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#73
Dec1205, 07:50 PM

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#74
Dec1305, 01:51 PM

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siddharth, you asked:



#75
Dec1305, 03:54 PM

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Stupid brain. Must read before posting.



#76
Dec1305, 11:16 PM

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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? 


#77
Dec1405, 12:52 AM

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[tex] f(x) = 0 [/tex] , [tex] f(x) = x [/tex] ,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? 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 


#78
Dec1405, 07:32 AM

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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. 


#79
Dec1405, 08:53 AM

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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.



#80
Dec1405, 08:59 AM

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We have a question with no known answer, that doens't really make it useful for this particular thread.



#81
Dec1405, 10:12 AM

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Besides, I wanted to revive interest in this thread. From the given condition [tex] f(\frac{x+y}{2}) = [f(x) f(1+\frac{y}{x})]/2 [/tex] so [tex] x \neq 0 [/tex] Here put y=0. So, [tex] f(\frac{x}{2})=\frac{f(x)f(1)}{2} [/tex] ie, [tex] 2f(x)=f(2x)f(1) [/tex] Now [tex] f'(x) = \lim_{h\rightarrow 0} \frac{f(x+h)  f(x)}{h} [/tex] [tex] = \lim_{h\rightarrow 0} \frac{f(\frac{2x+2h}{2})  f(x)}{h} [/tex] [tex]= \lim_{h\rightarrow 0} \frac{f(2x)f(1+\frac{h}{x})  2f(x)}{2h} [/tex] Since [tex] 2f(x)=f(2x)f(1) [/tex] we have, [tex]=(\frac{f(2x)}{2x}) \lim_{h\rightarrow 0} \frac{f(1+\frac{h}{x})  f(1)}{\frac{h}{x}} [/tex] So, we have [tex] f'(x) = \frac{f(2x)}{2x} f'(1) [/tex] Since [tex] 2f(x)=f(2x)f(1) [/tex] and [tex] f(1)=f'(1) [/tex] [tex] f'(x)=\frac{f(x)}{x} [/tex] [tex] f(x)=cx [/tex] Hence, by substituting back into the original condition, the only solutions are [tex] f(x) = x [/tex] and [tex] f(x) = 0 [/tex] 


#82
Dec1405, 10:32 AM

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Your 'solution' somhow omits the identity function and appears to claim that x is everywhere differentiable, and makes untold assumptions about f



#83
Dec1405, 10:38 AM

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



#84
Dec1405, 10:41 AM

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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!



#85
Dec1405, 10:47 AM

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#86
Dec1405, 10:55 AM

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But as I observed in my first post f(x)=x is a solution and not one of the ones you derive, or have you altered the conditions so that f(x)=x is not a solution?



#87
Dec1405, 11:09 AM

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That's a lesson then. Next time, I'll research the question thoroughly before I post it. 


#88
Dec1405, 01:19 PM

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You don't even need to assume f(x) is continuous to prove the result. All you need is f(xy/2)=f(x)f(y)/2 and f is differentiable at x=1 with f'(1)=f(1).
With just these, you can prove that f(x)=x or f(x)=x or f(x)=0. Siddharth had proved differentiability of f(x) at every nonzero x in post 81. You can use similar argument to show f(x) is continuous everywhere including x=0. 


#89
Dec2805, 08:00 AM

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How about another question. I heard this on the radio (Chris Maslanka, I think, presents a panel show with puzzles in on BBC radio 4) a few months ago.
There is a tribe of gnomes, each gnome, except the chief gnome, wears either a red or blue hat but doesn't know which (they're stuck on and there are no mirrors and whatever makes sense here). The chief gnome wants to conduct a census with the minimal opf fuss so he can count the number of red and blue hats. The gnomes being sensible people do this without even needing to communicate: one day they walk into the central clearing and form the red group and the blue group and the chief gnome can count them all. How do they do this? Remember, no communictation is needed (no gnome asks the colour of his hat). It's quite an interesting puzzle even if you don't think it's maths. 


#90
Dec2805, 12:20 PM

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Insertion sort: Gnomes arrive onebyone. The first two gnomes stand next to each other. The third gnome stands between the first two if the third gnome sees two different colored hats, otherwise the third gnome goes to the end of the line (after the second gnome). Each newly arriving gnome inserts himself between the two gnomes with different colored hats or goes to the end of the line if all of the gnomes in line have the same color hat. Some jostling is needed to make room for the newly arriving gnomes in the insertion sort method. This jostling is a kind of communication, just not talking. I assume that jostling and other physical contact is allowed. Therefore, Heap sort: The gnomes all meet in the clearing. The chief tells the gnomes they are to take any punches they receive like a gnome. He then tells the gnomes to pummel all the gnomes they see wearing red hats into unconsciousness and pile them in a heap. When the brawl ends, the chief gnomes counts all the standing gnomes (the blue group) and the unconscious gnomes (the red group). 


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