## 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 $$f(2009)$$.

1. Jhannybean

@AravindG

2. Jhannybean

Because 1 is a little bigger than 0?...

3. Jhannybean

Confused!!!

4. dan815

ah nvm scratch this stuff i dont think its the right way

5. beginnersmind

dan I think the math is just off. Use used f(2) = 2 :)

6. dan815

oh lets use this we can use that induction step after all

7. beginnersmind

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

edit: I meant 2009 = 49*41 of course.

9. Jhannybean

oohhh haha ok that makes a little bit more sense lol

10. dan815

|dw:1441425591667:dw|

11. dan815

|dw:1441425872198:dw|

12. dan815

u can get a base case and solve it

13. dan815

i think u can get a closed form from that too now

14. beginnersmind

your equations aren't right :( |dw:1441426034574:dw|

15. dan815

you are squaring and subtracing -1 every time u can definatley get a closed form for that given first term in seuqnce

16. dan815

wait how comes thats wrong f(x-y)=sqrt(f(x*y)+1) right

17. dan815

|dw:1441426260927:dw|

18. beginnersmind

Oh, sorry I thought the LHS was n*1. Better reread.

19. dan815

ah gotcha

20. dan815

can we do some telescoping stuff here though

21. dan815

|dw:1441426410790:dw|

22. dan815

we dont have f(1) or anythign so what can we do

23. beginnersmind

Still, you have 4 unknowns with 3 equations. You want k equations with k unknowns for the base case.

24. beginnersmind

Anyway, I have a solution strategy, but it's not pretty.

25. beginnersmind

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

oh cool that works

27. beginnersmind

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

28. dan815

we can end up solving for f(1) or something and we are done then we will have a complete closed formula

29. beginnersmind

There must be a simpler way than just brute forcing your way through.

30. dan815

ill work on finding closed form for this it looks kind of fun |dw:1441427086013:dw|

31. ganeshie8

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

oh serious

33. ganeshie8

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

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

assuming p1>q1, p2>q2 of course. BTW this is looking more and more like a number theory problem.

36. dan815

yea

37. dan815

like we know how many expreresion can exist for each f(k)

38. dan815

if k=xy i mean not the x-y part

39. beginnersmind

I think f(x) is just the constant function.

40. freckles

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

Close the thread :)

42. freckles

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

I was kind of playing with this more I think f is a constant function.

44. ganeshie8

thatn is simply brilliant !

45. ganeshie8

x - y = u xy = u does that allow us to get the closed form too

46. freckles

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

so solving that quadratic function will give us a closed form

48. ganeshie8

oh it does, nice..

49. freckles

well I might have a question...

50. freckles

$\sqrt{c^2+4c}>c \text{ is this true for all positive real } c \\$

51. ganeshie8

definitely

52. freckles

yeah you are right

53. freckles

I just wanted to verify the choice of y would always be positive for positive real c

54. beginnersmind

$$\sqrt{c^2} = c$$, and $$\sqrt(x)$$ is increasing, so yeah.

55. freckles

oh yeah duh thanks @beginnersmind