University of Florida/Egm4313/s12.team5.R6

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Report 6


R 6.1[edit]

Question[edit]

Find the (smallest) period of and .
Show that these functions also have a period p.
Show that the constant is also a periodic function with period p.

Solution[edit]

Given (1) p.9-2: for all and where p is the period.
The smallest period occurs when n=1 thus we can use the equation as:

For we have:

Thus,



, smallest period occurs when n=1 so, 



For we have:
< br /> Thus,



, smallest period occurs when n=1 so,  



From (1) p9-5, we know that
Then,
Thus, the period of both of these functions is also .



From (1) p.9-7, we know
Also,

Thus,  showing   is a periodic function with period . 

Author[edit]

This problem is solved and uploaded by Radina Dikova

R 6.2[edit]

Question[edit]

Is the given function even or odd or either even nor odd? Find its Fourier series. Show details of your work.



Author[edit]

Solved and uploaded by William Knapper

R 6.3[edit]

Question[edit]

K 2011 p.491 pbs 15,17
Find if the graph is for an even or odd problem and then find the fourier series. Graph the resulting equations.
Problem 15:
File:Problem 15.png
Problem 17:
File:Problem 17.png

Solution[edit]

For problem 15:
f(-x) = -f(x), therefore the graph shows an odd function. Similarly, L = .
Therefore we can use the Euler formula for an odd function:


During , f(x) = x. During ,
Therefore,
Using integration by parts shows to be:

when, for the first integral:

and for the second integral (where u, du, v, and dv are not the same as the first integral):

Therefore, when n is odd, and when n is even,
Therefore, the fourier series' are as follows:
For n=2 ->
For n=4 ->
For n = 8 ->
A graph of the fourier series' is shown below:
File:Graph 15.png

For Problem 17: Therefore, it is an even function and L = 1. Similarly, because it's an even function, the Euler equations for an even function can be used:



During [0,1],
Therefore:

However, the period is not 2, so in is replaced with to shift the function to a period of 2. Similarly, is replaced with since L = 1.
Therefore:

Letting where u, du, v, and dv are not the same as was used in number 15 above.
Integration by parts then yields:

and when n is even,
Therefore, the fourier series's are as follows:
For n = 2 ->
For n = 4 ->
For n = 8 ->

Author[edit]

This problem was solved and uploaded by John North

R 6.4[edit]

Question[edit]

Consider the L2-ODE-CC with the window function f(x) from p9-8 as excitation:

where r(x) = f(x)

and the initial conditions

1. Find such that:

with the same initial conditions as above.

Plot for n = 3, 6, 9 for x in [0, 10]

2. Use the matlab command ode45 to integrate the L2-ODE-CC and plot the numerical soln to compare with the analytical soln.

Level 1: n = 0,1

Solution[edit]

The Fourier series of a periodic function

cos(nωx) + sin(nωx)]

For an odd function the Fourier series will be the following:

sin(nωx)


The independent variable t will be used to shift x to the left

The period of oscillation and frequency of oscillation will be as follows:

ω = π/2

where

which comes to be: (1 - cosnπ)


= 1/2 + (2/nπ) sin(nπ/2}t

The homogeneous equation for Y will be as follows:

Author[edit]

This problem was solved and uploaded by Mike Wallace

R 6.5[edit]

Question[edit]

Redo R4.2 to redisplay the particular solution, the homogenous solution, and the exact solution for n = 3,5,9 over the interval [0,20π]

Redisplay the particular solution, the homogenous solution, and the exact solution. Superpose each solution with the exact solution.

Redo R4.3 with the TA code over the interval [0.10]. Zoom in about x = -0.5, 0, and 0.5 and comment on the accuracy of different approximations.

Redo R4.4 with the TA code over the interval [0.9,10] for n = 4, 7. Zoom in about x = 1, 1.5, 2, 2.5 and comment on the accuracy of different approximations.

Solution[edit]

R 4.2 redo[edit]

The re-displayed functions for the homogenous solutions are:

The re-displayed functions for the particular solutions are:

The re-displayed functions for the general solutions are:

Plot for

File:Part1a.png

Plot for

File:Part1b.png

Plot for

File:Part1c.png

R 4.3 Redo[edit]

Matlab code:

x = 0:0.01:10;

y = log(1+x);

EDU>> x1 = 0:0.01:10;

EDU>> y1 = zeros(1,1001);

EDU>> for i = 1:4

for j = 1:1001

y1(j) = y1(j) - ((-x1(j))^i)/i;

end

end

EDU>> y2 = zeros(1,1001);

EDU>> for i = 1:7

for j = 1:1001

y2(j) = y2(j) - ((-x1(j))^i)/i;

end

end

EDU>> y3 = zeros(1,1001);

EDU>> for i = 1:11

for j = 1:1001

y3(j) = y3(j) - ((-x1(j))^i)/i;

end

end

EDU>> y4 = zeros(1,1001);

EDU>> for i = 1:16

for j = 1:1001

y4(j) = y4(j) - ((-x1(j))^i)/i;

end

end

EDU>> h = plot(x,y);

orange = [1 0.5 0.2];

EDU>> set(h,'Color',orange);

EDU>> hold on;

EDU>> plot(x1,y1,'r');

EDU>> plot(x1,y2,'g');

EDU>> plot(x1,y3,'b');

EDU>> plot(x1,y4,'c');

legend('log(1+x)','T_4','T_7','T_1_1','T_1_6');

EDU>> grid on;

EDU>> axis([0 10 -10 10])

Captain general of the Air Force 2a.png

Zoom plot about 0:

File:2B.png

This plot makes it appear that only the 16 term approximation is in the window at point 0. This is, therefore, the most accurate of the other n solution approximations.

Zoom plot about 0.5:

File:2C.png

This plot makes it appear that most of the approximations are very accurate and close to the exact solution. The only approximation that appears to deviate a little after 0.5 is the n = 4 approximation.

R 4.4 redo[edit]

Matlab code:

syms x

EDU>> f = log(x+1);

EDU>> fT1 = taylor(f,5,1);

EDU>> fT2 = taylor(f,8,1);

X = 0.9:0.1:10;

Y(:,1) = subs(fT1,'x',X);

EDU>> Y(:,2) = subs(fT2,'x',X);

EDU>> Y(:,3) = log(1+X);

EDU>> figure

EDU>> plot(X,Y);

axis([0.9 10 -10 10])

Captain general of the Air Force 3a.png

Zoom in about 1:

File:3B.png

This shows that neither of the approximations are very close about x = 1 to the exact solution.

Zoom in about 1.5:

File:3C.png

This shows that the n = 7 approximation is close about x = 1.5 to the exact solution.

Zoom in about 2:

3d.png

This shows that all the approximations are very close to the exact solution about x = 2.

Zoom in about 2.5:

File:3e.png

This shows that the approximations are very close about x = 2.5, however, the n = 4 approximation is beginning to deviate and soon will not be a good approximation.

Author[edit]

This problem was solved and uploaded by David Herrick

R 6.6[edit]

Question[edit]

Given

(1) Simplify the first term on the lhs
(2) Simplify the second term and combine with the simplified first term
(3) Finally, add the third term

Solution[edit]

From Lecture Notes Sec.10 p.10-3:
Particular solution is of the form


From Lecture Notes Sec.10 p.10-3:


We want to substitute these derivatives back into the original ODE and verify that Wolfram Alpha's solution of is correct.


(1)


(2)


(3)



Comparing this to Wolfram Alpha's answer:



Author[edit]

Solved and uploaded by Joshua House

R 6.7[edit]

Question[edit]

(1) Find the separated ODEs for the heat equation

Solution[edit]

Assuming


Then:




Substituting partial derivatives back into original PDE:




* Where "c" is a constant, because if both sides were variables then they would never be equal one another (each side would be a function of a different variable.) Kreysig 2011, pp.546


Multiplying by the denominators to get two separate ODEs:




Author[edit]

Solved and uploaded by Joshua House

Contribution Summary[edit]

Problem 1 was solved and uploaded by: Radina Dikova

Problem 2 was solved and uploaded by: William Knapper

Problem 3 was solved and uploaded by: John North

Problem 4 was solved and uploaded by: Michael Wallace

Problem 5 was solved and uploaded by: David Herrick

Problem 6 was solved and uploaded by: Joshua House

Problem 7 was solved and uploaded by: Joshua House