f(k) = n*ln(k)-n*ln(1/n*sum(i=1 to n)xi^k+(k-1)sum of (1to n)ln(xi))-n

3 Kommentare

KSSV
KSSV am 13 Jul. 2023
What have you tried? You are facing any problem or error?
Taniya
Taniya am 13 Jul. 2023
i could not write the code for the sum series where x sum for 1 to n
Dyuman Joshi
Dyuman Joshi am 13 Jul. 2023
The expression you have written above is not clear. Please format it properly.

Melden Sie sich an, um zu kommentieren.

 Akzeptierte Antwort

Rahul
Rahul am 13 Jul. 2023

0 Stimmen

Hi Taniya,
Assuming you have k, n and xi, you can try the following code to find the Hessian Matrix:
f = n*log(k) - n*log(1/n * sum(xi^k, i, 1, n) + (k-1) * sum(log(xi), i, 1, n)) - n;
% Calculate the second partial derivatives
d2f_dk2 = diff(f, k, 2);
d2f_dxi_dk = diff(f, k, xi);
d2f_dk_dxi = diff(f, xi, k);
d2f_dxi2 = diff(f, xi, 2);
% Create the Hessian matrix
H = [d2f_dk2, d2f_dxi_dk; d2f_dk_dxi, d2f_dxi2];
Hope this helps.
Thanks.

Weitere Antworten (1)

Torsten
Torsten am 13 Jul. 2023
Bearbeitet: Torsten am 13 Jul. 2023

0 Stimmen

The vector of independent variables is (x1,x2,...,xn):
syms i k
n = 3;
x = sym('x',[1 n])
x = 
f = n*log(k) - n*log(1/n*sum(x.^k) + (k-1)*sum(log(x))) - n
f = 
df = gradient(f,x)
df = 
d2f = hessian(f,x)
d2f = 

Kategorien

Mehr zu Numerical Integration and Differential Equations finden Sie in Hilfe-Center und File Exchange

Gefragt:

am 13 Jul. 2023

Bearbeitet:

am 13 Jul. 2023

Community Treasure Hunt

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

Start Hunting!

Translated by