Anzeigen des Gesamtinhalts (oder Logo links anklicken) oder des Impressums.

Unterschiede

Hier werden die Unterschiede zwischen zwei Versionen angezeigt.

--- analyticalengine:bernoullinumbercalculation [2015-04-21 08:53] – created rainer
+++ analyticalengine:bernoullinumbercalculation [2015-05-06 18:17] – rainer
@@ Zeile 1: / Zeile 1: @@
-====== Note G and the calculation of Bernoulli numbers ======
+__false__
-See [[wp>Bernoulli numbers]] in Wikipedia for details on Bernoulli numbers.
-They may have been selected as an example because:
-  * They cannot be enumerated using the //Difference Engine//
-  * The values seem to be fairly random
-  * The values require a machine that can handle very large numbers
-  * In approximations of trigonometric functions, they allow less calculating steps than other approximations known at that time. Note that Tchebysheff's Approximation was published several years later (1859).
-  * They are a sequence where the n-th number can be calculated from the previous numbers
-Note that variants for the indexing and sign for Bernoulli numbers were proposed since that time; the later forms allow shorter writing of the formulas.
-In //Note G//, equation 8 shows the recursion formula used by AAL:
-<nowiki>
-`0 = -1/2 * (2n-1)/(2n+1) + B_1 ((2n)/2) + B_3 ((2n*(2n-1)*(2n-2))/(2*3*4)) + ... + B_(2n-1)`
-</nowiki>
-and  in terms of the number to be calculated:
-<nowiki>
-`-B_(2n-1) = -1/2 * (2n-1)/(2n+1) + B_1 ((2n)/2) + B_3 ((2n*(2n-1)*(2n-2))/(2*3*4)) + B_5 ((2n*(2n-1)... (2n-4))/(2*3*4*5*6)) +...`
-</nowiki>
-Using `A_i` as abbreviation for the factors gives
-<nowiki>
-`B_(2n-1) = A_0(n) + A_1(n) * B_1 + A_3(n) * B_3 + ... + A_5(n) * B_(2n-3)`
-</nowiki>
-In the text, AAL wrotes: //A<sub>1</sub>, A<sub>3</sub> &c. being ... functions of n ...//, but does not use a corresponding notation in the equation, which was done in the above equation.
-Thus, the coefficients -- which are not given in //Note G//, are:
-`A_0(n) = 1/2 * (2n-1)/(2n+1)`
-`A_1(n) = -(2n) / 2`
-`A_3(n) = A_1(n) * (2n-1)/3 * (2n-2)/4`
-`A_5(n) = A_3(n) * (2n-3)/5 * (2n-4)/6`
-This makes clear, that  for `A_5` and the following the calculation steps are structurally equal and thus can be calculated within a loop.
-Note the signs in contrast to //Note G//.
-(to be continued)

Unterschiede

Anmelden