Is this cool with you?

anonimnystefy · 2013-10-01 18:13:30

How did you get that?

bobbym · 2013-10-01 18:37:26

http://mathworld.wolfram.com/RiemannZetaFunction.html

Formula 19 - 20

But the important thing is why is it better?

anonimnystefy · 2013-10-02 05:57:12

Because it's alternating, of course. RRA is not working too fast, though. I cannot evaluate it quickly...

bobbym · 2013-10-02 09:35:33

Alternating series can be hard to accelerate too! Heard of the converger?

anonimnystefy · 2013-10-02 16:32:36

No. What is it?

bobbym · 2013-10-02 19:11:27

acc[n_]:=Module[{d,b,c,s},
d=(3+Sqrt[8])^n;
d=(d+1/d)/2;
b=-1;
c=-d;
s=0;
Table[c=b-c;s=s+c*a[k];b=(k+n)(k-n) b/((k+1/2)(k+1)),{k,0,n-1}];
s/d]

anonimnystefy · 2013-10-03 01:20:23

How does it work?

bobbym · 2013-10-03 02:21:58

I knew you were going to say that...

anonimnystefy · 2013-10-03 02:29:26

And you do not know?

bobbym · 2013-10-03 02:34:26

I knew you were going to ask that.

First, it is for alternating series only. So enter

a[k_]:=1 / ( k + 1 ) then

N[acc[250],200]

anonimnystefy · 2013-10-03 02:48:43

Hi

I am getting 0.69314718055994530941723212145817656807550013436025525412068000949339362196969471560586332699641868754200148102057068573368552023575813055703267075163507596193072757082837143519030703862389167347122161...

bobbym · 2013-10-03 02:50:10

Now enter:

N[Log[2],200]

what do you notice?

anonimnystefy · 2013-10-03 02:53:48

Wow, 196 digits!

bobbym · 2013-10-03 02:56:09

No what series you just did?

anonimnystefy · 2013-10-03 03:04:46

bobbym · 2013-10-03 03:06:25

The converger always starts at 0.

anonimnystefy · 2013-10-03 03:22:02

Same series.

bobbym · 2013-10-03 03:24:22

Yes, but it will affect what you enter into a[k_]

anonimnystefy · 2013-10-03 03:58:23

That is true, but you asked what the series was.

bobbym · 2013-10-03 04:04:05

Yes, but I meant in the form that the converger can use.

1) The series must be alternating.

2) It must start at 0.

The function acc has an input and a global variable a. These must be in the precise form required. That is why I did the transformation on the zeta series.

anonimnystefy · 2013-10-03 23:35:20

No, no, I got that it must start at 0. I was just saying that what I wrote and what you wrote are the same series.

So, you've got no idea how and why it works?

bobbym · 2013-10-04 01:41:09

None at all. Did you see how it works on zeta(3)?

anonimnystefy · 2013-10-04 01:52:50

Haven't tried. I will do it tomorrow.

What is tthe algorithm based on?

bobbym · 2013-10-04 01:58:51

The converger? I do not know,

anonimnystefy · 2013-10-04 21:24:14

Too bad.

Math Is Fun Forum

#1951 2013-10-01 18:13:30

Re: Is this cool with you?

#1952 2013-10-01 18:37:26

Re: Is this cool with you?

#1953 2013-10-02 05:57:12

Re: Is this cool with you?

#1954 2013-10-02 09:35:33

Re: Is this cool with you?

#1955 2013-10-02 16:32:36

Re: Is this cool with you?

#1956 2013-10-02 19:11:27

Re: Is this cool with you?

#1957 2013-10-03 01:20:23

Re: Is this cool with you?

#1958 2013-10-03 02:21:58

Re: Is this cool with you?

#1959 2013-10-03 02:29:26

Re: Is this cool with you?

#1960 2013-10-03 02:34:26

Re: Is this cool with you?

#1961 2013-10-03 02:48:43

Re: Is this cool with you?

#1962 2013-10-03 02:50:10

Re: Is this cool with you?

#1963 2013-10-03 02:53:48

Re: Is this cool with you?

#1964 2013-10-03 02:56:09

Re: Is this cool with you?

#1965 2013-10-03 03:04:46

Re: Is this cool with you?

#1966 2013-10-03 03:06:25

Re: Is this cool with you?

#1967 2013-10-03 03:22:02

Re: Is this cool with you?

#1968 2013-10-03 03:24:22

Re: Is this cool with you?

#1969 2013-10-03 03:58:23

Re: Is this cool with you?

#1970 2013-10-03 04:04:05

Re: Is this cool with you?

#1971 2013-10-03 23:35:20

Re: Is this cool with you?

#1972 2013-10-04 01:41:09

Re: Is this cool with you?

#1973 2013-10-04 01:52:50

Re: Is this cool with you?

#1974 2013-10-04 01:58:51

Re: Is this cool with you?

#1975 2013-10-04 21:24:14

Re: Is this cool with you?

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