Question

What is the Hessian for the function `f(x_1, fx_2) = (x_1 + x_2)^N`?

Answer 1

`| bb(H) f(x_1,x_2) | = 0`

Explanation:

The function is dependant upon two variables, assuming that `N` is a constant, so we should write:

`f(x_1, fx_2) = (x_1 + x_2)^N`

Then the first partial derivatives are:

`(partial f)/(partial x_1) = N(x_1+x_2)^(N-1)`
`(partial f)/(partial x_2) = N(x_1+x_2)^(N-1)`

Similarly, the second partial derivatives are:

`(partial^2 f)/(partial x_1^2) = N(N-1)(x_1+x_2)^(N-2)`
`(partial^2 f)/(partial x_2^2) = N(N-1)(x_1+x_2)^(N-2)`

(Note: In both cases the partial derivatives are the same due to the symmetry of the independent variables)

And the second partial cross derivatives are:

`(partial^2 f)/(partial x_1partial x_2) = N(N-1)(x_1+x_2)^(N-2)`
`(partial^2 f)/(partial x_2partial x_1) = N(N-1)(x_1+x_2)^(N-2)`

(Note: that the cross derivatives are identical due the the continuity of `f`)

Given these deviations, we then construct the Hessian Matrix.

`bb(H) f(x_1,x_2) = [ ((partial^2 f) / (partial x_1^2),(partial^2 f) / (partial x_1 partial x_2)), ((partial^2 f) / (partial x_2 partial x_2), (partial^2 f) / (partial x_2^2)) ]`

So the Hessian is:

`| bb(H) f(x_1,x_2) | = | ( N(N-1)(x_1+x_2)^(N-2), \ \ N(N-1)(x_1+x_2)^(N-2)) , (N(N-1)(x_1+x_2)^(N-2), \ \ N(N-1)(x_1+x_2)^(N-2)) |`

`\ \ \ = {N(N-1)(x_1+x_2)^(N-2)}^2 - {N(N-1)(x_1+x_2)^(N-2)}^2`

`\ \ \ = 0`