From f2be522089ced8d219acd27e9ec1d8867cf95ef3 Mon Sep 17 00:00:00 2001 From: rnhmjoj Date: Wed, 27 May 2020 15:29:59 +0200 Subject: [PATCH] ex-2: complete review - general review and proofread - complet the explanation of Lambert W based solutions - add section on 1 million digits computation - add references to GMP and mpmath --- notes/docs/bibliography.bib | 21 +- notes/sections/2.md | 460 +++++++++++++++++++----------------- notes/todo | 4 - 3 files changed, 267 insertions(+), 218 deletions(-) diff --git a/notes/docs/bibliography.bib b/notes/docs/bibliography.bib index 313c943..56ae8b8 100644 --- a/notes/docs/bibliography.bib +++ b/notes/docs/bibliography.bib @@ -4,7 +4,7 @@ journal={Journal of mathematics and physics}, volume={20}, number={1-4}, - pages={224 -- 230}, + pages={224--230}, year={1941}, publisher={Wiley Online Library} } @@ -198,3 +198,22 @@ computation of Euler's constant}, year={2017} } + +@manual{mpmath13, + key={mpmath}, + author={Fredrik Johansson and others}, + title={mpmath: a {P}ython library for arbitrary-precision floating-point arithmetic (version 0.18)}, + note={{\tt http://mpmath.org/}}, + month={December}, + year={2013}, +} + +@manual{gmp20, + key={gmp}, + author={Torbjörn Granlund and the GMP development team}, + title={GNU MP: The GNU Multiple Precision Arithmetic Library, + Edition 6.2.0 (version 0.18)}, + note={{\tt https://gmplib.org/}}, + month={January}, + year={2020}, +} diff --git a/notes/sections/2.md b/notes/sections/2.md index dadbd68..3fd348e 100644 --- a/notes/sections/2.md +++ b/notes/sections/2.md @@ -43,40 +43,40 @@ $$ | \gamma(n_{i+1}) - \gamma | > | \gamma(n_i) - \gamma| $$ -and $\gamma (n_i)$ was selected as the best result (see @tbl:1_results). +and $\gamma (n_i)$ was selected as the best result (see @tbl:naive-errs). ---------------------------------------------- -n $|\gamma(n)-\gamma|$ ----------------------- ---------------------- -\num{2e1} \num{2.48e-02} +---------------------------- +n $|γ(n)-γ|$ +----------- ---------------- +\num{2e1} \num{2.48e-02} -\num{2e2} \num{2.50e-03} +\num{2e2} \num{2.50e-03} -\num{2e3} \num{2.50e-04} +\num{2e3} \num{2.50e-04} -\num{2e4} \num{2.50e-05} +\num{2e4} \num{2.50e-05} -\num{2e5} \num{2.50e-06} +\num{2e5} \num{2.50e-06} -\num{2e6} \num{2.50e-07} +\num{2e6} \num{2.50e-07} -\num{2e7} \num{2.50e-08} +\num{2e7} \num{2.50e-08} -\num{2e8} \num{2.50e-09} +\num{2e8} \num{2.50e-09} -\num{2e9} \num{2.55e-10} +\num{2e9} \num{2.55e-10} -\num{2e10} \num{2.42e-11} +\num{2e10} \num{2.42e-11} -\num{2e11} \num{1.44e-08} ---------------------------------------------- +\num{2e11} \num{1.44e-08} +---------------------------- Table: Partial results using the definition of $\gamma$ with double -precision. {#tbl:1_results} +precision. {#tbl:naive-errs} The convergence is logarithmic: to fix the first $d$ decimal places, about -$10^d$ terms of the armonic series are needed. The double precision runs out at -the $10^{\text{th}}$ place, at $n=\num{2e10}$. +$10^d$ terms of the harmonic series are needed. The double precision runs out +at the $10^{\text{th}}$ place, at $n=\num{2e10}$. Since all the number are given with double precision, there can be at best 16 correct digits, since for a double 64 bits are allocated in memory: 1 for the sign, 8 for the exponent and 55 for the mantissa: @@ -87,43 +87,40 @@ $$ Only 10 digits were correctly computed: this means that when the terms of the series start being smaller than the smallest representable double, the sum of all the remaining terms gives a number $\propto 10^{-11}$. The best result is -shown in @tbl:first. +shown in @tbl:naive-res. ---------- ----------------------- -true: 0.57721\ 56649\ 01533 - -approx: 0.57721\ 56648\ 77325 - -diff: 0.00000\ 00000\ 24207 ---------- ----------------------- +------- -------------------- +exact 0.57721 56649 01533 +approx 0.57721 56648 77325 +diff 0.00000 00000 24207 +------- -------------------- Table: First method best result. From the top down: true value, best estimation -and difference between them. {#tbl:first} +and difference between them. {#tbl:naive-res} ### Alternative formula -As a first alternative, the constant was computed through the identity which -relates $\gamma$ to the $\Gamma$ function as follow [@davis59]: +As a first alternative, the constant was computed through the identity +[@davis59] which relates $\gamma$ to the $\Gamma$ function as follow: $$ \gamma = \lim_{M \rightarrow + \infty} \sum_{k = 1}^{M} \binom{M}{k} \frac{(-1)^k}{k} \ln(\Gamma(k + 1)) $$ Varying $M$ from 1 to 100, the best result was obtained for $M = 41$ (see -@tbl:second). It went sour: the convergence is worse than using the definition -itself. Only two places were correctly computed (@tbl:second). +@tbl:limit-res). This approximation gave an underwhelming result: the +convergence is actually worse than the definition itself. Only two decimal +places were correctly computed: ---------- ----------------------- -true: 0.57721\ 56649\ 01533 - -approx: 0.57225\ 72410\ 34058 - -diff: 0.00495\ 84238\ 67473 ---------- ----------------------- +-------- -------------------- +exact 0.57721 56649 01533 +approx 0.57225 72410 34058 +diff 0.00495 84238 67473 +-------- -------------------- Table: Best estimation of $\gamma$ using -the alternative formula. {#tbl:second} +the alternative formula. {#tbl:limit-res} Here, the problem lies in the binomial term: computing the factorial of a number greater than 18 goes over 15 places and so cannot be correctly @@ -146,69 +143,66 @@ $$ \prod_{k = 1}^{+ \infty} \left( 1 + \frac{z}{k} \right) e^{-z/k} \right) $$ -The execution stops when there is no difference between two consecutive therms +The execution stops when there is no difference between two consecutive terms of the infinite product (it happens for $k = 456565794 \sim \num{4.6e8}$, meaning that for this value of $k$ the term of the product is equal to 1 in terms of floating points). Different values of $z$ were checked, with $z_{i+1} = z_i + 0.01$ ranging from 0 to 20, and the best result was found for $z = 9$. ---------------------------------------------------------------- - z $|\gamma(z) - \gamma |$ z $|\gamma(z) - \gamma |$ ------ ------------------------ ------ ------------------------ - 1 \num{9.712e-9} 8.95 \num{9.770e-9} +----------------------------------------------- +z $|γ(z) - γ|$ z $|γ(z) - γ|$ +----- ---------------- ------ ---------------- + 1 \num{9.712e-9} 8.95 \num{9.770e-9} - 3 \num{9.320e-9} 8.96 \num{9.833e-9} + 3 \num{9.320e-9} 8.96 \num{9.833e-9} - 5 \num{9.239e-9} 8.97 \num{9.622e-9} + 5 \num{9.239e-9} 8.97 \num{9.622e-9} - 7 \num{9.391e-9} 8.98 \num{9.300e-9} + 7 \num{9.391e-9} 8.98 \num{9.300e-9} - 9 \num{8.482e-9} 8.99 \num{9.059e-9} + 9 \num{8.482e-9} 8.99 \num{9.059e-9} - 11 \num{9.185e-9} 9.00 \num{8.482e-9} + 11 \num{9.185e-9} 9.00 \num{8.482e-9} - 13 \num{9.758e-9} 9.01 \num{9.564e-9} + 13 \num{9.758e-9} 9.01 \num{9.564e-9} - 15 \num{9.747e-9} 9.02 \num{9.260e-9} + 15 \num{9.747e-9} 9.02 \num{9.260e-9} - 17 \num{9.971e-9} 9.03 \num{9.264e-9} + 17 \num{9.971e-9} 9.03 \num{9.264e-9} - 19 \num{10.084e-9} 9.04 \num{9.419e-9} ---------------------------------------------------------------- + 19 \num{10.084e-9} 9.04 \num{9.419e-9} +----------------------------------------------- Table: Differences between some obtained values of $\gamma$ and the exact one found with the reciprocal $\Gamma$ function formula. The values on the left are shown to give an idea of the $z$ large-scale behaviour; on the right, the values around the best -one ($z = 9.00$) are listed. {#tbl:3_results} +one ($z = 9.00$) are listed. {#tbl:recip-errs} -As can be seen in @tbl:3_results, the best value for $z$ is only by chance, +As can be seen in @tbl:recip-errs, the best value for $z$ is only by chance, since all $|\gamma(z) - \gamma |$ are of the same order of magnitude. The best -one is compared with the exact value of $\gamma$ in @tbl:third. +one is compared with the exact value of $\gamma$ in @tbl:recip-res. ---------- ----------------------- -true: 0.57721\ 56649\ 01533 +------- -------------------- +true 0.57721 56649 01533 +approx 0.57721 56564 18607 +diff 0.00000 00084 82925 +------- -------------------- -approx: 0.57721\ 56564\ 18607 +Table: Third method results for z = 9.00. {#tbl:recip-res} -diff: 0.00000\ 00084\ 82925 ---------- ----------------------- - -Table: Third method results for z = 9.00. {#tbl:third} - -This time, the convergence of the infinite product is fast enough to ensure the -$8^{th}$ place. +In this approximation, the convergence of the infinite product is fast enough +to reach the $8^{th}$ decimal place. ### Fastest convergence formula -The fastest known convergence belongs to the following formula, known as refined -Brent-McMillan formula[@yee19]: +The fastest known convergence to $\gamma$ belongs to the following formula, +known as refined Brent-McMillan formula [@yee19]: $$ \gamma_N = \frac{A(N)}{B(N)} -\frac{C(N)}{B^2(N)} - \ln(N) $$ {#eq:faster} - -with: +where \begin{align*} &A(N) = \sum_{k=1}^{+ \infty} \frac{N^k}{k!} \cdot H(k) \with H(k) = \sum_{j=1}^{k} \frac{1}{j} \\ @@ -217,95 +211,119 @@ with: \frac{((2k)!)^3}{(k!)^4 \cdot (16k)^2k} \\ \end{align*} -and the difference between $\gamma_N$ and $\gamma$ is given by [@demailly17]: +The asymptotic error of this estimation, given in [@demailly17], is: $$ - |\gamma_N - \gamma| < \frac{5 \sqrt{2 \pi}}{12 \sqrt{N}} e^{-8N} + |\gamma_N - \gamma| \sim \frac{5 \sqrt{2 \pi}}{12 \sqrt{N}} e^{-8N} = \Delta_N $$ {#eq:NeD} -This gives the value of $N$ which is to be used into the formula in order to get -$D$ correct decimal digits of $\gamma$. In fact, by imposing: +The error bound gives the value of $N$ to be used to get $D$ correct decimal +digits of $\gamma$. In fact, this is done by imposing: $$ \frac{5 \sqrt{2 \pi}}{12 \sqrt{N}} e^{-8N} < 10^{-D} $$ -With a bit of math, it can be found that the smallest integer which satisfies -the inequality is given by: +The inequality with the equal sign can be solved with the use of the Lambert +$W$ function. For a real number $x$, $W(x)$ is defined as the inverse function +of $x\exp(x)$, so the idea is to recast the equation into this form and take +$W$ both sides. + +\begin{align*}{3} +\frac{5 \sqrt{2 \pi}}{12 \sqrt{x}} e^{-8x} = 10^{-D} + & \thus \left(\frac{12}{5}\right)^2 \frac{x}{2 \pi} e^{16x} = 10^{2D} \\ + & \thus 16x e^{16x} = \left(\frac{5 \pi}{9}\right)^2 10^{2D + 1} \\ + & \thus x = \frac{1}{16} W\left(\left(\frac{5 \pi}{9}\right)^2 10^{2D + 1}\right) +\end{align*} + +The smallest integer which satisfies the inequality is then $$ N = 1 + \left\lfloor \frac{1}{16} W \left( \frac{5 \pi}{9} 10^{2D + 1}\right) \right\rfloor -$$ +$$ {#eq:refined-n} The series $A$ and $B$ were computed till there is no difference between two consecutive terms. Results are shown in @tbl:fourth. ---------- ------------------------------ -true: 0.57721\ 56649\ 01532\ 75452 +------- -------------------------- +exact 0.57721 56649 01532 75452 +approx 0.57721 56649 01532 53248 +diff 0.00000 00000 00000 33062 +------- -------------------------- -approx: 0.57721\ 56649\ 01532\ 53248 +Table: $\gamma$ estimation with the fastest known convergence formula +(@eq:faster). {#tbl:fourth} -diff: 0.00000\ 00000\ 00000\ 33062 ---------- ------------------------------ - -Table: $\gamma$ estimation with the fastest -known convergence formula (@eq:faster). {#tbl:fourth} - -Because of roundoff errors, the best result was obtained only for $D = 15$, for +Due to roundoff errors, the best result was obtained only for $D = 15$, for which the code accurately computes 15 digits and gives an error of \num{3.3e16}. For $D > 15$, the requested can't be fulfilled. + ### Arbitrary precision -To overcome the issues related to the double representation, one can resort to a -representation with arbitrary precision. In the GMP library (which stands for -GNU Multiple Precision arithmetic), for example, real numbers can be -approximated by a ratio of two integers (a fraction) with arbitrary precision: -this means a check is performed on every operation and in case of an integer -overflow, additional memory is requested and used to represent the larger result. -Additionally, the library automatically switches to the optimal algorithm -to compute an operation based on the size of the operands. +To overcome the issues related to the limited precision of the machine floating +points, one can resort to a software implementation with arbitrary precision. + +In the GMP [@gmp20] library (GNU Multiple Precision arithmetic), for +example, reals can be approximated by a rational or floating point numbers +with arbitrary precision. For integer and fractions this means a check is +performed on every operation that can overflow and additional memory is +requested and used to represent the larger result. For floating points it +means the size of the mantissa can be chosen before initialising the number. +Additionally, the library automatically switches to the optimal algorithm to +compute an operation based on the size of the operands. The terms in @eq:faster can therefore be computed with arbitrarily large -precision. Thus, a program that computes the Euler-Mascheroni constant within -a user controllable precision was implemented. +precision. Thus, a program that computes the Euler-Mascheroni constant within a +user controllable precision was implemented. + +To compute $N$ by @eq:refined-n, a trick must be used to avoid overflows in the +exponentiation, particularly when computing more than a few hundreds digits are +to be computed. The details are explained in @sec:optimised, below. Once the +number $N$ has been fixed, the series $A(N)$ and $B(N)$ are to be evaluated. +With rational numbers, a different criterion for the truncation must be +considered, because two consecutive term are always different. Brent and +McMillan[@brent00] prove that to reduce the partial sum to less than $\Delta_N$ +it is sufficient to compute $\alpha N$ terms, where $\alpha$ is the solution to +the equation: +\begin{align*} + \alpha\ln(\alpha) = 3 + \alpha && + \alpha = \exp(W(3e) - 1) \approx 4.97 +\end{align*} -According to [@brent00], the $A(N)$, $B(N)$ and $C(N)$ series were computed up -to $k_{\text{max}} = 5N$ \textcolor{red}{sure?}, since it guarantees the accuracy -od the result up to the $D$ decimal digit. The GMP library offers functions to perform some operations such as addition, multiplication, division, etc. However, the logarithm function is not implemented. Thus, most of the code carries out the $\ln(N)$ computation. -This is because the logarithm of only some special numbers can be computed +The reason is that the logarithm of only some special numbers can be computed with arbitrary precision, namely the ones of which a converging series is known. This forces $N$ to be rewritten in the following way: $$ N = N_0 \cdot b^e \thus \ln(N) = \ln(N_0) + e \cdot \ln(b) $$ -Since a fast converging series for $\ln(2)$ is known (it will be shwn shortly), -$b = 2$ was chosen. In case $N$ is a power of two, only $\ln(2)$ is to be -computed. If not, it must be handled as follows. -As well as for the scientific notation, in order to get the mantissa $1 +Since a fast converging series for $\ln(2)$ is known (it will be shown shortly), +$b = 2$ was chosen. If $N$ is a power of two, $N_0$ is 1 and only $\ln(2)$ is to be +computed. More generally, the problem reduces to the calculation of $\ln(N_0)$. + +Regarding the scientific notation, to find the mantissa $1 \leqslant N_0 < 2$, the number of binary digits of $N$ must be computed -(conveniently, a dedicated function `mpz_sizeinbase()` is found in GMP). If the +(conveniently, a dedicated function `mpz_sizeinbase()` exists in GMP). If the digits are $n$: $$ e = n - 1 \thus N = N_0 \cdot 2^{n-1} \thus N_0 = \frac{N}{2^{n - 1}} $$ -The logarithm of whichever number $N_0$ can be computed using the series of -$\text{atanh}(x)$, which converges for $|x| < 1$: +The logarithm of $N_0$ can be computed from the Taylor series of the hyperbolic +tangent, which is convergent for $|x| < 1$: $$ \text{atanh}(x) = \sum_{k = 0}^{+ \infty} \frac{x^{2k + 1}}{2x + 1} $$ -In fact: +The relation with the logarithm follows from the definition $$ \text{tanh}(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{e^{2x} - 1}{e^{2x} + 1} $$ - -with the change of variable $z = e^{2x}$, one gets: +and the change of variable $z = e^{2x}$: $$ \text{tanh} \left( \frac{\ln(z)}{2} \right) = \frac{z - 1}{z + 1} \thus \ln(z) = 2 \, \text{atanh} \left( \frac{z - 1}{z + 1} \right) @@ -322,19 +340,19 @@ $$ = 2 \sum_{k = 0}^{+ \infty} \frac{y^{2k + 1}}{2k + 1} $$ -But when to stop computing the series? -Given a partial sum $S_k$ of the series, it is possible to know when a digit is -definitely correct. The key lies in the following concept [@riddle08]. Letting -$S$ be the value of the series: +To estimate a series with a given precision some care must be taken: +different techniques apply to different series and are explained in +the paper [@riddle08]. In this case, letting $S$ be the value of the +series and $S_k$ the $k$-th partial sum the following bounds can be found: $$ S_k + a_k \frac{L}{1 -L} < S < S_k + \frac{a_{k+1}}{1 - \frac{a_{k+1}}{a_k}} $$ where $a_k$ is the $k^{\text{th}}$ term of the series and $L$ is the limiting -ratio of the series terms, which must be $< 1$ in order for it to converge (in +ratio of the series terms, which must be $\le 1$ in order for it to converge (in this case, it is easy to prove that $L = y^2 < 1$). The width $\Delta S$ of the interval containing $S$ gives the precision of the estimate $\tilde{S}$ if this -last is assumed to be the middle value of it, namely: +last is assumed to be its middle value, namely: $$ \tilde{S} = S_k + \frac{1}{2} \left( a_k \frac{L}{1 -L} + \frac{a_{k+1}}{1 - \frac{a_{k+1}}{a_k}} @@ -342,169 +360,185 @@ $$ \Delta S = \frac{a_{k+1}}{1 - \frac{a_{k+1}}{a_k}} - a_k \frac{L}{1 -L} $$ -In order to know when to stop computing the series, $\Delta S$ must be -evaluated. With a bit of math: -$$ - a_k = \frac{y^{2k + 1}}{2k + 1} \thus +For this series: +\begin{align*} + a_k = \frac{y^{2k + 1}}{2k + 1} && a_{k + 1} = \frac{y^{2k + 3}}{2k + 3} = a_k \, \frac{2k + 1}{2k + 3} \, y^2 -$$ - -hence: +\end{align*} +so, the right bound results: $$ \frac{a_{k+1}}{1 - \frac{a_{k+1}}{a_k}} = \frac{a_k}{\frac{2k + 3}{2k + 1}y^{-2} - 1} $$ - -from which: +and it finally follows: $$ - \Delta S = a_k \left( + \Delta S_k = a_k \left( \frac{1}{\frac{2k + 3}{2k + 1}y^{-2} - 1} - \frac{y^2}{1 - y^2} - \right) = \Delta S_k \with a_k = \frac{y^{2k + 1}}{2k + 1} + \right) $$ -By imposing $\Delta S < 10^{-D}$, where $D$ is the last correct decimal place -required, $k^{\text{max}}$ at which to stop can be obtained. This is achieved by -trials, checking for every $k$ whether $\Delta S_k$ is less or greater than -$10^{-D}$. +By imposing $\Delta S_k < 10^{-D}$, where $D$ is the number decimal places +required, $k^{\text{max}}$ at which to stop the summation can be obtained. This +is achieved by trials, checking for every $k$ whether $\Delta S_k$ is less or +greater than $10^{-D}$. -The same considerations holds for $\ln(2)$. It could be computed as the Taylor -expansion of $\ln(1 + x)$ with $x = 1$, but it would be too slow. A better idea -is to use the series with $x = -1/2$, which leads to a much more fast series: +Similar considerations can be made for $\ln(2)$. The number could be computed +from the MacLaurin series of $\ln(1 + x)$ with $x = 1$, but that yields a +slowly convergent series with alternating sign. A trick is to use $x = -1/2$, +which leads to a much faster series with constant sign: $$ - \log(2) = \sum_{k=1}^{+ \infty} \frac{1}{k \cdot 2^k} + \ln(2) = \sum_{k=1}^{+ \infty} \frac{1}{k \cdot 2^k} $$ -In this case the math siyplifies a lot: the ratio turns out to be $L = -1/2$ and therefore: +In this case the series ratio is found to be $L = 1/2$ and the error $$ - \Delta S = \frac{1}{k(k + 2) 2^{k-1}} + \Delta S_k = \frac{1}{k(k + 2) 2^{k-1}} $$ -Once computed the estimations of both logarithms and the series $A$, $B$ and -$C$ with the required precision, the program can be used to estimate $\gamma$ -up to whatever decimal digit. To give an idea of the time it takes to compute -it, 1000 digits were computed in \SI{0.63}{s}, with a \SI{3.9}{GHz} -2666mt al secondo -\textcolor{red}{Scrivere specifice pc che fa i conti} +Once the logarithms and the terms $A$, $B$ and $C$ have been computed @eq:faster +can finally be used to obtain $\gamma$ up to the given decimal place. + +The program was implemented with no particular care for performance but +was found to be relatively fast. On a \SI{3.9}{GHz} desktop computer with +\SI{2666}{MT\per\s} memory, it takes \SI{0.63}{\second} to compute the first +1000 digits. However, the quadratic complexity of the algorithm makes this +quite unpractical for computing more than a few thousands digits. For this +reason a more optimised algorithm was implemented. -### Faster implemented formula +### Optimised implementation {#sec:optimised} -When implemented, @eq:faster, which is the most fast known convergence formula, -is not fast at all. -The procedure can be accelerated by removing the $C(N)$ correction and keeping -only the two series $A(N)$ and $B(N)$. In this case, the equation can be -rewritten in a most convenient way [@brent00]: +The refined Brent-McMillan formula (@eq:faster) is theoretically the fastest +but is difficult to implement efficiently. In practice it turns out to be +slower than the standard formula even if its asymptotic error is better. + +The standard formula [@brent00] ignores the correction $C(N)$ and is rewritten +in a more convenient way: $$ - \gamma_k = \frac{U_k(N)}{V_k(N)} \with - U_k(N) = \sum_{j = 0}^k A_j(N) \et V_k(N) = \sum_{j = 0}^k B_j(N) + \gamma_k = \frac{\sum_{k = 0}^{k_\text{max}} A_k(N)} + {\sum_{k = 0}^{k_\text{max}} B_k(N)} $$ - where: +\begin{align*} + A_k &= \frac{1}{k} \left(\frac{A_{k-1} N^2}{k} + B_k \right) & A_0 &= - \ln(N) \\ + B_k &= \frac{B_{k-1} N^2}{k^2} & B_0 &= 1 \\ +\end{align*} + + +As said, the asymptotic error of the formula decreases slower $$ - A_j = \frac{1}{j} \left( \frac{A_{j-1} N^2}{j} + B_j \right) - \with A_0 = - \ln(N) + |\gamma - \gamma_k| \sim \pi e^{-4N} = \Delta_N $$ -and: +In order to avoid the expensive computation of $\ln(N)$, N was chosen to be a +power of 2, $N = 2^p$. As before: $$ - B_j = \frac{B_{j-1} N^2}{j^2} - \with B_0 = 1 + \Delta_p = \pi e^{-2^{p+2}} < 10^{-D} \thus + p > \log_2 \left( \ln(\pi) + D \ln(10) \right) - 2 $$ -This way, the error is given by: +and the smaller integer is found by taking the floor: $$ - |\gamma - \gamma_k| = \pi e^{-4N} + p = \left\lfloor \log_2 \left( \ln(\pi) + D \ln(10) \right) \right\rfloor + 1 $$ -which is greater with respect to the refined formula. Then, in order to avoid -computing the logarithm of a generic number $N$, instead of using @eq:NeD, a -number $N = 2^p$ of digits were computed, where $p$ was chosen to satisfy: -$$ - \pi e^{-4N} = \pi e^{-42^p} < 10^{-D} \thus - p > \log_2 \left[ \ln(\pi) + D \ln(10) \right] -2 -$$ +The number of terms to compute, $k_\text{max}$ is still proportional to $N$ +but by a different factor, $\beta$ such that: +\begin{align*} + \beta\ln(\beta) = 1 + \beta && + \beta = \exp(W(e^{-1}) + 1) \approx 3.59 +\end{align*} -and the smaller integer greater than it was selected, namely: -$$ - p = \left\lfloor \log_2 \left[ \ln(\pi) + D \ln(10) \right] \right\rfloor + 1 -$$ - -As regards the precision of the used numbers, this time the type `pmf_t` of GMP -was emploied. It consists of an arbitrary-precision floating point whose number -of bits is to be defined when the variable is initialized. +As regards the numbers representation, this time the GMP type `mpf_t` was +employed. It consists of an arbitrary-precision floating point with a fixed +exponent of 32 or 64 bits but arbitrary mantissa length. If a number $D$ of digits is required, the number of bits can be computed from: $$ 10^D = 2^b \thus b = D \log_2 (10) $$ -in order to the roundoff errors to not affect the final result, a security -range of 64 bits was added to $b$. +in order to avoid roundoff errors affecting the final result, a security range +of 64 bits was added to $b$. -Furthermore, the computation of the logarithm was made even faster by solving -@eq:delta instead -of finding $k^{\text{max}}$ by trials. +Furthermore, the computation of $\ln(2)$ was optimized by solving +@eq:delta analytically, instead of finding $k^{\text{max}}$ by trials. +The series estimation error was $$ - \Delta S = \frac{1}{k (k + 2) 2^{k - 1}} < 10^{-D} + \Delta S_k = \frac{1}{k (k + 2) 2^{k - 1}} < 10^{-D} $$ {#eq:delta} -This was accomplished as follow: +and the following condition should hold: $$ k (k + 2) 2^{k - 1} > 10^D \thus 2^{k - 1} [(k + 1)^2 - 1] > 10^D $$ -thus, for example, the inequality is satisfied for $k = x$ such that: +Asking for the equality with $k = x$ yields $$ 2^{x - 1} (x + 1)^2 = 10^D $$ - -In order to lighten the notation: $q := 10^D$, from which: +which can be again solved by the Lambert $W$ function as follows \begin{align*} - 2^{x - 1} (x + 1)^2 = q + 2^{x - 1} (x + 1)^2 = 10^D &\thus \exp \left( \ln(2) \frac{x - 1}{2} \right) (x + 1) - = \sqrt{q} \\ + = 10^{D/2} \\ &\thus \exp \left( \frac{\ln(2)}{2} (x + 1)\right) e^{- \ln(2)} (x + 1) - = \sqrt{q} \\ - &\thus \exp \left( \frac{\ln(2)}{2} (x + 1)\right) \frac{1}{2} (x + 1) - = \sqrt{q} \\ + = 10^{D/2} \\ &\thus \exp \left( \frac{\ln(2)}{2} (x + 1)\right) \frac{\ln(2)}{2} (x + 1) - = \ln (2) \sqrt{q} + = \ln (2) 10^{D/2} \end{align*} -Having rearranged the equation this way allows the use of the Lambert $W$ -function: +Taking $W$ both sides of the equation: $$ - W(x \cdot e^x) = x + \frac{\ln(2)}{2} (x + 1) = W ( \ln(2) 10^{D/2} ) $$ - -Hence, by computing the Lambert W function of both sides of the equation: -$$ - \frac{\ln(2)}{2} (x + 1) = W ( \ln(2) \sqrt{q} ) -$$ - which leads to: $$ - x = \frac{2}{\ln(2)} \, W ( \ln(2) \sqrt{q} ) - 1 + x = \frac{2}{\ln(2)} \, W ( \ln(2) 10^{D/2} ) - 1 $$ Finally, the smallest integer greater than $x$ can be found as: $$ k^{\text{max}} = - \left\lfloor \frac{2}{\ln(2)} \, W ( \ln(2) \sqrt{q} ) \right\rfloor - \with 10^D + \left\lfloor \frac{2}{\ln(2)} \, W (\ln(2) 10^{D/2}) \right\rfloor $$ -When a large number $D$ of digits is to be computed (but it also works for few -digits), the Lambert W function can be approximated by the first five terms of -its asymptotic value [@corless96]: +When a large number $D$ of digits is to be computed, the exponentiation can +easily overflow if working in double precision. However, $W$ grows like a +logarithm, so intuitively this operation could be optimized out. + +In fact this can be done by using the asymptotic expansion [@corless96] +at large $x$ of $W(x)$ : $$ W(x) = L_1 - L_2 + \frac{L_2}{L_1} + \frac{L_2 (L_2 - 2)}{2 L_1^2} - + \frac{L_2 (36L_2 - 22L_2^2 + 3L_2^3 -12)}{12L_1^4} + + \frac{L_2 (3L_2^3 - 22L_2^2 + 36L_2 - 12)}{12L_1^4} + + O\left(\left(\frac{L_2}{L_1}\right)^5\right) $$ - where $$ L_1 = \ln(x) \et L_2 = \ln(\ln(x)) $$ - -\textcolor{red}{Tempi del risultato e specifiche del pc usato} + +Now the usual properties of the logarithm can be applied to write: +\begin{align*} + L_1 &= \ln(\ln(2) 10^{D/2}) = \ln(\ln(2)) + \frac{D}{2}\ln(10) \\ + L_2 &= \ln(L_1) +\end{align*} + +In this way $k^{\text{max}}$ can be easily computed with standard double +precision floating points, with no risk of overflow or need of arbitrary +precision arithmetic. + + +### 1 million digits computation + +On the same desktop computer, the optimised program computes the first 1000 +digits of $\gamma$ in \SI{4.6}{\milli\second}: a $137\times$ improvement over +the previous program. This make it suitable for a more intensive computation. + +As a proof of concept, the program was then used to compute a million digits of +$\gamma$, which took \SI{1.26}{\hour} of running time. +The result was verified in \SI{6.33}{\hour} by comparing it with the +`mpmath.euler()` function from the mpmath [@mpmath13] multiple-precision +library with the Python integer backend. The library also uses the standard +Brent-McMillan algorithm and variable precision floating points but is not +based on GMP arithmetic types. diff --git a/notes/todo b/notes/todo index 59d7bd8..e69de29 100644 --- a/notes/todo +++ b/notes/todo @@ -1,4 +0,0 @@ -- aggiungere tempi di calcolo della γ - -simpy 1mln cifre: 6h 20' -fast 1mld cifre: 1h 37'