Week 6 Worksheet - Solutions
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Cauchy's Integral Formula
- We have and calculate
as well as for all . The Cauchy-Riemann equations are only satisfed at so is not holomorphic on any domain whatsoever.
- Note that when . It just so happens that
when so let's take .
- Now
and, since is closed, the existence of an antiderivative for on leaves us with
to be evaluated. The term in brackets is the Cauchy integral formula applied to and .
- Since is entire it is represented by a power series
centered at 0 with infinite radius of convergence. Moreover, for every we have
where on .
Thus
and the estimation lemma, together with the hypothesis, gives
which, for , converges to zero as . Thus all but finitely many of the coefficients are equal to zero and is a polynomial.
Taylor's Theorem
- We can differentiate power series. By induction
and plugging in 0 gives .
- First we verify that
so that
with infinite radius of convergence.
- Write
whenever using the geometric series. The radius of convergence is .
- We can plug into the power series of the exponential function
giving a Taylor series with infinite radius of convergence.
- We know that is holomorphic on . Thus is holomorphic on . It therefore has a Taylor on .
First .
Now
from the chain rule. But we know
on from the geometric series.
Thus
on .
- We can write
whenever and using the geometric series. Thus the series is valid on .
- Whenever we will have a Taylor series for on . The largest ball centered at and entirely contained within is the ball of radius .
Applications
- Plug into the quadratic formula
to get
as the roots.
- If has degree one then
as desired. We proceed by induction. Fix a polynomial
of degree . By the fundamental theorem of algebra for some . Polynomial long division therefore lets us write
where is a polynomial of degree . Plugging in gives so . By induction
and therefore has the desired form.
- Suppose there is not equal to any . Then the function is defined on all of and holomorphic there. Moreover, as in the proof of the fundamental theorem of algebra, the function is bounded. Liouville's theorem then implies and therefore is constant. This is a contradiction, as no polynomial of degree at least 1 is constant.