User:Egm6321.f09.team3.diggs/HW7

= 1 Hyperbolic Tangent Equivalent =

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

Show
The equivalence of the above statement.

Solution
In HW 6, we showed

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

Substituting in,

$$ x tanh^{-1}(x) - 1 = x tanh^{-1}(x) - 1$$

which is equivalent.

= 2 Qn as odd/even =

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

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 =

Given
$$P_n(x)$$ for n = 0, 1, 2, 3, 4

and

$$Q_n(x)$$ for n = 0, 1, 2, 3, 4

Find
Plot all functions.

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

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

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

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

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

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

$$ n=0 $$; $$ J=1 $$

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

$$ n=1 $$; $$ J=1 $$

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

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

$$ n=2 $$; $$ J=1 $$

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

$$Q_2(x) = \frac{1}{2}(3x^2 - 1) \cdot tanh^{-1}(x) - 2 \sum_{j=1}^1 \frac{2(2)-2(1)+1}{(2(2)-(1)+1)(1)} \cdot x $$

$$Q_2(x) = \frac{1}{2}(3x^2 - 1) \cdot tanh^{-1}(x) -  \frac{5}{2} x $$

$$ n=3 $$; $$ J=3 $$

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

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

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

$$ n=4 $$; $$ J=3 $$

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

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 ]

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

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

Plots:

P(x)



Q(x)



= 4 Proof of Pn and Qn =

Given
$$ = $$

Show
$$ = 0$$

Solution
Recall $$ = \int P_n(x) \cdot Q_n(x) dx$$

Since Pn(x) is even and Qn(x) is odd for any value of n, the integral is zero.

= 5 Distance Proof =

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

$$x_P^1 = r_P cos \theta_P cos \varphi_P$$

$$x_P^2 = r_P cos \theta_P sin \varphi_P$$

$$x_P^3 = r_P sin \theta_P$$

$$x_Q^1 = r_Q cos \theta_Q cos \varphi_Q$$

$$x_Q^2 = r_Q cos \theta_Q sin \varphi_Q$$

$$x_Q^3 = r_Q sin \theta_Q$$

$$(r_{PQ})^2 = (x_Q^1 - x_P^1)^2 + (x_Q^2 - x_P^2)^2 + (x_Q^3 - x_P^3)^2$$

$$(r_{PQ})^2 = (r_Q cos \theta_Q cos \varphi_Q - r_P cos \theta_P cos \varphi_P)^2 + (r_Q cos \theta_Q sin \varphi_Q - r_P cos \theta_P sin \varphi_P)^2 + (r_Q sin \theta_Q - r_P sin \theta_P)^2$$

$$(r_{PQ})^2 = r_Q^2 (cos \theta_Q)^2 (cos \varphi_Q)^2 - 2 r_Q r_P cos \theta_Q cos \theta_P cos \varphi_Q cos \varphi_P + r_P^2 (cos \theta_P)^2 (cos \varphi_P)^2 + r_Q^2 (cos \theta_Q)^2 (sin \varphi_Q)^2 - 2 r_Q r_P cos \theta_Q cos \theta_P sin \varphi_Q sin \varphi_P + r_P^2 (cos \theta_P)^2 (sin \varphi_P)^2 + r_Q^2 (sin \theta_Q)^2 - 2 r_Qr_P sin \theta_Q sin \theta_P + r_P^2 (sin \theta_P)^2$$

$$(r_{PQ})^2 = r_Q^2 (cos \theta_Q)^2 ((cos \varphi_Q)^2 + (sin \varphi_Q)^2) - 2 r_Q r_P cos \theta_Q cos \theta_P (cos \varphi_Q cos \varphi_P + sin \varphi_Q sin \varphi_P)+ r_P^2 (cos \theta_P)^2 ((cos \varphi_P)^2 + (sin \varphi_P)^2) + r_Q^2 (sin \theta_Q)^2 - 2 r_Qr_P sin \theta_Q sin \theta_P + r_P^2 (sin \theta_P)^2$$

$$(r_{PQ})^2 = r_Q^2((cos \theta_Q)^2 + (sin \theta_P)^2) + r_P^2((cos \theta_Q)^2 + (sin \theta_P)^2) - 2r_Q r_P [ cos \theta_Q cos \theta_P cos(\varphi_Q - \varphi_P) + sin \theta_P sin \theta_P ]$$

given the defintion of $$cos \gamma = [...] $$

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

= 6 Binomial Proof =

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

b. $$(r k) = \frac{r(r-1)...(r-k+1)}{k!}$$

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

where

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

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

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

$$y = 1$$, so $$y^k=1$$ for all values of k.

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

$$r = -\frac{1}{2}$$, Substitute in:

$$(x + 1)^{-\frac{1}{2}} = \sum \frac{-\frac{1}{2}(-\frac{1}{2}-1)...(-\frac{1}{2}-k+1)}{k!} \cdot x^{-\frac{1}{2}-k} $$

Now all for -x:

$$(-x + 1)^{-\frac{1}{2}} = \sum \frac{-\frac{1}{2}(-\frac{1}{2}-1)...(-\frac{1}{2}-k+1)}{k!} \cdot (-x)^{-\frac{1}{2}-k} $$

let $$i = -\frac{1}{2}-k$$ to get

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

Note that $$-x^i = (-1)^i \cdot x^i$$, which is combined into the fractional factorial.

therefore,

$$(1-x)^{-\frac{1}{2}} = \sum \alpha_i \cdot x^i $$

where $$\alpha_i = \frac{(2i-1)!}{(2i)!}$$, as expected.

= 7 Expressions for Pn =

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

$$(2 \mu \rho - \rho^2)^2 = 4 \mu^2 \rho^2 - 4 \mu \rho^3 + \rho^4$$

$$(2 \mu \rho - \rho^2)^3 = 8 \mu^3 \rho^3 - 8 \mu^2 \rho^4 + 2 \mu \rho^5 - 4 \mu^2 \rho^4 + 4 \mu \rho^5 + \rho^6$$

$$(2 \mu \rho - \rho^2)^4 = 16 \mu^4 \rho^4 - 16 \mu^3 \rho^5 + 4 \mu^2 \rho^6 - 8 \mu^3 \rho^5 + 8 \mu^2 \rho^6 + 2 \mu \rho^7 -    8 \mu^3 \rho^5 + 8 \mu^2 \rho^6 - 2 \mu \rho^7 + 4 \mu^2 \rho^6 - 4 \mu \rho^7 - \rho^8$$

$$(2 \mu \rho - \rho^2)^5 = 32 \mu^5 \rho^5 - 32 \mu^4 \rho^6 + 8 \mu^3 \rho^7 - 16 \mu^4 \rho^6 + 16 \mu^3 \rho^7 + 4 \mu^2 \rho^8 - 16 \mu^4 \rho^6 + 16 \mu^3 \rho^7 - 4 \mu^2 \rho^8 + 8 \mu^3 \rho^7 - 8 \mu^2 \rho^8 - 2 \mu \rho^9 -    16 \mu^4 \rho^6 + 16 \mu^3 \rho^7 - 4 \mu^2 \rho^8 + 8 \mu^3 \rho^7 - 8 \mu^2 \rho^8 + 2 \mu \rho^9 + 8 \mu^3 \rho^7 - 8 \mu^2 \rho^8 + 2 \mu \rho^9 - 4 \mu^2 \rho^8 + 4 \mu \rho^9 + \rho^10$$

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,

$$\alpha_0 = 1$$

$$\alpha_1 = \frac{1}{2}$$

$$\alpha_2 = \frac{3}{8}$$

$$\alpha_3 = \frac{5}{16}$$

$$\alpha_4 = \frac{35}{128}$$

$$\alpha_5 = \frac{63}{256}$$

Therefore,

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

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

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

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

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

$$P_5(x) = 6 \mu \alpha_3 - 24 \mu^3 \alpha_4 + 32 \mu^5 \alpha_5 = \frac{1}{8} (15 x - 70 x^3 + 63 x^5)$$

as expected.

= Contributing Team Members =

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