A question about definition of differentiability

Question

This question was asked to me by brother and I am not sure about its answer. So, I am asking it here.

As we know that a function $f :U $ (subset of $\mathbb{R}$) $ \to\mathbb{R} $ , defined on an open set $U$ is differentiable at $a\in U$ if

the derivative $\lim_{h\to 0}\frac{ f(x+h)- f(x) } { h }$ exists.

My question is can we change the definition of differentiability to $\lim_{h\to 0}\frac{ f(x+h) - f(x-h) } {h} $ exists.

Why or Why not?

This is often called the symmetric derivative (for googling purposes), and it can be shown (for example, see this answer) that if a function has a (finite) derivative at $x=a,$ then the function has a symmetric derivative at $x=a$ and the two derivatives are equal. However, as others have shown here (before I wrote this comment), it is possible for a function (even a continuous function) to have a symmetric derivative at a point without having an (ordinary) derivative at that point. — Dave L. Renfro, Aug 05 '20 at 09:46

score 2 · Answer 1 · answered Aug 05 '20 at 08:56

2

For every even function $f$ we have $\lim \frac {f(0+h)-f(0-h)} h =0$ but $f$ may not be differentiable at $0$. Example: $f(x)=|x|$.

answered Aug 05 '20 at 08:56

Kavi Rama Murthy

311,013

score 2 · Accepted Answer · answered Aug 05 '20 at 09:18

Well of course you can change it. Definitions can be whatever we want then to be. I think you mean are the 2 definitions any different. Yes they are different. The latter definition is symmetric about h and -h, meaning it doesn't matter if you take the limit from the right or the left, the result is the same, which means every point is differentiable. Even corners like y = abs(x) at the origin (the derivative is 0). A geometric interpretation is what you are doing is taking the limit as h goes to 0+ and the limit as h goes to 0- and computing the average between the two. For continuous functions, by definition the limits are the same, so you just get the normal derivative. But for non-continuous functions like y = abs(x), the normal derivative isn't defined at 0, because it's -1 coming from the left and +1 coming from the right. So this new definition says well let's just take the average of -1 and 1 which is 0 and say that's the derivative.

Nice extension of the derivative but doesn't have many further uses I think...

for non-continuous functions like y = abs(x) --- The absolute value function is continuous everywhere. — Dave L. Renfro, Aug 05 '20 at 09:38
Must've been drunk when I wrote this! I meant differentiable not continuous. — , Aug 05 '20 at 13:59

score 0 · Answer 3 · answered Aug 05 '20 at 09:41

Suppose that $f$ is differentiable at $x \in U.$ Then

$$\frac{ f(x+h) - f(x-h) } {h}=\frac{ f(x+h)-f(x)+f(x) - f(x-h) } {h}= \frac{ f(x+h) - f(x) } {h}+\frac{ f(x) - f(x-h) } {h}=\frac{ f(x+h) - f(x) } {h}+\frac{ f(x-h) - f(x) } {-h} \to 2 f'(x)$$

as $h \to 0.$

A question about definition of differentiability

3 Answers3