University of Florida/Egm6341/s10.Team2/HW4

Statement
P. 20-1

Develop higher-order corrected trapezoidal rules for $$ CT_{1}(n) \quad CT_{2}(n) \quad CT_{3}(n) $$

Then use these new corrected trapezoidal rules to find:

$$ I = \int_{0}^{1}\frac{\exp^{x}-1}{x}dx $$

until the error is of order of $$ 10^{-6} \quad $$ using $$\; n=2,4,8,16,32...$$

Solution
Higher Order corrected trapezoidal rules are of the form:

$$ \; CT_{k}(n) = CT_{k-1}(n) + a_{k}h^{2*k} $$

where the a coefficient is defined as follows:

$$ a_{i} = d_{i} \left[ f^{(2i-1)}(b) - f^{(2i-1)}(a) \right] $$

and $$ di=\frac{-B_{2i}}{(2i)!} $$

B are Bernoulli numbers. The first values for $$ d_{i} \quad $$ are:

$$ \; d_{1}=\frac{-1}{12} \qquad d_{1}=\frac{1}{720} \qquad d_{3}=\frac{-1}{30240} $$

These corrected trapezoidal rules define the integral as follows:

$$ \; I= CT_{k}(n) + \Theta (h^{(2*(k+1)}) $$

This means that higher order corrected trapezoidal rules can be developed as long as the function is continuous and its higher order derivatives can be calculated.

The first three corrected trapezoidal rules are found as follows:

$$\; CT_{1}(n) = CT_{0}(n) + a_{1}h^{2} $$

$$\; CT_{2}(n) = CT_{1}(n) + a_{2}h^{4} $$

$$\; CT_{3}(n) = CT_{2}(n) + a_{2}h^{6} $$

Using matlab to calculate the integrals, yields the results as follows:

It can be noticed that the corrected trapezoidal rules yield a solution to the integral much faster than the uncorrected Composite Trapezoidal Rule.

Matlab Code Used for Problem
 MATLAB Code: Finding the Corrected Trapezoidal Rules 

 MATLAB Code: Calculating the Values of the function 

 MATLAB Code: Calculating the Composite Trapezoidal Rule 

solved by--- Egm6341.s10.Team2.GV 21:56, 3 March 2010 (UTC)

checked by---Egm6341.s10.team2.patodon 20:53, 3 March 2010 (UTC)

Problem 2: Higher order error analysis of Trapezoidal rule for periodic functions

 * For which n does $$\; E_{n}^{1} \simeq 0$$

Statement
To prove that the value of the trapezoidal rule error: $$\; E_{n}^{1} \simeq 0$$ for a periodic function and find the effect of $$\; n$$(number of intervals or panels in trapezoidal rule) on the convergence of the function.

Solution
Let $$f\left ( x \right )$$ be a periodic function in the interval $$\left [ a,b \right ]$$

We know that for a periodic function with periodicity of $$\; b-a,$$ $$\; f\left ( a \right )= f\left ( a+\left ( b-a \right ) \right )=f\left ( b \right )$$

Differentiating $$\; f\left ( x \right )$$ w.r.t $$\; x$$ :

$$f'\left ( a \right ) = f'\left ( a +\left ( b-a \right )\right )=f'\left ( b \right )$$

similarly,

$$f^{k}\left ( a \right ) = f^{k}\left ( a +\left ( b-a \right )\right )=f^{k}\left ( b \right )$$

Therefore the coefficients of the expansion of $$\; E_{n}^{1}$$ i.e. $$\; a_{1},a_{2},a_{3},a_{4}$$ becomes zero due the following reason:

We know that:

$$a_{i}= d_{i}\times \left [ f^{\left (2i-1 \right )}\left ( b \right )-  f^{\left (2i-1  \right )}\left ( a \right )\right ]$$ where $$\; d_{i}= \frac{-B_{2i}}{\left (2i  \right )!}$$

$$f^{\left (2i-1 \right )}\left ( b \right ) \simeq  f^{\left (2i-1  \right )}\left ( a \right )$$ when periodicity of the function is $$\; b-a$$

Therefore, in the case of periodic function $$f\left ( x \right )$$ with periodicity $$\; b-a$$, $$\; a_{i}\simeq 0 \; \forall \; i=1,2,3...$$

Hence the error expressed as:

$$\; E_{n}^{1}= \sum_{i=1}^{\infty }a_{i} h^{2i} \simeq 0$$ as $$\; a_{i} \simeq 0 \; \forall \; i=1,2,3...$$

The same result is tested by using a Matlab program for trapezoidal rule and it proves to be true that the function converges rapidly for periodic functions using trapezoidal rule as the number of intervals 'n' increases and thus the error term $$\; E_{n}^{1}$$ tends to zero.

solved by-Srikanth Madala

checked by---Egm6341.s10.team2.patodon 20:55, 3 March 2010 (UTC)

Statement
Discuss the pros and cons of the following quadrature methods:


 * 1) Taylor Series
 * 2) Composite Trapezoidal rule
 * 3) Composite Simpson's rule
 * 4) Romberg table (including Richardson's extrapolation)
 * 5) Corrected Trapezoidal rule

Solution
 Taylor Series:   Pros: 


 * 1) Powerful tool that helps to approximate any type of simple or complex functions into a polynomial form $$\;P_{n}(x)$$ with an error in the form of remainder $$\;R_{n+1}(x)$$

 Cons: 


 * 1) The function $$\;f(x)$$ should be differentiable in the given limits 'a' and 'b' in which it is approximated as a polynomial.
 * 2) The product of two sub-functions in a given function $$\;f(x)$$ like for example $$e^{x}.\cos \left ( x \right )$$ would make it more difficult and more costly (in terms of computation) to evaluate successive derivatives.

 Composite Trapezoidal rule:   Pros: 
 * 1) Very intuitive and easy to use
 * 2) The numerical integral results obtained for periodic functions like trigonometric functions (sinx, cosx, tanx etc) is very accurate owing to the fact that the error is dependent on the difference between the odd derivatives of the limiting points - 'a' and 'b'.
 * 3) It yields better accuracy than simple trapezoidal rule in calculating the numerical integral of a function.

 Pros: 


 * 1) Needs many subintervals to converge and the convergence rate is low for non-periodic functions.
 * 2) Runge Phenomena.

 Composite Simpson's rule:   Pros: 
 * 1) It is still intuitive and easy to use
 * 2) The convergence rate is OK
 * 3) Exact for cubic polynomials

 Cons: 
 * 1) This rule can be applied to only even number of finite intervals between the limits- 'a' and 'b'. So the value of 'n' in Simpson's rule is always an even number. i.e. n=2i; where (i=1,2,3...)
 * 2) Runge Phenomena

 Romberg table (including Richardson's extrapolation):   Pros: 
 * 1) The successive computation of $$\;I(2n)$$ is cheaper and faster when $$\;I(n)$$ is already computed.
 * 2) The accuracy of the numerical integration results obtained are better than other methods. The higher the power of 'h' term in the trapezoidal error, the better the accuracy.

 Cons: 
 * 1) The success of Romberg integration is only justified if the integrand-'f' satisfies the hypotheses of the Euler–Maclaurin Theorem.For example, as illustrated in the | An Introduction to Numerical Analysis by Suli and Mayers Text Pg.218 The function $$\; x \to x^{\frac{1}{3}}$$ is not differentiable at $$\; x = 0$$, so the required conditions are not satisfied for any extrapolation.

solved by --- --Srikanth Madala

checked by --- --Egm6341.s10.Team2.GV 22:00, 3 March 2010 (UTC)

Statement
Given the expression


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

$$  \displaystyle \boldsymbol{E_n^1=\sum _{k=0}^{n-1}[\int_{x_k}^{x_{k+1}}f(x)dx-\frac{h}{2}[f(x_k)+f(x_{k+1})]]} $$     (1)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * 
 * }

where  $$\boldsymbol{x_k=a+kh}$$ and $$\boldsymbol{h=\frac{(b-a)}{n}}$$,

Show that by transformation of variables we get


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

$$  \displaystyle \boldsymbol{E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}[\int_{-1}^{1}g_k(t)dt-[g_k(-1)+g_k({+1})]]} $$     (2)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * 
 * }

where $$\boldsymbol{g_k(t)=f(x(t)); x\epsilon [x_k,x_{k+1}]}$$

Solution
Given the expression
 * {| style="width:100%" border="0"

$$  \displaystyle \boldsymbol{E_n^1=\sum _{k=0}^{n-1}[\int_{x_k}^{x_{k+1}}f(x)dx-\frac{h}{2}[f(x_k)+f(x_{k+1})]]} $$     (1)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * 
 * }

we use the transformation $$\displaystyle{x_k = a+kh}$$ to transform the interval from $$\displaystyle{[x_k,x_{k+1}]}$$       to      $$\displaystyle{[-1,1]}$$     by introducing  $$\displaystyle{[x_k,x_{k+1}]}$$.We get

$$\displaystyle{x(-1)=x_k}$$

$$x(0)=\frac{x_k+x_{k+1}}{2}$$

$$\displaystyle{x(1)=x_{k+1}}$$

substituting in (1) we get
 * {| style="width:100%" border="0"

$$  \displaystyle E_n^1=\sum _{k=0}^{n-1}[\int_{-1}^{+1}f(x(t))dt-\frac{h}{2}[f(x(-1))+f(x(+1)]] $$
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


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

$$  \displaystyle E_n^1=\sum _{k=0}^{n-1}[\int_{-1}^{+1}f(t\frac{h}{2}+0)dt-\frac{h}{2}[f(x(-1))+f(x(+1)]] $$
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


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

$$  \displaystyle E_n^1=\sum _{k=0}^{n-1}[\int_{-1}^{+1}f(t\frac{h}{2}+0)dt-\frac{h}{2}[f(x(-1))+f(x(+1)]] $$
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

taking $$\displaystyle{\frac{h}{2}}$$ out we get
 * {| style="width:100%" border="0"

$$  \displaystyle \boldsymbol{E_n^1=\frac{h}{2}\sum _{k=0}^{n-1}[\int_{-1}^{1}g_k(t)dt-[g_k(-1)+g_k({+1})]]} $$     (2) solved by---Egm6341.s10.team2.niki 17:46, 27 February 2010 (UTC)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * 
 * }

checked by--- --Srikanth Madala

Statement
Show the following relation via integration by parts

$$\int_{-1}^{1}(-t)\cdot g^{(1)}(t) = \int_{-1}^{1}g(t)dt - \left [ g(-1)+ g(1) \right ] $$

Solution
Integration by Parts

$$\begin{align} \int_{-1}^{1}g(t)dt &= \int_{-1}^{1}g(t)\cdot (1)dt\\ &= \left [ g(t)\cdot (t) \right ]_{-1}^{1} - \int_{-1}^{1}t \cdot g^{(1)}(t)\\ &= \left [ g(-1)+g(1) \right ] +\int_{-1}^{1}(-t)\cdot g^{(1)}(t)\\ \end{align}$$

Rearrangement yields

$$\int_{-1}^{1}(-t)\cdot g^{(1)}(t) = \int_{-1}^{1}g(t)dt - \left [ g(-1)+ g(1) \right ]$$

solved by---Egm6341.s10.team2.lee 01:45, 2 March 2010 (UTC)

checked by---Egm6341.s10.team2.niki 05:23, 2 March 2010 (UTC)

checked by---Egm6341.s10.team2.patodon 20:48, 3 March 2010 (UTC)

Statement
The following is defined:

$$ g_{k}(t)= f(x(t)) $$

where,

$$ x(t) = t\frac{h}{2} + \frac{x_{k}+x_{k+1}}{2} $$

Proof the following:

$$ g_{k}^{i} = \left( \frac{h}{2} \right) ^{i} f^{i}(x(t)) $$

Solution
The first step is to state the chain rule for a composite function which is the case in this situation,

$$ Y = F(G(x)) $$

$$ \frac{dY}{dx} = F^{'}(G(x))G^{'}(x) $$

The first derivative is found as follows:

$$ g_{k}(t) = f(x(t)) $$

$$ g_{k}^{'}(t) = f^{'}(x(t))x^{'}(t) $$

$$ x^{'}(t) = \frac{h}{2} $$

By this the first derivative is defined as:

$$ g_{k}^{(1)}(t) = \frac{h}{2}f^{'}(x(t)) $$

Using the chain rule again to find the second derivative as follows:

$$ g_{k}^{(2)}(t) = \frac{h}{2} \left[ \frac{d}{dt}f^{1}(x(t)) \right] $$

$$ g_{k}^{(2)}(t) = \left( \frac{h}{2} \right) ^{2} f^{(2)}(x(t)) $$

Apply the chain rule once more to find the third derivative as follows:

$$ g_{k}^{(3)}(t) = \left( \frac{h}{2} \right) ^{2} \frac{d}{dt}f^{(2)}(x(t)) $$

$$ g_{k}^{(3)}(t) = \left( \frac{h}{2} \right) ^{3} f^{(3)}(x(t)) $$

One notices that the value of $$ \frac{h}{2} $$ comes out of the derivative everytime it is applied.

One is able to then conclude a general expression for the derivative of the function as follows:

$$ g_{k}^{(i)}(t) = \left( \frac{h}{2} \right) ^{i} f^{(i)}(x(t)) $$

solved by---Egm6341.s10.Team2.GV 05:03, 27 February 2010 (UTC)

checked by---Egm6341.s10.team2.patodon 21:10, 3 March 2010 (UTC)

Statement
Show the following relation via integration by parts

$$\int_{-1}^{1}g^{(1)}(t)\cdot P_{1}(t)dt = \left [ P_{2(t)}\cdot g^{(1)}(t) \right ]_{-1}^{1} - \int_{-1}^{1}P_{2}(t)\cdot g^{(2)}(t)dt$$

Solution
Integration by Parts

$$\int_{-1}^{1}g^{(1)}(t)\cdot P_{1}(t)dt = \left [ P_{2(t)}\cdot g^{(1)}(t) \right ]_{-1}^{1} - \int_{-1}^{1}P_{2}(t)\cdot g^{(2)}(t)dt$$

$$\mbox{where,} \qquad P_{1}(t)=-t \qquad \mbox{and} \qquad P_{2}(t)=\int P_{1}(t)dt = -\frac{t^{2}}{2}+c$$

solved by---Egm6341.s10.team2.patodon 20:59, 3 March 2010 (UTC)

checked by---Egm6341.s10.team2.lee 02:15, 2 March 2010 (UTC)

Statement
Install Chebfun program (Tefethen et al. 2004)

Solution
Status - Successful

solved by---

checked by---Egm6341.s10.team2.patodon 21:27, 3 March 2010 (UTC)

Statement
For the integral $$\displaystyle{I=\int_{0}^{1}\frac{e^x-1}{x}dx}$$
 * 1) Optimise the code for Romberg table
 * 2) Compare second column of Romberg table with the results of Simpson's rule and derive any relationship.
 * 3) Compute $$\displaystyle{I_n}$$ to the order of $$\displaystyle{10^{-10}}$$ and find the computational time using


 * Code for Composite Trapezoidal Rule
 * Code for Composite Simpson's Rule
 * Code for Romberg table
 * Chebfun "sum" command
 * Built-in MATLAB function "trapz"
 * Built-in MATLAB function "quad"
 * Built-in MATLAB function "clencurt"

Solution
Exact value of the Integral from Mathematica is calculated by teh command "Integrate[((exp(x)-1)/x),+{x,+0,+1}]" the value is

1.3179021514544038948600088442492318379749012457927839928404611969976461077561394826119536468343922075 []]

Part1:Corrected code for Romberg table
 MATLAB Code: For Efficient Production of the Romberg Table 

Part2:Comaprison of Simpson's rule and Romberg table
The below table shows the values for the given integral computed using Composite Simpsons rule and the second column of the Romberg table for the same. By comparing the values of the Integral (or of the corresponding errors)) for any n, we see that the value from the composite Simpson's rule is the same as the value for n/2 (previous row) of $$\displaystyle{T_1(n)}$$.

For e.g : Denoting Simpson's as $$\displaystyle{S(n)}$$ we see that
 * {| style="width:100%" border="0"

$$  \displaystyle \displaystyle{S(16)=T_1(8)} $$
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }
 * {| style="width:100%" border="0"

$$  \displaystyle \displaystyle{S(32)=T_1(16)} $$
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


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

$$  \displaystyle \displaystyle{S(64)=T_1(32)} $$ etc. In general we can write
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


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

$$  \displaystyle \displaystyle{S(2^i)=T_1(2^{i-1})} $$
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

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

$$  \displaystyle T_1(2^i)=\frac{4T(2^i)-T(2^{i-1})}{3} $$
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }


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

$$  \displaystyle \therefore S(2^i)=\frac{4T(2^i)-T(2^{i-1})}{3} $$     (1)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * 
 * }

Part3:Codes
 Using Code for Composite trapezoidal rule  Gives n= 32768 ,In =1.31790215149321 and time 4.973 sec

Using Code for Composite Simpson's Rule Gives n= 128 ,In= 1.31790215146089 and time=0.017sec

 Using code for Romberg table 

Gives time =0.04sec

Using chebfun - sum: tic y = chebfun('(exp(x)-1)./x',[eps 1]); I=sum(y) toc

I = 1.317902151454404 and time = 0.009175 sec.

Using Trapz function: n=32768 ,In=1.31790215149321 and time = 0.006879 sec

Using Quad function Gives n=2 ,In= 1.31790215145441 and time = 0.004 sec

Using clencurt function tic [x,w] = clencurt(2^5); xx=(1+eps)/2+x.*(1-eps)/2; ww=w.*(1-eps)/2; yy=(exp(xx)-1)./xx; II=ww*yy toc

Gives II=1.317902151454404 and time = 0.000308 sec

Part3:Computational times
solved by---Egm6341.s10.team2.niki 18:20, 28 February 2010 (UTC)

checked by---Egm6341.s10.team2.lee 22:20, 2 March 2010 (UTC)

Statement
Repeat Problem 9 for integral $$\displaystyle{I=\int_{-5}^{5}\frac{1}{1+x^2}dx}$$

Solution
The exact value of the definite integral is atan(5)-atan(-5)=2.746801533890034...

own code - composite trapezoidal rule
Function Definition function I = trapzoid(f,a,b,n) h=(b-a)/n; x=linspace(a,b,n+1); fx=feval(f,x); I=h*(fx(1)/2+sum(fx(2:1:n))+fx(n+1)/2); Call Function clear;clc format long f = @(x) 1./(1+x.^2); tic I = trapzoid(f,-5,5,2^14) toc e = atan(5)-atan(-5)-I

Result I = 2.746801532971545 Elapsed time is 0.001745 seconds. e = 9.184866200939723e-010

own code - composite simpson's rule
Function Definition function In = simpson(f,a,b,n) % n must be a positive even integer h=(b-a)/n; %step size x=linspace(a,b,n+1); %(n+1) equally spaced points fx=feval(f,x); %function evaluations %Composite Simpson In=(h/3)*(fx(1)+4*sum(fx(2:2:n))+2*sum(fx(3:2:n-1))+fx(n+1));

Call Function clear;clc format long f = @(x) 1./(1+x.^2); tic I = simpson(f,-5,5,2^8) toc e = atan(5)-atan(-5)-I

Result I = 2.746801533727021 Elapsed time is 0.000343 seconds. e = 1.630109380812428e-010

Romberg quadrature rule
Function Definition function I_romb = rombquad(f,a,b,m) h = 2.^((1:m)-1)*(b-a)/2^(m-1);            % intervals used. k1 = 2.^((m-2):-1:-1)*2+1;                 % Index into intervals. f1 = feval(f,a:h(1):b);                    % Function evaluations. R = zeros(1,m);                            % Pre-allocation. % Define the starting vector: for ii = 1:m R(ii) = 0.5*h(ii)*(f1(1)+2*sum(f1(k1(end-ii+1):k1(end-ii+1)-1:end-1)) + f1(end)); end % Interpolations: for jj = 2:m jpower = (4^(jj-1)-1); for kk = 1:(m-jj+1) R(kk) = R(kk)+(R(kk)-R(kk+1))/jpower; end end I_romb = R(1);

Call Function clear;clc format long y = @(x) 1./(1+x.^2); tic I_romb=rombquad(y,-5,5,9) toc e=atan(5)-atan(-5)-I_romb

Result I_romb = 2.746801534346014 Elapsed time is 0.000356 seconds. e = -4.559819188898473e-010

matlab - trapz and quad
Call Function clear;clc format long x=linspace(-5,5,2^14+1); y = 1./(1+x.^2); I_exact=atan(5)-atan(-5) tic;I_ct=trapz(x,y);toc I_ct e_ct=I_exact-I_ct tic;I_cs=quad(@(x) 1./(1+x.^2),-5,5,1e-8);toc I_cs e_cs=I_exact-I_cs

Result I_exact = 2.746801533890032 Elapsed time is 0.003191 seconds. I_ct = 2.746801532971568 e_ct = 9.184639715442700e-010 Elapsed time is 0.009436 seconds. I_cs = 2.746801534322983 e_cs = -4.329510083778132e-010

matlab - clencurt
Call Function tic [xx,ww] = clencurt(2^10); xx=xx.*5; ww=ww.*5; yy=1./(1+xx.^2); II=ww*yy toc

Result II = 2.746801533890032 Elapsed time is 0.089293 seconds.

"sum" command of chebfun matlab toolbox
Call Function x = chebfun('x',[-5 5]); tic I=sum(1./(1+x.^2)) toc e=abs(atan(5)-atan(-5)-I)

Result I = 2.746801533890033 Elapsed time is 0.042520 seconds. e = 1.332267629550188e-015

solved by---Egm6341.s10.team2.lee 22:22, 2 March 2010 (UTC)

checked by-Egm6341.s10.team2.niki 20:45, 6 March 2010 (UTC)

Statement


Find the Arc Length of the ellipse where:

$$ r(\theta) = \frac{1 - e^{2}}{1 - e \cos(\theta)} $$

and

$$ eccentricity=e= \sin(\frac{\pi}{12}) $$

Using:

1)Composite Trapezoidal Rule

2)Composite Simpson's Rule

3)Romberg Table

4)"Sum" Command of the chebfun toolbox

5)"Trapz" Command of Matlab

6)"Quad" Command of Matlab

7)"Clencurt" Command of Matlab

Until the error is less than $$ 10^{-6} $$

Composite Simpson's Rule
 MATLAB Code: 

Romberg Table
Romberg Table for n=2

6.2832   6.0026    6.0722    6.0691    6.0691    6.0691    6.0727    6.0679    6.0692    6.0691    6.0691         0    6.0691    6.0691    6.0691    6.0691         0         0    6.0691    6.0691    6.0691         0         0         0    6.0691    6.0691         0         0         0         0    6.0691         0         0         0         0         0

Table for Error of Romberg Table

-0.2141   0.0665   -0.0031   -0.0000    0.0000   -0.0000   -0.0036    0.0012   -0.0001    0.0000   -0.0000         0   -0.0000    0.0000   -0.0000   -0.0000         0         0   -0.0000   -0.0000   -0.0000         0         0         0   -0.0000   -0.0000         0         0         0         0   -0.0000         0         0         0         0         0

Romberg Table for n=4

6.0727   6.0679    6.0692    6.0691    6.0691    6.0691    6.0691    6.0691    6.0691    6.0691    6.0691         0    6.0691    6.0691    6.0691    6.0691         0         0    6.0691    6.0691    6.0691         0         0         0    6.0691    6.0691         0         0         0         0    6.0691         0         0         0         0         0

Error for Romberg Table n=4

-0.0036   0.0012   -0.0001    0.0000   -0.0000   -0.0000   -0.0000    0.0000   -0.0000   -0.0000   -0.0000         0   -0.0000   -0.0000   -0.0000   -0.0000         0         0   -0.0000   -0.0000   -0.0000         0         0         0   -0.0000   -0.0000         0         0         0         0   -0.0000         0         0         0         0         0

Romberg Table for n=8

6.0691   6.0691    6.0691    6.0691    6.0691    6.0691    6.0691    6.0691    6.0691    6.0691    6.0691         0    6.0691    6.0691    6.0691    6.0691         0         0    6.0691    6.0691    6.0691         0         0         0    6.0691    6.0691         0         0         0         0    6.0691         0         0         0         0         0

Table for Error when n=8

1.0e-005 *

-0.1180   0.0281   -0.0109   -0.0084   -0.0085   -0.0085   -0.0085   -0.0085   -0.0085   -0.0085   -0.0085         0   -0.0085   -0.0085   -0.0085   -0.0085         0         0   -0.0085   -0.0085   -0.0085         0         0         0   -0.0085   -0.0085         0         0         0         0   -0.0085         0         0         0         0         0

"Sum" Command of Chebfun Matlab Toolbox
$$ In= 6.069090959564776 $$

$$ Error= I-In = -8.457248767967940e-008 $$

"Quad" Command of Matlab
The value found using the "Quad" Matlab Command was:

$$ I=6.069090875 $$

with an error smaller than $$10^{-6} $$

solved by--- --Egm6341.s10.Team2.GV 22:04, 3 March 2010 (UTC)

checked by---Egm6341.s10.team2.patodon 21:15, 3 March 2010 (UTC)

Signatures
solved problems 4 and 9, proofread problems 5 and 10:--Egm6341.s10.team2.niki 18:22, 28 February 2010 (UTC)

solved problems 5 and 10, proofread problems 7 and 9:--Egm6341.s10.team2.lee 22:26, 2 March 2010 (UTC)

solved problems 7, proofread problems 1, 2, 5, 6, 8 and 11 Egm6341.s10.team2.patodon 23:01, 2 March 2010 (UTC)

solved problems 1, 6 and 11. Proofread 3, 9. --Egm6341.s10.Team2.GV 22:06, 3 March 2010 (UTC)

solved problems 3 and 8. Proofread 4, 9--Srikanth Madala