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Let $R$ be a ring and given a power series $f \in R[[x]]$ , $f = \sum a_n x^n$:

Define $r(f)= \textrm{max} \{i : a_i \neq 0 \ \textrm{and} \ a_j=0 \ \textrm{for all} \ j \leq i\}$

Is it always true that $r(f)= \textrm{min} \{i: a_i \neq 0\}$?

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    For all $j\leq i$, implies $a_{i}=0$ contradiction.2012-05-24

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Of course. Let $i_0 = \max\{i\mid a_i\neq 0\text{ and }a_j=0\text{ for all }j\leq i\}$. Then $a_{i_0}\neq 0$, so $\min\{i\mid a_i\neq 0\}\leq i_0$. On the other hand, if $j\lt i_0$, then by definition of $i_0$ we must have $a_j=0$, so $\min\{i\mid a_i\neq 0\}\neq i_0$. As this holds for all $j\lt i_0$, this proves that $\min\{i\mid a_i\neq 0\}\geq i_0$, giving equality.

Note that the first definition is a bit silly, though: if there is at least one such $i$, there is at most one such $i$; because if $a_i\neq 0$, then for all $i'\gt i$ we have $i'\notin\{i\mid a_i\neq 0\text{ and }a_j=0\text{ for all }j\leq i\}$ (since $i\lt i'$ and $a_{i}\neq 0$. On the other hand, if $i$ is in the set, then no $j\lt i$ can be in the set (since the fact that $i$ is in the set implies that $a_j=0$ for all $j\lt i$. So the set in question is either empty or a singleton!

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    Thanks, yes I was confused by that definition because its a singleton if not all coefficients are zero.2012-05-24