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I don't understand how the definition for the generating function of Bell polynomials, i.e. $\Phi(t):=\exp(\sum_{k=1}^\infty x_k \frac{t^k}{k!})$, makes sense. If we write $F:=\sum_{k=1}^\infty x_k \frac{t^k}{k!}$, then $F\in \mathbb{C}[X_1,X_2,...][[t]]$ and $\Phi=\exp(F)$ but I don't get how one defines $\exp(F)$.

Let $\mathbb{C}[X_1,X_2,...][[t]]$ be equipped with the discrete topology. The above definition makes sense only if the sequence $\{\sum_{k=1}^n \frac{F^k}{k!}\}$ is convergent, but it is very unclear to me.

Or think of it in this way.

Define $\phi(u):=\sum_{n=0}^\infty \frac{(Fu)^n}{n!}$ so that $\phi \in \mathbb{C}[X_1,X_2,...][[t]][[u]]$.

$\Phi$ is well- define iff the evaluation of $\phi$ at $1$, i.e. $\phi(1)$ should be well-defined. But how?

Thank you in advance!

user26857
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Rubertos
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  • What makes you feel it is not well-defined? – lisyarus Apr 26 '17 at 13:37
  • @lisyarus I add that part ! – Rubertos Apr 26 '17 at 13:41
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    To compute $\Phi$ you need to compute its coefficient of $t^m$ for each $m$. For fixed $m$, show that the coefficient of $t^m$ in ${\sum_{k=1}^n \frac{F^k}{k!}}$ stays the same once $n$ is large enough. In that sense, the limit as $n \to \infty$ exists. – GEdgar Apr 26 '17 at 13:53
  • @GEdgar Is that still true if we replace $F$ by any arbitrary element $G$ in $\mathbb{C}[X_1,X_2,...]$? So that $exp(G)$ always make sense? – Rubertos Apr 26 '17 at 14:01
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    The important point here is that $F$ starts at $k=1$. If it started at $k=0$ we would be out of luck. – GEdgar Apr 26 '17 at 14:04
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    See this old answer and this related question. As GEdgar explained, the point is that the series you plug into the exponential series has no constant term. If you want to sound fancy, you can say that the resulting series converges $t$-adically. Basically saying that modulo any power of $t$, no matter how high, only finitely many terms will affect the "low degree part". – Jyrki Lahtonen Apr 26 '17 at 14:14

1 Answers1

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Here is a method (not original by me) to find $p(t) =\exp(f(t)) =\exp(\sum_{k=1}^\infty a_k \frac{t^k}{k!}) $.

If $p(t) =\exp(f(t)) $, differentiating, $p'(t) =f'(t)\exp(f(t)) =f'(t)p(t) $.

If $p(t) =\sum_{i=1}^{\infty} p_i \frac{ t^i}{i!} $ (we want to find the $p_i$), then $p'(t) =\sum_{i=1}^{\infty} ip_i \frac{t^{i-1}}{i!} =\sum_{i=0}^{\infty} p_{i+1} \frac{t^{i}}{i!} $ and $f'(t) =\sum_{k=1}^\infty ka_k \frac{t^{k-1}}{k!} =\sum_{k=0}^\infty a_{k+1} \frac{t^k}{k!} $.

Multiplying the power series,

$\begin{array}\\ p'(t) &=\sum_{n=0}^{\infty} p_{n+1} \frac{t^{n}}{n!}\\ \text{and}\\ f'(t)p(t) &=\sum_{k=0}^\infty a_{k+1} \frac{t^k}{k!}\sum_{i=0}^{\infty} p_i \frac{ t^i}{i!} \qquad\text{(with } p_0 = 0)\\ &=\sum_{k=0}^\infty\sum_{i=0}^{\infty} a_{k+1}p_i \frac{t^{k+i}}{k!i!}\\ &=\sum_{n=0}^\infty\sum_{k+i=n} a_{k+1}p_i \frac{t^{n}}{k!i!}\\ &=\sum_{n=0}^\infty\sum_{i=0}^n a_{n-i+1}p_i \frac{t^{n}}{(n-i)!i!} \qquad(k = n-i)\\ &=\sum_{n=0}^\infty\frac{t^n}{n!}\sum_{i=0}^n a_{n-i+1}p_i \frac{n!}{(n-i)!i!}\\ &=\sum_{n=0}^\infty\frac{t^n}{n!}\sum_{i=0}^n a_{n-i+1}p_i \binom{n}{i}\\ \end{array} $

Equating coefficients of $t^n$, $p_{n+1} =\sum_{i=0}^n a_{n-i+1}p_i \binom{n}{i} =\sum_{i=1}^n a_{n-i+1}p_i \binom{n}{i} $ since $p_0 = 0$.

To get $p_1$, set $t=0$ to get $p_1 =p'(0) =f'(0)p(0) =a_1\exp(f(0)) =a_1 $.

marty cohen
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