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Re: doing things on a procedural way and doing them on a functional way
*To*: mathgroup at smc.vnet.net
*Subject*: [mg47040] Re: doing things on a procedural way and doing them on a functional way
*From*: drbob at bigfoot.com (Bobby R. Treat)
*Date*: Mon, 22 Mar 2004 05:18:52 -0500 (EST)
*References*: <c3h165$b9o$1@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
Oops, I had that wrong a few minutes ago.
c[n_] = (2*n)!/(n!*n!*(n + 1));
FullSimplify[c[n + 1]/c[n]]
4 - 6/(2 + n)
Bobby
"Wolf, Hartmut" <Hartmut.Wolf at t-systems.com> wrote in message news:<c3h165$b9o$1 at smc.vnet.net>...
> A remark to add...
>
> >-----Original Message-----
> >From: Wolf, Hartmut
To: mathgroup at smc.vnet.net
> >Sent: Friday, March 19, 2004 7:36 AM
> >To: mathgroup at smc.vnet.net
> >Subject: [mg47040] Re: doing things on a procedural way
> >and doing them on a functional way
> >
> >
> >
> >>-----Original Message-----
> >>From: Paul Abbott [mailto:paul at physics.uwa.edu.au]
To: mathgroup at smc.vnet.net
> To: mathgroup at smc.vnet.net
> >>Sent: Thursday, March 18, 2004 10:38 AM
> >>To: mathgroup at smc.vnet.net
> >>Subject: [mg47040] Re: doing things on a procedural
> >>way and doing them on a functional way
> >>
> [...]
> >>
> >>I can't help but feel that there should be an elegant way to do this
> >>using Distribute ...
> >>
> >>Cheers,
> >>Paul
> >>
> [...]
> >
> >You're completely right,
> >the solution, I published yesterday, also works with
> >distribute (instead of Outer):
> >
> >In[32]:= nocnac[{arg_}, op_] := {arg}
> >In[33]:=
> >nocnac[{args__}, op_] :=
> > Flatten[ReplaceList[{args}, {a__, b__} :>
> > Distribute[op[nocnac[{a}, op], nocnac[{b}, op]], List]], 1]
> >
> >
> [...]
>
>
> Effectively both solutions are the same, as for Distribute you always can
> find an equivalent expression with Outer, e.g.:
>
> In[164]:= Distribute[List[{1, 2}, {{3, 4}, 5}, 6], List]
> Out[164]=
> {{1, {3, 4}, 6}, {1, 5, 6}, {2, {3, 4}, 6}, {2, 5, 6}}
>
> ...this is by definition the same as
>
> In[170]:= Distribute[List[{1, 2}, {{3, 4}, 5}, {6}], List]
> Out[170]=
> {{1, {3, 4}, 6}, {1, 5, 6}, {2, {3, 4}, 6}, {2, 5, 6}}
>
> ...functionally equivalent to
>
> In[166]:= Flatten[Outer[List, {1, 2}, {{3, 4}, 5}, {6}, 1], 2]
> Out[166]=
> {{1, {3, 4}, 6}, {1, 5, 6}, {2, {3, 4}, 6}, {2, 5, 6}}
>
>
> You may have observed that my first solution using Outer doesn't work with
> List, whereas that one above with Distribute does. This is not a problem of
> Outer, I just have been a bit careless with levels (at Flatten and Outer).
> This code also works with List:
>
> In[143]:=
> nocnac[{arg_}, op_] := {arg}
>
> nocnac[{args__}, op_] :=
> Flatten[ReplaceList[{args}, {a__, b__} :>
> Outer[op, nocnac[{a}, op], nocnac[{b}, op], 1]
> ], 2]
>
>
>
>
> >
> >
> >In[39]:= Length[nocnac[Range[#], op]] & /@ Range[10]
> >Out[39]=
> >{1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862}
> >
> >
> [...]
>
> >
> >I don't know, but assume that this can be expressed simply
> >through some sort of combinatorical numbers.
>
>
> Yes, indeed, it is given by
>
> In[171]:= Binomial[2 #, #]/(# + 1) & /@ Range[0, 9]
> Out[171]=
> {1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862}
>
> ...and the numbers
>
> c[n_] := Binomial[2n, n]/(n + 1)
>
>
> are called the Catalan numbers. See e.g.
> http://www.research.att.com/~njas/sequences/
> and my observation...
>
> >
> >In[57]:= nc[#]/nc[# - 1] & /@ Range[2, 30] // N
> >Out[57]=
> >{1., 2., 2.5, 2.8, 3., 3.14286, 3.25, 3.33333, 3.4,
> > 3.45455, 3.5, 3.53846, 3.57143, 3.6, 3.625, 3.64706,
> > 3.66667, 3.68421, 3.7, 3.71429, 3.72727, 3.73913,
> > 3.75, 3.76, 3.76923, 3.77778, 3.78571, 3.7931, 3.8}
> >
> [...]
> >
> >The conjecture of course is, that asymtotically nc[n] ~ 4^n.
> >
>
> is nothing else than the simple fact
>
> In[182]:= Limit[2n (2n - 1)/(n (n + 1)), n -> Infinity]
> Out[182]= 4
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