The Art and Craft of Problem Solving

9.4 POWER SERIES AND EULERIAN MATHEMATICS 345

for all tED. Integrating this with respect to t from a to x yields^12

• M 3 (x-a)::; l

x

fill (t)dt ::;M 3 (x-a).

By the fundamental theorem of calculus, the integral of fill is f", so the above becomes

• M 3 (x-a)::; f"(x) -f"(a)::; M 3 (x-a).
  Now let us replace x with the variable t and integrate with respect to t once again from
  a to x. We have
  (x-af '
  ()

'

() "()(

(x-a)^2

-M 3 2 ::; f x - f a -f a x-a) ::; M 3 2

Repeat the process once more to get

(x-a)^3 , ,, (x-a)^2 (x-a)^3

-M 3

2. 3 ::;

f(x)-f(a)-f(a)(x-a )-f (a) 2 ::;M (^3 2). 3.
Finally, we conclude that
(x a)^2
f(x) = f(a) + J
'
(a)(x-a) + f"(a) �
plus an error term that is at most equal to
in absolute magnitude.
M
(x-a)^3
(^3 6)

The general method is simple: just keep integrating the inequality If(n+^1 ) (x) I ::;
Mn until you get the nth-degree Taylor polynomial. The general formula is

where

n (x a)i

f(x)=f(a)+ �f(i)(a) � + Rn +l,

1= 1 l.

(x_a)n+l

IRn+ll ::;Mn +l

(n+l)!

(8)

From this remainder formula it is clear that if the bounds on the derivatives are

reasonable (for example, Mk does not grow exponentially in k), then the power series

will converge. And that is an amazing thing. For example, consider the familiar series

x^3 x?

sinx = x - -+ - -...

3! 5! '

which converges (verify!) for all real x. Yet the coefficients for this "global" series
come only from knowledge about the value of sin x and its derivatives at x = O. In

Example 9.4.3

(^12) We are using the fact that u(x) ::; v(x) implies J: u(x)dx ::; J: v(x)dx.