But
q
is a nonzero polynomial of degree 2
n
+ 2 exactly. Therefore,
x
1
and
x
n
+2
have to be simple zeros and so
x
1
=

1 and
x
n
+1
= 1. Note that the
polynomial
p
(
x
) = (1

x
2
)[
e
0
n
(
x
)]
2
∈
P
2
n
+2
has the same zeros as
q
and so
p
=
cq
, for some constant
c
. Comparing the coefficient of the leading order
term of
p
and
q
it follows that
c
= (
n
+ 1)
2
.
Therefore,
e
n
satisfies the
ordinary differential equation
(1

x
2
)[
e
0
n
(
x
)]
2
= (
n
+ 1)
2
k
e
n
k
2
∞

e
2
n
(
x
)
.
(2.61)
We know
e
0
n
∈
P
n
and its n zeros are the interior points
x
2
, . . . , x
n
+1
. There
fore,
e
0
n
cannot change sign in [

1
, x
2
].
Suppose it is nonnegative for
x
∈
[

1
, x
2
] (we reach the same conclusion if we assume
e
0
n
(
x
)
≤
0) then, taking
square roots in (2.61) we get
e
0
n
(
x
)
p
k
e
n
k
2
∞

e
2
n
(
x
)
=
n
+ 1
√
1

x
2
,
for
x
∈
[

1
, x
2
]
.
(2.62)
Using the trigonometric substitution
x
= cos
θ
, we can integrate to obtain
e
n
(
x
) =
k
e
n
k
∞
cos[(
n
+ 1)
θ
]
,
(2.63)
for
x
= cos
θ
∈
[

1
, x
2
] with 0
< θ
≤
π
, where we have chosen the constant of
integration to be zero so that
e
n
(1) =
k
e
n
k
∞
. Recall that
e
n
is a polynomial
of degree
n
+ 1 then so is cos[(
n
+ 1) cos

1
x
]. Since these two polynomials
agree in [

1
, x
2
], (2.63) must also hold for all
x
in [

1
,
1].
2.4.
CHEBYSHEV POLYNOMIALS
33
Definition 5.
The Chebyshev polynomial (of the first kind) of degree
n
,
T
n
is defined by
T
n
(
x
) = cos
nθ,
x
= cos
θ,
0
≤
θ
≤
π.
(2.64)
Note that (2.64) only defines
T
n
for
x
∈
[

1
,
1].
However, once the
coefficients of this polynomial are determined we can define it for any real
(or complex)
x
.
Using the trigonometry identity
cos[(
n
+ 1)
θ
] + cos[(
n

1)
θ
] = 2 cos
nθ
cos
θ,
(2.65)
we immediately get
T
n
+1
(cos
θ
) +
T
n

1
(cos
θ
) = 2
T
n
(cos
θ
)
·
cos
θ
(2.66)
and going back to the
x
variable we obtain the recursion formula
T
0
(
x
) = 1
,
T
1
(
x
) =
x,
T
n
+1
(
x
) = 2
xT
n
(
x
)

T
n

1
(
x
)
,
n
≥
1
,
(2.67)
which makes it more evident the
T
n
for
n
= 0
,
1
, . . .
are indeed polynomials
of exactly degree
n
. Let us generate a few of them.
T
0
(
x
) = 1
,
T
1
(
x
) =
x,
T
2
(
x
) = 2
x
·
x

1 = 2
x
2

1
,
T
3
(
x
) = 2
x
·
(2
x
2

1)

x
= 4
x
3

3
x,
T
4
(
x
) = 2
x
(4
x
3

3
x
)

(2
x
2

1) = 8
x
4

8
x
2
+ 1
T
5
(
x
) = 2
x
(8
x
4

8
x
2
+ 1)

(4
x
3

3
x
) = 16
x
5

20
x
3
+ 5
x.
(2.68)
From these few Chebyshev polynomials, and from (2.67), we see that
T
n
(
x
) = 2
n

1
x
n
+ lower order terms
(2.69)
and that
T
n
is an even (odd) function of
x
if
n
is even (odd), i.e.
T
n
(

x
) = (

1)
n
T
n
(
x
)
.
(2.70)
34
CHAPTER 2.
FUNCTION APPROXIMATION
Going back to (2.63), since the leading order coefficient of
e
n
is 1 and that
of
T
n
+1
is 2
n
, it follows that
k
e
n
k
∞
= 2

n
. Therefore
p
*
n
(
x
) =
x
n
+1

1
2
n
T
n
+1
(
x
)
(2.71)
is the best uniform approximation of
x
n
+1
in [

1
,
1] by polynomials of degree
at most
n
. Equivalently, as noted in the beginning of this section, the monic
polynomial of degree
n
with smallest infinity norm in [

1
,
1] is
˜
T
n
(
x
) =
1
2
n

1
T
n
(
x
)
.
(2.72)
Hence, for any other monic polynomial
p
of degree
n
max
x
∈
[

1
,
1]

p
(
x
)

>
1
2
n

1
.
(2.73)
The zeros and extrema of
T
n
are easy to find. Because
T
n
(
x
) = cos
nθ
and 0
≤
θ
≤
π
, the zeros occur when
θ
is an odd multiple of
π/
2. Therefore,
¯
x
j
= cos
(2
j
+ 1)
n
π
2
j
= 0
, . . . , n

1
.
(2.74)
The extrema of
T
n
(the points
x
where
T
n
(
x
) =
±
1) correspond to
nθ
=
jπ
for
j
= 0
,
1
, . . . , n
, that is
x
j
= cos
jπ
n
,
j
= 0
,
1
, . . . , n.
(2.75)