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.
\( D_t e^x = 0 \), 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.
