Is this cool with you?

bobbym · 2013-08-19 14:24:00

Hi;

A little more checking and it will be ready.

bobbym · 2013-08-19 19:14:57

Hi;

The formula is reasonably safe but who knows?!

The sum that generates the coefficients is:

Where

balls = total number of balls
urn = number of urns
max = maximum number of balls in any urn.

anonimnystefy · 2013-08-19 23:13:29

Have you tested it?

anonimnystefy · 2013-08-19 23:20:56

Oh, and, a general formula for the line and squares problem:

bobbym · 2013-08-20 00:01:49

I have tested it a great deal it will produce the coefficients of the gf.

anonimnystefy · 2013-09-27 07:24:00

bobbym wrote:

New Problem:
E says)
The sum

converges slowly. It takes 80,000 terms to get only 8 correct digits past the decimal. B can you speed up the convergence of this series? I know that Mathematica or Maple can get any number of digits for it but I would like to do it myself. What I am looking for is an acceleration of the series convergence.
A says) Hold on E, B isn't the only guy who can do things. I have never been wrong yet have I? Of course not. Just let me say there is no way. Now you can talk B.
B says) Thanks A you are too kind to me. Of course there is a simple way to produce a series that converges faster than that one. All you do is make use of...
C says) There you go again B. Contradicting A. There is a reason A goes first. He is the man! The best! He already said it can not be done.
D says) What is a series?
E says) A comes before B because of the dictionary. I want to hear what B has to say.
A says) I don't.
C says) Agreed!
D says) B, is it true that you are a distant cousin of Torricelli? Not Evangelista, the other one.
B says) Oh boy!?
Can you come up with a faster series that converges to the same value?

I have this

, but I don't how much faster it is...

bobbym · 2013-09-27 08:14:46

Hi;

anonimnystefy · 2013-09-27 08:19:45

Hi bobbym

I didn't know of any acceleration method besides the RRA, and that one works for alternating sequences only, so I used one I found on Wolfram MathWorld.

What do you have?

Last edited by anonimnystefy (2013-09-27 08:19:58)

bobbym · 2013-09-27 08:27:35

There are many other methods besides RRA. In addition, any series can be converted into an alternating one and then RRA or Euler applied.

A much better approach is this one mentioned in the Scheid book.

anonimnystefy · 2013-09-27 08:46:10

Which is in turn

, so it's 1/k^5 rate of convergence!

bobbym · 2013-09-27 08:49:32

That is true and the whole point but you had to see a trick first...

anonimnystefy · 2013-09-27 08:51:46

Which trick?

bobbym · 2013-09-27 08:53:11

The trick is how to sum that else you have replaced one cubic convergence with another. Do you see it?

anonimnystefy · 2013-09-27 09:06:15

I do. Partial fractions and telescoping. It is what the page I found used. http://mathworld.wolfram.com/Convergenc … ement.html

Also, I found this quintic convergence series:

bobbym · 2013-09-27 09:14:42

So you are only left with the sum on the extreme right and it has much faster convergence. Now you should numerically verify that.

anonimnystefy · 2013-09-27 09:24:11

Hi bobbym

I cheated and checked with M that it does actually get the right answer.

What I do not know is how do we estimate how many terms are needed for some accuracy?

bobbym · 2013-09-27 09:28:35

Which sum do you want to do?

anonimnystefy · 2013-09-27 09:29:59

All mentioned so far, starting with the first one (sum of 1/k^3)...

bobbym · 2013-09-27 09:33:51

Corny questions that come up in math courses which ask how many terms you need are replaced by what does the sum converge to and to how many digits can we get.

anonimnystefy · 2013-09-27 09:36:43

That does not answer my question of how to actually get the number of needed terms...

bobbym · 2013-09-27 09:43:04

I am not following you. It is a computational problem. To get the terms you have to use a computer and add them up. Then you need a tail analysis, remember most of the time you do not know what the sum is.

anonimnystefy · 2013-09-27 09:52:03

But, in the original problem, you said we need 80000 terms. How did you get that number?

bobbym · 2013-09-27 09:54:45

There are a couple of easy ways to back that statement up.First and simplest rule of thumb is the double rule.

anonimnystefy · 2013-09-27 09:57:58

Give me something concrete, please. I still haven't the slightest how to get that estimate.

bobbym · 2013-09-27 10:01:39

We can use the simplest command in M.

Sum[1/n^3, {n, 1, 40000}] // N

1.2020569028471022

Sum[1/n^3, {n, 1, 80000}] // N

1.2020569030814703

That is called the double method. Notice 8 digits passed the decimal point agree. You can expect the second answer is about accurate to 8 places.

Math Is Fun Forum

#1876 2013-08-19 14:24:00

Re: Is this cool with you?

#1877 2013-08-19 19:14:57

Re: Is this cool with you?

#1878 2013-08-19 23:13:29

Re: Is this cool with you?

#1879 2013-08-19 23:20:56

Re: Is this cool with you?

#1880 2013-08-20 00:01:49

Re: Is this cool with you?

#1881 2013-09-27 07:24:00

Re: Is this cool with you?

#1882 2013-09-27 08:14:46

Re: Is this cool with you?

#1883 2013-09-27 08:19:45

Re: Is this cool with you?

#1884 2013-09-27 08:27:35

Re: Is this cool with you?

#1885 2013-09-27 08:46:10

Re: Is this cool with you?

#1886 2013-09-27 08:49:32

Re: Is this cool with you?

#1887 2013-09-27 08:51:46

Re: Is this cool with you?

#1888 2013-09-27 08:53:11

Re: Is this cool with you?

#1889 2013-09-27 09:06:15

Re: Is this cool with you?

#1890 2013-09-27 09:14:42

Re: Is this cool with you?

#1891 2013-09-27 09:24:11

Re: Is this cool with you?

#1892 2013-09-27 09:28:35

Re: Is this cool with you?

#1893 2013-09-27 09:29:59

Re: Is this cool with you?

#1894 2013-09-27 09:33:51

Re: Is this cool with you?

#1895 2013-09-27 09:36:43

Re: Is this cool with you?

#1896 2013-09-27 09:43:04

Re: Is this cool with you?

#1897 2013-09-27 09:52:03

Re: Is this cool with you?

#1898 2013-09-27 09:54:45

Re: Is this cool with you?

#1899 2013-09-27 09:57:58

Re: Is this cool with you?

#1900 2013-09-27 10:01:39

Re: Is this cool with you?

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