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

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

Problem Statement: - For the Function f(x)=ex⋅cos(x)f(x)=ex⋅cos(x) derive the 4th Order Accurate Approximation of 2nd Order Derivative for the given function using Taylor Table Method for the following difference Methods: - 1. Central Difference 2. Skewed right-sided difference 3. Skewed left-sided difference Write A MATLAB…

  • MATLAB

  • Aditya Aanand

    updated on 09 Sep 2022

Problem Statement: -

For the Function f(x)=ex⋅cos(x)f(x)=ex⋅cos(x) derive the 4th Order Accurate Approximation of 2nd Order Derivative for the given function using Taylor Table Method for the following difference Methods: -

1. Central Difference 2. Skewed right-sided difference 3. Skewed left-sided difference

Write A MATLAB Code to Solve the function using the derived 4th Order Approximate 2nd Order Derivative and compare it with the 2nd Order Derivative of Analytical Solution

 

Solution: -

1. AIM: -

A. To Derive 4th Order Approximation for 2nd Order Derivative using: -

   a. Central Difference Method

   b. Skewed Right-Side Difference Method

   c. Skewed Left-Side Difference Method

B. Prove each Method to Be 4th Order Accurate

C. Write a MATLAB Code to Solve a function using all 3 Methods

D. Comment on the usefulness of Skewed Methods

 

2. Given: -

A. P = N - Q + ({1 for CDM} or {0 for Skewed Method})

    Here,

           P - Order of Accuracy of Approximation

           N - Number of Nodes in Numerical Stensil

           Q - Order of Derivative

           CDM - Central Difference Method

B.  Function→f(x)=ex⋅cos(x) Function→f(x)=ex⋅cos(x)

 

3. Assumed: -

A. Differentiation Point(x)=π3Differentiation Point(x)=π3

 

4. Analytical Derivative Formula: -

 second Derivative→∂2f(x)∂x2=-2ex⋅sin(x)

 

5. Formulas & Derivation of The Approximation Methods: -

A. Central Difference Method: -

 For calculating the 4th Order Approximation and 2nd Order Derivative

  (P=4) = N - (Q=2) + 1

 => Number of Nodes in Numerical Stensil(N) = 5

So,

       Numerical Stensil for Central Difference Method

 Num Stensil CDM

 

Let, (dx)2∂2f(x)∂x2=af(i-2)+bf(i-1)+cf(i)+df(i+1)+ef(i+2)

 

Using Taylor Table Method For Central Difference Method

Numerical Stensil ↓ f(i) (dx)∂f(i)∂x (dx)2∂2f(i)∂x2 (dx)3∂3f(i)∂x3 (dx)4∂4f(i)∂x4
    af(i-2)   a  -2a  4a2  -8a6  16a24
    bf(i-1)   b  -b  b2  -b6  b24
    cf(i)   c    0  0      0    0
   df(i+1)   d    d  d2    d6  d24
   ef(i+2)   e    2e  4e2    8e6  16e24
Sum=(dx)2∂2f(x)∂x2   0   0   1   0   0

 

On Solving,

    a=e=-112

    b=d=43

    c=52

Hence,

(dx)2∂2f(x)∂x2=-112f(i-2)+43f(i-1)-52f(i)+43f(i+1)-112f(i+2)

4th Order Approximation for 2nd Order Derivative of Central Difference Method

      ∂2f(x)∂x2=-112f(i-2)+43f(i-1)-52f(i)+43f(i+1)-112f(i+2)(dx)2

 

 B. Skewed Right-Side Difference Method: -

For Calculating 4th Order Approximation and 2nd Order Derivative

(P=4) = N - (Q=2) + 0

=> Number of Nodes in Numerical Stensil(N) = 6

So,

       Numerical Stensil for Skewed Right-Side Difference Method

Num Stencil Right Side Method

 

Let, (dx)2∂2f(x)∂x2=af(i)+bf(i+1)+cf(i+2)+df(i+3)+ef(i+4)+ff(i+5)

 

Numerical Stensil ↓ f(i) (dx)∂f(i)∂x (dx)2∂2f(i)∂x2 (dx)3∂3f(i)∂x3 (dx)4∂4f(i)∂x4 (dx)5∂5f(i)∂x5
    af(i)   a  0  0  0  0  0
    bf(i+1)   b  b  b2  b6  b24  b120
    cf(i+2)   c  2c  4c2  8c6  16c24  32c120
   df(i+3)   d  3d  9d2  27d6  81d24  729d120
   ef(i+4)   e  4e  16e2  64e6  256e24  1024e120
   ff(i+5)  f  5f  25f2  125f6  625f24  3125f120
Sum=(dx)2∂2f(x)∂x2   0   0   1   0   0   0

 

On Solving,

    a=154

    b=-776

    c=1076

    d=-13

    e=671132

    f=-56

Hence,

(dx)2∂2f(x)∂x2=154f(i)-776f(i+1)+1076f(i+2)-13f(i+3)+671132f(i+4)-56f(i+5)

4th Order Approximation for 2nd Order Derivative of Skewed Right-Side Difference Method

      ∂2f(x)∂x2=154f(i)-776f(i+1)+1076f(i+2)-13f(i+3)+671132f(i+4)-56f(i+5)(dx)2

 

C. Skewed Left-Side Difference Method: -

For Calculating 4th Order Approximation and 2nd Order Derivative

(P=4) = N - (Q=2) + 0

=> Number of Nodes in Numerical Stensil(N) = 6

So,

       Numerical Stensil for Skewed Right-Side Difference Method

Numerical Stencil Left Skewed

 

Let, (dx)2∂2f(x)∂x2=af(i-5)+bf(i-4)+cf(i-3)+df(i-2)+ef(i-1)+ff(i)

 

Numerical Stensil ↓ f(i) (dx)∂f(i)∂x (dx)2∂2f(i)∂x2 (dx)3∂3f(i)∂x3 (dx)4∂4f(i)∂x4 (dx)5∂5f(i)∂x5
    af(i-5)   a  -5a  25a2  -125a6  625a24  -3125a120
    bf(i-4)   b  -4b  16b2  -64b6  256b24  -1024b120
    cf(i-3)   c  -3c  9c2  -27c6  81c24  -729c120
   df(i-2)   d  -2d  4d2  -8d6  16d24  -32d120
   ef(i-1)   e  -e  e2  -e6  e24  -e120
   ff(i)  f  0  0  0  0  0
Sum=(dx)2∂2f(x)∂x2   0   0   1   0   0   0

 

On Solving,

    a=-56

    b=671132

    c=-13

    d=1076

    e=-776

    f=154

Hence,

(dx)2∂2f(x)∂x2=-56f(i-5)+671132f(i-4)-13f(i-3)+1076f(i-2)-776f(i-1)+154f(i)

4th Order Approximation for 2nd Order Derivative of Skewed Right-Side Difference Method

      ∂2f(x)∂x2=-56f(i-5)+671132f(i-4)-13f(i-3)+1076f(i-2)-776f(i-1)+154f(i)(dx)2

 

6. Proving That the Above Derived Formulas are 4th Order Accurate: -

A. Central Difference Method: -

As form proof the Coefficients of f(x),(dx)∂f(x)∂x,(dx)3∂3f(x)∂3xand(dx)4∂4f(x)∂x4 are equated to Zero(0) now for the Given Value of a, b, c, d, and e

Checking the Coefficient of(dx)5∂5f(x)∂x5

Coefficient of (dx)5∂5f(x)∂x5=-32a120-b120+0+d120+32e120

         For,a=e=-112andb=d=43

Coefficient of (dx)5∂5f(x)∂x5=0

Hence,

....+(coefficient of fIV(x))(dx)6∂6f(x)∂x6+(1)(dx)2∂2f(x)∂x2=-112f(i-2)+43f(i-1)-52f(i)+43f(i+1)-112f(i+2)

 

                                            Dividing Both-sides from(dx)2

⇒....+(coefficient of fIV(x))(dx)4∂6f(x)∂x6+∂2f(x)∂x2=-112f(i-2)+43f(i-1)-52f(i)+43f(i+1)-112f(i+2)(dx)2

 

Ignoring All multiples of dx4 (=> The Order of Accuracy is 4th Order)

∂2f(x)∂x2=-112f(i-2)+43f(i-1)-52f(i)+43f(i+1)-112f(i+2)(dx)2

 

B. Skewed Right-Side Difference Method: -

As form proof the Coefficients of f(x),(dx)∂f(x)∂x,(dx)3∂3f(x)∂3x,(dx)4∂4f(x)∂x4and(dx)5∂5f(x)∂x5 are equated to Zero(0)

Hence,

....+(coefficient of fIV(x))(dx)6∂6f(x)∂x6+(1)(dx)2∂2f(x)∂x2=154f(i)-776f(i+1)+1076f(i+2)-13f(i+3)+671132f(i+4)-56f(i+5)

 

 

                                            Dividing Both-sides from(dx)2

⇒....+(coefficient of fIV(x))(dx)4∂6f(x)∂x6+∂2f(x)∂x2=154f(i)-776f(i+1)+1076f(i+2)-13f(i+3)+671132f(i+4)-56f(i+5)(dx)2

 

Ignoring All multiples of dx4 (=> The Order of Accuracy is 4th Order)

∂2f(x)∂x2=154f(i)-776f(i+1)+1076f(i+2)-13f(i+3)+671132f(i+4)-56f(i+5)(dx)2

 

C. Skewed Left-Side Difference Method: -

As form proof the Coefficients of f(x),(dx)∂f(x)∂x,(dx)3∂3f(x)∂3x,(dx)4∂4f(x)∂x4and(dx)5∂5f(x)∂x5 are equated to Zero(0)

Hence,

....+(coefficient of fIV(x))(dx)6∂6f(x)∂x6+(1)(dx)2∂2f(x)∂x2=-56f(i-5)+671132f(i-4)-13f(i-3)+1076f(i-2)-776f(i-1)+154f(i)

 

 

                                            Dividing Both-sides from(dx)2

⇒....+(coefficient of fIV(x))(dx)4∂6f(x)∂x6+∂2f(x)∂x2=-56f(i-5)+671132f(i-4)-13f(i-3)+1076f(i-2)-776f(i-1)+154f(i)(dx)2

 

Ignoring All multiples of dx4 (=> The Order of Accuracy is 4th Order)

∂2f(x)∂x2=-56f(i-5)+671132f(i-4)-13f(i-3)+1076f(i-2)-776f(i-1)+154f(i)(dx)2

 

7. Explanation of Code: -

A. Section A(Defining Variables): -

    a. Defined x and a vector of different values of dx

    b. Calculated Analytical Value using Analytical Formula

    c. created "A" and "B" of Central and Skewed Forms ( "A" and "B" of AX = B)

    d. Calculated "X" of Central and Skewed Forms using X = AB

B. Section B(Differentiating): -

    a. Created A loop of Iteration of the length of dx vector

    b. Selected a dx at each iteration

    c. Created small functions for simplification of Calculations at each iteration

    d. Calculated Central Difference and Skewed Differences using the Function defined above and X calculated in Section A.

    e. Calculated Error generated per Iteration w.r.t Analytical Value Calculated in Section A and saved it in a Vector.

C. Section C(Plotting)

    a. Plotted a log Graph between The Value of Error vs The Value of dx for all 3 Methods

    b. Defined Proper Legends and Labels in the plot.

 

8. Code: -

 

% 4th Order 2nd Differentiation using Central and Both Skewed Schemes
clear all
close all
clc

%% Defining Variables
   x = pi/3;
   dx_all = x.*(10.^[-3:-1]);
   analytical_val = -(2*exp(x)*sin(x))
   A_crt_vec = [-2:1:2]';
   A_rgt_vec = [0:5]';
   A_lft_vec = [-5:0]';
   A_crt = [ones(5,1),(A_crt_vec.^[1:4]).*(1./cumprod([1:4]))]';
   A_rgt = [ones(6,1),(A_rgt_vec.^[1:5]).*(1./cumprod([1:5]))]';
   A_lft = [ones(6,1),(A_lft_vec.^[1:5]).*(1./cumprod([1:5]))]';
   B_crt = [0;0;1;0;0];
   B_skw = [0;0;1;0;0;0];
   X_crt = A_crtB_crt
   X_rgt = A_rgtB_skw
   X_lft = A_lftB_skw
   
%% Differentiating
   for i=1:length(dx_all)
       % Selecting h and Creating small Function for Ease
       dx = dx_all(i);
       f_mi_5 = exp(x-5*dx)*cos(x-5*dx);
       f_mi_4 = exp(x-4*dx)*cos(x-4*dx);
       f_mi_3 = exp(x-3*dx)*cos(x-3*dx);
       f_mi_2 = exp(x-2*dx)*cos(x-2*dx);
       f_mi_1 = exp(x-dx)*cos(x-dx);
       f_x = exp(x)*cos(x);
       f_pl_1 = exp(x+dx)*cos(x+dx);
       f_pl_2 = exp(x+2*dx)*cos(x+2*dx);
       f_pl_3 = exp(x+3*dx)*cos(x+3*dx);
       f_pl_4 = exp(x+4*dx)*cos(x+4*dx);
       f_pl_5 = exp(x+5*dx)*cos(x+5*dx);
       % Creating Formula and Executing it
       crt_fn_vec = [f_mi_2,f_mi_1,f_x,f_pl_1,f_pl_2];
       skw_rgt_fn_vec = [f_x,f_pl_1,f_pl_2,f_pl_3,f_pl_4,f_pl_5];
       skw_lft_fn_vec = [f_mi_5,f_mi_4,f_mi_3,f_mi_2,f_mi_1,f_x];
       dif_crt = (crt_fn_vec*X_crt)/(dx^2);
       dif_rgt = (skw_rgt_fn_vec*X_rgt)/(dx^2);
       dif_lft = (skw_lft_fn_vec*X_lft)/(dx^2);
       % Storing Errors
       err_crt(i) = abs(analytical_val - dif_crt);
       err_rgt(i) = abs(analytical_val - dif_rgt);
       err_lft(i) = abs(analytical_val - dif_lft);
   end

%% Plotting OUTPUT
loglog(dx_all,err_crt,dx_all,err_rgt,dx_all,err_lft)
xlabel('Value of dx','FontSize',15)
ylabel('Value of Error','FontSize',15)
legend('Central Difference','Skewed Right-Side Difference','Skewed Left-Side Difference')

 

9. OUTPUTS Generated on Running the Code: -

A. Command Window

Assign 3_5 Command Window Output

 

B. Plot OUTPUT

Assign 3_5 Graph

 

10. Usefulness of Skewed Difference Methods: -

Skewed Schemes are Generally useful where the Information is flowing from one direction then the Correct calculations can only be developed by points on one side, as the next value thus defined is required to be developed using the values just achieved. To explain this I'm using the example Below.

Example, 

Example for Skewed Scheme

In this example, if properties at points A and B are available and the Value of C is to be determined such that difference between each point is dx then,

Using Taylor Series, ϕC=ϕB+(∂ϕ∂x)Bdx

Now, As properties at points A and B are only Known So to determine (∂ϕ∂x)B

Skewed Left-Side Difference Method is used, (∂ϕ∂x)B=ϕB-ϕAdx

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