All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Fibonacci Fractions (Posted on 2005-03-09) Difficulty: 3 of 5
What is the sum of 0.1+ 0.01+ 0.002+ 0.0003+ 0.00005+ 0.000008+ ..., where each term is the n-th Fibonacci number, shifted n places to the right (that is, divided by 10^n)?

See The Solution Submitted by e.g.    
Rating: 3.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Extended precision using UBASIC | Comment 6 of 17 |

The following program evaluates the number numerically:
   10   F1=1:F2=1:Pwr=0.01
   15   Tot=0.11
   20   while Pwr>0
   30    F3=F1+F2
   35     PPwr=Pwr
   40    Pwr=Pwr/10
   50    Tot=Tot+F3*Pwr
   60    F1=F2:F2=F3
   70    print Tot
   80   wend
   90   print F1,int(-log(PPwr)/log(10)),int(-log(F1)/log(10)-log(PPwr)/log(10))

The line at the end is to determine how far we should trust the digits.  At the end the result came out to
 0.11235955056179775280898876404494382022471910112359550561797752808988764044943
82022471910112359550561797752808988764044943820224719101123595505617977528089887
64044943820224719101123595505617977528089887640449438202247191011235955056179775
28089887640449438202247191011235955056179775280898876404494382022471910112359550
56179775280898876404494382022471910112359550561797752808988764044943820224719101
12359550561797752808981346369378435976616688544876988124563923280800830811853429
4960708947265397738129336674012899278557170886817407
but the last Fibonacci number used was
 160315934185614122473327276500131314855604135915469629507118909263076087590471022644452597131283888029087109729
a 112-digit number so only the first 418 digits of the 529 shown can be relied upon.

But the sequence
1123595505617977528089887640449438202247191011235955056179775280898876404494382022471910
repeats every 88 digits, so this is a repeating decimal, or rational number. Its value can be obtained by dividing by a number with the same number of 9's:
1123595505617977528089887640449438202247191011235955056179775280898876404494382022471910
/ 9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999, which simplifies to 10/89.

So that seems to be the answer.  I can't imagine that its rationality is dependent on using the decimal system.  What if base 12 were used instead? Using the same procedure, but with base-12 instead (the following program):
   10   F1=1:F2=1:Pwr=1/144
   15   Tot=0.11
   20   while Pwr>0
   30    F3=F1+F2
   35     PPwr=Pwr
   40    Pwr=Pwr/12
   50    Tot=Tot+F3*Pwr
   60    F1=F2:F2=F3
   70    print Tot
   80   wend
   90   print F1,int(-log(PPwr)/log(10)),int(-log(F1)/log(10)-log(PPwr)/log(10))
we get (expressed in decimal)
 0.11132527565733672603901611535199321458863443596268023748939779474130619168787
10771840542832909245122985581000848176420695504664970313825275657336726039016115
35199321458863443596268023748939779474130619168787107718405428329092451229855810
00848176420695504664970313825275657336726039016115351993214588634435962680237489
39779474130619168787107718405428329092451229855810008481764206955046649703138252
75657336726039016115351993214539528242197440257200534106692895027679907764301385
4904136219838894999408607569446402474177970832755162

with 426 positions reliable.  After the 0.1113, the digits
2527565733672603901611535199321458863443596268023748939779474130
  619168787107718405428329092451229855810008481764206955046649703138
repeat, so we get


2527565733672603901611535199321458863443596268023748939779474130
619168787107718405428329092451229855810008481764206955046649703138
/ 9999999999999999999999999999999999999999999999999999999999999999
999999999999999999999999999999999999999999999999999999999999999999

= 298/1179

And we combine that with the first 0.1113 via: 298/11790000 + 1113/10000 = 52501/471600.

But that is the decimal representation when the number was formed in base-12.

The answer to the current problem is 10/89.


  Posted by Charlie on 2005-03-09 20:06:38
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information