1 The Numerator

Let us focus on $xf^{\prime}$. The first step is to compute $f^{\prime}(x)$.

$$ \begin{eqnarray} f^{\prime}(x) &=& \frac{d}{dx}\left( \sum\limits_{i=0}^{n} c_{i} x^{i} \right) \\ &=& \frac{d}{dx}\left( c_{0} x^{0} + \sum\limits_{i=1}^{n} c_{i} x^{i} \right) \\ &=& \frac{d}{dx}\left( c_{0} x^{0} \right) + \frac{d}{dx}\left( \sum\limits_{i=1}^{n} c_{i} x^{i} \right) \\ &=& \frac{d}{dx}\left( c_{0} \right) + \frac{d}{dx}\left( \sum\limits_{i=1}^{n} c_{i} x^{i} \right) \\ &=& \frac{d}{dx}\left( \sum\limits_{i=1}^{n} c_{i} x^{i} \right) \\ &=& \sum\limits_{i=1}^{n} \frac{d}{dx}\left(c_{i} x^{i} \right) \\ &=& \sum\limits_{i=1}^{n} i c_{i} x^{i-1} \\ \end{eqnarray} $$

The rest of the numerator can be computed by multiplying through by $x$.

$$ \begin{eqnarray} xf^{\prime}(x) &=& x\sum\limits_{i=1}^{n} i c_{i} x^{i-1} \\ &=& \sum\limits_{i=1}^{n} x^{1} i c_{i} x^{i-1} \\ &=& \sum\limits_{i=1}^{n} i c_{i} x^{i-1} x^{1} \\ &=& \sum\limits_{i=1}^{n} i c_{i} x^{i} \\ \end{eqnarray} $$

2 Plugging Everything In

$$ \begin{eqnarray} (cond f)(x) &=& \left | \frac{x f^{\prime}(x)}{f(x)} \right | \\ &=& \left| \frac{\sum\limits_{i=1}^{n} i c_{i} x^{i} } {\sum\limits_{i=0}^{n} c_{i} x^{i}} \right| \\ &=& \frac{\left|\sum\limits_{i=1}^{n} i c_{i} x^{i} \right|} {\left| \sum\limits_{i=0}^{n} c_{i} x^{i}\right|} \\ \end{eqnarray} $$

Now… how should we continue? How about adjusting the absolute values?

$$ \begin{eqnarray} (cond f)(x) &=& \frac{\left|\sum\limits_{i=1}^{n} i c_{i} x^{i} \right|} {\left| \sum\limits_{i=0}^{n} c_{i} x^{i}\right|} &\to& \frac{\sum\limits_{i=1}^{n} i \left|c_{i} x^{i} \right|} { \sum\limits_{i=0}^{n} \left|c_{i} x^{i}\right|} \\ \end{eqnarray} $$

Both the numerateor and denominator have been replaced with upper bounds. We need to be able to argue that the numerator and denominator were bounded by the “same amount” and that the expression has not been fundamentally changed.

1. The primary difference between the numerator and the denominator is the constant term. The denominator includes a $c_0$ term and the numerator does not.

2. The condition number of a monomial (i.e. $\alpha x^n$) is the degree of that monomial (i.e., $n$).

Let us continue with the condition number.

$$ \begin{eqnarray} (cond f)(x) &=& \frac{\sum\limits_{i=1}^{n} i \left|c_{i} x^{i} \right|} { \sum\limits_{i=0}^{n} \left|c_{i} x^{i}\right|} \\ &=& \frac{\sum\limits_{i=1}^{n} i \left|c_{i} x^{i} \right|} { \left| c_{0} \right| + \sum\limits_{i=1}^{n} \left|c_{i} x^{i}\right|} \\ \end{eqnarray} $$

3 Analyzing Behavior

Before analyzing the conditon number… let us expand the summations.

$$ \begin{eqnarray} (cond f)(x) &=& \frac{\sum\limits_{i=1}^{n} i \left|c_{i} x^{i} \right|} { \left| c_{0} \right| + \sum\limits_{i=1}^{n} \left|c_{i} x^{i}\right|} \\ &=& \frac{\left|c_{1} x \right| + 2\left|c_{2} x^2 \right| + 3\left|c_{3} x^3 \right| + \cdots n\left|c_{n} x^n \right|} {\left|c_{0}\right| + \left|c_{1} x \right| + \left|c_{2} x^2 \right| + \left|c_{3} x^3 \right| + \cdots \left|c_{n} x^n \right|} \end{eqnarray} $$

3.1 Large Values of x

Taking the limit as $x \to \infty$…

$$ \begin{eqnarray} \lim_{x \to \infty} \frac{\left|c_{1} x \right| + 2\left|c_{2} x^2 \right| + 3\left|c_{3} x^3 \right| + \cdots n\left|c_{n} x^n \right|} {\left|c_{0}\right| + \left|c_{1} x \right| + \left|c_{2} x^2 \right| + \left|c_{3} x^3 \right| + \cdots \left|c_{n} x^n \right|} &\to& \lim_{x \to \infty} \frac{n\left|c_{n} x^n \right|} {\left|c_{n} x^n \right|} &\to& n \end{eqnarray} $$

For large values of x… the condition number is the degree of the largest term, i.e., $n$. Note that the conditioning of a polynomial is the same for $x \to -\infty$.

3.2 Small Values of x

Taking the limit as $x \to 0$…

$$ \begin{eqnarray} \lim_{x \to 0} \frac{\left|c_{1} x \right| + 2\left|c_{2} x^2 \right| + 3\left|c_{3} x^3 \right| + \cdots n\left|c_{n} x^n \right|} {\left|c_{0}\right| + \left|c_{1} x \right| + \left|c_{2} x^2 \right| + \left|c_{3} x^3 \right| + \cdots \left|c_{n} x^n \right|} &\to& \lim_{x \to 0} \frac{\left| c_1 x \right|} {\left|c_{0} \right| + \left| c_1 x \right|} &<& 1 \end{eqnarray} $$

Note: This result assumes that $|c_0| \geq |c_1 x|$.