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analyticalengine:bernoullinumbercalculation

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analyticalengine:bernoullinumbercalculation [2015-04-21 22:23] – timanalyticalengine:bernoullinumbercalculation [2015-08-26 13:23] (aktuell) – compacted B_7 example calculation rainer
Zeile 15: Zeile 15:
 In //Note G//, equation 8 shows the recursion formula used by AAL: In //Note G//, equation 8 shows the recursion formula used by AAL:
  
-<nowiki> `0 = -1/2 * (2n-1)/(2n+1) + B_1 <sup>((1)) </sup> /(2*3*4)) + ... + B_(2n-1)` </nowiki>+<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: and  in terms of the number to be calculated:
Zeile 29: Zeile 29:
 Thus, the coefficients – which are not given in //Note G//   – are: Thus, the coefficients – which are not given in //Note G//   – are:
  
-`A_0(n) = 1/2 <nowiki>*</nowiki> (2n-1)/(2n+1)`+`A_0(n) = 1/2 * (2n-1)/(2n+1)`
  
 `A_1(n) = -(2n) / 2` `A_1(n) = -(2n) / 2`
  
-`A_3(n) = A_1(n) <nowiki>*</nowiki> (2n-1)/3 <nowiki>*</nowiki> (2n-2)/4`+`A_3(n) = A_1(n) * (2n-1)/3 * (2n-2)/4`
  
-`A_5(n) = A_3(n) <nowiki>*</nowiki> (2n-3)/5 <nowiki>*</nowiki> (2n-4)/6`+`A_5(n) = A_3(n) * (2n-3)/5 * (2n-4)/6`
  
 As each `A_i(n)` depends only on the previous  `A_(i-1)(n)` (for `i>3`), they can be enumerated within a loop, as the memory places for the earlier numbers can be reused. As each `A_i(n)` depends only on the previous  `A_(i-1)(n)` (for `i>3`), they can be enumerated within a loop, as the memory places for the earlier numbers can be reused.
Zeile 44: Zeile 44:
  
 Note that the signs are different to those in //Note G//. Note that the signs are different to those in //Note G//.
 +
 +====== Why not equation 2 or 3? ======
 +
 +On page 725, AAL shows two commonly used formulas in equation 2 and 3, that do calculate the n-th Bernoulli number just as a series without recursion to other Bernoulli numbers, i.e. as a function of n alone. Then she writes:
 +> As however our object is not simplicity or facility of computation, but the illustration of the powers of the engine, we prefer selecting the formula below, marked (8.)
 +
 +Thus she probably delibrately took into account the indexing problem.
 +
 +While equation (2.) looks fairly simple, it is an inifinite sum of terms, which by itself would rule out using the AE. (Still need to write down examples, as elements to be summed up are not integer numbers.)
 +
 +Equation (3.) looks like a finite sequence, but needs some rewriting to avoid arbitrary large numbers as intermediate results. (Would required deeper inspection too.)
 +
 +So the version in equation (8.) looks to me the most simple soulution that has a facility of computation; so why did AAL deny this fairly good argument? 
 +
  
 ====== The problem in line 21 ====== ====== The problem in line 21 ======
  
-The tabular programme in the //Diagram accompanying Translator's Note G//   is fairly straightforward upto line 12; the wrong indices in line 4 is evidently a typgraphic error.+The tabular programme in the //Diagram accompanying Translator's Note G//   is fairly straightforward upto line 12; the wrong indices in line 4 is evidently a typographic error.
  
 To calculate `B_7`, the lines 13 to 23 are executed once, then in line 24 and 25 the result is stored. To calculate `B_7`, the lines 13 to 23 are executed once, then in line 24 and 25 the result is stored.
Zeile 55: Zeile 69:
 Howerver, in line 21 during the second round, the value to be multiplied is not `V_22`, containing `B_5`, but `V_23`, contining the just calculated `B_7`. Howerver, in line 21 during the second round, the value to be multiplied is not `V_22`, containing `B_5`, but `V_23`, contining the just calculated `B_7`.
  
-AAL has evidently see that problem and writes on page 729 bottom: >The only exception to a //perfect identity//   in all the processes and columns used, for every repetition of Operations (13…23), is, that Operation 21 always requires one of its factors from a new column, and Operation 24 always puts its result on a new column. But as these variations follow the same law at each repetition (Operation 21 always requiring its factor from a column one in advance of that which it used the previous time, and Operation 24 always putting its result on the column one in advance of that which received the previous result), they are easily provided for in arranging the recurring group (or cycle) of Variable-cards.+AAL has evidently see that problem and writes on page 729 bottom:  
 +>The only exception to a //perfect identity//   in all the processes and columns used, for every repetition of Operations (13…23), is, that Operation 21 always requires one of its factors from a new column, and Operation 24 always puts its result on a new column. But as these variations follow the same law at each repetition (Operation 21 always requiring its factor from a column one in advance of that which it used the previous time, and Operation 24 always putting its result on the column one in advance of that which received the previous result), they are easily provided for in arranging the recurring group (or cycle) of Variable-cards.
  
 However, while it is correctly assumed that this may be solved, there is no hint of how this is //easily provided//, that the loom of variable cards are changed with each round. However, while it is correctly assumed that this may be solved, there is no hint of how this is //easily provided//, that the loom of variable cards are changed with each round.
Zeile 65: Zeile 80:
 However, to calculate millions of Bernoulli numbers, the AE would need a store of millions of numbers; this may be a place where she did not fully understand the problem. However, to calculate millions of Bernoulli numbers, the AE would need a store of millions of numbers; this may be a place where she did not fully understand the problem.
  
-Alan Bromley in his 1998 article on page 44 left column, second to last paragraph, writes: >With hindsight, we can note that in the Analytical Engine (at least until 1840), Babbage did not posess the variable-address concept; that is, there was no mechanism by which the machine could, as a result of a calculation, specify a particular variable in the store to be used of an operand for an instruction.+Alan Bromley in his 1998 article on page 44 left column, second to last paragraph, writes:  
 +>With hindsight, we can note that in the Analytical Engine (at least until 1840), Babbage did not posess the variable-address concept; that is, there was no mechanism by which the machine could, as a result of a calculation, specify a particular variable in the store to be used of an operand for an instruction.
  
 He refers this to solving matrices, and thus to a much much more complex problem. He refers this to solving matrices, and thus to a much much more complex problem.
Zeile 71: Zeile 87:
 ====== Actual Calculation ====== ====== Actual Calculation ======
  
-<nowiki>`A_0(1) = 1/2 * 1/3 = 1/6`</nowiki>+`A_0(1) = 1/2 * 1/3 = 1/6`
  
 `B_1 = A_0(1) = 1/6` `B_1 = A_0(1) = 1/6`
Zeile 77: Zeile 93:
 `A_0(2) = 3/10` `A_0(2) = 3/10`
  
-<nowiki>`A_1(2) = -2`</nowiki>+`A_1(2) = -2`
  
-`B_3 = A_0(2) + A_1(2) <nowiki>*</nowiki> B_1 = 3/10 - 2/6 = (9 - 10)/30 = - 1/30`+`B_3 = A_0(2) + A_1(2) * B_1 = 3/10 - 2/6 = (9 - 10)/30 = - 1/30`
  
 `A_0(3) = 5/14` `A_0(3) = 5/14`
Zeile 85: Zeile 101:
 `A_1(3) = -3` `A_1(3) = -3`
  
-`A_3(3) = -3 <nowiki>*</nowiki> 5/3 <nowiki>*</nowiki> 4/4 = -5`+`A_3(3) = -3 * 5/3 * 4/4 = -5`
  
 `B_5 = 5/14 - 3/6 + 5/30 = (5 - 7)/14 + 1/6 = -1/7 + 1/6 = 1/42` `B_5 = 5/14 - 3/6 + 5/30 = (5 - 7)/14 + 1/6 = -1/7 + 1/6 = 1/42`
Zeile 93: Zeile 109:
 `A_1(4) = -4` `A_1(4) = -4`
  
-`A_3(4) = -4 <nowiki>*</nowiki> 7/3 <nowiki>*</nowiki> 6/4 = -14`+`A_3(4) = -4 * 7/3 * 6/4 = -14`
  
-`A_5(4) = -14 <nowiki>*</nowiki> 5/5 <nowiki>*</nowiki> 4/6 = - 28/3`+`A_5(4) = -14 * 5/5 * 4/6 = - 28/3`
  
-`B_7 = 7/18 - 4/6 + 14/30 - 28/(3<nowiki>*</nowiki>42) = (7 - 12)/18 + 7/15 - 14/63 = -5/18 + 7/15 - 14/63 = (-25 + 28)/90 + 14/63 = 17/90 -14/63 = (119 - 140 ) / (7<nowiki>*</nowiki>9<nowiki>*</nowiki>10) = - 21/(7<nowiki>*</nowiki>9<nowiki>*</nowiki>10) = - 1/30`+`B_7 = 7/18 - 4/6 + 14/30 - 28/(3*42) = (7 - 12)/18 + 7/15 - 2/9 = -5/18 + (21-10)/45 = (-25 + 22)/90 = -3/90 = - 1/30`
  
analyticalengine/bernoullinumbercalculation.1429647834.txt.gz · Zuletzt geändert: 2015-04-21 22:23 von tim