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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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  1. JamesWolf Group Title
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    oh is that all? easy! one question, whats cassini's identity? To google!

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    • 2 years ago
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