Find the number of roots to $e^x = ax^2$ for all values of $a$. (x is real and so is $a$). I have tried some things but I am stuck.
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Why don't you tell us what you have tried? – Henrik supports the community Mar 22 '15 at 23:09
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@Henrik just basic things like applying log on each side but I dont get anywhere. – jdaUU Mar 22 '15 at 23:12
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These sort of equations are in general hard to solve. Often there isn't a closed form of the solution. A possible numerical strategy is to assume that one solution is close to 1. So you can expand $e^{x}$ in a Taylor series around 0. $e^{x} \approx 1+x+0.5x^2$. – abcdef Mar 22 '15 at 23:12
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One thing that should be clear is that for $a \leq 0$, there are no roots. So you can at least focus on the case of $a>0$. – graydad Mar 22 '15 at 23:15
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Easy enough with calculus, since they just want a count of the solutions $x.$ Without calculus, no. – Will Jagy Mar 22 '15 at 23:15
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See the technique. – science Mar 23 '15 at 01:28
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For $a>0$ the roots can be expressed in terms of Lambert $W$ function:
\begin{align} a x^2&=\exp(x) \\ x^2\exp(-x)&=\frac{1}{a} \\ x\exp(-\frac{x}{2})&=\pm\frac{1}{\sqrt{a}} \\ -\frac{x}{2}\exp(-\frac{x}{2})&=\mp\frac{1}{2\sqrt{a}} \\ -\frac{x}{2}&=\mathrm{W}\left( \mp\frac{1}{2\sqrt{a}} \right)\\ x&=-2\mathrm{W}\left( \mp\frac{1}{2\sqrt{a}} \right) \end{align}
So, $\forall a>0$ there is always one negative real root $x_1=-2\mathrm{W_0}\left(\frac{1}{2\sqrt{a}} \right)$. And since the Lambert $W(u)$ function has two real branches for $-\exp(-1)<u<0$, it would be two more (positive) real roots (three total):
\begin{align} x_2&=-2\mathrm{W_0}\left(-\frac{1}{2\sqrt{a}} \right), \\ x_3&=-2\mathrm{W_{-1}}\left(-\frac{1}{2\sqrt{a}} \right) \end{align} for $a>\exp(2)/4$.
Also, when $a=\exp(2)/4$ then $x_2=x_3$ and there are two real roots total.
Summarizing, the number $n$ of real roots to $\exp(x) = a x^2$: \begin{align} n&= \begin{cases} 0,\quad a\le0 \\ 1,\quad 0<a<\exp(2)/4 \\ 2,\quad a=\exp(2)/4 \\ 3,\quad a>\exp(2)/4 \\ \end{cases}. \end{align}
g.kov
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+1, but you left out the answer for $a\le 0$ and $0<a\le exp(2)/4$. I know these are obvious, but they should be stated. – Rory Daulton Mar 23 '15 at 00:18
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I am guessing you are looking for the roots over the whole real line. Note that $f(x)=e^x-ax^2$ is a continuous function on its domain but more importantly that the domain is a connected set. Does it sound like Intermediate value theorem?
Predrag Punosevac
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