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\title{Tutor Notes}
\author{A.G.M. Pietrow}

\begin{document}

\maketitle
\newpage
\tableofcontents
\newpage
\part{AN1NA}
\section{Complex Numbers}
Complex numnbers are factors of $i^2 = -1$. Complex numbers in the form of (z = a + bi)\footnote{Unless mentioned otherwise, this is {\bf{always}} the form that you should end up with.} can be added, subtracted and multiplied as any normal polynomial. Division however works slightly different, as can be seen in \eqref{1}. You multiply both sides of the fraction with the conjugate of the denominator.

\begin{equation}\frac{z}{w} = \frac{z}{w}\cdot \frac{\overline{w}}{\overline{w}} = \frac{z \overline{w}}{\abs{w}^2} \label{1}\end{equation}

In the last part of the equation, $w\cdot \overline{w}$ is the modulus of the complex number. This represents the length of the vector in the complex space. Next to that there is also the argument of the complex number, which represents the angle between the positive real axis and the complex number vector. Both of these quantities are given in \eqref{2} and \eqref{3}.

\begin{equation}\abs{z} = \sqrt{a^2 + b^2} =  r \label{2} \end{equation}

\begin{equation}arg(z) = \arctan{\rpar{\frac{b}{a}}} = \phi\footnote{Please note that the argument is not defined everywhere, and that sometimes the angle can be intuitively found.} \label{3}\end{equation}

The complex number w, is a vector drawn between the two axis of the system. However, now that we have the modulus and argument, we can represent the same vector in polar coordinates, namely with an angle $\phi$ and a length r. The complex number can then be represented in the form of a exponential.

\begin{equation}z = \abs{z}e^{i\phi}\label{4}\end{equation}

Alternatively the exponent representation can be reversed into a complex number by use of Euler's identity.

\begin{equation}re^{i\theta} = r\rpar{\cos\rpar{{\theta}} + i\sin{\rpar{\theta}}}\label{5}\end{equation}

This is important because it enables the user to jump back and forth between trigonometry and complex numbers.

Now, at this point, one may wonder why we would ever want to go back to those awful sines and cosines after discovering this beautiful representation. Unfortunately trigonometric functions are very powerful functions which in many instances are much easier to work with than exponentials. A good example of this is Moivre's theorem:

\begin{equation}\cos(n\theta) + i\sin(n\theta) = (\cos(\theta) + i\sin(\theta))^n\label{6}\end{equation}

However, we can translate this back into the complex form, making it even easier:

\begin{equation}z^n = \abs{z}^ne^{in\theta}\label{7}\end{equation}

Note that this also works for fractions, where n = $\frac{1}{2}$.\\
\newpage
\subsection{Examples}
\subsubsection{Basic Operations}
\fbox{\begin{minipage}{0.9\textwidth}
\begin{center}\bf 1) Solve: $z = 1 + 2i \oplus w = 3 + 4i$

for $\oplus$ = +,-,$*$,/\end{center}
For $\oplus$ = + and - we can simply add and subtract the coefficients like we would with vectors.
\begin{center}$\colvec{1}{2} + \colvec{3}{4} = \colvec{4}{6} = 4 + 6i$ and $\colvec{1}{2} - \colvec{3}{4} = \colvec{-2}{-2} = -2  -2i$\end{center}
For $\oplus = *$ we multiply the numbers just like a polynomial.
\begin{center}$zw = (1+2i)(3+4i) = 3 + 4i + 6i -8 = -5 + 10i$\end{center}
For $\oplus = /$ we trick the division into becoming a multiplication, by multiplying it with the complex conjugate.
\begin{center}$\frac{z}{w} = \frac{z}{w} \cdot \frac{\overline{w}}{\overline{w}} = \frac{z\overline{w}}{\abs{w}} = \frac{(1+2i)(3-4i)}{( 3 + 4i)( 3 - 4i)} = \frac{11 + 2i }{25} = \frac{11}{25} - \frac{2}{25}i$\end{center}
\end{minipage}}
\subsubsection{Exponentials}
\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 2) Find the modulus and argument of z and w and write down the exponential.\end{center}
For modulus we apply \eqref{2}:
\begin{center}$\abs{z} = \sqrt{1^2 + 2^2} = \sqrt{5}$ and $\abs{w} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$\end{center}
For the argument we apply \eqref{3}:
\begin{center}$\arg{z} = \arctan{\frac{2}{1}} \approx 63.3^o$ and $\arg{z} = \arctan{\frac{4}{3}} \approx 53.3^o$\end{center}
For the exponential we use \eqref{4}:
\begin{center}$z = \sqrt{5}e^{i\cdot \arctan{2/1}}$ and $z = 5e^{i\cdot \arctan{4/3}}$\end{center}
\end{minipage}}\\
\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 3) rewrite $\sqrt{1 + i}$ in the form of a+bi.\end{center}
We can use \eqref{4} to rewite $\sqrt{z}$:
$$\sqrt{\abs{z}e^{i\theta}}$$
Find the absolute value and modulus:
$$\abs{z} = \sqrt{2} \text{ and } \theta =  \arctan{\frac{1}{1}} = \frac{\pi}{4}$$
Now plug in the values and use that $\sqrt{x} = x^{\frac{1}{2}}$
$$\rpar{\sqrt{2}e^{\frac{i\pi}{4}}}^{\frac{1}{2}} = \sqrt[4]{2}e^{\frac{i\pi}{8}}$$
Now use \eqref{5}
$$\sqrt[4]{2}e^{\frac{i\pi}{8}} = \sqrt[4]{2}\rpar{\cos{\frac{\pi}{8}} + i\sin{\frac{\pi}{8}}}$$
The sine and cosine become something horrible at those points, so you can leave it like this.
\end{minipage}}

\newpage
\section{Limits and Continuity of Functions}
Limits are used to check how functions behave at certain points or at $\pm$ $\infty$. Informally, the limit is defined by \eqref{8}.
\begin{equation}\lim_{x \rightarrow a} f(x) = f(a)\label{8}\end{equation}
One can also have a certain preference for approaching a limit from a certain side. Below can be seen a example for the function $f(x) = \frac{1}{x}$
\begin{equation}\lim_{x \uparrow 0} \frac{1}{x} = -\infty\text{ and } \lim_{x \downarrow 0} \frac{1}{x} = \infty \label{9} \end{equation}
Next we will look at a few limit rules, showing that limits behave a lot like normal functions.\\\\
If the limits $\lim_{x \rightarrow a}$ f(x) and $\lim_{x \rightarrow a}$ g(x) both exist, then:\footnote{\url{http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/SandS/lHopital/limit_laws.html}}
\begin{enumerate}
\item $\lim_{x \rightarrow a} f(x) + g(x) = \lim_{x \rightarrow a} f(x) + \lim_{x \rightarrow a} g(x)$
\item $\lim_{x \rightarrow a} f(x) - g(x) = \lim_{x \rightarrow a} f(x) - \lim_{x \rightarrow a} g(x)$
\item $\lim_{x \rightarrow a} f(x)g(x) = \lim_{x \rightarrow a} f(x) \cdot \lim_{x \rightarrow a} g(x)$
\item $\lim_{x \rightarrow a} c f(x) = c \lim_{x \rightarrow a} f(x)$
\item $\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \frac{\lim_{x \rightarrow a} f(x)}{\lim_{x \rightarrow a} g(x)}  $ for $\lim_{x \rightarrow a} g(x) \neq 0$
\item $\lim_{x \rightarrow a} f(x)^n = \rpar{\lim_{x \rightarrow a} f(x)}^n  $
\end{enumerate}.
\\
Unfortunately there are no rules besides this, for both finite and infinite limits. This makes them tricky, requiring some experience and insight. We will try to build that up at the examples below. \\ \\
A function is considered continuous when there are no singularities in the whole thing. This means that at every point you can equate the limits from both sides, and get the same value.
\begin{equation}\lim_{x \uparrow a} f(x) = \lim_{x \downarrow a} f(x) = f(a) \label{10}\end{equation}
When there {\bf is} a discontinuity, usually the point {\bf can} be approached from one of the two sides. When this is possible from the left side (from below the value) then it is called left continuous and vice versa.
\\\\
Just like with the section about limits, there is not much to say here, as there are no clear rules on how to solve the problems in this section. \\\\
Finally we will cover l'Hopitals rule\footnote{If you have not yet covered derivatives, then skip this part.}, which can be applied to quotients of indeterminate limits. This being the form of $\frac{\infty}{\infty}$ or $\frac{0}{0}$. According to l'Hopital, in these cases you can take the derivative of both sides of the fraction, and keep doing so until the function becomes determinate. The then obtained value becomes the value of the limit.
\newpage
\subsection{Examples}
\subsubsection{Finite Limits}
\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 1) Calculate the limit of $\lim_{x \rightarrow 3}  \frac{1}{3-x}$\end{center}
We have to split this limit into a left and right limit, and approach the value from both sides.

\begin{center}$\lim_{x \rightarrow 3} = \frac{1}{3-x} \Rightarrow \lim_{x \uparrow 3} = \frac{1}{3-x} \text{ and } \lim_{x \downarrow 3} = \frac{1}{3-x}$\end{center}

These limits are identical to the ones in \eqref{9}, only translated onto 3. They give the result of $\infty$ and $-\infty$ respectively. This means that there is no limit at that point.\end{minipage}}
\\\\
\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 2) Calculate the limit of $\lim_{x \rightarrow 0}  \frac{\abs{x-2}}{x-2}$\end{center}
Here we simply fill in x =0 and get -1.
\end{minipage}}\\\\
\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 3) Calculate the limit of $\lim_{x \rightarrow 2} \rpar{\frac{1}{x-2} - \frac{1}{x^2 -4}}$\end{center}
Merge the two fractions
\begin{center}$\lim_{x \rightarrow 2} \frac{x + 2 - 1}{(x -  2)(x + 2)} = \lim_{x \rightarrow 2}\frac{x-1}{x^2 - 4}$\end{center}
From the top we get $\infty$ and from the bottom $-\infty$.
\end{minipage}}
\subsubsection{Infinite Limits}
Limits to infinity are usually easier than finite ones, because you can use the division rule. This is a rule where you simply divide every coefficient by the highest factor of x from the denominator. \\
 \fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 4) Calculate the limit of $\lim_{x \rightarrow \infty} \frac{x + x^3 +x^5}{1 + x^2 + x^3}$\end{center}
Here $x^3$ is the highest coefficient, so divide everything by that.
\begin{center}$\lim_{x \rightarrow \infty} \frac{x/x^3 + x^3/x^3 +x^5/x^3}{1/x^3 + x^2/x^3 + x^3/x^3} = \lim_{x \rightarrow \infty}\frac{1/x^2 + 1 + x^2}{1/x^3 + 1/x + 1}$ \end{center}
All fractions of $1/x^n$ go to zero when x goes to $\pm\infty$, so we are left with:
\begin{center}$\lim_{x \rightarrow \infty} \frac{1+x^2}{1} = \infty$ \end{center}
\end{minipage}}\\
\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 5) Calculate the limit of $\lim_{x \rightarrow \infty}\rpar{\sqrt{x^2 + 2x}- \sqrt{x^2 -2x}}$\end{center}
Multiply both sides of the root with its "conjugate":
\begin{center}$\lim_{x \rightarrow \infty}{\sqrt{x^2 + 2x}- \sqrt{x^2 -2x} \cdot \frac{\sqrt{x^2 + 2x}+ \sqrt{x^2 -2x}}{\sqrt{x^2 + 2x}+ \sqrt{x^2 -2x}}} = \lim_{x \rightarrow \infty} \frac{x^2 +2x -x^2 +2x}{\sqrt{x^2 + 2x}+ \sqrt{x^2 -2x}}$\end{center}
Now extract an x from the fraction and let x go to infinity:
\begin{center}$ \lim_{x \rightarrow \infty} \frac{4}{\sqrt{1 + 2/x}+ \sqrt{1 -2/x}} = \frac{4}{2} = 2$\end{center} \end{minipage}}
\newpage
\section{Differentiation}
The whole idea with differential calculus is to find the steepness of the slope of a function (or a part of a function), which in turn will give us information about the functions behaviour.

One of the most fundamental principles in this area is the Newton Quotient, which is a method of finding the slope of the function on a given point. Hence, the derivative at that one point.\\\\
 Note that this connects limits and derivatives, showing that if the limit at a point does not exist, there also is {\bf no} derivative at that point.
\begin{equation}a = \lim_{h \rightarrow 0} \frac{f(x_0 + h) - f(x_0)}{h}\label{11}\end{equation}
When taking the limit to 0, we find the slope of the tangent line of the function at that point. We can then put that slope into the equation of a straight line, to get the function of the tangent line.
\begin{equation}y = a(x-x_0) + y_0\label{12}\end{equation}
Note that the value of a {\bf has} to be equal on both sides of the function. So if the upper and lower limit differ from each other, then there is no tangent line at that point.
\\\\
Furthermore, in these notes I assume that the reader knows of basic derivatives, and hence that the following list needs no further explanation:
\begin{enumerate}
\item d/dx c = 0
\item d/dx x = 1
\item d/dx $x^n = nx^{n-1}$
\item d/dx $1/x^n$ = d/dx $x^{-n}$ = $-n x^{-n+1}$
\item d/dx $\sqrt{x}$ = d/dx $x^{0.5}$ = $0.5 x^{-0.5} = 1/(2\sqrt{x})$
\item d/dx $|x|$ = x/$|x|$ = $\text{sgn}x$
\end{enumerate}
The same goes for the following differentiation rules:\footnote{There are several notations to denote differentiation, the most general being d/dx f(x), followed by the shorthand f'(x) or f'. The other notation will be covered later. }
\begin{equation}(f \pm g)'(x) = f'(x) \pm g'(x)\label{13}\end{equation}
\begin{equation}(cf)'(x) = c\cdot f'(x)\label{14}\end{equation}
The chain rule:
\begin{equation}d/dx f(g(x)) = f'(g(x)) * g'(x)\label{16}\end{equation}
The product rule
\begin{equation}(fg)'(x) = f'(x)g(x) + f(x)g'(x) \text{ or } u'v + uv'\label{15}\end{equation}
The reciprocal rule: \footnote{Note that this is a combination of the chain rule and rule 4.}
\begin{equation}\rpar{\frac{1}{f}}'(x) = \frac{-f'(x)}{(f(x))^2} \text{ or } \frac{d}{dx}\frac{1}{u} = \frac{-u'}{u^2}\label{17}\end{equation}
The Quotient Rule
\begin{equation}\rpar{\frac{f}{g}}'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \text{ or } \frac{d}{dx}\frac{u}{v} = \frac{u'v -uv'}{v^2}\label{18}\end{equation}
\newpage
Next we will look at some trigonometric functions, and their derivatives.
\begin{enumerate}
\item d/dx $\sin x = \cos x$
\item d/dx $\cos x = -\sin x$
\item d/dx $\tan x = 1/\cos^2 x = \sec^2 x$\footnote{This derivative is based on the two derivatives above and the quotient rule. $\tan x = \frac{\sin x}{\cos x}$ Proving it is left as an exercise for the reader.}\footnote{the sec function is simply shorthand for 1/cos. Just like the csc = 1/sin and the cot = 1/tan = cos/sin.}
\end{enumerate}

Now that we know how derivatives work, we can look at higher order derivatives, or functions which are derived multiple times.
To denote these derivatives, we can repeat the d/dx operator or square it.
\begin{equation}d/dx \rpar{  d/dx \cdot f(x)} = d^2/dx^2 \cdot f(x) = f''(x)\label{19}\end{equation}
Obviously the square notation is much more compact then the other two, as if you want to find the $n^{th}$ derivative, you would have to put either n d/dx's or n apostrophe's above the f, as opposed to d$^{n}$/dx$^{n}$.
Beyond that they work exactly the same as normal derivatives, just taken multiple times.
\\
\\
Besides this we have a handy tool called the mean-value theorem, this is a method to look at the rough behaviour of a function. Seeing if it increases or decreases at a given interval.
 \begin{equation}f'(c) = \frac{f(b) - f(a)}{b-a}\label{20}\end{equation}
Here we take two points, a and b, and find the slope of the point, c,  in between the two.
\begin{enumerate}
\item If $f(x_2)$ $>$ $f(x_1)$ is called an {\bf increasing} interval
\item If $f(x_2)$ $<$ $f(x_1)$ is called a {\bf decreasing} interval
\item If $f(x_2)$ $\geq$ $f(x_1)$ is called a {\bf nondecreasing} interval
\item If $f(x_2)$ $\leq$ $f(x_1)$ is called a {\bf nonincreasing} interval
\end{enumerate}.\\
The more generalised version of this theorem is given by the next equation:
\begin{equation}\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}\label{21}\end{equation}
\newpage
\subsection{Examples}
\subsubsection{Newton Quotient}
\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 1) Find the tangent to the function $y = x^2$ at the point (1,1).\end{center}
Here we take f(x) to be $x^2$, so $x_0$ = 1 and $y_0$ = f(1) = 1:
\begin{center}$\lim_{h \rightarrow 0} \frac{(1+h)^2 -1}{h} = \lim_{h \rightarrow 0} \frac{2h+h^2}{h} = 2$ \end{center}
This gives us a tangent of:
\begin{center}$y = 2(x-1) + 1 = 2x-1$ \end{center}\end{minipage}}\\
\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 2) Find the tangent to the function $y = \frac{1}{\sqrt{x}}$ at x = 9.\end{center}
Just as we did above, we first calculate $x_0$ and $y_0$, which turn out to be 9 and $\frac{1}{3}$ respectively.
Then we plug that into the Newton quotient and get:
\begin{center}$\lim_{h \rightarrow 0} \frac{1/(9+h) - 1/3}{h} = \lim_{h \rightarrow 0} \frac{3 - \sqrt{9+h}}{3h\sqrt{9+h}} \cdot \frac{3+\sqrt{9+h}}{3+\sqrt{9+h}} =\lim_{h \rightarrow 0} \frac{9-9-h}{3h\sqrt{9+h}\rpar{3+\sqrt{9+h}}} = \lim_{h \rightarrow 0} \frac{-1}{27+3 h+9 \sqrt{9+h}} = - \frac{1}{54}$ \end{center}\end{minipage}}\\
\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 3) Find the tangent to the function $y = |x|$ at x = 0.\end{center}
Here we have $x_0 = y_0 = 0$, so:
\begin{center}$\lim_{h \rightarrow 0} \frac{|0+h| - |0|}{h} = \lim_{h \rightarrow 0} \frac{|h|}{h}$ \end{center}
This function is 1 when you take the top limit and -1 when you take the bottom one, so the limits don't match and the function has no tangent at (0,0).\end{minipage}}
\subsubsection{Differentials}
\fbox{\begin{minipage}{0.9\textwidth}
\begin{center}\bf 4) Solve $\frac{d}{dx} x\cos(x)$.\end{center}
This is a product of two functions, so use the product rule, with u=x and v=$\cos(x)$.
\begin{center}$ 1\cdot \cos(x) + x\cdot \sin(x)$\end{center}\end{minipage}}
\\
\fbox{\begin{minipage}{0.9\textwidth}
\begin{center}\bf 5) Solve $\frac{d}{dx} y^2 = x$.\end{center}
This is an odd function, because normally we differentiate functions in the form of "y =", so we have to fix that first by taking the root on both sides.
$$\frac{d}{dx} y^2 = x \Rightarrow \frac{d}{dx} y = \pm \sqrt{x}$$
This gives us two functions to derive:
$$\frac{d}{dx} \rpar{\pm \sqrt{x}} = \pm \frac{1}{2 \sqrt{x}}$$
\end{minipage}}
\newpage
\section{Integration}
Integrals are defined as the surface under a graph, they in turn are calculated by antiderivatives. These functions are the  inverse of derivatives, sometimes called antiderivatives. In this short chapter we will simply state the rules of integration and give a few examples.
\begin{enumerate}
\item $\int a dx = ax + c$
\item $\int af(x) dx = a\int f(x) dx$
\item $\int x^n dx = \frac{1}{n+1} x^{n+1} + c$ for $n \neq -1$
\item $\int \rpar{u \pm v \pm w} dx =  \int u dx \pm \int vdx \pm \int w dx$
\item $\int u dv = uv - \int v du $ (Integration by parts)
\end{enumerate}
\subsection{Examples}
\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 1) Calculate $\int x + \cos x dx$\end{center}
Split into two integrals and integrate:
\begin{center}$\int x dx + \int \cos x dx = \frac{1}{2}x^2 + \sin x + c$ \end{center}\end{minipage}}\\
\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 2) Calculate $\int \frac{1}{(1+x)^2} dx$\end{center}
Use integration rule 3:
\begin{center}$\frac{-1}{(1+x)^3} + c$ \end{center}\end{minipage}}\\
\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 3) Calculate $\int \cos^2 x dx$\end{center}
Use $\cos^2 x$ = $\frac{1}{2}(1 + \cos 2x)$
\begin{center}$\int \frac{1}{2}(1 + \cos 2x) dx =  \frac{1}{2}(x + \sin(2x)/2) + c$ \end{center}\end{minipage}}
\newpage
\section{Transcendental Functions}
Transcendental functions are functions that are not rational, meaning that they are not expressible as a fraction of polynomials.  Just like $\pi$ is not a rational number and cannot be expressed as one. However, you can combine sets of transcendental functions to get other transcendental functions or even cancel each other out. This last thing happens when you combine a function with its inverse. This simply means that you express one variable in the other, like in \eqref{22}.
\begin{equation}y = f(x) \Rightarrow x = f^{-1}(y)\label{22}\end{equation}
This process works the easiest for {\bf one to one} functions, these are functions which never cross the same number twice. (Like the function y=x, which never has a unique x for every y.) When a function is its own inverse we call it a {\bf Self inverse}.
\\
\\
Basically the inverse is a reflection over the line y=x, turning the range into the domain and vice versa.
\\
\\
Now one of the most important transcendental functions is the natural logarithm (ln(x)), known as the inverse of the exponential function ($e^x$).
The log has some very important rules, which will not be proven here, but rather just stated.\footnote{For proofs look in a mathematical textbook.}
\begin{enumerate}
\item d/dx ln(x) = $\frac{1}{x}$
\item $\int \frac{1}{x}$ = ln(x) + c
\item ln(xy) = ln(x) + ln(y)
\item ln(x/y) = ln(x) - ln(y)
\item ln(1/x) = -ln(x)
\item ln($x^r$) = r ln(x)
\item ln($e^x$) = x
\item $\lim_{x \uparrow 0} \ln(x)= - \infty$\footnote{Depending on if you are moving from above or below, so it is not differentiable at.}
\item $\lim_{x \rightarrow \infty} \ln(x) = \infty$
\end{enumerate}.\\
Because the $e^x$ is the inverse function, many of the same rules apply:
 \begin{enumerate}
\item d/dx $e^x = e^x$
\item $\int e^x = e^x  + c$
\item $(e^x)^y = e^{xy}$
\item $e^{x+y} = e^{x}e^{y}$
\item $e^{x-y} = \frac{e^{x}}{e^{y}}$
\item $e^{-x} = 1/e^x$
\item ln($e^x$) = x
\item $\lim_{x \rightarrow 0} e^x= 1$
\item $\lim_{x \rightarrow  \infty} e^x = \infty$
\item $\lim_{x \rightarrow  -\infty} e^x = 0$
\end{enumerate}.\\

Now you can design a very large scala of functions of this type, we call these functions log functions.
A well known log function is the $\log_{10}(x)$, which is the inverse of the function $10^x$. But you can create any type of log function that follows the same rules as above.
\begin{equation}\log_n(x) \text{ is the inverse of } n^x\label{23}\end{equation}
You can also convert any log into any other log by using \eqref{24}.
\begin{equation}\log_n(x) = \frac{\log_m(x)}{\log_m{n}}\label{24}\end{equation}
However, one must keep in mind that the integration and differential rules (1 and 2) do not apply to these logs. However, we can trick any log to work for these rules by rewriting it as a ln.
\begin{equation}\frac{d}{dx} \log_n{x} = \frac{d}{dx} \frac{\ln{x}}{\ln{n}}\label{25b}\end{equation}
Here $\ln{n}$ is a constant, so it can be ignored when integrated or differentiated. The same trick can be used for integration and the exponentials.
\\
\\
Next there is the question of power, if you would make a quotient of two functions, which one would "win"?
A exponential always wins from a power, a power always wins from a multitude of x, a multitude of x always wins from a logarithm and a logarithm always wins from a constant.

\begin{equation}\lim_{x \rightarrow \infty}\frac{x^a}{e^x} = 0\label{25}\end{equation}
\begin{equation}\lim_{x \rightarrow \infty}\frac{ax}{x^a} = 0\label{26}\end{equation}
\begin{equation}\lim_{x \rightarrow \infty}\frac{\ln{x}}{ax} = 0\label{27}\end{equation}
\begin{equation}\lim_{x \rightarrow \infty}\frac{a}{\ln{x}} = 0\label{28}\end{equation}
\newpage
\subsection{Examples}

\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 1) Find the inverse of f(x) = $\sqrt{x-1}$ \end{center}
Substitute f(x) with y, isolate the x and then replace x with f'(y).

\begin{center}$y^2 = x - 1 \Rightarrow x = y^2 + 1 \Rightarrow f'(y) = y^2 + 1$\end{center}
{\bf Note} that this only applies for x $\geq$ 0!\end{minipage}}\\
\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 2) Find the inverse of f(x) = $\frac{x}{\sqrt{x^2 +1}}$ \end{center}
Same substitutions as above:
\begin{align*}y\sqrt{x^2 +1} = x &\Rightarrow y^2(x^2 +1) = x^2 \\&\Rightarrow y^2 = (1-y^2)x^2 \\&\Rightarrow \sqrt{\frac{y^2}{1-y^2}} = x \\f'(y) &= \sqrt{\frac{y^2}{1-y^2}} \end{align*}\end{minipage}}\\
\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 3) Simplify the following expression: $e^3/\sqrt{e^5}$ \end{center}
Use exponent rule 5:
\begin{center}$e^{3-5/2} = e^{1/2} = \sqrt{e}$\end{center}\end{minipage}}\\
\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 4) Solve for x: $2^{x+1} = 3^x$ \end{center}
Use log rule 6 and 4:
\begin{align*}(x-1)\ln(2) = x\ln(3) &\Rightarrow -\ln(2) = (\ln(3) - \ln(2))x \\&\Rightarrow x = \frac{-\ln(2)}{\ln{(3/2)}} \\&\Rightarrow x = -\log_{3/2}(2)\end{align*}
The last step is not really nesesary, but fun to do.\end{minipage}}
\newpage
\section{Understanding and Drawing Functions}
Now having learnt some of the basics of function behaviour and differentiation, we can focus ourselves at function analysis. Which is where we learn how to describe and see the behaviour of functions. There are a few important terms in this section which will be used a lot, so try to learn these by heart.
\\\\
When a function has an {\bf absolute maximum} or {\bf absolute minimum}, it means that this maximum is the highest or lowest point of the function across the whole domain. For example the Bessel Function ($J_0$)\footnote{It's basically a cosine that gets more dampened the farther it is away from x=0. Google it!}, has an absolute maximum at x=0 and a lot of  local maxima and minima across the rest of its tail. A {\bf maximum local maximum} or {\bf minimum} is a maximum which does not surpass the absolute maximum.
\\\\
Next there are {\bf critical points}, which are points where the derivative f'(x) =0. {\bf Singular points}, where f'(x) is not defined and {\bf endpoints}, where the domain ends.
\\\\
Now, you may wonder; "How do I find these points?" Well, luckily there are theorems for that. One of the most used ones is called the {\bf first derivative test}, which basically is a test that compares the derivative at two intervals.
\begin{enumerate}
\item if f'(x) $<$ 0 for [a,$x_0$] and f'(x) $>$ 0 for [$x_0$, b], then you have a local maximum.
\item if f'(x) $>$ 0 for [a,$x_0$] and f'(x) $<$ 0 for [$x_0$, b], then you have a local minimum.
\item if  f'(x) $>$ 0 for (a,b), then it has a local minimum at a and a maximum at b.
\item if  f'(x) $<$ 0 for (a,b), then it has a local maximum at a and a minimum at a.
\end{enumerate}.\\
Besides that a function can have {\bf asymptotes}, which are places on a function where the function goes to infinity when approaching a point. (1/x has an asymptote at x=0, for example.) Now, these asymptotes come in a few different flavours, namely:
\begin{enumerate}
\item Vertical Asymptotes when $\lim_{x \uparrow a} f(x) = \pm \infty$ or $\lim_{x \downarrow a} f(x) = \pm \infty$
\item Horizontal Asymptotes when $\lim_{x \rightarrow \infty} f(x) = L$ or $\lim_{x \rightarrow \infty} f(x) = L$, with L a constant value.
\item Oblique Asymptotes when  $\lim_{x \rightarrow \pm \infty} \rpar{f(x) - (ax + b)} = 0$
\item if  f'(x) $<$ 0 for (a,b), then it has a local maximum at a and a minimum at a.
\end{enumerate}.\\

An oblique asymptote can only exist when the function is a quotient of two polynomials where the numerator is of exactly 1 higher degree than the denominator. Next you take the quotient of the two coefficients. So if we look at the next equation:
\begin{equation}f(x) = \frac{a_1x^{n+1} +  b_1x^{n} \cdots}{a_2x^{n} + b_2x^{n-1} \cdots}\label{A}\end{equation}
Then the slope is given by the long division of the two quotients.\footnote{You can ignore the remainder, as this term drops to 0 at large x's.}\\
\newpage
Now that we know all this we can make a checklist for plotting a graph:
\begin{enumerate}
\item Calculate f'(x) and f''(x)
\item Examine f(x) and look for asymptotes, symmetry and "obvious"\footnote{Basically anything that will help you get some information about the graph. The more points you know, the easier it becomes!} points.
\item Examine f'(x) and look for critical points, undefined points and local/absolute maxima/minima\footnote{Use derivative test 1 and 2.}
\item Examine f''(x) and look for points where it is 0, undefined and going up or down.\footnote{When you have an interval where there is a maximum in f''(x), then the function will be {\bf concave up} and the other way around when there is a minimum.}
\end{enumerate}.\\
It is best to make little tables where you set out f(x), f'(x) and f''(x) against some "obvious" points and the location of asymptotes, singular points, etc. Basically anything that could shed any information on the function. \\\\

When functions are too hard to estimate with the above methods, there are several methods to simplify it in such a way that they can be easily plotted and used in other ways. One of these methods is the {\bf Taylor polynomial}, which is a magical function, capable of expressing any continuous and differentiable function, around a chosen point, as a series of its derivatives. This is one of the most powerful and often used tools mathematical tools of the average physicist/astronomer.\footnote{This means that they are important, so learn it by heart!}
The function itself is given by \eqref{29}.
\begin{equation}P_n = \sum^{\infty}_{n=0} \frac{f(a)^{(n)}}{n!}(x-a)^n\label{29}\end{equation}
The series is infinite, but with only a few terms we can very quickly get a very accurate result. Below is the first 3 terms of the series:
 \begin{equation}P_2 = f(a) + \frac{f'(a)}{1!}(x-a) + \frac{f''(a)}{2!}(x-a)^2\label{30}\end{equation}
\\
There are a few very important series that can be Taylor developed:
\begin{enumerate}
\item $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$
\item $\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$
\item $\sin^x = x - \frac{x^2}{2!} + \frac{x^3}{3!} - \cdots$\footnote{Note that $\cos x + i\sin x =  e^{ix}$ from \eqref{5}.}
\item $\frac{1}{1-x} = 1 + x + x^2 + x^3 + \cdots$
\item $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$
\end{enumerate}.\\
\newpage
\subsection{Examples}
\subsubsection{Asymptotes}
\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 1) Find the domain and asymptotes of $\frac{x^2 + 3x +1}{4x^2 -9}$ \end{center}
Look for singularities in the denominator:
$$4x^2 -9 = 0 \Rightarrow \pm \sqrt{\frac{3}{2}}$$
This means that at $x=\pm \frac{3}{2}$ the function has a vertical asymptote and that that point is not in the functions domain.\\
\\
Next, try the limit to $\pm \infty$
$$\lim_{x \rightarrow \infty} f(x) = \frac{1}{4} \text{ and } \lim_{x \rightarrow -\infty} f(x) = \frac{1}{4}$$
This means that the function has a vertical asymptote on both sides that goes to $\frac{1}{4}$ and that all the other numbers are in the functions domain.
\end{minipage}}\\
\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 2) Find the domain and asymptotes of $\frac{-3x^2 }{x-1}$ \end{center}
We see an obvious singularity at x=1, this shows a vertical asymptote at that point. \\
\\
Next  we se that there is a oblique asymptote, so lets do long division:
\begin{align}x-1/&-3x^2\backslash -3x - 3 - \frac{1}{x-1}  \\&\underline{3x^2 +3x}_- \\&-3x\\&\underline{-3x + 1 }_-\\&1\end{align}
So our saymptote will be allong the line $-3x -3 - \frac{1}{x-1}$. However, we can see that the last term will go to 0 for large x's, so the line is y = -3x - 3. \\
\\
So the domain is $[-\infty, \infty]$ and does not exist on x=1.
\end{minipage}}
\subsubsection{Taylor}
\fbox{\begin{minipage}{0.9\textwidth}\begin{center}\bf 4) calculate $x^{\frac{1}{3}}$ about 8 and approximate $9^{\frac{1}{3}}$ \end{center}
We want to taylor expand f(x), around 8 and then fill in x=9:
$$P_2(x) = f(8) + \frac{f'(8)}{1!}(x-8) + \frac{f''(8)}{2!}(x-8)^2$$
Find f'(x) and f''(x)
$$f'(x) = \frac{1}{3}x^{\frac{-2}{3}} \text{ and } f''(x) = \frac{-2}{9}x^{\frac{-5}{3}}$$
Fill in the derivatives:
\begin{align*}P_2(x) &= 8^{\frac{1}{3}} + \frac{8^{\frac{-2}{3}}}{3}(x-8) + \frac{-2\cdot 8^{\frac{-5}{3}}}{9\cdot 2}(x-8)^2
& = 2 + \frac{1}{12}(x-8) - \frac{1}{288} (x-8)^2
\end{align*}
Fill in x=9 and get $\approx$ 2.07986
\end{minipage}}
\newpage
Session 1:
\begin{enumerate}
\item Express $|z|$ = 2 in the form a+bi and draw the result in the complex plane.
\item Solve $\sqrt[3]{-1}$ (for all 3 answers)
\item Find the solutions to: $z^4 + 1 - i\sqrt{3} =0 $
\item Calculate the limit of $\lim_{x \rightarrow \infty} \frac{x\sqrt{x+1}\rpar{1-\sqrt{2x+3}}}{7-6x+4x^2}$
\item Calculate the limit of $\lim_{x \rightarrow -\infty} x-\sqrt{x^2 + 1}$
\item Calculate the limit of $\lim_{x \rightarrow 0} 2xe^{x^8}/x^2$
\item Find the tangent of $x^3$ + x on point (2,10)
\item Find d/dx $\tan(x)$
\item Find d/dx \rpar{ y $\sin(x) = x^3 + \cos(y)$}
\item Find $\int x^3 dx$
\item Find $\int a^2 - x^2 dx$
\item Give $\lim_{x \rightarrow 0} x^a lnx$
\item Calculate $\frac{d}{dx}\log_\pi{x^2}$
\item Simplify $\ln{(x^2 + 6x + 9)}$
\item Simplify $e^{(3\ln{9})/2}$
\item Draw $y = \frac{x^3 -4x}{x^2 -1}$
\item Find $\sqrt{0.3}$ by Taylor expanding $\sqrt{x}$ around 1
\item Taylor expand $\gamma = \sqrt{\frac{1}{1-v^2/c^2}}$ so that you can calculate $\gamma$ for v=0.99c

\end{enumerate}


\end{document}