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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,991

Hi;

A little more checking and it will be ready.

**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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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,991

Hi;

The formula is reasonably safe but who knows?!

The sum that generates the coefficients is:

Where

balls = total number of balls

urn = number of urns

max = maximum number of balls in any urn.

**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: 14,883

Have you tested it?

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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,883

Oh, and, a general formula for the line and squares problem:

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: 82,991

I have tested it a great deal it will produce the coefficients of the gf.

**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: 14,883

bobbym wrote:

New Problem:

E says)

The sum

I have this

, but I don't how much faster it is...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: 82,991

Hi;

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: 14,883

Hi bobbym

I didn't know of any acceleration method besides the RRA, and that one works for alternating sequences only, so I used one I found on Wolfram MathWorld.

What do you have?

*Last edited by anonimnystefy (2013-09-27 08:19:58)*

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,991

There are many other methods besides RRA. In addition, any series can be converted into an alternating one and then RRA or Euler applied.

A much better approach is this one mentioned in the Scheid book.

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: 14,883

Which is in turn

, so it's 1/k^5 rate of convergence!Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,991

That is true and the whole point but you had to see a trick first...

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: 14,883

Which trick?

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,991

The trick is how to sum that else you have replaced one cubic convergence with another. Do you see it?

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: 14,883

I do. Partial fractions and telescoping. It is what the page I found used. http://mathworld.wolfram.com/Convergenc … ement.html

Also, I found this quintic convergence series:

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,991

So you are only left with the sum on the extreme right and it has much faster convergence. Now you should numerically verify that.

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: 14,883

Hi bobbym

I cheated and checked with M that it does actually get the right answer.

What I do not know is how do we estimate how many terms are needed for some accuracy?

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,991

Which sum do you want to do?

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: 14,883

All mentioned so far, starting with the first one (sum of 1/k^3)...

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,991

Corny questions that come up in math courses which ask how many terms you need are replaced by what does the sum converge to and to how many digits can we get.

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: 14,883

That does not answer my question of how to actually get the number of needed terms...

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,991

I am not following you. It is a computational problem. To get the terms you have to use a computer and add them up. Then you need a tail analysis, remember most of the time you do not know what the sum is.

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: 14,883

But, in the original problem, you said we need 80000 terms. How did you get that number?

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,991

There are a couple of easy ways to back that statement up.First and simplest rule of thumb is the double rule.

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: 14,883

Give me something concrete, please. I still haven't the slightest how to get that estimate.

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,991

We can use the simplest command in M.

Sum[1/n^3, {n, 1, 40000}] // N

1.2020569028471022

Sum[1/n^3, {n, 1, 80000}] // N

1.2020569030814703

That is called the double method. Notice 8 digits passed the decimal point agree. You can expect the second answer is about accurate to 8 places.

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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