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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,724

That is very good, you kept the dominant terms.

**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: 14,860

Exactly. They seem to do the job good enough. I will try getting the closed form analitically today.

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: 82,724

Uh wait. That is the analytical answer up there. So you will looking for a shorter one?

**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: 14,860

You didn't get it by analityc methods. You used M's FindSequenceFunction command. I want to see if I can get the answer analitically.

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: 82,724

All you provided me was with your original algorithm. I had to use the sequence. And why not just try induction and prove the a_n is correct?

**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: 14,860

Do you think there is a nicer reccurence for the sequence?

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: 82,724

Maybe. The weirdest thing is that sequences can have more than one difference equation.

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: 14,860

I have the difference equation. It's a[n+4]-4a[n+2]-a[n]+4a[n-2]=0. Don't ask how I got it!

*Last edited by anonimnystefy (2013-04-17 04:41:07)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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I am going to ask just 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: 14,860

It is a bit tough to explain. I played a little bit of spot-the-pattern...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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The solution to that will probably be longer than the other 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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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,860

I think you will get the same solution.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Possibly, or the trigonometric terms will be replaced by (-1)^n or something like 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: 14,860

The trigonometric terms will be replaced by i^n and (-i)^n.

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: 82,724

Yes, that is possible too but I would not exactly call that simpler.

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: 14,860

True. But the reduced form looks really nice, though!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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It is an asymptotic form yes? It is supposed to be shorter and easier to compute.

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: 14,860

Yes. It looks much much prettier than the nasty exact one.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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But you possibly missed the forest for the trees.

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: 14,860

How?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,724

It appears there might be a pattern worth investigating that is much simpler. It may not hold but it is worth a look. You have to get out of the box axiomatic math puts you in. It is small, dark and has little air.

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: 14,860

Which pattern? In the binary representation? That is the one I used to get the difference equation.

Hm, could you do asymptotic_a[n]-a[n] for me?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 82,724

One question at a time can be investigated well. This one is more important.

Which pattern? In the binary representation?

No, not that one. Look below, is there anything cooking?

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: 14,860

Well, they approach 1. That is why my formula is an asymptotic one. I also notice that every fourth term repeats two terms later. I do not notice anything else.

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: 82,724

Yes, the differences between the asymptotic form and the actual answer follow a pattern.

1,3, then 1,1,1,3...

This suggests 2 recurrences and just adding the appropriate constant. Now you would have a short exact answer.

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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