Sasho

Continuous functions and non-continuous derivatives...

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I am looking for case where there is continuous function, but its derivative is not-continuous...

Could you know any such example....?

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3 hours ago, Sasho said:

I am looking for case where there is continuous function, but its derivative is not-continuous...

Could you know any such example....?

A function with a corner or cusp such as y = mod(x) (The modulus function)

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5 hours ago, studiot said:

A function with a corner or cusp such as y = mod(x) (The modulus function)

Yes... I never thought for that.... THANK YOU!!!! :-)

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8 hours ago, studiot said:

A function with a corner or cusp such as y = mod(x) (The modulus function)

That's not the answer to the question the OP asked.

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Why isn't it?  The OP asked for an example of a function that is continuous but its derivative is not.  y= |x| is continuous for all x and its derivative, y'(x)= 1 for x> 0, y'(x)= -1 for x< 0, y'(0) not defined, is not continuous at x= 0.

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Posted (edited)
22 hours ago, HallsofIvy said:

Why isn't it?  The OP asked for an example of a function that is continuous but its derivative is not.  y= |x| is continuous for all x and its derivative, y'(x)= 1 for x> 0, y'(x)= -1 for x< 0, y'(0) not defined, is not continuous at x= 0.

Too strong. OP asked for a continuous function that has a derivative that is not continuous. |x| does not satisfy OP's requirement, since it does not have a derivative.

Edited by wtf

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I see your point- but that depends on exactly what you mean by "has a derivative".  I would say that f(x)= |x| certainly has a derivative.  That derivative does not happen to be defined at x= 0.  The original post did not say anything about the derivative being defined "for all x".

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I think the derived function of the modulus function has a discontinuity at x=0, where the slope changes abruptly from -1 to +1.

This you will find in all the standard textbooks as it is the bog standard example, and also if you look it up on the net.

What more is there to be said?

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On 3/29/2019 at 4:58 PM, Sasho said:

I am looking for case where there is continuous function, but its derivative is not-continuous...

Could you know any such example....?

1 hour ago, studiot said:

What more is there to be said?

An integral is just the reverse of a derivative.

Reference H.J.Keisler "Elementary Calculus an Infinitessimal Approach" (992 pages) PDF link: https://www.math.wisc.edu/~keisler/keislercalc-03-07-17.pdf

At a conceptual structural level improper integrals in physics can be piecewise continuous integrals, with limits from +infinity to -infinity, that converge. Refer H.J. Keisler,  p367,  Definition to p369,  examples 7, 8,  and 9. If they are continuous and don't converge then they are indefinite integrals which are entirely different. Refer H.J. Keisler,  p370, example 10, diagram 6.7.10 "It is tempting to argue that the positive area to the right of the origin and the negative area to the left exactly cancel each other out so that the improper integral is zero. But this leads to a paradox... So we do not give the integral ... the value 0, instead leave it undefined."

That doesn't mean that indefinite integrals don't play a part in our physics as an indefinite integral that cycles between +infinity and -infinity at its limits, as a sub function of a higher level function, is a valid proper use of indefinite integrals as definite integrals by change of variables. Refer H.J. Keisler, p224-5, Definition and example 8, diagram 4.4.6 second equation with u and substitute infinite limits. "We do not know how to find the indefinite integrals in this example. Nevertheless the answer is 0 because on changing variables both limits of integration become the same."

 

 

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9 hours ago, LaurieAG said:

An integral is just the reverse of a derivative.

Reference H.J.Keisler "Elementary Calculus an Infinitessimal Approach" (992 pages) PDF link: https://www.math.wisc.edu/~keisler/keislercalc-03-07-17.pdf

At a conceptual structural level improper integrals in physics can be piecewise continuous integrals, with limits from +infinity to -infinity, that converge. Refer H.J. Keisler,  p367,  Definition to p369,  examples 7, 8,  and 9. If they are continuous and don't converge then they are indefinite integrals which are entirely different. Refer H.J. Keisler,  p370, example 10, diagram 6.7.10 "It is tempting to argue that the positive area to the right of the origin and the negative area to the left exactly cancel each other out so that the improper integral is zero. But this leads to a paradox... So we do not give the integral ... the value 0, instead leave it undefined."

That doesn't mean that indefinite integrals don't play a part in our physics as an indefinite integral that cycles between +infinity and -infinity at its limits, as a sub function of a higher level function, is a valid proper use of indefinite integrals as definite integrals by change of variables. Refer H.J. Keisler, p224-5, Definition and example 8, diagram 4.4.6 second equation with u and substitute infinite limits. "We do not know how to find the indefinite integrals in this example. Nevertheless the answer is 0 because on changing variables both limits of integration become the same."

 

 

Thank you for the link to the pdf.
992 pages.

It will take some time to read!

However surely integrals, considered as antiderivatives or otherwise, are off topic in a thread devoted to the derived function?

In particular how is this derived function not discontinuous?

modf1.jpg.77b2759d4bcbdb25a0b3e085836f3639.jpg

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Posted (edited)
18 hours ago, studiot said:

Thank you for the link to the pdf.
992 pages.

It will take some time to read!

However surely integrals, considered as antiderivatives or otherwise, are off topic in a thread devoted to the derived function?

I identified the relevant pages (the actual PDF pages not the book pages) in my original post and hi lighted the relevant sections on those pages so you don't have to read the entire book.

A derivative in its most simplistic form is the application of n*x^(n-1) for all x in quadratic equations in the form of a*x^2 + b*x^1 + c*x^0 = 0 i.e. ax^2 + bx + c = 0.

The first derivative of this equation is 2*a*x^1 + 1*b*x^0 + 0*c*x^-1 = 2ax +b.

The first integral of 2ax +b is ax^2 + bx + c.

 

Edited by LaurieAG
re arrangement of sentences

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6 hours ago, LaurieAG said:

I identified the relevant pages (the actual PDF pages not the book pages) in my original post and hi lighted the relevant sections on those pages so you don't have to read the entire book.

A derivative in its most simplistic form is the application of n*x^(n-1) for all x in quadratic equations in the form of a*x^2 + b*x^1 + c*x^0 = 0 i.e. ax^2 + bx + c = 0.

The first derivative of this equation is 2*a*x^1 + 1*b*x^0 + 0*c*x^-1 = 2ax +b.

The first integral of 2ax +b is ax^2 + bx + c.

 

I fail to see any relevaence whatsoever to my example or to the topic in general.

In the first place the OP is learning standard analysis.

Keiseler in the section you refer to invokes non standard analysis and the hyper-reals.

In the second place the examples of improper integrals such as the integral are improper because the slope becomes infinite (eg the slope of y = 1/x near the y axis) are different from my example, which is bounded in both f(x) and f'(x) for all finite x.
The issue for my example is that

For f(x) the limits from the left always = the limits from the right

But for f'(x) the limits from the left = minus the linits from the right at the point x = 0

These are the standard analysi condition for continuity and non continuity.

 

So once again

Why is my example faulty?

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Posted (edited)
10 hours ago, studiot said:

 

Why is my example faulty?

If you think |x| has a derivative you failed freshman calculus. It's the classic elementary example of a continuous function that does NOT have a derivative. Because to "have a derivative" means to have a derivative at every point of its domain. 

Since |x| does not have a derivative, it certainly does not have a derivative that's discontinuous. That's because if I don't have a purple elephant, then I certainly don't have a purple elephant with wings.

Of course |x| does have a derivative that's undefined at one point of the domain of |x|. That MIGHT be what the OP meant, but it's NOT what they asked.

Edited by wtf

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Posted (edited)

ps ... found this good thread. https://math.stackexchange.com/questions/292275/discontinuous-derivative

One of the commenters notes that |x| does NOT answer the question of a function with discontinuous derivative, for exactly the reason I gave. 

The example given of such a beast is:

f(x) = x^2 sin(1/x) for x nonzero; and

f(0) = 0.

The details are in the link.

Edited by wtf

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18 hours ago, studiot said:

I fail to see any relevaence whatsoever to my example or to the topic in general.

A function is an equation and a derivative or an integral are both manipulations of the function.

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9 hours ago, LaurieAG said:

A function is an equation and a derivative or an integral are both manipulations of the function.

A function is more than an equation.

Functions have three parts, one of which is sometimes called the rule - I do not know of a posh mathematical term for it.

Equations can be rules, but not all equations.

for instance the equation

2 = 1 + 1

is not part of a function.

Equally not all functions have rules that are defined by an equation.
Sometimes this is actually impossible, as when the rule is defined by a process.

 

16 hours ago, wtf said:

If you think |x| has a derivative you failed freshman calculus. It's the classic elementary example of a continuous function that does NOT have a derivative. Because to "have a derivative" means to have a derivative at every point of its domain. 

Since |x| does not have a derivative, it certainly does not have a derivative that's discontinuous. That's because if I don't have a purple elephant, then I certainly don't have a purple elephant with wings.

Of course |x| does have a derivative that's undefined at one point of the domain of |x|. That MIGHT be what the OP meant, but it's NOT what they asked.

Since you are American, I will refer you to that well known american guru of these mateers - Tom Apostol.

Most writers concentrate on the definition of the continuous.
Few include a definition of the discontinuous.

Tom, bless him, includes several.

Specifically for my example he refers to a jump discontinuity.

 

This is mainstream not the guff on stackexchange you linked to which was not even about my example but the original question asked about

a very discontinuous function. My example has only one discontinuity and I have already outlined the proof (which incidentally I see looking it up is the way Tom presents it).

If you work it out you will find that (in Tom's notation) f(x) - A = 0 and therefore is always less than epsilon for any epsilon greater than zero.

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Posted (edited)
On 4/15/2019 at 7:27 AM, studiot said:

 

Most writers concentrate on the definition of the continuous.
Few include a definition of the discontinuous.]

You look silly doubling down on your error. You're wrong. I posted the correct example of a function with discontinuous derivative. 

Edited by wtf

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9 hours ago, wtf said:

You look silly doubling down on your error. You're wrong. I posted the correct example of a function with discontinuous derivative. 

 

What would be good would be if you could discontinue the ad hominem content and stick to mathematics.

 

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Posted (edited)
3 hours ago, studiot said:

 

What would be good would be if you could discontinue the ad hominem content and stick to mathematics.

 

Ok. You look perfectly sensible doubling down on your error. You look like a stud. A God. A man among boys. 

Happy now? Studiot you are the one not sticking to the math. I posted the correct answer. You're still trying to defend your incorrect point. What math should I stick to? I first noted that the "obvious" answer, which more than one person proposed, is wrong. I then looked up the right answer, which I was not able to work out for myself. I'm not claiming any special virtue here. So what more do you want? You're unhappy that I said you look silly? Ok you look terribly clever. 

I would also argue that "You look silly" is not an ad hominem. If I say you ARE silly that would be an ad hominem. But you're not always silly. Sometimes you display an impressive grasp of engineering math. Today, you are insisting you're right when it's perfectly clear that you're wrong. But if you want me to apologize for calling that behavior silly, I apologize. 

Edited by wtf

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51 minutes ago, wtf said:

. I posted the correct answer.

Is there a claim that there is one and only one correct answer ?

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8 minutes ago, studiot said:

Is there a claim that there is one and only one correct answer ?

No, there are many such functions. |x| is not one of them.

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