ex-2: written about the program fast.c
fancy, fancier and fast are now correctly described in the pdf. Only the computation time is missing.
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@ -179,3 +179,22 @@
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year={2000},
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publisher={Springer}
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}
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@article{corless96,
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title={On the Lambert W function},
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author={Corless, Robert M and Gonnet, Gaston H and Hare, David EG and Jeffrey,
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David J and Knuth, Donald E},
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journal={Advances in Computational mathematics},
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volume={5},
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number={1},
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pages={329--359},
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year={1996},
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publisher={Springer}
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}
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@article{demailly17,
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author={Jean-Pierre Demailly},
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title={Precise error estimate of the Brent-McMillan algorithm for the
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computation of Euler's constant},
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year={2017}
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}
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@ -196,9 +196,10 @@ $8^{th}$ place.
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### Fastest convergence formula
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The fastest known convergence belongs to the following formula [@yee19]:
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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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$$
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\gamma = \frac{A(N)}{B(N)} -\frac{C(N)}{B^2(N)} - \ln(N)
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\gamma_N = \frac{A(N)}{B(N)} -\frac{C(N)}{B^2(N)} - \ln(N)
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$$ {#eq:faster}
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with:
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@ -210,36 +211,48 @@ with:
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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 $B$ were computed till there is no difference between two
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consecutive terms. The number of desired correct decimals $D$ was given in
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input and $N$ was consequently computed through the formula:
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and the difference between $\gamma_N$ and $\gamma$ is given by [@demailly17]:
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$$
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N = \left\lfloor 2 + \frac{1}{4} \cdot \ln(10) \cdot D \right\rfloor
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|\gamma_N - \gamma| < \frac{5 \sqrt{2 \pi}}{12 \sqrt{N}} e^{-8N}
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$$ {#eq:NeD}
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given in [@brent00]. Results are shown in @tbl:fourth.
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Due to roundoff errors, the best results was obtained for $N = 10$. Up to 15
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places were correctly computed.
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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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$$
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\frac{5 \sqrt{2 \pi}}{12 \sqrt{N}} e^{-8N} < 10^{-D}
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$$
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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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$$
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N = 1 + \left\lfloor \frac{1}{16}
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W \left( \frac{5 \pi}{9} 10^{2D + 1}\right) \right\rfloor
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$$
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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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approx: 0.57721\ 56649\ 01532\ 86554
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approx: 0.57721\ 56649\ 01532\ 53248
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diff: 0.00000\ 00000\ 00000\ 11102
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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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which the code accurately computes 15 digits and gives an error of
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\SI{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 (fraction) with arbitrary precision:
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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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@ -249,30 +262,32 @@ 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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According to [@brent00], the $A(N)$, $B(N)$ and $C(N)$ series awere computed up
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to $k_{\text{max}} = 5N$, which was verified to compute the correct digits up to
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500 decimal places.
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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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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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This is because 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. As well as for the scientific notation, in order to get the
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mantissa $1 \leqslant N_0 < 2$, the number of binary digits of $N$ must be
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computed (conveniently, a dedicated function `mpz_sizeinbase()` can be found in
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GMP). If the digits are
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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 $1
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\leqslant N_0 < 2$, 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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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 od whichever number $N_0$ can be computed using the series of
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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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$$
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\text{atanh}(x) = \sum_{k = 0}^{+ \infty} \frac{x^{2k + 1}}{2x + 1}
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@ -290,26 +305,15 @@ $$
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\thus \ln(z) = 2 \, \text{atanh} \left( \frac{z - 1}{z + 1} \right)
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$$
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Then, by defining:
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The idea is to set $N_0 = z$ and define:
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$$
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y = \frac{z - 1}{z + 1} \thus z = \frac{1 + y}{1 - y}
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y = \frac{N_0 - 1}{N_0 + 1}
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$$
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from which:
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hence:
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$$
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\ln \left( \frac{1 + y}{1 - y} \right) = 2 \, \text{atanh}(y)
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= 2 \, \sum_{k = 0}^{+ \infty} \frac{y^{2k + 1}}{2y + 1}
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$$
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which is true if $|y| < 1$. The idea is to set:
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$$
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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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$$
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and therefore:
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$$
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\ln(N_0) = \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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\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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$$
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But when to stop computing the series?
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@ -322,7 +326,7 @@ $$
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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 $< 1$ in order for it to converge (in
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this case, it is easy to prove that $L = y^2$). The width $\Delta S$ of the
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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 the middle value of it, namely:
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$$
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@ -336,7 +340,7 @@ In order to know when to stop computing the series, $\Delta S$ must be
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evaluated. With a bit of math:
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$$
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a_k = \frac{y^{2k + 1}}{2k + 1} \thus
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a_{k + 1} = \frac{y^{2k + 3}}{2k + 3} = a_k \frac{2k + 1}{2k + 3} y^2
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a_{k + 1} = \frac{y^{2k + 3}}{2k + 3} = a_k \, \frac{2k + 1}{2k + 3} \, y^2
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$$
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hence:
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@ -352,9 +356,10 @@ $$
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\right) = \Delta S_k \with a_k = \frac{y^{2k + 1}}{2k + 1}
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$$
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By imposing $\Delta S < 1/10^D$, where $D$ is the correct decimal place required,
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$k$ at which to stop can be obtained. This is achieved by trials, checking for
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every $k$ whether $\Delta S_k$ is less or greater than $1/10^D$.
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By imposing $\Delta S < 10^{-D}$, where $D$ is the last correct decimal place
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required, $k^{\text{max}}$ at which to stop can be obtained. This is achieved by
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trials, checking for every $k$ whether $\Delta S_k$ is less or greater than
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$10^{-D}$.
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The same considerations holds for $\ln(2)$. It could be computed as the Taylor
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expansion of $\ln(1 + x)$ with $x = 1$, but it would be too slow. A better idea
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@ -376,6 +381,7 @@ it, 1000 digits were computed in \SI{0.63}{s}, with a \SI{3.9}{GHz}
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2666mt al secondo
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\textcolor{red}{Scrivere specifice pc che fa i conti}
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### Faster implemented formula
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When implemented, @eq:faster, which is the most fast known convergence formula,
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@ -397,17 +403,102 @@ $$
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and:
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$$
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B_j = \frac{B_{j-1} N^2}{j^2}
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\with B_0 = - \ln(N)
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\with B_0 = 1
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$$
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This way, the error is given by:
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$$
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|\gamma - \gamma_k| = \pi e^{4N}
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|\gamma - \gamma_k| = \pi e^{-4N}
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$$
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Then, in order to avoid computing the logarithm of a generic number $N$,
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instead of using @eq:NeD, a number $N = 2^p$ of digits were computed, where $p$
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was chosen in order to satisfy:
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which is greater with respect to the refined formula. Then, in order to avoid
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computing the logarithm of a generic number $N$, instead of using @eq:NeD, a
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number $N = 2^p$ of digits were computed, where $p$ was chosen to satisfy:
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$$
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\pi e^{-4N} = \pi e^{-42^p} < 10^{-D} \thus
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p > \log_2 \left[ \ln(\pi) + D \ln(10) \right] -2
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$$
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and the smaller integer greater than it was selected, namely:
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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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As regards the precision of the used numbers, this time the type `pmf_t` of GMP
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was emploied. It consists of an arbitrary-precision floating point whose number
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of bits is to be defined when the variable is initialized.
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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 the roundoff errors to not affect the final result, a security
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range of 64 bits was added to $b$.
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Furthermore, the computation of the logarithm was made even faster by solving
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@eq:delta instead
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of finding $k^{\text{max}}$ by trials.
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$$
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\Delta S = \frac{1}{k (k + 2) 2^{k - 1}} < 10^{-D}
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$$ {#eq:delta}
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This was accomplished as follow:
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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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thus, for example, the inequality is satisfied for $k = x$ such that:
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$$
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2^{x - 1} (x + 1)^2 = 10^D
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$$
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In order to lighten the notation: $q := 10^D$, from which:
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\begin{align*}
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2^{x - 1} (x + 1)^2 = q
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&\thus \exp \left( \ln(2) \frac{x - 1}{2} \right) (x + 1)
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= \sqrt{q} \\
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&\thus \exp \left( \frac{\ln(2)}{2} (x + 1)\right) e^{- \ln(2)} (x + 1)
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= \sqrt{q} \\
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&\thus \exp \left( \frac{\ln(2)}{2} (x + 1)\right) \frac{1}{2} (x + 1)
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= \sqrt{q} \\
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&\thus \exp \left( \frac{\ln(2)}{2} (x + 1)\right) \frac{\ln(2)}{2} (x + 1)
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= \ln (2) \sqrt{q}
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\end{align*}
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Having rearranged the equation this way allows the use of the Lambert $W$
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function:
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$$
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W(x \cdot e^x) = x
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$$
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Hence, by computing the Lambert W function of both sides of the equation:
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$$
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\frac{\ln(2)}{2} (x + 1) = W ( \ln(2) \sqrt{q} )
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$$
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which leads to:
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$$
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x = \frac{2}{\ln(2)} \, W ( \ln(2) \sqrt{q} ) - 1
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$$
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Finally, the smallest integer greater than $x$ can be found as:
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$$
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k^{\text{max}} =
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\left\lfloor \frac{2}{\ln(2)} \, W ( \ln(2) \sqrt{q} ) \right\rfloor
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\with 10^D
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$$
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When a large number $D$ of digits is to be computed (but it also works for few
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digits), the Lambert W function can be approximated by the first five terms of
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its asymptotic value [@corless96]:
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$$
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W(x) = L_1 - L_2 + \frac{L_2}{L_1} + \frac{L_2 (L_2 - 2)}{2 L_1^2}
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+ \frac{L_2 (36L_2 - 22L_2^2 + 3L_2^3 -12)}{12L_1^4}
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$$
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where
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$$
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L_1 = \ln(x) \et L_2 = \ln(\ln(x))
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$$
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\textcolor{red}{Tempi del risultato e specifiche del pc usato}
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