I am trying to calculate the Taylor series for the function:
$$f(x) = {\frac 1 x}(1 - \cos\sqrt{x})$$
How do I do it, if I know the Taylor series for $\cos(x)$?
$\cos x = {\Large \sum\limits_{k = 0}^{\infty}}\dfrac{(-1)^k x^{2k}}{(2k)!}$
It should be done just with algebraic modifications... Do I also need to know the Taylor series for $1/x$? Thank you!
First plug in $x \to \sqrt{x}$ to get $$ \cos(\sqrt{x}) = \sum_{k = 0}^\infty \frac{(-1)^k (\sqrt{x})^{2k}}{(2k)!} = \sum_{k = 0}^\infty \frac{(-1)^k x^{k}}{(2k)!} $$
Now if you look at $$ -\cos(\sqrt{x}) = -\sum_{k = 0}^\infty \frac{(-1)^k x^{k}}{(2k)!} = \sum_{k = 0}^\infty \frac{(-1)^{k+1} x^{k}}{(2k)!} $$ Do you see how to proceed from here?
EDIT:
Notice then that the $k = 0$ term is $-1$ so that when we add $1$ to the sum this term cancels out so $$ 1-\cos(\sqrt{x}) = \sum_{k = 1}^\infty \frac{(-1)^{k+1} x^{k}}{(2k)!} $$
Finally we have $$ \frac{1}{x} \left(1-\cos(\sqrt{x}) \right) = \frac{1}{x}\sum_{k = 1}^\infty \frac{(-1)^{k+1} x^{k}}{(2k)!} = \sum_{k = 1}^\infty \frac{(-1)^{k+1} x^{k}}{x(2k)!} = \sum_{k = 1}^\infty \frac{(-1)^{k+1} x^{k-1}}{(2k)!} $$