Numerical differentiation for function including vector
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I have implemented a numerical differentiation with central differences. I differentiate the function with respect to a. This appears to be relatively simple for functions containing scalars, e.g., . However, once the function also contains a vector, e.g., , where c is a 3 by 1 vector, I am struggling to implement this.
Below I share my code:
rng default
b = randn(1);
c = randn(3, 1);
f1 = @(a) a^4 * b^3;
%f2 = @(a) (a^4 * b^3 * c.').';
a = 1;
k = 3; % no. derivative
h = 0.1; % step size
cent_diff = zeros(k+1,1);
for i = 0:k
cent_diff(i+1, 1) = sum((-1)^i*nchoosek(k,i)*f1(a+(k/2-i)*h));
end
cent_diff = sum(cent_diff)/h^k;
cent_diff = 3.7304
% check result with symbolic differentiation
syms a
f1 = a^4 * b^3;
diff_3 = diff(f1, a, 3);
a = 1;
vpa(subs(d3)) = 3.7304
The differentiation of f2 can be easily achieved using the symbolic approach. How can I get the corresponding result using finite differences?
댓글 수: 2
Walter Roberson
2021년 6월 7일
cent_diff(i+1, 1) = sum((-1)^i*nchoosek(k,i)*f1(a+(k/2-i)*h));
What is the sum over? k i h are scalar. f1 returns a scalar for scalar input.
If you were looking ahead to the case where f1 might return a vector, then you would not want to sum over that vector.
채택된 답변
Walter Roberson
2021년 6월 7일
The wikipedia article shows .
for i = 0:k
cent_diff(i+1, 1) = sum((-1)^i*nchoosek(k,i)*f1(a+(k/2-i)*h));
end
inside that loop, i is scalar. There is no sum at that level. You are effectively recording the terms there
cent_diff = sum(cent_diff)/h^k;
and that is where you are doing the summation.
Now if you use a function such as f2 that returns a vector, then your stray sum() is adding together the individual terms for all of the vector elements. Which is not what you want.
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Walter Roberson
2021년 6월 7일
rng default
b = randn(1);
Nc = 3;
c = randn(Nc, 1);
f1 = @(a) a^4 * b^3;
f2 = @(a) (a^4 * b^3 * c.').';
a = 1;
k = 3; % no. derivative
h = 0.1; % step size
cent_diff = zeros(Nc, k+1);
for i = 0:k
cent_diff(:, i+1) = (-1).^i .* nchoosek(k,i) .* f2(a+(k/2-i)*h);
end
cent_diff
cent_diff = sum(cent_diff,2)./h^k;
cent_diff
추가 답변 (1개)
Sulaymon Eshkabilov
2021년 6월 7일
These answer discussions provide appropriate points how to apply the finite difference method:
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