Amitrano's formula (new)

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# Amitrano's formula (new)

Submitted by annagengi on Thu, 02/24/2005 - 15:45

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Binet's formula

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f(n) = f(n-2) + f(n-1)

f(0) = 0

f(1) = 1

x = [1 + sqrt(5)] / 2

y = [1 - sqrt(5)] / 2

f(n) = (x^n - y^n) / sqrt(5)

Amitrano's formula

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f(n) = a*f(n-2) + b*f(n-1)

For all a, b, f(0), f(1) are a member of R

x = [b + sqrt(b^2 + 4*a)] / 2

y = [b - sqrt(b^2 + 4*a)] / 2

f(n) = {a*f(0)*[x^(n-1) - y^(n-1)] + f(1)*[x^n - y^n]} / (x - y)

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## Re: Amitrano's formula (new)

Hi drini,

I have found a document in a math's italian site.

www.matematicamente.it (Section: Number theory).

Best regards

Anna

P.S.

I'm italian

## Re: Amitrano's formula (new)

so? all I'm claiming is that such result is not new, is just a condensed form of the characteristic method for solving linear homogeneous recurrences

f

G -----> H G

p \ /_ ----- ~ f(G)

\ / f ker f

G/ker f

## Re: Amitrano's formula (new)

what's so new about that formula? isn't that just an application of the characteristic method I used to derive binet's formula, but using variables for boundary conditions instead of 0 and 1 ?

f

G -----> H G

p \ /_ ----- ~ f(G)

\ / f ker f

G/ker f