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I’d like to solve following equation for $x.$

If it is not possible, why I can’t?

$$\sum_{i=0}^n (x-\mu_i)e^{-(x-\mu_i)^2} = 0$$

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  • $\begingroup$ Solve in what sense? $\endgroup$ Dec 14, 2018 at 8:35
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    $\begingroup$ there is no hope for an exact solution in closed form; also note that in general there are multiple solutions. $\endgroup$ Dec 14, 2018 at 8:49

4 Answers 4

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There always exists at least one solution: If $x >\mu_i$ for all $i$, then each of the terms in the sum is positive and if $x < \mu_i$ for all $i$, then each of the terms in the sum is negative. Thus, the function on the left hand side has at least one zero and all the zeroes lie strictly between the lowest of the $\mu_i$ and the highest.

Finding an explicit analytic solution is almost certainly not possible, but numerically, it will be easy. For example, starting with the smallest interval that contains the $\mu_i$, use the bisection method, which will converge exponentially fast to at least one root.

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It is simple to reduce to the case where $\sum_i\mu_i=0$. Then the simplest nontrivial case is $(x-\mu)e^{-(x-\mu)^2}+(x+\mu)e^{-(x+\mu)^2}=0$, which is equivalent to $(x+\mu)/(x-\mu)=e^{4x\mu}$. Here there is a solution of the form $$ x = \frac{1}{\sqrt{2}} \left(1 + \frac{1}{3}\mu^2 + \frac{11}{90}\mu^4 + \frac{17}{630}\mu^6 - \frac{281}{37800}\mu^8 - \frac{44029}{3742200}\mu^{10} + O(\mu^{12}) \right)$$ I have calculated further terms as far as $\mu^{240}$, and there is some rather interesting behaviour, with some features of the coefficient of $\mu^i$ apparently depending on $i$ mod $14$; I don't know how to explain that. The radius of convergence appears to be finite, perhaps close to $1.1$, but the evidence for that is not hugely compelling. Note that $281$ and $44029$ are prime; this means that various classes of simple formulae cannot produce the above list of coefficients. (The coefficient of $\mu^{12}$ is $-12147139/2043241200$, which also has prime numerator, but this pattern breaks down for $\mu^{14}$.)

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    $\begingroup$ Actually, because you missed a minus sign in your re-write of the equation (it should be $(x+\mu)/(x-\mu) = - e^{4x\mu}$), you have the roles of $x$ and $\mu$ reversed in your series solution. In fact, $x=0$ is always a root, and it's the only root unless $|\mu|>1/\sqrt{2}$, in which case, there are two additional roots, a positive one and its negative. $\endgroup$ Dec 15, 2018 at 6:31
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    $\begingroup$ In fact, allowing inversion of functions of one variable, one can solve the given equation explicitly: Multiply the original equation by $x$ and set $y=x^2$ and $z = \mu x$. Then $y= z\tanh (2z) = f(z^2)$ where $f'(w)>0$ for $w\ge 0$, so $f:[0,\infty)\to[0,\infty)$ has a smooth inverse $g:[0,\infty)\to[0,\infty)$. So $z^2 = g(y)$, or $\mu^2 = g(x^2)/x^2 = h(x^2)$ where $h:[0,\infty)\to [1/2,\infty)$ has a smooth inverse $k:[1/2,\infty)\to[0,\infty)$. Thus, $x^2 = k(\mu^2)$, or $$x = \pm\sqrt{k(\mu^2)}\,,$$ when $|\mu|\ge1/\sqrt2$ are the nonzero solutions (in addition to $x=0$, of course). $\endgroup$ Dec 15, 2018 at 14:51
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Here is why solutions do exist. $\newcommand{\bR}{\mathbb{R}}$ Consider the function $f:\bR\to\bR$ $$ f(x)=\sum_{i=1}^n e^{-(x-\mu_i)^2}. $$ Note that $$ 0\leq f(x)\leq n, \;\;\forall x\in\bR $$ and $$ \lim_{\vert x\vert\to\infty} f(x)= 0. \tag{$1$} $$ Set $$ M:=\sup_{x\in \bR} f(x). $$ If $(x_\nu)$ is a sequence of real numbers such that $$ \lim_{\nu\to\infty }f(x_\nu)=M, $$ then, due to (1), the sequence is bounded and thus admits a convergent subsequence. The limit $x^*$ of this subsequence satisfies $f(x^*) =M$ so $$ f'(x^*)=0. $$ The last equation is equivalent to yours $$ \sum_{i=1}^n (x^*-\mu_i) e^{-(x^*-\mu_i)^2}=0. $$

Using more sophisticated mathematics (namely $o$-minimality) one can say a bit more. $\newcommand{\eZ}{\mathscr{Z}}$

Consider the set $$\eZ:=\Big\{\; (x,\mu_1,\dotsc, \mu_n)\in\bR^{n+1}:\;\;\sum_{i=1}^n (x-\mu_i) e^{-(x-\mu_i)^2}=0\;\Big\}. $$ We have a natural projection $$ \pi:\eZ\to\bR^n,\;\;\eZ\ni (x,\mu_1,\dotsc,\mu_n)\mapsto (\mu_1,\dotsc,\mu_n). $$ For $\vec{\mu}\in\bR^n$ we set $$ \eZ(\vec{\mu}):=\pi^{-1}(\vec{\mu}). $$ The above discussion shows that $\eZ(\vec{\mu})$ is a nonempty discrete subset of $\bR$. A famous result of Alex Wilkie

Model completeness results for expansions of the ordered field of real numbers by restricted Pfaffian functions and the exponential function, J Amer Math Soc, 9, (4), 1996, 1051-1094.

implies that there exists a natural number $N$ such that

$$ 0<\# \eZ(\vec{\mu})\leq N,\;\;\forall \vec{\mu}\in\bR^n. $$ A more precise statement is true. There exist finitely many pairwise disjoint open subsets $\Omega_1,\dotsc, \Omega_\nu$ of the $\mu$ space with the following properties

  1. The union of the closures of the $\Omega_i$-s is the entire $\mu$ space.
  2. The boundaries of $\Omega_i$-s are piecewise $C^2$-hypersurfaces. (I won't elaborate on the precise meaning of "piecewise $C^2$".)
  3. The function $\mu\mapsto \#\eZ(\mu)$ is constant on each $\Omega_i$.

As others have indicated, your equation can have multiple solutions. E.g., for $n=5$ and $\mu_i=2i$ the graph of

$$ g(x)=\sum_{i=1}^n (x-\mu_i) e^{-(x-\mu_i)^2} $$ is depicted below (courtesy of MAPLE).

graph of g(x) for x in [-15,15]

As you can see, the graph of $g(x)$ intersects the $x$-axis in $8$ points.

Consider a point $(x_0,\mu)\in\eZ$ which is a regular point of the projection $\pi:\eZ\to\bR^n$. In other words $$ \partial_x G(x_0,\vec{\mu})\neq 0,\;\;G(x,\vec{\mu})=\sum_{i=1}^n (x-\mu_i) e^{-(x-\mu_i)^2}. $$ The implicit function theorem shows that near $(x_0,\vec{\mu})$ the solution $(x,\vec{\mu})$ of $G(x,\vec{\mu})$ depends smoothly on $\vec{\mu}$. Moreover $\newcommand{\pa}{\partial}$ $$ \frac{\pa G}{\pa x}\frac{\pa x}{\pa \mu_i}+\frac{\pa G}{\pa\mu_i}=0 $$ so $$ \frac{\pa x}{\pa\mu_i}=-\frac{G'_{\mu_i}}{G'_x} $$ We have $$G'_{\mu_i}= e^{-(x_0-\mu_i)^2}\Big(\; 2(x_0-\mu_i)^2-1\;\Big) $$ $$ G'_x= \sum_{j=1}^n e^{-(x_0-\mu_i)^2}\Big(1-2(x_0-\mu_j)^2\Big). $$ We deduce that $$ \sum_i \frac{\pa x}{\pa\mu_i}=1. $$ In particular $$ \frac{d}{dt}x(\mu_1+t,\dotsc, \mu_n+t)=1 $$ so $$ x(\mu_1+t, \dotsc, \mu_n+t)=x(\mu_1,\dotsc, \mu_n)+t. $$ Thus, as observed by Neil Strickland in his answer, it suffices to determine $x(\mu_1,\dotsc,\mu_n)$ only when $\sum_i\mu_i=0$. `

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  • $\begingroup$ Thanks for the nice explanation. I understand this equation has multiple solutions, and I can use a numerical method to get an answer. But, is it possible to get a closed form solution? (mu_i are constants here) $\endgroup$
    – Bugloss92
    Dec 18, 2018 at 2:22
  • $\begingroup$ The result of Wilkie says a bit more, namely that graph of the multivalued function $x(\vec{mu})$ belongs to a rather special class of sets that can be detected algorithmically in a finite number of steps. This is rather vague but the precie statement says that it belongs to an $o$-minimal collection of sets. $\endgroup$ Dec 18, 2018 at 10:57
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Since the question is how to solve the equation, you can look at the bisection method, which is a recursive application of the idea in @Rober answer.

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