Given $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + · · · + \frac{x^n}{n!} $$
I want to approximate $e^x$ for small $x$.
I'm having trouble understanding the difference between $o$ and $O$. I was wondering if someone could please work out this basic example for me and explain their steps.
Compute the value of $n$ (big o) and the value of $m$ (small o) in the following, and justify your calculations:
$$e^x = 1 + x + O(x^n ) \text{ as } x \rightarrow 0, \ \text{(Big O)}$$
$$e^x = 1 + x + o(x^m) \text{ as } x \rightarrow 0, \ \text{(Small o)}$$
Hint: write $f(x)$ as $e^x − (1 + x)$ and $g(x)$ as $x^n$ or as $x^m$.
Best Answer
For Big-O, you want to find an $n$ so that $$ \lim_{x\to 0}\frac{e^x-1-x}{x^n}=\lim_{x\to 0}\frac{x^2/2+x^3/3!+\cdots}{x^n}=c $$ for some constant $c$. It is clear that $n=2$ does the trick, since then $$ \lim_{x\to 0}\frac{e^x-1-x}{x^n}=\lim_{x\to 0}\frac{x^2/2+x^3/3!+\cdots}{x^2}\\ =\frac{1}{2}+\lim_{x\to 0}(x^2/3!+\cdots)=\frac12 $$ But of course, such a choice is not unique, which is why the use of the word "the" in this question is misleading. Taking $n=1$ or even $n=0$ will do.
For little-o, we would like that $c$ to be a $0$. Can you see from a very similar computation that taking $m=1$ will do, and $m=2$ will not? Can you find others?