ex-2: review
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@ -10,13 +10,11 @@ $$
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\sum_{k=1}^{n} \frac{1}{k}
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- \ln(n) \right)
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$$ {#eq:gamma}
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and represents the limiting blue area in @fig:gamma. The first 30 digits of
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$\gamma$ are:
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$$
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\gamma = 0.57721\ 56649\ 01532\ 86060\ 65120\ 90082 \dots
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$$ {#eq:exact}
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In complex analysis, this constant is related to many functions and can be
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evaluated through a variety of identities. In this work, five methods were
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implemented and their results discussed. In fact, evaluating $\gamma$ with a
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@ -43,7 +41,6 @@ worse, namely:
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$$
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| \gamma(n_{i+1}) - \gamma | > | \gamma(n_i) - \gamma|
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$$
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and $\gamma (n_i)$ was selected as the best result (see @tbl:naive-errs).
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----------------------------
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@ -79,12 +76,11 @@ The convergence is logarithmic: to fix the first $d$ decimal places, about
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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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correct digits. In fact, 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, hence:
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$$
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2^{55} = 10^{d} \thus d = 55 \cdot \log(2) \sim 16.6
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$$
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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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@ -108,7 +104,6 @@ $$
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\gamma = \lim_{M \rightarrow + \infty} \sum_{k = 1}^{M}
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\binom{M}{k} \frac{(-1)^k}{k} \ln(\Gamma(k + 1))
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$$
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Varying $M$ from 1 to 100, the best result was obtained for $M = 41$ (see
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@tbl: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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@ -124,7 +119,7 @@ Table: Best estimation of $\gamma$ using
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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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number greater than 18 goes over 16 places and so cannot be correctly
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represented. Furthermore, the convergence is slowed down by the logarithmic
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factor.
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@ -137,13 +132,11 @@ $$
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\frac{1}{\Gamma(z)} = z e^{yz} \prod_{k = 1}^{+ \infty}
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\left( 1 + \frac{z}{k} \right) e^{-z/k}
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$$
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which gives:
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$$
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\gamma = - \frac{1}{z} \ln \left( z \Gamma(z)
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\prod_{k = 1}^{+ \infty} \left( 1 + \frac{z}{k} \right) e^{-z/k} \right)
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$$
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The execution stops when there is no difference between two consecutive 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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@ -154,31 +147,30 @@ terms of floating points). Different values of $z$ were checked, with $z_{i+1}
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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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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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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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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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17 \num{9.971e-9} 9.03 \num{9.264e-9}
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19 \num{1.008e-8} 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:recip-errs}
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Table: Some results found with the reciprocal $\Gamma$ function formula.
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The values on the left are shown to give an idea of the $z$ large-scale
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behaviour; on the right, the values around the best one ($z = 9.00$) are
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listed. {#tbl:recip-errs}
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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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@ -193,7 +185,10 @@ diff 0.00000 00084 82925
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Table: Third method results for z = 9.00. {#tbl:recip-res}
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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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to reach the $8^{th}$ decimal place.
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With respect to the definition, this formula returns a worse estimation of
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$\gamma$ because the calculation of the infinite product leads to more
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round-off instabilities than the series.
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### Fastest convergence formula
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@ -216,31 +211,26 @@ The asymptotic error of this estimation, given in [@demailly17], is:
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$$
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|\gamma_N - \gamma| \sim \frac{5 \sqrt{2 \pi}}{12 \sqrt{N}} e^{-8N} = \Delta_N
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$$ {#eq:NeD}
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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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digits of $\gamma$. This is done by imposing:
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$$
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\frac{5 \sqrt{2 \pi}}{12 \sqrt{N}} e^{-8N} < 10^{-D}
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\Delta_N < 10^{-D}
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$$
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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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\begin{align*}
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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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$$
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N = 1 + \left\lfloor \frac{1}{16}
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W \left( \frac{5 \pi}{9} 10^{2D + 1}\right) \right\rfloor
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$$ {#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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@ -265,9 +255,9 @@ 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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with arbitrary precision. For integers 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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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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@ -277,19 +267,18 @@ 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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exponentiation, particularly when computing more than a few hundreds digits. The
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details are explained in @sec:optimised, below.
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Once the number $N$ has been fixed, the series $A(N)$ and $B(N)$ are to be
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evaluated. With rational numbers, a different criterion for the truncation must
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be considered, since 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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$$
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\alpha\ln(\alpha) = 3 + \alpha \thus
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\alpha = \exp(W(3e) - 1) \approx 4.97
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\end{align*}
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$$
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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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@ -299,27 +288,23 @@ 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 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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$b = 2$ was chosen. If $N$ is a power of two, $N_0$ is 1 and only $\ln(2)$ is to
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be computed. More generally, the problem lies in 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()` exists in GMP). If the
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digits are
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$n$:
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digits are $n$:
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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 $N_0$ 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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The relation with the logarithm follows from the definition
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The relation with the logarithm follows from the definition:
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$$
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\text{tanh}(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}
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= \frac{e^{2x} - 1}{e^{2x} + 1}
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@ -329,38 +314,33 @@ $$
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\text{tanh} \left( \frac{\ln(z)}{2} \right) = \frac{z - 1}{z + 1}
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\thus \ln(z) = 2 \, \text{atanh} \left( \frac{z - 1}{z + 1} \right)
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$$
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The idea is to set $N_0 = z$ and define:
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$$
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y = \frac{N_0 - 1}{N_0 + 1}
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$$
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hence:
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$$
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\ln(N_0) = 2 \, \text{atanh} (y)
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= 2 \sum_{k = 0}^{+ \infty} \frac{y^{2k + 1}}{2k + 1}
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$$
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To estimate a series with a given precision some care must be taken:
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different techniques apply to different series and are explained in
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the paper [@riddle08]. In this case, letting $S$ be the value of the
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series and $S_k$ the $k$-th partial sum the following bounds can be found:
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To estimate a series with a given precision, some care must be taken: different
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techniques apply to different series and are explained in the paper [@riddle08].
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In this case, letting $S$ be the value of the series and $S_k$ the $k$-th
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partial sum, the following bounds can be found:
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$$
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S_k + a_k \frac{L}{1 -L} < S < S_k + \frac{a_{k+1}}{1 - \frac{a_{k+1}}{a_k}}
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$$
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where $a_k$ is the $k^{\text{th}}$ term of the series and $L$ is the limiting
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ratio of the series terms, which must be $\le 1$ in order for it to converge (in
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this case, it is easy to prove that $L = y^2 < 1$). The width $\Delta S$ of the
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interval containing $S$ gives the precision of the estimate $\tilde{S}$ if this
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last is assumed to be its middle value, namely:
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$$
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\tilde{S} = S_k + \frac{1}{2} \left(
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\tilde{S} = S_k + \frac{1}{2} \left(
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a_k \frac{L}{1 -L} + \frac{a_{k+1}}{1 - \frac{a_{k+1}}{a_k}}
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\right) \et
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\Delta S = \frac{a_{k+1}}{1 - \frac{a_{k+1}}{a_k}} - a_k \frac{L}{1 -L}
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$$
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For this series:
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\begin{align*}
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a_k = \frac{y^{2k + 1}}{2k + 1} &&
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@ -377,7 +357,6 @@ $$
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\frac{1}{\frac{2k + 3}{2k + 1}y^{-2} - 1} - \frac{y^2}{1 - y^2}
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\right)
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$$
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By imposing $\Delta S_k < 10^{-D}$, where $D$ is the number decimal places
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required, $k^{\text{max}}$ at which to stop the summation can be obtained. This
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is achieved by trials, checking for every $k$ whether $\Delta S_k$ is less or
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@ -390,21 +369,20 @@ which leads to a much faster series with constant sign:
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$$
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\ln(2) = \sum_{k=1}^{+ \infty} \frac{1}{k \cdot 2^k}
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$$
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In this case the series ratio is found to be $L = 1/2$ and the error
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In this case the series ratio is found to be $L = 1/2$ and the error:
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$$
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\Delta S_k = \frac{1}{k(k + 2) 2^{k-1}}
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$$
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Once the logarithms and the terms $A$, $B$ and $C$ have been computed @eq:faster
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can finally be used to obtain $\gamma$ up to the given decimal place.
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Once the logarithms and the terms $A$, $B$ and $C$ have been computed,
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@eq:faster can finally be used to obtain $\gamma$ up to the given decimal
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place.
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The program was implemented with no particular care for performance but
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was found to be relatively fast. On a \SI{3.9}{GHz} desktop computer with
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\SI{2666}{MT\per\s} memory, it takes \SI{0.63}{\second} to compute the first
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1000 digits. However, the quadratic complexity of the algorithm makes this
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quite unpractical for computing more than a few thousands digits. For this
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reason a more optimised algorithm was implemented.
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1000 digits. However, this makes it quite unpractical for computing more than
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a few thousands digits. For this reason, a more optimised algorithm was
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implemented.
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### Optimised implementation {#sec:optimised}
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@ -413,43 +391,41 @@ The refined Brent-McMillan formula (@eq:faster) is theoretically the fastest
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but is difficult to implement efficiently. In practice it turns out to be
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slower than the standard formula even if its asymptotic error is better.
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The standard formula [@brent00] ignores the correction $C(N)$ and is rewritten
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in a more convenient way:
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The standard formula [@brent00] ignores the correction $C(N)$ and can be
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rewritten in a more convenient way:
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$$
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\gamma_k = \frac{\sum_{k = 0}^{k_\text{max}} A_k(N)}
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{\sum_{k = 0}^{k_\text{max}} B_k(N)}
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$$
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where:
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\begin{align*}
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A_k &= \frac{1}{k} \left(\frac{A_{k-1} N^2}{k} + B_k \right) & A_0 &= - \ln(N) \\
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B_k &= \frac{B_{k-1} N^2}{k^2} & B_0 &= 1 \\
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A_k &= \frac{1}{k} \left(\frac{A_{k-1} N^2}{k} + B_k \right)
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& A_0 &= - \ln(N) \\
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B_k &= \frac{B_{k-1} N^2}{k^2}
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& B_0 &= 1 \\
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\end{align*}
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As said, the asymptotic error of the formula decreases slower
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As said, the asymptotic error of the formula decreases slower:
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$$
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|\gamma - \gamma_k| \sim \pi e^{-4N} = \Delta_N
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$$
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In order to avoid the expensive computation of $\ln(N)$, N was chosen to be a
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power of 2, $N = 2^p$. As before:
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$$
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\Delta_p = \pi e^{-2^{p+2}} < 10^{-D} \thus
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p > \log_2 \left( \ln(\pi) + D \ln(10) \right) - 2
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$$
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and the smaller integer is found by taking the floor:
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$$
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p = \left\lfloor \log_2 \left( \ln(\pi) + D \ln(10) \right) \right\rfloor + 1
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$$
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The number of terms to compute, $k_\text{max}$ is still proportional to $N$
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but by a different factor, $\beta$ such that:
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The number of terms to compute, $k_\text{max}$, is still proportional to $N$
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but by a different factor $\beta$ such that:
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\begin{align*}
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\beta\ln(\beta) = 1 + \beta &&
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\beta = \exp(W(e^{-1}) + 1) \approx 3.59
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\end{align*}
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As regards the numbers representation, this time the GMP type `mpf_t` was
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employed. It consists of an arbitrary-precision floating point with a fixed
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exponent of 32 or 64 bits but arbitrary mantissa length.
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@ -457,27 +433,24 @@ If a number $D$ of digits is required, the number of bits can be computed from:
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$$
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10^D = 2^b \thus b = D \log_2 (10)
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$$
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in order to avoid roundoff errors affecting the final result, a security range
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In order to avoid roundoff errors affecting the final result, a security range
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of 64 bits was added to $b$.
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Furthermore, the computation of $\ln(2)$ was optimized by solving
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@eq:delta analytically, instead of finding $k^{\text{max}}$ by trials.
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The series estimation error was
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The series estimation error was:
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$$
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\Delta S_k = \frac{1}{k (k + 2) 2^{k - 1}} < 10^{-D}
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$$ {#eq:delta}
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and the following condition should hold:
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$$
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k (k + 2) 2^{k - 1} > 10^D \thus 2^{k - 1} [(k + 1)^2 - 1] > 10^D
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$$
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Asking for the equality with $k = x$ yields
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Asking for the equality, with $k = x$ it yields to:
|
||||
$$
|
||||
2^{x - 1} (x + 1)^2 = 10^D
|
||||
$$
|
||||
which can be again solved by the Lambert $W$ function as follows
|
||||
which can be again solved by the Lambert $W$ function as follows:
|
||||
\begin{align*}
|
||||
2^{x - 1} (x + 1)^2 = 10^D
|
||||
&\thus \exp \left( \ln(2) \frac{x - 1}{2} \right) (x + 1)
|
||||
@ -487,7 +460,6 @@ which can be again solved by the Lambert $W$ function as follows
|
||||
&\thus \exp \left( \frac{\ln(2)}{2} (x + 1)\right) \frac{\ln(2)}{2} (x + 1)
|
||||
= \ln (2) 10^{D/2}
|
||||
\end{align*}
|
||||
|
||||
Taking $W$ both sides of the equation:
|
||||
$$
|
||||
\frac{\ln(2)}{2} (x + 1) = W ( \ln(2) 10^{D/2} )
|
||||
@ -496,50 +468,45 @@ which leads to:
|
||||
$$
|
||||
x = \frac{2}{\ln(2)} \, W ( \ln(2) 10^{D/2} ) - 1
|
||||
$$
|
||||
|
||||
Finally, the smallest integer greater than $x$ can be found as:
|
||||
$$
|
||||
k^{\text{max}} =
|
||||
\left\lfloor \frac{2}{\ln(2)} \, W (\ln(2) 10^{D/2}) \right\rfloor
|
||||
$$
|
||||
|
||||
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 $W(x)$ :
|
||||
logarithm, so intuitively this operation could be optimized out.
|
||||
This can be done by using the asymptotic expansion [@corless96] at large $x$ of
|
||||
$W(x)$ :
|
||||
$$
|
||||
W(x) = L_1 - L_2 + \frac{L_2}{L_1} + \frac{L_2 (L_2 - 2)}{2 L_1^2}
|
||||
+ \frac{L_2 (3L_2^3 - 22L_2^2 + 36L_2 - 12)}{12L_1^4}
|
||||
+ O\left(\left(\frac{L_2}{L_1}\right)^5\right)
|
||||
$$
|
||||
where
|
||||
where:
|
||||
$$
|
||||
L_1 = \ln(x) \et L_2 = \ln(\ln(x))
|
||||
$$
|
||||
|
||||
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
|
||||
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.
|
||||
On the same desktop computer described before, 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 makes 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.
|
||||
$\gamma$, 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.
|
||||
This library uses the standard Brent-McMillan algorithm and variable precision
|
||||
floating points too but is not based on GMP arithmetic types.
|
||||
|
||||
Loading…
Reference in New Issue
Block a user