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
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@ -4,7 +4,7 @@
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journal={Journal of mathematics and physics},
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volume={20},
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number={1-4},
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pages={224 -- 230},
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pages={224--230},
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year={1941},
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publisher={Wiley Online Library}
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}
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@ -198,3 +198,22 @@
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computation of Euler's constant},
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year={2017}
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}
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@manual{mpmath13,
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key={mpmath},
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author={Fredrik Johansson and others},
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title={mpmath: a {P}ython library for arbitrary-precision floating-point arithmetic (version 0.18)},
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note={{\tt http://mpmath.org/}},
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month={December},
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year={2013},
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}
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@manual{gmp20,
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key={gmp},
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author={Torbjörn Granlund and the GMP development team},
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title={GNU MP: The GNU Multiple Precision Arithmetic Library,
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Edition 6.2.0 (version 0.18)},
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note={{\tt https://gmplib.org/}},
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month={January},
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year={2020},
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}
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@ -43,40 +43,40 @@ $$
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| \gamma(n_{i+1}) - \gamma | > | \gamma(n_i) - \gamma|
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$$
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and $\gamma (n_i)$ was selected as the best result (see @tbl:1_results).
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and $\gamma (n_i)$ was selected as the best result (see @tbl:naive-errs).
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---------------------------------------------
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n $|\gamma(n)-\gamma|$
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---------------------- ----------------------
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\num{2e1} \num{2.48e-02}
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----------------------------
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n $|γ(n)-γ|$
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----------- ----------------
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\num{2e1} \num{2.48e-02}
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\num{2e2} \num{2.50e-03}
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\num{2e2} \num{2.50e-03}
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\num{2e3} \num{2.50e-04}
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\num{2e3} \num{2.50e-04}
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\num{2e4} \num{2.50e-05}
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\num{2e4} \num{2.50e-05}
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\num{2e5} \num{2.50e-06}
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\num{2e5} \num{2.50e-06}
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\num{2e6} \num{2.50e-07}
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\num{2e6} \num{2.50e-07}
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\num{2e7} \num{2.50e-08}
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\num{2e7} \num{2.50e-08}
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\num{2e8} \num{2.50e-09}
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\num{2e8} \num{2.50e-09}
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\num{2e9} \num{2.55e-10}
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\num{2e9} \num{2.55e-10}
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\num{2e10} \num{2.42e-11}
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\num{2e10} \num{2.42e-11}
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\num{2e11} \num{1.44e-08}
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---------------------------------------------
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\num{2e11} \num{1.44e-08}
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----------------------------
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Table: Partial results using the definition of $\gamma$ with double
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precision. {#tbl:1_results}
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precision. {#tbl:naive-errs}
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The convergence is logarithmic: to fix the first $d$ decimal places, about
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$10^d$ terms of the armonic series are needed. The double precision runs out at
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the $10^{\text{th}}$ place, at $n=\num{2e10}$.
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$10^d$ terms of the harmonic series are needed. The double precision runs out
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at the $10^{\text{th}}$ place, at $n=\num{2e10}$.
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Since all the number are given with double precision, there can be at best 16
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correct digits, since for a double 64 bits are allocated in memory: 1 for the
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sign, 8 for the exponent and 55 for the mantissa:
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@ -87,43 +87,40 @@ $$
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Only 10 digits were correctly computed: this means that when the terms of the
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series start being smaller than the smallest representable double, the sum of
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all the remaining terms gives a number $\propto 10^{-11}$. The best result is
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shown in @tbl:first.
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shown in @tbl:naive-res.
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--------- -----------------------
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true: 0.57721\ 56649\ 01533
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approx: 0.57721\ 56648\ 77325
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diff: 0.00000\ 00000\ 24207
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--------- -----------------------
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------- --------------------
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exact 0.57721 56649 01533
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approx 0.57721 56648 77325
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diff 0.00000 00000 24207
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------- --------------------
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Table: First method best result. From the top down: true value, best estimation
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and difference between them. {#tbl:first}
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and difference between them. {#tbl:naive-res}
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### Alternative formula
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As a first alternative, the constant was computed through the identity which
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relates $\gamma$ to the $\Gamma$ function as follow [@davis59]:
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As a first alternative, the constant was computed through the identity
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[@davis59] which relates $\gamma$ to the $\Gamma$ function as follow:
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$$
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\gamma = \lim_{M \rightarrow + \infty} \sum_{k = 1}^{M}
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\binom{M}{k} \frac{(-1)^k}{k} \ln(\Gamma(k + 1))
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$$
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Varying $M$ from 1 to 100, the best result was obtained for $M = 41$ (see
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@tbl:second). It went sour: the convergence is worse than using the definition
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itself. Only two places were correctly computed (@tbl:second).
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@tbl:limit-res). This approximation gave an underwhelming result: the
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convergence is actually worse than the definition itself. Only two decimal
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places were correctly computed:
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--------- -----------------------
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true: 0.57721\ 56649\ 01533
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approx: 0.57225\ 72410\ 34058
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diff: 0.00495\ 84238\ 67473
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--------- -----------------------
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-------- --------------------
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exact 0.57721 56649 01533
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approx 0.57225 72410 34058
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diff 0.00495 84238 67473
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-------- --------------------
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Table: Best estimation of $\gamma$ using
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the alternative formula. {#tbl:second}
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the alternative formula. {#tbl:limit-res}
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Here, the problem lies in the binomial term: computing the factorial of a
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number greater than 18 goes over 15 places and so cannot be correctly
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@ -146,69 +143,66 @@ $$
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\prod_{k = 1}^{+ \infty} \left( 1 + \frac{z}{k} \right) e^{-z/k} \right)
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$$
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The execution stops when there is no difference between two consecutive therms
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The execution stops when there is no difference between two consecutive terms
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of the infinite product (it happens for $k = 456565794 \sim \num{4.6e8}$,
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meaning that for this value of $k$ the term of the product is equal to 1 in
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terms of floating points). Different values of $z$ were checked, with $z_{i+1}
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= z_i + 0.01$ ranging from 0 to 20, and the best result was found for $z = 9$.
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---------------------------------------------------------------
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z $|\gamma(z) - \gamma |$ z $|\gamma(z) - \gamma |$
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----- ------------------------ ------ ------------------------
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1 \num{9.712e-9} 8.95 \num{9.770e-9}
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-----------------------------------------------
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z $|γ(z) - γ|$ z $|γ(z) - γ|$
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----- ---------------- ------ ----------------
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1 \num{9.712e-9} 8.95 \num{9.770e-9}
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3 \num{9.320e-9} 8.96 \num{9.833e-9}
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3 \num{9.320e-9} 8.96 \num{9.833e-9}
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5 \num{9.239e-9} 8.97 \num{9.622e-9}
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5 \num{9.239e-9} 8.97 \num{9.622e-9}
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7 \num{9.391e-9} 8.98 \num{9.300e-9}
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7 \num{9.391e-9} 8.98 \num{9.300e-9}
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9 \num{8.482e-9} 8.99 \num{9.059e-9}
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9 \num{8.482e-9} 8.99 \num{9.059e-9}
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11 \num{9.185e-9} 9.00 \num{8.482e-9}
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11 \num{9.185e-9} 9.00 \num{8.482e-9}
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13 \num{9.758e-9} 9.01 \num{9.564e-9}
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13 \num{9.758e-9} 9.01 \num{9.564e-9}
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15 \num{9.747e-9} 9.02 \num{9.260e-9}
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15 \num{9.747e-9} 9.02 \num{9.260e-9}
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17 \num{9.971e-9} 9.03 \num{9.264e-9}
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17 \num{9.971e-9} 9.03 \num{9.264e-9}
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19 \num{10.084e-9} 9.04 \num{9.419e-9}
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---------------------------------------------------------------
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19 \num{10.084e-9} 9.04 \num{9.419e-9}
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-----------------------------------------------
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Table: Differences between some obtained values of $\gamma$ and
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the exact one found with the reciprocal $\Gamma$ function formula.
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The values on the left are shown to give an idea of the $z$
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large-scale behaviour; on the right, the values around the best
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one ($z = 9.00$) are listed. {#tbl:3_results}
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one ($z = 9.00$) are listed. {#tbl:recip-errs}
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As can be seen in @tbl:3_results, the best value for $z$ is only by chance,
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As can be seen in @tbl:recip-errs, the best value for $z$ is only by chance,
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since all $|\gamma(z) - \gamma |$ are of the same order of magnitude. The best
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one is compared with the exact value of $\gamma$ in @tbl:third.
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one is compared with the exact value of $\gamma$ in @tbl:recip-res.
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--------- -----------------------
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true: 0.57721\ 56649\ 01533
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------- --------------------
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true 0.57721 56649 01533
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approx 0.57721 56564 18607
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diff 0.00000 00084 82925
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------- --------------------
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approx: 0.57721\ 56564\ 18607
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Table: Third method results for z = 9.00. {#tbl:recip-res}
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diff: 0.00000\ 00084\ 82925
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--------- -----------------------
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Table: Third method results for z = 9.00. {#tbl:third}
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This time, the convergence of the infinite product is fast enough to ensure the
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$8^{th}$ place.
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In this approximation, the convergence of the infinite product is fast enough
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to reach the $8^{th}$ decimal place.
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### Fastest convergence formula
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The fastest known convergence belongs to the following formula, known as refined
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Brent-McMillan formula[@yee19]:
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The fastest known convergence to $\gamma$ belongs to the following formula,
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known as refined Brent-McMillan formula [@yee19]:
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$$
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\gamma_N = \frac{A(N)}{B(N)} -\frac{C(N)}{B^2(N)} - \ln(N)
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$$ {#eq:faster}
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with:
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where
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\begin{align*}
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&A(N) = \sum_{k=1}^{+ \infty} \frac{N^k}{k!} \cdot H(k)
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\with H(k) = \sum_{j=1}^{k} \frac{1}{j} \\
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@ -217,95 +211,119 @@ with:
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\frac{((2k)!)^3}{(k!)^4 \cdot (16k)^2k} \\
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\end{align*}
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and the difference between $\gamma_N$ and $\gamma$ is given by [@demailly17]:
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The asymptotic error of this estimation, given in [@demailly17], is:
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$$
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|\gamma_N - \gamma| < \frac{5 \sqrt{2 \pi}}{12 \sqrt{N}} e^{-8N}
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|\gamma_N - \gamma| \sim \frac{5 \sqrt{2 \pi}}{12 \sqrt{N}} e^{-8N} = \Delta_N
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$$ {#eq:NeD}
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This gives the value of $N$ which is to be used into the formula in order to get
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$D$ correct decimal digits of $\gamma$. In fact, by imposing:
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The error bound gives the value of $N$ to be used to get $D$ correct decimal
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digits of $\gamma$. In fact, this is done by imposing:
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$$
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\frac{5 \sqrt{2 \pi}}{12 \sqrt{N}} e^{-8N} < 10^{-D}
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$$
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With a bit of math, it can be found that the smallest integer which satisfies
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the inequality is given by:
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The inequality with the equal sign can be solved with the use of the Lambert
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$W$ function. For a real number $x$, $W(x)$ is defined as the inverse function
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of $x\exp(x)$, so the idea is to recast the equation into this form and take
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$W$ both sides.
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\begin{align*}{3}
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\frac{5 \sqrt{2 \pi}}{12 \sqrt{x}} e^{-8x} = 10^{-D}
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& \thus \left(\frac{12}{5}\right)^2 \frac{x}{2 \pi} e^{16x} = 10^{2D} \\
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& \thus 16x e^{16x} = \left(\frac{5 \pi}{9}\right)^2 10^{2D + 1} \\
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& \thus x = \frac{1}{16} W\left(\left(\frac{5 \pi}{9}\right)^2 10^{2D + 1}\right)
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\end{align*}
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The smallest integer which satisfies the inequality is then
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$$
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N = 1 + \left\lfloor \frac{1}{16}
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W \left( \frac{5 \pi}{9} 10^{2D + 1}\right) \right\rfloor
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$$
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$$ {#eq:refined-n}
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The series $A$ and $B$ were computed till there is no difference between two
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consecutive terms. Results are shown in @tbl:fourth.
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--------- ------------------------------
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true: 0.57721\ 56649\ 01532\ 75452
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------- --------------------------
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exact 0.57721 56649 01532 75452
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approx 0.57721 56649 01532 53248
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diff 0.00000 00000 00000 33062
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------- --------------------------
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approx: 0.57721\ 56649\ 01532\ 53248
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Table: $\gamma$ estimation with the fastest known convergence formula
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(@eq:faster). {#tbl:fourth}
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diff: 0.00000\ 00000\ 00000\ 33062
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--------- ------------------------------
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Table: $\gamma$ estimation with the fastest
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known convergence formula (@eq:faster). {#tbl:fourth}
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Because of roundoff errors, the best result was obtained only for $D = 15$, for
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Due to roundoff errors, the best result was obtained only for $D = 15$, for
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which the code accurately computes 15 digits and gives an error of
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\num{3.3e16}. For $D > 15$, the requested can't be fulfilled.
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### Arbitrary precision
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To overcome the issues related to the double representation, one can resort to a
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representation with arbitrary precision. In the GMP library (which stands for
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GNU Multiple Precision arithmetic), for example, real numbers can be
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approximated by a ratio of two integers (a fraction) with arbitrary precision:
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this means a check is performed on every operation and in case of an integer
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overflow, additional memory is requested and used to represent the larger result.
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Additionally, the library automatically switches to the optimal algorithm
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to compute an operation based on the size of the operands.
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To overcome the issues related to the limited precision of the machine floating
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points, one can resort to a software implementation with arbitrary precision.
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In the GMP [@gmp20] library (GNU Multiple Precision arithmetic), for
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example, reals can be approximated by a rational or floating point numbers
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with arbitrary precision. For integer and fractions this means a check is
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performed on every operation that can overflow and additional memory is
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requested and used to represent the larger result. For floating points it
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means the size of the mantissa can be chosen before initialising the number.
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Additionally, the library automatically switches to the optimal algorithm to
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compute an operation based on the size of the operands.
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The terms in @eq:faster can therefore be computed with arbitrarily large
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precision. Thus, a program that computes the Euler-Mascheroni constant within
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a user controllable precision was implemented.
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precision. Thus, a program that computes the Euler-Mascheroni constant within a
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user controllable precision was implemented.
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To compute $N$ by @eq:refined-n, a trick must be used to avoid overflows in the
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exponentiation, particularly when computing more than a few hundreds digits are
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to be computed. The details are explained in @sec:optimised, below. Once the
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number $N$ has been fixed, the series $A(N)$ and $B(N)$ are to be evaluated.
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With rational numbers, a different criterion for the truncation must be
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considered, because two consecutive term are always different. Brent and
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McMillan[@brent00] prove that to reduce the partial sum to less than $\Delta_N$
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it is sufficient to compute $\alpha N$ terms, where $\alpha$ is the solution to
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the equation:
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\begin{align*}
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\alpha\ln(\alpha) = 3 + \alpha &&
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\alpha = \exp(W(3e) - 1) \approx 4.97
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\end{align*}
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According to [@brent00], the $A(N)$, $B(N)$ and $C(N)$ series were computed up
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to $k_{\text{max}} = 5N$ \textcolor{red}{sure?}, since it guarantees the accuracy
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od the result up to the $D$ decimal digit.
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The GMP library offers functions to perform some operations such as addition,
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multiplication, division, etc. However, the logarithm function is not
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implemented. Thus, most of the code carries out the $\ln(N)$ computation.
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This is because the logarithm of only some special numbers can be computed
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The reason is that the logarithm of only some special numbers can be computed
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with arbitrary precision, namely the ones of which a converging series is known.
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This forces $N$ to be rewritten in the following way:
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$$
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N = N_0 \cdot b^e \thus \ln(N) = \ln(N_0) + e \cdot \ln(b)
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$$
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Since a fast converging series for $\ln(2)$ is known (it will be shwn shortly),
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$b = 2$ was chosen. In case $N$ is a power of two, only $\ln(2)$ is to be
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computed. If not, it must be handled as follows.
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As well as for the scientific notation, in order to get the mantissa $1
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Since a fast converging series for $\ln(2)$ is known (it will be shown shortly),
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$b = 2$ was chosen. If $N$ is a power of two, $N_0$ is 1 and only $\ln(2)$ is to be
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computed. More generally, the problem reduces to the calculation of $\ln(N_0)$.
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Regarding the scientific notation, to find the mantissa $1
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\leqslant N_0 < 2$, the number of binary digits of $N$ must be computed
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(conveniently, a dedicated function `mpz_sizeinbase()` is found in GMP). If the
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(conveniently, a dedicated function `mpz_sizeinbase()` exists in GMP). If the
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digits are
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$n$:
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$$
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e = n - 1 \thus N = N_0 \cdot 2^{n-1} \thus N_0 = \frac{N}{2^{n - 1}}
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$$
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The logarithm of whichever number $N_0$ can be computed using the series of
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$\text{atanh}(x)$, which converges for $|x| < 1$:
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The logarithm of $N_0$ can be computed from the Taylor series of the hyperbolic
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tangent, which is convergent for $|x| < 1$:
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$$
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\text{atanh}(x) = \sum_{k = 0}^{+ \infty} \frac{x^{2k + 1}}{2x + 1}
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$$
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In fact:
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The relation with the logarithm follows from the definition
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$$
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\text{tanh}(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}
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= \frac{e^{2x} - 1}{e^{2x} + 1}
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$$
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|
||||
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.
|
||||
|
||||
@ -1,4 +0,0 @@
|
||||
- aggiungere tempi di calcolo della γ
|
||||
|
||||
simpy 1mln cifre: 6h 20'
|
||||
fast 1mld cifre: 1h 37'
|
||||
Loading…
Reference in New Issue
Block a user