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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 \(f(2009)\).
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because 1 is a little bigger than 0?...

dan815
 one year ago
Best ResponseYou've already chosen the best response.1ah nvm scratch this stuff i dont think its the right way

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3dan I think the math is just off. Use used f(2) = 2 :)

dan815
 one year ago
Best ResponseYou've already chosen the best response.1oh lets use this we can use that induction step after all

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3The start is to rewrite \(f(x  y) = \sqrt{f(xy) + 1}\) as \[f(xy) = [f(xy)]^2  1\] In particular using that 2009 = 4941=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.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3edit: I meant 2009 = 49*41 of course.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oohhh haha ok that makes a little bit more sense lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.1u can get a base case and solve it

dan815
 one year ago
Best ResponseYou've already chosen the best response.1i think u can get a closed form from that too now

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3your equations aren't right :( dw:1441426034574:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.1you are squaring and subtracing 1 every time u can definatley get a closed form for that given first term in seuqnce

dan815
 one year ago
Best ResponseYou've already chosen the best response.1wait how comes thats wrong f(xy)=sqrt(f(x*y)+1) right

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3Oh, sorry I thought the LHS was n*1. Better reread.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1can we do some telescoping stuff here though

dan815
 one year ago
Best ResponseYou've already chosen the best response.1we dont have f(1) or anythign so what can we do

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3Still, you have 4 unknowns with 3 equations. You want k equations with k unknowns for the base case.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3Anyway, I have a solution strategy, but it's not pretty.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3You 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 = 32) That's 6 equations, 6 unknowns.

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3f(2) = [f(1)]^2  1, that is. BTW, from the last one at least we get f(5) = f(1)

dan815
 one year ago
Best ResponseYou've already chosen the best response.1we can end up solving for f(1) or something and we are done then we will have a complete closed formula

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3There must be a simpler way than just brute forcing your way through.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1ill work on finding closed form for this it looks kind of fun dw:1441427086013:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0good luck with that wolfram gave up already http://www.wolframalpha.com/input/?i=solve+a%28n%2B1%29+%3D+a%28n%29%5E21%2C+a%281%29%3D1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0there is a standard method for solving linear recurrence relations like \[a_{n+2} + a_{n+1} + a_n = f(n)\] but the relation \(a_{n+1} = {a_n}^21\) is nonlinear so it is kinda tough to get a closed form i think

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3I 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(p1q1) = f(p2q2)

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3assuming p1>q1, p2>q2 of course. BTW this is looking more and more like a number theory problem.

dan815
 one year ago
Best ResponseYou've already chosen the best response.1like we know how many expreresion can exist for each f(k)

dan815
 one year ago
Best ResponseYou've already chosen the best response.1if k=xy i mean not the xy part

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3I think f(x) is just the constant function.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4xy=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 \[f(2009)=\sqrt{f(2009)+1} \\ f^2(2009)f(2009)1=0\]

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3Close the thread :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.4By the way there does exist a solution to the system xy=2009 ,xy=2009 such that x>y>0 I have confirmed this :)

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I was kind of playing with this more I think f is a constant function.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0thatn is simply brilliant !

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0x  y = u xy = u does that allow us to get the closed form too

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

freckles
 one year ago
Best ResponseYou've already chosen the best response.4so solving that quadratic function will give us a closed form

freckles
 one year ago
Best ResponseYou've already chosen the best response.4well I might have a question...

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[\sqrt{c^2+4c}>c \text{ is this true for all positive real } c \\\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I just wanted to verify the choice of y would always be positive for positive real c

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.3\(\sqrt{c^2} = c\), and \(\sqrt(x)\) is increasing, so yeah.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4oh yeah duh thanks @beginnersmind
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