1549901369-Elements_of_Real_Analysis__Denlinger_

338 Chapter 6 • Differentiable Functions

Theorem 6.5.15 (nth Derivative Test for Maxima/Minima) Suppose
that n :'.'.: 2 and f, J', J", · · · , f(n-l) all exist in some neighborhood of a, f'(a) =
J"(a) = · · · = f(n-l)(a) = 0, and f(n)(a) exists but f(n)(a)-/-0.

(a) If n is even and f(n)(a) > 0, then f has a local minimum at a.

(b) If n is even and f(nl(a) < 0, then f has a local maximum at a.

(c) If n is odd, then f has neither local maximum nor local minimum at a.

Proof. See Exercise 18. •

*Theorem 6.5.16 (Irrationality of e) e is an irrational number.

a

Proof. For contradiction, suppose e is rational; say e = b' where a, b EN.

Let n be an arbitrary integer greater than b. By Taylor's theorem [see Example
6.5.13] we h ave

n 1 eCn

el= L k! + (n + 1)!

k=O

for some Cn E (0, 1). Multiplying both sides of this equation by n!, we h ave

a [ 1 1 1 1 1 ] n!ecn

n!b = n! O! +If+ 2! + 3! + ... + n! + (n + 1)!

= [ n!+n!+-+-+···+- +--.

n! n! n !] ecn

2! 3! n! n + 1

a

Since n > b, n! b is an integer. Also, the expression in square brackets in

the above equ ation is a n integer. Therefore, ~ must b e an integer. But,
n+l
eCn e
since Cn E (0, 1), 1 < ecn < e. Thus, 0 < --< --, so by the squeeze
n+l n+l

principle, Jim ~ = 0. Thus, we can take n sufficiently large so that
n->oo n + 1

eCn

0 <--1<1.

n +

This contradicts t he assertion that ~ is an integer. Therefore, e must
n+l
be an irrational number. •