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zmudz

  • one year ago

A sequence of real numbers (x_n) is defined recursively as follows: x_0=a and x_1=b are positive real numbers, and x_{n + 2} = (1 + x_{n + 1})/(x_n) for n = 0, 1, 2, .... so on. Find the value of x_{2012}, in terms of x_0 and x_1. Thanks!

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  1. danica518
    • one year ago
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    |dw:1437194441049:dw|

  2. danica518
    • one year ago
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    like this right?

  3. danica518
    • one year ago
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    |dw:1437194518228:dw|

  4. danica518
    • one year ago
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    |dw:1437194651461:dw|

  5. danica518
    • one year ago
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    im not seeing a trick, maybe its one of those continued fraction tricks?

  6. danica518
    • one year ago
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    |dw:1437194858049:dw|

  7. freckles
    • one year ago
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    the sequence starts to repeat itself starting at x_5

  8. danica518
    • one year ago
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    really

  9. freckles
    • one year ago
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    yep

  10. danica518
    • one year ago
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    should i simplify X3 and X4 or can i leave it all in that form

  11. freckles
    • one year ago
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    simplifying helps to see the pattern

  12. danica518
    • one year ago
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    |dw:1437195330961:dw|

  13. freckles
    • one year ago
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    \[x_4=\frac{1+\frac{a+b+1}{ab}}{\frac{1+b}{a}} \\ \text{ multiply top and bottom by } ab \\ \\ x_4=\frac{ab +a+b+1}{b(1+b)}=\frac{a(b+1)+(b+1)}{b(b+1)} \\ x_4=\frac{(b+1)(a+1)}{b(b+1)}=\frac{a+1}{b}\]

  14. zmudz
    • one year ago
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    That answer isn't right, and I'm not sure why because it makes sense to me... I don't know the answer but my homework is telling me it is wrong.

  15. anonymous
    • one year ago
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    the sequence is an example of one in rank-2 cluster algebras; it has a period of 5 for any initial values, so $$x_{2012}=x_{2012-2010}=x_2$$

  16. anonymous
    • one year ago
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    and we have $$x_2=\frac{1+x_1}{x_0}=\frac{1+b}a$$

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