University of Florida/Egm4313/s12.team11.R5

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Report 5

Intermediate Engineering Analysis
Section 7566
Team 11
Due date: March 30, 2012.

R5.1[edit | edit source]

Problem Statement[edit | edit source]

Given: Find for the following series:

1.
2.

Find for the Taylor series of

3. at

4. at

5. at

Solution[edit | edit source]

The radius of convergence is defined as



1.




                                                          

2.





However, in this problem, the series term is not , as is the general form.
Therefore, this implies:



                                                        


3. The Taylor series for is expressed as









Therefore:

                                                           

4. The Taylor series for at is expressed as











                                                             



5. The Taylor series for at is expressed as











For convergence:





Therefore,

                                                             



R5.2[edit | edit source]

Solved by: Andrea Vargas

Problem Statement[edit | edit source]

Part 1:Determine whether the following are linearly independent using the Wronskian

Part 2: Determine whether the following are linearly independent using the Gramian

Solution[edit | edit source]

Part 1[edit | edit source]

Using the Wronskian we check for linear independence.

We know from (1) and (2) in 7-35 that if

Then the functions are linearly independent.

Part 1.1[edit | edit source]




Taking the derivatives of each function:





                                           f(x) and g(x) are linearly independent
Part 1.2[edit | edit source]




Taking the derivatives of each function:





                                            f(x) and g(x) are linearly independent

Part 2[edit | edit source]

Using the Gramian we check for linear independence.
We know from the notes in (1) 7-34 that:

and that the Gramian is defined as:

Then f,g are linearly independent if

Part 2.1[edit | edit source]




Taking scalar products:




                                           f(x) and g(x) are linearly independent
Part 2.2[edit | edit source]




Taking scalar products:

We can use the trig identity for power reduction
Then we have,












                                           f(x) and g(x) are linearly independent

Conclusion[edit | edit source]

By both methods (the Wronskian and the Gramian) we obtain the same results.

R5.3[edit | edit source]

Problem Statement[edit | edit source]

Verify using the Gramian that the following two vectors are linearly independent.


Solution[edit | edit source]


We know from (3) 8-9 that:

We obtain,


Then,

                                            b_1 and b_2 are linearly independent

R5.4[edit | edit source]

Problem Statement[edit | edit source]

Show that is indeed the overall particular solution of the L2-ODE-VC with the excitation .

Discuss the choice of in the above table e.g., for why would you need to have both and in  ?

Solution[edit | edit source]

Because the ODE is a linear equation in y and its derivatives with respect to x, the superposition principle can be applied:

is a specific excitation with known form of and is a specific excitation with known form of

becomes

proving that

is indeed the overall particular solution of the L2-ODE-VC with the excitation

According to Fourier Theorem periodic functions can be represented as infinite series in terms of cosines and sines:

where the coefficients are the Fourier coefficients calculated using Euler formulas.

So even though the system is being excited by functions like the particular solution would still include both and in because the excitation is a periodic function that can be represented as the Fourier infinite series in terms of both and times the Fourier coefficients

R5.5[edit | edit source]

Part 1[edit | edit source]

Problem Statement[edit | edit source]

Show that and are linearly independant using the Wronskian and the Gramain (integrate over 1 period)

Solution[edit | edit source]


One period of
Wronskian of f and g

Plugging in values for



 They are linearly Independant using the Wronskian.







 They are linearly Independent using the Gramain.

Problem Statement[edit | edit source]

Find 2 equations for the 2 unknowns M,N and solve for M,N.

Solution[edit | edit source]




Plugging these values into the equation given () yields;

Simplifying and the equating the coefficients relating sin and cos results in;


Solving for M and N results in;

  

Problem Statement[edit | edit source]

Find the overall solution that corresponds to the initial conditions . Plot over three periods.

Solution[edit | edit source]

From before, one period so therefore, three periods is
Using the roots given in the notes , the homogenous solution becomes;

Using initial condtion ;


with

Solving for the constants;


Using the found in the last part;

 

R5 code.jpg

R5 plot.jpg

R5.6[edit | edit source]

solved by Luca Imponenti

Problem Statement[edit | edit source]

Complete the solution to the following problem

where

and

Find the overall solution corresponds to the initial condition:

Plot the solution over 3 periods.

Particular Solution[edit | edit source]

Taking the derivatives of the particular solution

Plugging these into the ODE yields

Equating like terms allows us to solve for M and N

So the particular solution is

Overall Solution[edit | edit source]

The overall solution in the sum of the homogeneous and particular solutions

To find A and B we apply the initial conditions

Taking the derivative

Giving us the overall solution

   

Plot[edit | edit source]

The period for is

Plotting the solution over 3 periods yields

Plot56.jpg

R5.7[edit | edit source]

Solved by Daniel Suh

Problem Statement[edit | edit source]





1. Find the components using the Gram matrix.
2. Verify the result by using and , and rely on the non-zero determinant matrix of and relative to the bases of and .

Part 1 Solution[edit | edit source]

Gram Matrix[edit | edit source]




Thus,






Defining c[edit | edit source]

Define:

If , then exists


Finding c[edit | edit source]

thus, exists



                                                     

Part 2 Solution[edit | edit source]






                                                      solution is correct

R5.8[edit | edit source]

Problem Statement[edit | edit source]

Find the integral

for and

Using integration by parts, and then with the help of of

General Binomial Theorem



Solution[edit | edit source]

For :



For substitution by parts,







Therefore:

                                     

Using the General Binomial Theorem:



Therefore:

Which we have previously found that answer as:

                                     




For :



Initially we use the following substitutions:



First let us consider the first term:

Next, we use the integration by parts:






Next let us consider the second term:

Again, we will use integration by parts:






Therefore:





Re-substituting for t:







Therefore:

                                     



Using the General Binomial Theorem for the integral with t substitution :



Therefore:

Which we have previously found that answer as:

                                     

R5.9[edit | edit source]

Solved by: Gonzalo Perez

Problem Statement[edit | edit source]

Consider the L2-ODE-CC (5) p.7b-7 with as excitation:

(5) p.7b-7

(1) p.7c-28

and the initial conditions

.

Part 1[edit | edit source]

Part A[edit | edit source]

Project the excitation on the polynomial basis

(1)

i.e., find such that

(2)

for x in , and for n = 3, 6, 9.

Plot and to show uniform approximation and convergence.

Note that:

(3)

Solution[edit | edit source]

To solve this problem, it is important to know that the scalar product is defined as the following:

.

Therefore, it follows that:

, where and .

We know that if are linearly independent, then by theorem on p.7c-37, the matrix is solvable.

According to this and (3)p.8-14:

If exists . (3)p.8-14

Now let's define the Gram matrix as a function of :

(1)p.8-13

Defining the "d" matrix as was done in (3)p.8-13, we get:

. (3)p.8-13

And according to (1)p.8-15: (1)p.8-15

Now, we can find the values to compare to .

Using Matlab, this is the code that was used to produce the results:

The Matlab code above produced the following graph:

Where is represented by the dashed line and the approximation,, is represented by the red line. This code can work for all n values.

Part B[edit | edit source]

In a seperate series of plots, compare the approximation of the function by Taylor series expansion about .

Where:

Solution[edit | edit source]

For n=1:

For n=2:

For n=3:

For n=4:

For n=5:

For n=6:

For n=7:

For n=8:

For n=9:

For n=10:

For n=11:

For n=12:

For n=13:

For n=14:

For n=15:

For n=16:

Using Matlab to plot the graph:

Part 2[edit | edit source]

Find such that:

(1) p.7c-27

with the same initial conditions as in (2) p.7c-28.

Plot for n = 3, 6, 9, for x in .

In a series of separate plots, compare the results obtained with the projected excitation on polynomial basis to those with truncated Taylor series of the excitation. Plot also the numerical solution as a baseline for comparison.

Solution[edit | edit source]

First, we find the homogeneous solution to the ODE:
The characteristic equation is:


Then,
Therefore the homogeneous solution is:


Now to find the particulate solution

For n=3:

We can then use a matrix to organize the known coefficients:



Then, using MATLAB and the backlash operator we can solve for these unknowns:
4 3n4.PNG
Therefore


Superposing the homogeneous and particulate solution we get


Differentiating:
Evaluating at the initial conditions:



We obtain:



Finally we have:


For n=6:

We can then use a matrix to organize the known coefficients:



Then, using MATLAB and the backlash operator we can solve for these unknowns:
4 3n7.PNG
Therefore


Superposing the homogeneous and particulate solution we get


Differentiating:
Evaluating at the initial conditions:



We obtain:



Finally


For n=9:


We can then use a matrix to organize the known coefficients:
N11.PNG

Then, using MATLAB and the backlash operator we can solve for these unknowns:
4 3n11.png
Therefore



Superposing the homogeneous and particulate solution we get



Differentiating:

Evaluating at the initial conditions:



We obtain:



Finally



Here is the graph for this problem using Matlab: