Please do not give a specific answer, just some guidance is what I'm looking for.
This is a problem from an analysis module I am currently taking that has left me scratching my head:
"Suppose that $g:I \to \mathbb{R}$ and that $g$ is differentiable at the point $x=x_0$ does the following limit exist? $\lim_{h \to 0} \frac{g(x_0+h)-g(x_0-h)}{2h}$
Edit: From testing of a few well-defined functions this appears to be some alternative way of expressing a derivative.
Here is guidance only. (You asked for guidance.)
This is how I would have posed the problem. I leave to others to solve it. You need context in order to understand why this is posed.
This is a more important problem than it might appear at first sight.
If $f:\mathbb R\to \mathbb R$ then the following "derivatives" at a point $x_0$ can be defined and are used extensively in analysis:
The ordinary derivative $$ f'(x_0) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}.$$
The symmetric derivative $$ SD\,f(x_0) = \lim_{h\to 0} \frac{f(x+h)-f(x-h)}{2h}.$$
The four Dini derivatives $$ D^+\,f(x_0) = \limsup_{h\to 0+} \frac{f(x+h)-f(x)}{h} \ \ \text{ and }\ \ D_+\,f(x_0) = \liminf_{h\to 0+} \frac{f(x+h)-f(x)}{h},$$ with similar definitions for $D^-\,f(x_0)$ and $D_-\,f(x_0)$.
Exercise 1. Show that if $f$ is differentiable at a point $x_0$ then $SD\,f(x_0)$ exists and is equal to $f'(x_0)$.
Exercise 2. Give an example of a function for which $SD\,f(x_0)$ exists but $f'(x_0)$ does not.
Exercise 3. Show that any function that has both right-hand and left-hand derivatives at a point must be symetrically differentiable there.
Exercise 4. Improve your example for #2 so that it is not explained by Ex 3.
Exercise 5. Show that a function is differentiable at a point if and only if all four Dini derivatives are the same real number.
Exercise 6. Give an example of a function with all four Dini derivatives finite and different from each other at $x=0$.
Exercise 7. Show that a continuous function with a positive ordinary derivative $f'(x)$ everywhere is monotone increasing.
Exercise 8. Show that a continuous function with a positive symmetric derivative $SD\,f(x)$ everywhere is monotone increasing.
Exercise 9. Show that a continuous function with a positive Dini derivative $D^+\,f(x)$ everywhere is monotone increasing.
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Commentary: The symmetric difference quotient $$ \frac{f(x+\Delta x )-f(x-\Delta x)}{2 \Delta x}$$ plays a small role in numerical differentiation. If the derivative exists at a point then this may be a better approximation to that derivative than the ordinary difference quotient. Some calculators use this to give their answers.
But for real analysis the interest in the symmetric derivative (or the Dini derivatives) is in situations where the function is not differentiable or where we do not know if it is differentiable. Thus these generalized derivatives are a replacement for the ordinary one.
Exercises 7, 8 & 9 illustrate this. Ex 7 is the standard monotonicity theorem all calculus students learn (proved by the mean-value theorem). But that theorem is way too weak. As you see in Ex 8 and Ex 9 the same result is available for these generalized derivatives too (although much harder to prove).
For real analysts the symmetric derivatives have attracted considerable attention, often because they play a role in the study of trigonometric series. The Russian mathematician Khintchine in 1927 showed that a measurable function with a symmetric derivative everywhere must have an ordinary derivative almost everywhere. For continued research that extends his ideas see the comments to this question:
If $\,\lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}\,$ exists for every $x$, what does this imply for $f$?
Often the first encounter a student has with the symmetric derivative is this simple question (which pops up all the time on StackExchange). Consider it an important preview of an interesting body of research, and not at all a meaningless exercise in limits.
The derivate is defined as the limit of the difference quotient. This limit has a geometric interpretation (try doing a sketch for different values getting closer to the limit). ¿Is there any interpretation to your limit?.
If the derivate could be interpeted as the instantaneous velocity, using your interpetation for this new limit, ¿can we make a physical interpretation?.
Does it matter here the value $h(x_0)$?
I think it exists because $$\lim_{h \rightarrow 0} \frac{g(x_0+h)-g(x_0 -h)}{2h}= \lim_{h \rightarrow 0} \frac{g(x_0+h)-g(x_0)+g(x_0)-g(x_0 -h)}{2h}$$ $$ = \lim_{h \rightarrow 0}\left[\frac{g(x_0+h)-g(x_0)}{2h}+\frac{g(x_0)-g(x_0 - h)}{2h}\right]= g^{'}(x_0)$$
Therefore,
$$\lim_{h \rightarrow 0} \frac{g(x_0+h)-g(x_0)}{2h} = \lim_{h\to 0}\frac{g(x_0)-g(x_0 - h)}{2h}.$$
– Adam Rubinson Jan 29 '22 at 22:49This is called the centered difference quotient. If $g$ is smooth enough, it converges to $g'(x_0)$ faster than the defining difference quotient $\frac{g(x_0 + h) - g(x_0)}{h}$. This can be shown via Taylor's theorem.
But you are only assuming that $g$ is differentiable at $x_0$. The centered difference quotients do converge to $g'(x_0)$. To prove this, use the definition of derivative: $$g(x_0 + h) = g(x_0) + g'(x_0)h + hR(h), \hspace{20pt}R(0) = 0, R(h) \to 0 \text{ as } h \to 0.$$
