Why does the Limit Process give an Exact Slope?

[WWGD]

It is not exact. The _actual_change in the values of $x^2$ between $x=1$ and $x=1.01$ is$\Delta f=(1.01)^2-1^2 =0.0201$. The approximate change $df=2xdx= 2(1)(0.01)=0.02 \neq 0.0201$.
Like you said, if the change was exact, we wouldn't need the Taylor series.
But I believe we're talking at cross from each other.

 

[jbriggs444]

It is not exact. The _actual_change in the values of $x^2$ between $x=1$ and $x=1.01$ is$\Delta f=(1.01)^2-1^2 =0.0201$. The approximate change $df=2xdx= 2(1)(0.01)=0.02 \neq 0.0201$.

I agree that the change predicted by the linear approximation is 0.02 while the actual change is 0.0201.
I disagree that the notation "$df$" properly denotes the linear approximation.

In particular, I do not agree that "$df$" denotes $f'(x) \times \Delta x$ and do not accept that "$df$" is intended to be an approximation for $f(x+\Delta x) - f(x)$.

But I believe we're talking at cross from each other.

Indeed. Your usage of notation is not standard.

 

[PeroK]

The derivative of the function $f(x) = x^2$ is $f'(x) = 2x$. This is exact. How you justify that mathematically is another question. Standard analysis uses limits - which are themselves exact.

This can also be written $f'(x) \equiv \frac{df}{dx} = 2x$, which leads to the concept and defining property of differentials, which are not real numbers: $$df = 2xdx$$More generally, if $y = f(x)$, then $$dy = f'(x)dx$$

 

[mathwonk]

you think this is confusing, how about this: a real number actually is a Cauchy sequence of rational numbers (modulo equivalence),

so if f is a rational polynomial like X^2, and X0 is a rational number then the slope, as a real number, of the graph y=f(X), at (X0,f(X0)), is equal to the (real number defined by the) Cauchy sequence of rational approximations:

An = [f(X0+1/n)-f(X0)]/(1/n), as n—> infinity.

I.e. the limit of the sequence {An} is actually the sequence itself!

Moral: of course, no finite part of a sequence ever gets to the limit, but we can still try to get there ourselves! (mentally).

Actually, I think the question in your title will be answered if you just decide for yourself in what sense the sequence 1, 1/2, 1/3, 1/4, ..., 1/n,... determines the exact number zero.

for more details, consult:
https://www.math.uga.edu/sites/default/files/inline-files/polynomial_tangents_without_limits.pdf

https://www.math.uga.edu/sites/default/files/inline-files/tangents_to_y.pdf

 

[WWGD]

The derivative of the function $f(x) = x^2$ is $f'(x) = 2x$. This is exact. How you justify that mathematically is another question. Standard analysis uses limits - which are themselves exact.

This can also be written $f'(x) \equiv \frac{df}{dx} = 2x$, which leads to the concept and defining property of differentials, which are not real numbers: $$df = 2xdx$$More generally, if $y = f(x)$, then $$dy = f'(x)dx$$

It depends on what you mean by ' exact'. Just how does this specifically counter anything Ive written here. Df is the differential, which is the change along the tangent line approximation, while $\ Delta f:=f(x+h)-f(x)$ is the actual change. The former is a limit, thus it may or may not exist. The latter is a difference of Real numbers, and will always exists. When f is a linear map, then $\Delta f==df$. Otherwise, this is not true. The best local linear map that approximates a linear function is that linear function itself.

Just what, specifically, have I said , that is wrong or misleading?

 

[jbriggs444]

It depends on what you mean by ' exact'.

There is no serious disagreement about the meaning of the word exact.

Just how does this specifically counter anything Ive written here. Df is the differential, which is the change along the tangent line approximation

$df$ is not the change along the tangent line approximation. That is a fiction of your own creation.

 

[WWGD]

There is no serious disagreement about the meaning of the word exact.


$df$ is not the change along the tangent line approximation. That is a fiction of your own creation.

No, it's a fiction of yours. Several sources use that, however you dislike it.
And, yes, I explained why, how I made the distinction.

 

[jbriggs444]

No, it's a fiction of yours. Several sources use that, however you dislike it.

Reference, please.

 

[WWGD]

Reference, please.

Will look it up. Why don't you provide sources with an alternative definition?

 

[WWGD]

Will look it up. Why don't you provide sources with an alternative definition?

For one, if you agree that $df=f'(x)dx$, this agrees with the change along the tangent line at $x=1$, with slope $2$, giving us $df=f'(x)dx=2dx=2(0.01)=0.02$, while the actual change, per one the prior posts, is 0.0201. The change Councidence? Since @fresh_42 may be unavailable, maybe @Mark44, your 44 colleague, can chime in.

 

[jbriggs444]

For one, if you agree that $df=f'(x)dx$

Indeed, I agree with this.

this agrees with the change along the tangent line at $x=1$, with slope $2$, giving us $df=f'(x)dx=2dx=2(0.01)=0.02$

0.01 is not an infinitesimal. To the extent that $df$ and $dx$ have values at all, they are infinitesimals.

There is a difference between "small" and "infinitesimal". It is not clear that 0.01 is even "small".

Let me be clear on how I understand the difference between $\Delta f$ and $df$. It is not that the one is an actual change in the function value while the other is the change in an associated linear map. It is that the one is the actual change in the function value across a finite change in $x$ while the other is the actual change in the function value across an infinitesimal change in $x$.

Still waiting on your references.

 

[WWGD]

Indeed, I agree with this.

0.01 is not an infinitesimal. To the extent that $df$ and $dx$ have values at all, they are infinitesimals.

There is a difference between "small" and "infinitesimal". It is not clear that 0.01 is even "small".

Still waiting on your references.

Will look it up. Maybe you can provide yours as well, regarding both the alternative definition and the requirement that dx be an infinitesimal. I don't remember infinitesimals being brought up when using these formulas in school. Do we also only compute Riemann sums against infinitesimals dx? Then just how is $\int dx=x$? Or do you disagree with this too?

 

[jbriggs444]

Maybe you can provide yours as well

Wikipedia has what seems to be a decent article.

I don't remember infinitesimals being brought up when using these formulas in school. Do we also only compute Riemann sums against infinitesimals dx? Then just how is $\int dx=x$?

When I went through school we stuck with the standard reals. There are no infinitesimals in the standard reals. My preferred understanding is that $df$ and $dx$ are not even values at all. They are notational placeholders. Handles on which one can hang an intuition.

But, if someone wants to treat them as having actual values, I am happy to adopt the idea of infinitesimals. It is when someone decides that they are finite that I call foul.

In integrals especially, the $dx$ is pure notation. It tells you what variable you are integrating over.

In a Riemann sum, one is taking a limit as the partition size goes to zero. That is where something akin to infinitesimals comes in -- the partition size gets small.

 

[fresh_42]

Maybe it's time to close this thread since it turned into a discussion about notation and wording which likely won't help the OP.

I once gathered some perspectives and wordings around derivatives and I found 10 different ones, and I did not even use the word slope. We are apparently short before discussing Pfaffian forms, surreal, and hyperreal numbers.

Infinitesimals have become obsolete in mathematics by the introduction of the epsilontic as a standard in teaching calculus. Unfortunately, they are still around in the language of physics as "arbitrary small changes". The d symbol is even more ambiguous. It stands for an abbreviation of a limit - which by the way is not a process, it is a number - or a vector, or a linear operator, or a derivation, or a piece of information, or a differential form. $\Delta x\neq dx \neq \displaystyle{\lim_{\Delta x \to 0}\Delta x =0}.$ I think it cannot even be called infinitesimal. $dx$ is context sensitive. In the context of this thread, it only makes sense as a quotient
$$
\left. \dfrac{dy}{dx}\right|_{x_0} =\lim_{\Delta x \to 0} \dfrac{y(x_0+\Delta x)-y(x_0)}{\Delta x}
$$
in which case it is the slope of the function $y(x)$ at $x_0,$ a limit, and a number.

 

[PeroK]

Differentials are often used informally in applied maths and calculus, without being formally defined. An example is an integral substitution. We start with the integral:
$$I = \int_a^b g(x) dx$$Then we let $x = f(u)$ hence $dx = f'(u) du$ and the integral becomes:
$$I = \int_{f^{-1}(a)}^{f^{-1}(b)} g(f(u))f'(u) du$$Differentials can be defined rigorously using linear maps on a manifold. But, for calculus 1, we can generally use differentials without worrying about a rigorous definition.

 

[pbuk]

However, what I don't know is why it is EXACTLY 2x and not just very close to 2x.

Because you can prove it.

1. Start by assuming the opposite: that the slope of the tangent line is not exactly $2x$, it is $2x + \varepsilon$.
2. Now you already know that for $f(x) = x^2$ the slope of the secant line between $x$ and $x + \Delta x$ is $2x + \Delta x$.
3. As you make $\Delta x$ smaller, the slope of the secant line will become closer to the slope of the tangent line - or rather it will never start getting further away (this is the definition of differentiability). So if you believe that the slope of the tangent is actually $2x + \varepsilon$ then the slope of the secant line can never be smaller than this.
4. Choose $\Delta x = \frac \varepsilon 2$. The slope of the secant line is $2 x + \Delta x = 2 x + \frac \varepsilon 2$. From 3. you know that the slope must be greater than or equal to $2 x + \varepsilon$. The only value of $\varepsilon$ for which $2 x + \frac \varepsilon 2 \ge 2 x + \varepsilon$ is $0$.
5. From 1. the slope of the tangent line is $2x + \varepsilon = 2x + 0 = 2x$.

Note that although this proof should be convincing at the 'Basic' level, it has a hole in it and to overcome this hole you need to study calculus more formally which is often called analysis.

@NoahsArk if you can spot the hole and it bothers you then you might find the book Calculus by Michael Spivak interesting. If you prefer learning from a combination of video lectures, notes and exercises then I recommend https://ocw.mit.edu/courses/18-01-single-variable-calculus-fall-2006/.

 

[Vanadium 50]

We've posted 32 messages since the OP was here last. Are we helping him? Or just piling on?

 

[pbuk]

We've posted 32 messages since the OP was here last. Are we helping him? Or just piling on?

Yes, there was a big diversion off-topic. I hope that my previous post (which ignores the diversion) is back on-topic and does actually help.

 

[fresh_42]

We've posted 32 messages since the OP was here last. Are we helping him? Or just piling on?

Indeed. We can discuss endlessly about differentiation. Some students once asked me to write a summary about this subject. I ended up with an article that had to be split into five parts to suit the length of typical, however yet long insight articles. I like to say that if you read two authors you will find four notations. There is a long way from a slope to the pullback of sections and it is paved with d's. It cannot be dealt with in a single thread, and by leaving the original limits of the question there is no restriction in place anymore.

I would like to see a discussion about the use of the word infinitesimal, its history, and its meaning in physics, especially as mathematicians have restricted its use to hyperreals and related concepts; but that would require a different thread.

This thread is closed now.