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

You are not logged in.

## #526 2013-12-05 18:15:32

gAr
Star Member

Offline

### Re: Series

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."

## #527 2013-12-05 21:09:13

bobbym

Online

### Re: Series

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.

## #528 2013-12-05 22:08:24

anonimnystefy
Real Member

Offline

### Re: Series

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

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #529 2013-12-05 22:14:41

bobbym

Online

### Re: Series

#### Code:

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.

## #530 2013-12-06 00:20:57

anonimnystefy
Real Member

Offline

### Re: Series

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.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #531 2013-12-06 00:24:05

bobbym

Online

### Re: Series

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.

## #532 2013-12-06 00:27:34

anonimnystefy
Real Member

Offline

### Re: Series

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

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #533 2013-12-06 00:31:01

bobbym

Online

### Re: Series

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

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.

## #534 2013-12-12 09:55:31

anonimnystefy
Real Member

Offline

### Re: Series

Hi bobbym

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

Last edited by anonimnystefy (2013-12-12 09:57:23)

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #535 2013-12-12 10:17:58

bobbym

Online

### Re: Series

Hi;

See post #527.

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.

## #536 2013-12-12 10:22:21

anonimnystefy
Real Member

Offline

### Re: Series

Hm, isn't romberg for integrals?

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #537 2013-12-12 10:25:46

bobbym

Online

### Re: Series

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

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.