University of Florida/Egm4313/s12.team5.R7

Report 7

R 7.1

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Question

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

 

(5):

 

Solution

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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

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This problem was solved and uploaded by: William Knapper

R 7.2

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Question

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Plot the truncated series  
with n=5 and for  
where  

Solution

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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

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This problem was solved and uploaded by: Josh House

R 7.3

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Question

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

  for  

and

  for  

Solution

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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

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This problem was solved and uploaded by: David Herrick

R 7.4

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Question

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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

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Problem 6: f(x) = |x|


MATLAB code used:



Problem 9:
  if  
  if  


MATLAB code used:



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

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The problem was solved and uploaded by John North.

R 7.5

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Question

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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

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Part 1

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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

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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:

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

Therefore through MATLAB's TRAPZ method,

 

Author

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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

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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