Math people:

I am looking for an upper bound on

$$g(x,t) = \frac{-\operatorname{erf}(t)+2\operatorname{erf}(\frac{1}{2}t(x+1))-\operatorname{erf}(xt)}{(x-1)^2},$$

where $x > 1$ and $t > 0$. Here, $\operatorname{erf}$ is the "error function" $\operatorname{erf}(x) = \frac{2}{\sqrt{\pi}}\int_0^x e^{-y^2}\,dy$. I want an actual bound, not just a proof that one exists. It doesn't have to be the least upper bound (though I am curious what it is), but I don't want something ridiculously large. I can do this myself, but I bet that it has been done before and I don't want to take credit for it or typeset a proof when I can quote someone else's finding.

Here is how I would do it: Obviously $g(x,t) \leq 2$ if $x \geq 2$. Maple gives $\lim_{x\to 1^+} g(x,t) = \frac{1}{\sqrt{\pi}}t^3 e^{-t^2}$. This can probably be easily done using L'Hopital's Rule twice, though I haven't actually done it. This quantity has a maximum value for $t > 0$, let's call it $K$. Extend $g$ to $x = 1$ using the limit. Fix a value of $t=t^*$, and calculate $\frac{\partial g}{\partial x}$ and $\frac{\partial^2 g}{\partial x^2}$ along the segment $\{[x,t^*] \mid 1 \leq x \leq 2 \}$. $\frac{\partial g}{\partial x}(1,t^*)=0$ and $\frac{\partial^2 g}{\partial x^2}$ is bounded along the segment, and the bound is independent of $t$. So you can use Taylor's Theorem with Remainder to find a bound for $g$ along the segment that is independent of $t$. This gives you a bound for the case $1 < x \leq 2$, and now you are done.

But I already said why I would like to see an existing proof.

Stefan (STack Exchange FAN)


You can show that $\frac{\partial g}{\partial x}(x,t) \leq 0$ for $t > 0$ and $x \geq 1$. Then just maximize $g(1,t)$ as you have done.

  • 1
    $\begingroup$ @StefanSmith, yes, I am sure. $\endgroup$ – Antonio Vargas Mar 19 '13 at 2:22
  • $\begingroup$ @StefanSmith try taking the derivative with respect to $x$. If you're going down the same path I did then it should look like the negative of something you already showed was positive. So, it's decreasing in $x$, and is then $\leq 0$. $\endgroup$ – Antonio Vargas Mar 19 '13 at 2:39
  • $\begingroup$ Got it. Thanks. This gives me the least upper bound too. $\endgroup$ – Stefan Smith Mar 19 '13 at 2:47
  • $\begingroup$ Glad to help, @Stefan! $\endgroup$ – Antonio Vargas Mar 19 '13 at 3:00
  • $\begingroup$ @Stefan, $g$ is decreasing in $x$ (regardless of what Maple tells you). The numerics are testy around $x=1$ because of the singularity in the denominator. Try the calculation again with more precision. $\endgroup$ – Antonio Vargas Mar 21 '13 at 2:54

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