University of Florida/Egm6341/s10.Team2/HW4
Problem 1: Development of higher order Corrected Trapezoidal rules and application
editStatement
editDevelop higher-order corrected trapezoidal rules for
Then use these new corrected trapezoidal rules to find:
until the error is of order of using
Solution
editHigher Order corrected trapezoidal rules are of the form:
where the a coefficient is defined as follows:
and
B are Bernoulli numbers. The first values for are:
These corrected trapezoidal rules define the integral as follows:
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:
Using matlab to calculate the integrals, yields the results as follows:
Integral Values using Corrected Trapezoidal Rules
| ||||||
n | Ct1 | Ct2 | Ct3 | Error Ct1 | Error Ct2 | Error Ct3 |
2 | 1.317875061148223 | 1.317908476816203 | 1.317908314863727 | 2.709080863771973e-005 | 6.324859342043965e-006 | 6.162906866391538e-006 |
4 | 1.317900452799411 | 1.317902541278660 | 1.317902538748152 | 1.699157449630562e-006 | 3.893217990214026e-007 | 3.867912916621918e-007 |
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
editMATLAB Code: Finding the Corrected Trapezoidal Rules
function [Ct1, Ct2, Ct3, Ect1, Ect2, Ect3] = cortrap (a,b,n)
tic
h=(b-a)/n;
x=linspace(a,b,n+1);
%declare the Function and its derivatives
F=(exp(x) - 1)./x;
F(1)=1;
F1=(exp(x)./x - (exp(x) - 1)./x.^2);
F1(1)=0.5;
F2=(exp(x)./x - (2*exp(x))./x.^2 + (2*(exp(x) - 1))./x.^3);
F2(1)=(1/3);
F3=(exp(x)./x - (3*exp(x))./x.^2 + (6*exp(x))./x.^3 - (6*(exp(x) - 1))./x.^4);
F3(1)=.25;
%singularity values were inserted
%calculating a1, a2, and a3 coefficients
r=n+1; %finds out the last term of the array
a1=(-1/12)*(F1(r)-F1(1)); %Bernoulli number d1=-1/12
a2=(1/720)*(F2(r)-F2(1)); %Bernoulli number d2=1/720
a3=(-1/30240)*(F3(r)-F3(1)); %Bernoulli number d3=-1/30240
%Calculating Ct0 using composite trapezoidal rule
Ct0=ctrapz(n);
Ct1=Ct0 + (a1*(h^2));
Ct2=Ct1 + (a2*(h^4));
Ct3=Ct2 + (a3*(h^6));
I=quad(@valus,0,1)
Ect1=abs(I-Ct1);
Ect2=abs(I-Ct2);
Ect3=abs(I-Ct3);
toc
MATLAB Code: Calculating the Values of the function
function F= valus(x)
F=(exp(x)-1)./x;
MATLAB Code: Calculating the Composite Trapezoidal Rule
function I=ctrapz(n)
i=0;
Itot=0;
It=0;
It2=0;
h=0;
while i<=n
if i==0
Itot1=1;
else if i<n
h=1/n;
It(i)=2*valus(h*i);
else
It2=valus(1);
end
end
Itot=Itot1+sum(It)+It2;
i=i+1;
end
I=Itot/(2*n);
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
edit- For which n does
Statement
editTo prove that the value of the trapezoidal rule error: for a periodic function and find the effect of (number of intervals or panels in trapezoidal rule) on the convergence of the function.
Solution
editLet be a periodic function in the interval
We know that for a periodic function with periodicity of
Differentiating w.r.t :
similarly,
Therefore the coefficients of the expansion of i.e. becomes zero due the following reason:
We know that:
where
when periodicity of the function is
Therefore, in the case of periodic function with periodicity ,
Hence the error expressed as:
as
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 tends to zero.
%As illustrated below in the Matlab code for a periodic function f=sin(x) between the periodicity (b-a)=(2*pi-0), the Error 'E'%
%tends to zero rapidly with the increase in the number of panels-'n'. In the below code the 'n' value varies from 1 through 10.%
%Error of trapezoidal rule as even power of 'h'%
tic;
clc;
clear;
syms x;
f=sin(x);
a=0;
b=(2*pi);
d=[-1/12;1/720;-1/30240];
for n=1:10
h=(b-a)/n;
E1=0;
for i=1:3
A(i)=d(i).*(subs(diff(f,2i-1),x,b)-subs(diff(f,2i-1),x,a));
E1=E1+A(i)*(h^(2*i));
end
E(n)=E1;
end
toc;
solved by-----Srikanth Madala
checked by---Egm6341.s10.team2.patodon 20:55, 3 March 2010 (UTC)
Problem 3: Discussions of pros and cons of various quadrature methods
editStatement
editDiscuss the pros and cons of the following quadrature methods:
- Taylor Series
- Composite Trapezoidal rule
- Composite Simpson's rule
- Romberg table (including Richardson's extrapolation)
- Corrected Trapezoidal rule
Solution
editTaylor Series:
Pros:
- Powerful tool that helps to approximate any type of simple or complex functions into a polynomial form with an error in the form of remainder
Cons:
- The function should be differentiable in the given limits 'a' and 'b' in which it is approximated as a polynomial.
- The product of two sub-functions in a given function like for example would make it more difficult and more costly (in terms of computation) to evaluate successive derivatives.
Composite Trapezoidal rule:
Pros:
- Very intuitive and easy to use
- 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'.
- It yields better accuracy than simple trapezoidal rule in calculating the numerical integral of a function.
Pros:
- Needs many subintervals to converge and the convergence rate is low for non-periodic functions.
- Runge Phenomena.
Composite Simpson's rule:
Pros:
- It is still intuitive and easy to use
- The convergence rate is OK
- Exact for cubic polynomials
Cons:
- 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...)
- Runge Phenomena
Romberg table (including Richardson's extrapolation):
Pros:
- The successive computation of is cheaper and faster when is already computed.
- 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:
- 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 is not differentiable at , 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)
Problem 4: Transformation of variables in the error expression of HO trapezoidal rule
editStatement
editGiven the expression
(1)
where and ,
Show that by transformation of variables we get
(2)
where
Solution
editGiven the expression
(1)
we use the transformation to transform the interval from to by introducing .We get
substituting in (1) we get
taking out we get
(2)
solved by---Egm6341.s10.team2.niki 17:46, 27 February 2010 (UTC)
checked by--- --Srikanth Madala
Problem 5: Integration by parts
editStatement
editShow the following relation via integration by parts
Solution
editIntegration by Parts
Rearrangement yields
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)
Problem 6: Derivative of as it relates to the proof to the theorem of higher order trapezoidal rules
editStatement
editThe following is defined:
where,
Proof the following:
Solution
editThe first step is to state the chain rule for a composite function which is the case in this situation,
The first derivative is found as follows:
By this the first derivative is defined as:
Using the chain rule again to find the second derivative as follows:
Apply the chain rule once more to find the third derivative as follows:
One notices that the value of 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:
solved by---Egm6341.s10.Team2.GV 05:03, 27 February 2010 (UTC)
checked by---Egm6341.s10.team2.patodon 21:10, 3 March 2010 (UTC)
Problem 7: Verifying the Error for Higher Order Trapezoidal Rule
editStatement
editShow the following relation via integration by parts
Solution
editIntegration by Parts
solved by---Egm6341.s10.team2.patodon 20:59, 3 March 2010 (UTC)
checked by---Egm6341.s10.team2.lee 02:15, 2 March 2010 (UTC)
Problem 8: Installation of chebfun program
editStatement
editInstall Chebfun program (Tefethen et al. 2004)
Solution
editStatus - Successful
solved by---
checked by---Egm6341.s10.team2.patodon 21:27, 3 March 2010 (UTC)
Problem 9: Comparison of codes
editStatement
editFor the integral
- Optimise the code for Romberg table
- Compare second column of Romberg table with the results of Simpson's rule and derive any relationship.
- Compute to the order of 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
editExact value of the Integral from Mathematica is calculated by teh command "Integrate[((exp(x)-1)/x),+{x,+0,+1}]" the value is
1.3179021514544038948600088442492318379749012457927839928404611969976461077561394826119536468343922075 [[((exp(x)-1)/x),+{x,+0,+1}]]
Part1:Corrected code for Romberg table
editMATLAB Code: For Efficient Production of the Romberg Table
function [table] = Rombergg(a,b,n)
tic
table=zeros(6,6);
%Generate 1st column of Romberg Table using built in composite trapezoidal
%function. Notice it calls the function 'Valus' which obtains the values
%for the function. The function is also corrected at initial point 0
for j=1:6
r=n;
x=linspace(a,b,r+1);
FX=valus(x);
FX(1)=1; %because of 0 start point
table(j,1)=trapz(x,FX);
n=2*r;
end
%Generate 2nd column of Romberg Table
for j=1:5
table(j,2)= ([(2^2)*table(j+1,1)] - [table(j,1)])/(2^2 - 1);
end
%Generate 3rd column of Romberg Table
for j=1:4
table(j,3) = ([(2^(2*2))*table(j+1,2)] - table(j,2))/(2^(2*2)-1);
end
%Generate 4th column of Romberg Table
for j=1:3
table(j,4) = ([(2^(2*3))*table(j+1,3)] - table(j,3))/(2^(2*3)-1);
end
%Generate 5th Column of Romberg Table
for j=1:2
table(j,5) = ([(2^(2*4))*table(j+1,4)] - table(j,4))/(2^(2*4)-1);
end
%Generate 6th column of Romberg Table
table(1,6) = ([(2^(2*5))*table(2,5)] - table(1,5))/(2^(2*5)-1);
toc
Part2:Comaprison of Simpson's rule and Romberg table
editThe 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 .
Comparison of Integral Values
| |||||
Number of Terms |
Values from |
Values from Romberg Table |
Error |
Error |
|
2 |
1.3180086656766800 |
1.3179089166831400 |
1.06514E-04 |
6.76523E-06 |
|
4 |
1.3179089166831400 |
1.3179025760028100 |
6.76523E-06 |
4.24548E-07 |
|
8 |
1.3179025760028100 |
1.3179021780157100 |
4.24548E-07 |
2.65613E-08 |
|
16 |
1.3179021780157100 |
1.3179021531149100 |
2.65613E-08 |
1.66051E-09 |
|
32 |
1.3179021531149100 |
1.3179021515581900 |
1.66051E-09 |
1.03790E-10 |
|
64 |
1.3179021515581900 |
- |
1.03790E-10 |
- |
For e.g: Denoting Simpson's as we see that
etc. In general we can write
wkt
(1)
Part3:Codes
editUsing Code for Composite trapezoidal rule
I =1.31790215145440389486; %Exact value of the integral
n = 1;
clc
format long;
tic %clock begins
while(true)
In = ctrapz(n); %calls ctrapz function which is the code used previously in the homework
if(abs((I - In)) < (10^(-10)))%check for error
break;
end
n = 2*n;
end
toc%clock stops
output = In
number = n %number of terms used
Gives n= 32768 ,In =1.31790215149321 and time 4.973 sec
Using Code for Composite Simpson's Rule
I = 1.31790215145; %Exact value of the integral
n = 1;
clc
format long;
tic%time starts
while(true)
In = simpb(0,1,n); %calls function simpb which is the code developed
if(abs((I - In)) < (10^(-10))) %check for order of error
break;
end
n = 2*n;
end
toc%clock stops
output = In % integral value calculated
number = n %gives number of terms
Gives n= 128 ,In= 1.31790215146089 and time=0.017sec
Using code for Romberg table
function [table] = Rombergg(a,b,n)
tic
table=zeros(6,6);
%Generate 1st column of Romberg Table using built in composite trapezoidal
%function. Notice it calls the function 'Valus' which obtains the values
%for the function. The function is also corrected at initial point 0
for j=1:6
r=n;
x=linspace(a,b,r+1);
FX=valus(x);
FX(1)=1; %because of 0 start point
table(j,1)=trapz(x,FX);
n=2*r;
end
%Generate 2nd column of Romberg Table
for j=1:5
table(j,2)= ([(2^2)*table(j+1,1)] - [table(j,1)])/(2^2 - 1);
end
%Generate 3rd column of Romberg Table
for j=1:4
table(j,3) = ([(2^(2*2))*table(j+1,2)] - table(j,2))/(2^(2*2)-1);
end
%Generate 4th column of Romberg Table
for j=1:3
table(j,4) = ([(2^(2*3))*table(j+1,3)] - table(j,3))/(2^(2*3)-1);
end
%Generate 5th Column of Romberg Table
for j=1:2
table(j,5) = ([(2^(2*4))*table(j+1,4)] - table(j,4))/(2^(2*4)-1);
end
%Generate 6th column of Romberg Table
table(1,6) = ([(2^(2*5))*table(2,5)] - table(1,5))/(2^(2*5)-1);
toc
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:
I = 1.31790215145; %Exact value of integral
n = 1;
tic %clock begins
while(true)
X = 0:1/n:1; %divides x into n equal parts
Y = (exp(X) - 1) ./ X; %given function
Y(1) = 1; %corrects for the singulairty at x=0
In = trapz(X,Y); %calls teh built-in function trapz
if(abs((I - In)) < (10^(-10))) %check order of error
break;
end
n = 2*n;
end
toc %clock ends
output = In
number = n
n=32768 ,In=1.31790215149321 and time = 0.006879 sec
Using Quad function
I =1.3179021514544038948600088; %Exact value of integral
n = 2;
clc
format long;
tic %clock begins
while(true)
X = 0:1/n:1; %divides the interval into n equal parts
F=@(X)(exp(X)-1)./X; %given function as an anonymous function
In=quad(F,0,1,1.e-10); %calls the built-in quad function
if(abs((I - In)) < ((10)^(10))) %check for error
break;
end
n = n+1;
end
toc
output = In
number = n
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
edit
Computational Times
| ||
|
|
|
Composite Trapezoidal |
32768 |
4.93 |
Composite Simpsons |
128 |
0.017 |
Romberg Table |
32 |
0.04 |
Chebfun-sum |
NA |
0.009175 |
Trapz |
32768 |
0.006879 |
Quad |
2 |
0.004 |
Clencurt |
16 |
0.000308
|
solved by---Egm6341.s10.team2.niki 18:20, 28 February 2010 (UTC)
checked by---Egm6341.s10.team2.lee 22:20, 2 March 2010 (UTC)
Problem 10: Another comparison of codes
editStatement
editRepeat Problem 9 for integral
Solution
editThe exact value of the definite integral is atan(5)-atan(-5)=2.746801533890034...
own code - composite trapezoidal rule
editFunction 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
editFunction 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
editFunction 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
editCall 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
editCall 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
editCall 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)
Problem 11: Finding The Arc Length of an Ellipse using Numerical Integration
editStatement
editFind the Arc Length of the ellipse where:
and
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
Solution
editComposite Trapezoidal Rule
edit
Composite Trapezoidal Rule
| ||
n | Value | Error |
2 | 6.283185307 | 2.E-01 |
4 | 6.072738504 | 4.E-03 |
8 | 6.069092055 | 1.E-06 |
16 | 6.06909096 | 8.E-08 |
function [I,E]=compostrap(a,b,n)
h=(b-a)/n;
pi2=2*pi();
i=0;
Itot=0;
It=0;
It2=0;
while i<=n
if i==0
Itot1=(1/2)*elrad(a);
else if i<n
u=h*i;
It(i)= elrad(u);
else
It2=(1/2)*elrad(b);
end
end
Itot=Itot1+sum(It)+It2;
i=i+1;
end
I=Itot*(h);
%Error Calculation
Ir=quad(@elrad,0,pi2);
E=Ir-I;
Composite Simpson's Rule
edit
Composite Simpson's Rule
| ||
n | Value | Error |
2 | 5.470081296 | 6.E-01
|
4 | 5.867072234 | 2.E-01 |
8 | 6.000117905 | 7.E-02 |
16 | 5.870432233 | 2.E-01 |
32 | 5.969761779 | 1.E-01 |
64 | 6.101815883 | 3.E-02 |
128 | 6.085453421 | 2.E-02 |
256 | 6.07727219 | 8.E-03 |
512 | 6.062882886 | 6.E-03 |
1024 | 6.071136267 | 2.E-03 |
2048 | 6.067538941 | 2.E-03 |
4096 | 6.06831495 | 8.E-04 |
8192 | 6.069346623 | 3.E-04 |
16384 | 6.068896957 | 2.E-04 |
32768 | 6.069154875 | 6.E-05 |
65536 | 6.069122917 | 3.E-05 |
131072 | 6.069106939 | 2.E-05 |
262144 | 6.069098949 | 8.E-06 |
524288 | 6.069084897 | 6.E-06 |
1048576 | 6.069092957 | 2.E-06 |
2097152 | 6.069091958 | 1.E-06 |
MATLAB Code:
function I = msimpb(a,b,w)
q=1;
i=a;
n=0;
Sum=0;
c=0;
e=sin(pi()/12);
while n<w
n=2^q;
h=(a+b)/n;
while i<=b
fx= (1-e^2)/(1-e*cos(i)); %(exp(i)-1)/(i); Declare Function Here
if i==a
Sum=1;
else if i==b
Sum=Sum+fx;
else if i==(b-h)
Sum=Sum+(4*fx);
else if rem(c,2)==0
Sum=Sum+(2*fx);
else
Sum=Sum+(4*fx);
end
end
end
end
c=c+1;
i=i+h;
end
n;
In=Sum*(h/3);
I=In;
i=a;
c=0;
q=q+1;
Sum=0;
end
Romberg Table
editRomberg 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
edit
>> chebfun('elrad(x)', [0 pi2])
ans =
chebfun column (1 smooth piece)
interval length endpoint values
( 0, 6.3) 43 1.3 1.3
vertical scale = 1.3
>> sum(F1)
ans =
6.069090959564776
>> I=quad(@elrad,0,pi2)
I =
6.069090874992289
>> Error=I-ans
Error =
-8.457248767967940e-008
"Trapz" Command of Matlab
edit
n-value |
Numerical Integral (In) |
Error (En) |
2 | 6.283185307 | 2.E-01 |
4 | 6.072738504 | 4.E-03 |
8 | 6.069092055 | 1.E-06 |
"Quad" Command of Matlab
editThe value found using the "Quad" Matlab Command was:
with an error smaller than
solved by--- --Egm6341.s10.Team2.GV 22:04, 3 March 2010 (UTC)
checked by---Egm6341.s10.team2.patodon 21:15, 3 March 2010 (UTC)
Contributing Authors and Assignments
editSignatures
editsolved 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
Solution Assignments and Reviewers
edit
Problem Assignments
| ||
Problem # | Solution | Reviewed |
Problem 1 | GV | PO |
Problem 2 | PO | JP |
Problem 3 | SM | GV |
Problem 4 | NN | SM |
Problem 5 | JP | NN |
Problem 6 | GV | PO |
Problem 7 | PO | JP |
Problem 8 | SM | GV |
Problem 9 | NN,GV | SM |
Problem 10 | JP | NN |
Problem 11 | GV | PO |