Advanced book on Mathematics Olympiad

Algebra 411

−( 2 n+ 1 )x

dn

dxn

( 1 −x^2 )n−

(^12)
−( 2 n+ 1 )

(

n

1

)

d

dx

x

dn−^1

dxn−^1

( 1 −x^2 )n−

(^12)
=−( 2 n+ 1 )x
dn
dxn
( 1 −x^2 )n−
(^12)
−n( 2 n+ 1 )
dn−^1
dxn−^1
( 1 −x^2 )n−
(^12)
.
So iftn(x)denotes the right-hand side, then
tn+ 1 (x)=xtn(x)−
(− 1 )n−^1 n
1 · 3 ···( 2 n− 1 )
dn−^1
dxn−^1
( 1 −x^2 )n−^1 +
(^12)
.
Look at the second identity from the statement! If it were true, then the last term would
be equal to

√

1 −x^2 Un− 1 (x). This suggests a simultaneous proof by induction. Call the
right-hand side of the second identityun(x).
We will prove by induction onnthattn(x) = Tn(x)/

√

√^1 −x^2 andun−^1 (x) =
1 −x^2 Un− 1 ( 2 x). Let us assume that this holds true for allk<n. Using the induction
hypothesis, we have

tn(x)=x

Tn− 1 (x)

√

1 −x^2

−

√

1 −x^2 Un− 2 (x).

Using the first of the two identities proved in the first problem of this section, we obtain
tn(x)=Tn(x)/

√

1 −x^2.

For the second half of the problem we show that

√

1 −x^2 Un− 1 (x)andun− 1 (x)are
equal by verifying that their derivatives are equal, and that they are equal atx=1. The
latter is easy to check: whenx=1 both are equal to 0. The derivative of the first is

−x

√

1 −x^2

Un− 1 (x)+ 2

√

1 −x^2 Un′− 1 (x).

Using the inductive hypothesis, we obtainu′n− 1 (x)=−nTn(x)/

√

1 −x^2. Thus we are
left to prove that

−xUn− 1 (x)+ 2 ( 1 −x^2 )Un′− 1 (x)=−nTn(x),

which translates to

−cosx

sinnx

sinx

+2 sin^2 x

ncosnxsinx−cosxsinnx

sin^2 x

·

1

sinx

=ncosnx.

This is straightforward, and the induction is complete.

Remark.These are called the formulas of Rodrigues.

199.IfM=A+iB, thenMt=At−iBt=A−iB.So we should take

A=

1

2

(

M+Mt

)

and B=

1

2 i

(

M−Mt

)

,

which are of course both Hermitian.