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

where

So we can rewrite the integral as

Since the power series defines an entire function we can exchange the integral and sum so that we have

where is the falling factorial.

This is just the newton series of f around 0.

TPID 13 is thus solved, as 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

where

So we can rewrite the integral as

Since the power series defines an entire function we can exchange the integral and sum so that we have

where is the falling factorial.

This is just the newton series of f around 0.

TPID 13 is thus solved, as 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?