Smooth Curve

Aug 2010
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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.
 

Drexel28

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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 \(\displaystyle \displaystyle f:[0,1]\to\mathbb{R}\) given by contiuously extending \(\displaystyle g: (0,1]\to\mathbb{R}:x\mapsto x^2\sin\left(\frac{1}{x}\right)\). You can easily prove, it's a classic exercise, that \(\displaystyle f\) is differentiable on \(\displaystyle [0,1]\) but that \(\displaystyle f'\) is discontinuous at \(\displaystyle 0\). The problem with constructing such a function, the reason why all the obvious examples don't work, is that if \(\displaystyle j\) is a function which is the derivative of another function (i.e. \(\displaystyle j=k'\) for some \(\displaystyle k:[a,b]\to\mathbb{R}\)) then \(\displaystyle j\) preserves connectedness. Or, said in a less highfalutin tongue, \(\displaystyle j\) will have the intermediate value property, that if \(\displaystyle x\in(j(c),j(d))\) for some \(\displaystyle c,d\in[a,b]\) then \(\displaystyle x=j(e)\) for some \(\displaystyle e\in[a,b]\). This is roughly what Darboux's theorem says.
 
Aug 2010
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No, and it's easy to construct one. Namely, how about \(\displaystyle \displaystyle f:[0,1]\to\mathbb{R}\) given by contiuously extending \(\displaystyle g: (0,1]\to\mathbb{R}:x\mapsto x^2\sin\left(\frac{1}{x}\right)\). You can easily prove, it's a classic exercise, that \(\displaystyle f\) is differentiable on \(\displaystyle [0,1]\) but that \(\displaystyle f'\) is discontinuous at \(\displaystyle 0\). The problem with constructing such a function, the reason why all the obvious examples don't work, is that if \(\displaystyle j\) is a function which is the derivative of another function (i.e. \(\displaystyle j=k'\) for some \(\displaystyle k:[a,b]\to\mathbb{R}\)) then \(\displaystyle j\) preserves connectedness. Or, said in a less highfalutin tongue, \(\displaystyle j\) will have the intermediate value property, that if \(\displaystyle x\in(j(c),j(d))\) for some \(\displaystyle c,d\in[a,b]\) then \(\displaystyle x=j(e)\) for some \(\displaystyle e\in[a,b]\). 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 \(\displaystyle lim_ {x \to \0}\) f'(x) = \(\displaystyle lim_ {x \to \0}\) 1/cosx doesn't exist.

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

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

Jose27

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*f'(0) = \(\displaystyle lim_ {x \to \0}\) (f(x) - f(0))/x = \(\displaystyle lim_ {x \to \0}\) (x^2sin(1/x)-0)/x = 0
Here you're proving that \(\displaystyle f'(0)\) exists, the point is that \(\displaystyle \lim_{x \to 0} f'(x)\) 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.
 
Aug 2010
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Here you're proving that \(\displaystyle f'(0)\) exists, the point is that \(\displaystyle \lim_{x \to 0} f'(x)\) 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.
 

HallsofIvy

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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 \(\displaystyle lim_ {x \to \0}\) f'(x) = \(\displaystyle lim_ {x \to \0}\) 1/cosx doesn't exist.

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

If f(x) = (x^3)sin(1/x), defining f'(0) = 0 makes sense because \(\displaystyle lim_ {x \to \0}\) 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.
 
Aug 2010
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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.
 
Feb 2008
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The derivative of the given example function does exist. Remember that the derivative of a function is given by

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

Given the function

\(\displaystyle \left\{\begin{array}{ll}x^2\sin(1/x),&0<x\leq 1\\0,&x=0\end{array}\right\}\)

we calculate the derivative at \(\displaystyle x_0=0\) by computing

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

Compute this and see what you find. Then compute (using standard differentiation techniques) the \(\displaystyle f'(x)\) for \(\displaystyle 0<x\leq 1\). Note that \(\displaystyle f'(x_0)\) is continuous if and only if \(\displaystyle \lim_{x\to x_0}f'(x)\) exists and is equal to \(\displaystyle f'(x_0)\). So does the limit exist ?

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

Drexel28

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EDIT: However I would be interested in knowing if it's possible to find an example of a function \(\displaystyle f\) which is differentiable on some domain \(\displaystyle D\) such that left- and right-sided limits of \(\displaystyle f'\) exist on all of \(\displaystyle D\), but such that \(\displaystyle f'\) is discontinuous in the particular sense of having the left- and right-sided limits unequal at some \(\displaystyle x_0\in D\). Perhaps that is impossible.
This was the effect of my first post. This can't happen really since derivatives have the intermediate value property.
 
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Aug 2010
961
101
The derivative of the given example function does exist. Remember that the derivative of a function is given by

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

Given the function

\(\displaystyle \left\{\begin{array}{ll}x^2\sin(1/x),&0<x\leq 1\\0,&x=0\end{array}\right\}\)

we calculate the derivative at \(\displaystyle x_0=0\) by computing

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

Compute this and see what you find. Then compute (using standard differentiation techniques) the \(\displaystyle f'(x)\) for \(\displaystyle 0<x\leq 1\). Note that \(\displaystyle f'(x_0)\) is continuous if and only if \(\displaystyle \lim_{x\to x_0}f'(x)\) exists and is equal to \(\displaystyle f'(x_0)\). So does the limit exist ?

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