text:
User:Jiangpx/6321 hw7

=Problem1 - Plot Legendre Functions for n=0,1,2,3=

Given
Legendre functions

$$\begin{align} & \left\{ {{P}_{n}}(x),n=0,1,2,3 \right\} \\ & \left\{ {{Q}_{n}}(x),n=0,1,2,3 \right\} \\ \end{align}$$

Find
Plot

$$\left\{ {{P}_{n}}(x),n=0,1,2,3 \right\}$$

Plot

$$\left\{ {{Q}_{n}}(x),n=0,1,2,3 \right\}$$

Plot

$$\left\{ \begin{align} & {{P}_{0}}(x),{{Q}_{0}}(x) \\ & {{P}_{1}}(x),{{Q}_{1}}(x) \\ & {{P}_{2}}(x),{{Q}_{2}}(x) \\ & {{P}_{3}}(x),{{Q}_{3}}(x) \\ \end{align} \right\}$$

Observe

$${{P}_{n}}(x)\And {{Q}_{n}}(x)as-1<x<1$$

Observe even-ness/odd-ness of

$$\begin{align} & \left\{ {{P}_{n}}(x),n=0,1,2,3 \right\} \\ & \left\{ {{Q}_{n}}(x),n=0,1,2,3 \right\} \\ \end{align}$$

Guess

$$\int_{-1}^{1}{{{P}_{n}}(x){{Q}_{n}}(x)}dx$$

Solution
Legendre polynomials:

$$\begin{align} & {{P}_{0}}(x)=1 \\ & {{P}_{1}}(x)=x \\ & {{P}_{2}}(x)=\frac{1}{2}(3{{x}^{2}}-1) \\ & {{P}_{3}}(x)=\frac{1}{2}(5{{x}^{3}}-3x) \\ \end{align}$$

and the associated Legendre functions:

$$\begin{align} & {{Q}_{0}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \\ & {{Q}_{1}}(x)=\frac{1}{2}x\log \left( \frac{1+x}{1-x} \right)-1 \\ & {{Q}_{2}}(x)=\frac{1}{4}(3{{x}^{2}}-1)\log \left( \frac{1+x}{1-x} \right)-\frac{3}{2}x \\ & {{Q}_{3}}(x)=\frac{1}{4}(5{{x}^{3}}-3x)\log \left( \frac{1+x}{1-x} \right)-\frac{5}{2}{{x}^{2}}+\frac{2}{3} \\ \end{align}$$

For x approaches 1 or -1, Pn(x) and Qn(x) both have the steepest slopes in the domain of (-1 1).

For n being even, Pn(x) is even while Qn(x) is odd;

For n being odd, Pn(x) is odd while Qn(x) is even;

Guess $$\int_{-1}^{1}{{{P}_{n}}(x){{Q}_{n}}(x)}dx=0$$

=Problem2 - Prove even-ness/odd-ness of general functions=

Given
even $$\left\{ {{g}_{i}}(x) \right\}$$ and odd $$\left\{ {{h}_{i}}(x) \right\}$$

Find
$$\sum\limits_{i}{{{g}_{i}}(x)}$$ is even, and

$$\sum\limits_{i}{{{h}_{i}}(x)}$$ is odd.

Solution
In general, for even function g, we have

$$g(-x)=g(x)$$

Given $${{g}_{1}}(x)$$ is even, then

$${{g}_{1}}(-x)={{g}_{1}}(x)$$

Given $${{g}_{2}}(x)$$ is even, then

$${{g}_{2}}(-x)={{g}_{2}}(x)$$

so

$${{g}_{1}}(-x)+{{g}_{2}}(-x)={{g}_{1}}(x)+{{g}_{2}}(x)$$

that is

function $$G={{g}_{1}}+{{g}_{2}}$$ is even

Therefore, in general, given even $$\left\{ {{g}_{i}}(x) \right\}$$

we have function $$\sum\limits_{i}{{{g}_{i}}(x)}$$ also being even.

By the same token, in general, given odd $$\left\{ {{h}_{i}}(x) \right\}$$

we have function $$\sum\limits_{i}{{{h}_{i}}(x)}$$ also being odd.

=Problem3 - Show the even-ness/odd-ness of Legendre polynomials=

Given
general expressions of Legendre polynomials

$${{P}_{n}}(x)=\sum\limits_{i=0}^{[n/2]}{{{(-1)}^{i}}\frac{(2n-2i)!{{x}^{n-2i}}}{{{2}^{n}}i!(n-i)!(n-2i)!}}$$

Find
(1)

$${{P}_{2k}}(x)$$ is even and

(2)

$${{P}_{2k+1}}(x)$$ is odd, for k=0,1,2,...

Solution
(1)

For general even function g, we have

$$g(-x)=g(x)$$

For n=2k, with k=0,1,2,...

$${{P}_{n}}(x)$$ becomes

$$ P_{2k}(x) = \sum_{0}^{k}(-1)^{i}\frac{(4k-2i)! \color{black}{(x)^{2k-2i}}}{2^{2k}i!(2k-i)!(2k-2i)!} $$

Consider $${{P}_{2k}}(-x)$$

$${{P}_{2k}}(-x)=\sum\limits_{0}^{k}\frac{(4k-2i)!{{(-x)}^{2k-2i}}}{{{2}^{2k}}i!(2k-i)!(2k-2i)!}={{P}_{2k}}(x)$$

Thus, it is clear that $${{P}_{2k}}(x)$$ is even for k=0,1,2,...

(2)

For general odd function h, we have

$$h(-x)=-h(x)$$

For n=2k+1, with k=0,1,2,...

$${{P}_{n}}(x)$$ becomes

$$ P_{2k+1}(x) = \sum_{0}^{[\frac{2k+1}{2}]}(-1)^{i}\frac{(4k+2-2i)! \color{black}{(x)^{2k+1-2i}}}{2^{2k+1}i!(2k+1-i)!(2k+1-2i)!} $$

Consider $${{P}_{2k+1}}(-x)$$

$${{P}_{2k+1}}(-x)=\sum\limits_{0}^{[\frac{2k+1}{2}]}\frac{(4k+2-2i)!{{(-x)}^{2k+1-2i}}}{{{2}^{2k+1}}i!(2k+1-i)!(2k+1-2i)!}=-{{P}_{2k+1}}(x)$$

Thus, it is clear that $${{P}_{2k+1}}(x)$$ is odd for k=0,1,2,...

=Problem4 - General polynomials vs Legendre polynomials=

Given
$$\begin{align} & \text{Polynomial} \\ & q(x)=\sum\limits_{i=0}^{5} \\ & \text{with} \\ & {{c}_{0}}=2,{{c}_{1}}=-5,{{c}_{2}}=-3,{{c}_{3}}=11,{{c}_{4}}=7,{{c}_{5}}=6 \\ & \\ \end{align}$$

Find
$$\begin{align} & \text{(1) }\left\{ {{a}_{i}} \right\}\text{ such that} \\ & q(x)=\sum\limits_{i=0}^{5}{{{a}_{i}}{{P}_{i}}(x)}\text{ } \\ & \text{where} \\ & {{P}_{i}}(x)-\text{Legendre Polynomials} \\ & \\  & \text{(2) Plot both expressions of }q(x)\text{ to verify} \\ \end{align}$$

Solution
Write c as

$$[c]=\left[ \begin{matrix} 2 & -5 & -3 & 11 & 7 & 6 \\ \end{matrix} \right]$$

Then

$$q=[c]{{[1\text{ }x\text{ }{{x}^{2}}\text{ }{{x}^{3}}\text{ }{{x}^{4}}\text{ }{{x}^{5}}]}^{T}}=[c]{{[{{x}^{i}}]}^{T}}=[c][X]$$

Also

$$q=[a]{{[{{P}_{0}}\text{ }{{P}_{1}}\text{ }{{P}_{2}}\text{ }{{P}_{3}}\text{ }{{P}_{4}}\text{ }{{P}_{5}}]}^{T}}=[a][P]$$

Express $${{[{{P}_{0}}\text{ }{{P}_{1}}\text{ }{{P}_{2}}\text{ }{{P}_{3}}\text{ }{{P}_{4}}\text{ }{{P}_{5}}]}^{T}}$$ as $$[A][X]$$

$$[P]=[A][X]=\left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 \\   0 & 1 & 0 & 0 & 0 & 0  \\   -1/2 & 0 & 3/2 & 0 & 0 & 0  \\   0 & -3/2 & 0 & 5/2 & 0 & 0  \\   3/8 & 0 & -15/4 & 0 & 35/8 & 0  \\   0 & 15/8 & 0 & -35/4 & 0 & 63/8  \\ \end{matrix} \right]\left[ \begin{matrix} 1 \\   x  \\ {{x}^{2}} \\ {{x}^{3}} \\ {{x}^{4}} \\ {{x}^{5}} \\ \end{matrix} \right]$$

From above derivations, we have

$$\begin{align} & q=[a][A][X] \\ & q=[c][X] \\ \end{align}$$

So for the above to be valid for all x, we must have

$$[a][A]=[c]$$

Therefore

$$[a]=[c]{{[A]}^{-1}}$$

With [c] and [A] known, [a] can be obtained as

$${{[a]}^{T}}=\left[ \begin{align} & 2.4000000000 \\ & 4.1714285714 \\  & 2.0000000000 \\  & 7.0666666666 \\  & 1.6000000000 \\  & 0.7619047619 \\ \end{align} \right]$$

The polynomial q(x) has been plot in terms of [c] and [a]

=Problem5 - Find The Coefficients An in Legendre-Fourier Series=

Given
Boundary condition

$$\begin{align} & (a)\text{ }\psi (r=1,\theta )=f(\theta )={{T}_{0}}{{\cos }^{6}}\theta  \\ & (b)\text{ }\psi (r=1,\theta )=f(\theta )={{T}_{0}}\exp \left( -\frac{4{{\theta }^{2}}} \right) \\ \end{align}$$

and expression of An

$${{A}_{n}}=\frac{2n+1}{2}$$

Find
(1) Properties of An, i.e.

$$\begin{align} & {{A}_{2k}}=0\text{ }? \\ & {{A}_{2k+1}}=0\text{  }? \\ & for\text{  }k=0,1,2,... \\ \end{align}$$

(2) Calculate non-zero

$$\left\{ {{A}_{n}} \right\}$$

Solution
(a).(1)

Define $$x=\sin \theta $$

Then $$f(x)={{T}_{0}}{{(1-{{x}^{2}})}^{3}}$$ and it is a even function

We know Legendre polynomial $${{P}_{n}}$$ is even with even n and odd with odd n

Hence, for n=2k+1, k=0,1,2,...

$$=0$$, i.e.,

$${{A}_{2k+1}}=0$$

Since

$$f(x)\in {{\Pi }_{6}},\text{ }a\text{  }set\text{  }of\text{  }all\text{  }polynomials\text{  }with\text{  }order\text{ }\le 6$$

and the linear independence of Legendre polynomials, we have

$$\begin{align} & {{P}_{n}}\bot f\text{ }for\text{  }n\ge 7,\text{  }i.e., \\ & =0 \\ & so\text{ } \\ & {{A}_{n}}=0,\text{ }for\text{  }n\ge 7 \\ \end{align}$$

$$Based\text{ }on\text{  }that,\text{  }only\text{  }{{A}_{0}},\text{  }{{A}_{2}},\text{  }{{A}_{4}},\text{  }and\text{  }{{A}_{6}}\text{  }are\text{  }non-zero$$

(a).(2)

We know that

$$\begin{align} & {{P}_{0}}(x)=1 \\ & {{P}_{2}}(x)=\frac{3}{2}{{x}^{2}}-\frac{1}{2} \\ & {{P}_{4}}(x)=\frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8} \\ & {{P}_{6}}(x)=\frac{231}{16}{{x}^{6}}-\frac{315}{16}{{x}^{4}}+\frac{105}{16}{{x}^{2}}-\frac{5}{16} \\ & \\ \end{align}$$

Then

$${{A}_{n}}=\frac{2n+1}{2}=\frac{2n+1}{2}\int_{-1}^{1}{f(x){{P}_{n}}(x)dx}$$

So $$\begin{align} & {{A}_{0}}=\frac{1}{2}=\frac{1}{2}\int_{-1}^{1}{{{T}_{0}}{{(1-{{x}^{2}})}^{3}}dx} \\ & {{A}_{2}}=\frac{5}{2}=\frac{5}{2}\int_{-1}^{1}{{{T}_{0}}{{(1-{{x}^{2}})}^{3}}\left( \frac{3}{2}{{x}^{2}}-\frac{1}{2} \right)dx} \\ & {{A}_{4}}=\frac{9}{2}=\frac{9}{2}\int_{-1}^{1}{{{T}_{0}}{{(1-{{x}^{2}})}^{3}}\left( \frac{35}{8}{{x}^{4}}-\frac{15}{4}{{x}^{2}}+\frac{3}{8} \right)dx} \\ & {{A}_{6}}=\frac{13}{2}=\frac{13}{2}\int_{-1}^{1}{{{T}_{0}}{{(1-{{x}^{2}})}^{3}}\left( \frac{231}{16}{{x}^{6}}-\frac{315}{16}{{x}^{4}}+\frac{105}{16}{{x}^{2}}-\frac{5}{16} \right)dx} \\ \end{align}$$

Evaluating the above definite integrals via Matlab, we may get

$$\begin{align} & {{A}_{0}}=\frac{16}{35}{{T}_{0}} \\ & {{A}_{2}}=-\frac{16}{21}{{T}_{0}} \\ & {{A}_{4}}=\frac{144}{385}{{T}_{0}} \\ & {{A}_{6}}=-\frac{16}{231}{{T}_{0}} \\ & \\ \end{align}$$

(b).(1) Define $$x=\theta $$

then $$f(x)={{T}_{0}}\exp \left( -\frac{4{{x}^{2}}} \right)$$

and

$$\begin{align} & \mu =\sin \theta =\sin x \\ & {{P}_{n}}(\mu )={{P}_{n}}(\sin x) \\ \end{align}$$

Note that $$f(x)$$ is an even function, along with the properties of $${{P}_{n}}(\sin x)$$, we have that

$$\begin{align} & =0 \\ & i.e.,\text{ }{{A}_{2k+1}}=0 \\ \end{align}$$

for k=0,1,2,...

Since $$f(x)$$ is not any kind of polynomials,

$$\begin{align} & \ne 0 \\ & so \\ & {{A}_{2k}}\text{ }exist\text{  }for\text{  }k=0,1,2,... \\ \end{align}$$

(b).(2) We have

$${{A}_{n}}=\frac{2n+1}{2}=\frac{2n+1}{2}\int_{-\pi /2}^{\pi /2}{f(x){{P}_{n}}(\sin x)dx}$$

Given

$$f(x)={{T}_{0}}\exp \left( -\frac{4{{x}^{2}}} \right)$$

and

$$\begin{align} & {{P}_{0}}(\sin x)=1 \\ & {{P}_{2}}(\sin x)=3/2{{\left( \sin \left( x \right) \right)}^{2}}-1/2 \\ & {{P}_{4}}(\sin x)=\frac{35}{8}{{\left( \sin \left( x \right) \right)}^{4}}-\frac{15}{4}{{\left( \sin \left( x \right) \right)}^{2}}+3/8 \\ & {{P}_{6}}(\sin x)=\frac{231}{16}{{\left( \sin \left( x \right) \right)}^{6}}-\frac{315}{16}{{\left( \sin \left( x \right) \right)}^{4}}+\frac{105}{16}{{\left( \sin \left( x \right) \right)}^{2}}-\frac{5}{16} \\ \end{align}$$

Evaluating the above definite integrals via Matlab, we may get

$$\begin{align} & {{A}_{0}}=1.1731086046{{T}_{0}} \\ & {{A}_{2}}=0.6393928815{{T}_{0}} \\ & {{A}_{4}}=0.7149018027{{T}_{0}} \\ & {{A}_{6}}=0.7261792127{{T}_{0}} \\ \end{align}$$

Note that we only calculated $${{A}_{0}}\to {{A}_{6}}$$ for demonstration.

As stated in (b).(1) that $${{A}_{2k}}\text{ }exist\text{  }for\text{  }k=0,1,2,...$$

=Problem6 - Derivation of Inverse hyperbolic tangent function=

Given
hyperbolic tangent function

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

Find
Inverse hyperbolic tangent function as

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

Solution
Give $$\tanh (x)=\frac{{{e}^{x}}-{{e}^{-x}}}{{{e}^{x}}+{{e}^{-x}}}$$

Consider

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

and it follows that

$$\begin{align} & \tanh \left[ \frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \right]=\tanh \left[ \log \sqrt{\frac{1+x}{1-x}} \right]=\frac{{{\left( \sqrt{\frac{1+x}{1-x}} \right)}^{2}}-1}{{{\left( \sqrt{\frac{1+x}{1-x}} \right)}^{2}}+1} \\ & =\frac{\frac{1+x}{1-x}-1}{\frac{1+x}{1-x}+1}=\frac{(1+x)-(1-x)}{(1+x)+(1-x)}=\frac{2x}{2}=x \\ \end{align}$$

Therefore

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

=Problem7 - Even-ness and odd-ness of Qn(x)=

Given
$$\begin{align} & {{P}_{n}}(x)=\sum\limits_{i=0}^{[n/2]}{{{(-1)}^{i}}\frac{(2n-2i)!{{x}^{n-2i}}}{{{2}^{n}}i!(n-i)!(n-2i)!}}\text{                                           (1)} \\ & {{Q}_{n}}(x)={{P}_{n}}(x){{\tanh }^{-1}}(x)-2\sum\limits_{j=1,3,5}^{J}{\frac{2n-2j+1}{(2n-j+1)j}}{{P}_{n-j}}(x)\text{                   (2)} \\ & J:=1+2\left[ \frac{n-1}{2} \right]\text{                                                                          (3)} \\ \end{align}$$

Find
Even-ness and odd-ness of Qn(x) depending on n

Solution
We know that $${{P}_{n}}(x)$$ is even when n is even (n=2k, k=0,1,2,...), and $${{P}_{n}}(x)$$ is odd when n is odd (n=2k+1, k=0,1,2,....)

Also note that $${{\tanh }^{-1}}(x)$$ is an odd function, because

$$\begin{align} & {{\tanh }^{-1}}(x)=\frac{1}{2}\log \left( \frac{1+x}{1-x} \right) \\ & {{\tanh }^{-1}}(-x)=\frac{1}{2}\log \left( \frac{1-x}{1+x} \right) \\ & =-\frac{1}{2}\log \left( \frac{1+x}{1-x} \right)=-{{\tanh }^{-1}}(x) \\ \end{align}$$

(1) Consider when n is odd, then

$${{P}_{n}}(x)\text{ }is\text{  }odd,\text{  }so\text{  }{{P}_{n}}(x){{\tanh }^{-1}}(x)\text{  }is\text{  }even$$,

n is odd and J is odd ---> n-j is even and $${{P}_{n-j}}(x)\text{ }is\text{  }even$$

Therefore, equation (2) is sum of even functions ---> $${{Q}_{n}}(x)$$ is even.

(2) Consider when n is even, then

$${{P}_{n}}(x)\text{ }is\text{  even},\text{  }so\text{  }{{P}_{n}}(x){{\tanh }^{-1}}(x)\text{  }is\text{  odd}$$,

n is even and J is odd ---> n-j is odd and $${{P}_{n-j}}(x)\text{ }is\text{  }odd$$

Therefore, equation (2) is sum of odd functions ---> $${{Q}_{n}}(x)$$ is odd.

Conclusion

(1) for n=2k, k=0,1,2,... $${{P}_{n}}(x)$$ is even and $${{Q}_{n}}(x)$$ is odd;

(2) for n=2k+1, k=0,1,2,... $${{P}_{n}}(x)$$ is odd and $${{Q}_{n}}(x)$$ is even.

= Problem 9 - Generate Legendre Polynomials Using Recurrence Relation=

Given
Reccurence relation of Legendre polynomials:

$$\left( n+1 \right){{P}_{n+1}}-\left( 2n+1 \right)x{{P}_{n}}+n{{P}_{n-1}}=0 (1)$$

Legendre polynomials for n=0,1:

$$\begin{align} & {{P}_{0}}(x)=1 \\ & {{P}_{1}}(x)=x \\ \end{align}$$

Find
Generate $$\displaystyle \left\{ {{P}_{2}},{{P}_{3}},{{P}_{4}},{{P}_{5}},{{P}_{6}} \right\}$$ using recurrence relation starting from $$\displaystyle {{P}_{0}}$$ and $$\displaystyle {{P}_{1}}$$ and compare with [[media:2010_11_09_15_00_14.djvu|Eq.4-6 in page-2 Mtg-36]]

Solution
(1) can be written as:

$${{P}_{n+1}}=\frac{\left( 2n+1 \right)x{{P}_{n}}-n{{P}_{n-1}}}{\left( n+1 \right)}$$

Generate $$\displaystyle {{P}_{2}}$$

Let n=1 in  (1), we get

$${{P}_{2}}=\frac{\left( 2\times 1+1 \right)x{{P}_{1}}-1\times {{P}_{0}}}{\left( 1+1 \right)}$$

$${{P}_{2}}=\frac{3x\cdot x-1\times 1}{2}$$

$${{P}_{2}}=\frac{1}{2}\left( 3{{x}^{2}}-1 \right)$$

Generate $$\displaystyle {{P}_{3}}$$

Let n=2 in  (1), we get

$${{P}_{3}}=\frac{\left( 2\times 2+1 \right)x{{P}_{2}}-2{{P}_{1}}}{\left( 2+1 \right)}$$

$${{P}_{3}}=\frac{5x\cdot \frac{1}{2}\left( 3{{x}^{2}}-1 \right)-2x}{3}$$

$${{P}_{3}}=\frac{1}{2}\left( 5{{x}^{3}}-3x \right)$$

Generate $$\displaystyle {{P}_{4}}$$

Let n=3 in  (1), we get

$${{P}_{4}}=\frac{\left( 2\times 3+1 \right)x{{P}_{3}}-3{{P}_{2}}}{\left( 3+1 \right)}$$

$${{P}_{4}}=\frac{7x\cdot \frac{1}{2}\left( 5{{x}^{3}}-3x \right)-3\times \frac{1}{2}\left( 3{{x}^{2}}-1 \right)}{4}$$

$${{P}_{4}}=\frac{35{{x}^{4}}-21{{x}^{2}}-9{{x}^{2}}+3}{8}$$

$${{P}_{4}}=\frac{1}{8}\left( 35{{x}^{4}}-30{{x}^{2}}+3 \right)$$

Generate $$\displaystyle {{P}_{5}}$$

Let n=4 in  (1), we get

$${{P}_{5}}=\frac{\left( 2\times 4+1 \right)x{{P}_{4}}-4{{P}_{3}}}{\left( 4+1 \right)}$$

$${{P}_{5}}=\frac{9x\cdot \frac{1}{8}\left( 35{{x}^{4}}-30{{x}^{2}}+3 \right)-4\times \frac{1}{2}\left( 5{{x}^{3}}-3x \right)}{5}$$

$${{P}_{5}}=\frac{63{{x}^{5}}}{8}-\frac{54{{x}^{3}}}{8}+\frac{27x}{8\times 5}-2{{x}^{3}}+\frac{6x}{5}$$

$${{P}_{5}}=\frac{1}{8}\left( 63{{x}^{5}}-70{{x}^{3}}+15x \right)$$

Generate $$\displaystyle {{P}_{6}}$$

Let n=5 in  (1), we get

$${{P}_{6}}=\frac{\left( 2\times 5+1 \right)x{{P}_{5}}-5{{P}_{4}}}{\left( 5+1 \right)}$$

$${{P}_{6}}=\frac{11x\cdot \frac{1}{8}\left( 63{{x}^{5}}-70{{x}^{3}}+15x \right)-5\times \frac{1}{8}\left( 35{{x}^{4}}-30{{x}^{2}}+3 \right)}{6}$$

$${{P}_{6}}=\frac{231{{x}^{6}}}{16}-\frac{770{{x}^{4}}}{48}+\frac{55{{x}^{2}}}{16}-\frac{175{{x}^{4}}}{48}+\frac{50{{x}^{2}}}{16}-\frac{5}{16}$$

$${{P}_{6}}=\frac{1}{16}\left( 231{{x}^{6}}-315{{x}^{4}}+105{{x}^{2}}-5 \right)$$

It is seen that $$\left\{ {{P}_{2}}\text{ }{{P}_{3}}\text{ }{{P}_{4}}\text{ }{{P}_{5}}\text{ }{{P}_{6}} \right\}$$ are the same as Eqs.4-6 in 36-2

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User:Eml5526.s11/Numerical computation like matlab

= Octave =

Octave: high-level language ... for numerical computations. ... command line interface for solving linear and nonlinear problems ... mostly compatible with Matlab. ... also ... used as a batch-oriented language. More about Octave from MacResearch: Octave, a free Matlab clone, and a bit more.

Comparison of Octave and Matlab commands
quad: Numerical integration. Comparison between Octave and Matlab; "In octave the quad command uses DQAGS and DQAGI from QUADPACK. In comparison with the quad command in Matlab, the quad command in Octave is to be preferred: it uses a global adaptive routine, is more reliable, handles integration over infinite intervals, returns an error estimate, has to possibility to specify relative and absolute error requests, etc... "

Clenshaw-Curtis (CC) quadrature: Octave has the quadcc command. There is no native matlab command for the CC quadrature, but matlab users could download the Fast CC matlab code by Greg von Winckel.

FEMOctave
Manual: A FEM algorithm in Octave. See also Finite element analysis.

OctMesh
A toolbox for the resolution of partial differential equations (PDE) by means of the ﬁnite element method on Octave. Presentation slides. See also Finite element analysis.

Octave interface to FEM code
OctaveFEMM: a Matlab toolbox that allows for the operation of Finite Element Method Magnetics (FEMM) via a set of Matlab functions. The toolbox works with Octave, a Matlab clone.

Quarteroni, Saleri 2010
Quarteroni, A., Saleri, F.,  Scientific Computing with MATLAB and Octave , Springer, 2nd ed, 2010. amazon

Eaton, Bateman, Hauberg 2009
Eaton, J.W., Bateman, D., Hauberg, S., Gnu Octave Version 3.0.1 Manual: A High-Level Interactive Language For Numerical Computations, CreateSpace, 2009. amazon Note: Under Ubuntu linux, the pdf file of the manual comes with the (free) installation of octave. For example, installing the package octave3.0-doc would provide the pdf file of the manual for octave 3.0 at /usr/share/doc/octave3.0-doc/octave.pdf.

= Scilab =

A free platform for numerical computation, similar to matlab / simulink.

Campbell, Chancelier, Nikoukhah 2010
Campbell, S.L., Chancelier, J.P., Nikoukhah, R., Modeling and Simulation in Scilab/Scicos with ScicosLab 4.4, Springer, 2010. amazon

= Sage =

A free open-source mathematics software system licensed under the GPL. It combines the power of many existing open-source packages into a common Python-based interface.

= Python(x,y) =

A free scientific and engineering development software for numerical computations, data analysis and data visualization based on Python programming language, Qt graphical user interfaces, Eclipse integrated development environment and Spyder interactive scientific development environment.

= SciPy = Open-source software for mathematics, science, and engineering. It is also the name of a ... conference on scientific programming with Python. The SciPy library depends on NumPy, which provides convenient and fast N-dimensional array manipulation. The SciPy library is built to work with NumPy arrays, and provides many ... numerical routines ... for numerical integration and optimization. Together, they run on all popular operating systems, ... and are free of charge.

= List of numerical analysis software (wikipedia) =

= Open-source software =


text:
User:Eml5526.s11/Finite element analysis

= Calculix =

A package designed to solve field problems. The method used is the finite element method. With CalculiX Finite Element Models can be build, calculated and post-processed. The pre- and post-processor is an interactive 3D-tool using the openGL API. The solver is able to do linear and non-linear calculations. Static, dynamic and thermal solutions are available. Both programs can be used independently. Because the solver makes use of the abaqus input format it is possible to use commercial pre-processors as well. In turn the pre-processor is able to write mesh related data for nastran, abaqus, ansys, code-aster and for the free-cfd codes duns, ISAAC and OpenFOAM. A vda CAD interface is available. The program is designed to run on Unix platforms like Linux and Irix computers but also on MS-Windows.

Calculix Tutorial by Ty Beede: Analysis of a beam using Calculix. Part 1: Buliding Geometry and Meshing Part 2: Exporting Mesh, Loads, and Boundary Conditions Part 3: Writing an Input File for the CCX Solver Part 4: Post Processing Results in CGX

= Octave and FEMOctave =

See Numerical computation like matlab.

= FEniCS project =

FEniCS project (wikipedia): a set of free software projects with the common goal to enable automated solution of differential equations. It provides software tools for working with computational meshes, finite element variational formulations of PDEs, ODE solvers and linear algebra. FEniCS project home page.

= FreeFEM++ =

FreeFEM++: is an implementation of a language dedicated to the finite element method, written in C++, with the FreeFem++ language being a C++ idiom. Problems involving PDE (2d, 3d) from several branches of physics such as fluid-structure interactions require interpolations of data on several meshes and their manipulation within one program. FreeFem++ includes a fast 2^d-tree-based interpolation algorithm and a language for the manipulation of data on multiple meshes.

= Elmer =

Elmer is an "Open Source Finite Element Software for Multiphysical Problems". A recent (as of 6 May 2011) discussion on the users group mailing list of Calculix (see also section Calculix here) indicated that there is a problem with the documentation of Elmer:

= List of finite element software packages (wikipedia) =

= Open-source software =


text:
User:Eml5526.s11/CAD modeling and meshing

= CAD modeling =

$$\displaystyle \color{black}{\spadesuit}$$ Salome: an open-source software that provides a generic platform for Pre- and Post-Processing for numerical simulation. It is based on an open and flexible architecture made of reusable components.

= Meshing =

$$\displaystyle \color{black}{\clubsuit}$$ Gmsh: a 3D finite element grid generator with a build-in CAD engine and post-processor. Its design goal is to provide a fast, light and user-friendly meshing tool with parametric input and advanced visualization capabilities. Gmsh is built around four modules: geometry, mesh, solver and post-processing. The specification of any input to these modules is done either interactively using the graphical user interface or in ASCII text files using Gmsh's own scripting language.

$$\displaystyle \color{black}{\clubsuit}$$ DistMesh: a simple MATLAB code for generation of unstructured triangular and tetrahedral meshes.  Paper:  P.-O. Persson, G. Strang, A Simple Mesh Generator in MATLAB. SIAM Review, Volume 46 (2), pp. 329-345, June 2004.

$$\displaystyle \color{black}{\clubsuit}$$ TOAST: a software suite for image reconstruction in optical diffusion tomography (ODT). It contains binary command line tools for numerical modelling of light transport and parameter recovery, but the main emphasis of development is currently shifting towards integration as a Matlab toolbox. Contains mesh-generation tools.

$$\displaystyle \color{black}{\clubsuit}$$ Mesh Generation & Grid Generation on the Web, maintained by Robert Schneider. Mesh-generation software

= Open-source software =


text:
User:Eml5526.s11/Plotting, image processing

= GNUPlot =

GNUPlot: a portable command-line driven graphing utility for ... many ... platforms. ... to visualize mathematical functions and data interactively ... used as a plotting engine by third-party applications like Octave ... supports many types of plots in either 2D and 3D... many different types of output ... to many file formats ... interactive terminal based on wxWidgets and the creation of mousable graphs for web display using the HTML5 canvas element. Book: Janert, P.K.,  Gnuplot in Action: Understanding Data with Graphs,  Manning Publications (2009). amazon

= Matplotlib =

matplotlib: a python 2D plotting library which produces publication quality figures in a variety of hardcopy formats and interactive environments across platforms. matplotlib can be used in python scripts, the python and ipython shell (ala MATLAB or Mathematica), web application servers, and six graphical user interface toolkits.

= PIL =

PIL: adds image processing capabilities to your Python interpreter. This library supports many file formats, and provides powerful image processing and graphics capabilities.

= Open-source software =


text:
User:Eml5526.s11/Symbolic computation

= wxMaxima =

wxMaxima: a cross platform GUI for the computer algebra system Maxima based on wxWidgets.

= SymPy =

SymPy: a Python library for symbolic mathematics. It aims to become a full-featured computer algebra system (CAS) while keeping the code as simple as possible in order to be comprehensible and easily extensible.

= List of numerical analysis software (wikipedia) =

= Open-source software =

title: User:Eml5526.s11/Symbolic computation


text:
User talk:Heri2000

 Hello Heri2000, 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

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User:Mateusz War.

Mateusz War.

Link

 * Ten użytkownik (Mateusz War.) ma swoje konto na Wikipedia PL.

title: User:Mateusz War.


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User:Nardaviel

High school + one year of college. Interests include linguistics (particularly English, Spanish, Japanese, and theoretical study), abnormal psychology, and philosophy.

title: User:Nardaviel


text:
Fast Fourier transform

This page is still under construction. Any contribution is appreciated.

Fast Fourier Transform
Fast fourier transform or FFT is an algorithm mainly developed for digital computing of a Discrete Fourier transform or DFT of a discrete signal. Computing of DFT of a digital signal (discrete-time signal) involves N2 Complex multiplications , where N is the number of points to which we consider taking DFT of the input periodic signal. Using an FFT algorithm, the number of complex multiplications can be reduced to a greater extent. Reduction is mainly due to decimation either in time domain or in frequency domain

Decimation in time FFT algorithm (Radix 2)
The N-point DFT of a sequence is given by

DFT\{x(n)\} = \sum_{n=0}^{N-1} x(n) w^{kn}_N \text{ ; } k = 0,1,\cdots N-1. $$

every time x(n) gets multiplied by $$w^{kn}_N$$, we get one complex multiplication. Therefore when the above summation is expanded for a given $$k$$, we get $$N$$ complex multiplications. Which means that, to compute Npoint DFT, we have $$N\times N = N^2$$ complex multiplications. Our aim is to reduce the number of complex multiplications. In the following presentation, The number of samples of $$x(n)$$ is assumed to be a power of 2. i.e, $$N = 2^v$$ where v is some fixed positive integer. In this approach, N point transforms are broken into $$\frac{N}{2}$$ point DFTs which are further broken into $$\frac{N}{4}$$ point DFTs. And this process is continued until we get 2-point DFT. This technique is known as Divide and Conquer approach. Let $$x(n)$$ be a finite length sequence of length equal to N. Given sequence is : $$x(n) = x(1),x(2),x(3),\cdots x(N-2),x(N-1).$$ Even indexed sequence is : $$x_e(n) = x(0),x(2),x(4)\cdots x(N-2).$$ and, odd indexed sequence is : $$x_o(n) = x(1),x(3),x(5)\cdots x(N-1).$$ Now we can split the above DFT equation into even and odd parts as below : $$X(k) = \underbrace{\sum_{n=0}^{N-2} x(n) w^{kn}_N}^\text{Even indexed} + \underbrace{\sum_{n=0}^{N-1} x(n) w^{kn}_N}^\text{Odd indexed}$$ Letting $$N = 2r$$ in the first summation and $$N = 2r + 1$$ in the second summation, we get : $$X(k) = \sum_{r=0}^{\frac{N}{2}-1} x(2r) w^{2kr}_N + \sum_{r=0}^{\frac{N}{2}-1} x(2r+1) w^{k(2r+1)}_N $$ since, $$w^{k2r)}_N = e^{-j\frac{2\pi}{N}k2r} = e^{-j\frac{2\pi}{\left(\frac{N}{2}\right)}kr} = w^{kr}_{\left(\frac{N}{2}\right)}$$ We get  $$X(k) = \sum_{r=0}^{\frac{N}{2}-1} x(2r) w^{kr}_{\left(\frac{N}{2}\right)} + w^{k}_{N)}\sum_{r=0}^{\frac{N}{2}-1} x(2r+1) w^{kr}_{\left(\frac{N}{2}\right)}$$ $$\implies X(k) = G(k) + w^k_N H(k) \text{ ; } k = 0,1,2\cdots \frac{N}{2}-1 \text{ or just } 0\leq k \leq \frac{N}{2}-1$$

Where $$G(k)$$ and $$H(k)$$ are $$\frac{N}{2}$$ point DFTs of even indexed and odd indexed sequences, respectively. For computing $$X(k)$$ for $$\frac{N}{2} \leq k \leq N-1$$, the periodicity of $$G(k)$$ and $$H(k)$$ are exploited. Since $$G(k)$$ and $$H(k)$$ are $$\frac{N}{2}$$-point DFTs, they are implicit periodic with a fundamental period $$\frac{N}{2}$$. Hence, we may write $$ X(k) = G(k) + w^k_N H(k) \text{ when } 0 \leq k \leq \frac{N}{2}-1 \text{   and }$$  $$ X(k) = G(k+\frac{N}{2}) + w^k_N H(k+\frac{N}{2}) \text{ when } \frac{N}{2}-1 \leq k \leq N-1 $$ Now, after first decimation, the total number of complex multiplications is, $$\eta _1 = 2\left({\frac{N}{2}}^2\right)+N$$ The same way will lead us to further decimation and thus a block diagram can be contructed as below :

Prepared from the notes of Dr. D. Ganesh Rao Tutorials, Bangalore
The derivation or the text in this article written by Sushruth Sastry 22:13, 19 November 2010 (UTC) is the content from the Tution notes of Dr. D. Ganesh rao, Prof. at MSRIT, bangalore.

title: Fast Fourier transform


text:
Practice: Expressing Linear Equations As Matrices

This page contains practice exercises for expressing linear equations as matrices.


text:
User talk:Jlester4

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Talk:Learning the basics of French/Learn the tenses

Where is the passé composé on this page? It's missing.


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User:Egm6321.f10.team4/HW7

= Problem 1 Plot Legendre Polynomials and Functions = From Meeting 37, p 37-1.

Given
The general form of the Legendre Equation:

And the solutions have the following format.

Definitions of the Legendre Polynomials:

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

Find
part.a Plot the given Legendre Polynomials and Infinite Series

With observing $$\displaystyle P_n(\mu)$$ and $$\displaystyle Q_n(\mu)$$ as $$\displaystyle \mu \rightarrow |\mu|=1 $$

part.b Plot each pairs of Legendre Polynomials and Functions

Observe even-ness and odd-ness of $$\displaystyle {P_i,Q_i} $$ where i=0 to 3 and make predictions

Solution for the part a

Solution for the part.b
Observation and prediction for Determining the function inside of the integral,

Egm6321.f10.team4.Yoon 00:00, 25 November 2010 (UTC) - Primary Author

Given
We have the following function f such that
 * {| style="width:100%" border="0"

\begin{align} f = \sum_{i}g_{i} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2-1)
 * }
 * }

Find
Prove that

if $$ \begin{align} \{g_i\} \end{align}$$ is odd, then function $$ \begin{align} f \end{align}$$ is odd.

if $$ \begin{align} \{g_i\} \end{align}$$ is even, then function $$ \begin{align} f \end{align}$$ is even.

Solution
A function $$ \begin{align} f(x) \end{align}$$ is odd, if and only if $$ \begin{align} f(x) \end{align}$$ = $$ \begin{align} -f(-x) \end{align}$$

Hence,

$$ \begin{align} f(x) \end{align}$$ = $$ \begin{align} \sum_{i}g_{i}(x) \end{align}$$ = $$ \begin{align} \sum_{i}-g_{i}(-x) \end{align}$$

$$ \begin{align} f(x) \end{align}$$ = $$ \begin{align} -\sum_{i}g_{i}(-x) \end{align}$$

when i = 3 $$ \begin{align} -\sum_{i=3}g_{i}(-x) \end{align}$$ = $$ \begin{align} -[g_{1}(-x)+g_{2}(-x)+g_{3}(-x)] \end{align}$$

let

$$ \begin{align} -[g_{1}(-x)+g_{2}(-x)+g_{3}(-x)] \end{align}$$ = $$ \begin{align} -f(-x) \end{align}$$

which then makes

$$ \begin{align} -f(-x) \end{align}$$ = $$ \begin{align} -\sum_{i}g_{i}(-x) \end{align}$$

then leads us to

$$ \begin{align} f(x) \end{align}$$ = $$ \begin{align} -f(-x) \end{align}$$ being an odd function

Now lets go to the even case:

A function $$ \begin{align} f(x) \end{align}$$ is even, if and only if $$ \begin{align} f(x) \end{align}$$ = $$ \begin{align} f(-x) \end{align}$$

Hence,

$$ \begin{align} f(x) \end{align}$$ = $$ \begin{align} \sum_{i}g_{i}(x) \end{align}$$ = $$ \begin{align} \sum_{i}g_{i}(-x) \end{align}$$

$$ \begin{align} f(x) \end{align}$$ = $$ \begin{align} \sum_{i}g_{i}(-x) \end{align}$$

when i = 3 $$ \begin{align} \sum_{i=3}g_{i}(-x) \end{align}$$ = $$ \begin{align} +[g_{1}(-x)+g_{2}(-x)+g_{3}(-x)] \end{align}$$

let

$$ \begin{align} +[g_{1}(-x)+g_{2}(-x)+g_{3}(-x)] \end{align}$$ = $$ \begin{align} f(-x) \end{align}$$

which then makes

$$ \begin{align} f(-x) \end{align}$$ = $$ \begin{align} \sum_{i}g_{i}(-x) \end{align}$$

then leads us to

$$ \begin{align} f(x) \end{align}$$ = $$ \begin{align} f(-x) \end{align}$$ being an even function

Egm6321.f10.team4.Corella.RL 18:39, 6 December 2010 (UTC) - Primary Author

= Problem 3* - Evenness and Oddness of Legendre Polynomials = Note: * The asterisk implies that this problem has been solved or partially solved in homework assignment for this course in a previous semester. Any sources used in developing these solutions will be cited accordingly.

3.1

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

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

 * where [n/2] is treated like floor(n/2), meaning that only the integer portion is used and the remainder is dropped. For example, [5/2] = [2.5] = 2.

3.2

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

 * $$\displaystyle q \in P_5 \Rightarrow q(x)=\sum_{i=0}^{5}c_i x^i $$
 * $$\displaystyle (Eq. 3-2) $$
 * }
 * }
 * }

where $$\displaystyle \begin{Bmatrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ c_4 \\ c_5 \end{Bmatrix} = \begin{Bmatrix} 2 \\ -5 \\ -3 \\ 11 \\ 7 \\ 6 \end{Bmatrix} $$

3.1
Show that $$\displaystyle P_{2k}(x) $$ is even and that $$\displaystyle P_{2k+1}(x) $$ is odd, for $$\displaystyle k=0,1,2,\dots $$.

3.2
Find $$\displaystyle {a_i} $$ such that $$\displaystyle q(x)= \sum_{i=0}^5 a_i P_i(x) $$.

Also, plot $$\displaystyle q=\sum_i c_i x^i = \sum_i a_i P_i $$.

3.1
Let us begin by finding the first several values of $$\displaystyle P_{2k}(x) $$ and $$\displaystyle P_{2k+1}(x) $$:

Case: k=0

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

 * $$\displaystyle P_{2 \cdot 0}(x)=P_{n=0}(x)=\frac{1 \cdot (0-0-1)}{2^0 \cdot 1 \cdot 1} (-1)^0 x^0 =1 $$
 * $$\displaystyle (Eq. 3-3) $$
 * }
 * }
 * }

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

 * $$\displaystyle P_{2 \cdot 0+1}(x)=P_{n=1}(x)= \frac{1 \cdot (2-0-1)}{2^0 \cdot 1 \cdot (1-0)!} (-1)^0 x^1 = x $$
 * $$\displaystyle (Eq. 3-4) $$
 * }
 * }
 * }

Note that when $$\displaystyle n $$ is odd in the summation in Eq. 3-1, the maximum value for $$\displaystyle i $$ in the summation in Eq. 3-1 is the same as it is when $$\displaystyle n=n-1 $$. For example, for:

 * $$\displaystyle P_n(x) = \sum_{i=0}^{[n/2]} \underbrace{\frac{1 \cdot 3 \cdot \dots \cdot \left(2n - 2i - 1\right)}{2^i \cdot i! (n-2i)!} (-1)^i x^{n-2i}}_{ \left( \cdot \right)} $$

When n=1:
 * $$\displaystyle P_1(x) = \sum_{i=0}^{[1/2]} \left( \cdot \right) = \sum_{i=0}^{[0.5]} \left( \cdot \right) = \sum_{i=0}^{0} \left( \cdot \right) $$

Case: k=1

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

 * $$\displaystyle P_{2 \cdot 1}(x)=P_{n=2}(x)=\frac{1 \cdot (4-0-1)}{2^0 \cdot 1 \cdot (2-0)!} (-1)^0 x^2 + \frac{1 \cdot (4-2-1)}{2^1 \cdot 1 \cdot (2-2)!} (-1)^1 x^0 =\frac{3}{2}x^2-\frac{1}{2} $$
 * $$\displaystyle (Eq. 3-5) $$
 * }
 * }
 * }

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

 * $$\displaystyle P_{2 \cdot 1+1}(x)=P_{n=3}(x)=\frac{1 \cdot 3 \cdot 5}{2^0 \cdot 1 \cdot (3)!} (-1)^0 x^3 + \frac{1 \cdot 3}{2^1 \cdot 1 \cdot (3-2)!} (-1)^1 x^1 =\frac{5}{2}x^3-\frac{3}{2}x $$
 * $$\displaystyle (Eq. 3-6) $$
 * }
 * }
 * }

For the sake of brevity, results of several cases are tabulated below:

Table 3-1: Even and Odd Legendre Polynomials

It can readily be observed that the polynomials $$\displaystyle P_{2k}(x)$$ have only even exponents on $$\displaystyle x$$, thus $$\displaystyle P_{2k}(x)=P_{2k}(-x)$$, so these are all even functions. Likewise, the polynomials $$\displaystyle P_{2k+1}(x)$$ have only odd exponents on $$\displaystyle x$$, so $$\displaystyle P_{2k+1}(x)=-P_{2k+1}(-x)$$, and these are all odd functions.

3.2
Let us expand the equation:

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

 * $$\displaystyle q(x)= \sum_{i=0}^5 a_i P_i(x) $$
 * $$\displaystyle = a_0 P_0(x) + a_1 P_1(x) +a_2 P_2(x) +a_3 P_3(x) +a_4 P_4(x) +a_5 P_5(x) $$
 * $$\displaystyle = a_0 P_0(x) + a_1 P_1(x) +a_2 P_2(x) +a_3 P_3(x) +a_4 P_4(x) +a_5 P_5(x) $$

 * $$\displaystyle (Eq. 3-7) $$
 * }
 * }

Substituting the Legendre Polynomials in Table 3-1:

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

 * $$\displaystyle q(x) = a_0 (1) + a_1 (x) +a_2 \left(\frac{3}{2}x^2-\frac{1}{2} \right) +a_3 \left(\frac{5}{2}x^3-\frac{3}{2}x \right) +a_4 \left( \frac{35}{8}x^4-\frac{15}{4}x^2+\frac{3}{8} \right) +a_5 \left(\frac{63}{8}x^5-\frac{35}{4}x^3+\frac{15}{8}x \right) $$
 * $$\displaystyle = \sum_{i=0}^5c_ix^i= \underbrace{\left(a_0-\frac{1}{2}a_2+\frac{3}{8}a_4 \right)}_{=c_0=2}x^0
 * $$\displaystyle = \sum_{i=0}^5c_ix^i= \underbrace{\left(a_0-\frac{1}{2}a_2+\frac{3}{8}a_4 \right)}_{=c_0=2}x^0

+ \underbrace{\left(a_1-\frac{3}{2}a_3+\frac{15}{8}a_5 \right)}_{=c_1=-5}x^1 + \underbrace{\left(\frac{3}{2}a_2-\frac{15}{4}a_4 \right)}_{=c_2=-3}x^2 + \underbrace{\left(\frac{5}{2}a_3-\frac{35}{4}a_5 \right)}_{=c_3=11}x^3 + \underbrace{\left(\frac{35}{8}a_4 \right)}_{=c_4=7}x^4 + \underbrace{\left(\frac{63}{8}a_5 \right)}_{=c_5=6}x^5 $$
 * $$\displaystyle (Eq. 3-8) $$
 * }
 * }

From Eq. 3-8 we can deduce the following:

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

$$\displaystyle a_4 = \frac{56}{35}=\frac{8}{5} $$ $$\displaystyle a_3 = \frac{2}{5} \left[11+\frac{35}{4}\left(\frac{16}{21}\right) \right] =\frac{2}{5}\left(\frac{33}{3}+\frac{20}{3}\right)=\frac{106}{15} $$ $$\displaystyle a_2 = \frac{2}{3} \left[-3+\frac{5}{4}\left(\frac{8}{5} \right)\right]=\frac{2}{3}\left(3\right)=2 $$ $$\displaystyle a_1 = 5+\frac{3}{2}\left(\frac{106}{15}\right)-\frac{15}{8}\left(\frac{16}{21}\right) =5+\frac{53}{5}-\frac{10}{7}=\frac{175}{35}+\frac{371}{35}-\frac{50}{35}=\frac{496}{35} $$ $$\displaystyle a_0 =2+ \frac{1}{2}\left(2\right)-\frac{3}{8}\left(\frac{8}{5}\right) =3-\frac{3}{5}=\frac{12}{5} $$
 * $$\displaystyle a_5 = \frac{48}{63}=\frac{16}{21} $$
 * $$\displaystyle a_5 = \frac{48}{63}=\frac{16}{21} $$
 * $$\displaystyle (Eq. 3-7) $$
 * }
 * }

Thus: $$\displaystyle \begin{Bmatrix} a_0 \\ a_1 \\ a_2 \\ a_3 \\ a_4 \\ a_5 \end{Bmatrix} = \begin{Bmatrix} \frac{12}{5} \\ \frac{496}{35} \\ 2 \\ \frac{106}{15} \\ \frac{8}{5} \\ \frac{16}{21} \end{Bmatrix} $$

Plots
The two definitions of q(x) are shown below and compared in plots, with both individual values and the summed values:

Figure 3-1: q(x), showing each individual $$ \displaystyle c_i x^i $$, for i=0,1,2,3,4, and 5. The sum $$ \displaystyle \sum_i c_i x^i $$ is shown in red.

Figure 3-1 was made in MATLAB using the following code:
 * clear all; close all; clc; hold on;
 * C=[2,-5,-3,11,7,6]
 * x=-1:0.1:1;
 * for i=0:5
 * plot(x,C(i+1)*x.^i)
 * end
 * plot(x,C(1)*x.^0+C(2)*x.^1+C(3)*x.^2+C(4)*x.^3+C(5)*x.^4+C(6)*x.^5,'r-')
 * xlabel('x'); ylabel('c(i)*x^i');

Figure 3-2: q(x), showing each individual $$ \displaystyle a_i P_i(x) $$, for i=0,1,2,3,4, and 5. The sum $$ \displaystyle \sum_i a_i P_i(x)$$ is shown in red.

Figure 3-2 was made in MATLAB using the following code: clear all; close all; clc; hold on; a=[12/5,496/35,2,106/15,8/5,16/21] x=-1:0.1:1; plot(x,a(1)*1) %a0*P0 plot(x,a(2)*x) %a1*P1 plot(x,a(3)*(1.5*x.^2-.5)) %a2*P2 plot(x,a(4)*(2.5*x.^3-1.5*x)) %a3*P3 plot(x,a(5)*((35/8)*x.^4-(15/4)*x.^2+3/8)) %a4*P4 plot(x,a(6)*((63/8)*x.^5-(35/4)*x.^3+(15/8)*x)) %a5*P5 % sum of the above in red: plot(x,a(1)*1+a(2)*x+a(3)*(1.5*x.^2-.5)+a(4)*(2.5*x.^3-1.5*x)+... a(5)*((35/8)*x.^4-(15/4)*x.^2+3/8)+a(6)*((63/8)*x.^5-(35/4)*x.^3+(15/8)*x),'r-') xlabel('x'); ylabel('a_i * P_i(x)');

Note that the blue curves in Figures 3-1 and 3-2 differ, because $$ \displaystyle a_i P_i(x) \neq c_ix^i $$. However, the red curves (the corresponding sums from i=0 to i=5) are identical.

Egm6321.f10.team4.ejm 15:17, 6 December 2010 (UTC) - Author

= Problem 4 =

From Meeting 38, p 37-5.

Given
And

Find
Part 1 Without calculation, find property of $$\displaystyle A_n(\mu)$$

i.e., $$\displaystyle A_{2k}=0?(\mu)$$, $$\displaystyle A_{2k+1}=0?(\mu)$$ for $$\displaystyle k=?(\mu)$$

Part 2 Evaluate non-zero coefficients

Solution for the Part 1
We know

$$\displaystyle f(\mu)$$ is an even function. i.e. for equation 4-1:

and we know $$\displaystyle P_n$$ is even for $$\displaystyle n=2k$$ as well as odd for $$\displaystyle n=2k+1$$

An integral of an even function is an odd function so for $$\displaystyle n=2k$$

and an integral of an odd function is an even function so for $$\displaystyle n=2k+1$$

Solution for the Part 2
For equation 4-1, start off by assuming

and we know

so

where

For $$\displaystyle n=0$$

using Wolfram Alpha with an input of: Integrate[(1-x^2)^3, x,0,1]

For $$\displaystyle n=2$$

using Wolfram Alpha with an input of: 5*Integrate[(1-x^2)^3*(3/2*x^2-1/2), x,0,1]

For $$\displaystyle n=4$$

using Wolfram Alpha with an input of: 9*Integrate[(1-x^2)^3*(35/8*x^4-15/4*x^2+3/8), x,0,1]

For $$\displaystyle n=6$$

using Wolfram Alpha with an input of: 13*Integrate[(1-x^2)^3*(231/16*x^6-315/16*x^4+105/16*x^2-5/16), x,0,1]

For equation 4-2, start off by assuming

so

For $$\displaystyle n=0$$

using Wolfram Alpha with an input of: Integrate[exp(-(2x/pi)^2), x,0,pi/2]

For $$\displaystyle n=2$$

using Wolfram Alpha with an input of: 5*Integrate[exp(-(2x/pi)^2)*(3/2*sin(x)^2-1/2), x,0,pi/2]

For $$\displaystyle n=4$$

using Wolfram Alpha with an input of: 9*Integrate[exp(-(2x/pi)^2)*(35/8*sin(x)^4-15/4*sin(x)^2+3/8), x,0,pi/2]

For $$\displaystyle n=6$$

using Wolfram Alpha with an input of: 13*Integrate[exp(-(2x/pi)^2)*(231/16*sin(x)^6-315/16*sin(x)^4+105/16*sin(x)^2-5/16),x,0,pi/2]

Egm6321.f10.team4.Auerbach 06:17, 7 December 2010 (UTC)Author

= Problem 5. Relationship Between Artanh(x) function And Legendre Functions $$\displaystyle {{Q}_{0}}(x),{{Q}_{1}}(x)$$ =

Given
Legendre functions are given as below,

Find
Reveal the relationship between Artanh(x) and Legendre Functions.

Solution for part.a
To find out the relationship between Legendre equation and artanh function, let's derivate artanh function from tanh fuction first.

let

Then, moving the denominator of the right hand side to the left hand side.

Multiplying e^y on the both side of the equation

Collecting the term of x yields

Deviding by (x-1) NoteSince the left land side is greater than 0, the condition of rigthand side is limited to |x|<1. Taking log on the both side of the equation,

Deviding by 2

In conclusion, artanh fuction is identical with Legendre series(Q0).

Solution for part.b
Multiplying x and substract 1 on both side of equation yield,

Egm6321.f10.team4.Yoon 16:00, 6 December 2010 (UTC) - Primary Author

= Problem 6 : Eveness and Oddness of the second solution to the Legendre Equation=

Given
The second solution to the Legendre Differential Equation

where

and

Find
Show that $$Q(x)$$ will be an even or odd function based on the integer value of $$ n $$

Solution
Noting that $$ acrtanh(x)$$ is an odd function and taking into account that $$j$$ can only take odd values we can create two tables.

The first shows that $$ Q(x)$$ is an odd function when $$n$$ takes even values

and the second shows that $$Q(x)$$ is an even function when $$n$$ takes odd values.

Egm6321.f10.team4.petralanda.n 21:06, 6 December 2010 (UTC)

= Problem 7 – Distance Between Two Points = From Meeting 40, p 40-2.

Given
To convert from rectangular to spherical coordinates (by the astronomic convention)

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

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

$$\displaystyle x_3 = z = r sin \theta $$

The expression for the distance between two points in space is given by

$$\displaystyle r_{PQ}^2 = \sum^{3}_{i = 1} ( x_Q^i - x_P^i )^2 $$ $$\displaystyle (Eq. 7-1) $$

Show
that the resulting expression for the distance between the points P and Q is

$$\displaystyle r_{PQ}^2 = r_Q^2 (1 + \rho^2 - 2 \rho cos \Gamma) $$ $$\displaystyle (Eq. 7-2) $$

where

$$\displaystyle \rho = \frac {r_P}{r_Q} $$

$$\displaystyle cos \Gamma = cos \theta_Q cos \theta_P cos (\phi_Q + \phi_P) + sin \theta_Q sin \theta_P $$

Solution
Starting from

$$\displaystyle r_{PQ}^2 = \sum^{3}_{i = 1} (x_Q^i - x_P^i)^2 $$ $$\displaystyle (Eq. 7-1) $$

$$\displaystyle = (x_{Q3} - x_{P3})^2 + (x_{Q2} - x_{P2})^2 + (x_{Q1} - x_{P1})^2 $$ $$\displaystyle (Eq. 7-3) $$

$$\displaystyle = (r_Q sin \theta_Q - r_P sin \theta_P)^2 + (r_Q sin \phi_Q cos \theta_Q - r_P sin\phi_P cos \theta_P)^2 + (r_Q cos \phi_Q cos \theta_Q - r_P cos \phi_P cos \theta_P)^2 $$
$$\displaystyle (Eq. 7-4) $$

$$\displaystyle = r_Q^2 sin^2 \theta_Q + r_P^2 sin^2 \theta_P - 2 r_Q r_P sin \theta_Q sin \theta_P + r_Q^2 sin^2 \phi_Q cos^2 \theta_Q + r_P^2 sin^2 \phi_P cos^2 \theta_P - 2 r_Q r_P sin \phi_Q sin \phi_P cos \theta_Q cos \theta_P + r_Q^2 cos^2 \phi_Q cos^2 \theta_Q + r_P^2 cos^2 \phi_P cos^2 \theta_P - 2 r_Q r_P cos \phi_Q cos \phi_P cos \theta_Q cos \theta_P $$
$$\displaystyle (Eq. 7-5) $$

Using the trigonometric identity

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

The expression simplifies to

$$\displaystyle = r_Q^2 + r_P^2 - 2 r_Q r_P (sin \theta_Q sin \theta_P + sin \phi_Q sin \phi_P cos \theta_Q cos \theta_P + cos \Phi_Q cos \phi_P cos \theta_Q cos \theta_P) $$
$$\displaystyle (Eq. 7-6) $$

Which can be regrouped

$$\displaystyle = r_Q^2 + r_P^2 - 2 r_Q r_P (sin \theta_Q sin \theta_P + cos \theta_Q cos \theta_P (sin \phi_Q sin \phi_P + cos \phi_Q cos \phi_P)) $$
$$\displaystyle (Eq. 7-7) $$

Dividing through the expression by $$\displaystyle r_{Q}^2 $$, yields

$$\displaystyle = r_Q^2 ( 1 + \frac {r_P^2}{r_Q^2} - 2 \frac {r_P}{r_Q} (sin \theta_Q sin \theta_P + cos \theta_Q cos \theta_P cos( \phi_Q - \phi_P))) $$
$$\displaystyle (Eq. 7-8) $$

Lastly, using our defintions of $$\displaystyle cos \Gamma $$ and $$\displaystyle \rho $$, our expression becomes

$$\displaystyle r_{PQ}^2 = r_Q^2 ( 1 + \rho^2 - 2 \rho cos \Gamma ) $$
$$\displaystyle (Eq. 7-2) $$

50.10.0.112 07:39, 5 December 2010 (UTC)Egm6321.f10.team4.osentowski - Author

Note: While this problem was assigned during the Fall 2009 term, it should be noted that the above solution was completed without reference to previously posted solutions.

= Problem 8 - Binomial Theorem Expansion= From Meeting 40, p 40-4.

Given
We have the following equation for the binomial theorem represented as
 * {| style="width:100%" border="0"

\begin{align} (x+y)^{r} = \sum_{k=0}^{\infty }\binom{r}{k}(x^{r-k})(y^{k}) \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8-1)
 * }
 * }

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

\begin{align} \binom{r}{k}=\frac{(r)(r-1)\cdots (r-k+1)}{k!} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8-2)
 * }
 * }

Find
Use Eq. 8-1 and Eq. 8-2 to obtain the following:

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

\begin{align} (1-x)^{-\frac{1}{2}} = \sum_{k=0}^{\infty }(\alpha _{i})(x^{i}) \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8-3)
 * }
 * }

and

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

\begin{align} \alpha _{i} = \frac{1\cdot 3\cdots (2i-1)}{2\cdot 4\cdots (2i)} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8-4)
 * }
 * }

Solution
Comparing Eq. 8-1 to Eq. 8-3 we can suggest that:

$$\displaystyle \begin{align} x=1, y=-x, r=-\frac{1}{2}, k=i \end{align} $$

Now plug-in the numbers from Eq. 8-3 to Eq. 8-1, we have

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

\begin{align} (1-x)^{-\frac{1}{2}} = \sum_{i=0}^{\infty }\binom{-\frac{1}{2}}{i}(1^{-\frac{1}{2}-i})(-x^{i}) \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8-5)
 * }
 * }

Since $$\displaystyle \begin{align} (1)^{-\frac{1}{2}}=1 \end{align} $$ we can cancel it out and rearrange it to be

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

\begin{align} (1-x)^{-\frac{1}{2}} = \sum_{i=0}^{\infty }\binom{-\frac{1}{2}}{i}(-1)^{i}(x)^{i} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8-6)
 * }
 * }

To obtain Eq. 8-4 we must set $$\displaystyle \begin{align} \alpha _{i} = \binom{-\frac{1}{2}}{i}(-1)^{i} \end{align} $$ and expand

By using Eq. 8-2 and doing the expansion we obtain

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

\begin{align} \alpha _{i} = \binom{-\frac{1}{2}}{i}(-1)^{i} =(-1)^{i} \frac{-\frac{1}{2}(-\frac{1}{2}-1)\cdots (-\frac{1}{2}-i+1)}{i!} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8-7)
 * }
 * }

Simplifying other terms we then get

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

\begin{align} \alpha _{i} = \binom{-\frac{1}{2}}{i}(-1)^{i} =(-1)^{i} \frac{-\frac{1}{2}(-\frac{3}{2})\cdots (\frac{1}{2}-i)}{i!} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8-8)
 * }
 * }

Redistributing $$\displaystyle \begin{align} (-1)^{i} \end{align} $$ and factoring out $$\displaystyle \begin{align}(-\frac{1}{2}) \end{align} $$we have our final equation

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

\begin{align} \alpha _{i} = \binom{-\frac{1}{2}}{i}(-1)^{i} =\frac{(1)\cdot (3)\cdots (2i-1)}{(2)\cdot (4)\cdots (2i)} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8-4)
 * }
 * }

We have now proven that we can derive Eq. 8-3 and Eq. 8-4 by using Eq. 8-1 and Eq. 8-2:

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

\begin{align} (x+y)^{r} = \sum_{k=0}^{\infty }\binom{r}{k}(x^{r-k})(y^{k})= \sum_{k=0}^{\infty }(\alpha _{i})(x^{i}) \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8-1)
 * }
 * }

where we showed from above that it is indeed

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

\begin{align} \alpha _{i} = \binom{-\frac{1}{2}}{i}(-1)^{i} =\frac{(1)\cdot (3)\cdots (2i-1)}{(2)\cdot (4)\cdots (2i)} \end{align} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 8-4)
 * }
 * }

Egm6321.f10.team4.Corella.RL 18:36, 6 December 2010 (UTC) - Primary Author

= Problem 9 – Generating Polynomials = From Meeting 41, p 41-2

Given
$$\displaystyle \mu P_n' - P_{n-1}' =  n P_n $$
$$\displaystyle (Eq. 9-1) $$

$$\displaystyle (n + 1) P_{n + 1} - (2n + 1) \mu P_n + n P_{n-1} =  0 $$
$$\displaystyle (Eq. 9-2) $$

$$\displaystyle P_0 =  1 $$

$$\displaystyle P_1 =  x $$

Find
$$\displaystyle P_2 $$ through $$\displaystyle P_6 $$, and compare the results to those derived using the formula

$$\displaystyle P_n (x) = \sum^{\frac {n}{2}}_{i =0} -1^i \frac {(2n - 2i)! x^{n-2i}}{2^n i! (n-1)! (n-2i)!} $$
$$\displaystyle (Eq. 9-3) $$

from Meeting 36 p. 36-2.

Solution
First, we must find $$\displaystyle \mu $$ using Equation 9-1, and the values for $$\displaystyle P_0 $$ and $$\displaystyle P_1 $$

$$\displaystyle \mu P_1' - P_0' = 1 P_1 $$
$$\displaystyle (Eq. 9-1) $$

$$\displaystyle \mu x' - 1' = x $$

$$\displaystyle \mu = x $$

Using this value for \mu, and Equation 9-2, the polynomials can be obtained.

$$\displaystyle P_2 $$ is found using $$\displaystyle n =  1 $$

$$\displaystyle (1+1) P_2 - (2*1 + 1) x P_1 + P_0 = 0 $$

$$\displaystyle 2 P_2 - 3 x (x) + 1 = 0 $$

$$\displaystyle 2 P_2 = 3 x^2 - 1 $$

$$\displaystyle P_2 = \frac {1}{2} (3x^2 - 1) $$
$$\displaystyle (Eq. 9-4) $$

$$\displaystyle P_3 $$ is found using $$\displaystyle n = 2 $$

$$\displaystyle (2 + 1) P_3 - (2*2 + 1) x P_2 + 2 P_1 = 0 $$

$$\displaystyle 3 P_3 - 5 x (\frac {1}{2} (3x^2 - 1)) + 2 x = 0 $$

$$\displaystyle 3 P_3 = 5 x (\frac {1}{2} (3x^2 - 1)) - 2 x $$

$$\displaystyle 3 P_3 = \frac {15}{2} x^3 - \frac{5}{2}x - 2x $$

$$\displaystyle P_3 =  \frac {1}{2} (5 x^3 - 3 x) $$
$$\displaystyle (Eq. 9-5) $$

$$\displaystyle P_4 $$ is found using $$\displaystyle n = 3 $$

$$\displaystyle (3 + 1) P_4 - (2*3 + 1) x P_3 + 3 P_2 =  0 $$

$$\displaystyle 4 P_4 - 7 x (\frac {1}{2} (5 x^3 - 3 x)) + 3 (\frac {1}{2} (3 x^2 - 1)) =  0 $$

$$\displaystyle 4 P_4 =  \frac {7}{2} x (5 x^3 - 3x) - \frac {3}{2} (3 x^2 - 1) $$

$$\displaystyle P_4 =  \frac {1}{8} (35 x^4 - 21 x^2 - 9 x^2 + 3) $$

$$\displaystyle P_4 =  \frac {35}{8} x^4 - \frac {15}{4} x^2 + \frac {3}{8} $$
$$\displaystyle (Eq. 9-6) $$

$$\displaystyle P_5 $$ is found using $$\displaystyle n = 4 $$

$$\displaystyle (4+1) P_5 - (2*4+1) P_4 + 4 P_3 =  0 $$

$$\displaystyle 5 P_5 - 9x P_4 + 4 P_3 =  0 $$

$$\displaystyle 5 P_5 - 9x (\frac {35}{8} x^4 - \frac {15}{4} x^2 + \frac {3}{8}) + 4 (\frac {1}{2} (5 x^3 - 3 x)) =  0 $$

$$\displaystyle 5 P_5 = \frac {315}{8} x^5 - \frac {135}{4} x^3 + \frac {27}{8} x - 10 x^3 + 6x $$

$$\displaystyle P_5 = \frac {63}{8} x^5 - \frac {35}{4} x^3 + \frac {15}{8} x $$
$$\displaystyle (Eq. 9-7) $$

$$\displaystyle P_6 $$ is found using $$\displaystyle n = 5 $$

$$\displaystyle (5 + 1) P_6 - (2*5 + 1) x P_5 - 5 P_4 =  0 $$

$$\displaystyle 6 P_6 - 11 x (\frac {63}{8} x^5 - \frac {35}{4} x^3 + \frac {15}{8} x) + 5 (\frac {35}{8} x^4 - \frac {15}{4} x^2 + \frac {3}{8}) $$

$$\displaystyle 6 P_6 =  \frac {693}{8} x^6 - \frac {385}{4} x^4 + \frac {165}{8} x^2 - \frac {175}{8} x^4 + \frac {75}{4} x^2 - \frac {15}{8} $$

$$\displaystyle 6 P_6 =  \frac {693}{8} x^6 - \frac {945}{8} x^4 + \frac {315}{8} x^2 - \frac {15}{8} $$

$$\displaystyle P_6 =  \frac {231}{16} x^6 - \frac {315}{16} x^4 + \frac {105}{16} x^2 - \frac {5}{16} $$
$$\displaystyle (Eq. 9-8) $$

The polynomials, represented in Equations 9-4 through 9-8, do indeed agree with the results presented in the Meeting 36 notes, developed using Equation 9-3. It can therefore be assumed that this is an acceptable method for generating Legendre Polynomials.

50.10.0.112 06:32, 6 December 2010 (UTC)Egm6321.f10.team4.osentowski - Author

128.227.11.220 19:00, 7 December 2010 (UTC) Egm6321.f10.team4.petralanda.n - Editor

= Problem 10: First solution to the Legendre Equation using Polynomial Expansion and Recurrence Method =

Find
$$P_2(x)$$ through $$P_6(x)$$ and compare the values with those obtained in Problem 7.9 using the second recurrence relation.

Solution
First we start by computing $$ P_5(x)$$ and $$P_6(x)$$.

$$n=5$$ and Eq (10-1) takes the following form

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

 * $$\displaystyle
 * $$\displaystyle

P_{5}(x)=\sum\limits_{i=0}^{5/2}{\frac{(-1)^{i}(2(5)-2i)!x^{5-2i}}{2^{5}i!(5-i)!(5-2i)!}}=\frac{(2(5))!x^{5}}{2^{5}(5)!(5)!}+\frac{(-1)^{1}(2(5)-2)!x^{5-2}}{2^{5}1!(5-1)!(5-2)!}+\frac{(-1)^{2}(2(5)-4)!x^{5-4}}{2^{5}2!(5-2)!(5-4)!}

$$

 * }
 * }
 * }

Which will yield

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

P_{5}(x)=\frac{63}{8}x^{5}-\frac{35}{4}x^{3}+\frac{15}{8}x $$
 * $$\displaystyle
 * $$\displaystyle
 *
$$\displaystyle (Eq. 10-2) $$
 * }
 * }

$$n=6$$ and Eq (10-1) will take the following form

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

P_{6}(x)=\sum\limits_{i=0}^{6/2}{\frac{(-1)^{i}(2(6)-2i)!x^{6-2i}}{2^{6}i!(6-i)!(6-2i)!}}=\frac{(2(6))!x^{6}}{2^{6}(6)!(6)!}-\frac{(2(6)-2)!x^{6-2}}{2^{6}(6-1)!(6-2)!}+\frac{(2(6)-4)!x^{6-4}}{2^{6}2!(6-2)!(6-4)!}-\frac{(2(6)-6)!}{2^{6}3!(6-3)!} $$
 * $$\displaystyle
 * $$\displaystyle

 * }
 * }
 * }

Which will result in

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

 * $$\displaystyle
 * $$\displaystyle

P_{6}(x)=\frac{231}{16}x^{6}-\frac{315}{16}x^{4}+\frac{105}{16}x^{2}-\frac{5}{16}$$

 *
$$\displaystyle (Eq. 10-3) $$
 * }
 * }

The following table compiles all the values generated using the polynomial expansion and the recurrence method.

Egm6321.f10.team4.petralanda.n 21:10, 6 December 2010 (UTC)

title: User:Egm6321.f10.team4/HW7


text:
Algebra/Powers

Laws of indices of all rational exponents
An index is of the form $$ a^x $$, and the laws on how to manipulate them is vital knowledge.

Power of one and zero
Any base raised to the power of one is simply the base.

For example,

$$ 1^1 = 1 $$

$$ 8^1 = 8 $$

$$ 0^1 = 0 $$

$$ a^1 = a $$

Zero raised to any positive, nonzero power is zero.

For example,

$$ 0^1 = 0 $$

$$ 0^2 = 0 $$

Any nonzero base raised to the power of zero is one.

For example,

$$ a^0 = 1 $$

$$ 5^0 = 1 $$

$$ 0^0 $$ is undefined, as it represents a mathematical singularity: raising zero to any other power is 0, but raising any number to the power of zero is 1.

Multiplication and division
When two indices are multiplied, as long as the bases are equal, the two indices are simply added together.

For example,

$$ a^x \times a^y = a^{x+y} $$

$$ 2^2 \times 2^3 = 2^{2+3} = 2^5 = 32 $$

For division the opposite is true, as long as the bases are equal, the indices are subtracted.

For example,

$$ a^x \div a^y = a^{x-y} $$

$$ 2^5 \div 2^2 = 2^{5-2} = 2^3 = 8 $$

Fractional indices
The denominator of a fractional index is the root that the base must be taken to.

For example,

$$ a^{\frac{1}{2}} = \sqrt{a^1}=\sqrt{a} $$

$$ a^{\frac{1}{3}} = \sqrt[3]{a^1}=\sqrt[3]{a} $$

The numerator of a fractional index is the power the base must be raised to.

For example,

$$ a^{\frac{2}{3}} = \sqrt[3]{a^2} $$

$$ 8^{\frac{2}{3}} = \sqrt[3]{8^2} = 2^2 = 4 $$

Negative indices
A negative sign for an index shows that the base is the denominator of a fraction, the index is the power it must be raised to. The numerator of this fraction is one (though if the term is multiplied by a constant, then the numerator is that constant).

For example,

$$ a^{-1} = \frac{1}{a} $$

$$ a^{-3} = \frac{1}{a^3} $$

$$ 2(a^{-1}) = \frac{2}{a} $$

Indices to the power of another index
When a number is raised to the power of an index, then this term is raised to the power of another index, the two indices are multiplied.

For example,

$$ \left ( a^x \right ) ^y = a^{x \times y} = a^{xy} $$

$$ \left ( a^2 \right ) ^3 = a^{2 \times 3} = a^6 $$

$$ \left ( \sqrt{a} \right ) ^4 = a^{\frac{1}{2} \times 4} = a^2 $$

It is tempting to think these mechanical rules always work as stated. However, this is not always the case. Consider

$$\sqrt{a^2} = (a^{2})^{1/2} = a^{2 \times \frac{1}{2}} = a$$

but this is actually wrong, because we have failed to account for any possible negative values of $$a$$. The correct way to "reduce" this is then

$$\sqrt{a^2} = |a|$$

Use and manipulation of surds
A surd is an irrational root of a whole number such as $$ \sqrt{2} $$ and $$ \sqrt{5} $$. Like indices, laws of use and manipulation of surds is vital knowlegde.

Multiplication and simplification
$$ \sqrt{a} \times \sqrt{b} = \sqrt{a \times b} $$

$$ \sqrt{2} \times \sqrt{8} = \sqrt{16} = 4 $$

Sometimes you will be asked to simplify your answer, this is done simply by finding a square number, like 4 or 9, that divides into the surd, then bringing it outside the square root.

For example,

$$ \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2} $$

Surds in fractions
$$ \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}} $$

$$ \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2} $$

Rationalising the denominator
Some times you may be given a fraction with a surd as the denominator and be asked to rationalise the denominator. If you are given a fraction with a single surd as a denominator you can simply multiply numerator and denominator by the surd to get rid of it. In effect you are always multiplying by 1.

For example,

$$ \frac{3}{\sqrt{2}} \times \left ( \frac{\sqrt{2}}{\sqrt{2}} \right ) = \frac{3\sqrt{2}}{2} $$

For a fraction that had a constant plus a surd, for example $$ a + \sqrt{b} $$, you need to multiply numerator and denominator by $$ a - \sqrt{b} $$.

For example,

$$ \frac{3}{3 + \sqrt{2}} \times \left ( \frac{3 - \sqrt{2}}{3 - \sqrt{2}} \right ) = \frac{3 \left ( 3 - \sqrt{2} \right )}{9 - 3 \sqrt{2} + 3 \sqrt{2} - 2} = \frac{9 - 3 \sqrt{2}}{7} $$

One of the reasons it is useful to rationalise a fraction is because a rationalised expression is easier to evaluate approximately by inspection. Consider the fraction $$1/\sqrt{2}$$, whose value is not easy to eyeball. However, the equivalent fraction $$\sqrt{2}/2$$ is easier to estimate, as we know that $$\sqrt{2} \approx 1.4$$, so $$\sqrt{2}/2 \approx 0.7$$. More precisely, we have $$1/\sqrt{2} = 0.70710678118 \ldots$$.

The discriminant of a quadratic function
We consider the following quadratic function, where $$A \neq 0$$:

f(x) = Ax^2 + Bx + C. $$

We define the discriminant of $$f$$ as the following quantity:

\text{Discriminant of } f = \text{Disc}(f) = B^2 - 4AC. $$

The discriminant allows us to classify the roots or zeroes of $$f$$. In particular, if $$\text{Disc}(f) > 0$$ we will have two distinct real roots; if $$\text{Disc}(f) < 0$$ we will have no real roots, and if $$\text{Disc}(f) = 0$$, we will have one (repeated) real root.

Questions
Here are some questions on the above topics to test your knowledge:

Laws of indices
Evaluate the following terms.

 * 1) $$ 2^2 \times 2^3 $$
 * 2) $$ 27^{\frac{2}{3}} $$
 * 3) $$ 4^{-\frac{3}{2}} $$
 * 4) $$ (-5)^{-3} $$
 * 5) $$ 9^{-0.5} $$
 * 6) $$ \left ( x^3 \right ) ^{\frac{2}{3}} $$

Surds
Express the following in the form $$ a\sqrt{b} $$.

 * 1) $$ \sqrt{45} $$
 * 2) $$ \sqrt{12} $$
 * 3) $$ \sqrt{32} $$

Rationalise the denominator of the following

 * 1) $$ \frac{4}{3 - \sqrt{5}} $$
 * 2) $$ \frac{11}{3 + \sqrt{11}} $$
 * 3) $$ \frac{3 - \sqrt{2}}{4 - \sqrt{5}} $$

title: Algebra/Powers


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User:Battered sav/E-portfolio

Week 1. Introduction
Generic Skills Covered in Unit
 * Communication: Create, express, present, listen, respond and negotiate ideas.
 * Information & Communication Technology: Select and use appropriate information and communication technology to retrieve, manipulate and present information.
 * Social responsibility: Work toward improvement in society, understand economic, political, social, and environmental systems with an international perspective, act in environmentally sustainable ways, accept service to the community as the primary purpose for professional life.

Introduction

 * Motivation is the process that gives behaviour energy and direction.
 * Energy: Behaviour is relatively strong, intense and persistent.
 * Direction: Behaviour is aimed toward achieving a particular purpose or goal.

Specific questions that constitute the core problems to be solved in motivation study
 * 1) What starts behaviour?
 * 2) How is behaviour sustained over time?
 * 3) Why is behaviour directed towards some ends but away from others?
 * 4) Why does behaviour change its direction?
 * 5) Why does behaviour stop?

History of Motivation

 * 1) Will
 * 2) Instinct
 * 3) Drive: Freud’s and Hull’s Drive Theory
 * 4) Incentive, Arousal, Discrepancy
 * 5) Rise of Mini-theories: The active nature of the person, the cognitive revolution and applied socially relevant research
 * 6) Contemporary Era

Summary of Freud’s Drive Theory
 * Drive’s Source: A bodily deficit occurs (e.g. Blood sugar drops and a sense of hunger emerges).
 * Drive’s Impetus: The intensity of the bodily deficit grows and emerges into consciousness as a psychological discomfort, which is anxiety.
 * Drive’s Object: Seeking to reduce anxiety and satisfy the bodily deficit, the person searches out and consumes a need satisfying environmental object (e.g. food)
 * Drives Aim: If the environmental object successfully satisfies the bodily deficit, satisfaction occurs and quiets anxiety, at least for a period of time.

Assessment Details
Textbook chapter Guildlines

Author a free, online, undergraduate\textbook chapter about a specific motivation and/or emotion topic.
 * 1) Topic
 * 2) Location
 * 3) Academic integrity
 * 4) Collaboration
 * 5) Word length
 * 6) Links
 * 7) Images
 * 8) Multimedia
 * 9) References
 * 10) Feedback

Multimedia

Create a five minute instructional video explaining the main content covered by your textbook chapter.

E-Portfolio

Develop an online e-portfolio sharing your reflections about the unit and its learning activities.

Tutorial 1
Our first tutorial of the unit motivation and emotion, involved our tutor, Dr James Neill asking or the participants to partake in a “get to know one another” exercise. This included divided the class in groups of height, thumb size, fast food preference and political party preference. Being a liberal supporter I was in the minority group of the class. Labour and The Greens with the majority of the support. This exercise helped us to pair up in groups with people of similar characteristics and views. After getting to know each other we formed small groups and were told by our tutor to individually define motivation and emotion. Our definitions included:

Motivation:

 * The inner drive behind our actions.
 * Why we do, what we do.
 * Driving force which determines our behaviour

Emotion:

 * State of mind which is determined by thoughts and feelings.
 * Having aroused, purposive or expressive feelings.
 * Our behaviour towards a particular situation.

We then discussed amongst our groups areas of motivation and emotion that are personally interesting. This was an effective approach to help us determine what subject to choose for our chapter summaries. I found intrinsic and extrinsic motivation very interesting because I could relate it to my day to day life. I am currently a part owner/ manager of a restaurant/bar and find it very useful in learning different strategies for motivating staff to do their job proficiently. Intrinsic and Extrinsic Motivation was an easy choice for my chapter summary, I believed I would be able to learn from it and put it into action, therefore becoming a better manager at my place of work.

Week 3. Brain and Physiological Needs
The Motivated and Emotional Brain

The brain carries out cognitive and intellectual functions, including thinking, learning, remembering, decision making and problem solving. The brain is also the centre of motivation and emotion. It generates cravings, needs, desires, pleasure, and the full range of the emotions. There are three general principles that guide research on the motivated and emotional brain:
 * 1) Specific Brain Structures Generate Specific Motivations: Different brain structures, when stimulated, give rise to specific motivational states. Stimulating one part of the hypothalamus, for instance, increases hunger, while stimulating a different part of the hypothalamus increases satiety.
 * 2) Biochemical Agents Stimulate Specific Brain Structures: Brain structures have receptor sites on them that endow them with the potential to be stimulated. The biochemical agents that stimulate these receptor sites are neurotransmitters and hormones. Neurotransmitters are the communication messengers of the nervous system, while hormones are the communication messengers of the endocrine system, allowing glands to communicate through the blood stream with bodily organs like the heart or lungs.
 * 3) Day to Day Events Stir Biochemical Agents Into Action: Day to day events stimulate the brain into action. If a person deprives themselves of food for a period of time, the ghrelin hormone is stirred into action. This biochemical agent then circulates the bloodstream and once it reaches the brain the stimulated hypothalamus creates the psychological experience of hunger.

Physiological Needs

Physiological needs occur with tissue and bloodstream deficits, as from water loss, nutrient deprivation, or physical injury. If neglected, bodily harm or pathology follows. Hence, physiological needs, when unmet and intense, represent life threatening emergencies. Examples of physiological needs include thirst, hunger and sex.

Week 4. Psychological and Social Needs/Tutorial 2
Psychological Needs

When an activity involves our psychological needs, we feel interest. When an activity satisfies our psychological needs, we feel enjoyment. Playing games, solving mysteries, and undertaking challenges are interesting and enjoyable things to do precisely because they provide an arena for involving and satisfying psychological needs.

Social Needs

Humans acquire social needs through experience, development and socialization.

Table: Incentive That Activate Each Social Need’s Emotional and Behaviour Potential

Tutorial 2.
In our second tutorial we were instructed to come up with a definition within our group on “what are needs”? With a little help from the textbook our group agreed on the most accurate description on:

What are Needs? Any condition within the person that is essential and necessary for life, growth and well being.

The organisation of needs was then shown in Maslow’s hierarchy in needs:

Brain Structure

We were instructed by our tutor to identify specific neural structures and their motivational function - in order to explain the neural bases of as many motivational states as you can. We did this individually then discussed it as a small group, and then shared key points with the class.
 * 1) 	Reticular formation
 * 2) 	Medial forebrain bundle: learning response outcome
 * 3) 	Hypothalamus: feeding, drinking and mating
 * 4) 	Amygdala: threat, danger
 * 5) 	Septal area: socialability, sexuality
 * 6) 	Hippocampus: threat, danger
 * 7) 	Cerebral cortex (Frontal lobes): planning, goals, intentions
 * 8)       Prefrontal Cerebral Cortex: motivational, emotional tendencies

Week 5. Intrinsic and Extrinsic Motivation / Goal Setting
Intrinsic Motivation

Intrinsic motivation is based in the innate, organismic needs for competence and self-determination. It energizes a wide variety of behaviours and psychological processes for which the primary rewards are the experiences of effectance and autonomy. Intrinsic motivation is the natural propensity to engage ones interests and exercise ones capacities, and in doing so, to seek and conquer optimal challenges. This motivation emerges spontaneously from internal tendencies and can motivate behaviour even without the aid of extrinsic rewards or environmental controls. Intrinsic needs differ from primary drives in that they are not based in tissue deficits and they do not operate cyclically, that is, breaking into awareness, pushing to be satisfied, and then when satisfied, receding into quiescence. Intrinsic motivation is an important motivator of the learning, adaption and growth in competencies that characterize human development.

Extrinsic Motivation

Extrinsically motivated behaviour has an external perceived locus of causality, the individual does it to get an extrinsic reward or to comply with an external constraint. Extrinsic motivation comes from an external source, such as money, good grades or the approval of others. With an external reward or constraint, an instrumentality develops such that the activity becomes a means to an end rather than an end in itself. The behaviour is no longer done because it is interesting, it is something that is done to get an external reward or to comply with an external constraint.The study of extrinsic motivation revolves around three central concepts of incentives, consequences and rewards. Incentives, consequences and rewards that are expected and tangible typically undermine motivation by decreasing autonomy, interfering with the learning process, and undermining peoples development of their own autonomous self regulation.

Goal Setting

A goal is whatever an individual is striving to accomplish. Goals generate motivation by focusing people’s attention on the discrepancy between their present level of accomplishment and their ideal level of accomplishment. Steps in an Effective Goal Setting Program

Sequential Steps within the Goal-Setting Process
 * 1) 	Specify the objective to be accomplished.
 * 2) 	Define goal difficulty.
 * 3) 	Clarify goal specificity.
 * 4) 	Specify the time span when performance will be assessed.

Sequential Steps within the Goal-Striving Process
 * 1) 	Check on goal acceptance.
 * 2) 	Discuss goal-attainment strategies.
 * 3) 	Create implementation intentions.
 * 4) 	Provide performance feedback.

Week 6. Personal Control Beliefs/Tutorial 3
Motivation to Exercise Personal Control

The desire to exercise personal control is predicted on a person’s belief that they have the power to produce favourable results. When people believe that they “have what it takes” to influence their environment, and the environment will be responsive to their influence attempts, then they will indeed try to make things happen for the better. They will be motivated to exercise personal control over life’s outcomes.

The Self

In a motivational analysis of the self and its strivings, four problems take centre stage:
 * 1) 	Defining or creating the self.
 * 2) 	Relating the self to society.
 * 3) 	Discovering and developing personal potential.
 * 4) 	Managing or regulating the self.

In the quest to discover and develop the self, we explore what does and what does not interest us, we internalize the values of those we respect, we strive to create meaning, we seek to discover and develop our talents, and we devote our time to developing some skills and relationships rather than others. In the quest to regulate the self, we reflect our capacities, monitor how well we are accomplishing our goals, and make the self-related adjustments that are needed to achieve enhanced self-functioning.

Six Dimensions of Psychological Well-Being
 * Self Acceptance: positive attitude towards the self, accepts multiple aspects of the self, feels positive about the past.
 * Positive Relations with Others: has warm, satisfying, trusting relationships with others, capable of strong empathy, affection and intimacy.
 * Autonomy: able to resist social pressures, regulates behaviour from within.
 * Environmental Mastery: controls complex array of external activities, makes effective use of surrounding opportunities.
 * Purpose in Life: has goals in life and a sense of directedness, feels there is a meaning to present and past life.
 * Personal Growth: has a feeling of continued development, is open to new experiences, has sense of realizing his or her potential.

Tutorial 3.
In this tutorial we were asked to develop a class map of the main underlying motivation sources for attending University. We found that there were six area’s in which motivation derived from:

 * 1) 	Career/Qualifications - for the degree, so I can get a better job etc.
 * 2) 	Self-Exploration/Learning - for the learning, curiosity, knowledge-seeking etc.
 * 3) 	Social Opportunities - to meet people, make and explore friendships, enjoy social environment
 * 4) 	Altruism - to become better able to help people, help society, help the planet etc.
 * 5) 	Social Pressure - expectations of family, friends, society etc.
 * 6) 	Rejection of Alternatives - better option than doing nothing, working etc.

We were then asked to individually rank your own personal motivational reasons for attending class and completed a University Student Motivation survey. We had to rank from 1-8, 1 being false, 8 being completely true a series of questions. The survey included questions such as:

I study at University...

 * To get a qualification, increase job opportunities, acquire valuable skills, and set up my future career.
 * Because I love learning and I want to explore ideas, improve myself, and gain personal growth by developing myself.
 * Because I want to help others, solve world problems, and contribute to society.
 * Because I enjoy the social environment, social opportunities, and can develop friendships.

Self Constructs
The class was discussed and distinguished the differences amongst self constucts:
 * Self-Esteem, Self Worth (Self Value)
 * Self Efficiacy (Behaviour)
 * Self Concept (Percepton of Self)
 * Self Identity (Groups, Roles, Social Norms)
 * Self Concordance (Agency Goals)

Week 9. Nature of Emotion
Emotions typically arise as a reaction to important life events. Once activated, emotions generate feelings, arouse the body to action, generate motivational states, and produce recognizable facial expressions.
 * 1) 	What is emotion?
 * 2) 	What causes an emotion?
 * 3) 	How many emotions are there?
 * 4) 	What good are the emotions?
 * 5) 	What is the difference between emotion and mood?

What Is An Emotion?

Emotions are multidimensional. They exist as subjective, biological, purposive, and a social phenomena. In part, emotions are subjective feelings, as they make us feel a particular way, such as angry or joyful. But emotions are also biological reactions, energy-mobilizing responses that prepare the body for adapting to whatever situation one faces.

What Causes An Emotion?

When we encounter a significant life event, an emotion comes to life. Encountering a significant life event activates cognitive and biological processes that collectively activate the critical components of emotion, including feelings, bodily arousal, goal directed purpose and expression.

How Many Emotions Are There?

Emotions come in two separate perspectives, biological and cognitive. Biological emphasize on primary emotions such as anger, fear, interest, surprise, distress and joy. The cognitive perspective asserts firmly that human beings experience a greater number of emotions. They point out that several emotions can arise from the same biological reaction. For instance, a single physiological response, such as a rapid rise in blood pressure, can serve as a biological basis for anger, jealousy or envy.

What Good Are The Emotions?

In Darwin’s research, he argues that emotions help animals adapt to their surroundings. Displays of emotion help adaption much in the same way that displays of physical characteristics (e.g. height) do. For example, the dog baring his teeth in defence of its territory helps it cope with hostile situations (by warding off opponents). Such expressiveness is functional, and emotions are therefore candidates for natural selection.

What Is The Difference Between Emotions And Mood?

Emotions emerge from significant life situations and from appraisals of their significance to our well-being. Moods emerge from processes that are ill-defined and are oftentimes unknown. Emotions mostly influence behaviour and direct specific courses of action. Moods on the other hand mostly influence cognition and direct what the person thinks about.

Week 10. Aspects of Emotion/Tutorial 4
Table: Biological and Cognitive Aspects of Emotion

Differential Emotions Theory

The Differential emotions theory takes its name from its emphasis on basic emotions serving unique or different, motivational purposes. The theory endorses the following:
 * 1) 	Ten emotions constitute the principal motivation system for human beings.
 * 2) 	Unique feeling: Each emotion has its own unique subjective, phenomenological quality.
 * 3) 	Unique expression: Each emotion has its own unique facial expression pattern.
 * 4) 	Unique neural activity: Each emotion has its own specific rate of neural firing that activates it.
 * 5) 	Unique purpose/motivation
 * Each emotion generates distinctive motivational properties and serves adaptive functions.

Izard’s 10 Fundamental Emotions Included in His Differential Emotions Theory

Positive Emotions: Interest, joy.

Neutral Emotions: Surprize.

Negative: Fear, anger, disgust, distress, contempt, shame, guilt.

Tutorial 4.
The class was divided into two groups and given a long list of emotions, approximately 150-200 words. We were then subjected to dividing these emotional definitions into sub-groups. After much discussion our group had divided all the emotions into nine different sub-groups. The sub-groups and definitions included:

Happy: cheerful, humorous, playful, grateful, hopeful, glad, perky and proud.

Sad: teary, doubtful, weary, livid, drained, sorrowful, miserable and self pity.

Fear: Qualm, horrified, panicked, rattled, powerless and paranoid.

Surprised: enlightened, distracted, amazed, thrall, fascinated and suspenseful.

Love: surgent, trusting, compassionate, lustful and desirous.

Angry: vexed, jealous, mad, combative, cursed and intrigued.

Cognitive: cheeky, overloaded, courageous, tenacious and eager.

Uncertain: baffled, aroused, puzzled, curious and intrigued.

Attitudes not Emotions: obnoxious, greedy, anomine and indifferent.

Below are some pictures of our emotions Q-sort.

Week 11. Personality, Motivation and Emotion
Individual Differences in Happiness, Arousal and Control

Happiness
 * Extraversion & happiness
 * Neuroticism & suffering
 * Extraverts & neurotics

Performance & emotion
 * Insufficient stimulation & underarousal
 * Excessive stimulation & overarousal
 * Credibility of the inverted-U hypothesis
 * Sensation seeking
 * Affect intensity

Control
 * Percieved control
 * Desire for control

Arousal’s Contribution to Motivation

 * 1) A person’s arousal level is mostly a function of how stimulating the environment is.
 * 2) People engage in behaviour to increase or decrease their level of arousal.
 * 3) When under aroused, people seek out opportunities to increase their arousal levels, because increases in environmental stimulation are pleasurable and enhance performance, whereas decreases are aversive and undermine performance.
 * 4) When over aroused, people seek out opportunities to decrease their arousal levels, because increases in environmental stimulation are aversive and undermine performance, whereas decreases are pleasurable and enhance performance.

Figure 2. Inverted U of arousal.

The Big 5 Personality Traits

 * Neuroticism
 * Extraversion
 * Openness to Experience
 * Agreeableness
 * Conscientiousness

Each of these is measured using 6 facets (traits), these traits are shown in the table below.

Week 12. Unconscious Motivation/Tutorial 5
Psychodynamic Perspective

The psychodynamic approach presents a largely deterministic and pessimistic image of human nature. Psychoanalysis is deterministic in that it holds that the ultimate cause of motivation and behaviour derives from biologically endowed and socially acquired impulses that determine our desires, thoughts, feelings, and behaviours, whether we like it or not. This view conveys that motivation comes across as something that has happened to us, rather than as something one chooses or creates

The Unconscious

Freud believed that the individual must express strong unconscious urges and impulses, though in a disguised form. Believing the unconscious constituted the “primary process” while consciousness was but a “secondary process,” Freud and his colleagues explored the contents and processes of the unconscious in a number of ways, including hypnosis, free association, dream analysis, humour, projective tests, errors and slips of the tongue, and so-called “accidents”. Due to this research, the idea that people have motives and intentions that lie outside of their everyday awareness is readily accepted by motivation researches.

Psychodynamics

Freud observed that people often engaged in behaviour that they clearly did not wish to do. Because people sometimes did what they did not want to do, he reasoned that motivation must be more complex than that which follows intentional violation. Conscious volition must have to wrestle with an unconscious counter-will. Freud’s depiction of the human mind was one of conflict, idea versus counter-idea, will versus counter-will, desire versus repression, excitation versus inhibition, and cathexis (sexual attraction) versus anti-cathexis (guilt). This clashing of the forces is what is meant by the term psychodynamics.

Ego Psychology The neo-Freudians saw ago functioning as the process of maturation, the ego was made independent from its id origins. For Hienz Hartman, the ego, unlike the id, developed through learning and experience. Learning occurred because the child engaged in a tremendous amount of manipulative, exploratory, and experimental activity (such as grasping, walking and thinking), all of which provided the ego with information about itself and its surroundings. With feedback from its manipulative, exploratory and experimental activity, the ego began to acquire ego properties of language, memory, intentions and complex ideas. This facilitated the ego’s ability to adapt successfully to the realities, demands and constraints of the world.

Tutorial 5.
I did not attend this tutorial, by checking on wikiversity i was able to find that the class was giving a Chapter Checklist to help out with our textbook chapter assessment. The Chapter Checklist included:

Does the chapter:
 * 1) 	have a title?
 * 2) 	have an introduction?
 * 3) 	set up some focus questions or specific topics/structure?
 * 4) 	explain relevant theory(ies)?
 * 5) 	explain relevant research?
 * 6) 	provide some examples?
 * 7) 	provide relevant links?
 * 8) 	illustrate the knowledge in an interesting way?
 * 9) 	provide a summary?
 * 10) 	use APA style (as much as reasonably possible in wiki-format)
 * 11) 	use and provide relevant peer-reviewed references

This chapter checklist helped me to format my textbook chapter and for any other questions involving the finer skills of wikiversity which i had never previously used, our tutor, Dr James Neill was always easily contactable and happy to help out.

Week 13. Growth and Motivation
Self Actualization

Self Actualization is an inherent developmental striving. It is a process of leaving behind timidity, defensive appraisals, and a dependence of others that is paired with the parallel process of moving toward courage to create, make realistic appraisals, and achieve autonomous self regulation. According to Roger’s Self Actualisation is “an underlying flow of movement toward constructive fulfilment of its inherent possibilities”. While Maslow believes it refers to an ever-fuller realization of one’s talents, capacities and potentialities.

Maslow: Six Behaviours That Encourage Self-Actualisation

 * 1) 	Make Growth Choices: See life as a series of choices, forever a choice toward progression and growth versus regression and fear.
 * 2) 	Be Honest: Dare to be different, unpopular and nonconformist. Take responsibility for your choices and the consequences of the choices.
 * 3) 	Situationally Position Yourself for Peak Experience: Set up conditions to make peak experiences more likely. Get rid of false notions and illusions.
 * 4) 	Give Up Defensiveness: Identify defences and find the courage to give them up.
 * 5) 	Let the Self Emerge: Perceive within yourself and see and hear the innate impulse voices. Instead of only looking to others to tell you who to become, also listen to your own personal interests and aspirations.
 * 6) 	Be Open to Experience: Experience fully, vividly, selflessly with full concentration and total absorption. Be spontaneous, original and open to experience.

Positive Psychology

Looks at people’s mental health and the quality of their lives to ask, “What could be?” Positive psychology also seeks to build people’s strengths and competencies.

Three Illustrative Personal Strengths

Optimism: A positive attitude or a good mood that is associated with what one expects to unfold in his or her immediate and long-term future. Related to better psychological & physical health, more health-promoting behaviors, greater persistence, and more effective problem solving.

Meaning: A sense of purpose, internalised values, and high efficacy are the motivational means to cultivate meaning in life. The act of creating meaning helps prevent future sickness.

Eudaimonic Well-Being: Eudaimonic well-being is self-realisation. Relatedness satisfaction and pursuit of self-endorsed goals forecast

Tutorial 6.
During our last tutorial of the subject we completed a “Sense of Coherence – Orientation to Life Questionnaire”. Sense of Coherence is “The extent to which one has a pervasive, enduring though dynamic, feeling of confidence that one’s environment is predictable and that things will work out as well as can reasonably be expected.”

It has three components – comprehensibility, manageability, and meaningfulness.

1. A sense of comprehensibility:
 * Do you feel that you can understand things, that things make sense and are not confusing?
 * Do you feel that things are predictable or can be expected? In other words, do you feel like you know what’s going to happen next, or that you know what’s coming?

2.A sense of manageability:
 * Do you feel that things are manageable or within your control, that things can be handled or taken care of?
 * Do you feel you have the skills or ability, the support, the help, or the resources necessary to take care of things?

3. A sense of meaningfulness:
 * Do you feel that things are interesting or fascinating, a source of pleasure or satisfaction?
 * Do you feel that things are really worth it, that there is good reason or purpose to care about what happens?

UC Student item means:

My score:
 * Overall SOC: 4.4
 * Comprehensibility: 4.6
 * Manageability: 4.5
 * Meaningfulness: 4

From these results I was able to conclude that my overall “Sense of Coherence” was slightly above the average mean, while on the other hand, my Meaningfulness was a whole standard deviation from the mean. Leaving the question in my mind whether or not my behaviours and actions towards life are meaningful and for the right cause???

Our tutor, Dr James Neill also discussed elements of our multimedia presentation. I was unfamiliar with the format of this assessment and after a quick chat with James he was able to point me in the right direction. After setting up a twitter account i was able to record my multimedia presetation successfully and without any problems.

title: User:Battered sav/E-portfolio