# How to find out if f and f' are continuous?

1. May 11, 2005

### twoflower

Hi,

Analyse continuity of f and f' on their domain, if
$$f(x) = \sum_{n=1}^{\infty} \arcsin x^{n}$$

No idea how to do that comes to my mind, I assume it will have something to do with uniform convergence but can't see any proper theorem, maybe More-Osgood, but it is rather confusing for me.

2. May 11, 2005

### saltydog

If $\{a_n\}$ is a set of positive numbers and:

$$|f_n(x)|\leq a_n \quad\text{for all n and}\quad$$

$$\sum_{n=1}^{\infty}a_n \quad\text{converges then: }\quad$$

$$\sum_{n=1}^{\infty}f_n(x) \quad\text{converges uniformly.}\quad$$

Now, what values of x would make these relations hold? I'm not sure about x values outside of this range.

3. May 11, 2005

### twoflower

Well, I can't see the connection between this and continuity of f and f' anyway...

4. May 11, 2005

### saltydog

Well here we go then: Right out of my Analysis text:

If $\{u_n\}$ is a sequence of continuous functions and if:

$$\sum_{n=1}^{\infty}u_n$$

converges uniformly to f (on some domain), then f is continuous on that domain.

5. May 11, 2005

### saltydog

Twoflower, I'm thinking the sum is:

$$\sum_{n=1}^{\infty}(Arcsin(x))^n$$

If that's the case, then we can compare it to a simple geometric series with Arcsin(x)<1. However, I'm thinking that's not your question but rather:

$$\sum_{n=1}^{\infty}Arcsin(x^n)$$

Although I suspect the same reasoning applies I'm not sure about it. Sorry.

6. May 11, 2005

### OlderDan

I may be missing the whole point, but I don't know the context of your problem. Is convergence really the issue here? There are values of x in the domain of $\arcsin x$ for which every term in the sum for f(x) will be > 1, and it will not converge. The domain of $\arcsin' x$ is the same as the domain of $\arcsin x$ except that the interval is open instead of closed. If you take the derivative term by term, I expect you can easily show convergence of f'(x), but all you need to show I think is continuity, and you can do that term by term.

7. May 11, 2005

### saltydog

Hello OlderDan. Suppose we restrict the domain to values of x in the range of:

$$(-\sqrt{Sin(1)}<x<\sqrt{Sin(1)})}$$

Thus $x^n$ will always be in this range and therefore, the Arcsin of the value will always be less than one. The function sequence quickly drops to zero. Does it converge uniformly when x>0 in the range? What about when x is negative in the range?. How to show that if so? Well, it's not my homework. Twoflower, don't you want to know?

8. May 11, 2005

### OlderDan

I don't see anything in the statement of the original problem to suggest restricting the domain. I see no problem with negative x. The even n terms are all even functions of x, and the odd n terms are odd functions of x. The terms certainly go quickly to zero if you stay away from |x| = 1. Maybe the point of the problem is that the high n terms abruptly change from 0 to pi/2 at x = 1, so maybe you need to identify the domain of both f and f' as the open interval (-1, 1). Like I said, I don't know the context, so maybe this is suppoed to be forrmulated as some kind of convergence problem. I don't think you have to approach it that way.

9. May 12, 2005

### twoflower

It's just standalone exercise from example analyse test I'm going to write. I thought it could have something to do with uniform convergence because we were just doing it and I know there are connections between continuity and uniform convergence. The right answer should be:

Functions f and f' are defined on (-1, 1) and are continuous there.

Anyway, f is defined on [-1, 1], isn't it? I know on this interval it isn't uniformly convergent, but that's all I know.

10. May 12, 2005

### saltydog

Hello guys. I'd like, if you don't mind, something definitive on the matter. You too Twoflower?

So we have:

$$f(x) = \sum_{n=1}^{\infty} \arcsin( x^{n})$$

Now, f(1) is not defined. It goes to infinity.

Now, f(-1) is not defined. It oscillates between 0 and $-\frac{\pi}{2}$ depending on whether n is even or odd.

So, this much seems clear: f(x) is defined on (-1,1).
(I was wrong about the square root stuf).

Now, is it uniformly convergent there? I suspect it is but I can't prove it. Can someone help?

11. May 12, 2005

### OlderDan

For |x|<1 it should not be hard to find a comparison test that will prove it. You might try looking at the ratio

$$\frac{ \arcsin( x^{n})}{1/x^n}$$

12. May 12, 2005

### saltydog

Thanks. I'll work with it. It's an interesting topic for me but I'm just not clear at all of how to approach convergence of function series on a "global level". That is, what determines when one series converges and the other doesn't. I would think there would exist some boundary that would separate them and that some algebraic structure of the function itself would be apparent to identify members of the each set. But that's good enough Dan. Don't want to distract you and others from the real homework help.

13. May 12, 2005

### OlderDan

There are a number of different tests for convergence. I've probably forgotten more of them than I remember, but if you just do a google on "Tests for convergence" you will find them.

14. May 12, 2005

### shmoe

Try comparing $$|\arcsin x^n|$$ with $$2|x|^n$$, this will handle uniform convergence by the Weierstrass test in post 2 (on what kind of interval?). Continuity of f follows from the theorem in post 4 (note that knowing the individual terms are contininuous does not imply that the limit will be a continuous function without uniform convergence).

You can show that the termwise derivatives converge uniformly to *something* in a similar way. Knowing that the arcsin(x^n) series converges will tell you that this *something* is in fact f' (standard kind of result in analysis texts, included is the result that f' actually exists) and continuity will follow (as the derivatives of arcsin(x^n) are continuuous).

15. May 12, 2005

### saltydog

That should do it Shmoe. Thanks. If you don't mind, I'd like to present a formal proof for my benefit as well as others who may be interested in this. My only concern is in the use of the Weierstrass test, the series to be compared to must be a positive sequence of numbers but below, I must specify that this sequence is such that |x|<b<1. I think that's Ok. Right?

First I'll state the Weierstrass M-Test:

If $u_n(x)$ is a set of functions on an interval M and if there exists a sequence $a_n$ of positive numbers such that:

$$\sum_{n=1}^{\infty}a_n$$

converges and:

$$|u_n(x)|\leq a_n\quad\text{for every}\quad x\in M$$

and every positive integer n. Then:

$$\sum_{n=1}^{\infty}u_n(x)\quad\text{converges uniformly on M}\quad$$

Now consider the first plot below which shows:

$$y(x)=arcsin(x)$$
$$g(x)=\frac{\pi}{2}x$$

The straight line is $g(x)=\frac{\pi}{2}x$.

This clearly shows:

$$|\arcsin(u)|\leq\frac{\pi}{2}|u| \quad\text{for}\quad u\in (-1,1)$$

Thus we can say:

$$|arcsin(x^n)|\leq \frac{\pi}{2}|x^n|\leq 2|b^n|\quad\text{for all}\quad x\in(-1,1)$$

with:

$$|x|<b<1$$

Now:

$$\sum_{n=1}^{\infty}2b^n$$

is a geometric series and converges.

Thus the function sequence:

$$\sum_{n=1}^{\infty}\arcsin(x^n)$$

Converges uniformly to a continuous function f(x) in the interval (-1,1).

QED (thanks to Shmoe)

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16. May 12, 2005

### OlderDan

Wouldn't that be "Try comparing $$|\arcsin x^n|$$ with $$\frac{2}{|x|^n}$$" or am I missing something?

I was holding out on the "2"

17. May 12, 2005

### OlderDan

DOH!! Yeah.. I was just driving home thinking what a dumb mistake that was hoping I could delete it before anyone saw it :uhh: