User:EGM6321.F10.TEAM1.HW7

Find
Plot $$ \left \{ P_0, Q_0 \right \} \ $$ for Figure 1.

Plot $$ \left \{P_1, Q_1 \right \} \ $$ for Figure 2.

Plot $$ \left \{ P_2, Q_2 \right \} \ $$ for Figure 3.

Plot $$ \left \{P_3, Q_3 \right \} \ $$ for Figure 4.

Plot $$ \left \{P_4, Q_4 \right \} \ $$ for Figure 5.

Observe evenness and oddness of $$ \left \{P_i, Q_i \right \} \ $$  as $$ i \ $$ goes to zero

"guess" $$ \int_{\mu=-1}^{+1}(P_i(\mu) Q_i(\mu))d\mu $$

Solution
From King:

The expression for $$\displaystyle Q_4(\mu)$$ is given by,



Author EGM6321.F10.TEAM1.WILKS 05:17, 23 November 2010 (UTC)EGM6321,F10.TEAM1.WILKS

Find
A) Show that if $$ \left \{ g_i \right \} $$ is odd, then $$ f \ $$ is odd. B) Show that if $$ \left \{ g_i \right \} $$ is even, then $$ f \ $$ is even.

Solution
A) $$ \left \{ g_i \right \} $$ is odd.

B) $$ \left \{ g_i \right \} $$ is even.

$$ \therefore \left \{ g_i \right \} $$ is odd, then $$ f \ $$ is odd.

$$ \therefore \left \{ g_i \right \} $$ is even, then $$ f \ $$ is even.

Author EGM6321.Taeho Kim

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)

Find
A) Without calculation, find property of $$ A_n \ $$. i.e. $$ A_{2k}= ? \ $$ $$ A_{2k+1}=? \ $$ for k=0 B) Evaluate non-zero coefficients (Use WA).

Solution
First solve with even value of n=2k. break this into two terms.

$$ f(\mu) $$ is even for any value by virtue of cos being even, while $$ P_{2k} $$ was shown to be an even polynomial function in a previous problem. This means that the second integral will become odd, once the integral is taken.

next, consider n=2k+1 Since P(2k+1) is known to be odd, the second integral will become even.

Solve the non-zero terms for k=0 and subbing in P and f

now, with k=1

Author EGM6321.f10.team1.allison

Find
A) Show that the second solution to Eq5.1 is:

A) Show that the second solution to Eq5.2 is:

Solution
by definition

so,

substituting in the given definition of $$Q_0 $$

the exponential and logarithms cancel to produce. use the multiplier

consider the solution from part a

substitute this back into 5.14

Author EGM6321.f10.team1.allison

Given
and for $$ n \ne \ m \ $$ and for $$ L_n=P_n \ $$

Find
A) Using Eq6.1, show $$ Q_n(x) \ $$ is even or odd depending on "n" B) Complete Eq6.4

Solution
$$\begin{align} & Q_{n}(x)=P_{n}(x)\tanh ^{-1}(x)-2\sum\limits_{j=1,3,5...}^{J}{\frac{2n-2j+1}{(2n-j+1)j}P_{n-j}(x)} \\ & \\  & \text{We can simplify this expression as} \\ & \\  & Q_{n}(x)=a_{n}(x)+b_{n}(x) \\ & \\  & \text{Where} \\ & \\  & \text{ }a_{n}(x)=P_{n}(x)\tanh ^{-1}(x)\text{ } \\ & \text{ }b_{n}(x)=-2\sum\limits_{j=1,3,5...}^{J}{\frac{2n-2j+1}{(2n-j+1)j}P_{n-j}(x)} \\ & \\  & \text{We know from the above Problem 3 that when n is an even number }P_{n}(x)\text{ } \\ & \text{is an even function}\text{. We also know that the product of an even function} \\ & \text{and an odd function is an odd function}\text{. This is easliy proven by defining} \\ & \text{ }f(x)=x^{n}\text{ as an even function when n is even and an odd function when} \\ & \text{n is an odd number}\text{. It can be seen that if }f(x)\text{ is an even function and } \\ & y(x)\text{ is an odd function, then their product is:} \\ & \\  & \text{ }f(x)\times y(x)=x^{2n}(x^{2n+1})=x^{2n+2n+1}=x^{4n+1} \\ & \\  & \text{Which is odd}\text{. Therefore, it directly follows that:} \\ & \tanh ^{-1}(x)=\cosh (x)(\frac{1}{\sinh (x)})=\text{even fuction}\times \text{odd function}=\text{odd function} \\ & \\  & \text{ This says that when n is an even number, we will have:} \\ & \\  & \text{ }a_{n}(x)=(\text{even function)}\times \text{(odd function)}=\text{odd function}\text{.} \\ \end{align}$$

$$\begin{align} & \text{Now looking at }b_{n}(x)=-2\sum\limits_{j=1,3,5...}^{J}{\frac{2n-2j+1}{(2n-j+1)j}P_{n-j}(x)},\text{ we notice that} \\ & P_{n-j}(x)\text{ will be an odd function when n is even, referencing both Problem 3} \\ & \text{above, and the fact that any odd number minus any even number is always } \\ & \text{an odd number}\text{. This is simply proven by expressing an odd number by } \\ & (2n+1)\text{ and and even number by (2k) with the following relation:} \\ & \\  & (2n+1)-(2k)=2(n-k)+1\text{ } \\ & \\  & \text{We see that odd minus even is always odd}\text{. Therefore when n is even, }P_{n-j}(x)\text{ } \\ & \text{will always be odd, and when n is even, we see that }b_{n}=\sum{\text{integer}\times \text{odd function}\text{. }} \\ & \text{It was shown in Problem 2 above that the summation of odd functions results } \\ & \text{in an odd function}\text{. Taking our above definition of an odd function, we see that } \\ & \text{if n is even, it results in both }a_{n}(x)\text{ and }b_{n}(x)\text{ being odd}\text{. Therefore, we see that } \\ & Q_{n}(x)\text{ will be odd}\text{.} \\ \end{align}$$

$$\begin{align} & \text{In looking at what happens to }Q_{n}(x)\text{ when n is odd we use the same above definitions}\text{.} \\ & \text{We know from the above Problem 3 that when n is an odd number }P_{n}(x)\text{ } \\ & \text{is an odd function}\text{. We already have shown in the above example of }Q_{2k}(x) \\ & \text{that}\tanh ^{-1}(x)\text{ is an odd function}\text{. We can easily see that the product of any 2 odd} \\ & \text{functions is an even function by the following simple relation:} \\ & \\  & f(x)\times y(x)=x^{2n+1}(x^{2k+1})=x^{2n+1+2k+1}=x^{2(n+k+1)} \\ & \\  & \text{ This says that when n is an odd number, we will have:} \\ & \\  & \text{ }a_{n}(x)=(\text{odd function)}\times \text{(odd function)}=\text{ even function}\text{.} \\ & \\  & \text{Now looking at }b_{n}(x)=-2\sum\limits_{j=1,3,5...}^{J}{\frac{2n-2j+1}{(2n-j+1)j}P_{n-j}(x)},\text{ we notice that} \\ & P_{n-j}(x)\text{ will be an even function when n is odd refrencing both Problem 3} \\ & \text{above, and the fact that any odd number minus any other odd number is always } \\ & \text{an even number}\text{. This is simply proven by expressing any 2 odd numbers by } \\ & (2k+1)\text{ and (2n+1) with the following relation:} \\ & \\  & (2k+1)-(2n+1)=2(k-n) \\ & \\  & \text{We see that odd minus odd is always even}\text{. Therefore when n is odd, }P_{n-j}(x)\text{ } \\ & \text{will always be even, and when n is odd, we see that }b_{n}=\sum{\text{integer}\times \text{even function}\text{. }} \\ & \text{It was shown in Problem 2 above that the summation of even functions results } \\ & \text{in an even function}\text{. Taking our above definition of an even function, we see that } \\ & \text{if n is odd, it results in both }a_{n}(x)\text{ and }b_{n}(x)\text{ being even}\text{. Therefore, we see that } \\ & Q_{n}(x)\text{ will be even}\text{.} \\ \end{align}$$

$$\begin{align} & \text{Referring the reader to the graphs in Problem }\!\!\#\!\!\text{ 1 above we would like to } \\ & \text{draw attention to the pairwise graphs of the functions P}_{n}(x)\text{ and} \\ & Q_{n}(x)\text{, which are known as the Legendre Functions of the First and Second} \\ & \text{Kind respectively}\text{. In particular, it is suggested the reader look } \\ & \text{at the similarities among the graphs when n is odd; i}\text{.e}\text{. when }P_{n}(x)\text{ is an odd} \\ & \text{function and }Q_{n}(x)\text{ is an even function as shown directly above}\text{.When }n\text{ is } \\ & \text{odd, it is seen the the graph of }P_{n}(x)\text{ vs }Q_{n}(x)\text{ on the interval }\!\![\!\!\text{ -1,1 }\!\!]\!\!\text{  intersects } \\ & \text{the }P_{n}(x)\text{ and }Q_{n}(x)\text{ axis at 0 every time}\text{. This suggests that the relationship between} \\ & P_{n}(x)\text{ and }Q_{n}(x)\text{ on the interval }\!\![\!\!\text{ -1,1 }\!\!]\!\!\text{  equals zero, but this is only by inspection} \\ & \text{and we should require a more concrete approach by looking at the generalized } \\ & \text{properties of the inner product between even and odd functions}\text{. } \\ &  \\  & \text{We can think of the inner product of 2 functions in the same manner} \\ & \text{as we do with the inner product of 2 vectors, because after all, a vector} \\ & \text{is nothing more than a function with direction}\text{. When we take the inner} \\ & \text{product of 2 vectors, we are actually projecting one vector onto the} \\ & \text{other; i}\text{.e}\text{. we are comparing the 2 in a manner that allows us to quantify } \\ & \text{similarities between the 2 vectors in terms of their directions}\text{. Therefore, the} \\ & \text{inner product has shown to have properties in itself}\text{. When the 2 vectors are perpendicular,} \\ & \text{ their inner product is zero}\text{. Vice versa, when they are parallel, their inner product is 1}\text{. This tells} \\ & \text{us that the inner product has some particular properties in its own right that} \\ & \text{at specific values, it goes to zero and 1}\text{. Let }\!\!'\!\!\text{ s search for these same type of } \\ & \text{characteristics between even and odd functions}\text{.} \\ & \\  & \text{When applying the inner product to 2 functions, it can be shown easily shown that} \\ & \text{the inner product between an even function and an odd function will be zero}\text{.} \\ & \text{Let }\!\!'\!\!\text{ s prove this with the following illustration by first considering the case }m\ne n\text{:} \\ & \\  & \sin (-mx)=-\sin (mx);\text{ therefore will be our odd function}\text{.} \\ & \\  & \cos (-nx)=\cos (nx);\text{ therefore will be our even function}\text{.} \\ & \\  & \langle \text{sin(nx),cos(nx)}\rangle =\int\limits_{-1}^{1}{\sin (nx)\cos (nx)} \\ & \text{Using the trigonometric identity sin(A)cos(B)=}\frac{1}{2}\{\sin (A-B)+\sin (A+B)\} \\ & \text{and plugging into the above integral we obtain the following:} \\ & \\  & \int\limits_{-1}^{1}{\sin (n\pi x)\cos (n\pi x)}=\frac{1}{2}\int\limits_{-1}^{1}{\sin [(n-n)\pi ]dx}+\frac{1}{2}\int\limits_{-1}^{1}{\sin [(n+n)\pi ]}dx \\ & \\  & =\frac{1}{2}\int\limits_{-1}^{1}{\sin [(0)\pi ]dx}+\frac{1}{2}\int\limits_{-1}^{1}{\sin [(2n)\pi ]}dx=0+\frac{1}{2}\int\limits_{-1}^{1}{\sin [(2n)\pi ]}dx \\ & \\  & =\frac{1}{2}\{\sin [(2)\pi ]-\sin [(-2)\pi ]\}=\frac{1}{2}\{\sin [(2)\pi ]+\sin [(2)\theta \pi ]\}=\sin (2\pi )=0 \\ & \\  & \text{Now let }\!\!'\!\!\text{ s explore the case were }m=n\text{:} \\ & \\  & \langle \text{sin(nx),cos(nx)}\rangle =\int\limits_{-1}^{1}{\sin (nx)\cos (nx)} \\ & \text{Again, using the identity sin(A)cos(A)=}\frac{1}{2}\{\sin (A-A)+\sin (A+A)\} \\ & \text{and plugging into the above integral we arrive at the following:} \\ & \\  & \frac{1}{2}\int\limits_{-1}^{1}{\sin [(2n)\pi ]}dx=0 \\ & \\  & \text{Therefore we see the the inner product of and even and odd function is always } \\ & \text{zero}\text{. Combined with prior result with Q}_{n}(x)'s\text{ even or oddness dependence on n,} \\ & \text{this is useful because we now then know the following to be true due to} \\ & \text{properties of the inner product between the Legendre Functions of the First and} \\ & \text{Second kind:} \\ \end{align}$$

Author EGM6321.Lang.Paul

Find
Show that Eq7.1 can be written equivalent to Eq7.2, i.e. prove:

Substitute (7.12),(7.13), and (7.14) into (7.11).

Solution
Author EGM6321.Taeho Kim

Find
Obtain Eq8.3 and Eq8.4 from Eq8.1 and Eq8.2

Solution
Author EGM6321.Taeho Kim

Given
From HW6.8:

Find
Generate the set $$ \left \{ P_2, P_3, P_4, P_5, P_6 \right \} \ $$ using the recurrence relation number 2 (RR2) $$ \left \{ P_0, P_1 \right \} \ $$

Solution
$$\begin{align} & \text{Knowing any 2 Legendre Polynomials, we would like to have a way of } \\ & \text{calculating other Legendre Polynomials to any order that we like}\text{. We will} \\ & \text{not derive this recurrence relation here, since it can be found in the lecture} \\ & \text{notes under meeting 44-2 (there it is referred to as RR2)}\text{. The result of the } \\ & \text{derivation the notes is the following:} \\ & \\  & \left( n+1 \right)P_{n+1}\left( x \right)-\left( 2n+1 \right)xP_{n}\left( x \right)+nP_{n-1}\left( x \right)=0 \\ & \\  & \text{Rearranging this equation to explicitly express }P_{n+1}\left( x \right)\text{ we get} \\ & \\  & P_{n+1}\left( x \right)=\frac{\left( 2n+1 \right)xP_{n}\left( x \right)-nP_{n-1}\left( x \right)}{\left( n+1 \right)} \\ & \text{From this we can calculate to the order of }n+1\text{ Legendre Polynomials as long as} \\ & \text{we know both }P_{n-1}\left( x \right)\text{ and }P_{n}\left( x \right).\text{ Referencing the given information, we have known} \\ & \text{values for up to the 4th order of }P_{n}\left( x \right).\text{ Let }\!\!'\!\!\text{ s take }P_{0}\left( x \right)\text{ and }P_{1}\left( x \right)\text{ and use the above } \\ & \text{recurrence relation to derive a set of Legendre Polynomials up to the order 6}\text{.} \\ & \\  & P_{0}\left( x \right)=1 \\ & P_{1}\left( x \right)=x \\ & \\  & \text{Plugging these 2 given polynomials into the recurrence relation we have:} \\ & \\  & P_{1+1}\left( x \right)=\frac{\left( 1\left( 1 \right)+1 \right)xP_{1}\left( x \right)-\left( 1 \right)P_{1-1}\left( x \right)}{\left( 1+1 \right)}=\frac{3xP_{1}\left( x \right)-P_{0}\left( x \right)}{2} \\ & P_{2}(x)=\frac{1}{2}\left( 3xP_{1}\left( x \right)-P_{0}\left( x \right) \right)=\frac{1}{2}\left( 3x^{2}-1 \right) \\ \end{align}$$

$$\begin{align} & \text{Now, to go further we just need to repeat the above steps all the way up to order 6}\text{.} \\ & P_{2+1}\left( x \right)=\frac{\left( 2\left( 2 \right)+1 \right)xP_{2}\left( x \right)-2P_{2-1}\left( x \right)}{\left( \left( 2 \right)+1 \right)}=\frac{5xP_{2}\left( x \right)-2P_{1}\left( x \right)}{3} \\ & P_{3}\left( x \right)=\frac{1}{3}\left[ 5x\left( \frac{1}{2}\left( 3x^{2}-1 \right) \right)-2\left( x \right) \right]=\frac{1}{3}\left( \frac{15}{2}x^{3}-\frac{5}{2}x-2x \right) \\ & P_{3}\left( x \right)=\frac{1}{3}\left( \frac{15}{2}x^{3}-\frac{9}{2}x \right)=\frac{1}{2}\left( 5x^{3}-3x \right) \\ \end{align}$$

$$\begin{align} & P_{3+1}\left( x \right)=\frac{\left( 2\left( 3 \right)+1 \right)xP_{3}\left( x \right)-3P_{3-1}\left( x \right)}{\left( 3+1 \right)}=\frac{7xP_{3}(x)-3P_{2}\left( x \right)}{4} \\ & P_{4}\left( x \right)=\frac{1}{4}\left[ 7x\left( \frac{1}{2}\left( 5x^{3}-3x \right) \right)-3\left( \frac{1}{2}\left( 3x^{2}-1 \right) \right) \right]=\frac{1}{8}\left[ \left( 35x^{4}-21x^{2} \right)-\left( 9x^{2}-3 \right) \right] \\ \end{align}$$

$$\begin{align} & P_{4+1}\left( x \right)=\frac{\left( 2\left( 4 \right)+1 \right)xP_{4}\left( x \right)-4P_{4-1}\left( x \right)}{\left( 4+1 \right)}=\frac{9xP_{4}\left( x \right)-4P_{3}\left( x \right)}{5} \\ & P_{5}\left( x \right)=\frac{1}{5}\left[ 9x\left( \frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8} \right)-4\left( \frac{1}{2}\left( 5x^{3}-3x \right) \right) \right]=\frac{1}{5}\left( \frac{315}{8}x^{5}-\frac{135}{4}x^{3}+\frac{27}{8}x-10x^{3}+6x \right) \\ & P_{5}\left( x \right)=\frac{63}{8}x^{5}-\frac{27}{4}x^{3}+\frac{27}{40}x-2x^{3}+\frac{6}{5}x=\frac{63}{8}x^{5}-\frac{35}{4}x^{3}+\frac{75}{40}x \\ \end{align}$$

$$\begin{align} & P_{5+1}\left( x \right)=\frac{\left( 2\left( 5 \right)+1 \right)xP_{5}\left( x \right)-5P_{5-1}\left( x \right)}{\left( 5+1 \right)}=\frac{11xP_{5}\left( x \right)-5P_{4}\left( x \right)}{6} \\ & P_{6}\left( x \right)=\frac{1}{6}\left[ 11x\left( \frac{63}{8}x^{5}-\frac{35}{4}x^{3}+\frac{15}{8}x \right)-5\left( \frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8} \right) \right] \\ & P_{6}\left( x \right)=\frac{1}{6}\left[ \left( \frac{693}{8}x^{6}-\frac{385}{4}x^{4}+\frac{165}{8}x^{2} \right)-\left( \frac{175}{8}x^{4}-\frac{75}{4}x^{2}+\frac{15}{8} \right) \right] \\ & P_{6}\left( x \right)=\frac{1}{6}\left[ \frac{693}{8}x^{6}-\frac{945}{8}x^{4}+\frac{315}{8}x^{2}-\frac{15}{8} \right]=\left( \frac{231}{16}x^{6}-\frac{315}{16}x^{4}+\frac{105}{16}x^{2}-\frac{5}{16} \right) \\ \end{align}$$

Author EGM6321.Lang.Paul

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

Signatures
Solved problem 1 -- EGM6321.F10.TEAM1.WILKS 05:30, 24 November 2010 (UTC)EGM6321,F10.TEAM1.WILKS

Solved problem 2 -- EGM6321.f10.team1.Kim.TH

Solved problem 3 -- EGM6321.f10.team1.russo 22:58, 22 November 2010 (UTC)

Solved problem 4 -- Egm6321.f10.team1.allison 16:48, 4 December 2010 (UTC)

Solved problem 5 -- Egm6321.f10.team1.allison 16:48, 4 December 2010 (UTC)

Solved problem 6 -- EGM6321.f10.team1.Lang.Paul70.185.113.180 04:56, 7 December 2010 (UTC)

Solved problem 7 -- EGM6321.f10.team1.Kim.TH

Solved problem 8 -- EGM6321.f10.team1.Kim.TH

Solved problem 9 -- EGM6321.f10.team1.Lang.Paul70.185.113.180 04:56, 7 December 2010 (UTC)

Solved problem 10 -- EGM6321.f10.team1.russo 08:13, 6 December 2010 (UTC)