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I asked Wolfram, and it gave me a complicated answer involving the ProductLog function $W$. Is there a simpler, approximate answer?

Erel Segal-Halevi
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1 Answers1

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Probably not possible without the $W$-function.

Here's how to get Wolfram's answer:

Writing $2^x=e^{x\log 2}$ taking the $d$th root, and doing some re-arranging, and you get:

$$xe^{-x\log 2/d} = d^{1/d}$$

Multiplying both sides by $-\frac{\log 2}d$ and letting $y=-\frac{x\log 2}d$ you get:

$$ye^y = -\frac{d^{1/d}\log 2}{d}$$

That means:

$$y=W\left(-\frac{d^{1/d}\log 2}{d}\right)$$

or:

$$x=\frac{-d}{\log 2}W\left(-\frac{d^{1/d}\log 2}{d}\right)$$

Since we are dealing with negative values, $W(u)$ has two real values for $-1/e<u<0$, called $W_0(u)$ and $W_{-1}(u)$. For $u$ small, we have that $W_0(u)=u-u^2+O(u^3)$. so we get:

$$\begin{align}x &= d^{1/d} +\frac{\log 2}{d^{1-2/d}} + O\left(\frac{1}{d^{2}}\right) \end{align}$$

The asymptotic for $W_{-1}(u)$ are more complicated, unfortunately, but they might be useful.

Thomas Andrews
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  • What is W here? Actually I don't know higher maths so please help me out – sgrmshrsm7 Apr 02 '17 at 15:50
  • $W(x)$ is defined as the solution $w$ of $we^w=x$. It is called the Lambert $W$-function, or the product logarithm function. @sgrmshrsm7 https://en.wikipedia.org/wiki/Lambert_W_function – Thomas Andrews Apr 02 '17 at 15:52
  • Wait... is the result negative? I was sure there must be a positive answer. For every $d$, $2^x$ is asymptotically larger than $x^d/d$, so the graph of $2^x$ must at some point (for some positive value of $x$) cross the graph of $x^d/d$. For example, this graph https://www.desmos.com/calculator/xxb2yjrqbc shows that, for $d=5$, the crossing point is at $x=18.8$. Is there an approximation to this positive solution? – Erel Segal-Halevi Apr 02 '17 at 17:18
  • @ErelSegal-Halevi Yep, I dropped a sign. I forgot the minus sign inside the $W$ call. Fixed – Thomas Andrews Apr 02 '17 at 18:18
  • Is there an approximation that holds at least asymptotically, using big-$O$ notation? – Erel Segal-Halevi Apr 03 '17 at 05:39
  • Don't know. Presumably, you mean for large $d$? – Thomas Andrews Apr 03 '17 at 15:34