Section 1.2.9: Generating Functions

Exercise 11. [M25] Equation (39) can also be used to express the $S$'s in terms of the $h$'s: We find $S_1=h_1$, $S_2=2h_2-h_1^2$, $S_3=3h_3-3h_1h_2+h_1^3$, etc. What is the coefficient of $h_1^{k_1}h_2^{k_2}\cdots h_m^{k_m}$ in this representation of $S_m$, when $k_1+2k_2+\cdots+mk_m=m$?

Verified: yes
Solve time: 7m

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Solution

Let

H(z)=\sum_{m\ge0}h_m z^m ] be the generating function for the complete homogeneous symmetric functions. By equation (39),

\ln H(z)=\sum_{j\ge1}\frac{S_j}{j}z^j.

\tag{1}

$$ Hence $$

S_m=m[z^m]\ln H(z),

\tag{2}

where $[z^m]$ denotes extraction of the coefficient of $z^m$. Since

H(z)=1+\sum_{j\ge1}h_j z^j,

$$ equation (24) gives $$

\ln H(z)

\sum_{r\ge1}\frac{(-1)^{r+1}}{r}

\left(\sum_{j\ge1}h_j z^j\right)^r.

\tag{3}

$$ To find the coefficient of the monomial $$

h_1^{k_1}h_2^{k_2}\cdots h_m^{k_m},

$$ assume $$

k_1+2k_2+\cdots+mk_m=m.

\tag{4}

$$ Let $$

r=k_1+k_2+\cdots+k_m.

\tag{5}

$$ In the expansion of $$

\left(\sum_{j\ge1}h_j z^j\right)^r,

$$ the term $$

h_1^{k_1}h_2^{k_2}\cdots h_m^{k_m}

z^{,k_1+2k_2+\cdots+mk_m}

h_1^{k_1}h_2^{k_2}\cdots h_m^{k_m}z^m

arises by choosing $k_j$ factors equal to $h_j z^j$ for each $j$. The number of such choices is the multinomial coefficient

\frac{r!}{k_1!k_2!\cdots k_m!}.

\tag{6}

$$ Therefore the coefficient of $$

h_1^{k_1}h_2^{k_2}\cdots h_m^{k_m}z^m

in $\ln H(z)$ is

\frac{(-1)^{r+1}}{r}

\frac{r!}{k_1!k_2!\cdots k_m!}

(-1)^{r+1}

\frac{(r-1)!}{k_1!k_2!\cdots k_m!}.

\tag{7}

Multiplying by $m$ according to (2), the coefficient of $h_1^{k_1}h_2^{k_2}\cdots h_m^{k_m}$ in $S_m$ is

m(-1)^{r+1}

\frac{(r-1)!}{k_1!k_2!\cdots k_m!},

\qquad

r=k_1+\cdots+k_m.

\tag{8}

$$ Hence the required coefficient is $$

\boxed{

(-1)^{k_1+\cdots+k_m+1},

m,

\frac{(k_1+\cdots+k_m-1)!}

{k_1!,k_2!\cdots k_m!}

}.

$$ This formula yields, for example, $$

S_2=2h_2-h_1^2,

$$$$

S_3=3h_3-3h_1h_2+h_1^3,

$$ in agreement with the values stated in the exercise. This completes the proof. ∎ $$