University of Florida/Egm4313/s12.team11.R4

Report 4


Intermediate Engineering Analysis
Section 7566
Team 11
Due date: March 14, 2012.

R4.1Edit

Problem StatementEdit

Obtain equations (2), (3), (n-2), (n-1), (n), and set up the matrix A as in (1) p.7-21 for the general case, with the matrix coefficients for rows 1, 2, 3, (n-2), (n-1), n, filled in, as obtained from equations (1), (2), (3), (n-2), (n-1), (n).

GivenEdit

As shown in p.7-21, the first equation is:

  (1) p.7-21

According to p.7-20, the general form of the series is:

  (2) p. 7-20

From (2) p.7-20, we can obtain n+1 equations for n+1 unknown coefficients  .

After referring to p.7-22, it can be determined that the matrix to be set up is of the following form:

 

where the rows signify the coefficients  , and the columns signify  .

SolutionEdit

Building the coefficient matrix as shown in p.7-22 of the class notes, we can begin to solve for the coefficients as follows:

Equation associated with  :

j=0:   (1)

Equation associated with  :

j=1:   (2)

Equation associated with  :

j=2:   (3)

Equation associated with  :

j=n-2:   (n-2)

Equation associated with  :

j=n-1:   (n-1)

Equation associated with  :

j=n:   (n)

Using all of the above equations, (1), (2), (3), (n-2), (n-1), (n), we can then determine the A matrix to be:


                            
                 

Solved by Gonzalo Perez

R4.2Edit

Part 1Edit

Solved by Jonathan Sheider

Problem StatementEdit

Given:  

With initial conditions:  

Using the Taylor series for sin(x), reproduce the graph in the lecture notes p.7-24.

SolutionEdit

The Taylor series expansion for sin(x) that is plotted with n = 13 is as follows:
 

Using MatLab, the following Taylor series expansion was plotted and coded:

 

The plot created with n = 13 (i.e. 13 terms) is as follows:

 

Part 2Edit

Solved by Jonathan Sheider

Problem StatementEdit

Given:  

With initial conditions:  

Letting   equal the truncated Taylor series of  , i.e.  

Find the overall solution   for   and plot these solutions on the interval from  

SolutionEdit

First finding the homogenous solution to the ODE:

The characteristic equation:

 

 

Therefore,  

And the homogenous solution is:

 

Next the particular solution will be evaluated.

For n = 3:

The excitation is therefore:  

Therefore the particular solution has a form:

 

Differentiating:

 

 

Plugging these values into the ODE and equating them to the excitation:

 
 
 

 
 
 

To solve this, we will use an upper triangular matrix, and solve using Matlab. The matrix is as follows:

 

The answer is calculated in Matlab:
 

Therefore:  

The final solution is then found to be:  

Evaluating at the initial conditions, we have:

 

 

Solving this system of equations, we find:

 

Therefore the final solution is:

 

Plotting this equation in Matlab yields:

 


For n = 5:

The excitation is therefore:  

Therefore the particular solution has a form:

 

Differentiating:

 

 

Plugging these values into the ODE and equating them to the excitation, and then solving them by using a upper triangular matrix as before (just as in the n=3 example) in Matlab:
 

Therefore:  
 

The final solution is then found to be:  
 

Evaluating at the initial conditions, we have:

 

 

Solving this system of equations, we find:

 

Therefore the final solution is:

 
 

Plotting this equation in Matlab yields:

 



For n = 9:

The excitation is therefore:  

Therefore the particular solution has a form:

 
 

Differentiating:

 
 

 
 

Plugging these values into the ODE and equating them to the excitation, and then solving them by using an upper triangular matrix as before (just as in the n=3 example) in Matlab:
 

Therefore:  
 
 

The final solution is then found to be:  
 
 


Evaluating at the initial conditions, we have:

 

 

Solving this system of equations, we find:

 

Therefore the final solution is:

 
 
 

Plotting this equation in Matlab yields:

 

Note: On the interval from   to  , the resulting plots of the solutions, on such a large scale (note that the Y-axis is forced to be on the order of  !) each plot looks almost exactly identical. It is not until you are able to zoom in a lot do you see the very slight change in the curve of each plot on this interval.

--Egm4313.s12.team11.sheider (talk) 01:09, 14 March 2012 (UTC)

Part 3Edit

Solved by Daniel Suh

Problem StatementEdit

Using the particular solution from Table 2.1 on Kreyszig, find the overall solution  , and plot it compared with   for n = 3,5,9

 
 
Initial Conditions:  ,  

Homogenous SolutionEdit

To find  ,
 
 
 
 
 

Particular SolutionEdit

To find  ,
Using Table 2.1, we find that for  ,
 
 
 

Plugging it back into the equation,
 
 

Solving for coefficients,
 
 
 
 
 

Overall SolutionEdit

 
 

Solving with initial conditions,
 
 

 
 
 
 

 
 

PlotEdit

Plotting in matlab yields:
 
As shown on the graph, the two plots are exactly the same.

Created by [Daniel Suh] 20:43, 13 March 2012 (UTC)

R4.3Edit

Part 1Edit

Solved by Francisco Arrieta

Problem StatementEdit
Develop   in Taylor Series about   to reproduce the figure on page 7-25
GivenEdit
 
SolutionEdit

For n=4:

 

For n=7:

 

For n=11:

 

For n=16:

 
 

 

Using MATLAB

--Egm4313.s12.team11.arrieta (talk) 00:44, 14 March 2012 (UTC)

Part 2Edit

Solved by: --Egm4313.s12.team11.vargas.aa (talk) 01:28, 14 March 2012 (UTC)

Problem StatementEdit

Given:  

where  

With initial conditions:  

Find the overall solution   for   and plot these solutions on the interval from  

SolutionEdit

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=4

 

 

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

 

Then, using MATLAB and the backlash operator we can solve for these unknowns:
 
Therefore
 

Superposing the homogeneous and particulate solution we get
 

Differentiating:
  Evaluating at the initial conditions:
 
 

We obtain:
 
 

Finally we have:
 

For n=7

 

 

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

 

Then, using MATLAB and the backlash operator we can solve for these unknowns:
 
Therefore
 

Superposing the homogeneous and particulate solution we get
 

Differentiating:
  Evaluating at the initial conditions:
 
 

We obtain:
 
 

Finally
 

For n=11  

 

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

Then, using MATLAB and the backlash operator we can solve for these unknowns:
 
Therefore
 
 

Superposing the homogeneous and particulate solution we get
 
 

Differentiating:
   
Evaluating at the initial conditions:
 
 

We obtain:
 
 

Finally
 
 

Part 3Edit

Plot for part 2 and part 3. Please note that because of the broad scale, it is almost impossible to distinguish between the graphs for n=4,7,11 nad the ODE45 operator. Only when zoomed in is there any noticeable difference
 

R4.4Edit

Part 1Edit

Problem StatementEdit

Find n sufficiently high so that   do not differ from the numerical solution by more than   at  

SolutionEdit

Using a program in MATLAB that iteratively added terms onto the taylor series of  , terms were added until the error between the exact answer and the series was less than  .

 

It was found after trial and error that   for the error to be of a magnitude of  . This error found was 9.7422e-005

Similarly, for  .

 

It was found after trial and error that   for the error to be of a magnitude of  . This error found was 9.3967e-005

Part 2Edit

Problem StatementEdit

Develop   in Taylor series about   for   and plot these truncated series vs. the exact function.
What is now the domain of convergence by observation?

SolutionEdit

A MATLAB program was created, which calculated the Taylor series of each n value, along with the exact function, then plotted these together to show the comparison of all the series.
Below is the Taylor series for   expanded at  .
 

 
 
It can be seen by observation that the domain of convergence has shifted to the right one unit.

--egm4313.s12.team11.gooding (talk) 03:48, 14 March 2012 (UTC)

Part 3=Edit

Solved by Luca Imponenti

Find   , for   such that:

 

for   in   with the initial conditions found.

Plot   for   for   in  .

Homogeneous SolutionEdit

The homogeneous case is shown below:

 

This equation has the following roots:

 

Which gives yields the homogeneous solution

 

General Solution, n=4Edit

Using the taylor series approximation from earlier with   we have

 

We know the particular solution,  , ve will have this form:

 

taking the derivatives of this solution

 

and

 

Plugging the above equations into the original ODE yields the following matrix equation:

 

The unknown vector   can be easily solved by forward substitution,the following values were calculated in matlab:

 

So the particular solution   is

 

We can now find the general solution for n=4,  .

 

   

Solving using the initial conditions yields;

         
           

General Solution, n=7Edit

Using the taylor series approximation from earlier with   we have

 

In a similar fashion we construct a matrix equation for n=7:

 

Solving:

 

So the particular solution   is

 

We can now find the general solution for n=7,  .

 

 

 

Solving using our initial conditions yields

        
  
                

General Solution, n=11Edit

Using the taylor series approximation from earlier with   we have

 

 

Finally, we write out the matrix equation for n=11:

 

 

Solving the system in matlab:

 

 

So the particular solution   is

 

 

We can now find the general solution for n=11,  .

 

 

 

 

Solving using our initial conditions yields

     
     
       
      
                 
   
     

PlotEdit

  shown in red

  shown in blue

  shown in green

 

Part 4Edit

Solved by Luca Imponenti

Use the matlab command ode45 to integrate numerically   with  

and the initial conditions from Part 3 to obtain the numerical solution for y(x).

Plot y(x) in the same figure as above.

Matlab SolutionEdit

The numerical solution calculated using the matlab ode45 command is shown below:

 ans =
   0.2788
   0.2854
   0.2923
   0.2997
   0.3074
   0.3229
   0.3401
   0.3592
   0.3804
   0.4040
   0.4302
   0.4595
   0.4921
   0.5285
   0.5691
   0.6145
   0.6651
   0.7218
   0.7850
   0.8557
   0.9346
   1.0228
   1.1213
   1.2313
   1.3542
   1.4914
   1.6445
   1.8155
   2.0063
   2.2193
   2.4569
   2.7219
   3.0175
   3.3471
   3.7146
   4.1243
   4.5809
   5.0898
   5.6568
   6.2885
   6.9921
   7.3442
   7.7142
   8.1032
   8.5119

PlotEdit

Plotting the aboved vector of y-values,along with the results from earlier yields the following graph:

 

where the answer calculated in matlab is shown in yellow.

Egm4313.s12.team11.imponenti (talk) 08:04, 14 March 2012 (UTC)