University of Florida/Egm4313/s12.team5.R7

R 7.1

Question

Verify (4)-(5) p.19-9.
(4):

$<{\mathrm{\Phi }}_i,{\mathrm{\Phi }}_j>=0,i\ne j$

(5):

$<{\mathrm{\Phi }}_i,{\mathrm{\Phi }}_j>=\frac{L}{2},i=j$

Solution

We know that:

${\mathrm{\Phi }}_j(x):=sin{\omega }_{j}x,j=1,2,...$

We also know by definition that:

$⟨\overline{f},\overline{g}⟩:=\underset{0}{\overset{L}{\int }}\overline{f}(x)\overline{g}(x)dx$

Therefore, we start with Eq. (4) and use the above definitions to obtain the following results:

$<{\mathrm{\Phi }}_i,{\mathrm{\Phi }}_j>=\underset{0}{\overset{L}{\int }}sin({\omega }_{i}x)sin({\omega }_{j}x)dx$

For simplicity's sake, we set L equal to $\pi$. This yields the following result:

$\underset{0}{\overset{\pi }{\int }}sin({\omega }_{i}x)sin({\omega }_{j}x)dx=\frac{({\omega }_i-{\omega }_j)sin(({\omega }_i+{\omega }_j)x)+(-{\omega }_i-{\omega }_j)sin(({\omega }_i-{\omega }_j)x)}{2{\omega }_i^2-2{\omega }_j^2}$

This last expression is evaluated from $\pi$ to zero. We know as a geometric property that the following two things are true:

$sin(C\pi )=0$
$sin(0)=0$

Where "C" is any positive integer. We also know that since L = $\pi$, p = 2L = $2\pi$, and $\omega =\frac{2\pi }{p}=\frac{2\pi }{2\pi }=1$, so therefore every sine term in the definite integral is 0 when both $\pi$ and 0 are evaluated. Therefore, we know that:

$<{\mathrm{\Phi }}_i,{\mathrm{\Phi }}_j>=0,i\ne j$

Now, for Eq. (5), we evaluate is as follows:

$<{\mathrm{\Phi }}_j,{\mathrm{\Phi }}_j>=\underset{0}{\overset{L}{\int }}sin({\omega }_{j}x{)}^{2}dx$

Again, we set L equal to $\pi$ to make the expression easier to evaluate. This yields:

$\underset{0}{\overset{\pi }{\int }}sin({\omega }_{j}x{)}^{2}dx=-(\frac{sin(2{\omega }_{j}x)-2{\omega }_{j}x}{4{\omega }_j})$

This integral is evaluated from $\pi$ to 0. Again, we know that the sine terms will be zero no matter which boundary we plug again, since $\omega =1$. Therefore, the evaluation simplifies to:

$-(-\frac{2{\omega }_j\pi }{4{\omega }_j}-0)=\frac{\pi }{2}$

Since we set L equal to $\pi$, we have thus proven Eq. (5).

Author

This problem was solved and uploaded by: William Knapper

R 7.2

Question

Plot the truncated series $u(x,t)=\sum _{j=1}^n{a}_{j}cos(c{w}_{j}t)sin({\omega }_{j}x)$

with n=5 and for $t=\alpha \frac{2L}{c}$

where $\alpha =0.5,1,1.5,2$

Solution

Matlab Code:

% Report 7 Problem 2
% Plot truncated series for n=5
% Given values
L = 2;
c = 3;
% Calculate alphas for use in t
alpha = (0.5:.5:2); % four alpha terms
t = ((alpha.*2*L)/c); % four "t" terms corresponding to the four alpha terms
% Calculate omegas (w)
j = (1:1:5); % series terms 1-5
w = ((j*pi)/L); % omega correspoding to series terms 1-5
% Calculate "a" coefficients
% Check if "a" is odd or even, then assign value
a = j;
for k=1:5;
if mod(k,2) == 0 % even term
   a(k) = 0;
else % odd term
   a(k) = -4/(pi^3*k^3);
end
end
% Calculate truncated series (note that a2=a4=0, so those terms in the
% series have no contribution
x=(0:0.01:2);
% For t=0.5
y1=zeros(size(x));
for k=1:5
   y1 = y1 + a(k)*cos(c*w(k)*t(1))*sin(w(k)*x);
end
% For t=1.0
y2=zeros(size(x));
for k=1:5
   y2 = y2 + a(k)*cos(c*w(k)*t(2))*sin(w(k)*x);
end
% For t=1.5
y3=zeros(size(x));
for k=1:5
  y3 = y3 + a(k)*cos(c*w(k)*t(3))*sin(w(k)*x);
end
% For t=2.0
y4=zeros(size(x));
for k=1:5
   y4 = y4 + a(k)*cos(c*w(k)*t(4))*sin(w(k)*x);
end
% Compare to original function
yorg = x.*(x-2);
plot(x,y1,x,y2,x,y3,x,y4,x,yorg)
legend('t=0.5','t=1.0','t=1.5','t=2.0','org function')

Plot of 4 different t's and original function:

Author

This problem was solved and uploaded by: Josh House

R 7.3

Question

Find the a) scalar product, b) the magnitude, and c) the angle between:

$1.f(x)=\cos (x),g(x)=x$ for $-2\le x\le 10$

and

$2.f(x)=\frac{1}{2}(3x^2-1),g(x)=\frac{1}{2}(5x^3-3x)$ for $-1\le x\le 1$

Solution

The scalar product between 2 functions on the interval [a,b] is defined by:

$<f,g>=\underset{a}{\overset{b}{\int }}f(x)g(x)dx$

The magnitude of a function is defined by:

$\Vert f\Vert =<f,f{>}^{\frac{1}{2}}={[\underset{a}{\overset{b}{\int }}{f}^2(x)dx]}^{\frac{1}{2}}$

The angle between 2 functions is defined by:

$\cos (\theta )=\frac{<f,g>}{\Vert f\Vert \Vert g\Vert }$

Problem 1:

$<f,g>=\underset{-2}{\overset{10}{\int }}x\cos (x)dx$

Using integration by parts: $u=x,du=dx,dV=\cos (x)dx,V=\sin (x)$

The integral becomes:  $x\sin (x)-\int \sin (x)dx=x\sin (x)+\cos (x)\underset{-2}{\overset{10}{\mid }}=-7.68$

$\Vert f\Vert =[\underset{-2}{\overset{10}{\int }}{\cos }^2(x)dx{]}^{\frac{1}{2}}=[\underset{-2}{\overset{10}{\int }}\frac{1+\cos (2x)}{2}dx{]}^{\frac{1}{2}}=[\frac{x}{2}+\frac{\sin (2x)}{4}\underset{-2}{\overset{10}{\mid }}{]}^{\frac{1}{2}}=2.46$

$\Vert g\Vert =[\underset{-2}{\overset{10}{\int }}{x}^{2}dx{]}^{\frac{1}{2}}=[\frac{x^3}{3}\underset{-2}{\overset{10}{\mid }}{]}^{\frac{1}{2}}=18.33$

$\cos (\theta )=\frac{-7.68}{18.33*2.46}\Rightarrow \theta =1.742$ rads

Problem 2:

$<f,g>=\underset{-1}{\overset{1}{\int }}\frac{1}{2}(3x^2-1)*\frac{1}{2}(5x^3-3x)dx=\underset{-1}{\overset{1}{\int }}(\frac{15}{4}{x}^5-\frac{7}{2}{x}^3+\frac{3}{4}x)dx$

The integral becomes: $\frac{5}{8}{x}^6-\frac{7}{8}{x}^4+\frac{3}{8}{x}^2\underset{-1}{\overset{1}{\mid }}=0$

$\Vert f\Vert =[\underset{-1}{\overset{1}{\int }}[\frac{1}{2}(3x^2-1){]}^{2}dx{]}^{\frac{1}{2}}=[\underset{-1}{\overset{1}{\int }}(\frac{9}{4}{x}^4-\frac{3}{2}{x}^2+\frac{1}{4})dx{]}^{\frac{1}{2}}=[\frac{9}{20}{x}^5-\frac{1}{2}{x}^3+\frac{1}{4}x\underset{-1}{\overset{1}{\mid }}{]}^{\frac{1}{2}}=0.632$

$\Vert g\Vert =[\underset{-1}{\overset{1}{\int }}[\frac{1}{2}(5x^3-3x){]}^{2}dx{]}^{\frac{1}{2}}=[\underset{-1}{\overset{1}{\int }}(\frac{25}{4}{x}^6-\frac{15}{2}{x}^4+\frac{9}{4}{x}^2)dx{]}^{\frac{1}{2}}=[\frac{25}{28}{x}^7-\frac{3}{2}{x}^5+\frac{3}{4}{x}^3\underset{-1}{\overset{1}{\mid }}{]}^{\frac{1}{2}}=0.535$

$\cos (\theta )=0\Rightarrow \theta =\pi /2$ rads.

Author

This problem was solved and uploaded by: David Herrick

R 7.4

Question

K. 2011 p482 pb. 6,9,12,13:

problems 6,9: Sketch or graph f(x) for $-\pi <x<\pi$.

problems 12,13: Find the fourier series for the f(x) in problems 6 and 9 respectively up to n = 5.

Problem 6: f(x) = |x| for $-\pi <x<\pi$

Problem 9: f(x) = x for $-\pi <x<0$

and f(x) = $\pi -x$ if $0<x<\pi$

Solution

Problem 6: f(x) = |x|

MATLAB code used:

Problem 9:
$f(x)=x$ if $-\pi <x<0$
$f(x)=\pi -x$ if $0<x<\pi$

MATLAB code used:

Problem 12:
f(x) = |x| breaks down into:
$f(x)=-x$ if $-\pi <x<0$ and
$f(x)=x$ if $0<x<\pi$
The formula for the fourier series is:
$f(x)=a_0+\sum _{n=1}^{\mathrm{\infty }}[a_n\cos (nx)+b_n\sin (nx)]$
The euler formulas are also:
$a_0=\frac{1}{2\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)dx$
$a_n=\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)\cos (nx)dx$
$b_n=\frac{1}{\pi }\underset{-\pi }{\overset{\pi }{\int }}f(x)\sin (nx)dx$
Therefore:
$a_0=\frac{1}{2\pi }[\underset{-\pi }{\overset{0}{\int }}-xdx+\underset{0}{\overset{\pi }{\int }}xdx]$
$a_n=\frac{1}{\pi }[\underset{-\pi }{\overset{0}{\int }}-x\cos (nx)dx+\underset{0}{\overset{\pi }{\int }}x\cos (nx)dx]$
$b_n=\frac{1}{\pi }[\underset{-\pi }{\overset{0}{\int }}-x\sin (nx)dx+\underset{0}{\overset{\pi }{\int }}x\sin (nx)dx]$
After integration for $a_0$ and integration by parts for $a_n$ and $b_n$:
$a_0=\frac{\pi }{2}$
$a_n=\frac{2}{\pi n^2}(\cos (-n\pi )-1)$
$b_n=0$
Similarly, for n = even, $a_n=0$ and for n = odd, $a_n=\frac{-4}{n^2\pi }$
Therefore, the fourier series out to n = 5 is:

$f_5(x)=\frac{\pi }{2}-\frac{4}{\pi }\cos (x)-\frac{4}{9\pi }\cos (3x)-\frac{4}{25\pi }\cos (5x)$

Problem 13:
$f(x)=x$ if $-\pi <x<0$
$f(x)=\pi -x$ if $0<x<\pi$
The euler formulas are the same as in problem 12. Plugging in problem 13's f(x):
$a_0=\frac{1}{2\pi }[\underset{-\pi }{\overset{0}{\int }}xdx+\underset{0}{\overset{\pi }{\int }}(\pi -x)dx]$
$a_n=\frac{1}{\pi }[\underset{-\pi }{\overset{0}{\int }}x\cos (nx)dx+\underset{0}{\overset{\pi }{\int }}(\pi -x)\cos (nx)dx]$
$b_n=\frac{1}{\pi }[\underset{-\pi }{\overset{0}{\int }}x\sin (nx)dx+\underset{0}{\overset{\pi }{\int }}(\pi -x)\sin (nx)dx]$
Therefore, after using integration for $a_0$ and integration by parts for $a_n$ and $b_n$:
$a_0=0$
$a_n=\frac{2}{n^2\pi }(1-\cos (n\pi ))$
$b_n=\frac{1}{n}(1-\cos (n\pi ))$
Therefore, when n is odd, $a_n=\frac{4}{n^2\pi }$ and $b_n=\frac{2}{n}$.
Similarly, when n is even, $a_n=0=b_n$.
Therefore, the fourier series out to n = 5 is:

$f_5(x)=\frac{4}{\pi }[\cos (x)+\frac{1}{9}\cos (3x)+\frac{1}{25}\cos (5x)]+2[\sin (x)+\frac{1}{3}\sin (3x)+\frac{1}{5}\sin (5x)]$

Author

The problem was solved and uploaded by John North.

R 7.5

Question

Consider (1)p.12-4:
$<{\mathrm{\Phi }}_{2j-1}{\mathrm{\Phi }}_{2k-1}>=\underset{0}{\overset{p}{\int }}\sin j\omega x\cdot \sin k\omega xdx$
where $p=2\pi ,j=2,k=3.$

1. Find the exact integration of (1)p.12-4 with the given data.
2. Confirm the result with matlab's trapz command for the trapezoidal rule as explained in (4)p.11-2.

Solution

Part 1

Using (3)p19-10 we get:
$\underset{0}{\overset{2\pi }{\int }}\sin j\omega x\cdot \sin k\omega xdx=\underset{0}{\overset{2\pi }{\int }}\frac{1}{2}[cos(j\omega -k\omega )x-cos(j\omega +k\omega )x]dx$

Integrating we get:
$\frac{sin((j\omega -k\omega )x)}{2(j\omega -k\omega )}-\frac{sin((j\omega +k\omega )x)}{2(j\omega +k\omega )}$

Plugging in the known values we have:
$\frac{sin((2\omega -3\omega )x)}{2(2\omega -3\omega )}-\frac{sin((2\omega +3\omega )x)}{2(2\omega +3\omega )}=\frac{sin(-1\omega x)}{-2\omega }-\frac{sin(5\omega x)}{10\omega }$

Solving from 0 to $2\pi$:
$[\frac{sin(-2\omega \pi )}{-2\omega }-\frac{sin(10\omega \pi )}{10\omega }]-[\frac{sin(0)}{-2\omega }-\frac{sin(0)}{10\omega }]$

We know that:
$sin(C\pi )=0$ if C is any integer
$sin(0)=0$

Thus, if $\omega$ is an an integer, the solution to the equation becomes:
$0-0=0$

Therefore:

$<{\mathrm{\Phi }}_3,{\mathrm{\Phi }}_5>=\underset{0}{\overset{2\pi }{\int }}\sin 2\omega x\cdot \sin 3\omega xdx=0$

Part 2

Using MATLAB's TRAPZ method, a variable X was created from 0 to $2\pi$ and the integral of

$\sin j\omega x\cdot \sin k\omega xdx$

was found from 0 to $2\pi$. The assumption of $\omega$ being an integer and in this case = 1 was made.

The following was the outcome in MATLAB:

-3.8317e-007 is used as an equivalent to 0.

Therefore through MATLAB's TRAPZ method,

$\underset{0}{\overset{2\pi }{\int }}\sin 2\omega x\cdot \sin 3\omega xdx=0$

Author

Part 1 of this problem was solved and uploaded by Radina Dikova.

Part 2 of this problem was solved and uploaded by Derik Bell.

Contribution Summary

Problem 1 was solved by: William Knapper

Problem 2 was solved by: Joshua House

Problem 3 was solved by: David Herrick

Problem 4 was solved by: John North

Problem 5 part 1 was solved by: Radina Dikova

Problem 5 part 2 was solved by: Derik Bell