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.

notes/docs/bibliography.bib

@@ -179,3 +179,22 @@
   year={2000},
   publisher={Springer}
 }
+
+@article{corless96,
+  title={On the Lambert W function},
+  author={Corless, Robert M and Gonnet, Gaston H and Hare, David EG and Jeffrey,
+          David J and Knuth, Donald E},
+  journal={Advances in Computational mathematics},
+  volume={5},
+  number={1},
+  pages={329--359},
+  year={1996},
+  publisher={Springer}
+}
+
+@article{demailly17,
+  author={Jean-Pierre Demailly},
+  title={Precise error estimate of the Brent-McMillan algorithm for the
+  computation of Euler's constant},
+  year={2017}
+}

notes/sections/2.md

@@ -196,9 +196,10 @@ $8^{th}$ place.
 
 ### Fastest convergence formula
 
-The fastest known convergence belongs to the following formula [@yee19]:  
+The fastest known convergence belongs to the following formula, known as refined
+Brent-McMillan formula[@yee19]:
 $$
-  \gamma = \frac{A(N)}{B(N)} -\frac{C(N)}{B^2(N)} - \ln(N)
+  \gamma_N = \frac{A(N)}{B(N)} -\frac{C(N)}{B^2(N)} - \ln(N)
 $$ {#eq:faster}
 
 with:
@@ -210,36 +211,48 @@ with:
           \frac{((2k)!)^3}{(k!)^4 \cdot (16k)^2k}         \\
 \end{align*}
 
-The series $A$ and $B$ were computed till there is no difference between two
+and the difference between $\gamma_N$ and $\gamma$ is given by [@demailly17]:
-consecutive terms. The number of desired correct decimals $D$ was given in
-input and $N$ was consequently computed through the formula: 
 $$
-  N = \left\lfloor 2 + \frac{1}{4} \cdot \ln(10) \cdot D \right\rfloor
+  |\gamma_N - \gamma| < \frac{5 \sqrt{2 \pi}}{12 \sqrt{N}} e^{-8N}
 $$ {#eq:NeD}
 
-given in [@brent00]. Results are shown in @tbl:fourth.  
+This gives the value of $N$ which is to be used into the formula in order to get
-Due to roundoff errors, the best results was obtained for $N = 10$. Up to 15
+$D$ correct decimal digits of $\gamma$. In fact, by imposing:
-places were correctly computed.
+$$
+  \frac{5 \sqrt{2 \pi}}{12 \sqrt{N}} e^{-8N} < 10^{-D}
+$$
 
+With a bit of math, it can be found that the smallest integer which satisfies
+the inequality is given by:
+$$
+  N = 1 + \left\lfloor \frac{1}{16}
+  W \left( \frac{5 \pi}{9} 10^{2D + 1}\right) \right\rfloor
+$$
+
+The series $A$ and $B$ were computed till there is no difference between two
+consecutive terms. Results are shown in @tbl:fourth.  
 
 --------- ------------------------------
 true:	     0.57721\ 56649\ 01532\ 75452
 
-approx:    0.57721\ 56649\ 01532\ 86554
+approx:    0.57721\ 56649\ 01532\ 53248
 
-diff:	     0.00000\ 00000\ 00000\ 11102
+diff:	     0.00000\ 00000\ 00000\ 33062
 --------- ------------------------------
 
 Table: $\gamma$ estimation with the fastest
 known convergence formula (@eq:faster). {#tbl:fourth}
 
+Because of roundoff errors, the best result was obtained only for $D = 15$, for
+which the code accurately computes 15 digits and gives an error of
+\SI{3.3e16}{}. For $D > 15$, the requested can't be fulfilled.
 
 ### Arbitrary precision
 
 To overcome the issues related to the double representation, one can resort to a
 representation with arbitrary precision. In the GMP library (which stands for
 GNU Multiple Precision arithmetic), for example, real numbers can be
-approximated by a ratio of two integers (fraction) with arbitrary precision: 
+approximated by a ratio of two integers (a fraction) with arbitrary precision: 
 this means a check is performed on every operation and in case of an integer
 overflow, additional memory is requested and used to represent the larger result.
 Additionally, the library automatically switches to the optimal algorithm
@@ -249,30 +262,32 @@ The terms in @eq:faster can therefore be computed with arbitrarily large
 precision.  Thus, a program that computes the Euler-Mascheroni constant within
 a user controllable precision was implemented.
 
-According to [@brent00], the $A(N)$, $B(N)$ and $C(N)$ series awere computed up
+According to [@brent00], the $A(N)$, $B(N)$ and $C(N)$ series were computed up
-to $k_{\text{max}} = 5N$, which was verified to compute the correct digits up to
+to $k_{\text{max}} = 5N$ \textcolor{red}{sure?}, since it guarantees the accuracy
-500 decimal places.  
+od the result up to the $D$ decimal digit.  
 The GMP library offers functions to perform some operations such as addition,
 multiplication, division, etc. However, the logarithm function is not
 implemented. Thus, most of the code carries out the $\ln(N)$ computation.
-First, it should be noted that the logarithm of only some special numbers can
+This is because the logarithm of only some special numbers can be computed
-be computed with arbitrary precision, namely the ones of which a converging
+with arbitrary precision, namely the ones of which a converging series is known.
-series is known. This forces $N$ to be rewritten in the following way:
+This forces $N$ to be rewritten in the following way:
 $$
   N = N_0 \cdot b^e \thus \ln(N) = \ln(N_0) + e \cdot \ln(b)
 $$
 
 Since a fast converging series for $\ln(2)$ is known (it will be shwn shortly),
-$b = 2$ was chosen. As well as for the scientific notation, in order to get the
+$b = 2$ was chosen. In case $N$ is a power of two, only $\ln(2)$ is to be
-mantissa $1 \leqslant N_0 < 2$, the number of binary digits of $N$ must be
+computed. If not, it must be handled as follows.  
-computed (conveniently, a dedicated function `mpz_sizeinbase()` can be found in
+As well as for the scientific notation, in order to get the mantissa $1
-GMP). If the digits are
+\leqslant N_0 < 2$, the number of binary digits of $N$ must be computed
+(conveniently, a dedicated function `mpz_sizeinbase()` is found in GMP). If the
+digits are
 $n$:
 $$
   e = n - 1 \thus N = N_0 \cdot 2^{n-1} \thus N_0 = \frac{N}{2^{n - 1}}
 $$
 
-The logarithm od whichever number $N_0$ can be computed using the series of
+The logarithm of whichever number $N_0$ can be computed using the series of
 $\text{atanh}(x)$, which converges for $|x| < 1$:
 $$
   \text{atanh}(x) = \sum_{k = 0}^{+ \infty} \frac{x^{2k + 1}}{2x + 1}
@@ -290,26 +305,15 @@ $$
   \thus \ln(z) = 2 \, \text{atanh} \left( \frac{z - 1}{z + 1} \right)
 $$
 
-Then, by defining:
+The idea is to set $N_0 = z$ and define:
 $$
-  y = \frac{z - 1}{z + 1} \thus z = \frac{1 + y}{1 - y}
+  y = \frac{N_0 - 1}{N_0 + 1}
 $$
 
-from which:
+hence:
 $$
-  \ln \left( \frac{1 + y}{1 - y} \right) = 2 \, \text{atanh}(y)
+  \ln(N_0) = 2 \, \text{atanh} (y)
-  = 2 \, \sum_{k = 0}^{+ \infty} \frac{y^{2k + 1}}{2y + 1}
+           = 2 \sum_{k = 0}^{+ \infty} \frac{y^{2k + 1}}{2k + 1}
-$$
-
-which is true if $|y| < 1$. The idea is to set:
-$$
-  N_0 = \frac{1 + y}{1 - y} \thus y = \frac{N_0 - 1}{N_0 + 1} < 1
-$$
-
-and therefore:
-$$
-  \ln(N_0) = \ln \left( \frac{1 + y}{1 - y} \right) =
-  2 \sum_{k = 0}^{+ \infty} \frac{y^{2k + 1}}{2k + 1}
 $$
 
 But when to stop computing the series?  
@@ -322,7 +326,7 @@ $$
 
 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
-this case, it is easy to prove that $L = y^2$). The width $\Delta S$ of the
+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:
 $$
@@ -336,7 +340,7 @@ In order to know when to stop computing the series, $\Delta S$ must be
 evaluated. With a bit of math:
 $$
   a_k = \frac{y^{2k + 1}}{2k + 1} \thus
-  a_{k + 1} = \frac{y^{2k + 3}}{2k + 3} = a_k \frac{2k + 1}{2k + 3} y^2
+  a_{k + 1} = \frac{y^{2k + 3}}{2k + 3} = a_k \, \frac{2k + 1}{2k + 3} \, y^2
 $$
 
 hence:
@@ -352,9 +356,10 @@ $$
   \right) = \Delta S_k \with a_k = \frac{y^{2k + 1}}{2k + 1}
 $$
 
-By imposing $\Delta S < 1/10^D$, where $D$ is the correct decimal place required,
+By imposing $\Delta S < 10^{-D}$, where $D$ is the last correct decimal place
-$k$ at which to stop can be obtained. This is achieved by trials, checking for
+required, $k^{\text{max}}$ at which to stop can be obtained. This is achieved by
-every $k$ whether $\Delta S_k$ is less or greater than $1/10^D$.
+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
@@ -376,6 +381,7 @@ 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}
 
+
 ### Faster implemented formula
 
 When implemented, @eq:faster, which is the most fast known convergence formula,
@@ -397,17 +403,102 @@ $$
 and:
 $$
   B_j = \frac{B_{j-1} N^2}{j^2}
-  \with B_0 = - \ln(N)
+  \with B_0 = 1
 $$
 
 This way, the error is given by:
 $$
-  |\gamma - \gamma_k| = \pi e^{4N}
+  |\gamma - \gamma_k| = \pi e^{-4N}
 $$
 
-Then, in order to avoid computing the logarithm of a generic number $N$,
+which is greater with respect to the refined formula. Then, in order to avoid
-instead of using @eq:NeD, a number $N = 2^p$ of digits were computed, where $p$
+computing the logarithm of a generic number $N$, instead of using @eq:NeD, a
-was chosen in order to satisfy:
+number $N = 2^p$ of digits were computed, where $p$ was chosen to satisfy:
+$$
+  \pi e^{-4N} = \pi e^{-42^p} < 10^{-D} \thus
+  p > \log_2 \left[ \ln(\pi) + D \ln(10) \right] -2
 $$
 
+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.  
+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 $b$.
+
+Furthermore, the computation of the logarithm was made even faster by solving
+@eq:delta instead
+of finding $k^{\text{max}}$ by trials.
+$$
+  \Delta S = \frac{1}{k (k + 2) 2^{k - 1}} < 10^{-D}
+$$ {#eq:delta}
+
+This was accomplished as follow:
+$$
+  k (k + 2) 2^{k - 1} > 10^D \thus 2^{k - 1} [(k + 1)^2 - 1] > 10^D
+$$
+
+thus, for example, the inequality is satisfied for $k = x$ such that:
+$$
+  2^{x - 1} (x + 1)^2 = 10^D
+$$
+
+In order to lighten the notation: $q := 10^D$, from which:
+\begin{align*}
+  2^{x - 1} (x + 1)^2 = q
+  &\thus \exp \left( \ln(2) \frac{x - 1}{2}  \right) (x + 1)
+    = \sqrt{q}                 \\
+  &\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} \\
+  &\thus \exp \left( \frac{\ln(2)}{2} (x + 1)\right) \frac{\ln(2)}{2} (x + 1)
+        = \ln (2) \sqrt{q}
+\end{align*}
+
+Having rearranged the equation this way allows the use of the Lambert $W$
+function:
+$$
+  W(x \cdot e^x) = x
+$$
+
+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:
+$$
+  x = \frac{2}{\ln(2)} \, W ( \ln(2) \sqrt{q} ) - 1
+$$
+
+Finally, the smallest integer greater than $x$ can be found as:
+$$
+  k^{\text{max}} =
+  \left\lfloor \frac{2}{\ln(2)} \, W ( \ln(2) \sqrt{q} ) \right\rfloor
+  \with 10^D
+$$
+
+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]:
+$$
+  W(x) = L_1 - L_2 + \frac{L_2}{L_1} + \frac{L_2 (L_2 - 2)}{2 L_1^2}
+       + \frac{L_2 (36L_2 - 22L_2^2 + 3L_2^3 -12)}{12L_1^4}
+$$
+
+where
+$$
+  L_1 = \ln(x) \et L_2 = \ln(\ln(x))
+$$
+ 
+\textcolor{red}{Tempi del risultato e specifiche del pc usato}