Comments on: MMM7 Solutions

By: samjshah

samjshah — Thu, 05 Jun 2008 13:15:40 +0000

hi john,
oh i see! makes sense. i love when problems have different ways of working them.
sam.

By: john

john — Wed, 04 Jun 2008 23:25:49 +0000

ahem…in my last i meant to say i prefer F_0=1. oops.

By: john

john — Wed, 04 Jun 2008 23:25:13 +0000

i tend to disagree with you on that. i prefer F_0=0. but that wasn’t the problem i had. when i was using the direct approach i had to sum two (convergent of course) geometric series. using the usual closed form for the fibonacci numbers resulted in those series having zero for their first terms…which i didn’t catch until your solution. i re-indexed the geometric series, and got the 25. then i found a general solution for fractions of the form n/m. the result is m^2/(m^2-mn-n^2), and holds for 0<(n/m)<(sqrt(5)-1)/2. fun stuff! thanks again for posting your solution!

By: samjshah

samjshah — Wed, 04 Jun 2008 18:41:29 +0000

Ahhhhhh. Makes sense. There was a whole discussion about that in the comment section in the Wild About Math post for the problem.

People say that it is more traditional to write F_0=0 and F_1=1, instead of F_0=1 and F_1=1.

I think that I’ve seen it both ways. But if forced to make a choice, I would come down on the F_0=0 and F_1=1 side.

By: John

John — Wed, 04 Jun 2008 17:31:42 +0000

hooray…i found my mistake. i was summing the closed form from zero, which gave an incorrect first term in my resulting geometric series. it feels good to figure this one out. good fun!

By: John

John — Wed, 04 Jun 2008 12:26:52 +0000

thanks for the updates. seems i made a mistake after-all. UnA, i used (phi^n – (1-phi)^n)/sqrt(5) with phi = (1+sqrt(5))/2 for my fibonacci closed form. i don’t recall seeing that the closed form required using the “negative” golden ratio,
(1-sqrt(5))/2 (although that is inseed 1-phi). i must have read things wrongly. anyway, thanks for clearing things up!

By: UnA

UnA — Wed, 04 Jun 2008 11:19:48 +0000

hi john,
I directly computed the sum too,using the golden ratio and i arrived at the common answer of 25. I did gotten the answer of 15 before but it was a careless mistake on my part that the polarity of the signs for one of the golden ratio terms was wrong. hmm could you have made that too? Cheers

By: samjshah

samjshah — Wed, 04 Jun 2008 03:02:37 +0000

Hi John,

I did enter some numbers in Excel and I’m getting a sum (after 401 terms) of 24.999827.

Sam.

By: samjshah

samjshah — Wed, 04 Jun 2008 02:40:06 +0000

How weird! I think I’ll probably try entering some numbers in Excel and see if they approach one or the other. Then we’ll know where to look.

Thanks for the heads up. I definitely won’t show this to my schools MathClub yet… Oh, wait, sad, there aren’t any more.

The year is just coming to a screeching halt! Sigh.

Sam.

By: john

john — Wed, 04 Jun 2008 02:07:26 +0000

i really like your solution. i solved the series on funmathblog as well, but arrived at a different answer than you. using a closed form for the fibonnaci series (involving the golden ratio), i directly computed the series and arrived at 15 as the answer. i’ve checked both my work, and yours, for convergence. my solution generates two geometric series, bit of which converge. your power series (relying on the closed form for fibonacci) certainly converges with x=1 (by the way, it isn’t the fraction that needs to be in the radius of convergence…the value of x needs to be in it. however the chosen fraction does influence the radius). i can’t see why our solutions result in different answers when both seem so right…weird…