hessian: 2x2 Hessian Matrix (Single-Choice)
hessianCompute the Hessian of the function \[ \begin{aligned} f(x_1, x_2) = 7 x_1^{2} + 5 x_1 x_2 + 3 x_2^{2} \end{aligned} \] at \((x_1, x_2) = (1, 4)\). What is the value of the upper left element?
The first-order partial derivatives are \[ \begin{aligned} f'_1(x_1, x_2) &= 14 x_1 + 5 x_2 \\ f'_2(x_1, x_2) &= 5 x_1 + 6 x_2 \end{aligned} \] and the second-order partial derivatives are \[ \begin{aligned} f''_{11}(x_1, x_2) &= 14\\ f''_{12}(x_1, x_2) &= 5\\ f''_{21}(x_1, x_2) &= 5\\ f''_{22}(x_1, x_2) &= 6 \end{aligned} \]
Therefore the Hessian is \[ \begin{aligned} f''(x_1, x_2) = \left( \begin{array}{rr} 14 & 5 \\ 5 & 6 \end{array} \right) \end{aligned} \] independent of \(x_1\) and \(x_2\). Thus, the upper left element is: \(f''_{11}(1, 4) = 14\).
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Compute the Hessian of the function \[ \begin{aligned} f(x_1, x_2) = -9 x_1^{2} + 4 x_1 x_2 + 7 x_2^{2} \end{aligned} \] at \((x_1, x_2) = (-5, 1)\). What is the value of the upper left element?
The first-order partial derivatives are \[ \begin{aligned} f'_1(x_1, x_2) &= -18 x_1 + 4 x_2 \\ f'_2(x_1, x_2) &= 4 x_1 + 14 x_2 \end{aligned} \] and the second-order partial derivatives are \[ \begin{aligned} f''_{11}(x_1, x_2) &= -18\\ f''_{12}(x_1, x_2) &= 4\\ f''_{21}(x_1, x_2) &= 4\\ f''_{22}(x_1, x_2) &= 14 \end{aligned} \]
Therefore the Hessian is \[ \begin{aligned} f''(x_1, x_2) = \left( \begin{array}{rr} -18 & 4 \\ 4 & 14 \end{array} \right) \end{aligned} \] independent of \(x_1\) and \(x_2\). Thus, the upper left element is: \(f''_{11}(-5, 1) = -18\).
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Compute the Hessian of the function \[ \begin{aligned} f(x_1, x_2) = -6 x_1^{2} -5 x_1 x_2 -2 x_2^{2} \end{aligned} \] at \((x_1, x_2) = (-2, 3)\). What is the value of the upper left element?
The first-order partial derivatives are \[ \begin{aligned} f'_1(x_1, x_2) &= -12 x_1 -5 x_2 \\ f'_2(x_1, x_2) &= -5 x_1 -4 x_2 \end{aligned} \] and the second-order partial derivatives are \[ \begin{aligned} f''_{11}(x_1, x_2) &= -12\\ f''_{12}(x_1, x_2) &= -5\\ f''_{21}(x_1, x_2) &= -5\\ f''_{22}(x_1, x_2) &= -4 \end{aligned} \]
Therefore the Hessian is \[ \begin{aligned} f''(x_1, x_2) = \left( \begin{array}{rr} -12 & -5 \\ -5 & -4 \end{array} \right) \end{aligned} \] independent of \(x_1\) and \(x_2\). Thus, the upper left element is: \(f''_{11}(-2, 3) = -12\).
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num_to_schoice() by providing the other three elements of the Hessian as typical errors. Further random answers are taken from a prespecified range, assuring only integer solutions. Although the Hessian of a quadratic function has only constant elements, an argument vector is given in the exercise because other types of functions might have non-constant Hessians.
(Note that the HTML output contains mathematical equations in MathML, rendered by MathJax using ‘mathjax = TRUE’. Instead it is also possible to use ‘converter = “pandoc-mathjax”’ so that LaTeX equations are rendered by MathJax directly.)
Demo code:
library("exams")
set.seed(403)
exams2html("hessian.Rmd", mathjax = TRUE)
set.seed(403)
exams2pdf("hessian.Rmd")
set.seed(403)
exams2html("hessian.Rnw", mathjax = TRUE)
set.seed(403)
exams2pdf("hessian.Rnw")