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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

Any other way? This one seems pretty impractical.

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Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,886

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 agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

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.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,886

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

http://www.youtube.com/watch?v=z4jeREy7Pbc

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 agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

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

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

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,886

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 agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

Ah, that one's nice.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,886

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.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,886

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.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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I know. It needs exponential convergence or some trick, I guess...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,886

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.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

Can't wait. Tomorrow, then!

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,886

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

**In mathematics, you don't understand things. You just get used to them.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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So, those sequence converters. What do they look like?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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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.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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Well, how does the first one work?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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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.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

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.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,886

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.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

That would be okay.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,886

Hi;

```
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.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

Does it work?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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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.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,673

Isn't it better to use recursion?

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