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## #26 2013-02-04 03:21:32

anonimnystefy
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### Re: Kummer and a tough integral.

That is not correct. With respect to t, 2pi*n is a constant, so it vanishes. You should have dt=dx.

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

## #27 2013-02-04 03:24:50

bobbym

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### Re: Kummer and a tough integral.

It does not vanish, multiplicative constants go to the front.

t = 2 n π x

dt/dx = 2 n π

dt = 2 n π dx

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.

## #28 2013-02-04 03:27:42

anonimnystefy
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### Re: Kummer and a tough integral.

You do not multiply x by it. You add it to x!

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

## #29 2013-02-04 03:30:34

bobbym

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### Re: Kummer and a tough integral.

Oh brother, I am doing the wrong problem. You said t=x+n*2pi and I am doing t=x*n*2pi

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.

## #30 2013-02-04 03:37:00

anonimnystefy
Real Member

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### Re: Kummer and a tough integral.

Yes. t=x+2pi*n.

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

## #31 2013-02-04 03:37:58

bobbym

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### Re: Kummer and a tough integral.

Okay thank you, I think I understand what he did there now.

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.

## #32 2013-02-04 04:23:30

anonimnystefy
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### Re: Kummer and a tough integral.

Great! So, do you think you will be able to expand it to 1000 digits with that?

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

## #33 2013-02-04 06:29:53

bobbym

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### Re: Kummer and a tough integral.

Hi;

Nope, objection 2 and 3 still apply.

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.

## #34 2013-02-04 07:36:47

anonimnystefy
Real Member

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### Re: Kummer and a tough integral.

Have you found any other ways?

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

## #35 2013-02-04 07:44:52

bobbym

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### Re: Kummer and a tough integral.

Nothing that could exceed about 100 digits. I used my idea with a few more bells and whistles.

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.

## #36 2013-02-04 09:48:26

anonimnystefy
Real Member

Offline

### Re: Kummer and a tough integral.

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

## #37 2013-02-04 10:02:34

bobbym

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### Re: Kummer and a tough integral.

I am not following you. Propositions? Do you mean other ideas?

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.

## #38 2013-02-04 10:32:22

anonimnystefy
Real Member

Offline

### Re: Kummer and a tough integral.

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

## #39 2013-02-04 10:44:58

bobbym

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### Re: Kummer and a tough integral.

Other problems like this 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.

## #40 2013-02-04 11:01:29

anonimnystefy
Real Member

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### Re: Kummer and a tough integral.

No. Did anyone tell you of another way to solve the integral problem?

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

## #41 2013-02-04 11:04:22

bobbym

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### Re: Kummer and a tough integral.

I am afraid not.

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.

## #42 2013-02-04 11:51:16

anonimnystefy
Real Member

Offline

### Re: Kummer and a tough integral.

Too bad. Be sure to tell me if you find something new.

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

## #43 2013-02-04 12:06:39

bobbym

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### Re: Kummer and a tough integral.

If I were to find something new I would tell immediately but every new idea just evaporates.

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.

## #44 2013-02-04 12:10:00

anonimnystefy
Real Member

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### Re: Kummer and a tough integral.

Great then!

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

## #45 2013-03-05 05:33:09

anonimnystefy
Real Member

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### Re: Kummer and a tough integral.

#### bobbym wrote:

To start, although any CAS known will gag on this problem they are able to do pieces of so it is to our advantage to split the integral like this.

The first integral is not too difficult and any CAS can do it to 100 digits.

Hi bobbym

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

## #46 2013-03-05 05:38:03

bobbym

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### Re: Kummer and a tough integral.

What does maxima provide for numeric integration that is built in?

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.

## #47 2013-03-05 05:48:43

anonimnystefy
Real Member

Offline

### Re: Kummer and a tough integral.

I don't think there is anything built in.

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

## #48 2013-03-05 05:55:39

bobbym

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### Re: Kummer and a tough integral.

There are numeric integration routines I think they are call quad_gags.

Also there is romberg and nint.

Last edited by bobbym (2013-03-05 06:02:50)

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.

## #49 2013-03-05 06:04:57

anonimnystefy
Real Member

Offline

### Re: Kummer and a tough integral.

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

## #50 2013-03-05 06:07:29

bobbym

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### Re: Kummer and a tough integral.

Does this run?

#### Code:

```(%i1) f(x) := (mode_declare (n, integer, x, float), n:n+1, exp(-x))\$
(%i2) translate(f)\$
Warning-> n is an undefined global variable.
(%i3) block ([rombergtol: 1.e-6, romberabs: 0.0], n:0, romberg (f, 0, 50));
(%o3)                   1.000000000488271
(%i4) n;
(%o4)                          257```

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.