Symmetric derivative

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

\( \lim _{{h\to 0}}{\frac {f(x+h)-f(x-h)}{2h}} \) .[1][2]

The expression under the limit is sometimes called the symmetric difference quotient.[3][4] 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.[3]

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.[1][5]

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,

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

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,

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

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 \( {\frac {|2|-|-1|}{2-(-1)}}={\frac {1}{3}} \) .[6]

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.[6]

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

\( f_{s}(x)\leq {\frac {f(b)-f(a)}{b-a}}\leq f_{s}(y). \) [6][7]

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):

\( f_{s}(z)={\frac {f(b)-f(a)}{b-a}}. \) [6]

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.[6]

Generalizations

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

It is defined as

\( \lim _{{h\to 0}}{\frac {f(x+h)-2f(x)+f(x-h)}{h^{2}}}. \) [2][8]

If the (usual) second derivative exists, then the second symmetric derivative equals it.[8] 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 {\displaystyle x=0} x=0:

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