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iterated derivation - Xorter - 06/09/2019
Hi, everyone! To be honest, iterated derivation as shifting may have been better, because as you will see, it is connected to shifting, but not known if it is just luck or what. As you can see in Dr. Peyam's vlog the exponential derivation is simply shifting the variable of function by the power of the derivation. So for example: (exp D f)(x) = f(x+1) ((exp D)^N f)(x) = f(x+N) And the so-called normal derivative of a function can also be represented as shifting if you draw the Taylor series of the function as in a "number" system: each coefficients are shifted to right (the radix is the variable x). What if all the function of the operator derivation is a kind of shifting of the function? Well, I have checked the (natural) logarithm of the operator, I do not want to make this post long and boring, so I am writing only the result. (If needed, it is okay for me to share the whies and hows.) So: (log D)(c x^n) = c Li2 x^n Which you can apply to all the coefficients of the Taylor series of a function. Would it be a kind of shifting? Furthermore I have checked also the double exponentiation of derivation, I have no so many results, but I know that it is related to numbers of stirling, but e. g. (exp exp D)(c x) = e(1 + c x). What is the rule of (exp^oN D f)(x) or more generally the rule of ((e [M] x)^oN o D f)(x)? Of course, (f^oN)(x) = (f o f o ... o f)(x) and D = d/dx, sorry, I should learn to use LaTex, again. |