C calculation of the gradient of a multidimensional function. (partial derivatives) - c

Question

I tried to implement an algorithm for calculating the gradient, but I can't even figure out whether it works correctly or not, because I don't know how to calculate it even with my hands.
I have attached the material on which I tried to make an implementation of the algorithm. I would be very grateful if someone could give an example of the input data and the result that should turn out, because I really want to figure it out. And also I would like to know what I have incorrectly prescribed in my program.
Thank you in advance. Because I'm really stuck on this and I can't move from a dead place.
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <locale.h>
double f(double x)
{
return (x * x );
}
int main(int argc, char** argv) {
double del = 0.1;
int n;
scanf_s("%d", &n);
double* x;
x = (double*)malloc(n * sizeof(double));
double* y;
y = (double*)malloc(n * sizeof(double));
double* fx;
fx = (double*)malloc(n * sizeof(double));
double* fy;
fy = (double*)malloc(n * sizeof(double));
double* grad;
grad = (double*)malloc(n * sizeof(double));
int i = 0;
for (i = 0; i < n; i++)
{
printf("x[%d] = ", i);
scanf_s("%lf", &x[i]);
y[i] = x[i];
fx[i] = f(x[i]);
}
double a = 1 / del;
for (i = 0; i < n; i++) {
y[i] = y[i] + del;
fy[i] = f(y[i]);
grad[i] = a * (fy[i] - fx[i]);
y[i] = x[i];
}
for (i = 0; i < n; i++) {
printf("%lf ", grad[i]);
}
}
UPD:
double f(double arr[], int size)
{
int a = sizeof(arr);
for (int i = 0; i < size; i++) {
arr[i] = arr[i] * arr[i];
}
return *arr;
}
int main(int argc, char** argv) {
double del = 0.1;
int n;
scanf_s("%d", &n);
double* x;
x = (double*)malloc(n * sizeof(double));
double* y;
y = (double*)malloc(n * sizeof(double));
double* fx;
fx = (double*)malloc(n * sizeof(double));
double* fy;
fy = (double*)malloc(n * sizeof(double));
double* grad;
grad = (double*)malloc(n * sizeof(double));
int i = 0;
for (i = 0; i < n; i++)
{
printf("x[%d] = ", i);
scanf_s("%lf", &x[i]);
y[i] = x[i];
fx[i] = f(x,n);
}
double a = 1 / del;
for (i = 0; i < n; i++) {
y[i] = y[i] + del;
fy[i] = f(y,n);
grad[i] = a * (fy[i] - fx[i]);
y[i] = x[i];
}
for (i = 0; i < n; i++) {
printf("%lf ", grad[i]);
}
}

Answer 1

I would suggest that you do this by hand for a single function before you worry about a vector of functions.
Assume you have one independent variable x and one function y = f(x):
Pick a value for x[i]
Calculate y(x[i]) = f(x[i])
Calculate x[i+1] = x[i] + delta
Calculate y(x[i+1]) = f(x[i+1])
The forward different approximation of the derivative at x[i] is (y(x[i+1]) - y(x[i])/delta.
There are other ways to approximate the derivative, but this is a start.
Once this works for a single function it's easy to see how to extend it to a vector of functions: turn all those subscripts to vectors.
You should understand that
independent variables x are a vector
dependent functions y are a vector
derivatives of y w.r.t. x are expressed as an m x n matrix, not a
vector:
dy[i]/dx[j] for i = 1 to m, j = 1 to n
There are n independent variables x and m dependent functions y.