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estudier

  • 2 years ago

Prove Cassini's identity using the matrix representation of the Fibonacci sequence

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  1. JamesWolf
    • 2 years ago
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    oh is that all? easy! one question, whats cassini's identity? To google!

  2. estudier
    • 2 years ago
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    Heh, I can open questions all over the shop now.....:-)

  3. JamesWolf
    • 2 years ago
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    http://planetmath.org/ProofofCassinisIdentity.html come across this which answeres your question, but certainly doesnt answer mine.

  4. JamesWolf
    • 2 years ago
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    out of interest. whats the formula for the fibonacci sequence

  5. estudier
    • 2 years ago
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    Where's the matrix?

  6. JamesWolf
    • 2 years ago
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    \[F_n = F_{n-2} + F_{n-1}\] ?

  7. JamesWolf
    • 2 years ago
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    oh sorry yes your right there is none there

  8. MrMoose
    • 2 years ago
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    You can represent the Fibonacci sequence as: \[\left[\begin{matrix}F_{n+1} & F_n \\ F_n &F_{n-1}\end{matrix}\right] = \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n\]

  9. MrMoose
    • 2 years ago
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    This is fairly easy to prove by induction

  10. MrMoose
    • 2 years ago
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    define F_0 as 0, F_1 as 1, and F_2 as 1

  11. MrMoose
    • 2 years ago
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    If the first equation is true, then\[\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^{n+1}=\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n*\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]\]: \[= \left[\begin{matrix}F_{n+1} & F_n \\ F_n & F_{n-1}\end{matrix}\right]*\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]\] \[= \left[\begin{matrix}F_{n+1}+F_n & F_{n+1} \\ F_n+F_{n-1} & F_n\end{matrix}\right]=\left[\begin{matrix}F_{n+2} & F_{n+1} \\ F_{n+1} & F_n\end{matrix}\right]\]

  12. MrMoose
    • 2 years ago
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    So: \[\left[\begin{matrix}F_{n+1} & F_n \\ F_n & F_{n-1}\end{matrix}\right] = \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n \forall n \in Z_+\]

  13. MrMoose
    • 2 years ago
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    now, take the determinant of both sides: \[F_{n+1}F_{n-1}-F_n^2=\det \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n\]

  14. MrMoose
    • 2 years ago
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    The determinant of the products of square matrices is the products of the determinants of the matrices: \[= (\det \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right])^n\]

  15. MrMoose
    • 2 years ago
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    =\[(0-1)^n=-1^n\]

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