6
$\begingroup$

There is a well-know quotation of Euler in a letter from 1746 to Goldbach:

Letztens habe ich gefunden, dass diese expressio $\sqrt{-1}^{\sqrt{-1}}$ einen valorem realem habe, welcher in fractionibus decimalibus $=0,2078795763$, welches mir merkwürdig zu seyn scheinet.

For the principal value of the logarithm the expresssion is $i^i=\exp(i \log(e^{i\pi/2}))= e^{-\pi/2}$.

My question is, how did Euler calculate this with such an accuracy?

$\endgroup$
3
  • $\begingroup$ Probably using Napier's tables and the series expansion of $e$. Or, maybe he used continued fraction of $e^x$ for more accuracy. $\endgroup$
    – vidyarthi
    Commented Jul 10, 2021 at 7:16
  • $\begingroup$ Another possibility is Newton's method for the equation $\log x = -\pi/2$. For better convergence, solve $\log x = -\pi/2^m$ for some $m$ instead, and then, $(m-1)$ times, square $x$. $\endgroup$ Commented Jul 10, 2021 at 9:27
  • $\begingroup$ See chronology of computation of pi. $\endgroup$
    – Lucian
    Commented Jul 10, 2021 at 9:49

1 Answer 1

10
$\begingroup$

On the numerical value of $i^i$ and Historical notes on the relation $e^{-\pi/2}=i^i$ describe how these accurate computations can be performed with logarithmic tables.

Euler described how he arrived at the identity in a paper read at the Berlin Academy in 1746, giving more decimals (13) than in the letter to Goldbach. A later calculation by Gauss computed 35 decimal places. Euler did not present his computation, but Gauss did [source].

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .