How accurate is this pandigital approximation to $e$?

Question

Given that: $$e \approx\displaystyle (1+9^{{-4}^{6\times 7}}) ^ {3^{2^{85}}}$$ But when I type the equation into wolfram alpha, I am unable to view up to how many decimal places this approximation is accurate up to.

How may I find out?

Answer 1

For a rough estimate of the accuracy note that $ n \ln \left(1 + \frac{1}{n}\right) = n \left( \frac{1}{n} + \frac{1}{2}\frac{1}{n^2} +\dots \right)$, taking $n=3^{2^{85}}$.

Answer 2

No direct calculation can't help us find the error in the computation, because $(1 - 9^{-4^{6\cdot 7}})^{3^{2^{85}}}$ approximates $e$ more closely than how closely we've actually computed digits of $e$.

To estimate the error, we have to figure out in general how good the approximation $e \approx (1 + \frac1n)^n$ is, in terms of $n$, and then set $n = 3^{2^{85}}$.

We can show that the following inequality holds (see the end of this answer) for integer $n>0$: $$\left(1 + \frac1n\right)^n < e < \left(1 + \frac1n\right)^{n+1}.$$ Therefore in approximating $e$ by $(1 + \frac1n)^n$, we're off by less than a factor of $1 + \frac1n$, which means we're off by an additive error of less than $\frac en$.

How small is this for $n = 3^{2^{85}}$? We can write this as $n = 10^{2^{85} \log_{10} 3}$, and the exponent is now small enough that even Wolfram Alpha can compute it: it is about $1.84577 \times 10^{25}$. That's how many correct digits we are guaranteed. (The factor of $e$ in $\frac en$ might reduce the number of correct digits by $1$, but we've calculated it less precisely than that anyway.)

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As promised, here is some inequality work. I will start from the inequality $e^x \ge 1 + x$ (with equality only if $x=0$), because I have to start from somewhere; how you prove this depends mainly on your definition of $e$, of which there are several.

Then $e > (1 + x)^{1/x}$ for $x>0$; setting $x = \frac1n$ proves the first half of the inequality we want.

To get the other half of the inequality, set $x = - \frac1{n+1}$, getting $e^{-1/(n+1)} > \frac{n}{n+1}$, or (taking the reciprocal) $e^{1/(n+1)} < \frac{n+1}{n} = 1 + \frac1n$. Finally, the $(n+1)$-th power of this gives us $e < (1 + \frac1n)^{n+1}$: the second half of the inequality.