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For what I know, $dy$-$dx$ aren't real numbers, exist as convenient notations to capture our intuitions about infinitesimal increments, while $\varepsilon$-$\delta$ are real distances, saying that $f$ can be as close as we want to $L$ if $x$ is sufficiently close to $c$. The latter has a formal statement:

$$ \lim_{x \to c} f(x) = L \iff (\forall \varepsilon > 0)(\exists \ \delta > 0) (\forall x \in D)(0 < |x - c | < \delta \ \Rightarrow \ |f(x) - L| < \varepsilon)$$

While the former denote for $\Delta f$ and $\Delta x$ respectively when we arrive at $\Delta f=A\Delta x+o(\Delta x)$, which comes from the definition of derivative $f'=\lim_{\Delta x\to0}\frac{\Delta f}{\Delta x}$, and a chain of denotations $\varepsilon(\Delta x)=\frac{\Delta f}{\Delta x}-f'$, $o(\Delta x)=\varepsilon(\Delta x)\Delta x$ and $A=f'(x)$, with a note that $\lim_{\Delta x\to0}\varepsilon(\Delta x)=0$. As I understand here $\varepsilon(\Delta x)$ is simply a random unimportant convenient notation and has nothing relates to the infinitesimal above.

Yet looking at the graph on Wikipedia: (ε, δ)-definition of limit, I can't help but thinking that they are just one thing:

So what is the difference, and more generally, the relation between these two notations? Can I use $\varepsilon$-$\delta$ in a integral? And if $df=A\Delta x,dx=\Delta x$, then they should be real, right?

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  • $\begingroup$ $\dfrac {dx} {dy}$ is a symbol, that has its origin from the Leibnizian approach but that now has been replaced by the $\epsilon - \delta$ definition. In this one, $\epsilon$ and $\delta$ are real numbers. $\endgroup$ – Mauro ALLEGRANZA Sep 7 '17 at 12:16
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    $\begingroup$ Apart from modern "infinitesimal" versions of calculus, we con recover the "infinitesimals" $dx$ and $dy$ also in a standard presentation (the $\epsilon-\delta$ one) as functions. See e.g. Vladimir Zorich, Mathematical Analysis I, Springer (2nd ed 2016), page 177. $\endgroup$ – Mauro ALLEGRANZA Sep 7 '17 at 12:28
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    $\begingroup$ @MauroALLEGRANZA Could you please post this as an answer? I can't see how you would do that- sure, you could define a translation algorithm from $dy-dx$ notation to $\epsilon-\delta$ notation, but how could you give $dx$ and $dy$ themselves a rigourous meaning? $\endgroup$ – Daniel Moskovich Sep 7 '17 at 12:55
  • $\begingroup$ @DanielMoskovich - can you open the page of Zorich's textbook referenced in the above comment ? $\endgroup$ – Mauro ALLEGRANZA Sep 7 '17 at 13:15
  • $\begingroup$ @MauroALLEGRANZA No- I wish I could! Google books doesn't display that page. $\endgroup$ – Daniel Moskovich Sep 7 '17 at 16:33
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To begin with, your graph is wrong (or at best, misleading). The standard $\epsilon$-$\delta$ definition does not say that $f(x-\delta) = f(x) - \epsilon$ and that $f(x+\delta) = f(x) + \epsilon,$ although your graph asserts both of those facts.

To make a more representative graph, increase $\epsilon$ so that the horizontal lines do not all intersect the curve at the same places as the vertical lines. Put the lower horizontal line below the lowest intersection and the upper horizontal line above the highest intersection. That will be more representative of how to think about $\epsilon$-$\delta$ definitions.

Now to take a concrete example, instead of your arbitrary function drawn as a red curve, let's redraw the graph using the function $f(x) = x+1.$ And let's specifically choose $c=3,$ $\delta = 1,$ and $\epsilon=2.$

Then if we take the lower difference, we have $\Delta x = 2 - 3 = -1$ and $\Delta f = f(2) - f(3) = 3-4 = -1.$ But $f'(3) = 1.$ so we have $$ \frac{\Delta f}{\Delta x} - f' = 1 - 1 = 0. $$ We get the same result for the upper difference.

In fact you will have $ \frac{\Delta f}{\Delta x} - f' = 0 $ whenever you have a function $f$ whose slope is constant over the interval $[x-|\Delta x|, x+|\Delta x|].$ But $0$ is never a suitable value for $\epsilon$ in an $\epsilon$-$\delta$ definition.

The notation $\epsilon(\Delta x)$ seems unfortunate to me, since it actually has very little to do with the $\epsilon$ in the figure.


Added notes:

There are "upper" and "lower" differences because when you're taking a limit at a point inside your domain, you have to be able to approach it "from above" and "from below." For any $x,$ the "difference" is $\Delta x = x-c.$ An upper difference $\Delta x$ is positive and occurs when you come at $c$ from above, that is, $x > c.$ A lower difference $\Delta x$ is negative and occurs when you come at $c$ from below, that is, $x < c.$

In the example I gave, $c=3,$ so $x=2$ makes $x<c.$ We then have $\Delta x = -1,$ so $\Delta x$ is a "lower" difference.

Remember, $\epsilon$ in the $\delta$-$\epsilon$ definition is related to $\Delta f$ (through $|f(x)-L|<\epsilon,$ and all we need to do to satisfy the definition is to make sure that $|f(x)-L|$ is less than $\epsilon$; we don't have to make them equal. I wanted simple numbers in the example I gave, so I chose a relatively large value of $\epsilon$ ($\epsilon=2$), which gave me plenty of room to choose $\delta.$ And with $c=3$ and $\delta=1,$ if $|x - c|<\delta$ (that is, if $2 < x < 4$), we have $3 < f(x) = x+1 < 5.$ The limit we're going for is $\lim_{x\to3} x+1 = 4,$ so $L=4$ and $|x - c|<\delta$ implies $|f(x)-L|<\epsilon=2$ as the definition requires.

On the other hand, $\delta=2$ and $\epsilon=1$ gives us $1<x<5,$ $2<f(x)<6,$ and therefore $|x - c|<\delta$ implies $|f(x)-L|<2.$ But in this case $|x - c|<\delta$ does not imply $|f(x)-L|<\epsilon$; consider what happens when $x = 4.5,$ for example.

It may also be worth pointing out that your presentation of the $\delta$-$\epsilon$ definition and your graph were relevant to a limit of the function $f,$ but to find $\frac{dy}{dx}$ you take a limit of $\frac{\Delta f}{\Delta x},$ which is not $f.$ So if you're trying to relate anything about $\lim_{\Delta x\to0} \frac{\Delta f}{\Delta x}$ to the definition you wrote or to your graph, the symbols are just not going to match up. To apply the $\delta$-$\epsilon$ definition to $\lim_{\Delta x\to0}\frac{\Delta f}{\Delta x}$ rather than $\lim_{x\to c}f,$ you can write it like this:

$$ \lim_{\Delta x\to0} \tfrac{\Delta f}{\Delta x} = L \iff (\forall \varepsilon> 0)(\exists \delta > 0) (\forall \Delta x \in D') \left(0 < |\Delta x - 0| < \delta \implies \left|\tfrac{\Delta f}{\Delta x} - L\right| < \varepsilon\right)$$

where $\Delta x = x-c$ for some constant $c,$ $D'$ is the set of possible values of $\Delta x$ where $x$ is allowed to be anywhere in the domain of $f,$ and $\Delta f = f(c+\Delta x) - f(c).$

And the notation $\varepsilon(\Delta x),$ suggesting that the variable $\varepsilon$ in the definition above is somehow a function of the quantity $\Delta x$ in the definition? That's just wrong, wrong, wrong. Look at the definition again; it doesn't work that way at all.

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  • $\begingroup$ What are the lower and upper differences? What is the role of $\delta$-$\epsilon$ in your example? Where is the number 2 in $\Delta x=3-2$? Do you mean $\delta=2, \epsilon=1$? $\endgroup$ – Ooker Sep 8 '17 at 8:01
  • $\begingroup$ I really did mean $\epsilon=2,\delta=1.$ The value $\epsilon=2$ does not occur at all in $\Delta x=3-2$; that's the point. I've added additional commentary to the answer. $\endgroup$ – David K Sep 8 '17 at 13:49
  • $\begingroup$ Before you said that the lower difference is $\Delta x=3-2=1$, but after the addendum it becomes $\Delta x=2-3=-1$. Which are right? The $\varepsilon(\Delta x)$ notation is insignificant, you can easily replace with with another notation; it is there to signify you that the $o(\Delta x)\rightarrow 0$ faster than $\Delta x$. And thanks to you I now understand the problem: $\varepsilon$-$\delta$ are to make sure that $f$ exists a limit at $L$, and does not tell whether it changes fast or slow at that point like $dx$-$dy$. Is it a requirement to have $f(x_0)$ exists to have its derivative? $\endgroup$ – Ooker Sep 9 '17 at 18:35
  • $\begingroup$ Thanks for spotting my inconsistent use of the $\Delta x$ notation. I see you read the answer carefully! The $-1$ example is the one I wanted, so I have made the first example match it. It's also true that technically it is fine to define a function of $\Delta x$ as in the formula in the question. I really only objected to the name of the function because I thought it could cause confusion. I'm glad the answer was helpful to you. $\endgroup$ – David K Sep 9 '17 at 20:28
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The $\frac{dy}{dx}$ notation was first introduced by Leibniz 350 years ago. To Leibniz, $dx$ and $dy$ were infinitesimals. Some centuries later, mathematicians following Cantor, Dedekind, and Weierstrass (CDW) attempted to eliminate infinitesimals from mathematical practice starting around 1870. This was a time of formalisation and rigorisation and since they CDW were unable to formalize infinitesimals in a satisfactory way they felt there was no choice but eliminating them altogether. Less than a century later, Robinson was able to formalize infinitesimals to an acceptable level of rigor so in a sense Leibniz's ideas were retroactively vindicated over three centuries later.

In most (infinitesimal!) calculus courses infinitesimals are still not being used, in my opinion to great detriment of the students. So unless you are studying at places like Bar Ilan university where calculus with infinitesimals is taught, your professor will consider it an error if you view $\frac{dy}{dx}$ as an actual fraction.

The epsilon-delta definitions you mentioned are long-winded paraphrases of the original infinisimal definitions. For example, Cauchy defined continuity of $y=f(x)$ as follows: each infinitesimal increment $\alpha$ necessarily produces an infinitesimal change $f(x+\alpha)-f(x)$ of the values of the function. Due to unavailability of infinitesimals in a traditional CDW framework, the kind of definition you are likely to hear in your course is as you noted: for every epsilon there exists a delta such that if $x-c$ is less than delta then $f(x)-f(c)$ is less than epsilon.

Keisler's textbook does a fine job treating calculus with infinitesimals. Once you understand the key concepts like continuity and derivative via the more intuitive infinitesimal approach, you will be in a better position to follow the long-winded paraphrases, as well (but don't tell your professor about it :-).

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    $\begingroup$ I don't understand the downvote of this answer. $\endgroup$ – P. Siehr Sep 7 '17 at 12:31
  • $\begingroup$ so basically the two are the same? That $\frac{\varepsilon}{\delta}=\frac{d}{dx}y, f=\int f'\delta$ and the like? $\endgroup$ – Ooker Sep 7 '17 at 12:38
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    $\begingroup$ @P.Siehr I don't know, but I'm questioning if it is a great detriment of the students actually. Of course this is marked as the opinion of mr Katz though. I'd guess that students think that they would be better of using infinitesimals instead of limits, but I'm not that sure. The fact that it took hundred years to sort that out and not nearly that time to sort limits out indicates that perhaps infinitesimals are not that easy after all - and we're talking about stuff that are to be tought to newbies that should be teached to be stringent. $\endgroup$ – skyking Sep 7 '17 at 12:42
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    $\begingroup$ @skyking Sure. If we want to teach to full rigour, the reals are simpler than the hyperreals. But modern universities so rarely do. If we're happy leaving some "black boxes", then I agree with Katz that the hyperreals (or another number system with infinitesimals) are much simpler. $\endgroup$ – Daniel Moskovich Sep 7 '17 at 12:45
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    $\begingroup$ @MauroALLEGRANZA OK- yes. The exact opposite (and now I can't edit). For fixed $\epsilon$ and $\delta$, the expression $\frac{\epsilon}{\delta}$ is just some real number, therefore not capturing the meaning of $\frac{dx}{dy}$. $\endgroup$ – Daniel Moskovich Sep 7 '17 at 12:48
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The $\epsilon$-$\delta$ notation and the $dy$-$dx$ notations live in different spaces.

Specifically, the $\epsilon$-$\delta$ notation is native to the real numbers $\mathbb{R}$, and has to do with how real numbers are constructed and defined. Therefore, working with real numbers necessarily entails $\epsilon$-$\delta$ work. One axiom of the real numbers $\mathbb{R}$ is completeness. One way of formulating this axiom is that every Cauchy sequence converges. The definition of Cauchy sequence, and the definition of convergence, both involve $\epsilon$-$\delta$. Intuitively (and I think it is fair to credit this observation to Bolzano), as real numbers can never we written to full precision (they are constructed as limits), they are defined, constructed, and thought as being infinite sequences of increasingly precise approximations. This captures the intuition that we can write a real number to within as small an error as we want (e.g. we can write $\pi$ to as many decimal places as we want, subject to limitations of time and writing space), but never to full precision.

Conversely, in the $dy$-$dx$ notation, $dy$ and $dx$ exist only if we have infinitesimals. Infinitesimals do not exist in $\mathbb{R}$, so when we are working over the real numbers, $dy$-$dx$ is used as a shorthand for $\epsilon$-$\delta$ and has no intrinsic meaning. Indeed, over the real numbers the $dy$-$dx$ notation is misleading and treating $\frac{dy}{dx}$ as a fraction doesn't make sense and is an error. To write something rigourous, you strictly need to rephrase your expression in terms of $\epsilon$-$\delta$. However, if we are working over another ordered field which does have infinitesimals, such as the hyperreals, the $dy$ and $dx$ are infinitesimal numbers.

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  • $\begingroup$ Daniel, it is true that "dy--dx notation is all about infinitesimals" in a way, but it is also used in the Weierstrassian approach, so this comment may be a little misleading. $\endgroup$ – Mikhail Katz Sep 7 '17 at 12:42
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    $\begingroup$ Okay. I rephrased it. "All about infinitesimals" in that it's formal notation for an epsilon-delta definition and nothing more, when we are working in a number system without infinitesimals. $\endgroup$ – Daniel Moskovich Sep 7 '17 at 12:53

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