User:Pjpline/sandbox

Introduction
I would like to suggest that formal power series can also be raised to any rational number, and that this follows from the fact of being able to raise them to integral powers.

When I first considered adding this concept to the formal power series page, I thought it wouldn't be allowed as it appears that it has not previously been published. However, it is a matter of fairly simple algebra to verify it, so I am asking for a ruling on whether it can be included.

I have also discovered in the History section of the Binomial series page, that Newton and Wallis did something similar for more restricted series. So, this more generalized result is not entirely new.

Reply
Thank you, that is a very interesting page, which is worthy of study.

I see that I wasn't clear enough. I am suggesting that it is possible to derive $$c_m$$ as a solution to

$$ \left( \sum_{k=0}^\infty a_k X^k \right)^{\!\frac{1}{n}} =\, \sum_{m=0}^\infty c_m X^m,$$

where n, k and m are integers, with similar restrictions to those specified for power series raised to powers. For example:

$$ \left( 1 + X + X^2 \right)^{\!\frac{1}{3}} =\ 1 + \frac{X}{3} + \frac{2X^2}{9} - \frac{13X^3}{81} + \frac{8X^4}{243} + \frac{37X^5}{729} - \frac{379X^6}{6561} + \cdots $$

Reply two
Restrictions:

$$c_0 = a_0^\frac{1}{n}$$ exists in the ring of coefficients.

$$c_0^{n-1}$$ exists and is invertible in the ring of coefficients.

n is invertible in the ring of coefficients.

Explanation
Since there isn't an immediate "No" here is an explanation of the method. It uses some concepts which appear complicated but which are actually quite simple.

The first is an extension to the partitions of an integer, making fixed length partitions by padding shorter partitions with zero(s). The partitions have an added restriction that no part is allowed to exceed a specified value. I use $$\overset{k}{\underset{n}{ \operatorname{pe}(m)}}$$ to represent the set of partitions of m with length n, where each part is limited to k.

$$\begin{align} \overset{k}{\underset{n}{ \operatorname{pe}(0)}} &= \{ 0^n \}, \ \ k \geq 0 , n \geq 1 && n \text{-partitions of zero}\\

\overset{k}{\underset{1}{ \operatorname{pe}(m)}} &= \{ m^1 \}, \ \ k \geq m , m \geq 0 && 1 \text{-partitions of } m \text{ with limit } k\\

\overset{k}{\underset{n}{ \operatorname{pe}(m)}} &= \bigcup_{i = \left \lceil \frac{m}{n} \right \rceil}^{\min(k,m)} \left ( \bigcup_{P \in \left ( \overset{i}{\underset{n - 1}{ \operatorname{pe}(m - i)}} \right ) } \{ i P \} \right ), \ \ n > 1 , m \leq kn && n \text{-partitions of } m \text{ with limit } k\\

\overset{3}{\underset{4}{ \operatorname{pe}(5)}} &= \{ 3^12^10^2, 3^11^20^1, 2^21^10^1, 2^11^3 \} && \text{example}\\

\end{align}$$

The second is a function that takes a partition set and an ordered list of coefficients and outputs a new coefficient:

$$\begin{align} tx \left ( \mathcal{P}, (a_k) \right ) &= \sum_{P \in \mathcal{P}} \left ( \left ( \sum_{p \in P} m(p) \right ) ! \prod_{p \in P} \frac{a_p^{m(p)}}{m(p)!} \right ) && m(p) \text{ is the count of part } p \\

tx \left ( \overset{3}{\underset{4}{ \operatorname{pe}(5)}}, (a_k) \right ) &= 4!\frac{a_3a_2a_0^2}{2!} + 4!\frac{a_3a_1^2a_0}{2!} + 4!\frac{a_2^2a_1a_0}{2!} + 4!\frac{a_2a_1^3}{3!} && \text{example} \\

\end{align}$$ Note that when $$\mathcal{P}$$ is a set of n-partitions, $$\sum_{p \in P} m(p)$$ is always n.

Using these, we have:

$$\begin{align} \left( \sum_{k=0}^\infty a_k X^k \right)^{\!n} &= \sum_{m=0}^\infty c_m X^m \\

c_m &= \operatorname{tx} \left ( \overset{m}{\underset{n}{ \operatorname{pe}(m)}}, (a_k) \right ) \\

\end{align}$$

Now, to find $$c_m$$ to satisfy

$$\begin{align} \left( \sum_{k=0}^\infty a_k X^k \right) ^\frac{1}{n} &= \sum_{m=0}^\infty c_m X^m, n>1 && \text{potentially multiple solutions} \\

\sum_{k=0}^\infty a_k X^k &= \left( \sum_{m=0}^\infty c_m X^m \right) ^{\!n} && \text{as a possibility} \\

a_m &= \operatorname{tx} \left ( \overset{m}{\underset{n}{ \operatorname{pe}(m)}}, (c_j) \right ) && (c_j) \text{ to be solved} \\

a_0 &= \operatorname{tx} \left ( \overset{0}{\underset{n}{ \operatorname{pe}(0)}}, (c_j) \right ) \\ &= n!\frac{c_0^n}{n!} \\ c_0 &= a_0^\frac{1}{n} && \text{possible multi-choice, obvious restrictions} \\

a_1 &= \operatorname{tx} \left ( \overset{1}{\underset{n}{ \operatorname{pe}(1)}}, (c_j) \right ) \\ &= n!\frac{c_1c_0^{n - 1}}{{(n - 1)}!} \\ c_1 &= \frac{a_1}{nc_0^{n-1}} && \text{obvious restrictions} \\

a_2 &= \operatorname{tx} \left ( \overset{2}{\underset{n}{ \operatorname{pe}(2)}}, (c_j) \right ) \\ &= n!\frac{c_2c_0^{n - 1}}{{(n - 1)}!} + n!\frac{c_1^2c_0^{n - 2}}{{2!(n - 2)}!} \\ c_2 &= \frac{a_2 - \frac{n(n - 1)c_1^2c_0^{n-2}}{2}}{nc_0^{n-1}} && \text{obvious restrictions} \\

\overset{m}{\underset{n}{ \operatorname{pe}(m)}} &= \{m0^{n-1}\} \cup \overset{m - 1}{\underset{n}{ \operatorname{pe}(m)}}, \ \ n \geq 1, m \geq 1 && \text{verify from definition}\\

a_m &= \operatorname{tx} \left ( \{m0^{n-1}\} \cup \overset{m - 1}{\underset{n}{ \operatorname{pe}(m)}}, (c_j) \right ) , \ \ n \geq 1, m  \geq 1 \\

&= \operatorname{tx} \left ( \{m0^{n-1}\}, (c_j) \right ) + \operatorname{tx} \left ( \overset{m - 1}{\underset{n}{ \operatorname{pe}(m)}}, (c_j) \right ) \\

&= nc_mc_0^{n - 1} + \operatorname{tx} \left ( \overset{m - 1}{\underset{n}{ \operatorname{pe}(m)}}, (c_j) \right ) \\

nc_mc_0^{n - 1} &= a_m - \operatorname{tx} \left ( \overset{m - 1}{\underset{n}{ \operatorname{pe}(m)}}, (c_j) \right ) \\

c_m &= \frac{a_m - \operatorname{tx} \left ( \overset{m - 1}{\underset{n}{ \operatorname{pe}(m)}}, (c_j) \right )}{nc_0^{n - 1}} && \text{exact for chosen } c_0 \\

\end{align}$$

The question is, would it be allowable to add a less verbose version of this to the Formal power series page?

Power series raised to rational powers
An alternative expression for a power peries raised to a natural number is as follows:

$$ \left( \sum_{k=0}^\infty a_k X^k \right)^{\!n} =\, \sum_{m=0}^\infty c_m X^m,$$ where $$\begin{align} c_m &= \operatorname{tx} \left ( \overset{m}{\underset{n}{ \operatorname{pe}(m)}}, (a_k) \right )\\

(a_k) &= a_0,a_1,a_2, \cdots \\

\overset{k}{\underset{n}{ \operatorname{pe}(0)}} &= \{ 0^n \}, \ \ k \geq 0 , n \geq 1 && n \text{ partitions of zero}\\

\overset{k}{\underset{1}{ \operatorname{pe}(m)}} &= \{ m^1 \}, \ \ k \geq m , m \geq 0 && 1 \text{ partitions of } m \text{ with limit } k\\

\overset{k}{\underset{n}{ \operatorname{pe}(m)}} &= \bigcup_{i = \left \lceil \frac{m}{n} \right \rceil}^{\min(k,m)} \left ( \bigcup_{P \in \left ( \overset{i}{\underset{n - 1}{ \operatorname{pe}(m - i)}} \right ) } \{ i P \} \right ), \ \ n > 1 , m \leq kn && n \text{-partitions of } m \text{ with limit } k\\

\overset{3}{\underset{4}{ \operatorname{pe}(5)}} &= \{ 3^12^10^2, 3^11^20^1, 2^21^10^1, 2^11^3 \} && \text{example}\\

tx \left ( \mathcal{P}, (a_k) \right ) &= \sum_{P \in \mathcal{P}} \left ( \left ( \sum_{p \in P} m(p) \right ) ! \prod_{p \in P} \frac{a_p^{m(p)}}{m(p)!} \right ) \\

tx \left ( \overset{3}{\underset{4}{ \operatorname{pe}(5)}}, (a_k) \right ) &= 4!\frac{a_3a_2a_0^2}{2!} + 4!\frac{a_3a_1^2a_0}{2!} + 4!\frac{a_2^2a_1a_0}{2!} + 4!\frac{a_2a_1^3}{3!} && \text{example} \\ \\ \left( \sum_{k=0}^\infty a_k X^k \right) ^\frac{1}{n} &= \sum_{m=0}^\infty c_m X^m && \text{solve for } (c_m) \\

\sum_{k=0}^\infty a_k X^k &= \left( \sum_{m=0}^\infty c_m X^m \right) ^{\!n} && \text{as a possibility} \\

a_m &= \operatorname{tx} \left ( \overset{m}{\underset{n}{ \operatorname{pe}(m)}}, (c_j) \right ) && (c_j) \text{ to be solved} \\

a_0 &= \operatorname{tx} \left ( \overset{0}{\underset{n}{ \operatorname{pe}(0)}}, (c_j) \right ) \\ &= n!\frac{c_0^n}{n!} \\ c_0 &= a_0^\frac{1}{n} && \text{possible multi-choice, obvious restrictions} \\

a_1 &= \operatorname{tx} \left ( \overset{1}{\underset{n}{ \operatorname{pe}(1)}}, (c_j) \right ) \\ &= n!\frac{c_1c_0^{n - 1}}{{(n - 1)}!} \\ c_1 &= \frac{a_1}{nc_0^{n-1}} && \text{obvious restrictions} \\

a_2 &= \operatorname{tx} \left ( \overset{2}{\underset{n}{ \operatorname{pe}(2)}}, (c_j) \right ) \\ &= n!\frac{c_2c_0^{n - 1}}{{(n - 1)}!} + n!\frac{c_1^2c_0^{n - 2}}{{2!(n - 2)}!} \\ c_2 &= \frac{a_2 - \frac{n(n - 1)c_1^2c_0^{n-2}}{2}}{nc_0^{n-1}} && \text{obvious restrictions} \\

\overset{m}{\underset{n}{ \operatorname{pe}(m)}} &= \{m0^{n-1}\} \cup \overset{m - 1}{\underset{n}{ \operatorname{pe}(m)}}, \ \ n \geq 1, m \geq 1 && \text{verify from definition}\\

a_m &= \operatorname{tx} \left ( \{m0^{n-1}\} \cup \overset{m - 1}{\underset{n}{ \operatorname{pe}(m)}}, (c_j) \right ) , \ \ n \geq 1, m  \geq 1 \\

&= \operatorname{tx} \left ( \{m0^{n-1}\}, (c_j) \right ) + \operatorname{tx} \left ( \overset{m - 1}{\underset{n}{ \operatorname{pe}(m)}}, (c_j) \right ) \\

&= nc_mc_0^{n - 1} + \operatorname{tx} \left ( \overset{m - 1}{\underset{n}{ \operatorname{pe}(m)}}, (c_j) \right ) \\

nc_mc_0^{n - 1} &= a_m - \operatorname{tx} \left ( \overset{m - 1}{\underset{n}{ \operatorname{pe}(m)}}, (c_j) \right ) \\

c_m &= \frac{a_m - \operatorname{tx} \left ( \overset{m - 1}{\underset{n}{ \operatorname{pe}(m)}}, (c_j) \right )}{nc_0^{n - 1}} \\

\end{align}$$

(This formula can only be used if the $$n^{\text{th}}$$ root of a0 exists, and n and c0 are invertible in the ring of coefficients.)

Note that if a0 is unity and all ak are rational, then all cm are rational.