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Functional Equation for Non negative Reals
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kunny
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#1
 Functional Equation for Non negative Reals
2009 Japan Mathematical Olympiad Finals, Problem 5
Find all functions f, defined on the non negative real numbers and taking non negative real numbers such that f(x^2)+f(y)=f(x^2+y+xf(4y)) for any non negative real numbers x,\ y.
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PostPosted: Sat Feb 21, 2009 6:44 pm
cyshine
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#2
Pretty cool problem.

I don't want to spoil the fun of solving this problem, so it's all hidden.

Click to reveal hidden content
First, notice that g(x) = x^2 + kx spans all non-negative reals for non-negative values of x, because g(0) = 0 and g(x) goes to infinity as x increases.

Then f is non-decreasing: if z = y + a, a > 0, then there is x > 0 such that x^2 + 4f(y)\cdot x = a. Choosing such x we have f(x^2) + f(y) = f(z) and since f can only assume non-negative values, f(z) \geq f(y).

Suppose that there is k > 0 such that f(k) = 0. Then, since f is non-decreasing and, by letting x = y = 0, f(0) = 0, f(x) = 0 for 0 \leq x \leq k. Letting x be \sqrt x and y = k/4, we obtain f(x) + f(k/4) = f(x + k/4)\iff f(x+k/4) = f(x). This implies that f is periodic with period k/4. Since f([0,k]) = \{0\}, f(x) = 0 for all non-negative x.

It remains to solve the case in which f(x) > 0 for all positive x. In this case, f is strictly increasing (f(z) \geq f(y) turns into a strict inequality) and, henceforth, f is injective. Substituting x by \sqrt x we obtain f(x) + f(y) = f(x + y + \sqrt xf(4y)). By symmetry, f(x + y + \sqrt xf(4y)) = f(x + y + \sqrt yf(4x)) \iff \frac{f(4x)}{\sqrt x} = \frac{f(4y)}{\sqrt y} This means that \frac{f(x)}{\sqrt x} is constant, so f(x) = c\cdot \sqrt x. Direct substitution yields c = 1 and we are done.
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PostPosted: Sat Feb 21, 2009 9:38 pm
xienohp
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#3
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1) Putting root(a) to be x and na to be y, for n to be non-negative real. Then we get
f(na)/f(a) = root(n)
By putting a = 1, we know that all values except 1 and 0 depends on f(1),
so we can directly manipulate f(x) = cx^(.5) which easily get c = 1 or 0 by substituting back to the original equation and we get f(x) = x^(.5) or 0 for x,yER+, f:R+|->R+
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PostPosted: Sat Aug 22, 2009 5:43 am
MichaelHo
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#4
This question from Japan Mathematical Olympiad Finals 2009
symmetry , black box
Japan-Mathematical_Olympiad_Finals-2009
NO. 5 : f(x^2)+f(y)=f(x^2+y+xf(4y))

f(x)=0, or f(x) progressively increase

swap x^2 with y,
f(y)+f(x^2)=f(y+x^2+\sqrt{y}f(4x^2))
so xf(4y) =\sqrt{y}f(4x^2)
\dfrac{f(x)}{x} = c

PostPosted: Sat Aug 22, 2009 5:50 am
FelixD
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#5
Re: Functional Equation for Non negative Reals
2009 Japan Mathematical Olympiad Finals, Problem 5
kunny wrote:
Find all functions f defined on the non negative real numbers and taking non negative real numbers such that
f(x^2) + f(y) = f(x^2 + y + xf(4y)) \quad (1)
for any non negative real numbers x, y.


Fix y and let g(x) = x^2 + ax, where f(4y) = a, which is constant. It's easy to see that g(x) spans all non-negative real numbers. Hence, f(y) \le f(x^2) + f(y) = f(y + g(x)), which implies that f is nondecreasing.
Setting x = y = 0 yields f(0) = 0. Assume that there exsits y_0 > 0 such that f(y_0) = 0. Then we have f(x) = 0 for x \in [0, y_0] as a consequence of the monotonicity. Plugging y = y_0/4 into (1) we get f(x^2) = f(x^2 + y_0/4). Construct the sequence x_n = \sqrt {x_{n - 1}^2 + y_0/4} for n \ge 1 with x_0 = 0. Hence f(x_n^2) = f(x_{n - 1}^2 + y_0/4) = f(0) = 0 and together with the monotonicity f(x) \equiv 0 for all non-negative x.
In the sequel, let f(x) \ne 0 for x \ne 0. Then we have f(y) < f(x^2) + f(y) = f(y + g(x)) for x, y \ne 0 and thus f is strictly increasing.
Plugging y \to y^2 into (1) we get f(x^2) + f(y^2) = f(x^2 + y^2 + xf(4y^2)). Switch x and y to get f(x^2 + y^2 + xf(4y^2)) = f(x^2 + y^2 + yf(4x^2)) and because f is strictly increasing xf(4y^2) = yf(4x^2) or equivalently \frac {f(4x^2)}{x} = \frac {f(4y^2)}{y} = c, where c is some constant. Hence f(x) = \frac {c\sqrt {x}}{2}. Plugging this into (1) we get c = 2 and hence f(x) = \sqrt {x} is the second solution.

PostPosted: Thu Aug 27, 2009 11:01 am
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