Another Fibonacci Contest
Tuesday, March 31, 2009 13:17
I am stuck on the Fibonacci series. A Wireless USP LP Evaluation Kit for the first to answer this.
If F(0) = 0 , F(1)= 1, etc
Solve for n
F(n-1)< googol <F(n)
When Fibonacci couldn't sleep,
He counted rabbits instead of sheep.
Wednesday, April 1 at 00:01 (GMT -8:00)
n=3
Wednesday, April 1 at 00:22 (GMT -8:00)
if you consider the fibonacci recurrence from negative no.s say ........-8,5,-3,2,-1,1,0,1,1,2,3,5,8
the answer might also include n=1,-1,-3,-5 and in postive side n>=3
therefore n=3 is the soluiton for your answer and if negative section of recurrence relation is included the values of n can be generalside as odd multiples.
Wednesday, April 1 at 08:12 (GMT -8:00)
Use Binet's formula for the n-th Fibonacci number:
F(n)=(phi^n-(-1/phi)^n)/sqrt(5)
Phi =(1+sqrt(5))/2 =1.618
as n will be very large (-1/phi)^n) will be negligible so we need to find n such that:
phi^(n-1)/sqrt5 < 10^100< phi^n/sqrt5
and it comes out to be n480.19
therefore solution is 480.19
Wednesday, April 1 at 08:13 (GMT -8:00)
n> 480.19
n< 481.19
solution is 480.19< n < 481.19
Wednesday, April 1 at 09:31 (GMT -8:00)
i made a calculation error. the answer is
n=476
Thursday, April 2 at 05:41 (GMT -8:00)
476
Thursday, April 2 at 06:02 (GMT -8:00)
i made an arithmetic error i believe while taking logarithm
the final answer is 480.17
n=481
Thursday, April 2 at 06:03 (GMT -8:00)
thank you sir for giving such problems it made me brush up my school mathematics i had solved this problem thrice now and i got the answer i suppose
Sunday, April 5 at 05:24 (GMT -8:00)
480.17
Tuesday, May 5 at 08:31 (GMT -8:00)
(solved with Mathematica)
F[958] =7.25562*10^99= 0.725562 googol< 1 googol
F[959] =1.17398*10^100= 1.17398 googol> 1 googol
F(n-1)< googol
Tuesday, May 5 at 08:32 (GMT -8:00)
(solved with Mathematica)
F[958] =7.25562*10^99= 0.725562 googol< 1 googol, F[959] =1.17398*10^100= 1.17398 googol> 1 googol
F(n-1)< googol
Tuesday, May 5 at 08:34 (GMT -8:00)
n=958 (solved with Mathematica), F[958] =7.25562*10^99= 0.725562 googol< 1 googol, F[959] =1.17398*10^100= 1.17398 googol> 1 googol, F(n-1)< googol
Wednesday, May 6 at 09:51 (GMT -8:00)
n=480.17
Friday, May 8 at 01:37 (GMT -8:00)
@vishal:
I don't know why you corrected your answer to "n=480.17", since when F(0)= 0 , F(1)= 1, ... the answer has to be an integer.
So my Excel says that googol is between F(480) and F(481). ;-)
Assuming that within the summing of 481 elements the mantissa(16)-exponent-error is marginal:
n = 481.
Friday, May 8 at 03:02 (GMT -8:00)
By the way:
"Asimov on numbers"
Portuguese version (page 34): http://tecnoeduka.110mb.com/documentos/matematica/numeros%20-%20asimov.pdf
"En efecto, F(481) es mayor que un googol."
... but who reads Asimov nowadys?
... have to go and get some Asimov stories for my kids ... and me :-)
Markus
Sunday, May 10 at 10:48 (GMT -8:00)
answer is 481 but i gave the range since solution lies in the interval 480.17 to 481.17
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