text:
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)

title: User:EGM6321.F10.TEAM1.HW7


text:
User talk:EGM6321.F10.TEAM1.HW7

 Hello EGM6321.F10.TEAM1.HW7, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 04:44, 12 November 2010 (UTC)

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

title: User talk:EGM6321.F10.TEAM1.HW7


text:
Carbon dioxide

Discovery
Carbon dioxide was discovered by Joseph Black, a Scottish chemist and physician, in 1754.

Quick Facts

 * Name: Carbon Dioxide
 * Chemical Formula: CO2
 * Melting Point: 194.7 K (-78°C, -109 °F) (Sublimes)
 * Boiling Point: 216.6 K (-57°C, -70°F) (Sublimes)
 * Appearance: Colourless gas
 * Molar Mass: 44.010 g/mol

Discussion questions and essay ideas

 * What role does carbon dioxide play in climate change on Earth?
 * What is the evidence that points to carbon dioxide emissions playing a significant role in climate change on Earth?

Carbon dioxide and climate change
Increases in atmospheric concentrations of carbon dioxide (CO2) and other long-lived greenhouse gases such as methane, nitrous oxide and ozone have strengthened their absorption and emission of infrared radiation, causing the rise in average global temperature since the mid-20th century. Carbon dioxide is of greatest concern because it exerts a larger overall warming influence than all of these other gases combined.

Wikipedia
As noted above, "Carbon dioxide is of greatest concern [of the various greenhouse gases] because it exerts a larger overall warming influence than all of these other gases combined."
 * List of countries by carbon dioxide emissions
 * Top contributors to greenhouse gas emissions
 * Greenhouse gas emissions by the United States
 * List of U.S. states and territories by carbon dioxide emissions
 * Greenhouse gas emissions by China
 * List of countries by greenhouse gas emissions
 * List of countries by carbon dioxide emissions per capita
 * List of countries by greenhouse gas emissions per person
 * Life-cycle greenhouse gas emissions of energy sources
 * Carbon dioxide in Earth's atmosphere
 * List of U.S. states and territories by carbon dioxide emissions

External

 * Just 100 companies responsible for 71% of global emissions, study says
 * Sources of carbon dioxide emissions

title: Carbon dioxide


text:
User talk:Teles

 Hello Teles, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 14:33, 12 November 2010 (UTC)

See my response on my page. --Majrjejm 03:48, 26 January 2011 (UTC)
 * Response on my page. --Majrjejm 04:03, 26 January 2011 (UTC)

title: User talk:Teles


text:
User:Michaelnelson/LearningGoals

Learning goals project
Todo: better name? (Goal Tracker is already in use - see below).

Frequently asked questions
* Why didn't other tools, like basecamp solve the problem?

Similar projects or products
Need to write up analysis of the following projects and how Learning Goals is different.

project-tools (aka. basecamp and other OSS variants)

www.goal-tracker.com

sbs's project

title: User:Michaelnelson/LearningGoals


text:
NYU Media, Culture, and Communication Master's Student Profiles

MCC Master's students are invited to insert their profiles alphabetically using the format below. Click the edit tab to add a bio. Note guidelines: 200 word max; content should focus on academic or professional profiles rather than personal information; bold name; italicize job title and company name; use the break tag to insert a line break; click the link icon to insert a hyperlink or email address. __________________________________________________________________________________

Stacy Rosenberg Assistant Director for External Relations and Adjunct Instructor Stacy manages web site content and edits print materials for the NYU Department of Media, Culture, and Communication (MCC) and the NYU Council for Media and Culture. She is the department's liaison to our alumni, internship employers and other external partners. Stacy teaches the courses Media and Identity at NYU Steinhardt and Business and Its Publics at the NYU Stern School of Business. She co-directed Media & Globalization: The European Experience in Amsterdam & Paris, and staffed Media & Globalization: The Asian Experience in Hong Kong and Beijing. Stacy also leads a financial planning seminar for young adults aging out of foster care. Previously, Stacy served as Associate Director at the NYU Tisch Center for Hospitality, Tourism, and Sports Management, was a strategic partners producer for the arts and entertainment guide Citysearch.com, and worked in television production management for The Walt Disney Company. Stacy earned her Master's in Culture and Communication from NYU and a Bachelors in Telecommunication with a concentration in Business from the University of Florida. visit my website | [mailto:stacy.rosenberg@nyu.edu email me] __________________________________________________________________________________

__________________________________________________________________________________


text:
User talk:MCC~enwikiversity

 Hello MCC, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
 * Visit a (kind of) random project.
 * Browse Wikiversity, or visit a portal corresponding to your educational level: pre-school, primary, secondary, tertiary, non-formal education.
 * Find out about research activities on Wikiversity.

 * Read an introduction for teachers and find out how to write an educational resource for Wikiversity.
 * Give feedback about your initial observations
 * Discuss Wikiversity issues or ask questions at the colloquium.
 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! If you're a twitter user, please follow http://twitter.com/Wikiversity. --Ottava Rima (talk) 15:48, 12 November 2010 (UTC)

Your account will be renamed
Hello,

The developer team at Wikimedia is making some changes to how accounts work, as part of our on-going efforts to provide new and better tools for our users like cross-wiki notifications. These changes will mean you have the same account name everywhere. This will let us give you new features that will help you edit and discuss better, and allow more flexible user permissions for tools. One of the side-effects of this is that user accounts will now have to be unique across all 900 Wikimedia wikis. See the announcement for more information.

Unfortunately, your account clashes with another account also called MCC. To make sure that both of you can use all Wikimedia projects in future, we have reserved the name MCC~enwikiversity that only you will have. If you like it, you don't have to do anything. If you do not like it, you can pick out a different name.

Your account will still work as before, and you will be credited for all your edits made so far, but you will have to use the new account name when you log in.

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:37, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:14, 19 April 2015 (UTC)

title: User talk:MCC~enwikiversity


text:
Template:Ambox/doc

This is the ambox or article message box meta-template.

It is used to build article message box templates such as wikify. It offers several different colours, uses default images if no image parameter is given and it has some other features.

This meta-template uses the ambox CSS classes in MediaWiki:Common.css. The classes can also be used directly in a wikitable if special functionality is needed. See the how-to guide for that.

Usage
Simple usage example:

Complex example:

But you are not really supposed to use red bold italic text.

Article message box types
The following examples use different type parameters but use no image parameters thus they use the default images for each type.

Other images
The default images shown above are mostly for convenience. In many cases it is more appropriate to use more specific images. These examples use the image parameter to specify an image other than the default images.

More examples
Some other parameter combinations.

Parameters
List of all parameters:

type
 * If no type parameter is given the template defaults to type notice. That means it gets a blue side bar.

image
 * No parameter = If no image parameter is given the template uses a default image. Which default image it uses depends on the type parameter.
 * An image = Should be an image with usual wiki notation. Widths of 40px - 50px are usually about right. (Images over 52 pixels wide will cause padding problems.)
 * Often an icon is purely decorative in the W3C sense that it repeats the text. To improve accessibility, it is desirable to not have it be announced by screen readers, as well as to avoid it linking to an irrelevant page. If (and only if) the image license allows this, it can be marked with "linkalt". For example:
 * Conversely, an icon that does not use "linkalt", and which therefore is announced to visually impaired readers, should use an "alt text" parameter that describes the icon. With no link parameter (using the default link), the alt text should describe the icon's visual appearance. For example:
 * With a nonempty "Page" the alt text should describe the icon's function. For example:
 * An icon whose license requires attribution may have alt text, but must keep the default link. Although public domain images do not require a link, many licenses do require one. Please see Purely decorative images for more information about licensing.
 * none = Means that no image is used.
 * blank = This parameter is now deprecated. If you see it in use, change it to "image=none".
 * An icon whose license requires attribution may have alt text, but must keep the default link. Although public domain images do not require a link, many licenses do require one. Please see Purely decorative images for more information about licensing.
 * none = Means that no image is used.
 * blank = This parameter is now deprecated. If you see it in use, change it to "image=none".

imageright
 * No parameter = If no imageright parameter is given then no image is shown on the right side.
 * An image = Should be an image with usual wiki notation. 40px - 50px width are usually about right depending on the image height to width ratio. (Images over 52 pixels width will cause padding problems.) For example:
 * Anything = Any other object that you want to show on the right side.
 * Anything = Any other object that you want to show on the right side.

style
 * Optional CSS values used by the entire message box table. Without quotation marks  but with the ending semicolons  . For example:

textstyle
 * Optional CSS values used by the text cell. For example:

text
 * The message body text.

The small parameters
The small article message boxes are meant for the top of sections. Normally they should only contain one or two lines of text.

small
 * left = Makes it a smaller left aligned message box. This also makes the default images smaller. Note that any data fed to the smallimage, smallimageright and smalltext parameters is only used if "small=left". To make it so your template also understands the small parameter you can use this code:

smallimage
 * No parameter = If no smallimage parameter is given then this template falls back to use the image parameter. If the image parameter also is empty then a small default image is used.
 * An image = Should be an image with usual wiki notation. 20px width is usually about right for boxes with one line of text, while 25px width is usually about right for boxes with two lines of text. For example:
 * none = Means that no image is used. This overrides any image fed to image, when "small=left".
 * none = Means that no image is used. This overrides any image fed to image, when "small=left".

smallimageright
 * No parameter = If no smallimageright parameter is given then this template falls back to use the imageright parameter. If the imageright parameter also is empty then no image is shown on the right side.
 * An image = Should be an image with usual wiki notation. 20px - 25px width is usually about right. For example:
 * Anything = Any other object that you want to show on the right side.
 * none = Means that no right side image is used. This overrides any image fed to imageright, when "small=left".
 * none = Means that no right side image is used. This overrides any image fed to imageright, when "small=left".

smalltext
 * A shorter version of the message body text. If no smalltext parameter is given then this template falls back to use the text parameter.

Technical details
If you need to use special characters in the text parameter then you need to escape them like this:

</nowiki>. And now a pipe and end braces <nowiki>|}}</nowiki>. }}

. And now a pipe and end braces |}}. }}

The  tags that surround the text in the example above are usually not needed. But if the text contains line breaks then sometimes we get weird line spacing. This especially happens when using vertical dotted lists. Then use the div tags to fix that.

This template uses CSS classes in MediaWiki:Common.css for most of its looks, thus it is fully skinnable.

This template calls ambox/core which holds most of the code for ambox, while ambox itself does parameter preprocessing.

Internally this meta-template uses HTML wikimarkup instead of wikimarkup for the table code. That is the usual way we make meta-templates since wikimarkup has several drawbacks. For instance it makes it harder to use parser functions and some special characters in parameters.

The default images for this meta-template are in png format instead of svg format. The main reason is that some older web browsers have trouble with the transparent background that MediaWiki renders for svg images. The png images here have hand optimised transparent background colour so they look good in all browsers. Note that svg icons only look somewhat bad in the old browsers, thus such hand optimisation is only worth the trouble for very widely used icons.

For more technical details see the talk page and the "See also" links below. Since this template works almost exactly like tmbox, imbox, cmbox and ombox their talk pages and related pages might also contain more details.

title: Template:Ambox/doc


text:
MediaWiki:Print.css

/*  */

title: MediaWiki:Print.css


text:
User:Egm6321.f10.team4/HW6

= Problem 1. The Legendre Equation when n=2 = From Meeting 33, p 33-1.

Given
The general form of the Legendre Equation:

Definitions of the Legendre Polynomials:

 * where $$\displaystyle m = \frac{n}{2}$$.

When n=2 and r=1, the first homogeneous solution, P2(x), is given below as:

Find
Show that the second homogeneous solution is:

using the Method of Variation of Parameters.

Solution
When n=2, the given Legendre Equation is as

Firstly, writing the equation in standard format as

On the assumption that we have one solution, say $$\displaystyle y(x) = u_1(x)$$. We will find the other solution by differentiating $$\displaystyle y(x) = U(x)\cdot {u}_{1}(x)$$ gives

Plugging the above deriviates into the (Eq. 1-6) gives

Since $$ \displaystyle {u}_{1}''+{a}_{1}{u}_{1}'+{a}_{0}{u}_{1} $$ is solution of (Eq. 1-6), the term is zero. We will use $$\displaystyle Z:=U'$$ to reduce the order of Eq. 1-8, via the missing-dependent-variable method,

After dividing the equation in terms of Z by the highest coefficient of Z, now we can find integrating factor h(x),

By definition, $$\displaystyle Z=U'=\frac{k_2}{h(x)}$$, we can go back to U(x) by integrating Z

Since we have U(x) now, we can solve y(x)

By plugging h(x) and u1(x) into above equation, we can get the desired solution Q2(x)

Egm6321.f10.team4.Yoon 00:00, 14 Nov 2010 (UTC) - Primary Author Egm6321.f10.team4.ejm 19:47, 17 November 2010 (UTC) Editor

= Problem 2. Non-homog. L2_ODE_VC using Var. of Paramerters =

From Meeting 33, p 33-1.

Given
Solve the Non-homog L2_ODE_VC with the condition $$\displaystyle f(x)=0 $$. K.p.28, Pb 1.1 ab '''Part. a'''
 * {| style="width:100%" border="0" align="left"

$$\displaystyle (x-1)y''-xy'+y=0 $$ $$
 * $$\displaystyle (Eq. 2-1)
 * }
 * }

'''Part. b'''
 * {| style="width:100%" border="0" align="left"

$$\displaystyle xy''+2y'+xy=0 $$ $$
 * $$\displaystyle (Eq. 2-2)
 * }
 * }

General trial solution is given on Lecture note p.32-1 as,
 * {| style="width:100%" border="0" align="left"

 * style="width:10%; padding:10px; border:2px solid #ff0000" |
 * style="width:10%; padding:10px; border:2px solid #ff0000" |

$$ \displaystyle y={x}^{c} {e}^{rx} $$ $$
 * $$\displaystyle (Eq. 2-3)
 * }
 * }

Find
Find solutions of the equations.

'''Part. a'''
 * {| style="width:100%" border="0" align="left"

$$\displaystyle (x-1)y''-xy'+y=0 $$ $$
 * $$\displaystyle (Eq. 2-1)
 * }
 * }

'''Part. b'''
 * {| style="width:100%" border="0" align="left"

$$\displaystyle xy''+2y'+xy=0 $$ $$
 * $$\displaystyle (Eq. 2-2)
 * }
 * }

Solution
Taking 1st and 2nd derivatives of the trial solution is

Solution for Part. a
Plugging the derivatives of trial solutions into the equation given (eq 2.1) yields,

Taking the common factor ahead of the equation, Collecting x terms by reduction order,

To make above equation into the format of trial solution, only one of the x term should be survived. Thus, the values for c and r by (Eq 2-7) are,

Substituting the values (Eq.2-1) into the trial solution, the homogeneous solutions are determined as

Solution for Part. b
Plugging the derivatives(from (eq.2-4) to (eq.2-6)) of trial solutions into the equation given (eq 2.2) yields,

Expanding the equation,

Collecting x terms by reduction order,

To make above equation into the format of trial solution, only one of the x terms should be survived. Thus, the values for c and r combinations by (Eq.2-11) are,

Substituting the values (Eq.2-2) into the trial solution, the homogeneous solutions are determined as

Egm6321.f10.team4.Yoon 00:00, 14 Nov 2010 (UTC) - Primary Author

= Problem 3: Direct method = From Meeting 33, p 33-1.

Given
The non-homogeneous Legendre equation with n=1:
 * {| style="width:100%" border="0" align="left"

$$\displaystyle \left( 1-x^{2} \right)y''+\left( -2x \right)y'+2y=\frac{1}{\left( 1-x^{2} \right)}

$$ $$
 * $$\displaystyle (Eq. 3-1)
 * }
 * }

Find
The complete solution $$y(x)$$ using the direct method

Solution
First modify Eq (3-1) to look as follows
 * {| style="width:100%" border="0"

\begin{align} y''+\underbrace{\frac{-2x}{(1-{x}^{2})}}_{=:a_1(x)}y'+\underbrace{\frac{2}{1-{x}^{2}}}_{=:a_0(x)}y=\frac{1}{\left( 1-x^{2} \right)^{2}} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3-2)
 * }
 * }

We are given the first and second homogeneous solutions [Mtg. 30 p. 2]:

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

$$\displaystyle u_1(x)=x $$ $$
 * $$\displaystyle (Eq. 3-3)
 * }
 * }

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

$$\displaystyle u_2(x)=\frac{x}{2}log \left( \frac{1+x}{1-x} \right)-1 $$ $$
 * $$\displaystyle (Eq. 3-4)
 * }
 * }

The next step is to find $$h(x)$$ using the definition below
 * {| style="width:100%" border="0" align="left"

$$h(x)=u^2_{1}(x) \cdot exp \left[{\int a_{1}(x)dx}\right]$$ $$
 * $$\displaystyle (Eq. 3-5)
 * }
 * }

Replace $$\displaystyle a_1(x)$$ from Eq. (3-2) and $$\displaystyle u_1(x)$$ from Eq. (3-3) into Eq. (3-5) to get the following

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

\begin{align} h(x) &=x^{2} \cdot exp\left[ \int{\frac{-2x}{1-x^{2}}dx}\right]\\ &=x^{2} \cdot (x^{2}-1)\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3-6)
 * }
 * }

The final step will be to compute the particular solution which is defined in the equation below
 * {| style="width:100%" border="0" align="left"

$$y_{p}(x)=u_1(x)\int{ \frac{1}{h(x)}\left[ \int{ h(x)f(x)dx} \right]}dx$$ $$
 * $$\displaystyle (Eq. 3-7)
 * }
 * }

from Eq. (3-2)
 * {| style="width:100%" border="0" align="left"

$$f(x)=\frac{1}{\left( 1-x^{2} \right)^{2}}$$
 * }
 * }

Substituting Eq.(3-3), Eq. (3-6) and $$f(x)$$ into Eq. (3-7), and using the help of Wolfram Alpha for solving the integrals:
 * {| style="width:100%" border="0"

\begin{align} y_{p}(x)&=x\int{\frac{1}{x^{2}(x^{2}-1)}\left[\int{x^2(x^2-1)\frac{1}{(1-x^{2})^{2}}dx}\right]}dx\\ &=-x\int{\frac{1}{x^2(x^2-1)}\underbrace{\left[\int{\frac{x^2}{1-x^2}}dx\right]}_{\frac{1}{2}\left[log\left(\frac{1+x}{1-x}\right)-2x\right]}}dx\\ &=\frac{-x}{2}\underbrace{\int{\frac{log \left(\frac{1+x}{1-x}\right)-2x}{x^2(x^2-1)}}dx}_{-\left[\frac{1}{2}log\left(\frac{1+x}{1-x} \right ) \right ]^2+\frac{1}{x}log\left(\frac{1+x}{1-x} \right )+log\left(\frac{1-x^2}{x^2} \right )}\\ &=\frac{x}{8}log^2\left(\frac{1+x}{1-x} \right )+\frac{1}{2}log\left(\frac{1-x}{1+x} \right )+\frac{x}{2}log\left(\frac{x^2}{1-x^2} \right )\\ \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3-8)
 * }
 * }

The complete solution to Eq. (3-1) will be a superposition of the two homogeneous solutions and the particular solution

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

\begin{align} y(x)&=k_1u_{1}(x)+k_2u_{2}(x)+y_{p}(x) \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3-9)
 * }
 * }

where $$\displaystyle k_1 $$ and $$\displaystyle k_2 $$ are constants. Our final solution is thus:
 * {| style="width:100%" border="0"

$$\begin{align} y(x)=k_1\underbrace{x}_{u_1(x)}+k_2\underbrace{\frac{x}{2}log \left( \frac{1+x}{1-x} \right)-1}_{u_2(x)}+\underbrace{\frac{x}{8}log^2\left(\frac{1+x}{1-x} \right )+\frac{1}{2}log\left(\frac{1-x}{1+x} \right )+\frac{x}{2}log\left(\frac{x^2}{1-x^2} \right)}_{y_p(x)} \end{align}$$
 * style="width:40%; padding:10px; border:2px solid #8888aa" |
 * style="width:40%; padding:10px; border:2px solid #8888aa" |
 *
 * }

Egm6321.f10.team4.petralanda.n 14:20, 16 November 2010 (UTC) Egm6321.f10.team4.ejm 17:26, 17 November 2010 (UTC) Editor

= Problem 4 = From Meeting 34, p 34-1,3

Find
4.1

Show

4.2

The Laplace equation

Solution
4.1 The infinitesimal segment $$\displaystyle ds $$ is defined as:

where

so

Substituting equations 4-3,4,5 into equation 4-1 and reducing,

or

4.2

The Laplace equation is defined as

where

thus

for $$\displaystyle i=1$$

for $$\displaystyle i=2$$

for $$\displaystyle i=3$$

Summing from $$\displaystyle i=1:3$$

Egm6321.f10.team4.Auerbach 05:20, 17 November 2010 (UTC) - Author

= Problem 5 = From Meeting 35, p 35-3,4

Find
5.1

Find $$\displaystyle \left \{ dx_i \right \}=\left \{dx_1,dx_2,dx_3\right \} $$ in terms of $$\displaystyle \left \{ d\xi_j \right \}=\left \{d\xi_1,d\xi_2,d\xi_3\right \} $$ and $$\displaystyle \left \{ d\xi_k \right \}$$

5.2

Find $$\displaystyle ds^2=dx_idx_i=\sum_{i=1}^{3}(dx_i)^2$$. Identify $$\displaystyle \left \{ dh_i \right \}$$ in terms of $$\displaystyle \left \{ d\xi_j \right \}$$

5.3

Find $$\displaystyle \Delta \psi $$ in cylindrical coordinates

5.4

Find the Bessel equation

Solution
5.1 The infinitesimal segment $$\displaystyle ds $$ is defined as:

where

so

5.2

Substituting equations 5-3,4,5 into equation 5-1 and reducing,

or

so

5.3

The Laplace equation is defined as

where

thus

5.4

Lets start off by assuming

and

so,

divide through by $$\displaystyle ABC $$

lets assume the last term to be

multiplying through by $$\displaystyle \xi_1^2 $$

lets assume the second term to be

multiplying through by $$\displaystyle A $$

The first term can be further reduced

If we now assume

and

We can achieve the Bessel equation

Egm6321.f10.team4.Auerbach 18:20, 17 November 2010 (UTC) - Author

= Problem 6 The Laplace Equation in Cylindrical Coordinates =

From [http://upload.wikimedia.org/wikiversity/en/f/fa/2010_11_05_07_46_02.djvu Mtg. 35]

Given
Converting from rectangular to spherical coordinates:

$$\displaystyle x_1 = x = r \sin \bar\theta \cos \phi $$

$$\displaystyle x_2 = y = r \sin \bar\theta \sin \phi $$

$$\displaystyle x_3 = z = r \cos\bar\theta $$

$$\displaystyle \xi_1 = r $$

$$\displaystyle \xi_2 = \bar\theta $$

$$\displaystyle \xi_3 = \phi $$

where $$\displaystyle \bar\theta $$ is equal to $$\displaystyle \pi / 2 - \theta $$ by the math/physics convention, as opposed to the astronomic convention

Find
the Laplacian, $$\displaystyle \Delta \Psi $$, in terms of spherical coordinates.

Solution
$$\displaystyle dx_1 = \frac {\partial x_1}{\partial xi_1} d \xi_1 + \frac {\partial x_1}{\partial xi_2} d \xi_2 + \frac {\partial x_1}{\partial xi_3} d \xi_3 $$

$$\displaystyle dx_1 = \sin \xi_2 \cos \xi_3 d \xi_1 + \xi_1 \cos \xi_2 \cos \xi_3 d \xi_2 - \xi_1 \sin \xi_2 \sin \xi_3 d \xi_3 $$

$$\displaystyle dx_2 = \frac {\partial x_2}{\partial xi_1} d \xi_1 + \frac {\partial x_2}{\partial xi_2} d \xi_2 + \frac {\partial x_2}{\partial xi_3} d \xi_3 $$

$$\displaystyle dx_2 = \sin\xi_2 \sin\xi_3 d\xi_1 + \xi_1 \cos\xi_2 \sin\xi_3 d\xi_2 + \xi_1 \sin\xi_2 \cos\xi_3 d\xi_3 $$

$$\displaystyle dx_3 = \frac {\partial x_3}{\partial xi_1} d \xi_1 + \frac {\partial x_3}{\partial xi_2} d \xi_2 + \frac {\partial x_3}{\partial xi_3} d \xi_3 $$

$$\displaystyle dx_3 = \cos\xi_2 d\xi_1 - \xi_1 \sin\xi_2 d\xi_2 + 0 $$

$$\displaystyle ds^2 = \sum dx_i^2 $$

$$\displaystyle $$

Using the trigonometric identity

$$\displaystyle \cos^2\alpha + \sin^2\alpha = 1 $$

and regrouping the terms in Eq. 6-1 yields

$$\displaystyle ds^2 = d\xi_1^2 + \xi_1^2 d\xi_2^2 + \xi_1^2 \sin^2 \xi_2 d\xi_3^2 $$

where $$\displaystyle h_i^2 $$ are the coefficients of each term in the above expression.

$$\displaystyle h_1 = 1 $$

$$\displaystyle h_2 = \xi_1 = r $$

$$\displaystyle h_3 = \xi_1 \sin\xi_2 = r \sin\bar\theta $$

For any set of orthogonal curvilinear coordinates, the Laplacian is given by

$$\displaystyle \Delta \Psi = \frac {1}{h_1h_2h_3} \sum \frac {\partial}{\partial \xi_i} (\frac {h_1h_2h_3}{h_i^2} \frac {\partial \Psi}{\partial \xi_i}) $$

for $$\displaystyle i = 1 $$

$$\displaystyle = \frac {1}{\xi_1} ( \frac {\partial}{\partial \xi_1} \frac {\xi_1}{h_1^2} \frac {\partial \Psi}{\partial \xi_1} ) $$

$$\displaystyle = \frac {1}{r^2 \sin\bar\theta} \frac {\partial}{\partial r} (r^2 \sin\bar\theta \frac {\partial \Psi}{\partial r} $$

for $$\displaystyle i = 2 $$

$$\displaystyle = \frac {1}{\xi_1} (\frac {\partial}{\partial \xi_2} \frac {\xi_1}{h_2^2} \frac {\partial \Psi}{\partial \xi_2}) $$

$$\displaystyle = \frac {1}{r^2 \sin\bar\theta} \frac{\partial}{\partial \bar\theta} (\frac {r^2 \sin \bar\theta}{r^2} \frac {\partial \Psi}{\partial \bar\theta}) $$

for $$\displaystyle i = 3 $$

$$\displaystyle = \frac {1}{\xi_1} (\frac {\partial}{\partial \xi_3} \frac {\xi_1}{h_3^2} \frac {\partial \Psi}{\partial \xi_3}) $$

$$\displaystyle = \frac {1}{r^2 \sin\bar\theta} \frac{\partial}{\partial phi} (\frac {r^2 \sin\bar\theta}{r^2 \sin^2 \bar\theta} \frac {\partial \Psi}{\partial \phi}) $$

Therefore, the expression for the Laplacian in terms of spherical coordinates becomes

128.227.51.176 17:51, 12 November 2010 (UTC)Egm6321.f10.team4.osentowski - Author Egm6321.f10.team4.ejm 20:09, 17 November 2010 (UTC) Editor

= Problem 7 - The Laplace Equation in Elliptical Coordinates=

From [http://upload.wikimedia.org/wikiversity/en/8/83/2010_11_09_15_00_14.djvu Mtg. 36]

Given
Converting from rectangular to elliptical coordinates:

$$\displaystyle x_1 = x = a * cosh \mu cos \nu$$

$$\displaystyle x_2 = y = a * sinh \mu sin \nu$$

$$\displaystyle \xi_1 = \mu $$

$$\displaystyle \xi_2 = \nu $$

Find
the Laplacian, $$\displaystyle \Delta \Psi $$, in terms of elliptical coordinates.

Solution
$$\displaystyle dx_1 = \frac {\partial x_1}{\partial \xi_1} d \xi_1 + \frac {\partial x_1}{\partial \xi_2} d \xi_2 $$

$$\displaystyle dx_1 = a \cdot sinh(\mu) \cdot cos(\nu) d\mu - a \cdot cosh(\mu) \cdot sin(\nu) d\nu $$

$$\displaystyle dx_2 = \frac {\partial x_2}{\partial \xi_1} d \xi_1 + \frac {\partial x_2}{\partial \xi_2} d \xi_2 $$

$$\displaystyle dx_2 = a * cosh \mu sin \nu d \mu + a * sinh \mu cos \nu d \nu $$

$$\displaystyle ds^2 = \sum dx_i^2 $$

Regrouping terms in Eq. 7-1 gives

$$\displaystyle = a^2 (sinh^2 \mu cos^2 \nu + cosh^2 \mu sin^2 \nu) d \mu^2 + a^2 (sinh^2 \mu cos^2 \nu + cosh^2 \mu sin^2 \nu) d \nu^2 $$

Using the trigonometric identity

$$\displaystyle cosh^2 \alpha = 1 + sinh^2 \alpha $$

and simplifying the expression gives

$$\displaystyle ds^2 = (a^2 sinh^2 \mu + a^2 sin^2 \nu) d \mu^2 + (a^2 sinh^2 \mu + a^2 sin^2 \nu) d \nu^2 $$

$$\displaystyle h^2 $$ is the coefficient of each term.

$$\displaystyle h_1^2 = h_2^2 = a^2 (sinh^2 \mu + sin^2 \nu) $$

The Laplacian can therefore be expressed as

128.227.51.176 17:51, 12 November 2010 (UTC)Egm6321.f10.team4.osentowski - Author

= Problem 8 - General Form Verification of Legendre Polynomials =

From Meeting 36, p 36-2 & 36-3

Given
With the following set of polynomials

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

\begin{align} P\left ( x \right ) = \sum_{i = 0}^{\left [ n/2 \right ]} (-1)^{i} \frac{(2n-2i)!(x^{n-2i})}{(2^{n})(i!)(n-i)!(n-2i)!} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8-1)
 * }
 * }

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

$$\displaystyle P\left ( x \right ) = \sum_{i = 0}^{\left [ n/2 \right ]} \frac{(1)(3)...(2n-2i-1)}{(2^{i})(i!)(n-2i)!}(-1)^{i}x^{n-2i} $$ $$
 *
$$\displaystyle (Eq. 8-2)
 * }
 * }

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

$$\displaystyle P_{0}\left ( x \right ) = 1 $$ $$
 *
$$\displaystyle (Eq. 8-3)
 * }
 * }

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

$$\displaystyle P_{1}\left ( x \right ) = x $$ $$
 *
$$\displaystyle (Eq. 8-4)
 * }
 * }

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

$$\displaystyle P_{2}\left ( x \right ) = \frac{1}{2}(3x^{2}-1) $$ $$
 *
$$\displaystyle (Eq. 8-5)
 * }
 * }

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

$$\displaystyle P_{3}\left ( x \right ) = \frac{1}{2}(5x^{3}-3x) $$ $$
 *
$$\displaystyle (Eq. 8-6)
 * }
 * }

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

$$\displaystyle P_{4}\left ( x \right ) = \frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8} $$ $$
 *
$$\displaystyle (Eq. 8-7)
 * }
 * }

Note: [n/2] = integer part of n/2 e.g. n = 5, n/2 = 2.5, [n/2] = 2

Find
Verify that the Legendre Polynomial Eqns. 8-3, 8-4, 8-5, 8-6, 8-7 can be written as Eqn. 8-1 and Eqn 8-2

Solution
Since Eqn 8-1 is equivalent to Eqn 8-2, we will only use Eqn 8-1 to demonstrate the solution

Using Eqn 8-1

At n = 0:

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

$$\displaystyle P_{0}\left ( x \right )= 1\cdot \frac{1\cdot 1}{1\cdot1\cdot1} $$

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

$$\displaystyle P_{0}\left ( x \right )= 1 $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }

At n = 1:

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

$$\displaystyle P_{1}\left ( x \right )= 1\cdot \frac{2!\cdot x}{2!\cdot1!} $$

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

$$\displaystyle P_{1}\left ( x \right )= x $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }

At n = 2:

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

$$\displaystyle P_{2}\left ( x \right )=(-1)^{0} \frac{(4-2(0))!(x^{2-2(0))})}{(4)((0))!)(2-(0))!(2-2(0)))!} + (-1)^{1} \frac{(4-2(1))!(x^{2-2(1))})}{(4)((1))!)(2-(1))!(2-2(1)))!} $$

$$\displaystyle P_{2}\left ( x \right )= \frac{4!\cdot x^2}{4\cdot 2!\cdot2!} -1 \frac{2!}{4\cdot 1!\cdot 1!\cdot 1!}= \frac{24\cdot x^2}{16} - \frac{1}{2} $$

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

$$\displaystyle P_{2}\left ( x \right )= \frac{1}{2}(3x^{2}-1) $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }

At n = 3:

$$\displaystyle P_{3}\left ( x \right )= \sum_{i = 0}^{\left [ 3/2 \right ]} (-1)^{i} \frac{(2(3)-2i)!(x^{(3)-2i})}{(2^{(3)})(i!)((3)-i)!((3)-2i)!}= \sum_{i = 0}^{\left [ 1.5 \right ]} (-1)^{i} \frac{(6-2i)!(x^{3-2i})}{(8)(i!)(3-i)!(3-2i)!} $$

$$\displaystyle P_{3}\left ( x \right )=(-1)^{0} \frac{(6-2(0))!(x^{3-2(0))})}{(8)((0))!)(3-(0))!(3-2(0)))!} + (-1)^{1} \frac{(6-2(1))!(x^{3-2(1))})}{(8)((1))!(3-(1))!(3-2(1)))!} $$

$$\displaystyle P_{3}\left ( x \right )= \frac{6!\cdot x^3}{8\cdot 3!\cdot3!} -\frac{4!}{8\cdot 2!\cdot 1!}= \frac{720\cdot x^2}{8\cdot 6\cdot 6} -\frac{24}{16} $$

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

$$\displaystyle P_{3}\left ( x \right )= \frac{1}{2}(5x^{3}-3x) $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }

At n = 4:

$$\displaystyle P_{4}\left ( x \right )= \sum_{i = 0}^{\left [ 4/2 \right ]} (-1)^{i} \frac{(2(4)-2i)!(x^{(4)-2i})}{(2^{(4)})(i!)((4)-i)!((4)-2i)!}= \sum_{i = 0}^{\left [ 2 \right ]} (-1)^{i} \frac{(8-2i)!(x^{4-2i})}{(16)(i!)(4-i)!(4-2i)!} $$

$$\displaystyle P_{4}\left ( x \right )=(-1)^{0} \frac{(8-2(0))!(x^{4-2(0))})}{(16)((0))!)(4-(0))!(4-2(0)))!} + (-1)^{1} \frac{(8-2(1))!(x^{4-2(1))})}{(16)((1))!(4-(1))!(4-2(1)))!}+(-1)^{2} \frac{(8-2(2))!(x^{4-2(2))})}{(16)((2))!(4-(2))!(4-2(2)))!} $$

$$\displaystyle P_{4}\left ( x \right )= \frac{8!\cdot x^4}{16\cdot 4!\cdot4!} -\frac{6!\cdot x^2}{16\cdot 3!\cdot 2!}+\frac{4!}{16\cdot 2!\cdot 2!}= \frac{1680\cdot x^4}{16\cdot 24} -\frac{120\cdot x^2}{16\cdot 2!}+\frac{24}{64} $$

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

$$\displaystyle P_{4}\left ( x \right )= \frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8} $$
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * style="width:5%; padding:10px; border:2px solid #8888aa" |
 * }
 * }
 * }

Egm6321.f10.team4.Corella.RL 12:54, 17 November 2010 (UTC) - Primary Author

= Problem 9 - Solution Verification of Legendre Equation =

From Meeting 36, p 36-3 Meeting 5, p 5-4 Meeting 24, p 24-1

Given
Equations 8.3-8.7 from Problem 8 and

Legendre Equation

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

$$\displaystyle (1-x^{2})\cdot y^{''}-2x\cdot y^{'}+n(n+1)\cdot y $$ $$
 *
$$\displaystyle (Eq. 9-1)
 * }
 * }

Find
Verify that the Legendre Polynomial Eqns. 8-3, 8-4, 8-5, 8-6, 8-7 are solutions of Eqn. 9-1

Solution
Let $$\displaystyle y_{n} = P_n\left ( x \right )$$, where $$\displaystyle y_{n}^{'} = P_{n}^{'}\left ( x \right ) $$ and $$\displaystyle y_{n}^{} = P_{n}^{}\left ( x \right ) $$

Use $$\displaystyle (1-x^{2})\cdot y_{n}^{''}-2x\cdot y_{n}^{'}+n(n+1)\cdot y_{n} $$

At n = 0

$$\displaystyle y_{0}=P_{0}\left ( x \right )= 1 $$,

$$\displaystyle y_{0}^{'} = 0 $$,

$$\displaystyle y_{0}^{''} = 0 $$

$$\displaystyle (1-x^{2})\cdot (0)-2x\cdot (0)+n(n+1)\cdot (1) = (1-x^{2})\cdot (0)-2x\cdot (0)+(0)((0)+1)\cdot (1) = 0 $$

Verified that Eq. 8-3 is a solution of Eq. 9-1

At n = 1

$$\displaystyle y_{1}=P_{1}\left ( x \right )= x $$,

$$\displaystyle y_{1}^{'} = 1 $$,

$$\displaystyle y_{1}^{''} = 0 $$

$$\displaystyle (1-x^{2})\cdot (0)-2x\cdot (1)+n(n+1)\cdot (x) = (1-x^{2})\cdot (0)-2x\cdot (1)+(1)((1)+1)\cdot (x) = x $$

Verified that Eq. 8-4 is not a solution of Eq. 9-1

At n = 2

$$\displaystyle y_{2}=P_{2}\left ( x \right )= \frac{1}{2}(3x^{2}-1) $$,

$$\displaystyle y_{2}^{'} = 3x $$,

$$\displaystyle y_{2}^{''} = 3 $$

$$\displaystyle (1-x^{2})\cdot (3)-2x\cdot (3x)+n(n+1)\cdot (\frac{1}{2}(3x^{2}-1)) = (1-x^{2})\cdot (3)-2x\cdot (3x)+(2)((2)+1)\cdot (\frac{1}{2}(3x^{2}-1)) = $$

$$\displaystyle 3-3x^{2}-6x^{2}+6\cdot (\frac{1}{2}(3x^{2}-1)) = 2 $$

Verified that Eq. 8-5 is not a solution of Eq. 9-1

At n = 3

$$\displaystyle y_{3}=P_{3}\left ( x \right )= \frac{1}{2}(5x^{3}-3x) $$,

$$\displaystyle y_{3}^{'} = \frac{15\cdot x^{2}}{2}-\frac{3}{2} $$,

$$\displaystyle y_{3}^{''} = 15x $$

$$\displaystyle (1-x^{2})\cdot (15x)-2x\cdot (\frac{15\cdot x^{2}}{2}-\frac{3}{2})+n(n+1)\cdot\frac{1}{2}(5x^{3}-3x)= $$ $$\displaystyle 18x-30x^{3}+(3)((3)+1)\cdot\frac{1}{2}(5x^{3}-3x) $$

$$\displaystyle 18x-30x^{3}+12\cdot\frac{1}{2}(5x^{3}-3x)= 18x-30x^{3}+30x^{3}-18x= 0 $$

Verified that Eq. 8-6 is a solution of Eq. 9-1

At n = 4

$$\displaystyle y_{4}=P_{4}\left ( x \right )= \frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8} $$,

$$\displaystyle y_{4}^{'} = \frac{35x^{3}}{2}-\frac{15x}{2} $$,

$$\displaystyle y_{4}^{''} = \frac{105x^{2}}{2}-\frac{15}{2} $$

$$\displaystyle (1-x^{2})\cdot (\frac{105x^{2}}{2}-\frac{15}{2})-2x\cdot (\frac{35x^{3}}{2}-\frac{15x}{2})+n(n+1)\cdot(\frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8}) $$

$$\displaystyle -\frac{245x^{4}}{2}+90x^{2}-\frac{15}{2}+(4)((4)+1)\cdot(\frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8}) $$

$$\displaystyle -\frac{245x^{4}}{2}+90x^{2}-\frac{15}{2}+\frac{175x^{4}}{2}-75x^{2}+\frac{15}{2}=-\frac{70x^{4}}{2}+15x^{2} $$

Verified that Eq. 8-7 is not a solution of Eq. 9-1

Egm6321.f10.team4.Corella.RL 21:29, 15 November 2010 (UTC) - Primary Author

title: User:Egm6321.f10.team4/HW6


text:
0/0

This page is about the indeterminate form 0/0 - a special case of the division by zero.

The files below show functions r(x), g(x) and their quotients m(x)=r(x)/g(x) , c(x)=g(x)/r(x).

Derivatives are drawn in the same color as their function, but thinner.

r(0)/g(0) = r'(0)/g'(0)

title: 0/0