Services & ResourcesWolfram ForumsMathGroup Archive

[Date Index] [Thread Index] [Author Index]

Re: I broke the sum into pieces

• To: mathgroup at smc.vnet.net
• Subject: [mg105089] Re: I broke the sum into pieces
• From: Roger Bagula <rlbagula at sbcglobal.net>
• Date: Fri, 20 Nov 2009 06:40:50 -0500 (EST)
• References: <4B0551F8.1050308@sbcglobal.net> <144987c90911190624s3f42b7e7g6545e37062631d1f@mail.gmail.com>
• Reply-to: rlbagulatftn at yahoo.com

Alexander Povolotsky wrote:
> Did you verify that with your friend - that guy who runs Mathematica group ?
> On 11/19/09, Roger Bagula <rlbagula at sbcglobal.net> wrote:
>   
>> Alexander Povolotsky
>>
>> This is the part Mathematica fails on:
>> Sum[(1/(64^n))*(1/(32*n + 333333)), {n, 0, Infinity}]
>> for some reason.
>> In[36]:=
>> N[333333/64,100]
>>
>> Out[36]=
>> 5208.3281250000000000000000000000000000000000000000000000000000000000000000000\
>> 00000000000000000000000
>>
>> In[39]:=
>> b[5]=Sum[(64^(-n-1))*(1/(n/2+5208.328125)),{n,0,Infinity}]
>>
>> Out[39]=
>> \!\(\(-2.05555574498401916904809250313`15.954589770191003*^18796\)\)
>> It just gives the wrong answer for that part.
>> Respectfully, Roger L. Bagula
>> 11759 Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
>> http://www.google.com/profiles/Roger.Bagula
>> alternative email: roger.bagula at gmail.com
>>
>>
>>
>>     
>
>   
Alexander Povolotsky,

No I haven't reported this.
Since the Mathematica guys ( Trott and Wolfram himself) put up my Beta 
cube fractal
without giving me any credit for it, they are on my list, ha, ha...

In trying to get a 5th term I tried:
$MaxExtraPrecision = 200
b[5] = Sum[(1/(64^n))*(1/(32*n + 3*1097)), {n, 0, Infinity}];
N[%, 100]
0.0003086349990279674310845177265991859681618864099951680507521205061977799618\
089309382536127817971331870245951164951`99.99999999999999
I'm pretty sure that is wrong too.

I'll cc this to him.
I can't use this approach to get closer to Pi at this rate.
Clear[b]
b[1]=Sum[(1/(64^n))*22/(32*n+7),{n,0,Infinity}];
b[2]=Sum[(1/(64^n))*(-1/(32*n+114)),{n,0,Infinity}];
b[3]=Sum[(1/(64^n))*(-1/(32*n+394)),{n,0,Infinity}];
b[4]=Sum[(1/(64^n))*(1/(32*n+781)),{n,0,Infinity}];
Sum[b[m],{m,1,4}];
N[%,100]
3.1415920104768860463819395769548831674912014189966869630288247319187743718795\
561171365118029140148452540476917807442`99.99999999999999
%-Pi
-6.431129071920807038063246197167059679803784188579461198603890420344066528814\
91523022428102222728100394732538`93.31113744211372*^-7
Respectfully, Roger L. Bagula
11759 Waterhill Road, Lakeside,Ca 92040-2905,tel: 619-5610814 :
http://www.google.com/profiles/Roger.Bagula
alternative email: roger.bagula at gmail.com

• Follow-Ups:
  • Re: Re: I broke the sum into pieces
    • From: Daniel Lichtblau <danl@wolfram.com>