Exact and Unique solution for base e^(1/e)
#4
Ok, I now understand your definition.
You take as initial function \( f : (-1,0)\to(0,1) \) the linear function \( f(x)=x+1 \) which just maps -1 to 0 and 0 to 1.
Hence the piecewise defined resulting function g defined by
\( g(x)=
\begin{cases}
f(x) &\text{for} &x\in (-1,0]\\
\exp_b(g(x-1)) &\text{for} &x>0\\
\log_b(f(x+1)) &\text{for} &x\in(-2,-1]
\end{cases}
\)
is continuous. And if we look at the first derivation at 0 from the left:
\( g'(0)=(x+1)'|_{x=0}=1 \) and from the right:
\( g'(+0)=\exp_b(f(x-1))'|_{x=0}=\exp_b'(0)=\exp_b(0)\log(b)=\log(b) \)

then we realize that both derivations are equal exactly for \( b=e \). By the recurrence that applies to all other joins of the piecewise defined function too and hence g is (once) continuously differentiable for \( b=e \) (but not several times continuously differentiable).

Ok, step by step we continue. Now we want to have g also differentiable for other bases b. It looks as if you change the initial interval for this purpose (though keeping the initial function f linear) in a way that provides for differentiability of g.
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Messages In This Thread
RE: Exact and Unique solution for base e^(1/e) - by bo198214 - 08/10/2007, 05:58 PM

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