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estudier
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Prove Cassini's identity using the matrix representation of the Fibonacci sequence
 one year ago
 one year ago
estudier Group Title
Prove Cassini's identity using the matrix representation of the Fibonacci sequence
 one year ago
 one year 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!
 one year ago

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

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

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

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

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

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

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

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

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

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

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