Nth Root of N

During my son’s math lesson today we got on the subject of \sqrt[n]n, which I prefer to write as n^{1/n}. We made a table of a few obvious values and limits:

    \[ \begin{tabular}{|r|c|} \hline $n$ & $n^{1/n}$\\ \hline $0\leftarrow$ & $0\leftarrow$ \\ 1 & 1 \\ 2 & $\sqrt{2} \approx 1.414$ \\ 3 & $\sqrt[3]{3} \approx 1.442$ \\ 4 & $\sqrt{2} \approx 1.414$ \\ $\rightarrow \infty$ & $\rightarrow 1$ \\ \hline \end{tabular}\]

So the values rise from n=0 \cdots 2, peak somewhere in n=2 \cdots 4, and asymptotically drop to 1 after that. So the maximum is probably between 2 and 4, and I asked my son how we could find it. And he suggested we find where the curve is flat and the derivative is zero.

And that’s when I realized I’d forgotten know how to find \frac{d}{dx}(x^{1/x}). My son is only 11 and we’ve only touched on calculus, so he couldn’t help. I figure the first step is to express it as \frac{d}{dx}(e^\frac{ln(x)}{x}), but I don’t know where to go from there.

So we turn to the internet, and I read Wikipedia’s entry for the nth root algorithm, which doesn’t answer my question but is nice to read because it’s a little gem. And then we get to WolframAlpha, which tells us everything we want to know. See for yourself, the WolframAlpha page for \frac{d}{dx}(x^{1/x}), which says the derivative is x^\frac{1}{x}(\frac{1}{x^2} - \frac{log(x)}{x^2}), which is zero when \frac{1}{x^2}=\frac{log(x)}{x^2} or 1=log(x) or x=e.

So the maximum of \sqrt[n]n is \sqrt[e]e\approx 1.4447. Not very surprising, but nice to have it confirmed. (When I described this problem to my neighbor he said the max would be at e as soon as he heard it was between 2 and 4.)

But I still don’t know how to calculate that derivative, except by asking WolframAlpha. Which brings up the question, is the internet making me stupid by making things easy? Or is it making me smarter by taking care of drudgery? I’d like to say it makes me smarter, but I know that, once in a while, what appears to be drudgery at first turns out to be important, providing unexpected insight.


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LaTeX Plugin

I’ve just installed Pavel Holoborodko’s amazing QuickLaTeX WordPress plugin, and so now I can show fancy equations like this: |y - x| < \delta

And this: w_i = \frac{2}{\left(1-\xi_i^2\right)\,\left[P'_n(\xi_i)\right]^2}

And even super complicated stuff like this:

( \langle \sqrt{ \sqrt[3]{ \langle \frac{ e ^ { |f(y) - f(x)| <  \pm \epsilon } } { |x - (y + z)| } \rangle  } ^ 3 } \rangle ) ^Q_m

LaTeX is pretty expressive, and these examples are kinda addictive.

|x| = \left\{ \begin{array}{ll}   x & \mbox{if $x \geq 0$};\\   -x & \mbox{if $x < 0$}.\end{array} \right.

Matrices look pretty good too. Let’s say I wanted to explain that the \emph{characteristic polynomial} \chi(\lambda) of the 3 \times 3 matrix

 \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right)
is given by the formula  \chi(\lambda) = \left| \begin{array}{ccc} \lambda - a & -b & -c \\ -d & \lambda - e & -f \\ -g & -h & \lambda - i \end{array} \right|

Looks awesome. Have a look at David R. Wilkins’ Getting Started with LaTeX for some excellent examples, including some of the ones above.


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Freeman Dyson and 1/19

I just read this article about Freeman Dyson and a math puzzle that asks you to find a number that is doubled when you tear off the rightmost digit and stick it on the left. For example, tearing the 2 off 12 and sticking it in front gets you 21, which isn’t 2×12, so 12 isn’t right. Moving the 1 in 7654321 gets you 1765432 which also dosn’t double it, so that’s not the number either.

Anyway, it turns out the number we’re looking for is 18 digits long. It’s 052,631,578,947,368,421 which doubles to 105,263,157,894,736,842. (You’re allowed to stick a zero in front.)

Well, my son loves this kind of base-10 number tomfoolery, so I asked him if he could figure it out. And he reeled off the answer without pause. All 18 digits.

“Uhh, how’d ya know?” I asked.

He said it’s because 1/19 is 0.052631578947368421… (repeated forever), and 2/19 is 0.105263157894736842… (just move the 1). And of course 2/19 is twice 1/19. So if you know your N/19 repeating decimals, this is apparently easy peasy. And all the N/19’s are buried in that infinite sequence. You just have to start at different places. 3/19 is 0.157894736842105263…, 4/19 is 0.210526315789473684…, 5/19 is 0.263157894736842105…, etc.

So the next time someone asks “what’s a number that’s quadrupled when you tear the last two digits off the right and stick them on the left” you can say “doy, it’s 052,631,578,947,368,421 of course” and then roll your eyes. Because you’ve seen 1/19 and 4/19 written out as decimals.

Anyway, my son also told me there’s something similar for 1/29, except you triple the number by tearing off the rightmost digit and putting it on the left. I think it’s 28 digits long.

And also for 1/39, except you quadruple (x4) the number. And with 1/49 you x5 the number, and 1/59 gets you x6. Even 1/99 (0.0101010101…) works (x10 gives you 0.1010101010…).

As for 1/109, I don’t know, I’ll bet there’s some kind of x11 trick there. My son’s in bed now, but I’ll ask him tomorrow when he gets up.

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