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.
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