03/28/2015, 06:46 AM
(This post was last modified: 03/28/2015, 06:47 AM by fivexthethird.)
It seems that this method is equivalent to the method of newton series.
To see this, note that
\( e^x \sum_{k=0}^{\infty}f(k) \frac{(-1)^k x^k}{k!} = (\sum_{k=0}^{\infty}\frac{x^k}{k!})(\sum_{k=0}^{\infty}f(k) \frac{(-1)^k x^k}{k!}) = \)
\( \sum_{k=0}^{\infty}x^k \sum_{j=0}^k\frac{(-1)^{j}f(j)}{(k-j)!j!} = \sum_{k=0}^{\infty}\frac{x^k}{k!} \sum_{j=0}^k(-1)^{j}f(j){k \choose j} = \)
\( \sum_{k=0}^{\infty}{\Delta}^kf(0) \frac{(-1)^{k}x^k}{k!} \)
where \( \Delta f(x) = f(x+1)-f(x) \)
So we can rewrite the integral as
\( \frac{1}{\Gamma(-z)}\int_0^{\infty} \sum_{k=0}^{\infty}{\Delta}^kf(0) \frac{(-1)^{k}x^{k-z-1}e^{-x}}{k! } dx \)
Since the power series defines an entire function we can exchange the integral and sum so that we have
\( \sum_{k=0}^{\infty}{\Delta}^kf(0)\frac{(-1)^{k}}{k!\Gamma(-z)} \int_0^{\infty} x^{k-z-1}e^{-x} dx = \)
\( \sum_{k=0}^{\infty}{\Delta}^kf(0)\frac{(-1)^{k}\Gamma(k-z)}{k!\Gamma(-z)} =\sum_{k=0}^{\infty}{\Delta}^kf(0)\frac{(z)_{k}}{k!} \)
where \( (z)_{k} = z(z-1)(z-2)...(z-(k-1)) \) is the falling factorial.
This is just the newton series of f around 0.
TPID 13 is thus solved, as \( x^{\frac{1}{x}} \) satisfies the bounds required for Ramanujan's master theorem to apply.
@tommy: I don't see how inverting around t helps us recover tetration from the interpolated super root... it gets us the slog, yes, but in either case we need to just invert around m to get tetration.
Or am I misinterpreting your post?
To see this, note that
\( e^x \sum_{k=0}^{\infty}f(k) \frac{(-1)^k x^k}{k!} = (\sum_{k=0}^{\infty}\frac{x^k}{k!})(\sum_{k=0}^{\infty}f(k) \frac{(-1)^k x^k}{k!}) = \)
\( \sum_{k=0}^{\infty}x^k \sum_{j=0}^k\frac{(-1)^{j}f(j)}{(k-j)!j!} = \sum_{k=0}^{\infty}\frac{x^k}{k!} \sum_{j=0}^k(-1)^{j}f(j){k \choose j} = \)
\( \sum_{k=0}^{\infty}{\Delta}^kf(0) \frac{(-1)^{k}x^k}{k!} \)
where \( \Delta f(x) = f(x+1)-f(x) \)
So we can rewrite the integral as
\( \frac{1}{\Gamma(-z)}\int_0^{\infty} \sum_{k=0}^{\infty}{\Delta}^kf(0) \frac{(-1)^{k}x^{k-z-1}e^{-x}}{k! } dx \)
Since the power series defines an entire function we can exchange the integral and sum so that we have
\( \sum_{k=0}^{\infty}{\Delta}^kf(0)\frac{(-1)^{k}}{k!\Gamma(-z)} \int_0^{\infty} x^{k-z-1}e^{-x} dx = \)
\( \sum_{k=0}^{\infty}{\Delta}^kf(0)\frac{(-1)^{k}\Gamma(k-z)}{k!\Gamma(-z)} =\sum_{k=0}^{\infty}{\Delta}^kf(0)\frac{(z)_{k}}{k!} \)
where \( (z)_{k} = z(z-1)(z-2)...(z-(k-1)) \) is the falling factorial.
This is just the newton series of f around 0.
TPID 13 is thus solved, as \( x^{\frac{1}{x}} \) satisfies the bounds required for Ramanujan's master theorem to apply.
@tommy: I don't see how inverting around t helps us recover tetration from the interpolated super root... it gets us the slog, yes, but in either case we need to just invert around m to get tetration.
Or am I misinterpreting your post?
