I am doing homework in PDE. One exercise says:
Set $P(x, D)$ a partial differential operator, show that $$P(x,\xi+\eta)=\sum_{\alpha}\frac{1}{\alpha!}\xi^\alpha P^{(\alpha)}(x,\eta)$$
I know what $P(x, D)$ is, but what about $P(x,\xi+\eta)$ and $P^{(\alpha)}(x,\eta)$? I check every section before this exercise and find nothing illustrating what $P(x,ξ)$ and $P^{(\alpha)}(x,\xi)$ mean. Nonetheless, I feel this is just some simple manipulation of binomial expansion...
Thanks to the comments. I think it makes since that $P(x,\cdot)$ means substitution of $x$ in a polynomial $P(x)$ by whatever is on the second slot.
The answer is a simple binomial expansion (as I have expected).
We have $$P(x,\xi+\eta)=\sum_{|\alpha|\le k}a_\alpha(x)(\xi+\eta)^\alpha$$ where $a_\alpha(x)\in C^{\infty}(\mathbb{R}^n)$ is a smooth function as a coefficient (I add it here because in our textbook an arbitrary smooth function is multiplied when defining the differential operator). Expand it by the binomial formula to get $$P(x,\xi+\eta)=\sum_{|\alpha|\le k}a_\alpha(x)\sum_{\beta\le\alpha}{\alpha\choose\beta}\xi^\beta\eta^{\alpha-\beta}$$ On the other hand we have $$\sum_{\alpha}\frac{1}{\alpha!}\xi^\alpha P^{(\alpha)}(x,\eta)=\sum_{|\alpha|\le k}\frac{1}{\alpha!}\xi^\alpha\sum_{|\beta| \le k}a_\beta(x)(\eta^\beta)^{(\alpha)}$$ obviously under the second summation, all terms such that $\beta<\alpha$ should vanish, and we get $$\sum_{\alpha}\frac{1}{\alpha!}\xi^\alpha P^{(\alpha)}(x,\eta)=\sum_{|\alpha|\le k}\frac{1}{\alpha!}\xi^\alpha\sum_{\beta \ge \alpha}a_\beta(x)(\eta^\beta)^{(\alpha)}=\sum_{|\alpha|\le k}{\beta\choose\alpha}\xi^\alpha\sum_{\beta \ge \alpha}a_\beta(x)(\eta^{\beta-\alpha})$$ which is identical to the previous one if we exchange the summation indices $\alpha$ and $\beta$.