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Material Type: Assignment; Class: Complex Analysis; Subject: Mathematics; University: University of California - Los Angeles; Term: Unknown 1989;
Typology: Assignments
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The reciprocal 1/Γ of the Gamma function is a holomorphic function having zeros at the negative integers. The purpose of this exercise is to introduce an alternative formula for this holomorphic function using a Weierstrass product.
γ = lim n→∞
( (^) n ∑
k=
k
) − log n
exists.
∏^ ∞
n=
( 1 +
z n
) e−^
z n
converges uniformly on compact sets on the entire complex plane.
1 Γ(z)
= zeγz
∏^ ∞
n=
( 1 +
z n
) e−^
z n
We prove some estimates on Γ.
|Γ(x + iy)| ≤ Cn,x|y|−n
(Hint: Cauchy’s integral formula and induction on n. )
e
)n ≤ Γ(n + 1) ≤ 2 πn
( (^) n
e
)n
Γ(n + 1) = (2πn)
(^12) ( n e
)n (1 + cn)
where cn → 0 as n → ∞. Hint: Prove Γ(n) = nne−n
∫ (^) ∞
−∞
e−n(e
y (^) − 1 −y) dy
and discuss the integral. You may use
∫ (^) ∞ −∞ e
−x^2 dx = √ 2 π.
Given an entire function f we shall consider the problem of finding a (natural) holo- morphic function F (z) which satisfies F (n) = f (n)(0). Let f denote an entire function which satisfies
|f (x + iy)| ≤ Cn(1 + |x|)−n
for all integers n, all x ∈ R and all |y| < 2. (For example P (z)e−z 2 for any polynomial P is such a function.) Consider the holomorphic branch of the logarithm defined on C \ {t : t ≤ 0 } which satisfies log(1) = 0 and define tz^ = ez^ log^ t. Consider
F+(f, z) :=
∫ (^) ∞
−∞
(x + i)z^ f (x + i) dx
F−(f, z) :=
∫ (^) ∞
−∞
(x − i)z^ f (x − i) dx
F (z) =
Γ(z + 1) 2 πi
(F−(f, − 1 − z) − F+(f, − 1 − z))
Prove: 1) F is a holomorphic function. 2) F (n) = f (n)(0) for all integers n.
Let r be a small real number. Consider the closed curve that starts at 1 − r in the complex plane, goes on a circle of radius r counter- clockwise about 1 back to 1 − r, then goes along a straight line to r , then counterclockwise on a circle of radius r about 0, back on a straight line to 1 − r, and then repeats the procedure but going clockwise about each of the points 1 and 0.