Prove that there exists a unique function f : [0, ∞)→[0,∞) such that

1.3.13. Prove that there exists a unique function f : [0, ∞)→[0,∞) such that

( f ◦ f )(x) = 6x − f (x) and f (x) > 0 , for all x >0 .

Solution. Fix x ≥ 0 and define the sequence (x n ) n ≥0 by x 0 = x and x n = f (x n −1 ), for all integers n ≥ 1. By hypothesis, we have x n ≥ 0 and

x n +2 +x n +1 − 6x n = ( f ◦ f )(x n ) + f (x n ) − 6x n = 0, for all n ≥ 0.

28 1 Sequences The roots of the associated characteristic equation r 2 + r − 6 = 0 are r 1 = 2 and

r 2 = −3. Thus, there exist real numbers a and b such that for all n ≥ 0,

(1.10) We intend to prove that b

n =a·2 + b · (−3) .

In this case, by (1.10), it follows that for any n sufficiently large (we choose n even if b < 0, resp. odd if b > 0) we have x n < 0, contradiction. So, b = 0 and hence

x 1 = f (x) = 2x 0 = 2x. Therefore the unique function satisfying our hypotheses is

f (x) = 2x. We recall that if (a n ) n ≥1 and (b n ) n ≥1 are sequences of positive real numbers, then

we use the notation

a n ∼b n as n →∞

if lim n → ∞ a n /b n = 1. ⊓ ⊔

be a sequence of real numbers such that e a n + na n =2 for all positive integers n . Compute lim n →∞ n (1 − na n ) .

1.3.14. Let (a n ) n ≥1

Teodora–Liliana R˘adulescu, Mathematical Reflections, No. 2 (2006) Solution. For any integer n ≥ 1, define f n (x) = e x + nx − 2. Then f n increases on

R, as a sum of increasing functions. So, by f n (0) = −1 < 0 and f n (ln 2) = n ln 2 > 0, we conclude that for all integers n ≥ 1, there exists a unique a n ∈ (0,ln2) such that

f n (a n ) = 0. Next, we observe that

f n +1 (a n )=e a n + na n −2+a n =f n (a n )+a n =a n >0. Since f n +1 (a n +1 ) = 0 and f n increases for any n ≥ 1, we deduce that a n >a n +1 , for

all n ≥ 1. So, there exists ℓ := lim n → ∞ a n ∈ [0,ln2). If ℓ > 0 then the recurrence rela- tion e a n + na n = 2 yields the contradiction +∞ = 2. Thus, ℓ = 0 and lim n → ∞ na n = 1, that is, a n ∼n −1 as n →∞. Using again the recurrence relation, we obtain

1 −1∼e /n − 1 as n→∞. Since e x − 1 = x + o(x) as x→0 we deduce that

It follows that lim n →∞ n (1 − na n ) = 1. ⊓ ⊔ With similar arguments as above (which are based on the asymptotic expansion of e x around x = 0) we can prove that lim n → ∞ n [n(1 − na n ) − 1] = 1/2 (exercise!).

In some cases, connections with trigonometry may be useful for computing lim- its. How can one solve the following problem without using trigonometric formulas?

1.3.15. Let −1 < a 0 <1 , and define recursively a n = [(1 + a n −1 )/2] 1 /2 , for all n >0 . Find the limits, as n →∞ , of b n =4 n (1 − a n ) and c n =a 1 a 2 ···a n .

M. Golomb, Amer. Math. Monthly, Problem E 2835

1.3 Recurrent Sequences 29 Solution. Let ϕ ∈ (0, π ) be such that cos ϕ =a 0 . Using the formula cos 2 x =

(1 + cos2x)/2 we find that a n = cos( ϕ /2 n ). Thus, since sin 2 x = (1 − cos2x)/2 and lim x → 0 sin x /x = 1, we obtain

=2 2n +1 sin 2 ϕ

ϕ 2 as n

2 n +1 −→ 2 →∞. Next, we observe that ϕ

b n =4 1 − cos

2 2 2 2 ···cos 2 n sin 2 n Hence

sin n c n = cos cos

sin ϕ

sin

→∞. ⊓ ⊔ Convergent linear combinations of the terms of a given sequence do not imply

as n

sin ( ϕ /2 n

that the sequence is convergent, too! We give below a complete response to a prob- lem of this type.

be a sequence of real numbers and −1 < α <1 such that the sequence (a n +1 + α a n ) n ≥1 is convergent. Prove that the sequence (a n ) n ≥1 con- verges, too.

1.3.16. Let (a n ) n ≥1

Solution. Set b n =a n +1 + α a n and ℓ := lim n → ∞ b n . We will prove that a n →ℓ/ (1 + α ) as n→∞. We first observe that if we set a ′ n =a n −ℓ/(1+ α ) and b ′ n =b n −ℓ, then the same recurrence relation is fulfilled (that is, b ′ =a ′

+ α a ′ ), but b n ′ n +1 n n →0 as n →∞. This shows that we can assume, without loss of generality, that ℓ = 0. It remains to prove that a n →0 as n→∞. Since the assertion is obvious for α = 0, we assume in what follows that α

n = (− α ) n x n . Then b n = (− α ) n +1 x n +1 + α (− α ) n x n and hence, for all n ≥ 1,

x n +1 −x n =− −

It follows that for all n ≥ 1,

n −1

x n =x 1 − ∑ − b

k =1

Since x n = (−1/ α ) n a n , we obtain, for all n ≥ 1,

n −1

a n = (− α ) n x 1 − (− α ) n ∑ − b .

k =1

But (− α ) n x α ) n ∑ 1 n →0 as n→∞. So, it remains to prove that (− −1 k =1 (−1/ α ) k b k →0 as n →∞. We first observe that by the Stolz–Ces`aro lemma,

∑ n −1 k =1 (−1/ α ) k b k

(−1/ α ) n b n

k =1 (−1/ α )

30 1 Sequences Hence

n −1

1 k −1 1 1 n

∑ − b k =o

(− α ) n

b k = (− α ) n

·o

= o(1) as n→∞.

k =1

This concludes the proof. ⊓ ⊔ Remarks. 1. The above statement is sharp, in the sense that the property does

not remain true if α

2. A “continuous” variant of the above “discrete” property is the following.

Assume that f is a real-valued continuously differentiable function on some int- erval [a, ∞) such that f ′ (x) + α f (x) tends to zero as x tends to infinity, for some α >0 . Prove that f (x) tends to zero as x tends to infinity.

A complete proof of this result will be given in Chapter 10.