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

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- Jhannybean

@AravindG

- Jhannybean

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

- Jhannybean

Confused!!!

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## More answers

- dan815

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

- beginnersmind

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

- dan815

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

- 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.

- beginnersmind

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

- Jhannybean

oohhh haha ok that makes a little bit more sense lol

- dan815

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- dan815

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- dan815

u can get a base case and solve it

- dan815

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

- beginnersmind

your equations aren't right :(
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- dan815

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

- dan815

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

- dan815

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- beginnersmind

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

- dan815

ah gotcha

- dan815

can we do some telescoping stuff here though

- dan815

|dw:1441426410790:dw|

- dan815

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

- beginnersmind

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

- beginnersmind

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

- 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.

- dan815

oh cool that works

- beginnersmind

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

- dan815

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

- beginnersmind

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

- dan815

ill work on finding closed form for this it looks kind of fun
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- 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

- dan815

oh serious

- 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

- 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)

- beginnersmind

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

- dan815

yea

- dan815

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

- dan815

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

- beginnersmind

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

- 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
\[f(2009)=\sqrt{f(2009)+1} \\ f^2(2009)-f(2009)-1=0\]

- beginnersmind

Close the thread :)

- 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 :)

- freckles

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

- ganeshie8

thatn is simply brilliant !

- ganeshie8

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

- 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.
\[x=c+y \\ y(c+y)=c \\ y^2+cy-c=0 \\ y=\frac{-c \pm \sqrt{c^2+4c}}{2} \\ \text{ now } y>0 \\ \text{ so } \\ y=\frac{-c + \sqrt{c^2+4c}}{2} \\ \text{ so } x=c+y=\frac{2c}{2}+\frac{-c+\sqrt{c^2+4c}}{2} =\frac{c + \sqrt{c^2+4c}}{2} \\ \text{ \it is obvious } \frac{c +\sqrt{c^2+4c}}{2}>\frac{-c + \sqrt{c^2+4c}}{2} \\ \text{ so } f^2(c)-f(c)-1=0 \\ \text{ will give us two solutions } \\ \text{ we take the positive one since } f(x)>0\]

- freckles

so solving that quadratic function will give us a closed form

- ganeshie8

oh it does, nice..

- freckles

well I might have a question...

- freckles

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

- ganeshie8

definitely

- freckles

yeah you are right

- freckles

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

- beginnersmind

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

- freckles

oh yeah duh
thanks @beginnersmind

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