Challenge

Week 3.5 - Deriving 4th order approximation of a 2nd order derivative using Taylor Table method

NNeetu Ruhela18 Oct 2023

Derivation:-

to derive the fourth order approximation for the second order derivative .

`(del^2 u)/(delx^2)=(af(i-2)+bf(i-1)+cf(i)+df(i+1)+ef(i+2))/dx-------------(1)`

taylor series derivation above eqaution.

`af(i-2)=af(i)-(2af'(i)dx)/(1!)+(4af''(i)dx^2) /(2!) +(8af'''(i)dx^3)/(3!)+(16af''''(i)dx^4)/(4!)`

Using taylor method express in table form

dx²f''(i) f(i) f'(i)dx f''(i)dx² f'''(i)dx³ f''''(i) dx⁴

af(i-2) a -2a 2a -4a/3 2a/3

bf(i-1) b -b b/2 -b/6 b/24

cf(i) c 0 0 0 0

df(i+1) d d d/2 d/6 d/24

ef(i+2) e 2e 2e 4e/3 2e/3

sum of 0 0 1 0 0

cofficient

euation conert into the matrix form

AX=B

x=inv(A)*B

respectively cofficient values.

Values put in equation (1):-

a=-0.8333

b=1.3333

c=-2.5000

d=1.3333

e=-0.0833

`(del^2 u)/(delx^2)=((0.0833)f(i-2)+(1.3333)f(i-1)+(-2.5000)f(i)+(1.3333)f(i+1)+(0.0833)f(i+2))/dx`

SKEWED RIGHT HAND DIFFERENCE SCHEME:

NO OF NODES:

The number of nodes = p+q

Where, p is order of derivative.

q is order of approximations.

The no of nodes = 2+4 =>6.

DERIVATION:

To derive a fourth order approximation for the second order derivative,we start with these approximation,

`(del^2u)/(delx^2)=(af(i)+bf(i+1)+cf(i+2)+df(i+3)+ef(i+4)+gf(i+5))/dx^2`-----------------------(2)

Using taylor method express in table form

`dx^2f''(i)`	f(i)	f'(i)dx	f''(i)dx²	f'''(i)dx³	f''''(i) dx⁴	`f'''''(i) dx^5`
af(i)	a	0	0	0	0	0
bf(i+1)	b	b	`b/2`	`b/6`	`b/24`	`b/120`
cf(i+2)	c	2c	`(4c)/2`	`(8c)/6`	`(16c)/24`	`(32c)/120`
df(i+3)	d	3d	`(9d)/2`	`(27d)/6`	`(81d)/24`	`(243d)/120`
ef(i+4)	e	4e	`(16e)/2`	`(64e)/6`	`(256e)/24`	`(1024e)/120`
gf(i+5)	g	5g	`(25g)/2`	`(125g)/6`	`(625g)/24`	`(3125g)/120`
sum of co-efficients	0	0	1	0	0	0

Linera eqaution convert into the mtrix form;-

AX=B

`[[1,1,1,1,1,1],[0,1,2,3,4,5],[0,1/2,4/2,9/2,16/2,25/2],[0,1/6,8/6,27/6,64/6,125/6],[0,1/24,16/24,81/24,256/24,625/24],[0,1/120,32/120,243/120,1024/120,3125/120]]`.`[(a),(b),(c),(d),(e),(g)]`=`[(0),(0),(1),(0),(0),(0)]`

x=inv(A)*B

a=3.7500

b=-12.8333

c=17.83333

d=13.0000

e=-12.8333

g=-0.8333

Subsistue these values in equation-(2)

`(del^2u)/(delx^2)=((3.7500)f(i)+(12.8333)f(i+1)+(-17.8333)f(i+2)+(-13.0000)f(i+3)+(5.0833)f(i+4)+(-0.8333)f(i+5))/dx^2`

skewed left hand differnce scheme:-

no of nodes

=p+q

Where, p is order of derivative.

q is order of approximations.

The no of nodes = 2+4 =>6.

derivation

fourth order approximation for the second order derivation:-

`(del^2u)/(delx^2)=(af(i-5)+bf(i-4)+cf(i-3)+df(i-2)+ef(i-1)+gf(i))/dx^2`------------------(3)

`dx^2f''(i)`	f(i)	f'(i)dx	f''(i)dx²	f'''(i)dx³	f''''(i) dx⁴	`f'''''(i) dx^5`
af(i-5)	a	-5a	`(25a)/2`	``-(125a)/6``	``-(625a)/25``	`-(3125a)/120`
bf(i-4)	b	-4b	`(16b)/2`	`-(64b)/6`	``(256b)/24``	`-(1024b)/120`
cf(i-3)	c	-3c	`(9c)/2`	`-(27c)/6`	`(81c)/24`	`-(243c)/120`
df(i-2)	d	-2d	`(4d)/2`	`-(8d)/6`	`(16d)/24`	`-(32d)/120`
ef(i-1)	e	-e	`(e)/2`	`-(e)/6`	`(e)/24`	`-(e)/120`
gf(i)	g	0	`0`	`0`	`0`	`0`
sum of co-efficients	0	0	1	0	0	0

linear Equation in matrix form:-

AX=B

``[[1,1,1,1,1,1],[-5,-4,-3,-2,-1,0],[25/2,16/2,9/2,4/2,1/2,0],[-125/6,-64/6,-27/6,-8/6,-1/6,0],[625/24,256/24,81/24,16/24,1/24,0],[-3125/120,-1024/120,-243/120,-32/120,-1/120,0]]``.`[(a),(b),(c),(d),(e),(g)]`=`[(0),(0),(1),(0),(0),(0)]`

x=inv(A)*B

a= -0.8333

b= 5.0833

c=-13.0000

d= 17.83333

g=3.7500

Subsistue these values in equation-(3)

`(del^2u)/(delx^2)=((-0.8333)f(i-5)+(5.0833)f(i-4)+(13.0000)f(i-3)+(17.8333)f(i-2)+(-12.8333)f(i-1)+(3.7500)f(i))/dx^2`

Matlab Code:-

FOURTH ORDER CENTRAL DIFFERNCING:-

function error_cental_differncing=fourth_central_differncing(x,dx)
% anyltic function =exp(x)*cos(x)
%analytic_derivative
% f'(x)=exp(x)*cos(x)-exp(x)*sin(x)
% f''(x)=-2*exp(x)*sin(x)
analytic_derivative=-2*exp(x)*sin(x);

%numerical_derivative
�ntral differncing
central_differncing=(((-0.0833)*((exp(x-(2*dx)))*(cos(x-(2*dx)))))+((1.3333)*((exp(x-dx))*(cos(x-dx))))-((2.5000)*((exp(x))*(cos(x))))+((1.3333)*((exp(x+dx))*(cos(x+dx))))-((0.0833)*((exp(x+(2*dx)))*(cos(x+(2*dx))))))/(dx^2);
error_cental_differncing=abs(central_differncing-analytic_derivative);
end

SKEWED RIGHT HAND DIFFERENCE SCHEME:

function error_skewed_right_hand=fourth_order_skewed_right_differnce(x,dx)
% anyltic function =exp(x)*cos(x)
%analytic_derivative
% f'(x)=exp(x)*cos(x)-exp(x)*sin(x)
% f''(x)=-2*exp(x)*sin(x)
analytic_derivative=-2*exp(x)*sin(x);

%numerical_derivative
�ntral differncing
right_side_skew=(((3.7500)*((exp(x))*(cos(x))))-((12.8333)*((exp(x+dx))*(cos(x+dx))))+((17.8333)*((exp(x+(2*dx)))*(cos(x+(2*dx)))))-((13.0000)*((exp(x+(3*dx)))*(cos(x+(3*dx)))))+((5.0833)*((exp(x+(4*dx)))*(cos(x+(4*dx)))))-((0.8333)*((exp(x+(5*dx)))*(cos(x+(5*dx))))))/(dx^2);
error_skewed_right_hand=abs(right_side_skew-analytic_derivative);
end

SKEWED LEFT HAND DIFFERENCE SCHEME:

function error_skewed_left_hand=fourth_order_skewed_left_differnce(x,dx)
% anyltic function =exp(x)*cos(x)
%analytic_derivative
% f'(x)=exp(x)*cos(x)-exp(x)*sin(x)
% f''(x)=-2*exp(x)*sin(x)
analytic_derivative=-2*exp(x)*sin(x);

%numerical_derivative
�ntral differncing
left_side_skew=(((-0.8333)*((exp(x-(5*dx)))*(cos(x-(5*dx)))))+((5.0833)*((exp(x-(4*dx)))*(cos(x-(4*dx)))))-((13.0000)*((exp(x-(3*dx)))*(cos(x-(3*dx)))))+((17.8333)*((exp(x-(2*dx)))*(cos(x-(2*dx)))))-((12.8333)*((exp(x-dx))*(cos(x-dx))))+((3.7500)*((exp(x))*(cos(x)))))/(dx^2);
error_skewed_left_hand=abs(left_side_skew-analytic_derivative);
end

MAIN CODE:-

close all
clear all
clc
% value of x
x=pi/3;
% range of dx
dx=linspace(pi/4,pi/4000,20);

for i=1:length(dx)
    central_differncing(i)=fourth_central_differncing(x,dx(i));
    right_side_skew(i) =fourth_order_skewed_right_differnce(x,dx(i));
   left_side_skew(i) =fourth_order_skewed_left_differnce(x,dx(i));
end

figure(1)
loglog(dx,central_differncing)
hold on
loglog(dx,right_side_skew,'r')
hold on
loglog(dx,left_side_skew,'g')
xlabel('range of dx')
ylabel('truncation error')
legend('central differncing','right side skew','left side skew')

LOGLOG GRAPH REPRESNTION

Conclusion:-

as per graph curve central differncing is very less error as compared to the right side skews scheme & left side skews scheme.ln the left and right skewed differnce method uses data only it gives the approximate and accurate values.

Why skewed scheme is useful:- the system is dose not

Week 3.5 - Deriving 4th order approximation of a 2nd order derivative using Taylor Table method

Ship work like this
in your own course.

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Ship work like thisin your own course.

Talk to a career counsellor

Ship work like this
in your own course.