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anonymous
 3 years ago
Prove Cassini's identity using the matrix representation of the Fibonacci sequence
anonymous
 3 years ago
Prove Cassini's identity using the matrix representation of the Fibonacci sequence

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Heh, I can open questions all over the shop now.....:)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0out of interest. whats the formula for the fibonacci sequence

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[F_n = F_{n2} + F_{n1}\] ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh sorry yes your right there is none there

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0This is fairly easy to prove by induction

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0define F_0 as 0, F_1 as 1, and F_2 as 1

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

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

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

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