- Suppose f is an entire function with f = 1. Let
Use Jensen’s equation to establish the following inequality.
where n is the number of zeros of
f in B
- The version of the Blaschke product presented above is that found in most complex
variable texts. However, there is another one in [?]. Instead of
Prove a version of Theorem 56.5.1 using this modification.
- The Weierstrass approximation theorem holds for polynomials of n variables on any
compact subset of ℝn. Give a multidimensional version of the Müntz-Szasz theorem
which will generalize the Weierstrass approximation theorem for n dimensions. You
might just pick a compact subset of ℝn in which all components are positive.
You have to do something like this because otherwise, tλ might not be
- Show cos =
- Recall sin =
. Use this to derive Wallis product,
- The order of an entire function, f is defined as
If no such a exists, the function is said to be of infinite order. Show the
order of an entire function is also equal to limsupr→∞ where
- Suppose Ω is a simply connected region and let f be meromorphic on Ω. Suppose
also that the set, S ≡ has a limit point in Ω
. Can you conclude
c for all z ∈ Ω?
- This and the next collection of problems are dealing with the gamma function.
with the convergence uniform on compact sets.
- ↑ Show ∏
converges to an analytic function on ℂ which has zeros
only at the negative integers and that therefore,
is a meromorphic function (Analytic except for poles) having simple poles at the
- ↑Show there exists γ such that if
then Γ = 1
. Thus Γ is a meromorphic function having simple poles at the
negative integers. Hint: ∏
e−1∕n = c = eγ.
- ↑Now show that
- ↑Justify the following argument leading to Gauss’s formula
- ↑ Verify from the Gauss formula above that Γ = Γ
z and that for n a
nonnegative integer, Γ =
- ↑ The usual definition of the gamma function for positive x is
n ≤ e−t for t ∈
. Then show
Use the first part to conclude that
Hint: To show
n ≤ e−t for t ∈
, verify this is equivalent to showing
n ≤ e−nu for u ∈.
- ↑Show Γ =
0∞e−ttz−1dt. whenever Rez > 0. Hint: You have already shown
that this is true for positive real numbers. Verify this formula for Rez > 0 yields an
- ↑Show Γ =
. Then find Γ
- Show that ∫
ds = .
Hint: Denote this integral by I and observe that
I2 = ∫
dxdy. Then change variables to polar coordinates,
x = r cos,
y = r sinθ.
- ↑ Now that you know what the gamma function is, consider in the formula for
Γ the following change of variables.
t = α + α1∕2s. Then in terms of the new
variable, s, the formula for Γ is
Show the integrand converges to e−
. Show that then
Hint: You will need to obtain a dominating function for the integral so that you
can use the dominated convergence theorem. You might try considering
s ∈ first and consider something like
on this interval. Then
look for another function for s > . This formula is known as Stirling’s
- This and the next several problems develop the zeta function and give a
relation between the zeta and the gamma function. Define for 0 < r < 2π
Show that Ir is an entire function. The reason 0 < r < 2π is that this prevents
− 1 from equaling zero. The above is just a precise description of the contour
dw where γ is the contour shown below.
in which on the integrals along the real line, the argument is different in going from
r to ∞ than it is in going from ∞ to r. Now I have not defined such contour
integrals over contours which have infinite length and so have chosen to
simply write out explicitly what is involved. You have to work with these
integrals given above anyway but the contour integral just mentioned is the
motivation for them. Hint: You may want to use convergence theorems from
real analysis if it makes this more convenient but you might not have
- ↑In the context of Problem 19 define for small δ > 0
where γrδ is shown below.
Show that limδ→0Irδ =
. Hint: Use the dominated convergence theorem if
it makes this go easier. This is not a hard problem if you use these theorems but
you can probably do it without them with more work.
- ↑ In the context of Problem 20 show that for r1 < r, Irδ
−Ir1δ is a contour
where the oriented contour is shown below.
In this contour integral, wz−1 denotes e
where log = ln
. Explain why this integral equals zero. From Problem 20
it follows that Ir = Ir1. Therefore, you can define an entire function,
≡ Ir for all
r positive but sufficiently small. Hint: Remember the
Cauchy integral formula for analytic functions defined on simply connected
regions. You could argue there is a simply connected region containing
- ↑ In case Rez > 1, you can get an interesting formula for I by taking the limit
r → 0. Recall that and now it is desired to take a limit in the case where Rez > 1. Show the first
integral above converges to 0 as r → 0. Next argue the sum of the two last integrals