MATH528

MATH528Due date: 12/05/18

Homework 12

1Exercises

Exercise. PS11.4.11

Solution. The Fourier series of the periodic rectangular wave function $f (x) = 2 k (u (π x) - . 5)$ , for $x \in [- π, π)$ , $f (x) = f (x + 2 π)$ , with $u (x)$ the unit step function is (Example 11.1.1, p. 478)

f (x) = \frac{4 k}{π} (\sin x + \frac{1}{3} \sin 3 x + \frac{1}{5} \sin 3 x + \dots) = \sum_{m = 1}^{\infty} b_{2 m - 1} \sin [(2 m - 1) x], b_{2 m - 1} = \frac{4 k}{π (2 m - 1)}

The Parseval equality states

\sum_{m = 1}^{\infty} b_{2 m - 1}^{2} = {(\frac{4 k}{π})}^{2} [1 + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \dots] = {|| f ||}^{2} = \int_{- π}^{π} f {(x)}^{2} d x = 2 π k^{2} \Rightarrow

1 + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \dots = \frac{π^{2}}{8}

In[3]:=

S[n_]:=Sum[1/(2m-1)^2,{m,1,n}]; Sinf=N[Pi^2/8.,16]

$1.2337$

In[11]:=

Table[N[S[n]],{n,10,100,10}]

${1.20872, 1.2212, 1.22537, 1.22745, 1.2287, 1.22953, 1.23013, 1.23058, 1.23092, 1.2312}$

In[13]:=

Table[Log10[Abs[N[(S[n]-Sinf)/Sinf]]],{n,10,100,10}]

${- 1.69363, - 1.99439, - 2.17043, - 2.29535, - 2.39225, - 2.47143, - 2.53838, - 2.59637, - 2.64752, - 2.69327}$

In[14]:=

Convergence is not rapid due to discontinuity in $f (x)$ at $x = \pm k π$

Exercise. PS11.5.7

Solution. The problem $y^{''} + λ y = 0$ , $y (0) = 0$ , $y (10) = 0$ is of Sturm-Liouville type, with the scalar product

(f, g) = \int_{0}^{10} f (x) g (x) d x .

For $λ = 0$ , applying boundary conditions to the general solution $y (x) = a x + b$ , $y (0) = b = 10$ , $y (10) = 10 a = 0 \Rightarrow a = 0$ , hence $y (x) = 0$ which is not an eigenfunction. For $λ > 0$ , applying boundary conditions to the general solution $y (x) = a \cos (\sqrt{λ} x) + b \sin (\sqrt{λ} x)$ , $y (0) = a = 0$ , $y (10) = b \sin (\sqrt{λ} 10) = 0 \Rightarrow \sqrt{λ} 10 = k π \Rightarrow λ_{k} = {(k π / 10)}^{2}$ are eigenvalues, with associated eigenfunctions $y_{k} (x) = \sin (k π x / \sqrt{10})$ . For $λ < 0$ , applying boundary conditions to the general solution $y (x) = a e^{\sqrt{λ} x} + b e^{- \sqrt{λ} x}$ , $y (0) = a + b = 0$ , $y (10) = a C + b / C = 0$ , leads to a homogeneous linear system with principal determinant

Δ = | \begin{array}{ll} 1 & 1 \\ C & 1 / C \end{array} | = \frac{1}{C} - C

that would have to be null in order to obtain a non-trivial solution. This occurs for $C = 1 = e^{\sqrt{λ} 10} \Rightarrow λ = 0$ , a contradiction. Hence the only eigenvalue, eigenfunction pairs are

λ_{k} = {(k π / 10)}^{2}, y_{k} (x) = \sin (k π x / \sqrt{10}) .

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Exercise. PS1.1.7

Solution.

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Exercise. PS1.1.8

Solution.

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Exercise. PS1.2.2

Solution.

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Exercise. PS1.2.3

Solution.

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Exercise. PS1.2.4

Solution.

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Exercise. PS1.3.5

Solution.

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2Problems

Problem. PS11.6.1

Solution. The function $f (x) = 63 x^{5} - 90 x^{3} + 35 x$ , has the Fourier-Legendre series expansion

f (x) = \sum_{m = 0}^{\infty} a_{m} P_{m} (x),

with

a_{m} = \frac{2 m + 1}{2} \int_{- 1}^{1} f (x) P_{m} (x) d x,

and $P_{m} (x)$ the Legendre polynomials

In[16]:=

p=Table[LegendreP[m,x],{m,0,5}]

${1, x, \frac{1}{2} (3 x^{2} - 1), \frac{1}{2} (5 x^{3} - 3 x), \frac{1}{8} (35 x^{4} - 30 x^{2} + 3), \frac{1}{8} (63 x^{5} - 70 x^{3} + 15 x)}$

In[17]:=

Since $f (x)$ has only odd powers less than $m = 5$ , the expansion is

f (x) = a_{1} P_{1} (x) + a_{3} P_{3} (x) + a_{5} P_{5} (x) .

One can compute $a_{m}$ directly, but simple observations lead to a quicker result

a_{5} = 8, f (x) - a_{5} P_{5} (x) =

In[17]:=

f[x_]=63 x^5 - 90 x^3 + 35 x;

Table[ (2m+1)/2 Integrate[f[x] LegendreP[m,x],{x,-1,1}],{m,1,5}]

${8, 0, - 8, 0, 8}$

In[18]:=

f[x]-8 LegendreP[5,x]

$20 x - 20 x^{3}$

In[20]:=

Expand[f[x]- 8 LegendreP[5,x] + 8 LegendreP[3,x]]

$8 x$

In[21]:=

Problem. PS1.1.17

Solution.

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Problem. PS1.1.18

Solution.

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Problem. PS1.3.22

Solution.

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1Exercises

2Problems

3Projects

3.1PS2.1.16