text:
University of Florida/Egm4313/s12.team12.R2

Given
$$ \lambda_1=-2, \lambda_2 = +5 $$

$$ y(0)=1, y'(0)=0 $$

For part one find the non-homogeneous L2-ODE-CC in standard form and the solution in terms of the initial conditions and the general excitation $$ r(x). $$

Consider no excitation and plot the solution

$$ r(x)=0 $$

For part two you need to generate 3 non-standard (and non-homogeneous) L2-ODE-CC that admit the two values $$ \lambda_1=-2, \lambda_2 = +5\ $$ as the two roots of the corresponding characteristic equation.

Solution
Find characteristic equation

$$ (\lambda-(-2))(\lambda-(5))=0 $$

$$ \lambda^2 - 3\lambda - 10 = 0 $$

The homogeneous solution takes the form

$$ y_h= C_1e^-2t + C_2e^5t $$

Take the derivative and subitute the intial conditions into the equations

$$ y'_h = -2C_1e^-2t + 5C_2e^5t $$

$$ y(0) >> 1 = C_1 + C_2 $$

$$ y'(0) >> 0 = -2C_1 + 5C_2 $$

$$ C_2 = 1-C_1 $$

$$ 0 = -2C_1 + 5 - 5C_1 $$

$$ -5 = -7C_1 $$

$$ C_1 = 5/7 $$

$$ C_2 = 2/7 $$

The Solution to the non homogeneous L2-ODE

$$ y(t) = 5/7C_1e^-2t + 2/7C_2e^5t + y_p $$

Part two:

Characteristic Equation

$$ (\lambda-3)(\lambda-(-6))=0 $$ $$ \lambda^2 + 3\lambda - 18 $$

Multiply equation by a variable, i chose m = 2,4,6 respectfully

$$ 2\lambda^2 + 6\lambda - 36 $$ $$ 4\lambda^2 + 12\lambda - 72 $$ $$ 6\lambda^2 + 18\lambda - 108 $$

Author
Patrick Greivell, 15:45 8 February 2012 (UTC)

Given
Given is an Initial Value Problem (IVP) to solve a second-order Ordinary Differential Equation (ODE) by finding a particular solution to the equation and its subsequent plot.

Initial Values:

$$y(0) = 1$$

$$y'(0) = 0$$

Differential Equation:

$$y'' - 10y' + 25y = r(x)$$

Assume that no excitation exists, i.e.

$$r(x) = 0$$

Therefore the ODE to be solved becomes:

$$y'' - 10y' + 25y = 0$$

Solution
With the given conditions, a homogenous linear second order ODE with constant coefficients is given to solve. In order to solve this ODE, a basis of solutions must be formed that meet the given conditions. Our goal is to find a solution to that will fit to the a general solution in the form of:

$$y = e^{\lambda x}$$

To do this, a quadratic equation in the form of the following is used to determine the method of how the basis of solutions will be formed.

$$\lambda^2 + a \lambda + b = 0$$

In order to do find the general solution, and ultimately the particular solution for this IVP, we will need to solve for λ using the characteristic equation above. Given that this is a quadratic equation, we can solve for λ by finding the roots of the equation. We know the variables a and b from our given ODE.

Let:

a = -10 b = 25

Plugging this into the characteristic equation, we yield the following:

$$\lambda^2 - 10 \lambda + 25 = 0$$

Now, instead of solving directly for λ, we can determine exactly how many and what kind of roots the equation will have based on the following cases given the value of the discriminant:

The discriminant is given by the following equation:

$$a^2 - 4b$$

If we evaluate the discriminant using the values of a and b from above, we can evaluate the discriminant to be:

$$(-10)^2 - 4(25) = 0$$

Therefore, we are in Case 2, where the value of our roots will be one real double root. Therefore, we can determine the value of λ by the following equation and by substituting the values into the equation subsequently.

From this, we can use the value of λ to form one equation of the basis as follows:

In order to form a basis, a second solution is needed. This is done by performing a reduction of order, yielding a first order differential equation. This process will produce the following general solution*:

Substituting the value for λ that was found:

In order to solve for the variables c1 and c2, we will use the initial conditions that were given in the problem statement. Therefore, we have two unknowns. We can solve for both of these variables by writing two equations, since one of the initial values was given as a value for the first derivative of the solution at a given point. The derivative of the general solution obtained (shown directly above) is obtained:

We can determine the value of c2 in terms of c1 by simply substituting in the known initial values as follows:

The second equation to solve for the unknown variable c1 can be determined by simply using the general solution equation and using the other initial value in the problem statement.

Therefore, we now know both of the unknown variables. Therefore the particular solution of the given ODE is:

It's subsequent plot is therefore:

Author
--Egm4313.s12.team12.stewart 04:35, 6 February 2012 (UTC)
 * Proof for form of general solution found in: Kreyszig, Erwin, Herbert Kreyszig, and E. J. Norminton. Advanced Engineering Mathematics. Hoboken, NJ: Wiley, 2011. Print.

 * Plot for particular solution plotted using Wolfram Alpha

Statement
K 2011 p.59 pbs. 3-4 Find a general solution. Check your answer by substitution.

Solution
Problem 3: Consider an ODE of the form: $$ \displaystyle {y}''+6{y}' +8.96y = 0$$ First we notice that the ODE is homogenous, linear, and has constant coefficients. Therefore, we find the auxiliary equation, which is of the form: $$ \displaystyle \lambda^2 + a\lambda + b = 0$$ The value of the discriminant is: $$ \displaystyle 6^2 - 4(8.96) = 0.16 > 0$$ We plug in a = 6 and b = 8.96 into the auxillary equation to find the roots. $$ \displaystyle \lambda ^{2} + 6\lambda + 8.96 = 0, \lambda _{1} = -3.2, \lambda _{2} = -2.8$$ The general solution will be of the form: $$ \displaystyle y=c_{1}e^{\lambda _{1}x}+c_{2}e^{\lambda _{2}x}$$ Therefore the solution will be $$ \displaystyle y=c_{1}e^{-3.2x} + c_{2}e^{-2.8x}$$

Problem 4: Consider an ODE of the form: $$ \displaystyle {y}'' + 4y' + (\pi ^{2}+4)y = 0$$ First we notice that the ODE is homogenous, linear, and has constant coefficients. Therefore, we find the auxiliary equation, which is of the form: $$ \displaystyle \lambda^2 + a\lambda + b = 0 $$ Looking at the coefficients we obtain: $$ \displaystyle \lambda^{2} + 4\lambda + (\pi^{2} + 4) = 0 $$ Next we obtain the roots of this equation with the quadratic formula: $$ \displaystyle \ \frac{-4 \pm \sqrt{(-4)^2 - 4(1)(\pi^2 + 4)}}{2(1)} $$ This can be simplified to: $$ \displaystyle -2\pm\frac{\sqrt{16 - 4\pi^2 - 16}}{2} $$ $$ \displaystyle -2\pm\frac{\sqrt{-4\pi^2}}{2} $$ We notice that by looking inside the radical, we will be obtaining complex roots. Continuing with simplification we get: $$ \displaystyle -2\pm{i\pi} $$ The general solution to the case of complex roots is: $$ \displaystyle y = e^{\frac{-ax}{2}}(Acos\omega{x} + Bsin\omega{x}) $$ Where complex roots are of the form: $$ \displaystyle -\frac{1}{2}a\pm{i\omega} $$ Knowing $$ \displaystyle a = 4 $$ and $$ \displaystyle \omega = \pi $$, we get the final general solution: $$ \displaystyle y = e^{-2x}(Acos\pi{x} + Bsin\pi{x}) $$

Author
Egm4313.s12.team12.sutcliffe 18:42, 8 February 2012 (UTC)

Statement
K 2011 p.59 pbs. 5-6 Find a general solution. Check your answer by substitution.

Solution
Problem 5: Consider an ODE of the form: $$ \displaystyle {y}'' + 2\pi y' + \pi ^{2}y = 0$$ This ODE is linear, and has constant coefficients. Therefore, its general solution is of the form: $$ \displaystyle {y}=e^{\lambda x} $$ Therefore: $$ \displaystyle {y}'=\lambda e^{\lambda x} $$ And: $$ \displaystyle {y}''={\lambda}^2 e^{\lambda x} $$ The ODE can be rearranged so that: $$ \displaystyle e^{\lambda x}(\lambda^2 + 2\pi \lambda + \pi^2)=0 $$ And since: $$ \displaystyle e^{\lambda x} \neq 0 $$ Then: $$ \displaystyle \lambda^2 + 2\pi \lambda + \pi^2=0 $$ $$ \displaystyle (\lambda + \pi)=0 $$ $$ \displaystyle \lambda =- \pi $$ Since: $$ \displaystyle \lambda_{1} = \lambda_{2} $$ The general solution then becomes: $$ \displaystyle y = c_{1}e^{\lambda x} + c_{2}xe^{\lambda x} $$ $$ \displaystyle y = e^{\lambda x} (c_{1} + c_{2}x) $$ Therefore: $$ \displaystyle y = e^{-\pi x} (c_{1} + c_{2}x) $$

Problem 6: Consider an ODE of the form: $$ \displaystyle 10{y}''-32{y}' +25.66y = 0$$ Divide through by 10 to get the equation into standard form where the leading coefficient is equal to 1. $$ \displaystyle {y}''-3.2{y}' +2.56y = 0$$ First we notice that the ODE is homogenous, linear, and has constant coefficients. Therefore, we find the auxiliary equation, which is of the form: $$ \displaystyle \lambda^2 + a\lambda + b = 0$$ The characteristic equation then becomes $$ \displaystyle \lambda^2 -3.2\lambda + 2.56 = 0$$ The value of the discriminant is: $$ \displaystyle (-3.2)^2 - 4(2.56) = 0$$ The 0 indicates that there is a real double root. The general solution will be of the form: $$ \displaystyle y=e^{-\frac{a}{2}x}(C_1+C_2x)$$ Where -a/2 = -(-3.2)/2 = 1.6 Therefore the solution will be $$ \displaystyle y=e^{1.6x}(C_1+C_2x)$$

Author
Egm4313.s12.team12.sutcliffe 18:43, 8 February 2012 (UTC)

Given
K 2011 p.59 pbs. 16-17 Find an ODE $$ \displaystyle y'' + ay' + by = 0 $$ for the given basis: $$ \displaystyle 16. e^{2.6x}, e^{-4.3x} $$ $$ \displaystyle 17. e^{-\sqrt{5}x}, xe^{-\sqrt{5}x} $$

Solution
Problem 16. Distinct real roots are of the form: $$ \displaystyle e^{\lambda_1x}, e^{\lambda_2x}$$ Therefore, $$ \displaystyle \lambda_1 = 2.6, \lambda_2 = -4.3 $$ The characteristic equation is: $$ \displaystyle (\lambda - \lambda_1)(\lambda - \lambda_2) = 0 $$ Therefore we have: $$ \displaystyle (\lambda - 2.6)(\lambda - (-4.3)) = 0 $$ Through simplification and distribution we get: $$ \displaystyle \lambda^{2} + 1.7\lambda - 11.18 = 0 $$ This is the auxiliary equation to the original linear, homogeneous ODE. Therefore our final solution is: $$ \displaystyle y'' + 1.7y' - 11.18y = 0 $$ Problem 17. Double real roots are of the form: $$ \displaystyle e^{-ax/2}, xe^{-ax/2} $$ Therefore, $$ \displaystyle \lambda = -\sqrt{5} $$ The characteristic equation is: $$ \displaystyle (\lambda - (-\sqrt{5}))^2 = \lambda^2 +2\sqrt{5}\lambda + 5 = 0 $$ This is the auxiliary equation to the original linear, homogeneous ODE. Therefore our final solution is: $$ \displaystyle y'' + 2\sqrt{5}y' + 5y = 0 $$

Author
Egm4313.s12.team12.stadick 24:00, 7 February 2012 (UTC)

Given

Find the parameters k, c, m of the the spring-dotplot-mass system with a double root $$ \lambda = -3 $$

Solution
The equation of motion for this system is

$$ m(y''_k + \frac{k}{c} y'_k)+ky_k = f(t) $$

or

$$ my''_k + m\frac{k}{c}y'_k + ky_k = f(t) $$

The homogeneous equation is

$$ Ay''_h + By'_h + Cy_h = 0 $$

Characteristic Equation of the double root

$$ (\lambda-(-3))^2 = \lambda^2 + 6\lambda + 9 $$

Then you find the homogeneous L2-ODE-CC for the characteristic eq.

$$ y'' + 6y' + 9y = 0 $$

Now compare the coeffients with the motion equation

$$ m = 1 $$

$$ k = 9 $$

$$ \frac{mk}{c} = 6 $$

$$ c = \frac{1*9}{6} $$

$$ c = \frac{3}{2} $$

Author
Patrick Greivell, 16:16 8 February 2012(UTC)

Given
Develop the Maclaurin Series (Taylor Series @ t = 0) for

a)  $$ e^t $$

b)  $$ cos(t) $$

c)  $$ sin(t) $$

Solution
The General form of a Taylor Series is

$$ \sum_{n=0}^{\infty} \frac{f^n(a)}{n!} (t-a)^n $$

with the expansion at point a being

$$ f(a) + \frac{f'(a)}{1!}(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f^3(a)}{3!}(x - a)^3 + ...$$

The Maclaurin Series (Taylor Series @ t=0) would be

a)

$$ e^t = \sum_{n=0}^{\infty} \frac{t^n}{n!} $$

which would equal

or

$$ 1 + t + \frac{x^2}{2} + \frac{x^3}{6} + ... $$ for all x

b)

$$ cos(t) = \sum_{n=0}^{\infty} \frac{(-1)^nt^{2n}}{(2n)!} $$

which would equal

$$ cos(0) + \frac{-sin(0)}{1!}(x) + \frac{-cos(0)}{2!}(x)^2 + \frac{sin(0)}{3!}(x)^3 + \frac{cos(0)}{4!}(x)^4 + ... $$

or

$$ 1 + 0 + \frac{x^2}{2} + 0 + \frac{x^4}{24} + ... $$ for all x

c)

$$ sin(t) = \sum_{n=0}^{\infty} \frac{(-1)^nt^{2n + 1}}{(2n + 1)!} $$

which would equal

$$ sin(0) + \frac{cos(0)}{1!}(x) + \frac{-sin(0)}{2!}(x)^2 + \frac{-cos(0)}{3!}(x)^3 + \frac{sin(0)}{4!}(x)^4 + ... $$

or

$$ 0 + x + 0 + \frac{x^3}{6} + 0 + ... $$ for all x

Author
Cpettigrew 21:15, 5 February 2012 (UTC)

Given
#8 P. 59 Given $$ \displaystyle y'' + y' + 3.25y = 0 $$ find a general solution and check your answer by substitution.

Solution
$$ \displaystyle 1y'' + 1y' + 3.25y = 0 $$ $$ \displaystyle a^2 - 4b = (1)^2 - (4)(1) = 1 - 4 = -3 $$ Since $$ \displaystyle a^2 - 4b < 0 $$ it is a case where the solutions are complex conjugate roots. A general solution for an ODE of this case can be given by the equation $$ \displaystyle y = e^{-ax/2} (A\cos \omega x + B\sin \omega x) $$ 1st Step The 1st step to finding the solutions for this ODE is to substitute the variable $$ \displaystyle \lambda $$ in for y and its derivatives. $$ \displaystyle \lambda = \frac{dy}{dx} $$ and $$ \displaystyle \lambda^2 = \frac{d^2y}{dx^2} $$ So $$ \displaystyle y'' + y' + 3.25y = 0 $$ becomes $$ \displaystyle \lambda^2 + \lambda + 3.25 = 0 $$ 2nd Step Now that the ODE has been made into a quadratic equation of form $$ \displaystyle ax^2 + bx + c = 0 $$ we can use the quadratic formula $$ \displaystyle x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Inputting the coefficients from the equation and solving for $$ \displaystyle \lambda $$ you get $$ \displaystyle \lambda = \frac {-1 \pm i \sqrt{12}}{2}$$ or $$ \displaystyle \lambda_1 = \frac {-1 + i \sqrt{12}}{2} $$ and $$ \displaystyle \lambda_2 = \frac {-1 - i \sqrt{12}}{2} $$ where $$ \displaystyle \lambda_1 = - \frac{1}{2} a + i \omega $$ and $$ \displaystyle \lambda_2 = - \frac{1}{2} a - i \omega $$ so $$\displaystyle \omega = \frac{\sqrt{12}}{2} $$ or $$ \displaystyle \omega = \sqrt{3} $$ 3rd Step Now if you plug all of this in to the general solution you will get $$ \displaystyle y = e^{- \frac {1}{2} x}(A \cos \sqrt{3} x + B \sin \sqrt{3} x) $$ Checking Solution with Substitution Using the general solution and taking the first and second derivatives of this general solution we can plug that in to the original equation and see if we in fact get zero. If so, then the general solution that was found is a solution for the ODE. $$ \displaystyle y = e^{- \frac{1}{2} x}(A \cos \sqrt{3} x + B \sin \sqrt{3} x) $$ The first derivative is $$ \displaystyle y' = - \frac{1}{2} e^{- \frac{1}{2} x}(A \cos \sqrt{3} x + B \sin \sqrt{3} x) + e^{- \frac{1}{2} x}(- \sqrt{3} A \sin \sqrt{3} x + \sqrt{3} B \cos \sqrt{3} x)$$ The second derivative is $$ \displaystyle y'' = \frac{1}{4} e^{- \frac{1}{2} x}(A \cos \sqrt{3} x + B \sin \sqrt{3} x) - \frac{1}{2} e^{- \frac{1}{2} x}(- \sqrt{3} A \sin \sqrt{3} x + \sqrt{3} B \cos \sqrt{3} x) - \frac{1}{2} e^{- \frac{1}{2} x}(- \sqrt{3} A \sin \sqrt{3} x + \sqrt{3} B \cos \sqrt{3} x)$$ $$ \displaystyle + e^{- \frac{1}{2} x}(-3 A \cos \sqrt{3} x -3 B \sin \sqrt{3} x)$$ Plugging all of this into the original equation you get $$ \displaystyle \frac{1}{4} e^{- \frac{1}{2} x}(A \cos \sqrt{3} x + B \sin \sqrt{3} x) - \frac{1}{2} e^{- \frac{1}{2} x}(- \sqrt{3} A \sin \sqrt{3} x + \sqrt{3} B \cos \sqrt{3} x) - \frac{1}{2} e^{- \frac{1}{2} x}(- \sqrt{3} A \sin \sqrt{3} x + \sqrt{3} B \cos \sqrt{3} x) $$ $$ \displaystyle + e^{- \frac{1}{2} x}(-3 A \cos \sqrt{3} x -3 B \sin \sqrt{3} x) - \frac{1}{2} e^{- \frac{1}{2} x}(A \cos \sqrt{3} x + B \sin \sqrt{3} x) + e^{- \frac{1}{2} x} (- \sqrt{3} A \sin \sqrt{3} x + \sqrt{3} B \cos \sqrt{3} x) $$ $$ \displaystyle + 3.25e^{- \frac{1}{2} x}(A \cos \sqrt{3} x + B \sin \sqrt{3} x) = 0$$ Simplifying all of these terms you end up getting $$ \displaystyle e^{- \frac{1}{2} x} (0) = 0$$

Author
--Egm4313.s12.team12.anders. 4:11, 7 February 2012 (UTC)

Given
#15 P.59 Given $$ \displaystyle y'' + 0.54y' + (0.0729 + \pi)y = 0 $$ find a general solution and check your answer by substitution.

Solution
$$ \displaystyle 1y'' + 0.54y' + (0.0729 + \pi)y = 0 $$ $$ \displaystyle a^2 - 4b = (1)^2 - (4)(0.54) = 1 - 2.16 = -1.16 $$ Since $$ \displaystyle a^2 - 4b < 0 $$ it is a case where the solutions are complex conjugate roots. A general solution for an ODE of this case can be given by the equation $$ \displaystyle y = e^{-ax/2} (A\cos \omega x + B\sin \omega x) $$ 1st Step The 1st step to finding the solutions for this ODE is to substitute the variable $$ \displaystyle \lambda $$ in for y and its derivatives. $$ \displaystyle \lambda = \frac{dy}{dx} $$ and $$ \displaystyle \lambda^2 = \frac{d^2y}{dx^2} $$ So $$ \displaystyle y'' + 0.54y' + (0.0729 + \pi)y = 0 $$ becomes $$ \displaystyle \lambda^2 + 0.54 \lambda + (0.0729 + \pi) = 0 $$ 2nd Step Now that the ODE has been made into a quadratic equation of form $$ \displaystyle ax^2 + bx + c = 0 $$ we can use the quadratic formula $$ \displaystyle x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Inputting the coefficients from the equation and solving for $$ \displaystyle \lambda $$ you get $$ \displaystyle \lambda = \frac {-1 \pm i \sqrt{12.56637061}}{2}$$ or $$ \displaystyle \lambda_1 = \frac {-1 + i \sqrt{12.56637061}}{2} $$ and $$ \displaystyle \lambda_2 = \frac {-1 - i \sqrt{12.56637061}}{2} $$ where $$ \displaystyle \lambda_1 = - \frac{1}{2} a + i \omega $$ and $$ \displaystyle \lambda_2 = - \frac{1}{2} a - i \omega $$ so $$\displaystyle \omega = \frac{\sqrt{12.56637061}}{2} $$ or $$ \displaystyle \omega = \sqrt{\pi} $$ 3rd Step Now if you plug all of this in to the general solution you will get $$ \displaystyle y = e^{-0.27 x}(A \cos \sqrt{\pi} x + B \sin \sqrt{\pi} x) $$ Checking Solution with Substitution Using the general solution and taking the first and second derivatives of this general solution we can plug that in to the original equation and see if we in fact get zero. If so, then the general solution that was found is a solution for the ODE. $$ \displaystyle y = e^{- 0.27 x}(A \cos \sqrt{\pi} x + B \sin \sqrt{\pi} x) $$ The first derivative is $$ \displaystyle y' = - 0.27 e^{- 0.27 x}(A \cos \sqrt{\pi} x + B \sin \sqrt{\pi} x) + e^{- 0.27 x}(- \sqrt{\pi} A \sin \sqrt{\pi} x + \sqrt{\pi} B \cos \sqrt{\pi} x)$$ The second derivative is $$ \displaystyle y'' = 0.0729 e^{- 0.27 x}(A \cos \sqrt{\pi} x + B \sin \sqrt{\pi} x) - 0.27 e^{- 0.27 x}(- \sqrt{\pi} A \sin \sqrt{\pi} x + \sqrt{\pi} B \cos \sqrt{\pi} x) - 0.27 e^{- 0.27 x}(- \sqrt{\pi} A \sin \sqrt{\pi} x + \sqrt{\pi} B \cos \sqrt{\pi} x)$$ $$ \displaystyle + e^{- 0.27 x}(- \pi A \cos \sqrt{\pi} x - \pi B \sin \sqrt{\pi} x)$$ Plugging all of this into the original equation you get $$ \displaystyle 0.0729 e^{- 0.27 x}(A \cos \sqrt{\pi} x + B \sin \sqrt{\pi} x) - 0.27 e^{- 0.27 x}(- \sqrt{\pi} A \sin \sqrt{\pi} x + \sqrt{\pi} B \cos \sqrt{\pi} x) - 0.27 e^{- 0.27 x}(- \sqrt{\pi} A \sin \sqrt{\pi} x + \sqrt{\pi} B \cos \sqrt{\pi} x) $$ $$ \displaystyle + e^{- 0.27 x}(- \pi A \cos \sqrt{\pi} x - \pi B \sin \sqrt{\pi} x) - 0.1458 e^{- 0.27 x}(A \cos \sqrt{\pi} x + B \sin \sqrt{\pi} x) + 0.54e^{- 0.27 x} (- \sqrt{\pi} A \sin \sqrt{\pi} x + \sqrt{\pi} B \cos \sqrt{\pi} x) $$ $$ \displaystyle + (0.0729 + \pi)e^{- 0.27 x}(A \cos \sqrt{\pi} x + B \sin \sqrt{\pi} x) = 0$$ Simplifying all of these terms you end up getting $$ \displaystyle e^{- 0.27 x} (0) = 0$$

Author
--Egm4313.s12.team12.anders. 4:11, 7 February 2012 (UTC)

Given
The ODE is: $$ \displaystyle y''+4y'+13y=r(x)     $$ The given initial conditions are: $$ \displaystyle  y(0) = 1    $$ $$ \displaystyle  y'(0) = 0    $$ $$ \displaystyle  r(x)=0      $$ The corresponding characteristic equation is: $$ \displaystyle \lambda^2 + 4\lambda +13=0     $$

Solution
Solving for $$ \lambda $$: $$ \displaystyle \lambda = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}     $$ $$ \displaystyle \lambda = \frac{-4 \pm \sqrt{4^2 -4(1)(13)}}{2(1)}     $$ $$ \displaystyle \lambda = \frac{-4 \pm \sqrt{-36}}{2}     $$ $$ \displaystyle \lambda = \frac{-4 \pm 2\sqrt{-9}}{2}     $$ $$ \displaystyle \lambda = {-2 \pm 3i}    $$ Therefore, the general solution will be: $$ \displaystyle y = C_1e^{a_1x}cosb_1x + C_2e^{a_2x}sinb_2x     $$ Where: $$ \displaystyle a_1=a_2=-2     $$ And: $$ \displaystyle b_1=b_2=3     $$ Thus: $$ \displaystyle  y = C_1e^{-2x}cos3x + C_2e^{-2x}sin3x   $$ Therefore: $$ \displaystyle y(0) = 1 = C_1e^{-2(0)}cos(3(0)) + C_2e^{-2(0)}sin(3(0)) = C_1(1)(1) + C_2(1)(0) = C_1 $$ $$ \displaystyle  C_1 = 1    $$ $$ \displaystyle y'(0) = 0 = -2C_1e^{-2x}cos(3x) - C_1e^{-2x}sin(3x)  -2C_2e^{-2x}sin(3x) + C_2e^{-2x}cos(3x) $$ $$ \displaystyle             0 = -2C_1 + 3C_2$$ $$ \displaystyle  C_2 = \frac{2}{3}    $$ Final Solution is as follows: $$ \displaystyle \  y = e^{-2x}cos(3x) + \frac{2}{3}e^{-2x}sin(3x)    \   $$

Author
Wagner Schulz 13:00, 8 February 2012 (UTC)


text:
Video journalism/Telling Stories

Genre (zhahn-ra) a kind or grouping or type of literary, musical, or artistic work.

Having, an understanding of genres or the groupings of video (film) stories is useful. In fact it can be a saviour when under pressure and producing a video news story everyday. The video worker intuitively starts designing their story by first structuring the ideas, the shooting script, writing the voice over wording and sequencing video. This is done in terms of their understanding of the type of story they are expected to make.

This understanding comes from a broad conception of the codes and conventions applying to specific genres. In video news for instance, a weekend shift journalist in Russia Today for instance, finding themselves in the middle of a staff shortage, could be expected to cut and voice main bulletin stories ranging from sport, court stories or international stories that have to be rebuilt and voiced from satellite feeds, to shooting and editing a weekend arts story. They would construct for each genre in direct response to pre determined templates that run every day in every major video news agency.

Knowledge of genre indicates a visual story telling language and set of conventions that belong to the particular genre or group in question. Recognition of a particular genre will help determine structure, style and depth that relates to the task of construction. The application of genre applies as much to in-depth investigative TV journalism on the intricacies and politics of international terrorism as it might to news.

Refer to the text below, by Gunther Kress, 1985, Linguistic processes in socio-cultural practice, Deakin Uni Press, pp. 46–47. Apply all the notions discussed here to your scripting, then to the process of editing your video news story.

Your video news story
Your video news story consists of many types of writing; print in the script and on screen to the camera’s writing of sound and visual text, while the editing as a writing, gives rise to new sequenced texts.

Writing has a structuring logic which differs fundamentally from that of speech. It is a logic of the nominal rather than of the verbal; of objects rather than processes; of abstraction rather than specificness/concreteness; a logic of hierarchy and of integration rather than a logic of sequence and addition. Western technological societies value the forms and logic of writing over the forms and logic of speaking. They present a kind of technology which is homologous with other technologies in our society. Writing also represents permanence and control rather than the impermanence and flux of speech. For these and other reasons writing is the medium of the domain of public social and political life while speaking is the medium of the domain of private life. The ‘public person has to adopt the modes of writing in speaking. For the powerful therefore, there is effectively only the one mode, that of writing; both in writing and in speech.

As you construct your script, a very public writing, keep these differences in mind. Be aware, when using the sometimes personal grabs from interviewees and then linking and describing the story sequences with the powerful and more public voice of the anchor, the journalist, you. Again Gunther Kress:

From what I have said so far it seems clear that mastery of speech, and of writing, confers power and effectiveness in different areas, in differing ways, in different domains. Command of writing gives access to certain cognitive, conceptual, social and political arenas. The person who commands both the forms of writing and of speech is therefore constructed in a fundamentally different way from the person who commands the forms of speech alone. Her or his range of cognitive, conceptual, social and political potential for being and acting differs fundamentally from that of the person who is confined to the forms of speech alone. Participation in public life and the power which that distributes depend on access to and mastery of the forms of writing. The possibility of being a certain kind of speaking and writing subject and therefore a certain kind of social and cultural agent depends on a person’s position in and relation to the forms and potentials of speech and of writing.

Authors, writers and originality
The general point of my discussion is to demonstrate the effect of social factors on the formation of a text. Discourse and genre both pre-exist the actions of the individual writer; so do the effects of the modes of speech and writing. Indeed, as I have attempted to show, writers are formed within discourse, genre, and the possibilities offered by speech and writing. What then is the role of the writer? And what are the demands for originality which can legitimately be made of child writers, or indeed of any writer? Notice that I have not included speakers here. It seems that we do not expect speakers to be original or creative in relation to the texts in which they are involved. Rather we seem to take it as assumed that speakers are largely obeying the demands of the occasion—of genre, of discourse, and of speech—when they speak. This may be due to the fact that spoken texts characteristically are multispeaker texts, so that no one speaker can be given responsibility for the construction of the text.

Moreover, in a multi-speaker text the speakers are not only fulfilling the demands of genre and discourse, but also seem to be under the reciprocal influence of each other as they take their turns in speaking. Written texts by contrast have the outward appearance of being the product of a single writer. However, if my argument is correct, then the function of the writer is not that of a creator of text, but of an assembler of text. That is, out of her or his experience of other texts, he or she creates a new text which meets the demands of a particular social occasion. My view stands in a fairly radical opposition to views held implicitly and explicitly in various linguistic or educational theories, which generally speaking regard the writer’s activity as one in which he or she draws on the resources of the linguistic system ‘as such’ (or draws on individual psychological, aesthetic resources) in freshly creating a text which fulfils the writer’s intentions or which meets the contingent demands of a particular situation. Such views lead on the one hand to demands for ‘originality’ end ‘creativity’ in children’s writing, and to tensions about plagiarism on the other hand in judgements made about children’s writing and other work.

In reality, all writers, poets, artists and journalists imitate and plagiarise - Kress again:

Therefore the task of any writer is importantly to understand the demands of generic form, the effects and meanings of discourses, and the forms of language in the written mode. The written texts of child learners everywhere bear the signs of the struggle to meet these demands, and everywhere bear the signs of their achievement in doing so.

The ‘stories’ written by children in primary school show the evidence of a (growing) understanding and mastery of narrative form, of narrative episodic structures, of knowledge of the syntax and the words appropriate to a genre, of formulae and of convention. All too frequently their developing understanding and mastery is classified and dismissed as copying, imitation, plagiarism.

As we have established, video-news and current affairs reports are visual ‘narratives’ or stories and they fall into distinct genres or groupings through their function, content, grammar, language and style. Your recognition of those genres and their visual conventions and codings will better serve your story telling ability.

The key insight here is that, over time, cultures establish certain relatively fixed ways of performing certain communicative functions such as arguing, explaining or story telling. Our concern is with the ways that our culture provides for the different forms of television-journalism story-telling.

Therefore, always start with an awareness of the intended audience, then visualise the genre into which you are producing. The next step is to acknowledge the fact that news and current affairs reports don’t simply reproduce reality. They don’t directly record what happened or what was said, rather video reportage is structured in production and editing so as to reflect key social values and to convey certain cultural meanings. These cultural meanings and messages are often built into journalistic practice to serve those in power – the networks and their advertisers. The visual language may also simply contribute to a news story’s entertainment values, making it more comfortable (less confronting) in viewing. A TV story that is easier to view, is more seamless, it is then more likely to convey information with clarity.

Video news story as narrative
Journalism and its culture can be viewed in many ways. The most reasonable of positions could involve taking a sceptical view of objectivity and reportage while also negotiating the negativity in journalism critique found in mass media and communications studies. The Nihilism in the critique of French media writer, Jean Baudrillard is to another extreme, while the unquestioning assertion of journalism producing nothing but objective truth is also as difficult to accept.

The Video News Story is but one of the ways of telling a story within the filmic text. Each story teller’s choice in content, depth and mode of delivery carries with it consequences as to meaning and rhetorical affect and this in turn affects, influences, viewers, conveys a political message. Documentary carries with it modes of representation that can be deeper, even less superficial than news, but within its longer story telling period, documentary has more opportunity to become indulgent and subjective, alienating the audience - ‘instead of citing the real (‘things’, ‘events’), it perpetually cites itself.’ [Baudrillard, J. In the Shadow of the Silent Majorities, New York: Semiotext(e) Foreign Agent Series, Published 1983, in Morris, M. Room 101 Or A Few Worst Things In The World - the lost referential. In Seduced And Abandoned, The Baudrillard Scene. Frankovits, A. Editor. Stonemoss Services, 1984, p. 112. ]

Throughout this Learning Schedule, you are asked to push the boundaries, experiment with the forms, experiment in your work and read up on the theory, which that might start to explain that experimentation. This practice is unusual in the global television industry, so enjoy the freedom in experimentation while it lasts. Here at least you’re free of the tyranny of the network agendas, producer styles and the packaging requirements in terms of advertising.

Across language and culture the video news story mimics itself. From a Cantonese language based Hong Kong television news-story to the news stories on Russia Today, to Fox news, to stories on air in Central Australia in several different Aboriginal Languages; they can all be seen holding similar lead-intro picture based form.

While there are subtle differences, audience screen literacy is something that is quite universal. This occurs through the globalisation of all modes of television, its strongest influence coming from the USA. Television is often cut to a literacy that is slightly below the average audience screen reading ability. Despite this, TV producers retain the assumption that all news programs are representing mainstream values.

International TV news packs all bulletins with encodings that purport news as natural, truthful, objective and universally understood. This Lead-dominated news structure is capable, of course, of serving the interests of non mainstream ideological perspectives. It can naturalise and construe as ‘objective’ the assumptions underlying alternative models of the social order. On the basis of this, Radio TV Hong Kong in the mid 1990s used to produce its Friday night program, ‘Headline’. According to Sze Wing Yuen, then Executive Producer at RTHK, the editor takes chunks of footage from the news throughout the week and re-edits the material into satirical, sarcastic and humorous vignettes that have a style akin to a music video. By Friday night the audience is already familiar with the news images and sound bites from throughout the week. This restructuring of the news footage through various dramatic and musical devices is pre-empted with a sophisticated audience knowledge of the stories and issues, gained throughout the week.

Programs like this play with and prosper on the news story structure and its intended function. With irony they subvert the news codes by making a mockery of the ideological positions and culturally-specific assumptions that underlay selection of news angles. Here, freedom of expression through comedy provides a questioning of the news story’s claim as arising naturally out the events themselves.

Seepage
A range of camera techniques, that a few years ago were confined to feature film or to the music video clip, have now seeped into documentary, current affairs and even news production. One popular example, that perhaps may have arisen from Oliver Stone’s Natural Born Killers, is the shooting of material already on the TV screen. This suggests the presence and affect of the screen’s persuasive ability on us all, it suggests the screen’s recent history and archival nature.

Another screen device now seeping even into news, is using a hand-held camera in a super-close up camera style. Producers know that this will most often evoke a sense of realism, as it did in the 1960s with the arrival of Sixteen Millimeter film cameras and Direct Cinema. But when this style is not edited skilfully, it will look clumsy and will not convey a seamlessness that is essential in structuring a reality of naturalness.

The adoption of codes and conventions from Hollywood iconography and films like Natural Born Killers, by TV-news stories is possibly subversive of the news and journalism culture itself. Television sequences and images can, in fact, get out of control and end up serving the interests of non mainstream ideological perspectives, despite their being manufactured in the mainstream. Such practice can naturalise and construe as ‘objective’ the assumptions underlying alternative models of the social order.

Television / video students who may be working and broadcasting in cultural perspectives and languages other than English should look to the journalistic traditions of their language to see what seepage might be occurring. You are encouraged to consider possible experimentation and alternatives that might be available within the form that you are working.

The public sphere — documentary, viewers and a current affair
Despite these blurred and rather suspect boundaries (as discussed above), both filmmaker and journalist have a privileged and powerful relationship with the lives they film. This involves the trading of information and the construction of film material for public consumption - as part of the Public Sphere. This notion of public sphere is encapsulated by Peter Dahlgren (editor) in his introduction to Communication and Citizenship, Journalism and the Public Sphere.

 * As the political and cultural significance of traditional and localised arenas continue to recede in the wake of social transformations and media developments, the notion of the public sphere moves to the fore and takes on a particularly normative valance. It becomes a focal point of our desire for a good society, the institutional site where political will should take form and citizens should be able to constitute themselves as active agents in the political process. How well the public sphere functions becomes a concrete manifestation of society’s democratic character and thus in a sense the most immediately visible indicator of our admittedly imperfect democracies. [Dahlgren, Peter ed. in Communication and Citizenship, Journalism and the Public Sphere Routledge Publ,1991, p. 1.]

In Australia, as in many social democracies – publicly funded documentary-filmmakers work for audiences hungry for entertainment in current affairs and documentary. Australian audiences in particular, have a sophisticated literacy in popular documentary. This ability enables complex and multiple readings and views of the language in a particular work. The journalist or film director’s observation and vision interacts through film with the audience. The assumed ‘naturalness’ and reputation of documentary is hinged on the assumption that it was filmed by a reputable professional, an artist, a well known filmmaker. These texts are packaged and submitted to the public forum as evidence or documented truth. In abuse of this, the filmmaker is sometimes caught out; with presumptuous arrogance, they are found assuming license or professional mandate to explore social reality and life’s casualties with bestowed levels of objectivity, balance and fairness by the mere fact of holding the job to make the film. This ‘naturalness’ and Public Sphere mandate in documentary, like that of the news story, is assumed and has been the embodiment of the profession since the documentary form began. Journalists and filmmakers could do the profession well by avoiding the assertion that they hold such privilege.

Highly criticised for their sensationalism, commercial current affairs programs like Australia’s, Channel-Nine Sixty-Minutes have until the 1990s, shot on film rather than video. This had the effect of creating a documentary look that in a stylised way is left suggesting authority and truthfulness to the audience. Now in the high definition video age, the same truthfulness is conveyed by the beautiful moving pictures of HD TV.

As documentary film methods were imported into the now mature television industry’s methodology, the use of film in mainstream TV current affairs sets up a look that purports an ethographic, privileged, reliable and tested form. Shot on the time tested format of HD video with a film look, a current affairs program like Sixty Minutes is assigned a documentary genre membership by nature of its deep (filmic) texture. Subliminally, this will assist the audience in making certain assumptions throughout their viewing. The documentary film look carries superior coding to that of video. Through its elitist and authoritative time tested culture and visual languages, documentary film, over earlier video technologies, can assert itself as being part of the time tested dominant paradigms.

The broad film genre – documentary, in the Public Sphere is packaged and marketed – presented as reliable, looking where others don’t, experimenting where others won’t, with camera, that since the days of its appearance, has wrongly been accepted as something akin to a scientific instrument. How the image and sound combinations are laid down in editing, is ultimately the journalist’s work, coming out of their thinking, their culture and views.

Experimental forms of video story telling are surely welcome to the scene. In Jean Baudrillard’s ‘Simulations’ this is given a context, an incentive to break away from the same:

 * Against the truth of the true, against the truer than true . . . against that unclean promiscuity with oneself called resemblance, we must re-make illusion, rediscover illusion, that both baleful and immoral power of tearing the same from the same called seduction. [Baudrillard, J. Simulations, New York: Semiotext(e) Foreign Agent Series, Published 1983, in Morris, M. Room 101 Or A Few Worst Things In The World - famous last words. In Seduced And Abandoned, The Baudrillard Scene, Frankovits, A. Editor. Stonemoss Services, 1984, p. 104.]

Or Bill Nichols:

 * The primary markers of stance, or space occupied, are the sound and image relayed to the viewer. To speak of the camera’s gaze is, in that one phrase, to mingle two distinct operations: the literal, mechanical operation of a device to reproduce images and the metaphorical, human process of gazing upon the
 * world. As a machine the camera produces an indexical record of what falls within its visual field. .................................... the camera reveals not only the world but its operator’s preoccupations, subjectivity, and values. The photographic (and aural) record provides an imprint of the visible surface of things. [Nichols, Bill. Axiographics in Representing reality. - Issues and Concepts in Documentary. Indiana University Press. 1991. 79.]

Activity
– In using video news examples on Vimeo, or Russia Today, or the like; study, deconstruct and discuss the origins and perspectives of the standard video News Story structure. Divide a story into sections containing the lead, the complicatiing action and the coda (the conclusion). – Does it purport objectivity? - What are the natural codes that sit in amongst the text to give it a tendency to convey truth, as discussed in this last section? – Throughout your regular video news viewings, what other structural or coding characteristics have you noticed in the typical video news story? – Could it be represented as having an atomic model rather than the inverted pyramid? The atomic model of story structure having a nucleus or centre part being the lead, intro and angle while the orbiting shells (the electrons), represent the dependant points in the story that follow in the complicating action and coda.

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

title: Video journalism/Telling Stories


text:
User talk:LllIIIllIIlIIllIIllIlIlI

Welcome

If you like you could invest your creativity in improving these learning resources:
 * A course on cabals
 * A course in troll sockpuppets
 * Mascot contest

your edits are easily revertable with tools, so just time waste, but if you are good with your ideas in above learning projects people might smile... Erkan Yilmaz uses the Chat + Identi.ca 02:30, 4 February 2012 (UTC)

title: User talk:LllIIIllIIlIIllIIllIlIlI


text:
Talk:Test and Quiz/Sandbox

Self-Assessment 1-5 Radio Buttons
I want to create a self-assessment where each question is answered on a 5 point scale ranging from 1 = Strongly Disagree through 5=Strongly Agree. There is no right or wrong answers. Is there some way to do this, perhaps with a horizontal set of 5 radio buttons that is clear, simple and compact? Thanks! --Lbeaumont 02:43, 4 February 2012 (UTC)
 * I am trying to simplify and improve the answer elements for What_Matters/assessment

title: Talk:Test and Quiz/Sandbox


text:
What Matters/assessment

This assessment is designed to determine if you are spending your time on what matters most.

Completing the Questionnaire
Based on your understanding of each topic gained from completing the course, your estimate of the time and effort you devote to each topic area, and the importance you place on each topic, respond to each of the following items by assigning a number 1-5 based on the scale provided below that best reflects your agreement or disagreement with each statement. Please record your answers on a separate piece of paper and not as edits to this page. It may be helpful to refer to the corresponding topic in the course to refresh your memory on what was covered as you complete this assessment. The questions are organized to align with the course topics. The list of statements is broad-ranging and not every statement will apply to all age groups. Skip any statements that are not appropriate for your age.

This is a self-assessment instrument; there are no right or wrong answers. The more accurate and honest you are in answering the more valuable it will be to you.

Especially if you are completing this assessment before completing the course it is perfectly acceptable to move quickly through making your best judgment based on your general knowledge and understanding of each statement. If a topic is unfamiliar to you, it may be best to score it 3 or below as a reminder to explore this topic in more detail later.

It may be helpful to print out a copy of this page so you can write down your answers and study the results.

1 Surviving
Score each of the following statements 1-5 using the following scale:

 * I get enough sleep.
 * My physiological needs are fully met.
 * My human rights are fully protected.
 * My psychological needs of autonomy, competency, and relatedness are each fully met.
 * I know what needs are currently met and which needs I am currently pursuing as described by Maslow’s hierarchy of needs.
 * My survival is not jeopardized by unmet needs.

2 Thriving
Score each of the following statements 1-5 using the following scale:
 * I am treated with dignity and respect.
 * I treat all others with dignity and respect.
 * My living space promotes my good health.
 * I am physically healthy.
 * I get regular exercise.
 * I am physically fit.
 * My wellness and vitality allow me to thrive physically.
 * My emotions are usually positive.
 * Often I am so engaged in work or other activities that I lose track of time.
 * I have close and positive relationships with other people.
 * I have a purpose in life that connects me with something meaningful beyond myself.
 * My accomplishments are a source of genuine pride.
 * I recognize, interpret, and respond constructively to emotions in myself and others.
 * I enjoy spending time with family members.
 * I am at peace with the members of my family.
 * I am pleased with my marital status.
 * I am pleased with my decision to have or not have children.
 * I am a good parent to my children.
 * My children are genuine a source of pride and joy.
 * I am working to pass on a beneficial legacy including helpful traditions to my offspring.
 * My mind is at peace.
 * I am content.
 * I am relaxed and not bothered by stress.
 * I notice and take time to enjoy many sources of beauty in my life.
 * I am curious and regularly enjoy learning new things.
 * I have a well-developed theory of knowledge.
 * I am good at solving problems.
 * I have mastered a body of knowledge in some domain or discipline pertaining to my work.
 * I have a dream.
 * I know who I am and I live my own life.
 * I have studied the virtues.
 * I understand what the virtuous good is.
 * I consistently strive toward that good.
 * I typically resist temptation and overcome fear to act according to my well-chosen values.
 * There are certain people I love.
 * There are certain people who love me.
 * I regularly enjoy safe sexual relations with my consenting adult partner.
 * I have friends I care about and who care about me.
 * I have developed a satisfying sense of spirituality.
 * I am thriving and flourishing.

3 Recreation
Score each of the following statements 1-5 using the following scale:
 * I spend adequate time playing and enjoying various forms of recreation.
 * I do not spend too much time playing.
 * I laugh enough.
 * I make sufficient time to enjoy various pleasures.
 * I don’t overindulge in pleasures to the detriment of other goals.
 * I enjoy a healthy and satisfying balance of recreation in my life.

4 Success
Score each of the following statements 1-5 using the following scale:
 * I have mastered one or more valuable skills
 * I am able to apply my skills to satisfying work.
 * People who know me and my work best think well of me.
 * I am creative.
 * I often have new and useful ideas.
 * I have a satisfying outlet for my creativity.
 * I am aware of the environmental impacts of my lifestyle
 * I take steps to conserve resources and reduce my environmental impact.
 * I am generally successful in my endeavors.

5 Significance
Score each of the following statements 1-5 using the following scale:
 * I demonstrate my sincere concern for the welfare of others.
 * I am impeccable with my word.
 * I do what I say.
 * I do my best.
 * I have reviewed my own values and carefully chosen them.
 * I live by a carefully chosen set of values.
 * I engage with the people who make up various communities I am a part of.
 * I feel I belong to those communities.
 * I genuinely trust many of the people in the various communities I belong to.
 * I evaluate decisions based on its global implications and long-term impacts.
 * I am engaged in meaningful work.
 * I have a positive impact and influence on others.

6 Transcendence
Score each of the following statements 1-5 using the following scale:
 * I am generous.
 * I enjoy helping others.
 * I resolve conflicts and disputes without resorting to violence or coercion.
 * I draw on my lifetime of experience and make decisions in the best long-term interest of all the humans impacted.
 * I use the well-being of all humans as my primary basis for decision making.
 * I am having a constructive and lasting impact on the world or universe at large.

7 Distractions and Dead Ends
Score each of the following statements 1-5 using the following scale:
 * I am humble and concerned with others rather than selfish and self-centered.
 * I am generous.
 * I wield power only while exercising great responsibility.
 * I am free of addictions such as tobacco use, alcohol abuse, drug abuse, gambling, unsafe sexual practices, eating disorders, etc.
 * I cope readily and constructively with hatred, anger, vengeance, envy, jealousy, or unfounded fears as they arise.
 * I am involved and engaged and rarely bored.
 * I am not bothered or distracted by ego, greed, addictions, destructive emotions, or boredom.

Examine your answers to identify the lowest scores in the areas that mean most to you.

Work to improve in those areas. Review the associated topics in this course. Read more of the materials in the suggested reading list for those topics. Repeat the assignments in those topics with a deeper sense of commitment. Spend more time in these areas and less time on activities that matter less.

title: What Matters/assessment


text:
User:Egm6322.s12.TEAM1.HW/HW4

Problem 1: Verification and Integration
 Solved without assistance

Find:
A. Verify Eq1.1 - Eq1.3 and Eq1.4 - Eq1.5 B. Intergrate both sides of Eq1.1 and use Eq1.2 to show the integral formula for weights, Eq1.6.

Solution:
A. verify equations First, according the definition, we have:

$$\displaystyle \pi (x):=\prod\limits_{k=1}^{n}{(x-{{x}_{k}})}$$.

The highest order of this equation won't be greater than n. Then,

$$\begin{array}{*{35}{l}} \pi (x)=(x-{{x}_{j}})\prod\limits_{k=1,k\ne j}^{n}{(x-{{x}_{k}})}\in {{\mathcal{P}}_{n}} \\ \end{array}$$ Moving the first term from right to the left side of this equation yields: $$\begin{array}{*{35}{l}} \frac{\pi (x)}{x-{{x}_{j}}}=\prod\limits_{k=1,k\ne j}^{n}{(x-{{x}_{k}})}\in {{\mathcal{P}}_{n-1}} \\ \end{array}$$ Taking the first derivative yields:

$$\begin{align} & {\pi }'(x)=(x-{{x}_{j}}{)}'\prod\limits_{k=1,k\ne j}^{n}{(x-{{x}_{k}})}+(x-{{x}_{j}}){{\left[ \prod\limits_{k=1,k\ne j}^{n}{(x-{{x}_{k}})} \right]}^{\prime }} \\ & =\prod\limits_{k=1,k\ne j}^{n}{(x-{{x}_{k}})}+(x-{{x}_{j}}){{\left[ \prod\limits_{k=1,k\ne j}^{n}{(x-{{x}_{k}})} \right]}^{\prime }} \end{align}$$

Substituting x_j into the above equation obtains:

$$\begin{align} & {\pi }'({{x}_{j}})=\prod\limits_{k=1,k\ne j}^{n}{({{x}_{j}}-{{x}_{k}})}+({{x}_{j}}-{{x}_{j}}){{\left[ \prod\limits_{k=1,k\ne j}^{n}{(x-{{x}_{k}})} \right]}^{\prime }} \\ & =\prod\limits_{k=1,k\ne j}^{n}{({{x}_{j}}-{{x}_{k}})} \end{align}$$ Because the highest order of polynomial P_n(x) won't be greater than n, so we can denote

$$\begin{array}{*{35}{l}} {{p}_{n}}(x)={{\alpha }_{n}}\pi (x) \\ \end{array}$$ in which $${{\alpha }_{n}}$$ is the leading coefficient, i.e. that of x^n. Then, the Lagrange basis function

$$\begin{array}{*{35}{l}} \begin{align} & {{\ell }_{j}}(x)=\frac{\pi (x)}{(x-{{x}_{j}}){\pi }'({{x}_{j}})} \\ & =\frac{{{P}_{n}}(x)/{{\alpha }_{n}}}{(x-{{x}_{j}}){{P}_}({{x}_{j}})/{{\alpha }_{n}}} \\ & =\frac{{{P}_{n}}(x)}{(x-{{x}_{j}}){{P}_}({{x}_{j}})}\in {{\mathcal{P}}_{n-1}} \\ \end{align} \\ \end{array}$$ B. integral formula for the weight

First, according to the equation (2)P45-6 and (1)P45-16C, we have: $$\begin{array}{*{35}{l}} {{I}_{n}}=\sum\limits_{j=1}^{n}{{{w}_{j}}f({{x}_{j}})}=\int\limits_{-1}^{+1}{{{f}^{h}}(x)dx} \\ \end{array}$$ Then, continued derivation yields: $$\begin{align} & \sum\limits_{j=1}^{n}{{{w}_{j}}f({{x}_{j}})}=\int\limits_{-1}^{+1}{\sum\limits_{j=1}^{n}{{{l}_{j}}(x)f({{x}_{j}})}dx} \\ & =\sum\limits_{j=1}^{n}{\int\limits_{-1}^{+1}{{{l}_{j}}(x)dx}f({{x}_{j}})} \end{align}$$

So, we can get: $$\begin{align} & {{w}_{j}}=\int\limits_{-1}^{+1}{{{l}_{j}}(x)dx} \\ & =\int\limits_{-1}^{+1}{\frac{{{P}_{n}}(x)}{(x-{{x}_{j}}){{P}_}({{x}_{j}})}dx} \end{align}$$

Problem 2: Christoffel-Darboux Identity
 Solved without assistance

Given: Equations 2.1 - 2.?
Where and and

with

Find
A.Show that the Christoffel-Darboux Identity, Eq2.1, with appropriate selection can be written as Eq2.5. B.Show that from the integral formula, Eq2.6 & Eq2.7 for the weight of the Gauss Legendre Quadrature, one can obtain the first formula for the weights, Eq2.8.

Solution:
A. When y is selected to be $${{x}_{j}}$$ , the CD identity can be shown as follows.

$$\displaystyle \sum\limits_{k=0}^{n}{\frac{{{P}_{k}}(x){{P}_{k}}({{x}_{j}})}}=\frac{{{P}_{n+1}}(x){{P}_{n}}({{x}_{j}})-{{P}_{n}}(x){{P}_{n+1}}({{x}_{j}})}{{{\beta }_{n}}{{h}_{n}}(x-{x}_{j})}$$ Because $${{x}_{j}}$$ is the root of n-order polynomial, so $${{P}_{n}}({{x}_{j}})=0$$. Then, CD identity becomes:

$$\displaystyle -\sum\limits_{k=0}^{n}{\frac{{{P}_{k}}(x){{P}_{k}}({{x}_{j}})}}=\frac{{{P}_{n}}(x){{P}_{n+1}}({{x}_{j}})}{{{\beta }_{n}}{{h}_{n}}(x-{x}_{j})}$$ The both sides of the above equation are multiplied by $${{P}_{0}}({{x}_{j}})$$, then integrate both side.

$$\displaystyle \begin{array}{*{35}{l}} \frac{{{P}_{n+1}}({{x}_{j}})}\frac{{{P}_{n}}(x){{P}_{0}}(x)}{(x-{{x}_{j}})}=-\sum\limits_{k=0}^{n}{\frac{{{P}_{k}}(x){{P}_{k}}({{x}_{j}})}} \\ \end{array}{{P}_{0}}(x)$$

$$\displaystyle \begin{align} & \int_{-1}^{+1}{\frac{{{P}_{n+1}}({{x}_{j}})}\frac{{{P}_{n}}(x){{P}_{0}}(x)}{(x-{{x}_{j}})}}dx=\int_{-1}^{+1}{-\sum\limits_{k=0}^{n}{\frac{{{P}_{k}}(x){{P}_{k}}({{x}_{j}})}}{{P}_{0}}(x)}dx \\ & =-\sum\limits_{k=0}^{n}{<{{P}_{0}},{{P}_{n}}>{{P}_{k}}({{x}_{j}})}=-{{P}_{0}}({{x}_{j}}) \end{align}$$ Since $${{P}_{0}}(x)=1$$, the above equation becomes:

$$\displaystyle \int_{-1}^{+1}{\frac{{{P}_{n}}(x)}{(x-{{x}_{j}})}}dx=-\frac{{{P}_{n+1}}({{x}_{j}})}$$

B. The weight of GL quadrature (1) P.45-9b can be further derived by using the integral derived in the above Part A. $$\displaystyle \begin{align} & \begin{array}{*{35}{l}} {{w}_{j}}=\int_{-1}^{+1}{\frac{{{P}_{n}}(x)}{(x-{{x}_{j}}){{P}_}({{x}_{j}})}}dx \\ \end{array} \\ & \begin{array}{*{35}{l}} =\frac{1}{{{P}_}({{x}_{j}})}\int_{-1}^{+1}{\frac{{{P}_{n}}(x)}{(x-{{x}_{j}})}}dx \\ \end{array} \\ & =\frac{-1}{{{P}_}({{x}_{j}})}\frac{{{P}_{n+1}}({{x}_{j}})} \end{align}$$

Using the general formula for Legendre polynomials in (2) P41-5 obtains: $$\displaystyle \begin{align} & \begin{array}{*{35}{l}} {{h}_{n}}=\frac{2}{2n+1} \\ \end{array} \\ & \begin{array}{*{35}{l}} {{\beta }_{n}}=\frac \\ \end{array}=\frac{1\centerdot 3\centerdot \cdots \left( 2n+1 \right)/(n+1)!}{1\centerdot 3\centerdot \cdots \left( 2n-1 \right)/(n)!}=\frac{2n+1}{n+1} \end{align}$$

Substituting the last two equation into $${{w}_{j}}$$ yields:

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

Problem 3: Polynomial Division
 Solved without assistance

Find
Prove Eq 3.4 and Eq3.5 are valid based on Equations 3.1-3.3

Solution
From Eq3.3, solve for $$ q(x)\ $$ to get the following Equation: Next, plug in values of Eq3.1, 3.2 and 3.5 to get the following equations: Next simplify Perform division to get: $$ \therefore \ $$ Eq3.4 and Eq3.5 are valid based on Equations 3.1, 3.2, 3.3

Problem 4: Pose Question to be solved
 Solved without assistance

Find
Use the table to verify Equations 4.1 and 4.2.

Solution:
Equation 4.1 states that the weights must be greater than zero. A quick scan of the third column verifies that all of the weights are >0. For the second part of this question, one must sum the weights, based on each number of points. Thus for $$ n=1 \ $$, $$ w_i=2 \ $$ Thus for $$ n=2 \ $$, $$ w_i=1 \ $$ in 2 places $$ \therefore \sum w_i = 1+1= 2 \ $$ Thus for $$ n=3 \ $$, $$ w_i=\frac{8}{9} \ $$ in one place and $$ w_i=\frac{5}{9} \ $$ in two places $$ \therefore \sum w_i = \frac{8}{9} + 2 \left ( \frac{8}{9} \right )= 2 \ $$ Thus for $$ n=4 \ $$, $$ w_i=\tfrac{18+\sqrt{30}}{36} \ $$ in 2 places and $$ \tfrac{18-\sqrt{30}}{36} \ $$ in two places $$ \therefore \sum w_i = 2*\tfrac{18+\sqrt{30}}{36} + 2*\tfrac{18-\sqrt{30}}{36}= 2 \ $$ Thus for $$ n=5 \ $$, $$ w_i=\frac{128}{225} \ $$ in one place and $$ w_i=\tfrac{322+13\sqrt{70}}{900} \ $$ in 2 places and $$ \tfrac{322-13\sqrt{70}}{900} \ $$ in two places $$ \therefore \sum w_i = \frac{128}{225} + 2* \tfrac{322-13\sqrt{70}}{900} + 2* \tfrac{322+13\sqrt{70}}{900} = 2 \ $$

Problem 5: Integrate and Find Number of Integration Points
 Solved without assistance

Find: A & B below
A) Integrate Equation 5.1 analytically (not numerically) exactly over $$ [-1,1] \ $$ B) Determine the least nmber of integration points to integrate Eq5.1 exactly with the GL Quadrature. Carry out the integration and compare to the exact result.

part A
$$\displaystyle p(x) = \frac{x^7}{8} + \frac{x^6}{7} + \frac{x^2}{3} + 5$$ $$\displaystyle \int_{x=-1}^{x=1} p(x)dx = \int_{x=-1}^{x=1} (\frac{x^7}{8} + \frac{x^6}{7} + \frac{x^3}{3} + 5)dx $$ $$\displaystyle = [\frac {x^8}{8^2} + \frac{x^7}{7^2} + \frac{x^3}{3^2} +5x]_{-1}^{1} = \frac{492}{4} $$

part B
p(x) is 7 order equation. If n = 4, $$\displaystyle \int_{-1}^{1} f[x]dx = w_1f[x_1] + w_2f[x_2] + w_3f[x_3]+w_4f[x_4] $$ Where $$\displaystyle f_1[x] =1,f_2[x] =x,f_3[x]=x^2, ...... ,f_8[x] = x^7$$ $$\displaystyle \int_{-1}^{1} f_1[x]dx = w_1f_1[x_1] + w_2f_1[x_2] + w_3f_1[x_3]+w_4f_1[x_4] $$ $$\displaystyle \int_{-1}^{1} f_2[x]dx = w_1f_2[x_1] + w_2f_2[x_2] + w_3f_2[x_3]+w_4f_2[x_4] $$ $$\displaystyle \int_{-1}^{1} f_3[x]dx = w_1f_3[x_1] + w_2f_3[x_2] + w_3f_3[x_3]+w_4f_3[x_4] $$ $$\displaystyle \int_{-1}^{1} f_4[x]dx = w_1f_4[x_1] + w_2f_4[x_2] + w_3f_4[x_3]+w_4f_4[x_4] $$ $$\displaystyle \int_{-1}^{1} f_5[x]dx = w_1f_5[x_1] + w_2f_5[x_2] + w_3f_5[x_3]+w_4f_5[x_4] $$ $$\displaystyle \int_{-1}^{1} f_6[x]dx = w_1f_6[x_1] + w_2f_6[x_2] + w_3f_6[x_3]+w_4f_6[x_4] $$ $$\displaystyle \int_{-1}^{1} f_7[x]dx = w_1f_7[x_1] + w_2f_7[x_2] + w_3f_7[x_3]+w_4f_7[x_4] $$ $$\displaystyle \int_{-1}^{1} f_8[x]dx = w_1f_8[x_1] + w_2f_8[x_2] + w_3f_8[x_3]+w_4f_8[x_4] $$ From the above equations, $$\displaystyle w_1 + w_2 + w_3 + w_4= 2$$

$$\displaystyle w_1x_1 + w_2x_2 + w_3x_3 + w_4x_4= 0$$

$$\displaystyle w_1x_1^2 + w_2x_2^2 + w_3x_3^2 + w_4x_4^2= \frac{2}{3}$$

$$\displaystyle w_1x_1^3 + w_2x_2^3 + w_3x_3^3 + w_4x_4^3= 0$$

$$\displaystyle w_1x_1^4 + w_2x_2^4 + w_3x_3^4 + w_4x_4^4= \frac{2}{5}$$

$$\displaystyle w_1x_1^5 + w_2x_2^5 + w_3x_3^5 + w_4x_4^5= 0$$

$$\displaystyle w_1x_1^6 + w_2x_2^6 + w_3x_3^6 + w_4x_4^6= \frac{2}{7}$$

$$\displaystyle w_1x_1^7 + w_2x_2^7 + w_3x_3^7 + w_4x_4^7= 0$$

As a result of the above equations, the solution is that $$\displaystyle x_1 = -0.86113$$ $$\displaystyle x_2 = -0.33998$$ $$\displaystyle x_3 = 0.33998$$ $$\displaystyle x_4 = 0.86113$$ $$\displaystyle w_1 = 0.34785$$ $$\displaystyle w_2 = 0.65214$$ $$\displaystyle w_3 = 0.65214$$ $$\displaystyle w_4 = 0.34785$$

These results show that $$\displaystyle \int_{x=-1}^{x=1} p(x)dx = \sum_{j=1}^{n}w_jp(x_j) $$ Actually, $$\displaystyle \int_{x=-1}^{x=1} p(x)dx = \frac{492}{49} = 10.0408 $$ However, using the above values, $$\displaystyle \sum_{j=1}^{n}w_jp(x_j) = 10.2630$$ I think it is round-off error and truncation error.

Problem 6: Plot Function and Find non-zero coefficients
 Solved without assistance

Find: A & B
A) Plot the function given in Eq6.1 B) Find the first three non-zero coefficients of the Fourier-Legendre Series, if not analytically then numerically using GL Quadrature with $$ 10^6 \ $$ accuracy.

Solution:

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\displaystyle f(\theta)=\frac{1}{1+\sin^2(sin^2 \theta)} $$
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\displaystyle \mu:=\sin \theta $$
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\displaystyle F(\mu)=f(\theta)=\frac{1}{1+\sin^2(\mu^2)} $$
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\displaystyle F(\mu)=\sum_{j=0}^{\infty}A_jP_j(\mu) $$
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\displaystyle A_j=\frac{\langle P_j,F \rangle}{\langle P_j,P_j \rangle}=\frac{2j+1}{2} \int \limits_{\mu=-1}^{\mu=+1}P_j(\mu)F(\mu)d\mu $$
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The first several Legendre polynomials are:
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\displaystyle P_0(\mu)=1, \, P_1(\mu)=\mu $$
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\displaystyle P_2(\mu)=\frac{1}{2}(3\mu^2-1) $$
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\displaystyle P_3(\mu)=\frac{1}{2}(5\mu^3-3\mu) $$
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\displaystyle P_4(\mu)=\frac{1}{8}(35\mu^4-30\mu^2+3) $$
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\displaystyle A_0=\frac{1}{2}\int_{-1}^11\cdot \frac{1}{1+\sin^2(\mu^2)}d\mu $$
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For the equation above can't be integrated analytically, then we use GL quadrature.
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\displaystyle A_0=\frac{1}{2}\cdot \sum_{j=1}^{n}w_j\cdot g(x_j) $$
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Here we use three points for accuracy.
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\displaystyle \left\{\begin{matrix} x_1=-\sqrt{\frac{3}{5}}&w_1=\frac{5}{9} \\ x_2=0&w_2=\frac{8}{9} \\ x_3=\sqrt{\frac{3}{5}}&w_3=\frac{5}{9} \end{matrix}\right. $$
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then:
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\displaystyle \begin{align} A_0 &=\frac{1}{2}[w_1\cdot g(x_1)+w_2\cdot g(x_2)+w_3\cdot g(x_3)]\\ &=0.8657 \end{align} $$
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\displaystyle \begin{align} A_1 &=\frac{3}{2}\int_{-1}^{1}\mu \cdot \frac{1}{1+\sin^2(\mu^2)}d\mu \\ &=\frac{3}{2}\sum_{j=1}^n w_j\cdot g(x_j) \\ &=\frac{3}{2}[w_1\cdot g(x_1)+w_2\cdot g(x_2)+w_3\cdot g(x_3)]\\ &=0 \end{align} $$
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\displaystyle \begin{align} A_2 &=\frac{5}{2}\int_{-1}^1 P_2(\mu)F(\mu)d\mu\\ &=\frac{5}{2}\sum_{j=1}^n w_j\cdot g(x_j)\\ &=-0.2686 \end{align} $$
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\displaystyle \begin{align} A_3 &=\frac{7}{2}\int_{-1}^1 P_3(\mu)F(\mu)d\mu\\ &=\frac{7}{2}\sum_{j=1}^n w_j\cdot g(x_j)\\ &=0 \end{align} $$
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From A4, we begin to use five points for accuracy.
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\displaystyle \begin{align} A_4 &=\frac{9}{2}\sum_{j=1}^n w_j\cdot g(x_j)\\ &=-0.0707 \end{align} $$
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\displaystyle F(\mu)=\sum_{j=0}^{\infty} A_jP_j(\mu) $$
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\displaystyle f(\theta)=\sum_{j=0}^{\infty}A_jP_j(\sin\theta) $$ Plot the results(Fourier-Legendre series of function $$f(\theta)$$) using Matlab: for only one term: for three terms: for five terms: Comparing with the originally exact one: from the above results, we see that with the increase of number of terms, it shows the convergence toward the original function.
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=Contributing Members & Referenced Lecture=

=Table of Contributions=

title: User:Egm6322.s12.TEAM1.HW/HW4


text:
User:Egm6322.s12.TEAM1.HW/HW6

Problem 1: Find the Continued Fraction for the Square Root of Five
 Solved without assistance

Find
Find the continued fraction for the $$ \sqrt{5} \ $$

Solution
Re-arrange to get To rid the denominator of the square root term, multiply numberator and denominator by $$ \sqrt{5}+2 \ $$ to get: Substituting the value for x in Eq 1.4 back into Eq 1.2: Values begin to repeat, because from Eq 1.2, the value of $$ \sqrt{5} \ $$ is already known: This may also be written in the following forms: $$ \sqrt{5} = 2+\frac{1}{4+}\frac{1}{4+}\frac{1}{4+}\frac{1}{4+}\frac{1}{4+}\frac{1}{\cdots} \ $$ $$ \sqrt{5} = \left \{ 2,4,4,4,\cdots \right \} \ $$ $$ \sqrt{5} = \left [ 2,\bar 4 \ \right ] \ $$

Problem 2: Approximate Pi using Euclid's Algorthm
 Solved without assistance

Given
$$ \pi\ \ $$ can be approximated successively by

Find
A) Use the method of successive subtractions or Euclid's Algorithm to show the given statement. B) Generate a sequence of rational numbers that approximates $$ \pi\ \ $$ from Wallis' Theorem: $$ \frac{ \pi\ }{2} = \frac{2*2*4*4*6*6*8*8 \cdots}{1*3*3*5*5*7*7*9\cdots} \ $$ C) Compare the rate of convergence of the continued fraction versus the Wallis Sequence.

Solution
A) Consider the shorthand values given in Eq2.1 and defining vectors as: Thus by the method of successive subtractions one must find the quotients given by above vectors $$ \overline{AB} \ $$, $$ \overline{CD} \ $$, $$ \overline{EF} \ $$, $$ \overline{GH} \ \ $$, beginning with the simplest, For Eq2.2 we have: For Eq2.3 we have: For Eq2.4 we have: For Eq2.5 we have: B) In order to generate a sequence of rational numbers similar to the Wallis sequence of: $$ \frac{ \pi\ }{2} = \frac{2*2*4*4*6*6*8*8 \cdots}{1*3*3*5*5*7*7*9\cdots} \ $$, I utilized an Excel Spreadsheet. I was able to use a sequence of 170 terms, at which point Excel stops computing terms any longer due to large numbers being generated in the sequence. See screen capture below (with several rows hidden to save space). One can see that Wallis' sequence slowly begins to approximate $$ \pi\ \ $$. C) The comparison of the speed of convergence between Wallis sequence versus the continued fraction is quite large. Using the terms found in this problem, one can see that it only required 2 terms to closely approximate $$ \pi\ \ $$ to the hundredths place using continued fractions, whereas I used 170 terms of Wallis' sequence and still only had an accuracy to the tenth place of a whole number.

Problem 3: Compare the convergence of power series and continued fraction method
 Solved without assistance

Given
Power series of tan x: $$\displaystyle x \in ]-\pi /2,\pi/2 [$$ $$\displaystyle tan x = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + ... + \frac{(-1)^{n-1}2^{2n}(2^{2n}-1)B_{2n}}{(2n)!}x^{2n-1} + ...$$ Continued fraction of tan x: $$\displaystyle tan x = \frac{x}{\displaystyle 1 -\frac{x^2}{\displaystyle 3 -\frac{x^2}{\displaystyle 5 -\frac{x^2}{\displaystyle 7 -\frac{x^2}{\ddots}}}}}$$

Find
Take $$x = \pi / 4$$ and compare the term -by-term convergence of the power series (2) p.48-25b and the continued fraction (1). Provide a plot to visualize the convergence.

Solution
the exact solution of $$\displaystyle tan\left(\frac{\pi}{4}\right )$$ is 1.0 x = $$\frac{\pi}{4}$$ For power series, when n = 1 $$\displaystyle tanx = x = \frac{\pi}{4}=0.7853$$ When n = 2, $$\displaystyle tanx = x + \frac{1}{3}x^3= 0.9467$$ When n = 3, $$\displaystyle tanx = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 = 0.9865$$ When n = 4, $$\displaystyle tanx = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \frac{17}{315}x^7 = 0.99649$$ When n =5, $$\displaystyle tanx = x + \frac{1}{3}x^3 + \frac{2}{15}x^5 + \frac{17}{315}x^7 + \frac{62}{2835}x^9 = 0.9989$$ For continued fraction method, n = 1 $$\displaystyle tan x = \frac{x}{\displaystyle 1} = \frac{\pi}{4} = 0.7853 $$ When n = 2, $$\displaystyle tan x = \frac{x}{\displaystyle 1 -{x^2}}= 2.0498$$ When n =3, $$\displaystyle tan x = \frac{x}{\displaystyle 1 -\frac{x^2}{\displaystyle 3 -{x^2}}} = 1.059685$$ When n =4, $$\displaystyle tan x = \frac{x}{\displaystyle 1 -\frac{x^2}{\displaystyle 3 -\frac{x^2}{\displaystyle 5 -{x^2}}}} = 1.00144$$ When n = 5, $$\displaystyle tan x = \frac{x}{\displaystyle 1 -\frac{x^2}{\displaystyle 3 -\frac{x^2}{\displaystyle 5 -\frac{x^2}{\displaystyle 7 -{x^2}}}}}= 1.000018$$ These results show that as n increases, the convergence speed of continued fraction method will be faster than power series.

Problem 4: proof of the conversion between cdf and pdf forms of normal distribution
 Solved without assistance

Given:
The definition of error function erf.

Find:
show that cdf of normal distribution can be obtained from pdf of normal distribution, and vice versa.

Solution:
From the defition of cdf (cumulative distribution function) we have: $$\displaystyle \Phi (x)=\int\limits_{-\infty }^{x}{f(t)dt}=\int\limits_{-\infty }^{x}{\frac{1}{\sqrt{2\pi }}{{e}^{-{{t}^{2}}/2}}dt}$$ Continued derivation based on the substitution of $$t=\sqrt{2}s$$,

$$\begin{align} & \Phi (x)=\frac{1}{\sqrt{2\pi }}\int\limits_{-\infty }^{x/\sqrt{2}}{{{e}^{-{{(\sqrt{2}s)}^{2}}/2}}d(\sqrt{2}s)} \\ & =\frac{1}{\sqrt{\pi }}\int\limits_{-\infty }^{x/\sqrt{2}}{{{e}^{-{{s}^{2}}}}ds} \\ & =\frac{1}{\sqrt{\pi }}\int\limits_{0}^{x/\sqrt{2}}{{{e}^{-{{s}^{2}}}}ds}+\frac{1}{\sqrt{\pi }}\int\limits_{-\infty }^{0}{{{e}^{-{{s}^{2}}}}ds} \\ & =\frac{1}{2}(\frac{2}{\sqrt{\pi }}\int\limits_{0}^{x/\sqrt{2}}{{{e}^{-{{s}^{2}}}}ds})+\frac{1}{2\sqrt{\pi }}\int\limits_{-\infty }^{\infty }{{{e}^{-{{s}^{2}}}}ds} \end{align}$$ Because $$\int\limits_{-\infty }^{\infty }{{{e}^{-{{s}^{2}}}}ds}=\sqrt{\pi }$$, which is proved in problem R13.5 and the definition of error function $$erf(x)=\frac{2}{\sqrt{\pi }}\int\limits_{0}^{x}{{{e}^{-{{s}^{2}}}}ds}$$ , we have: $$\begin{align} & \Phi (x)=\frac{1}{2}(\frac{2}{\sqrt{\pi }}\int\limits_{0}^{x/\sqrt{2}}{{{e}^{-{{s}^{2}}}}ds})+\frac{1}{2\sqrt{\pi }}\int\limits_{-\infty }^{\infty }{{{e}^{-{{s}^{2}}}}ds} \\ & =\frac{1}{2}(erf(x/\sqrt{2})+1) \end{align}$$ If we substitute the equation

$${x}'=\left( x-\mu \right)/\sigma $$ , into the above derivation, we obtain: $$\Phi (x)=\frac{1}{2}(erf(\left( x-\mu \right)/\sqrt{2}\sigma )+1)$$ The derivation from the cdf to pdf of normal distribution is the same procedure, just use the differential operator instead of integral operator.

Problem 5: integrate constant function with specific weight analytically and numerically
 Solved without assistance

Find:
Integrate the constant function 1 with the weight $${{e}^{-{{x}^{2}}}}$$ $$\displaystyle I(1)=\int\limits_{-\infty }^{\infty }{1\centerdot {{e}^{-{{x}^{2}}}}dx}$$ 1. Analytically 2. Verify with WolframAlpha 3. Numerically using Gauss-Hermite quadrature and plot the convergence of numerical value as a function of n.

Solution:
1.Analytically: First,square up the integral to convert the line integral into an integral over the plane $$\displaystyle {{I}^{2}}=\int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{{{e}^{-({{x}^{2}}+{{y}^{2}})}}dx}dy}$$ Then,convert the area integral to polar coordinates to integrate: $$\begin{align} & {{I}^{2}}=\int\limits_{-\infty }^{\infty }{\int\limits_{-\infty }^{\infty }{{{e}^{-({{x}^{2}}+{{y}^{2}})}}dx}dy} \\ & =\int\limits_{-\pi }^{\pi }{\int\limits_{0}^{\infty }{{{e}^{-({{(r\sin \theta )}^{2}}+{{(r\cos \theta )}^{2}})}}rdr}d\theta } \\ & =\int\limits_{-\pi }^{\pi }{\int\limits_{0}^{\infty }{{{e}^{-{{r}^{2}}}}\frac{1}{2}d\left( {{r}^{2}} \right)}d\theta } \\ & =\int\limits_{-\pi }^{\pi }{\frac{1}{2}\left( \int\limits_{0}^{\infty }{{{e}^{-t}}dt} \right)d\theta } \\ & =\int\limits_{-\pi }^{\pi }{\frac{1}{2}\left( -{{e}^{-\infty }}-(-{{e}^{-0}}) \right)d\theta } \\ & =\int\limits_{-\pi }^{\pi }{\frac{1}{2}d\theta } \\ & =\pi \end{align}$$ in which the substitution of $$t={{r}^{2}}$$ is used during derivation. So, we obtain: $$\displaystyle I=\int\limits_{-\infty }^{\infty }{{{e}^{-{{x}^{2}}}}dx}=\sqrt{\pi }$$ 2.Verify with WolframAlpha: $$\displaystyle \int\limits_{-\infty }^{\infty }{{{e}^{-{{x}^{2}}}}dx}=\sqrt{\pi }\approx 1.77245$$ So, the same result with analytical one is obtained. 3. Numerically: Use the Gauss-Hermite quadrature with the table for {x_i} and {H_i} at DLMF:Gauss-Hermite formula. $$\displaystyle {{I}_{n}}(1)=\sum\limits_{i=1}^{n}{{{H}_{i}}f({{x}_{i}})}=\sum\limits_{i=1}^{n}$$ n=2: $$0.\text{886227}*2=1.772454$$ n=3: $$\text{0}\text{.295409*2+1}\text{.18164}=1.772458$$ n=4: $$\text{0}\text{.804914*2+0}\text{.0813128*2}=1.7724536$$ n=5:

$$\text{0}\text{.94530872048294+0}\text{.393619323152241*2+0}\text{.0199532420590459*2=}1.7724538509055138$$ n=10:

$$\text{0}\text{.610862633735325799*2+0}\text{.240138611082314686*2+0}\text{.0338743944554810631*2+0}\text{.00134364574678123269*2+0}\text{.00000764043285523262063*2=}1.7724538509551602$$ And the plot is shown in the following figure:

Problem 6: Prove that a Polynomial belongs to a Real Set of Numbers
 Solved without assistance

Find:
Show that:

Solution:
For $$ p_n \in \mathcal P_n \  $$ and $$ p_n(x)= \sum_{i=0}^n k_i^nx^i \  $$ Consider the polynomial: $$ s(x):= p_{n+1}(x)-A_nxp_n(x)\in \mathcal P_{n+1} \ $$

$$ = \sum_{i=0}^{n+1}k_i^{n+1}x^i-A_n \sum_{i=0}^n k_i^nx^{i+1} \ $$

$$ = \sum_{j=-1}^n k_{j+1}^{n+1}x^{j+1}-A_n \sum_{i=0}^n k_i^nx^{i+1} \ $$

$$ = \sum_{j=-1}^{n-1}k_{j+1}^{n+1}x^{j+1}+k_{n+1}^{n+1}-A_n \sum_{i=0}^{n-1}k_i^nx^{i+1}-A_nk_n^nx^{n+1} \ $$

$$ = \sum_{j=-1}^{n-1}k_{j+1}^{n+1}x^{j+1}-A_n \sum_{i=0}^{n-1}k_i^nx^{i+1}+(k_{n+1}^{n+1}x^{n+1}-A_nk_n^nx^{n+1}) \ $$

If $$ A_n=\frac{k_{n+1}^{n+1}}{k_n^n}=:\frac{k_{n+1}}{k_n} \ $$

Then

$$ k_{n+1}^{n+1} x^{n+1} - A_n k_n^n x^{n+1} =(k_{n+1}^{n+1}- A_n k_n^n) x^{n+1} = 0 \ $$

$$ s(x) = \sum_{j=-1}^{n-1}k_{j+1}^{n+1}x^{j+1}-A_n \sum_{i=1}^{n-1}k_i^nx^{i+1}+0 = \sum_{i=0}^n k_i^{n+1}x^i-A_n \sum_{j=2}^nk_{j-1}^nx^j \in \mathcal P_n \ $$

Thus:

$$ s(x)=p_{n+1}(x)-A_nxp_n(x)=\sum_{i=0}^n d_ip_i(x) \ $$

Problem 7: Show Orthogonality using Scalar Products
 Solved without assistance

Find
Show that: for $$ j \ne n-1 \ $$ and for $$ j = n-1 \ $$ Also using Eqs 7.1 and 7.3, show that: for $$ j < \ n-1 \ $$

Solution:

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\displaystyle \langle p_j,p_{n+1}-A_nxp_n(x)\rangle= \sum_{i=0}^n d_i \underbrace{\langle p_j,p_i \rangle}_{\displaystyle h_j\delta_{ij} } =h_jd_j $$
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For the equation above to be satisfied
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\displaystyle j\leq n $$
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\displaystyle q_{j+1}:=xp_j \in \mathcal P_{j+1} $$
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When$$j=n$$
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\displaystyle q_{j+1}=q_{n+1}=xp_n=\sum _{i=0}^{n+1}d_ip_i(x) $$
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\displaystyle \langle q_{n+1},p_n \rangle =d_n\langle p_n,p_n \rangle =d_nh_n \neq 0 $$
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When$$j=n-1$$
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\displaystyle q_{j+1}=q_n=xp_{n-1}=\sum_{i=0}^nd_ip_i(x) $$
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\displaystyle \langle q_n,p_n \rangle =d_n \langle p_n,p_n \rangle =d_nh_n \neq 0 $$
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When$$j
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\displaystyle q_{j+1}\in \mathcal P_{j+1} $$ For $$p_n$$ is perpendicular to $$\mathcal P_{j+1}$$ So
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\displaystyle \langle q_{j+1},p_n\rangle =0 $$
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\displaystyle \begin{align} & \ \langle p_j, p_{n+1}-A_nxp_n(x) \rangle \\ &= \langle p_j, p_{n+1} \rangle - \langle p_j, A_nxp_n \rangle\\ &= 0-\langle A_n x p_j ,p_n \rangle \\ &= \langle q_{j+1}, p_n \rangle\\ &= 0 \end{align} $$
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=Contributing Members & Referenced Lecture=

=Table of Contributions=

title: User:Egm6322.s12.TEAM1.HW/HW6


text:
Video journalism/Editing Grammar

Grammar of the edit - become more conscious of what happens to meanings conveyed in a television story

The integrity of the cut - or in other words, don't use dissolves or other fancy transitions
We have established in previous sections of this Learning Schedule, that television products themselves provide a window on the process of communication. Audience responses to our products are constantly redefining our roles and approaches to making television. We can make these fine adjustments to production activities through a constant interrogation of the form. By analysing the form, gathering ideas about what audiences want and then responding through adusting editing grammar in a documentary, for instance, can sharpen product responsiveness to niche market. By fine tuning the rhythm of cuts and sequences one can suddenly make a film work for its audience. This involves anticipating contemporary desires and visual literacy levels while also anticipating what affect big feature films and other trends will have on audience expectations. This may sound impossible, but can be done by watching many variations of the form.

Re-write script: paper edit — rough cut — experiment with rhythm, grammar and pace
....... voices – such as those of the television anchor people and reporters – are attached to bodies that represent not personal witness, but institutional authority in anthropomorphic form.

This privileging of the world faces a crisis with every edit. With each cut the opportunity exists to re inscribe the filmmaker’s presence rather than excise it.

Each cut opens the gap between human agency and cinematic evidence only to anneal it again through continued exclusion. Documentary convention upholds the expectation of presence, of an ethic of witnessing, of a situated view, and yet exercises the bodily evidence of presence. [Nichols, Bill. Axiographics in Representing reality. – Issues and Concepts in Documentary. Indiana University Press, 1991, 90.]

Editing for associations — finding structure
Each piece of raw footage desires a final place in a film, news story, rock clip. Editing is a process that exists in both scriptwriting and in the actual cutting of film or videotape material. Editing in process chooses which sections are included and it sequences them with rules and conventions that constitute a film grammar.

Editing should ensure that the film story has seamlessness, rhythm and clarity. Experimental editing of the raw segments using low budget equpment like a computer based edit suite, should start to suggest new associations and story to your camera footage. By starting with a strong idea and working from that as the lead, like one might in a news story, the rough cut editing process of documentary or current affairs can present and build its own rhythm and grammatical structure. This process is better served by working with two people, the editor and the director or journalist, over a realistic period of time.

To indicate the time consuming nature of editing, it is felt in the industry, that the finishing of a minute of final sequencing in each day of editing is about the right rate of progress for a thoughtful film to emerge. A thoughtful one minute news-story can take the best part of the day to cut, a current affairs story might take a week to edit, while a half hour documentary a month – and so on.

Bill Nichols suggests that while the film segments are raw representations of reality they are also something that is not actual. ‘If I am to analyse this film properly, I must not mistake it for reality; but if I do not mistake it for reality, I cannot analyse it properly.’ [Bill Nichols, Ideology and the Image (Bloomington, Ind., 1981), 250.]

If David MacDougall’s ideas were applied, then one could proceed with editing so that each social actor (subject) had a distinct vantage point within the film. Their positions holding certain perspectives that are given through their narratives.

 * The experiences of individual social actors are seen increasingly as vantage points within a society which, in the complex dynamics of their engagement, structure particular ‘readings’ of social phenomena. The portrayal of subjective experience becomes a strategy for resituating anthropological understanding within the textual richness of a society. The goal is not simply to present the ‘indigenous view’, nor to invade voyeuristically the consciousness of other individuals, but to see social behaviour, and indeed culture, as a continuous process of interpretation and reinvention. [David MacDougal, Ethnographic Film in Fields of Vision, Essays in Film Studies, Visual Anthropology, and Photography. Leslie Devereaux and Roger Hillman – eiditors. University of California Press. 1995. ISBN 0- 520-08522-1 (alk. paper).-ISBN 0-520-08524-8 (pbk. :alk. paper), 219.]

Editing aims at finding the best dramatic elements and sequence orders. Pace and the emotive and telling sound-image relationships are found by editing and sequencing outwards from the angle, argument or central theme. In the news story, the central theme is positioned and defined through the Lead or the Intro.

Sequences are built in blocks of simular theme, to provide ‘distinct vantage’ points around the social actors (subjects) until one theme converges with and enhances the next. The subject’s grabs are sequenced so that the story is told using the best grabs from each speaker.

 * It is impossible not to talk of sequences, of associated relationships between shots, and yet the cutting has the effect of isolating each shot, giving it its own weight and unity, and not restricting the associative process to the relationship between immediately neighbouring shots. Because each shot sinks into memory, the possibility for association, for waiting, for letting things happen, is an extended, unproscribed one. [Susan Dermody. – The Subjective Voice in Documentary The Pressure of the Unconscious upon the Image. In Fields of Vision, Essays in Film Studies, Visual Anthropology, and Photography. Leslie Devereaux and Roger Hillman – editors. University of California Press, 1995, 296.]

Film segments or ‘factual referents’ in their sequences and blocks of ‘distinct vantage points’ are built outwards into something like overlapping concentric circles. Their growth provides the opportunity for screening or trialing blocks in different sequences, combinations and permutations. Experiments can be re-tested by relaying sound with newly matched scenes and edited sequences while pace and rhythm of edit cuts can be changed.

Trying these different permutations demands uncomfortable screenings with friends and small audiences. These screenings leave you wondering about the different and culturally determined interpretations that people make of television programs. Interpretations occur in ways that are often different to expected outcomes held by filmmakers, academics or producers.

Sometimes audience readings can be contrary and subversive to the culture that produced the material in the first instance. Susan Dermody signals a process which is both necessary and permissible in searching for those ‘best fit’ permutations that will ultimately constitute a film.

 * One purpose of this paper is to try to identify something about a mode of film that is not quite documentary nor yet quite fiction, not essay, not poetry, not anything yet satisfyingly described by a single word. I am interested in trying to tease some of these things out, because I’m in the midst of trying to write a film like this, to get it right.
 * It is a mode that may use images collected or created from the real world, documentary’ evidence from the real world, but is free to use them to speak of something other than primarily their ‘factual’ referents as documentary tends to do. It uses images to be truthful rather than necessarily factual; and sometimes too be truthful toward private rather than necessarily public realities.
 * But all of this is in one way or another in service of an engagement with reality, a communal sharing of reality, a commutas such as the best documentaries always provide at some level. [Susan Dermody. – The Subjective Voice in Documentary The Pressure of the Unconscious upon the Image. In Fields of Vision, Essays in Film Studies, Visual Anthropology, and Photography. Leslie Devereaux and Roger Hillman – editors. University of California Press, 1995, 296.]
 * But all of this is in one way or another in service of an engagement with reality, a communal sharing of reality, a commutas such as the best documentaries always provide at some level. [Susan Dermody. – The Subjective Voice in Documentary The Pressure of the Unconscious upon the Image. In Fields of Vision, Essays in Film Studies, Visual Anthropology, and Photography. Leslie Devereaux and Roger Hillman – editors. University of California Press, 1995, 296.]
 * But all of this is in one way or another in service of an engagement with reality, a communal sharing of reality, a commutas such as the best documentaries always provide at some level. [Susan Dermody. – The Subjective Voice in Documentary The Pressure of the Unconscious upon the Image. In Fields of Vision, Essays in Film Studies, Visual Anthropology, and Photography. Leslie Devereaux and Roger Hillman – editors. University of California Press, 1995, 296.]

title: Video journalism/Editing Grammar


text:
Video journalism/Ethics of the Gaze

This topic is a work in progress...

Discussions around journalistic representation and fairness, or notions of privacy verses public interest, or comment on the subjectivity-objectivity balance; often become bogged down in their own problematics. The contradictions and problematics within journalism filmmaking are the centre of this work-in progress discussion which deliberately exposes ethical dilemmas in filmmaking. This topic reviews questions of representation and ethical practice in video journalism that employs the new, portable and unobtrusive digital camera. In particular, it looks at questions of ethics surrounding the camera gazing in on the very private.

Altered world-views
It is often said that television has altered our world. In the same way, people often speak of a new world, a new society, a new phase of history, being created – ‘brought about’ – by this or that new technology...

Most of us know what is generally implied when such things are said. But this may be the central difficulty: that we have got so used to statements of this general kind, in our most ordinary discussions, that we fail to realise their specific meanings. [Raymond Wiliams Television – Technology and Cultural Form Fontana. UK, 1974, 9.]

Read this essay for some back-grounding: Ethics in the Immersive Documentary by John Dentino December 2013

http://sensesofcinema.com/2013/feature-articles/ethics-in-the-immersive-documentary/

Academic deliberation
Writing about these issues works at finding a place in culture – for journalism, TV news and documentary. It also allows for a critique of the accountability in news and documentary in the context of the new world information order and perhaps most importantly, in the context of the documentary subject (the social actor).

Film making ethics provides the opportunity to analyse and critique genre and write about changes, trends and processes within documentary journalism and film culture. In this context, one can provide a reflexivity that is difficult to maintain in academic writing on documentary when there is no direct contact or experience with the medium.

title: Video journalism/Ethics of the Gaze


text:
User talk:Zhaomeng

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Hello
Hi, thanks for your edits to Invariant Tasks: Skill Builder Practice today. Looking forward to your contributions on Wikiversity. --HappyCamper 06:18, 4 February 2012 (UTC)

Copyright problems on photos
Hello. There is a problem with one or more files you have uploaded. Every file you upload must have a clear statement of the copyright position. If you took the photos yourself, add PD-self if you are placing it unreservedly in the public domain. If you wish to retain minimal rights, add Cc-by-sa-3.0. If you did not take the photo, please add an explanation of why it can be hosted here. I am afraid that if you do not do this, the file wil be deleted. Let me know if you have any problems.--Collingwood (talk) 11:50, 2 July 2012 (UTC)
 * After you have uploaded the photo, edit the page. If the photo is somewhere on the web and that place explains the copyright situation, just add a link to that page.  Otherwise, you can say something like "the author has given me permission".  If you made the photo yourself, then as I say add PD-self if you are placing it unreservedly in the public domain, otherwise add  .  We work here on the principle of good faith, so we will believe whatever you put.--Collingwood (talk) 21:58, 6 October 2012 (UTC)

title: User talk:Zhaomeng