Advanced book on Mathematics Olympiad

3.2 Continuity, Derivatives, and Integrals 139

Rolle’s theorem.Iff:[a, b]→Ris continuous on[a, b], differentiable on(a, b), and
satisfiesf(a)=f(b), then there existsc∈(a, b)such thatf′(c)= 0.

Its standard use was on problems like the following.

Example.Prove that the Legendre polynomial

Pn(x)=

dn

dxn

(x^2 − 1 )n

hasndistinct zeros in the interval(− 1 , 1 ).

Solution.Consider the polynomial functionQn(x)=(x^2 − 1 )n. Its zerosx=1 and
x=−1 have multiplicityn. Therefore, for everyk<n, thekth derivativeQ(k)n (x)has
1 and−1 as zeros. We prove by induction onkthat for 1<k≤n,Q(k)n (x)haskdistinct
zeros in(− 1 , 1 ).
By Rolle’s theorem this is true fork=1. Assume that the property is true fork<n,
and let us prove it fork+1. The polynomialQ(k)n (x)hask+2 zerosx 0 =− 1 <
x 1 <···xk<xk+ 1 =1. By Rolle’s theorem, between any two consecutive zeros of the
function there is a zero of the derivativeQ(kn+^1 )(x). HenceQ(kn+^1 )(x)hask+1 distinct
zeros between−1 and 1. This completes the induction.
In particular,Q(n)n (x)=Pn(x)hasndistinct zeros between−1 and 1, as desired. 

Rolle’s theorem applied to the functionφ:[a, b]→R,

φ(x)=

∣

∣∣

∣∣

∣

f (x) g(x) 1

f (a) g(a) 1

f (b) g(b) 1

∣

∣∣

∣∣

∣

,

yields the following theorem.

Cauchy’s theorem.Iff, g:[a, b]→Rare two functions, continuous on[a, b]and
differentiable on(a, b), then there exists a pointc∈(a, b)such that

(f (b)−f (a))g′(c)=(g(b)−g(a))f′(c).

In the particular caseg(x)=x, we have the following.

The mean value theorem (Lagrange).Iff:[a, b]→Ris a function that is continu-
ous on[a, b]and differentiable on(a, b), then there existsc∈(a, b)such that

f′(c)=

f(b)−f(a)

b−a

.

We use the mean value theorem to solve a problem of D. Andrica.