User:Egm6321.f09.Team2/HW7

Homework Assignment #7 - due Wednesday, 12/9, 22:00 UTC

Problem 1
p. 37-1 Show $$Q_1(x)={1\over 2}xln\Bigg({{1+x}\over {1-x}}\Bigg)-1=xtanh^{-1}x-1$$

$$\frac{1}{2}xln(\frac{1+x}{1-x})-1=xtanh^{-1}x-1$$

Add 1 to both sides and divide by x: $$\frac{1}{2}ln(\frac{1+x}{1-x}) = tanh^{-1}x$$

Assume: $$y = tanh^{-1}(x)$$

Then: $$x = tanh (y) = \frac{e^y - e^{-y}}{e^y + e^{-y}}$$

Cross-multiply: $$ xe^y + xe^{-y} = e^y - e^{-y}$$

Multiply through by e^y: $$ x(e^y)^2 + x = (e^y)^2 - 1 $$

Re-arrange terms: $$ e^{2y} = \frac{1+x}{1-x}$$

Square root of both sides: $$ e^y = \sqrt{\frac{1+x}{1-x}}$$

To solve for y, take the natural log: $$ y = ln{(\sqrt{\frac{1+x}{1-x}})}$$ $$ y = \frac{1}{2}ln(\frac{1+x}{1-x})$$

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

Problem 2
p. 37-1 1) Using $$Q_ \left( x \right) =P_ \left( x \right) {\it tanh^{-1}} \left( x \right) -2\,\sum _{j=1,3,5}^{J}{\frac { \left( 2\,n-2\,j+1 \right) P_ \left( x \right) }{ \left( 2\,n-j+1 \right) j}} $$ show when $$Q_n$$ is even or odd depending on "n"

2) Plot $$[P_0,\ P_1,\ ...,\ P_4]\ and\ [Q_0,\ Q_1,\ ...,\ Q_4]$$

1) Using the properties of even and odd number and functions, one can determine if $$Q_n \!$$ is even or odd. The product of an odd and even function is odd. The product of two odd functions is even. The sum of two odd functions is odd. The sum of two even functions is even. The sum of an even and odd number is odd. The sum of two odd numbers is even. Constant multiples of a function do not change its evenness or oddness.

For $$n=2k\!$$ (even number) $$P_{2k}(x)\!$$ is even and $$tanh^{-1}(x)\!$$ is odd. Therefore, the left-hand side of $$Q_{2k}(x)\!$$ is odd. On the right side, $$2k-j\!$$ is odd because $$j\!$$ is odd. This means $$P_{2k-j}(x)\!$$ is odd, so the summation is odd. Therefore, the right-hand side of $$Q_{2k}(x)\!$$ is odd. The end result is that $$Q_{2k}(x)\!$$ is odd.

For $$n=2k+1\!$$ (odd number) $$P_{2k+1}(x)\!$$ is odd. Since $$tanh^{-1}(x)\!$$ is still odd, the left-hand side of $$Q_{2k+1}(x)\!$$ is even. On the right side, $$2k+1-j\!$$ is even because $$j\!$$ is odd. This means $$P_{2k+1-j}(x)\!$$ is even, so the summation is even. Therefore, the right-hand side of $$Q_{2k+1}(x)\!$$ is even. The end result is that $$Q_{2k+1}(x)\!$$ is even.

2) $$P_n\!$$



$$Q_n\!$$



Problem 3
Describe why $$=0\!$$

Since $$Pn\!$$ has opposite parity to $$Qn\!$$ for the same choice of $$n\!$$ and the product of an odd and even function is odd, this means the inner product will be equal to zero. The inner product is zero due to the property of odd functions, i.e. the integral of an odd function over a discrete boundary [-A,A] is zero.

Problem 4
p. 37-3 Given $$-\int_{-1}^{+1}L_m\Big[(1-x^2)L'_n\Big]'dx+n(n+1)\int_{-1}^{+1}L_mL_ndx=0$$

Using integration by parts show

$$-\int_{-1}^{+1}(1-x^2)L'_nL'_mdx+n(n+1)=0$$

Integration by parts follows the form,


 * $$\int_{b}^{b}u\, dv=[uv]_{a}^{b}-\int_{a}^{b}v\, du$$

Taking the first term on the left-hand side and separating it by,


 * $$u=L_m \!$$
 * $$dv = [(1-x^2)L_n']'\, dx$$

we can find that,


 * $$du=L_m'\, dx \!$$
 * $$v=[(1-x^2)L_n'] \!$$

This yields,


 * $$\int_{-1}^{1}L_m[(1-x^2)L_n']'\, dx = [L_mL_n'(1-x^2)]_{-1}^{1}-\int_{-1}^{1}(1-x^2)L_n'L_m'\, dx$$

Since $$1-x^2$$ is zero at $$-1$$ and $$1$$, the result is,


 * $$\int_{-1}^{1}L_m[(1-x^2)L_n']'\, dx = -\int_{-1}^{1}(1-x^2)L_n'L_m'\, dx$$

Using $$\!$$ to denote the inner product, the original equation can now be written as,


 * $$-\int_{-1}^{1}(1-x^2)L_n'L_m'\, dx+n(n+1)=0$$

Problem 5
p. 38-2 Given $$(r_{PQ})^2=\sum_{i=1}^3(x_Q^i-x_P^i)^2$$ Where $$x_P^1=r_Pcos\theta_Pcos\phi_P$$, $$x_P^2=r_Pcos\theta_Psin\phi_P$$, $$x_P^3=r_Psin\theta_P$$

$$x_Q^1=r_Qcos\theta_Qcos\phi_Q$$, $$x_Q^2=r_Qcos\theta_Qsin\phi_Q$$, $$x_Q^3=r_Qsin\theta_Q$$

Show $$(r_{PQ})^2=r_p^2+r_Q^2-2r_Pr_Qcos\gamma$$

where $$ cos\gamma=cos\theta_Qcos\theta_Pcos(\phi_Q-\phi_P)+sin\theta_Qsin\theta_P$$

$$(r_{PQ})^2=\sum_{i=1}^3(x_Q^i-x_P^i)^2$$

$$(r_{PQ})^2= \left( x_{1{Q}}-x_{1{P}} \right) ^{2}+ \left( x_{2{Q}}-x_{2{P}}\right) ^{2}+ \left( x_{3{Q}}-x_{3{P}} \right) ^{2}$$

$$(r_{PQ})^2={x_{1{Q}}}^{2}-2\,x_{1{Q}}x_{1{P}}+{x_{1{P}}}^{2}+{x_{2{Q}}}^{2}-2\,x_{2{Q}}x_{2{P}}+{x_{2{P}}}^{2}+{x_{3{Q}}}^{2}-2\,x_{3{Q}}x_{3{P}}+{x_{3{P}}}^{2}$$


 * $$={r_}^{2} \left( \cos \left( \theta_ \right) \right) ^{2}

\left( \cos \left( \phi_ \right) \right) ^{2}-2\,r_\cos \left( \theta_ \right) \cos \left( \phi_ \right) r_ \cos \left( \theta_ \right) \cos \left( \phi_ \right)$$
 * $$ +{r_}^{2} \left( \cos \left( \theta_ \right) \right) ^{2} \left(\cos \left( \phi_ \right)  \right) ^{2}+{r_}^{2} \left( \cos \left( \theta_ \right)  \right) ^{2} \left( \sin \left( \phi_ \right)  \right) ^{2}-2\,r_\cos \left( \theta_ \right)

\sin \left( \phi_ \right) r_\cos \left( \theta_ \right) \sin \left( \phi_ \right)$$
 * $$ +{r_}^{2} \left( \cos \left( \theta_ \right) \right) ^{2} \left( \sin \left( \phi_ \right)  \right) ^{2}+{r_}^{2} \left( \sin \left( \theta_ \right)  \right) ^{2}-2\,r_\sin \left( \theta_ \right) r_\sin \left( \theta_ \right) +{r_}^{2} \left( \sin \left( \theta_ \right)  \right) ^{2}$$

Combine terms;


 * $$={r_}^{2}+ \left (\,\cos \left( \theta_ \right) \cos

\left( \theta_ \right) \cos \left( \phi_ \right) \cos \left( \phi_ \right)-2\, \cos \left( \theta_ \right) \cos \left( \theta_ \right) \sin \left( \phi_ \right) \sin \left( \phi_ \right) -2\,\sin \left( \theta_ \right) \sin \left( \theta_ \right) \right)r_r_+{r_}^{2} $$


 * $$={r_}^{2}+{r_}^{2}-2r_r_(\,\cos \left( \theta_ \right) \cos

\left( \theta_ \right) \cos \left( \phi_ \right) \cos \left( \phi_ \right)+\, \cos \left( \theta_ \right) \cos \left( \theta_ \right) \sin \left( \phi_ \right) \sin \left( \phi_ \right) +\,\sin \left( \theta_ \right) \sin \left( \theta_ \right)) $$


 * $$={r_}^{2}+{r_}^{2}-2r_r_[\,\cos \left( \theta_ \right) \cos

\left( \theta_ \right)[ \cos \left( \phi_ \right) \cos \left( \phi_ \right)+ \sin \left( \phi_ \right) \sin \left( \phi_ \right)] +\,\sin \left( \theta_ \right) \sin \left( \theta_ \right)]$$

Using the identity: $$\cos \left( a \right) \cos \left( b \right) +\sin \left( a \right) \sin \left( b \right)=\cos \left( a-b \right)$$

We can further reduce
 * $$={r_}^{2}+{r_}^{2}-2r_r_[\,\cos \left( \theta_ \right) \cos

\left( \theta_ \right)[ \cos \left( \phi_-\phi_ \right)] +\,\sin \left( \theta_ \right) \sin \left( \theta_ \right)] $$

Substitute $$ cos\gamma=cos(\theta_Q)cos(\theta_P)cos(\phi_Q-\phi_P)+sin(\theta_Q)sin(\theta_P)$$

Yields  $$(r_{PQ})^2=r_p^2+r_Q^2-2r_Pr_Qcos\gamma$$

Problem 6
p. 38-4

Use $$(x+y)^r=\sum_{k=0}^\infty{n\choose k}x^{r-k}y^k$$ and $${n\choose k}={r(r-1)...(r-k+1)\over k!}$$

to show

$$(1-x)^{-1/2}=\sum_{i=0}^\infty\alpha_ix^i$$ and $$\alpha_i={1*3*...(2i-1)\over 2*4*...(2i)}$$

Start by making variables match-up:

$$(1-x)^{-1/2}=\sum_{k=0}^\infty\alpha_kx^k$$ and $$\alpha_k={1*3*...(2k-1)\over 2*4*...(2k)}$$

$$\frac{1}{(1-x)^{1/2}} = \sum_{k=0}^\infty\alpha_kx^k$$

Working from: $$(x+y)^r=\sum_{k=0}^\infty{r\choose k}x^{r-k}y^k$$

Assume: $$x = 1$$, $$ y = -x $$ ,  $$ r = \frac{1}{2}$$

As 1 to any power is still 1, we are left with:

$$\frac{1}{(1-x)^{1/2}} = \sum_{k=0}^\infty{r\choose k}-x^k$$

$${r\choose k} = {r-k+1\choose k}$$

$$\frac{1}{(1-x)^{1/2}}=\sum_{k=0}^\infty{r-k+1\choose k}-x^k$$

$$\frac{1}{(1-x)^{1/2}}=\sum_{k=0}^\infty{\frac{1}{2}+k-1\choose k}x^k$$

$$\frac{1}{(1-x)^{1/2}}=\sum_{k=0}^\infty{k-\frac{1}{2}\choose k}x^k$$

Simplfying the complex fraction / choose function:

$$\frac{1}{(1-x)^{1/2}}=\sum_{k=0}^\infty{2k-1\choose 2k}x^k$$

Defining: $$\alpha_i={(2i-1)\choose (2i)} = {(2k-1)\choose (2k)}$$, with k defined from 0 to $$\infty$$

When $$k=0$$, $$\alpha_0=1$$; $$k=1,\alpha_1=\frac{1}{2}$$; $$k=2,\alpha_2=\frac{3}{8}$$, etc.

 $$(1-x)^{-1/2}=\sum_{i=0}^\infty\alpha_ix^i$$

Problem 7
p.39-2

Continue the power series development of $${1\over\sqrt{A(\mu,\rho)}}=\alpha_0+\alpha_1(2\mu\rho-\rho^2)+\alpha_2(2\mu\rho-\rho^2)^2+....$$ to find $$P_3(\mu),\ P_4(\mu),\ P_5(\mu)$$

Compare to the results obtained on p.31-3.

Contributing Team Members
--Joe Gaddone 20:16, 30 November 2009 (UTC) --Matt Walker --Egm6321.f09.Team2.sungsik 02:22, 4 December 2009 (UTC) --Kumanchik 17:58, 9 December 2009 (UTC)