coef <- sample(c(2:9, -(2:9)), 3, replace = TRUE) x <- sample(c(-5:5), 2, replace = TRUE) H <- matrix(c(2 * coef[1], coef[2], coef[2], 2 * coef[3]), nrow = 2, ncol = 2) ix <- sample(1:4, 1, prob=c(0.35, 0.15, 0.15, 0.35)) ixt <- c("upper left", "upper right", "lower left", "lower right")[ix] ixn <- c("11", "12", "21", "22")[ix] sol <- H[ix] err <- unique(H[-ix]) err <- err[err != sol] sc <- num_to_schoice(sol, wrong = err, range = -25:25, method = "delta", delta = 1, digits = 0) plus <- ifelse(coef < 0, "", "+")
Compute the Hessian of the function
$$
\begin{aligned}
f(x_1, x_2) = r coef[1]
x_1^{2} r plus[2]
r coef[2]
x_1 x_2 r plus[3]
r coef[3]
x_2^{2}
\end{aligned}
$$
at $(x_1, x_2) = (r x[1]
, r x[2]
)$.
What is the value of the r ixt
element?
answerlist(sc$questions, markup = "markdown")
The first-order partial derivatives are
$$
\begin{aligned}
f'1(x_1, x_2) &= r H[1,1]
x_1 r plus[2]
r H[1,2]
x_2 \
f'_2(x_1, x_2) &= r H[2,1]
x_1 r plus[3]
r H[2,2]
x_2
\end{aligned}
$$
and the second-order partial derivatives are
$$
\begin{aligned}
f''{11}(x_1, x_2) &= r H[1,1]
\
f''{12}(x_1, x_2) &= r H[1,2]
\
f''{21}(x_1, x_2) &= r H[2,1]
\
f''_{22}(x_1, x_2) &= r H[2,2]
\end{aligned}
$$
Therefore the Hessian is
$$
\begin{aligned}
f''(x_1, x_2) = r toLatex(H, escape = FALSE)
\end{aligned}
$$
independent of $x_1$ and $x_2$. Thus, the r ixt
element is:
$f''_{r ixn
}(r x[1]
, r x[2]
) = r sol
$.
answerlist(ifelse(sc$solutions, "True", "False"), markup = "markdown")
extype: schoice
exsolution: r mchoice2string(sc$solutions)
exname: Hessian
Add the following code to your website.
For more information on customizing the embed code, read Embedding Snippets.