<s>[INST] How to find n for $n = 2^{\frac {n^2}{16}}$ I have this equation $n = 2^{\frac{n^2}{16}}$
I need to find n.
I tried using the Lambert W function, but I don't know how to get one side into the form $W(xe^x)=x$ without having an $n$ on the other side.
I have done:
$n = e^{\frac{n^2ln2}{16}}$
$1 = \frac{1}{n}e^{\frac{n^2ln2}{16}}$
I'm stuck because I am multiplying $\frac{1}{n}$ by $\frac{n^3}{16}ln2$ but that leaves n on the other side.
 [/INST] $n = 2^{(\frac {n}{4})^2}$
We need to get the right hand side into the form ${f(n)}e^{f(n)} = c$
Then $f(n) = W(c)$
$1 = n^{-1}e^{(\frac {n}{4})^2\ln 2}$
$1^{-1} = ne^{-(\frac {n}{4})^2\ln 2}$
$\frac 14 = \frac {n}{4}e^{-(\frac {n}{4})^2\ln 2}$
$\frac 1{16} = (\frac {n}{4})^2e^{-2(\frac {n}{4})^2\ln 2}$
$\frac {\ln 2}{8} = (2(\frac {n}{4})^2\ln 2)e^{-2(\frac {n}{4})^2\ln 2}$
$-\frac {\ln 2}{8} = (-2(\frac {n}{4})^2\ln 2)e^{-2(\frac {n}{4})^2\ln 2}$
Now we have the right-hand side in the correct form.
$-2(\frac {n}{4})^2\ln 2 = W(-\frac {\ln 2}{8})$
And unwind to find $n$
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