home work ii

1 | P a g e Richardson's Extrapolation

Numerical Methods (COSC 607)

Home Work II

Richardson's Extrapolation First and second derivatives approximation

Group No 1 Group Members:-

1. Biruk Ambachewu 2. Ayalnesh Tigabie 3. Admasu Chane 4. Hidya Nurhusen 5. Wasie Legesse 6. Abebe Alemu Submitted by:-

Name Abebe Alemu Balcha ID No. GSR/3003/08

Submitted to:- Teacher Prof. Okey Oseloka Onyejekwe July , 2016


Table of Contents

1. Introduction .......................................................................................................................................... 3

2. Overview ............................................................................................................................................... 3

2.1. Taylor's series ................................................................................................................................ 3

2.2. Extrapolation ................................................................................................................................. 7

3. Scope of work........................................................................................................................................ 8

4. Problem Definition One ........................................................................................................................ 8

4.1. Richardson's' Extrapolation using centered formula for First Derivatives .................................... 9

4.1.1. Part I: - Working By hand ...................................................................................................... 9

4.1.2. Analysis on Problem one ..................................................................................................... 15

4.1.3. PART II: - Computer Program using MATLAB ...................................................................... 16

4.2. Richardson's' Extrapolation using centered formula for Second Derivatives ............................. 22

4.2.1. PART I:- Working by Hand .................................................................................................. 22


5. Problem definition Two ...................................................................................................................... 32

5.1. Richardson's Extrapolation first derivatives for even power Order of h .................................... 32

5.1.1. PART I :- working by hand ................................................................................................... 32


5.2. Richardson's Extrapolation Using three point Second Derivatives ............................................. 46

5.2.1. PART I: - working by hand ................................................................................................... 46


6. Observation ......................................................................................................................................... 53

Richardson extrapolation algorithm for central differences ................................................................ 53


1. Introduction This paper is prepared for numerically finding Richardson's extrapolation using centered

formula for First derivatives and second derivatives, in first, second and third orders, by

changing and dividing equal fractions, the step size h towards zero. This helps us to

extrapolate the first and second derivatives to approximate the first and second derivatives at

the reference point x.

2. Overview

2.1. Taylor's series

Taylor series is a representation of a function as an infinite sum of terms that are

calculated from the values of the function's derivatives at a single point. A Taylor series

is a series expansion of a function about a point. A one-dimensional Taylor series is an

expansion of a real function about a point is given by

'' (3) ( )' 2 3( ) ( ) ( )

( ) ( ) ( )( ) ( ) ( ) ... ( )2! 3! !

nnf a f a f a

f x f a f a x a x a x a x an

A series expansion is a representation of a particular function as a sum of powers in one

of its variables, or by a sum of powers of another (usually elementary) function .

Taylors series use for

Evaluating definite Integrals: Some functions have no antiderivative which can be

expressed in terms of familiar functions. This makes evaluating definite integrals of

these functions difficult because the Fundamental Theorem of Calculus cannot be

used. If we have a polynomial representation of a function, we can oftentimes use that

to evaluate a definite integral.

Understanding asymptotic behavior: Sometimes, a Taylor series can tell us useful

information about how a function behaves in an important part of its domain.

Understanding the growth of functions

Solving differential equations

The applications of Taylor series is mainly to approximate ugly functions into nice

ones (polynomials)!

Taylor series provide the basic method for computing transcendental functions such

as ex, sinx, and cosx.


All of computational science is built on Taylor's theorem

'' (3) ( )2 3

'

( ) ( ) ( )( ) ( ) ( ) ( ) ... ( )

2! 3! !( )

( )

nnf a f a f a

f x f a x a x a x an

f ax a

,

We can use Taylor series to find approximation to the first and second derivatives of f(x)

at the point x (called reference point)

For a given step size h we can have the following formula that derived from Taylors

series

I.

'' 2 3 3'

'' 2 3 3

'

'' 2 3 3'

( ) ( )( ) ( ) ( ) ...

2! 3!

( ) ( )( ) ( ) ( ...)

2! 3!( )

( ) ( ) ( ) ( )( ) , ( ...)

2! 3!

f x h f x hf x h f x f x h

f x h f x hf x h f x

f xh

f x h f x f x h f x hf x where error

h

' ( ) ( )( )

f x h f xf x

h

-------------------------- ---------eq (1)

Called FORWARD DIFFERENCE FORMULA

II.

'' 2 3 3'

'' 2 3 3

'

'' 2 3 3'

( ) ( )( ) ( ) ( ) ...

2! 3!

( ) ( )( ) ( ) ...)

2! 3!( )

( ) ( ) ( ) ( )( ) , ...)

2! 3!

f x h f x hf x h f x f x h

f x h f x hf x f x h

f xh

f x f x h f x h f x hf x where error

h

' ( ) ( )( )

f x f x hf x

h

-------------------------------------------eq (2)


Called backward difference formula

III. Add the last statement of the first and the second expressions we will have the

following expression called Central Difference Formula for estimating first

derivatives

'

'

'

( ) ( )( )

( ) ( )( ) ,

( ) ( ) ( ) ( )2 ( )

f x h f xf x

h

f x f x hf x

h

f x h f x f x f x hf x

h

' ( ) ( )( )

2

f x h f x hf x

h

-------------------------------------- eq(3)

IV. To get the Second derivatives and lets rewrite the first and second expressions as

follows

'

'

( ) ( ) ''( )( )

2

( ) ( ) ''( )( )

2

f x h f x f x hf x

h

f x f x h f x hf x

h

Let's subtract the first one from the second one and we will have


( ) ( ) ''( ) ( ) ( ) ''( )( )

2 2

( ) ( ) ''( ) ( ) ( ) ''( )( )

2 2

( ) ( ) ( ) ( ) ''( ) ''( )0

2 2

''( ) ( ) 2 ( ) ( )

2

f x h f x f x h f x f x h f x h

h h

f x h f x f x h f x f x h f x h

h h

f x h f x f x f x h f x f xh h

h h h h

f x f x h f x f x hh

h h h

Thus 2

( ) 2 ( ) ( )''( )

f x h f x f x hf x

h

------------------------------------ eq(4)

V. Order of the error:- to get the order of the error O(h), let's start again form the

forward, backward difference method and finally we can deduce the central

difference methods. Which is

VI. Forward Difference

2

2

( ) ( ) '( ) ( )

'( ) ( ) ( ) ( )

( ) ( )'( ) ( )

f x h f x f x h O h

f x h f x h f x O h

f x h f xf x O h

h

so that here

the order is in the order of h

VII. Back ward difference

2

2

( ) ( ) '( ) ( )

'( ) ( ) ( ) ( )

( ) ( )'( ) ( )

f x h f x f x h O h

f x h f x f x h O h

f x f x hf x O h

h

here is also

it is in the order of h or which is call is first order

VIII. Central difference

(2) 2 (3) 3 (4) 4

(2) 2 (3) 3 (4) 4

( ) ( ) ( )( ) ( ) '( ) ...

2! 3! 4!

( ) ( ) ( )( ) ( ) '( ) ...

2! 3! 4!

f x h f x h f x hf x h f x f x h

f x h f x h f x hf x h f x f x h

subtract From the above expression s a and b, we have down and we will get the

following expressions


(3) 3

2

( )( ) ( ) 2 '( ) 2 ...

3!

( ) ( )'( ) ( )

2

f x hf x h f x h f x h

f x h f x hf x O h

h

Here it is in the order of two, for small h, or if the step size h decreases, we will

have more accurate value, or the error will be very small. So that, the central

difference formula is more accurate than other two formulas.

IX. Let's make the forward difference to have second order

' ( ) ( ) ''( )( )

2

f x h f x f x hf x

h

Taylor series and equation 1

Substitute equation 4 in the second derivatives, and we would have

2 12

1

( ) 2 ( ) ( )( ) ( )

( )2

i i i

i ii

f x f x f xf x f x hf x h

h

2 1( ) 4 ( ) 3 ( )( )

2i i i

i

f x f x f xf x

h

--------------------------eq (5)

Where xi is reference point

Here by substitution of the second derivatives we improve the accuracy y

to order two O(h2)

2.2. Extrapolation Extrapolation: is the process of estimating, beyond the original observation range, the

value of a variable on the basis of its relationship with another variable. It is similar to

interpolation, which produces estimates between known observations, but extrapolation is

subject to greater uncertainty and a higher risk of producing meaningless results.

Extrapolation may also mean extension of a method, assuming similar methods will be

applicable. Extrapolation may also apply to human experience to project, extend, or

expand known experience into an area not known or previously experienced so as to

arrive at a (usually conjectural) knowledge of the unknown

Numerical methods approximates or (extrapolates) to some value, for example from

Taylors first derivatives and central difference formula


F(h) will approximate to ( ) ( )

( ) '( )2

f x h f x hF h f x

h

-----------------eq(6)

2 4 61 1 2 3( ) ...F h R C h C h C h Where

0lim ( )h

R F h

There for Richardson's Extrapolation formula for even power error series is given by

1

( ) ( )2( ) ( ) , 1,2,...

2 4 1

i i

i i i

hF F h

hF h F wherei

--------------------------------------------eq(7)

And for All power error series the Extrapolation Formula is given by

1

( ) ( )2( ) ( ) , 1,2,...

2 2 1

i i

i i i

hF F h

hF h F where i

-----------------------------------------eq(8)

3. Scope of work In Richardson's Extrapolation we can do first and second derivatives in Forward,

Backward and Centered formula for this paper is working only with centered formula. So

that the scope of this work is for the given problem, finding first and second derivatives

using Richardson's extrapolation using centered formula, to determining how the error is

minimized by approximating the step size h to zero. Here the work will be done first by

hand calculation; next I will write a MATLAB code and Plot a graph to compare the

derivative of the function at the given reference point with respect to the step size.

4. Problem Definition One Using Richardson's Extrapolation Centered formula for the function

3( ) cos( ),f x x x With reference point x=2.3

a. Calculate by hand and compare the result with the following given table

b. Prepare an algorithm based on Richardson's Extrapolation centered formula

c. Write a Matlab code and analyze the result with that has done by hand

Extrapolation Table h F_1's F_2's F_3's F-4's 1 -18.03738 1/2 -19.35862 -19.79904 1/4 -19.58246 -19.65707 -19.64761 1/8 -19.63120 -19.64745 -19.64681 -19.64680


Solution

4.1. Richardson's' Extrapolation using centered formula for First Derivatives

4.1.1. Part I: - Working By hand

4.1.1.1. First Order F_1's

Step 1- For step size h=1, and we have reference point x=2.3 the first derivative with order h

From eq(5) we have

2 41 1 2

( ) ( )( ) '( ) ...

2

f x h f x hF h f x C h C h

h

Substitute x and h value in the equation

1

(2.3 1) (2.3 1)(1)

2(1)

f fF

1

(3.3) (1.3)(1)

2(1)

f fF

3 3

1

3.3 cos(2.3) 1.3 cos(1.3)(1)

2(1)F

1

35.937cos(2.3) 2.197cos(1.3)(1)

2(1)F

1

35.937( 0.98747977) 2.197(0.26749883)(1)

2(1)F

1

35.48706049 0.58769493(1)

2F

1

35.48706049 0.58769493 36.07475542(1)

2 2F

There for we have 1(1) 18.037377F ------------------------------------exp (1)

Step 2- For step size h=0.5, and we have reference point x=2.3 the first derivative with order h

x+h=2.3+0.5=2.8 and x-h=2.3-0.5=1.8

3 3

1

2.8 cos(2.8) 1.8 cos(1.8)0.5

2 0.5F

1

21.952cos(2.8) 5.892cos(1.8)0.5

2 0.5F

1

20.68366482 ( 1.32504262)0.5

2 0.5F

1

20.68366482 1.32504262 19.358622200.5

2 0.5 1F

1 0 .5 1 9 .3 5 8 6 2F --------------------------------------------------------exp(2)


Step 3- For step size h=0.25, reference point x=2.3 the first derivative with order h

Here x+h=2.3+0.25=2.55 and x-h=2.3-0.25=2.05, substituting to the equation we have

1

(2.55) (2.05)14 12

4

f fF

3 3

1

2.55 cos(2.55) 2.05 cos(2.05)0.25

2 0.25F

1

16.581375cos(2.55) 8.615125cos(2.05)0.25

2 0.25F

1

13.76342894 3.97219887 9.781230070.25

2 0.25 0.5F

1 0.25 19.58246F ------------------------------------------------------------------exp(3)

Step 4- For step size h=0.125, reference point x=2.3 the first derivative with order h

Here x+h=2.3+0.125=2.425 and x-h=2.3-0.125=2.175

1

(2.425) (2.175)18 12

8

f fF

3 3

1

2.425 cos(2.425) 2.175 cos(2.175)0.125

2 0.125F

1

14.26051556cos(2.425) 10.28910934cos(2.175)0.125

2 0.125F

1

14.26051556( 0.75404811) 10.28910934( 0.56810692)0.125

2 0.125F

1

10.75311489 ( 5.84531419)0.125

2 0.125F

1

10.75311489 ( 5.84531419) 4.907800700.125

2 0.125 0.25F

1

4.907800700.125 19.63120

0.25F

1 0.125 19.63120F ------------------------------------------------------------exp(4)


4.1.1.2. Second Order F_2's

From equation 7 above we have 1

( )2

( )2 4 1

j j

j j j

hF F h

hF h F

1, 2,3,...j

Step 1:- at j=1 and step size h=0.5

From above calculation we do have 12

hF

= -19.358622 and 1( )F h = -18.037377

1 1

1 1 1

( )2

( )2 4 1j

hF F h

hF h F

2 1

19.358622 ( 18.037377)(0.5) 19.358622

4 1F

2

19.358622 18.037377(0.5) 19.358622

4 1F

2

1.32124(0.5) 19.358622

3F

2 (0.5) 19.358622 0.44041F

2 (0.5) 19.79903F ---------------------------------------------------------------exp(5)


1 1

2 1

1 4 2

4 4 4 1j

h hF F

hF F

for j=2 and h=1/4 or h=0.25

14

hF

=-19.58246, 12

hF

=-19.35862

1 1

2 1 1

1 4 2

4 4 4 1

h hF F

hF F

2 1

19.58246 ( 19.358622)0.25 19.58246

4 1F

2 1

19.58246 19.3586220.25 19.58246

4 1F


2

0.223840.25 19.58246

3F

2 0.25 19.58246 ( 0.07461)F

2 0.25 19.58246 0.07461F

2 0.25 19.65707F -------------------------------------------------------------exp(6)


1 1

2 1

1 8 4

8 8 4 1j

h hF F

hF F

for j=1 and h=1/8=0.125 18

hF

=-19.63120

1 1

2 1 1

8 40.125

8 4 1

h hF F

hF F

2

19.63120 19.582460.125 19.63120

4 1F

2

0.048740.125 19.63120

3F

2 0.125 19.63120 ( 0.01625)F

2 0.125 19.63120 0.01625F

2 0.125 19.64745F ------------------------------------------------------------exp(7)


4.1.1.3. Third Order F_3's

From equation 7 above 1

( )2

( )2 4 1

j j

j j j

hF F h

hF h F

1, 2,3,...j we have

1 1

2 1

( )4 2

( )4 4 1

j j

j j j

h hF F

hF h F


24

hF

-19.65707 and 2 ( )2

hF -19.79904

2 2

3 2 2

( )4 2

( )4 4 1

h hF F

hF h F

2 2

3 2 2

4 2

4 4 4 1

h hF F

h hF F

3 2

1 19.65707 ( 19.79904)19.65707

4 4 1F

3

19.65707 19.799040.25 19.65707

16 1F

3

0.141970.25 19.65707

15F

3 0.25 19.65707 0.00946F

3 0.25 19.64761F ------------------------------------------------------------exp(8)


2 2

3 2 2

8 4

8 8 4 1

h hF F

h hF F

2 2

3 2 2

1 8 4

8 8 4 1

h hF F

hF F


3

1 19.64745 ( 19.65707)19.64745

8 16 1F

3

19.64745 19.657070.125 19.64745

15F

3

0.009620.125 19.64745

15F

3 0.125 19.64745 0.00064F

3 0.125 -19.646808F ------------------------------------------------------------exp(9)


3 3

3 1 3 3

8 4

8 8 4 1

h hF F

h hF F

3 3

4 3 3

8 4

8 8 4 1

h hF F

h hF F

4

1 19.64681 ( 19.64761)19.64681

8 64 1F

4

19.64681 19.647610.125 19.64681

63F

4

0.00080.125 19.64681

63F

4 0.125 19.64681 0.00001F

4 0.125 19.64680F ------------------------------------------------------------exp(10)


4.1.2. Analysis on Problem one

Here we have done by hand calculation and we have got the following solutions that

shows from exp (1) to exp (10)

H F_1's F_2's F_3's F_4's

1 -18.03737

0.5 -19.35862 -19.79903

0.25 -19.58246 -19.65707 -19.64761

0.125 -19.63120 -19.64745 -19.646808 -19.64680

This result from the given table is the same, let's find analytically the exact derivative

value of the function

3( ) cos( ),f x x x at the point of x=2.3

3 2 3( cos( )) 3* cos( ) ( sin( ))df dy

x x x x x xdx dx

at x=2.3

2 3'(2.3) 3*(2.3) cos(2.3) 2.3 ( sin(2.3))f = -19.64680

The exact derivative value of the function at the point x=2.3. is equal to -19.64680

Now from the table and the exact value we can deduce that as the order increase, the error

becomes near zero (NOT EQUAL) and we have good numerical approximation. At the

fourth order we have almost the same result as of the exact value


4.1.3. PART II: - Computer Program using MATLAB

4.1.3.1. Algorithm

Step 1:- Set the number of step sizes n

Step 2:- set h(i) as h/2^-(i-1), where i=1 to n

Step 3:-Set F (1,n) for the first order values

Step 4:- evaluate for j--> to n,

F(1,j)=((x+h(j))3cos(x+h(j)) - ( x-h(j))3cos(x-h(j))) /2*h(j)

Step 5:- Calculate for the second, third and fourth order, i=1,..,n

F(i+1,j)=F(i,j)+(F(i,j)-F(i,j-1))/4j-1

Step 6: Get the result and plot

4.1.3.2. MATLAB Program Code

Richardson's Extrapolation using Centered formula for first derivative MATLAB Program

function RichardSonExtraP() % %================================================================== % % This Program is Richardson's' Extrapolation for approximating % % the first derivative % % Written By Abebe Alemu, % % ID GSR/3003/08 % % Addis Ababa University, Faculty of Natural Science % % Computational Science Department % % Graduate Program, -------------- Dated on July 2016 % % ===================================================================== format short; %==============Definition ================ h=zeros(50); h1=zeros(50); h2=zeros(50); h3=zeros(50); F1=zeros(50,50); F2=zeros(50,50); A1=zeros(10); A2=zeros(10); A3=zeros(10); %======Enter Input size from key board ============= fprintf('Please Enter the number of Step sizes n \n n='); n=input(''); A=zeros(n,n); % Assining A1=zeros(n); A2=zeros(n); A3=zeros(n); h=zeros(n);


h1=zeros(n); h2=zeros(n); h3=zeros(n); x=2.3;% ------------Reference point %===========define Step size ============================================== for i=1:n h(i)=2^-(i-1); % Calculating the step size h end %============End ========================================================== %========================================================================== % Richardson's Extrapolation of first order using centered formula for f'(x) %========================================================================== for j=1:n h(j); F1(1,j)=(((x+h(j))^3)*(cos(x+h(j)))-((x-h(j)))^3*cos(x-h(j)))/(2*h(j)); A(j,1)= F1(1,j); A1(j)=A(j,1); h1(j)=h(j); end % % ---- Second order -------------------------------- for k=2:n F2(k,2)=F1(1,k)+((F1(1,k)-F1(1,k-1))/((4^1)-1)); A(k,2)= F2(k,2); A2(k)=A(k,2); h2(k)=h(k); end % % ------------Third, fourth up to n order ----------------- for m=3:n for k=2:m-1 F2(m,k+1)=F2(m,k)+((F2(m,k)-F2(m-1,k))/((4^k)-1)); A(m,k+1)=F2(m,k+1); A3(k)=A(m,k+1); h3(k)=h(k); end end %======================Dispaly====================================== % Display the result of first derivative value of f(x) at x=1.25 % numerically approximating using Richardson extrapolation disp(' f(x)=(x^3)*cos(x), its first derivative approximate value at x=1.25'); disp('==========================================================='); disp(' No h F_1s | F_2s | F_3s | F_4s |' ); disp('==========================================================='); for i=1:n msg1=sprintf('%0.0f %0.4f %0.6f %0.6f %0.6f %0.6f ',... i, h(i), A(i,1), A(i,2), A(i,3), A(i,4) ); disp([' ',msg1,' ']); end % %===================================================================


% % The diagram for n step size and its respective value of f'(x) % %=================================================================== figure; % plot on the same figure windows in to four sub plots %--------------- First order plot----------------------------------- subplot(2,2,1) plot(h1,A1,'--rs'); legend('First Order'); axis([0.0 1.0 -20.0 -17.0]); ylabel('f''(x)'); xlabel('Step size h(n)'); title('First Order Richardson''s Extrapolation using Centered Formula'); %------------- Second order plot------------------------------------ subplot (2,2,2); plot(h2,A2,'*'); legend('Second order'); axis([0.0 1.0 -20.0 -17.0]); ylabel('f''(x)'); xlabel('Step size h(n)'); title('Second Order Richardson''s Extrapolation using Centered Formula'); %--------------Third order plot------------------------------------- subplot(2,2,3); plot(h3,A3,'d'); legend('Third Order'); axis([0.0 1.0 -20.0 -17.0]); ylabel('f''(x)'); xlabel('Step size h(n)'); title('Third Order Richardson''s Extrapolation using Centered Formula'); %----------------- on this quadrant I try to show all three plots------ subplot(2,2,4); plot(h1,A1,'--rs'); hold on; plot(h2,A2,'*'); hold on; plot(h3,A3,'d'); hold off; lgnd=legend('Richardson''s Extrapolation using Centered Formula'); axis([0.0 1.0 -20.0 -17.0]); ylabel('f''(x)'); xlabel('Step size h(n)'); title(lgnd,' Title'); % ----------------All in one new figure windows --------------- figure plot(h1,A1,'--rs'); hold on; plot(h2,A2,'*'); hold on; plot(h3,A3,'d');


hold off; lgnd=legend('Richardson''s Extrapolation using Centered Formula'); axis([0.0 1.0 -20.0 -17.0]); ylabel('f''(x)'); xlabel('Step size h(n)'); title(lgnd,'Title') % % ------------------------END of PROGRAM -------------------- end

4.1.3.3. MATLAB Program Out put

Please Enter the number of Step sizes n

n=4

f(x)=(x^3)*cos(x), its first derivative approximate value at x=1.25

=============================================================

No h F_1s | F_2s | F_3s | F_4s |

=============================================================

1 1.0000 -18.037378 0.000000 0.000000 0.000000

2 0.5000 -19.358622 -19.799037 0.000000 0.000000

3 0.2500 -19.582460 -19.657073 -19.647608 0.000000

4 0.1250 -19.631203 -19.647450 -19.646809 -19.646796

>>


4.1.3.4. Richardson's Extrapolation step size h vs first order derivative Plot Out

put


Plot for Richardson extrapolation with for first order derivatives


4.2. Richardson's' Extrapolation using centered formula for Second Derivatives

4.2.1. PART I:- Working by Hand

From Equation 4 above we have for second derivatives

1 2

( ) 2 ( ) ( )f x h f x f x hF

h

3( ) cos( )f x x x with x=2.3

4.2.1.1. First Order F_1's

Step 1:- at step size h=1

1 2

(2.3 1) 2 (1) (2.3 1)

1

f f fF

3 3 3

1 2

3.3 cos(3.3) 2(2.3 cos(2.3)) 1.3 cos(1.3)

1F

1 2

35.937 cos(3.3) 2(12.167 cos(2.3)) 2.197 cos(1.3)

1F

1 2

35.48706049 2( 8.10650351) 0.58769493

1F

1 2

35.48706049 16.21316070 0.58769493

1F

1

18.68620486

1F

1 18.68620486F

Step 2:- at step size h=0.5

1 2

( ) 2 ( ) ( )f x h f x f x hF

h

1 2

(2.3 0.5) 2 (2.3) (2.3 0.5)

0.5

f f fF

1 2

(2.8) 2 (2.3) (1.8)

0.5

f f fF

3 3 3

1 2

2.8 cos(2.8) 2(2.3 cos(2.8)) 1.8 cos(1.8)

0.5F

1

21.952cos(2.8) 2(12.167 cos(2.3)) 5.832cos(1.8)

0.25F

1

20.68366482 2( 8.10658035) ( 1.32504262)

0.25F


1

20.68366482 16.2131607 1.32504262

0.25F

1

5.79554674

0.25F

1 23.18218696F

Step 3:- at step size h=0.25

1 2

( ) 2 ( ) ( )f x h f x f x hF

h

1 2

(2.3 0.25) 2 (2.3) (2.3 0.25)

0.25

f f fF

3 3 3

1 2

2.55 cos(2.55) 2(2.3 cos(2.3)) 2.05 cos(2.05)

0.25F

1 2

16.581375*cos(2.55) 2(12.167cos(2.3)) 8.615125*cos(2.05)

0.25F

1 2

13.76342894 2( 8.10658035) 3.97219887

0.25F

1

13.76342894 16.2131607 3.97219887

0.0625F

1

1.52246711

0.0625F

1 24.35947376F

Step 4:- At step size h=0.125

1 2

( ) 2 ( ) ( )f x h f x f x hF

h

h=1/8=0.125

1 2

(2.3 0.125) 2 (2.3) (2.3 0.125)

0.125

f f fF

1 2

(2.425) 2 (2.3) (2.175)

0.125

f f fF

3 3 3

1 2

2.425 cos(2.425) 2(2.3 cos(2.3)) (2.175 cos(2.175))

0.125F

1 2

14.260515625cos(2.425) 2(12.167 cos(2.3)) (10.289109375cos(2.175))

0.125F


1

10.75311449 2( 8.10658035) ( 5.84531419)

0.015625F

1

10.75311449 16.2131607 5.84531419

0.015625F

1

0.38526798

0.015625F

1 24.65715072F

4.2.1.2. Second Order F_2's

12

hF

=-23.18218696 and 1F h =-18.68620486


1 1

1 1 1

( )2

( )2 4 1j

hF F h

hF h F

2 1

23.18218696 ( 18.68620486)( ) 23.18218696

4 1F h

2

23.18218696 18.68620486( ) 23.18218696

4 1F h

2

4.4959821( ) 23.18218696

3F h

2 ( ) 23.18218696 ( 1.4986607)F h

2 ( ) 23.18218696 1.4986607F h

2 ( ) 24.68084766F h


1( )2

hF -23.18218696 1( )

4

hF -24.35947376


1 1

2 1 1

4 2

4 4 4 1

h hF F

h hF F

2 1

24.35947376 ( 23.18218696)24.35947376

4 4 1

hF

2

24.35947376 23.1821869624.35947376

4 4 1

hF

2

1.177286824.35947376

4 3

hF

2 24.35947376 ( 0.39242894)4

hF

2 24.35947376 0.499553574

hF

2 24.751902694

hF


1( )4

hF -24.35947376 and 1( )

8

hF -24.65715072

1 1

2 1 1

8 4

8 8 4 1

h hF F

h hF F

2 1

24.65715072 ( 24.35947376)24.65715072

8 4 1

hF

2 1

24.65715072 24.3594737624.65715072

8 4 1

hF

2

0.2976769624.65715072

8 3

hF

2 24.65715072 ( 0.09922565)8

hF

2 24.756376378

hF


4.2.1.3. Third Order F_3's at j=2

Step 1:- At step size of h=0.25

2 ( )2

hF -24.68084766 and 2 ( )

4

hF -24.75190269

2 2

3 2 2

4 2

4 4 4 1

h hF F

h hF F

3 2

24.75190269 ( 24.68084766)24.75190269

4 4 1

hF

3

24.75190269 24.6808476624.75190269

4 16 1

hF

3

0.0710550324.75190269

4 15

hF

3 24.756639694

hF


2 ( )4

hF -24.75190269 and 2 ( )

8

hF -24.75637637

2 2

3 2 2

8 4

8 8 4 1

h hF F

h hF F

3 2

24.75637637 ( 24.75190269)24.75637637

8 4 1

hF

3 2

24.75637637 24.7519026924.75637637

8 4 1

hF

3

0.0044736824.75637637

8 16 1

hF

3

0.0044736824.75637637

8 15

hF

3 24.75637637 ( 0.00029825)8

hF


3 24.756674628

hF

4.2.1.4. Fourth Order F_4's

at j=3

3( )4

hF 24.75663969 and 3( )

8

hF -24.75667462

3 3

3 1 3

( )8 4

( )8 8 4 1j

h hF F

h hF F

for j=3 and h=1/8=0.125

3 2

4 3 3

8 4

8 8 4 1

h hF F

h hF F

4 3

24.75667462 ( 24.75663969)24.75667462

8 4 1

hF

4

24.75667462 24.7566396924.75667462

8 64 1

hF

4

0.0000349324.75667462

8 63

hF

4 24.75667462 ( 0.00000055)8

hF

4 24.75667462 0.000000558

hF

4 24.756675178

hF



4.2.2.1. Algorithm



Step 3:-Set F (1,n) for the first order values


F(1,j)=((x+h(j))3cos(x+h(j)) -2*(x)3*cos(x)+( x-h(j))3cos(x-h(j))) /2*h(j)


F(i+1,j)=F(i,j)+(F(i,j)-F(i,j-1))/4j-1


4.2.2.2. Computer MATLAB Program function SecRichardSonExtraP() % %================================================================== % % This Program is Extrapolation for approximating the first derivative % % Written By Abebe Alemu, % % ID GSR/3003/08 % % Addis Ababa University, Faculty of Natural Science % % Computational Science Department % % Graduate Program, -------------- Dated on July 2016 % % ===================================================================== format shorte; %==============Definition ================ h=zeros(50); h1=zeros(50); h2=zeros(50); h3=zeros(50); F1=zeros(50,50); F2=zeros(50,50); A1=zeros(10); A2=zeros(10); A3=zeros(10); n=10; %======Enter Input size from key board ============= fprintf('Please Enter the number of Step sizes n \n n='); n=input(''); A1=zeros(n); A2=zeros(n); A3=zeros(n); h=zeros(n); h1=zeros(n); h2=zeros(n); h3=zeros(n); x=2.3;% ------------Reference point


%===========define Step size ============================================== for i=1:n h(i)=2^-(i-1); end A=zeros(n,n); %============End ========================================================== %====================================================================== % Richardson's Extrapolations for second order derivatives %====================================================================== for j=1:n h(j); F1(1,j)=(((x+h(j))^3)*(cos(x+h(j)))-2*x^3*cos(x)+((x-h(j)))^3*cos(x-h(j)))/(h(j)^2); A(j,1)= F1(1,j); A1(j)=A(j,1); h1(j)=h(j); end for k=2:n % ---- Second order F2(k,2)=F1(1,k)+((F1(1,k)-F1(1,k-1))/((4^1)-1)); A(k,2)= F2(k,2); A2(k)=A(k,2); h2(k)=h(k); end for m=3:n % Third to n th order for k=2:m-1 F2(m,k+1)=F2(m,k)+((F2(m,k)-F2(m-1,k))/((4^k)-1)); A(m,k+1)=F2(m,k+1); A3(k)=A(m,k+1); h3(k)=h(k); end end %=======================End ======================================== %======================Dispaly====================================== % Display the approximate first derivative value of f(x) at x=1.25 disp(' f(x)=x^3cos(x), its first derivative approximate value at x=1.25'); disp('=============================================================='); disp(' No h F_1s | F_2s | F_3s | F_4s |' ); disp('=============================================================='); for i=1:n msg1=sprintf('%0.0f %0.4f %0.8f %0.8f %0.8f %0.8f ',... i, h(i), A(i,1), A(i,2), A(i,3), A(i,4) ); disp([' ',msg1,' ']); end % %=================================================================== % % Plot % %=================================================================== figure; %--------------- First order plot------------------------------------- subplot(2,2,1) plot(h1,A1,'--rs'); title('Richardson''s Extrapolation First Order Derivatives'); xlabel('Step Size h'); legend('First Order'); ylabel('f''(x)');


axis([0.0 1.0 -25.0 -17.0]); %------------- Second order plot--------------------------------------- subplot (2,2,2); plot(h2,A2,'*'); title('Richardson''s Extrapolation Second Order Derivatives'); xlabel('Step Size h'); legend('Second Order'); axis([0.0 1.0 -25.0 -17.0]); %--------------Third order plot-------------------------------------- subplot(2,2,3); plot(h3,A3,'d'); title('Richardson''s Extrapolation Third Order Derivatives'); xlabel('Step Size h'); legend('Third Order'); axis([0.0 1.0 -25.0 -17.0]); subplot(2,2,4); plot(h1,A1,'--rs',h2,A2,'*',h3,A3,'d'); lgnd=legend('Richardson''s Extrapolation '); ylabel('f''(x)'); xlabel('Step size h(n)'); title(lgnd,' Title'); axis([0.0 1.0 -25.0 -17.0]); % % ------------------------END of PROGRAM -------------------- end

4.2.2.3. Computer MATLAB Output


n=6

f(x)=x^3cos(x), its first derivative approximate value at x=1.25

============================================================

No h F_1s | F_2s | F_3s | F_4s |

============================================================

1 1.0000 -18.68620486 0.00000000 0.00000000 0.00000000

2 0.5000 -23.18218695 -24.68084764 0.00000000 0.00000000

3 0.2500 -24.35947370 -24.75190262 -24.75663962 0.00000000

4 0.1250 -24.65717631 -24.75641052 -24.75671104 -24.75671218

5 0.0625 -24.73181406 -24.75669331 -24.75671216 -24.75671218

6 0.0313 -24.75048676 -24.75671100 -24.75671218 -24.75671218

>>


4.2.2.4. Richardson's extrapolation plot step size for second derivatives


5. Problem definition Two Given f(x)=exp(x2)cos(3x), reference point x=1.25

Use Richardson's Extrapolation and the central difference results of the first derivatives and

second derivatives,

The exact value is given as which is equal to exact= -1.606374738, To check this is true, let's find analytically by finding the first derivatives of the function f( f(x)=exp(x2)cos(3x)

2

2 2

2 2

( ) (exp( ) cos(3 )) '

exp( )*2 *cos(3 ) exp( )( 3*sin(3 ))

'(1.25) exp(1.25 )*2(1.25)*cos(3*1.25) exp(1.25 )*3*sin(3*1.25)

'(1.25) -1.606374738

dfx x x

dx

x x x x x

f

f

Yes analytically we have '(1.25) -1.606374738f

Solution

5.1. Richardson's Extrapolation first derivatives for even power Order of h

5.1.1. PART I :- working by hand

20 00

0 00

( ) ( )( ) ( )

2

( ) ( )'( )

2

f x h f x hdfx O h

dx h

f x h f x hf x

h

, ----------------------------------------------------eq(1)

Find the truncation error O(h2) we use 20( ) '( )O h exact f x ,-------------------------eq(2)

Now let's find first order derivatives, with step size h=1


5.1.1.1. Second order First derivative

Step 1:- Step Size h=0.4,

1

2 2

2

2

2

(2.3 1) (2.3 1)( )

2(1)

(3.3) (1.3)(1)

2

(3.3) cos(3.3) (1.3) cos(1.3)

2

(35.0*0.2353814) (2.05957572)( 0.8300535)

2

6.61458451

(2)

( ) 1.606374738 ( )

( ) 1.606374738 6.61458451

( )

f fF h

f fF

From eq

O h F h

O h

O h

8.220959248

Step 2:- At Step Size h/2=0.2

2 2

( ) 0.2

(1)

(1.25 0.2) (1.25 0.2)( )

2(0.2)

(1.45) (1.05)(0.2)

0.4

exp(1.45 )cos(3*1.45) exp(1.05 )cos(3*1.05)(0.2)

0.4

8.18661088*( 0.35450907) (3.01168583* 0.99996466)

0.4

0.2733789

step size h is

From eq

f fF h

f fF

F

2

2

2

675

(2)

( ) 1.606374738 ( )

( ) 0.2

( ) 1.606374738 0.2733789675

( ) 1.8797537055

From eq

O h F h

a step size h is

O h

O h


Step 3:- At Step Size h/4= 0.1

2 2

, ( ) 0.1

(1)

(1.25 0.1) (1.25 0.1)( )

2(0.1)

(1.35) (1.15)(0.1)

0.2

exp(1.35 )cos(3*1.35) exp(1.15 )cos(3*1.15)(0.1)

0.2

6.1873074*( 0.61500238) (3.75279164*( 0.95281821)

0.2

1.

when a step size h is

From eq

f fF h

f fF

F

14740285

2

2

2

(2)

( ) 1.606374738 ( )

, ( ) 0.1

( ) 1.606374738 1.14740285

( ) 0.45897189

From eq

O h F h

when a step size h is

O h

O h

Step 4:- At step size h/8=0.05

2 2

2

(1)

(1.25 0.05) (1.25 0.05)( )

2(0.05)

(1.3) (1.2)(0.05)

0.1

exp(1.3 ) cos(3*1.3) exp(1.2 )cos(3*1.2)(0.05)

0.1

5.41948071*( 0.725932304) (4.22069582*( 0.89675842)

0.1

1.49231606

(2)

(

From eq

f fF h

f fF

F

From eq

O h

2

2

) 1.606374738 ( )

( ) 1.606374738 1.49231606

( ) 0.11405864

F h

O h

O h


Step 5 :- At Step Size h/16=0.025

2 2

(1)

(1.25 0.025) (1.25 0.025)( )

2(0.025)

(1.(1.275) (1.225)(0.025)

0.05

exp((1.275) ) cos(3*1.275) exp((1.225) )cos(3*1.225)(0.025)

0.05

5.08159404*( 0.77542569) (4.484491*( 0.86107954)

0.05

From eq

f fF h

f fF

F

1.5779022

2

2

2

(2)

( ) 1.606374738 ( )

( ) 1.606374738 1.5779022

( ) 0.02847254

From eq

O h F h

O h

O h


2 2

(1)

(1.25 0.0125) (1.25 0.0125)( )

2(0.0125)

(1.2652) (1.2375)(0.0125)

0.05

exp((1.2625) ) cos(3*1.2625) exp((1.2375) ) cos(3*1.2375)(0.0125)

0.05

4.92294166*( 0.79855394) (4.6246757*( 0.84141

From eq

f fF h

f fF

F

2

2

2

0995)

0.025

1.5992592

(2)

( ) 1.606374738 ( )

( ) 1.606374738 1.5992592

( ) 0.00711554

From eq

O h F h

O h

O h



2 2

(1)

(1.25 0.0125) (1.25 0.0125)( )

2(0.00625)

(1.25652) (1.24375)(0.00625)

0.0125

exp((1.25625) )cos(3*1.2625) exp((1.24375) ) cos(3*1.24375)(0.00625)

0.0125

4.846050593*( 0.80969898) (4.6969

From eq

f fF h

f fF

F

2

2

2

53291*( 0.83113127)

0.0125

1.6045976

(2)

( ) 1.606374738 ( )

( ) 1.606374738 1.6045976

( ) 0.0001777138

From eq

O h F h

O h

O h

5.1.1.2. Fourth Order first derivative

1

( ) ( )2( ) ( ) ( ), 1, 2,3,

2 4 1

j j

j j

hF F h

hF h Fj j

-----------------------------------eq(3)


1 1 1

2

2

(0.2) (0.4)(0.2) (0.2)

4 1

0.27337907 6.6145854401(0.2) 0.27337907+( )

3

(0.2) -1.8403563833

F FF F

F

F

4

4

4

4

(2) ( )

( ) 1.606374738 ( )

( ) 1.60637473 (-1.8403563833)

( ) 0.2339816453

from eq O h

O h F h

O h

O h



1 1 1

2

2

(3)

(0.1) (0.2)(0.1) (0.1)

4 1

-1.1474026294 0.2733790725(0.1) -1.1474026294+( )

3

(0.1) -1.6209965300

From eq

F FF F

F

F

4

4

4

4

(2) ( )

( ) 1.606374738 ( )

( ) 1.60637473 (-1.6209965301)

( ) 0.01462179201

from eq O h

O h F h

O h

O h

Step 3:- At Step size h/8=0.05

1 1 1

2

2

(3),

(0.05) (0.1)(0.05) (0.05)

4 1

-1.4923161916+1.1474026294(0.05) -1.4923161916+( )

3

(0.05) -1.6072873807

From eq

F FF F

F

F

4

4

4

4

(2) ( )

( ) 1.606374738 ( )

( ) 1.60637473 (-1.6072873807)

( ) 0.0009126427

from eq O h

O h F h

O h

O h


1 1 1

2

2

(3)

(0.025) (0.05)(0.025) (0.025)

4 1

-1.5779028639+1.4923161916(0.025) -1.5779028639+( )

3

(0.025) -1.6064317547

From eq

F FF F

F

F


4

4

4

4

(2) ( )

( ) 1.606374738 ( )

( ) 1.606374738 (-1.6064317547)

( ) 0.0000570167

from eq O h

O h F h

O h

O h


1 1 1

2

2

(3)

(0.0125) (0.025)(0.0125) (0.0125)

4 1

-1.59925944+1.57790286(0.0125) -1.59925944+( )

3

(0.0.125) -1.60637830

From eq

F FF F

F

F

4

4

4

4

(2) ( )

( ) 1.60637473 ( )

( ) 1.60637473 (-1.60637830)

( ) 0.00000356

from eq O h

O h F h

O h

O h

Step6:- At Step size h/64=0.00625

1 1 1

2

2

(3),

(0.00625) (0.0125)(0.00625) (0.00625)

4 1

-1.60459608+1.59925944(0.00625) -1.60459608+( )

3

(0.00625) 1.60637496

From eq

F FF F

F

F

4

4

4

4

(2) ( )

( ) 1.60637473 ( )

( ) 1.60637473 ( 1.60637496)

( ) 0.00000022

from eq O h

O h F h

O h

O h


5.1.1.3. Sixth Order first derivative

Step 1:- At the step size of h/4=0.1

2 1 2

3

3

(3),

(0.1) (0.2)(0.1) (0.1)

4 1

- 1.62099653-(-1.84035638)(0.1) -1.62099653+( )

15

(0.1) 1.60637253

From eq

F FF F

F

F

6

6

6

6

(2) ( )

( ) 1.606374738 ( )

( ) 1.60637473 ( 1.60637254)

( ) 0.0000022

from eq O h

O h F h

O h

O h


2 1 2

3

3

(3),

(0.05) (0.1)(0.05) (0.05)

4 1

- 1.60728738-(- 1.62099653)(0.05) -1.60728738+( )

15

(0.05) 1.60637344

From eq

F FF F

F

F

6

6

6

6

(2) ( )

( ) 1.606374738 ( )

( ) 1.60637473 ( 1.60637343)

( ) 0.000001301

from eq O h is

O h F h

O h

O h


2 1 2

3

3

(3)

(0.025) (0.05)(0.025) (0.025)

4 1

-1.60643175-(-1.60728738)(0.025) -1.60643175+( )

15

(0.025) 1.60637471

From eq

F FF F

F

F


6

6

6

6

(2) ( )

( ) 1.60637473 ( )

( ) 1.60637473 ( 1.60637471)

( ) 0.00000003

from eq O h

O h F h

O h

O h


2 1 2

3

3

(3),

(0.0125) (0.025)(0.0125) (0.025)

4 1

-1.60637830-(-1.60643175)(0.0125) -1.60637830+( )

15

(0.0125) 1.60637474

From eq

F FF F

F

F

6

6

6

6

(2) ( )

( ) 1.60637473 ( )

( ) 1.60637473 ( 1.60637471)

( ) 0.0000000004

from eq O h

O h F h

O h

O h


2 1 2

3

3

(3),

(0.00625) (0.0125)(0.00625) (0.025)

4 1

-1.60637496-(-1.60637830)(0.00625) -1.60637496+( )

15

(0.00625) 1.606374738

From eq

F FF F

F

F

6

6

6

6

(2) ( )

( ) 1.606374738 ( )

( ) 1.60637473 ( 1.60637471)

( ) 0.000000000009

from eq O h

O h F h

O h

O h


5.1.1.4. Eighth Order first derivative


3 1 3

4

4

(3),

(0.05) (0.1)(0.05) (0.05)

4 1

-1.6063734374-(-1.6063725398)(0.05) -1.60637343+( )

63

(0.05) 1.6063734516

From eq

F FF F

F

F

8

8

8

8

(2) ( )

( ) 1.606374738 ( )

( ) 1.606374738 ( 1.60637345167)

( ) 0.00000128632

from eq O h

O h F h

O h

O h


3 1 3

4

4

(3),

(0.025) (0.05)(0.025) (0.025)

4 1

-1.606374713-(-1.60637343742)(0.025) -1.6063747129+( )

63

(0.025) 1.6037473324

From eq

F FF F

F

F

8

8

8

8

(2) ( )

( ) 1.606374738 ( )

( ) 1.606374738 ( 1.60637345167)

( ) 0.0000000047

from eq O h

O h F h

O h

O h


3 1 3

4

4

(3)

(0.0125) (0.025)(0.0125) (0.0125)

4 1

-1.6063747376-(-1.6063747)(0.0125) -1.6063747376+( )

63

(0.0125) 1.606374737995

From eq

F FF F

F

F


8

8

8

8

(2) ( )

( ) 1.606374738 ( )

( ) 1.606374738 ( 1.60637345167)

( ) 0.00000000000479299

from eq O h

O h F h

O h

O h


5.1.2.1. Algorithm



Step 3:-Set F (1,n) for the Second order values


F(1,j)=(exp(x+h(j))2cos(3*(x+h(j)) )-( exp( x-h(j))2 cos(3*(x-h(j)))) /2*h(j)

Step 5:- Calculate for the foruth, sixth and eighth order, as well i=1,..,n

F(i+1,j)=F(i,j)+(F(i,j)-F(i,j-1))/4j-1


5.1.2.2. Computer MATLAB Program function OrderH() % %================================================================== % % This Program is Richardson's Extrapolation and central difference formula % % for the first derivative % % Written By Abebe Alemu, % % ID GSR/3003/08 % % Addis Ababa University, Faculty of Natural Science % % Computational Science Department % % Graduate Program, -------------- Dated on July 2016 % % ===================================================================== format shorte; %==============Definition ================ h=zeros(50); h1=zeros(50); h2=zeros(50); e=zeros(50); e2=zeros(50); e3=zeros(50); F1=zeros(50,50); F2=zeros(50,50); x=1.25; % ------------Reference point %======Enter Input size from key board ============= fprintf('Please Enter the number of Step sizes n \n n='); n=input('');


% % ============================================ % % f(x)=exp(x^2)cos(3x) given function % % ================================================ % Analytical Exact value of the first derivative of the function at x=1.25 exact=-1.606374738; %===========define Step size ============= for i=1:n h(i)=0.4*(2^-(i-1)); end %A=zeros(n,n); Oh2=zeros(n,n); %============End ============================= %=========================================== % Richardson's Extrapolation, centered difference for first derivative %============================================== for j=1:n h(j); F1(1,j)=(((exp((x+h(j))^2))*(cos(3*(x+h(j)))))- ... ((exp((x-h(j))^2))*cos(3*(x-h(j)))))/(2*h(j)); Oh2(j,1)=exact-F1(1,j); e(j)=Oh2(j,1); end for k=2:n F2(k,2)=F1(1,k)+((F1(1,k)-F1(1,k-1))/((4^1)-1)); Oh2(k,2)= exact-F2(k,2); e2(k)=Oh2(k,2); end for m=3:n for k=3:m F2(m,k)=F2(m,k-1)+((F2(m,k-1)-F2(m-1,k-1))/((4^(k-1))-1)); Oh2(m,k)=exact-F2(m,k); e3(k)=Oh2(m,k); end end % %=============================End ============================= % %===========================Display============================ disp('=============================================================='); disp(' No h O(h^2) O(h^4) O(h^6) O(h^8) '); disp('=============================================================='); for i=1:n % 0.10f ten digit precision msg1=sprintf('%0.0f %0.4f %0.10f %0.10f %0.10f %0.10f ', ... i, h(i), Oh2(i,1), Oh2(i,2), Oh2(i,3), Oh2(i,4) ); disp([' ',msg1,' ']); end %========================================================================= for i=2:n h1(i)=h(i); end for j=3:n h2(j)=h(j); end % %=======================================================


% % Plotting the graph using step size h and the error % %======================================================= figure; subplot(2,2,1)% ----- Plot the first order in the first quadrant plot(h,e,'--rs') title('Richardson''s Extrapolation Second Derivatives'); xlabel('Step Size h'); legend('Secodn order O(h^2)'); ylabel('f''(x)'); axis([0.0 0.5 -10.0 0.0]); grid on; subplot (2,2,2); % ----- Plot the second order in the second quadrant plot(h1,e2,'*'); title('Richardson''s Extrapolation Second Derivatives'); xlabel('Step Size h'); legend('Secodn order O(h^4)'); ylabel('f''(x)'); axis([0.0 0.3 0.0 0.25]); grid on; subplot(2,2,3); % -----Plot the third order on the third quadrant plot(h2,e3,'--rs'); title('Richardson''s Extrapolation Second Derivatives'); xlabel('Step Size h'); legend('Second order O(h^6)'); ylabel('f''(x)'); axis([0.0 0.510 -0.10 0.0]); grid on; subplot(2,2,4); % -----Plot the three figures in the same quadrant plot(h,e) hold on; plot(h1,e2,'*'); hold on; plot(h2,e3); title('Richardson''s Extrapolation Secodn Derivatives'); xlabel('Step Size h'); legend('Secodn order O(h^2)'); ylabel('f''(x)'); axis([0.0 0.5 -10.0 0.0]); grid on; end


5.1.2.3. Computer MATLAB Output


n=6

===========================================================

No h O(h^2) O(h^4) O(h^6) O(h^8)

===========================================================

1 0.4000 -8.2209601781 0.0000000000 0.0000000000 0.0000000000

2 0.2000 -1.8797538106 0.2339816453 0.0000000000 0.0000000000

3 0.1000 -0.4589721086 0.0146217921 -0.0000021981 0.0000000000

4 0.0500 -0.1140585463 0.0009126411 -0.0000013023 -0.0000012881

5 0.0250 -0.0284718740 0.0000570167 -0.0000000249 -0.0000000046

6 0.0125 -0.0071152961 0.0000035632 -0.0000000004 0.0000000000

>>

5.1.2.4. Computer MATLAB Plot Ext.Polation for first derivatives


5.2. Richardson's Extrapolation Using three point Second Derivatives

Given 2

20 0 002 2

( ) 2 ( ) ( )( ) ( )

f x h f x f x hfx o h

x h

where 0x =1.25, and exact analytical

value is given 43.8375005482.

'' 0 0 00 2

( ) 2 ( ) ( )( )

f x h f x f x hf x

h

-----------------------------------------eq(1)

From above solution A, we have got the first derivative of the function

2( ) exp( ) cos(3 )f x x x , we can find easily the second derivatives annalistically

2

2 2

2 2 2 2

2 2

( ) (exp( ) cos(3 )) '

'( ) exp( )*2 *cos(3 ) exp( )( 3*sin(3 ))

''( ) [exp( )*4 *cos(3 ) exp( )*2*cos(3 ) exp( )*2 *( sin(3 ))]

[exp( )*2 *( 3*sin(3 )) exp( )( 9*cos(3 ))]

dfx x x

dx

f x x x x x x

f x x x x x x x x x

x x x x x

Substituting the value of x=1.25 at the expression, and we have got 43.8375005482 this is

exact value analytically; then let's find the relative error by finding in each step size

error of order nth of h is O(hn)=exact - f''(x) --------------------------------eq(2)

Solution

5.2.1. PART I: - working by hand

Reference value 0x =1.25, 2( ) exp( ) cos(3 )f x x x

Step 1:- Order two O(h2), at h=0.4

''

2

(1.25 0.4) 2 (1.25) (1.25 0.4)(1.25)

0.4

f f ff

=

(1.65) 2 (1.25) (0.85)

0.16

f f f

2 2 2'' exp((1.65) ) cos(3*1.65) 2*exp(1.25 )cos(3*1.25) exp((0.85) )cos(3*0.85)(1.25)

0.16f

'' (1.25) 60.6368227893f

from the above equation 2 second order error 2( )o h

2

2

( ) 43.8375005482 60.6368227893

( ) 16.7993222411

o h

o h


Step 2:- At Step size h=0.2

'' 0 0 02

( ) 2 ( ) ( )( )

f x h f x f x hf x

h

''

2

(1.25 0.2) 2 (1.25) (1.25 0.2)(1.25)

0.2

f f ff

=

(1.45) 2 (1.25) (1.05)

0.04

f f f

2 2 2'' exp((1.45) )cos(3*1.45) 2*exp(1.25 )cos(3*1.25) exp((1.05) ) cos(3*1.05)(1.25)

0.04f

'' (1.25) 47.8883085539f

Find second order error 2( )o h for h/2=0.2 is from eq(2)

2

2

( ) 43.8375005482 47.8883085539

( ) 4.0508080057

o h

o h

Step 3:- at Step size h/4=0.1

''

2

(1.25 0.1) 2 (1.25) (1.25 0.1)(1.25)

0.1

f f ff

=

(1.35) 2 (1.25) (1.15)

0.01

f f f


0.04f

'' (1.25) 44.8402524014f

Find second order error 2( )o h at step size 0.1

2

2

( ) 43.8375005482 44.8402524014

( ) 1.0027518532

o h

o h

Step 4:- at Step size h/8=0.05

''

2

(1.25 0.05) 2 (1.25) (1.25 0.05)(1.25)

0.05

f f ff

=

(1.30) 2 (0.05) (1.2)

0.0025

f f f



0.0025f

'' (1.25) 44.0875580582f

Find second order error 2( )o h fro step size 0.05

2

2

( ) 43.8375005482 44.0875580582

( ) 0.2500575100

o h

o h


''

2

(1.25 0.025) 2 (1.25) (1.25 0.025)(1.25)

0.025

f f ff

=(1.275) 2 (1.25) (1.2225)

0.000625

f f f

2 2 2'' exp((1.275) )cos(3*1.275) 2*exp(1.25 )cos(3*1.25) exp((1.2225) )cos(3*1.2225)(1.25)

0.000625f

'' (1.25) 43.8999753936f

Find second order error 2( )o h for step size 0.025

2

2

( ) 43.8375005482 43.8999753936

( ) 0.0624748454

o h

o h


''

2

(1.25 0.0125) 2 (1.25) (1.25 0.0125)(1.25)

0.0125

f f ff

=(1.2625) 2 (1.25) (1.2375)

0.00015625

f f f

2 2 2'' exp((1.2625) )cos(3*1.2625) 2*exp(1.25 )cos(3*1.25) exp((1.2375) )cos(3*1.2375)(1.25)

0.00015625f

'' (1.25) 43.8531167568f


Find second order error 2( )o h at step size h/32=0.0125

2

2

( ) 43.8375005482 43.8531167568

( ) 0.0156162386

o h

o h

Generally after working second derivative with three point' using second derivative by

hand we have the following order of errors

h 2( )o h 4( )o h 6( )o h

0.4 -16.7993222411 ##### ####

0.2 -4.0508080057 0.1986967394 ####

0.1 -1.0027518532 0.0132668641 0.0009048724

0.05 -0.2500575100 0.0008406044 0.0000121871

0.025 -0.0624748454 0.0000527095 0.0000001832

0.0125 -0.0156162386 0.0000032970 0.0000000028


5.2.2.1. Algorithm


Step 2:- set h (i) as h/2^-(i-1), where i=1 to n

Step 3:-Set F (1,n) for the Second order values


F(1,j)=(exp(x+h(j))2cos(3*(x+h(j)) )-2*(exp(x)2*cos(3x)+(exp( x-h(j))2

cos(3*(x-h(j)))) /h(j)2


Step 6:- Get the result and plot

5.2.2.2. Computer MATLAB Program function Order2H() % %================================================================== % % This Program is Extrapolation for approximating the first derivative % % Written By Abebe Alemu, % % ID GSR/3003/08 % % Addis Ababa University, Faculty of Natural Science % % Computational Science Department % % Graduate Program, -------------- Dated on July 2016 % % ===================================================================== format short; %==============Defintion ================


h=zeros(50); F1=zeros(50,50); F2=zeros(50,50); h1=zeros(50); h2=zeros(50); e=zeros(50); e2=zeros(50); e3=zeros(50); %======Enter Input size from key board ============= fprintf('Please Enter the number of Step sizes n \n n='); n=input(''); x=1.25;% ------------Refernce point exact=43.8375005482; %===========Define Step size ============= for i=1:n h(i)=0.4*(2^-(i-1)); h(i); A=zeros(n,n); Oh2=zeros(n,n); %============End ============================= %=========================================== % Calucation %f(x)=exp(x^2)cos(3x) %============================================== for j=1:n h(j); F1(1,j)=((exp((x+h(j))^2)*(cos(3*(x+h(j)))))-(2*(exp(x^2))*cos(3*x))+((exp((x-h(j))^2))*cos(3*(x-h(j)))))/(h(j)^2); Oh2(j,1)=exact-F1(1,j); e(j)=Oh2(j,1); % A(j,1)= F1(1,j); end for k=2:n F2(k,2)=F1(1,k)+((F1(1,k)-F1(1,k-1))/((4^1)-1)); Oh2(k,2)= exact-F2(k,2); e2(k)=Oh2(k,2); end for m=3:n for k=2:m-1 F2(m,k+1)=F2(m,k)+((F2(m,k)-F2(m-1,k))/((4^k)-1)); Oh2(m,k+1)=exact-F2(m,k+1); e3(k)=Oh2(m,k); end end % for i=2:n % h1(i)=h(i); % end % for j=3:n % h2(j)=h(j); % end % %=======================End ==================================== %===============Dispaly=========================== disp('=============================================================='); disp(' F_1s F_2s F_3s F_4s '); disp('=============================================================='); %Oh2(i,j)=Oh2(i,j);


for i=1:n %for j=1:i %err=Oh2(i,j); msg1=sprintf('%0.0f %0.4f %0.10f %0.10f %0.10f %0.10f ', ... i, h(i), Oh2(i,1), Oh2(i,2), Oh2(i,3), Oh2(i,4) ); disp([' ',msg1,' ']); %end end figure; subplot(2,2,1)% ----- Plot the first order in the first quadrant plot(h,e,'--rs') title('Richardson''s Extrapolation Second Derivatives'); xlabel('Step Size h'); legend('Secodn order O(h^2)'); ylabel('f''(x)'); axis([0.0 0.5 -10.0 0.0]); grid on; subplot (2,2,2); % ----- Plot the second order in the second quadrant plot(h1,e2,'*'); title('Richardson''s Extrapolation Second Derivatives'); xlabel('Step Size h'); legend('Secodn order O(h^4)'); ylabel('f''(x)'); axis([0.0 0.3 0.0 0.25]); grid on; subplot(2,2,3); % -----Plot the third order on the third quadrant plot(h2,e3,'--rs'); title('Richardson''s Extrapolation Second Derivatives'); xlabel('Step Size h'); legend('Second order O(h^6)'); ylabel('f''(x)'); axis([0.0 0.510 -0.10 0.0]); grid on; subplot(2,2,4); % -----Plot the three figures in the same quadrant plot(h,e) hold on; plot(h1,e2,'*'); hold on; plot(h2,e3); title('Richardson''s Extrapolation Secodn Derivatives'); xlabel('Step Size h'); legend('Secodn order O(h^2)'); ylabel('f''(x)'); axis([0.0 0.5 -10.0 0.0]); grid on; %================================================ end


5.2.2.3. Computer MATLAB Output for second derivatives

=============================================================

No o(h^n) F_1s F_2s F_3s F_4s

=============================================================

1 0.4000 -16.7993222411 0.0000000000 0.0000000000 0.0000000000

2 0.2000 -4.0508080057 0.1986967395 0.0000000000 0.0000000000

3 0.1000 -1.0027518532 0.0132668643 0.0009048726 0.0000000000

4 0.0500 -0.2500575100 0.0008406043 0.0000121870 -0.0000019826

5 0.0250 -0.0624748454 0.0000527095 0.0000001832 -0.0000000073

6 0.0125 -0.0156162386 0.0000032970 0.0000000028 -0.0000000000

>>

5.2.2.4. Computer MATLAB Plot


6. Observation Recursive difference formulas for derivatives can be obtained by canceling the truncation

error at each order of numerical approximation. This method is called Richardson

extrapolation. It can be used only if the data values are equally spaced with step size h.

1. From Taylors series '' (3) ( )

2 3

'

( ) ( ) ( )( ) ( ) ( ) ( ) ... ( )

2! 3! !( )

( )

nnf a f a f a

f x f a x a x a x an

f ax a

One can deduce that

2. Forward difference formula

' ( ) ( )( )

f x h f xf x

h

for first derivatives

Denote the two-point forward difference approximation as F1(h). It has the truncation

error of order O(h). By canceling the O(h) error, we define a new forward difference:

F2(h) = 2F1(h) - F1(2h)

The new approximation F2(h) is in fact the three-point forward difference:

The three-point forward difference is more accurate than the two-point difference: it

has the truncation error of order O(h2)!

3. Backward formula ' ( ) ( )( )

f x f x hf x

h

for first derivatives

4. Centered difference formula,

' ( ) ( )( )

2

f x h f x hf x

h

for first

derivatives Richardson extrapolation algorithm for central differences

Construct the following mapping from Fk(h) to Fk+1(h):

Fk+1 = Fk(h) + (Fk(h) - Fk(2h)) / (4k - 1) And we have

1 2

( ) 2 ( ) ( )f x h f x f x hF

h

, centered formula for second derivatives


5. Generally

a. Centered difference formula is preferable to get more accurate approximation

derivative result

b. Making with same order having more equally spaced size h tends to accurate

approximation result (which means if h 0 , the error tends to zero, but

not equal to zero)

c. Use higher order error for getting good approximation of first and second

derivatives of a given function f(x).

d. MATLAB programming is useful for easily calculating the derivatives and

determining the order of the error

e. From the above plot we can see that for higher order the graph tends to zero

asymptotically.

home work ii

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