I saw the following identities being used for my Complex Variables class:
$e^{x} = 1 + x + O(x)^{2}$
$\cos(y+y') - \cos(y) = -(\sin(y))y' + O(y')^2$
$\sin(y+y') - \sin(y) = i(\cos(y))y' + O(y')^2$
Where $i$ denotes the imaginary unit. I'm not sure how the big-$O$ notation would work in these particular cases. What would it mean in this case, and how would one derive the above properties? I'm not sure where to start.
Thanks for the help!
For the first one, for small $x$, $e^x =\sum_{n=0}^{\infty} \frac{x^n}{n!} =1+x+\sum_{n=2}^{\infty} \frac{x^n}{n!} $, so
$\begin{array}\\ |e^x-1-x| &=|\sum_{n=2}^{\infty} \frac{x^n}{n!}|\\ &\le\sum_{n=2}^{\infty} |\frac{x^n}{n!}|\\ &=|x^2|\sum_{n=2}^{\infty} |\frac{x^{n-2}}{n!}|\\ &=|x^2|\sum_{n=0}^{\infty} |\frac{x^{n}}{(n+2)n!}|\\ \end{array} $
If $|x| \le \frac12$, $\sum_{n=0}^{\infty} |\frac{x^{n}}{(n+2)n!}| \le \sum_{n=0}^{\infty} |\frac{1}{2^n(n+2)n!}| < \sum_{n=0}^{\infty} |\frac{1}{2^n}| =1 $, so $|e^x-1-x| \le |x^2| = O(x^2) $.
Note that almost exactly the same argument will show that, for any $m > 0$, as $x \to 0$, $e^x =\sum_{n=0}^{m} \frac{x^n}{n!}+O(x^{m+1}) $.