Is this cool with you?

anonimnystefy · 2013-09-27 10:07:50

Any other way? This one seems pretty impractical.

bobbym · 2013-09-27 10:11:13

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?

anonimnystefy · 2013-09-27 10:17:30

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.

bobbym · 2013-09-27 10:22:16

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.

anonimnystefy · 2013-09-27 10:55:49

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

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

bobbym · 2013-09-27 11:05:35

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.

anonimnystefy · 2013-09-27 11:37:34

Ah, that one's nice.

bobbym · 2013-09-27 11:47:18

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

anonimnystefy · 2013-09-27 13:36:30

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

bobbym · 2013-09-27 13:46:19

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

anonimnystefy · 2013-09-27 14:25:26

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

bobbym · 2013-09-27 14:47:24

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

anonimnystefy · 2013-09-27 14:58:50

Can't wait. Tomorrow, then!

bobbym · 2013-09-27 21:34:52

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

anonimnystefy · 2013-09-28 01:36:18

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

bobbym · 2013-09-28 03:18:40

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

anonimnystefy · 2013-09-28 03:49:47

Well, how does the first one work?

bobbym · 2013-09-28 03:57:31

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.

anonimnystefy · 2013-09-28 04:38:12

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.

bobbym · 2013-09-28 07:29:28

Hi;

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

anonimnystefy · 2013-09-28 10:53:29

That would be okay.

bobbym · 2013-09-28 11:02:58

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]]];

anonimnystefy · 2013-09-28 12:07:38

Does it work?

bobbym · 2013-09-28 12:11:42

Yes, try it on this simple one:

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

anonimnystefy · 2013-09-28 13:00:17

Isn't it better to use recursion?

Math Is Fun Forum

#1901 2013-09-27 10:07:50

Re: Is this cool with you?

#1902 2013-09-27 10:11:13

Re: Is this cool with you?

#1903 2013-09-27 10:17:30

Re: Is this cool with you?

#1904 2013-09-27 10:22:16

Re: Is this cool with you?

#1905 2013-09-27 10:55:49

Re: Is this cool with you?

#1906 2013-09-27 11:05:35

Re: Is this cool with you?

#1907 2013-09-27 11:37:34

Re: Is this cool with you?

#1908 2013-09-27 11:47:18

Re: Is this cool with you?

#1909 2013-09-27 13:36:30

Re: Is this cool with you?

#1910 2013-09-27 13:46:19

Re: Is this cool with you?

#1911 2013-09-27 14:25:26

Re: Is this cool with you?

#1912 2013-09-27 14:47:24

Re: Is this cool with you?

#1913 2013-09-27 14:58:50

Re: Is this cool with you?

#1914 2013-09-27 21:34:52

Re: Is this cool with you?

#1915 2013-09-28 01:36:18

Re: Is this cool with you?

#1916 2013-09-28 03:18:40

Re: Is this cool with you?

#1917 2013-09-28 03:49:47

Re: Is this cool with you?

#1918 2013-09-28 03:57:31

Re: Is this cool with you?

#1919 2013-09-28 04:38:12

Re: Is this cool with you?

#1920 2013-09-28 07:29:28

Re: Is this cool with you?

#1921 2013-09-28 10:53:29

Re: Is this cool with you?

#1922 2013-09-28 11:02:58

Re: Is this cool with you?

#1923 2013-09-28 12:07:38

Re: Is this cool with you?

#1924 2013-09-28 12:11:42

Re: Is this cool with you?

#1925 2013-09-28 13:00:17

Re: Is this cool with you?

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