4 D . J. NEWMA N

being similar. Writin g then,

(1.12) x = cos(0/2), 0 0 7T,

we find the "Fourier

expansion5*

of ** = (cos(0/2)) * is given directly b y th e binomial ex-

pansion oi(e

iej2

+ e~

ie,2)kf2

an d the result is

We may immediately rea d off ou r upper bound. Sinc e cos/0 = T 2£x) i t follow s tha t

A0 n/2

while

||JC*-P(JC)|| Z A

l%

fn/2

which, by its very definition , is Pk

n

.

The lower bound, on the other hand, is something of a struggle and we need a few

lemmas.

LEMMA

1.1 . Suppose that F(6) has the Fourier Series a0/2 + 2J11af- co s JO where aj

is nonnegative and nonincreasing. Suppose also that P(6) is an nth degree trigonometric poly-

nomial, and that v is a positive integer Then sup0 \F(0) - P(d)\ (v + l)a u+n/4.

PROOF.

Conside r the integral

(1.14) J = i - F (F{e) - P(0))e*

n + 1

*(l + e

ie

+ e

2id

+ • • • + e

{v~l )W )2

dO.

2nJ~n

On the one hand, since P(8) has degree n we see that

fn

P(0)e

i(n +

'

)d

(\ + e

ie

+ • • •

)2

dd = 0

so that

(1.15) J = i- r F(d)e* n + ' )0 (1+ e " + • •• + e ("-' ''9)2 49,

and by using the explicitly given Fourie r Series for F(x), we may read off fro m (1.15) th e

expression

(1.16)

Furthermore sinc e we are assuming that th e a} are nonnegative an d nonincreasing thi s

results in

(1.17) 2J(\ + 2 + 3 + — 4 - ^n

+ ir

= (*

2

+ i0* n+l,/2.