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Miklós Schweitzer
1958 Miklós Schweitzer
10
10
Part of
1958 Miklós Schweitzer
Problems
(1)
Miklós Schweitzer 1958- Problem 10
Source:
10/23/2015
10. Prove that the function
f
(
x
)
=
∫
−
∞
∞
(
sin
θ
θ
)
2
k
cos
(
2
x
θ
)
d
θ
f(x)= \int_{-\infty}^{\infty} \left (\frac{\sin\theta}{\theta} \right )^{2k}\cos (2x\theta) d\theta
f
(
x
)
=
∫
−
∞
∞
(
θ
s
i
n
θ
)
2
k
cos
(
2
x
θ
)
d
θ
where
k
k
k
is a positive integer, satisfies the following conditions:(i)
f
(
x
)
=
0
f(x)=0
f
(
x
)
=
0
if
∣
x
∣
≥
k
\mid x \mid \geq k
∣
x
∣≥
k
and
f
(
x
)
≥
0
f(x) \geq 0
f
(
x
)
≥
0
elsewhere; (ii) in interval
(
l
,
l
+
1
)
(l,l+1)
(
l
,
l
+
1
)
(
l
=
−
k
,
−
k
+
1
,
…
,
k
−
1
)
(l= -k, -k+1, \dots , k-1)
(
l
=
−
k
,
−
k
+
1
,
…
,
k
−
1
)
the function
f
(
x
)
f(x)
f
(
x
)
is a polynomial of degree
2
k
−
1
2k-1
2
k
−
1
at most. (R. 7)
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