Erik McClure

Leap Motion Impressions, Input Sanitation, and 3D Gesture Ideas


**Pros:**
  • For the most part, does what it claims it does, and gives you extremely precise, fast tracking of fingers.

Cons:

  • Really hates thumbs for some reason.
  • Has a lot of trouble with pens or other implements.
  • Fingers must be separated.
  • Fairly easy to get positions that break the camera because it can’t see the fingers.
  • No one has any idea how to write software for it.

I just got my Leap Motion device today, and for the most part, I like it. It does most of the things I expected it to do. It seems like something that could become very useful down the line. Unfortunately, right now, I wouldn’t recommend getting one if you wanted something that’s more than a toy.

This isn’t entirely the fault of the device itself, but more a combination of driver limitations and boneheaded software that doesn’t know what it’s doing. The device is so new, no one knows how to properly utilize it, so almost all the apps that exist right now are either garbage or ridiculously sensitive to your setup. There are, however, a few more serious problems that should be addressed by the Leap Motion team.

First of all, the driver hates thumbs. It seems like it’s trying to edit them out when they aren’t positioned like a finger, which makes any sort of pinching gesture impossible to do. In addition, it doesn’t like it when you try to use a pen, despite the fact that it’s specifically designed to let you do so. Both of these problems seem more like driver analysis problems, so I expect they’ll be corrected soon. I know these aren’t hardware limitations, because the camera can see the stupid pen, the driver just has to recognize it as a tool and disregard the hand holding it, which is really doesn’t want to do - It’s obsessed with hands. I should call it the Lyra driver from now on.

A more fundamental and concerning issue is the ease in which I can assume a position that blocks the camera’s view of some or most of my fingers. Pretty much anything that involves my fingers being vertically aligned breaks the driver, along with everything from a roughly 90 degree to 135 degree angle, pointed away from the device. This appears to be an inherent issue with the device itself, and its the same problem the kinect would have, because the camera’s view of the fingers gets blocked. How much more effective would a leap motion device with two cameras, one on either side of the monitor, be? Or just a wider camera angle (since it’s using dual emitters anyway to get depth information). As it is right now, it’s cute and futuristic looking, but any minority-report-esque gestures completely break it.

A much more unexpected issue is the fact that the device makes no attempt to deal with fingers that are held together, despite the fact that the shape of two touching fingers is fairly easy to figure out. If you’ve been tracking one finger and suddenly it meets up with another one and you end up with one really wide “finger”, it should be trivial to send this information as a “double finger” pointing, because this is an incredibly useful gesture (see below for details). It’s not like the finger is going to change into a hippopotamus all of a sudden.

3D Gestures

Most of the other problems are software oriented, not hardware or driver related. The simple fact is that no one has any idea how to properly utilize the device. Most apps seem to assume there’s only going to be one finger pointed at the screen, and will get confused if you have any other fingers hanging off even if they’re not pointed at the screen. The Leap Motion driver gives you a rough estimate of what it thinks the hand orientation is, and using this information its fairly trivial to figure out which finger someone is pointing with, even if their other fingers are only slightly below that finger. The camera is on a flat surface at a 90 degree angle from the screen, which means the finger you want is almost always going to be the finger that is the “highest” in relation to the hand orientation.

The rest of the apps either don’t do much processing on the raw data, or do an incredibly bad job of it. The touch screen recreation had this horrible tendency to start sliding off, and the cursor would sometimes jump around. I knew it shouldn’t be doing that because I’ve been watching the diagnostic information and it’s just getting confused by ghost fingers popping in and out. Guess what? A finger can’t teleport. If you’ve been tracking one finger, chances are you should keep tracking it no matter what else happens.

In addition, no one’s really come up with effective 3D gestures. All the apps try to use the 3D plane by turning it into a virtual touchscreen and it doesn’t work very well. Instead, we should be using gestures that are independent of a 2D plane. The apps need to be paying attention to the angle the finger is pointing in instead of trying to rely on moving your entire hand around. It seems like everyone is using the position information instead of the angle information, which the driver provides. If I’m pointing up and to the right, my cursor should be in the corner of the screen no matter where my hand is.

Furthermore, no one seems to be trying very hard to deal with noise. When the driver starts losing precision, you can tell because the angle information or position information will start to jitter. The more jittering, the less accurate the reading is. A simple way to quantify this is by calculating the variance of the recent data. The higher the variance, the less accurate your reading is. You should then scale how heavily you average out the points based on how much variance there is in the data. This allows you to avoid destroying precision while preventing the cursor from exploding when you lose accuracy. There’s a lot of algorithms dedicated to solving this exact problem, so it’s kind of ridiculous that no one is using them. In a more specific scenario, humans do precise manipulations with their fingers, not their hands. Small pertubations in the input data that occur in the hands should be removed from the finger positions, because it’s just the hand trembling, not the actual finger.

Instead of trying to recreate “pushing a button” in 3D, we should use a gesture interface that lets you move the cursor around with your index finger, and then when you bring up your middle finger against your index finger (so you are now pointing at the screen with two fingers), this counts as a “click” or a mouse-down. You can then drag things around with two fingers, or separate them to generate a mouse-up event. To right-click, bring your two fingers together again, and then rotate your hand clockwise. This could be combined with a more traditional “pushing a button” gesture for easier clicks, but provides a much more robust way of moving things around that isn’t awkward or difficult to use.

Since the driver currently doesn’t let you track two fingers held together, this same behavior can be emulated by simply keeping track of two fingers and detecting when both of them accelerate towards each other until one of them vanishes. On this same note, because fingers can drop out of tracking, programs can’t rely on the driver to be perfect. If you lose track of a finger, it’s not that hard to guess where it might be based on the orientation of the hand, which is necessary if we’re going to have good gesture recognition. The recognizer needs to be able to interpolate likely positions of lost fingers if it’s going to make an effective guess about what gesture was intended.

Leap Motion is really cool. Leap Motion is definitely the future. Unfortunately, we haven’t gotten to the future yet. We need a better software framework to build gesture interfaces on top of, and we need a new standard set of gestures for 3D that aren’t simply 2D ripoffs. Incidentally, I’d love to try building an open-source library that addresses a lot of the input sanitation problems I outlined in this article, but I’m a bit busy with the whole “finding a job so I don’t starve to death” thing. Hopefully, someone else will take up the cause, but until they do, don’t expect to be very productive with the Leap Motion.


Aurora Theory Released!


Aurora Theory has been released! Buy it on bandcamp for $9, or $10 on iTunes, Amazon, and Google Play. The album is also available on Spotify, last.fm, other online radios, and can be previewed on YouTube.

Aurora Theory has been 4 years in the making, a compilation of all the songs I managed to make in the middle of attending university. The earlier songs have been extensively improved, and all songs have been remastered for the album’s release. This album represents nothing less than the highest quality work I can possibly manage at this point in my career. I’ve acquired a fair number of fans over the years, and I want to thank you for supporting my musical endeavors so far.

Track List:

  1. Soar (Original Mix) [5:49]
  2. Aurora Theory (Redux) [4:10]
  3. Cosminox [5:41]
  4. Tendril [5:32]
  5. Birefringence [3:57]
  6. Critical Damage (Extended Mix) [3:45]
  7. Starstruck [4:22]
  8. Dream Among The Stars [4:06]
  9. At The Nationals [4:02]
  10. The Cloud River [5:38]
  11. The Piano And The Cello [3:53]
  12. Snowflower [5:53]
  13. Spiral Nebula [5:44]
  14. Next Year [5:31]

What I Learned In College


*"In times of change, learners inherit the earth, while the learned find themselves beautifully equipped to deal with a world that no longer exists."* ― Eric Hoffer

Yesterday, the University of Washington finally mailed me my diploma. A Bachelor of Science in Applied Computational Math and Science: Discrete Math and Algorithms. I learned a lot of things in college. I learned how to take tests and how to pinpoint exactly what useless crap a particular final needed me to memorize. I learned that math is an incredibly beautiful thing that has been butchered so badly I hated it all the way until my second year of college. I learned that creativity is useless and all problems have one specific right answer you can find in the back of a textbook somewhere, because that’s all I was ever graded on. I learned that getting into the CSE major is more about fighting an enormous, broken bureaucratic mess than actually being good at computer science. But most of all, I learned that our educational system is so obsessed with itself it can’t even recognize it’s own shortcomings.

The first accelerated program I was accepted into was the Gifted program in middle school. I went from getting As in everything to failing every single one of my core classes. Determined to prove myself, I managed to recover my grades to Bs and Cs by the end of 7th grade, and by the end of 8th grade I was back up to As and Bs. I didn’t do this by getting smarter, I did it by getting better at following directions. I got better at taking tests. I became adept at figuring out precisely what the teacher wanted me to do, and then doing only that, so I could maximize both my free time and my grades. By the time I reached high school, I would always meticulously go over the project requirements, systematically satisfying each bullet point in order to maximize my score. During tests, I not only skipped over difficult questions, I would actively seek out hints in the later questions to help me narrow down possible answers. My ability to squeeze out high grades had more to do with my aptitude at filling in the right bubbles on a piece of paper then actually understanding the material.

I fantasized about attending college, where I would be judged on my intellectual prowess, and not on my test taking skills. I longed for the pursuit of knowledge in it’s purest form, only for this dream to be completely and utterly crushed. Instead of a place free from the endless battery of tests I had been subjected to during high school, I quickly realized that college was nothing but tests. I once had a math course where 95% of my grade was split between a first midterm, a second midterm, and a final. By the end of my second year of college, I simply stopped attending lectures. I could teach myself the material out of the textbook, and went to class only to take a test or turn in homework. I earned my degree by becoming incredibly adept at memorizing precisely which useless, specific facts were needed to properly answer questions. I was never asked nor told how to apply these to real world scenarios.

Thankfully, in one of the last classes I actually attended lecture in, the TA teaching the class said something that sparked a epiphany in me: “Math is simply repeated abstraction and generalization.” Suddenly, I was able to connect math and programming, and began to realize that I had loved math all my life. What I hated about math was the trivial nonsense they taught in middle school. I signed up for the most advanced math classes I could get away with, even when I could barely pass them. I began to realize that the most important thing these classes taught me was what I didn’t know. Once I knew what I didn’t know, I could teach it to myself, but only after I found the holes in my knowledge. You can’t fill a hole if you don’t know where it is. I didn’t know what combinatorics was until it was mentioned to me by that TA; Chrome still doesn’t think combinatorics is even a word.

Everyone finds the beauty of math in their own way, but we teach it like an automated assembly line of cars. Math is taught as some kind of rigid tool, when it is really a language for expressing logic, one with multiple dialects, each with their own personality. We invented music to express emotions that cannot be described; we invented math to express logical abstractions that defy explanation. Every tool in math is like another instrument in a grand orchestra, each note echoing off the others, reflecting a whole that is greater than the sum of its parts. Some composers prefer the string section, others prefer the woodwinds. There is no single right answer, only different ones. Instead of giving our children a brush and telling them to use their imagination, we give them a coloring book and grade them on how well they stay inside the lines.

I mean, we all know creativity is overrated. It must be, since we systematically destroy it even when we try to encourage it. It doesn’t matter how many programs we fund for encouraging things like art and music when the kids are still ultimately judged on how well they follow instructions and fill in little scantron bubbles. Kids are not stupid. I cannot believe how many adults still think they can get away with telling kids one thing and then doing another. They know what you’re up to. They know the only thing the school system cares about is their grades, and that their grades are based entirely on how well they follow directions. They know that answering a question “almost right” doesn’t matter. They know all problems have one answer in the back of the teacher’s textbook, and their job is to figure out what it is. The vast majority of them have absolutely no idea how to approach a problem that has no correct answer. They don’t know how to judge the correctness of a solution, because in school, everything is either right or wrong. All they know how to do is guess how likely it is that their solution is the solution the teacher wants, not how well the solution would actually work.

This is, of course, completely contradictory to everything in life. Life does not have answers in the back of the book. Life does not have a single correct answer to any problem. There is no right way to do anything, there are simply pros and cons. The fact that many people continue to delude themselves into thinking otherwise is a sad symptom of this issue. Our obsession with tests has trained a generation of robots, not engineers. They’re more skilled at working their way through a bureaucracy than designing rockets. Then again, considering that colleges have now turned into enormous, obstructive bureaucracies, perhaps this isn’t entirely a bad thing.

After all, with only 160 (now 200) spots open in its CSE major program each year when it has over 27000 undergraduate students enrolled[1], the University of Washington gets mighty picky about who they let in. After getting a 3.7 and 3.4 in my first two calculus classes, I slipped and got a 2.8 in my third math class, so despite the fact that I got a perfect 5 on the AP Computer Science AB exam and was qualified to skip both introductory programming courses, they rejected my application and demanded I take Matrix Algebra before letting me in. So I got a 3.9 in Matrix Algebra (a grade that was exceptionally good, according to one professor), and then… they still didn’t let me in. They complained that my entrance essay sounded “too cocky” and had me take the second introductory programming course even though I already had credit for it. When I failed to get an exceptionally good grade in that class for all the wrong reasons (like being graded down for having both too few comments and too many comments in my code), I simply could not bring myself to compete in a hyper-competitive environment where the only thing I was judged on was how many irrelevant details I could regurgitate. So, I majored in Applied Mathematics and simply took all the condensed, non-major CSE courses instead.

This obsession with tests extends into the evaluation of the educational system itself. One of the reasons nothing is getting better is because we use the very thing that is wrong with the educational system to judge it. We fill out ridiculous polls made out of those same interminable bubbles that are destroying the curriculum. We introduce more standardized testing. The entire system is completely obsessed with tests, and yet the only place that tests actually exist is… inside the system itself. Education has become so enraptured with this imaginary world it has constructed, it’s completely forgotten about the reality it’s supposed to be teaching kids about.

Kids know this imaginary world has nothing to do with reality. We lament about how to teach kids math when they refuse to understand it, without realizing that they are simply applying the same method of learning they use in everything else - memorize useless facts, then regurgitate them on a test. The reason our math curriculum is failing so badly is because in math, you can’t simply memorize things, you need to understand them. Consequently, Math acts as a canary in the coal mine for our entire educational system. Kids make no effort to understand anything, because they aren’t graded on how well they understand concepts, they are graded on how well they memorize random, useless details and follow directions.

We live in a world being overrun by automation. Any task that can be reduced to simply following a set of instructions over and over is being done by robots and software. This constant attrition of jobs involving menial work and physical labor will continue at a steady pace for the foreseeable future. We are teaching our kids skills that are being made irrelevant in a modern economy. We are preparing our children for a world that no longer exists. At the same time, while I could write a series of blog posts outlining an effective educational system, it will never be implemented in public schools. The establishment is too deeply entrenched. Foolish startups repeatedly and continually attempt to “disrupt” the educational system without realizing just how laughably outmatched they are. This is not something you can fix with a cute website. People complain about global warming, space travel, all sorts of adorable little problems, but miss the elephant in the room.

The greatest challenge our species has ever faced is the educational system itself.


Course Notes


It's standard procedure at the University of Washington to allow a single sheet of handwritten notes during a Mathematics exam. I started collecting these sheets after I realized how useful it was to have a reference that basically summarized all the useful parts of the course on a single sheet of paper. Now that I've graduated, it's easy for me to quickly forget all the things I'm not using. The problem is that, when I need to say, develop an algorithm for simulating turbulent airflow, I need to go back and re-learn vector calculus, differential equations and nonlinear dynamics. So I've decided to digitize all my notes and put them in one place where I can reference them. I've uploaded it here in case they come in handy to anyone else.

The earlier courses listed here had to be reconstructed from my class notes because I'd thrown my final notesheet away or otherwise lost it. The classes are not listed in the order I took them, but rather organized into related groups. This post may be updated later with expanded explanations for some concepts, but these are highly condensed notes for reference, and a lot of it won't make sense to someone who hasn't taken a related course.

Math 124 - Calculus I
lost

Math 125 - Calculus II
lost

Math 126 - Calculus III
lost

Math 324 - Multivariable Calculus I
\[ r^2 = x^2 + y^2 \]
\[ x= r\cos\theta \]
\[ y=r\sin\theta \]

\[ \iint\limits_R f(x,y)\,dA = \int_\alpha^\beta\int_a^b f(r\cos\theta,r\sin\theta)r\,dr\,d\theta=\int_\alpha^\beta\int_{h_1(\theta}^{h_2(\theta)} f(r\cos\theta,r\sin\theta)r\,dr\,d\theta \]

\[ m=\iint\limits_D p(x,y)\,dA \begin{cases} \begin{aligned} M_x=\iint\limits_D y p(x,y)\,dA & \bar{x}=\frac{M_y}{m}=\frac{1}{m}\iint x p(x,y)\,dA \end{aligned} \\ \begin{aligned} M_y=\iint\limits_D x p(x,y)\,dA & \bar{y}=\frac{M_x}{m}=\frac{1}{m}\iint y p(x,y)\,dA \end{aligned} \end{cases} \]

\[ Q = \iint\limits_D \sigma(x,y)\,dA \]
\[ I_x = \iint\limits_D y^2 p(x,y)\,dA \]
\[ I_y = \iint\limits_D x^2 p(x,y)\,dA \]
\[ I_0 = \iint\limits_D (x^2 + y^2) p(x,y)\,dA \]

\[ \iiint f(x,y,z) dV = \lim_{l,m,n\to\infty}\sum_{i=1}^l\sum_{j=1}^m\sum_{k=1}^n f(x_i,y_j,z_k) \delta V \]
\[ \iiint\limits_B f(x,y,z)\,dV=\int_r^s\int_d^c\int_a^b f(x,y,z)\,dx\,dy\,dz = \int_a^b\int_r^s\int_d^c f(x,y,z)\,dy\,dz\,dx \]

$$E$$ = general bounded region
Type 1: $$E$$ is between graphs of two continuous functions of $$x$$ and $$y$$.
\[ E=\{(x,y,z)|(x,y)\in D, u_1(x,y) \le z \le u_2(x,y)\} \]

$$D$$ is the projection of E on to the xy-plane, where
\[ \iiint\limits_E f(x,y,z)\,dV = \iint\limits_D\left[\int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z)\,dz \right]\,dA \]

$$D$$ is a type 1 planar region:
\[ \iiint\limits_E f(x,y,z)\,dV = \int_a^b \int_{g_1(x)}^{g_2(x)} \int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z)\,dz\,dy\,dx \]

$$D$$ is a type 2 planar region:
\[ \iiint\limits_E f(x,y,z)\,dV = \int_d^c \int_{h_1(y)}^{h_2(y)} \int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z)\,dz\,dx\,dy \]


Type 2: $$E$$ is between $$y$$ and $$z$$, $$D$$ is projected on to $$yz$$-plane
\[ E=\{(x,y,z)|(y,z)\in D, u_1(y,z) \le x \le u_2(y,z)\} \]
\[ \iiint\limits_E f(x,y,z)\,dV = \iint\limits_D\left[\int_{u_1(y,z)}^{u_2(y,z)} f(x,y,z)\,dx \right]\,dA \]

Type 3: $$E$$ is between $$x$$ and $$z$$, $$D$$ is projected on to $$xz$$-plane
\[ E=\{(x,y,z)|(x,z)\in D, u_1(x,z) \le y \le u_2(x,z)\} \]
\[ \iiint\limits_E f(x,y,z)\,dV = \iint\limits_D\left[\int_{u_1(x,z)}^{u_2(x,z)} f(x,y,z)\,dy \right]\,dA \]

Mass
\[ m = \iiint\limits_E p(x,y,z)\,dV \]
\[ \bar{x} = \frac{1}{m}\iiint\limits_E x p(x,y,z)\,dV \]
\[ \bar{y} = \frac{1}{m}\iiint\limits_E y p(x,y,z)\,dV \]
\[ \bar{z} = \frac{1}{m}\iiint\limits_E z p(x,y,z)\,dV \]
Center of mass: $$(\bar{x},\bar{y},\bar{z})$$
\[ Q = \iiint\limits_E \sigma(x,y,z)\,dV \]
\[ I_x = \iiint\limits_E (y^2 + z^2) p(x,y,z)\,dV \]
\[ I_y = \iiint\limits_E (x^2 + z^2) p(x,y,z)\,dV \]
\[ I_z = \iiint\limits_E (x^2 + y^2) p(x,y,z)\,dV \]

Spherical Coordinates:
\[ z=p\cos\phi \]
\[ r=p\sin\phi \]
\[ dV=p^2\sin\phi \]
\[ x=p\sin\phi\cos\theta \]
\[ y=p\sin\phi\sin\theta \]
\[ p^2 = x^2 + y^2 + z^2 \]
\[ \iiint\limits_E f(x,y,z)\,dV = \int_c^d\int_\alpha^\beta\int_a^b f(p\sin\phi\cos\theta,p\sin\phi\sin\theta,p\cos\phi) p^2\sin\phi\,dp\,d\theta\,d\phi \]

Jacobian of a transformation $$T$$ given by $$x=g(u,v)$$ and $$y=h(u,v)$$ is:
\[ \begin{aligned} \frac{\partial (x,y)}{\partial (u,v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\[0.1em] \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u} \end{aligned} \]

Given a transformation T whose Jacobian is nonzero, and is one to one:
\[ \iint\limits_R f(x,y)\,dA = \iint\limits_S f\left(x(u,v),y(u,v)\right)\left|\frac{\partial (x,y)}{\partial (u,v)}\right|\,du\,dv \]

Polar coordinates are just a special case:
\[ x = g(r,\theta)=r\cos\theta \]
\[ y = h(r,\theta)=r\sin\theta \]

\[ \frac{\partial (x,y)}{\partial (u,v)} = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta + r\sin^2\theta=r(\cos^2\theta + \sin^2\theta) = r \]

\[ \iint\limits_R f(x,y)\,dx\,dy = \iint\limits_S f(r\cos\theta, r\sin\theta)\left|\frac{\partial (x,y)}{\partial (u,v)}\right|\,dr\,d\theta=\int_\alpha^\beta\int_a^b f(r\cos\theta,r\sin\theta)|r|\,dr\,d\theta\]

For 3 variables this expands as you would expect: $$x=g(u,v,w)$$ $$y=h(u,v,w)$$ $$z=k(u,v,w)$$
\[ \frac{\partial (x,y,z)}{\partial (u,v,w)}=\begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & \frac{\partial z}{\partial v} & \frac{\partial z}{\partial w} \end{vmatrix} \]

\[ \iiint\limits_R f(x,y,z)\,dV = \iiint\limits_S f(g(u,v,w),h(u,v,w),k(u,v,w)) \left|\frac{\partial (x,y,z)}{\partial (u,v,w)} \right| \,du\,dv\,dw\]

Line Integrals
Parameterize: $$r(t)=\langle x(t),y(t),z(t) \rangle$$ where $$r'(t)=\langle x'(t),y'(t),z'(t) \rangle$$ and $$|r'(t)|=\sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2} $$
\[ \int_C f(x,y,z)\,ds = \int_a^b f(r(t))\cdot |r'(t)|\,dt = \int_a^b f(x(t),y(t),z(t))\cdot\sqrt{x'(t)^2 + y'(t)^2 + z'(t)^2}\,dt\;\;\;a<t<b \]

For a vector function $$\mathbf{F}$$:
\[ \int_C \mathbf{F}\cdot dr = \int_a^b \mathbf{F}(r(t))\cdot r'(t)\,dt \]

Surface Integrals
\[ \]

Parameterize: $$r(u,v) = \langle x(u,v),y(u,v),z(u,v) \rangle$$
\[ \begin{matrix} r_u=\frac{\partial x}{\partial u}\vec{\imath} + \frac{\partial y}{\partial u}\vec{\jmath} + \frac{\partial z}{\partial u}\vec{k} \\ r_v=\frac{\partial x}{\partial v}\vec{\imath} + \frac{\partial y}{\partial v}\vec{\jmath} + \frac{\partial z}{\partial v}\vec{k} \end{matrix}\]
\[ r_u \times r_v = \begin{vmatrix} \vec{\imath} & \vec{\jmath} & \vec{k} \\ \frac{\partial x}{\partial u} & \frac{\partial y}{\partial u} & \frac{\partial z}{\partial u} \\ \frac{\partial x}{\partial v} & \frac{\partial y}{\partial v} & \frac{\partial z}{\partial v} \end{vmatrix} \]

\[ \iint\limits_S f(x,y,z) dS = \iint\limits_D f(r(t))|r_u \times r_v|\,dA \]

For a vector function $$\mathbf{F}$$:
\[ \iint\limits_S \mathbf{F}\cdot dr = \iint\limits_D \mathbf{F}(r(u,v))\cdot (r_u \times r_v)\,dA) \]

Any surface $$S$$ with $$z=g(x,y)$$ is equivalent to $$x=x$$, $$y=y$$, and $$z=g(x,y)$$, so
$$xy$$ plane:
\[ \iint\limits_S f(x,y,z)\,dS = \iint\limits_D f(x,y,g(x,y))\sqrt{\left(\frac{\partial z}{\partial x}\right)^2+\left(\frac{\partial z}{\partial y}\right)^2+1}\,dA \]

$$yz$$ plane:
\[ \iint\limits_S f(x,y,z)\,dS = \iint\limits_D f(x,h(x,z),z)\sqrt{\left(\frac{\partial y}{\partial x}\right)^2+\left(\frac{\partial y}{\partial z}\right)^2+1}\,dA \]

$$xz$$ plane:
\[ \iint\limits_S f(x,y,z)\,dS = \iint\limits_D f(g(y,z),y,z)\sqrt{\left(\frac{\partial x}{\partial y}\right)^2+\left(\frac{\partial x}{\partial z}\right)^2+1}\,dA \]

Flux:
\[ \iint\limits_S\mathbf{F}\cdot dS = \iint\limits_D\mathbf{F}\cdot (r_u \times r_v)\,dA \]

The gradient of $$f$$ is the vector function $$\nabla f$$ defined by:
\[ \nabla f(x,y)=\langle f_x(x,y),f_y(x,y)\rangle = \frac{\partial f}{\partial x}\vec{\imath} + \frac{\partial f}{\partial y}\vec{\jmath} \]

Directional Derivative:
\[D_u\,f(x,y) = f_x(x,y)a + f_y(x,y)b = \nabla f(x,y)\cdot u \text{ where } u = \langle a,b \rangle \]
\[ \int_C\,ds=\int_a^b |r'(t)|=L \]
\[ \]

If $$\nabla f$$ is conservative, then:
\[ \int_{c_1} \nabla f\,dr=\int_{c_2} \nabla f\,dr \]

This means that the line integral between two points will always be the same, no matter what curve is used to go between the two points - the integrals are path-independent and consequently only depend on the starting and ending positions in the conservative vector field.
A vector function is conservative if it can be expressed as the gradient of some potential function $$\psi$$: $$\nabla \psi = \mathbf{F}$$
\[ curl\,\mathbf{F} =\nabla\times\mathbf{F}\]
\[ div\,\mathbf{F} =\nabla\cdot\mathbf{F}\]


Math 326 - Multivariable Calculus II
$$f(x,y)$$ is continuous at a point $$(x_0,y_0)$$ if
\[ \lim\limits_{(x,y)\to(0,0)} f(x,y) = f(x_0,y_0) \]

$$f+g$$ is continuous if $$f$$ and $$g$$ are continuous, as is $$\frac{f}{g}$$ if $$g \neq 0$$
A composite function of a continuous function is continuous
\[ \frac{\partial f}{\partial x} = f_x(x,y) \]
\[ \frac{\partial f}{\partial x}\bigg|_{(x_0,y_0)}=\left(\frac{\partial f}{\partial x}\right)_{(x_0,y_0)}=f_x(x_0,y_0) \]

To find $$\frac{\partial z}{\partial x} F(x,y,z)$$, differentiate $$x$$ as normal, hold y constant, and differentiate $$z$$ as a function (such that $$z^2 = 2z \frac{\partial z}{\partial x}$$ and $$2z = 2 \frac{\partial z}{\partial x}$$)
Ex:
\[ F(x,y,z) = \frac{x^2}{16} + \frac{y^2}{12} + \frac{z^2}{9} = 1 \]
\[ \frac{\partial z}{\partial x} F = \frac{2x}{16} + \frac{2z}{}\frac{\partial z}{\partial x} = 0 \]

The tangent plane of $$S$$ at $$(a,b,c): z-c = f_x(a,b)(x-a) + f_y(a,b)(y-b)$$ where $$z=f(x,y)$$, or $$z =f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)$$
Note that
\[ f_x(a,b)=\frac{\partial z}{\partial x}\bigg|_{(a,b)} \]
which enables you to find tangent planes implicitly.
Set $$z=f(x,y)$$. $$f_x=f_y=0$$ at a relative extreme $$(a,b)$$.
Distance from origin:
\[ D^2 = z^2 + y^2 + x^2 = f(a,b)^2 + y^2 + x^2 \]

Minimize $$D$$ to get point closest to the origin.
The differential of $$f$$ at $$(x,y)$$:
\[ df(x,y;dx,dy)=f_x(x,y)\,dx + f_y(x,y)\,dy \]
\[ dz=\frac{\partial f}{\partial x}\,dx + \frac{\partial f}{\partial y}\,dy \]

$$f$$ is called differentiable at $$(x,y)$$ if it $$s$$ defined for all points near $$(x,y)$$ and if there exists numbers $$A$$,$$B$$ such that
\[ \lim_{(n,k)\to(0,0)}\frac{|f(x+h,y+k) - f(x,y) - Ah - Bk|}{\sqrt{h^2 + k^2}}=0 \]

If $$f$$ is differentiable at $$(x,y)$$ it is continuous there.
\[ A = f_x(x,y) \]
and
\[ B=f_y(x,y) \]

If $$F_x(x,y)$$ and $$F_y(x,y)$$ are continuous at a point $$(x_0,y_0)$$ defined in $$F$$, then $$F$$ is differentiable there.
Ex: The differential of $$f(x,y)=3x^2y+2xy^2+1$$ at $$(1,2)$$ is $$df(1,2;h,k)=20h+11k$$
\[ d(u+v)=du+dv \]
\[ d(uv)=u\,dv + v\,du \]
\[ d\left(\frac{u}{v}\right)=\frac{v\,du -u\,dv}{v^2} \]

Taylor Series:
\[ f^{(n)}(t)=\left[\left(h\frac{}{} + k\frac{}{})^n F(x,y) \right) \right]_{x=a+th,y=b+tk} \]
Note:
\[ f''(t)=h^2 F_{xx} + 2hk F_{xy} + k^2 F_{yy} \]

\[ \begin{matrix} x=f(u,v) \\ y=g(u,v) \end{matrix} \]
\[ J=\frac{\partial(f,g)}{\partial(u,v)} \]
\[ \begin{matrix} u=F(x,y) \\ v=G(x,y) \end{matrix} \]
\[ j = J^{-1} = \frac{1}{\frac{\partial(f,g)}{\partial(u,v)}} \]

\[ \begin{matrix} x = u-uv \\ y=uv \end{matrix} \]
\[ \iint\limits_R\frac{dx\,dy}{x+y} \]
R bounded by
\[ \begin{matrix} x+y=1 & x+y=4 \\ y=0 & x=0 \end{matrix}\]
\[ \int_0^1\int_0^x \frac{du\,dv}{u-uv+ux}\left|\frac{\partial(x,y)}{\partial(u,v)}\right| \]

\[ \frac{\partial(F,G)}{\partial(u,v)}=\begin{vmatrix} \frac{\partial F}{\partial u} & \frac{\partial F}{\partial v} \\ \frac{\partial G}{\partial u} & \frac{\partial G}{\partial v} \end{vmatrix} \]
\[ \nabla f = \langle f_x, f_y, f_z \rangle \]
\[ G_x(s_0,t_0) =F_x(a,b)f_x(s_0,t_0) + F_y(a,b)g_x(s_0,t_0) \]
\[ U\times V = U_xV_y - U_yV_x \text{ or } A\times B = \left\langle \begin{vmatrix} a_y & a_z \\ b_y & b_z \end{vmatrix}, -\begin{vmatrix} a_x & a_z \\ b_x & b_z \end{vmatrix}, \begin{vmatrix} a_x & a_y \\ b_x & b_y \end{vmatrix} \right\rangle \]

Given $$G(s,t)=F(f(s,t),g(s,t))$$, then:
\[ \begin{matrix} \frac{\partial G}{\partial s} = \frac{\partial F}{\partial x}\frac{\partial f}{\partial s} + \frac{\partial F}{\partial y}\frac{\partial g}{\partial s} \\ \frac{\partial G}{\partial t} = \frac{\partial F}{\partial x}\frac{\partial f}{\partial t} + \frac{\partial F}{\partial y}\frac{\partial g}{\partial t} \end{matrix} \]

Alternatively, $$ u=F(x,y,z)=F(f(t),g(t),h(t))$$ yields
\[ \frac{du}{dt}=\frac{\partial u}{\partial x}\frac{dx}{dt} + \frac{\partial u}{\partial y}\frac{dy}{dt} + \frac{\partial u}{\partial z}\frac{dz}{dt} \]

Examine limit along line $$y=mx$$:
\[ \lim_{x\to 0} f(x,mx) \]

If $$g_x$$ and $$g_y$$ are continuous, then $$g$$ is differentiable at that point (usually $$0,0$$).
Notice that if $$f_x(0,0)=0$$ and $$f_y(0,0)=0$$ then $$df(0,0;h,l)=0h+0k=0$$

The graph of $$y(x,y)$$ lies on a level surface $$F(x,y,z)=c$$ means $$f(x,y(x,z),z)=c$$. So then use the chain rule to figure out the result in terms of $$F$$ partials by considering $$F$$ a composite function $$F(x,y(x,z),z)$$.

Fundamental implicit function theorem: Let $$F(x,y,z)$$ be a function defined on an open set $$S$$ containing the point $$(x_0,y_0,z_0)$$. Suppose $$F$$ has continuous partial derivatives in $$S$$. Furthermore assume that: $$F(x_0,y_0,z_0)=0, F_z(x_0,y_0,z_0)\neq 0$$. Then $$z=f(x,y)$$ exists, is continuous, and has continuous first partial derivatives.
\[ f_x = -\frac{F_x}{F_z} \]
\[ f_y = -\frac{F_y}{F_z} \]

Alternatively, if
\[ \begin{vmatrix} F_x & F_y \\ G_x & G_y \end{vmatrix} \neq 0 \]
, then we can solve $$x$$ and $$y$$ as functions of $$z$$. Since the cross-product is made of these determinants, if the $$x$$ component is nonzero, you can solve $$y,z$$ as functions of $$x$$, and therefore graph it on the $$x-axis$$.

To solve level surface equations, let $$F(x,y,z)=c$$ and $$G(x,y,z)=d$$, then use the chain rule, differentiating by the remaining variable (e.g. $$\frac{dy}{dx}$$ and $$\frac{dz}{dx}$$ means do $$x$$)
\[ \begin{matrix} F_x + F_y y_x + F_z z_x = 0 \\ G_x + G_y y_x + G_z z_x = 0 \end{matrix} \]
if you solve for $$y_x$$,$$z_x$$, you get
\[ \left[ \begin{matrix} F_y & F_z \\ G_y & G_z \end{matrix} \right] \left[ \begin{matrix} y_x \\ z_x \end{matrix} \right] = \left[\begin{matrix}F_x \\ G_x \end{matrix} \right] \]

Mean value theorem:
\[ f(b) - f(a) = (b-a)f'(X) \]
\[ a < X < b \]

or:
\[ f(a+h)=f(a)+hf'(a + \theta h) \]
\[ 0 < \theta < 1 \]

xy-plane version:
\[ F(a+h,b+k)-F(a,b)=h F_x(a+\theta h,b+\theta k)+k F_y(a+\theta h, b+\theta k) \]
\[ 0 < \theta < 1 \]

Lagrange Multipliers: $$\nabla f = \lambda\nabla g$$ for some scale $$\lambda$$ if $$(x,y,z)$$ is a minimum:
\[ f_x=\lambda g_x \]
\[ f_y=\lambda g_y \]
\[ f_z=\lambda g_z \]
Set $$f=x^2 + y^2 + z^2$$ for distance and let $$g$$ be given.

Math 307 - Introduction to Differential Equations
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Math 308 - Matrix Algebra
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Math 309 - Linear Analysis
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AMath 353 - Partial Differential Equations

Fourier Series:
\[ f(x)=b_0 + \sum_{n=1}^{\infty} \left(a_n \sin\frac{n\pi x}{L} + b_n\cos\frac{n\pi x}{L} \right) \]
\[ b_0 = \frac{1}{2L}\int_{-L}^L f(y)\,dy \]
\[ a_m \frac{1}{L}\int_{-L}^L f(y) \sin\frac{m\pi y}{L}\,dy \]
\[ b_m = \frac{1}{L}\int_{-L}^L f(y)\cos\frac{m\pi y}{L}\,dy \]
\[ m \ge 1 \]

\[ u_t=\alpha^2 u_{xx} \]
\[ \alpha^2 = \frac{k}{pc} \]
\[ u(x,t)=F(x)G(t) \]

Dirichlet: $$u(0,t)=u(L,t)=0$$
Neumann:$$u_x(0,t)=u_x(L,t)=0$$
Robin:$$a_1 u(0,t)+b_1 u_x(0,t) = a_2 u(L,t) + b_2 u_x(L,t) = 0$$
Dirichlet:
\[ \lambda_n=\frac{n\pi}{L}\;\;\; n=1,2,... \]
\[ u(x,t)=\sum_{n=1}^{\infty} A_n \sin\left(\frac{n\pi x}{L} \right)\exp\left(-\frac{n^2\alpha^2\pi^2 t}{L^2}\right) \]
\[ A_n = \frac{2}{L} \int_0^L f(y) \sin\frac{n\pi y}{L}\,dy = 2 a_m\text{ for } 0\text{ to } L \]

Neumann:
\[ \lambda_n=\frac{n\pi}{L}\;\;\; n=1,2,... \]
\[ u(x,t)=B_0 + \sum_{n=1}^{\infty} B_n \cos\left(\frac{n\pi x}{L} \right)\exp\left(-\frac{n^2\alpha^2\pi^2 t}{L^2}\right) \]
\[ B_0 = \frac{1}{L} \int_0^L f(y)\,dy \]
\[ B_n = \frac{2}{L} \int_0^L f(y) \cos\frac{n\pi y}{L}\,dy = 2 b_m\text{ for } 0\text{ to } L \]

Dirichlet/Neumann:
\[ \lambda_n=\frac{\pi}{2L}(2n + 1)\;\;\; n=0,1,2,... \]
\[ u(x,t)=\sum_{n=0}^{\infty} A_n \sin\left(\lambda_n x\right) \exp\left(-\alpha^2 \lambda_n^2 t\right) \]
\[ A_n = \frac{2}{L} \int_0^L f(y) \sin\left(\lambda_n y\right)\,dy \]

Neumann/Dirichlet:
\[ \lambda_n=\frac{\pi}{2L}(2n + 1)\;\;\; n=0,1,2,... \]
\[ u(x,t)=\sum_{n=0}^{\infty} B_n \cos\left(\lambda_n x\right) \exp\left(-\alpha^2 \lambda_n^2 t\right) \]
\[ B_n = \frac{2}{L}\int_0^L f(y)\cos(\lambda_n y)\,dy \]

\[ v_2(x,y)=\sum_{n=1}^{\infty} C_n(t)[\sin/\cos](\lambda_n x) \]
Replace $$[\sin/\cos]$$ with whatever was used in $$u(x,t)$$
\[ C_n(t)=\int_0^t p_n(s) e^{\lambda_n^2\alpha^2 (s-t)}\,ds \]
\[ p_n(t)=\frac{2}{L}\int_0^L p(y,t)[\sin/\cos](\lambda_n y)\,dy \]
Replace $$[\sin/\cos]$$ with whatever was used in $$u(x,t)$$. Note that $$\lambda_n$$ for $$C_n$$ and $$p_n(t)$$ is the same as used for $$u_1(x,t)$$
Sturm-Liouville:
\[ p(x)\frac{d^2}{dx^2}+p'(x)\frac{d}{dx}+q(x) \]
\[ a_2(x)\frac{d^2}{dx^2} + a_1(x)\frac{d}{dx} + a_0(x) \rightarrow p(x)=e^{\int\frac{a_1(x)}{a_2(x)}\,dx}\]
\[ q(x)=p(x)\frac{a_0(x)}{a_2(x)} \]

Laplace's Equation:
\[ \nabla^2 u=0 \]
\[ u=F(x)G(y) \]
\[ \frac{F''(x)}{F(x)} = -\frac{G''(y)}{G(y)} = c \]
for rectangular regions
\[ F_?(x)=A =\sinh(\lambda x) + B \cosh(\lambda x) \]
use odd $$\lambda_n$$ if $$F$$ or $$G$$ equal a single $$\cos$$ term.
\[ G_?(y)=C =\sinh(\lambda y) + D \cosh(\lambda y) \]
\[ u(x,?)=G(?) \]
\[ u(?,y)=F(y) \]

Part 1: All $$u_?(x,?)=0$$
\[ \lambda_n=\frac{n\pi}{L y}\text{ or }\frac{(2n+1)\pi}{2L y} \]
\[ u_1(x,y)=\sum_{n=0}^{\infty}A_n F_1(x)G_1(y) \]
\[ A_n = \frac{2}{L_y F_1(?)}\int_0^{L_y} u(?,y)G_1(y)\,dy \]
\[ u(?,y)=h(y) \]
\[ ? = L_x\text{ or }0 \]

Part 2: All $$u_?(?,y)=0$$
\[ \lambda_n=\frac{n\pi}{L x}\text{ or }\frac{(2n+1)\pi}{2L x} \]
\[ u_2(x,y)=\sum_{n=1}^{\infty}B_n F_2(x)G_2(y) \]
\[ B_n = \frac{2}{L_x G_2(?)}\int_0^{L_x} u(x,?)F_2(x)\,dx \]
\[ u(x,?)=q(x) \]
\[ ? = L_y\text{ or }0 \]

\[ u(x,y)=u_1(x,y)+u_2(x,y) \]

Circular $$\nabla^2 u$$:
\[ u_{rr} + \frac{1}{r} u_r + \frac{1}{r^2}u_{\theta \theta} = 0 \]
\[ \frac{G''(\theta)}{G(\theta)}= - \frac{r^2 F''(r) + r F(r)}{F(r)} = c\]

\[ \left\langle f,g \right\rangle = \int_a^b f(x) g(x)\,dx \]
\[ \mathcal{L}_s=-p(x)\frac{d^2}{dx^2}-p'(x)\frac{d}{dx} + q(x) \]
\[ \mathcal{L}_s\phi(x)=\lambda r(x) \phi(x) \]
\[ \left\langle\mathcal{L}_s y_1,y_2 \right\rangle =\int_0^L \left(-[p y_1']'+q y_1\right)y_2\,dx \]

$$\mathcal{L}_s = f(x)$$, then $$\mathcal{L}_s^{\dagger} v(x) = 0$$, where if $$v=0$$, $$u(x)$$ is unique, otherwise $$v(x)$$ exists only when forcing is orthogonal to all trivial solutions.
if $$c=0$$,
\[ G(\theta)=B \]
\[ F(r)=C\ln r + D \]

if $$c=-\lambda^2 <0$$,
\[ G(\theta)=A\sin(\lambda\theta) + B\cos(\lambda\theta) \]
\[ F(r) = C\left(\frac{r}{R} \right)^n + D\left( \frac{r}{R} \right)^{-n}\]
\[ u(r,\theta)=B_0 + \sum_{n=1}^{\infty} F(r) G(\theta) = B_0 + \sum_{n=1}^{\infty}F(r)[A_n\sin(\lambda\theta) + B_n\cos(\lambda\theta)] \]
\[ A_n = \frac{1}{\pi}\int_0^{2\pi} f(\theta)\sin(n\theta)\,d\theta \]
\[ B_n = \frac{1}{\pi}\int_0^{2\pi} f(\theta)\cos(n\theta)\,d\theta \]
\[ B_0 = \frac{1}{2\pi}\int_0^{2\pi} f(\theta)\,d\theta \]

Wave Equation:
\[ u_{tt}=c^2 u_{xx} \]
\[ u = F(x)G(t) \]
\[ \frac{F''(x)}{F(x)}=\frac{G''(t)}{G(t)}=k \text{ where } u(t,0)=F(0)=u(t,L)=F(L)=0 \]
\[ F(x) = A\sin(\lambda x) + B\cos(\lambda x) \]
\[ u_t(x,0)=g(x)=\sum_{n=1}^{\infty} A_n\lambda_n c \sin(\lambda_n x) \]
\[ A_n = \frac{2}{\lambda_n c L}\int_0^L g(y)\sin(\lambda_n y)\,dy \]

\[ G(t) = C\sin(\lambda t) + D\cos(\lambda t) \]
\[ u(x,0)=f(x)=\sum_{n=1}^{\infty}B_n\sin(\lambda_n x) \]
\[ B_n = \frac{2}{L}\int_0^L f(y)\sin(\lambda_n y)\,dy \]

\[ u(x,t)=\sum_{n=1}^{\infty}F(x)G(t)=\sum_{n=1}^{\infty}F(x)\left[C\sin(\lambda t) + D\cos(\lambda t)\right] \]
\[ \lambda_n=\frac{n\pi}{L}\text{ or }\frac{(2n+1)\pi}{2L} \]

Inhomogeneous:
\[ u(0,t)=F(0)=p(t) \]
\[ u(L,t)=F(L)=q(t) \]
\[ u(x,t)=v(x,t) + \phi(x)p(t) + \psi(x)q(t) \]

Transform to forced:
\[ v(x,y) = \begin{cases} v_{tt}=c^2 v_{xx} - \phi(x)p''(x) - \psi(x)q''(t) = c^2 v_{xx} + R(x,t) & \phi(x)=1 - \frac{x}{L}\;\;\;\psi(x)=\frac{x}{L}\\ v(0,t)=v(L,t)=0 & t>0 \\ v(x,0)=f(x)-\phi(x)p(0) - \psi(x)q(0) & f(x)=u(x,0) \\ v_t(x,0)=g(x)-\phi(x)p'(0) - \psi(x)q'(0) & g(x)=u_t(x,0) \end{cases} \]

Then solve as a forced equation.
Forced:
\[ u_{tt}=c^2 u_{xx} + R(x,t) \]
\[ u(x,t)=u_1(x,t)+u_2(x,t) \]
$$u_1$$ found by $$R(x,t)=0$$ and solving.
\[ u_2(x,t)=\sum_{n=1}^{\infty} C_n(t)\sin\left(\frac{n\pi x}{L}\right) \]
where $$\sin\frac{n\pi x}{L}$$ are the eigenfunctions from solving $$R(x,t)=0$$
\[ R(x,t)=\sum_{n=1}^{\infty} R_n(t)\sin\left(\frac{n\pi x}{L}\right) \]
\[ R_n(t)=\frac{2}{L}\int_0^L R(y,t)\sin\left(\frac{n\pi x}{L}\right)\,dy \]
\[ C_n''(t) + k^2 C_n(t)=R_n(t) \]
\[ C_n(t)=\alpha\sin(k t) + \beta\cos(k t) + sin(k t)\int_0^t R_n(s)\cos(k s)\,ds - \cos(k t)\int_0^t R_n(s)\sin(k s)\,ds \]

where $$C_n(0)=0$$ and $$c_n'(0)=0$$
Fourier Transform:
\[ \mathcal{F}(f)=\hat{f}(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} f(\xi) e^{-i k \xi}\,d\xi \]
\[ \mathcal{F}^{-1}(f)=f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \hat{f}(k) e^{ikx}\,dk \]
\[ \mathcal{F}(f+g)=\widehat{f+g}=\hat{f} + \hat{g} \]
\[ \widehat{fg}\neq\hat{f}\hat{g} \]
\[ \widehat{f'}=ik\hat{f}\text{ or }\widehat{f^{(n)}}=(ik)^n\hat{f} \]

\[ \widehat{u_t} = \frac{\partial \hat{u}}{\partial t} = \hat{u}_t \]
\[ \widehat{u_{tt}}=\frac{\partial^2\hat{u}}{\partial t^2}=\hat{u}_{tt} \]
\[ \widehat{u_{xx}}=(ik)^2\hat{u}=-k^2\hat{u} \]

\[ u(x,t)=\mathcal{F}^{-1}(\hat{u})=\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \hat{f}(k) e^{\alpha^2 k^2 t} e^{ikx}\,dk \]
\[ u(x,t)=\mathcal{F}^{-1}\left(\hat{f}(k) e^{-\alpha^2 k^2 t}\right) \]

Semi-infinite:
\[ \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \hat{F}(k) e^{\alpha^2 k^2 t} e^{ikx}\,dk \]
where
\[ \hat{F}(k)=-i\sqrt{\frac{2}{\pi}}\int_0^{\infty} f(\xi)\sin(k\xi)\,d\xi \]
$$f(x)$$$$\hat{f}(k)$$
\[1\text{ if } -s < x < s, 0\text{ otherwise }\]\[ \sqrt{\frac{2}{\pi}}\frac{\sin(ks)}{k} \]
\[ \frac{1}{x^2 + s^2} \]\[ \frac{1}{s}\sqrt{\frac{\pi}{2}}e^{-s|k|} \]
\[ e^{-sx^2} \]\[ \frac{1}{\sqrt{2s}}e^{\frac{k^2}{4s}} \]
\[ \frac{\sin(sx)}{x}\]\[ \sqrt{\frac{\pi}{2}}\text{ if } |k|<s, 0\text{ otherwise} \]
\[ e^{isx}\text{ if } a < x < b, 0\text{ otherwise} \]\[ \frac{i}{\sqrt{2\pi}}\left(\frac{1}{s-k} \right)\left(e^{ea(s-k)} - e^{ib(s-k)} \right) \]


AMath 402 - Dynamical Systems and Chaos
Dimensionless time
\[ \tau = \frac{t}{T} \]
where T gets picked so that
\[\frac{}{}\]
and
\[\frac{}{}\]
are of order 1.
Fixed points of $$x'=f(x)$$: set $$f(x)=0$$ and solve for roots
Bifurcation: Given $$x'=f(x,r)$$, set $$f(x,r)=0$$ and solve for $$r$$, then plot $$r$$ on the $$x-axis$$ and $$x$$ on the $$y-axis$$.
Saddle Node
Saddle Node
Transcritical Node
Transcritical
Subcritical Pitchfork
Subcritical Pitchfork
Supercritical Pitchfork
Supercritical Pitchfork

\[ \begin{matrix} x'=ax+by \\ y'=cx+dy \end{matrix} \]
\[ A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\]
\[ \begin{vmatrix} a-\lambda & b \\ c & d-\lambda \end{vmatrix}=0 \]
\[ \begin{matrix} \lambda^2 - \tau\lambda + \delta=0 & \lambda = \frac{\tau \pm \sqrt{\tau^2 - 4\delta}}{2} \\ \tau = a+d & \delta=ad-bc \end{matrix} \]
\[ x(t)=c_1 e^{\lambda_1 t}v_1 + c_2 e^{\lambda_2 t}v_2 \]
$$v_1,v_2$$ are eigenvectors.
Given $$\tau = \lambda_1 + \lambda_2$$ and $$\delta=\lambda_1\lambda_2$$:
if $$\delta < 0$$, the eigenvalues are real with opposite signs, so the fixed point is a saddle point.
otherwise, if $$\tau^2-4\delta > 0$$, its a node. This node is stable if $$\tau < 0$$ and unstable if $$\tau > 0$$
if $$\tau^2-4\delta < 0$$, its a spiral, which is stable if $$\tau < 0$$, unstable if $$\tau > 0$$, or a center if $$\tau=0$$
if $$\tau^2-4\delta = 0$$, its degenerate.
\[ \begin{bmatrix} x'=f(x,y) \\ y'=g(x,y) \end{bmatrix} \]
Fixed points are found by solving for $$x'=0$$ and $$y'=0$$ at the same time.
nullclines are when either $$x'=0$$ or $$y'=0$$ and are drawn on the phase plane.
\[ \begin{bmatrix} \frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y} \end{bmatrix} \]


$$\leftarrow$$ For nonlinear equations, evaluate this matrix at each fixed point, then use the above linear classification scheme to classify the point.

A basin for a given fixed point is the area of all trajectories that eventually terminate at that fixed point.
Given $$x'=f(x)$$, $$E(x)$$ is a conserved quantity if $$\frac{dE}{dt}=0$$
A limit cycle is an isolated closed orbit in a nonlinear system. Limit cycles can only exist in nonlinear systems.
If a function can be written as $$\vec{x}'=-\nabla V$$, then its a gradient function and can't have closed orbits.
The Liapunov function $$V(x)$$ for a fixed point $$x^*$$ satisfies $$V(x) > 0 \forall x \neq x^*, V(x^*)=0, V' < 0 \forall x \neq x^*$$
A Hopf bifurcation occurs as the fixed points eigenvalues (in terms of $$\mu$$) cross the imaginary axis.

Math 300 - Introduction to Mathematical Reasoning
PQ$$P \Rightarrow Q$$$$\neg P \vee Q$$
TTTT
TFFF
FTTT
FFTT

All valid values of $$x$$ constitute the Domain.
$$f(x)=y$$ The range in which $$y$$ must fall is the codomain
The image is the set of $$y$$ values that are possible given all valid values of $$x$$
So, $$\frac{1}{x}$$ has a domain $$\mathbb{R} - \{0\}$$ and a codomain of $$\mathbb{R}$$. However, no value of $$x$$ can ever produce $$f(x)=0$$, so the Image is $$\mathbb{R}-\{0\}$$

Injective: No two values of $$x$$ yield the same result. $$f(x_1)\neq f(x_2)$$ if $$x_1 \neq x_2$$ for all $$x$$
Surjective: All values of $$y$$ in the codomain can be produced by $$f(x)$$. In other words, the codomain equals the image.
Bijective: A Bijective function is both Injective and Surjective - all values of $$y$$ are mapped to exactly one value of $$x$$. A simple way to prove this is to solve $$f(x)$$ for $$x$$. If this can be done without creating multiple solutions (a square root, for example, yields $$\pm$$, not a single answer), then it's a bijective function.
If $$f(x)$$ is a bijection, it is denumerable. Any set that is equivalent to the natural numbers is denumerable.
$$\forall x \in \mathbb{R}$$ means "for all $$x$$ in $$\mathbb{R}$$", where $$\mathbb{R}$$ can be replaced by any set.
$$\exists y \in \mathbb{R}$$ means "there exists a $$y$$ in $$\mathbb{R}$$", where $$\mathbb{R}$$ can be replaced by any set.
\[ A \vee B = A\text{ or }B \]
\[ A \wedge B = A\text{ and }B \]
\[ P \Leftrightarrow Q\text{ means }P \Rightarrow Q \wedge Q\Rightarrow P \text{ or } P\text{ iff }Q \text{ (if and only if)} \]

$$A \cup B$$ = Union - All elements that are in either A or B, or both.
\[ \{x|x \in A \text{ or } x \in B \} \]

$$A \cap B$$ = Intersection - All elements that are in both A and B
\[ \{x|x \in A \text{ and } x \in B \} \]

$$A\subseteq B$$ = Subset - Indicates A is a subset of B, meaning that all elements in A are also in B.
\[ \{x \in A \Rightarrow x \in B \} \]

$$A \subset B$$ = Strict Subset - Same as above, but does not allow A to be equal to B (which happens if A has all the elements of B, because then they are the exact same set).
$$A-B$$ = Difference - All elements in A that aren't also in B
\[ \{x|x \in A \text{ and } x \not\in B \} \]

$$A\times B$$ = Cartesian Product - All possible ordered pairs of the elements in both sets:
\[ \{(x,y)|x \in A\text{ and }y\in B\} \]

ProofWe use induction on $$n$$
Base Case:[Prove $$P(n_0)$$]
Inductive Step:Suppose now as inductive hypothesis that [$$P(k)$$ is true] for some integer $$k$$ such that $$k \ge n_0$$. Then [deduce that $$P(k+1)$$ is true]. This proves the inductive step.
Conclusion:Hence, by induction, [$$P(n)$$ is true] for all integers $$n \ge n_0$$.
ProofWe use strong induction on $$n$$
Base Case:[Prove $$P(n_0)$$, $$P(n_1)$$, ...]
Inductive Step:Suppose now as inductive hypothesis that [$$P(n)$$ is true for all $$n \le k$$] for some positive integer $$k$$, then [deduce that $$P(k+1)$$ is true]. This proves the inductive step.
Conclusion:Hence, by induction [$$P(n)$$ is true] for all positive integers $$n$$.
Proof:Suppose, for contradiction, that the statement $$P$$ is false. Then, [create a contradiction]. Hence our assumption that $$P$$ is false must be false. Thus, $$P$$ is true as required.

The composite of $$f:X\rightarrow Y$$ and $$g:Y\rightarrow Z$$ is $$g\circ f: x\rightarrow Z$$ or just $$gf: X\rightarrow Z$$. $$g\circ f = g(f(x)) \forall x \in X$$
\[ a\equiv b \bmod m \Leftrightarrow b\equiv a \bmod m \]
If
\[ a \equiv b \bmod m\text{ and }b \equiv c \bmod m,\text{ then }a \]

Negation of $$P \Rightarrow Q$$ is $$P \wedge (\neg Q)$$ or $$P and (not Q)$$
If $$m$$ is divisible by $$a$$, then $$a b_1 \equiv a b_2 \bmod m \Leftrightarrow b_1 \equiv b_2 \bmod\left(\frac{m}{a} \right)$$
Fermat's little theorem: If $$p$$ is a prime, and $$a \in \mathbb{Z}^+$$ which is not a multiple of $$p$$, then $$a^{p-1}\equiv 1 \bmod p$$.
Corollary 1: If $$p$$ is prime, then $$\forall a \in \mathbb{Z}, a^p = a \bmod p$$
Corollary 2: If $$p$$ is prime, then $$(p-1)! \equiv -1 \bmod p$$

Division theorem: $$a,b \in \mathbb{Z}$$, $$b > 0$$, then $$a = bq+r$$ where $$q,r$$ are unique integers, and $$0 \le r < b$$. Thus, for $$a=bq+r$$, $$\gcd(a,b)=\gcd(b,r)$$. Furthermore, $$b$$ divides $$a$$ if and only if $$r=0$$, and if $$b$$ divides $$a$$, $$\gcd(b,a)=b$$

Euclidean Algorithm: $$\gcd(136,96)$$
$$136=96\cdot 1 + 40$$
\[ 290x\equiv 5\bmod 357 \rightarrow 290x + 357y = 5 \]

$$96=40\cdot 2 + 16$$
\[ ax\equiv b \bmod c \rightarrow ax+cy=b \]

$$40=16\cdot 2 + 8$$
$$16=8\cdot 2 + 0 \leftarrow$$ stop here, since the remainder is now 0
$$\gcd(136,96)=8$$

Diophantine Equations (or linear combinations):
$$140m + 63n = 35$$ $$m,n \in \mathbb{Z}$$ exist for $$am+bn=c$$ iff $$\gcd(a,b)$$ divides $$c$$
$$140=140\cdot 1 + 63\cdot 0$$
$$63=140\cdot 0 + 63\cdot 1$$ dividing $$am+bn=c$$ by $$\gcd(a,b)$$ always yields coprime $$m$$ and $$n$$.
$$14=140\cdot 1 - 63\cdot 2$$
$$7=63\cdot 1 - 14\cdot 4 = 63 - (140\cdot 1 - 63\cdot 2)\cdot 4 = 63 - 140\cdot 4 + 63\cdot 8 = 63\cdot 9 - 140\cdot 4 $$
$$0=14 - 7\cdot 2 =140\cdot 1 - 63\cdot 2 - 2(63\cdot 9 - 140\cdot 4) = 140\cdot 9-63\cdot 20 $$
So $$m=9$$ and $$n=-20$$

Specific: $$7=63\cdot 9 - 140\cdot 4 \rightarrow 140\cdot(-4)+63\cdot 9=7\rightarrow 140\cdot(-20)+63\cdot45=35$$
So for $$140m+63n=35, m=-20, n=45$$
Homogeneous: $$140m+63n=0 \; \gcd(140,63)=7 \; 20m + 9n = 0 \rightarrow 29m=-9n$$
$$m=9q, n=-20q$$ for $$q\in \mathbb{Z}$$ or $$am+bn=0 \Leftrightarrow (m,n)=(bq,-aq)$$
General: To find all solutions to $$140m+63n=35$$, use the specific solution $$m=-20, n=45$$
$$140(m+20) + 63(n-45) = 0$$
$$(m+20,n-45)=(9q,-20q)$$ use the homogeneous result and set them equal.
$$(m,n)=(9q-20,-20q+45)$$ So $$m=9q-20$$ and $$n=-20q+45$$
\[ [a]_m = \{x\in \mathbb{Z}| x\equiv a \bmod m \} \Rightarrow \{ mq + a|q \in \mathbb{Z} \} \]

$$ax\equiv b \bmod m$$ has a unique solution $$\pmod{m}$$ if $$a$$ and $$m$$ are coprime. If $$\gcd(a,b)=1$$, then $$a$$ and $$b$$ are coprime. $$[a]_m$$ is invertible if $$\gcd(a,m)=0$$.

Math 327 - Introduction to Real Analysis I
Cauchy Sequence: For all $$\epsilon > 0$$ there exists an integer $$N$$ such that if $$n,m \ge N$$, then $$|a_n-a_m|<\epsilon$$

Alternating Series: Suppose the terms of the series $$\sum u_n$$ are alternately positive and negative, that $$|u_{n+1}| \le |u_n|$$ for all $$n$$, and that $$u_n \to 0$$ as $$n\to\infty$$. Then the series $$\sum u_n$$ is convergent.

Bolzano-Weierstrass Theorem: If $$S$$ is a bounded, infinite set, then there is at least one point of accumulation of $$S$$.

$$\sum u_n$$ is absolutely convergent if $$\sum|u_n|$$ is convergent
If $$\sum u_n$$ is absolutely convergent, then $$\sum u_n = \sum a_n - \sum b_n$$ where both $$\sum a_n$$ and $$\sum b_n$$ are convergent. If $$\sum u_n$$ is conditionally convergent, both $$\sum a_n$$ and $$\sum b_n$$ are divergent.

\[ \sum u_n = U = u_0 + u_1 + u_2 + ... \]
\[ w_0 = u_0 v_0 \]
\[ w_1 = u_0 v_1 + u_1 v_0 \]

\[ \sum v_n = V = v_0 + v_1 + v_2 + ... \]
\[ w_n = u_0 v_n + u_1 v_{n-1} + ... + u_n v_0 \]

\[ \sum w_n = UV = w_0 + w_1 + w_2 + ... \]
Provided
\[ \sum u_n, \sum v_n \]
are absolutely convergent.

$$\sum a_n x^n$$ is either absolutely convergent for all $$x$$, divergent for all $$x\neq 0$$, or absolutely convergent if $$-R < x < R$$ (it may or may not converge for $$x=\pm R$$ ).
$$-R < x < R$$ is the interval of convergence. $$R$$ is the radius of convergence.
\[ R=\lim_{n \to \infty}\left|\frac{a_n}{a_{n+1}}\right| \]
if the limit exists or is $$+\infty$$
Let the functions of the sequence $$f_n(x)$$ be defined on the interval $$[a,b]$$. If for each $$\epsilon > 0$$ there is an integer $$N$$ independent of x such that $$|f_n(x) - f_m(x)| < \epsilon$$ where $$N \le m,n$$.

You can't converge uniformly on any interval containing a discontinuity.
\[ |a|+|b|\le|a+b| \]
\[ \lim_{n \to \infty}(1+\frac{k}{n})^n = e^k \]
\[ lim_{n \to \infty} n^{\frac{1}{n}}=1 \]
\[ \sum_{k=0}^{\infty}=\frac{a}{1-r} \]
if $$|r| < 1$$
\[\sum a_n x^n\]
has a radius of convergence
\[R = \lim_{n \to \infty}\left|\frac{a_n}{a_{n+1}} \right|\]
if it exists or is $$+\infty$$. The interval of convergence is $$-R < x < R$$
\[ \sum^{\infty}\frac{x^2}{(1+x^2)^n} = x^2\sum\left(\frac{1}{1+x^2}\right)^n = x^2\left(\frac{1}{1-\frac{1}{1+x^2}}\right) = x^2\left( \frac{1+x^2}{1+x^2-1}\right) = x^2\left( \frac{1+x^2}{x^2} \right) = 1+x^2 \]

Math 427 - Complex Analysis
0$$\frac{\pi}{6}$$$$\frac{\pi}{4}$$$$\frac{\pi}{3}$$$$\frac{\pi}{2}$$
$$\sin$$0$$\frac{1}{2}$$$$\frac{\sqrt{2}}{2}$$$$\frac{\sqrt{3}}{2}$$1
$$\cos$$1$$\frac{\sqrt{3}}{2}$$$$\frac{\sqrt{2}}{2}$$$$\frac{1}{2}$$0
$$\tan$$0$$\frac{1}{\sqrt{3}}$$1$$\sqrt{3}$$$$\varnothing$$
\[ \frac{z_1}{z_2} = \frac{x_1 x_2 + y_1 y_2}{x_2^2 + y_2^2} + i\frac{y_1 x_1 - x_1 y_2}{x_2^2 + y_2^2} \]
\[ |z| = \sqrt{x^2 + y^2} \]
\[ \bar{z} = x-i y \]
\[ |\bar{z}|=|z| \]

\[ \text{Re}\,z=\frac{z+\bar{z}}{2} \]
\[ \text{Im}\,z=\frac{z-\bar{z}}{2i} \]
\[ \text{Arg}\,z=\theta\text{ for } -\pi < \theta < \pi \]
\[ z=re^{i\theta}=r(\cos(\theta) + i\sin(\theta)) \]

\[ \lim_{z \to z_0} f(z) = w_0 \iff \lim_{x,y \to x_0,y_0} u(x,y) = u_0 \text{ and }\lim_{x,y \to x_0,y_0} v(x,y)=v_0 \text{ where } w_0=u_0+iv_0 \text{ and } z_0=x_0+iy+0 \]

\[ \lim_{z \to z_0} f(z) = \infty \iff \lim_{z \to z_0} \frac{1}{f(z)}=0 \]

\[ \lim_{z \to \infty} f(z) = w_0 \iff \lim_{z \to 0} f\left(\frac{1}{z}\right)=w_0\]
\[ \text{Re}\,z\le |\text{Re}\,z|\le |z| \]
\[ \text{Im}\,z\le |\text{Im}\,z| \le z \]

\[ \lim_{z \to \infty} f(z) = \infty \iff \lim_{z\to 0}\frac{1}{f\left(\frac{1}{z}\right)}=0\]
\[ \left| |z_1| - |z_2| \right| \le |z_1 + z_2| \le |z_1| + |z_2| \]

Roots:
\[ z = \sqrt[n]{r_0} \exp\left[i\left(\frac{\theta_0}{n} + \frac{2k\pi}{n} \right)\right] \]
Harmonic:
\[ f_{xx} + f_{yy} = 0 \text{ or } f_{xx}=-f_{yy} \]

Cauchy-Riemann:
\[ f(z) = u(x,y)+iv(x,y) \]
\[ \sinh(z)=\frac{e^z-e^{-z}}{2}=-i\sin(iz) \]
\[ \frac{d}{dz}\sinh(z)=\cosh(z) \]

\[ u_x = v_y \]
\[ u_y = -v_x \]
\[ \cosh(z) = \frac{e^z + e^{-z}}{2} = cos(iz) \]
\[ \frac{d}{dz}\cosh(z)=\sinh(z) \]

\[ f(z) = u(r,\theta)+iv(r,\theta) \]
\[ \sin(z)=\frac{e^{iz}-e{-iz}}{2i}=i-\sinh(iz)=\sin(x)\cosh(y) + i\cos(x)\sinh(y) \]

\[ r u_r = v_{\theta} \]
\[ u_{\theta}=-r v_r \]
\[ \cos(z)=\frac{e^{iz}+e^{-iz}}{2}=\cosh(iz)=\cos(x)\cosh(y) - i \sin(x)\sinh(y) \]

\[ \sin(x)^2 + \sinh(y)^2 = 0 \]
\[ |\sin(z)|^2 = \sin(x)^2 + \sinh(y)^2 \]
\[ |\cos(z)|^2 = \cos(x)^2 + \sinh(y)^2 \]

$$e^z$$ and Log:
\[ e^z=e^{x+iy}=e^x e^{iy}=e^x(\cos(y) + i\sin(y)) \]
\[ e^x e^{iy} = \sqrt{2} e^{i\frac{\pi}{4}} \]
\[ e^x = \sqrt{2} \]
\[ y = \frac{\pi}{4} + 2k\pi \]
\[ x = \ln\sqrt{2} \]
So,
\[ z = \ln\sqrt{2} + i\pi\left(\frac{1}{4} + 2k\right) \]
\[ \frac{d}{dz} log(f(z)) = \frac{f'(z)}{f(z)} \]

Cauchy-Gourgat: If $$f(z)$$ is analytic at all points interior to and on a simple closed contour $$C$$, then
\[ \int_C f(z) \,dz=0 \]

\[ \int_C f(z) \,dz = \int_a^b f[z(t)] z'(t) \,dt \]
\[ a \le t \le b \]
$$z(t)$$ is a parameterization. Use $$e^{i\theta}$$ for circle!
Maximum Modulus Principle: If $$f(z)$$ is analytic and not constant in $$D$$, then $$|f(z)|$$ has no maximum value in $$D$$. That is, no point $$z_0$$ is in $$D$$ such that $$|f(z)| \le |f(z_0)|$$ for all $$z$$ in $$D$$.
Corollary: If $$f(z)$$ is continuous on a closed bounded region $$R$$ and analytic and not constant through the interior of $$R$$, then the maximum value of $$|f(z)|$$ in $$R$$ always exists on the boundary, never the interior.
Taylor's Theorem: Suppose $$f(z)$$ is analytic throughout a disk $$|z-z_0|<R$$. Then $$f(z)$$ has a power series of the form
\[ f(z) = \sum_{n=0}^{\infty}\frac{f^{(n)}(z_0)}{n!}(z-z_0)^n \]
where $$|z-z_0| < R_0$$ and $$n=0,1,2,...$$
Alternatively,
\[ \sum_{n=0}^{\infty} a_n(z-z_0)^n \text{ where } a_n=\frac{f^{(n)}(z_0)}{n!} \]

For $$|z|<\infty$$:
\[ e^z = \sum_{n=0}^{\infty}\frac{z^n}{n!} \]
\[ \sin(z) = \sum_{n=0}^{\infty}(-1)^n\frac{z^{2n+1}}{(2n+1)!} \]
\[ \cos(z) = \sum_{n=0}^{\infty}(-1)^n\frac{z^{2n}}{(2n)!} \]

Note that
\[ \frac{1}{1 - \frac{1}{z}}=\sum_{n=0}^{\infty}\left(\frac{1}{z}\right)^n \text{ for } \left|\frac{1}{z}\right| < 1\]
, which is really $$1 < |z|$$ or $$|z| > 1$$
For $$|z| < 1$$:
\[ \frac{1}{1-z} = \sum_{n=0}^{\infty} z^n \]
\[ \frac{1}{1+z} = \sum_{n=0}^{\infty} (-1)^n z^n \]
For $$|z-1|<1$$:
\[ \frac{1}{z} = \sum_{n=0}^{\infty} (-1)^n (z-1)^n \]

Analytic: $$f(z)$$ is analytic at point $$z_0$$ if it has a derivative at each point in some neighborhood of $$z_0$$.
$$f(z)=u(x,y)+iv(x,y)$$ is analytic if and only if $$v$$ is a harmonic conjugate of $$u$$.
If $$u(x,y)$$ and $$v(x,y)$$ are harmonic functions such that $$u_{xx}=-u{yy}$$ and $$v_{xx}=-v_{yy}$$, and they satisfy the Cauchy-Reimann conditions, then $$v$$ is a harmonic conjugate of $$u$$.

Differentiable: $$f(z)=u(x,y)+iv(x,y)$$ is differentiable at a point $$z_0$$ if $$f(z)$$ is defined within some neighborhood of $$z_0=x_0+iy_0$$, $$u_x$$ $$u_y$$ $$v_x$$ and $$v_y$$ exist everywhere in the neighborhood, those first-order partial derivatives exist at $$(x_0,y_0)$$ and satisfy the Reimann-Cauchy conditions at $$(x_0,y_0)$$.
Cauchy Integral Formula:
\[ \int_C \frac{f(z)}{z-z_0}\,dz = 2\pi i f(z_0) \]
where $$z_0$$ is any point interior to $$C$$ and $$f(z)$$ is analytic everywhere inside and on $$C$$ taken in the positive sense. Note that $$f(z)$$ here refers to the actual function $$f(z)$$, not $$\frac{f(z)}{z-z_0}$$
Generalized Cauchy Integral Formula:
\[ \int_C \frac{f(z)}{(z-z_0)^{n+1}}\,dz = \frac{2\pi i}{n!} f^{(n)}(z_0) \]
(same conditions as above)
Remember, discontinuities that are roots can be calculated and factored:
\[ \frac{1}{z^2-w_0}=\frac{1}{(z-z_1)(z-z_2)} \]

Residue at infinity:
\[ \int_C f(z)\,dz = 2\pi i \,\mathop{Res}\limits_{z=0}\left[\frac{1}{z^2} f\left(\frac{1}{z}\right) \right] \]

A residue at $$z_0$$ is the $$n=-1$$ term of the Laurent series centered at $$z_0$$.
A point $$z_0$$ is an isolated singular point if $$f(z_0)$$ fails to be analytic, but is analytic at some point in every neighborhood of $$z_0$$, and there's a deleted neighborhood $$0 < |z-z_0| < \epsilon$$ of $$z_0$$ throughout which $$f$$ is analytic.
\[ \mathop{Res}\limits_{z=z_0}\,f(z)=\frac{p(z_0)}{q'(z_0)}\text{ where }f(z) = \frac{p(z)}{q(z)}\text{ and } p(z_0)\neq 0 \]

\[ \int_C f(z)\,dz = 2\pi i \sum_{k=1}^n\,\mathop{Res}\limits_{z=z_k}\,f(z) \]
where $$z_0$$, $$z_1$$, $$z_2$$,... $$z_k$$ are all the singular points of $$f(z)$$ (which includes poles).

If a series has a finite number of $$(z-z_0)^{-n}$$ terms, its a pole ($$\frac{1}{z^4}$$ is a pole of order 3). If a function, when put into a series, has no $$z^{-n}$$ terms, it's a removable singularity. If it has an infinite number of $$z^{-n}$$ terms, its an essential singularity (meaning, you can't get rid of it).

AMath 401 - Vector Calculus and Complex Variables
$$||\vec{x}||=\sqrt{x_1^2+x_2^2+ ...}$$
$$\vec{u}\cdot\vec{v} = ||\vec{u}||\cdot||\vec{v}|| \cos(\theta)=u_1v_1+u_2v_2+ ...$$
$$||\vec{u}\times\vec{v}|| = ||\vec{u}||\cdot||\vec{v}|| \sin(\theta) = \text{ area of a parallelogram}$$
\[ \vec{u}\times\vec{v} = \begin{vmatrix} \vec{\imath} & \vec{\jmath} & \vec{k} \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} \]
\[ \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad-bc \]
\[ \vec{u}\cdot\vec{v}\times\vec{w} = \vec{u}\cdot(\vec{v}\times\vec{w})=\text{ volume of a parallelepiped} \]

\[ h_?\equiv\left|\left|\frac{\partial\vec{r}}{\partial ?}\right|\right| \]
\[ \hat{e}_r(\theta)=\frac{1}{h_r}\frac{\partial\vec{r}}{\partial r} \]
\[ \hat{e}_r'(\theta)=\frac{d\hat{e}_r}{d\theta}=\hat{e}_{\theta}(\theta) \]

\[ \vec{r}=r(t)\hat{e}_r(\theta(t)) \]
\[ \vec{r}'(t)=r'(t)\hat{e}_r + r(t)\frac{d\hat{e}_r}{dt}\]
\[ \vec{r}'(t)=r'(t)\hat{e}_r + r(t)\frac{d\hat{e}_r}{d\theta}\frac{d\theta}{dt}=r'(t)\hat{e}_r + r(t)\hat{e}_{\theta} \theta'(t) \]

Projection of $$\vec{u}$$ onto $$\vec{v}$$:
\[ ||\vec{u}|| \cos(\theta) = \frac{\vec{u}\cdot\vec{v}}{||\vec{v}||} \]
Arc Length:
\[ \int_a^b \left|\left| \frac{d\vec{r}}{dt} \right|\right| \,dt = \int_a^b ds \]

Parametrization:
\[ s(t) = \int_a^t \left|\left|\frac{d\vec{r}}{dt}\right|\right| \,dt \]
\[ \frac{dx}{dt}=x'(t) \]
\[ dx = x'(t)\,dt \]
\[ \oint x \,dy - y \,dx = \oint\left(x \frac{dy}{dt} - y \frac{dx}{dt}\right)\,dt = \oint x y'(t) - y x'(t) \,dt \]
\[ \nabla = \frac{\partial}{\partial x}\vec{\imath} + \frac{\partial}{\partial y}\vec{\jmath} + \frac{\partial}{\partial z}\vec{k} \]
\[ \nabla f = \left\langle \frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z} \right\rangle \]
\[ D_{\vec{u}} f(\vec{P})=f'_{\vec{u}}=\nabla f \cdot \vec{u} \]
\[ \nabla(fg)=f\nabla g + g \nabla f \]

\[ \vec{F}(x,y,z)=F_1(x,y,z)\vec{\imath} + F_2(x,y,z)\vec{\jmath} + F_3(x,y,z)\vec{k} \]

\[ \text{div}\, \vec{F} = \nabla \cdot \vec{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \]

\[ \text{curl}\, \vec{F} = \nabla \times \vec{F} = \left\langle \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \right\rangle \]

Line integral:
\[ \int_{\vec{r}(a)}^{\vec{r}(b)}\vec{F}\cdot \vec{r}'(t)\,dt = \int_{\vec{r}(a)}^{\vec{r}(b)}\vec{F}\cdot d\vec{r} \]
Closed path:
\[ \oint \vec{F}\cdot d\vec{r} \]
\[ d\vec{r} = \nabla\vec{r}\cdot \langle du_1,du_2,du_3 \rangle = h_1 du_1 \hat{e}_1 + h_2 du_2 \hat{e}_2 + h_3 du_3 \hat{e}_3 \]
\[ h_?= \left|\left| \frac{\partial\vec{r}}{\partial ?} \right|\right| = \frac{1}{||\partial ?||} \]
\[ \hat{e}_?=\frac{1}{h_?}\frac{\partial\vec{r}}{\partial?}=h_?\nabla?(x,y,z)=\frac{\nabla?}{||\nabla?||} \]
If $$\vec{r}=\vec{r}(s)$$ then
\[ \int_a^b\vec{F}\cdot \frac{d\vec{r}}{ds} \,ds = \int_a^b\vec{F}\cdot\vec{T} \,ds\]

\[ \nabla f \cdot \hat{e}_?=\frac{1}{h_?}\frac{\partial f}{\partial ?} \]
\[ \nabla f = (\nabla f \cdot \hat{e}_r)\hat{e}_r + (\nabla f \cdot \hat{e}_{\theta})\hat{e}_{\theta} = \frac{\partial f}{\partial r}\hat{e}_r + \frac{1}{r}\frac{\partial f}{\partial \theta}\hat{e}_{\theta} \]
\[ ds = \sqrt{d\vec{r}\cdot d\vec{r}} = \left|\left| \frac{d\vec{r}}{dt}dt \right|\right| = \sqrt{h_1^2 du_1^2 + h_2^2 du_2^2+h_3^2 du_3^2} \]

\[ Spherical: h_r=1, h_{\theta}=r, h_{\phi}=r\sin(\theta) \]

\[ \vec{r}=\frac{\nabla f}{||\nabla f||} \]
\[ \vec{r}=\vec{r}(u,v) \]
\[ \vec{n}= \frac{\frac{\partial\vec{r}}{\partial u}\times\frac{\partial\vec{r}}{\partial v}}{\left|\left| \frac{\partial\vec{r}}{\partial u}\times\frac{\partial\vec{r}}{\partial v} \right|\right|} \]

$$\vec{F}$$ is conservative if:
\[ \vec{F}=\nabla\phi \]
\[ \nabla\times\vec{F}=0 \]
\[ \vec{F}=\nabla\phi \iff \nabla\times\vec{F}=0 \]

\[ \iint\vec{F}\cdot d\vec{S} \]
\[ d\vec{S}=\left(\frac{\partial\vec{r}}{\partial u} \times \frac{\partial\vec{r}}{\partial v} \right)\,du\,dv \]
\[ d\vec{S}=\vec{n}\,dS \]
Surface Area:
\[ \iint \,dS \]

Flux:
\[ \iint\vec{F}\cdot d\vec{S} \]

Shell mass:
\[ \iint p\cdot dS \]

Stokes:
\[ \iint(\nabla\times\vec{F})\cdot d\vec{S} = \oint\vec{F}\cdot d\vec{r} \]

Divergence:
\[ \iiint\limits_V\nabla\cdot\vec{F}\,dV=\iint\limits_{\partial v} \vec{F}\cdot d\vec{S} \]

If $$z=f(x,y)$$, then
\[ \vec{r}(x,y) = \langle x,y,f(x,y) \rangle \]

\[ \nabla\cdot\vec{F}=\frac{1}{h_1 h_2 h_3}\left[\frac{\partial}{\partial u_1}(F_1 h_2 h_3) + \frac{\partial}{\partial u_2}(F_2 h_1 h_3) + \frac{\partial}{\partial u_3}(F_3 h_1 h_2) \right] \]
\[ dV = \left| \frac{\partial\vec{r}}{\partial u_1}\cdot\left(\frac{\partial\vec{r}}{\partial u_2}\times\frac{\partial\vec{r}}{\partial u_3} \right) \right| du_1 du_2 du_3 \]
If orthogonal,
\[ I = \iiint\limits_V f(u_1,u_2,u_3)h_1 h_2 h_3\,du_1\,du_2\,du_3 \]
\[ (x-c_x)^2 + (y-c_y)^2 = r^2 \Rightarrow \vec{r}(t)=\langle c_x + r\cos(t),c_y+r\sin(t) \rangle \]

Ellipse:
\[ \Rightarrow \vec{r}(t)=\langle c_x+a\cos(t), c_y+b\sin(t) \rangle \]

For spherical, if $$\theta$$ is innermost, its max value is $$\pi$$, or if its on the z value.
\[ \vec{r}(\theta,\phi)=\langle \sin(\theta)\cos(\phi), \sin(\theta)\sin(\phi),\cos(\theta) \rangle \]


Laplace Transform:
\[ \mathcal{L}\left[f(t)\right]=F(s)=\int_0^{\infty}e^{-st}f(t)\,dt \]
\[ \mathcal{L}[f'(t)] = s\mathcal{L}[f(t)] - f(0) = s\cdot F(s)-f(0)\]
\[ \mathcal{L}[f''(t)] = s^2\mathcal{L}[f(t)] - s\cdot f(0) - f'(0) = s^2\cdot F(s) - s\cdot f(0) - f'(0)\]

\[ \mathcal{L}[0] = 0 \]
\[ \mathcal{L}[1] = \frac{1}{s} \]
\[ \mathcal{L}[k] = \frac{k}{s} \]
\[ \mathcal{L}[e^{at}] = \frac{1}{s-a} \]
\[ \mathcal{L}[\cos(\omega t)] = \frac{s}{s^2 + \omega^2} \]
\[ \mathcal{L}[\sin(\omega t)] = \frac{\omega}{s^2 + \omega^2} \]


Math 461 - Combinatorial Theory I
General Pigeon: Let $$n$$,$$m$$, and $$r$$ be positive integers so that $$n>rm$$. Let us distribute $$n$$ balls into $$m$$ boxes. Then there will be at least 1 box in which we place at least $$r+1$$ balls.



Base Case:Prove $$P(0)$$ or $$P(1)$$
Inductive Step:Show that by assuming the inductive hypothesis $$P(k)$$, this implies that $$P(k+1)$$ must be true.
Strong Induction:Show that $$P(k+1)$$ is true if $$P(n)$$ for all $$n < k+1$$ is true.

There are $$n!$$ permutations of an $$n$$-element set (or $$n!$$ linear orderings of $$n$$ objects)
$$n$$ objects sorted into $$a,b,c,...$$ groups have $$\frac{n!}{a!b!c!...}$$ permutations. This is also known as a $$k$$-element multiset of $$n$$.
Number of $$k$$-digit strings in an $$n$$-element alphabet: $$n^k$$. All subsets of an $$n$$-element set: $$2^n$$
Let $$n$$ and $$k$$ be positive integers such that $$n \ge k$$, then the number of $$k$$-digit strings over an $$n$$-element alphabet where no letter is used more than once is $$\frac{n!}{(n-k)!}=(n)_k$$
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} = \frac{(n)_k}{k!} \rightarrow \]
This is the number of $$k$$-element subsets in $$[n]$$, where $$[n]$$ is an $$n$$-element set.
\[ \binom{n}{k} = \binom{n}{n-k} \]
\[ \binom{n}{0} = \binom{n}{n} = \binom{0}{0} = 1 \]

Number of $$k$$-element multisets in $$[n]$$:
\[ \binom{n+k-1}{k} \]

Binomial Theorem:
\[ (x+y)^n = \sum_{k=0}^n\binom{n}{k}x^k y^{n-k} \text{ for } n \ge 0 \]

Multinomial Theorem:
\[ (x_1 + x_2 + ... + x_k)^n = \sum_{a_1,a_2,...,a_k} \binom{n}{a_1,a_2,...,a_k} x_1^{a_1} x_2^{a_2} ... x_k^{a_k} \]
\[ \binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1} \text{ for } n,k \ge 0 \]
\[ \binom{k}{k} + \binom{k+1}{k} + ... + \binom{n}{k} = \binom{n+1}{k+1} \text{ for } n,k \ge 0 \]
\[ \binom{n}{k} \le \binom{n}{k} \text{ if and only if } k \le \frac{n-1}{2} \text{ and } n=2k+1 \]
\[ \binom{n}{k} \ge \binom{n}{k} \text{ if and only if } k \ge \frac{n-1}{2} \text{ and } n=2k+1 \]
\[ \binom{n}{a_1,a_2,...,a_k} = \frac{n!}{a_1!a_2!...a_k!} \]
\[ \binom{n}{a_1,a_2,...,a_k} = \binom{n}{a_1}\cdot \binom{n-a_1}{a_2} \cdot ... \cdot \binom{n-a_1-a_2-...-a_{k-1}}{a_k} \text{ if and only if } n=\sum_{i=1}^k a_i\]
\[ \binom{m}{k} = \frac{m(m-1)...(m-k+1)}{k!} \text{ for any real number } m \]
\[ (1+x)^m = \sum_{n\ge 0}^{\infty} \binom{m}{n}x^n \]
\[ \sum_{k=0}^n (-1)^k\binom{n}{k} = 0 \text{ for } n \ge 0 \]
\[ 2^n = \sum_{k=0}^n \binom{n}{k} = 0 \text{ for } n \ge 0 \]
\[ \sum_{k=1}^n k\binom{n}{k} = n 2^{n-1} \text{ for } n \ge 0 \]
\[ \binom{n+m}{k} = \sum_{i=0}^k \binom{n}{i} \binom{n}{k-i} \text{ for } n,m,k \ge 0 \]

\[ |E| = \frac{1}{2}\sum_{v\in V} deg(v) \]
or, the number of edges in a graph is half the sum of its vertex degrees.
Compositions:$$n$$ identical objects
$$k$$ distinct boxes
\[\binom{n-1}{k-1}\]$$n$$ identical objects
any number of distinct boxes
$$2^{n-1}$$
Weak Compositions:
(empty boxes allowed)
$$n$$ identical objects
$$k$$ distinct boxes
\[ \binom{n+k-1}{k-1} \]$$n$$ distinct objects
$$k$$ distinct boxes
$$k^n$$

Split $$n$$ people into $$k$$ groups of $$\frac{n}{k}$$:
Unordered:
\[ \frac{n!}{\left(\left(\frac{n}{k}\right)!\right)^k k!} \]
Ordered:
\[ \frac{n!}{\left(\left(\frac{n}{k}\right)!\right)^k} \]

Steps from (0,0) to (n,k) on a lattice:
\[\binom{n+k}{k}\]

Ways to roll $$n$$ dice so all numbers 1-6 show up at least once: $$6^n - \binom{6}{5}5^n + \binom{6}{4}4^n + \binom{6}{3}3^n + \binom{6}{2}2^n + \binom{6}{1}$$
The Sieve: $$|A_1| + |A_2| - |A_1\cap A_2|$$ or $$|A_1| + |A_2| + |A_3| - |A_1\cap A_2| - |A_1\cap A_3| - |A_2\cap A_3| + |A_1\cap A_2\cap A_3|$$
Also, $$|S - (S_A\cap S_B\cap S_C)| = |S| - |S_A| - |S_B| - |S_C| + |S_A\cap S_B| + |S_A\cap S_C| + |S_B\cap S_C| - |S_A\cap S_B\cap S_C|$$

Graphs: A simple graph has no loops (edges connecting a vertex to itself) or multiple edges between the same vertices. A walk is a path through a series of connected edges. A trail is a walk where no edge is traveled on more than once. A closed trail starts and stops on the same vertex. A Eulerian trail uses all edges in a graph. A trail that doesn't touch a vertex more than once is a path. $$K_n$$ is the complete graph on $$n$$-vertices.
If one can reach any vertex from any other on a graph $$G$$, then $$G$$ is a connected graph.
A connected graph has a closed Eulerian trail if and only if all its vertices have even degree. Otherwise, it has a Eulerian trail starting on S and ending on T if only S and T have odd degree.
In a graph without loops there are an even number of vertices of odd-degree. A cycle touches all vertices only once, except for the vertex it starts and ends on. A Hamiltonian cycle touches all vertices on a graph.
Let $$G$$ be a graph on $$n \ge 3$$ vertices, then $$G$$ has a Hamiltonian cycle if all vertices in $$G$$ have degree equal or greater than $$n/2$$. A complete graph $$K_n$$ has $$\binom{n}{k}\frac{(k-1)!}{2}$$ cycles of $$k$$-vertices.
Total Cycles:
\[ \sum_{k=3}^n\binom{n}{k}\frac{(k-1)!}{2} \]
Hamiltonian cycles:
\[ \frac{(n-1)!}{2} \]

Two graphs are the same if for any pair of vertices, the number of edges connecting them is the same in both graphs. If this holds when they are unlabeled, they are isomorphic.
A Tree is a minimally connected graph: Removing any edge yields a disconnected graph. Trees have no cycles, so any connected graph with no cycles is a tree.
All trees on $$n$$ vertices have $$n-1$$ edges, so any connected graph with $$n-1$$ edges is a tree.
Let $$T$$ be a tree on $$n \ge 2$$ vertices, then T has at least two vertices of degree 1.
Let $$F$$ be a forest on $$n$$ vertices with $$k$$ trees. Then F has $$n-k$$ edges.
Cayley's formula: The number of all trees with vertex set $$[n]$$ is $$A_n = n^{n-2}$$
A connected graph $$G$$ can be drawn on a plane so that no two edges intersect, then G is a planar graph.
A connected planar graph or convex polyhedron satisfies: Vertices + Faces = Edges + 2
$$K_{3,3}$$ and $$K_5$$ are not planar, nor is any graph with a subgraph that is edge-equivalent to them.
A convex Polyhedron with V vertices, E edges, and F faces satisfies: $$3F \le 3E$$, $$3V \le 3E$$, $$E \le 3V-6$$, $$E \le 3F-6$$, one face has at most 5 edges, and one vertex has at most degree 5.
$$K_{3,3}$$ has 6 vertices and 9 edges
A graph on $$n$$-vertices has $$\binom{n}{2}$$ possible edges.
If a graph is planar, then $$E \le 3V - 6$$. However, some non-planar graphs, like $$K_{3,3}$$, satisfy this too.
Prüfer code: Remove the smallest vertex from the graph, write down only its neighbor's value, repeat.

Math 462 - Combinatorial Theory II
All labeled graphs on $$n$$ vertices:
\[ 2^{\binom{n}{2}} \]

There are $$(n-1)!$$ ways to arrange $$n$$ elements in a cycle.
Given $$h_n=3 h_{n-1} - 2 h_{n-2}$$, change to $$h_{n+2}=3 h_{n+1} - 2 h_n$$, then $$h_{n+2}$$ is $$n=2$$ so multiply by $$x^{n+2}$$
\[ \sum h_{n+2}x^{n+2} = 3 \sum h_{n+1}x^{n+2} - 2 \sum h_n x^{n+2} \]
Then normalize to
\[ \sum_{n=2}^{\infty}h_n x^n \]
by subtracting $$h_0$$ and $$h_1$$
\[ \sum h_n x^n - x h_1 - h_0 = 3 x \sum h_{n+1}x^{n+1} - 2 x^2 \sum h_n x^n \]

\[ G(x) - x h_1 - h_0 = 3 x (G(x)-h_0) - 2 x^2 G(x) \]
Solve for:
\[ G(x) = \frac{1-x}{2x^2-3x+1} = \frac{1}{1-2x} = \sum (2x)^n \]

\[ G(x) = \sum(2x)^n = \sum 2^n x^n = \sum h_n x^n \rightarrow h_n=2^n \]

\[ \sum_{n=0}^{\infty} \frac{x^n}{n!} = e^x \]
\[ \sum_{n=0}^{\infty} (cx)^n = \frac{1}{1-cx} \]

Also,
\[ 2e_1 + 5e_2 = n \rightarrow \sum h_n x^n = \sum x^{2e_1+5e_2} = \sum x^{2e_1} x^{5e_2} = \left(\sum^{\infty} x^{2e_1}\right)\left(\sum^{\infty} x^{5e_2}\right)=\frac{1}{(1-x^2)(1-x^5)}\]


$$S(n,k)$$: A Stirling number of the second kind is the number of nonempty partitions of $$[n]$$ into k blocks where the order of the blocks doesn't matter. $$S(n,k)=S(n-1,k-1)+k S(n-1,k)$$, $$S(n,n-2) = \binom{n}{3} + \frac{1}{2}\binom{k}{2}\binom{n+2}{2}$$
Bell Numbers:
\[ B(n) = \sum_{i=0}^n S(n,i) \]
, or the number of all partitions of $$[n]$$ into nonempty parts (order doesn't matter).
Catalan Number $$C_n$$:
\[ \frac{1}{n+1}\binom{2n}{n} \]
derived from
\[ \sum_{n \ge 1} c_{n-1} x^n = x C(x) \rightarrow C(x) - 1 = x C(x)\cdot C(x) \rightarrow C(x) = \frac{1 - \sqrt{1-4x}}{2x} \]

Products: Let $$A(n) = \sum a_n x^n$$ and $$B(n) = \sum b_n x^n$$. Then $$A(x)B(x)=C(x)=\sum c_n x^n$$ where
\[ c_n = \sum_{i=0}^n a_i b_{n-i} \]


Cycles: The number of $$n$$-permuatations with $$a_i$$ cycles of length $$i \in [n]$$ is \frac{n!}{a_1!a_2!...a_n!1^{a_1}2^{a_2}...n^{a_n}}
The number of n-permutations with only one cycle is $$(n-1)!$$
$$c(n,k)$$: The number of $$n$$-permutations with $$k$$-cycles is called a signless stirling number of the first kind.
$$c(n,k) = c(n-1,k-1) + (n-1) c(n-1,k)$$
\[ c(n,n-2)= 2\binom{n}{3} + \frac{1}{2}\binom{n}{2}\binom{n-2}{2} \]

$$s(n,k) = (-1)^{n-k} c(n,k)$$ and is called a signed stirling number of the first kind.

Let $$i \in [n]$$, then fro all $$k \in [n]$$, there are $$(n-1)!$$ permutations that contain $$i$$ in a $$k$$-cycle.

\[ T(n,k)=\frac{k-1}{2k}n^2 - \frac{r(k-r)}{2k} \]
Let Graph $$G$$ have $$n$$ vertices and more than $$T(n,k)$$ edges. Then $$G$$ contains a $$K_{k+1}$$ subgraph, and is therefore not $$k$$-colorable.

$$N(T)$$ = all neighbors of the set of vertices $$T$$
$$a_{s,d} = \{ s,s+d,s+2d,...,s+(n-1)d \}$$
$$\chi(H)$$: Chromatic number of Graph $$H$$, or the smallest integer $$k$$ for which $$H$$ is $$k$$-colorable.
A 2-colorable graph is bipartite and can divide its vertices into two disjoint sets.
A graph is bipartite if and only if it does not contain a cycle of odd length.
A bipartite graph has at most $$\frac{n^2}{4}$$ edges if $$n$$ is even, and at most $$\frac{n^2 - 1}{4}$$ edges if $$n$$ is odd.
If graph $$G$$ is not an odd cycle nor complete, then $$\chi(G) \le$$ the largest vertex degree $$\ge 3$$.
A bipartite graph on $$n$$ vertices with a max degree of $$k$$ has at most $$k\cdot (n-k)$$ edges.
A tree is always bipartite (2-colorable).

Philip Hall's Theorem: Let a bipartite graph $$G=(X,Y)$$. Then $$X$$ has a perfect matching to $$Y$$ if and only if for all $$T \subset X, |T| \le |N(T)|$$

$$R(k,l):$$ The Ramsey Number $$R(k,l)$$ is the smallest integer such that any 2-coloring of a complete graph on $$R(k,l)$$ vertices will contain a red $$k$$-clique or blue $$l$$-clique. A $$k$$-clique is a complete subgraph on $$k$$ vertices.
\[ R(k,l) \le R(k,l-1) + R(k-1,l) \]
\[ R(k,k) \le 4^{k-1} \]
\[ R(k,l) \le \binom{k+l-2}{l-1} \]


Let $$P$$ be a partially ordered set (a "poset"), then:
1) $$\le$$ is reflexive, so $$x \le x$$ for all $$x \in P$$
2) $$\le$$ is transitive, so that if $$x \le y$$ and $$y \le z$$ then $$x \le z$$
3) $$\le$$ is anti-symmetric, such that if $$x\le y$$ and $$y \le x$$, then $$x=y$$

Let $$P$$ be the set of all subsets of $$[n]$$, and let $$A \le B$$ if $$A \subset B$$. Then this forms a partially ordered set $$B_n$$, or a Boolean Algebra of degree $$n$$.

\[ E(X)=\sum_{i \in S} i\cdot P(X=i) \]

A chain is a set with no two incomparable elements. An antichain has no comparable elements.
Real numbers are a chain. $$\{ (2,3), (1,3), (3,4), (2,4) \}$$ is an antichain in $$B_4$$, since no set contains another.
Dilworth: In a finite partially ordered set $$P$$, the size of any maximum antichain is equal to the number of chains in any chain cover.

Weak composition of n into 4 parts: $$a_1 + a_2 + a_3 + a_4 = n$$
Applying these rules to the above equation: $$a_1 \le 2, a_2\text{ mod }2 \equiv 0, a_3\text{ mod }2 \equiv 1, a_3 \le 7, a_4 \ge 1$$
Yields the following:
\[ a_1 + 2a_2 + (2a_3 + 1) + a_4 = n \]
\[(\sum_0^2 x^{a_1})(\sum_0^{\infty} x^{2a_2})(\sum_0^3 x^{2a_3 + 1})(\sum_1^{\infty} x^{a_4})=\frac{1+x+x^2}{1-x^2}(x+x^3+x^5+x^7)\left(\frac{1}{1-x} - 1\right)\]


Math 394 - Probability I
both E and F:
\[ P(EF) = P(E\cap F) \]
If E and F are independent, then
\[ P(E\cap F) = P(E)P(F) \]


$$P(F) = P(E) + P(E^c F)$$ $$P(E\cup F) = P(E) + P(F) - P(EF)$$
$$P(E\cup F \cup G) = P(E) + P(F) + P(G) - P(EF) - P(EG) - P(FG) + P(EFG)$$

E occurs given F:
\[P(E|F)=\frac{P(EF)}{P(F)} \]
$$P(EF)=P(FE)=P(E)P(F|E)$$
Bayes formula:
\[ P(E)=P(EF)+P(EF^c) \]
\[P(E)=P(E|F)P(F) + P(E|F^c)P(F^c) \]
\[P(E)=P(E|F)P(F) + P(E|F^c)[1 - P(F)] \]

\[ P(B|A)=P(A|B)\frac{P(B)}{P(A)} \]

The odds of A:
\[ \frac{P(A)}{P(A^c)} = \frac{P(A)}{1-P(A)} \]
\[P[\text{Exactly } k \text{ successes}]=\binom{n}{k}p^k(1-p)^{n-k} \]

\[ P(n \text{ successes followed by } m \text{ failures}) = \frac{p^{n-1}(1-q^m)}{p^{n-1}+q^{m-1}-p^{n-1}q^{m-1}} \]
where $$p$$ is the probability of success, and $$q=1-p$$ for failure.

\[ \sum_{i=1}^{\infty}p(x_i)=1 \]
where $$x_i$$ is the $$i^{\text{th}}$$ value that $$X$$ can take on.
\[ E[X] = \sum_{x:p(x) > 0}x p(x) \text{ or }\sum_{i=1}^{\infty} x_i p(x_i) \]
\[ E[g(x)]=\sum_i g(x_i)p(x_i) \]
\[ E[X^2] = \sum_i x_i^2 p(x_i) \]

\[ Var(X) = E[X^2] - (E[X])^2 \]
\[ Var(aX+b) = a^2 Var(X) \]
for constant $$a,b$$
\[ SD(X) = \sqrt{Var(X)} \]


Binomial random variable $$(n,p)$$:
\[ p(i)=\binom{n}{i}p^i(1-p)^{n-i}\; i=0,1,...,n\]
where $$p$$ is the probability of success and $$n$$ is the number of trials.
Poisson:
\[ E[X] = np \]
\[ E[X^2] = \lambda(\lambda + 1) \]
\[ Var(X)=\lambda \]
\[ P[N(t)=k)] = e^{-\lambda t}\frac{(\lambda t)^k}{k!} \: k=0,1,2,... \]
where $$N(t)=[s,s+t]$$
\[ E[X] = \int_{-\infty}^{\infty} x f(x) \,dx \]
\[ P\{X \le a \} = F(a) = \int_{-\infty}^a f(x) \,dx \]
\[ \frac{d}{da} F(g(a))=g'(a)f(g(a))\]
\[ E[g(X)]=\int_{-\infty}^{\infty} g(x) f(x) \,dx \]
\[ P\{ a \le X \le b \} = \int_a^b f(x) \,dx \]

Uniform:
\[ f(x) = \frac{1}{b-a} \text{ for } a\le x \le b \]
\[ E[X] = \frac{a+b}{2} \]
\[ Var(X) = \frac{(b-a)^2}{12}\]

Normal:
\[ f(x) = \frac{1}{\sqrt{2\pi} \sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \text{ for } -\infty \le x \le \infty \]
\[ E[X] = \mu \]
\[ Var(X) = \sigma^2 \]
\[ Z = \frac{X-\mu}{\sigma} \]

\[ P\left[a \le X \le b\right] = P\left[ \frac{a - \mu}{\sigma} < \frac{X - \mu}{\sigma} < \frac{b - \mu}{\sigma}\right] = \phi\left(\frac{b-\mu}{\sigma}\right) - \phi\left(\frac{a-\mu}{\sigma}\right) \]



\[ \phi(x) = GRAPH \]
\[ P[X \le a]=\phi\left(\frac{a-\mu}{\sigma}\right) \]
\[ P[Z > x] = P[Z \le -x] \]
\[ \phi(-x)=1-\phi(x) \]

\[ Y=f(X) \]
\[ F_Y=P[Y\le a]= P[f(X) \le a] = P[X \le f^{-1}(a)]=F_x(f^{-1}(a)) \]
\[ f_Y=\frac{d}{da}(f^{-1}(a))f_x(f^{-1}(a)) \]
\[ Y = X^2 \]
\[ F_Y = P[Y \le a] = P[X^2 \le a] = P[-\sqrt{a} \le X \le \sqrt{a}] = \int_{-\sqrt{a}}^{\sqrt{a}} f_X(x) dx \]
\[ f_Y = \frac{d}{da}(F_Y) \]


\[N(k) \ge x \rightarrow 1 - N(k) < x \]
\[ P(A \cap N(k) \ge x) = P(A) - P(A \cap N(k) < x) \]




Discrete:
\[ P(X=1) = P(X \le 1) - P(X < 1) \]
Continuous:
\[ P(X \le 1) = P(X < 1) \]


Exponential:
\[ f(x) = \lambda e^{-\lambda x} \text{ for } x \ge 0 \]
\[ E[X] = \frac{1}{\lambda} \]
\[ Var(X) = \frac{1}{\lambda^2} \]


Gamma:
\[ f(x) = \frac{\lambda e^{-\lambda x} (\lambda x)^{\alpha - 1}}{\Gamma(\alpha)} \]
\[ \Gamma(\alpha)=\int_0^{\infty} e^{-x}x^{\alpha-1} \,dx \]
\[ E[X] = \frac{\alpha}{\lambda} \]
\[ Var(X) = \frac{\alpha}{\lambda^2} \]

\[ E[X_iX_j] = P(X_i = k \cap X_j=k) = P(X_i = k)P(X_j = k|X_i=k) \]


Stat 390 - Probability and Statistics for Engineers and Scientists
\[ p(x) = \text{categorical (discrete)} \]
\[ f(x) = \text{continuous} \]
$$\mu_x=E[x]$$$$\sigma_x^2 = V[x]$$
Binomial$$n\pi$$$$n\pi (1-\pi)$$
Normal$$\mu$$$$\sigma^2$$
Poisson$$\lambda$$$$\lambda$$
Exponential$$\frac{1}{\lambda}$$$$\frac{1}{\lambda^2}$$
Uniform$$\frac{b+a}{2}$$$$\frac{(b-a)^2}{12}$$
Binomial\[ p(x) = \binom{n}{x} \pi^x (1 - \pi)^{n-x}\]
Poisson\[ p(x) = \frac{e^{-\lambda} \lambda^x}{x!} \]
Normal\[ f(x) = \frac{1}{\sqrt{2 \pi \sigma^2}} e^{-\frac{1}{2} \left(\frac{x - \mu}{\sigma} \right)}\]
Exponential\[ f(x) = \lambda e^{-\lambda x} \]
Uniform\[ f(x) = \frac{1}{b-a}\]
$$\pi = p =$$ proportion
$$n\pi = C = \lambda =$$ mean
$$\mu = n \pi$$
$$\sigma^2 = n \pi (1 - \pi)$$
Sample mean: $$\bar{x} =$$\[ \frac{1}{n} \sum_{i=1}^n x_i \]Sample median:\[\tilde{x} = \text{if } n \text{ is odd,} \left(\frac{n+1}{2}\right)^{\text{th}} \text{ value} \] \[ \text{if } n \text{ is even, average} \frac{n}{2} \text{ and } \frac{n}{2}+1\]
Sample variance:\[s^2 = \frac{1}{n-1}\sum (x_i - \bar{x})^2 = \frac{S_{xx}}{n-1} = \sum x_i^2 - \frac{1}{n} \left( \sum x_i \right)^2\]Standard deviation: $$s = \sqrt{s^2}$$

low/high quartile: median of lower/upper half of data. If $$n$$ is odd, include median in both.
low = 1st quartilehigh = 3rd quartilemedian = 2nd quartile

IQR: high - low quartiles
range: max - min
Total of something: $$\bar{x} n$$
An outlier is any data point outside the range defined by IQR $$\cdot 1.5$$
An extreme is any data point outside the range defined by IQR $$\cdot 3$$

\[ \mu_{\bar{x}} = \mu = \bar{x} \]
\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]
\[ \mu_p = p = \pi \]
\[ \sigma_p = \sqrt{\frac{p(1-p)}{n}} \]

$$p(x)$$ distribution
[discrete]
mean: $$\mu_x = \sum x \cdot p(x) = E[x]$$
variance: $$\sigma_x^2 = \sum(x-\mu_x)^2 p(x) = V[x]$$
$$f(x)$$ distribution
[continuous]
mean:
\[ \mu_x = \int_{-\infty}^{\infty} x \cdot f(x) \]

median:
\[ \tilde{\mu} \rightarrow \int_{-\infty}^{\tilde{\mu}} f(x) = \int_0^{\tilde{\mu}} f(x) = 0.5 \]

variance:
\[ \sigma^2 = \int_{-\infty}^{\infty} (x =\mu_x)\cdot f(x) \]
Normal
\[ k = \frac{\mu + k \sigma - \mu}{\sigma} \]
standardize:
\[ \frac{x - \mu}{\sigma} \]
\[ -k = \frac{\mu - k \sigma - \mu}{\sigma} \]
upper quartile:
\[ \mu + 0.675\sigma \]
lower quartile:
\[ \mu - 0.675\sigma \]
Exponential
\[ -ln(c) \cdot \frac{1}{\lambda} \text{ where c is the quartile }(0.25,0.5,0.75) \]
\[ S_{xx} = \sum x_i^2 - \frac{1}{n}\left(\sum x_i\right)^2 = \sum(x_i - \bar{x})^2 \]
\[ S_{yy} = \sum y_i^2 - \frac{1}{n}\left(\sum y_i\right)^2 = \sum(y_i - \bar{y})^2 \]
\[ S_{xy} = \sum{x_i y_i} - \frac{1}{n}\left(\sum x_i\right)\left(\sum y_i\right) \] \[ \text{SSResid} = \text{SSE (error sum of squares)} = \sum(y_i - \hat{y}_i)^2 = S_{yy} - b S_{xy} \]
\[ \text{SSTo} = \text{SST} = S_{yy} = \text{Total sum of squares} = \text{SSRegr} + \text{SSE} = \text{SSTr} + \text{SSE} = \sum_i^k \sum_j^n (x_{ij} - \bar{\bar{x}})^2 \]
\[ \text{SSRegr} = \text{regression sum of squares} \]
\[ r^2 = 1 - \frac{\text{SSE}}{\text{SST}} = \frac{\text{SSRegr}}{\text{SST}} = \text{coefficient of determination} \]

\[ r = \frac{S_{xy}}{\sqrt{S_{xx}}\sqrt{S_{yy}}} \]
\[ \hat{\beta} = \frac{S_{xy}}{S_{xx}} \]
\[ \hat{\alpha} = \bar{y} - \beta\bar{x} \]
prediction:
\[ \hat{\alpha} = \bar{y} - \beta\bar{x} \]
Percentile ($$\eta_p$$):
\[ \int_{-\infty}^{\eta_p} f(x) = p \]

MSE (Mean Square Error) =
\[ \frac{1}{n} SSE = \frac{1}{n}\sum (y_i-\hat{y}_i)^2 = \frac{1}{n} \sum (y_i - \alpha - \beta x_i)^2 \]

MSTr (Mean Square for treatments)

ANOVA
\[ SST = SS_{\text{explained}} + SSE \]
\[ R^2 = 1 - \frac{SSE}{SST} \]
\[ \sum(y_i - \bar{y})^2 = \sum(\hat{y}_i - \bar{y})^2 + \sum(y_i - \hat{y}_i)^2 \]
\[ s_e^2 = \frac{SSE}{n-2} \text{ or } \frac{SSE}{n-(k+1)} \]
\[ R_{adj}^2 = 1-\frac{SSE(n-1)}{SST(n-(k+1))}\]
\[ H_0: \mu_1 = \mu_2 = .. = \mu_k \]
\[ \bar{\bar{x}} = \left(\frac{n_1}{n}\right)\bar{x}_1 + \left(\frac{n_2}{n}\right)\bar{x}_2 + ... + \left(\frac{n_k}{n}\right)\bar{x}_k \]
\[ SSTr = n_1(\bar{x}_1 - \bar{\bar{x}})^2 + n_2(\bar{x}_2 - \bar{\bar{x}})^2 + ... + n_k(\bar{x}_k - \bar{\bar{x}})^2 \]

SourcesdfSSMSF
Between Samples$$k-1$$SSTrMSTr$$\frac{\text{MSTr}}{\text{MSE}}$$
Within Samples$$n-k$$SSEMSE
Total$$n-1$$SST

One-sample t interval:
\[ \bar{x} \pm t^* \frac{s}{\sqrt{n}} \]
(CI)
Prediction interval:
\[ \bar{x} \pm t^* s\sqrt{1+\frac{1}{n}} \]
(PI)
Tolerance interval:
\[ \bar{x} \pm k^* s \]

Difference t interval:
\[ \bar{x}_1 - \bar{x}_2 \pm t^*\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]

Adjusted
\[ R^2 = 1 - \left(\frac{n-1}{n-(k+1)}\right) \frac{SSE}{SST} \]

Type I error: Reject $$H_0$$ when true; If the F-test p-value is small, it's useful.
Type II error: Don't reject $$H_0$$ when false.
B(Type II) =
\[ z^* \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \]


Simple linear regression: $$y = \alpha + \beta x$$
General multiple regression: $$y = \alpha + \beta_1 x_1 + ... + \beta_k x_k$$
Prediction error: $$\hat{y} - y^* = \sqrt{s_{\hat{y}}^2 + s_e^2} \cdot t$$
\[ t = \frac{\hat{y}-y^*}{\sqrt{s_{\hat{y}}^2 + s_e^2}} \]

\[ P(\hat{y} - y^* > 11) = P\left(\sqrt{s_{\hat{y}}^2 + s_e^2} \cdot t > 11\right) = P\left(t > \frac{11}{\sqrt{s_{\hat{y}}^2 + s_e^2}}\right) \]

\[ \mu_{\hat{x}_1 - \hat{x}_2} = \mu_{\hat{x}_1} - \mu_{\hat{x}_2} = \mu_1 - \mu_2 \]
\[ \sigma_{\hat{x}_1 - \hat{x}_2} = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \]
\[ \hat{x}_1 - \hat{x}_2 \pm z^* \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]


To test if $$\mu_1 = \mu_2$$, use a two-sample t test with $$\delta = 0$$
If you're looking for the true average, it's a CI, not the standard deviation about the regression.
A large n value may indicate a z--test, but you must think about whether or not its paired or not paired.
A hypothesis is only valid if it tests a population parameter. Do not extrapolate outside of a regression analysis unless you use a future predictor.

F-Test
\[ F = \frac{MSRegr}{MSE} \]
\[ MSRegr = \frac{SSRegr}{k} \]
\[ MSResid = \frac{SSE}{n - (k+1)} \]
\[ H_0: \beta_1=\beta_2=...=\beta_k=0 \]


Chi-Squared ($$\chi^2$$)
\[ H_0: \pi_1 = \frac{\hat{n}_1}{n_1},\pi_2 = \frac{\hat{n}_2}{n_2}, ..., \pi_k = \frac{\hat{n}_k}{n_k} \]
\[ \chi^2 = \sum_{i=1}^k \frac{(n_i - \hat{n}_i)^2}{\hat{n}_i} = \sum \left(\frac{\text{observed - expected}}{\text{expected}}\right) \]


Linear Association Test
\[ H_0: p=0 \]
\[ t = \frac{r\sqrt{n-2}}{\sqrt{1-r^2}} \]
\[ \sigma_{\hat{y}} = \sigma \sqrt{\frac{1}{n} + \frac{(x^* - \bar{x})^2}{S_{xx}} } \]
\[ \mu_{\hat{y}} = \hat{y} = \alpha + \beta x^* \]
\[ \hat{y} \pm t^* s_{\hat{y}} df = n-2 \text{ for a mean y value (population)} \]
\[ \hat{y} \pm t^* \sqrt{s_e^2 + s_{\hat{y}}^2} df=n-2 \text{ for single future y-values (PI)} \]


Paired T-test
\[ \bar{d} = \bar{(x-y)} \]
\[ t = \frac{\bar{d} - \delta}{\frac{s_d}{\sqrt{n}}} \]
\[ \sigma_d = \sigma \text{ of x-y pairs } = s_d \]


Large Sample:
\[ z = \frac{\bar{x} - 0.5}{\frac{s}{\sqrt{n}}} \]
\[ \text{P-value } \le \alpha \]

Small Sample:
\[ z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}} \]
\[ df = n-1 \]


Difference:
\[ H-0: \mu_1 - \mu_2 = \delta \]
\[ t = \frac{\bar{x}_1 - \bar{x}_2 - \delta}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]
\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)}{\frac{1}{n_1-1} \left(\frac{s_1^2}{n_1}\right)^2 + \frac{1}{n_2-1} \left(\frac{s_2^2}{n_2}\right)^2 } \]

Contact


*"We must be the change we want to see in the world." - Mahatma Gandhi*
Given the enormous influence of The Matrix on a lot of what I make, many people assume that it is my favorite movie. My favorite movie of all time is actually [Contact](https://en.wikipedia.org/wiki/Contact_(film)).

Many people assume the message behind Contact is that Religion and Science aren’t mutually exclusive, but this is only part of the equation. Contact is really about faith. It’s about never losing faith in your own passion, regardless of what it is. It can be science, religion, art, music, it doesn’t matter. All that matters is that you find your passion, and you have faith in it, and no matter what happens, no matter what stands in your way, you fight for it.

Naturally, in real life, we don’t have a Dues Ex Hadden Industries to magic away all our problems, but the movie does an excellent job of making sure the drama that plays out on screen does not undermine it’s purpose. Even with virtually unlimited funding, Ellie finds out she’s being forcefully kicked out of the Very Large Array. No matter what anyone says to her, she refuses to let it go. At that point in the movie, it is abundantly clear that absolutely nothing will stop this woman from pursuing her passion, no matter how insane or misunderstood it is, no matter what obstacles are put in her path, no matter what friends she may lose along the way. At this darkest hour, after everything seems to be falling apart, she goes to a cliffside to think.

And then she comes back.

She comes back, and she starts listening again, because she isn’t going to give up, and that is when she hears the signal. That is when our hollywood movie is yanked out of reality and brings us on a fantastic journey through the stars. It happens after Ellie refuses to give up, even after the universe itself seems to be conspiring to ruin her.

Then, the project is stolen from her repeatedly, and she has to fight to keep it. Then, she is denied her desire to be the one to ride through the machine. Then the machine is completely destroyed and everything she fought for is taken away from her.

Then, our little Dues Ex Hadden comes back. Hadden gave Ellie her funding because of her emotional outburst at the investor meeting. Hadden saw the passion she had, the unstoppable fire and unshakable faith she had in her science project, and that is when he decided to fund her. Once again, Hadden moves us into the realm of fiction, and gives Ellie a second chance.

Contact is brilliant because it begins firmly anchored in reality, and only later, with a few nudges in the right direction, throws us into a world of make-believe. Carl Sagan is trying to make us ask, what if? What if we could fly? What if we could go to the moon? What if we colonized Mars? What if aliens contacted us? Contact shows us how beautiful life can be. It reminds us of all the amazing things that human beings are capable of when we ask that simple question, What If?

To answer it, we just need a little faith.


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