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When I was a senior in high school in 2001, as I took calculus, I made the following conjecture that proves resistive to attack. It goes like this: For every positive integer $n$, there are exactly $n$ positive real zeroes of $$\frac {d^n}{dx^n}x^{1/x}$$ and no two derivatives has a common root.

A few things to keep in mind are that it is not hard to show that $$\lim_{x\rightarrow 0^+}\frac {d^n}{dx^n}x^{1/x}=\lim_{x\rightarrow \infty}\frac {d^n}{dx^n}x^{1/x}=0$$ for all $n\geq 1$ and that $x^{1/x}$ is smooth and then to use these to show that the $n$th derivative has at least $n$ positive real zeroes. The proof of smoothness uses induction on $n$.

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    I assume you want the limits to be as $x\to 0^+$ and $x\to\infty$, correct?2012-08-30
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    Or pointwise convergence?2012-08-30
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    I wonder if it's easier to work with the logarithm, $(\log x)/x$.2012-08-30
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    Ok, what I said didn't make sense since $n \to 0^+$ doesn't make sense.2012-08-30
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    If you want to, you can try the proof of the weaker result, which will obviously include proving the above limits, and I will tell you if you are right. For the main conjecture, though, I wouldn't be surprised if you get stumped, as it stumped the teachers who know me and also the professors at UMass Boston where I did my college work. Also, I am changing the typo2012-08-30
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    The $n^{th}$ derivative will be an exponential mutiplied by a two variable polynomial in $x$ and $\log x$. The highest power of $\log x$ will be $n$, so we can look at this as a polynomial in $\log x$ of degree $n$, but there are these extra other $x$'s hanging around. If we could ignore these other $x$'s since they don't move two much around zero, which might be possible using some variable changes and complex analysis stuff, we could show that this polynomial has exactly $n$ zeros, and none outside of some disk. However, to show everything is distinct seems extremely difficult to me.2012-08-30

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