University of Florida/Egm4313/s12.team5.R7

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Report 7


R 7.1[edit]

Question[edit]

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



(5):




Solution[edit]

We know that:



We also know by definition that:



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



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



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




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



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



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



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



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

Author[edit]

This problem was solved and uploaded by: William Knapper

R 7.2[edit]

Question[edit]

Plot the truncated series
with n=5 and for
where

Solution[edit]

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:

File:R7prob2.jpg

Author[edit]

This problem was solved and uploaded by: Josh House

R 7.3[edit]

Question[edit]

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

for

and

for

Solution[edit]

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

The magnitude of a function is defined by:

The angle between 2 functions is defined by:

Problem 1:

Using integration by parts:

The integral becomes:  


 rads

Problem 2:

The integral becomes: 


 rads.

Author[edit]

This problem was solved and uploaded by: David Herrick

R 7.4[edit]

Question[edit]

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

problems 6,9: Sketch or graph f(x) for .

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

Problem 9: f(x) = x for

and f(x) = if

Solution[edit]

Problem 6: f(x) = |x|
File:Problem 6.png
MATLAB code used:
File:Problem 6 mcode.png

Problem 9:
if
if
File:Problem 9.png
MATLAB code used:
File:Problem 9 mcode.png

Problem 12:
f(x) = |x| breaks down into:
if and
if
The formula for the fourier series is:

The euler formulas are also:



Therefore:



After integration for and integration by parts for and :



Similarly, for n = even, and for n = odd,
Therefore, the fourier series out to n = 5 is:




Problem 13:
if
if
The euler formulas are the same as in problem 12. Plugging in problem 13's f(x):



Therefore, after using integration for and integration by parts for and :



Therefore, when n is odd, and .
Similarly, when n is even, .
Therefore, the fourier series out to n = 5 is:


Author[edit]

The problem was solved and uploaded by John North.

R 7.5[edit]

Question[edit]

Consider (1)p.12-4:

where

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[edit]

Part 1[edit]

Using (3)p19-10 we get:


Integrating we get:


Plugging in the known values we have:


Solving from 0 to :


We know that:
if C is any integer


Thus, if is an an integer, the solution to the equation becomes:


Therefore:




Part 2[edit]

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



was found from 0 to . The assumption of being an integer and in this case = 1 was made.


The following was the outcome in MATLAB:



File:7.5part2.jpg



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

Therefore through MATLAB's TRAPZ method,



Author[edit]

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[edit]

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