Calculating 2*2 with high precision

The definition of derivative,

\displaystyle    f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h) - f(x)}{h}   \ \ \ \ \ \ \ \ \ \ (1)

is not such a great way to actually find the derivative numerically. Its symmetric version,

\displaystyle    f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h) - f(x-h)}{2h}   \ \ \ \ \ \ \ \ \ \ (2)

performs much better in computations. For example, consider the derivative {f(x)=e^x } at the point {x=1}. We know that {f'(1)=2.718281828\dots}. Numerically, with {h=0.001}, we get

\displaystyle    \frac{f(1+h) - f(1)}{h} \approx \mathbf{2.71}9 \dots

(error {>10^{-3}}) versus

\displaystyle    \frac{f(1+h) - f(1-h)}{2h} \approx \mathbf{2.71828}2 \dots

(error {<10^{-6}}).

Considering this, why don’t we ditch (1) altogether and adopt (2) as the definition of derivative? Just say that by definition,

\displaystyle    f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x-h)}{2h}

whenever the limit exists.

This expands the class of differentiable functions: for example, {f(x)=|x|} becomes differentiable with {f'(0)=0}. Which looks more like a feature than a bug: after all, {f} has a minimum at {0}, and the horizontal line through the minimum is the closest thing to the tangent line that it has.

Another example: the function

\displaystyle     f(x) = \begin{cases} x ,\quad & x\le 0 \\ 3x,\quad & x>0 \end{cases}

has {f'(0)=2} under this definition, because

\displaystyle    \lim_{h\rightarrow 0^+}\frac{f(x+h)-f(x-h)}{2h} = \lim_{h\rightarrow 0^+}\frac{3h-(-h)}{2h} = 2

and

\displaystyle    \lim_{h\rightarrow 0^-}\frac{f(x+h)-f(x-h)}{2h} = \lim_{h\rightarrow 0^-}\frac{h-3(-h)}{2h} = 2

This example also makes sense: since {f(x)=|x|+2x}, getting {f'(0)=0+2} is expected. In fact, with the new definition we still have basic derivative rules: if {f,g} are differentiable, then {f+g}, {f-g}, {fg}, {f/g} are also differentiable (with the usual caveat about {g\ne 0}) and the familiar formulas hold.

Let’s test the chain rule on the function {g = f\circ f}. The rule says

\displaystyle     g'(0) = f'(f(0)) f'(0)    \ \ \ \ \ \ \ \ \ \ (3)

Since {f(0)=0}, the product on the right is {2\cdot 2}. On the other hand,

\displaystyle     g(x) = \begin{cases} x ,\quad & x\le 0 \\ 9x,\quad & x>0 \end{cases}

which implies, by a computation similar to the above, that {g'(0)=5}. So, if we want to have the chain rule (3), we must accept that

\displaystyle     \mathbf{2\cdot 2 = 5}

This is where the desire for high numerical precision leads.

Plenty of other things go wrong with the symmetric definition:

  • Maximum or minimum of {f} may be attained where {f'} exists and is nonzero.
  • A differentiable function may be discontinuous.
  • Having {f'>0} everywhere does not imply that {f} is increasing.
  • Mean Value Theorem fails.

4 thoughts on “Calculating 2*2 with high precision”

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