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ex-2: revised section 2.2.1

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Giù Marcer 2020-05-26 09:58:16 +02:00 committed by rnhmjoj
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notes/sections/2.md
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@ -43,44 +43,50 @@ $$
| \gamma(n_{i+1}) - \gamma | > | \gamma(n_i) - \gamma|
$$
and $\gamma (n_i)$ was selected as the result (see @tbl:1_results).
and $\gamma (n_i)$ was selected as the best result (see @tbl:1_results).
-----------------------------------------------
n sum $|\gamma(n)-\gamma|$
----------- ------------- ---------------------
\SI{2e1}{} \SI{2.48e-02}{}
---------------------------------
n $|\gamma(n)-\gamma|$
----------- ---------------------
\SI{2e1}{} \SI{2.48e-02}{}
\SI{2e2}{} \SI{2.50e-03}{}
\SI{2e2}{} \SI{2.50e-03}{}
\SI{2e3}{} \SI{2.50e-04}{}
\SI{2e3}{} \SI{2.50e-04}{}
\SI{2e4}{} \SI{2.50e-05}{}
\SI{2e4}{} \SI{2.50e-05}{}
\SI{2e5}{} \SI{2.50e-06}{}
\SI{2e5}{} \SI{2.50e-06}{}
\SI{2e6}{} \SI{2.50e-07}{}
\SI{2e6}{} \SI{2.50e-07}{}
\SI{2e7}{} \SI{2.50e-08}{}
\SI{2e7}{} \SI{2.50e-08}{}
\SI{2e8}{} \SI{2.50e-09}{}
\SI{2e8}{} \SI{2.50e-09}{}
\SI{2e9}{} \SI{2.55e-10}{}
\SI{2e9}{} \SI{2.55e-10}{}
\SI{2e10}{} \SI{2.42e-11}{}
\SI{2e10}{} \SI{2.42e-11}{}
\SI{2e11}{} \SI{1.44e-08}{}
-----------------------------------------------
\SI{2e11}{} \SI{1.44e-08}{}
---------------------------------
Table: Partial results using the definition of $\gamma$ with double
precision. {#tbl:1_results}
The convergence is logarithmic: to fix the first $d$ decimal places, about
$10^d$ terms are needed. The double precision runs out at the
10\textsuperscript{th} place, $n=\SI{2e10}{}$.
Since all the number are given with double precision, there can be at best 15
correct digits but only 10 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 give a number $\propto 10^{-11}$.
$10^d$ terms of the armonic series are needed. The double precision runs out at
the $10^{\text{th}}$ place, at $n=\SI{2e10}{}$.
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:
$$
2^{55} = 10^{d} \thus d = 55 \cdot \log(2) \sim 16.6
$$
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 $\propto 10^{-11}$.
Best result in @tbl:first.
--------- -----------------------
@ -106,7 +112,7 @@ $$
Varying $M$ from 1 to 100, the best result was obtained for $M = 41$ (see
@tbl:second). It went sour: the convergence is worse than using the definition
itself. Only two places were correctly computed (#@tbl:second).
itself. Only two places were correctly computed (@tbl:second).
--------- -----------------------
true: 0.57721\ 56649\ 01533