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Re: Bernoulli Numbers

  • To: mathgroup at smc.vnet.net
  • Subject: [mg122633] Re: Bernoulli Numbers
  • From: DrMajorBob <btreat1 at austin.rr.com>
  • Date: Fri, 4 Nov 2011 05:59:31 -0500 (EST)
  • Delivered-to: l-mathgroup@mail-archive0.wolfram.com
  • References: <j8r9gv$3jd$1@smc.vnet.net> <201111030844.DAA15137@smc.vnet.net>
  • Reply-to: drmajorbob at yahoo.com

Much faster is:

Clear[b]
b[0] = 1;
b[n_] := b[n] = -Sum[Binomial[n + 1, k]*b[k], {k, 0, n - 1}]/(n + 1)
Array[b, 17, 0] // Timing

{0.000679, {1, -(1/2), 1/6, 0, -(1/30), 0, 1/42, 0, -(1/30), 0, 5/66,
   0, -(691/2730), 0, 7/6, 0, -(3617/510)}}

whereas before, the timing was:

Clear[b]
b[0] = 1;
b[n_] := -Sum[Binomial[n + 1, k]*b[k], {k, 0, n - 1}]/(n + 1)
Array[b, 17, 0] // Timing

{0.717417, {1, -(1/2), 1/6, 0, -(1/30), 0, 1/42, 0, -(1/30), 0, 5/66,
   0, -(691/2730), 0, 7/6, 0, -(3617/510)}}

Bobby

On Thu, 03 Nov 2011 03:44:09 -0500, Dr. Wolfgang Hintze <weh at snafu.de>  
wrote:

> Why not solve by hand for b[n] (and preferrably use lower case symbols
> for your own ones), i.e.
>
> b[n_] := -Sum[Binomial[n + 1, k]*b[k], {k, 0, n - 1}]/(n + 1)
>
> and then calculate the values recursively in a table
> b[0] = 1;
>
> Table[{n, b[n]}, {n, 0, 16}]
>
> {{0, 1}, {1, -(1/2)}, {2, 1/6}, {3, 0}, {4, -(1/30)},
> {5, 0}, {6, 1/42}, {7, 0}, {8, -(1/30)}, {9, 0},
> {10, 5/66}, {11, 0}, {12, -(691/2730)}, {13, 0},
> {14, 7/6}, {15, 0}, {16, -(3617/510)}}
>
> Regards,
> Wolfgang
>
> "David Turner" <DTurner at faulkner.edu> schrieb im Newsbeitrag
> news:j8r9gv$3jd$1 at smc.vnet.net...
>> Hello,
>>
>> I wish to compute several Bernoulli numbers, say B0 through B20.  The
>> Bernoulli numbers are defined recursively by
>>
>> B0 = 1, and Solve[Sum[Binomial[n,k]*Bk,{k,0,n-1}]==0,Bn-1] for n > 1
>>
>> I am trying to compute these numbers in some type of loop, and
>> display them in a table.  Any help is greatly appreciated.
>>
>> Thanks,
>>
>> David
>>
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-- 
DrMajorBob at yahoo.com



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