Value for Function with 2nd order Central difference scheme

I am trying to write code for the above problem but getting wrong answer, Kindly help me to find the error in the code or suggest if there is any better alternate way to write code for the problem.

Right answer is 2.3563

c=1.5;
h=0.1;
x=(c-h):h:(c+h);
Fun=@(x) exp(x)-exp(-x)/2;
dFun=@(x) 2*exp(x)+2*exp(-x)/2;
F=Fun(x);
n=length(x);

dx= (F(:,end)-F(:,1))/(2*h)

dx = 4.6009

Answer 1

[Star Strider]

See First and Second Order Central Difference and add enclosing parentheses to the numerator of your implementation of the cosh function.

[VBBV]

c=1.5;
h=0.1;
x=(c-h):h:(c+h);
Fun=@(x) (exp(x)-exp(-x))/2;  % parenthesis 
dFun=@(x) 2*(exp(x)+exp(-x))/2; % parenthesis 
F=Fun(x);
n=length(x);

dx= (F(:,end)-F(:,1))/(2*h)

dx = 2.3563

[Anu]

c = 1.5;

h = 0.1;

x = (c - h):h:(c + h);

Fun = @(x) (exp(x) - exp(-x)) / 2;

F = Fun(x);

n = length(x);

dx = (F(3) - F(1)) / (2 * h); % Corrected calculation of derivative at x=c

Answer 2

[Anu]

• c is the central point.
• h is the step size.
• x is a vector of values around c.
• Fun is the function you want to calculate the derivative for.
• F is the function values at the points in x.
• dx calculates the derivative at the central point c using finite differences.