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## Math Q&A Game

It doesn't work for arbitrary $c$.

If $f(x)=cx$ then $f(\frac{xy}{2}) = c \frac{xy}{2}$ and $\frac{f(x)f(y)}{2} = c^2 \frac{ xy}2$. Equating these two yields $c=c^2$ so the only solutions for $c$ are the two Matt has already found.

 Recognitions: Gold Member Homework Help Yes, I see that. Can you prove that the only possible solutions are $$f(x)=x$$ and $$f(x)=0$$ from the given conditions? Sorry for the poorly worded question.

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 Quote by D H It doesn't work for arbitrary $c$. If $f(x)=cx$ then $f(\frac{xy}{2}) = c \frac{xy}{2}$ and $\frac{f(x)f(y)}{2} = c^2 \frac{ xy}2$. Equating these two yields $c=c^2$ so the only solutions for $c$ are the two Matt has already found.
What doesn't work for arbitrary c?

 Recognitions: Homework Help Science Advisor From f(2x/2) = f(x), we get that either: a) f = 0, OR b) f(2) = 2 From case b), using that f(0) = f(0)f(x)/2 for all x, we get either: b1) f(0) = 0, OR b2) f(x) = 2 for all x b2) is impossible given the condition that f'(1) = f(1), so we have two cases overall: a) f = 0 b) f(0) = 0 and f(2) = 2 In general, it holds that f(x) = +/- f(-x) since f(xx/2) = f((-x)(-x)/2). In fact, by looking at f((-x)y/2) = f(x(-y)/2), we can make an even stronger claim that either for all x, f(x) = f(-x) or for all x, f(x) = -f(-x). Note that this gives a solution f(x) = |x| which satisfies the criteria (it is not required that f be differentiable everywhere, only at 1 is necessary) but is neither 0 nor identity. Suppose f(x) = 0 for some non-zero x. Then for all y, f(y) = f(x(2y/x)/2) = 0, so f(0) = 0, and either f(x) is non-zero for all other x, or f(x) is zero for all other x: a) f = 0 b) f(0) = 0, f(2) = 2, f is either odd or even, and f(x) is non-zero for x non-zero.

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 Quote by AKG Note that this gives a solution f(x) = |x| which satisfies the criteria (it is not required that f be differentiable everywhere, only at 1 is necessary) but is neither 0 nor identity.
I followed what you said up until this part, but I don't understand what you are saying here. Can you explain it in more detail so that I can understand?

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 Can you prove that the only possible solutions are f(x)=x and f(x)=0
I'm simply saying that f(x) = |x| satisfies the given criteria [i.e. |xy/2| = |x||y|/2 for all real x, y and f'(1) = 1 = |1| = f(1)] but f is neither of those two solutions you proposed above. The stuff I wrote in brackets in my post #72 was to acknowledge that f(x) = |x| is not differentiable at x=0, but the given criteria only require that f be diff'able at 1, not necessarily everywhere.

 So, other than f(x)=x, f(x)=|x| and f(x)=0, is there a forth solution? In general, we are not even assuming continuity except at x=1, at which point f is also differentiable?

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 Quote by AKG I'm simply saying that f(x) = |x| satisfies the given criteria [i.e. |xy/2| = |x||y|/2 for all real x, y and f'(1) = 1 = |1| = f(1)] but f is neither of those two solutions you proposed above. The stuff I wrote in brackets in my post #72 was to acknowledge that f(x) = |x| is not differentiable at x=0, but the given criteria only require that f be diff'able at 1, not necessarily everywhere.
Ok then, my question was incorrect. So the only solutions are
$$f(x) = 0$$ , $$f(x) = |x|$$ ,ie ( f(x) = x for x>0 and f(x)=-x for x<0 )

I hope this is the final edit to the question. Can you prove that the only solutions are f(x) = 0 and f(x) = |x| from the given conditions?

 Quote by chingkui In general, we are not even assuming continuity except at x=1, at which point f is also differentiable?
I picked this question up from a calculus book and no information is given in the question in the book about it's differentiability at points other than x=1.

But I think it is necessary for the condition that f(x) is differentiable at all points except x=0 to be given.
So you are given that f(x) is differentiable at all points except x=0.

I hope that the question is still clear and I did not confuse everyone too much

 Recognitions: Homework Help Science Advisor I thought the setter was supposed to know the answer so that they could agree upon the first correct response? Without imposing some more conditions I don't see that you can conclude there won't be some more answers. And as you keep changing the question we presume that you don't know the answer, which makes it a little unfair on those attempting it. If we were to require that the function had a taylor series with radius of convergence infinity about 0 then you could get the answer but that would necessarily exclude AKG's answer. Didn't the original question ask for f(x) for positive x only? Because then, assuming that the function is smooth in some neighborhood of 1, you can get the answer since every number is nth the power of a number close to 1.
 I thought we have a much more interesting question... it would be a lot more fun to assume nothing about continuity and differentiability (except at x=1)... I think we can prove continuity everywhere and differentiability except possibly at x=0 with the assumption of differentiability at x=1.
 Recognitions: Homework Help Science Advisor We have a question with no known answer, that doens't really make it useful for this particular thread.

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 Quote by matt grime I thought the setter was supposed to know the answer so that they could agree upon the first correct response? Without imposing some more conditions I don't see that you can conclude there won't be some more answers. And as you keep changing the question we presume that you don't know the answer, which makes it a little unfair on those attempting it.
I understand that I messed up this question, so I'll provide the answer which I was expecting. The reason I had to change the question was that the book I picked it up from did not give the correct answer.
Besides, I wanted to revive interest in this thread.

From the given condition
$$f(\frac{x+y}{2}) = [f(x) f(1+\frac{y}{x})]/2$$ so $$x \neq 0$$

Here put y=0.
So,

$$f(\frac{x}{2})=\frac{f(x)f(1)}{2}$$

ie,

$$2f(x)=f(2x)f(1)$$

Now
$$f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h}$$

$$= \lim_{h\rightarrow 0} \frac{f(\frac{2x+2h}{2}) - f(x)}{h}$$

$$= \lim_{h\rightarrow 0} \frac{f(2x)f(1+\frac{h}{x}) - 2f(x)}{2h}$$

Since $$2f(x)=f(2x)f(1)$$
we have,
$$=(\frac{f(2x)}{2x}) \lim_{h\rightarrow 0} \frac{f(1+\frac{h}{x}) - f(1)}{\frac{h}{x}}$$

So, we have
$$f'(x) = \frac{f(2x)}{2x} f'(1)$$

Since $$2f(x)=f(2x)f(1)$$ and $$f(1)=f'(1)$$

$$f'(x)=\frac{f(x)}{x}$$

$$f(x)=|cx|$$

Hence, by substituting back into the original condition, the only solutions are

$$f(x) = |x|$$ and $$f(x) = 0$$

 Recognitions: Homework Help Science Advisor Your 'solution' somhow omits the identity function and appears to claim that |x| is everywhere differentiable, and makes untold assumptions about f

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That's why I changed the question, so that f(x) is differentiable everywhere except x=0 was given

 Quote by siddharth So you are given that f(x) is differentiable at all points except x=0.
May I know what untold assumptions are being made?

 Recognitions: Homework Help Science Advisor Firstly I must admit I've given up on rereading the new assumptions you keep making, and secondly you conclude f(x)=|x| is a solution. That is not an everywhere differentiable function, and you *do not conclude* that f(x)=x is a solution. Read your own proof and its conclusion that states the solutions are f(x)=|x| and f(x)=0!

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 Quote by matt grime Firstly I must admit I've given up on rereading the new assumptions you keep making, and secondly you conclude f(x)=|x| is a solution. That is not an everywhere differentiable function, and you *do not conclude* that f(x)=x is a solution. Read your own proof and its conclusion that states the solutions are f(x)=|x| and f(x)=0!
I agree, and that is why I messed up the question. Sorry. I changed the question in post #77 so that the solutions are f(x)=0 and f(x)=|x|