# 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.

• Any reason to avoid numerical methods?
– lulu
Commented Sep 28, 2022 at 23:00
• Hi, thanks for answering. But I cannot see how wolfram got to that answer. Commented Sep 28, 2022 at 23:06
• Presumably by using Newton's method, or the like. Easy enough to bound the possible real answers, after which any numerical method you favor should go through.
– lulu
Commented Sep 28, 2022 at 23:07
• Thanks! I'd appreciate it if you could make an answer showing how newton's method works for this problem. If not, thanks for steering me in the right direction! Commented Sep 28, 2022 at 23:19
• Just try it, it goes very quickly. Guessing $10$ gives you one root in about six steps. Guessing $0$ gives you the other even more quickly.
– lulu
Commented Sep 28, 2022 at 23:21

$$n=2^{n^2/16}$$ $$n^2=2^{n^2/8}=e^{n^2(\ln2)/8}$$ $$-\frac{\ln2}8n^2e^{-n^2(\ln2)/8}=-\frac{\ln2}8$$ Now we can apply Lambert W: $$-\frac{\ln2}8n^2=W\left(-\frac{\ln2}8\right)$$ $$n=\sqrt{\frac{W(-(\ln2)/8)}{-(\ln2)/8}}$$ The argument to $$W$$ falls in the range where both real branches $$W_0$$ and $$W_{-1}$$ have real output, so we get two solutions $$n=1.0488074\dots$$ and $$n=6.5999439\dots$$

• Thanks! I'm curious about step three. Is n/e^x == ne^-x? Commented Sep 29, 2022 at 16:46
• @BenAlan Exactly. Moving $e^x$ to the other side negates the exponent. Commented Sep 29, 2022 at 16:49

It is interesting to notice that $$f(x)=2^{\frac{x^2}{16}}$$ is contraction map for $$x$$ in the interval $$[-4.6,4.6]$$ and a solution for the fixed point $$x=f(x)$$ can be found by the sequence $$x_{i+1}=f(x_i)$$, $$i=0,1,2,...$$, starting at any $$x_0\in [-4.6,4.6]$$. This sequence will converge to one of the solutions ($$x\approx 1.048807$$).

• Alan wants a solution with Lambert W function. But, I love the fixed point theorem too. All bad things in life happen because of that theorem. E.g. Nash. Commented Sep 29, 2022 at 10:22
• @Bob Dobbs, Sure :). I love the fixed point theorem too. I could not resist to post a note on that just to show an alternative solution route for the question. Commented Sep 29, 2022 at 10:57

$$n = 2^{(\frac {n}{4})^2}$$

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 $$n$$

In the definition of Lambert W function, put $$x=\ln(u^{-1})$$ then we have the property: $$W(-\frac{\ln u}{u})=-\ln u$$.

If $$n=2^{\frac{n^2}{16}}$$ then $$\ln n=\frac{n^2\ln 2}{16}$$. Multiplying by 2 we have $$\ln (n^2)=\frac{n^2\ln 2}{8}$$ and $$-\frac{\ln (n^2)}{n^2}=-\frac{\ln 2}{8}.$$

Now, taking $$W$$ of both sides and using the property above we get $$-\ln(n^2)=W(-\frac{\ln 2}{8})$$. Hence, $$n=\sqrt{\exp(-W(-\frac{\ln 2}{8}))}.$$ Since, $$W(-\frac{\ln 2}{8})$$ has two negative real roots, (Use the calculator here, with $$x\approx -0.08664339756999$$: https://www.had2know.org/academics/lambert-w-function-calculator.html), we have two solutiıns for Alan's equation.