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zmudz

  • one year ago

Let \(f(x)\) be a function defined for all positive real numbers satisfying the conditions \(f(x) > 0\) for all \(x > 0\) and \(f(x - y) = \sqrt{f(xy) + 1}\) for all \(x > y > 0\). Determine \(f(2009)\).

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  1. Jhannybean
    • one year ago
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    @AravindG

  2. Jhannybean
    • one year ago
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    Because 1 is a little bigger than 0?...

  3. Jhannybean
    • one year ago
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    Confused!!!

  4. dan815
    • one year ago
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    ah nvm scratch this stuff i dont think its the right way

  5. beginnersmind
    • one year ago
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    dan I think the math is just off. Use used f(2) = 2 :)

  6. dan815
    • one year ago
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    oh lets use this we can use that induction step after all

  7. beginnersmind
    • one year ago
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    The start is to rewrite \(f(x - y) = \sqrt{f(xy) + 1}\) as \[f(xy) = [f(x-y)]^2 - 1\] In particular using that 2009 = 49-41=8 \[f(2009) = [f(8)]^2 - 1\] So we're not looking for a closed formula. Just a few values like f(1) and f(2) would be enough.

  8. beginnersmind
    • one year ago
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    edit: I meant 2009 = 49*41 of course.

  9. Jhannybean
    • one year ago
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    oohhh haha ok that makes a little bit more sense lol

  10. dan815
    • one year ago
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    |dw:1441425591667:dw|

  11. dan815
    • one year ago
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    |dw:1441425872198:dw|

  12. dan815
    • one year ago
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    u can get a base case and solve it

  13. dan815
    • one year ago
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    i think u can get a closed form from that too now

  14. beginnersmind
    • one year ago
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    your equations aren't right :( |dw:1441426034574:dw|

  15. dan815
    • one year ago
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    you are squaring and subtracing -1 every time u can definatley get a closed form for that given first term in seuqnce

  16. dan815
    • one year ago
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    wait how comes thats wrong f(x-y)=sqrt(f(x*y)+1) right

  17. dan815
    • one year ago
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    |dw:1441426260927:dw|

  18. beginnersmind
    • one year ago
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    Oh, sorry I thought the LHS was n*1. Better reread.

  19. dan815
    • one year ago
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    ah gotcha

  20. dan815
    • one year ago
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    can we do some telescoping stuff here though

  21. dan815
    • one year ago
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    |dw:1441426410790:dw|

  22. dan815
    • one year ago
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    we dont have f(1) or anythign so what can we do

  23. beginnersmind
    • one year ago
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    Still, you have 4 unknowns with 3 equations. You want k equations with k unknowns for the base case.

  24. beginnersmind
    • one year ago
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    Anyway, I have a solution strategy, but it's not pretty.

  25. beginnersmind
    • one year ago
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    You can start with f(2) = [f(1)]^2 + 1 f(3) = [f(2)]^2 + 1 , , , f(6) = [f(5)]^2 + 1 f(6) = [f(1)]^2 + 1 (6=3*2, 1 = 3-2) That's 6 equations, 6 unknowns.

  26. dan815
    • one year ago
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    oh cool that works

  27. beginnersmind
    • one year ago
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    f(2) = [f(1)]^2 - 1, that is. BTW, from the last one at least we get f(5) = f(1)

  28. dan815
    • one year ago
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    we can end up solving for f(1) or something and we are done then we will have a complete closed formula

  29. beginnersmind
    • one year ago
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    There must be a simpler way than just brute forcing your way through.

  30. dan815
    • one year ago
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    ill work on finding closed form for this it looks kind of fun |dw:1441427086013:dw|

  31. ganeshie8
    • one year ago
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    good luck with that wolfram gave up already http://www.wolframalpha.com/input/?i=solve+a%28n%2B1%29+%3D+a%28n%29%5E2-1%2C+a%281%29%3D1

  32. dan815
    • one year ago
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    oh serious

  33. ganeshie8
    • one year ago
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    there is a standard method for solving linear recurrence relations like \[a_{n+2} + a_{n+1} + a_n = f(n)\] but the relation \(a_{n+1} = {a_n}^2-1\) is nonlinear so it is kinda tough to get a closed form i think

  34. beginnersmind
    • one year ago
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    I found something. Let n be a positive integer such that it can be written as n = p1*q1 and n = p2*q2 then we have f(p1-q1) = f(p2-q2)

  35. beginnersmind
    • one year ago
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    assuming p1>q1, p2>q2 of course. BTW this is looking more and more like a number theory problem.

  36. dan815
    • one year ago
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    yea

  37. dan815
    • one year ago
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    like we know how many expreresion can exist for each f(k)

  38. dan815
    • one year ago
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    if k=xy i mean not the x-y part

  39. beginnersmind
    • one year ago
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    I think f(x) is just the constant function.

  40. freckles
    • one year ago
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    x-y=2009 xy=2009 we should be able to find x and y where x>y>0 and if so we just have to solve this quadratic for f(2009) and take the positive value \[f(2009)=\sqrt{f(2009)+1} \\ f^2(2009)-f(2009)-1=0\]

  41. beginnersmind
    • one year ago
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    Close the thread :)

  42. freckles
    • one year ago
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    By the way there does exist a solution to the system x-y=2009 ,xy=2009 such that x>y>0 I have confirmed this :)

  43. freckles
    • one year ago
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    I was kind of playing with this more I think f is a constant function.

  44. ganeshie8
    • one year ago
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    thatn is simply brilliant !

  45. ganeshie8
    • one year ago
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    x - y = u xy = u does that allow us to get the closed form too

  46. freckles
    • one year ago
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    Say we have xy=c x-y=c since x>y>0 then c>0 Now I think there will always be a solution such that x>y>0 is true. \[x=c+y \\ y(c+y)=c \\ y^2+cy-c=0 \\ y=\frac{-c \pm \sqrt{c^2+4c}}{2} \\ \text{ now } y>0 \\ \text{ so } \\ y=\frac{-c + \sqrt{c^2+4c}}{2} \\ \text{ so } x=c+y=\frac{2c}{2}+\frac{-c+\sqrt{c^2+4c}}{2} =\frac{c + \sqrt{c^2+4c}}{2} \\ \text{ \it is obvious } \frac{c +\sqrt{c^2+4c}}{2}>\frac{-c + \sqrt{c^2+4c}}{2} \\ \text{ so } f^2(c)-f(c)-1=0 \\ \text{ will give us two solutions } \\ \text{ we take the positive one since } f(x)>0\]

  47. freckles
    • one year ago
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    so solving that quadratic function will give us a closed form

  48. ganeshie8
    • one year ago
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    oh it does, nice..

  49. freckles
    • one year ago
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    well I might have a question...

  50. freckles
    • one year ago
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    \[\sqrt{c^2+4c}>c \text{ is this true for all positive real } c \\\]

  51. ganeshie8
    • one year ago
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    definitely

  52. freckles
    • one year ago
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    yeah you are right

  53. freckles
    • one year ago
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    I just wanted to verify the choice of y would always be positive for positive real c

  54. beginnersmind
    • one year ago
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    \(\sqrt{c^2} = c\), and \(\sqrt(x)\) is increasing, so yeah.

  55. freckles
    • one year ago
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    oh yeah duh thanks @beginnersmind

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