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

You are not logged in.

- Topics: Active | Unanswered

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

Hi;

Here it is:

**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.**

Offline

**zetafunc.****Guest**

I'm confused how you got that... how are you getting that?

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

Repeated integration by parts is how that is generated. Most asymptotic series are generated by IBP.

**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.**

Offline

**zetafunc.****Guest**

Just tried it, I get the same answer. That is cool, I always thought that if you got an infinite series through IBP it was just poor form. I will try this with some other integrals...

Wait, is repeated IBP therefore another way of integrating something for which there is no closed form? So like an alternative to integrating the Taylor series?

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

Sometimes IBP will get the integral exactly, as you have seen. Sometimes it can get an asymptotic form like the one above. Sometimes it can get an ordinary power series. Each one has uses if used correctly.

**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.**

Offline

**zetafunc.****Guest**

You will get an asymptotic form for the integral of sinx / x too, right? (through repeated IBP)

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

Yes, that is an asymptotic form for any x.

*Last edited by bobbym (2013-02-21 15:20:15)*

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.

Offline

**zetafunc.****Guest**

Thank you, this was useful.

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

To compute with them you only use a finite number of terms rather than an infinite number like with a power series.

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.

Offline

**zetafunc.****Guest**

Because the terms quickly get relatively smaller enabling you to justify the accuracy of your sum?

I also realised that continued fractions can produce interesting results for things like sqrt(pi), rather than just the simple surds.

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

Yes, continu3d fractions is nice for approximating constants and they also solve pell equations.

Because the terms quickly get relatively smaller enabling you to justify the accuracy of your sum?

You keep using the terms that are getting smaller. The first term that gets larger you stop there.

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.

Offline

**zetafunc.****Guest**

Yes, I remember, using the convergents. I think you showed me a post of yours about it...

Why do you stop there, and not at the term right before that one?

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

That is what I meant. The term right before that 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.

Offline

**zetafunc.****Guest**

Oh, okay. Also, I have a question about continued fractions again. About 4 months ago you told me that the continued fraction of root 11 was {3; 3, 6, 3, 6, 3, 6, ...} with period 2.

I am getting it to be {1; 0.2, 0.2, 0.2, ...) with period 1, however. Or, in other words,

But the numerators must be 1 so I just divided through by 5 to get 0.2 for each quotient apart from the zeroth one.

Can you show me how you got a different continued fraction? Is mine not valid? It is a problem because I am getting different convergents, so I cannot solve Pell equations with my current method.

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

That is not the continued fraction for 11.

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.

Offline

**zetafunc.****Guest**

so

Where am I going wrong, if I truncate it after a bit it seems to converge well?

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

Hi;

That appears to be working. There is more than one way to represent a constant with cf's.

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.

Offline

**zetafunc.****Guest**

Could you show me how you got your continued fraction?

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

Do you remember the algorithm I showed you?

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.

Offline

**zetafunc.****Guest**

Yes, but it is impractical without a calculator/CAS...

Can't convergents be found from my fraction, somehow?

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

The method used in post #8266 seems okay but much more difficult than with a calculator.

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.

Offline

**zetafunc.****Guest**

I should really learn to like calculators...

So is my method not the typical way one would derive a continued fraction?

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

To answer that I suggest you do another √19.

I should really learn to like calculators...

Yes!!!!!! If you hope to get to CERN or any other installation in the world they will demand you use a calculator or a CAS. Start now, especially since there is someone around to teach you.

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.

Offline

**zetafunc.****Guest**

The convergence will probably be slower for larger numbers, but from my method, doesn't it follow that

where p is prime? So for your example p = 19...

I guess I don't mind calculators but I've never used a CAS before, I don't think. Aside from WolframAlpha or something... as for CERN, I don't know if I want to do that. That was back when I loved physics... at the moment I'm undecided.

**zetafunc.****Guest**

Actually ignore the 'p is prime' comment, that is irrelevant...