Using the standard multi-index notation, suppose we have the two Taylor series $$ f(\theta) := \sum_{|\alpha|=0}^{\infty} a_{\alpha} \theta^{\alpha} $$ and $$ g(\theta) := \sum_{|\alpha|=0}^{\infty} b_{\alpha} \theta^{\alpha} $$ where $a_{\alpha} \in \mathbb{R}$, $\theta := (\theta_1, \theta_2)^T \in \mathbb{R}^2$ and $\alpha \in \left( \mathbb{Z}_{\geq 0} \right)^2$.

Question: On the convergence domain of $fg$, does $$ (fg)(\theta) = \sum_{|\alpha|=0}^{\infty} c_{\alpha} \theta^{\alpha} $$ where $$ c_{\alpha} := \sum_{\substack{\beta_1 + \gamma_1 = \alpha_1 \\ \beta_2 + \gamma_2 = \alpha_2\\}} a_{\beta} b_{\gamma}? $$


The proof is the same as in the one-variable case.

Suppose $f,g$ both converge at a point $(S_1,S_2)$ with $S_1,S_2\neq 0$ (if convergence requires $S_2=0$ then you are in a one-variable case). Let $C$ be such that $|a_\alpha.S^\alpha|,|b_\alpha.S^\alpha| \leq C$ for all $\alpha$ (this exists by convergence).

Now let $0<r<1$ and let $T_i=r|S_i|$. We show that the series for $fg$ converges uniformly absolutely in $[-T_1,T_1]\times[-T_2,T_2]$.

For this, consider the series

$$(*)\quad\sum_{\alpha,\beta} a_\alpha b_\beta \theta^\alpha \theta^\beta.$$

Putting absolute values we get the series

$$\sum_{\alpha,\beta} |a_\alpha b_\beta \theta^\alpha \theta^\beta| \leq \sum_{\alpha,\beta} |a_\alpha S^\alpha||b_\beta S^\beta|r^{|\alpha+\beta|}$$ $$\leq C^2 \sum_{\alpha,\beta} r^{|\alpha+\beta|} = C^2 \left(\sum_{n=0}^\infty r^n\right)^4 = \frac{C^2}{(1-r)^4}.$$

It follows that the series $(*)$ converges absolutely and uniformly in the stated domain. To see that it converges to $fg$ note that $$\left(\sum_{|\alpha|\leq N}a_\alpha \theta^\alpha\right) \left(\sum_{|\beta|\leq N}b_\beta \theta^\beta\right) $$ is a sequence of partial sums of $(*)$ which exhausts all summands. Now take $N\to\infty$. On the one hand you get $fg$ and on the other hand the sum of $(*)$.

Finally, given that the series $(*)$ converges absolutely to $fg$, we can arbitrarily reorder it without changing either the fact of convergence or the value of the sum. In particular, we can order the terms by the value of $\alpha+\beta$. Then the partial sums of your $\sum_\alpha c_\gamma \theta^\gamma$ are a subsequence of the partial sums of the rearrangement, so again they converge and to the same value.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.