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 What is the convergence radius of this power series? mike3 Long Time Fellow Posts: 368 Threads: 44 Joined: Sep 2009 07/04/2011, 04:14 AM (This post was last modified: 07/04/2011, 04:20 AM by mike3.) @JmsNxn, I fed the given series $e^x \ln(x) \stackrel{\mathrm{supposedly}}{=} \sum_{n=0}^{\infty} \frac{\psi_0(n + 1)}{n!} x^n$ to the Wolfram Alpha calculator. It says that it does not give the left-hand side, but instead, it gives $e^x (\ln(x) + \Gamma(0, x)) = \sum_{n=0}^{\infty} \frac{\psi_0(n + 1)}{n!} x^n$ (I have no idea how it managed to derive that formula -- any suggestions?) But this shows the problem. There is an extra term present, which is the upper incomplete gamma function. It decays to 0 as $x \rightarrow \infty$. So, your series is asymptotic to, but not equal to, $e^x \ln(x)$. I.e. the first "equation" is really $e^x \ln(x) \sim \sum_{n=0}^{\infty} \frac{\psi_0(n + 1)}{n!} x^n \quad (x \rightarrow \infty)$. So obviously, there must be something wrong in the derivation. I think I found it. In the beginning, you assume $D_t e^x = 0$, which, given the definition of your $D_t$ operator, would mean that $\frac{d^t}{dx^t} e^x = e^x$. But that's a problem. There's a catch when working with fractional derivatives. Namely, that they are "non-local". This is similar to how the integral requires a "lower (or upper) bound". Another way to think of choosing the bound is "choosing a branch" of the inverse function. The fractional derivative is a continuous and real-indexed iteration of the derivative, which means it must also include all negative iterates as well -- and those are integrals, so the "non-local" property of the integrals must show up somewhere, and it shows up at every non-nonnegative-integer order of differentiation. In general, fractional-iterate functions that can do real and complex iterates will be multi-valued functions if whatever is being iterated is not injective, and derivative is definitely not injective (differentiate a constant function). In the theory of fractional derivatives, when you took $\frac{d^t}{dx^t} e^x = e^x$, you took the lower bound as $-\infty$. (See how you have to take this as the lower bound for an integral of the exponential function if you want it to return the exponential function back? Mmm-hmm. Same here.) But, when you took $\frac{d^t}{dx^t} x^n = \frac{\Gamma(n+1)}{\Gamma(n+1-t)} x^{n-t}$ which you used to build $D_t x^n$, and thus the power series, you were taking the lower bound to be 0. That's a funny thing about these formulas -- and in some cases it is not said what the lower bound is. When you then took the power series, it was kind of like saying $\int_{-\infty}^{x} e^t dt = \int_{0}^{x} \left(\sum_{n=0}^{\infty} \frac{t^n}{n!}\right) dt$ and here, the error becomes clear. « Next Oldest | Next Newest »