Maclaurin expansion series at $x=0$

Question

I have the function $$f(x)=\left\{\begin{array}{cl} \dfrac{e^{x}-1}{x}, & x\neq0 \\ 1, & x=0 \end{array}\right.$$ And I gotta find its Maclaurin expansion series.

I know how to find the one from $\dfrac{e^{x}-1}{x}$. But how do get a function that $f(0)=1$?

score 2 · Accepted Answer · answered Feb 25 '19 at 01:11

2

$$e^x - 1 = \sum_{k\ge 1} {x^k\over k!}.$$ Now divide by $x$ and you are done.

answered Feb 25 '19 at 01:11

ncmathsadist

49,383

This function has a removable singularity at 0. – ncmathsadist Feb 25 '19 at 01:12
So the answer is $$\sum\frac{x^{k}}{(k+1)!}$$ for $k\geq0$? – mvfs314 Feb 25 '19 at 01:23
2

Just re-index the sum! – ncmathsadist Feb 25 '19 at 01:39

score 1 · Answer 2 · answered Feb 25 '19 at 01:08

1

Your constant is equal to 1, as your function approaches 1 when x=0. So set your constant to 1 and just use derivatives to approximate $\dfrac{e^{x}-1}{x}$ around x=0.

answered Feb 25 '19 at 01:08

Sina Babaei Zadeh

2,357

Maclaurin expansion series at $x=0$

2 Answers2