User:Egm6321.f09.team3/HW7

= 1 Hyperbolic Tangent Equivalent =

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

Given

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

Show
The equivalence of the above statement.

Solution
In HW 6, we showed


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

Substituting in,


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


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


 * }
 * }
 * }

which is equivalent.


 * }

= 2 Qn as odd/even =

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

Given

 * $$Q_n(x) = P_n(x) \cdot \tanh^{-1}(x) - 2 \cdot \sum \frac{2n-2j+1}{(2n-j+1)j} \cdot P_{n-j}(x)$$

Find
The values for :$$n (2k, 2k+1)$$ for which :$$Q_n(x)$$ is odd or even.

Solution
Recall that


 * $$\tanh^{-1}(x)$$

is odd. and for


 * $$n = 2k,P_n(x)$$

is even and for


 * $$n=2k+1,P_n(x)$$

is odd.

Therefore, for $$n = 2k$$, $$Q_n(x)$$ is odd (because odd * even = odd)

Alternately, for $$n = 2k+1$$, $$Q_n(x)$$ is even (because odd * odd = even)


 * }

= 3 Plots of Pn and Qn =

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Given

 * $$P_n(x)\quad \text{for} \quad n = 0, 1, 2, 3, 4 $$

and


 * $$Q_n(x)\quad \text{for} \quad n = 0, 1, 2, 3, 4 $$

Find
Plot all functions.

Solution

 * $$P_0(x) = \alpha_0 = 1$$


 * $$P_1(x) = 2 \mu \alpha_1 = x$$


 * $$P_2(x) = (- \alpha_1 + 4) = \frac{1}{2}(3x^2 - 1) $$


 * $$P_3(x) = -4 \mu \alpha_2 + 8 \mu^3 \alpha_3 = \frac{1}{2}(-3x + 5x^3) $$


 * $$P_4(x) = \alpha_2 - \mu^2 \alpha_3 + 16 \mu^4 \alpha_4 = \frac{1}{8}(3 - 30 x^2 + 35 x^4)$$


 * $$Q_n(x) = P_n(x) \cdot \tanh^{-1}(x) - 2\cdot\sum_{j=1,3,5}^J \frac{2n-2j+1}{(2n-j+1)j} \cdot P_{n-j}(x) $$

where


 * $$ J:=1+2\left[\frac{n-1}{2}\right]$$


 * $$ n=0 $$; :$$ J=1 $$


 * $$Q_0(x) = P_0(x) \cdot \tanh^{-1}(x) - 2 \cdot \sum_{j=1}^1 \frac{2(0)-2(1)+1}{(2(0)-(1)+1)1} \cdot P_{0-1}(x) $$


 * $$Q_0(x) =1\cdot \tanh^{-1}(x) - 2 (0)= \tanh^{-1}(x) $$


 * $$ n=1 $$; :$$ J=1 $$


 * $$Q_1(x) = P_1(x) \cdot \tanh^{-1}(x) - 2 \cdot \sum_{j=1}^1 \frac{2 (1) -2(1)+1}{(2(1)-(1)+1)(1)} \cdot P_{1-1}(x)$$


 * $$Q_1(x) = x \cdot \tanh^{-1}(x) - 2 \cdot \sum_{j=1}^1 \frac{2 (1) -2(1)+1}{(2(1)-(1)+1)(1)} \cdot (1) = x \tanh^{-1}(x)-1 $$


 * $$ n=2 $$; :$$ J=1 $$


 * $$Q_2(x) = P_2(x) \cdot \tanh^{-1}(x) - 2 \cdot \sum_{j=1}^1 \frac{2(2)-2(1)+1}{(2(2)-(1)+1)(1)} \cdot P_{(2)-(1)}(x) $$


 * $$Q_2(x) = \frac{1}{2}(3x^2 - 1)

\cdot \tanh^{-1}(x) - 2 \sum_{j=1}^1 \frac{2(2)-2(1)+1}{(2(2)-(1)+1)(1)} \cdot x $$


 * $$Q_2(x) = \frac{1}{2}(3x^2 - 1)

\cdot \tanh^{-1}(x) -  \frac{5}{2} x $$


 * $$ n=3 $$; :$$ J=3 $$


 * $$Q_3(x) = P_3(x) \cdot \tanh^{-1}(x) - 2 \cdot \sum_{j=1,3}^3 \frac{2(3)-2j+1}{(2(3)-j+1)j} \cdot P_{3-j}(x) $$


 * $$Q_3(x) = P_3(x) \cdot \tanh^{-1}(x) - 2 \cdot \left [ \frac{2(3)-2(1)+1}{(2(3)-(1)+1)(1)} \cdot P_{3-1}(x) + \frac{2(3)-2(3)+1}{(2(3)-(3)+1)(3)} \cdot P_{3-3}(x)\right] $$


 * $$Q_3(x) = \frac{1}{2}(-3x + 5x^3) \cdot \tanh^{-1}(x) - 2 \left [ \frac{2(3)-2(1)+1}{(2(3)-(1)+1)(1)} \cdot \frac{1}{2}(3x^2 - 1) (x) + \frac{2(3)-2(3)+1}{(2(3)-(3)+1)(3)} \cdot 1 \right]  $$


 * $$Q_3(x) = \frac{1}{2}(-3x + 5x^3) \cdot \tanh^{-1}(x) - \left(\frac{5}{2}x^3+\frac{1}{3}\right) $$


 * $$ n=4 $$; :$$ J=3 $$


 * $$Q_4(x) = P_4(x) \cdot \tanh^{-1}(x) - 2 \cdot \sum_{j=1,3}^3 \frac{2(4)-2j+1}{(2(4)-j+1)j} \cdot P_{4-j}(x) $$


 * Q_4(x) = P_4(x) \cdot \tanh^{-1}(x) - 2 \left [ \frac{2(4)-2(1)+1}{(2(4)-(1)+1)(1)} \cdot P_{4-1}(x) +  \frac{2(4)-2(3)+1}{(2(4)-(3)+1)(1)} \cdot P_{4-3}(x) \right ]


 * $$Q_4(x) = \frac{1}{8}(3 - 30 x^2 + 35 x^4) \cdot \tanh^{-1}(x) - 2 \left [ \frac{2(4)-2(1)+1}{(2(4)-(1)+1)(1)} \cdot \frac{1}{2}(-3x + 5x^3) +  \frac{2(4)-2(3)+1}{(2(4)-3+1)(1)} \cdot x \right] $$


 * $$Q_4(x) = \frac{1}{8}(3 - 30 x^2 + 35 x^4) \cdot \tanh^{-1}(x) - \left(\frac{35}{8} x^3-\frac{55}{24}x\right) $$

Plots:

P(x)



Q(x)




 * }

= 4 Proof of Pn and Qn =

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

Given

 * $$ = $$

Show

 * $$ = 0$$

Solution
Recall


 * $$ = \int P_n(x) \cdot Q_n(x) dx$$

since


 * $$P_n(x)$$

is even and


 * $$Q_n(x)$$

is odd for any value of n, the integral is zero.


 * }

= 5 Distance Proof =

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

Given

 * $$(r_{PQ})^2 = (r_P)^2 + (r_Q)^2 - 2r_Pr_Q \cos \gamma$$

where


 * $$ \cos \gamma = \cos \theta_Q \cos \theta_P \cos(\varphi_Q - \varphi_P) + \sin \theta_P \sin \theta_P $$

Show
The above relation is true.

Solution
Starting with,


 * $$(r_{PQ})^2 = \sum (x_Q^i - x_P^i)^2$$

where



\begin{align} x_P^1& = r_P \cos \theta_P \cos \varphi_P\\ x_P^2& = r_P \cos \theta_P \sin \varphi_P\\ x_P^3& =r_P \sin \theta_P\\ x_Q^1& = r_Q \cos \theta_Q \cos \varphi_Q\\ x_Q^2& = r_Q \cos \theta_Q \sin \varphi_Q\\ x_Q^3& = r_Q \sin \theta_Q\\ (r_{PQ})^2& = (x_Q^1 - x_P^1)^2 + (x_Q^2 - x_P^2)^2 + (x_Q^3 - x_P^3)^2\\ (r_{PQ})^2& = (r_Q \cos \theta_Q \cos \varphi_Q - r_P \cos \theta_P \cos \varphi_P)^2 + (r_Q \cos \theta_Q \sin \varphi_Q - r_P \cos \theta_P \sin \varphi_P)^2 + (r_Q \sin \theta_Q - r_P \sin \theta_P)^2\\ (r_{PQ})^2& = r_Q^2 (\cos \theta_Q)^2 (\cos \varphi_Q)^2 - 2 r_Q r_P \cos \theta_Q \cos \theta_P \cos \varphi_Q \cos \varphi_P + r_P^2 (\cos \theta_P)^2 (\cos \varphi_P)^2 + r_Q^2 (\cos \theta_Q)^2 (\sin \varphi_Q)^2 \\ &\quad -2 r_Q r_P \cos \theta_Q \cos \theta_P \sin \varphi_Q \sin \varphi_P + r_P^2 (\cos \theta_P)^2 (\sin \varphi_P)^2 + r_Q^2 (\sin \theta_Q)^2 - 2 r_Qr_P \sin \theta_Q \sin \theta_P + r_P^2 (\sin \theta_P)^2\\ (r_{PQ})^2& = r_Q^2 (\cos \theta_Q)^2 ((\cos \varphi_Q)^2 + (\sin \varphi_Q)^2) - 2 r_Q r_P \cos \theta_Q \cos \theta_P (\cos \varphi_Q \cos \varphi_P + \sin \varphi_Q \sin \varphi_P)+ r_P^2 (\cos \theta_P)^2 ((\cos \varphi_P)^2 \\ &\quad + (\sin \varphi_P)^2) + r_Q^2 (\sin \theta_Q)^2 - 2 r_Qr_P \sin \theta_Q \sin \theta_P + r_P^2 (\sin \theta_P)^2\\ (r_{PQ})^2& = r_Q^2((\cos \theta_Q)^2 + (\sin \theta_P)^2) + r_P^2((\cos \theta_Q)^2 + (\sin \theta_P)^2) - 2r_Q r_P [ \cos \theta_Q \cos \theta_P \cos(\varphi_Q - \varphi_P) + \sin \theta_P \sin \theta_P ] \end{align} $$

given the defintion of


 * $$\cos \gamma = \cos \theta_Q \cos \theta_P \cos(\varphi_Q - \varphi_P) + \sin \theta_P \sin \theta_P $$
 * {| style="width:100%" border="0" align="left"


 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * $$(r_{PQ})^2 = r_Q^2 + r_P^2 - 2r_Q r_P \cos \gamma $$
 * $$(r_{PQ})^2 = r_Q^2 + r_P^2 - 2r_Q r_P \cos \gamma $$


 * }
 * }
 * }


 * }

= 6 Binomial Proof =

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

Given

 * $$a.)\quad(x + y)^r = \sum (r k) x^{r-k}y^k$$


 * $$b.)\quad(r k) = \dfrac{r(r-1)...(r-k+1)}{k!}$$

Find

 * $$(1-x)^{-1/2} = \sum \alpha_i x^i$$

where


 * $$\alpha_i = \dfrac{1 \cdot 3 \cdot ... \cdot (2i-1)}{2 \cdot 4 \cdot ... \cdot (2i)}$$

Solution
Substitute b into a, since r is not an integer, but is any real value.


 * $$(x + y)^r = \sum \dfrac{r(r-1)...(r-k+1)}{k!} \cdot x^{r-k}y^k$$


 * $$y = 1$$, so :$$y^k=1$$

for all values of k.


 * $$(x + 1)^r = \sum \dfrac{r(r-1)...(r-k+1)}{k!} \cdot x^{r-k} $$


 * $$r = -\dfrac{1}{2}$$,

Substitute in:


 * $$(x + 1)^{-\dfrac{1}{2}} = \sum \dfrac{-\dfrac{1}{2}\left(-\dfrac{1}{2}-1\right)...\left(-\dfrac{1}{2}-k+1\right)}{k!} \cdot x^{(-1/2)-k} $$

Now all for -x:


 * $$(-x + 1)^{-\dfrac{1}{2}} = \sum \dfrac{-\dfrac{1}{2}\left(-\dfrac{1}{2}-1\right)...\left(-\dfrac{1}{2}-k+1\right)}{k!} \cdot (-x)^{(-1/2)-k} $$

let


 * $$i = -\dfrac{1}{2}-k$$

to get


 * $$(1-x)^{-\dfrac{1}{2}} = \sum \dfrac{(2i-1)!}{(2i)!} \cdot x^i $$

Note that


 * $$-x^i = (-1)^i \cdot x^i$$,

which is combined into the dfractional factorial.

therefore,


 * {| style="width:20%" border="0" align="left"


 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * $$(1-x)^{-\dfrac{1}{2}} = \sum \alpha_i \cdot x^i $$
 * $$(1-x)^{-\dfrac{1}{2}} = \sum \alpha_i \cdot x^i $$

where


 * $$\alpha_i = \dfrac{(2i-1)!}{(2i)!}$$,

as expected.
 * }
 * }
 * }


 * }

= 7 Expressions for Pn =

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

Given

 * $$\frac{1}{\sqrt{A(\mu, \rho)}} = \sum \alpha_i (2\mu \rho - \rho^2)^i$$

Find
The expressions for :$$P_3$$, :$$P_4$$, :$$P_5$$ using the expansion

Solution

 * $$\frac{1}{\sqrt{A(\mu, \rho)}} = \alpha_0 + \alpha_1 \cdot (2 \mu \rho - \rho^2) + \alpha_2 \cdot (2 \mu \rho - \rho^2)^2 + \alpha_3 \cdot (2 \mu \rho - \rho^2)^3 + \alpha_4 \cdot (2 \mu \rho - \rho^2)^4 + \alpha_5 \cdot (2 \mu \rho - \rho^2)^5$$

where



\begin{align} (2 \mu \rho - \rho^2)^2& = 4 \mu^2 \rho^2 - 4 \mu \rho^3 + \rho^4\\ (2 \mu \rho - \rho^2)^3& = 8 \mu^3 \rho^3 - 8 \mu^2 \rho^4 + 2 \mu \rho^5 - 4 \mu^2 \rho^4 + 4 \mu \rho^5 + \rho^6\\ (2 \mu \rho - \rho^2)^4& = 16 \mu^4 \rho^4 - 16 \mu^3 \rho^5 + 4 \mu^2 \rho^6 - 8 \mu^3 \rho^5 + 8 \mu^2 \rho^6 + 2 \mu \rho^7\\ &\quad - 8 \mu^3 \rho^5 + 8 \mu^2 \rho^6 - 2 \mu \rho^7 + 4 \mu^2 \rho^6 - 4 \mu \rho^7 - \rho^8\\ (2 \mu \rho - \rho^2)^5& = 32 \mu^5 \rho^5 - 32 \mu^4 \rho^6 + 8 \mu^3 \rho^7 - 16 \mu^4 \rho^6 + 16 \mu^3 \rho^7 + 4 \mu^2 \rho^8\\ &\quad - 16 \mu^4 \rho^6 + 16 \mu^3 \rho^7 - 4 \mu^2 \rho^8 + 8 \mu^3 \rho^7 - 8 \mu^2 \rho^8 - 2 \mu \rho^9 -    16 \mu^4 \rho^6  \\ &\quad + 16 \mu^3 \rho^7 - 4 \mu^2 \rho^8 + 8 \mu^3 \rho^7 - 8 \mu^2 \rho^8 + 2 \mu \rho^9 + 8 \mu^3 \rho^7 - 8 \mu^2 \rho^8 \\ &\quad + 2 \mu \rho^9 - 4 \mu^2 \rho^8 + 4 \mu \rho^9 + \rho^10 \end{align} $$

gather by :$$\rho$$:


 * $$\frac{1}{\sqrt{A(\mu, \rho)}} = \alpha_0 + 2 \mu \alpha_1 \cdot \rho + (- \alpha_1 + 4) \rho^2 + (-4 \mu \alpha_2 + 8 \mu^3 \alpha_3) \cdot \rho^3 + (\alpha_2 - \mu^2 \alpha_3 + 16 \mu^4 \alpha_4) \cdot \rho^4 + (6 \mu \alpha_3 - 24 \mu^3 \alpha_4 + 32 \mu^5 \alpha_5) \cdot \rho^5 + ...$$

From Eqn 7, pg 38-4,


 * $$\alpha_0 = 1\quad \alpha_1 = \frac{1}{2} \quad \alpha_2 = \frac{3}{8} \quad \alpha_3 = \frac{5}{16} \quad \alpha_4 = \frac{35}{128} \quad \alpha_5 = \frac{63}{256}$$

Therefore,


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


 * style="width:2%; padding:10px; border:2px solid #1888aa" |
 * style="width:2%; padding:10px; border:2px solid #1888aa" |

\begin{align} P_0(x)& = \alpha_0 = 1\\ P_1(x)& = 2 \mu \alpha_1 = x\\ P_2(x)& = (- \alpha_1 + 4) = \frac{1}{2}(3x^2 - 1)\\ P_3(x)& = -4 \mu \alpha_2 + 8 \mu^3 \alpha_3 = \frac{1}{2}(-3x + 5x^3)\\ P_4(x)& = \alpha_2 - \mu^2 \alpha_3 + 16 \mu^4 \alpha_4 = \frac{1}{8}(3 - 30 x^2 + 35 x^4)\\ P_5(x)& = 6 \mu \alpha_3 - 24 \mu^3 \alpha_4 + 32 \mu^5 \alpha_5 = \frac{1}{8} (15 x - 70 x^3 + 63 x^5) \end{align} $$
 * }
 * }
 * }

as expected.


 * }

= Contributing Team Members =

Angela Diggs Egm6321.f09.team3.diggs 20:21, 6 December 2009 (UTC)

Catherine Snow EGM6321.f09.Team3.cgs 15:24, 7 December 2009 (UTC)

Francisco Kuljevan --Egm6321.f09.team3.Kuljevan.Francisco 05:22, 8 December 2009 (UTC)

Barrett Mooney Egm6321.f09.team3.mooney 13:38, 8 December 2009 (UTC)