Value for Function with 2nd order Central difference scheme
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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)
采纳的回答
Star Strider
2023-8-12
0 个投票
See First and Second Order Central Difference and add enclosing parentheses to the numerator of your implementation of the cosh function.
2 个评论
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)
Anu
2023-9-30
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
更多回答（1 个）
Anu
2023-9-30
0 个投票
- 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.
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