Webpage of Richard Williams

MTH 121

Deductive Reasoning is General$\rightarrow$Specific, Inductive Reasoning is Specific$\rightarrow$General
Truth Tables for NOT, AND, OR, and IMPLIES

$P$	$Q$	$\neg P$	$P\wedge Q$	$P\vee Q$	$P\rightarrow Q$
T	T	F	T	T	T
T	F	F	F	T	F
F	T	T	F	T	T
F	F	T	F	F	T

Found a great guide on two-way tables here.

When you convert units...
- Start by making a path from the units you have to the units you want. For example, when going from seconds to years,$$\text{second }\rightarrow\text{ minute }\rightarrow\text{ hour }\rightarrow\text{ day }\rightarrow\text{ year}$$
- Take the units you have and, using the conversion factor, multiply by a fraction with the unit you want on top (according to our path) and your current units on the bottom. $$30\text{ sec} \left(\frac{1 \text{ min}}{60\text{ sec}}\right)\left(\frac{1\text{ hr}}{60\text{ min}}\right)\left(\frac{1\text{ day}}{24\text{ hr}}\right)\dotsc$$

A great place to practice scientific notation can be found here. While it is a chemistry article, the guide of working with powers of 10 is incredibly useful.

MTH 127

Function transformations

Work inside out! Turn into $$af(b(x-c))+d,$$ Inner Translation → Inner Coefficient → Outer Coefficient → Outer Translation

Midpoint of $(x_0,y_0),(x_1,y_1)$ is given by $$\left(\frac{x_0+x_1}{2},\frac{y_0+y_1}{2}\right)$$ and the distance $d$ is given by $$d=\sqrt{(x_0-x_1)^2+(y_0-y_1)^2}.$$

Intervals are lenghts described by their endpoints. Whenever the endpoint is included (a filled dot), we use square brackets. Whenever the endpoint is not included (an open dot), we use parenthesis. When one endpoint is $\infty$ or $-\infty$, we generally use parenthesis since we cannot realistically reach that endpoint.
The average rate of range of a function $f$ in the interval $[a,b]$ is given by $$\frac{f(b)-f(a)}{b-a}.$$

Even functions have the property that $f(-x)=f(x)$

Odd functions have the property that $f(-x)=-f(x)$.

When given $f(x)=ax^2+bx+c$, you can turn it into vertex with the process below and get a formula! \begin{align*} f(x) &= ax^2+bx+c \\ &= a\left(x^2 + \frac{b}{a}x\right) + c \\ &= a\left( x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 \right) + c \\ &= a\left( \left(x+\frac{b}{2a} \right)^2 - \left(\frac{b}{2a}\right)^2 \right)+c \\ &= a\left(x+\frac{b}{2a}\right)^2 - \frac{ab^2}{4a^2}+c \\ &= a\left(x+\frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c \end{align*}
Notice the equation is now in vertex form, $f(x)=a(x-h)^2 + k$ where $(h,k)$ is the vertex and $$h=-\frac{b}{2a}\quad\quad\text{and}\quad\quad k=-\frac{b^2}{4a}+c.$$

When you're given the vertex form $f(x)=a(x-h)^2+k$ and need to produce the standard form, just multiply it out! \begin{align*} f(x) &= a(x-h)^2 + k \\ &= a(x^2-2hx+h^2)+k \\ &= ax^2 -2ahx + ah^2 + k \\ &= ax^2 -2ahx + (ah^2+k)\end{align*}

Here's the steps to find the quadratic formula if you're interested! \begin{align*} ax^2+bx+c&=0 \\ x^2+\frac{b}{a}x+\frac{c}{a}&=0 \\ x^2+\frac{b}{a}x&=-\frac{c}{a} \\ x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2 &= -\frac{c}{a} + \left(\frac{b}{2a}\right)^2 \\ \left(x+\frac{b}{2a}\right)^2 &= -\frac{c}{a}+\frac{b^2}{4a^2} = \frac{b^2-4ac}{4a^2} \\ x+\frac{b}{2a} &= \pm\sqrt{\frac{b^2-4ac}{4a^2}} \\ &= \pm\frac{\sqrt{b^2-4ac}}{2a} \\ x &= -\frac{b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a} \\ &= \frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{align*}

MTH 140

To find the derivative of a function at a point $c$, you evaluate the limit $$f'(c)=\lim_{h\rightarrow 0} \frac{f(c+h)-f(c)}{h}.$$

Example: Let $f(x)=3x^2-5x+7$. Find $f'(11)$.

We start with the limit definition and substitute $x=11+h$ and $x=11$ into $f(x)$. \begin{align*} f(11+h) &= 3(11+h)^2 - 5(11+h)+7 \\ &= 3(11^2+2(11)(h)+h^2) - 5(11+h)+7 \\ &= 3(121+22h+h^2)-5(11+h)+7\\ &= 363+66h+3h^2-55-5h+7 \\ &= 315+61h+3h^2 \\[5mm] f(11) &= 3(11)^2-5(11)+7 \\ &= 3(121)-55+7 \\ &= 363-55+7 \\ &= 315.\end{align*}Now lets plug that result into our limit! \begin{align*} f'(11) &= \lim_{h\rightarrow 0} \frac{f(11+h)-f(11)}{h} \\ &= \lim_{h\rightarrow 0} \frac{(315+61h+3h^2)-(315)}{h} \\ &= \lim_{h\rightarrow 0} \frac{61h+3h^2}{h} \\ &= \lim_{h\rightarrow 0} 61+3h \\ &= 61 \end{align*}

To avoid plugging in values for every point we want a derivative of, we often compute it without plugging in a specific $x$.

Example: $g(x)=x^2+3x$. Find $g'(x)$.

We use the same limit definition and directly substitute. \begin{align*}g'(x) &= \lim_{h\rightarrow 0} \frac{g(x+h)-g(x)}{h} \\ &= \lim_{h\rightarrow 0} \frac{\big((x+h)^2 + 3(x+h)\big) - (x^2+3x)}{h} \\ &= \lim_{h\rightarrow 0} \frac{(x^2+2xh+h^2+3x+3h) -x^2-3x}{h} \\ &= \lim_{h\rightarrow 0} \frac{2xh+h^2+3h}{h} \\ &= \lim_{h\rightarrow 0} 2x+h+3 \\ &= 2x+3 \end{align*}We could now use this to find the derivative of $g$ at any $x$ value we like. Try verifying it works for $g'(-3/2)$.

I prefer to do it this way because, as seen in the previous example, putting in explicit values can lead to numbers in the equation that are not so easy to deal with when you cannot use a calculator.

Here is the problem you wanted me to put here from our meeting on Oct 6!

Find the derivative of $f(x)=3x-8x^2+8$. \begin{align*} f'(x) &= \lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ &= \lim_{h\rightarrow 0}\frac{\left[3(x+h)-8(x+h)^2+8\right]-(3x-8x^2+8)}{h} \\ &= \lim_{h\rightarrow 0} \frac{\left[3x+3h-8(x^2+2xh+h^2)+8\right]-3x+8x^2-8}{h} \\ &= \lim_{h\rightarrow 0} \frac{3x+3h-8x^2-16xh-8h^2+8-3x+8x^2-8}{h} \\ &= \lim_{h\rightarrow 0} \frac{3h-16xh-8h^2}{h} \\ &= \lim_{h\rightarrow 0} 3-16x-8h \\ &= 3-16x \end{align*}

MTH 220

Nothing here yet!

MTH 229

The derivative of a function $f$ at $x=a$ is $$f'(a)=\lim_{h\rightarrow 0}\frac{f(a+h)-f(a)}{h}$$

We talked about funny stuff that happens with the axiom of choice during one meeting for entertainment. The axiom of choice "loosely" states $$\text{For any nonempty set, we can make a choice function } f \text{ that maps this set to an element of that set, our "chosen element".}$$ Here are some problems when you accept this axiom:
- You can prove that there are disjoint subsets $A,B$ of $\mathbb{R}$ where the "length" (more specifically, the outer measure) of $A\cup B$ is not equal to the sum of the length of $A$ and length of $B$. You learn more about this in the topic of measure theory. Almost all of measure theory can be thrown out the window without choice, since we jump through many hoops of definitions and criteria to have $\mu(A\cup B)=\mu(A)+\mu(B)$
- Often, it is used to prove that things exist that cannot explicitly be constructed. We cannot explicitly construct $A,B$ in above, but we know they exist! This is seen by many as a philisophical problem.
- The Banach-Tarski Paradox is a consequence of the axiom of choice that says that we can cut up a sphere into some finite number of pieces and move them around to make two copies of the sphere of equal volume.
Similarly, when you reject this axiom:
- You can no longer choose certain items from a set casually. It becomes mathematically not guaranteed that you can make a function to describe which sock from a pair you want to put on first one morning.
- Without the axiom of choice, you can split up a set into more patitions than the original had elements! "How to have more things by forgetting how to count them", see Lemma 6.10. In summary of the proof, even though $A$ has no countably infinite subset (i.e, it's Dedekind-finite), its finite subsets can be enough in number that the set $[A]^{<\omega}$ is Dedkind-infinite, informally saying "there are more finite subsets than elements". Big words and big definitions, but very interesting and unintuitive claim!
- You can make an infinite cardinality smaller than $\aleph_0$, the number of integers.
Many more neat consequences of rejecting/accepting can be found here. If you're interested seeing excerpts from the book or learning more, let me know!

MTH 300

Nothing here yet!

MTH 335

An ordinary differential equation with solution $y(x)$ is linear if it can be written as $$\sum_{i=0}^n a_i(x)y^{(i)}=g(x).$$

A linear ordinary differential equation of order $n$ has a unique solution on the interval $I$ when it can be written in the form $$y^{(n)} + \sum_{i=1}^{n-1}a_i(t)y^{(i)}=f(t)$$ when all $a_i(t)$ and $f(t)$ are continuous on $I$. The interval of existence/uniqueness, when given an initial value $y(t_0)=y_0$, is the maximal interval of existence $I_0$ where $t_0\in I_0$.

An ordinary differential equation with solution $y(t)$ is called separable if it can be written as $$\frac{\dd y}{\dd t}=g(t)h(y).$$

THE FIRST STEPS OF SOLVING A DIFFERENTIAL EQUATION: Identify identify identify! Once you identify the type of ODE, you get the solution method. Practice just identifying different ODEs - you need to be at the point where it becomes almost immediate recognition to succeed in the course!

If you have a separable ODE $$\frac{\dd y}{\dd t}=g(t)h(y),\quad y(t_0)=y_0$$, you use the following method: \begin{align*} \dd y&=g(t)h(y) \dd t \\[3mm] \frac{\dd y}{h(y)} &= g(t)\dd t \\[3mm] \int \frac{\dd y}{h(y)} &= \int g(t)\dd t \end{align*} Solving for $y$ from here with give you a general solution, then use your initial condition to solve for you constant term.

If you have a nonseparable linear 1st-order ODE $$y'+p(t)y=q(t),\quad y(t_0)=y_0,$$ the solution method is to find an integrating factor $$\mu(t) = e^{\int p(t)\dd t} = \exp\left(\int p(t) \dd t\right).$$ Notice that $\mu'(t)=p(t)\mu(t)$. Now we multiply by $\mu(t)$, \begin{align*} y'+p(t)y&= q(t) \\ \mu(t)y'+ p(t)\mu(t)y &= \mu(t)q(t) \\ \mu(t)y' + \mu'(t)y &=\mu(t)q(t) \\ \frac{\dd}{\dd t}\left[\mu(t)y\right] &= \mu(t)q(t) \\ \mu(t)y &= \int \mu(t)q(t)\dd t \\ y &= \frac{1}{\mu(t)}\int \mu(t)q(t)\dd t. \end{align*}Plugging in initial conditions for the general solution produced here solves the ODE.

If you have a linear 2nd-order homogenous ODE $$ay'' + by' + cy = 0,\quad y(t_0)=y_0,\quad y'(t_1)=y_1$$ where $a,b,c$ are constant, the solution method is to "assume what the solution looks like", called an ansatz. For these problems, our ansatz is $$y(t)=e^{\lambda t}.$$ Taking derivatives of $y$, we also get \begin{align*} y'(t) &= \lambda e^{\lambda t} \\ y''(t) &= \lambda^2 e^{\lambda t}. \end{align*} By substituting our $y, y', y''$, we have \begin{align*} ae^{\lambda t} + b\lambda e^{\lambda t} + c\lambda^2 e^{\lambda t}&=0 \\ a + b\lambda + c\lambda^2 &= 0 \end{align*} This a quadratic! Generally you'll only have real $\lambda$, but if you get a complex $\lambda$, worry ONLY about the real parts of $\lambda$. There are two cases - you have 1 or 2 unique $\lambda$.

For one unique $\lambda$, the solution $y$ takes the form $$y(t)=\left(C_1 + C_2\lambda\right)e^{\lambda t}.$$ Use your initial conditions to find $C_1, C_2$ with simultaneous systems of equations.

For two unique $\lambda$, the solution $y$ takes the form of a linear combination of your ansatz, being $$y(t)=C_1 e^{\lambda_1 t} + C_2 e^{\lambda_2 t}.$$ Use initial conditions to find your $C_1, C_2$ with simultaneous systems of equations.

The Laplace transform is defined as $$\mathcal{L}\{f\}(s) = \int_0^\infty f(t)e^{-st} \dd t.$$ You should make sure you know how to compute these! The inverse transform requires contour integration and I don't expect this to be covered in class. Singularities as describe there are, simply put, complex $s$ that make $F(s)$ undefined. Thus, to compute and invert the most common functions, refer to the table below. The hardest part is making the form of the function match!