Readers,
I would like to propose a challenge.
What follows ...
a) 1 - 2 - 6 - 15 - 40 - 104 - 273 - ?
b) If the sequence in a) is generated by a function f: N->R then what is f[n] ? Or... what is the closed form for the sequence in a).
c) Modify the polynomial involved in creating f such that at least one power of ten becomes a member of the image of f.
Remarks:
a) Should be simple for those with a pass on MS221;
b) Not solvable with the standard MS221 / M208 tools but I'll expect 221+ folk will accept the challenge;
c) No hints as it is the bonus question;
Solvers are nominated for the gallery of excellence.
Update:
( 1/5-'10)
What sort of a person are you? Do you give it a try? Or don't you think it isn't worth the time? Then may I ask, why read my blog in the first place? Do you only like the " TMA-sort-of-exercices " which are easy to solve and get marked? Then... mathematics is not for you. Think about it. It concerns: -you-.
Anyway, I designed parts b) and c) of the exercise for myself. I haven't find the answer yet. It's more difficult than I thought. But I am learning... I'll solve it.l
2-2024 Quran and mathematics
2 weeks ago
u1 = 1 x 1
ReplyDeleteu2 = 1 x 2
u3 = 2 x 3
u4 = 3 x 5
u5 = 5 x 8
u6 = 8 x 13
u7 = 13 x 21
We recognise the 2 columns of factors as two instances of the Fibonacci sequence.
Giving:
u8 = Fib(8)Fib(9) = 714
un = Fib(n)Fib(n+1)
From here we can use the Binet formula (to get something horrible for the closed form)
That answer is worth 33% of the marks. A resit I am afraid. ;-)
ReplyDeleteThe closed form is rather sexy.
1/10 (-2 (-1)^n-(Sqrt[5]-1) (1/2 (3-Sqrt[5]))^n+(1+Sqrt[5]) (1/2 (3+Sqrt[5]))^n)
I had the unalluring:
ReplyDeletephi = (1 + sqr(5))/2
Un = (1/5)((phi)^(2n+1) + (1-phi)^(2n+1) + (-1)^(n+1))