# How did Hermite calculate eπ√163e^{\pi\sqrt{163}} in 1859?

Pretend you are in 1859. What is a fast, efficient, and accurate way to numerically evaluate constants like that to, say, 20 decimal places, using ONLY pen and paper?

The French paper of Hermite (1859) ‘Sur la théorie des equations modulaires’ is available freely at Google books, it begins at page 29.

You’ll find there the value of $e^{\pi\sqrt{43}}$ with all the correct digits but concerning $e^{\pi\sqrt{163}}$ there is only the indication that the fractional part should begin with twelve consecutive $9$. He clearly used modular properties to deduce this as explained by Matt. Wikipedia’s page concerning Heegner numbers could help too.

At first glance Hermite doesn’t seem to provide his computations’ secrets concerning $e^{\pi\sqrt{43}}$, polynomials of order $48$ of page 67 (he speaks of “quite long but not at all impractical calculation” about them) and so on… Clearly he didn’t fear tedious computations!

So let’s see what we can do with logarithms or exponential tables. The ‘state of the art’ for these (1859) days may perhaps be found in this paper of 1881 from Alexander Ellis : Gray published Tables for twelve-place logarithms in 1845 and Thoman a “Tables de logarithmes à 27 décimales” in 1867. Twelve digits seem enough for the nine digits of the integer part of $e^{\pi\sqrt{43}}$ and Hermite probably deduced the fractional part by inversion and multiplication of this result.

Let’s suppose arbitrary that he used exponential tables (he could as well have used logarithm tables or $log_{10}$ tables or something more subtle ‘AGM like’ who knows…)

I’ll start with $\pi \sqrt{43} \approx 20.6008006943$ ($\sqrt{43}$ is obtained quickly by iterations of $n'=n^2+43 d^2$,$d'=2nd$) and try to get $e^{\pi \sqrt{43}}\approx 884736744.000$

I used the ‘relatively recent’ Abramowitz and Stegun tables that I’ll ’round’ at 12 digits for my ‘H-emulation’.

$e^{20}$ is tabulated page 138 as ( 8)4.85165 19540 98 and this value of $485165195.410$ will be our reference. We will have to multiply this by $e^{0.6008006943}$ (nearly $1.82357834477$). Let’s try to get this one :

• by interpolation between $e^{0.600}\approx 1.82211880039$ and $e^{0.601}\approx 1.82394183055$ from the tables we get :
$1.82357849025$ with the final result $884736814.567$ clearly insufficient…
To get something more exact we need more tabulated values or expand $\displaystyle e^{0.601-0.6008006943}$ in Taylor series getting $e^{0.601}e^{-0.0002}e^{0.0000007}e^{-0.000000005}=1.82357834605$ with the final result $884736744.607$ not to far from our target (the idea is that the Taylor series for $e^{-0.0002}$, $e^{0.0000007}$… are fast)

• or by multiplication by $e^{0.6}$, $e^{0.0001}$ (8 times), $e^{0.0000001}$ (6 times), $e^{0.00000001}$ (9 times), $e^{0.000000001}$ (4 times) at this point the final result is $884736743.722$ and we are nearly there…

Let’s note that the second method allows to get high precision just by ‘pre-evaluation’ of some terms. In fact $\displaystyle e^{10^k}$ (for $k=1,0,-1,-2\cdots -18$) evaluated with high precision should have worked for $e^{\pi\sqrt{163}}$ !
EDIT2: evaluating directly $e^{0.0000006}$, $e^{0.00000009}$, $e^{0.0000000043}$ to 18 digits would probably be more efficient and the result could be evaluated this way : $\displaystyle a e^{\epsilon}= a + (a)\frac{\epsilon}{1} + \left(\frac{a\epsilon}{1}\right)\frac{\epsilon}2+\cdots$ (each time the previous term is mutiplied/divided by $\frac{\epsilon}n$ and added to the result)
some excellent mental calculators of these days could have contributed too!

Of course Hermite’s method could have be much more subtle. Vladimir Arnold pointed out that many techniques like elliptic functions were better known in the nineteenth century than now. I think that the same could be said about quite some methods to solve Diophantine equations before the time of computers as you may confirm! 🙂

OTHER METHODS : Let’s try to apply the general method proposed by Apostol (in tzs’ answer) or rather the idea of another Caltech professor for computing ‘exponentials in your head’ : Richard Feynman (see J.M.’s extract ‘Lucky Numbers’ from Feynman’s very funny book “Surely You’re Joking, Mr. Feynman!”).

Let’s look at one of his favorite examples : he was asked to compute $e^3$ ‘in his head’ and was able to answer quickly $20.085$. He needed only the values of $\ln 2$ and $\ln 10$ in this case. Let’s detail this (to 6 digits here) : $\ln 10\approx 2.302585$ and $\ln 2\approx 0.693147$ but their sum is $\approx 2.995732= 3-\epsilon$ with $\epsilon\approx 0.004268$ so that :

If asked for $e^4$ he would simply multiply this by $2.7182818$ to get : $e^4\approx 54.59816$ (the relation used is $4 \approx 1 +\ln 2 +\ln 10$).

A generalization of Feynman’s trick to compute $e^a$ would consist in searching linear relations between $a$, $1$, $\ln 2$ and $\ln 10$ (we may accept other constants especially logarithms since $e^c$ should be easy to evaluate). This is easier to do now with algorithms like PSLQ or LLL, in Hermite’s days I think you could only use continued fractions or guess…

For $\pi \sqrt{43}$ we may get excellent approximations like $2^{18} 15^3$, $5^{64/5}$.
For $\pi \sqrt{163}$ approximations are $2^{30}\cdot 5^{12}$, $2^{10}\cdot 3^7\cdot 5^{11}\cdot 7^4$, $2^7\cdot 3^{20}\cdot 5\cdot 7^6$, $e^{40} 4\cdot 3^8\cdot 5/7^6$ (I suppose that we have a table of logarithms and of exponentials of integers given with high precision).

But to be honest I think that none of these tricks were used by Hermite himself for these evaluations because he knew, as you may find in the previous links, ‘Heegner numbers’ for example, that the nearest integer was given by $960^3+744=884736744$ for $e^{\pi \sqrt{43}}$ and by $640320^3+744=262537412640768744$ for $e^{\pi \sqrt{163}}$ so that he needed no evaluation of exponential at all !!