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Mean value theorem for derivatives

Anonymous Mon 30 Sep 18:33:13 2019 No.11020304 View Reply Original Report

Quoted By: >>11020329 >>11020382 >>11020406 >>11020440 >>11022318 >>11023049

The mean value theorem is always defined as working for differentiable functions, but I think it should require a slightly weaker property: being able to draw a tangent line at every point. This allows in functions like cuberoot(x), which work fine with the MVT

Thoughts?

Anonymous

Anonymous Mon 30 Sep 2019 18:38:02 No.11020315 Report

Quoted By: >>11020328

How would you draw that tangent line?

Anonymous

Anonymous Mon 30 Sep 2019 18:41:03 No.11020328 Report

Quoted By: >>11020342

>>11020315
It's a vertical tangent line: x=0

Anonymous

Anonymous Mon 30 Sep 2019 18:41:12 No.11020329 Report

Quoted By: >>11020440

>>11020304
Please define "drawing a tangent line". How would you prove it's possible without invoking the concept of differentiability? Mate, the two ideas are so intertwined that vectors/derivations on a manifold can be defined as equivalence classes of curves (!) that are first-order identical.

Anonymous

Anonymous Mon 30 Sep 2019 18:44:58 No.11020342 Report

Quoted By: >>11020357

>>11020328
Which is the limit of the derivative for x -> 0

Anonymous

Anonymous Mon 30 Sep 2019 18:50:05 No.11020357 Report

Quoted By: >>11020360

>>11020342
Right, but it still can't be differentiated there.

Anonymous

Anonymous Mon 30 Sep 2019 18:52:30 No.11020360 Report

Quoted By:

>>11020357
And you can't find a mean there

Anonymous

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Anonymous Mon 30 Sep 2019 18:52:47 No.11020362 Report

Quoted By:

https://en.wikipedia.org/wiki/Constructive_analysis#The_intermediate_value_theorem

Anonymous

Anonymous Mon 30 Sep 2019 19:01:37 No.11020382 Report

Quoted By: >>11020400

>>11020304
that's already the way it's stated, differentiability isn't required at the endpoints.

an example that violates your sense is the function y=1/|x|
you cannot find a point that has a tangent line y=0, though you can find abundant secants that fulfill this criterion.

Anonymous

Anonymous Mon 30 Sep 2019 19:14:00 No.11020400 Report

Quoted By:

>>11020382
>y=0
*y=c for some constant

Anonymous

Anonymous Mon 30 Sep 2019 19:19:13 No.11020406 Report

Quoted By: >>11023204

>>11020304
and if you like, one reason the mvt applies to x^(1/3) is that every tangent value you take could be found on one everywhere differentiable half of the graph. so no test cases exist that require drawing a secant across the line x=0

Anonymous

Anonymous Mon 30 Sep 2019 19:30:01 No.11020440 Report

Quoted By: >>11020445 >>11022024

>>11020304
>>11020329
formal definition of "being able to draw a tangent line" is that the graph can be parametrized as a regular curve (this is possible for the cuberoot for example by (t^3,t)). and a mean value theorem for this kind of situation of course already exists, it's the cauchy mean value theorem.

Anonymous

Anonymous Mon 30 Sep 2019 19:31:30 No.11020445 Report

Quoted By: >>11020452

>>11020440
>regular curve
You did it again.

Anonymous

Anonymous Mon 30 Sep 2019 19:32:47 No.11020452 Report

Quoted By: >>11021866

>>11020445
what ?

Anonymous

Anonymous Tue 01 Oct 2019 03:24:17 No.11021866 Report

Quoted By:

>>11020452
He's a cultist, don't pay him any mind

Anonymous

Anonymous Tue 01 Oct 2019 04:35:57 No.11022024 Report

Quoted By: >>11022143

>>11020440
Regularity requires differentiability.

Anonymous

Anonymous Tue 01 Oct 2019 05:33:50 No.11022143 Report

Quoted By: >>11022154

>>11022024
It doesn't. OP is completely right that there are functions which satisfy the mean value property, have graphs with well defined tangent lines everywhere, and are not differentiable everywhere. I'm saying that this is however already covered by a much more general theorem.

Anonymous

Anonymous Tue 01 Oct 2019 05:43:43 No.11022154 Report

Quoted By: >>11022158

>>11022143
>It doesn't.
Of course it does, by definition.

Anonymous

Anonymous Tue 01 Oct 2019 05:46:01 No.11022158 Report

Quoted By: >>11022190

>>11022154
the parametrization of the GRAPH needs to be differentiable (obviously), but the FUNCTION itself need not be. that's what the thread is about.

Anonymous

Anonymous Tue 01 Oct 2019 05:59:06 No.11022190 Report

Quoted By: >>11022197

>>11022158
The one implies the other. It's a simple exercise for 1st year students.

Anonymous

Anonymous Tue 01 Oct 2019 06:02:23 No.11022197 Report

Quoted By: >>11022292

>>11022190
f(x) = cuberoot(x) is a function which is not differentiable at 0, but its graph y = cuberoot(x) can be parametrized as a regular curve by c(t) = (t^3,t).

Anonymous

Anonymous Tue 01 Oct 2019 06:55:52 No.11022292 Report

Quoted By: >>11022307

>>11022197
On what basis are you treating those as equivalent with respect to differentiability (or even continuity) at zero? It is trivial to construct a function for which your genuis method will break. How the fuck did you finish an analysis course?

Anonymous

Anonymous Tue 01 Oct 2019 07:14:05 No.11022307 Report

Quoted By:

>>11022292
Read the whole thread, or at least the original post. OP is saying that the Lagrange's mean value theorem doesn't need the function to be differentiable everywhere in the interior, it's only necessary that "we can draw tangent line everywhere". And he's right. His assertion about drawing tangent lines can be formalized by saying that the graph can be parametrized as a regular curve, then his idea follows from the Cauchy mean value theorem.

f(x) = cuberoot is a demonstration. It's a function which is not differentiable, but it satisfies the mean value property: we can parametrize the graph as (t^3,t) and apply the Cauchy mean value theorem.

Anonymous

Anonymous Tue 01 Oct 2019 07:34:11 No.11022318 Report

Quoted By: >>11022322

>>11020304
Isn't this Cauchy's mean value theorem? Pretty sure it is.

Anonymous

Anonymous Tue 01 Oct 2019 07:35:51 No.11022322 Report

Quoted By: >>11022996

>>11022318
of course it is. finally someone else who isn't a complete retard

Anonymous

Anonymous Tue 01 Oct 2019 15:06:05 No.11022996 Report

Quoted By: >>11023012

>>11022322
>However Cauchy's theorem does not claim the existence of such a tangent in all cases where (f(a), g(a)) and (f(b), g(b)) are distinct points, since it might be satisfied only for some value c with f(c) = g(c) = 0, in other words a value for which the mentioned curve is stationary; in such points no tangent to the curve is likely to be defined at all.

Anonymous

Anonymous Tue 01 Oct 2019 15:15:03 No.11023012 Report

Quoted By: >>11023021

>>11022996
we're assuming that there are no stationary points ("...require a slightly weaker property: being able to draw a tangent line at every point")

Anonymous

Anonymous Tue 01 Oct 2019 15:20:38 No.11023021 Report

Quoted By: >>11023030

>>11023012
>we're assuming
no, we're not. OP specifically mentioned cube root of x

Anonymous

Anonymous Tue 01 Oct 2019 15:25:18 No.11023030 Report

Quoted By: >>11023195 >>11023204

>>11023021
yes, cuberoot(x) is not differentiable at 0, BUT we are still able to draw a tangent line to its graph even at 0 (that's what OP assumes), because the graph y = cuberoot(x) as a subset of the plane admits a parametrization as a curve without stationary points. the mean value property then follows from cauchy.

Anonymous

Anonymous Tue 01 Oct 2019 15:38:05 No.11023049 Report

Quoted By:

>>11020304
There is a version for parametric equations.

Anonymous

Anonymous Tue 01 Oct 2019 17:02:43 No.11023195 Report

Quoted By: >>11023204 >>11023211

>>11023030
>the graph y = cuberoot(x) as a subset of the plane admits a parametrization as a curve without stationary points
Untrue. there is a stationary point at x=0. Show me a parametrization for which this is not true.

Anonymous

Anonymous Tue 01 Oct 2019 17:06:19 No.11023204 Report

Quoted By:

>>11023195
>>11023030
or do you mean the subset of the plane excluding x=0? because then we are talking past each other and you should read >>11020406

Anonymous

Anonymous Tue 01 Oct 2019 17:12:08 No.11023211 Report

Quoted By: >>11023234

>>11023195
(t^3,t)

Anonymous

Anonymous Tue 01 Oct 2019 17:24:11 No.11023234 Report

Quoted By:

>>11023211
gotcha, I'm a dummy, sorry

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