User:Egm6341.s11.team3/sub5

= 5.1 =

Given: A number of functions with their finite bounds
I)$$ I = \int_0^6 \frac{dx}{1+(x-3)^2}, x \in [0, 6] \ $$

II)$$ I = \int_0^1 x^{\frac{1}{3}}dx, x \in [0, 1] \ $$

Find: Integrate them by different integration methods
1. Find the integral by using Composite Trapezoid, Composite Simpson's rules and plot the functions

a) By using uniform discretization b) By using non-uniform discretization

2. Use Gauss-Lagrange method to approximate the function

Composite Trapezoid rule is given by Vu Quoc's notes
$$ I_n^t = h[\frac{1}{2}f_0 + f_1 +f_2 +.... + f_{n-1} + \frac{1}{2}f_n] $$

Matlab code was written to calculate the generic functions (need to input function and B.C. and interval and the code generates number of elements, approximate integral value, and the error.

For:

I)$$ I = \int_0^6 \frac{dx}{1+(x-3)^2} \ $$

Exact versus approximate function plots, blue is exact/analytical and red points are approximate function at 100 points

II)$$ I = \int_0^1 x^{\frac{1}{3}}dx \ $$

Matab output

Exact versus approximate function plot

Composite Simpson rule is given by Vu Quoc's notes is
$$ I_n^t = \frac{h}{3}[f_0 + 4f_1 +2f_2 +4f_3 +2f_4 +.... + 2f_{n-2} + 4f_{n-1} + f_n] $$

Matlab code was written to calculate the generic functions (need to input function and B.C. and interval and the code generates number of elements, approximate integral value, and the error.

For:

I)$$ I = \int_0^6 \frac{dx}{1+(x-3)^2} \ $$

Exact versus approximate function plots, blue is exact/analytical and red points are approximate function at 100 points

II)$$ I = \int_0^1 x^{\frac{1}{3}}dx \ $$

Matab output

Exact versus approximate function plot

Gauss - Legendre quadrature rule
For this method we had to use the variable transformation since the defined Gauss-Legendre quadrature is in the interval $$ [-1, +1] \ $$

I)$$ I = \int_0^6 \frac{dx}{1+(x-3)^2} \in x = [0, 6] \ $$

Therefore we converted to new value "z", that is $$ x \in [0, 6] \rightarrow z \in  [-1, +1] \ $$

Therefore, the conversion became:

$$ z(x) = \frac{x}{3} - 1 \ $$

For check, we checked that z is linear in x as well as boundary conditions.

$$ x = min = 0 \rightarrow z(0) = \frac{0}{3} - 1 = -1 \ $$

$$ x = max = 6 \rightarrow z(6) = \frac{6}{3} - 1 = 1 \ $$

Therefore, it's consistent

Note that 1st thing was done in conversion from "x" to "z(x)" was adjustment of length "z" = 2, while "x" = 6, therefore it was divided by 3 1st and then adjusted to fit the $$ [-1, +1] \ $$ interval.

Finally, since

$$ 3*z(x) = x \ $$

the final integral of Gauss - Legendre quadrature had to be multiplied by factor of "3" to represent the integrated function.

Below is attached matlab code for Gauss - Legendre quadrature for n = 2, 3, 4, 5

and the Gauss - Legendre quadrature integration output on that function

For

II)$$ I = \int_0^1 x^{\frac{1}{3}}dx \ $$

$$ z = 2x - 1 \ $$

and the Gauss - Legendre quadrature integration output on that function

--Egm6341.s11.team3.JA 03:21, 24 March 2011 (UTC) and Kim

= 5.2 =

Given

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

$$\underline{F} = \underline{I} + \triangle \underline{A} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 5.2.1)
 * }

$$\underline{I} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\quad \underline{A} = \begin{pmatrix} -0.2& 1 \\ -1 & -0.2 \end{pmatrix}, \quad \triangle = 0.02 \quad \underline{x_0}= \dbinom{{{x}_{0}}^{1}}{{{x}_{0}}^{2}} $$


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

$$LSSM: \quad \underline{x}_{k+1} = \underline{F} \quad \underline{x}_{k}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 5.2.2)
 * }

Find
$$ Run \quad LSSM \quad and \quad plot \quad({x}_{j}, j=0,1,...)$$

Solution
$$ By \quad (Eq 5.2.1) \quad $$ $$ \underline{F}=\begin{pmatrix} 0.9960 & 0.200 \\ -0.200 & 0.9960 \end{pmatrix}$$

$$ By \quad (Eq 5.2.2) \quad build \quad code \quad$$

Result is



Given

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

$$\lim_{k \to \infty}\underline{x}_{k}=\lim_{k \to \infty} \underline^{k+1}\underline{{x}_{0}} = \underline{\hat x}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 5.2.3)
 * }

Find
Find Equilibrium   Point plot $$\underline{\hat x} $$ as big red dot $$\underline{{x}_{0}} $$as big blue dot

Solution
By (Eq5.2.3)

Result is



Given

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

$$ LSSMRN: \quad \underline{{x}_{k+1}} = \underline{F} \quad \underline{x_k} + \underline{G} \quad \underline{w}_{k+1}$$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 7.0.1)
 * }

$$\underline{G}= \dbinom{1}{1} \alpha\ $$ where $$\alpha\ = 0.5, 1, 2$$

Find
Plot $$ ( \underline{{x}_{j}}, \quad j=0,1,2,...)$$

Solution
By Eq(5.2.4)

Result is when $$\alpha\ = 0.5$$



when $$\alpha\ = 1$$

when $$\alpha\ = 2$$

Given
Cauchy Random Noise

Find
Plot $$ ( \underline{{x}_{j}}, \quad j=0,1,2,...)$$ in single-slit diffraction

Solution
by (Eq5.2.4) with Cauchy Random Noise

Result is Egm6341.s11.team3.sungsikkim 20:43, 22 March 2011 (UTC)

= 5.3 =

Given:
Given a pdf function $$ f(x) \quad $$ and the defination of Quartile points: $$ Q_1, Q_3 \quad $$ 
 * {| style="width:100%" border="0"

$$ F(Q_1) = P(x $$ \displaystyle
 * }


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

$$ F(Q_3) = P(x $$ \displaystyle (Eq. 5.3.1)
 * }

The Cauchy pdf $$ C(x_0,r) \quad $$ and Gauss (normal) pdf $$ N(\mu,\sigma) \quad $$ are defined as following:


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

$$ C(x_0,r) := \frac{r}{\pi[r^2+(x-x_0)^2]} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 5.3.2)
 * }


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

$$ N(\mu,\sigma) := \frac{1}{\sigma\sqrt{2\pi}} e^{\frac{-(x-\mu)^2}{2\sigma^2}} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 *  $$ \displaystyle (Eq. 5.3.3)
 * }

with the half width represented by $$ r^C \quad $$ and $$ r^G \quad $$ respectively.

Find:
1) Find $$\quad Q_1, Q_3 $$ for $$ C(x_0,r) \quad $$.

2) Find $$\quad Q_1, Q_3 $$ for $$ N(\mu,\sigma) \quad $$.

3) Let $$ x_0 = \mu = 0 \quad $$ and $$ r^C = 1 \quad $$. Find $$ \sigma^1 \quad $$ so that $$ r^G =1 \quad $$. Plot $$ C(x_0,r) \quad $$ and $$ N(\mu,\sigma) \quad $$.

4) Find: $$ Q_1^C, Q_3^C \quad $$ for $$ C(x_0,r) \quad $$ and $$ C(0,1) \quad $$, and $$ Q_1^G, Q_3^G \quad $$ for $$ N(\mu,\sigma) \quad $$ and $$ N(0,\sigma^1) \quad $$.

Plot: $$ Q_1^C, Q_3^C \quad $$ for $$ C(0,1) \quad $$,and $$ Q_1^G, Q_3^G \quad $$ for $$ N(0,\sigma^1) \quad $$. Comment on results.

Solution:
1)

Integrate Cauchy pdf function, we get

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \int C(x_0,r)dx = \int \frac{r}{\pi[r^2+(x-x_0)^2]} dx = \frac{1}{\pi} tan^{-1}(\frac{x-x_0}{r}) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 5.3.4)
 * }

For $$ Q_1 \quad $$, we have

<span id="(1)">
 * {| style="width:100%" border="0"

$$ F(Q_1) = \int_{-\infty}^{Q_1} \frac{r}{\pi[r^2+(x-x_0)^2]} dx = \frac{1}{\pi} [tan^{-1}(\frac{Q_1-x_0}{r}) - tan^{-1}(\frac{-\infty-x_0}{r})] = 0.25 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Rightarrow tan^{-1}(\frac{Q_1-x_0}{r}) = -\frac{\pi}{4} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Rightarrow Q_1 = x_0 - r $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

For $$ Q_3 \quad $$, we have

<span id="(1)">
 * {| style="width:100%" border="0"

$$ F(Q_3) = \int_{-\infty}^{Q_3} \frac{r}{\pi[r^2+(x-x_0)^2]} dx = \frac{1}{\pi} [tan^{-1}(\frac{Q_3-x_0}{r}) - tan^{-1}(\frac{-\infty-x_0}{r})] = 0.75 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Rightarrow tan^{-1}(\frac{Q_3-x_0}{r}) = \frac{\pi}{4} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Rightarrow Q_3 = x_0 + r $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

2)

Integrate Gauss pdf function with the help from WolframAlpha, we get

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \int N(\mu,\sigma)dx = \int \frac{1}{\sigma\sqrt{2\pi}} e^{\frac{-(x-\mu)^2}{2\sigma^2}} dx = \frac{1}{2} erf(\frac{x-\mu}{\sqrt{2}\sigma}) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 5.3.5)
 * }

For $$ Q_1 \quad $$, we have

<span id="(1)">
 * {| style="width:100%" border="0"

$$ F(Q_1) = \int_{-\infty}^{Q_1} N(\mu,\sigma)dx = \frac{1}{2} [erf(\frac{Q_1-\mu}{\sqrt{2}\sigma}) - erf(\frac{-\infty-\mu}{\sqrt{2}\sigma})] = 0.25 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Rightarrow erf(\frac{Q_1-\mu}{\sqrt{2}\sigma}) = -0.5 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

solve from WolframAlpha, we get

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \frac{Q_1-\mu}{\sqrt{2}\sigma} = -0.476936 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Rightarrow Q_1 = \mu - 0.674489\sigma $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

For $$ Q_3 \quad $$, we have

<span id="(1)">
 * {| style="width:100%" border="0"

$$ F(Q_3) = \int_{-\infty}^{Q_3} N(\mu,\sigma)dx = \frac{1}{2} [erf(\frac{Q_3-\mu}{\sqrt{2}\sigma}) - erf(\frac{-\infty-\mu}{\sqrt{2}\sigma})] = 0.75 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Rightarrow erf(\frac{Q_3-\mu}{\sqrt{2}\sigma}) = 0.5 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

solve from WolframAlpha, we get

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \frac{Q_3-\mu}{\sqrt{2}\sigma} = 0.476936 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Rightarrow Q_3 = \mu + 0.674489\sigma $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

3)

From defination of half width: <span id="(1)">
 * {| style="width:100%" border="0"

$$ \frac{N}{N_{max}} = e^{\frac{-(r^G)^2}{2(\sigma^1)^2}} = 0.5 \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

solve from WolframAlpha ,

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Rightarrow \frac{-1}{2(\sigma^1)^2} = ln(0.5) = -0.693147 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \Rightarrow \sigma^1 = 0.849322 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 5.3.6)
 * }

The plot of $$ C(x_0,r) \quad $$ and $$ N(\mu,\sigma) \quad $$ are shown in part 4).

4)

From part 1) and part 2)

For $$ C(x_0,r) \quad $$

<span id="(1)">
 * {| style="width:100%" border="0"

$$ Q_1^C = x_0 - r $$ and $$ Q_3^C = x_0 + r $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

For $$ C(0,1) \quad $$

<span id="(1)">
 * {| style="width:100%" border="0"

$$ Q_1^C = -1 $$ and $$ Q_3^C = 1 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

For $$ N(\mu,\sigma) \quad $$

<span id="(1)">
 * {| style="width:100%" border="0"

$$ Q_1^G = \mu - 0.674489\sigma $$ and $$ Q_3^G = \mu + 0.674489\sigma $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

For $$ N(0,\sigma^1) \quad $$, with $$ \sigma^1 \quad $$ given in (Eq. 5.3.6)

<span id="(1)">
 * {| style="width:100%" border="0"

$$ Q_1^G = -0.572858 $$ and $$ Q_3^G = 0.572858 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

The plot is shown in the following figure:



Comments: By comparing the probability density fuctions of Cauchy and Gauss with the same width, a heavy tail of the Cauchy pdf is observed. Furthermore, because of the heavy tail, the distance between the 1st and 3rd quartile points of Cauchy pdf are larger than that of Gauss pdf.

Egm6341.s11.team3.Xia 2:00, 22 March 2011 (UTC)

Given
Applications and implementation logic for 5 different numerical methods given below ,

1. Taylor Series 2. Composite Trapezoidal rule 3. Composite Simpson's rule 4. Romberg Richardson's rule 5. CTk

Find
The Pros and Cons of the 5 different techniques

Solution
 Taylor Series : 

 Pros: 

1. Approximates simple or complex functions as a polynomial with remainder representing the error.

2. Higher order T-Series can solve for oscillatory functions as well.

 Cons: 

1. One of the main drawback is that the function should be differentiable in the given limits in which it is approximated as a polynomial.

2. Number of iterations can be large depending on the problem. But depending on the problem in-hand we could choose a smaller or larger interval ,which in turn would decrease or increase the processing time / resources respectively.

Composite Trapezoidal rule:

 Pros: 

1. Periodic functions can be analyzed with ease and accuracy because the error is dependent on the difference between the odd derivatives of the limits.

2. Better flexibility/operational procedure than simple trapezoidal rule and more accurate.

 Cons: 

1. Convergence time lapse is very slow when compared to other techniques.

2. We need to make sure to have sufficiently large number of intervals in the domain.

Composite Simpson's rule:

 Pros: 

1. Exactly solves for polynomials of order 3 or less.

2. It is often advantageous to use sub-intervals of different lengths, and concentrate on the places where the integrand is not well-behaved.

 Cons: 

1. It is applicable to only even number of intervals in the domains. So the value of 'n' in Composite Simpson's rule is always even. i.e. n=2i; where (i=1,2,3,4....)

2. Not possible to choose odd number of intervals because of the even domain.

 Romberg Richardson's Rule: 

 Pros: 

1. The successive computation of $$\;I(2n)$$ is less resource consuming and less computational time when $$\;I(n)$$ is already available.

2. Relatively higher accuracy when compared to the other methods and is pretty robust.

3. Applied for energy cascade in turbulence and in weather predictions

4. Applicable to numerical integration of ODEs to estimate the error.

 Cons: 

1. Romberg integration is successful only when integrand satisfies the hypotheses of the Euler–Maclaurin.

2. The integrand must have continuous derivatives, though fairly good results may be obtained if only a few derivatives exist

 CTk: 

 Pros: 

1. Similar to their counter part, it is useful for solving periodic functions exactly

 Cons: 

1. The function derivatives have to be solved for first ,at the end points of the domain which is given.

2. Consumes considerably more memory resources compared to Composite rule.

= 5.8  =

Given:
<span id="(1)">
 * {| style="width:100%" border="0"

$$ E_n^T = \sum_{k=0}^{n-1} [\int_{x_k}^{x_{k+1}} f(x)dx - \frac{h}{2}(f(x_k) + f(x_{k+1}))] \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 5.8.1)
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ x(t) := \frac{x_k+x_{k+1}}{2} + \frac{h}{2} t \quad $$, $$ t \in [-1,1] \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 5.8.2)
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ g_k(t) := f(x(t)) \quad $$, $$ x \in [x_k,x_{k+1}] \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 5.8.3)
 * }

Find:
1) Derive the following equation:

<span id="(1)">
 * {| style="width:100%" border="0"

$$ E_n^T = \frac{h}{2} \sum_{k=0}^{n-1} [\int_{-1}^{1} g_k(t)dt - (g_k(-1) + g_k(1))] \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

2) Derive the following equation:

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \int_{-1}^{1} (-t)g^{(1)}(t)dt = \int_{-1}^{1} g(t)dt - [g(-1) + g(1)] \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

3) Derive the following equation:

<span id="(1)">
 * {| style="width:100%" border="0"

$$ g_k^{(i)}(t) = (\frac{h}{2})^if^{(i)}(x(t)) \quad $$, $$ x \in [x_k,x_{k+1}] \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

Solution:
1) From (Eq. 5.8.1), (Eq. 5.8.2) and (Eq. 5.8.3)

<span id="(1)">
 * {| style="width:100%" border="0"

$$ E_n^T = \sum_{k=0}^{n-1} [\int_{x_k}^{x_{k+1}} f(x(t))dx(t) - \frac{h}{2}(g_k(-1) + g_k(1))] \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = \sum_{k=0}^{n-1} [\int_{x_k}^{x_{k+1}} g_k(t)d(\frac{x_k+x_{k+1}}{2} + \frac{h}{2} t) - \frac{h}{2}(g_k(-1) + g_k(1))] \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = \sum_{k=0}^{n-1} [\int_{-1}^{1} g_k(t) \frac{h}{2}dt - \frac{h}{2}(g_k(-1) + g_k(1))] \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = \frac{h}{2} \sum_{k=0}^{n-1} [\int_{-1}^{1} g_k(t)dt - (g_k(-1) + g_k(1))] \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

2) Integration by parts:

<span id="(1)">
 * {| style="width:100%" border="0"

$$ \int_{-1}^{1} (-t)g^{(1)}(t)dt \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = [(-t) \int g(t)^{(1)}dt]_{-1}^{1} - \int_{-1}^{1} (-t)^{(1)} (\int g(t)^{(1)}dt) dt \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = [(-t) g(t)]_{-1}^{1} - \int_{-1}^{1} (-t)^{(1)} g(t) dt \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = [(-1) g(1)-(1) g(-1)] - \int_{-1}^{1} (-1) g(t) dt \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = \int_{-1}^{1} g(t) dt - [g(1)+ g(-1)] \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

3) We can build from i=1:

<span id="(1)">
 * {| style="width:100%" border="0"

$$ g_k^{(1)}(t) = f^{(1)} x^{(1)}(t) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = f^{(1)} \frac{h}{2} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

so

<span id="(1)">
 * {| style="width:100%" border="0"

$$ g_k^{(2)}(t) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = (g_k^{(1)}(t))^{(1)} \frac{h}{2} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = (f^{(1)} \frac{h}{2})^{(1)} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = f^{(2)} x^{(1)}(t) \frac{h}{2} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = f^{(2)} (\frac{h}{2})^{2} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle (Eq. 5.8.4)
 * }

From observation of (Eq. 5.8.4), it is implied:

<span id="(1)">
 * {| style="width:100%" border="0"

$$ g_k^{(i-1)}(t) = f^{(i-1)} (\frac{h}{2})^{i-1} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

so

<span id="(1)">
 * {| style="width:100%" border="0"

$$ g_k^{(i)}(t) \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = (f^{(i-1)} (\frac{h}{2})^{i-1})^{(1)} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = f^{(i)} x^{(1)}(t) (\frac{h}{2})^{i-1} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

<span id="(1)">
 * {| style="width:100%" border="0"

$$ = f^{(i)} (\frac{h}{2})^{i} \quad $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right"> $$ \displaystyle
 * }

Egm6341.s11.team3.Xia 20:00, 23 March 2011 (UTC)