# Understanding Transfer Principle About Formal Power Series

The theorem I am referring to is from "The Concrete Tetrahedron" by Kauers and Paule, which states:

Theorem 2.8 (Transfer Principle) Let $$a(z) = \sum_{n=0}^{\infty} a_nz^n$$ and $$b(z) = \sum_{n=0}^{\infty} b_nz^n$$ be real or complex functions analytic in a non-empty open neighbourhood of zero. If $$a(z) = b(z)$$ for all $$z \in U$$, then $$a_n = b_n$$ for all $$n \in \mathbb{N}$$.

An example given in the book to illustrate this principle is:

"The transfer principle can be used for obtaining simple proofs of identities. For instance, to demonstrate that $$\exp(\log(1+x)) = 1 + x$$ as formal power series, it is sufficient to observe that this relation is true for the corresponding analytic functions. Proving this without the transfer principle would necessitate elaborate calculations."

My interpretation is that the Transfer Principle implies that the Taylor series of the analytic function $$\exp(\log(1+x))$$ is simply $$1 + x$$.

However, when we regard $$\exp(\log(1+x))$$ as a formal power series (generating functions), we use the definitions: $$\exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!}$$ and $$\log(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}.$$ Then, by the definition of the composition of power series, we get: $$\exp(\log(1+x)) = \sum_{n=0}^\infty \frac{\left(\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}\right)^n}{n!}.$$ My confusion arises here: How can we assert that the above power series, $$\sum_{n=0}^\infty \frac{\left(\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}\right)^n}{n!},$$ is equivalent to the Taylor series of $$\exp(\log(1+x))$$ and therefore equals $$1+x$$?

I am unable to see how this conclusion is derived directly from the Transfer Principle.

Update: I don't have a problem with the theorem itself. But I am not sure it implies the identity given as an example. More specifically, if we consider $$\exp(x)$$ and $$\log(1+x)$$ as short-hands to write the corresponding power series, how do we know that $$\exp(\log(1+x))$$ as a composition of two power series, has the same coefficients as $$\exp(\log(1+x))$$ when considered as a composition of two analytic functions.

• I think you are reading too much into it. The point is that since we know the identify is true (from the definition of $\log,\exp$) then the expansions of the left and right are identical. So you avoid the computation. Jan 30 at 7:36
• I know what is the point. I just don't see the details. There is a gap between composing to power series and composing the corresponding analytic functions. I think it's been hidden and there needs to be lemma somewhere which says that if $A(B(x)) = C(x)$ as functions, then $A(B(x)) = C(x)$ as power series. Jan 30 at 7:46
• If we think of $a(z)$ and $b(z)$ as analytic functions (near $0$) rather than just formal power series, then the coefficients of the power series for $a(z)$ and $b(z)$ are determined uniquely as residues: $a_n=\mathrm{Res}(a(z)/z^{n+1};0)$. So, if there is an identity for analytic functions, then it holds for their power series as well. Also, by the way, I think the combinatorial content of your example is that there is the same number of permutations of length $n\ge 2$ with an even number of cycles and with an odd number of cycles. Jan 30 at 11:41
• Note that $\exp(\log(1+x)) = \sum_{n=0}^\infty \frac{\left(\sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}\right)^n}{n!}$ is not a power series. The terms need to be rearranged formally to write this as a power series. Then their Theorem 2.8 asserts that the two power series are equal. Presumably they provide a proof of Theorem 2.8, which may not be entirely trivial. Jan 30 at 12:39
• Of course, I agree with you, that it can be seen quite easily that the two types of composition yield the same power series, but being an introductory book this should at least be mentioned at some point. Jan 31 at 6:46