Discussion
In the Taylor expansion of \( f \) about \( x = a \), the coefficient of \( (x-a)^{n} \) is
Answer: C
Writing \( f(x) = \sum c_{n}(x-a)^{n} \) and differentiating \( n \) times at \( x = a \) leaves only the \( n \)th term, giving \( f^{(n)}(a) = n!\,c_{n} \) and so
\[ c_{n} \;=\; \frac{f^{(n)}(a)}{n!}. \]
The factorial divisor is what makes the series converge for well-behaved functions; omitting it produces a series that diverges for all but the smallest arguments
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