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

• chickensandwich
In summary: So the question still remains.In summary, the Taylor series for the function F(x) = cos(\sqrt{x}) about a=0 is given by 1 - \frac{x}{2!} + \frac{x^2}{4!} - \frac{x^3}{6!} \ldots. This series is valid for x \ge 0 and can be obtained by substituting \sqrt{x} into the Taylor series of cos(x) about a=0. The function F(x) = cos(\sqrt{x}), f : \mathbb{R} \geq 0 \to \mathbb{R} is differentiable at x=0 and therefore, functions on closed intervals can
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 $$\cosh(\sqrt{-x})$$ 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?

chickensandwich said:
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 $$\cosh(\sqrt{-x})$$ 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.

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.

Last edited:
chickensandwich said:
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.

Last edited:
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,∞).

Last edited:
Delta² said:
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 ##

PeroK said:
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.

## Question 1: What is a Taylor series expansion?

A Taylor series expansion is a mathematical representation of a function as an infinite sum of terms, where each term is a polynomial function of the variable x. It is used to approximate the value of a function at a certain point by using its derivatives at that point.

## Question 2: What is the significance of having a valid Taylor series expansion at a=0?

Having a valid Taylor series expansion at a=0 means that the function can be approximated at that point using its derivatives. This is important because it allows us to calculate the value of the function at that point without having to evaluate the function directly.

## Question 3: Does every function have a valid Taylor series expansion at a=0?

No, not every function has a valid Taylor series expansion at a=0. The function must be infinitely differentiable at a=0 in order for its Taylor series expansion to be valid.

## Question 4: How do we determine if cos(sqrt(x)) has a valid Taylor series expansion at a=0?

We can determine if cos(sqrt(x)) has a valid Taylor series expansion at a=0 by checking if the function is infinitely differentiable at a=0. This means that all of its derivatives exist at a=0 and can be calculated.

## Question 5: What is the Taylor series expansion of cos(sqrt(x)) at a=0?

The Taylor series expansion of cos(sqrt(x)) at a=0 is 1 - (x/2) + (x^2/24) - (x^3/720) + ... , where each term is calculated using the derivatives of the function at a=0.

• Calculus and Beyond Homework Help
Replies
1
Views
990
• Calculus and Beyond Homework Help
Replies
7
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
26
Views
1K
• Calculus and Beyond Homework Help
Replies
27
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
443
• Calculus and Beyond Homework Help
Replies
6
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
462
• Calculus and Beyond Homework Help
Replies
1
Views
351
• Calculus and Beyond Homework Help
Replies
5
Views
1K