Fix an integer n ≥1 . (i) Prove that the equation

4.3.5. Fix an integer n ≥1 . (i) Prove that the equation

x n +x n −1 + ··· + x − 1 = 0

has a unique positive solution. Let a n denote this solution. Show that a n ≤1 . (ii) Prove that the sequence (a n ) n ≥1 is decreasing.

(iii) Show that

a n +1 n − 2a n +1=0 and deduce that lim n → ∞ a n = 1/2 .

(iv) Prove that

+o

as n →∞ .

4 ·2 n

150 4 Continuity Solution. (i) Set f n (x) = x n +x n −1 +···+x−1. Then f n is continuous, f n (0) =−1,

and f n (1) = n − 1 ≥ 0. Thus, by the intermediate value property, there exists a n ∈ (0, 1] such that f n (a n ) = 0. Since f n is strictly increasing on [0, ∞), it follows that

the equation x n +x n −1 + ··· + x − 1 = 0 has a unique positive solution a n . (ii) We have f n (a n +1 ) = −a n +1 n +1 < 0. Since f n increases in [0, ∞) and

0 =f n (a n )>f n (a n +1 ), we deduce that a n >a n +1 . (iii) We observe that

(x − 1)(x n +x n −1 n + ··· + x − 1) = x +1 − 2x + 1. Thus, a n +1 n − 2a n + 1 = 0. Therefore

1 a n +1

a n +1

0 <a n − = n < 2 →0 as n→∞,

because 0 <a 2 < 1. So, by the squeezing principle, we deduce that lim n → ∞ a n = 1/2.

(iv) Set b n =a n −1/2. Then ln(2b n ) = (n + 1) lna n . But ln a n = −ln2+ln(1+2b n ) and ln (1 + 2b n ) ∼ 2b n = o(1/n). We deduce that

ln (2b n ) = −(n + 1)ln2 + o(1) as n→∞,

4.3.6. Let f :R →R

be a surjective continuous function that takes any value at most twice. Prove that f is strictly monotone.

Solution. Fix a , b ∈ R with a < b and assume, without loss of generality, that

f (a) ≤ f (b). Since f is continuous, f ([a,b]) is a bounded interval, so there exist x 0 ,x 1 ∈ [a,b] such that

f (x 0 ) ≤ f (x) ≤ f (x 1 ) for all x ∈ [a,b]. We first claim that f (x 0 ) < f (x 1 ). Indeed, if f (x 0 ) = f (x 1 ), then f is constant in

[a, b], which contradicts our assumption that f takes any value at most twice. Thus,

f (x 0 ) < f (x 1 ). Next, we prove that f (x 0 ) ≥ f (a) and f (x 1 ) ≤ f (b). We prove only the first inequality, the second one being left to the reader as an exercise with similar argu- ments. Arguing by contradiction, we assume that f (x 0 ) < f (a). Then there exists x 2 ∈ R such that f (x 2 ) < f (x 0 ). Then either x 2 < a or x 2 > b. Let us assume that

x 2 < a. Since f has the intermediate value property, it follows that f takes a cer-

tain value at least three times on the interval [x 2 , b], a contradiction. This shows that

f (x 0 ) ≥ f (a). It follows that for any real numbers a < b and for all x ∈ [a,b], f (a) ≤ f (x) ≤

f (b). Thus, f is monotone. Moreover, since f −1 {y} has at most two elements for

4.4 Types of Discontinuities 151 all y ∈ R, we conclude that f cannot take the same value at different points; hence

f is strictly monotone. ⊓ ⊔