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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

repeat step 2.

*Last edited by bobbym (2013-02-25 00:18:28)*

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

Doing it by hand, I mean.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

Hi;

I do not know of a hand method. I think I remember one though but do not remember where I saw it.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

Do you remember what it was?

**zetafunc.****Guest**

Also, why does this method solve Pell equations?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

For that we would have to ask Mr. Fermat and Mr. Legendre.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

For which question?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

How taking the cf and convergents solves a Fermat-Pell equation.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

It's an unsolved problem?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

I do not think so. This is all covered in number theory books that do a little computation.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

I will see if I can find out about it somewhere, then...

Nothing from adriana yet, 16:41. Usually on Mondays she contacts me while she is at school.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

The method I showed is very important because it finds a simple continued fraction. This simple continued fraction's convergents are the best rational approximations of the constant. Other cf's do not have that property.

Look here:

http://en.wikipedia.org/wiki/Continued_fraction

*Last edited by bobbym (2013-02-25 05:01:36)*

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

I have skimmed that page for days but I've still not been able to find a method for finding a simple continued fraction that differs from yours (I believe they use your method to find the continued fraction for pi). However, I did notice that on the Wiki page for sqrt(3), they said this:

The square root of 3 can be expressed by generalised continued fractions such as [2; -4, -4, -4, ...] which is identical to [1; 1, 2, 1, 2, 1, 2, ...] evaluated at every second term.

The trouble is, I don't understand what they mean by that... this seems to be the important bit because they can get the simple CF from a generalised one. Do you know what they mean by 'evaluated at every second term'?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

I think that is a misprint.

Generalized ones a generated using other methods some of them tricky, than the floor method.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

But, are the two continued fractions I mentioned identical?

They do both converge to root 3, but I am unsure how I show that they're actually the same fraction...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

I think that is a misprint. The convergents of the two are different.

There is no 30 / 17 in the second group of convergents.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Hmm... so, do you remember the other way to get the simple continued fraction? Every page I go to seems to use your method but it is difficult to do by hand...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

First things first.

does not converge to the √3

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Yes it does... I evaluated it after a few terms and I got root 3 with an error of only 0.0000274 %.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

I am getting convergence to something close to 1.76393202250021

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

And, you are using the CF

?

I am getting root 3...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

Yes that cf is correct but is that

{2,-4,-4,-4,-4,-4...}?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Yes, they are just distributing the negative through the denominator, I think.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,481

I do not think that is correct.

{2,-4,-4,-4,-4,-4...} is

which is not converging to the √3

*Last edited by bobbym (2013-02-25 05:59:28)*

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Hmm, I agree, that one does not converge...