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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,274

RRA: Riemann Rearrangement Algorithm

First shown to me by the courtesy of Thomas E. Gantner

This was supposedly used by Riemann himself to show the folly of rearranging series irresponsibly. Few people are aware of its real usefulness.

For any convergent alternating series

we can usually generate a faster converging series with the transformation

Notice that the series inside the curly brackets is itself an alternating series and the algorithm can be reapplied on 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: 15,230

Hi bobbym

Good morning from me!

Thanks for creating a new thread! I assume you are going to expand your post so that we can later talk about it.

*Last edited by anonimnystefy (2012-04-10 18:17:18)*

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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: 84,274

Hi;

This thread will be devoted to it entirely. Examples and theory.

**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: 15,230

Cool! Again,thank you!

I figured out immideately why it works,but I am not sure how it applies to finite series or how many terms to take...

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: 84,274

Hi;

I think the purpose is geared more to an infinite series which is being estimated by a finite one to some desired accuracy.

**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: 15,230

But how many terms do we take for a 10 digits approximation?

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: 84,274

You can usually get the general term of the new alternating series

that is produced. From that you use the Leibniz 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: 15,230

Aha...Just what I thought. I just thought you had something better.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,274

It is enough to do the job, why do we need something else?

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: 15,230

Dunno,but it seems that every time I try to reuse a method,you have a better one.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,274

There are always other methods. In numerical analysis there are thousands of methods. Experience helps in the choice of the correct one. Someday, you might read "Numerical Methods that Usually Work," by the great Forman S. Acton, my teacher. You will then learn that the greatest numerical mathematician is a chemist!

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: 15,230

But,for now we should use Leibniz?

Can you give me a series to apply Leibniz to?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,274

Any alternating series that converges. What kind of example do you want?

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: 15,230

An easy one,for starters.Then maybe we can gradually go to some unusual series.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,274

I am not following you. All that Leibniz does is bound the tail of an alternating series. Is that what 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
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No.Give me a simple alternating function to estimate to a certain number of digits and then lower the number of terms needed to get that estimate.I want to do the process you showed me.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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How many terms are needed to estimate the above sum with 11 correct 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.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Well,our first approximation using Leibniz rules says we need 38 terms.But we can do better than that!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,274

Hi;

38 is correct! That is approximately what is going to be needed with a straight summation.

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
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I applied the RRA I got

When I used Leibniz again it told me I needed 37 terms.

*Last edited by anonimnystefy (2012-04-11 00:02:13)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,274

That is essentially what I got. You know what happened?

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: 15,230

Hi bobbym

No I don't.Did it make a faster converging series (obviously)?

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: 84,274

From post #1

we can usually generate a faster converging series with the transformation

In this case it does not accelerate the convergence.

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**

**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,230

Then what?

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: 84,274

Numerical techniques have conditions on them. When you violate those conditions strange things can occur. Take the simplest and most familiar numerical technique, Newtons iteration. People think they understand it but they do not. It is a very temperamental beast like the RRA.

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**