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 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 [counter]example 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. Symmetric Derivative, however exists for the function at x=0.

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

$\begin{align} f_s(0)&= \lim_{h \to 0}\frac{f(0+h) - f(0-h)}{2h} \\ &= \lim_{h \to 0}\frac{f(h) - f(-h)}{2h} \\ &= \lim_{h \to 0}\frac{\left\vert h \right\vert - \left\vert -h \right\vert}{2h} \\ &= \lim_{h \to 0}\frac{h-(-(-h))}{2h} \\ &= 0 \\ \end{align}$

only, where remember that $h>0$ and $h\longrightarrow 0$, and hence $\left\vert -h \right\vert$ is equal to $-(-h)$ only! So, we observe that the symmetric derivative of the modulus function exists at $x=0$,and is equal to zero, even if its ordinary derivative won't 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 exits, but they are unequal (one is -1 and the other is 1); their average is 0, as expected.

x^-2

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

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

$\begin{align} f_s(0)&= \lim_{h \to 0}\frac{f(0+h) - f(0-h)}{2h} \\ &= \lim_{h \to 0}\frac{f(h) - f(-h)}{2h} \\ &= \lim_{h \to 0}\frac{1/h^2 - 1/(-h)^2}{2h} \\ &= \lim_{h \to 0}\frac{1/h^2-1/h^2}{2h} \\ &= 0 \\ \end{align}$

only, where again, $h>0$ and $h\longrightarrow 0$. See that again, for this function, its symmetric derivative exists at $x=0$, its ordinary derivative does not occur at $x=0$, due to discontinuity in the curve at $x=0$. Furthermore, neither the left nor the right derivatives are finite at 0, i.e. this is an essential discontinuity.

The Dirichlet function

The Dirichlet function, defined as:

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

may be analysed to realize that it has symmetric derivatives $\forall x \in \mathbb{Q}$ but not $\forall x \in \mathbb{R}-\mathbb{Q}$, i.e. 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 by in 1967 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 f_s(x) ≥ 0 and f_s(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, f_s(z) ≥ 0. Analogously, if f(b) < f(a), then there exists a point z in (a, b) where f_s(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 $\sgn(x)$ which is defined through

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

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 $x=0$:

$\begin{align} \lim_{h \to 0} \frac{\sgn(0+h) - 2\sgn(0) + \sgn(0-h)}{h^2} &= \lim_{h \to 0} \frac{1 - 2\cdot 0 + (-1)}{h^2} \\ &= \lim_{h \to 0} \frac{0}{h^2} \\ &= 0 \end{align}$

See also

• Central differencing scheme
• density point
• generalized derivative
• Generalizations of the derivative
• Symmetrically continuous function

References

1. Peter R. Mercer (2014). More Calculus of a Single Variable. Springer. p. 173. ISBN 978-1-4939-1926-0.
2. Thomson, p. 1
3. Peter D. Lax; Maria Shea Terrell (2013). Calculus With Applications. Springer. p. 213. ISBN 978-1-4614-7946-8.
4. Shirley O. Hockett; David Bock (2005). Barron's how to Prepare for the AP Calculus. Barron's Educational Series. p. 53. ISBN 978-0-7641-2382-5.
5. Thomson, p. 6
6. Prasanna Sahoo; Thomas Riedel (1998). Mean Value Theorems and Functional Equations. World Scientific. pp. 188–192. ISBN 978-981-02-3544-4.
7. Thomson, p. 7
8. A. Zygmund (2002). Trigonometric Series. Cambridge University Press. pp. 22–23. ISBN 978-0-521-89053-3.

• Thomson, Brian S. (1994). Symmetric Properties of Real Functions. Marcel Dekker. ISBN 0-8247-9230-0.
• A.B. Kharazishvili (2005). Strange Functions in Real Analysis, Second Edition. CRC Press. p. 34. ISBN 978-1-4200-3484-4.
• Aull, C.E.: "The first symmetric derivative". Am. Math. Mon. 74, 708–711 (1967)

External links

• Approximating the Derivative by the Symmetric Difference Quotient (Wolfram Demonstrations Project)