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User:G-man~enwikiversity

Problem 1
Problem Statement: Find the Euler Integrating Factor h(y) for a non-linear 1st order ODE for the particular case where we assume hxN=0

Two conditions must be satisfied in order for a nonlinear 1st order ODE to be exact.

The first condition is that $$F(x,y,y')=0$$ must be in the form $$M(x,y)+N(x,y)y'=0$$

The second condition is that $$M_y=N_x$$. This example will show how to satisfy the second condition.

Start with equation $$M(x,y)+N(x,y)y'=0$$ which already satisfies the first condition.

If we assume the equation is not exact we can multiply it by an unkown factor $$h(x,y)$$ (Euler Integrating Factor) to attempt to find a factor that will make the equation exact.

The equation becomes:

$$h(x,y)[M(x,y)dx+N(x,y)dy]=0$$

$$(hM)dx+(hN)dy=0, where\ hM=\overline M, and\ hN=\overline N$$

$$\overline M_y=(hM)_y=h_yM+hM_y$$

$$\overline N_x=(hN)_y=h_xN+hN_x$$

Set $$\overline M_y=\overline N_x$$ and rearrange,

$$h_xN-h_yM+h(N_x-M_y)=0$$

To solve for $$h_y$$ we will let $$h_xN=0$$ which implies that $$h_x=0$$ since $$\scriptstyle M\ne 0$$

$$-h_yM+h(N_x-M_y)=0$$, rearrage terms

$${h_y\over h}dy={1\over M}(N_x-M_y)$$ take the integral of both sides,

$$\int {h_y\over h}dy=\int {1\over M}(N_x-M_y)dy$$

If the right hand side of the integral is only a function of y then it can be written as $$g(y)$$

$${1\over M}(N_x-M_y)=g(y)$$

$$ln\mid h\mid=\int g(y)dy$$

$$h(y)=exp\int_{}^{y}g(s)ds$$

Problem 2
Problem Statement: Given the Non-homogeneous linear 1st order ODE with VC $$y'+{1\over x}y=x^2$$ show the steps to obtain $$y={x^3\over 4}+{C\over x}$$ where C=Const.

The first step is to check if the given equation is in the proper form:

$$P_{(x)}y'+Q_{(x)}y=R_{(x)}$$ ; Yes [pg.(8-1) Eg.(1)]

If $$P_{(x)}\ne 0$$ for all (x) then:

$$y'+{Q_{(x)}\over P_{(x)}}y={R_{(x)}\over P{(x)}}$$

In this form we can use the Integration Factor Method to solve for y

$$N_{(x,y)}{dy\over dx}+M_{(x,y)}=0$$

Multiply by $$h_{(x,y)}$$

$$h_{(x,y)}[N_{(x,y)}dy+M_{(x,y)}dx]$$

$$(hM)dx+(hN)dy=0$$ where $$(hM)=\overline M;\ (hN)=\overline N$$

Satisfy the exactness condition $$\overline M_y=\overline N_x$$ where:

$$\overline M_y=(hM)_y=h_yM+hM_y$$

$$\overline N_x=(hN)_x=h_xN+hN_x$$

$$h_xN-h_yM+h(Nx-My)=0$$

To solve for h(x,y) we must make an assumption that h is a function of (x) only.

Assume hyM=0, so our equation becomes;

$$h_xN+h(N_x-M_y)=0$$

$${h_x\over h}=-{1\over N_{(x,y)}}(N_x-M_y)$$

Let $$N_{(x,y)}=1$$ and $$M_{(x,y)}={Q_{(x)}\over P_{(x)}}y={1\over x}y$$

$$N_x=0$$ and $$M_y={1\over x}$$

Substitute back into the right hand side of the equationto verify h is a function of (x) only,

$${1\over 1}(0-{1\over x})={1\over x}=f(x)$$

Using that result we can now solve for $$h$$

$$\int {h_x\over h}dx=\int {1\over x}dx$$

$$ln|h|=ln|x|$$

$$h_x=x$$

Now that we have $$h_x$$ we can multiply through our original equation to obtain,

$$(x)y'+(x){1\over x}y=(x)x^2\ =xy'+y=x^3$$

Integrating yields:

$$xy={x^4\over 4}+C$$ divide by x,

 $$y={x^3\over 4}+{C\over x}$$

Problem 1
Problem Statement : Given a L2_ODE_VC $$\sqrt{x}y''+2xy'+3y=0$$

Find (m,n) from the integrating factor (xm,yn) that makes the equation exact.

A first integral is $$\phi(x,y,p)=xp+\big(2x^{3/2}\big)y+k_1=k_2$$

 * $$\phi_p=x$$
 * $$\phi_x=p+3x^{1/2}$$
 * $$\phi_y=2x^{3/2}-1$$

$$\phi(x,y,p)=h(x,y)+\int_{}f(x,y,p)dp, f=\phi_p$$

$$\phi(x,y,p)=h(x,y)+\int_{}xdp={h(x,y)+xp}$$

$$g(x,y,p)=\phi_x+\phi_yp=h_x+\phi_x+(h_y+\phi_y)p$$

$$g(x,y,p)=p+3x^{1/2}+(2x^{3/2}-1)p=h_x+x+(h_y+0)p$$

 * $$h_y=2x^{3/2}$$
 * $$h_x=3x^{1/2}-x$$

$$y^n=h_y\Rightarrow lny^n=ln(2x^{3/2})$$  $$n=ln(2x^{3/2})$$ $$x^m=h_x\Rightarrow lnx^m=ln(3x^{1/2}-x)$$  $$m=ln(3x^{1/2}-x)$$

Problem 2
Problem Statement: Given a first integral$$\phi$$ of a L2_ODE_VC, solve for $$y(x)$$.

$$\phi(x,y,p)=xp+(2x^{3\over 2}-1)y+k_1=k_2$$ (1)

where k1 and k2 are const, and $$p=y'$$

Eq. (1) is in the form $$M(x,y)+N(x,y)y'$$ where

$$M(x,y)=(2x^{3\over 2}-1)y$$

$$N(x,y)=x$$

so it satisfies the 1st condition of exactness.

Check if $$M_y=N_x$$ for the 2ndcondition of exactness

$$M_y=2x^{3\over2}-1$$

$$N_x=1$$

$$M_y\ne N_x$$ so we do not satisfy the 2nd condition of exactness.

We must apply the integrating factor method for a L1_ODE_VC.

$$xp+(2x^{3\over 2}-1)y=k_2-k_1$$, divide by x to obtain the form:

$$y'+a_0(x)y=b(x)$$ where:

$$a_o(x)={1\over x}(2x^{3\over 2}-1)=2\sqrt{x}-{1\over x}$$

$$b(x)={k_2-k_1\over x}$$

From our solution of a general non-homogeneous L1_ODE_VC p.8-1

$$h(x)=exp\int_{}^{x}a_0(s)ds$$

$$h(x)=exp\int_{}^{x}2\sqrt{x}-{1\over x}=exp\bigg({4\over 3}x^{3\over 2}-ln\left|x\right|\bigg)\bigg({{k_2-k_1}\over x}\bigg)=exp{4\over 3}x^{3\over 2}-x$$

From p.8-2 Eq. (4)

$$y(x)={1\over h(x)}\int_{}^{x}h(s)b(s)ds$$

Use the product rule of integration $$\int_{} ab=a\int_{} b-\int_{}a'\int_{}b$$

$$y(x)={1\over h(x)}\bigg[h(x)\int_{}^{x}b(x)-\int_{}^{x}h'(x)\int_{}^{x}b(x)\bigg]$$

In our example $$\int_{}^{x}h'(x)=h(x)$$ so,

$$y(x)={1\over h(x)}\bigg[h(x)\int_{}^{x}b(x)-h(x)\int_{}^{x}b(x)\bigg]$$

$$y(x)=0$$

Problem 3
Problem Statement: From ([[media:Egm6321.f09.p13-1.png|p.13-1]]), find the mathematical structure of $$\Phi$$ that yields the above class of ODE. A class of exact L2_ODE_VC.

$$F(x,y,y',y'')={d\phi\over dx}=\phi_x(x,y,p)+\phi_y(x,y,p)p+\phi_p(x,y,p)p'\ where\ p=y'$$

$$\phi=h(x,y)+\int_{}\phi_pdp=h(x,y)+\phi_pp$$

$$\phi_x=h_x+\phi_{px}p$$

$$\phi_y=h_y+\phi_{py}p$$

$$g=\phi_x+\phi_yp=h_x+\phi_{px}p+(h_y+\phi_{py}p)p$$

$$h_y=\phi_y-\phi_{px}$$

Take the integral of $$h_y$$

$$h(x,y)=(\phi_y-\phi_{px})y+k_1$$

Substitute back into the equation for $$\phi$$

$$\phi=(\phi_y-\phi_{px})y+k_1+\phi_pp$$

Rearrange the terms to obtain

$$\phi(x,y,p)=\phi_pp+(\phi_y-\phi_{px})y+k$$ where,

 * $$P(x)=\phi_p$$
 * $$T(x)=(\phi_y-\phi_{px})y$$
 * $$k=k_1$$

$$\phi(x,y,p)=P(x)p+T(x)y+k$$

Problem 4
Problem Statement: Given a N1_ODE, for the case n=1 $$F(x,y,y')=0\Leftrightarrow {d\phi\over dx}(x,y)$$

Show that $$f_0-{df_1\over df_x}=0\Leftrightarrow\phi_{xy}=\phi_{yx},$$ Hint:$$f_1=\phi_y$$

$$F={d\phi\over dx}(x,y^{(0)},....y^{(n-1)}=\phi_x+\phi_y6{(0)}y^{(1)}$$

$$F=\phi_x+\phi_yy'$$

$$f_i:={\partial F\over \partial y^i}$$

4.1
Find $$f_0$$ in terms of $$\phi$$.

$$f_0={\partial F\over \partial y}={\partial (\phi_x+\phi_yy')\over \partial y}=\phi_{xy}$$  $$f_0=\phi_{xy}$$

4.2
Find $$f_1$$ in terms of $$\phi_y$$

$$f_1={\partial F\over \partial y'}={\partial (\phi_x+\phi_yy')\over \partial y'}=\phi_y$$  $$f_1=\phi_y$$

4.3
Show that $$f_0-{df_1\over df_x}=0\Leftrightarrow\phi_{xy}=\phi_{yx},$$

$$\phi_{xy}-{d\phi_y\over df_x}=\phi_{xy}-\phi_{yx}=0$$  $$\phi_{xy}=\phi_{yx}$$

Problem 9
$$y_{xxx}=(d/dx)\bigg[{d/dx\over (d/dx)y}\bigg]=(dt/dx)(d/dt)(dt/dx)(d/dt)(dt/dx)(d/dt)y$$

Replace $$(dt/dx)\ with\ e^{-t}\ and\ (d/dt)y\ with\ y_t.$$

$$y_{xxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)((e^{-t})y_t)$$ 'Chain Rule'

$$y_{xxx}=(e^{-t})(d/dt)(e^{-t})(-e^{-t}(y_t)+e^{-t}(t_{tt}))$$

$$y_{xxx}=(e^{-t})(d/dt)(-e^{-2t}(y_t)+e^{-2t}(y_{tt}))$$

$$y_{xxx}=(e^{-t})(2e^{-2t}(y_t)-e^{-2t}(y_{tt})-2e^{-2t}(y_{tt})+e^{-2t}(y_{ttt}))$$

$$y_{ttt}=2e-3t(y_t)-e^{-3t}(y_{tt})-2e^{-3t}(y_{tt})+e^{-3t}9y_{ttt})$$

Factor out $$e^{-3t}$$ and re-arrange terms in ordre of derivative,

$$y_{xxx}=(e^{-3t})(y_{ttt}-3y_{tt}+2y_t)$$

$$y_{xxxx}=(d/dx)(d/dx)(d/dx)(d/dx)y$$

$$y_{xxxx}=(dt/dx)(d/dt)[(dt/dx)(d/dt)]((dt/dx)(d/dt))\langle(dt/dx)(d/dt)y\rangle$$

Replace $$(dt/dx)\ with\ e^{-t}\ and\ (d/dt)y\ with\ y_t.$$

$$y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)(e^{-t})(d/dt)((e^{-t})y_t)$$

$$y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)(e^{-t})(-e^{-t}(y_t)+e^{-t}(y_{tt}))$$

$$y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(d/dt)(e^{-2t}(y_t)+e^{-2t}(y_{tt}))$$

$$y_{xxxx}=(e^{-t})(d/dt)(e^{-t})(-2e^{-2t}(y_t)+e^{-2t}(y_{tt})-2e^{-2t}(y_{tt})+e^{-2t}(y_{ttt}))$$

$$y_{xxxx}=(e^{-t})(d/dt)(-2e^{-3t}(y_t)+e^{-3t}(y_{tt})-2e^{-3t}(y_{tt})+e^{-3t}(y_{ttt}))$$

$$y_{xxxx}=(e^{-t})(6e^{-3t}(y_t)-2e^{-3t}(y_{tt})-3e^{-3t}(y_{tt})+e^{-3t}(y_{ttt})+6e^{-3t}(y_{tt})-2e^{-3t}(y_{ttt})-3e^{-3t}(y_{ttt})+e^{-3t}(y_{tttt}))$$

$$y_{xxxx}=6e^{-4t}(y_t)+2e^{-4t}(y_{tt})+3e^{-4t}(y_{tt})-e^{-4t}(y_{ttt})+6e^{-4t}(y_{tt})-2e^{-4t}(y_{ttt})-3e^{-4t}(y_{ttt})+e^{-4t}(y_{tttt}))$$

Factor out $$e^{-4t}$$ and re-arrange terms in order of derivative.

$$y_{xxxx}=(e^{-4t})(y_{tttt}-6y_{ttt}+11y_{tt}-6y_t)$$

Problem #1
Given the Legendra differential equation $$(1-x^2)y''-2xy'+n(n+1)y=0$$ with $$n=0$$ and given the homogeneous solution $$u_1(x)=1$$. use the reduction of order method 2 (undetermined factor) to find $$u_2(x)$$, the second homgeneous solution.

$$(1-x^2)y''-2xy'+n(n+1)y=0\ at\ n=0\ yields,$$

$$(1-x^2)y''-2xy'=0$$

From the form $$y''+a_1y'+a_0y=0$$

$$a_1(x)=-{2x\over 1-x^2}$$

The homogeneous solution is represented by:

$$y_H(x)=k_1u_1(x)+k_2u_2(x)$$

The full homogeneous solution is assumed to be:

$$y(x)=U(x)u_1(x)\ where\ U\ is\ unknown.$$

 * $$y'=Uu'_1+U'u_1$$ multiply by $$a_1(x)$$

 * $$y=Uu_1+2U'u'_1+U''u_1$$ multiply by $$a_2(x)=1$$

Adding yields: $$a_1y'+y=U[a_1u'_1+u_1]+U'[a_1u_1+2u'_1]+U''u_1$$

Since $$u_1$$ is a homogeneous solution $$[a_1u'_1+u_1]=0$$ and $$a_1y'+y=0$$ we get:

$$U'[a_1u_1+2u'_1]+U''u_1=0$$

Let $$Z=U'$$

$$Z[a_1u_1+2u'_1]+Z'u_1=0$$, solve for Z by direct integration.

$${Z'\over Z}+(a_1+{2u'_1\over u_1})=0$$, solve for Z by direct integration.

$$Z(x)={e^k\over (u_1)^2}exp[-\int_{}^ta_1(s)ds]$$

$$U(x)=\int_{}^x{e^k\over (u_1(t))^2}exp[-\int_{}^ta_1(s)ds]dt+k_1$$

 * $$\int_{}^ta_1(s)ds=\int_{}^t-{2s\over (1-s^2)}ds=ln(1-t^2)$$

$$y_H(x)=k_1u_1(x)+k_2u_2(x)=U(x)u_1(x)=e^ku_1\int_{}^x{1\over (u_1(t))^2}exp[-ln(1-t^2)]dt+k_1u_1(x)$$

Solve for $$u_2(x)$$

$$u_2(x)=u_1\int_{}^x{1\over (u_1(t))^2}exp[-ln(1-t^2)]dt=x\int_{}^x{1\over (x)^2}exp[-ln(1-t^2)]dt=x\int_{}^x{dt\over (x)^2(1-t^2)}$$

$$u_2(x)={x\over 2}ln\bigg({1+x\over 1-x}\bigg)-1$$

Problem #2
Problem 1.1 part (b) from King, Differential Equations page 28.

Show that the function $$u_1=x^{-1}sinx$$ is a solution of the differential equation $$ xy''+2y'+xy=0$$. Use the reduction of order method to find the second independent solution $$u_2$$.

Arrange the differential equation into the form $$y''+a_1y'+a_0y=0$$

$$y''+{2\over x}y'+y=0$$

$$a_1(x)=2/x$$

$$\int_{}^ta_1(s)ds=\int_{}^t{2\over s}ds=2ln(t)$$

$$u_2(x)=u_1(x)\int_{}^x{1\over u^2_1(t)}exp\bigg[-\int_{}^ta_1(s)ds\bigg]dt$$ [King Eq.(1.3)p.6]

$$u_2(x)={sinx\over x}\int_{}^x{1\over {sin^2t\over t^2}}exp[-2ln(t)]dt={sinx\over x}\int_{}^x{t^2\over sin^2t}exp[-2ln(t)]dt$$

$$u_2(x)=\Bigg({sinx\over x}\Bigg)\Bigg({x^3\over 3}\Bigg)\Bigg({1\over {1\over 2}x-{1\over 4}sin2x}\Bigg)exp[-2ln(x)]$$

Problem 1
p.29-4 Complete equations (3) & (4) for i=2 and i=3 respectively.

Problem Statement: From the Laplace equation $$\mathcal {4}\psi={1\over h_1h_2h_3}\sum_{i=1}^3{\partial \over \partial \xi_i}\Big[{h_1h_2h_3\over (h_i)^2}{\partial \psi \over \partial \xi_i}\Big]$$ where $$h_1(\xi)=1$$ $$h_2(\xi)=rcos\theta=\xi_1cos\xi_2$$ $$h_3(\xi)=r=\xi_1$$ $$h_1h_2h_3=r^2cos\theta$$ $$\xi_1=r, \xi_2=\psi, \xi_3=\theta$$

Solve for when $$i=2, i=3$$

When $$i=2$$ $$\mathcal {4}\psi={1\over h_1h_2h_3}\sum_{i=1}^3{\partial \over \partial \xi_i}\Big[{h_1h_2h_3\over (h_i)^2}{\partial \psi \over \partial \xi_i}\Big]$$ becomes

$$\mathcal {4}\psi_{i=2}={1\over r^2cos\theta}{\partial \over \partial \phi}\Big[{r^2cos\theta\over (rcos\theta)^2}{\partial \psi \over \partial \phi}\Big]$$

$$\mathcal {4}\psi_{i=2}={1\over r^2cos\theta}{\partial \over \partial \phi}\Big[{r^2cos\theta\over (rcos\theta)^2}{\partial \psi \over \partial \phi}\Big]={1\over r^2cos\theta}{\partial \over \partial \phi}\Big[{r^2cos\theta\over r^2cos^2\theta}{\partial \psi \over \partial \phi}\Big]$$

$$\mathcal {4}\psi_{i=2}={1\over r^2cos\theta}{r^2cos\theta\over r^2cos^2\theta}{\partial \over \partial \phi}\Big[{\partial \psi \over \partial \phi}\Big]$$

 $$\mathcal {4}\psi_{i=2}={1\over r^2cos^2\theta}\Big[{\partial^2 \psi \over \partial \phi^2}\Big]$$

When $$i=3$$ $$\mathcal {4}\psi={1\over h_1h_2h_3}\sum_{i=1}^3{\partial \over \partial \xi_i}\Big[{h_1h_2h_3\over (h_i)^2}{\partial \psi \over \partial \xi_i}\Big]$$ becomes

$$\mathcal {4}\psi_{i=3}={1\over r^2cos\theta}{\partial \over \partial \theta}\Big[{r^2cos\theta\over (r)^2}{\partial \psi \over \partial \theta}\Big]$$

$$\mathcal {4}\psi_{i=3}={1\over r^2cos\theta}{\partial \over \partial \theta}\Big[{r^2cos\theta\over r^2}{\partial \psi \over \partial \phi}\Big]$$

 $$\mathcal {4}\psi_{i=3}={1\over r^2cos\theta}{\partial \over \partial \theta}\Big[cos\theta{\partial \phi\over \partial\theta}\Big]$$

Problem 2
p.30-2 Solve eq.(1) using the trial solution method where $$R(r)=r^\lambda$$

Problem Statement:

Solve $${d\over dr}\Bigg(r^2{dR\over dr}\Bigg)=kR$$ by trial solution: $$R(r)=r^\lambda$$

where,

$$R'={d\over dr}R=\lambda r^{\lambda-1}$$

$$R''=\lambda(\lambda-1)r^{\lambda-2}$$

$$r^2{dR\over dr}=r^2R'=r^2(\lambda r^{\lambda-1})=\lambda r^{\lambda-1+2}$$

$${d\over dr}\Bigg(r^2{dR\over dr}\Bigg)={d\over dr}\Big(\lambda r^{\lambda-1+2}\Big)=kR$$

$$\lambda (\lambda-1+2)r^{\lambda-2+2}=kr^\lambda$$

$$\lambda (\lambda+1)r^\lambda=kr^\lambda$$

Divide by $$r^\lambda$$

$$\lambda (\lambda+1)=k$$

=Homework #6=

Problem 9
p. 33-3 For $$f=\sum_i g_i$$ 1) show if {gi} is odd then f is odd 2) show if {gi} is even then f is even

A continuous function f can be expressed as $$f(x)=\sum_{n=0}^\infty A_nP_n(x)$$

We have observed that when n is EVEN $$P_n$$ is EVEN and when n is ODD $$P_n$$ is ODD

If we say $$g_i=A_nP_n$$

Then $$g_i$$ will be EVEN for all EVEN values of n.

And $$g_i$$ will be ODD for all ODD values of n.

Problem 10
p. 33-3 Show $$P_{2k}(x)$$ is even for k=0,1,2,... Show $$P_{2k+1}(x)$$ is odd for k=0,1,2,... Use equation (6) from p.31-3

For $$P_{2k}(x)$$ 'n' will always be an even number for all values of 'k' i.e. k=0,1,2,3,4,5......; n=0,2,4,6,8,10....

For $$P_{2k+1}(x)$$ 'n' will always be an odd number for all values of 'k' i.e. k=0,1,2,3,4,5......; n=1,3,5,7,9,11....

From the equation $$P_n(x)=\sum_{i=0}^{[n/2]}(-1)^i{(2n-2i)!x^{n-2i}\over 2^ni!(n-i)!(n-2i)!}$$ where [n/2] is only the integer part of the quotient i.e. [3/2]=1.5=1

$$\ x^{n-2i}$$ will always be an even number for all even values of 'n'.

$$\ x^{n-2i}$$ will always be an odd number for all odd values of 'n'.

Problem 11
p. 33-4 For $$q(x)=\sum_{i=0}^4c_ix^i$$ where $$c_0=3,\ c_1=10,\ c_2=15,\ c_3=-1,\ c_4=5$$

Find $$(a_i)$$ such that $$q=\sum_{i=0}^4a_iP_i$$

Plot $$q=\sum_ic_ix^i\ and\ q=\sum_ia_iP_i$$

$$q(x)=\sum_{i=0}^4c_ix^i$$ can be shown as a column matrix:

$$\begin{bmatrix} c_0x^0\\c_1x^1\\c_2x^2\\c_3x^3\\c_4x^4 \end{bmatrix}=\begin{bmatrix} 3x^0\\10x^1\\15x^2\\-1x^3\\5x^4 \end{bmatrix}$$

$$q(x)=\sum_{i=0}^4a_iP_i=\begin{bmatrix} a_0&a_1&a_2&a_3&a_4 \end{bmatrix}\begin{bmatrix} P_0\\P_1\\P_2\\P_3\\P_4 \end{bmatrix}$$

Setting $$q(x)=\sum_{i=0}^4c_ix^i=\sum_{i=0}^4a_iP_i(x)$$ yields

$$\begin{bmatrix} a_0&a_1&a_2&a_3&a_4 \end{bmatrix}\begin{bmatrix} P_0\\P_1\\P_2\\P_3\\P_4 \end{bmatrix}=\begin{bmatrix} 3x^0\\10x^1\\15x^2\\-1x^3\\5x^4 \end{bmatrix}$$

$$a_0P_0=3x^0 \Rightarrow a_0={3/P_0}$$

$$a_1P_1=10x^1 \Rightarrow a_1={10x/P_1}$$

$$a_2P_1=15x^2 \Rightarrow a_2={15x^2/P_2}$$

$$a_3P_3=-1x^3 \Rightarrow a_3={-x^3/P_3}$$

$$a_4P_4=5x^4 \Rightarrow a_4={5x^4/P_4}$$

From lecture p.31-3

$$P_0(x)=1$$

$$P_1(x)=x$$

$$P_2(x)=1/2(3x^2-1)$$

$$P_3(x)=1/2(5x^3-3x)$$

$$P_4(x)=25/8x^4-15/4x^2+3/8$$

 $$a_0=3$$

$$a_1={10x\over x}=10$$

$$a_2={15x^2\over 1/2(3x^2-1)}$$

$$a_3={-x^3\over 1/2(5x^3-3x)}$$

$$a_4={5x^4\over 25/8x^4-15/4x^2+3/8}$$

Fig. 1 Fig. 2

Problem 14
p35-3 Verify the table for the Gaus-Legendre quadrature in wikepedia, analytic expression of {xj} and {wj} for j=1,...,n and n=1,...,5. You must first verify the expression for Pn for n=5,6. Evaluate numerically {xj} and {wj} and compare results with Abramovitz & Stegun, Table 25.4 p.916.

First we will verify $$P_n\ when\ n=5,6$$

$$P_n(x)=\sum_{i=0}^{[n/2]}(-1)^i{(2n-2i)!x^{n-2i}\over 2^ni!(n-i)!(n-2i)!}$$

When n=5

$$P_{5(i=0)}(x)=(-1)^0{10!x^5\over 2^5(0!)(5!)(5!)}={(1)3628800x^5\over 460800}={63\over 8}x^5$$

$$P_{5(i=1)}(x)=(-1)^1{8!x^3\over 2^5(1!)(4!)(3!)}={(-1)40320x^3\over 4608}={-70\over 8}x^3$$

$$P_{5(i=2)}(x)=(-1)^2{6!x\over 2^5(2!)(3!)(1!)}={(1)720x\over 384}={15\over 8}x$$

$$P_5(x)=P_{5(i=0)}+P_{5(i=1)}+P_{5(i=2)}={63\over 8}x^5-{70\over 8}x^3+{15\over 8}x$$
 $$P_5(x)={1\over 8}(63x^5-70x^3+15x)$$ When n=6

$$P_{6(i=0)}(x)=(-1)^0{12!x^6\over 2^6(0!)(6!)(6!)}={(1)479001600x^6\over 33177600}={231\over 16}x^6$$

$$P_{6(i=1)}(x)=(-1)^1{10!x^3\over 2^6(1!)(5!)(4!)}={(-1)3628800x^4\over 184320}={-315\over 16}x^4$$

$$P_{6(i=2)}(x)=(-1)^2{8!x^2\over 2^6(2!)(4!)(2!)}={(1)40320x^2\over 6144}={105\over 16}x^2$$

$$P_{6(i=3)}(x)=(-1)^3{6!x^0\over 2^6(3!)(3!)(0!)}={(-1)720\over 2304}=-5$$

$$P_6(x)=P_{6(i=0)}+P_{6(i=1)}+P_{6(i=2)}+P_{6(i=3)}={231\over 16}x^6-{315\over 16}x^4+{105\over 16}x^2-5$$
 $$P_6(x)={1\over 16}(231x^6-315x^4+105x^2-5)$$

$$P_1,\ P_2,\ P_3,\ P_4$$ have been shown earlier in problem #6 of HW#6

Solve for the roots of $$(x_i)\ for\ n=1,..,5$$

$$P_1(x)=x\Rightarrow roots\ x=0$$

$$P_2(x)=1/2(3x^2-1)\Rightarrow x^2=1/3\Rightarrow roots\ x=+/-\sqrt {1/3}$$

$$P_3(x)=1/2(5x^3-3x)\Rightarrow (5/2)x^3=(3/2)x\Rightarrow x^2=3/5\Rightarrow roots\ x=0,\ x=+/-\sqrt{3/5}$$

$$P_4(x)=1/8(35x^4-30x^2+3)\Rightarrow roots\ x=+/-\sqrt {(3-2\sqrt 6/5)\over 7},\ x=+/-\sqrt {(3+2\sqrt 6/5)\over 7}$$

$$P_5(x)=1/8(63x^5-70x^3-15x)\Rightarrow roots\ x=0, x=+/-1/3\sqrt {5-2\sqrt 10/7},\ x=0'\ x=+/-1/3\sqrt {5+2\sqrt 10/7}$$

Solve for {wj} for j=1,...n and n=1,...5

where $$w_j={-2\over (n+1)P'_n(x_j)P_{n+1}(x_j)}$$

n=1

$$P_1=x;\ P'_1=1;\ P_2=1/2(3x_j^2-1)$$
 $$w_1={-2\over (2)(1)(-1/2)}=2$$

n=2

$$P_2=1/2(3x_j^2-1);\ P'_2=3x_j;\ P_3=1/2(5x_j^3-3x)$$
 $$w_{2+}={-2\over (3)(3)\sqrt{1/3}(1/2)[5\sqrt{1/3}^3-3\sqrt{1/3}]}={-2\over 9(1/3)^{1/2}[5/2(1/3)^{3/2}-3/2(1/3)^{1/2}]}={-2\over -2}=1$$

$$w_{2-}={-2\over (3)(3)(-\sqrt{1/3})(1/2)[5(-\sqrt{1/3}^3)-3(-\sqrt{1/3})]}={-2\over -9(1/3)^{1/2}[5/2(-1/3)^{3/2}-3/2(-1/3)^{1/2}]}={-2\over -2}=1$$

n=3

$$P_3=1/2(5x_j^3-3x);\ P'_3=15/2x_j^2-3/2;\ P_4=1/8(35x^4-30x^2+3)$$
 $$w_{3-0}={-2\over (4)[(15/2)(0)-3/2](1/8)[35(0)-30(0)+3]}={-2\over (4)(-3/2)(3/8)}={-2\over -9/4}=8/9$$

$$w_{3+}={-2\over (4)[(15/2)\sqrt{3/5}^2-3/2](1/8)[35\sqrt{3/5}^4-30\sqrt{3/5}^2+3]}={-2\over (4)(3)(-18/5)}={-2\over -3.6}=5/9$$

$$w_{3-}={-2\over (4)[(15/2)(-\sqrt{3/5}^2)-3/2](1/8)[35(-\sqrt{3/5}^4)-30(-\sqrt{3/5}^2)+3]}={-2\over (4)(3)(-18/5)}={-2\over -3.6}=5/9$$

n=4

$$P_4=1/8(35x^4-30x^2+3);\ P'_4=1/8(140x^3-60x);\ P_5=1/8(63x^5-70x^3+15x)$$
 $$w_{4+-}={-2\over (5)\Bigg[1/8\Bigg(140\Bigg(\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)^3-60\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)\Bigg]\Bigg[1/8\Bigg(63\Bigg(\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)^5-70\Bigg(\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)^3+15\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)\Bigg]}={18+\sqrt 30\over 36}$$

$$w_{4--}={-2\over (5)\Bigg[1/8\Bigg(140\Bigg(-\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)^3-60-\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)\Bigg]\Bigg[1/8\Bigg(63\Bigg(-\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)^5-70\Bigg(-\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)^3+15\Bigg(-\sqrt {(3-2\sqrt 6/5)\over 7}\Bigg)\Bigg)\Bigg]}={18+\sqrt 30\over 36}$$

$$w_{4++}={-2\over (5)\Bigg[1/8\Bigg(140\Bigg(\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)^3-60\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)\Bigg]\Bigg[1/8\Bigg(63\Bigg(\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)^5-70\Bigg(\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)^3+15\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)\Bigg]}={18-\sqrt 30\over 36}$$

$$w_{4-+}={-2\over (5)\Bigg[1/8\Bigg(140\Bigg(-\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)^3-60-\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)\Bigg]\Bigg[1/8\Bigg(63\Bigg(-\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)^5-70\Bigg(-\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)^3+15\Bigg(-\sqrt {(3+2\sqrt 6/5)\over 7}\Bigg)\Bigg)\Bigg]}={18-\sqrt 30\over 36}$$

n=5

$$P_5=1/8(63x^5-70x^3+15x);\ P'_5=1/8(315x^4-210x^2+15);\ P_6=1/16(231x^6-315x^4+105x^2-5)$$

 $$w_{5-0}={-2\over (6)\Big[1/8\Big(315(0)-210(0)+15\Big)\Big]\Big[1/16\Big(231(0)-315(0)-5\Big)\Big]}={-2\over (6)(15/8)(-5/16)}=128/225$$

$$w_{5+-}={-2\over (6)\Bigg[1/8\Bigg(315\Bigg(1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^4-210\Bigg(1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^2+15\Bigg)\Bigg]\Bigg[1/16\Bigg(231\Bigg(1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^6-315\Bigg(1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^4+105\Bigg(1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^2-5\Bigg)\Bigg]}={332+13\sqrt70\over 900}$$

$$w_{5--}={-2\over (6)\Bigg[1/8\Bigg(315\Bigg(-1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^4-210\Bigg(-1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^2+15\Bigg)\Bigg]\Bigg[1/16\Bigg(231\Bigg(-1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^6-315\Bigg(-1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^4+105\Bigg(-1/3\sqrt{5-2\sqrt{10/7}}\Bigg)^2-5\Bigg)\Bigg]}={332+13\sqrt70\over 900}$$

$$w_{5++}={-2\over (6)\Bigg[1/8\Bigg(315\Bigg(1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^4-210\Bigg(1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^2+15\Bigg)\Bigg]\Bigg[1/16\Bigg(231\Bigg(1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^6-315\Bigg(1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^4+105\Bigg(1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^2-5\Bigg)\Bigg]}={332-13\sqrt70\over 900}$$

$$w_{5-+}={-2\over (6)\Bigg[1/8\Bigg(315\Bigg(-1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^4-210\Bigg(-1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^2+15\Bigg)\Bigg]\Bigg[1/16\Bigg(231\Bigg(-1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^6-315\Bigg(-1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^4+105\Bigg(-1/3\sqrt{5+2\sqrt{10/7}}\Bigg)^2-5\Bigg)\Bigg]}={332-13\sqrt70\over 900}$$

Numerical values of {xj} and {wj}

$$\begin{bmatrix} {} & x_i & w_i \\ n=2 &

0.577350269189626 & 1.000000000000000 \\ n=3 & 0.000000000000000 & 0.888888888888889 \\ n=3 & 0.774596669241483 & 0.555555555555556 \\ n=4 & 0.339981043584856 & 0.652145154862546 \\ n=4 & 0.861136311594053 & 0.347854845137454\\ n=5 & 0.000000000000000 & 0.568888888888889 \\ n=5 & 0.538469310105683 & 0.478628670499366 \\ n=5 & 0.906179845938664 & 0.236926885056189 \end{bmatrix}$$

=Homework #7=

Problem 4
p. 38-2 Given $$(r_{PQ})^2=\sum_{i=1}^3(x_Q^i-x_P^i)^2$$ Where $$x_P^1=r_Pcos\theta_Pcos\phi_P$$, $$x_P^2=r_Pcos\theta_Psin\phi_P$$, $$x_P^3=r_Psin\theta_P$$

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

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

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

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

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

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

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

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

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

Combine terms;

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

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

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

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

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

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

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

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

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

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

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

title: User:G-man~enwikiversity


text:
User talk:Dean.jenkins~enwikiversity

 Hello Dean.jenkins, 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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Connectivism and Connectivism glossary
Thanks for your edits, Dean, much appreciated. Am hitting the sack in Oztralia and will listen to CCK09 recording later. -- Jtneill - Talk - c 17:17, 17 September 2009 (UTC)
 * You may not have edited wiki much before (!?), but you're picking it up v. quick - anyone who can edit a wiki table like that knows a trick or so ! Let me know if I can help out. Small tip re signatures below. Sincerely, James -- Jtneill - Talk - c 01:38, 18 September 2009 (UTC)

School:Medicine
Just letting you know about this rather neglected outpost: School:Medicine. There's not really anyone active in this area on Wikiversity. A few people have jumped in at times (enthusiastic medical students) and then disappeared. -- Jtneill - Talk - c 13:56, 21 September 2009 (UTC)

Your account will be renamed
Hello,

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

title: User talk:Dean.jenkins~enwikiversity


text:
Connectivism/Glossary

About this glossary
This is a glossary of terms used in connectivism to help define how they they differ from the dictionary definitions and other learning theories. This glossary is currently under development as part of the CCK09 and CCK11 course. Please help to improve.

amplification
A central concept of connectivism. "learning, knowledge and understanding [can be amplified] through the extension of a personal network" (Siemens 2005) the connection of one concept or skill set with another complementary concept or skill set that produces a greater impact than each element could produce on its own. (Siemens, 2011)

capacity to know
This is the potential that exists in a learner and their connections. It "is more critical than what is currently known" (Siemens 2005)

centrality
A measure of the importance of a node in a network. Degree centrality - the number of direct connections a node has - a well connected node - a busy hub. Betweenness centrality - a calculation of how 'between' constituencies or groups of nodes a particular node is - a broker node or 'boundary spanner'. Closeness centrality - nodes with short connections - nodes best placed to monitor information flow. (Krebs 2008)

connected specialization
In complex systems, individual agents/nodes become increasingly specialized. In order to enact new domains of knowledge (see above), we need to connect specialized nodes. Understanding how and why nodes form and connect may help us to understand why we have an iPad but not a Windows tablet (as promised by Balmer in 2010). Connections have an impact – but we don’t want random connections for connections sake. We need connections that increase the capacity of a network of individuals to create and grow knowledge. ([ ])

connections
"A connection is a link between two entities [in a network] such that a change of state in one entity may result in a change of state in the second entity." (Downes 2009) These connections are where knowledge and understanding are represented and they may be at a neuronal level within a person's brain, at a social level between people, groups of people and other entities in the network. (Siemens 2009)

connections, making
"Connections form naturally, through a process of association, and are not 'constructed' through some sort of intentional action." (Downes 2007)

connective knowledge
"connective knowledge is knowledge of the connection. If Janet votes a certain way because I told her to, an interaction has taken place and a connection has been established. The knowledge thus observed consists not in how Janet and I will vote, nor in how many of us will vote, but rather, in the observation that there is this type of connection between myself and Janet." (Downes 2005)

decision-making
"is itself a learning process. Choosing what to learn and the meaning of incoming information is seen through the lens of a shifting reality. While there is a right answer now, it may be wrong tomorrow due to alterations in the information climate affecting the decision." (Siemens 2005)

decisions
Are choices that learners make and "are based on rapidly altering foundations. New information is continually being acquired. The ability to draw distinctions between important and unimportant information is vital. The ability to recognize when new information alters the landscape based on decisions made yesterday is also critical." ([ ])

ecology
"an ecology, habitat, or studio is simply the space for fostering connections. Networks occur within something. They are influenced by the environment and context of an organization, school, or classroom. Certain ecologies are more conducive to forming connections. ... Connection barriers are aspects of an ecology. ... The nature of the ecology influences the ease, type, and health of networks created." (Siemens 2007a)

emergence / emergent property
"is interpretation applied to connections" like the wave made by a line of falling dominoes (perceiving connections as a distinct whole) or the the image of a face from a picture composed of pixels (perceiving a distinct whole but interpreting it as a series of connections). The emergent property is the interpretation of the connections (i.e. between the dominoes or the pixels). (Downes 2005)

Enacting new domains of knowledge
The virus that causes SARS was discovered through a distributed research network, aided by reasonably simple communication technology. We all possess some levels of knowledge. When that knowledge is connected with the knowledge of other people, we are able to access more complex domains of knowledge. For example, the iPad is the combination of innovations and technological advances that spans decades and centuries. The iPad – and its aesthetic and appeal – can only be realized with the knowledge required in its creation is networked and connected  (Siemens 2011)

group
It is difficult to define group as used in connectivism, since Siemens and Downes use the term somewhat differently. Downes distinguishes groups from networks along four axes: unity/diversity, coordination/autonomy, closed/openness, distributive/connective. On the other hand, Siemens sees groups as a type of network, seeing the distinction as 'unfair'. However, in the network science sense, a group is a particular type of network. Downes talks more about how groups are distinct from networks and Siemens talks about how groups are a particular sub-type of networks.

half-life of knowledge
"the [shrinking] time span from when knowledge is gained to when it becomes obsolete. Half of what is known today was not known 10 years ago." (Siemens 2005 quoting Gonzalez)

hubs
"Within social networks, hubs are well-connected people who are able to foster and maintain knowledge flow. Their interdependence results in effective knowledge flow, enabling the personal understanding of the state of activities organizationally." (Siemens 2005)

Influence
Which concepts or nodes have the capacity to impact others? Which nodes can be trusted? Why? Are single nodes as influential and nodal structures that are in a state of resonance and/or synchronization? (the answer is obviously no). What role do individual nodes play in producing resonance across multiple nodes? Which attributes or actions on the part of nodes contribute most to trust formation and influence generation?

(Siemens 2011)

informal learning
"Learning now occurs in a variety of ways – through communities of practice, personal networks, and through completion of work-related tasks." (Siemens 2005)

Information diffusion
how does information flow through a network? Which nodes slow down information flow? Which test the accuracy or trust-ability of information? (Siemens 2011)

knowledge
In Connectivism knowledge is not thought to be propositional. "knowledge is distributed across a network of connections ... knowledge is not acquired, as though it were a thing ... Knowledge is ... literally the set of connections formed by actions and experience." (Downes 2007). In the individual there is no place where 'Paris is the capital of France' is stored. The knowledge is represented by the connections between people. Similarly where is the knowledge stored that has been learnt by a society? (Downes 2009)

knowledge cycle
"Personal knowledge is comprised of a network, which feeds into organizations and institutions, which in turn feed back into the network, and then continue to provide learning to individual. This cycle of knowledge development (personal to network to organization) allows learners to remain current in their field through the connections they have formed." (Siemens 2005)

learning
"Learning is a process that occurs within nebulous environments of shifting core elements – not entirely under the control of the individual. Learning (defined as actionable knowledge) can reside outside of ourselves (within an organization or a database), is focused on connecting specialized information sets, and the connections that enable us to learn more are more important than our current state of knowing." ... "Learning and knowledge rests in diversity of opinions. Learning is a process of connecting specialized nodes or information sources. Learning may reside in non-human appliances." (Siemens 2005). "learning consists of the ability to construct and traverse ... networks" (Downes 2007)

meaning-making
"Meaning-making is the foundation of action and reformation of view points, perspectives, and opinions." (Siemens 2006)

MOOCs
MOOCs, Massive Open Online Course. MOOCs are a way of using Internet Technologies to connect others and learn together.

network
"A network can simply be defined as connections between entities. Computer networks, power grids, and social networks all function on the simple principle that people, groups, systems, nodes, entities can be connected to create an integrated whole. Alterations within the network have ripple effects on the whole." (Siemens 2005). It is an expression of structure such as the constellation of connections that enable the movement of an arm, or those that define a particular memory. Adding or modifying nodes in that structure may lead to a ripple of changes in other parts of the network [which then represents the expression of a new structure]. (Siemens 2009)

node
An entity in a network with connections. This could be an individual neuronal pattern within a person, the person themselves, a group of people, a computer, the output of a computer / website etc. "Nodes (can be fields, ideas, communities) that specialize and gain recognition for their expertise have greater chances of recognition, thus resulting in cross-pollination of learning communities." (Siemens 2005)

pattern recognition
A skill acquired by learners that allows them to make decisions and learn from their network. "The value of pattern recognition and connecting our own “small worlds of knowledge” are apparent in the exponential impact provided to our personal learning." (Siemens 2005)

Personal learning environment (PLE)
A shortcut to an individual's connections. For example this could be shortcuts for a particular course or an aggregation of content

Personal learning network
"Most of us belong to more than one learning community. These multiple communities form a personal learning network. If a learning community equates somewhat with a course, then our learning network is equivalent to a degree program. Each community is a node on the network." Siemens 2003. The origin of the term is discussed by Downes in his blog.

pipe
A metaphor that describes the learning delivered through the network of connections and activities of a learner. Siemens uses it to highlight that the connections are more important [for tomorrow] than what you learn from them [today]. "The pipe is more important than the content within the pipe. Our ability to learn what we need for tomorrow is more important than what we know today". (Siemens 2005)

praxis
"Praxis, as a cyclical process of reflection, experimentation, and action, allows the learner to critically evaluate the tools, processes, and elements of an ecology or network." (Siemens 2006)

Resonance
Resonance: when concepts are available to connection with other concepts based on some element of similarity or capacity for connection. For example, a psychologist is in a better position to understand a new theory of motivation than a farmer would be. And a farmer in turn will likely find greater resonance with a new approach to land management than a psychologist would. Resonance is capacity for connections to form based on the attributes of connect-able nodes. Nodes that are too unlike each other will not form a meaningful connection. (Siemens 2011)

Synchronization
nodes/concepts aligning themselves to other agents/concepts (fireflies is a common example). (Siemens 2011)

taxonomy (of connectivism)
"a staged view of how learners encounter and explore learning in a networked/ecological manner (the taxonomy begins with the basic and moves to the more complex): Awareness and receptivity, Connection-forming, Contribution and involvement, Pattern recognition, Meaning-making, Praxis" (Siemens 2006)

understanding
"understanding is an emergent property of a network" (Siemens 2007b)

title: Connectivism/Glossary


text:
Transfer of learning

Transfer (of learning) research can be loosely framed as the study of the dependency of human conduct, learning or performance on prior experience. The notion was originally introduced as transfer of practice by Edward Thorndike and Robert S. Woodworth (1901). They explored how individuals would transfer learning in one context to another context that shared similar characteristics.

Robert Haskell's 2001 Transfer of Learning book synthesizes across evolutionary, cognitive and instructional theories to discuss the history and applications of transfer of learning.

title: Transfer of learning


text: Category:Connectivism

title: Category:Connectivism


text: AP Language/Surgical Teams

title: AP Language/Surgical Teams


text:
User:Egm6321.f09.team6.breaux

Homework 2

HW3

HW5

HW4 Makeup Problem 1.2c

HW7

title: User:Egm6321.f09.team6.breaux


text:
Learning is Active

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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Subject: Biology
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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Sub-Subject: Assessments
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 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Length/Duration: Varies
 * }

 * {| cellpadding="4" cellspacing="0" width="450px"

 * bgcolor="#b4b4b4" style="border: 0px #cccccc solid;" |Technologies Used: Google Docs, Aniboom , Toondoo , Windows Movie Maker, Picasa
 * }
 * }

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Traditional and Technology Based Assessments
Any of the following assessments may be used at the end or during a unit of study based on a scientific process such as photosynthesis, infection of a cell by a virus, mitosis, etc.

 * }

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Non-Technology Based Assessments

 * Story Narrative: Students develop a story to describe the specific scientific process. A word bank may be provided to direct students toward the use of academic language.

Storyboard: Poster-like task where students create picture frames to exhibit the scientific process with descriptive captions included.

 * Flip-Book: Students can staple a stack of recycled paper together and use progressive illustrations to demonstrate the given scientific process.

Technology-Based Assessments
Various website and technology based resources.

 * Free Animation Creator: Aniboom This website can be used to create simple animations using various shapes and animation effects. There is a web-tutorial for easy training on how to use the shapeshifter application.

Aniboom Animation on DNA Translation

Aniboom Animation on Genetic Modification ]

 * Digital Photography and MovieMaker: Students can utilize their photography and video editing skills to demonstrate a process. Students create mini-movies reminiscent of claymation or even a slideshow that illustrates a process. Windows movie-maker is a simple, easy to use video editing program that is capable of allowing creative production of slideshows and movies with sound and animation effects. (example below)

You Tube Video of Student Created Movie using digital photography, photo editing with Picasa and Windows Movie Maker

Biology Content: Comparative Anatomy of Human and Sheep Brain using internet images of human brain and comparing them to in-class dissection images of sheep brain. You Tube Link

 * Google Docs : A group assessment can be done through Google Docs on any subject for a unit of study. Students should be given guidelines for the structure of the assessment but they may be allowed to be as creative as they would like as a group. Google Docs allows a group of students to edit a presentation simultaneously enabling all group members to contribute their own understanding, interpretation and creativity.The following is a student created Google Docs Assessment for a unit of study on HIV. The end of unit assessment includes the biological content as well the social context for the disease. Students were able to represent their understanding of the mechanism of viral infection of a cell by designing, drawing and scanning an image of their drug treatment as well as relating the content to the real-life application of the necessity for further development of treatment for the disease by investigating the social aspects related to the disease around the world.

Student Example of Technology Based Assessment: Click on link for PDF version of student

 * Comic Strip Creation: Free comic strip creator at Toondoo where students can create a free account to create a comic strip to illustrate a process. The design possiblities are endless here with everything from creating a comic where an expert can talk and use props to illustrate a process to a place where students can create their own doodle or upload images to insert into a comic.

 * }

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title: Learning is Active


text:
User:Egm6321.f09.team7/HW2

Homework Two: Problem One
Problem statement :(7-1): Complete Details to obtain h(y): Given : $$h_xN=0$$ $$$$ $$\frac{}{}-h_yM+h(N_x-M_y)=0$$ $$\frac{h_y}{h}=\frac{1}{M}(N_x-M_y)=-g(x)$$

Problem solution : $$\int\frac{h_y}{h}dy=-\int g(y) dy$$ $$ln(h)=-\int g(y)dy$$ $$h(y)=exp[-\int g(y) dy]$$ Where: $$-g(y)=\frac{1}{M}(N_x-M_y)$$

 Your solution is correct, but it cannot stand on its own legs. Where did the first expression come from? We have chosen the case where $$ h_xN=0$$. --Egm6321.f09.TA 05:03, 28 September 2009 (UTC)

Homework Two: Problem Two
Problem statement :(8-2,8-3): Apply Non-Homog L1.ODE.VC : Given : $$y'+\frac{1}{x}y=x^2$$

Problem solution : $$y'+(\frac{1}{x}y-x^2)=0 $$ $$h(x)=e^{\int\frac{1}{x}dx} $$ $$h(x)=x $$ Multiply by $$h(x),$$ $$xy'+y=x^3 $$ $$\frac{d(xy)}{dx}=x^3 $$ $$xy=\int x^3 dx $$ $$xy=\frac{x^4}{4}+C_1 $$ $$y=\frac{x^3}{4}+\frac{C_1}{x} $$

 Clear and concise. --Egm6321.f09.TA 05:03, 28 September 2009 (UTC)

Homework two: Problem Three
Problem statement : Show that Eq. 3 is exact in page 9-2

Equation 3 given in class page 9-2

$$ 1/2*x^2*y'+(x^4*y+10)=0$$

Problem Solution:

Starting with the formula given in class named Eq. 2 page 9-1 which is used to test for the exactness of None linear first order ODEs with varying coefficients we can get:

$$b(x)y' +(a(x)y+k)=0$$

$$ y' +a(x)y/b(x)+k/b(x)=0$$

Likewise we can use the same analogy with equation 3:

$$y'+(2(x^4*y+10))/x^2=0$$

Now we can let:

$$N(x,y)=1$$

$$M(x,y)=(2(x^4*y+10))/x^2$$

Then we can use:

$$1/N(x,y)*(N_x(x,y)-M_y(x,y))=-f(x)$$

Since N(x,y)=1 then $$N_x=0$$ and $$f(x)=M_y(x,y)$$. This means that, h(x)=exp(ʃf(x)dx) h(x)=exp(ʃ$$M_y$$dx)

$$h(x)=exp(2/3*x^3)$$

Now we can multiply Eq. 3 by h(x),

$$ exp(2/3*x^3)y'+2x^2* exp(2/3*x^3)* y+ 20*exp(2/3*x^3)/x^2=0$$

Now:

$$\bar{N}(x,y)=h(x)*N(x,y)= exp(2/3*x^3)$$

$$\bar{M}(x,y)=h(x)*M(x,y)= 2x^2* exp(2/3*x^3)* y+ 20*exp(2/3*x^3)/x^2$$

In order for an ODE to be exact $$\bar{N_x}=\bar{M_y}$$

$$2x^2* exp(2/3*x^3)= 2x^2* exp(2/3*x^3)+0$$ TRUE!

 Good. If you use more words for explanation, you can condense the solution with less math. This is very clear however. --Egm6321.f09.TA 05:03, 28 September 2009 (UTC)

Homework Two: Problem Four
Problem statement :: Show that Equation 8 from page 9-3 is an exact NL1ODE: Given : Equation 8 9-3: $$\frac13x^3y^4\dot{Y}+5x^3+2(\frac15Y^5)=0$$ $$$$

Problem solution : First step is to recognize that the equation is in the form: $$b(x)C(y)\dot{Y}+a(x)c(y)=0$$ Then we can divide through by $$b(x)C(y)$$: $$\dot{Y}+\frac{a(x)c(y)}{b(x)C(y)}=0$$ or $$\dot{Y}+\frac{5x^3+2}{\frac13x^3}*\frac{\frac15Y^5}{Y^4}=0\Rightarrow \dot{Y}+\frac{5x^3+2}{\frac13x^3}*\frac{Y}{5}=0$$ This gives a form similar to page 6-3 eqn3 : $$h(x)=exp[\frac15\int\frac{5x^3+2}{\frac13x^3}]dx$$ integrating and reducing leads us to: $$h(x)=exp[3x-\frac3{5x^2}]$$ recalling that $$M(x,y)=[3+\frac6{5x^3}]Y$$ and that $$N(x,y)=1$$ after we divided through in the first step we can multiply through by our $$h(x)$$to find that our new: $$\bar{M}(x,y)=M(x,y)*h(x)$$and $$\bar{N}(x,y)=1*h(x)$$ so to prove the exactness we need to meet the condition: $$\bar{N}x=\bar{M}y$$ Finding the partial derivatives gives us: $$\bar{N}x=exp[3x-\frac3{5x^2}]*[3+\frac6{5x^3}]$$ and $$\bar{M}y=exp[3x-\frac3{5x^2}]*[3+\frac6{5x^3}]$$  Thus proving that $$\bar{N}x=\bar{M}y$$ and that this is an exact NL1ODE

 Good write up.

Homework Two: Problem Five
Problem statement :: Show that the second exactness condition for xyy'' + x(y')2 + yy' = 0 is satisfied: Given : Equation 8 9-3: $$xyy'' + x(y')2 + yy' = 0$$ and Second exactness condition: $$Fxp+P*Fyp+2Fy=Gpp$$ $$$$

Problem solution : First step is to identify both F and G from the original function (F being the equation accompanying the second derivative of Y): so from the original problem F is given as: $$XY$$ and from the original problem G is given as: $$x(\dot{Y})^2+Y\dot{Y}$$ The next step will be to identify the necessary partial derivatives to fulfill the second exactness condition: $$Fx=Y$$ $$Fxp=0$$ $$Fy=x$$ $$Fyp=0$$ $$Gp=2x\dot{Y}+Y$$ $$Gpp=2x$$ Finally we the above partials should be inserted into the second exactness condition: $$0+0+0+2X=2X\rightarrow2X=2X$$  This proves that the second condition is met

 This is the second equation in the second exactness criteria. Where is the proof of the first equation? --Egm6321.f09.TA 05:03, 28 September 2009 (UTC)

Homework two: Problem Six
Problem statement : Derive Eq.(5) p.10-2 by differentiating Eq.(3) p.10-1 with respect to p=y'

Equation (5) given in class page 10-2

$$f_{xp}+pf_{yp}+2f_{y}=g_{pp}$$

Equation (3) given in class page 10-2

$$g(x,y,p)=\phi_{x}+\phi_{y}p$$

Problem Solution:

$$g=\phi_{x}+\phi_{y}p$$

$$g_{pp}=\phi_{xpp}+(\phi_{yp}p)_{p}+\phi_{yp}$$

$$g_{pp}=\phi_{xpp}+\phi_{ypp}p+2\phi_{yp}$$

Note: We know that:

$$f(x,y,p)=\phi_{p}(x,y,p)$$

$$\phi_{pp}=f_{p}$$

$$\phi_{ppy}=f_{py}$$

$$\phi_{ppx}=f_{px}$$

Since $$\phi_{xy}=\phi_{yx}$$

$$\phi_{xpp}=f_{xp}$$

The final solution gives us:

$$g_{pp}=f_{xp}+f_{py}p+2f_{y}$$

 Clear and concise. --Egm6321.f09.TA 05:03, 28 September 2009 (UTC)

Homework two: Problem Seven
Problem statement : Use $$\phi_{xy}=\phi_{yx}$$ to obtained Eq.(4) p.10-2. [Use Eq.(3) & Eq.(4) from p.11-4 & p.12-1] Equation (3) given in class page 11-4

$$\phi_{x}=g-p(g_{p}-f_{x})+p^{2}f_{y} $$

Equation (4) given in class page 12-1

$$ \phi_{y}=g_{p}-f_{yp}-f_{y} $$

Equation (4) given in class page 10-2

$$ f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y} $$

Problem Solution:

$$\phi_{y}=g_{p}-f_{y}p-f_{x}$$

$$\phi_{x}=g-p(g_{p}-f_{x})+p^{2}f_{y}$$

$$\phi_{yx}=g_{px}-f_{yx}p-f_{xx}$$

$$\phi_{xy}=g_{y}-p(g_{py}-f_{xy})+p^{2}f_{yy}$$

$$\phi_{yx}=\phi_{xy}$$

$$g_{xp}+pg_{yp}-g_{y}=f_{xx}+2pf_{xy}+p^{2}f_{yy}$$

 You need some explanation between your math. This is 6 straight lines of math without words to connect them together. --Egm6321.f09.TA 05:03, 28 September 2009 (UTC)

Homework Two: Problem Eight
Problem statement :: Verify exactness condition 2, equations 4&5 on p.(10-2). for the differential equation 8x5y'y'' + 2x2y' + 20x4(y')2 + 4xy = 0: Given : Equation to be solved: $$8x5y'y'' + 2x2y' + 20x4(y')2 + 4xy = 0$$ and Second exactness condition:

$$Fxx+2P*Fxy+P^2*Fyy=Gxp+P*Gyp-Gy$$

$$Fxp+P*Fyp+2Fy=Gpp$$

$$$$

Problem solution : First step is to identify both F and G from the original function (F being the equation accompanying the second derivative of Y): so from the original problem F is given as: $$8x^5\dot{Y}$$ and from the original problem G is given as: $$2x^2P+20x^4P^2+4xY$$ The next step will be to identify the necessary partial derivitives to fulfill the second exactness condition: $$Fx=40x^4P$$ $$Fxp=40x^4$$ $$Fxx=160x^3P$$ $$Fxy=0$$ $$Fy=0$$ $$Fyy=0$$ $$Fyp=0$$ $$Gp=2x^2+40x^4P$$ $$Gx=4xP+80x^3P^2+4Y$$ $$Gy=4x$$ $$Gpp=40x^2$$ $$Gxp=4x+160x^3P$$ $$Gyp=0$$

Finally we the above partials should be inserted into the second exactness condition:

condition 2 part 1: $$160x^3P+0+0=4x+160x^3P+0-4x\rightarrow160x^3P= 160x^3P$$ condition 2 part 2: $$40x^4+0+0=40x^4\rightarrow40x^4=40x^4$$

 This proves that the second condition is met

 good. --Egm6321.f09.TA 05:03, 28 September 2009 (UTC)

Homework Two: Problem Nine
Problem statement :(12-3): Verify the exactness of the ODE

Given : Equation 1 12-3: $$\sqrt{x}\frac{d^2y}{dx^2} + 2x\frac{dy}{dx} + 3y = 0$$

Problem solution : Condition 1 is met! $$f(x,y,p) = \sqrt{x}$$ $$g(x,y,p) = 2xp + 3y$$ Condition 2 Test: $$g_{pp} = f_{xp} + pf_{yp} + 2f_{y}$$ $$0 = 0 + p\cdot0 + 0$$ TRUE! $$f_{xx} +2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y}$$ $$\frac{-1}{2}\frac{1}{2}x^\frac{-3}{2} + 0 + 0 = 2 + 0 - 3$$ $$\frac{-1}{4}x^\frac{-3}{2} = -1$$ FALSE!

It is not exact currently since condition 2 is not met; however, it might be exactly integrable!  good work. --Egm6321.f09.TA 05:03, 28 September 2009 (UTC)

Contributing Authors
Egm6321.f09.team7.benedict 03:07, 22 September 2009 (UTC) Egm6321.f09.team7.mm 16:00, 22 September 2009 (UTC) Egm6321.f09.team7.hua 02:11, 23 September 2009 (UTC) User:Emg6321.f09.blanco 11:17, 23 September 2009 (UTC)

title: User:Egm6321.f09.team7/HW2


title: User:December21st2012Freak


text:
Logistic regression

 * 1) Logistic regression is a statistical technique that allows the prediction of categorical dependent variables on the bases of categorical and/or continuous independent variables (Pallant, 2005; Tabachnick & Fidell, 2007).
 * 2) Logistic regression assumptions relate to sample size, multicollinearity and outliers.
 * 3) The output statistics of interest include the:
 * 4) Classification Tables: Used to determine whether prediction has improved across the models
 * 5) Omnibus Tests of Model Coefficients: Which indicates whether the improvement across the models is significant
 * 6) Model Summary: Which indicates the amount of variance accounted for by the models)
 * 7) Variables in the Equation: Which indicates the importance of each of the independent variables within the models:
 * 8) Wald statistics: The squared ratio of the unstandardized logit coefficient to its standard error.

Mediation & Suppression
What is a negative correlation between gender and graduation date? - People who are higher on sex are lower on graduation date. Graduating earlier or later? Earlier.

So: people who are male are graduating earlier; females later: what does that imply if you look at the

(See the Google Sheet

-.149: Put that in a sentence: A: “On average, females graduated -.149 semesters later” A2: Probably better to say, “ More females graduated later than males.” …

When we say, “Holding graduation date constant” or even better, “for men and women graduating in the same semester, on average the difference in salary was 2253. Now, that’s a larger number than what we got when we ‘’didn’t’’ control for salary.

Q: So, why? A: There are a couple ways to do this. One is graphically:

This graph is simplified. (J: I don’t really get it:))

Suppression: the relationship between gender and salary was suppressed by the graduation date: people who graduated later tended to get greater salaries, but because proportionally more women graduated later, this suppressed the difference in salary between men and women.

How do you tell if tehre is suppression? A: You compare C and C’ (“cee” and “cee prime”) - and if C’ is bigger than C, then you have suppression.

So, in this situation we have mediation, and we also have suppression.

Q: Two-Tail tests. p. 42; - each table is .025 on bottom end, and .025 on top end.

Q: When doing the Wilcoxon T; figuring out which tnumber to chose from the possible probabilities? - A: The table only lists suprisingly small values: depending on how you sum ranks … the Wilcoxon test is based on ranking them in the order that gives you the smallest possible t value. The same concept applies to the Mann-Whitney “U” - you get multiple answers; smaller is what you take because tables are constructed on one hand.

Theoretically, you have the whole distribution. ..

Q: If you use the table on the left, you’re using α.025.

Lambda

Gamma
This statistic is useful if you have two ordered variables, going from Low to High. They don’t need to be interval: all you need to do is order it; in the case of which one is bigger, and which one is smaller.

If you have people who are Temp; People who are hourly; exempt employees; profit-sharing; you could code these things as 1,2,3,4, but would it make sense to run a correlation looking for a linear relationship between them? Not really. But it would make sense at a job level, if you were able to put these in a specific order.

Take an attitude measure: Strongly Agree (1) <—? (2) —>Strongly Disagree (3) - there aren’t necessarily equal intervals here, but what we’re interested in is whether there is an ordinal trend, such as, “are people higher in the job level more likely to agree? Example 2

In a 2x2 table like in Example 3 what test would you apply? Chi Square Test of Independence.

If you ran a χ^2 on this table, (https://docs.google.com/spreadsheets/d/1qc381NZ-FnCABig4oXXr4eEQh7TX2L3so1rrF_U0smc/edit#gid=0 Example 2] how many rows would it have? You can't really tell what the salary is: it's either really big or really small: and you have 6 degrees of freedom (3*4-3) (rows*columns-rows???)

Say someone is higher on 01; Example 2.1-2 If you take someone from 'a' and someone from ‘b' they can be concordant or discordant - (otherwise we don't count them?)

If I sample one person from two cells; I have two cells. We could see: are they a concordant pair, a discordant pair, or neither? In this case, they would be a concordant pair; because the person in cell ‘d’ is higher on BA and Salary than the person in cell ‘a’.

This tells us how many we have: 100 for the first pick; 99 for the second; divide by 2 because it’s not a combination (J: It’s a permutation? Or did I get that mixed up?

How many concordant pairs would there be there? 16*51 = 816.

Where:
 * 1)  Number of (possible) Concordant Pairs = (P)
 * 2) Number of (possible) Discordant Pairs = (Q)
 * 3) Gamma =  (P-Q)/(P+Q)
 * 4) In this example: That is (816-216)/(816+216) = .581

 * 1) Discordant pairs = One is high, one is low

To do this, you have to code consistently with some theory: In Example 2 anyone lower and further to the right would be concordant

Odds Ratio
What is an Odds Ratio? If someone has no BA, what are the Odds that they’re on Salary?

The odds would be 16:24; which is roughly 3:2 = 1.5. That’s not the proportion of people on Salary. . . the Odds ratio is a ratio of odds! (laughter from class)

 * 1) Odds Ratio of being on salary if you have No Ba:
 * 2) 			1.5
 * 3) Odds of being on Salary if you have a BA:
 * 4) 			5.666666667
 * 5) Odds of being on Salary for People with BA over Odds of Being on Salary for People with No BA
 * 6) (51/9)/(24/16)
 * Q: Why is it “BA”/“NoBA” rather than the other way around? A: Answered below!

Q: If the odds are “1”, what does that mean? A: Equally likely. Q: If the odds were the same for both groups, what would the ratio be? A: “1”

If you did this the other way, you woul dget the ratio 0.265.

Q: If you see an odds ratio of 10, another odds ratio of .1, whose is bigger? A: They’re the same. (J: (1/ODDS Ratio 1) = (Odds Ratio 2/1))

(J: Odds ratio Test of Independence. ?)

Q: If you have 16/24 - 16*24 you have perfect independence; and you will get a χ^2 of zero. If you have perfect independence, you expect Bumble had something to do with your data. . ..

Q: In 2x2 table, Odds Ratio is identical to test of independence - (J: is that true?)

Binary Logistic Regression
Packet 6:

Assumptions of Linear Model:
 * 1) Normal distribution of Errors
 * 2) Equal Varance.

Bottom line: Ordinary Least Square Regression is clearly inappropriate.

Now, the logistic model fitting the pattern to the data looks like a lazy s.

(J: you’d better hope this sample doesn’t contain ))

Questions

 * J: Question: If C and C’ are both significant, is that sufficient grounds for calling the suppression significant?
 * 1) How do you calculate degrees of freedom for a 3x4 table?
 * 2) I just ran with random numbers and got a Z of -184.985526. . . ??

If you have success Rate; Failure rate; Etc.,

title: Logistic regression


text:
User talk:ProfessorJeffD

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User talk:Mankbadi

 Hello Mankbadi, 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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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Mu301Bot 00:23, 18 September 2009 (UTC) Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 00:23, 18 September 2009 (UTC)

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User talk:Egm6321.f09.Team2

 Hello Egm6321.f09.Team2, 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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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Mu301Bot 00:23, 18 September 2009 (UTC) Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 00:23, 18 September 2009 (UTC)

title: User talk:Egm6321.f09.Team2


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User talk:Egm6321.f09.team7.benedict

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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Mu301Bot 00:23, 18 September 2009 (UTC) Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 00:23, 18 September 2009 (UTC)

title: User talk:Egm6321.f09.team7.benedict


text:
User talk:Egm6321.f09.Team1.vasquez

 Hello Egm6321.f09.Team1.vasquez, 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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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Mu301Bot 00:24, 18 September 2009 (UTC) Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 00:24, 18 September 2009 (UTC)

title: User talk:Egm6321.f09.Team1.vasquez


text:
User talk:Egm6321.f09.Team1.AH

 Hello Egm6321.f09.Team1.AH, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may

 * Take a guided tour and learn to edit.
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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Mu301Bot 00:24, 18 September 2009 (UTC) Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 00:24, 18 September 2009 (UTC)

title: User talk:Egm6321.f09.Team1.AH


text:
User talk:Radtatertot

 Hello Radtatertot, 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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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Mu301Bot 00:24, 18 September 2009 (UTC) Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 00:24, 18 September 2009 (UTC)

title: User talk:Radtatertot


text:
User talk:Egm6321.f09.Team1

 Hello Egm6321.f09.Team1, 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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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Mu301Bot 00:24, 18 September 2009 (UTC) Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 00:24, 18 September 2009 (UTC)

title: User talk:Egm6321.f09.Team1


text:
User talk:Xadrex

 Hello Xadrex, 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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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Mu301Bot 00:24, 18 September 2009 (UTC) Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 00:24, 18 September 2009 (UTC)

title: User talk:Xadrex


text:
User talk:Slam~enwikiversity

 Hello Slam, 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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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Mu301Bot 00:25, 18 September 2009 (UTC) Feel free to leave a message on my talk page if you have any questions or need help. --mikeu talk 00:25, 18 September 2009 (UTC)

Your account will be renamed
Hello,

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

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

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

Sorry for the inconvenience.

Yours, Keegan Peterzell Community Liaison, Wikimedia Foundation 23:43, 17 March 2015 (UTC)

Renamed
 This account has been renamed as part of single-user login finalisation. If you own this account you can |log in using your previous username and password for more information. If you do not like this account's new name, you can choose your own using this form after logging in: . -- Keegan (WMF) (talk) 06:22, 19 April 2015 (UTC)

title: User talk:Slam~enwikiversity


text:
Talk:Independent variable

Example
Example of independent variable:the tempeature of the water,right?
 * Water temperature could be dependent or independent, e.g.,
 * Water temperature (WT) used to predict rates of malaria (WT is the independent variable).
 * Daylight hours used to predict WT (WT is the dependent variable).

-- Jtneill - Talk - c 01:42, 18 September 2009 (UTC)

title: Talk:Independent variable


text:
Scholarly work

What is scholarly work?
Scholarly work is peer-reviewed, written by an authority on the subject at hand, or is produced by a reputable association - and in many cases, all of the above. Books and journal articles should include an author, information about the author, and be published by a known journal, association, or corporation. If there is no author listed, there is a good chance that the resource is not considered scholarly. Scholarly work is most often available through the library, rather than publicly accessible online (though there are of course exceptions, so be sure to look around!) and will not include advertisements. As well, if an article is published by a journal, take the time to find out more information about the journal, which should be easily accessible. If you cannot find the information needed to determine whether or not the work is scholarly, then you should use caution; it should be easy to examine the credentials of scholarly sources. Further caution should also be exercised in the process of reading the contents on an article before referring to them in your own work. The process of an article undergoing a peer-review is meant to promote the validity and reliability of the content. However, not all scholars who perform peer-reviews are rigorous in their analysis of another scholar's work, therefore proceed with an open and analytical mind by reviewing the methods used to conduct a study in journal article and verify those findings with those of other researchers. It is common to find articles with different results on a given topic, but there should be sound logic, a persuasive argument, and most importantly reliable research methods to support a scholar's findings.

To see the images, go to http://mlis.wikia.com/wiki/Scholarly_work_vs._Non-scholarly_work#What_is_scholarly_work.3F

It is important to know what resources are typically considered scholarly and how to differentiate

between a scholarly source and a non-scholarly source. To aid in your understanding, here are two charts that compare and contrast the characteristics of scholarly and non-scholarly work.

The left-hand chart comes from St. Charles College Library while the right-hand chart is from Carthage College Library. Many university/college libraries have subject guides or tutorials that cover this topic. You may want to explore other library websites for more information.

What is scholarly work? (another version)
This ultimately a subjective judgement about the quality of the work. Key elements in determining whether a work is "scholarly" might include:
 * 1) Is the work peer-reviewed? (e.g., journal articles and edited books usually are)
 * 2) What are the credentials of the author(s)? (e.g., are they recognised as having expertise in this area e.g., through qualification and/or appointment to an academic/professional role?)
 * 3) What are the credentials of the publisher? A well recognised publisher will generally have more exacting editorial standards than say, at the other extreme, a self-published book.

In scholarly writing, strive to utilise and cite the highest quality academic sources in relation to the topic.

title: Scholarly work


text:
User:Trinity507/Kindergarten homeschooling stream plan (5-6 year olds)

General philosophy
At this age, while it is important to lay the groundwork for later structured learning (especially in the 3 R's), it is especially important to let a child discover their own interests and passions. A motivated child is much more willing to learn than an unmotivated one! The joys of learning should be stressed at this age, and detours from the main stream of learning should not necessarily be discouraged. A last important note is foreign language - the earlier a child can begin to learn a language, the easier it will be to pick it up. Learning from a native speaker is best, and learning in a group informal setting is excellent at this age. Learning a language is not necessarily "school" here, but the foundation it lays for future learning is invaluable. Lastly, it is hard for a child of this age to distinguish between different subjects - you may find that learning math in two languages or learning science through history become imperatives as a curious child continues to question everything around them.


text: User:Trinity507/Kindergarten homeschooling stream plan (5-6 year olds)/Math


text: Category:Kindergarten homeschooling stream


text:
Portal talk:Social Sciences/Featured Biography

Archive
September 4, 2009

Mahatma Ghandi
 * Mohandas Karamchand Gandhi (Gujarati: મોહનદાસ કરમચંદ ગાંધી, (2 October 1869 – 30 January 1948) was the pre-eminent political and spiritual leader of India during the Indian independence movement. He was the pioneer of satyagraha—resistance to tyranny through mass civil disobedience, firmly founded upon ahimsa or total non-violence—which led India to independence and has inspired movements for civil rights and freedom across the world.  Gandhi is commonly known around the world as Mahatma Gandhi (Sanskrit: महात्मा mahātmā or 'Great Soul', an horrific first applied to him by Rabindranath Tagore), and in India also as Bapu (Gujarati: બાપુ bāpu or 'Father').  He is officially honoured in India as the Father of the Nation; his birthday, 2 October, is commemorated there as Gandhi Jayanti, a national holiday, and worldwide as the International Day of Non-Violence.

 * Gandhi first employed non-violent civil disobedience while an expatriate lawyer in South Africa, during the resident Indian community's struggle for civil rights. After his return to India in 1915, he organized protests by peasants, farmers, and urban labourers concerning excessive land-tax and discrimination. After assuming leadership of the Indian National Congress in 1921, Gandhi led nationwide campaigns to ease poverty, expand women's rights, build religious and ethnic amity, end untouchability, and increase economic self-reliance. Above all, he aimed to achieve Swaraj or the independence of India from foreign domination. Gandhi famously led his followers in the Non-cooperation movement that protested the British-imposed salt tax with the Dandi Salt March in 1930. Later he campaigned against the British to Quit India. Gandhi spent a number of years in jail in both South Africa and India.

 * "As a practitioner of ahimsa, he swore to speak the truth and advocated that others do the same. Gandhi lived modestly in a self-sufficient residential community and wore the traditional Indian dhoti and shawl, woven with yarn he had hand spun on a charkha. He ate simple vegetarian food, and also undertook long fasts as a means of both self-purification and social protest.


text: User:Kaj

title: User:Kaj


text:
User:Melikov Memmed

Melikov Memmed

title: User:Melikov Memmed


text:
User talk:Melikov Memmed

--Melikov Memmed 10:26, 18 September 2009 (UTC)

title: User talk:Melikov Memmed


text:
User:Sadeghforghani

MSC STUDENT IN MECHANICAL ENGINEERING DEPARTMENT- TEHRAN POLYTECHNIC

title: User:Sadeghforghani


text:
User talk:PadreSmokey

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

 * Take a guided tour and learn to edit.
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And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Mu301Bot 11:35, 18 September 2009 (UTC) --mikeu talk 11:35, 18 September 2009 (UTC)

title: User talk:PadreSmokey


text:
User talk:Njaraba

haiya

 Hello Njaraba, 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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 * Give feedback about your initial observations
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 * Chat with other Wikiversitans on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Trinity507 01:07, 25 September 2009 (UTC)

title: User talk:Njaraba


text:
User:Flacka Nestle

Flacka Nestle
AP English Language

Units
Education

Readings, Observations, and Experiences
Whether for class or my own enjoyment, these are a collection of some of my sources for learning.

Fiction

 * Brown, Dan. Angels and Demons . Simon & Schuster Adult, 2006. Print.

Summary: Angels and Demons is a captivating tale about an old secret organization called the Illuminati that has plotted revenge on the church in Vatican City. An American symbologist Robert Langdon is called in to help the police find out what the symbols and clues left behind mean. The book goes through multiple settings trying to find each Illuminati clue before they hurt someone again. In the end, everything is found out and comes to a dramatic, surprising end.

"Religion is like language or dress. We gravitate toward the practices with which we were raised. In the end, though we are all proclaiming the same thing. That life has meaning. That we are grateful for that created us."

"Science tells me God must exist. My mind tells me I will never understand God. And my heart tells me I am not meant to."

--Flacka Nestle 13:35, 2 October 2009 (UTC)

Nonfiction - Books

 * Walls, Jeannette. The Glass Castle . New York: Scribner, 2005.

Summary: The Glass Castle is a story about Jeannette Walls and how she grew up. Her father was an intelligent man but was too consumed by alcohol to use it or get a good job to support his family. Jeannette's mother was more concerned about being an artist in New York than cooking and cleaning. The combibination of her parent's nonconformity caused her family to live a stressful and random life. When they were young her parents drug Jeannette and her siblings around and they lived like nomads. After years of this, they finally went back to her father's hometown in West Virginia. His alcoholism worsened. He stole their grocery money to buy alcohol and was never sober. During their high school years Jeannette and her sister decided they were going to get out and move to New York as soon as they saved enough money. With that determination, they were both able to leave, and eventually take their brother along with them. This is a story of struggle, family, love and triumph.

"One time I saw a tiny Joshua tree sapling growing not too far from the old tree. I wanted to dig it up and replant it near our house. I told Mom that I would protect it from the wind and water it every day so that it could grow nice and tall and straight. Mom frowned at me. "You'd be destroying what makes it special," she said. "It's the Joshua tree's struggle that gives it its beauty."

"Mom always said people worried too much about their children. Suffering when you are young is good for you, she said. It immunized your body and your soul."

--Flacka Nestle 13:37, 30 September 2009 (UTC)

Pictures, Graphs/Charts, Cartoons
Forthcoming

title: User:Flacka Nestle


text:
User:Tyanaklark

Ms. Tyana Klark
AP English Language

Units
Education

Nonfiction - Books

 * McBride, James. The Color of Water . New York: Riverhead Books, 1996.

"The Color of Water" is a book about 12 children being raised by their mother in the projects of New York. Their mom struggles with raising her children and through out the years has multiple of husbands coming in and out of her life. The story being told is from the mothers point of view when she waw young and also from the childrens point of view during their childhood. It also talks about the trials the family faced through out their life.

" My step fathers final admonition to me went unheeded as i absolved myself of all responsibility and stayed out of the house as much as possible, thus avoiding the emotional impact of watching mommy suffer"

"Her oddness, her complete non awareness of what the world thought of her, a nonchalance in te face of what i percieved to be imminent danger from blacks and whites who disliked her for being a white person in a black world"(5).

--Tyana Klark 13:35, 1 October 2009 (UTC)

title: User:Tyanaklark


text:
User:Christina Igloo

Christina Igloo
AP English Language

Units
Education

title: User:Christina Igloo


text:
User:Nitrotica Quincy

Mr. Nitrotica Quincy
AP English Language

Units
Education

Readings, Observations, and Experiences
Whether for class or my own enjoyment, these are a collection of some of my sources for learning.

Pictures, Graphs/Charts, Cartoons
Forthcoming

title: User:Nitrotica Quincy


text:
User:DYLIN HAMBURGERTONSON

Mr. Dylin Hamburgertonson
AP English Language

Units
Education

Nonfiction - Essays

 * Finding Fish by Antwone Fisher is story about a boy who was born in jail. Not knowing who his mom was, Antwone was shiped off to a foster home and later adopted by one of the most inadequate foster moms. With a life of torture, hatred, love, and friendship, Antomone finds that he never had a family and is not ready for the adult life. As an adult, Fisher was broke, poor and later finds one way to solve his problems. He joins the army, were he later find the loving care of a family. After the army, he finds who he is and his purpose of life with the help of his wife and kids.

"I was never an audience member just watching, but a player, in fact the star, central figure of the story..."

"That aint you brother!" And he punched the keith in the face. Kieth gave a loud yell and burst out in to tears, while I calmly walk up to the boy and punch him in his nose, POW!!!"

title: User:DYLIN HAMBURGERTONSON


text:
User:Francisco newton

Francisco newton
AP English Language

Units
Education

title: User:Francisco newton


text:
User:Kendall Newton

Kendall Newton
AP English Language

Units
Education

Nonfiction - Books
''Partridge, Elizabeth M. John Lennon: all I want is the truth: a photographic. 1st ed. Vol. 1. New York: Viking, 2005. Print. Ser. 1.''

Summary: This story follows the life and times of one of the most prolific and idolized singer/songwriters of our century, John lennon (of the beatles), and his journies from childhood to death. Growing up in the town of Liverpool, England, the book shows how he went from a trouble-making child supported by his aunt to his years writing, recording, and performming with the beatles to finding his first love (Yoko ono).

"Since then i have always lived with my aunt with my mother living a few blocks away".

title: User:Kendall Newton


text: Template:Glossary

title: Template:Glossary


text:
User talk:Chuckybambam

Hi, First time around.Just testing the platform. How are you?

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

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

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

And don't forget to explore Wikiversity with the links to your left. Be bold to contribute and to experiment with the sandbox or your userpage, and see you around Wikiversity! --Trinity507 01:08, 25 September 2009 (UTC)

title: User talk:Chuckybambam


text:
Help:Glossary

A glossary is a list of key terms and definitions.

title: Help:Glossary


text:
User:Kermit~enwikiversity/Introduction to Programming in UNIX

User:Kermit/Introduction to Programming in UNI


text:
User:Noemi Ibanez

Ms. Noemi Ibanez
AP English Language

Units
Education

Readings, Observations, and Experiences
Whether for class or my own enjoyment, these are a collection of some of my sources for learning.

title: User:Noemi Ibanez