1. ## Smooth Curve

If x=x(t) has a derivative at every point of [a,b], isn't that derivative automatically continuous? If not, please give an example of a derivative which exists at every point but which is not continuous. All I can think of is a spike in the curve which has different (one sided) derivatives on either side, but the derivative doesn't exist at the spike.

This question has its origin in the definition of a smooth curve, which requires that the derivatives of x=x(t), y=y(t), and z=z(t) exist AND be continuous.

2. ## Re: Smooth Curve

Originally Posted by Hartlw
If x=x(t) has a derivative at every point of [a,b], isn't that derivative automatically continuous? If not, please give an example of a derivative which exists at every point but which is not continuous. All I can think of is a spike in the curve which has different (one sided) derivatives on either side, but the derivative doesn't exist at the spike.

This question has its origin in the definition of a smooth curve, which requires that the derivatives of x=x(t), y=y(t), and z=z(t) exist AND be continuous.
No, and it's easy to construct one. Namely, how about $\displaystyle f:[0,1]\to\mathbb{R}$ given by contiuously extending $g: (0,1]\to\mathbb{R}:x\mapsto x^2\sin\left(\frac{1}{x}\right)$. You can easily prove, it's a classic exercise, that $f$ is differentiable on $[0,1]$ but that $f'$ is discontinuous at $0$. The problem with constructing such a function, the reason why all the obvious examples don't work, is that if $j$ is a function which is the derivative of another function (i.e. $j=k'$ for some $k:[a,b]\to\mathbb{R}$) then $j$ preserves connectedness. Or, said in a less highfalutin tongue, $j$ will have the intermediate value property, that if $x\in(j(c),j(d))$ for some $c,d\in[a,b]$ then $x=j(e)$ for some $e\in[a,b]$. This is roughly what Darboux's theorem says.

3. ## Re: Smooth Curve

Originally Posted by Drexel28
No, and it's easy to construct one. Namely, how about $\displaystyle f:[0,1]\to\mathbb{R}$ given by contiuously extending $g: (0,1]\to\mathbb{R}:x\mapsto x^2\sin\left(\frac{1}{x}\right)$. You can easily prove, it's a classic exercise, that $f$ is differentiable on $[0,1]$ but that $f'$ is discontinuous at $0$. The problem with constructing such a function, the reason why all the obvious examples don't work, is that if $j$ is a function which is the derivative of another function (i.e. $j=k'$ for some $k:[a,b]\to\mathbb{R}$) then $j$ preserves connectedness. Or, said in a less highfalutin tongue, $j$ will have the intermediate value property, that if $x\in(j(c),j(d))$ for some $c,d\in[a,b]$ then $x=j(e)$ for some $e\in[a,b]$. This is roughly what Darboux's theorem says.
Thanks for interesting example. However:

Let f(x) = (x^2)sin(1/x). f'(0) doesn't exist because 1/0 doesn't exist. Defining f'(0) = 0* makes no sense because $lim_ {x \to \0}$ f'(x) = $lim_ {x \to \0}$ 1/cosx doesn't exist.

*f'(0) = $lim_ {x \to \0}$ (f(x) - f(0))/x = $lim_ {x \to \0}$ (x^2sin(1/x)-0)/x = 0

If f(x) = (x^3)sin(1/x), defining f'(0) = 0 makes sense because $lim_ {x \to \0}$ f'(x) = 0. In this case existence of f'(0) implies continuity at 0.

4. ## Re: Smooth Curve

Originally Posted by Hartlw

*f'(0) = $lim_ {x \to \0}$ (f(x) - f(0))/x = $lim_ {x \to \0}$ (x^2sin(1/x)-0)/x = 0
Here you're proving that $f'(0)$ exists, the point is that $\lim_{x \to 0} f'(x)$ doesn't exist, meaning the derivative isn't continous (defining the derivative at 0 as 0 makes sense because the limit defining it exists) which is what you want.

In fact, a more elaborate example gives a function everywhere differentiable with the deerivative having a set of discontinuities of positive measure, see here.

5. ## Re: Smooth Curve

Originally Posted by Jose27
Here you're proving that $f'(0)$ exists, the point is that $\lim_{x \to 0} f'(x)$ doesn't exist, meaning the derivative isn't continous (defining the derivative at 0 as 0 makes sense because the limit defining it exists) which is what you want.

In fact, a more elaborate example gives a function everywhere differentiable with the deerivative having a set of discontinuities of positive measure, see here.
f'(0) doesn't exist because 1/0 (f(0)) doesn't exist. You can define it to be zero, but there's no point if lim f'(x) as x-> zero doesn't exist. You're simply defining a discontinuity which is pointless except as a contrived example.

I'll pass on the more elaborate example. Not my cup of tea. I'm really just trying to get a working feeling for what a curve looks like with derivatives existing everywhere on a continuous curve but not continuous. I don't have the ability to remember abstract derivations.

6. ## Re: Smooth Curve

Originally Posted by Hartlw
Thanks for interesting example. However:

Let f(x) = (x^2)sin(1/x). f'(0) doesn't exist because 1/0 doesn't exist. Defining f'(0) = 0* makes no sense because $lim_ {x \to \0}$ f'(x) = $lim_ {x \to \0}$ 1/cosx doesn't exist.

*f'(0) = $lim_ {x \to \0}$ (f(x) - f(0))/x = $lim_ {x \to \0}$ (x^2sin(1/x)-0)/x = 0

If f(x) = (x^3)sin(1/x), defining f'(0) = 0 makes sense because $lim_ {x \to \0}$ f'(x) = 0. In this case existence of f'(0) implies continuity at 0.
What is your point? You asked if having a derivative implied that the derivative must be continuous. Drexel28 gave an example showing that the derivative does NOT have to be continuous. The fact that there exist derivatives that are continuous is not relevant to your question.

7. ## Re: Smooth Curve

Originally Posted by HallsofIvy
What is your point? You asked if having a derivative implied that the derivative must be continuous. Drexel28 gave an example showing that the derivative does NOT have to be continuous. The fact that there exist derivatives that are continuous is not relevant to your question.
My point was that I felt the example was contrived in a manner that didn't make sense to me.

For example, sinx/x is undefined at x=0. But since lim sinx/x -> 0 exists, it makes sense to define it as 0. If limit didn't exist, defining it as 0 would be meaningless.

Suppose f'(x) = x^2 except at x=1 where I define it to be 10. Then clearly existence of derivative everywhere does not imply continuity of f'(x). But the example is contrived in a manner that doesn't make sense. I'm not saying existence of f'(x) implies continuity, just trying to get a feel for why it isn't true. f(x) = (x^3) sin(1/x) made sense to me because definition of f(0) and limit f'(x) were the same.

Perhaps it would help to elaborate on my original post. I didn't understand the requirement of a smooth curve that f'(x) exist everywhere AND be continuous. I have made some progress. I believe it relates to the definition of line integral. If you set up partitions to define a line integral by summation and use the derivative and intermediate value theorem along the way, then intuitiveley I see that in order for the limit to exist independent of definition of partition just as long as all partitions go to zero, f'(t) has to be continuous.

But out of curiousity I wondered what a continuous (part of the requirement) curve with existing derivatives that weren't continuous would look like. It's easy enough to imagine what a continuous curve looks like though I vaqueley recollect some non-intuitive examples.

8. ## Re: Smooth Curve

The derivative of the given example function does exist. Remember that the derivative of a function is given by

$f'(x_0)=\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$.

Given the function

$\left\{\begin{array}{ll}x^2\sin(1/x),&0

we calculate the derivative at $x_0=0$ by computing

$f'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0}\frac{f(h)-f(0)}{h}$.

Compute this and see what you find. Then compute (using standard differentiation techniques) the $f'(x)$ for $0. Note that $f'(x_0)$ is continuous if and only if $\lim_{x\to x_0}f'(x)$ exists and is equal to $f'(x_0)$. So does the limit exist ?

EDIT: However I would be interested in knowing if it's possible to find an example of a function $f$ which is differentiable on some domain $D$ such that left- and right-sided limits of $f'$ exist on all of $D$, but such that $f'$ is discontinuous in the particular sense of having the left- and right-sided limits unequal at some $x_0\in D$. Perhaps that is impossible.

9. ## Re: Smooth Curve

Originally Posted by hatsoff
EDIT: However I would be interested in knowing if it's possible to find an example of a function $f$ which is differentiable on some domain $D$ such that left- and right-sided limits of $f'$ exist on all of $D$, but such that $f'$ is discontinuous in the particular sense of having the left- and right-sided limits unequal at some $x_0\in D$. Perhaps that is impossible.
This was the effect of my first post. This can't happen really since derivatives have the intermediate value property.

10. ## Re: Smooth Curve

Originally Posted by hatsoff
The derivative of the given example function does exist. Remember that the derivative of a function is given by

$f'(x_0)=\lim_{h\to 0}\frac{f(x_0+h)-f(x_0)}{h}$.

Given the function

$\left\{\begin{array}{ll}x^2\sin(1/x),&0

we calculate the derivative at $x_0=0$ by computing

$f'(0)=\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0}\frac{f(h)-f(0)}{h}$.

Compute this and see what you find. Then compute (using standard differentiation techniques) the $f'(x)$ for $0. Note that $f'(x_0)$ is continuous if and only if $\lim_{x\to x_0}f'(x)$ exists and is equal to $f'(x_0)$. So does the limit exist ?

EDIT: However I would be interested in knowing if it's possible to find an example of a function $f$ which is differentiable on some domain $D$ such that left- and right-sided limits of $f'$ exist on all of $D$, but such that $f'$ is discontinuous in the particular sense of having the left- and right-sided limits unequal at some $x_0\in D$. Perhaps that is impossible.
I did all that and explained why f'(0) didn't exist in posts 3) and 4). Can't address your edit. Once again, f'(0) doesn't exist because 1/0 doesn't exist. But you can find f'(0) using definition of derivative and the limiting process and then DEFINE this as the derivative at 0. It doesn't make sense to do so because limit f'(x) as x->0 doesn't exist. In my opinion it is a contrived example which doesn't make sense. f(x) = (x^3)sin(1/x) does make sense because f'(0) obtained by a limiting process using definition of derivative (although f'(0) doesn't exist because 1/0 doesn't exist) can be defined consistent with lim f'(x) as x ->0.

Still looking for a non-contrived example of a continuous function on a closed interval which has a derivative defined at each point but which (the derivative) isn't continuous. Kaplan mentions that a curve can have jump discontinuities in the derivative but doesn't elaborate. Yes, yes, I know- Drexel28's example in first post. Oh well, i beat my objection to death and since nobody is buying it or maybe doesn't understand it, I'll let it ride without a feel for what a curve with discontinuous derivatives looks like.

Thanks for the example Drexel28.

PS I missed edit on a previous post so let me correct a typo. sinx/x, x=0 isn't defined. But BECAUSE lim sinx/x as x->0 =1 (not 0), it makes sense to define it as 1. So-so analogy.

What I am saying may make more sense if you read my last post in "show that f'(c) exists" nearby.

11. ## Re: Smooth Curve

Originally Posted by Hartlw
I did all that and explained why f'(0) didn't exist in posts 3) and 4). Can't address your edit. Once again, f'(0) doesn't exist because 1/0 doesn't exist. But you can find f'(0) using definition of derivative and the limiting process and then DEFINE this as the derivative at 0. It doesn't make sense to do so because limit f'(x) as x->0 doesn't exist. In my opinion it is a contrived example which doesn't make sense. f(x) = (x^3)sin(1/x) does make sense because f'(0) obtained by a limiting process using definition of derivative (although f'(0) doesn't exist because 1/0 doesn't exist) can be defined consistent with lim f'(x) as x ->0.

Still looking for a non-contrived example of a continuous function on a closed interval which has a derivative defined at each point but which (the derivative) isn't continuous. Kaplan mentions that a curve can have jump discontinuities in the derivative but doesn't elaborate. Yes, yes, I know- Drexel28's example in first post. Oh well, i beat my objection to death and since nobody is buying it or maybe doesn't understand it, I'll let it ride without a feel for what a curve with discontinuous derivatives looks like.

Thanks for the example Drexel28.

PS I missed edit on a previous post so let me correct a typo. sinx/x, x=0 isn't defined. But BECAUSE lim sinx/x as x->0 =1 (not 0), it makes sense to define it as 1. So-so analogy.

What I am saying may make more sense if you read my last post in "show that f'(c) exists" nearby.
I'll try to say one more thing, and if you don't buy it, that's fine. The property I stated about what functions which are the derivatives of other functions satisfy (i.e. the intermediate value property) forces any example to be 'contrived'. Indeed, morally one would like to create a function which just has a jump discontinuity when one differentiates--this can't happen. The only real way you are going to create such a function is to create one that DOESN'T HAVE A LIMIT. It's just a fact of life. Now, if you were to say to me "Yo, dude, those kind of functions are stupid, I wouldn't consider them anyways." then I'd go "Ok, fine, that makes sense to me. So, just don't include such examples in your theory." But, you asked why they make this caveat and we gave you an example, as stupid as it may be.

12. ## Re: Smooth Curve

Originally Posted by Drexel28
This was the effect of my first post. This can't happen really since derivatives have the intermediate value property.
Ah, cool, thanks !

13. ## Re: Smooth Curve

let's try to compute the derivative of $f(x) = x^2sin(1/x)$, at x = 0 (remember we are defining f(0) = 0, this is ok, since this defintion makes f continuous), since its existence seems to be in doubt:

$\lim_{h \to 0} \frac{f(h+0) - f(0)}{h}$

$= \lim_{h \to 0} \frac{f(h)}{h} = \lim_{h \to 0}\frac{h^2sin(1/h)}{h}$

$= \lim_{h \to 0} hsin(1/h)$.

now, here, we have a problem, because we can't evaluate sin(1/0).

but...let's try doing the left-hand and right-hand limits separately:

for small enough h > 0, we can ensure that sin(1/h) > 0, thus for these same h,

hsin(1/h) > 0, and we know that sin(1/h) ≤ 1, so by the squeeze theorem:

$0 = \lim_{h \to 0^+} 0h \leq \lim_{h \to 0^+}hsin(1/h) \leq \lim_{h \to 0^+}1h = 0$ hence

$\lim_{h \to 0^+}hsin(1/h) = 0$.

similarly, for h < 0 and |h| sufficiently small,

-1 ≤ sin(1/h) < 0, so

$0 = \lim_{h \to 0^-} -1h \leq \lim_{h \to 0^-}hsin(1/h) \leq \lim_{h \to 0^-}0h = 0$ so we conclude that:

$\lim_{h \to 0^-}hsin(1/h) = 0$.

since the left-hand and right-hand limits are equal,

$\lim_{h \to 0}hsin(1/h) = 0$, so f'(0) exists.