4.2 KiB
Math416 Lecture 10
Fast reload on Power Series
Suppose \sum_{n=0}^\infty a_n converges absolutely. (\sum_{n=0}^\infty |a_n|<\infty)
Then rearranging the terms of the series does not affect the sum of the series.
For any permutation \sigma of the set of positive integers, \sum_{n=0}^\infty a_{\sigma(n)}=\sum_{n=0}^\infty a_n.
Proof:
Let \epsilon>0, then \exists N\in\mathbb{N} such that \forall n\geq N,
\sum_{n=N}^\infty |a_n|<\epsilon
So there exists N_0 such that if M\geq N_0, then
\sum_{n=N_0}^M |a_n|<\epsilon
for any first M terms of \sigma, we choose N_0 such that all the terms (no overlapping with the first M terms) on the tail is less than $\epsilon$.
\sum_{n=1}^{\infty} a_n=\sum_{n=1}^{M} a_n+\sum_{n=M+1}^\infty a_n
Let K>N, L>N_0,
\left|\sum_{n=1}^{K}a_n-\sum_{n=1}^{L}a_{\sigma(n)}\right|<2\epsilon
QED
Chapter 4 Complex Integration
Complex Integral
Definition 6.1
If \phi(t) is a complex function defined on [a,b], then the integral of \phi(t) over [a,b] is defined as
\int_a^b \phi(t) dt = \int_a^b \text{Re}\{\phi(t)\} dt + i\int_a^b \text{Im}\{\phi(t)\} dt
Theorem 6.3 (Triangle Inequality)
If \phi(t) is a complex function defined on [a,b], then
\left|\int_a^b \phi(t) dt\right| \leq \int_a^b |\phi(t)| dt
Proof:
Let \lambda(t)=\frac{\left|\int_a^t \phi(t) dt\right|}{\int_a^t |\phi(t)| dt}, then \left|\lambda(t)\right|=1.
\begin{aligned}
\left|\int_a^b \phi(t) dt\right|&=\lambda\int_a^b \phi(t) dt\\
&=\int_a^b \lambda(t)\phi(t) dt\\
&=\text{Re} \{\int_a^b \lambda(t)\phi(t) dt\}\\
&\leq\int_a^b |\lambda(t)\phi(t)| dt\\
&=\int_a^b |\phi(t)| dt
\end{aligned}
Assume \phi is continuous on [a,b], the equality means \lambda(t)\phi(t) is real and positive everywhere on [a,b], which means \arg \phi(t) is constant.
QED
Definition 6.4 Arc Length
Let \gamma be a curve in the complex plane defined by \gamma(t)=x(t)+iy(t), t\in[a,b]. The arc length of \gamma is given by
\Gamma=\int_a^b |\gamma'(t)| dt=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} dt
N.B. If \int_{\Gamma} f(z) dz depends on orientation of \Gamma, but not the parametrization.
We define
\int_{\Gamma} f(z) dz=\int_{\Gamma} f(\gamma(t))\gamma'(t) dt
Example:
Suppose \Gamma is the circle centered at z_0 with radius R
\int_{\Gamma} \frac{1}{z-z_0} dz
Parameterize the unit circle:
\gamma(t)=z_0+Re^{it}\quad
\gamma'(t)=iRe^{it}, t\in[0,2\pi]
f(z)=\frac{1}{z-z_0}
f(\gamma(t))=\frac{1}{(z_0+Re^{it})-z_0}
\int_{\Gamma} f(z) dz=\int_0^{2\pi} f(\gamma(t))\gamma'(t) dt=\int_0^{2\pi} \frac{1}{Re^{-it}}iRe^{it} dt=2\pi i
Theorem 6.11 (Uniform Convergence)
If f_n(z) converges uniformly to f(z) on \Gamma, assume length of \Gamma is finite, then
\lim_{n\to\infty} \int_{\Gamma} f_n(z) dz = \int_{\Gamma} f(z) dz
Proof:
Let \epsilon>0, since f_n(z) converges uniformly to f(z) on \Gamma, there exists N\in\mathbb{N} such that for all n\geq N,
\left|f_n(z)-f(z)\right|<\epsilon
\begin{aligned}
\left|\int_{\Gamma} f_n(z) dz - \int_{\Gamma} f(z) dz\right|&=\left|\int_{\Gamma} (f_n(\gamma(t))-f(\gamma(t)))\gamma'(t) dt\right|\\
&\leq \int_{\Gamma} |f_n(\gamma(t))-f(\gamma(t))||\gamma'(t)| dt\\
&\leq \int_{\Gamma} \epsilon|\gamma'(t)| dt\\
&=\epsilon\text{length}(\Gamma)
\end{aligned}
QED
Theorem 6.6 (Integral of derivative)
Suppose \Gamma is a closed curve, \gamma:[a,b]\to\mathbb{C} and \gamma(a)=\gamma(b).
\begin{aligned}
\int_{\Gamma} f'(z) dz &= \int_a^b f'(\gamma(t))\gamma'(t) dt\\
&=\int_a^b \frac{d}{dt}f(\gamma(t)) dt\\
&=f(\gamma(b))-f(\gamma(a))\\
&=0
\end{aligned}
QED
Example:
Let R be a rectangle \{-a,a,ai+b,ai-b\}, \Gamma is the boundary of R with positive orientation.
Let \int_{R} e^{-z^2}dz.
Is e^{-z^2}=\frac{d}{dz}f(z)?
Yes, since
e^{z^2}=1-\frac{z^2}{1!}+\frac{z^4}{2!}-\frac{z^6}{3!}+\cdots=\frac{d}{dz}\left(\frac{z}{1!}-\frac{1}{3}\frac{z^3}{2!}+\frac{1}{5}\frac{z^5}{3!}-\cdots\right)
This is polynomial, therefore holomorphic.
So
\int_{R} e^{z^2}dz = 0
with some limit calculation, we can get
\int_{R} e^{-z^2}dz = 2\pi i