How to solve $n$ for $n^n = 2^c$?
What's the numerical method?
I don't get this.
For $c=1000$, $n$ should be approximately $140$, right?
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Srivatsan
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andandandand
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try something like solve(n^n = 2^1000,n) instead of what you did. – Myself Mar 17 '11 at 13:13
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Out of curiosity. Where does this equation come up? Or is it just an exercise? – Mar 17 '11 at 16:57
3 Answers
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Hint: Consider this.
Hint 2: First take $\log$ on both sides.
And explicitly: The solution to your question is given by $$ n = e^{W(c\log 2)} = \frac{{c\log 2}}{{W(c\log 2)}}. $$ For $c=1000$, this gives $n \approx 140.2217$. The function $W$ is standard (ProductLog in Wolfram Mathematica).
EDIT: For large $c$, a rough but very simple approximation to the solution $n$ of $n^n = 2^c$ can be obtained as follows (cf. this, also for improvement of the approximation): $$ n \approx (c\log 2)[\log (c\log 2)]^{1/\log (c\log 2) - 1} . $$ For example, for $c=1000$ this gives $n \approx 141.2083$, not far from the exact value of about $140.2217$.
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The wolframalpha link in the question actually has the $n = \exp(W(c\log 2))$ solution. :) – leif Mar 17 '11 at 15:08
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Alpha sometimes goes off into the complex plane when what you want is only the reals. I agree with you and get about 140.222. If you ask to solve n(ln(n))=1000 ln(2) you get what you want.
Ross Millikan
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Yes, take the $log_2$ of both sides gives you:
$$n log_2(n) = c$$
You can use Newton's method to solve this:
$$x_0 = c$$ $$x_{k+1} = x_k - (x_k log(x_k) - c log(2))/(1 + log(x_k))$$
where now "log" is the natural logarithm.
This gives the solution $n \sim x_4 = 140.221667$.
Starting with a better $x_0$, like $x_0=c / log(c)$ gives you even faster convergence. With c=1000 or c=1000000, the value $x_3$ is correct with an error of $10^{-8}$.
Thomas Andrews
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