Numeric derivative with multiple variables

Hello,

given a function f = @(a) [exp(a(1)) - exp(a(2))] with a= [0,0 ]I want to calculate the numeric deriative based on (f(a+h) - f(a-h)) / (2*h);

Calling f(a+h) results in the function getting evaluated with both a values.

But what I want is extracting the result of the first a(1) and a(2) in two different calls.

How can this be achieved?

Thanks in advance

Answer 1

[John D'Errico]

Here is a simple trick to remember. See how I used da1 and da2 below.

f = @(a) sin(a(1)) + cos(a(2)) + a(1) - 2*a(2);

a0 = [1 3];
h = 1e-8;

da1 = [1 0];
da2 = [0 1];

format long g
(f(a0 + da1*h) - f(a0 - da1*h))/(2*h) % numerical partial drivative df/dx

ans = 
          1.54030233012747

(f(a0 + da2*h) - f(a0 - da2*h))/(2*h) % numerical df/dy

ans = 
         -2.14111999241595

Are those partial derivatives correct? We can check them, by an analytical computation.

syms x y
subs(diff(f([x,y]),x),[x,y],[1 3]) % analytical derivative wrt x, at the point (1,3)

ans = 

vpa(ans)

ans = 

1.540302305868139717400936607443

Looks good to me.

subs(diff(f([x,y]),y),[x,y],[1 3]) % analytical derivative wrt y

ans = 

vpa(ans)

ans = 

The numerical finite difference seems to have worked well enough. They are not exact, but h was only 1e-8. That is about what I would expect for accuracy from a central finite difference approximation.

[Clueless]

Thank you John for the answer.

How exactly does da1 & da2 work in this example? If i have a(3) would I achieve this by da3 = [0 0 1] ?