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We need to use the Maclaurin series of $\cosh(x)$.

RobPratt
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noname
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  • It is a classic exercise to prove that $a_{n} = 2^{n-1}$ (for example) - does that help? – preferred_anon Feb 26 '21 at 23:46
  • It’s quite easy to derive this exponential generating function without any reference to the hyperbolic cosine. In what sense are you required to use its Maclaurin series? – Brian M. Scott Feb 26 '21 at 23:51
  • I have to use the Product formula, exponential version. Where I have to assume a_n be the number of ways to build a certain structure on an n-th element set, and b_n be the number of ways to build another structure on an n-th element. And c_n be the number of ways to seperate [n] into the disjoint subsets S and T, such that the union of S and T = [n]. – noname Feb 27 '21 at 00:04
  • and then build a structure of the first kind on S, and a structure of the second kind on T. Let A(x), B(x) and C(x) be the respective exponential generating functions of the sequences {a_n}, {b_n}, and c_{n}. Then A(x)B(x) = C(x) – noname Feb 27 '21 at 00:06

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HINT: The Maclaurin series for $\cosh x$ is

$$\cosh x=\sum_{n\ge 0}\frac{x^{2n}}{(2n)!}\,,$$

so it corresponds to the sequence $\langle e_n:n\ge 0\rangle$, where

$$e_n=\begin{cases} 1,&\text{if }n\text{ is even}\\ 0,&\text{if }n\text{ is odd.} \end{cases}$$

The number of even-sized subsets of $\{1,\ldots,n\}$ is

$$\sum_{k\ge 0}\binom{n}{2k}=\sum_{k\ge 0}\binom{n}ke_k\cdot 1\,.$$

This last expression is the coefficient of $\frac{x^n}{n!}$ in the product of the power series $\sum_{n\ge 0}e_n\frac{x^n}{n!}$ and $\sum_{n\ge 0}1\cdot\frac{x^n}{n!}$ corresponding to the sequences $\langle e_n:n\ge 0\rangle$ and $\langle 1,1,1,\ldots\rangle$. Now all you have to do is identify the generating function for that last sequence and put the pieces together.

You can check your result be deriving it in a different way. The empty set has one even-sized subset, and $\{1,\ldots,n\}$ has $2^{n-1}$ even-sized subsets for $n\ge 1$, so the power series that you want is

$$1+\sum_{n\ge 1}2^{n-1}\frac{x^n}{n!}=1+\frac12\sum_{n\ge 1}\frac{(2x)^n}{n!}\,.$$

Write this in terms of an exponential function, and compare that result with the ways of expressing $\cosh x$ in terms of exponential functions.

Brian M. Scott
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