Is this cool with you?

bobbym · 2013-09-28 13:16:49

Recursion in my opinion is rarely the best way to go.

It performs well for the tasks it is given. The lists are rarely longer than 50 and usually shorter.

When we begin you will see why this program is more than fast enough. But if you want to optimize it then go ahead.

anonimnystefy · 2013-09-28 19:38:43

Okay, what now?

bobbym · 2013-09-28 21:49:55

Hi;

Lets start with the way a dummy would approach this problem.

Make a table:

s=Table[Sum[1/n^3, {n, 1, 2^m}], {m, 0, 17}]

That is really stupid code!

s//N

{1., 1.125, 1.177662037037037, 1.1951602435617104, \
1.2002220387226148, 1.2015836423576125, 1.2019367252957784, \
1.2020266230687449, 1.202049303509178, 1.2020549995326137, \
1.20205642678787, 1.2020567840084981, 1.2020568733645467, \
1.2020568957099231, 1.202056901297063, 1.2020569026939472, \
1.2020569030431807}

Using the double rule we can see that we probably have about 8 places.

Now run

N[romberg[s], 50]

1.2020569031595942853997381615114499907649857486888

we see we have about 41 places! A great acceleration.

anonimnystefy · 2013-09-30 06:26:19

Hi

Here's my code for romberg:

romberg[l_] := Module[{x, a}, a = Dimensions[l][[1]]; x = 1/2;
   Clear[f];
   f[n_, 1] := f[n, 1] = l[[n]];
   f[i_, j_] := f[i, j] = (f[i, j - 1] - x^(j - 1)*f[i - 1, j - 1])/(1 - x^(j - 1));
   f[a, a]];

Hm, let me try it on another one.

bobbym · 2013-09-30 06:33:48

Did you write it?

anonimnystefy · 2013-09-30 06:36:00

Write what?

bobbym · 2013-09-30 06:39:52

That Romberg or all of Shakespeare?

anonimnystefy · 2013-09-30 06:44:12

I wrote the romberg function. Why?

bobbym · 2013-09-30 06:46:46

That is good. Did you get the point of it in post 1928?

anonimnystefy · 2013-09-30 06:48:31

Hi

I did. I tried it on both of our accelerated sequences, and it worked worse than on the original sequence...

If you were wondering about the stuff I used, like f[i,j]:=f[i,j]=...; it's something I found a while ago while searching for a way to do memoization in recursion in M.

Last edited by anonimnystefy (2013-09-30 06:49:44)

bobbym · 2013-09-30 06:56:39

Yes, I am familiar with memoization.

What do you mean by worse?

anonimnystefy · 2013-09-30 07:00:20

Table[Sum[(11 i + 6)/(i^3 (i + 1) (i + 2) (i + 3)), {i, 1, n}], {n, 1, 50}];

N[romberg1[%], 30]
0.785389896574820021713652114058

5/12 + % - Zeta[3]
-3.39918107597019419380786*10^-7

Last edited by anonimnystefy (2013-09-30 07:01:25)

bobbym · 2013-09-30 07:10:18

Sequence acceleration has rules and the tail does not apply.

anonimnystefy · 2013-09-30 07:11:27

Could you elaborate?

bobbym · 2013-09-30 07:13:11

Notice how I constructed s.

bobbym · 2013-09-30 07:16:23

Sequences that are decreasing geometrically and have steep curves seem to accelerate better. The tail is almost flat.

anonimnystefy · 2013-09-30 07:16:49

Oh! It goes from 1 to 2^m. I didn't see that one. Sorry! I am getting it now. Thanks!

bobbym · 2013-09-30 07:18:21

The point is we only needed a couple of terms to get 41 digits. Continuing that table would have required about 10^20.

anonimnystefy · 2013-09-30 07:24:27

Yeah, I know. It's a good one. Only about 130 000 is much more doable.

bobbym · 2013-09-30 07:25:55

Romberg is working fine for about 50 digits but for 100 or so another idea is necessary.

anonimnystefy · 2013-09-30 08:02:39

What do you suggest then?

bobbym · 2013-09-30 08:08:39

I tried shanks, it did not work.

There are 3 areas open yet. Winjgaarden transformation,Euler Mclaurin and the new accelerator I have.

anonimnystefy · 2013-09-30 08:42:30

Does any of them work on this one?

bobbym · 2013-09-30 09:15:50

I do not know yet, as soon as I finish another problem I will start on this one.

bobbym · 2013-10-01 12:06:32

Hi anonimnystefy;

A simple idea is to make use of the fact:

it is now easy to get 100+ digits.

Math Is Fun Forum

#1926 2013-09-28 13:16:49

Re: Is this cool with you?

#1927 2013-09-28 19:38:43

Re: Is this cool with you?

#1928 2013-09-28 21:49:55

Re: Is this cool with you?

#1929 2013-09-30 06:26:19

Re: Is this cool with you?

#1930 2013-09-30 06:33:48

Re: Is this cool with you?

#1931 2013-09-30 06:36:00

Re: Is this cool with you?

#1932 2013-09-30 06:39:52

Re: Is this cool with you?

#1933 2013-09-30 06:44:12

Re: Is this cool with you?

#1934 2013-09-30 06:46:46

Re: Is this cool with you?

#1935 2013-09-30 06:48:31

Re: Is this cool with you?

#1936 2013-09-30 06:56:39

Re: Is this cool with you?

#1937 2013-09-30 07:00:20

Re: Is this cool with you?

#1938 2013-09-30 07:10:18

Re: Is this cool with you?

#1939 2013-09-30 07:11:27

Re: Is this cool with you?

#1940 2013-09-30 07:13:11

Re: Is this cool with you?

#1941 2013-09-30 07:16:23

Re: Is this cool with you?

#1942 2013-09-30 07:16:49

Re: Is this cool with you?

#1943 2013-09-30 07:18:21

Re: Is this cool with you?

#1944 2013-09-30 07:24:27

Re: Is this cool with you?

#1945 2013-09-30 07:25:55

Re: Is this cool with you?

#1946 2013-09-30 08:02:39

Re: Is this cool with you?

#1947 2013-09-30 08:08:39

Re: Is this cool with you?

#1948 2013-09-30 08:42:30

Re: Is this cool with you?

#1949 2013-09-30 09:15:50

Re: Is this cool with you?

#1950 2013-10-01 12:06:32

Re: Is this cool with you?

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