= Solution
Minimize the square of the <Euclidean distance>, using <minimizers under strictly increasing transformations>. For $s=x+y>0$,
$$
x^2+y^2+\ln^2(x+y)=\frac{s^2}{2}+\frac{(x-y)^2}{2}+\ln^2s.
$$
For fixed $s$ the unique minimum has $x=y=s/2$. Put $x=y=\alpha>0$ and $J(\alpha)=2\alpha^2+\ln^2(2\alpha)$. Its <derivative> is
$$
J'(\alpha)=\frac{2}{\alpha}\left(2\alpha^2+\ln(2\alpha)\right).
$$
Let $h(\alpha)=2\alpha^2+\ln(2\alpha)$. Since $h'(\alpha)=4\alpha+1/\alpha>0$, it is <strictly increasing>, with limits $-\infty$ and $+\infty$ at the two ends of $(0,\infty)$. It therefore has exactly one zero. The sign of $J'$ changes there from negative to positive, proving the unique <global minimum>. The nearest point is \b[$P=(\alpha,\alpha,\ln(2\alpha))$] at that zero.<Discrete mathematics, continuous mathematics, and probability examination paper A10100W1 (2022)>[Question 4(d), p. 4: the logarithm is interpreted as ln(2α)]{ref}
The <Newton root-finding iteration> is
$$
\alpha_{k+1}=\alpha_k-\frac{2\alpha_k^2+\ln(2\alpha_k)}{4\alpha_k+1/\alpha_k}.
$$
Starting at $0.4$ gives $0.376376476$, $0.376544572$, and $0.376544582$. Moreover $h(0.3765)<0<h(0.3766)$ certifies the rounding. Thus \b[$\alpha=0.377$ to three significant figures]; more accurately $P\approx(0.3765445825,0.3765445825,-0.2835716452)$.
Back to article page
