Why does Maclaurin's series for this function exist:
$$y=\cos{\sqrt{x}}$$
even though the second derivative of the function at $x=0$ is undefined?
I know that you can use the standard series of $\cos{x}$ and replacing $x$ with $\sqrt{x}$ to find its Maclaurin's expansion, but why can't I use the standard Maclaurin's theorem to expand the function as a power series?
Best Answer
It leads to the same results. Call
$$ f(x) = \cos\sqrt{x} $$
$$ f'(x) = -\frac{\sin\sqrt{x}}{2\sqrt{x}} $$
So that
$$ \lim_{x\to 0 }f'(x) = -\frac{1}{2} $$
where I used the well known limit $\lim_{t\to 0}\sin t/t = 1$.
$$ f''(x) = \frac{1}{4 x}\left(-\cos \sqrt{x} + \frac{\sin \sqrt{x}}{\sqrt{x}}\right) $$
And the limit is
$$ \lim_{x\to0} f''(x) = \frac{1}{12} $$
So the first two terms lead to
$$ \cos\sqrt{x} = 1 - \frac{x}{2} + \frac{x^2}{24} + \cdots\tag{1} $$
For comparison, this is the expansion for $\cos (t)$
$$ \cos t = 1 -\frac{t^2}{2} + \frac{t^4}{24} + \cdots \tag{2} $$
You can obtain (1) from (2) just by replacing $t = \sqrt{x}$