08/22/2009, 05:50 PM
(This post was last modified: 08/22/2009, 11:57 PM by Base-Acid Tetration.)
Uh Oh I saw a problem with the f-logarithm terminology. What if it can mean a function/operator being plugged into the exp's Taylor series (powers meaning iteration)? Someone might call THAT a functional exponential, in analog to the matrix exponential...
consider this example (just an example, don't say it's circular because i use exp's taylor series) this statement of analyticity of a function:
Iff the following statement holds true for all \( x, x_0 \in U: \)
\( f(x) = e^{(x-x_0)\frac{d}{dx}}[f](x_0), \)
then the function is analytic in \( U. \)
Can you call such a usage an operator exponential?
Annotation: if you're confused about my example:
(x-x0) is scalar multiplication by x-x0, and d/dx is the differentiation operator ...
so \( e^{(x-x_0)\frac{d}{dx}} = \sum^\infty_{n=0}\frac{1}{n!} (x-x_0)^n \frac{d^n}{dx^n} \)
Apply that to f, and you get: \( e^{(x-x_0)\frac{d}{dx}}[f] = \sum^\infty_{n=0}\frac{(x-x_0)^n}{n!} \frac{d^n f}{dx^n} \)
"Plug in" x_0 to this resulting function: \( e^{(x-x_0)\frac{d}{dx}}[f](x_0) = \sum^\infty_{n=0}\frac{(x-x_0)^n}{n!} f^{(n)} (x_0)= \sum^\infty_{n=0} \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n \).
consider this example (just an example, don't say it's circular because i use exp's taylor series) this statement of analyticity of a function:
Iff the following statement holds true for all \( x, x_0 \in U: \)
\( f(x) = e^{(x-x_0)\frac{d}{dx}}[f](x_0), \)
then the function is analytic in \( U. \)
Can you call such a usage an operator exponential?
Annotation: if you're confused about my example:
(x-x0) is scalar multiplication by x-x0, and d/dx is the differentiation operator ...
so \( e^{(x-x_0)\frac{d}{dx}} = \sum^\infty_{n=0}\frac{1}{n!} (x-x_0)^n \frac{d^n}{dx^n} \)
Apply that to f, and you get: \( e^{(x-x_0)\frac{d}{dx}}[f] = \sum^\infty_{n=0}\frac{(x-x_0)^n}{n!} \frac{d^n f}{dx^n} \)
"Plug in" x_0 to this resulting function: \( e^{(x-x_0)\frac{d}{dx}}[f](x_0) = \sum^\infty_{n=0}\frac{(x-x_0)^n}{n!} f^{(n)} (x_0)= \sum^\infty_{n=0} \frac{f^{(n)}(x_0)}{n!} (x-x_0)^n \).