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Is there any particular reason "infinite" derivatives are not well-defined? For example, $x \mapsto x^{\frac 13}$ at $x=0$. More precisely, what is wrong with the following definition of differentiability?

Let $f:I \to \mathbb{R}$ be a real function and let $c$ be an interior point of $I$. If $f$ is continuous at $c$ and the limit $$\lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$$ exists and is equal to $L$, where $L \in \mathbb{R} \cup \{+\infty, -\infty\}$, $f$ is said to be differentiable at $c$.

There doesn't seem to be anything obviously wrong with this.

The most important theorem, the mean value theorem, still holds with this definition, according to Wikipedia. Furthermore, unless I'm mistaken, with this definition there is the nice property that if $f:(a,b) \to \mathbb{R}$ is injective and differentiable, $f^{-1}$ is as well.

I am assuming, of course, that there are problems since we use the stricter definition:
what are they?

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  • $\begingroup$ $f(x+\varepsilon)=f(x)+\varepsilon f'(x)$ is axiomatic in smooth infinitesimal analysis. It also 'works' if $\varepsilon$ is finite. $\endgroup$ – user301988 May 30 '16 at 1:45
  • $\begingroup$ I don't think you can define a $+\infty$ derivative at one point,you'll need probably that $f$ is (continuously) differentiable on an open, where the continuity is meant in $\mathbb{R} \cup \{ \pm \infty\}$. I guess this will be not so different to the definition of meromorphic / holomorphic on an open $U$ except at some isolated points. $\endgroup$ – reuns May 30 '16 at 2:12
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    $\begingroup$ You might be interested in Dini derivatives. (Royden's Real Analysis is a well-written source for details.) $\endgroup$ – Andrew D. Hwang May 30 '16 at 2:17
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    $\begingroup$ One possible reason I've thought of (without an explicit example) is that that standard "linearity" properties of the derivative (ie. $(cf + cg)' = cf' + cg'$) as well as the chain rule might not work. I'm not sure this is entirely satisfying though, because the aforementioned theorems can just be formulated with the additional proviso that the derivative is finite. Perhaps there is some "deeper" reason we want derivative to be finite (or, maybe it's something not-so-deep which I'm not seeing). I have placed a bounty on this question. $\endgroup$ – MathematicsStudent1122 Jun 2 '16 at 2:42
  • $\begingroup$ So what is $\lim_{x\to 0}\dfrac 1x?$ $\endgroup$ – steven gregory Jun 3 '16 at 16:26
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It is a matter of convention and agreement between mathematicians.

For me, there is no problem if you say that the function is differentiable at some point $c$ even if $\lim_{h \to 0} \frac{f(c+h) - f(c)}{h}$ is equal to $+\infty$ or $- \infty$. This would only extend differentiability of some functions to some larger set, for example your example $x \mapsto x^{\frac 13}$ would with your definition be differentiable on $\mathbb R$ and not only on $\mathbb R \setminus \{0\}$. So I would not say as you say that derivatives of some functions in some points that are equal to $+\infty$ or $- \infty$ are not well-defined, they are well-defined, it is just that we usually take by definition that the derivative of some function at some point is a finite number.

Situation is, viewen in this way, similar to that of series.

You could define that the series $\sum_{i=1}^{\infty}a_i$ of real numbers is convergent if the limit $\lim_{n \to \infty}\sum_{i=1}^{n}a_i$ exists and is equal to some member of the set $\mathbb{R} \cup \{+\infty, -\infty\}$. With this definition, for example, the harmonic series $\sum_{i=1}^{\infty}\frac {1}{n}$ would be a convergent series.

The only "problem" that I see with these extended definitions of the derivative at some point and convergence of the series is that maybe we would have to, when proving some theorems, replace the assumption Suppose that $f$ is differentiable at some point $c$... with the assumption Suppose that $f$ is differentiable at some point $c$ and that the derivative at that point is not equal to $+ \infty$ or $- \infty$... (and similarly for the series(and integrals)).

So, I would say that there is nothing wrong with your extended definition.

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You would lose the sum, product, and quotient rules for derivatives. You would lose the chain rule. You would lose the fact that a derivative at a point implies continuity at that point. The intermediate value theorem would no longer apply to differentiable functions. You lose the Darboux property of derivatives. Say goodbye to Taylor. Our freshmen are going to love it!

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  • $\begingroup$ Everything you're saying seems correct, but I overlooked continuity. Now it seems a function like $\frac 1x$ with $f(0) = 0$ would be differentiable, which is definitely not what I wanted. I was thinking more along the lines of continuous functions with vertical tangents. I will edit this, but this is my fault so I'll +1 your answer. $\endgroup$ – MathematicsStudent1122 Jun 3 '16 at 17:22
  • $\begingroup$ Well OK, but now differentiable implies continuous is vapid, since continuity is assumed in the definition. $\endgroup$ – zhw. Jun 3 '16 at 21:21
  • $\begingroup$ A "practical" reason why its a bad idea Newton--Raphson method may "converge" to the wrong solution, for functions $f$ that have $|f'|\geq c>0$. E.g., $x\mapsto x^{1/3}+1=0$ with initial guess $x_0=0$. $\endgroup$ – Oskar Limka Jun 8 '16 at 13:35
  • $\begingroup$ Can you give examples? What's an example of the sum-rule failing, for example? $\endgroup$ – goblin Jul 9 '17 at 5:08
  • $\begingroup$ @goblin If $f'(0) = \infty, g'(0)= -\infty,$ then $(f+g)'(0) = \infty - \infty?$ $\endgroup$ – zhw. Jul 9 '17 at 14:49
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I agree with the above answer of @Farewell.

Another aspect worth considering in my opinion is the inverse function theorem.

If a function with "derivative" $\pm \infty$ has an inverse, then in many cases the derivative of the inverse at the point will be $0$. (Basically we get a vertical line with a horizontal line when we switch the dependent and independent variables.)

First let's try to consider some geometric problems associated with having "derivative" $\pm \infty$ of a one-dimensional function. At that point, the associated tangent line would clearly be vertical.

But therein lies a major issue: how can one consistently define the slope of a vertical line? One can't -- it's impossible because both $\infty$ and $-\infty$ will be equally reasonable choices -- this none-uniqueness problem doesn't happen for any other type of tangent line by the way.

Sure, in the case of $x^{-1/3}$ one could argue that "by continuity" the slope should be defined to be $+\infty$. But what about $\sqrt{x}$ and $-\sqrt{x}$? By continuity the derivative of one at 0 would be $+\infty$ and the other would have derivative $-\infty$ at 0, but both would correspond to the same tangent line of the curve $x=y^2$.

In more than one dimension, the geometric problems associated with trying to define an "infinite derivative" are even worse. Specifically, an "infinite derivative" would correspond to the non-existent inverse of a singular matrix, and there are literally uncountably many ways in which a matrix can be singular (i.e. not be invertible and have determinant zero), so any attempt to find a reasonably small number of "pseudoinverses" for all singular matrices would not be tractable.

(Moreover the space of invertible matrices has a nice property called "openness" which is similar to the idea of an open interval that non-invertible matrices simply do not have. Think of it this way: the set of real numbers which have well-defined reciprocals is $(-\infty,0) \cup (0,\infty)$ -- two open intervals, whereas the set of real numbers that don't have a well-defined reciprocal, $\{0\}$ is a point (points have the property of being "closed"). A similar situation exists in higher dimensions.)

In the proof of the inverse function (for a general number of dimensions, including $n=1$), we rely on the derivative being "non-zero" (in a generalized sense) in order to show that we can find a local inverse for the function centered at that point.

The proof doesn't go through when the derivative is "zero" because we can't define a unique value for the derivative of the local inverse function at that point (again, this is even true for $n=1$ as I mentioned above).

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  • $\begingroup$ What about the Reimann projective line where $-\infty = \infty$? I would say that derivatives at infinity aren't wrong and make sense for some problems but that the machinery for working with infinities is annoying to work with and so many mathematicians don't bother with the case. $\endgroup$ – Steven Stewart-Gallus Jun 3 '16 at 17:15
  • $\begingroup$ But a single infinity for the real line does NOT lead to useful analysis (because of its total order, which we use extensively), since $\infty$ means essentially "without upper bound" and $-\infty$ means "without lower bound". In higher-dimensional spaces, where there is no total order, people are more willing to consider a single "point at infinity", but using such a concept to try to define a single inverse for uncountably many singular matrices still seems futile. $\endgroup$ – Chill2Macht Jun 3 '16 at 19:12

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