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

AIM

To derive the fourth order approximations of a second order derivative using central differencing scheme, skewed right sided difference and skewed left sided difference with the help of Taylor table method and to compare the analytical derivative with obtained values for the given function.

 

OBJECTIVE

 

1. Derive the following 4th order approximation of the second-order derivative using:- 
2. Central difference
3. Skewed right-sided difference
4. Skewed left-sided difference
5. Also, prove that your skewed schemes are fourth-order accurate.  

• Write a program in Matlab to evaluate the second-order derivative of the analytical function exp(x)*cos(x) and compare it with the 3 numerical approximations that you have derived. 

1. Provide a plot that compares the absolute error between the above-mentioned schemes
2. Describe why a skewed scheme is useful? What can a skewed scheme do that a CD scheme cannot do?

 

INTRODUCTION

            In this challenge we will be dealing with Taylor Table method for the approximation.  Taylor table is a general method for choosing the coefficients of a finite difference formula to ensure the highest possible order of accuracy. "Taylor Table" will determine the coefficients of the terms which minimizes the Taylor series error. It will also determine the Taylor series error and the order of accuracy of the method.

 

THEORY

 

Taylor series

The Taylor series of a function is an infinite sum of terms that are expressed in terms of the function's derivatives at a single point. For most common functions, the function and the sum of its Taylor series are equal near this point.

`f(x)=f(a)+(f'(a))/(1!)(x-a)+(f''(a))/(2!) (x-a)^2+(f'''(a))/(3!) (x-a)^3+...`

`f(x+a)=f(x)+(f'(x))/(1!)(a)+(f''(x))/(2!) (a)^2+(f'''(x))/(3!) (a)^3+...`

 

As the degree of the Taylor polynomial rises, it approaches the correct function. This image shows sin x and its Taylor approximations by polynomials of degree 1, 3, 5, 7, 9, 11, and 13 at x = 0.

 

Approximations

 

We can use the first few terms of a Taylor Series to get an approximate value for a function.

Here we show better and better approximations for cos(x). The red line is cos(x), the blue is the approximation.

 

Deriving the 4th order approximations of the second-order derivative

Let's consider a numerical stencile with a set of grid points as shown in the fig below

right-sided difference

 

left -sided difference

 

Above shown both numerical stencils are Skewed 2nd order right & left -sided difference scheme  for fourth order approximation.

central-difference

And this is symmetric central difference for 4th order opproximation

 

DERIVATION

 

1. CENTRAL DIFFERENCE

 

Define the no.of nodes & the coefficients to find:

     for central  difference approximation we use

                   Number of nodes= P +Q - 1

               P = Order of the approximation = 4

              Q = Order of the derivative = 2

                Number of nodes= 4 +2 – 1 = 5

So we will find up to 5 nodes for 2nd derivative of the 4th order approximation central difference approximation.

 

Since we’re deriving using central difference; our numerical stencil will look like.

Here, we’re computing 4th order approximation at point (i), where we take information from the four-point surrounding (i); i.e. we’re going to take two points on the left of (i) and two points on the right of (i). Let those points be (i-2), (i-1), (i+1), and (i+2) and the distance between each of them is (Δx). Our numerical stencil will look like as shown in the above figure. 

Based on the above stencil our hypothesis equation will be:- 

`(∂^2 f)/(∂x^2 )=af(i-2)+bf(i-1)+cf(i)+df(i+1)+ef(i+2).........(1)`

In order to find the values of coefficient(a,b,c,d,e), we’re going to use the Taylor table method. Writing Taylor series expansion for each term of equation (1) as:- 

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

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

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

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

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

 

By arranging the above Taylor series expansions in Taylor table; we get:- 

 

 

Note: Only for the first five terms we assume our hypothesis. That’s why we can’t say about the last two terms. 

Here we have five unknowns, so, we have to solve five equations and get values of coefficients (a,b,c,d,e). We are going to form these equations by adding terms for the individual columns. The summation of coefficients of `f''(i) is equal to 1, and f(i), f′(i), f′'(i), f′''(i), f′'''(i) are equal to 0; because we want to calculate the second-order derivative, f′'(i).

`a+b+c+d+e=0`

`-2a-b+0+d+2e=0`

`(4a)/2+(b/2)+0+d/2+(4e)/2=1`

`-(8a)/6-b/6+0+d/6+(8e)/6=0`

`(16a)/24-b/24+0+d/24+(16e)/24=1`

we can easily find out the coefficient values if we convers the above eq's into a matrix form :

`AX=B`

`X=[[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]]`

X=[[a],[b],[c],[d],[e]]B=[[0],[0],[1],[0],[0]]`

`X=A^(-1)⋅B`

`X=[[a],[b],[c],[d],[e]]=[[-0.08333],[1.3333],[-2.5000],[1.3333],[-0.08333]]`

 

Using mat lab we will solve the equation

A=[1, 1, 1, 1, 1;-2, -1, 0, 1, 2;4/2, 1/2, 0, 1/2, 4/2;-8/6, -1/6, 0, 1/6, 8/6;16/24, 1/24, 0, 1/24, 16/24]
B=[0;0;1;0;0]
X=A^-1*B

 

 

Substitute the values of X matrix in equation and find the second derivative using central difference method.

Here, we observed that a= -0.0833; b=1.3333; c=-2.5; d=1.3333; e= -0.8333;

a=e;b=d;

Substitute in the table for `f^v (i)dx^5` and we find that their sum of coefficient = 0;
After that, the term is `f^(vi) (i)dx^6`

Divide `f^(vi) (i)dx^6`, we get the coefficient in the order of `dx^4` . We are neglecting that terms, so we can say that our approximation is fourth order accurate.

 

2)Using Skewed Right-Sided Difference:

 

Define the no.of nodes & the coefficients to find:

     for one sided / skewed schemes approximation we use

                   Number of nodes= P +Q

               P = Order of the approximation = 4

              Q = Order of the derivative = 2

                Number of nodes= 4 +2  = 6

So we will find up to 6 nodes for 2nd derivative of the 4th order approximation for one sided / skewed schemes approximation.

To derive 4th order skewed right-sided difference at point (i) of function. Since we’re using the skewed right-sided method, here we have to define five points on the right of (i). Let those points be (i+1), (i+2), (i+3), (i+4), and (i+5); with a spacing of Δx each. 

 

According to the above stencil, our hypothesis equation will be: 

`(∂^2 f)/(∂x^2 )=af(i)+bf(i+1)+cf(i+2)+df(i+3)+ef(i+4)+gf(i+5)`

 

In order to find the values of coefficients (a,b,c,d,e,g), we’re going to use the Taylor table method. 

Writing Taylor series expansion for each term of equation as:- 

`(a)f(i)=(a)f(i)`

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

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

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

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

`(g)f(i+5)=(g)f(i)+((g)f'(i)5Δx)/(1!)+((g)f''(i)(5Δx)^2)/(2!)+((g)f'''(i)(5Δx)^3)/(3!)+((g)f''''(i)(5Δx)^4)/(4!)+((g)f'''''(i)(5Δx)^5)/(5!)+((g)f''''''(i)(5Δx)^6)/(6!)+....`

 

By arranging the above Taylor series expansions in Taylor table; we get:- 

Only for the first six terms, we assume our hypothesis. That’s why we can’t say about the last term. 

Here we have six unknowns, so, we have to solve six equations and get values of coefficients (a,b,c,d,e,g). We are going to form these equations by adding terms for the individual columns. The summation of coefficients of f′'(i) is equal to 1, and f(i), f′(i), f′'(i), f′''(i), f′'''(i), f'''''(i) is equal to 0; because we want to calculate the second-order derivative, f''(i).

 

Therefore, the six equations are:- 

`a+b+c+d+e+g=0`

`0+b+2c+3d+4e+5g=0`

`0+b/2+(4c)/2+(9d)/2+(16e)/2+(25g)/2=1`

`0+b/6+(8c)/6+(27d)/6+(64e)/6+(125g)/6=0`

`0+b/24+(16c)/24+(81d)/24+(256e)/24+(625g)/24=0`

`0+b/120+(32c)/120+(243d)/120+(1024e)/120+(3125g)/120=0`

 

we can easily find out the coefficient values if we convers the above eq's into a matrix form :

`AX=B`

`X=[[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],[0,1/120,32/120,243/120,1024/120,3125/24]]`

`X=[[a],[b],[c],[d],[e],[g]]``B=[[0],[0],[1],[0],[0],[0]]`

`X=A^(-1)⋅B`

`X=[[a],[b],[c],[d],[e],[g]]=[[3.7500],[-12.8333],[17.8333],[-13.0000],[5.0833],[-0.8333]]`

 

Using mat lab we will solve the equation

A=[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]
  B=[0;0;1;0;0;0]
  X=inv(A)*B

Substitute the values of X matrix in equation and find the second derivative using central difference method.

Here, we observed that a=3.75;  b=-12.8333;  c=17.8333;   d=-13;   e=5.0833;   g=-0.8333;

We find the coefficient, upto the term`f^v (i)dx^5`

Divide `f^(vi) (i)dx^6`, we get the coefficient in the order of `dx^4` . We are neglecting that terms, so we can say that our approximation is fourth order accurate.

 

3)Using Skewed Left-Sided Difference:

 

Define the no.of nodes & the coefficients to find:

     for one sided / skewed schemes approximation we use

                   Number of nodes= P +Q

               P = Order of the approximation = 4

              Q = Order of the derivative = 2

                Number of nodes= 4 +2  = 6

So we will find up to 6 nodes for 2nd derivative of the 4th order approximation for one sided / skewed schemes approximation.

To derive 4th order skewed left-sided difference at point (i) of function. Since we’re using the skewed left-sided method, here we have to define five points on the left of (i). Let those points be (i-1), (i-2), (i-3), (i-4), and (i-5); with a spacing of Δx^2 each. 

 

According to the above stencil, our hypothesis equation will be: 

`(∂^2 f)/(∂x^2 )=af(i)+bf(i-1)+cf(i-2)+df(i-3)+ef(i-4)+gf(i-5)`

 

To find the values of coefficients (a,b,c,d,e,g), we’re going to use the Taylor table method. 

Writing Taylor series expansion for each term of equation as:-

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

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

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

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

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

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

 

By arranging the above Taylor series expansions in Taylor table; we get:- 

Only for the first six terms, we assume our hypothesis. That’s why we can’t say about the last term. 

Here we have six unknowns, so, we have to solve six equations and get values of coefficients (a,b,c,d,e,g). We are going to form these equations by adding terms for the individual columns. The summation of coefficients of f′'(i) is equal to 1, and f(i), f′(i), f′'(i), f′''(i), f′'''(i), f'''''(i)  is equal to 0; because we want to calculate the second-order derivative, f''(i).

Therefore, the six equations are:- 

`a+b+c+d+e+g=0`

`-5a-4b-3c-2d-e+0=0`

`(25a)/2+8b+(9c)/2+2d+e/2+0=1`

`-(125a)/6-(64b)/6-(27c)/6-(8d)/6-e/6+0=0`

`(625a)/24+(256b)/24+(81c)/24+(16d)/24+e/24+0=0`

`-(3125a)/120-(1024b)/120-(243c)/120-(32d)/120-e/120+0=0`

 

we can easily find out the coefficient values if we convers the above eq's into a matrix form :

`AX=B`

`X=[[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]]`

`X=[[a],[b],[c],[d],[e],[g]]``B=[[0],[0],[1],[0],[0],[0]]`

`X=A^(-1)⋅B`

`X=[[a],[b],[c],[d],[e],[g]]=[[3.7500],[-12.8333],[17.8333],[-13.0000],[5.0833],[-0.8333]]`

 

Using mat lab we will solve the equation

A=[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]
  B=[0;0;1;0;0;0]
  x_r=inv(A)*B

 

Substitute the values of X matrix in equation and find the second derivative using central difference method.

Here, we observed that a=3.75;  b=-12.8333;  c=17.8333;   d=-13;   e=5.0833;   g=-0.8333;

We find the coefficient, upto the term`f^v (i)dx^5`

Divide `f^(vi) (i)dx^6`, we get the coefficient in the order of `dx^4` . We are neglecting that terms, so we can say that our approximation is fourth order accurate.

 

we have rewrite the equations here ;

 central difference approximation :

`(∂^2 f)/(∂x^2 )=(a⋅f(x-2dx)+b⋅f(x-dx)+c⋅f(x)+d⋅f(x+dx)+e⋅f(x+2dx))/(Δx^2 )`

 

 skewed forward difference approximation :

`(∂^2 f)/(∂x^2 )=(a⋅f(x)+b⋅f(x+dx)+c⋅f(x+2dx)+d⋅f(x+3dx)+e⋅f(x+4dx)+g⋅f(x+5dx))/(Δx^2 )`

 

skewed backward difference approximation :

`(∂^2 f)/(∂x^2 )=(a⋅f(x)+b⋅f(x-dx)+c⋅f(x-2dx)+d⋅f(x-3dx)+e⋅f(x-4dx)+g⋅f(x-5dx))/(Δx^2 )`

 

 

FUNCTION CODE:

 

CENTRAL DIFFERENCE APPROXIMATION

function central_diff_error=central_difference_approximation(x,dx)
%function=exp(x)*cos(x)
%f ''(x)=-2*exp(x)*sin(x)
analytical_derivative=-2*exp(x)*sin(x);
%numerical derivative
a=-0.083333;
 
b=1.333333;
 
c=-2.500000;
 
d=1.333333;
 
e=-.083333;
%central_diff_approx=(a*f(x-2*dx)+b*f(x-dx)+c*f(x)+d*f(x+dx)+e*f(x+2*dx))/(dx^2)
central_diff_approx=(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);
 
central_diff_error=abs(central_diff_approx-analytical_derivative)
end

 

BACKWARD DIFFERENCE APPROXIMATION

function skewed_backward_error=backward_difference_approximation(x,dx)
%function=exp(x)*cos(x)
%f ''(x)=-2*exp(x)*sin(x)
analytical_derivative=-2*exp(x)*sin(x);
%numerical derivative
a=-0.83333;
 
b=5.08333;
 
c=-13.00000;
 
d=17.83333;
 
e=-12.83333;
 
g=3.750000;
% skewed left sided approx=(a*f(x-5*dx)+b*f(x-4*dx)+c*f(x-3*dx)+d*f(x-2*dx)+e*f(x-dx)+g*f(x))/(dx^2)
skewed_backward_approx=(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);
skewed_backward_error=abs(skewed_backward_approx-analytical_derivative);
end

 

FORWARD DIFFERENCE APPROXIMATION

function skewed_forward_error=forward_difference_approximation(x,dx)
%function=exp(x)*cos(x)
%f ''(x)=-2*exp(x)*sin(x)
analytical_derivative=-2*exp(x)*sin(x);
%numerical derivative
a=3.75000;
 
b=-12.83333;
 
c=17.83333;
 
d=-13.00000;
 
e=5.08333;
 
g=-0.83333;
%skewed rightsided approx=a*f(x)+b*f(x+dx)+c*f(x+2*dx)+d*f(x+3*dx)+e*f(x+4*dx)+g*f(x+5*dx)
skewed_forward_approx=(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);
skewed_forward_error=abs(skewed_forward_approx-analytical_derivative);
end

 

MAIN CODE

clear all
close all
clc
 
x=pi/3;
dx=linspace(pi/4,pi/40000,40);
for i=1:length(dx)
    central_diff_error(i)=central_difference_approximation(x,dx(i));
    skewed_backward_error(i)=backward_difference_approximation(x,dx(i));
    skewed_forward_error(i)=forward_difference_approximation(x,dx(i));
end
figure(1)
plot(dx,central_diff_error,"Color",'r','LineWidth',2)
hold on
plot(dx,skewed_backward_error,"Color",'b','LineWidth',2)
plot(dx,skewed_forward_error,"Color",'g','LineWidth',2)
legend('Central Difference Approximation','Backward Difference Approximation','Forward Difference Approximation')
grid on
xlabel('dx')
ylabel('error')
figure(2)
loglog(dx,central_diff_error,'color','r','linewidth',2)
hold on
loglog(dx,skewed_backward_error,'color','b','LineWidth',2)
loglog(dx,skewed_forward_error,'color','g','linewidth',2)
legend('Central Difference Approximation','Backward Difference Approximation','Forward Difference Approximation')
grid on
xlabel('dx')
ylabel('error')

 

Explanation

1. clear all, close all, clc is used to clear all the variables and delet the graph, plot which are created already.
2. Input value is assigned to variables x and range operator is used to assign dx value (dx range = [pi/4000,pi/40]).
3. Use for loop with i ranges from 1 to length of dx and assign the central and skewed right differance and skewed left differance function which is created.
4. A function is created for 4th order error with  skewed right side  difference error as the function definition and forth order skewed right side difference as the function name and x,dx as the input and assign the skewed right side  difference approximation and analytical differentiate formulas and end the function.
5. Create next function for 4th order error with central difference error as the function definition and forth order central differance as the function name and x,dx as the input and assign the central difference approximation and analytical differentiate formulas and end the function.
6. Then create another function for 4th order error with skewed left side  difference error as the function definition and forth order skewed left side differenceas the function name and x,dx as the input and assign the skewed left side  difference approximation and analytical differentiate formulas and end the function.
7. To plot the graph loglog command is used for (dx vs  central difference error , dx vs skewed right side  difference error and dx vs skewed left side  difference error) which is used to compare the curve in detaily and xlabel and ylabel is added and legend is also added.

 

RESULT

From the above plot we can conclude that cemtral difference scheme has lesser error with decreasing dx as compared to the skewed right sided and skewed left sided schemes.

CD uses information from both the sides from the point of consideration to give approximative and accurate values.

As the left and right skewed difference methods uses data only from one side of point of interest it gives less approximative and accurate values about the point.

 

 

The 4th order approximations of the second-order derivative of Central difference , Skewed right-sided difference, Skewed left-sided difference and  the program in Matlab to evaluate the second-order derivative of the analytical function exp(x)*cos(x) and compare it with the 3 numerical approximations that are derived are successfully completed.

 

1.why a skewed scheme is useful?

Skewed scheme is useful when there aren't sufficient information available, in such cases the numerical stencil might not be symmetric. Thus we use skewed scheme.

 

2.What can a skewed scheme do that a CD scheme cannot?

Since there are no boundaries or sufficient information available CD scheme cannot be used to find the value at the specified point thus we use skewed scheme in place of CD scheme even though CD scheme is more accurate than the skewed scheme and has lesser error as compared to skewed scheme.

 

CONCLUSION

 

1. As we can see from the above figure the error for the central difference method is very less compared to the skewed methods.
2. Skewed right and left difference methods almost gives the same results in terms of error for a small value of time step.

1. When data from the grid points is only available from either the left or right end i.e. at nodes closer to the boundaries,for computations that requires a high order approximations we make use of skewed schemes.Having a symmetric stencil for computation is not possioble in this scenarios like we do for central differencing schemes or having a low order accuracy.For better consistency of scheme to be maintained the values of dx are kept as low as possible for providing proper computational plots.
2. A skewed scheme is useful when we have to predict using the boundary condition as we have only one direction to go. Hence the skewed scheme will be useful and we cannot use the central differencing scheme for such an application because it required information from both directions of the point.