Download due March 22, 2006 - Harvard Math Department

Math 55b, Assignment #5, March 15, 2006
(due March 22, 2006)
Problem 1 (Expansions of Four Trigonometric Functions in Partial Fractions Derived From the Double Angle Formula for the Tangent and Cotangent Functions). Please precisely specify the meaning of convergence for the
series in your answer to this problem and rigorously justify your arguments
involving the manipulation and the convergence of series.
(a) Verify the following double angle formula for the cotangent function
2 cot 2x = cot x − tan x
from the double angle formulae for the sine and cosine functions.
(b) Iterate n times the following variation
µ
¶
π(x − 1)
1
πx
+ cot
cot
,
cot πx =
2
2
2
µ
¶
1
πx
π(x + 1)
cot πx =
cot
+ cot
2
2
2
of the formula of Part (a) to derive
πx
πx cot πx = n
2
Ã
n−1
¶
2X
−1 µ
π(x + k)
πx
π(x − k)
πx
cot
−
tan
+
cot
cot n +
2
2n
2n
2n
k=1
!
(c) Use
ξ
1
1
cot n =
n
n→∞ 2
2
ξ
lim
and Part (b) to derive the following expansion in partial fractions of the
cotangent function
∞
1 X
π cot πx = +
x n=1
µ
1
1
+
x+n x−n
¶
∞
X
1
1
.
= + 2x
2
x
x − n2
n=1
(d) Use Part (c) and
π cot
πx
πx
− 2π cot πx = π tan
2
2
1
.
to derive the following expansion in partial fractions of the tangent function
∞
πx X
4x
π tan
.
=
2
(2n + 1)2 − x2
n=0
(e) Use Part (d) and
1
x
=
2
sin x
to derive the following expansion in partial fractions of the cosecant function
∞
X
π
1
(−1)n−1
= + 2x
.
sin πx
x
n2 − x 2
n=1
cot x + tan
(f) Use Part (e) and
cos x = sin
³π
−x
´
2
to derive the following expansion in partial fractions of the secant function
∞
X
2n − 1
π
=4
.
(−1)n−1
2 − 4x2
cos πx
(2n
−
1)
n=1
Problem 2 (Euler’s Reflection Formula for the Gamma Function Derived
From the Expansion in Partial Fractions of the Cotangent Function). Use
the expansion in partial fractions of the cotangent function and
µ ¶
√
1
= π
Γ
2
and the comparison of
d
log (Γ(x)Γ (1 − x))
dx
and
d
π
log
dx
sin πx
to verify the following reflection formula of Euler for the Gamma function
π
.
Γ(x)Γ(1 − x) =
sin πx
Problem 3 (Infinite Product Expansion of the Sine Function). Use the expansion in partial fractions of the cotangent function to show
¶
Yµ
x2
sin πx = πx
1− 2
n
n∈N
for x ∈ R.
2
Problem 4 (Values of Riemann’s Zeta Function at Positive Even Integers Derived From the Power Series Expansion and the Expansion in Partial Fractions of the Cotangent Function ). The Bernoulli’s numbers Bk are defined
by the formula
∞
X
x
Bk xk
=
.
ex − 1 k=0 k!
They satisfy the recurrent relation
µ ¶
µ ¶
µ ¶
µ
¶
n
n
n
n
B0 +
B1 +
B2 + · · · +
Bn−1 = 0,
0
1
2
n−1
or the symbolic equation
(B + 1)n − B n = 0
when B k is interpreted as Bk . Use
eix + e−ix
cot x = i ix
e − e−ix
and the expansion in partial fractions of the cotangent function to show that
∞
2k
X
1
k−1 B2k (2π)
= (−1)
n2k
2(2k)!
n=1
for any positive integer k.
Problem 5 (Gauss’s Multiplication Theorem for the Gamma Function). Prove
that
¶ µ
¶
µ
¶
µ
n−1
1
2
n−1
1
Γ x+
···Γ x +
= (2π) 2 n 2 −nx Γ (nx)
Γ (x) Γ x +
n
n
n
for x > 0.
Hint: Use Stirling’s formula
lim
Γ(x + 1)
√
= 1.
2πx
x→∞ xx e−x
and the Euler formula
(m − 1)! my
m→∞ y(y + 1) · · · (y + m − 1)
Γ(y) = lim
3
for y = x + nk (0 ≤ k ≤ n − 1) and with m replaced by nm for y = nx to
verify that the quotient
¡
¢ ¡
¢
¡
¢
nnx Γ (x) Γ x + n1 Γ x + n2 · · · Γ x + n−1
n
Γ (nx)
is independent of x.
Problem 6 (Evaluation of Definite Integrals by Using the Beta Function and
Euler’s Reflection Formula for the Gamma Function). Use the formula
Z 1
Γ(x)Γ(y)
tx−1 (1 − t)y−1 dt
=
Γ(x + y)
0
for the Beta function and Euler’s Reflection Formula for the Gamma function
to show that
Z ∞ a−1
π
x
dx =
1+x
sin πa
0
for 0 < a < 1.
4