User:EGM6321.f10.team1.russo

This is my Principles of Engineering Analysis 1 wiki page. Below are the contributions that I made to my team's assignments. I took this course Fall semester of 2010.

Given
given the boundary values:

$$\begin{align} & y(a)=\alpha \\ & y(b)=\beta \\ \end{align}$$.

Find
Solve for variables c and d in the following equation:


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$$y(x)=cy_{H}^{1}(x)+dy_{H}^{2}(x)+{{y}_{p}}(x)$$

(p5-3)
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Solution
Substituting the boundary point data into the given equation yields:


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$$y(a)= \alpha = cy_{H}^{1}(a)+dy_{H}^{2}(a)+{{y}_{p}}(a)$$ (4.1),(4.2)
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$$y(b)= \beta = cy_{H}^{1}(b)+dy_{H}^{2}(b)+{{y}_{p}}(b)$$ (4.3),(4.4)
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Using algebra to solve equations (4.2) and (4.4) for variable c yields:


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$$c = \frac{1}{y_{H}^{1}(a)}[\alpha - dy_{H}^{2}(a)-{{y}_{p}}(a)]$$ (4.5)
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$$c = \frac{1}{y_{H}^{1}(b)}[\beta - dy_{H}^{2}(b)-{{y}_{p}}(b)]$$ (4.6)
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Setting equations (4.5) and (4.6) equal to one another by recognizing that c=c yields:


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$$\frac{1}{y_{H}^{1}(a)}[\alpha - dy_{H}^{2}(a)-{{y}_{p}}(a)]= \frac{1}{y_{H}^{1}(b)}[\beta - dy_{H}^{2}(b)-{{y}_{p}}(b)]$$ (4.7)
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Solving equation (4.7) for variable d using algebra:


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$$\frac{y_{H}^{1}(b)}{y_{H}^{1}(a)}[\alpha - dy_{H}^{2}(a)-{{y}_{p}}(a)]= \beta - dy_{H}^{2}(b)-{{y}_{p}}(b)$$ (4.8)
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$$\frac{y_{H}^{1}(b)}{y_{H}^{1}(a)}[\alpha -{{y}_{p}}(a)]+{{y}_{p}}(b)-\beta = \frac{y_{H}^{1}(b)}{y_{H}^{1}(a)}dy_{H}^{2}(a) - dy_{H}^{2}(b) = d(\frac{y_{H}^{1}(b)}{y_{H}^{1}(a)}y_{H}^{2}(a) - y_{H}^{2}(b))$$ (4.9),(4.10)
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$$d = \frac{\frac{y_{H}^{1}(b)}{y_{H}^{1}(a)}[\alpha -{{y}_{p}}(a)]+{{y}_{p}}(b)-\beta}{\frac{y_{H}^{1}(b)}{y_{H}^{1}(a)}y_{H}^{2}(a) - y_{H}^{2}(b)}$$ (4.11)
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Equation (4.11) is the final answer for variable d.

Now, variable c may be found by substituting equation (4.11) into equation (4.5):

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$$c = \frac{1}{y_{H}^{1}(a)}[\alpha - \frac{\frac{y_{H}^{1}(b)}{y_{H}^{1}(a)}[\alpha -{{y}_{p}}(a)]+{{y}_{p}}(b)-\beta}{\frac{y_{H}^{1}(b)}{y_{H}^{1}(a)}y_{H}^{2}(a) - y_{H}^{2}(b)}y_{H}^{2}(a)-{{y}_{p}}(a)]$$ (4.12)
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Equation (4.12) is the final answer for variable c, and so the solution is now complete.

EGM6321.f10.team1.russo 22:23, 13 September 2010 (UTC)

Given
<span id="(1)">
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$$ \varnothing(x,y)= x^2y^\frac{3}{2}+log(x^3y^2)=k$$ (4.1)
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Find
1)$$ F(x,y,y')=\frac{d\varnothing}{dx}(x,y)$$ 2) Verify that F is exact N1_ODE 3) Invent three more N1_ODE's

Solution
This problem asks for F(x,y,y') and for the verification that F is exactly N1_ODE. In other words, verify that there exists a function $$ \phi(x,y)\ $$ such that: <span id="(1)">
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$$ F = \frac{d\phi}{dx}$$ (4.2)
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In other terms, verify that:

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$$F(x,y,y') = \frac{d}{dx}\phi(x,y) = \frac{d}{dx}k$$ (4.3, 4.4)
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Solving for the partial derivatives in equations 2 and 3 yields:

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$$F(x,y,y') = \frac{d\phi}{dx}(x,y) + \frac{d\phi}{dx}(x,y)\frac{dy}{dx} = M(x,y) + N(x,y)\frac{dy}{dx} = 0$$ (4.5, 4.6 and 4.7)
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Where:

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$$M(x,y) = \frac{d\phi}{dx}(x,y) = (2)(y^{3/2})x + \frac{3}{x}$$ (4.8, 4.9)
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$$N(x,y) = \frac{d\phi}{dx}(x,y) = \frac{3}{2}(x^2)(y^{1/2})x + \frac{2}{y}$$ (4.10, 4.11)
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1) and 2) Substituting equations 4.9 and 4.11 into equation 4.5 yields the final answer for F(x,y,y') (4.12) and also verifies that it is exactly N1_ODE (4.13)

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$$F(x,y,y') = ((2)(y^{3/2})x + \frac{3}{x}) + (\frac{3}{2}(x^2)(y^{1/2})x + \frac{2}{y})y' = 0$$ (4.12, 4.13)
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3) Below are 3 examples of N1_ODEs

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$$ F(x,y,y') = 22x^2y + y^5y'\ $$ (4.15) <span id="(1)">
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$$ F(x,y,y') = sin(x^{7/2}) - \frac{4}{3}y^6y'\ $$ (4.16) <span id="(1)">
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$$ F(x,y,y') = x^8 + y^4y'\ $$ (4.17)
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Author: User:EGM6321.f10.team1.russo

Find
Use the integrating factor method to obtain the result of eq3.1 directly

Solution
The given solution is one of a first order linear ODE. The derivative of such an equation has the form:

Where $$\ \tau\, $$ is the constant of integration. We will assume the case of constant coefficients, so this equation becomes:

Equation 3.3 can be rearranged as:

We desire a solution for $$\ x(\tau\,) $$, which may be found using the integration by factors method. First, the function $$\ h(\tau\,) $$ must be found:

Multiplying both sides of equation 3.4 by $$\ h(\tau\,) $$ yields:

This equation may be simplified by recalling that $$\ h(\tau\,)a(\tau\,) = \dot{h}(\tau\,) $$:

Substituting equation 3.5 into equation 3.7 for $$h(\tau\,)$$ yields:

It is now possible to find an expression for x(t) by integrating both sides of this equation with respect to $$\ \tau\, $$:

Performing this integration yields:

By performing some algebra on this equation, x(t) is found:

Author: EGM6321.f10.team1.russo 07:15, 6 October 2010 (UTC)

Find
Find $$ y_{xxxxx}\ $$ in terms of the derivative of y with respect to t

Solution
Author Egm6321.f10.team1.allison 17:08, 21 October 2010 (UTC), EGM6321.f10.team1.russo 16:31, 2 November 2010 (UTC)

Given
and

Find
Show that eq 6.1 agrees with with expression in King, page 8, eq 1.6:

Where:

Solution
Equation (6.2) may be rearranged as follows:

Taking the derivative of both sides of equation (6.5) yields:

Carrying out the derivative of the left side of equation (6.6) yields:

Upon inspection, one can notice a relationship between equations (6.7) and (6.4), which may be written as:

It is now possible to develop an expression for $$ h(x)\ $$ by setting equations (6.6) and (6.8) equal to one another and rearranging:

King's expression (equation (6.3)) may be rearranged, and its integration variables substituted, yielding: Where the two homogeneous solutions, namely $$ Au_1(x)\ $$ and $$ Bu_2(x)\ $$ have been omitted so that only the particular solution is considered.

The objective now is to use rearrangement and substitution to put the particular solution in a form equivalent to equation (6.10). The rule of Integration by Parts may be written as follows:

Rearranging this rule into an applicable form yields:

It can easily be seen that the left side of equation (6.12) is of nearly the same form as the particular solution, equation (6.1).

To apply the Integration by Parts rule to the particular solution, the following substitutions may be made: and

Substituting equations (6.13),(6.14) and (6.15) into equation (6.12) yields:

Multiplying both sides of equation (6.15) by $$ u_1(x)\ $$ and substituting equation (6.9) for $$ h(x)\ $$ in the right side of (6.16) yields:

By inspection, the left side of equation (6.17) is exactly the particular solution from equation (6.1). Thus, after rearranging the right side of equation (6.17) and canceling out like terms, the particular solution may be written as:

It is my belief that the given particular solution from lecture 30-1 equation 2 is a misprint. If the particular equation had been given as:

Then equation (6.18) would instead have become:

Which is equivalent to King's expression for y(x) as stated in equation (6.10).

Author EGM6321.f10.team1.russo 00:58, 3 November 2010 (UTC)

Given
Where $$\ [n/2] $$ = the integer part of $$\ n/2 $$

Find
Verify that Eq8.3 thru Eq8.7 can be written in the form of Eq8.1 or Eq8.2

Solution
Because equations 8.1 and 8.2 are equivalent, it is only necessary to prove that equations 8.3 through 8.7 may be written in the form of equation 8.1.

1) Verify that equation 8.3 may be written in the form of equation 8.1.

Equation 8.3 is given as:

By observing the subscript in equation 8.3, the two required parameters may be found: and

Equation 8.1 now becomes:

Performing the summation in equation 8.10 yields:

After performing the necessary calculations, equation 8.11 becomes:

This expression of equation 8.1 is equivalent to equation 8.3.

Therefore, equation 8.3 may be written as equation 8.1

2) Verify that equation 8.4 may be written in the form of equation 8.1.

Equation 8.4 is given as:

By observing the subscript in equation 8.4, the two required parameters may be found: and

Equation 8.1 now becomes:

Performing the summation in equation 8.15 yields:

After performing the necessary calculations, equation 8.16 becomes:

This expression of equation 8.1 is equivalent to equation 8.4.

Therefore, equation 8.4 may be written as equation 8.1

3) Verify that equation 8.5 may be written in the form of equation 8.1.

Equation 8.5 is given as:

By observing the subscript in equation 8.5, the two required parameters may be found: and

Equation 8.1 now becomes:

Performing the summation in equation 8.20 yields:

After performing the necessary calculations, equation 8.21 becomes:

This expression of equation 8.1 is equivalent to equation 8.5.

Therefore, equation 8.5 may be written as equation 8.1

4) Verify that equation 8.6 may be written in the form of equation 8.1.

Equation 8.6 is given as:

By observing the subscript in equation 8.6, the two required parameters may be found: and

Equation 8.1 now becomes:

Performing the summation in equation 8.25 yields:

After performing the necessary calculations, equation 8.26 becomes:

This expression of equation 8.1 is equivalent to equation 8.6.

Therefore, equation 8.6 may be written as equation 8.1

5) Verify that equation 8.7 may be written in the form of equation 8.1.

Equation 8.7 is given as:

By observing the subscript in equation 8.7, the two required parameters may be found: and

Equation 8.1 now becomes:

Performing the summation in equation 8.30 yields:

After performing the necessary calculations, equation 8.31 becomes:

This expression of equation 8.1 is equivalent to equation 8.7.

Therefore, equation 8.7 may be written as equation 8.1

It has thus been verified that equations 8.3 through 8.7 may be written as such.

Therefore, equations 8.3 through 8.7 may be written in the form of equation 8.1 or 8.2

Author EGM6321.f10.team1.russo 02:40, 12 November 2010 (UTC)

Find
Verify that Eq9.3 thru Eq9.7 are solutions of Legendre Equations as given in Eq9.1 or Eq9.2

Solution
Since equations 9.2 and 9.1 are identical, it is only necessary to prove that equations 9.3 through 9.7 are solutions to equation 9.1. This will be accomplished by taking the first and second derivatives of each potential solution equation, substituting the results into equation 9.1, and then verifying that there exists an integer $$\ n $$ for which the Legendre differential equation is satisfied:

1) Verify that equation 9.3 is a solution of equation 9.1:

Begin with the potential solution:

Taking the first derivative of equation 9.3 yields:

and taking the second derivative of equation 9.3 yields:

Substituting $$\ P_0(x)$$, $$\ P_0'(x)$$, and $$\ P_0''(x) $$ into equation 9.1 yields:

Carrying out the multiplication yields:

It can be seen that there are two values of the parameter n that satisfy this condition.

When $$\ n=0 $$, equation 9.11 becomes:

And when $$\ n=-1 $$ equation 9.11 becomes:

when $$\ n=0 $$ or when $$\ n=-1 $$

2) Verify that equation 9.4 is a solution of equation 9.1:

Begin with the potential solution:

Taking the first derivative of equation 9.4 yields:

and taking the second derivative of equation 9.4 yields:

Substituting $$\ P_1(x)$$, $$\ P_1'(x)$$, and $$\ P_1''(x) $$ into equation 9.1 yields:

Carrying out the multiplication yields:

Dividing both sides of equation 9.18 by $$\ x $$ and rearranging the remaining terms yields: It can be seen that there are two values of the parameter n that satisfy this condition.

When $$\ n=1 $$, equation 9.19 becomes:

And when $$\ n=-2 $$ equation 9.19 becomes:

when $$\ n=1 $$ or when $$\ n=-2 $$

3) Verify that equation 9.5 is a solution of equation 9.1:

Begin with the potential solution:

Taking the first derivative of equation 9.5 yields:

and taking the second derivative of equation 9.5 yields:

Substituting $$\ P_2(x)$$, $$\ P_2'(x)$$, and $$\ P_2''(x) $$ into equation 9.1 yields:

Carrying out the multiplication yields:

Rearranging equation 9.26 yields:

Dividing both sides of equation 9.27 by $$\ \frac{1}{2}(3x^2 - 1) $$ yields: It can be seen that there are two values of the parameter n that satisfy this condition.

When $$\ n=2 $$, equation 9.28 becomes:

And when $$\ n=-3 $$ equation 9.28 becomes:

when $$\ n=2 $$ or when $$\ n=-3 $$

4) Verify that equation 9.6 is a solution of equation 9.1:

Begin with the potential solution:

Taking the first derivative of equation 9.6 yields:

and taking the second derivative of equation 9.6 yields:

Substituting $$\ P_3(x)$$, $$\ P_3'(x)$$, and $$\ P_3''(x) $$ into equation 9.1 yields:

Carrying out the multiplication yields:

Rearranging equation 9.35 yields:

Dividing both sides of equation 9.36 by $$\ \frac{1}{2}(5x^3 - 3x) $$ yields: It can be seen that there are two values of the parameter n that satisfy this condition.

When $$\ n=3 $$, equation 9.37 becomes:

And when $$\ n=-4 $$ equation 9.37 becomes:

when $$\ n=3 $$ or when $$\ n=-4 $$

4) Verify that equation 9.7 is a solution of equation 9.1:

Begin with the potential solution:

Taking the first derivative of equation 9.7 yields:

and taking the second derivative of equation 9.7 yields:

Substituting $$\ P_4(x)$$, $$\ P_4'(x)$$, and $$\ P_4''(x) $$ into equation 9.1 yields:

Carrying out the multiplication yields:

Rearranging equation 9.44 yields:

Dividing both sides of equation 9.45 by $$\ \left(\frac{35}{8}x^4 - \frac{15}{4}x^2 + \frac{3}{8}\right) $$ yields: It can be seen that there are two values of the parameter n that satisfy this condition.

When $$\ n=4 $$, equation 9.46 becomes:

And when $$\ n=-5 $$ equation 9.46 becomes:

when $$\ n=4 $$ or when $$\ n=-5 $$

It has thus been verified that equations 9.3 through 9.7 are solutions to equation 9.1.

Therefore, equations 9.3 through 9.7 are valid solutions to Legendre's differential equation as shown in equation 9.1

Author EGM6321.f10.team1.russo 23:08, 11 November 2010 (UTC)

Given
Using either: or:

Find
A) Prove that Eq 3.3 is even for k=0,1,2...

B} Prove that Eq 3.4 is odd for k=0,1,2...

Solution
A) Prove that Equation 3.3 is even for k=0,1,2,... using Equation 3.1:

To prove that Equation 3.3 is even, it must be shown that it fulfills the definition of an even function, which is as follows:

Let f(x) be a real-valued function of a real variable. Then f is even if the following equation holds for all x in the domain of f:

Therefore, it must be shown that the following condition is satisfied:

This will be shown using Equation 3.1. First, it must be put into the proper form. Upon inspection, it can be seen that $$\ n = 2k $$ and that $$\ x = \mu $$ for equation 3.3.

Substituting these parameters into equation 3.1 yields:

Performing some of the algebra in equation 3.6 yields:

If $$ \mu $$ were replaced with $$\ -\mu $$, then equation 3.1 would instead have become:

The only term in equation 3.7 that contains the variable $$\ \mu $$ is $$\ (\mu^2)^{k - i} $$, while the only term in equation 3.8 containing $$\ \mu $$ is $$\ ((-\mu)^2)^{k - i} $$. Since $$\ \mu $$ and $$\ k $$ are integers, $$\ \mu^2 $$ must equal $$\ (-\mu)^2 $$.

This means that 3.7 must equal equation 3.8, and so condition 3.5 is satisfied.

B) Prove that Equation 3.4 is odd for k=0,1,2,... using Equation 3.1:

To prove that Equation 3.4 is odd, it must be shown that it fulfills the definition of an odd function, which is as follows:

Again, let f(x) be a real-valued function of a real variable. Then f is odd if the following equation holds for all x in the domain of f:

Therefore, it must be shown that the following condition is satisfied:

This will be shown using Equation 3.1. First, it must be put into the proper form. Upon inspection, it can be seen that $$\ n = 2k+1 $$ and that $$\ x = \mu $$ for equation 3.4.

Substituting these parameters into equation 3.1 yields:

Performing some of the algebra in equation 3.10 and multiplying both sides by -1 yields:

If $$ \mu $$ were replaced with $$\ -\mu $$, then equation 3.1 would instead have become:

By bringing the negative sign from the term $$\ (-\mu) $$ to the front of equation 3.12, this may be rewritten as:

Recalling the fact that $$\ \mu^2 $$ must equal $$\ (-\mu)^2 $$ as discussed in part A above, equation 3.11 must equal equation 3.13, and so condition 3.9 is satisfied.

The work of Team 1 from the Fall 2009 term was referenced during this solution. Specifically, the team's solution to Problem 1 of Homework 7 was viewed to find out how the definition of an even or odd function corresponds to equation 3.1. This reference may be found at the following link:

While the referenced material simply mentions the correlations between equation 3.1 and the properties of even and odd functions, the above solution begins with these properties and proves that they are followed by equations 3.3 and 3.4.

Author EGM6321.f10.team1.russo 22:54, 22 November 2010 (UTC)

Given
From the solution to Problem 7.8 above, we are given equations 8.7 and 8.8, which are simply renamed below:

And

Find
Continue Power Series expansion of Eq10.1 and use equations 10.4-10.6 to find $$ \left \{ P_3, P_4, P_5, P_6 \right \} \ $$ and compare results with those obtained by a) Solutions from HW6.8

b) Solutions from HW7.9

Solution
$$\begin{align} & \text{The generating function for the Legendre Polynomials was derived during meeting} \\ & \text{40 so therefore we will just present the result as follows:} \\ & \\  & \left[ A\left( \mu ,\rho  \right) \right]^{-\frac{1}{2}}=\frac{1}{\sqrt{1-2\mu \rho +\rho ^{2}}}=\frac{1}{\sqrt{1-x}}\text{              where }x=2\mu \rho -\rho ^{2} \\ & \\  & \text{We would like to expand the generating function as a power series in terms of }\rho \text{ to the} \\ & \text{order 6}\text{. This is done by the following:} \\ & \\  & \left[ A\left( \mu ,\rho  \right) \right]^{-\frac{1}{2}}=\sum\limits_{i=0}^{6}{\alpha _{i}x^{i}}=\alpha _{0}+\alpha _{1}x+\alpha _{2}x^{2}+\alpha _{3}x^{3}+\alpha _{4}x^{4}+\alpha _{5}x^{5}+\alpha _{6}x^{6} \\ & \\  & \text{We can then use the given information, as well as equation 10}\text{.8 to solve for the }\alpha _{i}\text{ }\!\!'\!\!\text{ s}\text{.} \\ & \\  & \alpha _{0}=1 \\ & \alpha _{1}=\frac{1}{2} \\ & \alpha _{2}=\frac{1\cdot 3}{2\cdot 4}=\frac{3}{8} \\ & \alpha _{3}=\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}=\frac{15}{48}=\frac{5}{16} \\ & \alpha _{4}=\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}=\frac{105}{384}=\frac{35}{128} \\ & \alpha _{5}=\frac{1\cdot 3\cdot 5\cdot 7\cdot 9}{2\cdot 4\cdot 6\cdot 8\cdot 10}=\frac{945}{3840}=\frac{63}{256} \\ & \alpha _{6}=\frac{1\cdot 3\cdot 5\cdot 7\cdot 9\cdot 11}{2\cdot 4\cdot 6\cdot 8\cdot 10\cdot 12}=\frac{10395}{46080}=\frac{231}{1024} \\ & \\  & \text{Now we use our definition for }x\text{ to organize the correct }\alpha _{i}'s\text{ with the correct powers of }\rho. \\ & \text{To we use the integer form of the binomial expansion to expand all }x\text{ expressions}\text{. The} \\ & \text{binomial theorem is given as follows:} \\ & \\  & \left( a+b \right)^{n}=\sum\limits_{r=0}^{n}{\frac{n!}{r!\left( n-r \right)!}a^{n-r}b^{r}} \\ & \\  & x=2\mu \rho -\rho ^{2} \\ & \\  & x^{2}=\left( 2\mu \rho -\rho ^{2} \right)^{2}=4\mu ^{2}\rho ^{2}-4\mu \rho ^{3}+\rho ^{4} \\ & \\  & x^{3}=\left( 2\mu \rho -\rho ^{2} \right)^{3}=\sum\limits_{r=0}^{3}{\frac{3!}{r!\left( 3-r \right)!}\left( 2\mu \rho  \right)^{3-r}\left( -\rho ^{2} \right)^{r}} \\ & \text{  }=\frac{3\cdot 2\cdot 1}{1\left( 3\cdot 2\cdot 1 \right)}\cdot 8\mu ^{3}\rho ^{3}-\frac{3\cdot 2\cdot 1}{1\left( 2\cdot 1 \right)}\cdot 4\mu ^{2}\rho ^{4}+\frac{3\cdot 2\cdot 1}{2\cdot 1\left( 1 \right)}\cdot 2\mu \rho ^{5}-\frac{3\cdot 2\cdot 1}{3\cdot 2\cdot 1\left( 1 \right)}\rho ^{6} \\ & \therefore x^{3}=8\mu ^{3}\rho ^{3}-12\mu ^{2}\rho ^{4}+6\mu \rho ^{5}-\rho ^{6} \\ & \\  & x^{4}=\left( 2\mu \rho -\rho ^{2} \right)^{4}=\sum\limits_{r=0}^{4}{\frac{4!}{r!\left( 4-r \right)!}\left( 2\mu \rho  \right)^{4-r}\left( -\rho ^{2} \right)^{r}} \\ & \text{  }=\frac{4\cdot 3\cdot 2\cdot 1}{1\left( 4\cdot 3\cdot 2\cdot 1 \right)}16\mu ^{4}\rho ^{4}-\frac{4\cdot 3\cdot 2\cdot 1}{1\left( 3\cdot 2\cdot 1 \right)}8\mu ^{3}\rho ^{5}+\frac{4\cdot 3\cdot 2\cdot 1}{2\cdot 1\left( 2\cdot 1 \right)}4\mu ^{2}\rho ^{6}-\frac{4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot 1\left( 1 \right)}2\mu \rho ^{7}+\frac{4\cdot 3\cdot 2\cdot 1}{4\cdot 3\cdot 2\cdot 1\left( 1 \right)}\rho ^{8} \\ & \text{  }=16\mu ^{4}\rho ^{4}-32\mu ^{3}\rho ^{5}+24\mu ^{2}\rho ^{6}-8\mu \rho ^{7}+\rho ^{8} \\ & \\  & x^{5}=\left( 2\mu \rho -\rho ^{2} \right)^{5}=\sum\limits_{r=0}^{5}{\frac{5!}{r!\left( 5-r \right)!}\left( 2\mu \rho  \right)^{5-r}\left( -\rho ^{2} \right)^{r}} \\ & \text{  }=\frac{5\cdot 4\cdot 3\cdot 2\cdot 1}{1\left( 5\cdot 4\cdot 3\cdot 2\cdot 1 \right)}32\mu ^{5}\rho ^{5}-\frac{5\cdot 4\cdot 3\cdot 2\cdot 1}{1\left( 4\cdot 3\cdot 2\cdot 1 \right)}16\mu ^{4}\rho ^{6}+\frac{5\cdot 4\cdot 3\cdot 2\cdot 1}{2\cdot 1\left( 3\cdot 2\cdot 1 \right)}8\mu ^{3}\rho ^{7}-\frac{5\cdot 4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot 1\left( 2\cdot 1 \right)}4\mu ^{2}\rho ^{8}+ \\ & \text{     }\frac{5\cdot 4\cdot 3\cdot 2\cdot 1}{4\cdot 3\cdot 2\cdot 1\left( 1 \right)}2\mu \rho ^{9}-\frac{5\cdot 4\cdot 3\cdot 2\cdot 1}{5\cdot 4\cdot 3\cdot 2\cdot 1\left( 1 \right)}\rho ^{10} \\ & \\  & \therefore x^{5}=32\mu ^{5}\rho ^{5}-80\mu ^{4}\rho ^{6}+80\mu ^{3}\rho ^{7}-40\mu ^{2}\rho ^{8}+10\mu \rho ^{9}-\rho ^{10} \\ & \\  & x^{6}=\left( 2\mu \rho -\rho ^{2} \right)^{6}=\sum\limits_{r=0}^{6}{\frac{6!}{r!\left( 6-r \right)!}\left( 2\mu \rho  \right)^{6-r}\left( -\rho ^{2} \right)^{r}} \\ & \text{  }=\frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{1\left( 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1 \right)}64\mu ^{6}\rho ^{6}-\frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{1\left( 5\cdot 4\cdot 3\cdot 2\cdot 1 \right)}32\mu ^{5}\rho ^{7}+\frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{2\cdot 1\left( 4\cdot 3\cdot 2\cdot 1 \right)}16\mu ^{4}\rho ^{8}-\frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{3\cdot 2\cdot 1\left( 3\cdot 2\cdot 1 \right)}8\mu ^{3}\rho ^{9}+ \\ & \text{   }\frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{4\cdot 3\cdot 2\cdot 1\left( 2\cdot 1 \right)}4\mu ^{2}\rho ^{10}-\frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{5\cdot 4\cdot 3\cdot 2\cdot 1\left( 1 \right)}2\mu \rho ^{11}+\frac{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1}{6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1\left( 1 \right)}\rho ^{12} \\ & \text{ } \\ & \therefore x^{6}=64\mu ^{6}\rho ^{6}-192\mu ^{5}\rho ^{7}+240\mu ^{4}\rho ^{8}-160\mu ^{3}\rho ^{9}+60\mu ^{2}\rho ^{10}-12\mu \rho ^{11}+\rho ^{12} \\ & \\  & \text{After plugging in the values for the respective }x\text{ }\!\!'\!\!\text{ s and }\alpha \text{ }\!\!'\!\!\text{ s, we group the terms together which have common } \\ & \text{factors of }\rho ^{i}.\text{ Each of these factors represent an order of expansion and their respective coeffiecents are} \\ & \text{Legendre polynomials of their respective order}\text{. We were asked to calculate up to order 6, so after } \\ & \text{rearranging and factoring the expansion looks as follows}\text{.} \\ & \\  & \frac{1}{\sqrt{1-2\mu \rho +\rho ^{2}}}=1+\mu \rho +\frac{1}{2}\left( 3\mu ^{2}-1 \right)\rho ^{2}+\frac{1}{2}\left( 5\mu ^{3}-3\mu  \right)\rho ^{3}+\left( \frac{35}{8}\mu ^{4}-\frac{15}{4}\mu ^{2}+\frac{3}{8} \right)\rho ^{4}+\left( \frac{63}{8}\mu ^{5}-\frac{35}{4}\mu ^{3}+\frac{15}{8}\mu  \right)\rho ^{5} \\ & \text{                       +}\left( \frac{231}{16}\mu ^{6}-\frac{315}{16}\mu ^{4}+\frac{105}{16}\mu ^{2}-\frac{5}{16} \right)\rho ^{6} \\ \end{align}$$

PART A)

In Lecture 36, we were given the following two identical equations:

For homework problem 6.8, the results of these two equations were found for n=[0,4] as follows:

Using equation 10.21 to find $$\ P_5(x) $$ yields:

Likewise, equation 10.21 may be used to find $$\ P_6(x) $$, yielding:

Upon inspection, it is obvious that equation 10.18 is equivalent to 10.26, 10.19 is equivalent to 10.27, 10.20 is equivalent to 10.29, and 10.21 is equivalent to 10.31.

Therefore, the power series expansion of equation 10.1 yields the same results as equations 10.21 and 10.22

PART B)

$$\begin{align} & \text{We can see just by inspection the Legendre Polynomials which were derived by the recurrence } \\ & \text{relation in Problem 6 above, and the Legendre Polynomials which were dervied by power series} \\ & \text{expansion in the first part of Problem 10 here are equal}\text{.} \\ & \text{The Legendre polynomials found in Problem 6 are as follows:} \\ & \\  & P_{3}\left( \mu  \right)=\frac{1}{2}\left( 5\mu ^{3}-3\mu  \right) \\ & P_{4}\left( \mu \right)=\frac{35}{8}\mu ^{4}-\frac{15}{4}\mu ^{2}+\frac{3}{8} \\ & P_{5}\left( \mu \right)=\frac{63}{8}\mu ^{5}-\frac{35}{4}\mu ^{3}+\frac{15}{8}\mu  \\ & P_{6}\left( \mu \right)=\frac{1}{16}\left( 231\mu ^{6}-315\mu ^{4}+105\mu ^{2}-5 \right) \\ & \\  & \text{The Legendre Polynomials found in the first part of Problem 10 are equal to the above}\text{.} \\ \end{align}$$

$$\begin{align} & \text{Because we see that we get the same exact solutions by expanding the generating function by} \\ & \text{power series as we do using the 2 general equations for the Legendre Polynomials, there must be} \\ & \text{a way to go directly from the generating function to the general expression}\text{. We know from the } \\ & \text{above power series expansion the following is true}\text{.} \\ & \\  & \frac{1}{\sqrt{1-2\mu \rho +\rho ^{2}}}=\sum\limits_{n=0}^{\infty }{P_{n}\left( \mu  \right)\rho ^{n}} \\ & \\  & \text{The goal is to seek a path to directly derive a general expression for }P_{n}\left( \mu  \right)\text{ by using the binomial} \\ & \text{expansion on the Legendre generating function}\text{. The binomial expansion is defined as follows:} \\ & \\  & \left( a+b \right)^{n}=\sum\limits_{r=0}^{n}{\frac{n!}{r!\left( n-r \right)!}}a^{n-r}b^{r} \\ & \\  & \text{We know that a factoral is defined for only positive integers, but it }\!\!'\!\!\text{ s obvious if we are to move directly} \\ & \text{from the generating function into an expansion we need to be creative in our approach}\text{. In this light let }\!\!'\!\!\text{ s} \\ & \text{assume for a second that we can have a factorial of a negative non-integer}\text{. If we can manipulate the } \\ & \text{factorial in such a way that in the end, all the undefined characteristics of the factorial; i}\text{.e}\text{. the negative} \\ & \text{non-integer parts are removed, then we have a direct method of expressing the generating function in} \\ & \text{a general sense and should be able to derive the following equation}\text{.} \\ \end{align}$$

$$\begin{align} & \\  & P_{n}\left( \mu  \right)=\sum\limits_{i=0}^{\frac{n}{2}}{\left( -1 \right)^{i}\frac{\left( 2n-2i \right)!x^{n-2i}}{2^{n}i!\left( n-i \right)!\left( n-2i \right)!}} \\ & \\  & \text{Therefore let }\!\!'\!\!\text{ s assume the following is valid and see what pops out}\text{.} \\ \end{align}$$

$$\begin{align} & \\  & \left( 1-x \right)^{-\frac{1}{2}}=\sum\limits_{i}^{\infty }{\frac{\left( -\frac{1}{2} \right)!\left( -1 \right)^{i}}{i!\left( -\frac{1}{2}-i \right)!}x^{i}} \\ \end{align}$$

$$\begin{align} & \\  & \text{We see the problem up front lies in the factor of the binomial coefficient, } \\ & \\  & \frac{\left( -\frac{1}{2} \right)!}{\left( -\frac{1}{2}-i \right)!} \\ & \\ \end{align}$$

$$\begin{align} & \text{We know right away we can deal with the negative number issue by multiplying through by 1}^{i}.\text{ So let }\!\!'\!\!\text{ s } \\ & \text{expand out the first few terms and see if some common factors will pop out of both the numerator and } \\ & \text{the denominator}\text{. Remember, because these expressions do not exist in a sense that there is not a factoral } \\ & \text{definition for them, we can not make any assumption about the tail end of these expressions either in the } \\ & \text{numerator or the denominator}\text{. We must leave them trailing off; i}\text{.e}\text{. they just keep going and going to infinity}\text{.} \\ \end{align}$$

$$\begin{align} & \\  & \frac{\left( -\frac{1}{2} \right)\left( -\frac{1}{2}-1 \right)\left( -\frac{1}{2}-2 \right)\left( -\frac{1}{2}-3 \right)............................................}{\left( -\frac{1}{2}-i \right)\left( -\frac{1}{2}-i-1 \right)\left( -\frac{1}{2}-i-2 \right)\left( -\frac{1}{2}-i-3 \right).........................} \\ & \\  & \text{Now let }\!\!'\!\!\text{ s get rid of }i\text{ number of negative terms from the numerator}\text{.} \\ & \\  & \frac{\left( \frac{-1}{-1} \right)^{i}\left( -\frac{1}{2} \right)\left( -\frac{1}{2}-1 \right)\left( -\frac{1}{2}-2 \right)\left( -\frac{1}{2}-3 \right)............................................}{\left( -\frac{1}{2}-i \right)\left( -\frac{1}{2}-i-1 \right)\left( -\frac{1}{2}-i-2 \right)\left( -\frac{1}{2}-i-3 \right)...................................} \\ & \\  & \frac{\left( -1 \right)^{i}\left( \frac{1}{2} \right)\left( \frac{1}{2}+1 \right)\left( \frac{1}{2}+2 \right)\left( \frac{1}{2}+3 \right)..........\left( \frac{1}{2}+i-1 \right)\left( -\frac{1}{2}-i \right)\left( -\frac{1}{2}-i-1 \right)\left( -\frac{1}{2}-i-2 \right)\left( -\frac{1}{2}-i-3 \right).......}{\left( -\frac{1}{2}-i \right)\left( -\frac{1}{2}-i-1 \right)\left( -\frac{1}{2}-i-2 \right)\left( -\frac{1}{2}-i-3 \right)........................................................................} \\ & \\  & \text{What we find is that the entire denominator (all terms to infinity) divide out all the negative terms} \\ & \text{in the numerator, therefore leaving the fraction free of negative terms spanning all the way to infinity}\text{.} \\ & \text{Even though we know nothing about these terms in the beginning, we know under these manipulations} \\ & \text{they do they are irrelevant to our solution}\text{. } \\ \end{align}$$

$$\begin{align} & \\  & \left( -1 \right)^{i}\left( \frac{1}{2} \right)\left( \frac{1}{2}+1 \right)\left( \frac{1}{2}+2 \right)\left( \frac{1}{2}+3 \right)........\left( \frac{1}{2}+i-1 \right) \\ & \\  & \left( -1 \right)^{i}\left( \frac{1}{2} \right)\left( \frac{3}{2} \right)\left( \frac{5}{2} \right)\left( \frac{7}{2} \right)........\left( \frac{2i-1}{2} \right) \\ & \\  & \frac{\left( -1 \right)^{i}1\cdot 3\cdot 5\cdot 7.........\left( 2i-1 \right)}{2^{i}} \\ & \\  & \text{Because we are looking for a way to expand in terms of the binomial expansion, we want to derive an} \\ & \text{expression that will be factorial in nature}\text{. Therefore we multiply the numerator by the missing even } \\ & \text{numbers}\text{.} \\ & \\  & \frac{\left( -1 \right)^{i}1\cdot 3\cdot 5\cdot 7.........\left( 2i-1 \right)}{2^{i}}\cdot \frac{2\cdot 4\cdot 6........2i}{2\cdot 4\cdot 6........2i} \\ & \\  & \frac{\left( -1 \right)^{i}1\cdot 3\cdot 5\cdot 7.........\left( 2i-1 \right)\cdot \left( 2\cdot 4\cdot 6........2i \right)}{2^{i}\cdot \left( 2\cdot 4\cdot 6........2i \right)} \\ & \\  & \frac{\left( -1 \right)^{i}1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7.........\left( 2i-1 \right)\cdot 2i}{2^{i}\cdot \left[ 2^{i}\left( 1\cdot 2\cdot 3\cdot 4........\text{ }i \right) \right]} \\ & \\  & \frac{\left( -1 \right)^{i}2i\cdot \left( 2i-1 \right).........3\cdot 2\cdot 1}{2^{2i}\left( i \right)!} \\ & \\  & \frac{\left( -1 \right)^{i}\left( 2i \right)!}{2^{2i}i!} \\ & \\  & \text{Therefore, we have successfully derived an expression for our negative, non-integer factorial}\text{.} \\ & \\  & \left( 1-x \right)^{-\frac{1}{2}}=\sum\limits_{i}^{\infty }{\frac{\left( -\frac{1}{2} \right)!\left( -1 \right)^{i}}{i!\left( -\frac{1}{2}-i \right)!}x^{i}}=\sum\limits_{i}^{\infty }{\frac{\left( -1 \right)^{2i}\left( 2i \right)!}{2^{2i}\left( i! \right)^{2}}x^{i}} \\ & \\  & \text{Using this form, it is easy to see that we can apply this to our generating function for }P_{n}\left( \mu  \right). \\ &  \\  & \frac{1}{\sqrt{1-2\mu \rho +\rho ^{2}}}=\frac{1}{\sqrt{1-\left( 2\mu \rho -\rho ^{2} \right)}}=\sum\limits_{i}^{\infty }{\frac{\left( -1 \right)^{2i}\left( 2i \right)!}{2^{2i}\left( i! \right)^{2}}\left( 2\mu \rho -\rho ^{2} \right)^{i}}=\sum\limits_{i}^{\infty }{\frac{\left( -1 \right)^{2i}\left( 2i \right)!}{2^{2i}\left( i! \right)^{2}}\left( \rho \left( 2\mu -\rho  \right) \right)^{i}} \\ & \\  & \sum\limits_{i}^{\infty }{\frac{\left( -1 \right)^{2i}\left( 2i \right)!}{2^{2i}\left( i! \right)^{2}}\left( \rho \left( 2\mu -\rho  \right) \right)^{i}}=\sum\limits_{i}^{\infty }{\frac{\left( -1 \right)^{2i}\left( 2i \right)!}{2^{2i}\left( i! \right)^{2}}\rho ^{i}\left( 2\mu -\rho  \right)^{i}} \\ & \\  & \text{We again want to expand }\left( 2\mu -\rho  \right)^{i}\text{ using the binomial expansion}\text{.} \\ & \\  & \left( 2\mu -\rho  \right)^{i}=\sum\limits_{k=0}^{i}{\frac{i!}{k!\left( i-k \right)!}}\left( 2\mu  \right)^{i-k}\left( -\rho  \right)^{k} \\ & \\  & \sum\limits_{i}^{\infty }{\frac{\left( -1 \right)^{2i}\left( 2i \right)!}{2^{2i}\left( i! \right)^{2}}\rho ^{i}\left( 2\mu -\rho  \right)^{i}}=\sum\limits_{i}^{\infty }{\frac{\left( -1 \right)^{2i}\left( 2i \right)!}{2^{2i}\left( i! \right)^{2}}\rho ^{i}}\sum\limits_{k=0}^{i}{\frac{i!}{k!\left( i-k \right)!}}\left( 2\mu  \right)^{i-k}\left( -\rho  \right)^{k} \\ \end{align}$$

$$\begin{align} & \sum\limits_{i=0}^{\infty }{\sum\limits_{k=0}^{i}{\frac{\left( -1 \right)^{2i}\left( 2i \right)!}{2^{2i}\left( i! \right)^{2}}\rho ^{i}\frac{i!}{k!\left( i-k \right)!}}}\left( 2\mu \right)^{i-k}\left( -\rho  \right)^{k} \\ & \\  & \text{Now we make the following simplifications:} \\ & \left( -1 \right)^{2i}=1 \\ & \left( -\rho \right)^{k}=(-1)^{k}\rho ^{k} \\ & \frac{i!}{\left( i! \right)^{2}}=\frac{1}{i!} \\ & \\  & \text{This leaves us with the following:} \\ & \\  & \sum\limits_{i=0}^{\infty }{\sum\limits_{k=0}^{i}{\frac{\left( 2i \right)!\left( -1 \right)^{k}\left( 2\mu  \right)^{i-k}\rho ^{i+k}}{2^{2i}i!\left( i-k \right)!k!}}} \\ & \\  & \text{Now let }\!\!'\!\!\text{ s look again at our vey first relationship between the generating function and the form of the} \\ & \text{expansion}\text{.} \\ & \\  & \frac{1}{\sqrt{1-2\mu \rho +\rho ^{2}}}=\sum\limits_{n=0}^{\infty }{P_{n}\left( \mu  \right)\rho ^{n}} \\ & \\  & \text{We see that if we are to have our expression match up to the series expansion, we need to make }\rho \text{ } \\ & \text{independent of }k.\text{ We can easily do this by asking ourself, if we let }k\text{ go to a new variable, say n,} \\ & \text{what do we allow }i\text{ to go to, or do we leave }i\text{ alone? The logic is easily seen in the following:} \\ & \\  & \text{If }k\to n \\ & \text{Then }i+n+?=i\text{ } \\ & \text{We see that we let }i\to i-n \\ & \\  & \text{This then allows us to express }\rho \text{ as being independent of the second summation as can be seen}\text{.} \\ & \\  & \sum\limits_{i=0}^{\infty }{\sum\limits_{k\to n=0}^{i\to \left( i-n \right)}{\frac{\left( 2i\to 2\left( i-n \right) \right)!\left( -1 \right)^{k\to n}\left( 2\mu  \right)^{i\to \left( i-n \right)-k\to \left( -n \right)}\rho ^{i\to \left( i-n \right)+k\to n}}{2^{2i\to 2\left( i-n \right)}\left( i\to i-n \right)!\left( i\to \left( i-n \right)-k\to \left( -n \right) \right)!\left( k\to n \right)!}}}=\sum\limits_{i=0}^{\infty }{\left\{ \sum\limits_{n=0}^{i-n}{\frac{\left( 2i-2n \right)!\left( -1 \right)^{n}\left( 2\mu  \right)^{i-2n}}{2^{2i-2n}\left( i-n \right)!\left( i-2n \right)!n!}} \right\}}\rho ^{i} \\ & \\  & \text{This is exactely the form that we are looking for and it can be seen that the upper} \\ & \text{limit of the inner summation be }\frac{i}{2},\text{ not }i-n.\text{ } \\ & \\ \end{align}$$

$$\begin{align} & \text{Therefore, our derivation of the general expression for }P_{n}\left( \mu \right)\text{ is complete and we have shown } \\ & \text{mathematically why on Problem }\!\!\#\!\!\text{ 6 and Problem  }\!\!\#\!\!\text{ 10 Part A above we got the same answer as } \\ & \text{in the beginning of Problem }\!\!\#\!\!\text{ 10}\text{.} \\ \end{align}$$

$$\begin{align} & \text{Proof of }(i-n)\to \frac{i}{2} \\ & \text{Keep LHS variables the same and change RHS variables}\text{. This should allow a relationship to be } \\ & \text{derived on what happens to }i\text{ and }k\text{ when I set them to }i-n\text{ and }n\text{ respectively}\text{.} \\ & \\  & \frac{\left( 2i \right)!(-1)^{k}\left( 2\mu  \right)^{i-k}\rho ^{k+i}}{2^{2i}i!\left( i-k \right)!k!}=\frac{\left( 2i \right)!(-1)^{k}\left( 2\mu  \right)^{i-k}\rho ^{k+i}}{2^{2i}i!\left( i-k \right)!k!} \\ & \\  & \frac{\left( 2 \right)!\left( i \right)!(-1)^{k}\left( 2 \right)^{i-k}\left( \mu  \right)^{i-k}\rho ^{k+i}}{2^{2i}i!\left( i-k \right)!k!}=\frac{\left( 2 \right)!\left( i \right)!(-1)^{k}\left( 2 \right)^{i-k}\left( \mu  \right)^{i-k}\rho ^{k+i}}{2^{2i}i!\left( i-k \right)!k!} \\ & \\  & \rho \text{ will be independent of }P_{n}\left( \mu  \right)\text{ so we can divide it out}\text{.} \\ & \\  & \frac{\left( 2 \right)!\left( i \right)!(-1)^{k}\left( 2 \right)^{i-k}\left( \mu  \right)^{i-k}}{2^{2i}i!\left( i-k \right)!k!}=\frac{\left( 2 \right)!\left( i \right)!(-1)^{k}\left( 2 \right)^{i-k}\left( \mu  \right)^{i-k}}{2^{2i}i!\left( i-k \right)!k!} \\ & \\  & \text{Change LHS variables; i}\text{.e}\text{. let }i\to i-n\text{ and }k\to n. \\ & \\  & \frac{\left( 2 \right)!\left( i \right)!(-1)^{k}\left( 2 \right)^{i-k}\left( \mu  \right)^{i-k}}{2^{2i}i!\left( i-k \right)!k!}=\frac{\left( 2 \right)!\left( i-n \right)!(-1)^{n}\left( 2 \right)^{i-n-n}\left( \mu  \right)^{i-n-n}}{2^{2(i-n)}\left( i-n \right)!\left( i-n-n \right)!n!} \\ & \\  & \frac{\left( 2 \right)!\left( i \right)!(-1)^{k}\left( 2 \right)^{i-k}\left( \mu  \right)^{i-k}}{2^{2i}i!\left( i-k \right)!k!}=\frac{\left( 2 \right)!\left( i-n \right)!(-1)^{n}\left( 2 \right)^{i-2n}\left( \mu  \right)^{i-2n}}{2^{2i-2n}\left( i-n \right)!\left( i-2n \right)!n!} \\ & \\  & \text{Expand expontential terms on RHS and LHS}\text{.} \\ & \\  & \frac{\left( 2 \right)!\left( i \right)!(-1)^{k}\left( 2 \right)^{i}\left( \mu  \right)^{i-k}}{2^{2i}i!\left( i-k \right)!2^{k}k!}=\frac{2^{2n}\left( 2 \right)!\left( i-n \right)!(-1)^{n}\left( 2 \right)^{i}\left( \mu  \right)^{i-2n}}{2^{2i}2^{2n}\left( i-n \right)!\left( i-2n \right)!n!} \\ & \\  & \text{Divide out terms on RHS and LHS}\text{.} \\ & \\  & \frac{\left( \mu  \right)^{i-k}}{2^{i}\left( i-k \right)!2^{k}k!}=\frac{\left( 2 \right)^{i}\left( \mu  \right)^{i-2n}}{2^{2i}\left( i-2n \right)!n!} \\ & \\  & \frac{\left( \mu  \right)^{i-k}}{\left( i-k \right)!2^{k}k!}=\frac{\left( 2 \right)^{2i}\left( \mu  \right)^{i-2n}}{2^{2i}\left( i-2n \right)!n!} \\ & \\  & \frac{\left( \mu  \right)^{i-k}}{\left( i-k \right)!2^{k}}=\frac{\left( \mu  \right)^{i-2n}}{\left( i-2n \right)!} \\ & \\  & \frac{\left( \mu  \right)^{i-k}}{2^{k}\left( i-k \right)!}=\frac{\left( \mu  \right)^{2\left( \frac{i}{2}-n \right)}}{\left( 2\left( \frac{i}{2}-n \right) \right)!} \\ & \\  & \text{The above relationship shows that as }k\to n\text{ we see }i\to \frac{i}{2} \\ \end{align}$$

Author EGM6321.f10.team1.russo 08:10, 6 December 2010 (UTC) 70.185.113.180 10:34, 7 December 2010 (UTC)EGM6321.f10.team1.Lang.Paul