• 0 Vote(s) - 0 Average
• 1
• 2
• 3
• 4
• 5
 What is the convergence radius of this power series? JmsNxn Long Time Fellow Posts: 291 Threads: 67 Joined: Dec 2010 05/26/2011, 02:50 AM (This post was last modified: 05/26/2011, 04:48 PM by JmsNxn.) This proof starts out by considering the differential operator $D_t f(x) = \frac{d}{dt}\frac{d^t}{dx^t} f(x)$ which is spreadable across addition $D_t [f(x) + g(x)] = D_t f(x) + D_t g(x)$. And: $D_t e^x = 0$, which is important for this proof. And next, using traditional fractional calculus laws $\frac{d^t}{dx^t} x^n = \frac{\Gamma(n+1)}{\Gamma(n+1-t)} x^{n-t}$: $D_t x^n = \frac{d}{dt} \frac{\Gamma(n+1)}{\Gamma(n+1-t)} x^{n-t}$ which comes to, (if you want me to show you the long work out just ask, I'm trying to be brief) $D_t x^n = x^{n-t}\frac{\Gamma(n+1)}{\Gamma(n+1-t)}(\psi_0(n+1-t) - \ln(x))$ where $\psi_0(x)$ is the digamma function. So now we do the fun part: $D_t e^x = 0\\\\ D_t (\sum_{n=0}^{\infty} \frac{x^n}{n!}) = 0\\\\ \sum_{n=0}^{\infty} D_t \frac{x^n}{n!} = 0\\\\$ So we just plug in our formula for $D_t x^n$ and divide it by n!: $0 = \sum_{n=0}^{\infty} x^{n-t}\frac{\Gamma(n+1)}{n!\Gamma(n+1-t)}(\psi_0(n+1-t) - \ln(x))$ (divide $\Gamma(n+1)$ by n!.) we expand these and seperate and rearrange: $0 = \sum_{n=0}^{\infty} (\frac{x^{n-t}}{\Gamma(n+1-t)}\psi_0(n+1-t) - \frac{x^{n-t}}{\Gamma(n+1-t)}\ln(x))\\\\ 0 = (\sum_{n=0}^{\infty} \frac{x^{n-t}}{\Gamma(n+1-t)}\psi_0(n+1-t)) - (\sum_{c=0}^{\infty} \frac{x^{c-t}}{\Gamma(c+1-t)}\ln(x))\\\\ \ln(x)(\sum_{c=0}^{\infty} \frac{x^{c-t}}{\Gamma(c+1-t)}) = (\sum_{n=0}^{\infty} \frac{x^{n-t}}{\Gamma(n+1-t)}\psi_0(n+1-t))\\\\$ And now if you're confused what t represents, you'll be happy to hear we eliminate it now by setting it to equal 0. therefore, all our gammas are factorials and the left hand side becomes e^x by ln(x) and the since digamma function for integers arguments can be expressed through harmonic numbers $\psi_0(n+1) = \sum_{k=1}^{n} \frac{1}{k} - \gamma$ where $\gamma$ is the euler/mascheroni constant: $\sum_{n=0}^{\infty} \frac{x^n}{n!}\psi_0(n+1) =\ln(x)e^x\\\\ \ln(x) = e^{-x}(1 - \gamma + \sum_{n=1}^{\infty} \frac{x^n}{n!} (\sum_{k=1}^{n} \frac{1}{k} - \gamma))$ I've been unable to properly do the ratio test but using Pari gp it seems to converge for values x >e, but failed at 1000, worked for 900. I decided to multiply the infinite series and I got: $ln(x) = \sum_{n=0}^{\infty} x^n (\sum_{k=0}^{n} (-1)^k \frac{\sum_{c=1}^{n-k}\frac{1}{c} - \gamma}{k!(n-k)!})$ Using Pari gp nothing seems to converge, but that may be fault to may coding. I'm wondering, has anybody seen this before? And I am also mainly wondering how I can prove the radius of convergence of this. Also, I wonder if this could further suggest the gamma function as the natural extension of the factorial function, since these values do converge. « Next Oldest | Next Newest »