Why is the value of the coefficient $a_0=-1/2$ in this product of two series?

Question

I've read this answer here and know why $a_{-1}=1$ but I really don't get how $a_0=-\frac{1}{2}$.

Here is the equation in question: $$\left(\sum_{n\ge0}a_{n-1}z^n\right)\left(\sum_{n\ge0}\frac{z^n}{(n+1)!}\right)=1.$$ I can only see that $a_{-1}\frac{z^2}{6}+a_0z\cdot \frac{z}{2}+a_1z^2 = 0$.

Please, use descriptive titles. "Why is $a_0=−1/2$?" says nothing about the subject of the question. — jjagmath, Nov 08 '22 at 23:32
@jjagmath What would you write as title? As it relates to another question I'm not sure what to add — Quotenbanane, Nov 08 '22 at 23:35
How about something like "the value of a coefficient in the product of two series" or something like that? — jjagmath, Nov 08 '22 at 23:37

score 1 · Accepted Answer · answered Nov 08 '22 at 23:33

1

The coefficient of $z$ on the left side is $\frac {a_{-1}} {2!}+a_0$ and this must be $0$. This gives $a_0=-\frac 1 2$.

answered Nov 08 '22 at 23:33

Kavi Rama Murthy

359,332

Ah okay. So I'm looking at $z$ instead of $z^2$. Oops. – Quotenbanane Nov 08 '22 at 23:37

Why is the value of the coefficient $a_0=-1/2$ in this product of two series?

1 Answers1