From: keyintl@pop.iquest.net (Chris Adams)

>Harry,

>

>WOW! What a great web page. I spent several hours Sunday looking through

>your web pages. Unfortunately, for my wife, I spent time in your site

>rather than completing her "honey dos", ie. painting. I shared your site

>with a friend of mine, since it is simply the best, and developed two

>questions.

>

>You started your series with 0 and 1, indeed this is the classic case. But

>in general, you can start with any two integers to define such a sequence.

>In particular, the next number in the sequence is always the sum of the

>last two. E.g. if you start with 3 and 7 you get the Fibonacci sequence:

> 3, 7, 10, 17, 27, 44, 71, etc.

>

>With the classic start, it turns out that you can write a closed form for

>the nth term without having to calculate all of the middle terms. In the

>classic case the value of the nth term is

>

> F(n) = (Phi^n - PhiP^n)/Sqrt(5),

> where Phi = (1 + Sqrt(5))/2 = the Golden Ratio,

> and PhiP = Phi Prime = (1 - Sqrt(5))/2 = 1 - Phi = -1/Phi,

>as you have provided. Please note that in the classic Fibonacci sequence

>

>that the first term, 0, is the zeoth term or F(0) here.

>

>FIRST QUESTION, I would like to ask about a closed form solution for the

>general start.

>

>Also, for the classic series the ratio F(n+1)/F(n) gets closer and closer

>to the golden ratio as n gets larger and larger (If I recall, the limit

>from my calc days.) SECOND QUESTION, I would like to inquire about the

>limit of the ratio for the general series. (Something tells me that it is

>not very hard to establish that it also gets closer and closer to the

>golden ratio, about 1.618...).

Chris:

Thanks for buttering me up about my web page.

Good questions.

1) The sequence S(0), S(1), S(0) + S(1), S(0) + 2S(1), 2S(0) + 3S(1),

3S(0) + 5s(1), ...

has a closed form for the nth term of

S(n) = S(0) F(n-1) + S(1) F(n).

2) The ratio of S(n)/S(n-1) is

[S(0) F(n-1) + S(1) F(n)] / [S(0) F(n-2) + S(1) F(n-1)].

For large n, replace F(n) in the numerator with Phi F(n-1). Then the ratio is

[S(0) Phi F(n-2) + S(1) Phi F(n-1)] / [S(0) F(n-2) + S(1) F(n-1)] = Phi.

So, the limit of the ratio is always Phi.

-Harry

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