1
$\begingroup$

This is a generalization of a previous MO question, "Reducing system of equations involving Erf, Error Function".

Consider the system of equations: $$1/2 + {\rm Erf}(x) - \alpha {\rm Erf}(\frac{x+y}{2})=0$$ $$-1/2 + {\rm Erf}(y) - \alpha {\rm Erf}(\frac{x+y}{2})=0,$$ where $x \le y$ and ${\rm Erf}$ is the Error Function. By ansatz it is clear that one solution is $(x,y)=(-{\rm Erf}^{-1}(1/2),{\rm Erf}^{-1}(1/2))$, independent of $\alpha$. The question is to prove the uniqueness of this solution, for any $\alpha$.

(Update: A visual inspection shows how there are clearly multiple solutions for a robust interval of $\alpha \in (0.5, \approx 0.8)$, and so the general $\alpha$ formulation was not helpful. I am revising my question to specifically focus on uniqueness in the case of $\alpha =1/2$.)

I am particularly interested in a proof for $\alpha = 1/2$, but I hope that by phrasing the problem in terms of a general $\alpha$, someone may see an elegant elementary proof.

Observe that proving that $x+y=0$ is sufficient. My previous MO question proves uniqueness in this way for the case where $\alpha=1$, with two proofs (one by Noam Elkies, one by myself). Unfortunately these proofs do not appear to generalize to other values of $\alpha$ (Noam's proposition is false for $\alpha = 1/2$, my approach does not appear to generalize), even though I strongly suspect the general statement.

Again tagged with probability because of the relation to the Normal distribution CDF.

Numerics: I have embarked on some numerical exploration. If one considers the equation $${\rm Erf}(x) + {\rm Erf}(y) = 2\alpha {\rm Erf}(\frac{x+y}{2}),$$ For $\alpha=1$ we know that $y=-x$ and $y=x$ are solutions. Numerically, it appears as through $y=-x$ is always a solution, while the second solution is $\alpha$-dependent, and not always so nice. For $\alpha=1/2$ the second equation appears to be ${\rm Erf}(c_{1/2} y) - {\rm Erf}(c_{1/2}x) =1$ where $c_{1/2} \approx 0.57285884$. For other values of $\alpha$ the equation appears to be of the form ${\rm Erf}(c_\alpha y - d_\alpha) - {\rm Erf}(c_\alpha x +d_\alpha) =1$. Any insight on what the value of $c_{1/2}$ might be analytically or why $d_{1/2}$ might be zero would be much appreciated.

$\endgroup$
2
  • $\begingroup$ In the interest of full disclosure, I have just cross-posted the question to math.stackexchange (cross-posting also disclosed there): math.stackexchange.com/questions/137000/… $\endgroup$ Apr 25, 2012 at 22:23
  • $\begingroup$ A useful illustration has been added over at the math.SE question, but a proof of uniqueness for $\alpha=1/2$ still escapes me. Using implicit differentiation I have attempted to show that one derivative is strictly greater than the other (a visual suggests this), but the analysis of the resulting implicit expressions quickly becomes unwieldy. $\endgroup$ Apr 26, 2012 at 16:24

1 Answer 1

3
$\begingroup$

It has lots of other solutions, at least for some alphas. Try $x = -.225312055012178104725014013952$, $y = 0.813419847597618541690289359893$, $\alpha = 0.775232509215700110280368495370$.

In general, choose $x,y$ such that $Erf(x)=Erf(y)+1=0$ and set $$\alpha = \frac{1+2 Erf(x)}{2 Erf(x/2+y/2)}.$$ This value of $\alpha$ follows from the first equation, then $Erf(x)=Erf(y)+1=0$ follows by substituting it in the second equation. Alternatively, $Erf(x)=Erf(y)+1=0$ holds by subtracting the equations.

$\endgroup$
3
  • $\begingroup$ $Erf(x)=Erf(y)+1=0$ implies that $x=0$ and $y=\infty$, and I guess I'm looking for bounded solutions. Substracting the equations gives $Erf(x)-Erf(y)+1=0$, is that what you meant? $\endgroup$ Apr 26, 2012 at 1:00
  • $\begingroup$ I meant $x=0, y=-\infty$ $\endgroup$ Apr 26, 2012 at 1:01
  • $\begingroup$ I see now what you meant. Yes, There are multiple solutions when $\alpha \in (0.5, \approx 0.8)$. I have revised the question to focus on $\alpha=1/2$. Thanks for the pointer. $\endgroup$ Apr 26, 2012 at 2:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.