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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,933

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.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

Okay, what now?

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,933

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.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

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.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,933

Did you write it?

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

Write what?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,933

That Romberg or all of Shakespeare?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

I wrote the romberg function. Why?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,933

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

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

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)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,933

Yes, I am familiar with memoization.

What do you mean by worse?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

```
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)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,933

Sequence acceleration has rules and the tail does not apply.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

Could you elaborate?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Notice how I constructed s.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,933

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

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,933

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

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,933

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

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

What do you suggest then?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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I tried shanks, it did not work.

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

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,861

Does any of them work on this one?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,933

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

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,933

Hi anonimnystefy;

A simple idea is to make use of the fact:

it is now easy to get 100+ digits.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**