1
$\begingroup$

I am having a problem with the final question of this exercise.

Show that $e$ is irrational (I did that). Then find the first $5$ digits in a decimal expansion of $e$ ($2.71828$).

Can you approximate $e$ by a rational number with error $< 10^{-1000}$ ?

Thank you in advance

  • 0
    I think that a more interesting question is whether there is a good short rational approximation to e, analogous to 355/113 for $\pi$. Approximating e to absurd accuracy is just donkeywork for a computer. What puzzles me here is: if you are a good enough mathematician to prove that e is irrational, why would you ask such a naive question?2012-11-14

3 Answers 3

4

In the standard proof that $e$ is irrational, one first proves that $ 0 < e -s_n < \frac1{n!n} \qquad\mbox{where}\qquad s_n = \sum_{k=0}^n \frac1{k!} $ So you only need to find $n$ such that $\frac1{n!n}< 10^{-1000}$ or $n!>10^{1000}$. You can use Stirling's approximation for that I guess. Wolfram Alpha says $n=450$ suffices.

  • 0
    I would think that stating that such an n exists would probably suffice for this question. One could show the Taylor expansion for e has an Nth term such that all n > N terms will be smaller than 10^{-1000}.2012-11-14
3

I don't think we have to really know anything about $e$ to say that we can approximate it with a rational number with an error less then $10^{-1000}$.

Say $e=a_0.a_1 a_2 \ldots$

There is clearly a rational number $q=a_0.a_1 \ldots a_{1001}$ and $|q-e|\leq10^{-1000}$ (note that the difference is bounded by a geometric sum)

  • 0
    @robjohn - thanks for pointing that out, I edited accordingly.2012-11-16
0

If 2.71828 are the first few digits of $e$ the we have $2.71828 < e < 2.71829$ Put $q = 271828/100000$ we deduce $0 < e-q < 0.00001 = 10^{-5}.$

  • 1
    @Belgi: the question is "Can you approximate $e$..." not "Approximate $e$..." No answer actually computes a rational approximation of $e$ to an error of less than $10^{-1000}$.2012-11-17