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ganeshie8

  • one year ago

Identify a reasonable pattern and find a formula for \(n\) th term of the sequence : \[\large -3,~21,~3,~129,~147,\ldots\]

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  1. amistre64
    • one year ago
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    we can build a degree 4 (maybe?) polynomial :)

  2. ganeshie8
    • one year ago
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    haha sure guess we can do that with any sequence, including \[1,~1,~2,~3,~5,~8,~13,~\cdots\] ;p

  3. amistre64
    • one year ago
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    i actually reworked the closed form of the fibber today :)

  4. ganeshie8
    • one year ago
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    the sequence in question is fibonacci like one.. but identifying the pattern might be tough as there is no uniq answer yeah

  5. amistre64
    • one year ago
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    if i toss in a switch function, then for all points in the ... it can be zero or even some other function to turn on ;)

  6. amistre64
    • one year ago
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    so your pattern is what, an alternating sign fibber?

  7. amistre64
    • one year ago
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    well, not precise

  8. ganeshie8
    • one year ago
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    was hoping to work it using \[u_n = Au_{n-1}+Bu_{n-2}\] where \(A\) and \(B\) are constants

  9. amistre64
    • one year ago
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    what have you determined to be your recursive form?

  10. ganeshie8
    • one year ago
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    exercise 15B / last problem http://www.cimt.plymouth.ac.uk/projects/mepres/alevel/discrete_ch15.pdf

  11. ganeshie8
    • one year ago
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    \[u_n = u_{n-1}+6u_{n-2}\] http://www.wolframalpha.com/input/?i=solve+3+%3D+21x-3y%2C+129%3D3x%2B21y

  12. amistre64
    • one year ago
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    ok, you already have the recursion ... i just figured it out lol

  13. amistre64
    • one year ago
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    -3 21 3 129 147 -3x + 21y = 3 21x + 3y = 129 6x+y 6(3)+129 = 147

  14. amistre64
    • one year ago
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    might be easier to work like this tho f(n+2) = 6 f(n+1) + f(n) f(n+2) - 6 f(n+1) - f(n) = 0 let f(n) = x^n x^(n+2) - 6 x^(n+1) - x^(n) = 0 x^(n) (x^2 - 6x - 1) = 0 what are the roots of our quadratic?

  15. amistre64
    • one year ago
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    i think my 6 is in the wrong spot tho ... didnt double check to see if my fingers were behaving

  16. ganeshie8
    • one year ago
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    f(n+2) = f(n+1) + 6f(n) right

  17. ganeshie8
    • one year ago
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    x^2 - x - 6 = 0 x = 3, -2

  18. amistre64
    • one year ago
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    since 3^n and (-2)^n are linearly independant, then we can form a ... linear combination of the solutions as: A3^n + B(-2)^n and setup a system of 2 equations in 2 unknowns do we like to start with n=0 or n=1?

  19. ganeshie8
    • one year ago
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    0 i guess, the trick part is identifying a pattern and after that finding the general solution is pretty straightforward!

  20. amistre64
    • one year ago
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    :) yeah A+B = -3 3A -2B = 21

  21. amistre64
    • one year ago
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    2A+2B = -6 3A -2B = 21 ----------- 5A = 15 A = 3, so B = -9

  22. amistre64
    • one year ago
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    we can chk with the wolf: table [3(3^n)-9(-2)^n,{n,10}]

  23. ganeshie8
    • one year ago
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    *B = -6 \[u_n = 3^{n+1}-6(-2)^{n}\]

  24. amistre64
    • one year ago
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    lol ... fine we will use -6 instead :)

  25. amistre64
    • one year ago
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    ... wow, he can solve a closed form for some random fibber sequence, but that addition really throws him for a loop ;)

  26. ganeshie8
    • one year ago
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    checks out wid wolf ! http://www.wolframalpha.com/input/?i=Table%5B3%5E%28n%2B1%29-6%28-2%29%5En%2C+%7Bn%2C0%2C4%7D%5D i use to do these problems using generating functions, they're long but they gave nice practice with power series

  27. amistre64
    • one year ago
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    i cant say that i have played with generating functions much, if at all.

  28. ikram002p
    • one year ago
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    is it unique ?

  29. amistre64
    • one year ago
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    well i can tell you using -9 doesnt work .. so im gonna say yes :)

  30. ikram002p
    • one year ago
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    well :P

  31. ganeshie8
    • one year ago
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    using generating functions \[\begin{align}G(x) &= u_0+u_1x+u_2x^2+u_3x^3+\cdots\\~\\ &=-3 + 21x + (u_1+6u_0)x^2 + (u_2+6u_1)x^3 + \cdots \\~\\ &=-3+21x+x(u_1x+u_2x^2+\cdots) + 6x^2(u_0+u_1x+\cdots)\\~\\ &=-3+21x+x(G(x)-u_0) + 6x^2G(x)\\~\\ \end{align}\] \(\implies G(x)[1-x-6x^2]=-3+21x-u_0x\\~\\ \begin{align}\implies G(x) &= \dfrac{-3+24x}{1-x-6x^2} \\~\\ &= \dfrac{3}{1-3x}-\dfrac{6}{1+2x}\\~\\ &=3\sum\limits_{n=0}^{\infty} (3x)^n-6\sum\limits_{n=0}^{\infty} (-2x)^n\\~\\ &=\sum\limits_{n=0}^{\infty} \left[\color{blue}{3^{n+1}-6(-2)^n}\right] x^n \end{align}\)

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