# Symmetric derivative

(Redirected from Symmetric difference quotient)

In mathematics, the symmetric derivative is an operation generalizing the ordinary derivative. It is defined as:

$\lim _{h\to 0}{\frac {f(x+h)-f(x-h)}{2h}}.$ The expression under the limit is sometimes called the symmetric difference quotient. A function is said to be symmetrically differentiable at a point x if its symmetric derivative exists at that point.

If a function is differentiable (in the usual sense) at a point, then it is also symmetrically differentiable, but the converse is not true. A well-known counterexample is the absolute value function f(x) = |x|, which is not differentiable at x = 0, but is symmetrically differentiable here with symmetric derivative 0. For differentiable functions, the symmetric difference quotient does provide a better numerical approximation of the derivative than the usual difference quotient.

The symmetric derivative at a given point equals the arithmetic mean of the left and right derivatives at that point, if the latter two both exist.

Neither Rolle's theorem nor the mean value theorem hold for the symmetric derivative; some similar but weaker statements have been proved.

## Examples

### The modulus function

Graph of the modulus function. Note the sharp turn at x=0, leading to non-differentiability of the curve at x=0. The function hence possesses no ordinary derivative at x=0. The symmetric derivative, however, exists for the function at x=0.

For the modulus function, $f(x)=\left\vert x\right\vert$ , we have, at $x=0$ ,

{\begin{aligned}f_{s}(0)&=\lim _{h\to 0}{\frac {f(0+h)-f(0-h)}{2h}}\\&=\lim _{h\to 0}{\frac {f(h)-f(-h)}{2h}}\\&=\lim _{h\to 0}{\frac {\left\vert h\right\vert -\left\vert -h\right\vert }{2h}}\\&=\lim _{h\to 0}{\frac {h-(-(-h))}{2h}}\\&=0,\\\end{aligned}}

where since $h>0$  we have $\left\vert -h\right\vert$  = $-(-h)$ . So, we observe that the symmetric derivative of the modulus function exists at $x=0$ , and is equal to zero, even though its ordinary derivative does not exist at that point (due to a "sharp" turn in the curve at $x=0$ ).

Note in this example both the left and right derivatives at 0 exist, but they are unequal (one is -1 and the other is 1); their average is 0, as expected.

### The function x−2

Graph of y=1/x². Note the discontinuity at x=0. The function hence possesses no ordinary derivative at x=0. The symmetric derivative, however, exists for the function at x=0.

For the function $f(x)=1/x^{2}$ , we have, at $x=0$ ,

{\begin{aligned}f_{s}(0)&=\lim _{h\to 0}{\frac {f(0+h)-f(0-h)}{2h}}\\&=\lim _{h\to 0}{\frac {f(h)-f(-h)}{2h}}\\&=\lim _{h\to 0}{\frac {1/h^{2}-1/(-h)^{2}}{2h}}\\&=\lim _{h\to 0}{\frac {1/h^{2}-1/h^{2}}{2h}}\\&=0,\\\end{aligned}}

where again, $h>0$ . Again, for this function the symmetric derivative exists at $x=0$ , while its ordinary derivative does not exist at $x=0$ , due to discontinuity in the curve there. Furthermore, neither the left nor the right derivative is finite at 0; i.e. this is an essential discontinuity.

### The Dirichlet function

The Dirichlet function, defined as

$f(x)={\begin{cases}1,&{\text{if }}x{\text{ is rational}}\\0,&{\text{if }}x{\text{ is irrational}}\end{cases}}$

has symmetric derivatives $\forall x\in \mathbb {Q}$  but not $\forall x\in \mathbb {R} -\mathbb {Q}$ ; i.e. the symmetric derivative exists for rational numbers but not for irrational numbers.

## Quasi-mean value theorem

The symmetric derivative does not obey the usual mean value theorem (of Lagrange). As counterexample, the symmetric derivative of f(x) = |x| has the image {-1, 0, 1}, but secants for f can have a wider range of slopes; for instance, on the interval [-1, 2], the mean value theorem would mandate that there exist a point where the (symmetric) derivative takes the value ${\frac {|2|-|-1|}{2-(-1)}}={\frac {1}{3}}$ .

A theorem somewhat analogous to Rolle's theorem but for the symmetric derivative was established in 1967 by C.E. Aull, who named it Quasi-Rolle theorem. If f is continuous on the closed interval [a, b] and symmetrically differentiable on the open interval (a, b) and f(b) = f(a) = 0, then there exist two points x, y in (a, b) such that fs(x) ≥ 0 and fs(y) ≤ 0. A lemma also established by Aull as a stepping stone to this theorem states that if f is continuous on the closed interval [a, b] and symmetrically differentiable on the open interval (a, b) and additionally f(b) > f(a) then there exist a point z in (a, b) where the symmetric derivative is non-negative, or with the notation used above, fs(z) ≥ 0. Analogously, if f(b) < f(a), then there exists a point z in (a, b) where fs(z) ≤ 0.

The quasi-mean value theorem for a symmetrically differentiable function states that if f is continuous on the closed interval [a, b] and symmetrically differentiable on the open interval (a, b), then there exist x, y in (a, b) such that

$f_{s}(x)\leq {\frac {f(b)-f(a)}{b-a}}\leq f_{s}(y)$ .

As an application, the quasi-mean value theorem for f(x) = |x| on an interval containing 0 predicts that the slope of any secant of f is between -1 and 1.

If the symmetric derivative of f has the Darboux property, then the (form of the) regular mean value theorem (of Lagrange) holds, i.e. there exists z in (a, b):

$f_{s}(z)={\frac {f(b)-f(a)}{b-a}}$ .

As a consequence, if a function is continuous and its symmetric derivative is also continuous (thus has the Darboux property), then the function is differentiable in the usual sense.

## Generalizations

The notion generalizes to higher-order symmetric derivatives and also to n-dimensional Euclidean spaces.

### The second symmetric derivative

It is defined as

$\lim _{h\to 0}{\frac {f(x+h)-2f(x)+f(x-h)}{h^{2}}}.$ 

If the (usual) second derivative exists, then the second symmetric derivative equals it. The second symmetric derivative may exist however even when the (ordinary) second derivative does not. As example, consider the sign function $\operatorname {sgn}(x)$  which is defined through

$\operatorname {sgn}(x)={\begin{cases}-1&{\text{if }}x<0,\\0&{\text{if }}x=0,\\1&{\text{if }}x>0.\end{cases}}$

The sign function is not continuous at zero and therefore the second derivative for $x=0$  does not exist. But the second symmetric derivative exists for $x=0$ :

{\begin{aligned}\lim _{h\to 0}{\frac {\operatorname {sgn}(0+h)-2\operatorname {sgn}(0)+\operatorname {sgn}(0-h)}{h^{2}}}&=\lim _{h\to 0}{\frac {1-2\cdot 0+(-1)}{h^{2}}}\\&=\lim _{h\to 0}{\frac {0}{h^{2}}}\\&=0\end{aligned}}