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  1. Home/
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  3. Week 3.5 - Deriving 4th order approximation of a 2nd order derivative using Taylor Table method

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

  USING THE CENTRAL DIFFERENCE METHOD: IN central difference method numerical stencil will be    (i-2)  (i-1)  (i)  (i+1)  (i+2) for fourth order approximation at point (i) we have to take information from both side right& left and the distance between them will be ∆x. our hypothesis…

    • Arun Reddy

      updated on 01 Jan 2022

     

    USING THE CENTRAL DIFFERENCE METHOD:

    IN central difference method numerical stencil will be

       (i-2)  (i-1)  (i)  (i+1)  (i+2)

    for fourth order approximation at point (i) we have to take information from both side right& left and the distance between them will be ∆x.

    our hypothesis equation will be :

        ∂²u/∂x² = af(i-2)+bf(i-1)+cf(i)+df(i+1)+ef(i+2)...............(1)

    to find the value of (a,b,c,d,e) we are gone a use taylor table series method.

    (a)f(i-2)= (a)f(i)-(a)f'(i)(2∆x)/1! +(a)f''(i)(2∆x)^2/2! - (a)f'''(i)(2∆x)^3/3! +(a)f''''(i)(2∆x)^4/4! - (a)f'''''(i)(2∆x)^5/5! + (a)f''''''(i)(2∆x)^6/6!+.............

     

    (b)f(i-1)= (b)f(i)-(b)f'(i)(∆x)/1! + (b)f''(i)(∆x)^2/2! - (b)f'''(i)(∆x)^3/3! + (b)f''''(i)(∆x)^4/4! - (b)f'''''(i)(∆x)^5/5! + (b)f''''''(i)(∆x)^6/6!+ ..............

     

    (c)f(i)=(c)f(i)

     

    (d)f(i+1)=(d)f(i)+(d)f'(i)(∆x)/1! - (d)f''(i)(∆x)^2/2! + (d)f'''(i)(∆x)^3/3! - (d)f''''(i)(∆x)^4/4! + (d)f'''''(i)(∆x)^5/5! - (d)f''''''(i)(∆x)^6/6!+..........

     

    (e)f(i+2)= (e)f(i)+(e)f'(i)(2∆x)/1! - (e)f''(i)(2∆x)^2/2! + (e)f'''(i)(2∆x)^3/3! - (e)f''''(i)(2∆x)^4/4! + (e)f'''''(i)(2∆x)^5/5! - (e)f''''''(i)(2∆x)^6/6!+........

     

    arranging above terms in taylor tyabel:

      f(i) ∆xf(i) ∆x^2f''(i) ∆xf^3'''(i) ∆x^4f''''(i) ∆x^5f'''''(i) ∆x^6f''''''(i)
    (a)f(i-2) a -2a 4a/2 -8a/6 16a/24 -32a/120 64a/240

    (b)f(i-1)

    		 b -b b/2 -b/6 b/24 -b/120 b/240
    (c)f(i) c 0 0 0 0 0 0
    (d)f(i+1) d d d/2 d/6 d/24 d/120 d/240
    (e)f(i+2) e 2e 4e/2 8e/6 16e/24 -32e/120 64e/240
    sum of above all 0 0 1 0 0 ? ?

     

    from above this we can write in matrix form AX=B

    [1 1 1 1 1 ; -2 -1 0 1 2 ; 2 1/2 0 1/2 2 ; -8/6 -1/6 0 1/6 8/6 ; 16/24 1/24 0 1/24 16/24] [a b c d e ]=[0 0 1 0 0 ]

    after solving above matrix the values we get

    a=-0.083333

    b=1.333333

    c=-2.500000

    d=1.333333

    e=-0.83333

    dividing both the side by equation (1) by (∆x^2) we get

    ∂²u/∂x² = af(i-2)+bf(i-1)+cf(i)+df(i+1)+ef(i+2) /  (∆x^2)      %we can put the above vlues(a b c d e) in it

    after this to get the second order deravative the coeficient of f''(i) has to be 1 so as to get 1 we have to ignore others or set it as 0.

    here we can also see that a is equal to e and b is equal to d we can aslso deravie that and solve the equation and get to the final equation.

     

    USING THE SKEWED RIGHT SIDED DIFFERENCE:

    In this skewed right hand side difference method we have to define the five points on the right of (i).

    the numerical stencil will be

      (i)  (i+1)  (i+2)  (i+3)  (i+4)   (i+5)

    the hypothesis eqquation will be

        ∂²u/∂x² = af(i)+bf(i+1)+cf(i+2)+df(i+3)+ef(i+4)+gf(i+5)

    writing taylor series expansion for each term

     af(i)=af(i)

    (b)f(i+1)= (b)f(i)+(b)f'(i)(∆x)/1! + (b)f''(i)(∆x)^2/2! + (b)f'''(i)(∆x)^3/3! + (b)f''''(i)(∆x)^4/4! + (b)f'''''(i)(∆x)^5/5! + (b)f''''''(i)(∆x)^6/6!+.........

    (c)f(i+2)=(c)f(i)+(c)f'(i)(2∆x)/1! + (c)f''(i)(2∆x)^2/2! + (c)f'''(i)(2∆x)^3/3! + (c)f''''(i)(2∆x)^4/4! + (c)f'''''(i)(2∆x)^5/5! + (c)f''''''(i)(2∆x)^6/6!+.......

    (d)f(i+3)=(d)f(i)+(d)f'(i)(3∆x)/1! + (d)f''(i)(3∆x)^2/2! + (d)f'''(i)(3∆x)^3/3! + (d)f''''(i)(3∆x)^4/4! + (d)f'''''(i)(3∆x)^5/5! + (d)f''''''(i)(3∆x)^6/6!+..........

    (e)f(i+4)= (e)f(i)+(e)f'(i)(4∆x)/1! + (e)f''(i)(4∆x)^2/2! + (e)f'''(i)(4∆x)^3/3! + (e)f''''(i)(4∆x)^4/4! + (e)f'''''(i)(4∆x)^5/5! + (e)f''''''(i)(4∆x)^6/6!+........

    (g)f(i+5)= (g)f(i)+(g)f'(i)(5∆x)/1! + (g)f''(i)(5∆x)^2/2! + (g)f'''(i)(5∆x)^3/3! - (g)f''''(i)(5∆x)^4/4! + (g)f'''''(i)(5∆x)^5/5! + (g)f''''''(i)(5∆x)^6/6!+........

     

    arranging taylor series expansion in taylor term:

      f(i) ∆xf(i) ∆x^2f''(i) ∆x^3f'''(i) ∆x^4f''''(i) ∆x^5f'''''(i) ∆x^6f''''''(i)
     af(i) a 0 0 0 0 0 0
    (b)f(i+1) b b b/2 b/6 b/24 b/120 b/720
    (c)f(i+2) c 2c 4c/2 8c/6 16c/24 32c/20 64c/720
    (d)f(i+3) d 3d 9d/2 27d/6 81d/24 243d/120 729d/720
    (e)f(i+4) e 4e 16e/2 64e/6 256e/24 1024e/120 4096e/720
    (g)f(i+5) g 5g 25g/2 125g/6 625g/24 3125g/120 15625g/720
    sum of above all 0 0 1 0 0 0 ?

     

    from above this we can write in matrix form AX=B

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

    after solving the matrix the value we get:

    a=3.75000

    b=-12.83333

    c= 17.83333

    d=-13.00000

    e=5.08333

    g=-0.83333

    dividing both the side by equation (1) by (∆x^2) we get

      ∂²u/∂x² = af(i)+bf(i+1)+cf(i+2)+df(i+3)+ef(i+4)+gf(i+5)      %we can put the above vlues(a b c d e g) in it

    putting the above all values we can get the equation.

     

    USING SKEWED LEFT SIDED DIFFERENCE:

    In this skewed left hand side difference method we have to define the five points on the left of (i).

    the numerical stencil will be:

      (i-5)  (i-4)  (i-3)  (i-2)  (i-1)   (i)

    the hypothesis eqquation will be

        ∂²u/∂x² = af(i-5)+bf(i-4)+cf(i-3)+df(i-2)+ef(i-1)+gf(i)

    writing taylor series expansion for each term

    (a)f(i-5)= (a)f(i)-(a)f'(i)(5∆x)/1! + (a)f''(i)(5∆x)^2/2! - (a)f'''(i)(5∆x)^3/3! + (a)f''''(i)(5∆x)^4/4! - (a)f'''''(i)(5∆x)^5/5! + (a)f''''''(i)(5∆x)^6/6!+.........

    (b)f(i-4)=(b)f(i)-(b)f'(i)(4∆x)/1! + (b)f''(i)(4∆x)^2/2! - (b)f'''(i)(4∆x)^3/3! + (b)f''''(i)(4∆x)^4/4! - (b)f'''''(i)(4∆x)^5/5! + (b)f''''''(i)(4∆x)^6/6!+.......

    (c)f(i-3)=(c)f(i)-(c)f'(i)(3∆x)/1! + (c)f''(i)(3∆x)^2/2! - (c)f'''(i)(3∆x)^3/3! + (c)f''''(i)(3∆x)^4/4! - (c)f'''''(i)(3∆x)^5/5! + (c)f''''''(i)(3∆x)^6/6!+..........

    (d)f(i-2)= (d)f(i)-(d)f'(i)(2∆x)/1! + (d)f''(i)(2∆x)^2/2! - (d)f'''(i)(2∆x)^3/3! + (d)f''''(i)(2∆x)^4/4! - (d)f'''''(i)(2∆x)^5/5! + (d)f''''''(i)(2∆x)^6/6!+........

    (e)f(i-1)= (e)f(i)-(e)f'(i)(∆x)/1! + (e)f''(i)(∆x)^2/2! - (e)f'''(i)(∆x)^3/3! + (e)f''''(i)(∆x)^4/4! - (e)f'''''(i)(∆x)^5/5! + (e)f''''''(i)(∆x)^6/6!+........

    (g)f(i)=(g)f(i)

    arranging taylor series expansion in taylor term:

      f(i) ∆xf(i) ∆x^2f''(i) ∆x^3f'''(i) ∆x^4f''''(i) ∆x^5f'''''(i) ∆x^6f''''''(i)
     af(i-5) a -5a 25/a -125a/6 625a/24 -3125a/120 15625a/720
    (b)f(i-4) b -4b 8b -64b/6 256b/24 -1024b/120 4096b/720
    (c)f(i-3) c -3c 9c/2 -27c/6 81c/24 -243c/120 729c/120
    (d)f(i-2) d -2d 2d -8d/6 16d/24 32d/120 64d/720
    (e)f(i-1) e -e e/2 -e/6 e/24 -e/120 e/720
    (g)f(i) g 0 0 0 0 0 0
    sum of above all 0 0 1 0 0 0 ?

    from above this we can write in matrix form AX=B

    [1 1 1 1 1 1 ; -5 -4 -3 -2 -1 0 ; 25/2 8/2 9/2 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]

     

    solving the above matrix we get the values

    a= -0.83333

    b= 5.08333

    c=-13.00000

    d=17.83333

    e=-12.83333

    g=3.75000

     

    dividing both the side by equation (1) by (∆x^2) we get

    ∂²u/∂x² = af(i-5)+bf(i-4)+cf(i-3)+df(i-2)+ef(i-1)+gf(i)        %we can put the above vlues(a b c d e g) in it

    putting the above all values we can get the equation.

     

    % function for central difference error

    % creating the fuction for 4order approximation
function  difference_central_order =central_difference(x,dx)

%analytical function f(x)= exp(x)*cos(x);
% f''(x)=-2*exp(x)*sin(x);

analytical_deravative= -2*exp(x)*sin(x);
% values of coefficients from  taylor table
a=-0.083333;
b=1.333333;
c=-2.500000;
d=1.333333;
e=-0.83333;

%central differencing
%((a*f(x-2*dx))+(b*f(x-dx))+(c*f(x))+(d*f(x+dx))+(e*f(x+2*dx)))/dx^2;

central_differencing_fourth_order=((a*exp(x-2*dx)*cos(x-2*dx))+(b*exp(x-dx)*cos(x-dx))+(c*exp(x)*cos(x))+(d*exp(x+dx)*cos(x+dx))+(e*exp(x+2*dx)*cos(x+2*dx)))/dx^2;
difference_central_order= abs(central_differencing_fourth_order-analytical_deravative)
end

     

    % function for skewed right side

    % creating 4 order approximation using 2 order deravative by skewed right side difference
function error_skewed_right_side=skewed_right_side(x,dx)

%analytical function f(x)= exp(x)*cos(x);
% f''(x)=-2*exp(x)*sin(x);

analytical_deravative= -2*exp(x)*sin(x);
% values of coefficients from  taylor table
a=3.75000;
b=-12.83333;
c= 17.83333;
d=-13.00000;
e=5.08333;
g=-0.83333;

skewed_right_side_fourth_order=((a*exp(x)*cos(x))+(b*exp(x+dx)*cos(x+dx))+(c*exp(x+2*dx)*cos(x+2*dx))+(d*exp(x+3*dx)*cos(x+3*dx))+(e*exp(x+4*dx)*cos(x+4*dx))+(g*exp(x+5*dx)*cos(x+5*dx)))/dx^2;

error_skewed_right_side=abs(skewed_right_side_fourth_order-analytical_deravative)
end

     

    %function for skewed left side

     % creating 4 order approximation using 2 order deravative by skewed left side difference
function error_skewed_left_sided= skewed_left_side(x,dx)
%analytical function f(x)= exp(x)*cos(x);
% f''(x)=-2*exp(x)*sin(x);

analytical_deravative= -2*exp(x)*sin(x);
% values of coefficients from  taylor table
a= -0.83333;
b= 5.08333;
c=-13.00000;
d=17.83333;
e=-12.83333;
g=3.75000;

skewed_left_side_fourth_order=((a*exp(x-5*dx)*cos(x-5*dx))+(b*exp(x-4*dx)*cos(x-4*dx))+(c*exp(x-3*dx)*cos(x-3*dx))+(d*exp(x-2*dx)*cos(x-2*dx))+(e*exp(x-dx)*cos(x-dx))+(g*exp(x)*cos(x)))/dx^2;
error_skewed_left_sided=abs(skewed_left_side_fourth_order-analytical_deravative);
end

     

    %programing

    clear all
clc
close all
 x=pi/3;
 dx=linspace(pi/4,pi/4000,30);

 for i=1:length(dx)
    central_difference_error (i) =central_difference(x,dx(i));
     skewed_right_side_error (i) =skewed_right_side(x,dx(i));
     skewed_left_side_error (i) =skewed_left_side(x,dx(i));
 end
 % plotting graph of dx vs error
 figure(1)
 plot(dx,central_difference_error,'LineWidth',2)
 hold on
 plot(dx,skewed_right_side_error,"Color",'r','LineWidth',2)
 hold on
 grid on
 plot(dx,skewed_left_side_error,"Color",'g','LineWidth',2)
 xlabel('dx')
 ylabel('error')
 legend('central difference error','skewed right sided','skewed left side')

 figure(2)
  loglog(dx,central_difference_error,'LineWidth',2)
 hold on
 loglog(dx,skewed_right_side_error,"Color",'r','LineWidth',2)
 hold on
 grid on
 loglog(dx,skewed_left_side_error,"Color",'g','LineWidth',2)
 xlabel('dx')
 ylabel('error')
 legend('central difference error','skewed right sided','skewed left side')

    %result graph

    here we can easily notice central difference approximation produce more accurate solution.

     

     

     

     

     

     

     

     

     

     

     

     

     

     

     

     

     

     

     

     

     

     

     

     

     

     

     

     

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