(New page: We recall that <math>V_0^x</math> is monotone increasing, thus f is monotone decreasing, and <math>V_0^x = f(0)-f(x)</math>. Then we have <math>f(0)-f(x) = (f(0)-f(x))^{\frac{1}{2}} \Ri...) |
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\Rightarrow (f(0)-f(x))^2 = (f(0)-f(x)) </math> | \Rightarrow (f(0)-f(x))^2 = (f(0)-f(x)) </math> | ||
| − | + | Factoring yields <math>f(x) = f(0)</math> or <math>f(x) = f(0)-1</math>, with two compatible functions: | |
<math>f_1(x) = \frac{1}{3}</math>, | <math>f_1(x) = \frac{1}{3}</math>, | ||
Latest revision as of 13:44, 22 July 2008
We recall that $ V_0^x $ is monotone increasing, thus f is monotone decreasing, and $ V_0^x = f(0)-f(x) $.
Then we have
$ f(0)-f(x) = (f(0)-f(x))^{\frac{1}{2}} \Rightarrow (f(0)-f(x))^2 = (f(0)-f(x)) $
Factoring yields $ f(x) = f(0) $ or $ f(x) = f(0)-1 $, with two compatible functions:
$ f_1(x) = \frac{1}{3} $,
$ f_2(x) = \frac{1}{3}, \ 0<x\leq 1, f_2(0) = \frac{4}{3} $
-pw
