Proposition 13

Tag 06CJ, permanent. Cite it as [Bourbaki, Tag 06CJ].

— In $Q[[X, Y]]$ we have $e^{X+Y} = e^X e^Y$ .

For the binomial formula gives

\frac{(X+Y)^n}{n!} = \sum_{i+j=n} \frac{X^i}{i!} \frac{Y^j}{j!}

Hence

e^X e^Y = \left( \sum_{i \geq 0} \frac{X^i}{i!} \right) \left( \sum_{j \geq 0} \frac{Y^j}{j!} \right) = \sum_{i,j \geq 0} \frac{X^i}{i!} \frac{Y^j}{j!} = \sum_{n \geq 0} \sum_{i+j=n} \frac{X^i}{i!} \frac{Y^j}{j!}

= \sum_{n \geq 0} \frac{(X+Y)^n}{n!} = e^{X+Y}.

We shall define two elements $e(X)$ , I(X) of $Q[[X]]$ by

e(X) = e^X - 1 = \sum_{n \geq 1} \frac{X^n}{n!}

(34)

l(X) = \sum_{n \geq 1} (-1)^{n-1} \frac{X^n}{n}.

We have (35)

e(X+Y) = e(X) + e(Y) + e(X)e(Y)

(36)

D(e^X) = D(e(X)) = e^X

(37)

D(l(X)) = \sum_{n \geq 0} (-X)^n = (1+X)^{-1}.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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