I've got Schur polynomials of multivariables $x=x_1,x_2,x_3,\ldots$defined as
$$\exp\left(\sum_{k=1}^\infty x_k z^k\right) = \sum_{k=0}^\infty P_k(x) z^k$$
So, how can i get $ P_0(x)=1 $ ? And how can I get the formula for, for example $ P_{30}(x) $ ? How can i get recursive formula for this?
When I tried to expand this equation I get something like this:
$$\exp(x_1 z^1+x_2 z^2+x_3 z^3+\cdots) = P_0(x) $$
which isn't even similar to 1. And if I try to cut this expression in the way that I don't see the reason it would be correct, I get something like this:
$$\exp(x_1z^1) = P_0(x).$$
Still not 1. And I don't think it is even correct... Is it an assumption that $P_0=1$? How would You get recurisive formula for Schur polynomials then?
Since $\exp(u)=\sum_{i\geq0}u^i$, we have \begin{align} \exp(\sum_{k=1}^{\infty} x_k z^k)&=\sum_{i\geq0}\frac{1}{i!}\left(\sum_{k=1}^{\infty} x_k z^k\right)^i\\ &=1+z\left(\sum_{k=1}^{\infty} x_k z^{k-1}\right)+\frac{z^2}{2!}\left(\sum_{k=1}^{\infty} x_k z^{k-1}\right)^2+\cdots \end{align} As every term after the first one has $z$ as a factor, the constant term in the sum is $1$, and that is $P_0$. Similarly, all terms from the third one on are divisible by $z^2$, so the term involving $z$ comes from the secodn term and he coefficient of $z$ in the sum is $x_1$. All terms after the fourth one are divisible by $z^2$, so the coefficient of $z^2$ in the whole sum is the sam as the coefficient of $z^2$ in $$1+z\left(\sum_{k=1}^{\infty} x_k z^{k-1}\right)+\frac{z^2}{2!}\left(\sum_{k=1}^{\infty} x_k z^{k-1}\right)^2,$$ which is $x_2+x_1^2/2$.
The coefficient of $z^3$ is $\frac{x_1^3}{6}+x_2 x_1+x_3$ and comes fromthe first four terms, and so on.