- MacTutor History of Mathematics

Solution: Day 3, problem 2

We will say that a polynomial with real coefficients splits if all its roots are real.
The first observation is that, if

h(x)

splits, then so does

({{d}\over{dx}} - a)h(x)

for any

a

in R.
To see this, suppose that

a

is non-zero (the case a = 0 can be done by a similar argument, or by continuity), and consider the polynomial

h_{1}(x) = h(x) - a^{-1} h'(x)

.
If

h

has degree

n

and

n

distinct roots

x_{1} < ... < x_{n}

, then

h

has alternating signs in the

n + 1

intervals into which these roots divide R, and has a maximum or minimum in each of these intervals (at infinity for the two end intervals). But

h_{1}

has the same asymptotics as

h

for

x

going to infinity (because it has the same degree and the same leading coefficient as

h

), and has the same value and hence the same sign as

h

at all maxima or minima of

h

, so it also has

n

sign changes and therefore splits.
If there are repeated roots, we modify the argument slightly (if

h

has an

r

- fold root at

x_{i}

with

r

> 1, then

h_{1}

has an

(r - 1)

- fold root there and at least one further sign change between

x_{i}

and the nearest maximum or minimum of

h

on the left or right, because

r - 1

has the opposite parity from

r

, so we still get as many real roots as we need) or argue by continuity.
The next step is: if

g(x)

and

h(x)

both split, then so does

g({{d}\over{dx}})(h(x))

. Indeed,

g(x)

can be written as

c (x - a_{1}) ... (x - a_{r})

with some non-zero real number

c

and real

a_{i}

, so the result follows from the previous one by induction on

r

. Now assume that

b_{n}x^{n} + b_{n-1}x^{n-1} + ... + b_{0}

splits and apply the assertion just proved to

g(x) = b_{n}x^{N-n} + b_{n-1}x^{N-n+1} + ... + b_{0}x^{N}

and

h(x) = x^{N}/N!

, which both split.