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need help about sequence of fibonacci

if you cant see or that picture seem not clear , please feel free to ask me, thank you

Note by NurFitri Hartina
4 years, 3 months ago

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11 votes

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The generating function for the Fibonacci numbers is \(G(x) = \frac{x}{1-x-x^2}\). (Think of the infinite geometric series and what terms combine to make \(x^n\)...). So, you want to solve \(\frac{a}{1-a-a^2} = 2\) or \(2a^2 +3a -2 = 0\). Since we must have \(0<a<\frac{\sqrt{5}-1}{2}\), this leaves \(a=\frac{1}{2}\).

Eric Edwards - 4 years, 3 months ago

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Could you elaborate on how you found that closed form?

Tim Vermeulen - 4 years, 3 months ago

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Here is an interesting idea. Try to follow these steps carefully:

\( G(x) = x+x^2+2 x^3+3 x^4+5 x^5+8 x^6+13 x^7+\text{...} \)

\( \frac{G(x)}{x} = 1+x+2 x^2+3 x^3+5 x^4+8 x^5+13 x^6+\text{...} \)

\( \frac{G(x)}{x}-G(x) = 1+x^2+x^3+2 x^4+3 x^5+5 x^6+8 x^7+\text{...} \)

\( x G(x) = x^2+x^3+2 x^4+3 x^5+5 x^6+8 x^7+13 x^8+\text{...} \)

And finally,

\( \frac{G(x)}{x}-G(x)-x G(x)=1 \)

Solve this for \( G(x) \) to get \( G(x) = \frac{x}{1−x−x^2} \)

Ivan Stošić - 4 years, 3 months ago

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@Ivan Stošić Wow, that's amazing. It's like the first time I saw a proof of the sum of a geometric series, where suddenly all the terms canceled out.

Tim Vermeulen - 4 years, 3 months ago

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I highly recommend Wilf's Generatingfunctionology, which deals with formal generating functions. The second edition is available online for free on the author's site. Using this, you can actually generate a formal proof.

George Williams - 4 years, 3 months ago

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@George Williams That's looking really good, thank you.

Tim Vermeulen - 4 years, 3 months ago

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@George Williams that's very nice book, thanks

NurFitri Hartina - 4 years, 3 months ago

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