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ThudnBlunder
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Radically Constant
« on: Aug 1st, 2010, 9:24am » |
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Find numbers a and b (a < b) and the value of the constant such that
[x + 2 (x - 1)]1/2 + [x - 2(x - 1)]1/2 is constant for a  x b.
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| « Last Edit: May 19th, 2011, 4:42am by ThudnBlunder » |
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towr
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Re: Radically Constant
« Reply #1 on: Aug 2nd, 2010, 1:26am » |
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[x + 2 (x - 1)] + [x - 2 (x - 1)] = c
[x + 2 (x - 1)] = c - [x - 2 (x - 1)]
x + 2 (x - 1) = c2 - 2c [x - 2 (x - 1)] + x - 2 (x - 1)
4 (x - 1) = c2 - 2c [x - 2(x - 1)]
4 (x - 1) - c2 = - 2c [x - 2(x - 1)]
16 (x - 1) - 8c2 (x - 1) + c2 = 4c2 [x - 2(x - 1)]
16 (x - 1) + c^4 = 4 c2 x
take c2 = y
y2 - 4 y x + 16 (x - 1) = 0
y = [4x +/- (16x2 - 64 (x - 1)) ]/2
y = 2(x +/- ((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.
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| « Last Edit: Aug 2nd, 2010, 1:40am by towr » |
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0.999...
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Re: Radically Constant
« Reply #2 on: Aug 2nd, 2010, 5:43am » |
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If we have a real function f(x) = [x+2(x-1)] + [x-2(x-1)] then it is defined only for values of x  1.
Let m = (x-1). Hence, f(x) = (m2+2m+1) + (m2-2m+1) = |m+1|+|m-1| = 2(x-1) if x 2.
On the other hand, if 1  x 2, then m+1 0 and m-1 0. So f(x) = m+1-(m-1) = 2.
Hence a = 1, b = 2, f(x) = 2 in this interval.
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| « Last Edit: Aug 2nd, 2010, 6:31am by 0.999... » |
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towr
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Re: Radically Constant
« Reply #3 on: Aug 2nd, 2010, 7:02am » |
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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.
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0.999...
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Re: Radically Constant
« Reply #4 on: Aug 2nd, 2010, 7:12am » |
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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.
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towr
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Re: Radically Constant
« Reply #5 on: Aug 2nd, 2010, 7:24am » |
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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.
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| « Last Edit: Aug 2nd, 2010, 7:25am by towr » |
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ThudnBlunder
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Re: Radically Constant
« Reply #6 on: Aug 2nd, 2010, 7:37am » |
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I believe the intention is that the square root is meant to be real.
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rmsgrey
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Re: Radically Constant
« Reply #7 on: Aug 2nd, 2010, 8:59am » |
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on Aug 2nd, 2010, 7:12am, 0.999... wrote:| 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. |
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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.
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