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Here's a cute little problem involving the Lambert-$W$ function. For the reader's convenience, I've included some of my own working.

Let $t$ and $x$ be non-negative reals, and let $\alpha \in (0,1/2]$. Find the smallest $y$ greater than $x$ that satisfies $$ \frac{y + t^\alpha}{t}\exp\{ - (y + t^\alpha)^2/(2 t) \} \leq \frac{x }{t}\exp\{ - (x + t^\alpha)^2/(2 t) \}, $$ In other words, find $y$ defined by $$ y = \inf \left\{ z > x : \frac{z + t^\alpha}{t}\exp\{ - (z + t^\alpha)^2/(2 t) \} \leq \frac{x }{t}\exp\{ - (x + t^\alpha)^2/(2 t) \} \right\}. $$ By continuity of the LHS and RHS in $y$ and $x$ respectively, we simply need to solve the equation $$ \frac{y + t^\alpha}{t}\exp\{ - (y + t^\alpha)^2/(2 t) \} = \frac{x }{t}\exp\{ - (x + t^\alpha)^2/(2 t) \}. $$ Squaring both sides, then multiplying with $-t$ gives $$ -\frac{(y + t^\alpha)^2}{t}\exp\{ - (y + t^\alpha)^2/ t \} = - \frac{x^2 }{t}\exp\{ - (x + t^\alpha)^2/ t \}. $$ Now the LHS expression can simply be recognised as the inverse of the Lambert-$W$ function $\mathcal{W}$, i.e. $$ \mathcal{W}(LHS) = -\frac{(y + t^\alpha)^2}{t}. $$ The RHS doesn't simplify so easily, but with some restrictions on the range of $x$ and $t$, we can use Taylor expansion of $\mathcal{W}$. \begin{align} LHS &= - \frac{x^2 }{t}\exp\{ - (x + t^\alpha)^2/ t \} \\ &= - \frac{(x + t^\alpha)^2 - 2 x t^\alpha - t^{2 \alpha}}{t}\exp\{ - (x + t^\alpha)^2/ t \} \\ &= - \frac{(x + t^\alpha)^2 - 2 x t^\alpha - t^{2 \alpha}}{t}\exp\{ - (x + t^\alpha)^2/ t \} + \frac{ 2 x t^\alpha + t^{2 \alpha}}{t}\exp\{ - (x + t^\alpha)^2/ t \} \\ &=: A + B \quad \text{(labelling the two terms)}. \end{align}

Now as long as the LHS stays away from a ball around $-1/e$, we can use Taylor expansion of $\mathcal{W}$ at the term $A = - \frac{(x + t^\alpha)^2 }{t}\exp\{ - (x + t^\alpha)^2/ t \}$ and say \begin{align} \mathcal{W}(LHS) &= \mathcal{W}(A + B) \\ &= \mathcal{W}(A) + \mathcal{W}'(A)(B - A) \\ &= \mathcal{W}\left( - \frac{(x + t^\alpha)^2 }{t}\exp\{ - (x + t^\alpha)^2/ t \} \right) + O\left( - \frac{(x + t^\alpha)^2 }{t}\exp\{ - (x + t^\alpha)^2/ t \} \right) \\ &= .- \frac{(x + t^\alpha)^2 }{t} + O\left( - \frac{(x + t^\alpha)^2 }{t}\exp\{ - (x + t^\alpha)^2/ t \} \right) \end{align} Thus $$ y = \sqrt{ (x + t^\alpha)^2 + O\left( (x + t^\alpha)^2 \exp\{ - (x + t^\alpha)^2/ t \} \right) - t^\alpha. } $$

For $z < 0$, $\mathcal{W}$ is a two-valued function. Should there be two roots to the equation?

Next, how does $y$ behave when $t \to 0$ and $x \to 0$?

Any problems with sending $x \to 0$ independently of $t$? What if $x \asymp \sqrt{t}$ and $t \to 0$ (i.e., we end up at the one point where $\mathcal{W}$ is not differentiable, and Taylor doesn't work)?

I currently don't have access to any computational software (Matlab, Mathematica etc) or I'd just plot the damn thing. Would anyone be kind enough to plot examples of LHS and RHS to just to see where they intersect? (Fix a $t$ and $\alpha$, and plot the two functions of $x$ and $y$ to see where they intersect).

Many thanks in advance.

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    $\begingroup$ Please no, I'm allergic to cuteness! -asdf $\endgroup$ – Simply Beautiful Art Dec 17 '16 at 15:59
  • $\begingroup$ Hah. There are some not so cute bits near the end as antidote - if you look hard ;-) $\endgroup$ – zab Dec 17 '16 at 16:06
  • $\begingroup$ desmos.com has online graphing. It accepts Latex input so you should be able to cut'n'paste pretty easily from MSE. Here's a link to your problem: desmos.com/calculator/qlu5kg9ks4 . I had to change $\alpha$ to $a$. $\endgroup$ – Χpẘ May 4 '17 at 21:52

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