312 lines
11 KiB
Markdown
312 lines
11 KiB
Markdown
# Exercise 2
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## Euler-Mascheroni constant
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The Euler-Mascheroni constant is defined as the limiting difference between the
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partial sums of the harmonic series and the natural logarithm:
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$$
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\gamma = \lim_{n \rightarrow +\infty} \left( \sum_{k=1}^{n} \frac{1}{k}
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- \ln(n) \right)
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$$ {#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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limited precision due to floating points number representation entails limited
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precision on the estimation of $\gamma$ itself due to roundoff errors. All the
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known methods involve sums, subtractions or products of very big or small
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numbers, packed in series, partial sums or infinite products. Thus, the
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efficiency of the methods lies on how quickly they converge to their limit.
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{#fig:gamma width=7cm}
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## Computing the constant
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### Definition
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First, in order to have a quantitative idea of how hard it is to reach a good
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estimation of $\gamma$, it was naively computed using the definition given in
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@eq:gamma. The difference was computed for increasing value of $n$, with
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$n_{i+1} = 10 \cdot n_i$ and $n_1 = 20$, till the approximation starts getting
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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 result (see @tbl:1_results).
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-----------------------------------------------
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n sum $|\gamma(n)-\gamma|$
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----------- ------------- ---------------------
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\SI{2e1}{} \SI{2.48e-02}{}
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\SI{2e2}{} \SI{2.50e-03}{}
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\SI{2e3}{} \SI{2.50e-04}{}
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\SI{2e4}{} \SI{2.50e-05}{}
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\SI{2e5}{} \SI{2.50e-06}{}
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\SI{2e6}{} \SI{2.50e-07}{}
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\SI{2e7}{} \SI{2.50e-08}{}
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\SI{2e8}{} \SI{2.50e-09}{}
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\SI{2e9}{} \SI{2.55e-10}{}
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\SI{2e10}{} \SI{2.42e-11}{}
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\SI{2e11}{} \SI{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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The convergence is logarithmic: to fix the first $d$ decimal places about
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$10^d$ terms are needed. The double precision runs out at the
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10\textsuperscript{th} place, $n=\SI{2e10}{}$.
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Since all the number are given with double precision, there can be at best 15
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correct digits but only 10 were correctly computed: this means that when the
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terms of the series start being smaller than the smallest representable double,
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the sum of all the remaining terms give a number $\propto 10^{-11}$.
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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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Table: First method results. {#tbl:first}
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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:
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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.
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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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Table: Second method results. {#tbl:second}
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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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represented. Furthermore, the convergence (even if this is not a series
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and, consequently, it is not properly a "convergence") is slowed down by
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the logarithmic factor.
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### Reciprocal $\Gamma$ function
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A better result was found using the well known reciprocal $\Gamma$ function
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formula:
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$$
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\frac{1}{\Gamma(z)} = z e^{yz} \prod_{k = 1}^{+ \infty}
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\left( 1 + \frac{z}{k} \right) e^{-z/k}
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$$
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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 therms
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of the infinite product (it happens for $k = 456565794$, meaning that for this
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value of $k$, the term of the product is equal to 1). Different values of $z$
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were checked, with $z_{i+1} = z_i + 0.01$, ranging from 0 to 20 and the best
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result was found for $z = 9$. As can be seen in @tbl:3_results, it's only by
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chance, since all $|\gamma(z) - \gamma |$ are of the same order of magnitude.
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The best one is compared with the exact value of $\gamma$ in @tbl:third.
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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 \SI{9.712e-9}{} 8.95 \SI{9.770e-9}{}
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3 \SI{9.320e-9}{} 8.96 \SI{9.833e-9}{}
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5 \SI{9.239e-9}{} 8.97 \SI{9.622e-9}{}
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7 \SI{9.391e-9}{} 8.98 \SI{9.300e-9}{}
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9 \SI{8.482e-9}{} 8.99 \SI{9.059e-9}{}
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11 \SI{9.185e-9}{} 9.00 \SI{8.482e-9}{}
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13 \SI{9.758e-9}{} 9.01 \SI{9.564e-9}{}
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15 \SI{9.747e-9}{} 9.02 \SI{9.260e-9}{}
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17 \SI{9.971e-9}{} 9.03 \SI{9.264e-9}{}
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19 \SI{10.084e-9}{} 9.04 \SI{9.419e-9}{}
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---------------------------------------------------------------
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Table: Differences between the obtained values of $\gamma$ and the exact
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one. {#tbl:3_results}
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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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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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### Fastest convergence formula
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The fastest known convergence belongs to the following formula:
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(source: http://www.numberworld.org/y-cruncher/internals/formulas.html):
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$$
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\gamma = \frac{A(N)}{B(N)} -\frac{C(N)}{B^2(N)} - \ln(N)
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$$ {#eq:faster}
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with:
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\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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&B(N) = \sum_{k=1}^{+ \infty} \frac{N^k}{k!} \\
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&C(N) = \frac{1}{4N} \sum_{k=0}^{2N}
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\frac{((2k)!)^3}{(k!)^4 \cdot (16k)^2k} \\
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\end{align*}
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The series $A$ and $B$ are computed till there is no difference between two
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consecutive terms.
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The number of desired correct decimals is given in input and $N$ is
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consequently computed through a formula given in the same article
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above-mentioned. Results are shown in @tbl:fourth.
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--------- ------------------------------
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true: 0.57721\ 56649\ 01532\ 75452
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approx: 0.57721\ 56649\ 01532\ 86554
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diff: 0.00000\ 00000\ 00000\ 11102
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--------- ------------------------------
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Table: Fourth method results. {#tbl:fourth}
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Due to roundoff errors, the best results is obtained for $N = 10$. Since up to
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15 places were correctly computed, an approximation of $\gamma$ better than the
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one reached with the definition in @eq:gamma was obtained.
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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 (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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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 has been implemented. Unlike the previously
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mentioned programs, this one was more carefully optimized.
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The $A$ and $B$ series are computed up to an arbitrary limit $k_{\text{max}}$.
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Different values of $k_{\text{max}}$ were tested but, obviously, they all
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eventually reach a point where the approximation cannot guarantee the
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requested precision; the solution turned out to be to let $k_{\text{max}}$ depends on
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$N$. Consequently, $k_{\text{max}}$ was chosen to be $5N$, after it has been
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verified to produce the correct digits up to 500 decimal places.
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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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First, it should be noted that the logarithm of only some special numbers can
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be computed with arbitrary precision, namely the ones of which a converging
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series is known. 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 $b = 2$ was chosen. As
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well as for the scientific notation, in order to get the mantissa $1 \leqslant
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N_0 < 2$, the number of binary digits of $N$ must be computed (conveniently, a
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dedicated function `mpz_sizeinbase()` can be found in GMP). If the digits are $n$:
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$$
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e = b - 1 \thus N_0 = \frac{N}{2^{n - 1}}
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$$
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Then, by defining:
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$$
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N_0 = \frac{1 + y}{1 - y} \thus y = \frac{N_0 - 1}{N_0 + 1} < 1
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$$
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and the following series (which is convergent for $y < 1$) can therefore be
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used:
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$$
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\ln \left( \frac{1 + y}{1 - y} \right) =
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2 \sum_{k = 0}^{+ \infty} \frac{y^{2k + 1}}{2k + 1}
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$$
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But when to stop computing the series?
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Given a partial sum $S_k$ of the series, it is possible to know when a digit is
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definitely correct, meaning that no matter how large $k$ can be,
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it will not affect that decimal place. The key lies in the following concept.
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Letting $S$ the value of the series:
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$$
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S_k + \frac{a_{k+1}}{1 - \frac{a_{k+1}}{a_k}} < S < S_k + a_k \frac{L}{1 -L}
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$$
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where $L$ is the limiting ratio of the series terms, which must be $< 1$ in
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order for it to converge (in this case, it is easy to prove that $L = y^2$).
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The width $\Delta S$ of the interval containing $S$, for a given $S_k$, gives
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the precision of the partial sum with respect to $S$. By imposing $\Delta S <
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1/10^D$ where $D$ is the correct decimal place required, the desired precision
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is obtained. This is achieved by trials, checking at every step whether $\Delta
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S$ is less or greater than $1/10^D$.
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The same holds for $\ln(2)$, which is given by the following series:
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$$
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\log(2) = \sum_{k=1}^{+ \infty} \frac{1}{k \cdot 2^k}
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$$
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In this case the ratio is $L = 1/2$.
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