Approximation Theorem & Linear Interpolation Explained

Let $f (x)$ be a function having at least $n$ derivatives f', f'', \dots, f^{(n)} in an interval. Let $x_{0}$ be a point of this interval. Let $f (x)$ have a zero of multiplicity $n$ of $f (x)$ if $f (x_{0}) = 0$ , f'(x_0) = 0, $\dots$ , $f^{(n - 1)} (x_{0}) = 0$ , but $f^{(n)} (x_{0}) \neq 0$ . Thus if

f(x_0) = 0 = f'(x_0) = \dots = f^{(n-1)}(x_0),

we can say that $x_{0}$ is a zero of $f (x)$ of multiplicity at least $n$ .

If $f (x)$ has zeros (of multiplicity $\geq 1$ ) at $x_{1}$ and $x_{2}$ , then by Rolle's Theorem f'(x) has a zero (of multiplicity at least $1$ ) between $x_{1}$ and $x_{2}$ . If $f (x)$ has zeros of multiplicity at least $2$ at $x_{1}$ and of multiplicity at least $1$ at $x_{2}$ and, say, $x_{1} < x_{2}$ , then f'(x) has a zero of multiplicity at least $1$ at $x_{1}$ and a zero $t$ of multiplicity at least $1$ such that $x_{1} < t < x_{2}$ . Then f''(x) has a zero of multiplicity at least $1$ between $x_{1}$ and $t$ . Likewise, if $f (x)$ has zeros at $x_{1}, x_{2}, x_{3}$ , and $x_{1} < x_{2} < x_{3}$ , then f'(x) has a zero between $x_{1}$ and $x_{2}$ and one between $x_{2}$ and $x_{3}$ . Therefore f''(x) has a zero between these zeros of f'(x). Now let us make a general statement.

Let $f (x)$ have $n$ zeros, counting multiplicities, in an interval; we show that f'(x) has at least $(n - 1)$ zeros in the interval, counting multiplicities. If the zeros of $f (x)$ are $x_{1} < x_{2} < \dots < x_{r}$ , with respective multiplicities $m_{1}, m_{2}, \dots, m_{r}, m_{1} + m_{2} + \dots + m_{r} = n$ , then $x_{1}$ is a zero of f'(x) of multiplicity $m_{1} - 1$ , $x_{2}$ a zero of multiplicity $m_{2} - 1$ , and so on. (Zeros of multiplicity $0$ are simply not zeros.) Moreover, f'(x) has a zero between $x_{1}$ and $x_{2}$ , one between $x_{2}$ and $x_{3}$ , etc., giving at least $r - 1$ more zeros. The total number of zeros, counting multiplicities, of f'(x) in the interval is thus at least

\begin{aligned} (m_1 - 1) &+ (m_2 - 1) + \dots + (m_r - 1) + (r - 1)\\ &= (m_1 + m_2 + \dots + m_r) - r + (r - 1) \\ &=m_1+m_2+\cdots+m_r-1\\ &= n - 1 \cdot \end{aligned}

a key theorem which will lead us to several interesting forms of approximations.

Theorem 1

Suppose that $f (x)$ has zeros at $x_{1}, x_{2}, \dots, x_{r}$ , of respective multiplicities at least $m_{1}, m_{2}, \dots, m_{r}$ . Let $x_{0}$ be such that $f (x)$ has $n = m_{1} + m_{2} + \dots + m_{r}$ derivatives in the smallest interval containing $x_{0}, x_{1}, x_{2}, \dots, x_{r}$ . Let

H (x) = (x - x_{1})^{m_{1}} (x - x_{2})^{m_{2}} \dots (x - x_{r})^{m_{r}} .

Then there is a $ξ$ in this interval such that

f (x_{0}) = \frac{H (x_{0})}{n!} f^{(n)} (ξ) \cdot

PROOF. Suppose first that $x_{0}$ is one of $x_{1}, x_{2}, \dots, x_{r}$ . Then $f (x_{0}) = 0 = H (x_{0})$ , and we can take $ξ$ to be any point of the interval. It remains to consider the case where $x_{0}$ is different from all of $x_{1}, x_{2}, \dots, x_{r}$ . Then consider

g (x) = f (x) - C H (x),

where $C$ is a constant such that $g (x_{0}) = 0$ . Note that to be able to choose such a $C$ requires that we can solve for $C$ in $f (x_{0}) - C H (x_{0}) = 0$ , or simply that $H (x_{0}) \neq 0$ , which we know to be the case. Of course we could write down explicitly what $C$ is, but this is not essential.

Now $g (x)$ has $x_{1}$ as a zero of multiplicity at least $m_{1}$ , $x_{2}$ as a zero of multiplicity at least $m_{2}, \dots, x_{r}$ as a zero of multiplicity at least $m_{r}$ , and $x_{0}$ is also a zero of $g (x)$ . Thus $g (x)$ has at least $n + 1$ zeros, counting multiplicities, in our interval. Consequently there is a $ξ$ in this interval such that $g^{(n)} (ξ) = 0$ . But $g^{(n)} (x) = f^{(n)} (x) - C H^{(n)} (x)$ . Observe that when we expand $H (x)$ , we get a polynomial with highest term $x^{n}$ and other terms with lower exponents. But when we take $n$ derivatives, all these other terms become $0$ and $(x^{n})^{(n)} = n! \cdot$ Thus

g^{(n)} (x) = f^{(n)} (x) - C \cdot n!,

0 = g^{(n)} (ξ) = f^{(n)} (ξ) - C \cdot n!,

and

C = \frac{f^{(n)} (ξ)}{n!} \cdot

From the fact that $g (x_{0}) = 0 = f (x_{0}) - C H (x_{0})$ , we see that

f (x_{0}) = C H (x_{0}) = \frac{f^{(n)} (ξ)}{n!} H (x_{0}),

which was to be proved.

As a first application, let us consider the problem of interpolating from tables, for instance from logarithm tables. If we have tables for the function $f (x)$ and we read off $f (x_{1})$ and $f (x_{2})$ for $x_{1} < x_{2}$ , we want to estimate the error introduced by using the straight line through $(x_{1}, f (x_{1}))$ and $(x_{2}, f (x_{2}))$ in place of the curve for finding $f (x)$ when $x_{1} < x < x_{2}$ . Let $y = a x + b$ be the approximating line. Then we consider the function $f (x) - (a x + b)$ . This has zeros of multiplicity at least $1$ at $x_{1}$ and $x_{2}$ . Therefore we have $n = 2$ , and

H (x) = (x - x_{1}) (x - x_{2}) \cdot

By our theorem,

f(x) - (ax + b) = \frac{(x-x_1)(x-x_2)}{2!} f''(\xi)

for some $ξ$ between $x_{1}$ and $x_{2}$ . Now $H (x) \leq 0$ , since $x - x_{1}$ and $x - x_{2}$ have opposite signs, and the difference between $f (x)$ and $a x + b$ will be, in absolute value, less than or equal to the absolute value of $H (x)$ at its minimum times the largest possible absolute value for f''(x) in the interval, divided by $2$ . But H'(x) = (x - x_1) + (x - x_2), which is $0$ at the midpoint $x = \frac{1}{2} (x_{1} + x_{2})$ . This is the minimum for $H (x)$ , and the value of $H (x)$ there is

\frac{(x_{2} - x_{1}) (x_{1} - x_{2})}{4} \cdot

Thus

|f(x) - (ax + b)| \le \frac{(x_2 - x_1)^2}{8} \cdot \text{Max } |f''(\xi)| \cdot \tag{*}

For a table of logarithms, with $x_{1} = 1.01, x_{2} = 1.02, the error is at most$

\frac{(.01)^{2}}{8} \cdot \underset{1.01 \leq ξ \leq 1.02}{Max} (\frac{1}{ξ^{2}}) = \frac{.0001}{8} \cdot \frac{1}{1.0201} < .00002 \cdot

EXERCISES

Exercise 1.

Give a rule of the form (*) for extrapolation (where $x$ lies outside the interval $[x_{1}, x_{2}]$ ).

Exercise 2.

$\log 70.0 = 4.24850$ ; $\log 80.0 = 4.38203$ . Extrapolate to get a value for $\log 82.0$ and estimate your error.

Exercise 3.

$\log \tan 1.00 = 0.1924$ ; $\log \tan 1.10 = 0.2933$ . Estimate the error in interpolating for any angle between 1.100 and 1.10.

An Approximation Theorem. Linear Interpolation

EXERCISES

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