Use vector input to anonymous function and append output to a vector (to plot the results).

I have written a short code to calculate the forward difference and relative error of a function at a point with a variable step size "h". I can succesfully calculate the forward difference at a single step size and have created a vector of values of "h" . When I try to use the vector of step sizes in my equation for the forward difference it only calculates the difference using the first value in the step size vector. I don'y understand why.

My intent is to calculate the value of the forward difference for all values of "h" and append the results to a single vector so that I can plot them on a graph with relative error. I will need to use the same approach to make a relative error plot. Here is my code:

% clear the command window, workspace, and any plots to avoid error in future calcs.

clc

clear

clf

% declare symbolic variables

syms x1 x2 h

% Declare the function

F = (x1+10).^2 + (x2+5).^2 + exp(-(x1.^2)-(x2.^2));

% find the symbolic gradient

g = gradient(F,[x1,x2]);

V = double(subs(g,{x1,x2},{-8,-8}));

grad_f = sum(V);

% set point of interest

x1=-8;

x2=-8;

% setup given function

A = @(x1,x2) (((x1)+10).^2) + (((x2)+5).^2) + exp(-(((x1).^2)-((x2).^2)));

F = A(x1,x2);

a = [1:1:25];

h = [10.^(-a)];

FWD = [(A(x1+h,x2+h) - F)/h]

Answer 1

[Voss]

Use "./" rather than "/" in your calculation of FWD, in order to do element-wise division.

% clear the command window, workspace, and any plots to avoid error in future calcs.
clc 
clear
clf
% declare symbolic variables 
syms x1 x2 h
% Declare the function
F = (x1+10).^2 + (x2+5).^2 + exp(-(x1.^2)-(x2.^2));
% find the symbolic gradient
g = gradient(F,[x1,x2]);
V = double(subs(g,{x1,x2},{-8,-8}));
grad_f = sum(V);
% set point of interest
x1=-8;
x2=-8;
% setup given function
A = @(x1,x2) (((x1)+10).^2) + (((x2)+5).^2) + exp(-(((x1).^2)-((x2).^2)));
F = A(x1,x2);
a = [1:1:25];
  
h = [10.^(-a)];
FWD = [(A(x1+h,x2+h) - F)./h]

FWD = 1×25

-1.8000 -1.9800 -1.9980 -1.9998 -2.0000 -2.0000 -2.0000 -2.0000 -2.0000 -2.0000 -2.0000 -2.0002 -2.0073 -1.9540 -1.7764 0 0 0 0 0 0 0 0 0 0

Answer 2

[Francis Pantuso]

Fantastic. This works perfectly for my application.

[Voss]

Great. Please mark my answer as "Accepted" if you don't mind.