Odd converging power series for ln(x) centered at zero

JmsNxn · **Posted:** Sat, 2 Jul 2011 00:58:19 UTC

This proof starts out by considering the differential operator $\frac{D_t}{Dx} f(x) = \frac{d}{dt}\frac{d^t}{dx^t} f(x)$ which is spreadable across addition. I call it the super derivative. For this proof we require:
$\frac{D_t}{Dx} e^x = 0$

And next, using traditional fractional calculus laws:

$\frac{D_t}{Dx} x^n = \frac{d}{dt} \frac{\Gamma(n+1)}{\Gamma(n+1-t)} x^{n-t}$

which comes to, (if you want me to show you the long work out just ask, I'm trying to be brief)

$\frac{D_t}{Dx} x^n = x^{n-t}\frac{\Gamma(n+1)}{\Gamma(n+1-t)}(\psi_0(n+1-t) - \ln(x))$ where $\psi_0(x)$ is the digamma function.

So now we do the fun part:

$\frac{D_t}{Dx} e^x = 0\\\\ \frac{D_t}{Dx} (\sum_{n=0}^{\infty} \frac{x^n}{n!}) = 0\\\\ \sum_{n=0}^{\infty} \frac{D_t}{Dx} \frac{x^n}{n!} = 0\\\\$

So we just plug in our formula for $\frac{D_t}{Dx} x^n$ 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 and the digamma function for integers arguments can be expressed through harmonic numbers :
$\sum_{n=0}^{\infty} \frac{x^n}{n!}\psi_0(n+1) =\ln(x)e^x\\\\ \ln(x) = e^{-x}(-\gamma + \sum_{n=1}^{\infty} \frac{x^n}{n!} \psi_0(n+1))$

I've been unable to properly do the ratio test but using Pari gp it seems to converge for values x >[4,5], but failed at around 1000, worked for 900.

I also multiplied the series but it gave a smaller convergence radius.

$ln(x) = \sum_{n=0}^{\infty} x^n (\sum_{k=0}^{n} (-1)^k \frac{\psi_0(n-k+1)}{k!(n-k)!})$

which converges for x > [4,5] and fails at around 35 or 36.

The two very odd things about this series is that a) it's a power series of ln(x) centered about zero, and secondly, the first series has an infinite convergence radius through the ratio test and through another proof by another forum member that I showed this to. So my question is, logically. What is the proper convergence radius. Or exactly when this series is equal to ln(x) and when it is not.

Any questions comments would be much appreciated, thanks for reading if you got this far.

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Odd converging power series for ln(x) centered at zero

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