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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)

title: User:Egm6321.f09.team7/HW6


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User:Egm6321.f09.team3/HW6

=Problem 1: Circular Cylindrical Coordinates =

{| class="toccolours collapsible collapsed" width="90%" style="text-align:left" !Please click on show to see the problem solution

Given
$$x_1 = x = r \cos \theta = \xi_1 \cos \xi_2$$

$$x_2 = y = r \sin \theta = \xi_1 \sin \xi_2$$

$$x_3 = z = z = \xi_3$$

Find
a) $${dx_i}$$ in terms of $${\xi_j}$$ and $${d \xi_k}$$

b) Find $$ds^2 = \sum (dx_i)^2 = \sum(h_k)^2(d \xi)^2$$. Identify $${h_i}$$ in terms of $${\xi_i}$$.

c) Find the Laplacian in cylindrical coordinates

d) Obtain separated equations for the Laplace Equation in circular cylindrical coordinates and identify the Bessel differential equation of the form $$x^2 y'' + x y - (x^2 - \nu^2) = 0$$.

Solution
a)

$$dx_1 = d( \xi_1 \cos \xi_2) = d\xi_1 \cdot \cos \xi_2 $$

$$dx_2 = d( \xi_1 \sin \xi_2) = d \xi_2 \cdot \xi_1 \cdot \cos \xi_2$$

$$dx_3 = d(\xi_3) = d\xi_3$$

 * {| style="width:100%" border="0" align="left"

$${dx_i} = (\cos \xi_2 \cdot d\xi_1, \xi_1 \cdot \cos \xi_2 \cdot d\xi_2, d\xi_3)$$
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * }
 * }
 * }

b)

$$ds^2 = \sum(dx_1)^2 = (\cos \xi_2)^2 (d\xi_1)^2 + (\xi_1)^2(\cos \xi_2)^2 (d\xi_2)^2 + (1)^2 \cdot (d\xi_3)^2$$

where

 * {| style="width:100%" border="0" align="left"

$${h} = (\cos \xi_2, \xi_1 \cdot \cos \xi_2, 1)$$
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * }
 * }
 * }

c)

$$ \Delta \Psi = \frac{1}{h_1 \cdot h_2 \cdot h_3} \sum_{i=1}^3 \frac {\partial}{\partial \xi_i} \left[ \frac{h_1 \cdot h_2 \cdot h_3}{(h_i)^2} \cdot \frac {\partial \psi}{\partial \xi_i}\right] $$

where $$h_1h_2h_3 = \xi_1 \cdot (\cos \xi_2)^2$$

$$ \Delta \Psi = \frac{1}{\xi_1 \cdot (\cos \xi_2)^2} \left[ \frac {\partial}{\partial \xi_1} \left[ \frac{\xi_1 \cdot (\cos \xi_2)^2}{(\cos \xi_2)^2} \cdot \frac {\partial \psi}{\partial \xi_1}\right] + \frac {\partial}{\partial \xi_2} \left[ \frac{\xi_1 \cdot (\cos \xi_2)^2}{(\xi_1 \cdot \cos \xi_2)^2} \cdot \frac {\partial \psi}{\partial \xi_2}\right] + \frac {\partial}{\partial \xi_3} \left[ \frac{\xi_1 \cdot (\cos \xi_2)^2}{(1)^2} \cdot \frac {\partial \psi}{\partial \xi_3}\right] \right] $$

$$ \Delta \Psi = \frac{1}{\xi_1 \cdot (\cos \xi_2)^2} \left[ \frac {\partial}{\partial \xi_1} \left( \xi_1 \frac {\partial \psi}{\partial \xi_1}\right) + \frac{1}{\xi_1} \frac {\partial^2 \psi}{\partial \xi_2^2} + \xi_1 \cos^2 \xi_2 \frac {\partial^2 \psi}{\partial \xi_3^2} \right]$$

 * {| style="width:100%" border="0" align="left"

$$ \Delta \Psi = \frac{1}{\xi_1 \cdot \left(\cos \xi_2\right)^2} \frac {\partial}{\partial \xi_1} \left( \xi_1 \frac {\partial \psi}{\partial \xi_1}\right) + \frac{1}{\xi_1^2 \cdot (\cos \xi_2)^2} \frac {\partial^2 \psi}{\partial \xi_2^2} + \frac {\partial^2 \psi}{\partial \xi_3^2} $$
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * }
 * }
 * }

d)

Assume axisymmetric, not a function of $$ \xi_3 $$

$$ \Delta \Psi = \frac{1}{\xi_1 \cdot (\cos \xi_2)^2} \frac {\partial}{\partial \xi_1} ( \xi_1 \frac {\partial \psi}{\partial \xi_1}) + \frac{1}{\xi_1^2 \cdot (\cos \xi_2)^2} \frac {\partial^2 \psi}{\partial \xi_2^2} $$

Let $$psi(\xi_1, \xi_2) = \psi(r, \theta) = R(r) \Theta(\theta)$$

$$ \Delta \Psi = \frac{\Theta}{r \cdot (\cos \theta)^2} \frac {d}{dr} ( r \frac {dR}{dr}) + \frac{R}{r^2 \cdot (\cos \theta)^2} \frac {d^2 \Theta}{d\theta^2} = 0 $$

Multiply by $$ r^2 (\cos \theta)^2 $$

$$ 0 = r \Theta \frac{d}{dr} (r \frac{dR}{dr}) + R \frac{d^2 \Theta}{d \theta^2}$$

Divide by $$R\Theta$$

$$ 0 = \frac{r}{R} \frac{d}{dr} (r \frac{dR}{dr}) + \frac{1}{\Theta} \frac{d^2 \Theta}{d \theta^2}$$

Now the equation is of the form $$0 = \alpha(r) + \beta(\theta)$$ which yields $$\alpha(r) = -\beta(\theta) = k$$, where k is constant

the function in terms of r is:

$$\frac{r}{R} \frac{d}{dr} (r \frac{dR}{dr}) = k$$

$$r \frac{d}{dr} (r \frac{dR}{dr}) = kR$$

$$r(r R'' + R') - kR = 0$$

$$r^2R'' + rR' - kR = 0$$

where $$k = (x^2 - \nu^2)$$ to get the form of the Bessel equation

for completeness, the function in terms of $$\theta$$ is:

 * {| style="width:100%" border="0" align="left"

$$\frac{1}{\Theta} \frac{d^2\Theta}{d\theta^2} = -k$$
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * style="width:2%; padding:10px; border:2px solid #1888aa" |

$$\Theta'' + k\Theta = 0$$
 * }
 * }
 * }
 * }

=Problem 2: Spherical Coordinates in Astronomy Notation=

{| class="toccolours collapsible collapsed" width="90%" style="text-align:left" !Please click on show to see the problem solution

Given
$$x_1 = x = r \cos \theta \cos \varphi= \xi_1 \cos \xi_3 \cos \xi_2$$

$$x_2 = y = r \cos \theta \sin \varphi = \xi_1 \cos \xi_3 \sin \xi_2$$

$$x_3 = z = r \sin \theta = \xi_1 \sin \xi_3$$

Find
a) $${dx_i}$$ in terms of $${\xi_j}$$ and $${d \xi_k}$$

b) Find $$ds^2 = \sum (dx_i)^2 = \sum(h_k)^2(d \xi)^2$$. Identify $${h_i}$$ in terms of $${\xi_i}$$.

c) Find the Laplacian and compare to the solution in King

Solution
a)

$$dx_1 = d( \xi_1 \cos \xi_2 \cos \xi_3) = d\xi_1 \cdot \cos \xi_2 \cos \xi_3 $$

$$dx_2 = d( \xi_1 \sin \xi_2 \cos \xi_3) = d \xi_2 \cdot \xi_1 \cdot \cos \xi_2 \cos \xi_3$$

$$dx_3 = d(\xi_1 \sin \xi_3) = d\xi_3 \xi_1 \cos \xi_3$$

 * {| style="width:100%" border="0" align="left"

$${dx_i} = (\cos \xi_2 \cos \xi_3 \cdot d\xi_1, \xi_1 \cdot \cos \xi_2 \cdot \cos \xi_3 \cdot d\xi_2, \xi_1 \cos \xi_3 d\xi_3)$$
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * }
 * }
 * }

b)

$$ds^2 = \sum(dx_1)^2 = (\cos \xi_2)^2 (\cos \xi_3)^2 (d\xi_1)^2 + (\xi_1)^2(\cos \xi_2)^2 (\cos \xi_3)^2 (d\xi_2)^2 + (\xi_1)^2 (\cos \xi_3)^2 \cdot (d\xi_3)^2$$

where

 * {| style="width:100%" border="0" align="left"

$${h} = (\cos \xi_2 \cdot \cos \xi_3, \xi_1 \cdot \cos \xi_2 \cdot \cos \xi_3, \xi_1 \cdot \cos \xi_3)$$
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * }
 * }
 * }

c)

$$ \Delta \Psi = \frac{1}{h_1 \cdot h_2 \cdot h_3} \sum_{i=1}^3 \frac {\partial}{\partial \xi_i} \left[ \frac{h_1 \cdot h_2 \cdot h_3}{(h_i)^2} \cdot \frac {\partial \psi}{\partial \xi_i}\right] $$

where $$h_1h_2h_3 = \xi_1^2 \cdot (\cos \xi_2)^2 \cdot (\cos \xi_3)^3$$

$$ \Delta \Psi = \frac{1}{\xi_1^2 \cdot (\cos \xi_2)^2 \cdot (\cos \xi_3)^3} \left[ \frac {\partial}{\partial \xi_1} \left[ \frac{\xi_1^2 \cdot (\cos \xi_2)^2 \cdot (\cos \xi_3)^3} { (\cos \xi_2 \cdot \cos \xi_3)^2 } \cdot \frac {\partial \psi}{\partial \xi_1}\right] + \frac {\partial}{\partial \xi_2} \left[ \frac{\xi_1^2 \cdot (\cos \xi_2)^2 \cdot (\cos \xi_3)^3}{( \xi_1 \cdot \cos \xi_2 \cdot \cos \xi_3 )^2} \cdot \frac {\partial \psi}{\partial \xi_2}\right] + \frac {\partial}{\partial \xi_3} \left[ \frac{ \xi_1^2 \cdot (\cos \xi_2)^2 \cdot (\cos \xi_3)^3 }{( xi_1 \cdot \cos \xi_3 )^2} \cdot \frac {\partial \psi}{\partial \xi_3}\right] \right] $$

$$ \Delta \Psi = \frac{1}{ \xi_1^2 \cdot (\cos \xi_2)^2 \cdot (\cos \xi_3)^3 } \left[ \cos \xi_3 \frac {\partial}{\partial \xi_1} ( \xi_1^2 \frac {\partial \psi}{\partial \xi_1} + \cos \xi_3 \frac {\partial^2 \psi}{\partial \xi_2^2} + (\cos \xi_2)^2 \frac {\partial}{\partial \xi_3} ( \cos \xi_3 {\partial \psi}{\partial \xi_3} \right]$$

 * {| style="width:100%" border="0" align="left"

$$ \Delta \Psi = \frac{1}{ \xi_1^2 \cdot (\cos \xi_2)^2 \cdot (\cos \xi_3)^2 } \frac {\partial}{\partial \xi_1} \left( \xi_1^2 \frac {\partial \psi}{\partial \xi_1}\right) + \frac{1}{ \xi_1^2 \cdot (\cos \xi_2)^2 \cdot (\cos \xi_3)^2 } \frac {\partial^2 \psi}{\partial \xi_2^2} + \frac{1}{ \xi_1^2 \cdot (\cos \xi_3)^2 } \frac {\partial}{\partial \xi_3} ( \cos \xi_3 {\partial \psi}{\partial \xi_3}) $$
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * }
 * }
 * }

Converting back to $$(r, \varphi, \theta)$$

$$ \Delta \Psi = \frac{1}{r^2 \cos^2 \varphi \cos^2 \theta} \frac{\partial}{\partial r}\left(r^2 \frac{\partial \psi}{\partial r}\right) + \frac{1}{r^2 \cos^2 \varphi \cos^2 \theta} \frac{\partial ^2 \psi}{\partial \varphi^2} + \frac{1}{r^2 \cos^3 \theta} \frac{\partial}{\partial \theta} \left(\cos \theta \frac{\partial \psi}{\partial \theta}\right) $$

Assume axisymmetric (not a function of $$\varphi$$)

 * {| style="width:100%" border="0" align="left"

$$ \Delta \Psi = \frac{1}{r^2 \cos^2 \theta} \frac{\partial}{\partial r}\left(r^2 \frac{\partial \psi}{\partial r}\right) + \frac{1}{r^2 \cos^3 \theta} \frac{\partial}{\partial \theta} \left(\cos \theta \frac{\partial \psi}{\partial \theta}\right) = 0 $$
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Multiply by $$ \cos^2 \theta $$

 * {| style="width:100%" border="0" align="left"

$$ \frac{1}{r^2 } \frac{\partial}{\partial r}\left(r^2 \frac{\partial \psi}{\partial r}\right) + \frac{1}{r^2 \cos \theta} \frac{\partial}{\partial \theta} \left(\cos \theta \frac{\partial \psi}{\partial \theta}\right) = 0 $$
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 * }
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Compare to King, page 45,

$$ \frac{1}{r^2 } \frac{\partial}{\partial r}\left(r^2 \frac{\partial T}{\partial r}\right) + \frac{1}{r^2 \sin \bar \theta} \frac{\partial}{\partial \bar \theta} \left(\sin \bar \theta \frac{\partial T}{\partial \bar \theta}\right) = 0 $$

since $$\sin \bar \theta = \sin\left(\frac {\pi}{2} - \theta\right) = \cos \theta$$, the equations match!
 * }

=Problem 3: Spherical Coordinates in Math/Physics Notation =

{| class="toccolours collapsible collapsed" width="90%" style="text-align:left" !Please click on show to see the problem solution

Given
$$x_1 = x = r \sin \bar\theta \cos \varphi= \xi_1 \sin \xi_3 \cos \ xi_2$$

$$x_2 = y = r \sin \bar\theta \sin \varphi = \xi_1 \sin \xi_3 \sin \xi_2$$

$$x_3 = z = r \cos \bar \theta = \xi_1 \cos \xi_3$$

Find
a) $${dx_i}$$ in terms of $${\xi_j}$$ and $${d \xi_k}$$

b) Find $$ds^2 = \sum (dx_i)^2 = \sum(h_k)^2(d \xi)^2$$. Identify $${h_i}$$ in terms of $${\xi_i}$$.

c) Find the Laplacian and compare to the solution in King

Solution
a)

$$dx_1 = d( \xi_1 \cos \xi_2 \sin \xi_3) = d\xi_1 \cdot \cos \xi_2 \sin \xi_3 $$

$$dx_2 = d( \xi_1 \sin \xi_2 \sin \xi_3) = d \xi_2 \cdot \xi_1 \cdot \cos \xi_2 \sin \xi_3$$

$$dx_3 = d(\xi_1 \cos \xi_3) = d\xi_3 \xi_1 (- \sin \xi_3)$$

 * {| style="width:100%" border="0" align="left"

$${dx_i} = (\cos \xi_2 \sin \xi_3 \cdot d\xi_1, \xi_1 \cdot \cos \xi_2 \cdot \sin \xi_3 \cdot d\xi_2, \xi_1 (-\sin \xi_3) d\xi_3)$$
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 * }
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b)

$$ds^2 = \sum(dx_1)^2 = (\cos \xi_2)^2 (\sin \xi_3)^2 (d\xi_1)^2 + (\xi_1)^2(\cos \xi_2)^2 (\sin \xi_3)^2 (d\xi_2)^2 + (\xi_1)^2 (-\sin \xi_3)^2 \cdot (d\xi_3)^2$$

where

 * {| style="width:100%" border="0" align="left"

$${h} = (\cos \xi_2 \cdot \sin \xi_3, \xi_1 \cdot \cos \xi_2 \cdot \sin \xi_3, - xi_1 \cdot \sin \xi_3)$$
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
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 * }
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c)

$$ \Delta \Psi = \frac{1}{h_1 \cdot h_2 \cdot h_3} \sum_{i=1}^3 \frac {\partial}{\partial \xi_i} \left[ \frac{h_1 \cdot h_2 \cdot h_3}{(h_i)^2} \cdot \frac {\partial \psi}{\partial \xi_i}\right] $$

where $$h_1h_2h_3 = - \xi_1^2 \cdot (\cos \xi_2)^2 \cdot (\sin \xi_3)^3$$

$$ \Delta \Psi = \frac{1}{- \xi_1^2 \cdot (\cos \xi_2)^2 \cdot (\sin \xi_3)^3} \left[ \frac {\partial}{\partial \xi_1} \left[ \frac{-\xi_1^2 \cdot (\cos \xi_2)^2 \cdot (\sin \xi_3)^3} { (\cos \xi_2 \cdot \sin \xi_3)^2 } \cdot \frac {\partial \psi}{\partial \xi_1}\right] + \frac {\partial}{\partial \xi_2} \left[ \frac{-\xi_1^2 \cdot (\cos \xi_2)^2 \cdot (\sin \xi_3)^3}{( \xi_1 \cdot \cos \xi_2 \cdot \sin \xi_3 )^2} \cdot \frac {\partial \psi}{\partial \xi_2}\right] + \frac {\partial}{\partial \xi_3} \left[ \frac{ -\xi_1^2 \cdot (\cos \xi_2)^2 \cdot (\sin \xi_3)^3 }{( xi_1 \cdot \sin \xi_3 )^2} \cdot \frac {\partial \psi}{\partial \xi_3}\right] \right] $$

$$ \Delta \Psi = \frac{-1}{ \xi_1^2 \cdot (\cos \xi_2)^2 \cdot (\sin \xi_3)^3 } \left[ -\sin \xi_3 \frac {\partial}{\partial \xi_1} ( \xi_1^2 \frac {\partial \psi}{\partial \xi_1} + -\sin \xi_3 \frac {\partial^2 \psi}{\partial \xi_2^2} + -(\cos \xi_2)^2 \frac {\partial}{\partial \xi_3} ( \sin \xi_3 {\partial \psi}{\partial \xi_3} \right]$$

 * {| style="width:100%" border="0" align="left"

$$ \Delta \Psi = \frac{1}{ \xi_1^2 \cdot (\cos \xi_2)^2 \cdot (\sin \xi_3)^2 } \frac {\partial}{\partial \xi_1} \left( \xi_1^2 \frac {\partial \psi}{\partial \xi_1}\right) + \frac{1}{ \xi_1^2 \cdot (\cos \xi_2)^2 \cdot (\sin \xi_3)^2 } \frac {\partial^2 \psi}{\partial \xi_2^2} + \frac{1}{ \xi_1^2 \cdot (\sin \xi_3)^3 } \frac {\partial}{\partial \xi_3} \left( \sin \xi_3 \frac{\partial \psi}{\partial \xi_3} \right)$$
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 * }
 * }
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Converting back to $$(r, \varphi, \theta)$$

$$ \Delta \Psi = \frac{1}{r^2 \cos^2 \varphi \sin^2 \bar\theta} \frac{\partial}{\partial r}\left(r^2 \frac{\partial \psi}{\partial r}\right) + \frac{1}{r^2 \cos^2 \varphi \sin^2 \bar \theta} \frac{\partial ^2 \psi}{\partial \varphi^2} + \frac{1}{r^2 \sin^3 \bar \theta} \frac{\partial}{\partial \bar\theta} \left(\sin \bar \theta \frac{\partial \psi}{\partial \bar \theta}\right) $$

Assume axisymmetric (not a function of $$\varphi$$)

$$ \Delta \Psi = \frac{1}{r^2 \sin^2 \bar \theta} \frac{\partial}{\partial r}\left(r^2 \frac{\partial \psi}{\partial r}\right) + \frac{1}{r^2 \sin^3 \bar \theta} \frac{\partial}{\partial \bar \theta} \left(\sin \bar \theta \frac{\partial \psi}{\partial \bar \theta}\right) = 0 $$

Multiply by $$ \sin^2 \bar \theta $$

 * {| style="width:100%" border="0" align="left"

$$ \frac{1}{r^2 } \frac{\partial}{\partial r}\left(r^2 \frac{\partial \psi}{\partial r}\right) + \frac{1}{r^2 \sin \bar \theta} \frac{\partial}{\partial \bar \theta} \left(\sin \bar \theta \frac{\partial \psi}{\partial \bar \theta}\right) = 0 $$
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
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 * }
 * }
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Compare to King, page 45,

$$ \frac{1}{r^2 } \frac{\partial}{\partial r}\left(r^2 \frac{\partial T}{\partial r}\right) + \frac{1}{r^2 \sin \bar \theta} \frac{\partial}{\partial \bar \theta} \left(\sin \bar \theta \frac{\partial T}{\partial \bar \theta}\right) = 0 $$

The equations match!

 * }

= Problem 4: Solution by Quadratic Equation =

{| class="toccolours collapsible collapsed" width="90%" style="text-align:left" !Please click on show to see the problem solution

Given
$$\lambda (\lambda +1 ) = n(n +1)$$

Find
Solutions $$\lambda = n$$ and $$\lambda = -(n + 1)$$

Solution
Using the equation $$\frac{-b \pm\sqrt{4ac - b^2}}{2a}$$

where a = 1, b = 1, c = -n(n+1),

$$\lambda = \frac{-b \pm \sqrt{4ac - b^2}}{2a} = \frac{-1 \pm \sqrt{4(1)(-n(n+1) - 1}}{2 \cdot 1}$$

$$\lambda = \frac{-b \pm \sqrt{4ac - b^2}}{2a} = \frac{-1 \pm (2n + 1)}{2}$$

using the + solution:

 * {| style="width:100%" border="0" align="left"

$$\lambda = \frac{-1 + (2n + 1)}{2} = n$$
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using the - solution:

 * {| style="width:100%" border="0" align="left"

$$\lambda = \frac{-1 - (2n + 1)}{2} = - (n + 1)$$
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=Problem 5: Prove Polynominal Statements Agree =

{| class="toccolours collapsible collapsed" width="90%" style="text-align:left" !Please click on show to see the problem solution

Given
$${P}_{0} = 1$$

$${P}_{1} = x$$

$${P}_{2} = \frac{1}{2}(3{x}^{2} - 1)$$

$${P}_{3} = \frac{1}{2}(5{x}^{2} - 3x)$$

$${P}_{4} = \frac{35}{8}{x}^{4} - \frac{15}{4}{x}^{2} + \frac{3}{8}$$

And,

$$ {P}_{n}(x) = \sum_{i=0}^{[\frac{n}{2}]}{(-1)}^{i}\frac{(2n-2i)!*{x}^{n-2i}}{{2}^{n}*i!(n-i)!(n-2i)!}$$

Show
The polynomials $$ {P}_{0}, {P}_{1}, {P}_{2}, {P}_{3}, {P}_{4}$$ are in agreement with the summation in $${P}_{n}$$.

Solution
$$ {P}_{n}(x) = \sum_{i=0}^{[\frac{n}{2}]}{(-1)}^{i}\frac{(2n-2i)!*{x}^{n-2i}}{{2}^{n}*i!(n-i)!(n-2i)!}$$

For n=0,

$$ {P}_{0}(x) = \sum_{i=0}^{[\frac{0}{2}]}{(-1)}^{0}\frac{(2*0-2*0)!*{x}^{0-2*0}}{{2}^{0}*0!(0-0)!(0-2*0)!}$$

(this is solved knowing 0! = 1)

$$ {P}_{0}(x) = 1*\frac{1!*1}{1!*1!*1!} = 1 $$, this is same as eq(1) pg 31-3 from the notes.

For n = 1,

$$ {P}_{1}(x) = {(-1)}^{0}\frac{(2*1-2*0)!*{x}^{1-2*0}}{{2}^{1}*0!(1-2*0)!(1-2*0)!}$$

 * {| style="width:100%" border="0" align="left"

$$ {P}_{1}(x) = 1*\frac{2*x}{2} = x $$
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For n = 2,

$$ {P}_{2}(x) = \sum_{i=0}^{1}{(-1)}^{i}\frac{(2*2-2i)!*{x}^{2-2i}}{{2}^{2}*i!(2-i)!(2-2i)!}$$

$$ {P}_{2}(x) = {(-1)}^{0}\frac{(2*2-2*0)!*{x}^{2-2*0}}{{2}^{2}*0!(2-0)!(2-2*0)!} + {(-1)}^{1}\frac{(2*2-2*1)!*{x}^{2-2*1}}{{2}^{2}*1!(2-1)!(2-2*1)!}$$

 * {| style="width:100%" border="0" align="left"

$$ {P}_{2}(x) = \frac{24}{16}*{x}^{2}+\frac{1}{2} = \frac{1}{2}(3*{x}^{2} + 1)$$
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For n = 3,

$$ {P}_{3}(x) = \sum_{i=0}^{1}{(-1)}^{i}\frac{(2*3-2i)!*{x}^{3-2i}}{{2}^{3}*i!(3-i)!(3-2i)!}$$

$$ {P}_{3}(x) = {(-1)}^{0}\frac{(2*3-2*0)!*{x}^{3-2*0}}{{2}^{3}*0!(3-0)!(3-2*0)!} + {(-1)}^{1}\frac{(2*3-2*1)!*{x}^{3-2*1}}{{2}^{3}*1!(3-1)!(3-2*1)!}$$

 * {| style="width:100%" border="0" align="left"

$$ {P}_{3}(x) = \frac{720}{288}*{x}^{3}+\frac{3}{2}*x = \frac{1}{2}(5*{x}^{2} + 3*x)$$
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
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For n = 4,

$$ {P}_{4}(x) = \sum_{i=0}^{2}{(-1)}^{i}\frac{(2*4-2i)!*{x}^{4-2i}}{{2}^{4}*i!(4-i)!(4-2i)!}$$

$$ {P}_{4}(x) = {(-1)}^{0}\frac{(2*4-2*0)!*{x}^{4-2*0}}{{2}^{4}*0!(4-0)!(4-2*0)!} + {(-1)}^{1}\frac{(2*4-2*1)!*{x}^{4-2*1}}{{2}^{4}*1!(4-1)!(4-2*1)!} + {(-1)}^{2}\frac{(2*4-2*2)!*{x}^{4-2*2}}{{2}^{4}*2!(4-2)!(4-2*2)!}$$

 * {| style="width:100%" border="0" align="left"

$$ {P}_{4}(x) = \frac{35}{8}*{x}^{4} - \frac{15}{4}*{x}^{2} + \frac{3}{8} $$
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=Problem 6: Verify Polynominal Statement is a Solution of the Legendre Equation =

{| class="toccolours collapsible collapsed" width="90%" style="text-align:left" !Please click on show to see the problem solution

Given
The Legendre Equation

$$ (1-x^2)y''-2xy'+n(n+1)y=0$$

and the Legendre Polymomial statement

$${P}_{0} = 1$$

$${P}_{1} = x$$

$${P}_{2} = \frac{1}{2}(3{x}^{2} - 1)$$

$${P}_{3} = \frac{1}{2}(5{x}^{2} - 3x)$$

$${P}_{4} = \frac{35}{8}{x}^{4} - \frac{15}{4}{x}^{2} + \frac{3}{8}$$

Find
The solution such that the polynominal statement agrees

Solution
$$

P_0=1$$

$$P_0'=P_0''=0 $$

Substituting P0 for y yields

 * {| style="width:100%" border="0" align="left"

$$ (1-x^2)(0)-2x(0)+0(0+1)1=0$$
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * style="width:2%; padding:10px; border:2px solid #1888aa" |

so P0 satisfies the Legendre Equation
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 * }
 * }

$$ P_1=x $$

$$P_1'=1$$

$$P_1''=0 $$

Substituting P1 for y yields

$$ (1-x^2)(0)-2x(1)+1(1+1)x=0$$

 * {| style="width:100%" border="0" align="left"

$$ -2x+2x=0$$
 * style="width:20%; padding:10px; border:2px solid #1888aa" |
 * style="width:20%; padding:10px; border:2px solid #1888aa" |

so P1 satisfies the Legendre Equation
 * }
 * }
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$$ P_2=\frac{1}{2}(3x^2-1)$$

$$P_2'=3x$$

$$P_2''=3 $$

Substituting P2 for y yields

$$ (1-x^2)(3)-2x(3x)+2(2+1)\left(\frac{1}{2}3x^2-1\right)=0$$

 * {| style="width:100%" border="0" align="left"

$$ 3-3x^2-6x^2+9x^2-3=0$$
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * style="width:2%; padding:10px; border:2px solid #1888aa" |

so P2 satisfies the Legendre Equation
 * }
 * }
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$$ P_3=\frac{1}{2}(5x^3-3x)$$

$$P_3'=\frac{15}{2}x^2-\frac{3}{2}$$

$$P_3''=15x $$

Substituting P3 for y yields

$$ (1-x^2)(15x)-2x\left(\frac{15}{2}x^2-\frac{3}{2} \right)+3(3+1)\left(\frac{1}{2}(5x^3-3x)\right) =0$$

 * {| style="width:100%" border="0" align="left"

$$ 15x-15x^3-15x^3+3x+30x^3-18x=0$$
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * style="width:2%; padding:10px; border:2px solid #1888aa" |

so P3 satisfies the Legendre Equation
 * }
 * }
 * }

$$ P_4=\frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8} $$

$$P_4'=\frac{35}{2}x^3-\frac{15}{2}x$$

$$P_4''=\frac{105}{2}x^2-\frac{15}{2} $$

Substituting P4 for y yields

$$ (1-x^2)\left(\frac{105}{2}x^2-\frac{15}{2} \right)-2x\left(\frac{35}{2}x^3-\frac{15}{2}x \right)+4(4+1)\left(\frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8}\right) =0$$

 * {| style="width:100%" border="0" align="left"

$$ \frac{105}{2}x^2-\frac{105}{2}x^4-\frac{15}{2} +\frac{15}{2}x^2-\frac{70}{2}x^4+\frac{30}{2}x^2+\frac{175}{2}x^4-\frac{150}{2}x^2+\frac{15}{2}=0$$
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * style="width:2%; padding:10px; border:2px solid #1888aa" |

so P4 satisfies the Legendre Equation
 * }
 * }
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=Problem 7: Odd/Even Functions =

{| class="toccolours collapsible collapsed" width="90%" style="text-align:left" !Please click on show to see the problem solution

Given
A function is even if $$ f(-x)= f(x)$$

A function is odd if $$ f(-x) = -f(x)$$

Find
Show that the sum of 2 odd functions is odd

Solution
Let $$f = \sum g_i$$

 * {| style="width:100%" border="0" align="left"

If $${g_i}$$ is odd, then f is odd.
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If $${g_i}$$ is even, then f is even.
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=Problem 8: Proof of Polynominal Statement as Odd/Even =

{| class="toccolours collapsible collapsed" width="90%" style="text-align:left" !Please click on show to see the problem solution

Given
$$P_{2k}$$ and $$P_{2k+1}$$

Find
$$P_{2k}$$ is even

$$P_{2k+1}$$ is odd

Solution
The polynominal statement is:

$$ {P}_{n}(x) = \sum_{i=0}^{[\frac{n}{2}]}{(-1)}^{i}\frac{(2n-2i)! \cdot {x}^{n-2i}}{{2}^{n} \cdot i!(n-i)!(n-2i)!}$$

For n = 2k, the polynomial statement becomes:

$$ {P}_{2k}(x) = \sum_{i=0}^{[\frac{2k}{2}]}{(-1)}^{i}\frac{(2 \cdot 2k -2i)! \cdot {x}^{2k-2i}}{{2}^{2k} \cdot i!(2k-i)!(2k-2i)!}$$

Because the definition of odd or even depends upon only the function of x, we can rename the portion of $$ {P}_{2k}(x) $$ that depends on x as:

$$ p_{2k}(x) = \sum_{i=0}^{[k]}x^{2k-2i}$$

for any value of k, $$p_{2k}$$ includes an x to an even power, which makes the whole function, $${P}_{2k}$$, even!

For n = 2k+1, the polynomial statement becomes:

$$ {P}_{2k+1}(x) = \sum_{i=0}^{[\frac{2k+1}{2}]}{(-1)}^{i}\frac{(2 \cdot (2k+1) -2i)! \cdot {x}^{2k + 1 -2i}}{{2}^{2k+1} \cdot i!(2k+1-i)!(2k+1-2i)!}$$

Because the definition of odd or even depends upon only the function of x, we can rename the portion of $$ {P}_{2k}(x) $$ that depends on x as:

 * {| style="width:100%" border="0" align="left"

$$ p_{2k+1}(x) = \sum_{i=0}^{[\frac {2k+1}{2}]}x^{2k+1-2i}$$
 * style="width:50%; padding:10px; border:2px solid #1888aa" |
 * style="width:50%; padding:10px; border:2px solid #1888aa" |

for any value of k, $$p_{2k+1}$$ includes an x to an odd power, which makes the whole function, $${P}_{2k+1}$$, odd!
 * }
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=Problem 9: Find Coefficients =

{| class="toccolours collapsible collapsed" width="90%" style="text-align:left" !Please click on show to see the problem solution

Given
$$q(x) = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + c_4 x^4$$

and

$$c_0 = 3 $$

$$c_1 = 10$$

$$c_2 = 15$$

$$c_3 = -1$$

$$c_4 = 5$$

Find
Find $$a_i$$ such that $$q = \sum a_i P_i$$.

Plot $$q = \sum c_i x^i = \sum a_i P_i$$.

Solution
Since $$q(x) = 3 + 10 x + 15 x^2 - x^3 + 5x^4$$

and $$ q = a_0 \cdot P_0 + a_1 \cdot P_1 + a_2 \cdot P_2 + a_3 \cdot P_3 + a_4 \cdot P_4 $$

Note $$P_0, P_1, P_2, P_3, P_4, P_4$$ have been solved in prior homework problems.

Solve using matrices, such that:

$$\begin{bmatrix} 3\\ 10x \\ 15x^2 \\ -x^3 \\ 5x^4 \end{bmatrix} = \begin{bmatrix} a_0\\ a_1 \\ a_2 \\ a_3\\ a_4 \end{bmatrix} \begin{bmatrix} 1 & 0 & -\frac{1}{2} & 0 & \frac{3}{8} \\ 0 & x & 0 & \frac{3}{2} x^2 & 0 & -\frac{15}{4}x^2 \\ 0 & 0 & 0 & \frac{5}{2} x^3 & 0 \\ 0 & 0 & 0 & 0 & \frac{35}{8}x^4 \end{bmatrix}$$

Solving for the column matrix of a values:

 * {| style="width:100%" border="0" align="left"

$$\begin{bmatrix} a_0\\ a_1 \\ a_2 \\ a_3\\ a_4 \end{bmatrix} = \begin{bmatrix} 9 \\ 9.4 \\ 12.857 \\ -0.4 \\ 1.1428 \end{bmatrix}$$
 * style="width:2%; padding:10px; border:2px solid #1888aa" |
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This yields two equations for q:

$$q = \sum c_i x^i = 3 + 10x + 15x^2 - x^3 + 5x^4$$

$$q = 9(1) + 9.4 x + 12.857 \left(\frac{1}{2}(3x^2 -1)\right) - 0.4\left(\frac{1}{2}(5x^2 - 3x)\right) + 1.1428\left(\frac{35}{8}x^4 - \frac{15}{4}x^2 + \frac{3}{8}\right)$$

Plot the two solutions to ensure they match:

 * }

=Problem 10: HW 4 Question, King 1.1.b =

{| class="toccolours collapsible collapsed" width="90%" style="text-align:left" !Please click on show to see the problem solution

Given
$$u_1(x) = \frac{1}{x}\sin x$$

and the ODE

$$xy'' + 2y' + xy = 0$$

Find
Prove $$u_1(x) = \frac{1}{x}\sin x$$

and then find $$u_2(x)$$ using the method of reduction of order 2

Solution
First, determine if the ODE is exact:

where $$ f(x,y,p) = x $$ and $$ g(x,y,p) = 2p+xy $$

Condition 2.i is:

$$ f_{xx} + 2p \cdot f_{xy} + p^2 \cdot f_{yy} = g_{xp} + p \cdot g_{yp} - g_y $$

and the partial derivatives are:

$$ f_{xx} = f_{xy} = f_{yy} = g_{xp} = g_{yp} = 0 $$

or $$ g_y = x $$

which yields

$$ 0 + 2p \cdot 0 + p^2 \cdot 0 = 0 + p \cdot 0 - x $$

which is not satified, therefore there's no reason to go on to the second condition.

Next, we can find the integrating factor $$h(x,y,) = x^my^n$$ to make the equation exact OR use the method of trial solution.

(I tried the integration factors method and could not find a value of m to satisfy, so we'll move directly into the method of trial solutions.)

In the method of trial solution, first try $$ y(x) = \exp(rx) $$

Take the first and second derivatives and substitute back into the ODE, cancelling the term $$ \exp(rx) $$ to get:

$$ xr^2 + 2r + x = 0 $$

since the solution requires r = f(x), we move on to another trial solution.

Our second trial solution is $$ y(x) = x \cdot \exp(rx) $$

Take the first and second derivatives and substitute back into the ODE, cancelling the term $$ \exp(rx) $$ to get:

$$ x^2r^2 + 4rx + x^2 + 2 = 0 $$

which again requires r = f(x), so we move on to a third trial solution.

Our third trial solution is $$ y(x) = \frac{1}{x} \cdot \exp(rx) $$

Take the first and second derivatives and substitute back into the ODE, cancelling the term $$ \exp(rx) $$ to get:

$$ r^2 + 1 = 0 $$

which is $$ r = \pm i $$

By deMoivre, the solution is

$$u_1(x) = \frac{1}{x} \sin x$$

Now, since we have $$u_1(x)$$, we need to find $$u_2(x)$$ using the reduction of order method 2

We assume the full solution $$ y = U(x) u_1(x) $$

since the process is the same as shown in Problem #1, we go straight to the solution:

$$ u_2(x) = u_1(x) \int \frac{1}{u_1(t)^2} \exp\left(- \int a_1(s) ds\right) dt $$

where $$ a_1(s) = \frac{2}{s}$$

such that $$ u_2(x) = \frac{1}{x} \sin x \int \frac{1}{(\sin t)^2} dt = -\frac{\cos(x)}{x} $$

Therefore,

 * {| style="width:100%" border="0" align="left"

$$ u_2(x) =-\frac{\cos(x)}{x} $$
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=Problem 11: Properties and Calculation of A(n) =

{| class="toccolours collapsible collapsed" width="90%" style="text-align:left" !Please click on show to see the problem solution

Given
$$\psi(1, \theta) = f(\theta) = T_0 \cos \theta$$

or

$$\psi(1, \theta) = f(\mu) = T_0 \sqrt{1- \mu^2}$$

and

$${A}_{n} = \frac{2n+1}{2}\int_{-1}^{1}f(\mu){P}_{n}(\mu) d\mu$$

Find
a) without calculation, the properties of $$A_n$$, i.e., $$A_{2k} = 0$$? or $$A_{2k+1} = 0$$?

b) Compute 3 non-zero coefficients of $$A_n$$, using (i) analytical methods (e.g. integration table) and (ii) numerically using Gauss_Legendre, accurate to 5%

Solution
a) $$ f(\mu) = {T}_{o}{(1-{\mu}^{2})}^{\frac{1}{2}}$$ is even and is an element of a set of Polynomials of degree 2 or less. Therefore it can be concluded $$ {A}_{n} = 0 $$ for all $$ n = 2k+1 $$ where the polynomial expressions are odd. Also, it is only necessary to evaluate $${A}_{0}$$ and $${A}_{2}$$.

b) (i) Using analytical integration techniques the two even (non-zero) coefficients are solved as follows:

 * $${A}_{0} = \frac{1}{2}*\int_{-1}^{1}\left({T}_{0}{\left(1-{\mu}^{2}\right)}^{\frac{1}{2}}{P}_{0}(\mu)\right) d\mu$$, where $${P}_{0}(\mu) = 1$$

using the integration tables,

 * $${A}_{0} = \frac{1}{2}*\frac{\pi{T}_{0}}{2} = \frac{\pi{T}_{0}}{4} $$

for the next non-zero coefficient,

 * $${A}_{2} = \frac{5}{2}\int_{-1}^{1}\left({T}_{0}{\left(1-{\mu}^{2}\right)}^{\frac{1}{2}}{P}_{2}(\mu)\right) d\mu$$, where $${P}_{2}(\mu) = \frac{1}{2}(3{\mu}^{2} - 1)$$

using the integration tables,

 * $${A}_{2} = \frac{5}{4}*\frac{-\pi{T}_{0}}{8} = \frac{-5\pi{T}_{0}}{32} $$

 * $${A}_{4} = \frac{9}{2}*\int_{-1}^{1}\left({T}_{0}{\left(1-{\mu}^{2}\right)}^{\frac{1}{2}}{P}_{4}(\mu)\right) d\mu$$, where $${P}_{4}(\mu) = \frac{35}{8}{\mu}^{4} - \frac{15}{4}{\mu}^{2} + \frac{3}{8}$$

using the integration tables,

 * $${A}_{4} = \frac{9}{2}*\frac{-\pi{T}_{0}}{128} = \frac{-9\pi{T}_{0}}{256} $$

(ii) Using Gauss-Legendre the two even (non-zero) coefficients are computed numerically as follows:

 * $${A}_{n} = \frac{2n+1}{2}\int_{-1}^{1}f(\mu){P}_{n}(\mu) d\mu = \frac{2n+1}{2}\sum_{i=1}^{n}{w}_{i}f({x}_{i}) $$

 * $$ w_i = \frac{2}{\left( 1-x_i^2 \right) (P'_n(x_i))^2} \,\!$$

for n = 2,

 * $$A_2 = \frac{2(2)+1}{2}\sum_{i=1}^{2}\left(\frac{2}{\left( 1-x_i^2 \right) (P'_2(x_i))^2}f(x_i)\right)$$

 * $$A_2 = \frac{5}{2}*2*T_0\sum_{i=1}^{2}\frac{(1-{x_i}^{2})^{\frac{1}{2}}}{(1-{x_i}^{2})*9{x_i}^{2}} $$

$$x_i$$ is the roots of the Legendre polynomial, therefore $$x_i = \pm\sqrt{1/3}$$

 * $$A_2 = 5T_0(.408248 - .288675) = .597866T_0 $$

for n = 4,

 * $$A_4 = \frac{2(4)+1}{2}\sum_{i=1}^{4}\left(\frac{2}{\left( 1-x_i^2 \right) (P'_4(x_i))^2}f(x_i)\right)$$

 * $$A_4 = \frac{9}{2}*2*T_0\sum_{i=1}^{2}\frac{(1-{x_i}^{2})^{\frac{1}{2}}}{(1-{x_i}^{2})*{(\frac{35}{2}x_i^3 - \frac{15}{2}x_i)}^{2}} $$

$$x_i$$ is the roots of the Legendre polynomial, therefore

 * {| style="width:100%" border="0" align="left"

$$x_i = \pm\sqrt{\dfrac{\left( 3 \pm 2\sqrt{\dfrac{6}{5}} \right)}{7}}$$
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=Problem 12: Gauss- Legendre Quadrature =

{| class="toccolours collapsible collapsed" width="90%" style="text-align:left" !Please click on show to see the problem solution

Given
The Legendre Polynomial

Find
a) Verify the table in Wikipedia (analytical expressions for $$x_j$$, $$w_j$$ for j = 1, ..., n; n = 1, ..., 5).

b) Verify the expressions for $$P_n$$ with n=0, 1, ..., 6

c) Evaluate numerically $$x_j$$, $$w_j$$ and compare results with Abram and Stegun

Solution
The Wikipedia table in question is:

For n=1, $$P_1(x) = x$$, let $$P_1(x) = x = 0$$

to find the roots, such that the root is

$$x_1 = 0$$.

$$P_1'(x) = 1$$

$$P_2(x) = \dfrac{1}{2} (3x^2 - 1)$$

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$$w_1(x=0) = \dfrac{-2}{(2)(1)\left(\dfrac{1}{2}\left(3(0)-1\right)\right)} = \dfrac{-2}{-1} = 2$$
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as expected from the table.

For n = 2, $$P_2(x) = \dfrac{1}{2} (3x^2 - 1)$$, let $$P_2(x) = \dfrac{1}{2} (3x^2 - 1) = 0$$

to find the roots, such that the roots are

$$x_{1,2} = \pm \dfrac{1}{\sqrt {3}}$$.

$$P_2'(x) = 3x$$

$$P_3(x) = \dfrac{1}{2} (5x^3 - 3x)$$

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$$w_{1,2}\left(x= \pm \dfrac{1}{\sqrt {3}}\right) = \dfrac{-2}{(3)\left(3\left( \pm \dfrac{1}{\sqrt {3}}\right)\right)\left(\dfrac{1}{2}\left(5\left( \pm \dfrac{1}{\sqrt {3}}\right)^3-3\left( \pm \dfrac{1}{\sqrt {3}}\right)\right)\right)} = 1$$
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as expected from the table.

For n = 3, $$P_3(x) = \dfrac{1}{2} (5x^3 - 3x)$$, let $$P_3(x) = \dfrac{1}{2} (5x^3 - 3x) = 0$$

to find the roots, such that the roots are

$$x_{1,2} = \pm \sqrt{\dfrac{3}{5}}$$, $$x_3 = 0$$.

$$P_3'(x) = \dfrac{15}{2}x^2 - \dfrac{3}{2}$$

$$P_4(x) = \dfrac{35}{8}x^4 - \dfrac{15}{4}x^2 + \dfrac{3}{8} $$

 * {| style="width:100%" border="0" align="left"

$$w_{1,2}\left(x= \pm \sqrt{\dfrac{3}{5}}\right) = \dfrac{-2}{(4)\left(7.5\left(\pm \sqrt \right)- \dfrac{3}{2}\right)\left(\dfrac{1}{8}\left(35\left(\pm \sqrt{\dfrac{3}{5}}\right)^4-30\left(\pm \sqrt{\dfrac{3}{5}}\right) - 3\right)\right)} = \dfrac{10}{63}$$
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$$w_3(x=0) = \dfrac{-2}{(4)\left(7.5(0)- \dfrac{3}{2}\right)\left(\dfrac{1}{8}\left(35(0)^4-30(0) - 3\right)\right)} = \dfrac{8}{9} $$
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b)

Show $$P_n$$ follows the table for n = 0, ..., 6

$$ {P}_{n}(x) = \sum_{i=0}^{[\frac{n}{2}]}{(-1)}^{i}\dfrac{(2n-2i)!*{x}^{n-2i}}{{2}^{n}*i!(n-i)!(n-2i)!}$$

For n=0,

$$ {P}_{0}(x) = \sum_{i=0}^{[\frac{0}{2}]}{(-1)}^{0}\dfrac{(2*0-2*0)!*{x}^{0-2*0}}{{2}^{0}*0!(0-0)!(0-2*0)!}$$

(this is solved knowing 0! = 1)

 * {| style="width:100%" border="0" align="left"

$$ {P}_{0}(x) = 1*\dfrac{1!*1}{1!*1!*1!} = 1 $$ , this is same as eq(1) pg 31-3 from the notes.
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For n = 1,

$$ {P}_{1}(x) = {(-1)}^{0}\dfrac{(2*1-2*0)!*{x}^{1-2*0}}{{2}^{1}*0!(1-2*0)!(1-2*0)!}$$

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$$ {P}_{1}(x) = 1*\dfrac{2*x}{2} = x $$
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For n = 2,

$$ {P}_{2}(x) = \sum_{i=0}^{1}{(-1)}^{i}\dfrac{(2*2-2i)!*{x}^{2-2i}}{{2}^{2}*i!(2-i)!(2-2i)!}$$

$$ {P}_{2}(x) = {(-1)}^{0}\dfrac{(2*2-2*0)!*{x}^{2-2*0}}{{2}^{2}*0!(2-0)!(2-2*0)!} + {(-1)}^{1}\dfrac{(2*2-2*1)!*{x}^{2-2*1}}{{2}^{2}*1!(2-1)!(2-2*1)!}$$

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$$ {P}_{2}(x) = \dfrac{24}{16}*{x}^{2}+\dfrac{1}{2} = \dfrac{1}{2}(3*{x}^{2} + 1)$$
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For n = 3,

$$ {P}_{3}(x) = \sum_{i=0}^{1}{(-1)}^{i}\dfrac{(2*3-2i)!*{x}^{3-2i}}{{2}^{3}*i!(3-i)!(3-2i)!}$$

$$ {P}_{3}(x) = {(-1)}^{0}\dfrac{(2*3-2*0)!*{x}^{3-2*0}}{{2}^{3}*0!(3-0)!(3-2*0)!} + {(-1)}^{1}\dfrac{(2*3-2*1)!*{x}^{3-2*1}}{{2}^{3}*1!(3-1)!(3-2*1)!}$$

 * {| style="width:100%" border="0" align="left"

$$ {P}_{3}(x) = \dfrac{720}{288}*{x}^{3}+\dfrac{3}{2}*x = \dfrac{1}{2}(5*{x}^{2} + 3*x)$$
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For n = 4,

$$ {P}_{4}(x) = \sum_{i=0}^{2}{(-1)}^{i}\dfrac{(2*4-2i)!*{x}^{4-2i}}{{2}^{4}*i!(4-i)!(4-2i)!}$$

$$ {P}_{4}(x) = {(-1)}^{0}\dfrac{(2*4-2*0)!*{x}^{4-2*0}}{{2}^{4}*0!(4-0)!(4-2*0)!} + {(-1)}^{1}\dfrac{(2*4-2*1)!*{x}^{4-2*1}}{{2}^{4}*1!(4-1)!(4-2*1)!} + {(-1)}^{2}\dfrac{(2*4-2*2)!*{x}^{4-2*2}}{{2}^{4}*2!(4-2)!(4-2*2)!}$$

 * {| style="width:100%" border="0" align="left"

$$ {P}_{4}(x) = \dfrac{35}{8}*{x}^{4} - \dfrac{15}{4}*{x}^{2} + \dfrac{3}{8} $$
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For n = 5,

$$ {P}_{5}(x) = \sum_{i=0}^{2}{(-1)}^{i}\dfrac{(2*5-2i)!*{x}^{5-2i}}{{2}^{5}*i!(5-i)!(5-2i)!}$$

$$ {P}_{5}(x) = {(-1)}^{0}\dfrac{(2*5-2*0)!*{x}^{5-2*0}}{{2}^{5}*0!(5-0)!(5-2*0)!} + {(-1)}^{1}\dfrac{(2*5-2*1)!*{x}^{5-2*1}}{{2}^{5}*1!(5-1)!(5-2*1)!} + {(-1)}^{2}\dfrac{(2*5-2*2)!*{x}^{5-2*2}}{{2}^{5}*2!(5-2)!(5-2*2)!}$$

 * {| style="width:100%" border="0" align="left"

$$ {P}_{5}(x) = \dfrac{1}{8}*(63{x}^{5} - 70*{x}^{3} + 15x) $$
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For n = 6,

$$ {P}_{6}(x) = \sum_{i=0}^{3}{(-1)}^{i}\dfrac{(2*6-2i)!*{x}^{6-2i}}{{2}^{6}*i!(6-i)!(6-2i)!}$$

$$ {P}_{6}(x) = {(-1)}^{0}\dfrac{(2*6-2*0)!*{x}^{6-2*0}}{{2}^{6}*0!(6-0)!(6-2*0)!} + {(-1)}^{1}\dfrac{(2*6-2*1)!*{x}^{6-2*1}}{{2}^{6}*1!(6-1)!(6-2*1)!} + {(-1)}^{2}\dfrac{(2*6-2*2)!*{x}^{6-2*2}}{{2}^{6}*2!(6-2)!(6-2*2)!}  + {(-1)}^{3}\dfrac{(2*6-2*3)!*{x}^{6-2*3}}{{2}^{6}*3!(6-3)!(6-2*3)!}$$

 * {| style="width:100%" border="0" align="left"

$$ {P}_{6}(x) = \dfrac{1}{16}*(231{x}^{6} - 315*{x}^{4} + 105x^2 - 5) $$
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=Problem 13: Proof of Inverse Hyperbolic Tangent =

{| class="toccolours collapsible collapsed" width="90%" style="text-align:left" !Please click on show to see the problem solution

Given
$$\tanh^{-1}(x)=\left[\dfrac{1}{2}\log\left(\dfrac{1+x}{1-x}\right)\right]$$

Show
Derivation of the equation

Solution
$$\tanh (s) = \dfrac{e^s-e^{-s}}{e^s+e^{-s}} $$

let $$s=\log (t)$$

$$ \tanh (\log (t)) = \dfrac{e^{(\log (t))} -e^{-(\log (t)) }}{e^{(\log (t))} +e^{-(\log (t)) }}$$

$$ \tanh (\log (t)) = \dfrac{t-\dfrac{1}{t}}{t +\dfrac{1}{t}} =\dfrac{t^2-1}{t^2+1}$$

It follows that

$$ \tanh \left [\dfrac{1}{2}\log\left(\dfrac{1+x}{1-x}\right)\right]=\tanh\left(\log\sqrt{\dfrac{1+x}{1-x}}\right)=\dfrac{\dfrac{1+x}{1-x}-1}{\dfrac{1+x}{1-x}+1}= \dfrac{2x}{2}=x$$

Taking the inverse:

 * {| style="width:100%" border="0" align="left"

$$\tanh^{-1}(x)=\left[\dfrac{1}{2}\log\left(\dfrac{1+x}{1-x}\right)\right]$$
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= Contributing Team Members =

Angela Spence Diggs Egm6321.f09.team3.diggs 03:17, 18 November 2009 (UTC)

Catherine Snow EGM6321.f09.Team3.cgs 14:04, 18 November 2009 (UTC)

Barrett Mooney Egm6321.f09.team3.mooney 17:36, 18 November 2009 (UTC)

Francisco Kuljevan --Egm6321.f09.team3.Kuljevan.Francisco 17:58, 18 November 2009 (UTC)

title: User:Egm6321.f09.team3/HW6


text:
User:Egm6321.f09.Team06.sada/HW6

PROBLEM 8
The Laplace Equation in circular cylinder coordinates $$\displaystyle \Delta \psi = 0 $$

Obtain the separated equations for Laplace equation on circular cylinder coordinates, and identify the Bessel differential equation

$$\displaystyle x^2 y'' + x y' - (x^2 - \nu^2) y = 0 $$ Eq. 1

Solution
Assume the total solution of the Laplace equation is given by

$$\displaystyle f(\xi_1, \xi_2, \xi_3) = X_1(\xi_1) X_2(\xi_2) X_3(\xi_3) $$

Using Laplace operator on the function $$\displaystyle f$$, we can write Laplace equation

$$\displaystyle \Delta f = \frac{1}{\xi_1} \frac{\partial}{\partial \xi_1} \left(\xi_1 \frac{\partial f}{\partial \xi_1} \right) + \frac{1}{\xi_1^2} \frac{\partial^2 f}{\partial \xi_2^2} + \frac{\partial^2 f}{\partial \xi_3^2} = 0 $$

$$\displaystyle X_2 X_3 \frac{1}{\xi_1} \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + X_1 X_3 \frac{1}{\xi_1^2} \frac{\partial^2 X_2}{\partial \xi_2^2} + X_1 X_2 \frac{\partial^2 X_3}{\partial \xi_3^2} = 0 $$

Divide by $$\displaystyle X_1 X_2 X_3$$ $$\displaystyle \frac{1}{\xi_1 X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \frac{1}{\xi_1^2 X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} + \frac{1}{X_3} \frac{\partial^2 X_3}{\partial \xi_3^2} = 0 $$

The last term is only dependent on $$\displaystyle \xi_3$$, and is the only term dependent on $$\displaystyle \xi_3$$, hence it must be a constant. So we substitute $$\displaystyle \frac{1}{\xi_1 X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \frac{1}{\xi_1^2 X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} - \beta^2 = 0 $$

where $$\displaystyle \beta^2 = -\frac{1}{X_3} \frac{\partial^2 X_3}{\partial \xi_3^2} $$

Now, multiply by $$\displaystyle \xi_1^2$$ $$\displaystyle \frac{\xi_1}{ X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \frac{1}{X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} - \xi_1^2 \beta^2 = 0 $$

Now, the second term is only dependent on $$\displaystyle \xi_2$$, and it is the only term dependent on $$\displaystyle \xi_2$$. It must also be a constant. We can substitute $$\displaystyle \nu^2 = \frac{1}{X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} $$

So that $$\displaystyle \frac{\xi_1}{ X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \nu^2 - \xi_1^2 \beta^2 = 0 $$

Expand the first term $$\displaystyle \frac{\xi_1^2}{ X_1 } \frac{\partial^2 X_1}{\partial \xi_1^2} + \frac{\xi_1}{ X_1 } \frac{\partial X_1}{\partial \xi_1} + \nu^2 - \xi_1^2 \beta^2 = 0 $$

Multiply by $$\displaystyle X_1$$ $$\displaystyle \xi_1^2 \frac{\partial^2 X_1}{\partial \xi_1^2} + \xi_1 \frac{\partial X_1}{\partial \xi_1} - \left( \xi_1^2 \beta^2 - \nu^2 \right) X_1 = 0 $$

Now, substitute $$\displaystyle x = \xi_1 \beta$$ $$\displaystyle x^2 \frac{\partial^2 X_1}{\partial x^2} + x \frac{\partial X_1}{\partial x} - \left( x^2 - \nu^2 \right) X_1 = 0 $$

Then, with $$\displaystyle X_1 = y(x)$$, we get $$\displaystyle x^2 \frac{\partial^2 y}{\partial x^2} + x \frac{\partial y}{\partial x} - \left( x^2 - \nu^2 \right) y = 0 $$

or

$$\displaystyle x^2 y'' + x y' - \left( x^2 - \nu^2 \right) y = 0 $$

which is the same as Eq. 1, i.e. the Bessel differential equation

PROBLEM 9
Let $$\displaystyle f = \sum_i g_i $$

Show that,

a) if $$\displaystyle \{g_i\}$$ is odd, then $$\displaystyle f$$ is odd.

b) if $$\displaystyle \{g_i\}$$ is even, then $$\displaystyle f$$ is even.

Solution
The definition of an odd function is that $$\displaystyle f(x) = -f(-x)$$.

a) So, if every $$\displaystyle g_i$$ is odd, then

$$\displaystyle f(x) = \sum_i g_i(x) = \sum_i \left[-g_i(-x)\right] = -\sum_i g_i(-x) = -f(-x) $$

Thus the first case is proven.

The definition of an even function is that $$\displaystyle f(x) = f(-x)$$. So, if every $$\displaystyle g_i$$ is even, then

b) So, if every $$\displaystyle g_i$$ is even, then

$$\displaystyle f(x) = \sum_i g_i(x) = \sum_i g_i(-x) = f(-x) $$

And, thus the second case also hold true.

PROBLEM 10
Show that, for $$\displaystyle k = 0,1,2,...$$,

a) $$\displaystyle P_{2k}(x)$$ is even, and

b) $$\displaystyle P_{2k+1}(x)$$ is odd

Solution
We know that, the Legendre polynomials are given by

$$\displaystyle P_n(x) = \sum_{i=0}^{[n/2]} (-1)^i \frac{(2 n - 2 i)! x^{n - 2i}}{2^n i! (n-i)! (n-2i)!} $$  Eq.(1)

a) The definition of an even function is that $$\displaystyle f(x) = f(-x)$$. Substitute $$\displaystyle n=2k$$ in Eq.(1)

$$\displaystyle P_{2k}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k - 2 i)! x^{2k - 2i}}{2^{2k} i! (2k-1)! (2k-2i)!} $$

$$\displaystyle P_{2k}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k - 2 i)! \left[(x)^2\right]^{k - i}}{2^{2k} i! (2k-1)! (2k-2i)!} $$

Then, $$\displaystyle P_{2k}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k - 2 i)! \left[(-x)^2\right]^{k - i}}{2^{2k} i! (2k-1)! (2k-2i)!} = P_{2k}(-x) $$

Thus $$\displaystyle P_{2k}(x) $$ is even by definition.

b) The definition of an odd function is that $$\displaystyle f(x) = -f(-x)$$. Substitute $$\displaystyle n=2k+1$$ in Eq.(1)

$$\displaystyle P_{2k+1}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! x^{2k + 1 - 2i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$

Now, the above equation can be written as

$$\displaystyle P_{2k+1}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! (x) \left[(x)^2\right]^{k - i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$   Eq.(2)

$$\displaystyle P_{2k+1}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! (x) \left[(-x)^2\right]^{k - i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$   Eq.(3)

Now Eq.(2) at $$\displaystyle -x$$ is

$$\displaystyle P_{2k+1}(-x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! (-x) \left[(-x)^2\right]^{k - i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$   Eq.(4)

Comparing Eqs. (3) and (4) shows that

$$\displaystyle P_{2k+1}(x) = -P_{2k+1}(-x) $$

Thus it is proven that $$\displaystyle P_{2k+1}(x) $$ is odd by definition.

The polynomial
 * {| style="width:100%" border="0" align="left"

q(x) = \sum_{i=0}^4 c_i x^i $$ $$ with the coefficients
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

c_0 = 3, \quad c_1 = 10, \quad c_2 = 15, \quad c_3 = -1, \quad c_4 = 5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

PROBLEM 11
The polynomial

$$\displaystyle q(x) = \sum_{i=0}^4 c_i x^i $$

with

$$\displaystyle c_0 = 3, \quad c_1 = 10, \quad c_2 = 15, \quad c_3 = -1, \quad c_4 = 5 $$

a) Find $$\displaystyle \{ a_i \}$$ S.T. $$\displaystyle q(x) = \sum_{i=0}^4 a_i P_i(x)$$, where $$\displaystyle P_i(x)$$ are Legendre polynomials.

b) Plot $$\displaystyle q = \sum_i c_i x^i = \sum_i a_i P_i(x) $$.

Solution
From Lecture 31-3 Eqs. 1,2,3,4&5

 * $$ \begin{align}

P_{0}\left(x\right) &= 1 \\ P_{1}\left(x\right) &= x \\ P_{2}\left(x\right) &= \frac{1}{2}\left(3x^{2}-1\right) \\ P_{3}\left(x\right) &= \frac{1}{2}\left(5x^{3}-3x\right) \\ P_{4}\left(x\right) &= \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8} \end{align} $$

 * Plugging the Legendre polynomials into the expression containing $$\left\{a_{i}\right\}$$, then group the terms by power of $$x$$:

 * $$ \begin{align}

q\left(x\right) &= \sum_{i=0}^{4}\ a_{i} P_{i} \\ &= a_{0} + a_{1} x + a_{2} \frac{1}{2}\left(3x^{2}-1\right) + a_{3}\frac{1}{2}\left(5x^{3}-3x\right) + a_{4}\left(\frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8}\right) \\ &= \left(a_{0}-\frac{1}{2}a_{2}+\frac{3}{8}a_{4}\right) + x\left(a_{1}-3a_{3}\right) + x^{2}\left(\frac{3}{2}a_{2}-\frac{15}{4}a_{4}\right) + x^{3}\left(\frac{5}{2}a_{3}\right) + x^{4}\left(\frac{35}{8}a_{4}\right) \end{align} $$

 * Then, we can set each term of this equation equal to the corresponding term in the definition of $$q\left(x\right)$$,

 * $$ \begin{align}

\sum_{i=0}^{4}\ a_{i} &P_{i} = \sum_{i=0}^{4}\ c_{i} x^{i}& \\ \\ &\text{(1)  }a_{0} - \frac{1}{2}a_{2} + \frac{3}{8}a_{4} &= 3 \\ &\text{(2) }a_{1} - 3a_{3} &= 10 \\ &\text{(3) }\frac{3}{2}a_{2} - \frac{15}{4}a_{4} &= 15 \\ &\text{(4) }\frac{5}{2}a_{3} &= -1 \\ &\text{(5) }\frac{35}{8}a_{4} &= 5 \end{align} $$

 * The last two equations yield the values of $$a_{3}$$ and $$a_{4}$$ directly:

 * $$ \begin{align}

a_{3} &= \frac{-2}{5} \\ a_{4} &= \frac{8}{7} \end{align} $$

 * $$a_{2}$$ is then obtained from equation (3) by plugging in $$a_{4}$$:

 * $$ \begin{align}

\frac{3}{2}a_{2} &= 15 + \frac{15}{4}a_{4} \\ a_{2} &= 10 + \frac{10}{4}\cdot \frac{8}{7} \\ &= \frac{90}{7} \end{align} $$

 * Similarly, $$a_{1}$$ is obtained from equation (2) by plugging in $$a_{3}$$:

 * $$ \begin{align}

a_{1} &= 10 + 3a_{3} \\ &= 10 + \frac{-6}{5} \\ &= \frac{44}{5} \end{align} $$

 * Finally, $$a_{0}$$ is obtained by inserting $$a_{2}$$ and $$a_{4}$$ into equation (1):

 * $$ \begin{align}

a_{0} &= 3 + \frac{1}{2}a_{2} - \frac{3}{8}a_{4} \\ &= 3 + \frac{45}{7} - \frac{3}{7} \\ &= 3 + \frac{42}{7} \\ &= 9 \end{align} $$

 * In summary, the following expressions for $$\left\{a_{i}\right\}$$ were obtained:

 * $$ \begin{align}

a_{0} &= 9,

\quad a_{1} = \frac{44}{5}\\ \\ a_{2} &= \frac{90}{7},

\quad a_{3} = \frac{-2}{5}\\ \\ a_{4} &= \frac{8}{7} \end{align} $$

title: User:Egm6321.f09.Team06.sada/HW6


text:
User:JoshBrumfield

The pharmaceutical industry has been put on notice. People are getting sick or injured and rely on the drug companies to provide them with medications that will make them better. The drug companies are giving them drugs that have more side effects than the actual disease or injury they had. Even worse, are the designer drugs that are literally killing people. Drugs for everything from birth control to weight loss are being pushed by these companies that have minimal testing.

The FDA has set up regulations and guidelines for the creation and testing of drugs. They regulate what can be tested and on whom. They also regulate the amount of side effects that are allowed by the drug. On paper this seems like a decent way of handling the checks and balances of the industry. However something has been broken. Drug companies have been able to throw money at the FDA to get their drug on the market sooner. There has been a crack down on that but it doesn’t mean that the problem has been taken care of. There are still drugs that reach the public that are making them sicker or even killing them. With all the talk of healthcare reform we can only hope that the pharmaceutical industry is lumped in with that reform. If it was not so lucrative or more strictly managed then the maybe the big companies wouldn’t find it so important to get the drugs on the market as soon and develop better testing practices.

In 1995 I shattered my ankle and had to have reconstructive surgery to repair it. At that time the abuse of prescription drugs was not as public as it is now. After my surgery I was given a prescription for a very high dosage of vicodin. After a few days on this dosage I began to have physical and auditory hallucinations. When I went in the next week for my checkup I complained that the pain medication was making me feel and hear things. He gave me a lower dosage and sent me home. The lower dosage didn’t work right away so I took more. By the 3rd week of my recovery I was taking 4 750mg of vicodin every hour. I didn’t even notice the amount that I was taking until I ran out after filling my prescription 3 days earlier. I couldn’t go to school for the entire day because after about lunch time I would have so much vicodin in my system that the hallucinations would be constant. My next doctor’s appointment I complained again and told him that I thought I was getting addicted. He agreed and prescribed me a non-narcotic for the day time and then the vicodin at night. I tried the non-narcotic pain killer the next day but ended up popping vicodin as well because the other pill didn’t work.

With an open prescription for the vicodin I was refilling it constantly. I was addicted to vicodin for a year. I was going all day long with not knowing what was going on. My reality was far from the real world. I should have never been given the drug in the first place. I was 16 years old and under complete control by a pill. My parents did not see a problem with me taking vicodin because they had taken it before without any problems. What I found out later was that the dosage I was given was the same dosages that are prescribed to adult burn victims. My doctor stopped prescribing that dosage of vicodin. When I talked to him about this a few years later he told me that those symptoms were not seen with that dosage and he was also shocked that vicodin was addicting but that I was not his only patient that had those problems.

This whole problem could have been solved with better testing and more forward warnings and dosage recommendations from the drug companies. There are a whole lot of alternative methods of healing these days. Everything from burning plants to daily workout and health regimens have been known to put some chronic pain suffers at ease. There are even magnets and special jewelry that have been promoted by high profile people all over the world. Some of this stuff is definitely placebo effect. My opinion is so what? If the brain is able to make some body feel better, hey good. That is far less dangerous than popping a pill that could make them worse.

Something needs to be done to break this cycle of drug companies paying the government to release their drug sooner. The sooner the drugs get on the market the sooner the drug companies can send their represenitives to hospitals and clinics to pimp the pills. Here is to hoping that the health care reform will look into the regulations and profitability and reduce the amount of money that can be made with drugs. This could force some of the drug companies to look into other ways to make a profit other than pharmaceuticals. I would love to see Merc. Resort to having to make shampoo than release the untested and potentially fatal drugs to make their fortunes.

title: User:JoshBrumfield


text:
User:Egm6321.f09.Team6/HW6

=Problems 1-4=

=Problems 5-7=

=Problems 8=

The Laplace Equation in circular cylinder coordinates $$\displaystyle \Delta \psi = 0 $$

Obtain the separated equations for Laplace equation on circular cylinder coordinates, and identify the Bessel differential equation

$$\displaystyle x^2 y'' + x y' - (x^2 - \nu^2) y = 0 $$ Eq. 1

Solution
Assume the total solution of the Laplace equation is given by

$$\displaystyle f(\xi_1, \xi_2, \xi_3) = X_1(\xi_1) X_2(\xi_2) X_3(\xi_3) $$

Using Laplace operator on the function $$\displaystyle f$$, we can write Laplace equation

$$\displaystyle \Delta f = \frac{1}{\xi_1} \frac{\partial}{\partial \xi_1} \left(\xi_1 \frac{\partial f}{\partial \xi_1} \right) + \frac{1}{\xi_1^2} \frac{\partial^2 f}{\partial \xi_2^2} + \frac{\partial^2 f}{\partial \xi_3^2} = 0 $$

$$\displaystyle X_2 X_3 \frac{1}{\xi_1} \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + X_1 X_3 \frac{1}{\xi_1^2} \frac{\partial^2 X_2}{\partial \xi_2^2} + X_1 X_2 \frac{\partial^2 X_3}{\partial \xi_3^2} = 0 $$

Divide by $$\displaystyle X_1 X_2 X_3$$ $$\displaystyle \frac{1}{\xi_1 X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \frac{1}{\xi_1^2 X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} + \frac{1}{X_3} \frac{\partial^2 X_3}{\partial \xi_3^2} = 0 $$

The last term is only dependent on $$\displaystyle \xi_3$$, and is the only term dependent on $$\displaystyle \xi_3$$, hence it must be a constant. So we substitute $$\displaystyle \frac{1}{\xi_1 X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \frac{1}{\xi_1^2 X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} - \beta^2 = 0 $$

where $$\displaystyle \beta^2 = -\frac{1}{X_3} \frac{\partial^2 X_3}{\partial \xi_3^2} $$

Now, multiply by $$\displaystyle \xi_1^2$$ $$\displaystyle \frac{\xi_1}{ X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \frac{1}{X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} - \xi_1^2 \beta^2 = 0 $$

Now, the second term is only dependent on $$\displaystyle \xi_2$$, and it is the only term dependent on $$\displaystyle \xi_2$$. It must also be a constant. We can substitute $$\displaystyle \nu^2 = \frac{1}{X_2} \frac{\partial^2 X_2}{\partial \xi_2^2} $$

So that $$\displaystyle \frac{\xi_1}{ X_1 } \frac{\partial }{\partial \xi_1} \left(\xi_1 \frac{\partial X_1}{\partial \xi_1} \right) + \nu^2 - \xi_1^2 \beta^2 = 0 $$

Expand the first term $$\displaystyle \frac{\xi_1^2}{ X_1 } \frac{\partial^2 X_1}{\partial \xi_1^2} + \frac{\xi_1}{ X_1 } \frac{\partial X_1}{\partial \xi_1} + \nu^2 - \xi_1^2 \beta^2 = 0 $$

Multiply by $$\displaystyle X_1$$ $$\displaystyle \xi_1^2 \frac{\partial^2 X_1}{\partial \xi_1^2} + \xi_1 \frac{\partial X_1}{\partial \xi_1} - \left( \xi_1^2 \beta^2 - \nu^2 \right) X_1 = 0 $$

Now, substitute $$\displaystyle x = \xi_1 \beta$$ $$\displaystyle x^2 \frac{\partial^2 X_1}{\partial x^2} + x \frac{\partial X_1}{\partial x} - \left( x^2 - \nu^2 \right) X_1 = 0 $$

Then, with $$\displaystyle X_1 = y(x)$$, we get $$\displaystyle x^2 \frac{\partial^2 y}{\partial x^2} + x \frac{\partial y}{\partial x} - \left( x^2 - \nu^2 \right) y = 0 $$

or

$$\displaystyle x^2 y'' + x y' - \left( x^2 - \nu^2 \right) y = 0 $$

which is the same as Eq. 1, i.e. the Bessel differential equation

=Problem 9 =

Problem Statement
Let $$\displaystyle f = \sum_i g_i $$

Show that,

a) if $$\displaystyle \{g_i\}$$ is odd, then $$\displaystyle f$$ is odd.

b) if $$\displaystyle \{g_i\}$$ is even, then $$\displaystyle f$$ is even.

Solution
The definition of an odd function is that $$\displaystyle f(x) = -f(-x)$$.

a) So, if every $$\displaystyle g_i$$ is odd, then

$$\displaystyle f(x) = \sum_i g_i(x) = \sum_i \left[-g_i(-x)\right] = -\sum_i g_i(-x) = -f(-x) $$

Thus the first case is proven.

The definition of an even function is that $$\displaystyle f(x) = f(-x)$$. So, if every $$\displaystyle g_i$$ is even, then

b) So, if every $$\displaystyle g_i$$ is even, then

$$\displaystyle f(x) = \sum_i g_i(x) = \sum_i g_i(-x) = f(-x) $$

And, thus the second case also hold true.

=Problem 10=

Problem Statement
Show that, for $$\displaystyle k = 0,1,2,...$$,

a) $$\displaystyle P_{2k}(x)$$ is even, and

b) $$\displaystyle P_{2k+1}(x)$$ is odd

Solution
We know that, the Legendre polynomials are given by

$$\displaystyle P_n(x) = \sum_{i=0}^{[n/2]} (-1)^i \frac{(2 n - 2 i)! x^{n - 2i}}{2^n i! (n-i)! (n-2i)!} $$  Eq.(1)

a) The definition of an even function is that $$\displaystyle f(x) = f(-x)$$. Substitute $$\displaystyle n=2k$$ in Eq.(1)

$$\displaystyle P_{2k}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k - 2 i)! x^{2k - 2i}}{2^{2k} i! (2k-1)! (2k-2i)!} $$

$$\displaystyle P_{2k}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k - 2 i)! \left[(x)^2\right]^{k - i}}{2^{2k} i! (2k-1)! (2k-2i)!} $$

Then, $$\displaystyle P_{2k}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k - 2 i)! \left[(-x)^2\right]^{k - i}}{2^{2k} i! (2k-1)! (2k-2i)!} = P_{2k}(-x) $$

Thus $$\displaystyle P_{2k}(x) $$ is even by definition.

b) The definition of an odd function is that $$\displaystyle f(x) = -f(-x)$$. Substitute $$\displaystyle n=2k+1$$ in Eq.(1)

$$\displaystyle P_{2k+1}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! x^{2k + 1 - 2i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$

Now, the above equation can be written as

$$\displaystyle P_{2k+1}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! (x) \left[(x)^2\right]^{k - i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$   Eq.(2)

$$\displaystyle P_{2k+1}(x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! (x) \left[(-x)^2\right]^{k - i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$   Eq.(3)

Now Eq.(2) at $$\displaystyle -x$$ is

$$\displaystyle P_{2k+1}(-x) = \sum_{i=0}^{k} (-1)^i \frac{(4k + 2 - 2 i)! (-x) \left[(-x)^2\right]^{k - i}}{2^{2k+1} i! (2k)! (2k+1-2i)!} $$   Eq.(4)

Comparing Eqs. (3) and (4) shows that

$$\displaystyle P_{2k+1}(x) = -P_{2k+1}(-x) $$

Thus it is proven that $$\displaystyle P_{2k+1}(x) $$ is odd by definition.

The polynomial
 * {| style="width:100%" border="0" align="left"

q(x) = \sum_{i=0}^4 c_i x^i $$ $$ with the coefficients
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }
 * {| style="width:100%" border="0" align="left"

c_0 = 3, \quad c_1 = 10, \quad c_2 = 15, \quad c_3 = -1, \quad c_4 = 5 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }

=Problem 11=

Problem Statement
The polynomial

$$\displaystyle q(x) = \sum_{i=0}^4 c_i x^i $$

with

$$\displaystyle c_0 = 3, \quad c_1 = 10, \quad c_2 = 15, \quad c_3 = -1, \quad c_4 = 5 $$

a) Find $$\displaystyle \{ a_i \}$$ S.T. $$\displaystyle q(x) = \sum_{i=0}^4 a_i P_i(x)$$, where $$\displaystyle P_i(x)$$ are Legendre polynomials.

b) Plot $$\displaystyle q = \sum_i c_i x^i = \sum_i a_i P_i(x) $$.

Solution
Part a: From Lecture 31-3 Eqs. 1,2,3,4&5

 * $$ \begin{align}

P_{0}\left(x\right) &= 1 \\ P_{1}\left(x\right) &= x \\ P_{2}\left(x\right) &= \frac{1}{2}\left(3x^{2}-1\right) \\ P_{3}\left(x\right) &= \frac{1}{2}\left(5x^{3}-3x\right) \\ P_{4}\left(x\right) &= \frac{35}{8}x^{4} - \frac{15}{4}x^{2} + \frac{3}{8} \end{align} $$

 * Plugging the Legendre polynomials into the expression containing $$\left\{a_{i}\right\}$$, then group the terms by power of $$x$$:

 * $$ \begin{align}

q\left(x\right) &= \sum_{i=0}^{4}\ a_{i} P_{i} \\ &= a_{0} + a_{1} x + a_{2} \frac{1}{2}\left(3x^{2}-1\right) + a_{3}\frac{1}{2}\left(5x^{3}-3x\right) + a_{4}\left(\frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8}\right) \\ &= \left(a_{0}-\frac{1}{2}a_{2}+\frac{3}{8}a_{4}\right) + x\left(a_{1}-3a_{3}\right) + x^{2}\left(\frac{3}{2}a_{2}-\frac{15}{4}a_{4}\right) + x^{3}\left(\frac{5}{2}a_{3}\right) + x^{4}\left(\frac{35}{8}a_{4}\right) \end{align} $$

 * Then, we can set each term of this equation equal to the corresponding term in the definition of $$q\left(x\right)$$,

 * $$ \begin{align}

\sum_{i=0}^{4}\ a_{i} &P_{i} = \sum_{i=0}^{4}\ c_{i} x^{i}& \\ \\ &\text{(1)  }a_{0} - \frac{1}{2}a_{2} + \frac{3}{8}a_{4} &= 3 \\ &\text{(2) }a_{1} - 3a_{3} &= 10 \\ &\text{(3) }\frac{3}{2}a_{2} - \frac{15}{4}a_{4} &= 15 \\ &\text{(4) }\frac{5}{2}a_{3} &= -1 \\ &\text{(5) }\frac{35}{8}a_{4} &= 5 \end{align} $$

 * The last two equations yield the values of $$a_{3}$$ and $$a_{4}$$ directly:

 * $$ \begin{align}

a_{3} &= \frac{-2}{5} \\ a_{4} &= \frac{8}{7} \end{align} $$

 * $$a_{2}$$ is then obtained from equation (3) by plugging in $$a_{4}$$:

 * $$ \begin{align}

\frac{3}{2}a_{2} &= 15 + \frac{15}{4}a_{4} \\ a_{2} &= 10 + \frac{10}{4}\cdot \frac{8}{7} \\ &= \frac{90}{7} \end{align} $$

 * Similarly, $$a_{1}$$ is obtained from equation (2) by plugging in $$a_{3}$$:

 * $$ \begin{align}

a_{1} &= 10 + 3a_{3} \\ &= 10 + \frac{-6}{5} \\ &= \frac{44}{5} \end{align} $$

 * Finally, $$a_{0}$$ is obtained by inserting $$a_{2}$$ and $$a_{4}$$ into equation (1):

 * $$ \begin{align}

a_{0} &= 3 + \frac{1}{2}a_{2} - \frac{3}{8}a_{4} \\ &= 3 + \frac{45}{7} - \frac{3}{7} \\ &= 3 + \frac{42}{7} \\ &= 9 \end{align} $$

 * In summary, the following expressions for $$\left\{a_{i}\right\}$$ were obtained:

 * $$ \begin{align}

a_{0} &= 9,

\quad a_{1} = \frac{44}{5}\\ \\ a_{2} &= \frac{90}{7},

\quad a_{3} = \frac{-2}{5}\\ \\ a_{4} &= \frac{8}{7} \end{align} $$ Part b:

=Problem 12=

Part 1
Problem Statement Without calculating, find the properties of $$A_n$$. Solution First, $$A_n$$ is defined by $$A_n=\frac{2n+1}{2}\int^{+1}_{-1}T_0(1-\mu^2)^2P_n(\mu)d\mu,   n=0,1,2,...$$ where $$\mu=sin(\theta), T_0=const., k=1,2,...$$ One of the properties of Legendre polynomials is that $$P_{2k}$$ is an even function and $$P_{2k}$$ is odd. Since $$(1-\mu^2)^2$$ is even and $$P_{2k}(\mu)$$ is also even, the product must therefore be even. It follows that the integral from -1 to 1 is equal to twice the integral from 0 to 1. $$A_{2k}$$ can therefore be written as $$A_{2k}=(4k+1)\int^{+1}_{0}T_0(1-\mu^2)^2P_{2k}(\mu)d\mu$$ $$A_{2k+1}$$ can be written by $$A_{2k+1}=(4k+1)\int^{1}_{-1}T_0(1-\mu^2)^2P_{2k+1}(\mu)d\mu$$ Since $$(1-\mu^2)^2$$ is even and $$P_{2k+1}(\mu)$$ is odd, the product must be even. The integral of an odd function from -1 to 1 is zero. Therefore, $$A_{2k+1}=0$$

Part 2
Problem Statement Compute three non-zero coefficients of $$A_n$$ analytically, using either $$\theta$$ or $$\mu$$ as integrating variables. Solution First, $$P_n(x)$$ is defined by $$P_n(x)=\sum_{r=0}^{m}(-1)^r\frac{(2n-2r)!x^{n-2r}}{2^nr!(n-r)!(n-2r)!}$$ Therefore, the following Legendre Polynomials are given $$P_0(x)=1$$ $$P_1(x)=x$$ $$P_2(x)=\frac{1}{2}(3x^2-1)$$ $$P_3(x)=\frac{1}{2}(5x^3-3x)$$ $$P_4(x)=\frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8}$$ The first three non-zero coefficients are $$A_0$$, $$A_2$$ and $$A_4$$. First, we solve for $$A_0$$. $$A_0=\int^{1}_{0}T_0(1-\mu^2)^2d\mu=T_0\left[\mu-\frac{2}{3}\mu^3+\frac{1}{5}\mu^5\right]^{1}_{0}=\frac{8}{15}T_0$$ Next, $$A_2=5\int^{1}_{0}T_0(1-\mu^2)^2\left[\frac{1}{2}(3\mu^2-1)^2\right]d\mu=\frac{5}{2}T_0\left[\mu^9-\frac{24}{7}\mu^7+\frac{22}{5}\mu^5-\frac{8}{3}\mu^3+\mu\right]^{1}_{0}=-\frac{16}{21}T_0$$ Finally, $$A_4=9\int^{1}_{0}T_0(1-\mu^2)^2\left(\frac{35}{8}\mu^4-\frac{15}{4}\mu^2+\frac{3}{8}\right)d\mu=9T_0\left[\frac{35}{72}\mu^9-\frac{25}{14}\mu^7+\frac{19}{8}\mu^5-\frac{3}{2}\mu^3+\frac{3}{8}\mu\right]^{1}_{0}=\frac{8}{35}T_0$$ =Problem 13=

Problem Statement
Verify the values for $${x_i}$$, $${w_i}$$ and $$P_n(x)$$ in the table for Gauss-Legendre quadrature on the Gaussian Quadrature Wikipedia page.

Solution
For $$P_1=x$$, letting $$x=0$$ yields $$P_1=0$$. Similarly, if we substitute $$x=/pm\sqrt{\frac{1}{3}}$$ into the equation for $$P_2$$, we have $$P_2=\frac{1}{2}(3\frac{1}{3}-1)=0$$. We then perform the same substitution for the remainder of the Legendre Polynomials. $$P_3=\frac{1}{2}=(5(0)^3-3(0))=0$$ and $$P_3=\frac{1}{2}\left(5\left(\frac{3}{5}\right)^{3/2}-3\sqrt{\frac{3}{5}}\right)=0$$ $$P_4=\frac{35}{8}\left(\sqrt{(3+2\sqrt{6/5})/7}\right)^4-\frac{15}{4}\left(\sqrt{(3+2\sqrt{6/5})/7}\right)^2+\frac{3}{8}=0$$ Additionally, $$P_4=\frac{35}{8}\left(\sqrt{(3-2\sqrt{6/5})/7}\right)^4-\frac{15}{4}\left(\sqrt{(3-2\sqrt{6/5})/7}\right)^2+\frac{3}{8}=0$$ Finally, we substitute the roots for the 5th degree Legendre Polynomial, $$P_5$$. $$P_5=\frac{1}{8}\left(63\left(\frac{1}{3}\sqrt{5-2\sqrt{10/7}}\right)^5-70\left(\frac{1}{3}\sqrt{5-2\sqrt{10/7}}\right)^3+15\left(\frac{1}{3}\sqrt{5-2\sqrt{10/7}}\right)\right)=0$$ and $$P_5=\frac{1}{8}\left(63\left(\frac{1}{3}\sqrt{5+2\sqrt{10/7}}\right)^5-70\left(\frac{1}{3}\sqrt{5+2\sqrt{10/7}}\right)^3+15\left(\frac{1}{3}\sqrt{5+2\sqrt{10/7}}\right)\right)=0$$ Next, the weight $${w_i}$$ is given as $$ w_i = \frac{2}{\left( 1-x_i^2 \right) (P'_n(x_i))^2}$$ The first derivatives of the Legendre Polynomials are: $$P'_1=1$$ $$P'_2=3x$$ $$P'_3=\frac{1}{2}(15x^2-3)$$ $$P'_4=\frac{35}{2}x^3-\frac{15}{2}x$$ $$P'_5=\frac{1}{8}\left(\frac{63}{5}x^4-\frac{70}{3}x^2+15\right)$$ Substituting the roots and Legendre Polynomial derivatives into the weight equation yields the following results: $$w_1=2$$ $$w_2=1$$ $$w_{3,1}=\frac{8}{9}$$ $$w_{3,2}=\frac{5}{9}$$ $$w_{4,1}=\frac{8+\sqrt{30}}{36}$$ $$w_{4,2}=\frac{8-\sqrt{30}}{36}$$ $$w_{5,1}=\frac{128}{225}$$ $$w_{5,2}=\frac{322+13\sqrt{70}}{900}$$ $$w_{5,3}=\frac{322-13\sqrt{70}}{900}$$ =Problem 14=

Part 1
Problem From Lecture p.22-2, King p.28 problem 1.1.b: Verify the exactness of the following differential equation using the Integrating Factor Method $$xy''+2y'+xy=0$$ Solution In order to satisfy the first condition of exactness, the differential equation must be able to be expressed using the following functions $$F=f(x,y,p)y''+g(x,y,p)$$ The equation meets the first condition for exactness where $$f(x,y,p)=x$$ $$g(x,y,p)=2p+xy$$ Now, we check to see if the equation satisfies the second condition for exactness where $$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$ $$f_{xp}+pf_{yp}+2f_{y}=g_{pp}$$ Since $$f_{xx}=f_{xy}=f_{yy}=g_{xp}=g_{yp}=0$$ and $$g_y=x$$ the first equation is not satisfied and the differential equation is not exact.

Part 2
Problem Solve the differential equation using the following trial solutions:
 * 2.1) $$y(x)=e^{rx}$$
 * 2.2) $$y(x)=xe^{rx}$$
 * 2.3) $$y(x)=\frac{1}{x}e^{rx}$$

Solution 2.1) By substitution, we find that $$xr^2e^{rx}+2re^{rx}+xe^{rx}=0$$  dividing by $$e^{rx}$$, we have  $$xr^2+2r+x=0$$  Letting $$x=1$$ , we have  $$r^2+2r+x=0$$  Therefore,  $$r_{1,2}=1$$  and the particular solution is  $$y_p=c_1e^{-t}+c_2te^{-t}$$

2.2) By substitution, we have $$x(2re^{rx}+r^2xe^{rx})+2(e^{rx}+rxe^{rx})+x^2e^{rx}=0$$  Dividing by $$e^{rx}$$ and letting $$x=0$$  $$r^2+4r+3=0$$  This yields the following particular solution  $$y_p=c_1e^{-t}+c_2e^{-3t}$$

2.3) By substitution, we have $$x\left(\frac{2}{x^3}e^{rx}-\frac{2r}{x^2}e^{rx}+\frac{r^2}{x}e^{rx}\right)+r\left(-\frac{1}{x^2}e^{rx}+\frac{r}{x}e^{rx}\right)+x\left(\frac{1}{x}e^{rx}\right)=0$$  Dividing by $$e^{rx}$$ and letting $$x=0$$  $$r^2+1=0$$  This yields the following particular solution  $$y_p=c_1cos(t)+c_2sin(t)$$

Part 3
Problem Use the Undetermined Coefficient Method to find $$u_2(x)$$, knowing $$u_1(x)=\frac{sin(x)}{x}$$. Compare to above results.

Solution Part 1: Showing that $$u_1$$ is a solution to the differential equation.

$$y = x^{-1}sinx$$

$$\frac{dy}{dx}=\frac{d}{dx}\left(x^{-1}sinx\right)=-x^{-2}sinx+x^{-1}cosx$$

$$\frac{d^2y}{dx^2}=\frac{d^2}{dx^2}\left(x^{-1}sinx\right)=2x^{-3}sinx-2x^{-2}cosx-x^{-1}sinx$$

Substituting into the original equation, we have:

$$x\left(2x^{-3}sinx-2x^{-2}cosx-x^{-1}sinx\right)+2\left(-x^{-2}sinx+x^{-1}cosx\right)+x\left(x^{-1}sinx\right)=0$$

The left-hand side of the equation above reduces to zero, proving that $$u_1$$ is a solution to the differential equation.

Solution Part 2: Solving for $$u_2$$. We will look for a solution of the form $$y(x)=U(x)u_1(x)$$. First, we represent the first and second derivative as follows:

$$\frac{dy}{dx}=\frac{dU}{dx}u_1+U\frac{du_1}{dx}$$

$$\frac{d^2y}{dx^2}=\frac{d^2U}{dx^2}u_1+2\frac{dU}{dx}\frac{du_1}{dx}+U\frac{d^2u_1}{dx^2}$$

Next, we substitute the above equations into the differential equation.

$$x\left(\frac{d^2U}{dx^2}u_1+2\frac{dU}{dx}\frac{du_1}{dx}+U\frac{d^2u_1}{dx^2}\right)+2\left(\frac{dU}{dx}u_1+U\frac{du_1}{dx}\right)+x\left(U(x)u_1(x)\right)=0$$

We can now collect the terms.

$$U\left(\frac{d^2U}{dx^2}+2\frac{du_1}{dx}+xu_1\right)+xu_1\frac{d^2U}{dx^2}+\frac{dU}{dx}\left(2x\frac{du_1}{dx}+2u_1\right)=0$$

Since $$u_1$$ is a solution of the differential equation, the term multiplying by $$U$$ is zero. Next, we define $$Z:=\frac{dU}{dx},$$ and $$\frac{dZ}{dx}=\frac{d^2U}{dx^2}$$. By removing the $$u_1$$ terms, which are equal to zero and substituting the new definitions, we arrive at the following first-order, linear differential equation:

$$xu_1\frac{dZ}{dx}+Z\left(2x\frac{du_1}{dx}+2u_1\right)=0$$

Dividing by $$Zxu_1$$, we have

$$\frac{1}{Z}\frac{dZ}{dx}+\frac{2}{u_1}\frac{du_1}{dx}+\frac{2}{x}=0$$

By integration, we have

$$log|Z|+2log|u_1|+\int^x\frac{2}{s}ds=C$$

This can be used to solve for $$Z$$.

$$Z=\frac{k}{u_{1}^{2}}exp\left(-\int^x\frac{2}{s}ds\right)=\frac{dU}{dx}$$, where $$k=e^C$$

Integrating the above equation gives us an equation for $$U(x)$$.

$$U(x)=\int^x\frac{c}{u_{1}^{2}(t)}exp\left(-\int^{t}\frac{2}{s}ds\right)dt+\tilde{c}$$

This gives us an equation for $$y(x)$$.

$$y(x)=u_1(x)\int^x\frac{c}{u_{1}^{2}(t)}exp\left(-\int^{t}\frac{2}{s}ds\right)dt+\tilde{c}u_1(x)$$

Finally, we arrive at the equation for $$u_2(x)$$.

$$u_2(x)=u_1(x)\int^{x}\frac{1}{u_{1}^{2}(t)}exp\left(-\int^{t}\frac{2}{s}ds\right)dt$$

By solving the above equation, we arrive at the solution for $$u_2$$.

$$u_2(x)=x^{-1}cosx$$

=Contributing Members= Egm6321.f09.Team6.PaulMoore 10:21, 18 November 2009 (UTC)

Egm6321.f09.team6.breaux 17:19, 18 November 2009 (UTC)

Egm6321.f09.Team6.dianafoster 18:08, 18 November 2009 (UTC)

Egm6321.f09.Team06.sada 18:44, 18 November 2009 (UTC)

title: User:Egm6321.f09.Team6/HW6


text:
User:Lwitiko mwakatuma

I never used to work with Adobe photoshops and illastrution, and I feel to learn more and more so as I can be come a good technician in designing things and just like pictures designing webs or business cards as well I need to have the materials in such way I can do all or try with me alone at home practizing. Thank you--Lwitiko mwakatuma 07:49, 18 November 2009 (UTC)lwitiko metron mwakatuma, BOx 63283. Dar es Salaam. Tanian-East Africa.

title: User:Lwitiko mwakatuma


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User talk:JPLeRouzic

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User talk:Drhalat

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User:Dry.eats/Solving Engineering Problems with Numerical Tools

Many engineering problems are solved using numerical tools. Programming languages such as MATLAB, C++, Python, and FORTRAN include packages to solve engineering problems numerically. The purpose of this wiki is to provide examples for instructors (and students) to understand how to numerically simulate systems from various engineering fields. Below is the current linked list of examples of solving engineering problems numerically. Hopefully this list will grow over time.

Heat Transfer

 * Heat conduction on a spoon in hot tea

Control Theory

 * Proportional-integral controller in a flow system
 * MATLAB system dynamics primer
 * LQR Control for an Inverted Pendulum on a Cart

Numerical Analysis

 * Using MATLAB to show the Lagrange Interpolation

128.32.14.19 17:18, 24 November 2009 (UTC)Devin