5,...\}$?

7 events

when toggle format	what		by	license	comment
Feb 6, 2023 at 19:23	history	edited	Tom Copeland	CC BY-SA 4.0	Missing polynomial added. Links added. Caveat added to avoid a common mistake by some.
Feb 3, 2023 at 17:46	comment	added	Tom Copeland		To avoid problems in umbral eval in other contexts, note $(c+a.x)^m = \sum_{n=0}^m \binom{m}{n} c^{m-n} (a.x)^n=\sum_{n=0}^m \binom{m}{n} c^{n} (a.x)^{m-n} $ umbrally evaluated gives $\sum_{n=0}^m \binom{m}{n} c^{m-n} a_nx^n =\sum_{n=0}^m \binom{m}{n} c^{n} a_{m-n}x^{m-n} $. Then generalizing, $\frac{x}{1+ax} = x(1+ax)^{-1} = x\sum_{n \geq 0} \binom{-1}{n} a^n x^n =x\sum_{n \geq 0} (-1)^n a^n x^n = a^0x - a^1x^2 + a^2x^3 -\cdots $ umbrally evaluated gives $a_0 x- a_1x^2 + a_2x^3 -\cdots$. With $a_0= 1$, this detail can be ignored as is certainly the case for $a_n=1$ for all $n \geq 0$.
Feb 3, 2023 at 15:52	history	edited	Tom Copeland	CC BY-SA 4.0	Overlooked bar added
Feb 3, 2023 at 4:49	history	edited	Tom Copeland	CC BY-SA 4.0	Missing bar and argument summand added
Feb 3, 2023 at 2:24	history	edited	Tom Copeland	CC BY-SA 4.0	Added action of reciprocal polynomials
Feb 2, 2023 at 18:00	history	edited	Tom Copeland	CC BY-SA 4.0	Missing argument provided. Cut-and-paste error eliminated.
Feb 2, 2023 at 16:44	history	answered	Tom Copeland	CC BY-SA 4.0