Re: Functions That Remember Values They Have Found
- To: mathgroup at smc.vnet.net
- Subject: [mg127610] Re: Functions That Remember Values They Have Found
- From: Bob Hanlon <hanlonr357 at gmail.com>
- Date: Tue, 7 Aug 2012 03:02:06 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- Delivered-to: l-mathgroup@wolfram.com
- Delivered-to: mathgroup-newout@smc.vnet.net
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- References: <20120806083728.AC1FE683B@smc.vnet.net>
Clear[Hgeom, Hgeom2, Hgeom3]
Hgeom[n_] := H[n] = Sum[
(1 - 0.5)^i/(1 - 0.5^i) Hgeom[n - i],
{i, 1, n}]/n
Hgeom[0] = 1;
Hgeom[4]
0.203175
The memory is not effective when stored under a separate name
?? Hgeom
Use the same function name for the memory
Hgeom2[n_Integer] := Hgeom2[n] =
Sum[(1 - 0.5)^i/(1 - 0.5^i)
Hgeom2[n - i], {i, 1, n}]/n;
Hgeom2[0] = 1;
Hgeom2[4]
0.203175
?? Hgeom2
You could also use exact values
Hgeom3[n_Integer] := Hgeom3[n] =
Sum[2^-i/(1 - 2^-i)
Hgeom3[n - i], {i, 1, n}]/n;
Hgeom3[0] = 1;
Hgeom3[4]
64/315
?? Hgeom3
Note the timing differences
m = 18;
{Timing[Hgeom[m];][[1]],
Timing[Hgeom2[m];][[1]],
Timing[Hgeom3[m];][[1]]}
{3.09407, 0.001196, 0.002241}
Demonstrating that the results are equal
And @@ Table[
Hgeom[n] == Hgeom2[n] == Hgeom3[n],
{n, 0, m}]
True
Bob Hanlon
On Mon, Aug 6, 2012 at 4:37 AM, Esteban Gonz=E1lez Morales
<yo8231 at gmail.com> wrote:
> Hi, I was trying to create a recursive function and I read the help about=
it, and wrote this code
>
> Hgeom[n_] :=
> H[n] = Sum[(1 - 0.5)^i/(1 - 0.5^i) Hgeom[n - i], {i, 1, n}]/n
> Hgeom[0] = 1;
>
> However, when I calculate Hgeom[10] it gives me the right value, and then=
I ask for the information about Hgeom and get that it has values calculate=
d.
>
> Have you any idea what could have gone wrong?
>
- References:
- Functions That Remember Values They Have Found
- From: Esteban González Morales <yo8231@gmail.com>
- Functions That Remember Values They Have Found