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estudier
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Prove Cassini's identity using the matrix representation of the Fibonacci sequence
 2 years ago
 2 years ago
estudier Group Title
Prove Cassini's identity using the matrix representation of the Fibonacci sequence
 2 years ago
 2 years ago

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JamesWolf Group TitleBest ResponseYou've already chosen the best response.0
oh is that all? easy! one question, whats cassini's identity? To google!
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
Heh, I can open questions all over the shop now.....:)
 2 years ago

JamesWolf Group TitleBest ResponseYou've already chosen the best response.0
http://planetmath.org/ProofofCassinisIdentity.html come across this which answeres your question, but certainly doesnt answer mine.
 2 years ago

JamesWolf Group TitleBest ResponseYou've already chosen the best response.0
out of interest. whats the formula for the fibonacci sequence
 2 years ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
Where's the matrix?
 2 years ago

JamesWolf Group TitleBest ResponseYou've already chosen the best response.0
\[F_n = F_{n2} + F_{n1}\] ?
 2 years ago

JamesWolf Group TitleBest ResponseYou've already chosen the best response.0
oh sorry yes your right there is none there
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.1
You can represent the Fibonacci sequence as: \[\left[\begin{matrix}F_{n+1} & F_n \\ F_n &F_{n1}\end{matrix}\right] = \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n\]
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.1
This is fairly easy to prove by induction
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.1
define F_0 as 0, F_1 as 1, and F_2 as 1
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.1
If the first equation is true, then\[\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^{n+1}=\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n*\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]\]: \[= \left[\begin{matrix}F_{n+1} & F_n \\ F_n & F_{n1}\end{matrix}\right]*\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]\] \[= \left[\begin{matrix}F_{n+1}+F_n & F_{n+1} \\ F_n+F_{n1} & F_n\end{matrix}\right]=\left[\begin{matrix}F_{n+2} & F_{n+1} \\ F_{n+1} & F_n\end{matrix}\right]\]
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.1
So: \[\left[\begin{matrix}F_{n+1} & F_n \\ F_n & F_{n1}\end{matrix}\right] = \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n \forall n \in Z_+\]
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.1
now, take the determinant of both sides: \[F_{n+1}F_{n1}F_n^2=\det \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n\]
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.1
The determinant of the products of square matrices is the products of the determinants of the matrices: \[= (\det \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right])^n\]
 2 years ago

MrMoose Group TitleBest ResponseYou've already chosen the best response.1
=\[(01)^n=1^n\]
 2 years ago
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