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ex-2: review

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notes/sections/2.md
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@ -10,13 +10,11 @@ $$
\sum_{k=1}^{n} \frac{1}{k}
- \ln(n) \right)
$$ {#eq:gamma}
and represents the limiting blue area in @fig:gamma. The first 30 digits of
$\gamma$ are:
$$
\gamma = 0.57721\ 56649\ 01532\ 86060\ 65120\ 90082 \dots
$$ {#eq:exact}
In complex analysis, this constant is related to many functions and can be
evaluated through a variety of identities. In this work, five methods were
implemented and their results discussed. In fact, evaluating $\gamma$ with a
@ -43,7 +41,6 @@ worse, namely:
$$
| \gamma(n_{i+1}) - \gamma | > | \gamma(n_i) - \gamma|
$$
and $\gamma (n_i)$ was selected as the best result (see @tbl:naive-errs).
----------------------------
@ -79,12 +76,11 @@ The convergence is logarithmic: to fix the first $d$ decimal places, about
$10^d$ terms of the harmonic series are needed. The double precision runs out
at the $10^{\text{th}}$ place, at $n=\num{2e10}$.
Since all the number are given with double precision, there can be at best 16
correct digits, since for a double 64 bits are allocated in memory: 1 for the
sign, 8 for the exponent and 55 for the mantissa:
correct digits. In fact, for a double 64 bits are allocated in memory: 1 for the
sign, 8 for the exponent and 55 for the mantissa, hence:
$$
2^{55} = 10^{d} \thus d = 55 \cdot \log(2) \sim 16.6
$$
Only 10 digits were correctly computed: this means that when the terms of the
series start being smaller than the smallest representable double, the sum of
all the remaining terms gives a number $\propto 10^{-11}$. The best result is
@ -108,7 +104,6 @@ $$
\gamma = \lim_{M \rightarrow + \infty} \sum_{k = 1}^{M}
\binom{M}{k} \frac{(-1)^k}{k} \ln(\Gamma(k + 1))
$$
Varying $M$ from 1 to 100, the best result was obtained for $M = 41$ (see
@tbl:limit-res). This approximation gave an underwhelming result: the
convergence is actually worse than the definition itself. Only two decimal
@ -124,7 +119,7 @@ Table: Best estimation of $\gamma$ using
the alternative formula. {#tbl:limit-res}
Here, the problem lies in the binomial term: computing the factorial of a
number greater than 18 goes over 15 places and so cannot be correctly
number greater than 18 goes over 16 places and so cannot be correctly
represented. Furthermore, the convergence is slowed down by the logarithmic
factor.
@ -137,13 +132,11 @@ $$
\frac{1}{\Gamma(z)} = z e^{yz} \prod_{k = 1}^{+ \infty}
\left( 1 + \frac{z}{k} \right) e^{-z/k}
$$
which gives:
$$
\gamma = - \frac{1}{z} \ln \left( z \Gamma(z)
\prod_{k = 1}^{+ \infty} \left( 1 + \frac{z}{k} \right) e^{-z/k} \right)
$$
The execution stops when there is no difference between two consecutive terms
of the infinite product (it happens for $k = 456565794 \sim \num{4.6e8}$,
meaning that for this value of $k$ the term of the product is equal to 1 in
@ -174,11 +167,10 @@ z $|γ(z) - γ|$ z $|γ(z) - γ|$
19 \num{1.008e-8} 9.04 \num{9.419e-9}
-----------------------------------------------
Table: Differences between some obtained values of $\gamma$ and
the exact one found with the reciprocal $\Gamma$ function formula.
The values on the left are shown to give an idea of the $z$
large-scale behaviour; on the right, the values around the best
one ($z = 9.00$) are listed. {#tbl:recip-errs}
Table: Some results found with the reciprocal $\Gamma$ function formula.
The values on the left are shown to give an idea of the $z$ large-scale
behaviour; on the right, the values around the best one ($z = 9.00$) are
listed. {#tbl:recip-errs}
As can be seen in @tbl:recip-errs, the best value for $z$ is only by chance,
since all $|\gamma(z) - \gamma |$ are of the same order of magnitude. The best
@ -194,6 +186,9 @@ Table: Third method results for z = 9.00. {#tbl:recip-res}
In this approximation, the convergence of the infinite product is fast enough
to reach the $8^{th}$ decimal place.
With respect to the definition, this formula returns a worse estimation of
$\gamma$ because the calculation of the infinite product leads to more
round-off instabilities than the series.
### Fastest convergence formula
@ -216,31 +211,26 @@ The asymptotic error of this estimation, given in [@demailly17], is:
$$
|\gamma_N - \gamma| \sim \frac{5 \sqrt{2 \pi}}{12 \sqrt{N}} e^{-8N} = \Delta_N
$$ {#eq:NeD}
The error bound gives the value of $N$ to be used to get $D$ correct decimal
digits of $\gamma$. In fact, this is done by imposing:
digits of $\gamma$. This is done by imposing:
$$
\frac{5 \sqrt{2 \pi}}{12 \sqrt{N}} e^{-8N} < 10^{-D}
\Delta_N < 10^{-D}
$$
The inequality with the equal sign can be solved with the use of the Lambert
$W$ function. For a real number $x$, $W(x)$ is defined as the inverse function
of $x\exp(x)$, so the idea is to recast the equation into this form and take
$W$ both sides.
\begin{align*}{3}
\begin{align*}
\frac{5 \sqrt{2 \pi}}{12 \sqrt{x}} e^{-8x} = 10^{-D}
& \thus \left(\frac{12}{5}\right)^2 \frac{x}{2 \pi} e^{16x} = 10^{2D} \\
& \thus 16x e^{16x} = \left(\frac{5 \pi}{9}\right)^2 10^{2D + 1} \\
& \thus x = \frac{1}{16} W\left(\left(\frac{5 \pi}{9}\right)^2 10^{2D + 1}\right)
\end{align*}
The smallest integer which satisfies the inequality is then
$$
N = 1 + \left\lfloor \frac{1}{16}
W \left( \frac{5 \pi}{9} 10^{2D + 1}\right) \right\rfloor
$$ {#eq:refined-n}
The series $A$ and $B$ were computed till there is no difference between two
consecutive terms. Results are shown in @tbl:fourth.
@ -265,9 +255,9 @@ points, one can resort to a software implementation with arbitrary precision.
In the GMP [@gmp20] library (GNU Multiple Precision arithmetic), for
example, reals can be approximated by a rational or floating point numbers
with arbitrary precision. For integer and fractions this means a check is
with arbitrary precision. For integers and fractions, this means a check is
performed on every operation that can overflow and additional memory is
requested and used to represent the larger result. For floating points it
requested and used to represent the larger result. For floating points, it
means the size of the mantissa can be chosen before initialising the number.
Additionally, the library automatically switches to the optimal algorithm to
compute an operation based on the size of the operands.
@ -277,19 +267,18 @@ precision. Thus, a program that computes the Euler-Mascheroni constant within a
user controllable precision was implemented.
To compute $N$ by @eq:refined-n, a trick must be used to avoid overflows in the
exponentiation, particularly when computing more than a few hundreds digits are
to be computed. The details are explained in @sec:optimised, below. Once the
number $N$ has been fixed, the series $A(N)$ and $B(N)$ are to be evaluated.
With rational numbers, a different criterion for the truncation must be
considered, because two consecutive term are always different. Brent and
exponentiation, particularly when computing more than a few hundreds digits. The
details are explained in @sec:optimised, below.
Once the number $N$ has been fixed, the series $A(N)$ and $B(N)$ are to be
evaluated. With rational numbers, a different criterion for the truncation must
be considered, since two consecutive term are always different. Brent and
McMillan[@brent00] prove that to reduce the partial sum to less than $\Delta_N$
it is sufficient to compute $\alpha N$ terms, where $\alpha$ is the solution to
the equation:
\begin{align*}
\alpha\ln(\alpha) = 3 + \alpha &&
$$
\alpha\ln(\alpha) = 3 + \alpha \thus
\alpha = \exp(W(3e) - 1) \approx 4.97
\end{align*}
$$
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.
@ -299,27 +288,23 @@ 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 shown shortly),
$b = 2$ was chosen. If $N$ is a power of two, $N_0$ is 1 and only $\ln(2)$ is to be
computed. More generally, the problem reduces to the calculation of $\ln(N_0)$.
$b = 2$ was chosen. If $N$ is a power of two, $N_0$ is 1 and only $\ln(2)$ is to
be computed. More generally, the problem lies in the calculation of $\ln(N_0)$.
Regarding the scientific notation, to find the mantissa $1
\leqslant N_0 < 2$, the number of binary digits of $N$ must be computed
(conveniently, a dedicated function `mpz_sizeinbase()` exists in GMP). If the
digits are
$n$:
digits are $n$:
$$
e = n - 1 \thus N = N_0 \cdot 2^{n-1} \thus N_0 = \frac{N}{2^{n - 1}}
$$
The logarithm of $N_0$ can be computed from the Taylor series of the hyperbolic
tangent, which is convergent for $|x| < 1$:
$$
\text{atanh}(x) = \sum_{k = 0}^{+ \infty} \frac{x^{2k + 1}}{2x + 1}
$$
The relation with the logarithm follows from the definition
The relation with the logarithm follows from the definition:
$$
\text{tanh}(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}}
= \frac{e^{2x} - 1}{e^{2x} + 1}
@ -329,26 +314,22 @@ $$
\text{tanh} \left( \frac{\ln(z)}{2} \right) = \frac{z - 1}{z + 1}
\thus \ln(z) = 2 \, \text{atanh} \left( \frac{z - 1}{z + 1} \right)
$$
The idea is to set $N_0 = z$ and define:
$$
y = \frac{N_0 - 1}{N_0 + 1}
$$
hence:
$$
\ln(N_0) = 2 \, \text{atanh} (y)
= 2 \sum_{k = 0}^{+ \infty} \frac{y^{2k + 1}}{2k + 1}
$$
To estimate a series with a given precision some care must be taken:
different techniques apply to different series and are explained in
the paper [@riddle08]. In this case, letting $S$ be the value of the
series and $S_k$ the $k$-th partial sum the following bounds can be found:
To estimate a series with a given precision, some care must be taken: different
techniques apply to different series and are explained in the paper [@riddle08].
In this case, letting $S$ be the value of the series and $S_k$ the $k$-th
partial sum, the following bounds can be found:
$$
S_k + a_k \frac{L}{1 -L} < S < S_k + \frac{a_{k+1}}{1 - \frac{a_{k+1}}{a_k}}
$$
where $a_k$ is the $k^{\text{th}}$ term of the series and $L$ is the limiting
ratio of the series terms, which must be $\le 1$ in order for it to converge (in
this case, it is easy to prove that $L = y^2 < 1$). The width $\Delta S$ of the
@ -360,7 +341,6 @@ $$
\right) \et
\Delta S = \frac{a_{k+1}}{1 - \frac{a_{k+1}}{a_k}} - a_k \frac{L}{1 -L}
$$
For this series:
\begin{align*}
a_k = \frac{y^{2k + 1}}{2k + 1} &&
@ -377,7 +357,6 @@ $$
\frac{1}{\frac{2k + 3}{2k + 1}y^{-2} - 1} - \frac{y^2}{1 - y^2}
\right)
$$
By imposing $\Delta S_k < 10^{-D}$, where $D$ is the number decimal places
required, $k^{\text{max}}$ at which to stop the summation can be obtained. This
is achieved by trials, checking for every $k$ whether $\Delta S_k$ is less or
@ -390,21 +369,20 @@ which leads to a much faster series with constant sign:
$$
\ln(2) = \sum_{k=1}^{+ \infty} \frac{1}{k \cdot 2^k}
$$
In this case the series ratio is found to be $L = 1/2$ and the error
In this case the series ratio is found to be $L = 1/2$ and the error:
$$
\Delta S_k = \frac{1}{k(k + 2) 2^{k-1}}
$$
Once the logarithms and the terms $A$, $B$ and $C$ have been computed @eq:faster
can finally be used to obtain $\gamma$ up to the given decimal place.
Once the logarithms and the terms $A$, $B$ and $C$ have been computed,
@eq:faster can finally be used to obtain $\gamma$ up to the given decimal
place.
The program was implemented with no particular care for performance but
was found to be relatively fast. On a \SI{3.9}{GHz} desktop computer with
\SI{2666}{MT\per\s} memory, it takes \SI{0.63}{\second} to compute the first
1000 digits. However, the quadratic complexity of the algorithm makes this
quite unpractical for computing more than a few thousands digits. For this
reason a more optimised algorithm was implemented.
1000 digits. However, this makes it quite unpractical for computing more than
a few thousands digits. For this reason, a more optimised algorithm was
implemented.
### Optimised implementation {#sec:optimised}
@ -413,43 +391,41 @@ The refined Brent-McMillan formula (@eq:faster) is theoretically the fastest
but is difficult to implement efficiently. In practice it turns out to be
slower than the standard formula even if its asymptotic error is better.
The standard formula [@brent00] ignores the correction $C(N)$ and is rewritten
in a more convenient way:
The standard formula [@brent00] ignores the correction $C(N)$ and can be
rewritten in a more convenient way:
$$
\gamma_k = \frac{\sum_{k = 0}^{k_\text{max}} A_k(N)}
{\sum_{k = 0}^{k_\text{max}} B_k(N)}
$$
where:
\begin{align*}
A_k &= \frac{1}{k} \left(\frac{A_{k-1} N^2}{k} + B_k \right) & A_0 &= - \ln(N) \\
B_k &= \frac{B_{k-1} N^2}{k^2} & B_0 &= 1 \\
A_k &= \frac{1}{k} \left(\frac{A_{k-1} N^2}{k} + B_k \right)
& A_0 &= - \ln(N) \\
B_k &= \frac{B_{k-1} N^2}{k^2}
& B_0 &= 1 \\
\end{align*}
As said, the asymptotic error of the formula decreases slower
As said, the asymptotic error of the formula decreases slower:
$$
|\gamma - \gamma_k| \sim \pi e^{-4N} = \Delta_N
$$
In order to avoid the expensive computation of $\ln(N)$, N was chosen to be a
power of 2, $N = 2^p$. As before:
$$
\Delta_p = \pi e^{-2^{p+2}} < 10^{-D} \thus
p > \log_2 \left( \ln(\pi) + D \ln(10) \right) - 2
$$
and the smaller integer is found by taking the floor:
$$
p = \left\lfloor \log_2 \left( \ln(\pi) + D \ln(10) \right) \right\rfloor + 1
$$
The number of terms to compute, $k_\text{max}$ is still proportional to $N$
but by a different factor, $\beta$ such that:
The number of terms to compute, $k_\text{max}$, is still proportional to $N$
but by a different factor $\beta$ such that:
\begin{align*}
\beta\ln(\beta) = 1 + \beta &&
\beta = \exp(W(e^{-1}) + 1) \approx 3.59
\end{align*}
As regards the numbers representation, this time the GMP type `mpf_t` was
employed. It consists of an arbitrary-precision floating point with a fixed
exponent of 32 or 64 bits but arbitrary mantissa length.
@ -457,27 +433,24 @@ If a number $D$ of digits is required, the number of bits can be computed from:
$$
10^D = 2^b \thus b = D \log_2 (10)
$$
in order to avoid roundoff errors affecting the final result, a security range
In order to avoid roundoff errors affecting the final result, a security range
of 64 bits was added to $b$.
Furthermore, the computation of $\ln(2)$ was optimized by solving
@eq:delta analytically, instead of finding $k^{\text{max}}$ by trials.
The series estimation error was
The series estimation error was:
$$
\Delta S_k = \frac{1}{k (k + 2) 2^{k - 1}} < 10^{-D}
$$ {#eq:delta}
and the following condition should hold:
$$
k (k + 2) 2^{k - 1} > 10^D \thus 2^{k - 1} [(k + 1)^2 - 1] > 10^D
$$
Asking for the equality with $k = x$ yields
Asking for the equality, with $k = x$ it yields to:
$$
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 $W(x)$ :
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.