vu uu t2 − vu ut2 + s 2 + r 2

Problem 10973

(American Mathematical Monthly, Vol.109, November 2002) Proposed by L. D. Servi (USA).

WithRk(n) defined as below, prove that limk→∞Rk(2)/Rk(3) = 3/2.

Rk(n) =

k square roots

z }| {

vu uu t2 −

vu ut2 +

s 2 +

r

2 + · · · + q

2 +√ n

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

Let f (x) =√

2 + x then for any starting point x ∈ (−2, 2) the iterates f^k(x) is an increasing sequence which converges to the attracting fixed point 2. Moreover

Rk(n) = q

2 − f^k−1(n − 2) and lim

k→∞

Rk(2) Rk(3) = lim

k→∞

s

2 − f^k(0) 2 − f^k(1). First note that

2 − f^k−1(x) = 4 − (f^k−1(x))²

2 + f^k−1(x) = 4 − (2 + f^k−2(x))²

2 + f^k−1(x) = 2 − f^k−2(x) (f^k(x))² and therefore going backward

2 − f^k−1(x) = 2 − f^k−2(x)

(f^k(x))² = 2 − f^k−3(x)

(f^k(x)f^k−1(x))² = · · · = 2 − f⁰(x)

(f^k(x)f^k−1(x) · · · f²(x))². Since 2f^′(x) = 1/√

2 + x = 1/f (x) then

2^k(f^k(x))^′f (x) = 1

f^k(x)f^k−1(x) · · · f²(x) and we obtain the precious identity

4 − (f^k(x))²= 2 − f^k−1(x) = 2 − x

(f^k(x)f^k−1(x) · · · f²(x))² = (4 − x²) · (2^k(f^k(x))^′)². Let yk(x) = f^k(x) and solve the following differential equation under the condition yk(2) = 2

p4 − (y^k(x))²=p

4 − x²· 2^ky^′k(x).

Then

2^k Z yk(x)

2

dy p4 − y² =

Z x 2

√ dx 4 − x², and integrating we find that

2^karccos yk(x) 2



= arccos x 2

.

So we have an explicit formula for the k-th iterate of f

f^k(x) = yk(x) = 2 cos arccos(^x₂) 2^k

 .

Now it is easy to compute our limit

k→∞lim s

2 − f^k(0) 2 − f^k(1) = lim

k→∞

vu uu t

1 − cos_π/2

2^k

 1 − cos

π/3 2^k

 = 3 2.

 Remark: for x ∈ [−2, 2] the map f(x) has an inverse

f⁻¹(x) = 2 T2(^x₂) for x ∈ [0, 2]

where T2(x) = 2x²− 1 is the Tchebycheff polynomial of degree 2.

It is well known that

T₂^k(x) = T₂^k(x) and Tn(x) = cos(n arccos(x)).

Hence

f^−k(x) = 2 T₂^k(^x₂) = 2 T2^k x 2

= 2 cos 2^karccos ^x₂

, and therefore we find again the formula

f^k(x) = 2 cos arccos(^x₂) 2^k

 .