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        riddles >> medium >> Radically Constant
        
(Message started by: ThudanBlunder on Aug 1^st, 2010, 9:24am)

Title: Radically Constant
Post by ThudanBlunder on Aug 1^st, 2010, 9:24am

Find numbers a and b (a < b) and the value of the constant such that
[x + 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x - 1)]^1/2 + [x - 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x - 1)]^1/2 is constant for a http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/leqslant.gif x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/leqslant.gif b.

Title: Re: Radically Constant
Post by towr on Aug 2^nd, 2010, 1:26am

[hide]http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif[x + 2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x - 1)] + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif[x - 2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x - 1)] = c
http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif[x + 2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x - 1)] = c - http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif[x - 2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x - 1)]
x + 2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x - 1) = c² - 2c http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif[x - 2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x - 1)] + x - 2 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x - 1)
4 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x - 1) = c² - 2c http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif[x - 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x - 1)]
4 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x - 1) - c² = - 2c http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif[x - 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x - 1)]
16 (x - 1) - 8c² http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x - 1) + c² = 4c² [x - 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x - 1)]
16 (x - 1) + c^4 = 4 c² x

take c² = y
y² - 4 y x + 16 (x - 1) = 0
y = [4x +/- http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(16x² - 64 (x - 1)) ]/2
y = 2(x +/- http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif((x-2)²)))
y = 2(x +/- (x-2))
If y is constant, than y must be 4, so c is +/- 2

From the graph it's easy to see that a = -inf, b=2, c=2, but I'm going to have to take a raincheck on trying to proof it.[/hide]

Title: Re: Radically Constant
Post by 0.999... on Aug 2^nd, 2010, 5:43am

[hide]If we have a real function f(x) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif[x+2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x-1)] + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif[x-2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x-1)] then it is defined only for values of x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif1.
Let m = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x-1). Hence, f(x) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(m²+2m+1) + http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(m²-2m+1) = |m+1|+|m-1| = 2http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/surd.gif(x-1) if x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif2.
On the other hand, if 1 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gifx http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif2, then m+1 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ge.gif0 and m-1 http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/le.gif0. So f(x) = m+1-(m-1) = 2.
Hence a = 1, b = 2, f(x) = 2 in this interval.[/hide]

Title: Re: Radically Constant
Post by towr on Aug 2^nd, 2010, 7:02am

[hide]But the function as a whole is both real and 2 for x < 1.
The only problem there is if you want sqrt to be a real function.[/hide]

Title: Re: Radically Constant
Post by 0.999... on Aug 2^nd, 2010, 7:12am

[hide]I held the belief that (for analogy, inform me if the analogy doesn't apply) if g(x) = (x-1)/(x-1), then g would not have a value at x = 1.[/hide]

Title: Re: Radically Constant
Post by towr on Aug 2^nd, 2010, 7:24am

[hide]I'm not really sure.
If one insists sqrt has to be real, then you can't calculate the values of the function for x < 1; but if you don't insist on that and allow complex values for sqrt then the overall function exists everywhere and is real for every x.[/hide]

Title: Re: Radically Constant
Post by ThudanBlunder on Aug 2^nd, 2010, 7:37am

I believe the intention is that the square root is meant to be real.

Title: Re: Radically Constant
Post by rmsgrey on Aug 2^nd, 2010, 8:59am

on 08/02/10 at 07:12:41, 0.999... wrote:

[hide]I held the belief that (for analogy, inform me if the analogy doesn't apply) if g(x) = (x-1)/(x-1), then g would not have a value at x = 1.[/hide]

In your example, g has a well-defined value everywhere except x=1, and the same well-defined limit as x tends to 1 from both directions, so there is a unique, well-defined function, everywhere continuous, that is equal to g whenever x is not 1, where g is not necessarily defined, so, by including continuity as a condition on g, you can treat it as having a value at x=1.

My default interpretation would be to say that g has a value at x=1 and only pay attention to the distinction between the continuous version of g, and the, mostly discontinuous, family of well-defined functions that are equal except at x=1, only paying attention to that distinction if it's brought up by context.