清空筛选条件
清空筛选条件

Absolute Error and Relative Error, Help please!!

25 次查看（过去 30 天）
SB
SB 2012-11-2
评论： Torsten 2023-3-7
So I've found one approximation of ln(x) to be 2* the sum of 1/(2n+1)*((x-1)/(x+1))^(2n+1) and I need to find the absolute and relative errors at every new x and n value and then graph them in a 3d Surface Plot.
Here's what I have so far, I'm not sure where to go from here:
%
function [abserror,relerror]=error_comparison(x,n)
sum1=0;
for i=0:n
sum1=sum1+ 1./(2.*i+1).*(((x-1)./(x+1)).^(2*i+1)); % an approximation of ln(x)
end
sum=2*sum1;
abserror= log(x)-sum
relerror=abserror./log(x)
end
  9 个评论
显示 7更早的评论
Mahmoud
Mahmoud 2023-3-7
Writing the negative exponent to give the same result as the calculator. Example: 0.3000 * 10^-3 gives a result of 3 * 10^-4
But in MATLAB it gives a very different result. What is the problem here or how to write the equation correctly to give the same value as the calculator
Torsten
Torsten 2023-3-7
Ran in:
I don't know what you mean.
0.3000 * 10^-3
ans = 3.0000e-04
3 * 10^-4
ans = 3.0000e-04

请先登录，再进行评论。

请先登录，再回答此问题。

回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Calculus 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by