cn in(0, 1) such that n Y k=1 f′(ck

Problem 11753

(American Mathematical Monthly, Vol.121, January 2014) Proposed by Prapanpong Pongsriiam (Thailand).

Letf be a continuous map from [0, 1] to R that is differentiable on (0, 1), with f (0) = 0 and f (1) = 1.

Show that for each positive integern there exist distinct numbers c1, . . . , cⁿ in(0, 1) such that

n

Y

k=1

f^′(c^k) = 1.

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

Assume that f is not identically equal to the identity map (otherwise the required property holds trivially), then |f (x) − x| attains a positive maximum value in some point t ∈ (0, 1).

Let us consider the set D = f^′((0, 1)). By the Mean Value Theorem,

b := f (t)

t =f (t) − f (0)

t − 0 ∈ D and a := 1 − f (t) − t

1 − t = 1 − t − (f (t) − t)

1 − t = f (1) − f (t) 1 − t ∈ D.

Note that b > 1 and a < 1 if f (t) > t, and b < 1 and a > 1 if f (t) > t. Hence by Darboux’s Theorem, there exists r > 0 such that (1 − r, 1 + r) ⊂ D. Moreover, for any y ∈ (1, 1 + r), we have that 1/y ∈ (1/(1 + r), 1) ⊂ (1 − r, 1).

Finally select m := ⌊n/2⌋ different numbers y1, y2, . . . , y^m ∈ (1, 1 + r) and let ym+k := 1/y^k for k = 1, . . . , m. Take also yⁿ = 1 if n is odd. Then y1, y2, . . . , yⁿ are n different values in D and therefore there exist n different numbers c1, c2, . . . , cⁿ∈ (0, 1), such that f^′(c^k) = y^k for k = 1, . . . , n

and _n

Y

k=1

f^′(c^k) =

m

Y

k=1

 y^k· 1

yk



= 1.