 2

Periodic Function

Definition

Answer 2

-the function f(x) is said to be periodic, with (minimum) period 2L, if f(x+2L)=f(x) for all x in the domain of f

Question 3

Linear Superposition

Answer 3

-clearly if f and g are both periodic functions with same period 2L, then any linear combination of them, c1f(x)+c2g(x) is also periodic with the same period

Question 4

Fourier Series

Theorem

Answer 4

-given a function such that, f(x+2L)=f(x) we can find an, bn such that:
fN(x) = ao/2 + Σ[ancos(nπx/L) + bnsin(nπx/L)]
-where sum is from 1 to N and 1≤N
-and the Fourier coefficients are given by:
an = 1/L * ∫ f(x)cos(nπx/L) dx
bn = 1/L * ∫ f(x)sin(nπx/L) dx
-both integrals taken between -L and +L

Question 5

Continuity

Definition

Answer 5

-a function f(x) is continuous at xo, if
lim f(x) = lim f(x)
-where the first limit is taken as x tends to xo from below and the second limit is taken as x tends to xo from above
-both limits must exist and be equal

Question 6

Jump Discontinuity

Definition

Answer 6

-f(x) has a jump discontinuity at x=xo if both limits (x->xo-)lim f(x) and (x->xo+)lim f(x) exist but are not equal

Question 7

Piecewise Continuity

Definition

Answer 7

-f(x) is piecewise continuous on interval [a,b] if it only has a finite number of jumps in the interval [a,b]

Question 8

Piecewise Smooth

Definition

Answer 8

• f(x) is piecewise smooth on interval [a,b] if both f(x) and f’(x) are piecewise continuous on [a,b]
• this means that wherever the function jumps, the gradient also changes

Question 9

Integral Over a Period

Answer 9

-when f(x) is piecewise continuous and has period T:
(0,T)∫ f(x) dx = (a,a+T) ∫ f(x) dx
-i.e. integral of f(x) between 0 and T is equal to the integral of f(x) between a and a+T

Question 10

Orthogonality Relations

sine-sine

Answer 10

∫sin(mπx/L)sin(nπx/L)dx = {0, m≠n or L, m=n

Question 11

Orthogonality Relations

cosine-cosine

Answer 11

∫cos(mπx/L)cos(nπx/L)dx = {0, m≠n or L, m=n≠0 or 2L, m=n=0

Question 12

Orthogonality Relations

sine-cosine

Answer 12

∫sin(mπx/L)cos(nπx/L)dx = {0, for all m,n

Question 13

Double Angle Identites

Answer 13

sinAsinB = 1/2[cos(A-B)-cos(A+B)]
cosAcosB = 1/2[cos(A-B)+cos(A+B)]
-in particular
sin²A = 1/2[1-cos(2A)]
cos²A = 1/2[1+cos(2A)]

Question 14

Fourier Series

Inner Product

Answer 14

⟨f,g⟩ = ∫f(x) g(x) dx

-where the integral is taken between -L and +L

Answer 15

**check last paragraph from L16

Question 16

Fourier Series

Differentiation Theorem

Answer 16

• given f(x), a continuous periodic function with piecewise continuous derivative f’(x)
• the Fourier series for f’(x) can be obtained by term by term differentiation of the Fourier Series of f(x)

Question 17

Fourier Series

Convergence Theorem

Answer 17

-let f(x) be a piecewise smooth periodic function
-then, for all x, we have the pointwise convergence:
1/2[f(x+)+f(x-)] = ao/2 + Σ[ancos(nπx/L)+bnsin(nπx/L)]
-where an and bn are given by the Fourier coefficient formulae
-therefore, at a jump discontinuity, the Fourier series converges to the average value whilst at a point of continuity it just converges to the common value

Question 18

Gibbs Phenomenon

Answer 18

• each finite expansion is an analytic function, so cannot truly represent a discontinuous function
• whilst the graph passes through the average at the precise point of discontinuity, the graph overshoots on either side
• this is known as the Gibbs phenomenon

Question 19

Odd Function

Definition

Answer 19

-an odd function satisfies:

f(-x) = -f(x)

Question 20

Even Function

Definition

Answer 20

-an even function satisfies:

f(-x) = f(x)

Question 21

Odd or Even and Linear Combinations

Answer 21

-the property of being either even or odd is preserved under linear combinations
af(-x)+bg(-x) = ±(af(-x)+bg(-x))

Question 22

Odd and Even Functions Under Multiplication

Answer 22

f(-x)g(-x) = {f(x)g(x) if f and g are both even or both odd OR -f(x)g(x) if one is odd and one is even}

Question 23

Odd and Even Functions

Fourier Series

Answer 23

f(-x)=f(x) => bn=0 for all n
=> an = 1/L ∫ f(x)*cos(nπx/L) dx
-integral btwn -1 and 1
=> an = 2/L ∫ f(x)*cos(nπx/L) dx
-integral btwn 0 and 1

f(-x)=-f(x) => an=0 for all n
=> bn = 1/L ∫ f(x)*sin(nπx/L) dx
-integral btwn -1 and 1
=> bn = 2/L ∫ f(x)*sin(nπx/L) dx
-integral btwn 0 and 1

f(-x)=f(x) => f(x)cos(nπx/L) is even
f(-x)=-f(x) => f(x)sin(nπx/L) is odd

Question 24

Even and Odd Parts of a Function

Answer 24

-given any function f(x), the even and odd parts are defined:
fe(x) = 1/2[f(x)+f(-x)]
fo(x) = 1/2[f(x)-f(-x)]
-and we have:
f(x) = fe(x) + fo(x)

Question 25

What is a half-range Fourier series?

Answer 25

-in applications to separation of variables for partial differential equations we are often given a function f(x) on only the half-range 0

Question 26

Even Extension

Answer 26

f(x) = ao/2 + Σancos((nπx/L)
-where the sum is between 1 and infinity and:
an = 2/L ∫ f(x)cos(nπx/L) dx

Question 27

Odd Extension

Answer 27

f(x) = Σbnsin((nπx/L)
-where the sum is between 1 and infinity and:
bn = 2/L ∫ f(x)sin(nπx/L) dx