Revisiting the Reciprocal

Here we come back to the function $f:x\rightarrow ax-\log x.$

We start with a new majorization of the logarithm. In other contexts the logarithm, which is concave, has been majorized by a linear function. Since our $f$ uses the negative logarithm, which is convex, that will not work in our case. By Taylor $$\log(x) = \log (y) + \frac{1}{y}(x-y)-\frac12\frac{1}{z^2}(x-y)^2,$$ where $z$ is between $x$ and $y$. Thus if we define $$h(x,y)\mathop{=}\limits^{\Delta}\log (y) + \frac{1}{y}(x-y)-\frac12\begin{cases} \frac{(x-y)^2}{x^2}&\text{if }x\leq y,\\ \frac{(x-y)^2}{y^2}&\text{otherwise}. \end{cases}$$ then, for all $x>0$ and $y>0$, $$\log(x)\geq h(x,y),$$ and thus, with $g(x,y)=ax-h(x,y)$, we have $f(x)\leq g(x,y).$ with equality if and only if $x=y$.

Now $g$, as a function of $x$, is differentiable on the positive reals for all $y$. In fact $$\mathcal{D}_1g(x,y)= a-\frac{1}{y}+\begin{cases} \frac{y}{x^3}(x-y)&\text{if }x\leq y,\\ \frac{1}{y^2}(x-y)&\text{otherwise}. \end{cases}$$ Let us find the solutions of $\mathcal{D}_1g(x,y)=0$. First check if $$a-\frac{1}{y}+\frac{1}{y^2}(x-y)=0$$ has a root $x\geq y$. The unique root is $x=2y-ay^2$. Thus if $2y-ay^2\geq y$, i.e. if $y\leq\frac{1}{a}$, this gives a solution to $\mathcal{D}_1g(x,y)=0$. Note that the update in this case is the same update as Newton's update for the reciprocal.

Matters are a bit more complicated for finding a solution of $$a-\frac{1}{y}+\frac{y}{x^3}(x-y)=0.$$ with $x\leq y$. The equation can be written as the cubic equation in $x$ $$(a-\frac{1}{y})x^3+yx-y^2=0.$$ If $y>\frac{1}{a}$ then the cubic has only one real root. Because if there are two, then by Rolle the derivative should vanish somewhere on the interval between them, but the derivative is always positive. Since the cubic is negative for $x=0$ and positive for $x=y$, the unique root is between zero and $y$, and thus satisfies $\mathcal{D}_1g(x,y)=0$.