Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**gAr****Member**- Registered: 2011-01-09
- Posts: 3,478

I don't mind having the discussion.

"Believe nothing, no matter where you read it, or who said it, no matter if I have said it, unless it agrees with your own reason and your own common sense" - Buddha?

"Data! Data! Data!" he cried impatiently. "I can't make bricks without clay."

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,605

Thanks gAr.

Can you post the accelerator Borwein found?

I am sure I showed it to you already. I know we discussed it. It is also useless for this sequence. I used Romberg and it worked well.

Want the code for Borwien's?

**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.**

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,397

Yes. Is it the one with all the square roots and stuff?

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,605

```
accelerate[n_]:=Module[{d,b,c,s},
d=(3+Sqrt[8])^n;
d=(d+1/d)/2;
b=-1;
c=-d;
s=0;
Table[c=b-c;s=s+c*a[k];b=(k+n)(k-n) b/((k+1/2)(k+1)),{k,0,n-1}];
s/d]
```

Remember how to use 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.**

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,397

That's the one I was thinking of.

I set a[n] to be the array of the series terms. Then I do N[acc[number_of_terms],number_of_digits.

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,605

I am pretty sure it is just for alternating sequences.

**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.**

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,397

Yes, I know. It also has to start at 0.

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,605

You can adjust the index to handle that. Anyways it is not useful on this problem.

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.

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,397

Hi bobbym

You didn't tell me how we actually get the numerical answer here.

*Last edited by anonimnystefy (2013-12-11 10:57:23)*

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,605

Hi;

See post #527.

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.

Offline

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,397

Hm, isn't romberg for integrals?

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

Offline

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,605

It is for sequences that are the result of numerical integrations on an integral. Actually it is a bunch of sequence accelerators.

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.

Offline