Surface states

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

It is defined as:

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

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

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

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 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 $x=0$,

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

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 $x=0$ (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:

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

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 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