The Extended Mean Value Theorem

Following the method of the last section, let $R$ be defined by the equation

$$f(b) - f(a) - (b - a)f'(a) - \frac{1}{2} (x - a)^2R = 0.$$ (13.7)

Let $F(x)$ be a function formed by replacing $b$ by $x$ in the left-hand member of (13.1); that is,

$$F(x) = f(x) - f(a) - (x - a) f'(a) - \frac{1}{2} (x - a)^2 R.$$ (13.8)

From (13.7), $F(b) = 0$; and from (13.8), $F(a) = 0$; therefore, by Rolle's Theorem, at least one value of $x$ between $a$ and $b$, say $x_1$ will cause $F'(x)$ to vanish. Hence, since

$$F'(x) = f'(x) - f'(a) - (x - a)R,$$

we get

$$F'(x_1) = f'(x_1) - f'(a) - (x_1 - a)R = 0.$$

Since $F'(x_1) = 0$ and $F'(a) = 0$, it is evident that $F'(x)$ also satisfies the conditions of Rolle's Theorem, so that its derivative, namely $F''(x)$, must vanish for at least one value of $x$ between $a$ and $x_1$, say $x_2$, and therefore $x_2$ also lies between $a$ and $b$. But $F''(x) = f''(x) - R$; therefore $F''(x_2) = f''(x_2) - R = 0$, and $R = f''(x_2)$. Substituting this result in (13.7), we get

$$f(b) = f(a) + (b - a)f'(a) + \frac{1}{2!} (b - a)^2 f''(x_2), \ \ \ \ a < x_2 < b.$$

In the same manner, if we define $S$ by means of the equation

$$f(b) -f(a) - (b - a) f'(a) - \frac{1}{2!} (b - a)^2 f''(a) - \frac{1}{3!}(b - a)^2 f''(a) S = 0,$$

we can derive the equation

$$\begin{array}{ll} f(b) = f(a) & + (b - a)f'(a) + \frac{1}{2!}... ...& + \frac{1}{3!} (b - a)^3 f'''(x_3), a < x_3 < b , \end{array}$$ (13.9)

where $x_3$ lies between $a$ and $b$. By continuing this process we get the general result,

$$\begin{array}{ll} f(b) = f(a) & + \frac{(b - a)}{1!}f'(a) ... ...{(b - a)^n}{n!}f^{(n)} (x_1), \ \ \ \ a < x_1 < b, \end{array}$$

where $x_1$ lies between $a$ and $b$. This equation is called the Extended Theorem of Mean Value^13.2, or Taylor's formula.