User:Egm6321.f09.team7/HW6

Homework Six: Problem 1
Problem Statement:  '''See attached file contains the Problem Statement!  Solution:''' '''See attached file contains the solution! '''

Homework Six: Problem 2
Problem Statement:  '''See attached file contains the Problem Statement!  Solution:''' '''See attached file contains the solution! '''

Homework Six: Problem 3
Problem Statement:  '''See attached file contains the Problem Statement!  Solution:''' '''See attached file contains the solution! '''



Homework Six: Problem 4
Problem Statement:  '''See attached file contains the Problem Statement!  Solution:''' '''See attached file contains the solution! '''

Problem Statement
Pg. 33-3 []

Show that $$P_{(2k+1)}$$ is odd

Problem Solution
first we need the definition of $$P_{(n)}$$ it is as follows from Pg. 31-3

$$P_{n}(x)= \sum\nolimits_{i=0}^{[n/2]} (-1)^i \frac{(2n-2i)! x^{n-2i}}{2^ni!(n-i)!(n-2i)!}$$

As shown in the previous problem for summation polynomials the coefficient of a number does not make a difference in whether it is even or odd. Therefore we are only concerned with the portion of the above equation concerning x and its exponent:

$$x^{n-2i}$$

The first step will be to separate the above statement into:

$$x^n*x^{-2i}$$

We then realize that what we have is a product of two functions, for n = 2k+1 the functions are

$$x^{(2k+1)}$$ and  $$x^{-2i}$$

if we realize that any number multiplied by 2 is an even number we can state that the first function $$x^{(2k+1)}$$ will always be an odd function because an even number plus 1 will be an odd number

The second function $$x^{-2i}$$ will always be even because as stated previously for any number i when multiplied by 2 it will give an even number.

Therefore we are left with the product of the first function $$x^{(2k+1)}$$ which will always be odd and the second function $$x^{-2i}$$ which will always be even.

By observing the statements of the previous problem we can see that any time we have the product of an even and an odd function we are left with an odd function. Also from the previous problem the summation of odd functions will result in an odd function.

For this reason the polynomial $$P_{(2k+1)}$$ will always be an odd function.

Problem Statement
Pg. 31-3 []

Show that Equation 6 and equation 7 from 31-3 are equal

i.e. $$\sum\nolimits_{i=0}^{[n/2]} (-1)^i \frac{(2n-2i)! x^{n-2i}}{2^ni!(n-i)!(n-2i)!}$$   =     $$\sum\nolimits_{i=0}^{[n/2]}  \frac{1.3...(2n-2i-1)}{2^ii!(n-2i)!}(-1)^ix^{n-2i}$$

Problem Solution
The first step is to realize that some of terms are common to both equations specifically $$(-1)^i$$ and  $$x^{n-2i}$$ and therefore can be canceled out. This leaves us with:

$$\sum\nolimits_{i=0}^{[n/2]} \frac{(2n-2i)! }{2^ni!(n-i)!(n-2i)!}$$   =     $$\sum\nolimits_{i=0}^{[n/2]}  \frac{1.3...(2n-2i-1)}{2^ii!(n-2i)!}$$

In the first equation we will rewrite the $$2^n$$ terms as $$2^n-i*2^i$$ to help match to the second equations denominator

Next we can redefine our new term in the following manner$$ 2^{n - i} = \prod\limits_{k = 1}^{n - i} 2 $$

Using a similar method $$(n - i)! = \prod\limits_{k = 1}^{n - i} k $$

Leaving the left hand equation in the form:

$$\sum\nolimits_{i=0}^{[n/2]} \frac{(2n-2i)! }{\prod\limits_{k = 1}^{n - i} 2i! 2^i\prod\limits_{k = 1}^{n - i} k (n-2i)!}$$

The two new terms in the denominator can be combined:

$$\sum\nolimits_{i=0}^{[n/2]} \frac{(2n-2i)! }{2^i \prod\limits_{k = 1}^{n - i} 2k (i!)((n-2i)!)}$$

which leads us to:

$$\sum\nolimits_{i=0}^{[n/2]} \frac{(2n-2i)! }{2^i \prod\limits_{k = 1}^{n - i} 2k (i!)((n-2i)!)}$$ = $$\sum\nolimits_{i=0}^{[n/2]} \frac{1.3...(2n-2i-1)}{2^ii!(n-2i)!}$$

to further simplify we can recognize that $$(n-2i)!$$ and $$(i!)$$ are common to both equations and are eliminated to obtain:

$$\sum\nolimits_{i=0}^{[n/2]} \frac{(2n-2i)! }{2^i \prod\limits_{k = 1}^{n - i} 2k }$$ = $$\sum\nolimits_{i=0}^{[n/2]} \frac{1.3...(2n-2i-1)}{2^i}$$

from here the first equation can be manipulate to give us:

$$\sum\nolimits_{i=0}^{[n/2]} \frac{(2n - 2i - 1){2( n - i)}}{2^i{2(n-i)} }$$

Simply Canceling the $${2( n - i)}$$ terms makes this equation equal to the other equation and proves the problem

Homework Six: Problem 7
Problem Statement:  '''See attached file contains the Problem Statement!  Solution:''' '''See attached file contains the solution! '''

Homework Six: Problem 8
Problem Statement:  '''See attached file contains the Problem Statement!  Solution:''' '''See attached file contains the solution! '''

Problem Statement
Pg. 33-3 []

let f be the sum of functions $$g_i$$ $$f=\sum\nolimits_{i=0}^{n} g_i$$

if all $$g_i$$ are even show that f is even

if all $$g_i$$ are odd then show that f is odd

Problem Solution
To prove the statement above it is important to understand the properties of even and odd functions.

First to define a function as even it needs to have the following property:

$$Q(-x) = Q(x)$$

In other words a function Q of x is equal to the same function Q even if the argument x is taken as negative.

Some good examples of even functions are:

cos(x), x^2, |x|

Conversely to define a function as even it needs to have the following property:

$$Q(-x) = -Q(x)$$

In other words a function Q of x is negative to the same function Q if the argument x is taken as negative.

Some good examples of even functions are:

sin(x), x, and x^3

now some other properties for odd and even functions. For multiplication functions have the following properties taking an even function as f(+) and an odd as f(-):

f(+) * f(+) = f(+), an even times an even equals an even f(-) * f(-) = f(+), an odd times an odd equals an even f(-) * f(+) or f(+) * f(-) = f(-) an even times an odd or vice versa equals an odd

finally when taking the summation:

f(+) + f(+) = f(+)

f(-) + f(-) = f(-)

Therefore from the problem statement the sum of all odd or even functions will come to an odd or even function respectiviley to exemplify this feature consider the following:

if we take the definition from above f(-x) = f(x) and express it as f(-1*x) then make $$f(x) = x^n$$ then we get $$(-1*x)^n$$ which can be expressed as $$(-1)^n*(x)^n$$ now $$(x)^n$$ is just the original f and it is known that $$(-1)^n$$ will be positive for all n that are even and negative for all n that are odd so: $$ f(x) = x^n$$ = odd for n = odd or even for n = even

now if we take the sum of even functions like the following:

$$x^4+x^6+x^8$$ none of the exponents will change and there fore even if you make x negative the entire function will be the same and therefore stay even.

Conversely for the sum of odd functions:

$$ x+x^3+x^5$$ the exponents are un-altered by summing and therefore by making x negative each value will be negative and exactly equal to the negative of the original summation.

Problem Statement
[] From King et al., p.28, problem 1.1 b Use various methods to try to determine the first homogeneous solution u1 for the following equation:


 * $$xy''+2y'+xy=0 \!$$

The following methods should be used:

Check exactness of equation

Check if it possible to use the integrating factor method?

Attempt trial solutions:


 * 1) $$y(x)=e^{rx} \!$$
 * 2) $$y(x)=xe^{rx} \!$$
 * 3) $$y(x)=\frac{1}{x}e^{rx} \!$$

Problem Solution
Verifying Exactness

From previous homework and lectures we can define the conditions:

Condition 1:

equation is of the form f(x,y,p)Y" + g(x,y,p)

Condition 2: do the following relations hold,

$$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y\!$$ $$f_{xp}+pf_{yp}+2f_y=g_{pp} \!$$

If we define f and g in the following ways: f = x g = 2p+xy

It is easy to see that the first condition is satisfied

for the second condition

$$f_{xx}=0 \!$$ $$f_{xy}=0 \!$$ $$f_{yy}=0 \!$$ $$g_{xp}=0 \!$$ $$g_{yp}=0 \!$$ $$g_{y}=x \!$$

for the second equation of the second condition this leads to:

$$ 0 = -x$$

Which is not true and therefore the equation is not exact

Integrating factor method

Since the equation is not exact we need to Multiply through by $$x^my^n \!$$ and solve for the $$m \!$$ and $$n \!$$

The second equation simplifies to 0=0 no matter what m and n are.

The first equation has factors of both m and n and therefore cannot be solved since we have one equation and two unknowns

Trial Solutions

First attempt:

$$y=e^{rx} \!$$

$$y'=re^{rx} \!$$

$$y''=r^2e^{rx} \!$$

Plugging into the equation yields


 * $$xr^2+2+x=0 \!$$

Since r is by definition a constant we cannot factor out a solution

Second attempt:

$$y=xe^{rx} \!$$

$$y'=e^{rx}+xre^{rx} \!$$

$$y''=2re^{rx}+xr^2e^{rx} \!$$

Plugging into the equation yields


 * $$x^2r^2+4xr+2+x^2 \!$$

Again r is defined as a constant and cannot be a function of x therefore r cannot be factored out

Third attempt:

$$y = \frac{1}{x}e^{rx} \!$$

$$y' = -\frac{1}{x^2}e^{rx}+\frac{1}{x}re^{rx} \!$$

$$y'' = 2\frac{1}{x^3}e^{rx}-2\frac{1}{x^2}re^{rx}+\frac{1}{x}r^2e^{rx} \!$$

Plugging into the equation yields


 * $$r^2+1=0 \!$$

This solution can be factored when we keep in mind that imaginary i is a constant and not a function of x

This yields the solutions, $$r=\sqrt{-1}=i \!$$.

and therefore:

$$y(x)=\frac{e^{ix}}{x} \!$$

employing the moivre relationship tells us that:

$$y(x) = \frac{cos(x)}{x} + \frac{isin(x)}{x}$$

Lastly this can be compared with the undetermined factor method which was completed in Team 7's homework 4

Problem Statement
Given: $$ q(x)=\sum_{i=0}^4 c_ix^i$$ $$c_0=3$$ $$c_1=10$$ $$c_2=15$$ $$c_3=-1$$ $$c_4=5$$ Find the coefficients ($$a_i$$)to the Legendre polynomials($$P_i$$) and plot the solution. $$ q(x)=\sum_{i=0}^4 c_ix^i=\sum_{i=0}^4 a_iP_i$$

Problem Solution
The Legendre Polynomials are: $$P_0(x)=1$$ $$P_1(x)=x$$ $$P_2(x)=\frac{3}{2}x^2-\frac{1}{2}$$ $$P_3(x)=\frac{5}{2}x^3-\frac{3}{2}x$$ $$P_4(x)=\frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8}$$

$$q=3+10x+15x^2+-1x^3+5x^4=a_0(1)+a_1(x)+a_2(\frac{3}{2}x^2-\frac{1}{2})+a_3(\frac{5}{2}x^3-\frac{3}{2}x)+a_4(\frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8})$$

$$P(x)=\left[\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ -\frac{1}{2} & 0 & \frac{3}{2} & 0 & 0 \\ 0 & -\frac{3}{2} & 0 & \frac{5}{2} & 0 \\ \frac{3}{8} & 0 & -\frac{15}{4} & 0 & \frac{35}{8} \end{matrix}\right]

\left[\begin{matrix} 1\\ x \\ x^2 \\ x^3\\ x^4 \end{matrix}\right]=Ax$$

Solving for the coefficients: $$\left[\begin{matrix} a_0 & a_1 & a_2 & a_3 & a_4 \\ \end{matrix}\right]= \left[\begin{matrix} c_0 & c_1 & c_2 & c_3 & c_4 \\ \end{matrix}\right]* \left[\begin{matrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ -\frac{1}{2} & 0 & \frac{3}{2} & 0 & 0 \\ 0 & -\frac{3}{2} & 0 & \frac{5}{2} & 0 \\ \frac{3}{8} & 0 & -\frac{15}{4} & 0 & \frac{35}{8} \end{matrix}\right]^{-1}$$ The coefficients equal: $$\left[\begin{matrix} a_0 & a_1 & a_2 & a_3 & a_4 \\ \end{matrix}\right]= \left[\begin{matrix} 9 & 9.4 & 12.857 & -.4 & 1.143 \\ \end{matrix}\right]$$

$$q=3+10x+15x^2+-1x^3+5x^4=9(1)+9.4(x)+12.857(\frac{3}{2}x^2-\frac{1}{2})-.4(\frac{5}{2}x^3-\frac{3}{2}x)+1.143(\frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8})$$

The plot shows that the two equations are equal.

Homework Six: Problem 13
Problem Statement:  '''See attached file contains the Problem Statement!  Solution:''' '''See attached file contains the solution! '''

Homework Six: Problem 12
Problem Statement:  '''See attached file contains the Problem Statement!  Solution:''' '''See attached file contains the solution! '''

Ref1. Ref2.

Contributing Authors
Egm6321.f09.team7.mm 05:16, 17 November 2009 (UTC) Egm6321.f09.team7.benedict 01:10, 7 November 2009 (UTC) Egm6321.f09.team7.hua 10:57, 18 November 2009 (UTC) User:Emg6321.f09.blanco 11:17, 2 November 2009 (UTC)