Maclaurin's series for $\cos{\sqrt{x}}$

Question

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?

Answer 1

It leads to the same results. Call

$$ f(x) = \cos\sqrt{x} $$

• $f'(0)$

$$ f'(x) = -\frac{\sin\sqrt{x}}{2\sqrt{x}} $$

So that

$$ \lim_{x\to 0 }f'(x) = -\frac{1}{2} $$

where I used the well known limit $\lim_{t\to 0}\sin t/t = 1$.

• $f''(0)$

$$ f''(x) = \frac{1}{4 x}\left(-\cos \sqrt{x} + \frac{\sin \sqrt{x}}{\sqrt{x}}\right) $$

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 $t = \sqrt{x}$

Answer 2

Are you sure the second derivative is not defined at $x=0$? The first derivative is $$ f'(x) = \begin{cases} -\frac{\sin{\sqrt{x}}}{2\sqrt{x}} & x > 0 \\ -\frac{1}{2} & x = 0 \end{cases}. $$ The second derivative at zero is $$ f''(0) = \lim_{h \to 0} \frac{1}{h} \left( \frac{1}{2} - \frac{\sin{\sqrt{h}}}{2\sqrt{h}} \right) = \lim_{h \to 0} \frac{\sqrt{h}-\sin{\sqrt{h}}}{2h^{3/2}} = \frac{1}{12}. $$ Just because $g'(a)$ does not exist, doesn't mean that $(f \circ g)'(a)$ doesn't exist: the chain rule may only work if $g'(a)$ is defined, but the derivative is calculated as the limit of the quotient $(f(g(a+h))-f(g(a)))/h$, so, as here, can exist even if $g'(a)$ does not.

Answer 3

You seem to misinterpret the nature of the chain rule $$(f\circ g)'(x) = f'(g(x))g'(x).$$ What it means is that if $g$ is differentiable at $x_0$ and $f$ is differentiable at $g(x_0)$, then $f\circ g$ is differentiable at $x_0$ and the derivative is given by the above formula.

Basically, a crude way to say it is "conditions xyz are true $\implies$ formula works" and its contrapositive is "formula doesn't work $\implies$ conditions xyz are not true". The contrapositive of the chain rule doesn't say that composition isn't differentiable if the formula doesn't work.

For example, $(\cos\sqrt x)' = -\frac{\sin\sqrt x}{2\sqrt x}$ for $x\neq 0$ by the chain rule. As you say, it is not defined at $0$, so the chain rule doesn't work at $0$ and from that we can conclude that either $\cos$ or $\sqrt{\,\cdot\,}$ is not differentiable at $0$. And take a look at that, indeed we know that $\sqrt{\,\cdot\,}$ is not differentiable at $0$.

However, that still doesn't mean that $\cos\sqrt x$ isn't differentiable at $0$. We can check the definition of the derivative, in this case we must verify that the limit $$\lim_{x\to 0}\frac{\cos\sqrt x - 1}{x}$$ exists. It does and is equal to $-\frac 12$. Not only that, $$\lim_{x\to 0}\frac{-\sin\sqrt x}{2\sqrt x} = -\frac 12$$ as well, so not only is $\cos\sqrt x$ differentiable everywhere, its first derivative is continuous.

Checking for the second and higher derivatives leads to similar conclusion.