# Symmetric derivative

(Redirected from Symmetric difference quotient)

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

${\displaystyle \lim _{h\to 0}{\frac {f(x+h)-f(x-h)}{2h}}.}$[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, ${\displaystyle f(x)=\left\vert x\right\vert }$ , we have, at ${\displaystyle x=0}$ ,

{\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 ${\displaystyle h>0}$  we have ${\displaystyle \left\vert -h\right\vert }$  = ${\displaystyle -(-h)}$ . So, we observe that the symmetric derivative of the modulus function exists at ${\displaystyle x=0}$ , 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 ${\displaystyle x=0}$ ).

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 ${\displaystyle f(x)=1/x^{2}}$ , we have, at ${\displaystyle x=0}$ ,

{\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, ${\displaystyle h>0}$ . Again, for this function the symmetric derivative exists at ${\displaystyle x=0}$ , while its ordinary derivative does not exist at ${\displaystyle x=0}$ , 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

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

has symmetric derivatives ${\displaystyle \forall x\in \mathbb {Q} }$  but not ${\displaystyle \forall x\in \mathbb {R} -\mathbb {Q} }$ ; 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 ${\displaystyle {\frac {|2|-|-1|}{2-(-1)}}={\frac {1}{3}}}$ .[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

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

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

${\displaystyle \lim _{h\to 0}{\frac {f(x+h)-2f(x)+f(x-h)}{h^{2}}}.}$ [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 ${\displaystyle \operatorname {sgn}(x)}$  which is defined through

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

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

{\displaystyle {\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}}}

## Notes

1. ^ a b Peter R. Mercer (2014). More Calculus of a Single Variable. Springer. p. 173. ISBN 978-1-4939-1926-0.
2. ^ a b Thomson, p. 1
3. ^ a b 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. pp. 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 b A. Zygmund (2002). Trigonometric Series. Cambridge University Press. pp. 22–23. ISBN 978-0-521-89053-3.

## References

• 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)