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

Using Central differencing method:

The stencil will look like this: ….(i-2)….(i-1)….(i)….(i+1)…..(i+2)

Equation from above stencil: δu^2/δx^2 = af(i − 2) + bf(i − 1) + cf(i) + df(i + 1) + ef(i +2)

Taylors Table :

	f(i)	△xf(i)	△x^2f''(i)	△x^3f'''(i)	△x^4f''''(i)	△x^5f'''''(i)	△x^6f'''''(i)
f(i-2)	a	-2a	4a/2	-8a/6	16a/4	-32a/120	64a/240
f(i-1)	b	-b	b/2	-b/6	b/24	-b/120	b/240
f(i)	c	0	0	0	0	0	0
f(i+1)	d	d	d/2	d/6	d/24	d/120	d/240
f(i+2)	e	2e	4e/2	8e/6	16e/24	-32e/120	64e/120
af(i − 2) + bf(i − 1) + cf(i) + df(i + 1) + ef(i +2)	0	0	1	0	0	?	?

a=-020833 b=1.33 c=-2.500 d=1.33 e=-0.08333

using skewed right difference:

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

a = 3.75 b=-12.8333 c=17.833 d=-13.00 e=5.0833 g=-0.833

using skewed left-sided difference:

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

we get a=0.8333 b=5.0833 c=-13.00 d=17.8333 e=-12.83333 g=3.7500

% 4th order centraldiffrence
function error_central_differencing = central difference (x, dx)

% Analytical function f(x) = exp(x)*cos(x)

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

%  Taylor table;
a=-0.083333;
b=1.333333;
c=-2.500000; 
d=1.333333;
e=0.083333;

central_differencing_fourth_order_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;
error_central_differencing_differencing = abs(central_differencing_fourth_order_approx - analytical_derivative);
end

%right_skewed_difference

function error_skewed_right_sided=skewed_right_sided(x,dx)

% analytical function f(x)=exp(x) *cos(x);

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

%values of coefficients calculated from our Taylor table;
a=3.75000;
b=-12.83333;
c=17.83333;
d=-13.00000;
e=5.08333;
g=-8.83333;


%skewed right-sided differencing=((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)))/dx^2;

skewed_right_sided_fourth_order_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;

error_skewed_right_sided=abs(skewed_right_sided_fourth_order_approx-analytical_derivative);
end

%skewed left diffrence

function error_skewed_left_sided=skewed_left_sided(x, dx)

% analytical function f(x)=exp(x)*cos(x); 

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

%values of coefficients calculated from our Taylor table;
a=-0.83333;
b=5.08333;
c=-13.00000;
d=17.83333;
e=-12.83333;
g=3.75000;



%skewed left-sided differencing=((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 left-sided fourth order approximation;

skewed_left_sided_fourth_order_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;

error_skewed_left_sided=abs(skewed_left_sided_fourth_order_approx-analytical_derivative);
end