Surface states

In mathematics, the symmetric derivative is an operation related to the ordinary derivative.

It is defined as:

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

A function is symmetrically differentiable at a point x if its symmetric derivative exists at that point. It can be shown that if a function is differentiable at a point, it is also symmetrically differentiable, but the converse is not true. The best known 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. It can also be shown that the symmetric derivative at a point is the mean of the one-sided derivatives at that point, if they both exist.

Examples

1. The modulus function,$f(x)=\left|x\right|$
For absolute value function, or the modulus function, we have, at $x=0$,

$\begin{array}{c}\\ f_s(0)=\underset{h\to 0}{lim}\frac{f(0+h)-f(0-h)}{2h}\\ \\ f_s(0)=\underset{h\to 0}{lim}\frac{f(h)-f(-h)}{2h}\\ \\ f_s(0)=\underset{h\to 0}{lim}\frac{\left|h\right|-|-h|}{2h}\\ \\ f_s(0)=\underset{h\to 0}{lim}\frac{h-(-(-h))}{2h}\\ \\ f_s(0)=0\end{array}$

only, where remember that $h>0$ and $h⟶0$, and hence $|-h|$ 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 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.

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

$\begin{array}{c}\\ f_s(0)=\underset{h\to 0}{lim}\frac{f(0+h)-f(0-h)}{2h}\\ \\ f_s(0)=\underset{h\to 0}{lim}\frac{f(h)-f(-h)}{2h}\\ \\ f_s(0)=\underset{h\to 0}{lim}\frac{1/h^2-1/(-h{)}^2}{2h}\\ \\ f_s(0)=\underset{h\to 0}{lim}\frac{1/h^2-1/h^2}{2h}\\ \\ f_s(0)=0\end{array}$

only, where again, $h>0$ and $h⟶0$. 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 $x=0$ (i.e. essential discontinuity).

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.

3. The Dirichlet function, defined as:

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

may be analysed to realize that it has symmetric derivatives $\mathrm{\forall }x\in \mathbb{Q}$ but not $\mathrm{\forall }x\in ℝ-\mathbb{Q}$, i.e. symmetric derivative exists for rational numbers bur not for irrational numbers.

See also

• Symmetrically continuous function

References

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External links

• Approximating the Derivative by the Symmetric Difference Quotient (Wolfram Demonstrations Project)
• Dirichlet Function
• Dirichlet function and its modifications
• Dirichlet function-Wolfram Alpha