Problem with the concept of differentiation

[WWGD]

You mean like v=32t. Here the v at 5sec is 160ft/s and it will be approximately 160 around 5sec with some error. So we will be able to calculate approximate actual changes between two nearby points. we say the velocity of body from say 5sec to 5.000001 sec is approximately 160ft/sec. In this way we can calculate the approximate change of position in between them as 0.00016 ft.
We can take any value of t and approximate the actual change between t and some near point assuming nothing unexpected happens.

because the velocity keeps on changing. Right? That’s why “approximate change” and not exact change.

Let’s say 160.016 ft/s is good enough. I got that speed by taking my other point as 5.001sec. So now this approximation works for many points near 5.001 hoping there is no any sudden major changes in speed. Points like 5.01sec,5.03sec,etc
In the same way if we take the limit we get precisely 160 at 5. This is very exact than 160.016 ft/s. Now this works for points near 5. Right!

Just a small caveat that velocity ( nor the derivative in general) is not necessarily changing. Speed may be constant. Please ignore if you're aware of this.

 

[bob012345]

Summary:: What is the change actually at some time $t_0$?

We define differentiation as the limit of $\frac{f(x+h)-f(x)}h$ as $h->0$. We find the instantaneous velocity at some time $t_0$ using differentiation and call it change at $t_0$. We show tangent on the graph of the function at $t_0$. But after taking h or time interval as zero to find the instantaneous change there is no two points left for change to occur. Change at $t_0$ is meaningless in the same way as the slope at a point. How can there be a change at say 5 sec? We have to take two points to talk about change in between them.

I want to reiterate what was said in post #13. The definition is not about change, it is about the rate of change. In the limit, the rate of change at a point is exact.

Let $v(t)= 32t$, then $\frac {dv}{dt} = 32$.

We can get $v(5.001)$ two ways 1) exactly compute is $v(5.001)=32(5.001) = 160.032$

or 2) estimate it with the derivative;

$v_{approx}(5.001) ≈ v(5) + \frac{dv}{dt} \Delta{t} = 160 + 32*0.001 = 160.032$

It is exact only because we are using a linear function with a constant slope.

 

[diegogarcia]

Summary:: What is the change actually at some time $t_0$?

We define differentiation as the limit of $\frac{f(x+h)-f(x)}h$ as $h->0$. ... Change at $t_0$ is meaningless in the same way as the slope at a point. How can there be a change at say 5 sec? We have to take two points to talk about change in between them.

We do. As h gets closer and closer to zero the curve more and more resembles a linear curve, i.e. a straight line with a definite slope.

Think about magnifying the curve at x0 billions, trillions, septillions of times. The magnified portion will resemble a straight line with a definite slope. It is this linear slope that defines the derivative at x0.

The differential, dx, which used to be defined as an infinitesimal distance, is now defined as that distance over which the curve becomes linear (or a surface becomes a flat plane).

So there are essentially two points, x0 and x0+h, that define a slope which is considered the derivative at x0.

There are points at which a derivative does not exist. See, for example, the absolute value function:

https://en.wikipedia.org/wiki/File:Absolute_value.svg

The AV function has no derivative at x=0 because there is no interval, however small, around x=0 that would closely approximate a linear curve or straight line. (We could however consider left- and right-sided derivatives.)

So we are, in effect, considering two points, x0 and x0+h, in the definition of the derivative.

There is also a difference between the mathematical and the physical interpretation of the differential. In math, the differential is truly vanishingly small. It is a limit. But in physics, the differential is something that is small enough to be experimentally undetectable and how small depends on the particular experiment.

Again, the key idea is that we can "magnify" any curve (or surface) so that it becomes essentially linear (planar) and thus will have a definite slope (or tangent plane). This slope is the derivative.

 

[Mark44]

Again, the key idea is that we can "magnify" any curve (or surface) so that it becomes essentially linear (planar) and thus will have a definite slope (or tangent plane). This slope is the derivative.

Not so for just any old curve. A counterexample is the Weierstrass function, a particularly pathological function the is continuous everywhere, but differentiable nowhere (see https://en.wikipedia.org/wiki/Weierstrass_function).

 

[benorin]

I find the picture usually helps.

 

[diegogarcia]

Not so for just any old curve. A counterexample is the Weierstrass function, a particularly pathological function the is continuous everywhere, but differentiable nowhere (see https://en.wikipedia.org/wiki/Weierstrass_function).

So what? That's not the issue nor does it constitute a counterexample to anything I have stated.

The OP asked what I consider to be a legimate question. The status quo in current mathematics is to obscure all hope of intuition with endless "epsilon delta" proofs. Mathematics need not be so visually sterile.

Differentiation implies continuity but continuity does not imply differentiation. The Weirstrauss function demonstrates this but this is also a completely different issue.

The OP wants to know how a "slope," or derivative, can be defined using only a single point, which is the OP's
interpretation. I simply illustrated the fact that two points, in effect, are actually used to define the derivative.

 

[vela]

That's not the issue nor does it constitute a counterexample to anything I have stated.

You said you could magnify any curve and it will eventually look linear. @Mark44 gave you an example of a curve where you can't do that. How is that not a counterexample?

You don't even need that weird of a function. No matter how much you magnify f(x)=|x| around x=0, it will never look linear.

 

[diegogarcia]

You said you could magnify any curve and it will eventually look linear. @Mark44 gave you an example of a curve where you can't do that. How is that not a counterexample?

Since the topic is differentiation it should have been clear that I meant any differentiable curve. In fact I even provided my own counterexample with the absolute value function which is not differentiable at a single point. The Weirstrauss function is not differentiable at any point.

Furthermore, in this case, the Weirstrauss function is a bad pedagogical example because it does not combine both differentiablity and non-differentiability in the same curve, as does the absolute value and other functions.

 

[sysprog]

Since the topic is differentiation it should have been clear that I meant any differentiable curve. In fact I even provided my own counterexample with the absolute value function which is not differentiable at a single point. The Weirstrauss function is not differentiable at any point.

By "is not differentiable at a single point" you presumably meant "is non-differentiable at only a single point".

Furthermore, in this case, the Weirstrauss function is a bad pedagogical example because it does not combine both differentiablity and non-differentiability in the same curve, as does the absolute value and other functions.

As @Mark44 observed, it's continuous everywhere, and differentiable nowhere. Whether that makes it a good or bad example depends on what it's being presented as an example of. In this instance, the context was as a counterexample to your statement that you only later qualified as meant to be referential only to differentiable curves. Consequently, I disagree with the contention that the fact that the function "does not combine differentiability and non-differentiability in the same curve" makes it "a bad pedagogical example".

 

[benorin]

I mean this in the nicest, most non-condescending way possible: you guys please remember the purpose of this is thread to help the OP, is there some pressing need for a counter example still? I can breakout with my ample Counterexamples in Analysis text by Gelbaum & Olmstead if need be, it's got a cornucopia non-differentiable functions for every occasion, just tell me the specific need?

 

[Mark44]

is there some pressing need for a counter example still?

Probably not any additional ones, but the reason for them was to counter an incorrect statement earlier on, that

we can "magnify" any curve (or surface) so that it becomes essentially linear (planar) and thus will have a definite slope (or tangent plane). This slope is the derivative.

(Emphasis added by me.)

 

[sysprog]

@benorin, your reference to Counterexamples in Analysis by Gelbaum & Olmstead is appreciated $\dots$