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Diffstat (limited to 'content')
| -rw-r--r-- | content/know/concept/binomial-distribution/index.pdc | 72 | ||||
| -rw-r--r-- | content/know/concept/convolution-theorem/index.pdc | 26 | ||||
| -rw-r--r-- | content/know/concept/curvilinear-coordinates/index.pdc | 154 | ||||
| -rw-r--r-- | content/know/concept/dirac-delta-function/index.pdc | 41 | ||||
| -rw-r--r-- | content/know/concept/heaviside-step-function/index.pdc | 27 | ||||
| -rw-r--r-- | content/know/concept/holomorphic-function/index.pdc | 71 | ||||
| -rw-r--r-- | content/know/concept/parsevals-theorem/index.pdc | 32 |
7 files changed, 271 insertions, 152 deletions
diff --git a/content/know/concept/binomial-distribution/index.pdc b/content/know/concept/binomial-distribution/index.pdc index 70cc897..e644164 100644 --- a/content/know/concept/binomial-distribution/index.pdc +++ b/content/know/concept/binomial-distribution/index.pdc @@ -22,7 +22,7 @@ that $n$ out of the $N$ trials succeed: $$\begin{aligned} \boxed{ - P_N(n) = \binom{N}{n} \: p^n (1 - p)^{N - n} + P_N(n) = \binom{N}{n} \: p^n q^{N - n} } \end{aligned}$$ @@ -41,8 +41,20 @@ $$\begin{aligned} The remaining factor $p^n (1 - p)^{N - n}$ is then just the probability of attaining each microstate. -To find the mean number of successes $\mu$, -the trick is to treat $p$ and $q$ as independent: +The expected or mean number of successes $\mu$ after $N$ trials is as follows: + +$$\begin{aligned} + \boxed{ + \mu = N p + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-mean"/> +<label for="proof-mean">Proof</label> +<div class="hidden"> +<label for="proof-mean">Proof.</label> +The trick is to treat $p$ and $q$ as independent until the last moment: $$\begin{aligned} \mu @@ -54,16 +66,26 @@ $$\begin{aligned} = N p (p + q)^{N - 1} \end{aligned}$$ -By inserting $q = 1 - p$, we find the following expression for the mean: +Inserting $q = 1 - p$ then gives the desired result. +</div> +</div> + +Meanwhile, we find the following variance $\sigma^2$, +with $\sigma$ being the standard deviation: $$\begin{aligned} \boxed{ - \mu = N p + \sigma^2 = N p q } \end{aligned}$$ -Next, we use the same trick to calculate $\overline{n^2}$ -(the mean of the squared number of successes): +<div class="accordion"> +<input type="checkbox" id="proof-var"/> +<label for="proof-var">Proof</label> +<div class="hidden"> +<label for="proof-var">Proof.</label> +We use the same trick to calculate $\overline{n^2}$ +(the mean squared number of successes): $$\begin{aligned} \overline{n^2} @@ -79,7 +101,7 @@ $$\begin{aligned} &= N p + N^2 p^2 - N p^2 \end{aligned}$$ -Using this and the earlier expression for $\mu$, we find the variance $\sigma^2$: +Using this and the earlier expression $\mu = N p$, we find the variance $\sigma^2$: $$\begin{aligned} \sigma^2 @@ -88,18 +110,26 @@ $$\begin{aligned} = N p (1 - p) \end{aligned}$$ -Once again, by inserting $q = 1 - p$, we find the following expression for the variance: +By inserting $q = 1 - p$, we arrive at the desired expression. +</div> +</div> + +As $N \to \infty$, the binomial distribution +turns into the continuous normal distribution: $$\begin{aligned} \boxed{ - \sigma^2 = N p q + \lim_{N \to \infty} P_N(n) = \frac{1}{\sqrt{2 \pi \sigma^2}} \exp\!\Big(\!-\!\frac{(n - \mu)^2}{2 \sigma^2} \Big) } \end{aligned}$$ -As $N$ grows to infinity, the binomial distribution -turns into the continuous normal distribution. -We demonstrate this by taking the Taylor expansion of its -natural logarithm $\ln\!\big(P_N(n)\big)$ around the mean $\mu = Np$: +<div class="accordion"> +<input type="checkbox" id="proof-normal"/> +<label for="proof-normal">Proof</label> +<div class="hidden"> +<label for="proof-normal">Proof.</label> +We take the Taylor expansion of $\ln\!\big(P_N(n)\big)$ +around the mean $\mu = Np$: $$\begin{aligned} \ln\!\big(P_N(n)\big) @@ -108,7 +138,7 @@ $$\begin{aligned} D_m(n) = \dv[m]{\ln\!\big(P_N(n)\big)}{n} \end{aligned}$$ -We use Stirling's approximation to calculate all these factorials: +We use Stirling's approximation to calculate the factorials in $D_m$: $$\begin{aligned} \ln\!\big(P_N(n)\big) @@ -163,7 +193,7 @@ $$\begin{aligned} = - \frac{1}{\sigma^2} \end{aligned}$$ -The higher-order derivatives tend to zero for large $N$, so we discard them: +The higher-order derivatives tend to zero for $N \to \infty$, so we discard them: $$\begin{aligned} D_3(n) @@ -184,13 +214,9 @@ $$\begin{aligned} = \ln\!\Big( \frac{1}{\sqrt{2\pi \sigma^2}} \Big) - \frac{(n - \mu)^2}{2 \sigma^2} \end{aligned}$$ -Thus, as $N$ goes to infinity, the binomial distribution becomes a Gaussian: - -$$\begin{aligned} - \boxed{ - \lim_{N \to \infty} P_N(n) = \frac{1}{\sqrt{2 \pi \sigma^2}} \exp\!\Big(\!-\!\frac{(n - \mu)^2}{2 \sigma^2} \Big) - } -\end{aligned}$$ +Taking $\exp$ of this expression then yields a normalized Gaussian distribution. +</div> +</div> ## References diff --git a/content/know/concept/convolution-theorem/index.pdc b/content/know/concept/convolution-theorem/index.pdc index 9d1a666..1454cc0 100644 --- a/content/know/concept/convolution-theorem/index.pdc +++ b/content/know/concept/convolution-theorem/index.pdc @@ -32,7 +32,12 @@ $$\begin{aligned} } \end{aligned}$$ -To prove this, we expand the right-hand side of the theorem and +<div class="accordion"> +<input type="checkbox" id="proof-fourier"/> +<label for="proof-fourier">Proof</label> +<div class="hidden"> +<label for="proof-fourier">Proof.</label> +We expand the right-hand side of the theorem and rearrange the integrals: $$\begin{aligned} @@ -45,8 +50,8 @@ $$\begin{aligned} = A \cdot (f * g)(x) \end{aligned}$$ -Then we do the same thing again, this time starting from a product in -the $x$-domain: +Then we do the same again, +this time starting from a product in the $x$-domain: $$\begin{aligned} \hat{\mathcal{F}}\{f(x) \: g(x)\} @@ -57,6 +62,8 @@ $$\begin{aligned} &= B \int_{-\infty}^\infty \tilde{g}(k') \tilde{f}(k - k') \dd{k'} = B \cdot (\tilde{f} * \tilde{g})(k) \end{aligned}$$ +</div> +</div> ## Laplace transform @@ -76,9 +83,14 @@ $$\begin{aligned} \boxed{\hat{\mathcal{L}}\{(f * g)(t)\} = \tilde{f}(s) \: \tilde{g}(s)} \end{aligned}$$ -We prove this by expanding the left-hand side. Note that the lower -integration limit is 0 instead of $-\infty$, because we set both $f(t)$ -and $g(t)$ to zero for $t < 0$: +<div class="accordion"> +<input type="checkbox" id="proof-laplace"/> +<label for="proof-laplace">Proof</label> +<div class="hidden"> +<label for="proof-laplace">Proof.</label> +We expand the left-hand side. +Note that the lower integration limit is 0 instead of $-\infty$, +because we set both $f(t)$ and $g(t)$ to zero for $t < 0$: $$\begin{aligned} \hat{\mathcal{L}}\{(f * g)(t)\} @@ -98,6 +110,8 @@ $$\begin{aligned} &= \int_0^\infty \tilde{f}(s) g(t') \exp(- s t') \dd{t'} = \tilde{f}(s) \: \tilde{g}(s) \end{aligned}$$ +</div> +</div> diff --git a/content/know/concept/curvilinear-coordinates/index.pdc b/content/know/concept/curvilinear-coordinates/index.pdc index e1c0465..925eda3 100644 --- a/content/know/concept/curvilinear-coordinates/index.pdc +++ b/content/know/concept/curvilinear-coordinates/index.pdc @@ -50,7 +50,7 @@ and [parabolic cylindrical coordinates](/know/concept/parabolic-cylindrical-coor In the following subsections, we derive general formulae to convert expressions -from Cartesian coordinates in the new orthogonal system $(x_1, x_2, x_3)$. +from Cartesian coordinates to the new orthogonal system $(x_1, x_2, x_3)$. ## Basis vectors @@ -93,7 +93,26 @@ $$\begin{aligned} ## Gradient -For a given direction $\dd{\ell}$, we know that +In an orthogonal coordinate system, +the gradient $\nabla f$ of a scalar $f$ is as follows, +where $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$ +are the basis unit vectors respectively corresponding to $x_1$, $x_2$ and $x_3$: + +$$\begin{gathered} + \boxed{ + \nabla f + = \vu{e}_1 \frac{1}{h_1} \pdv{f}{x_1} + + \vu{e}_2 \frac{1}{h_2} \pdv{f}{x_2} + + \vu{e}_3 \frac{1}{h_3} \pdv{f}{x_3} + } +\end{gathered}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-grad"/> +<label for="proof-grad">Proof</label> +<div class="hidden"> +<label for="proof-grad">Proof.</label> +For a direction $\dd{\ell}$, we know that $\dv*{f}{\ell}$ is the component of $\nabla f$ in that direction: $$\begin{aligned} @@ -104,7 +123,7 @@ $$\begin{aligned} \end{aligned}$$ Where $\vu{u}$ is simply a unit vector in the direction of $\dd{\ell}$. -We can thus find an expression for the gradient $\nabla f$ +We thus find the expression for the gradient $\nabla f$ by choosing $\dd{\ell}$ to be $h_1 \dd{x_1}$, $h_2 \dd{x_2}$ and $h_3 \dd{x_3}$ in turn: $$\begin{gathered} @@ -112,49 +131,59 @@ $$\begin{gathered} = \vu{e}_1 \dv{x_1}{\ell} \pdv{f}{x_1} + \vu{e}_2 \dv{x_2}{\ell} \pdv{f}{x_2} + \vu{e}_3 \dv{x_3}{\ell} \pdv{f}{x_3} - \\ - \boxed{ - \nabla f - = \vu{e}_1 \frac{1}{h_1} \pdv{f}{x_1} - + \vu{e}_2 \frac{1}{h_2} \pdv{f}{x_2} - + \vu{e}_3 \frac{1}{h_3} \pdv{f}{x_3} - } \end{gathered}$$ - -Where $\vu{e}_1$, $\vu{e}_2$ and $\vu{e}_3$ -are the basis unit vectors respectively corresponding to $x_1$, $x_2$ and $x_3$. +</div> +</div> ## Divergence -Consider a vector $\vb{V}$ in the target coordinate system -with components $V_1$, $V_2$ and $V_3$: +The divergence of a vector $\vb{V} = \vu{e}_1 V_1 + \vu{e}_2 V_2 + \vu{e}_3 V_3$ +in an orthogonal system is given by: + +$$\begin{aligned} + \boxed{ + \nabla \cdot \vb{V} + = \frac{1}{h_1 h_2 h_3} + \Big( \pdv{(h_2 h_3 V_1)}{x_1} + \pdv{(h_1 h_3 V_2)}{x_2} + \pdv{(h_1 h_2 V_3)}{x_3} \Big) + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-div"/> +<label for="proof-div">Proof</label> +<div class="hidden"> +<label for="proof-div">Proof.</label> +As preparation, we rewrite $\vb{V}$ as follows +to introduce the scale factors: $$\begin{aligned} \vb{V} - &= \vu{e}_1 V_1 + \vu{e}_2 V_2 + \vu{e}_3 V_3 - \\ &= \vu{e}_1 \frac{1}{h_2 h_3} (h_2 h_3 V_1) + \vu{e}_2 \frac{1}{h_1 h_3} (h_1 h_3 V_2) + \vu{e}_3 \frac{1}{h_1 h_2} (h_1 h_2 V_3) \end{aligned}$$ -We take only the $\vu{e}_1$-component of this vector, -and expand its divergence using a vector identity, -where $f = h_2 h_3 V_1$ is a scalar -and $\vb{U} = \vu{e}_1 / (h_2 h_3)$ is a vector: +We start by taking only the $\vu{e}_1$-component of this vector, +and expand its divergence using the following vector identity: $$\begin{gathered} \nabla \cdot (\vb{U} \: f) = \vb{U} \cdot (\nabla f) + (\nabla \cdot \vb{U}) \: f - \\ +\end{gathered}$$ + +Inserting the scalar $f = h_2 h_3 V_1$ +the vector $\vb{U} = \vu{e}_1 / (h_2 h_3)$, +we arrive at: + +$$\begin{gathered} \nabla \cdot \Big( \frac{\vu{e}_1}{h_2 h_3} (h_2 h_3 V_1) \Big) = \frac{\vu{e}_1}{h_2 h_3} \cdot \Big( \nabla (h_2 h_3 V_1) \Big) + \Big( \nabla \cdot \frac{\vu{e}_1}{h_2 h_3} \Big) (h_2 h_3 V_1) \end{gathered}$$ -The first term is straightforward to calculate -thanks to our preceding expression for the gradient. +The first right-hand term is easy to calculate +thanks to our expression for the gradient $\nabla f$. Only the $\vu{e}_1$-component survives due to the dot product: $$\begin{aligned} @@ -162,8 +191,8 @@ $$\begin{aligned} = \frac{\vu{e}_1}{h_1 h_2 h_3} \pdv{(h_2 h_3 V_1)}{x_1} \end{aligned}$$ -The second term is a bit more involved. -To begin with, we use the gradient formula to note that: +The second term is more involved. +First, we use the gradient formula to observe that: $$\begin{aligned} \nabla x_1 @@ -177,7 +206,7 @@ $$\begin{aligned} \end{aligned}$$ Because $\vu{e}_2 \cross \vu{e}_3 = \vu{e}_1$ in an orthogonal basis, -we can get the vector whose divergence we want: +these gradients can be used to express the vector whose divergence we want: $$\begin{aligned} \nabla x_2 \cross \nabla x_3 @@ -196,15 +225,9 @@ $$\begin{aligned} \end{aligned}$$ After repeating this procedure for the other components of $\vb{V}$, -we arrive at the following general expression for the divergence $\nabla \cdot \vb{V}$: - -$$\begin{aligned} - \boxed{ - \nabla \cdot \vb{V} - = \frac{1}{h_1 h_2 h_3} - \Big( \pdv{(h_2 h_3 V_1)}{x_1} + \pdv{(h_1 h_3 V_2)}{x_2} + \pdv{(h_1 h_2 V_3)}{x_3} \Big) - } -\end{aligned}$$ +we get the desired general expression for the divergence. +</div> +</div> ## Laplacian @@ -229,31 +252,55 @@ $$\begin{aligned} ## Curl -We find the curl in a similar way as the divergence. -Consider an arbitrary vector $\vb{V}$: +The curl of a vector $\vb{V}$ is as follows +in a general orthogonal curvilinear system: + +$$\begin{aligned} + \boxed{ + \begin{aligned} + \nabla \times \vb{V} + &= \frac{\vu{e}_1}{h_2 h_3} \Big( \pdv{(h_3 V_3)}{x_2} - \pdv{(h_2 V_2)}{x_3} \Big) + \\ + &+ \frac{\vu{e}_2}{h_1 h_3} \Big( \pdv{(h_1 V_1)}{x_3} - \pdv{(h_3 V_3)}{x_1} \Big) + \\ + &+ \frac{\vu{e}_3}{h_1 h_2} \Big( \pdv{(h_2 V_2)}{x_1} - \pdv{(h_1 V_1)}{x_2} \Big) + \end{aligned} + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-curl"/> +<label for="proof-curl">Proof</label> +<div class="hidden"> +<label for="proof-curl">Proof.</label> +The curl is found in a similar way as the divergence. +We rewrite $\vb{V}$ like so: $$\begin{aligned} \vb{V} - = \vu{e}_1 V_1 + \vu{e}_2 V_2 + \vu{e}_3 V_3 = \frac{\vu{e}_1}{h_1} (h_1 V_1) + \frac{\vu{e}_2}{h_2} (h_2 V_2) + \frac{\vu{e}_3}{h_3} (h_3 V_3) \end{aligned}$$ -We expand the curl of its $\vu{e}_1$-component using a vector identity, -where $f = h_1 V_1$ is a scalar and $\vb{U} = \vu{e}_1 / h_1$ is a vector: +We expand the curl of its $\vu{e}_1$-component using the following vector identity: $$\begin{gathered} \nabla \cross (\vb{U} \: f) = (\nabla \cross \vb{U}) \: f - \vb{U} \cross (\nabla f) - \\ +\end{gathered}$$ + +Inserting the scalar $f = h_1 V_1$ +and the vector $\vb{U} = \vu{e}_1 / h_1$, we arrive at: + +$$\begin{gathered} \nabla \cross \Big( \frac{\vu{e}_1}{h_1} (h_1 V_1) \Big) = \Big( \nabla \cross \frac{\vu{e}_1}{h_1} \Big) (h_1 V_1) - \frac{\vu{e}_1}{h_1} \cross \Big( \nabla (h_1 V_1) \Big) \end{gathered}$$ -Previously, when calculating the divergence, +Previously, when proving the divergence, we already showed that $\vu{e}_1 / h_1 = \nabla x_1$. Because the curl of a gradient is zero, -the first term thus disappears, leaving only the second, -which contains a gradient turning out to be: +the first term disappears, leaving only the second, +which contains a gradient that turns out to be: $$\begin{aligned} \nabla (h_1 V_1) @@ -273,20 +320,9 @@ $$\begin{aligned} \end{aligned}$$ If we go through the same process for the other components of $\vb{V}$ -and add the results together, we get the following expression for the curl $\nabla \cross \vb{V}$: - -$$\begin{aligned} - \boxed{ - \begin{aligned} - \nabla \times \vb{V} - &= \frac{\vu{e}_1}{h_2 h_3} \Big( \pdv{(h_3 V_3)}{x_2} - \pdv{(h_2 V_2)}{x_3} \Big) - \\ - &+ \frac{\vu{e}_2}{h_1 h_3} \Big( \pdv{(h_1 V_1)}{x_3} - \pdv{(h_3 V_3)}{x_1} \Big) - \\ - &+ \frac{\vu{e}_3}{h_1 h_2} \Big( \pdv{(h_2 V_2)}{x_1} - \pdv{(h_1 V_1)}{x_2} \Big) - \end{aligned} - } -\end{aligned}$$ +and add up the results, we get the desired expression for the curl. +</div> +</div> ## Differential elements diff --git a/content/know/concept/dirac-delta-function/index.pdc b/content/know/concept/dirac-delta-function/index.pdc index 76b6e97..9eecefd 100644 --- a/content/know/concept/dirac-delta-function/index.pdc +++ b/content/know/concept/dirac-delta-function/index.pdc @@ -21,7 +21,7 @@ defined to be 1: $$\begin{aligned} \boxed{ - \delta(x) = + \delta(x) \equiv \begin{cases} +\infty & \mathrm{if}\: x = 0 \\ 0 & \mathrm{if}\: x \neq 0 @@ -56,12 +56,10 @@ following integral, which appears very often in the context of [Fourier transforms](/know/concept/fourier-transform/): $$\begin{aligned} - \boxed{ - \delta(x) - %= \lim_{n \to +\infty} \!\Big\{\frac{\sin(n x)}{\pi x}\Big\} - = \frac{1}{2\pi} \int_{-\infty}^\infty \exp(i k x) \dd{k} - \:\:\propto\:\: \hat{\mathcal{F}}\{1\} - } + \delta(x) + = \lim_{n \to +\infty} \!\Big\{\frac{\sin(n x)}{\pi x}\Big\} + = \frac{1}{2\pi} \int_{-\infty}^\infty \exp(i k x) \dd{k} + \:\:\propto\:\: \hat{\mathcal{F}}\{1\} \end{aligned}$$ When the argument of $\delta(x)$ is scaled, the delta function is itself scaled: @@ -72,18 +70,22 @@ $$\begin{aligned} } \end{aligned}$$ -*__Proof.__ Because it is symmetric, $\delta(s x) = \delta(|s| x)$. Then by -substituting $\sigma = |s| x$:* +<div class="accordion"> +<input type="checkbox" id="proof-scale"/> +<label for="proof-scale">Proof</label> +<div class="hidden"> +<label for="proof-scale">Proof.</label> +Because it is symmetric, $\delta(s x) = \delta(|s| x)$. +Then by substituting $\sigma = |s| x$: $$\begin{aligned} \int \delta(|s| x) \dd{x} &= \frac{1}{|s|} \int \delta(\sigma) \dd{\sigma} = \frac{1}{|s|} \end{aligned}$$ +</div> +</div> -*__Q.E.D.__* - -An even more impressive property is the behaviour of the derivative of -$\delta(x)$: +An even more impressive property is the behaviour of the derivative of $\delta(x)$: $$\begin{aligned} \boxed{ @@ -91,16 +93,21 @@ $$\begin{aligned} } \end{aligned}$$ -*__Proof.__ Note which variable is used for the -differentiation, and that $\delta'(x - \xi) = - \delta'(\xi - x)$:* +<div class="accordion"> +<input type="checkbox" id="proof-dv1"/> +<label for="proof-dv1">Proof</label> +<div class="hidden"> +<label for="proof-dv1">Proof.</label> +Note which variable is used for the +differentiation, and that $\delta'(x - \xi) = - \delta'(\xi - x)$: $$\begin{aligned} \int f(\xi) \: \dv{\delta(x - \xi)}{x} \dd{\xi} &= \dv{x} \int f(\xi) \: \delta(x - \xi) \dd{x} = f'(x) \end{aligned}$$ - -*__Q.E.D.__* +</div> +</div> This property also generalizes nicely for the higher-order derivatives: diff --git a/content/know/concept/heaviside-step-function/index.pdc b/content/know/concept/heaviside-step-function/index.pdc index 0471acf..dbbca6f 100644 --- a/content/know/concept/heaviside-step-function/index.pdc +++ b/content/know/concept/heaviside-step-function/index.pdc @@ -50,7 +50,23 @@ $$\begin{aligned} \end{aligned}$$ The [Fourier transform](/know/concept/fourier-transform/) -of $\Theta(t)$ is noteworthy. +of $\Theta(t)$ is as follows, +where $\pv{}$ is the Cauchy principal value, +$A$ and $s$ are constants from the FT's definition, +and $\mathrm{sgn}$ is the signum function: + +$$\begin{aligned} + \boxed{ + \tilde{\Theta}(\omega) + = \frac{A}{|s|} \Big( \pi \delta(\omega) + i \: \mathrm{sgn}(s) \pv{\frac{1}{\omega}} \Big) + } +\end{aligned}$$ + +<div class="accordion"> +<input type="checkbox" id="proof-fourier"/> +<label for="proof-fourier">Proof</label> +<div class="hidden"> +<label for="proof-fourier">Proof.</label> In this case, it is easiest to use $\Theta(0) = 1/2$, such that the Heaviside step function can be expressed using the signum function $\mathrm{sgn}(t)$: @@ -77,15 +93,10 @@ $$\begin{aligned} &= A \pi \delta(s \omega) + \frac{A}{2} \pv{\int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t}} = \frac{A}{|s|} \pi \delta(\omega) + i \frac{A}{s} \pv{\frac{1}{\omega}} \end{aligned}$$ +</div> +</div> The use of $\pv{}$ without an integral is an abuse of notation, and means that this result only makes sense when wrapped in an integral. Formally, $\pv{\{1 / \omega\}}$ is a [Schwartz distribution](/know/concept/schwartz-distribution/). -We thus have: -$$\begin{aligned} - \boxed{ - \tilde{\Theta}(\omega) - = \frac{A}{|s|} \Big( \pi \delta(\omega) + i \: \mathrm{sgn}(s) \pv{\frac{1}{\omega}} \Big) - } -\end{aligned}$$ 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:* +<div class="accordion"> +<input type="checkbox" id="proof-int-theorem"/> +<label for="proof-int-theorem">Proof</label> +<div class="hidden"> +<label for="proof-int-theorem">Proof.</label> +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}$$ -*Since $f(z)$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann -equations, such that the integrands disappear and the final result is zero.* -*__Q.E.D.__* +Since $f(z)$ is holomorphic, $u$ and $v$ satisfy the Cauchy-Riemann +equations, such that the integrands disappear and the final result is zero. +</div> +</div> An interesting consequence is **Cauchy's integral formula**, which states that the value of $f(z)$ at an arbitrary point $z_0$ is @@ -109,11 +114,15 @@ $$\begin{aligned} } \end{aligned}$$ -*__Proof__*. -*Thanks to the integral theorem, we know that the shape and size +<div class="accordion"> +<input type="checkbox" id="proof-int-formula"/> +<label for="proof-int-formula">Proof</label> +<div class="hidden"> +<label for="proof-int-formula">Proof.</label> +Thanks to the integral theorem, we know that the shape and size of $C$ is irrelevant. Therefore we choose it to be a circle with radius $r$, such that the integration variable becomes $z = z_0 + r e^{i \theta}$. Then -we integrate by substitution:* +we integrate by substitution: $$\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.__* +</div> +</div> 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:* +<div class="accordion"> +<input type="checkbox" id="proof-diff-formula"/> +<label for="proof-diff-formula">Proof</label> +<div class="hidden"> +<label for="proof-diff-formula">Proof.</label> +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 $n$. +</div> +</div> ## Residue theorem @@ -205,24 +219,29 @@ $$\begin{aligned} } \end{aligned}$$ -*__Proof__*. *From the definition of a meromorphic function, +<div class="accordion"> +<input type="checkbox" id="proof-res-theorem"/> +<label for="proof-res-theorem">Proof</label> +<div class="hidden"> +<label for="proof-res-theorem">Proof.</label> +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.__* +</div> +</div> This theorem might not seem very useful, but in fact, thanks to some clever mathematical magic, diff --git a/content/know/concept/parsevals-theorem/index.pdc b/content/know/concept/parsevals-theorem/index.pdc index 824afa6..9f440f2 100644 --- a/content/know/concept/parsevals-theorem/index.pdc +++ b/content/know/concept/parsevals-theorem/index.pdc @@ -17,24 +17,24 @@ markup: pandoc and the inner product of their [Fourier transforms](/know/concept/fourier-transform/) $\tilde{f}(k)$ and $\tilde{g}(k)$. There are two equivalent ways of stating it, -where $A$, $B$, and $s$ are constants from the Fourier transform's definition: +where $A$, $B$, and $s$ are constants from the FT's definition: $$\begin{aligned} \boxed{ - \braket{f(x)}{g(x)} = \frac{2 \pi B^2}{|s|} \braket*{\tilde{f}(k)}{\tilde{g}(k)} - } - \\ - \boxed{ - \braket*{\tilde{f}(k)}{\tilde{g}(k)} = \frac{2 \pi A^2}{|s|} \braket{f(x)}{g(x)} + \begin{aligned} + \braket{f(x)}{g(x)} &= \frac{2 \pi B^2}{|s|} \braket*{\tilde{f}(k)}{\tilde{g}(k)} + \\ + \braket*{\tilde{f}(k)}{\tilde{g}(k)} &= \frac{2 \pi A^2}{|s|} \braket{f(x)}{g(x)} + \end{aligned} } \end{aligned}$$ -For this reason, physicists like to define the Fourier transform -with $A\!=\!B\!=\!1 / \sqrt{2\pi}$ and $|s|\!=\!1$, because then it nicely -conserves the functions' normalization. - -To prove the theorem, we insert the inverse FT into the inner product -definition: +<div class="accordion"> +<input type="checkbox" id="proof-fourier"/> +<label for="proof-fourier">Proof</label> +<div class="hidden"> +<label for="proof-fourier">Proof.</label> +We insert the inverse FT into the defintion of the inner product: $$\begin{aligned} \braket{f}{g} @@ -54,7 +54,7 @@ $$\begin{aligned} \end{aligned}$$ Where $\delta(k)$ is the [Dirac delta function](/know/concept/dirac-delta-function/). -Note that we can equally well do the proof in the opposite direction, +Note that we can equally well do this proof in the opposite direction, which yields an equivalent result: $$\begin{aligned} @@ -73,6 +73,12 @@ $$\begin{aligned} &= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x} = \frac{2 \pi A^2}{|s|} \braket{f}{g} \end{aligned}$$ +</div> +</div> + +For this reason, physicists like to define the Fourier transform +with $A\!=\!B\!=\!1 / \sqrt{2\pi}$ and $|s|\!=\!1$, because then it nicely +conserves the functions' normalization. |