Laplacian of the value function

Jeffrey Eiyike
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Homework Statement



Laplacian of the function V(x,t)=-1/2* x' D x + h' *x + D

Homework Equations

The Attempt at a Solution


is equals D.
 
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Your answer is wrong, please explain your reasoning. Note that D is a matrix and the Laplacian of a scalar function should be a scalar.

Edit: Note that I am assuming your x is a vector and x' its transpose. You really have not made this point clear in your post.
 
Laplacian of the function V(x,t)=-1/2* x' D x + h' *x - Z

x is a vector D is a matrix which depends on time h is a vector which depends on time Z is also a vector depends on time
 
D is a square matrix..
 
Jeffrey Eiyike said:

Homework Statement



Laplacian of the function V(x,t)=-1/2* x' D x + h' *x + D

Homework Equations

The Attempt at a Solution


is equals D.

PF rules require you to show your work. What is preventing you from just going ahead and actually computing the Laplacian?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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