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Linear Algebra : Gradients and Hessians of Quadratic and Linear Functions

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Example Question #11 : Gradients And Hessians Of Quadratic And Linear Functions

$\begin{align*}&\text{Calculate the Hessian matrix of the function}\\&f(x,y)=2cos(2x)cos(5y) + 6\cdot 2^{(6x)}tan(4y)\end{align*}$

Possible Answers:

$\begin{bmatrix}216\cdot 2^{(6x)}tan(4y)ln(2)^{2} - 8cos(2x)cos(5y)&20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)\\20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)&48\cdot 2^{(6x)}tan(4y)\cdot (4tan(4y)^{2} + 4) - 50cos(2x)cos(5y)\end{bmatrix}$

$\begin{bmatrix}20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)&48\cdot 2^{(6x)}tan(4y)\cdot (4tan(4y)^{2} + 4) - 50cos(2x)cos(5y)\\216\cdot 2^{(6x)}tan(4y)ln(2)^{2} - 8cos(2x)cos(5y)&20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)\end{bmatrix}$

$\begin{bmatrix}20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)&216\cdot 2^{(6x)}tan(4y)ln(2)^{2} - 8cos(2x)cos(5y)\\48\cdot 2^{(6x)}tan(4y)\cdot (4tan(4y)^{2} + 4) - 50cos(2x)cos(5y)&20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)\end{bmatrix}$

$\begin{bmatrix}48\cdot 2^{(6x)}tan(4y)\cdot (4tan(4y)^{2} + 4) - 50cos(2x)cos(5y)&20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)\\20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)&216\cdot 2^{(6x)}tan(4y)ln(2)^{2} - 8cos(2x)cos(5y)\end{bmatrix}$

Correct answer:

Explanation:

$\begin{align*}&\text{The Hessian of a function is a matrix of second derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\\...&...&...\\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\\&\text{Considering our function: }f(x,y)=2cos(2x)cos(5y) + 6\cdot 2^{(6x)}tan(4y)\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[cos(u)]=-sin(u)du\\&d[a^u]=a^uduln(a)\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\\&Hf=\begin{bmatrix}216\cdot 2^{(6x)}tan(4y)ln(2)^{2} - 8cos(2x)cos(5y)&20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)\\20sin(2x)sin(5y) + 36\cdot 2^{(6x)}ln(2)\cdot (4tan(4y)^{2} + 4)&48\cdot 2^{(6x)}tan(4y)\cdot (4tan(4y)^{2} + 4) - 50cos(2x)cos(5y)\end{bmatrix}\end{align*}$

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Example Question #61 : Matrix Calculus

$\begin{align*}&\text{Determine the Hessian, }H\text{, of the function}\\&f(x,y,z)=9ln(2x)ln(2z)e^{(y)}\end{align*}$

Possible Answers:

$\begin{bmatrix}-\frac{(9ln(2z)e^{(y)})}{x^{2}}&\frac{(9ln(2z)e^{(y)})}{x}&\frac{(9e^{(y)})}{(xz)}\\\frac{(9ln(2z)e^{(y)})}{x}&9ln(2x)ln(2z)e^{(y)}&\frac{(9ln(2x)e^{(y)})}{z}\\\frac{(9e^{(y)})}{(xz)}&\frac{(9ln(2x)e^{(y)})}{z}&-\frac{(9ln(2x)e^{(y)})}{z^{2}}\end{bmatrix}$

$\begin{bmatrix}9ln(2x)ln(2z)e^{(y)}&\frac{(9ln(2z)e^{(y)})}{x}&\frac{(9ln(2x)e^{(y)})}{z}\\\frac{(9ln(2z)e^{(y)})}{x}&-\frac{(9ln(2z)e^{(y)})}{x^{2}}&\frac{(9e^{(y)})}{(xz)}\\\frac{(9ln(2x)e^{(y)})}{z}&\frac{(9e^{(y)})}{(xz)}&-\frac{(9ln(2x)e^{(y)})}{z^{2}}\end{bmatrix}$

$\begin{bmatrix}\frac{(9e^{(y)})}{(xz)}&\frac{(9ln(2x)e^{(y)})}{z}&-\frac{(9ln(2x)e^{(y)})}{z^{2}}\\\frac{(9ln(2z)e^{(y)})}{x}&9ln(2x)ln(2z)e^{(y)}&\frac{(9ln(2x)e^{(y)})}{z}\\-\frac{(9ln(2z)e^{(y)})}{x^{2}}&\frac{(9ln(2z)e^{(y)})}{x}&\frac{(9e^{(y)})}{(xz)}\end{bmatrix}$

$\begin{bmatrix}\frac{(9e^{(y)})}{(xz)}&\frac{(9ln(2z)e^{(y)})}{x}&-\frac{(9ln(2z)e^{(y)})}{x^{2}}\\\frac{(9ln(2x)e^{(y)})}{z}&9ln(2x)ln(2z)e^{(y)}&\frac{(9ln(2z)e^{(y)})}{x}\\-\frac{(9ln(2x)e^{(y)})}{z^{2}}&\frac{(9ln(2x)e^{(y)})}{z}&\frac{(9e^{(y)})}{(xz)}\end{bmatrix}$

Correct answer:

Explanation:

$\begin{align*}&\text{The Hessian of a function is a matrix of second derivatives taken with}\\&\text{respect to its constituent variables. The form is as follows:}\\&\begin{bmatrix}\frac{d^2f}{d_{x_1}^2}&...&\frac{d^2f}{d_{x_1}d_{x_n}}\\...&...&...\\\frac{d^2f}{d_{x_n}d_{x_1}}&...&\frac{d^2f}{d_{x_n}^2}\end{bmatrix}\\&\text{Considering our function: }f(x,y,z)=9ln(2x)ln(2z)e^{(y)}\\&\text{And utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[e^u]=e^udu\\&Hf=\begin{bmatrix}-\frac{(9ln(2z)e^{(y)})}{x^{2}}&\frac{(9ln(2z)e^{(y)})}{x}&\frac{(9e^{(y)})}{(xz)}\\\frac{(9ln(2z)e^{(y)})}{x}&9ln(2x)ln(2z)e^{(y)}&\frac{(9ln(2x)e^{(y)})}{z}\\\frac{(9e^{(y)})}{(xz)}&\frac{(9ln(2x)e^{(y)})}{z}&-\frac{(9ln(2x)e^{(y)})}{z^{2}}\end{bmatrix}\end{align*}$

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Linear Algebra : Gradients and Hessians of Quadratic and Linear Functions

Study concepts, example questions & explanations for Linear Algebra

All Linear Algebra Resources

Example Questions

Example Question #11 : Gradients And Hessians Of Quadratic And Linear Functions

Example Question #61 : Matrix Calculus

All Linear Algebra Resources

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