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## #351 2013-03-31 11:21:36

anonimnystefy
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### Re: PSLQ and LLL?

How?

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #352 2013-03-31 11:23:52

bobbym

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### Re: PSLQ and LLL?

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.

## #353 2013-03-31 11:25:41

anonimnystefy
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### Re: PSLQ and LLL?

Oh, that. I thought you meant that PSLQ can solve the problem from scratch.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #354 2013-03-31 11:26:19

bobbym

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### Re: PSLQ and LLL?

It can evaluate the integral, isn't that from scratch?

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.

## #355 2013-03-31 11:29:37

anonimnystefy
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### Re: PSLQ and LLL?

The numerical integration evaluates it, and the PSLQ just finds a nice form. The PSLQ cannot get the answer on its own, so that is not fron scratch.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #356 2013-03-31 11:31:47

bobbym

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### Re: PSLQ and LLL?

The PSLQ finds the correct form using the basis vector you provide. It is correct inasmuch as it will agree to the digits you provide for that vector. The other method requires help from residues and equation solvers so it can not do it by itself either.

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.

## #357 2013-03-31 11:33:41

anonimnystefy
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### Re: PSLQ and LLL?

It does do it on its own, because finding the residues is the part of tthe method. But solving the integral is not a part of the PSLQ.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #358 2013-03-31 11:35:13

bobbym

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### Re: PSLQ and LLL?

The roots of the equation are a separate process. And it requires an argand diagram.

You do not solve the integral, you get a numerical approximation.

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.

## #359 2013-03-31 11:39:42

anonimnystefy
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### Re: PSLQ and LLL?

Getting the roots is still a part of the contour integration method. Numerical integration is not a part of the PSLQ.

Anyway, it's just semantics. We can just agree to disagree.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #360 2013-03-31 11:42:15

bobbym

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### Re: PSLQ and LLL?

Does it matter whether or not what is part or what is not part? It is another method, that is what I am saying.

For instance, if I tell you that the integral we are now working on is equal to π / 3, how do you know that is correct? Are you going to trust the manipulations only?

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.

## #361 2013-03-31 11:48:35

anonimnystefy
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### Re: PSLQ and LLL?

I can check in another way. What does that have to do with anything?

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #362 2013-03-31 11:50:27

bobbym

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### Re: PSLQ and LLL?

Yes but the PSLQ is still another method. I never said it was the only other method.

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.

## #363 2013-03-31 11:53:07

anonimnystefy
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### Re: PSLQ and LLL?

But, it is not a method for solving the problem. It doesn't tell you all the steps necessary to do a problem. Contour integration method does.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #364 2013-03-31 11:55:37

bobbym

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### Re: PSLQ and LLL?

Is that a requirement for a method? Especially for a check? The steps of the program you typed in show you clearly the method.

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.

## #365 2013-03-31 12:14:47

anonimnystefy
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### Re: PSLQ and LLL?

It depends on what you expect it to do. PSLQ is a method for turning approximations into exact anwers, but it is not a method for solvibg integrals.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #366 2013-03-31 12:18:13

bobbym

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### Re: PSLQ and LLL?

I disagree, it is a method of proving an approximation is being well represented by a closed form to the the number of digits.

There is no known method for all integrals. As a matter of fact I believe the integral I showed you in that post was only done thanks to a PSLQ as are many others.

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.

## #367 2013-03-31 12:21:48

anonimnystefy
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### Re: PSLQ and LLL?

#### bobbym wrote:

I disagree, it is a method of proving an approximation is being well represented by a closed form to the the number of digits.

Yes, that is true. But, as I said, it sai, it is not a method for solving integrals.

It might be a part of a method for solving integrals, but it itself is not a method for solving integrals.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #368 2013-03-31 12:23:06

bobbym

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### Re: PSLQ and LLL?

It refines numerical integration which is a method to get an integral.

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.

## #369 2013-03-31 12:26:27

anonimnystefy
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### Re: PSLQ and LLL?

I do not see how that statement contradicts what I said in my last post.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #370 2013-03-31 12:36:51

bobbym

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### Re: PSLQ and LLL?

But as I said there are no cut and dry methods to solve an integral. More integrals can not be done than can be done by the standard methods. Numerical integration is a robust technique which theoretically can get a definite integral to any degree of precision. The PSLQ then determines whether that number has a simple relationship with the constants provided. Sounds like a method we all should be using and know.

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.

## #371 2013-03-31 12:41:29

anonimnystefy
Real Member

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### Re: PSLQ and LLL?

That still doesn't contradict what I said. Numerical integration+PSLQ is a method for evaluating integrals, PSLQ on its own is not a method for evaluating integrals.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #372 2013-03-31 12:47:26

bobbym

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### Re: PSLQ and LLL?

Neither is a contour. It requires, limits, derivatives and solving a tough polynomial.

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.

## #373 2013-03-31 13:00:03

anonimnystefy
Real Member

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### Re: PSLQ and LLL?

But it is! The solving of the equation, the limits and derivatives are all things that the method explicitly demands, and with all that is enough to solve an integral (not every though).

The PSLQ doesn't explicitly demand for numerical integration, and it itself is not enough to solve a problem.

Thos discussion is obviously pointless and I will not continue any further.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #374 2013-03-31 13:05:30

bobbym

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### Re: PSLQ and LLL?

Hi;

I am not angry at the discussion and you should not be either. If you do not want to discuss it anymore, I am okay with that. See you in the other thread.

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.

## #375 2013-03-31 13:07:28

anonimnystefy
Real Member

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### Re: PSLQ and LLL?

I am not angry. I am just tired of being misunderstood.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment