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You can treat derivation as a linear operator on vector-space of functions, let's name it D. Now second and third derivatives as operators are quite simple sinc
by leni536 11y ago
You can treat derivation as a linear operator on vector-space of functions, let's name it D. Now second and third derivatives as operators are quite simple since they are just D^2 and D^3. The question is if you can well define sqrt(D), or saying that if there is one and only one operator D_half such that D_half^2 is D (existence and unicity). It breaks at unicity since there are multiple half derivative constructions (frequency domain approach, Taylor-series approach,...).
On the other hand sqrt is well defined for positive real numbers (however not well defined for general complex numbers). There is a similar definition only for positive operators on Hilbert spaces (you pick the positive operator from all the possible square roots). However derivation is not a positive operator on the most used Hilbert-spaces.
- carlob 11y agoI remember reading a paper that claimed that integrating on a Grassmann variable space was equivalent to integrating in a space with -2 dimensions of the usual commuting variables.
- ngoldbaum 11y agoAt some point in my college Linear Algebra class I realized that the exponential function is an eigenvector of the derivative operator. Hilbert spaces are fun.
- c_lebesgue 11y agoAnd the Gaussian distribution is an eigenvector of the Fourier transform. Not surprisingly both of these facts correlate with how fundamental is the exponents function to solving linear differential equations, and how fundamental is the Gaussian distribution in probability theory.
- tome 11y agoYour approach seems to be simpler than the blog post. Can't you just note that the pth derivative of exp ax is a^p exp ax and then just extend via a fourier series? Maybe the non-uniqueness is too unsatisfying.
- leni536 11y agoI assume you are thinking the pth derviative of exp(jkx) which is (jk)^p exp(jkx) and extending this to Fourier-transform (or series). However (j*k)^(1/2) is not so well-defined, you can pick arbitrarily from two complex values for each k. This approach leads to infinite half derivative constructions.
- deleted 11y ago[deleted]
- cxseven 11y agoYes, and a single irrational power of a non-zero number has an infinite set of possible values to choose from.
- xamuel 11y agoNitpick: the act of taking derivatives is not "derivation", it is "differentiation"
- madcaptenor 11y agoNitpick squared: http://en.wikipedia.org/wiki/Derivation_(differential_algebra) http://en.wikipedia.org/wiki/Derivation_(differential_algebr... When I first learned this in a graduate abstract algebra class, however, I was convinced that my professor, who was not a native English speaker, was simply mistranslating from whatever language he was used to talking about these things in. (He was, IIRC, a Romanian, educated in Germany.)