Does cos(sqrt(x)) have a valid Taylor series expansion at a=0?

[chickensandwich]

Find the Taylor series about a=0 for the function F(x) = \cos(\sqrt{x}).

Taylor series expansion of a function f(x) about a
\sum^{\infty}_0 \frac{f^{(n)}(a)}{n!}(x-a)^n

Taylor series of \cos{x} about a=0 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} \ldots

From these expansions I could easily find a solution by a simple substitution of \sqrt{x} into the Taylor series of \cos{x} about a=0 which produces the series
1 - \frac{x}{2!} + \frac{x^2}{4!} - \frac{x^3}{6!} \ldots
I found (through the use of a graphics calculator and the equality cos(z) = cosh(iz)) that this series converges to \cos(\sqrt{x}) for x \ge 0 and \cosh(\sqrt{-x}) <br /> for x \le 0
My question is, is this Taylor series about a valid at a=0,\, as for a Taylor series to exist about a point a the function has to have the property that it is infinitely differentiable at a and \cos(\sqrt{x}) is not differentiable at all at 0 (at least if you don't extend the function \cos{x} to take imaginary values or my understanding of differentiability of endpoints is incorrect).

I know that a function's continuity is defined differently at endpoints (for example the function F(x) = \sqrt{x} is not defined as x \to 0^- but is continuous at x=0), is a function's differentiability also defined differently at endpoints?

I think the main point of my question is, is the function F(x) = \cos(\sqrt{x}),\, f : \mathbb{R} \geq 0 \to \mathbb{R} differentiable at x=0? Or in a more general form, are functions on closed intervals differentiable at their endpoints?

 

[Dick]

Find the Taylor series about a=0 for the function F(x) = \cos(\sqrt{x}).

Taylor series expansion of a function f(x) about a
\sum^{\infty}_0 \frac{f^{(n)}(a)}{n!}(x-a)^n

Taylor series of \cos{x} about a=0 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} \ldots

From these expansions I could easily find a solution by a simple substitution of \sqrt{x} into the Taylor series of \cos{x} about a=0 which produces the series
1 - \frac{x}{2!} + \frac{x^2}{4!} - \frac{x^3}{6!} \ldots
I found (through the use of a graphics calculator and the equality cos(z) = cosh(iz)) that this series converges to \cos(\sqrt{x}) for x \ge 0 and \cosh(\sqrt{-x})<br /> for x \le 0
My question is, is this Taylor series about a valid at a=0,\, as for a Taylor series to exist about a point a the function has to have the property that it is infinitely differentiable at a and \cos(\sqrt{x}) is not differentiable at all at 0 (at least if you don't extend the function \cos{x} to take imaginary values or my understanding of differentiability of endpoints is incorrect).

I know that a function's continuity is defined differently at endpoints (for example the function F(x) = \sqrt{x} is not defined as x \to 0^- but is continuous at x=0), is a function's differentiability also defined differently at endpoints?

I think the main point of my question is, is the function F(x) = \cos(\sqrt{x}),\, f : \mathbb{R} \geq 0 \to \mathbb{R} differentiable at x=0? Or in a more general form, are functions on closed intervals differentiable at their endpoints?

I think this is a question for the lawyers. There is a concept of a one-sided derivative and don't see any reason why you can't Taylor expand based on that. The point about the form of the function for $x \lt 0$ is interesting, but I don't think that's very important. The point is that your series does converge for $x \ge 0$. My say.

 

[Delta2]

Though you found the series by a smart and easy trick, i think it would be beneficial to prove that there is the n-th derivative of the function F(x) (as you defined it) at the point x=0 and it is equal to F^{n}(0)=\frac{(-1)^n}{(n+1)...(2n)}

Notice we care only for the right derivative as the function is only defined in the right of x=0.

Functions on close intervals are not necessarily differentiable in the end points, but for a function f/[a,b] we care only for the right derivatives at a, in order to apply taylor's theorem around point a.

 

[PeroK]

My question is, is this Taylor series about a valid at a=0,\, as for a Taylor series to exist about a point a the function has to have the property that it is infinitely differentiable at a and \cos(\sqrt{x}) is not differentiable at all at 0 (at least if you don't extend the function \cos{x} to take imaginary values or my understanding of differentiability of endpoints is incorrect).

Using the substitution of $\sqrt{x}$ into the Taylor series for $cos(x)$ is perfectly valid and produces a series expansion for $cos(\sqrt{x})$ valid for $x \ge 0$.

The way to look at it is that the Taylor series for $cos(x)$ is valid for whatever numerical value you choose for $x$. For example, you could also get a perfectly valid series expansion for $cos(|x|)$ or even $cos(f(x))$ where $f(x)$ is not continuous anywhere.

In this case, as Delta2 points out, $cos(\sqrt{x})$ may be infinitely (right) differentiable at 0. Remember that a derivative is a limit, not a formula.

In fact, you could use your valid series expansion for $cos(\sqrt{x})$ to see that this function is infinitely differentiable and calculate its nth derivatives at 0.

 

[Delta2]

There is a slight problem though, it seems to me its not so easy to find the generic form of the n-th derivative of the function f(x)=cos(\sqrt x) at (0,∞) which we need in order to find f^n(0) as the limit \lim_{x \to 0^{+}}\frac{f^{n-1}(x)-f^{n-1}(0)}{x-0} which can be simplified using De Hospital rule to f^n(0)=\lim_{x \to 0^+}f^n(x) since we know the n-th derivative exists at (0,∞).

 

[PeroK]

There is a slight problem though, it seems to me its not so easy to find the generic form of the n-th derivative of the function f(x)=cos(\sqrt x) at (0,∞) which we need in order to find f^n(0) as the limit \lim_{x \to 0^{+}}\frac{f^{n-1}(x)-f^{n-1}(0)}{x-0} which can be simplified using De Hospital rule to f^n(0)=\lim_{x \to 0^+}f^n(x) since we know the n-th derivative exists at (0,∞).

It's easy if you use:

$cos(\sqrt{x}) = 1 - \frac{x}{2!} + \frac{x^2}{4!} - \frac{x^3}{6!} \ldots$

 

[Delta2]

It's easy if you use:

$cos(\sqrt{x}) = 1 - \frac{x}{2!} + \frac{x^2}{4!} - \frac{x^3}{6!} \ldots$

Oh well yes but that's the thing we want to prove (i mean that we ignore the proof that has been given though it is 100% correct) with an alternative method.