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

You are not logged in.

## #1901 2013-09-28 08:07:50

anonimnystefy
Real Member

Online

### Re: Is this cool with you?

Any other way? This one seems pretty impractical.

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

## #1902 2013-09-28 08:11:13

bobbym

Online

### Re: Is this cool with you?

Hmmm, impractical? In 135 years of working with that rule it failed only once! Now that may not be good enough for a load of mathematicians like A that never calculated anything but it is a workhorse for us.

Sometimes we can do a tail analysis which can help us know how close we are. You are aware of the methods of numerical integration? Know some of the theory?

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.

## #1903 2013-09-28 08:17:30

anonimnystefy
Real Member

Online

### Re: Is this cool with you?

I'm not sure. I have heard of a few formulas, including the trapezoidal, simpsons, romberg,... Of these, I know only what the trapezoidal is. I am willing to learn it though.

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

## #1904 2013-09-28 08:22:16

bobbym

Online

### Re: Is this cool with you?

You know of a simpler one than those. You know about stuffing rectangles to get an area?

By the way, when you say things like you said in the previous post I am reminded of what pappym told me.

bobbym, everything you know is wrong! - pappym

Everything they taught you is wrong. Think for yourself, do not opine the words of others. The double method is highly practical and we can improve it.

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.

## #1905 2013-09-28 08:55:49

anonimnystefy
Real Member

Online

### Re: Is this cool with you?

Oh, yes, Riemann integration. I also remember Euler-Mclaurin...

So, what kind of tail analysis do you have in mind?

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

## #1906 2013-09-28 09:05:35

bobbym

Online

### Re: Is this cool with you?

The simplest one, If areas ( integrals ) can be approximated by rectangles then rectangles can be approximated by integrals. Each 1 x n rectangle can be thought of as a term in the sum.

Testing this we see that we are off by

by actual computation. There are some conclusions, notice that the double rule was predicting 8 and we really had about 10. Showing that it is on the conservative side and can be trusted.

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.

## #1907 2013-09-28 09:37:34

anonimnystefy
Real Member

Online

### Re: Is this cool with you?

Ah, that one's nice.

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

## #1908 2013-09-28 09:47:18

bobbym

Online

### Re: Is this cool with you?

With it you can answer your question of how many terms you would need for 100 digits of accuracy.

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.

## #1909 2013-09-28 11:36:30

anonimnystefy
Real Member

Online

### Re: Is this cool with you?

10^50 with the original series! That is really a lot!!

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

## #1910 2013-09-28 11:46:19

bobbym

Online

### Re: Is this cool with you?

For that many digits none of the methods looked at so far will suffice.

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.

## #1911 2013-09-28 12:25:26

anonimnystefy
Real Member

Online

### Re: Is this cool with you?

I know. It needs exponential convergence or some trick, I guess...

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

## #1912 2013-09-28 12:47:24

bobbym

Online

### Re: Is this cool with you?

We will get into a few sequence converters that might help tomorrow.

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.

## #1913 2013-09-28 12:58:50

anonimnystefy
Real Member

Online

### Re: Is this cool with you?

Can't wait. Tomorrow, 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

## #1914 2013-09-28 19:34:52

bobbym

Online

### Re: Is this cool with you?

I did a little preliminary work. It was not very encouraging.

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.

## #1915 2013-09-28 23:36:18

anonimnystefy
Real Member

Online

### Re: Is this cool with you?

So, those sequence converters. What do they look like?

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

## #1916 2013-09-29 01:18:40

bobbym

Online

### Re: Is this cool with you?

I use a couple of them. Shanks, Romberg, Aitken, Euler, RRA, and a new 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.

## #1917 2013-09-29 01:49:47

anonimnystefy
Real Member

Online

### Re: Is this cool with you?

Well, how does the first one work?

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

## #1918 2013-09-29 01:57:31

bobbym

Online

### Re: Is this cool with you?

Romberg acceleration worked best so far. We will start there.

Romberg integrates by taking a sequence of values and using Richardson extrapolation on them. This means it can accelerate many sequences. Have you ever programmed a Romberg integration?

If you have not and do not know how to write it I will give you mine.

See you a little later, chores are calling.

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.

## #1919 2013-09-29 02:38:12

anonimnystefy
Real Member

Online

### Re: Is this cool with you?

I haven't programmed it, but I remember you helping me find it among the Maxima packages and use it for the 1/(cos x+x^2) 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

## #1920 2013-09-29 05:29:28

bobbym

Online

### Re: Is this cool with you?

Hi;

I can give you mine if you want and we can go on.

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.

## #1921 2013-09-29 08:53:29

anonimnystefy
Real Member

Online

### Re: Is this cool with you?

That would be okay.

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

## #1922 2013-09-29 09:02:58

bobbym

Online

### Re: Is this cool with you?

Hi;

#### Code:

```romberg[l_] := Module[{x, a, o, l1}, a = Dimensions[l][[1]]; x = 1/2;
o = Table[0, {a}, {a - 1}];
l1 = augment[Transpose[{l}], o];
For[i = 2, i <= a, ++i,
For[j = 2, j <= a, ++j,
l1[[i, j]] = (l1[[i, j - 1]] -
x^(j - 1)*l1[[i - 1, j - 1]])/(1 - x^(j - 1));];];
l1[[a, a]]];

augment[a_, b_] := Transpose[Join[Transpose[a], Transpose[b]]];```

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.

## #1923 2013-09-29 10:07:38

anonimnystefy
Real Member

Online

### Re: Is this cool with you?

Does it work?

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

## #1924 2013-09-29 10:11:42

bobbym

Online

### Re: Is this cool with you?

Yes, try it on this simple one:

romberg[{1, 5/4, 205/144, 1077749/705600,
822968714749/519437318400}]

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.

## #1925 2013-09-29 11:00:17

anonimnystefy
Real Member

Online

### Re: Is this cool with you?

Isn't it better to use recursion?

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