# 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”

1. Are you trying to imply that a desire for high numerical precision is bad?

1. I suppose it’s just the irony of the situation 🙂

2. mathwonk says:

But $2\cdot 2 = 5$ for large values of 2 and 5, right?

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