Re: Bernoulli Numbers

*To*: mathgroup at smc.vnet.net*Subject*: [mg122637] Re: Bernoulli Numbers*From*: "Alexander Elkins" <alexander_elkins at hotmail.com>*Date*: Fri, 4 Nov 2011 06:00:14 -0500 (EST)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*References*: <j8r9gv$3jd$1@smc.vnet.net>

Mathematica provides for the computation of Bernoulli numbers: In[1]:= expect=Table[BernoulliB[k], {k, 0, 20}] Out[1]= {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), 0, 43867/798, 0, -(174611/330)} For one way to display them in a table use: Grid[Table[{Subscript[B, k], BernoulliB[k]}, {k, 0, 20}], Alignment -> {{Left, Center}}] We can check the recursive definition you gave: In[2]:= Sum[Binomial[n, k]*BernoulliB[k], {k, 0, n - 1}] == 0 Out[2]= True Mathematica's result is incorrect, since this is true if and only if n > 1, as is shown here when n = 1: In[3]:= With[{n = 1}, Sum[Binomial[n, k]*BernoulliB[k], {k, 0, n - 1}] == 0] Out[3]= False The simplest way to do what you ask is to use Nest to build the list of Bernoulli numbers using Solve on the recursive definition every time: In[4]:= expect == Nest[With[{B = #, n = Length[#] + 1}, Join[B, b /. Solve[Sum[Binomial[n, k]*Part[Append[B, b], k + 1], {k, 0, n - 1}] == 0, b]]] &, {1}, 20] Out[4]= True However, with a little effort it is possible solve for BernoulliB[n - 1] this way: In[5]:= Solve[Sum[Binomial[n, k]*BernoulliB[k], {k, 0, n - 1}] == Sum[Binomial[n, k]*BernoulliB[k], {k, 0, n - 2}] + Sum[Binomial[n, k]*BernoulliB[k], {k, n - 1, n - 1}], BernoulliB[n - 1]] Out[5]= {{BernoulliB[-1 + n] -> -(Sum[BernoulliB[k] Binomial[n, k], {k, 0, -2 + n}]/n)}} Now, still using Nest to build the list of Bernoulli numbers, we have: In[6]:= expect == Nest[With[{B = #, n = Length[#] + 1}, Append[B, Sum[Binomial[n, k]*B[[k + 1]], {k, 0, n - 2}]/-n]] &, {1}, 20] Out[6]= True It is also possible to use Dot instead of Sum giving in an even simpler expression: In[7]:= expect == Nest[With[{B = #, n = Length[#] + 1}, Append[B, Binomial[n, Range[0, n - 2]].B/-n]] &, {1}, 20] Out[7]= True Note that since Binomial[n, k] == n!/k!/(n-k)! then Binomial[n, k]/-n can also be written as -(n-1)!/k!/(n-k)! Enjoy! "David Turner" <DTurner at faulkner.edu> wrote in message 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 > > _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ > > This email and any files transmitted with it are confidential and intended solely for the use of the individual or entity to whom they are addressed. This message contains confidential information and is intended only for the individual named. If you are not the named addressee you should not dissem inate, distribute or copy this e-mail. Please notify the sender immediately by e-mail if you have received this e-mail by mistake and delete this e-mail from your system. If you are not the intended recipient you are notified that disclosing, copying, distributing or taking any action in reliance on the contents of this information is strictly prohibited.