1 Condition Number Properties

Can one claim that $(cond f \times g)(x) \leq (cond \phantom{1} f)(x) + (cond \phantom{1} g)(x)$ for the general form of the condition number?

1.1 What is Known?

The general form of the condition number…

$$ (cond \phantom{1} \lambda)(x) = \left| \frac{x\lambda’(x)}{\lambda(x)} \right| $$

We know from the general form of the condition number that $\lambda(x) \neq 0$ and $x \neq 0$.

1.2 Notation and Pieces

Let us define $h(x)$.

$$ \begin{eqnarray} h(x) &=& f(x) \times g(x) \\ &=& f(x) g(x) \end{eqnarray} $$

To define $h’(x)$… we need to apply the product rule.

$$ \begin{eqnarray} h’(x) &=& f(x) g’(x) + f’(x) g(x) \\ \end{eqnarray} $$

1.3 Solving the Problem

Let us rewrite the problem… plug in $h(x)$… and $h’(x)$.

$$ \begin{eqnarray} (cond h)(x) &\leq& (cond \phantom{1} f)(x) + (cond \phantom{1} g)(x) \\ \left| \frac{xh’(x)}{h(x)} \right| & \leq & \left| \frac{xf’(x)}{f(x)} \right| + \left| \frac{xg’(x)}{g(x)} \right| \\ \end{eqnarray} $$

Now that we have plugged everything in… we need to pick a side. Let us focus on the left side. The right side must be left alone.

$$ \begin{eqnarray} (cond h)(x) &\leq& (cond \phantom{1} f)(x) + (cond \phantom{1} g)(x) \\ \left| \frac{xh’(x)}{h(x)} \right| & \leq & \left| \frac{xf’(x)}{f(x)} \right| + \left| \frac{xg’(x)}{g(x)} \right| \\ \left| x\left(\frac{f(x) g’(x) + f’(x)g(x)}{f(x)g(x)}\right) \right| &\leq& \ldots \\ \left| x\left(\frac{f(x) g’(x)}{f(x)g(x)} + \frac{f’(x)g(x)}{f(x)g(x)}\right) \right| &\leq& \ldots \\ \left| x\left(\frac{g’(x)}{g(x)} + \frac{f’(x)}{f(x)}\right) \right| &\leq& \ldots \\ \left| \frac{xg’(x)}{g(x)} + \frac{xf’(x)}{f(x)} \right| &\leq& \ldots \\ \left| \frac{xf’(x)}{f(x)} + \frac{xg’(x)}{g(x)} \right| &\leq& \left| \frac{xf’(x)}{f(x)} \right| + \left| \frac{xg’(x)}{g(x)} \right| \\ \end{eqnarray} $$

The claim holds by the triangle inequality (i.e., $|a + b| \leq |a| + |b|$).

Alternatively… we could split the result into cases…

Case 1: pre absolute value

$(cond \phantom{1}f)(x)$ and $(cond \phantom{1}g)(x)$ are both positive or both negative. That would mean $lhs = rhs$.

Case 2: pre absolute value

$(cond \phantom{1}f)(x)$ and $(cond \phantom{1}g)(x)$ have opposite signs. That would mean $lhs < rhs$.

But… this is just a more verbose statement of the triangle inequality.

2 Extending h(x) to n Functions

What if we made the same claim for 3 functions… or 4 functions… or $n$ functions? Would the claim still hold. Consider defining z(x) for 3 functions…

$$ z(x) = f(x) \times g(x) \times h(x) $$

The derivative of z(x) requires a recursive application of the product rule.

$$ \begin{eqnarray} z’(x) &=& f(x) g(x) h(x) \\ &=& \frac{d}{dx}\left(f(x)g(x)h(x)\right) \\ &=& f(x)\frac{d}{dx}\Big(g(x)h(x)\Big) + f’(x)g(x)h(x) \\ &=& f(x) \left(g(x) h’(x) + g’(x) h(x)\right) + f’(x)g(x)h(x) \\ &=& f(x) g(x) h’(x) + f(x) g’(x) h(x) + f’(x)g(x)h(x) \\ &=& f’(x) g(x) h(x) + f(x) g’(x) h(x) + f(x)g(x)h’(x) \\ \end{eqnarray} $$

After a little algebra… we have…

$$ \begin{eqnarray} \frac{z’(x)}{z(x)} &=& \frac{f’(x)}{f(x)} + \frac{g’(x)}{g(x)} + \frac{h’(x)}{h(x)} \end{eqnarray} $$

I will leave the algebra as an exercise to the reader (i.e. you).