[spec] inverse p-value and distance integral
[notes.git] / spec / spec-collated.tex
index 785a7fdcb6da91ca2a2b82ef125294daf4e1b9a4..d66ece975983a2cf40ff524d6d65e934a4a16a08 100644 (file)
       \hline
     \end{tabularx}
 
+    \begin{theorembox}{title=Factor theorem}
+      If \(\beta z + \alpha\) is a factor of \(P(z)\), \\
+      \-\hspace{1em}then \(P(-\dfrac{\alpha}{\beta})=0\).
+    \end{theorembox}
+
     \subsection*{\(n\)th roots}
 
     \(n\)th roots of \(z=r\operatorname{cis}\theta\) are:
                       \addplot[gray, dotted, thick, domain=-35:35] {-1.5708} node [black, font=\footnotesize, above left, pos=1] {\(y=-\frac{\pi}{2}\)};
                     \end{axis}
                   \end{tikzpicture}
-\columnbreak
+
+                  \subsection*{Mensuration}
+
+                  \begin{tikzpicture}[draw=blue!70,thick]
+                    \filldraw[fill=lblue] circle (2cm);
+                    \filldraw[fill=white] 
+                    (320:2cm) node[right] {} 
+                    -- (220:2cm) node[left] {} 
+                    arc[start angle=220, end angle=320, radius=2cm] 
+                    -- cycle;
+                    \node {Major Segment};
+                    \node at (-90:2) {Minor Segment};
+
+                    \begin{scope}[xshift=4.5cm]
+                      \draw circle (2cm);
+                      \filldraw[fill=lblue] 
+                      (320:2cm) node[right] {}
+                      -- (0,0) node[above] {}
+                      -- (220:2cm) node[left] {} 
+                      arc[start angle=220, end angle=320, radius=2cm]
+                      -- cycle;
+                      \node at (90:1cm) {Major Sector};
+                      \node at (-90:1.5) {Minor Sector};
+                    \end{scope}
+                  \end{tikzpicture}
+
+                  \subsubsection*{Sectors}
+
+                  \begin{align*}
+                    A &= \pi r^2 \dfrac{\theta}{2\pi} \\
+                    &= \dfrac{r^2 \theta}{2}
+                  \end{align*}
+
+                  \subsubsection*{Segments}
+
+                  \[ A = \dfrac{r^2}{2} \left( \theta = \sin \theta \right) \]
+
+                  \subsubsection*{Chords}
+
+                  \begin{align*}
+                    \operatorname{crd} \theta &= \sqrt{(1 - \cos\theta)^2 + \sin^2 \theta} \\
+                    &= \sqrt{2 - 2\cos\theta} \\
+                    &= 2 \sin \left(\dfrac{\theta}{2}\right)
+                  \end{align*}
+
                   \section{Differential calculus}
 
+                  \[f^\prime(x) = \lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={\frac{dy}{dx}}\]
+
                   \subsection*{Limits}
 
                   \[\lim_{x \rightarrow a}f(x)\]
                       \(f(x)\) is continuous \(\iff L^-=L^+=f(x) \> \forall x\)
                   \end{enumerate}
 
-                  \subsection*{Gradients of secants and tangents}
+                  \subsection*{Gradients}
 
                   \textbf{Secant (chord)} - line joining two points on curve\\
                   \textbf{Tangent} - line that intersects curve at one point
 
-                  \subsection*{First principles derivative}
-
-                  \[f^\prime(x) = \lim_{\delta x \rightarrow 0}{\delta y \over \delta x}={\frac{dy}{dx}}\]
-
-                  \subsubsection*{Logarithmic identities}
-
-                  \(\log_b (xy)=\log_b x + \log_b y\)\\
-                  \(\log_b x^n = n \log_b x\)\\
-                  \(\log_b y^{x^n} = x^n \log_b y\)
-
-                  \subsubsection*{Index identities}
-
-                  \(b^{m+n}=b^m \cdot b^n\)\\
-                  \((b^m)^n=b^{m \cdot n}\)\\
-                  \((b \cdot c)^n = b^n \cdot c^n\)\\
-                  \({a^m \div a^n} = {a^{m-n}}\)
-
-                  \subsection*{Reciprocal derivatives}
-
-                  \[\frac{1}{\frac{dy}{dx}} = \frac{dx}{dy}\]
-
-                  \subsection*{Differentiating \(x=f(y)\)}
-                  Find \(\dfrac{dx}{dy}\), then \(\dfrac{dy}{dx} = \dfrac{1}{\left(\dfrac{dx}{dy}\right)}\)
-
-                  \subsection*{Second derivative}
-                  \begin{align*}f(x) \longrightarrow &f^\prime (x) \longrightarrow f^{\prime\prime}(x)\\
-                  \implies y \longrightarrow &\frac{dy}{dx} \longrightarrow \frac{d^2 y}{dx^2}\end{align*}
-
-                  \noindent Order of polynomial \(n\)th derivative decrements each time the derivative is taken
-
                   \subsubsection*{Points of Inflection}
 
                   \emph{Stationary point} - i.e.
                   \end{warning}
 
 
+                  \subsection*{Second derivative}
+                  \begin{align*}f(x) \longrightarrow &f^\prime (x) \longrightarrow f^{\prime\prime}(x)\\
+                  \implies y \longrightarrow &\frac{dy}{dx} \longrightarrow \frac{d^2 y}{dx^2}\end{align*}
+
+                  \noindent Order of polynomial \(n\)th derivative decrements each time the derivative is taken
+
+
+                  \subsection*{Slope fields}
+
+                  \begin{tikzpicture}[declare function={diff(\x,\y) = \x+\y;}]
+                    \begin{axis}[axis equal, ymin=-4, ymax=4, xmin=-4, xmax=4, ticks=none, enlargelimits=true, ]
+                      \addplot[thick, orange, domain=-4:2] {e^(x)-x-1};
+                      \pgfplotsinvokeforeach{-4,...,4}{%
+                        \draw[gray] ( {#1 -0.1}, {4 - diff(#1, 4) *0.1}) --  ( {#1 +0.1}, {4  + diff(#1, 4) *0.1});
+                        \draw[gray] ( {#1 -0.1}, {3 - diff(#1, 3) *0.1}) --  ( {#1 +0.1}, {3  + diff(#1, 3) *0.1});
+                        \draw[gray] ( {#1 -0.1}, {2 - diff(#1, 2) *0.1}) --  ( {#1 +0.1}, {2  + diff(#1, 2) *0.1});
+                        \draw[gray] ( {#1 -0.1}, {1 - diff(#1, 1) *0.1}) --  ( {#1 +0.1}, {1  + diff(#1, 1) *0.1});
+                        \draw[gray] ( {#1 -0.1}, {0 - diff(#1, 0) *0.1}) --  ( {#1 +0.1}, {0  + diff(#1, 0) *0.1});
+                        \draw[gray] ( {#1 -0.1}, {-1 - diff(#1, -1) *0.1}) --  ( {#1 +0.1}, {-1  + diff(#1, -1) *0.1});
+                        \draw[gray] ( {#1 -0.1}, {-2 - diff(#1, -2) *0.1}) --  ( {#1 +0.1}, {-2  + diff(#1, -2) *0.1});
+                        \draw[gray] ( {#1 -0.1}, {-3 - diff(#1, -3) *0.1}) --  ( {#1 +0.1}, {-3  + diff(#1, -3) *0.1});
+                        \draw[gray] ( {#1 -0.1}, {-4 - diff(#1, -4) *0.1}) --  ( {#1 +0.1}, {-4  + diff(#1, -4) *0.1});
+                      }
+                    \end{axis}
+                  \end{tikzpicture}
+
                   \begin{table*}[ht]
                     \centering
                     \begin{tabularx}{\textwidth}{|r|Y|Y|Y|}
                       \-\hspace{1em}\texttt{impDiff(y\^{}2+ax=5,\ x,\ y)}
                   \end{cas}
 
-                  \subsection*{Slope fields}
+                  \subsection*{Function of the dependent
+                  variable}
 
-                  \begin{tikzpicture}[declare function={diff(\x,\y) = \x+\y;}]
-                    \begin{axis}[axis equal, ymin=-4, ymax=4, xmin=-4, xmax=4, ticks=none, enlargelimits=true, ]
-                      \addplot[thick, orange, domain=-4:2] {e^(x)-x-1};
-                      \pgfplotsinvokeforeach{-4,...,4}{%
-                        \draw[gray] ( {#1 -0.1}, {4 - diff(#1, 4) *0.1}) --  ( {#1 +0.1}, {4  + diff(#1, 4) *0.1});
-                        \draw[gray] ( {#1 -0.1}, {3 - diff(#1, 3) *0.1}) --  ( {#1 +0.1}, {3  + diff(#1, 3) *0.1});
-                        \draw[gray] ( {#1 -0.1}, {2 - diff(#1, 2) *0.1}) --  ( {#1 +0.1}, {2  + diff(#1, 2) *0.1});
-                        \draw[gray] ( {#1 -0.1}, {1 - diff(#1, 1) *0.1}) --  ( {#1 +0.1}, {1  + diff(#1, 1) *0.1});
-                        \draw[gray] ( {#1 -0.1}, {0 - diff(#1, 0) *0.1}) --  ( {#1 +0.1}, {0  + diff(#1, 0) *0.1});
-                        \draw[gray] ( {#1 -0.1}, {-1 - diff(#1, -1) *0.1}) --  ( {#1 +0.1}, {-1  + diff(#1, -1) *0.1});
-                        \draw[gray] ( {#1 -0.1}, {-2 - diff(#1, -2) *0.1}) --  ( {#1 +0.1}, {-2  + diff(#1, -2) *0.1});
-                        \draw[gray] ( {#1 -0.1}, {-3 - diff(#1, -3) *0.1}) --  ( {#1 +0.1}, {-3  + diff(#1, -3) *0.1});
-                        \draw[gray] ( {#1 -0.1}, {-4 - diff(#1, -4) *0.1}) --  ( {#1 +0.1}, {-4  + diff(#1, -4) *0.1});
-                      }
-                    \end{axis}
-                  \end{tikzpicture}
+                  If \({\frac{dy}{dx}}=g(y)\), then
+                  \(\frac{dx}{dy} = 1 \div \frac{dy}{dx} = \frac{1}{g(y)}\). Integrate both sides to solve equation. Only add \(c\) on one side. Express
+                  \(e^c\) as \(A\).
+
+                  \subsection*{Reciprocal derivatives}
+
+                  \[\frac{1}{\frac{dy}{dx}} = \frac{dx}{dy}\]
+
+                  \subsection*{Differentiating \(x=f(y)\)}
+                  Find \(\dfrac{dx}{dy}\), then \(\dfrac{dy}{dx} = \dfrac{1}{\left(\dfrac{dx}{dy}\right)}\)
 
                   \subsection*{Parametric equations}
 
 
                 \[\int f(x) \cdot dx = F(x) + c \quad \text{where } F^\prime(x) = f(x)\]
 
-                  \subsubsection*{Definite integrals}
-
-                  \[\int_a^b f(x) \cdot dx = [F(x)]_a^b=F(b)-F(a)\]
-
-                  \begin{itemize}
-
-                    \item
-                      Signed area enclosed by\\
-                      \(\> y=f(x), \quad y=0, \quad x=a, \quad x=b\).
-                    \item
-                      \emph{Integrand} is \(f\).
-                  \end{itemize}
-
                   \subsubsection*{Properties}
 
                   \begin{align*}
                       \(\sin 2x = 2 \sin x \cos x\)
                   \end{itemize}
 
+                  \subsection*{Separation of variables}
+
+                  If \({\frac{dy}{dx}}=f(x)g(y)\), then:
+
+                  \[\int f(x) \> dx = \int \frac{1}{g(y)} \> dy\]
+
                   \subsection*{Partial fractions}
 
                   To factorise \(f(x) = \frac{\delta}{\alpha \cdot \beta}\):
                     To reverse, use \texttt{combine(...)}
                   \end{cas}
 
+                  \subsection*{Integrating \(\boldsymbol{\dfrac{dy}{dx} = g(y)}\)}
+
+                  \[ \text{if } \dfrac{dy}{dx} = g(y), \text{ then } x = \int{\dfrac{1}{g(y)}} \> dy \]
+
                   \subsection*{Graphing integrals on CAS}
 
                   \begin{cas}
                     For restrictions, \texttt{Define\ f(x)=...} then \texttt{f(x)\textbar{}x\textgreater{}...}
                   \end{cas}
 
-                  \subsection*{Applications of antidifferentiation}
-
-                  \begin{itemize}
-
-                    \item
-                      \(x\)-intercepts of \(y=f(x)\) identify \(x\)-coordinates of
-                      stationary points on \(y=F(x)\)
-                    \item
-                      nature of stationary points is determined by sign of \(y=f(x)\) on
-                      either side of its \(x\)-intercepts
-                    \item
-                      if \(f(x)\) is a polynomial of degree \(n\), then \(F(x)\) has degree
-                      \(n+1\)
-                  \end{itemize}
-
-                  To find stationary points of a function, substitute \(x\) value of given
-                  point into derivative. Solve for \({\frac{dy}{dx}}=0\). Integrate to find
-                  original function.
-
                   \subsection*{Solids of revolution}
 
                   Approximate as sum of infinitesimally-thick cylinders
                     \end{enumerate}
                   \end{cas}
 
+                  \subsection*{Applications of antidifferentiation}
+
+                  \begin{itemize}
+
+                    \item
+                      \(x\)-intercepts of \(y=f(x)\) identify \(x\)-coordinates of
+                      stationary points on \(y=F(x)\)
+                    \item
+                      nature of stationary points is determined by sign of \(y=f(x)\) on
+                      either side of its \(x\)-intercepts
+                    \item
+                      if \(f(x)\) is a polynomial of degree \(n\), then \(F(x)\) has degree
+                      \(n+1\)
+                  \end{itemize}
+
+                  To find stationary points of a function, substitute \(x\) value of given
+                  point into derivative. Solve for \({\frac{dy}{dx}}=0\). Integrate to find
+                  original function.
+
                   \subsection*{Rates}
 
                   \subsubsection*{Gradient at a point on parametric curve}
                     To verify solutions, find \(\frac{dy}{dx}\) from \(y\) and substitute into original
                   \end{warning}
 
-                  \subsubsection*{Function of the dependent
-                  variable}
-
-                  If \({\frac{dy}{dx}}=g(y)\), then
-                  \(\frac{dx}{dy} = 1 \div \frac{dy}{dx} = \frac{1}{g(y)}\). Integrate both sides to solve equation. Only add \(c\) on one side. Express
-                  \(e^c\) as \(A\).
-
 
 
                   \subsubsection*{Mixing problems}
 
                   \[\left(\frac{dm}{dt}\right)_\Sigma = \left(\frac{dm}{dt}\right)_{\text{in}} - \left(\frac{dm}{dt}_{\text{out}}\right)\]
 
-                  \subsubsection*{Separation of variables}
+                  \subsection*{Euler's method}
 
-                  If \({\frac{dy}{dx}}=f(x)g(y)\), then:
+                  \[\dfrac{f(x+h) - f(x)}{h} \approx f^\prime (x) \quad \text{for small } h\]
 
-                  \[\int f(x) \> dx = \int \frac{1}{g(y)} \> dy\]
+                  \[\implies f(x+h) \approx f(x) + hf^\prime(x)\]
 
-                  \subsubsection*{Euler's method for solving DEs}
+                  \begin{theorembox}{}
+                    If \(\dfrac{dy}{dx} = g(x)\) with \(x_0 = a\) and \(y_0 = b\), then:
+                    \begin{align*}
+                      x_{n+1} &= x_n + h \\
+                      y_{n+1} &= y_n + hg(x_n)
+                    \end{align*}
+                  \end{theorembox}
 
-                  \[\frac{f(x+h) - f(x)}{h} \approx f^\prime (x) \quad \text{for small } h\]
 
-                  \[\implies f(x+h) \approx f(x) + hf^\prime(x)\]
 
                   \include{calculus-rules}
 
         \end{align*}
 
         \noindent \textbf{Distance travelled between \(t=a \rightarrow t=b\):}
-        \[= \int^b_a \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \cdot dt \]
+        \begin{align*}
+          &= \int^{b}_{a}{\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}} \> dt \tag{2D} \\
+          &= \int^{t=b}_{t=a}{\dfrac{dx}{dt}} \> dt \tag{linear}
+        \end{align*}
 
         \noindent \textbf{Shortest distance between \(\boldsymbol{r}(t_0)\) and \(\boldsymbol{r}(t_1)\):}
         \[ = |\boldsymbol{r}(t_1) - \boldsymbol{r}(t_2)| \]