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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.

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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/… –  Johan Ugander Apr 25 '12 at 22:23
    
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. –  Johan Ugander Apr 26 '12 at 16:24

1 Answer 1

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.

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$Erf(x)=Erf(y)+1=0$ 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? –  Johan Ugander Apr 26 '12 at 1:00
    
I meant $x=0, y=-\infty$ –  Johan Ugander Apr 26 '12 at 1:01
    
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. –  Johan Ugander Apr 26 '12 at 2:18

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