# Showing that a function is C infinity?

1. Jul 3, 2009

### Skynt

I'm currently reading a book, The Road To Reality by Roger Penrose, and trying to tackle some of the exercises in the process. My knowledge in mathematics is limited, but broad enough to complete some of the exercises. Anyway, one of them wants you to consider the one function $$e^{\frac{-1}{x^{2}}}$$ and show that it is $$C^{\infty}$$ but not analytic..

The entire concept of C-infty functions has been hard to grasp... can anyone explain? (such as proving a function to be C-infty)

Thanks!

2. Jul 3, 2009

### Hurkyl

Staff Emeritus
Start small. Can you prove it's C0? C1?

3. Jul 3, 2009

### Skynt

I have no clue where to begin. I suppose a C1 is differentiable once? This section really lost me.

4. Jul 3, 2009

### D H

Staff Emeritus
Differentiability is not quite right. A function is C1 if its derivative is continuous. A function is C-infinity if derivatives of all order are continuous.

What is the derivative of $\exp(-x^2)$? Is it continuous? How about the second derivative, and so on?

5. Jul 3, 2009

### Hurkyl

Staff Emeritus
Of course, he should first worry if the function is continuous before he worries if its derivative is continuous!

(for the OP: if the function isn't continuous, it doesn't have a derivative. I assume you know that, but I felt I should say it anyways)

6. Jul 3, 2009

### Skynt

$$y = e^{-x^{2}}$$
It took me a little bit to realize that
$$ln \ y = -x^{2}$$
$$\frac{y'}{y} = -2x$$
$$y' = -2x ( e^{-x^{2}} )$$

It is continous and so forth if I did the same thing again... so that would prove it is $$C^{\infty}$$ ?

What would be a more formal way of proving that (if what I did was correct) in a more generalized fashion like I've noticed in hard to read proofs?

7. Jul 3, 2009

### D H

Staff Emeritus
That's not the function in the original post. You should be looking at

\aligned y &= e^{-1/x^{2}} \\ \ln y &= -1/x^2

As Hurkyl mentioned, the first (zeroth) thing you have to do is show that the function itself is continuous. Is it? Is it even defined at x=0?

To get around this issue, some people define the function in question to be

$$f(x) = \Bigl{\lbrace}\;\; \begin{matrix} 0 & \quad \text{x=0} \\ \exp(-1/x^2) & \quad\text{otherwise} \end{matrix}$$

8. Jul 3, 2009

### Skynt

Oh gosh I'm sorry I was going off your previous post ( exp(-x^2) )

$$f(x) = \Bigl{\lbrace}\;\; \begin{matrix} 0 & \quad \text{x=0} \\ \exp(-1/x^2) & \quad\text{otherwise} \end{matrix}$$

I'm guessing if its defined like that, it would be continuous, otherwise it would be undefined in some manner at x=0

But it sort of works out the same way it did with $$e^{-x^{2}}$$ if you derive $$e^{-1/x^{2}}$$

9. Jul 3, 2009

### D H

Staff Emeritus
My bad; what I wrote in post #4 was a typo. The function in question is that posted in your first post in this thread.

Don't just guess. Is the function as clarified in post #7 continuous? How do you go about showing that? How about the derivatives?

10. Jul 5, 2009

### Preno

Which holds iff they all exist, so you just have to check that they do.

11. Jul 5, 2009

### n!kofeyn

This is the correct definition that you need. To prove that f is $C^\infty$ (smooth), use induction. For f to be smooth, $f^{(k)}$ must exist and be continuous for all k=0,1,2,... To do induction, prove that for k=0, $f^{(0)}$, which is just f, is continuous. Then assume that $f^{(k)}$ exists and is continuous. Use this information to show that $f^{(k+1)}$ exists and is continuous. You will get a feel of this if you do take a few of the first derivatives as suggested.

A function is real-analytic at a point p if it is equal to its Taylor series about p in a neighborhood of p. To show that f is not analytic, show that around zero f is not equal to its Taylor series about zero. (You will find that $f^{(k)}(0) = 0$ for all k, so that the Taylor series of f(x) about zero is zero. But f(x) is definitely not zero in a neighborhood of zero, so it can't be equal to its Taylor series.)

Another way to show that it isn't analytic is to use the power series expansion of the exponential function, which is unique. That is
$$e^y = 1+y+\frac{y^2}{2!} + \frac{y^3}{3!} + \cdots$$
Now substitute y=-1/x^2. You will then get a series with negative terms in the exponents, which is not the Taylor series of the function.