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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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 \(f(2009)\).

Mathematics
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Because 1 is a little bigger than 0?...
Confused!!!

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ah nvm scratch this stuff i dont think its the right way
dan I think the math is just off. Use used f(2) = 2 :)
oh lets use this we can use that induction step after all
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.
edit: I meant 2009 = 49*41 of course.
oohhh haha ok that makes a little bit more sense lol
|dw:1441425591667:dw|
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u can get a base case and solve it
i think u can get a closed form from that too now
your equations aren't right :( |dw:1441426034574:dw|
you are squaring and subtracing -1 every time u can definatley get a closed form for that given first term in seuqnce
wait how comes thats wrong f(x-y)=sqrt(f(x*y)+1) right
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Oh, sorry I thought the LHS was n*1. Better reread.
ah gotcha
can we do some telescoping stuff here though
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we dont have f(1) or anythign so what can we do
Still, you have 4 unknowns with 3 equations. You want k equations with k unknowns for the base case.
Anyway, I have a solution strategy, but it's not pretty.
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.
oh cool that works
f(2) = [f(1)]^2 - 1, that is. BTW, from the last one at least we get f(5) = f(1)
we can end up solving for f(1) or something and we are done then we will have a complete closed formula
There must be a simpler way than just brute forcing your way through.
ill work on finding closed form for this it looks kind of fun |dw:1441427086013:dw|
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
oh serious
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
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)
assuming p1>q1, p2>q2 of course. BTW this is looking more and more like a number theory problem.
yea
like we know how many expreresion can exist for each f(k)
if k=xy i mean not the x-y part
I think f(x) is just the constant function.
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 \[f(2009)=\sqrt{f(2009)+1} \\ f^2(2009)-f(2009)-1=0\]
Close the thread :)
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 :)
I was kind of playing with this more I think f is a constant function.
thatn is simply brilliant !
x - y = u xy = u does that allow us to get the closed form too
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. \[x=c+y \\ y(c+y)=c \\ y^2+cy-c=0 \\ y=\frac{-c \pm \sqrt{c^2+4c}}{2} \\ \text{ now } y>0 \\ \text{ so } \\ y=\frac{-c + \sqrt{c^2+4c}}{2} \\ \text{ so } x=c+y=\frac{2c}{2}+\frac{-c+\sqrt{c^2+4c}}{2} =\frac{c + \sqrt{c^2+4c}}{2} \\ \text{ \it is obvious } \frac{c +\sqrt{c^2+4c}}{2}>\frac{-c + \sqrt{c^2+4c}}{2} \\ \text{ so } f^2(c)-f(c)-1=0 \\ \text{ will give us two solutions } \\ \text{ we take the positive one since } f(x)>0\]
so solving that quadratic function will give us a closed form
oh it does, nice..
well I might have a question...
\[\sqrt{c^2+4c}>c \text{ is this true for all positive real } c \\\]
definitely
yeah you are right
I just wanted to verify the choice of y would always be positive for positive real c
\(\sqrt{c^2} = c\), and \(\sqrt(x)\) is increasing, so yeah.
oh yeah duh thanks @beginnersmind

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