From cc295b5da8e3db4417523a507caf106d5839d989 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Wed, 2 Jun 2021 13:28:53 +0200
Subject: Introduce collapsible proofs to some articles
---
.../know/concept/holomorphic-function/index.pdc | 71 ++++++++++++++--------
1 file changed, 45 insertions(+), 26 deletions(-)
(limited to 'content/know/concept/holomorphic-function')
diff --git a/content/know/concept/holomorphic-function/index.pdc b/content/know/concept/holomorphic-function/index.pdc
index 1077060..1c2f092 100644
--- a/content/know/concept/holomorphic-function/index.pdc
+++ b/content/know/concept/holomorphic-function/index.pdc
@@ -77,8 +77,12 @@ $$\begin{aligned}
}
\end{aligned}$$
-*__Proof__*.
-*Just like before, we decompose $f(z)$ into its real and imaginary parts:*
+
+
+
+
+
+Just like before, we decompose $f(z)$ into its real and imaginary parts:
$$\begin{aligned}
\oint_C f(z) \:dz
@@ -88,16 +92,17 @@ $$\begin{aligned}
&= \oint_C u \dd{x} - v \dd{y} + i \oint_C v \dd{x} + u \dd{y}
\end{aligned}$$
-*Using Green's theorem, we integrate over the area $A$ enclosed by $C$:*
+Using Green's theorem, we integrate over the area $A$ enclosed by $C$:
$$\begin{aligned}
\oint_C f(z) \:dz
&= - \iint_A \pdv{v}{x} + \pdv{u}{y} \dd{x} \dd{y} + i \iint_A \pdv{u}{x} - \pdv{v}{y} \dd{x} \dd{y}
\end{aligned}$$
-*Since $f(z)$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann
-equations, such that the integrands disappear and the final result is zero.*
-*__Q.E.D.__*
+Since $f(z)$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann
+equations, such that the integrands disappear and the final result is zero.
+
+
An interesting consequence is **Cauchy's integral formula**, which
states that the value of $f(z)$ at an arbitrary point $z_0$ is
@@ -109,11 +114,15 @@ $$\begin{aligned}
}
\end{aligned}$$
-*__Proof__*.
-*Thanks to the integral theorem, we know that the shape and size
+
+
+
+
+
+Thanks to the integral theorem, we know that the shape and size
of $C$ is irrelevant. Therefore we choose it to be a circle with radius $r$,
such that the integration variable becomes $z = z_0 + r e^{i \theta}$. Then
-we integrate by substitution:*
+we integrate by substitution:
$$\begin{aligned}
\frac{1}{2 \pi i} \oint_C \frac{f(z)}{z - z_0} \dd{z}
@@ -121,15 +130,15 @@ $$\begin{aligned}
= \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta}
\end{aligned}$$
-*We may choose an arbitrarily small radius $r$, such that the contour approaches $z_0$:*
+We may choose an arbitrarily small radius $r$, such that the contour approaches $z_0$:
$$\begin{aligned}
\lim_{r \to 0}\:\: \frac{1}{2 \pi} \int_0^{2 \pi} f(z_0 + r e^{i \theta}) \dd{\theta}
&= \frac{f(z_0)}{2 \pi} \int_0^{2 \pi} \dd{\theta}
= f(z_0)
\end{aligned}$$
-
-*__Q.E.D.__*
+
+
Similarly, **Cauchy's differentiation formula**,
or **Cauchy's integral formula for derivatives**
@@ -143,16 +152,20 @@ $$\begin{aligned}
}
\end{aligned}$$
-*__Proof__*.
-*By definition, the first derivative $f'(z)$ of a
-holomorphic function $f(z)$ exists and is given by:*
+
+
+
+
+
+By definition, the first derivative $f'(z)$ of a
+holomorphic function exists and is:
$$\begin{aligned}
f'(z_0)
= \lim_{z \to z_0} \frac{f(z) - f(z_0)}{z - z_0}
\end{aligned}$$
-*We evaluate the numerator using Cauchy's integral theorem as follows:*
+We evaluate the numerator using Cauchy's integral theorem as follows:
$$\begin{aligned}
f'(z_0)
@@ -166,7 +179,7 @@ $$\begin{aligned}
\oint_C \frac{f(\zeta) (z - z_0)}{(\zeta - z)(\zeta - z_0)} \dd{\zeta}
\end{aligned}$$
-*This contour integral converges uniformly, so we may apply the limit on the inside:*
+This contour integral converges uniformly, so we may apply the limit on the inside:
$$\begin{aligned}
f'(z_0)
@@ -174,9 +187,10 @@ $$\begin{aligned}
= \frac{1}{2 \pi i} \oint_C \frac{f(\zeta)}{(\zeta - z_0)^2} \dd{\zeta}
\end{aligned}$$
-*Since the second-order derivative $f''(z)$ is simply the derivative of $f'(z)$,
-this proof works inductively for all higher orders $n$.*
-*__Q.E.D.__*
+Since the second-order derivative $f''(z)$ is simply the derivative of $f'(z)$,
+this proof works inductively for all higher orders $n$.
+
+
## Residue theorem
@@ -205,24 +219,29 @@ $$\begin{aligned}
}
\end{aligned}$$
-*__Proof__*. *From the definition of a meromorphic function,
+
+
+
+
+
+From the definition of a meromorphic function,
we know that we can decompose $f(z)$ like so,
-where $h(z)$ is holomorphic and $p$ are all its poles:*
+where $h(z)$ is holomorphic and $p$ are all its poles:
$$\begin{aligned}
f(z) = h(z) + \sum_{p} \frac{R_p}{z - z_p}
\end{aligned}$$
-*We integrate this over a contour $C$ which contains all poles, and apply
-both Cauchy's integral theorem and Cauchy's integral formula to get:*
+We integrate this over a contour $C$ which contains all poles, and apply
+both Cauchy's integral theorem and Cauchy's integral formula to get:
$$\begin{aligned}
\oint_C f(z) \dd{z}
&= \oint_C h(z) \dd{z} + \sum_{p} R_p \oint_C \frac{1}{z - z_p} \dd{z}
= \sum_{p} R_p \: 2 \pi i
\end{aligned}$$
-
-*__Q.E.D.__*
+
+
This theorem might not seem very useful,
but in fact, thanks to some clever mathematical magic,
--
cgit v1.2.3