Thus far

Allow me to clear the air about graduate school.  Everything you hear is absolutely true!

Only 10 hours is required each week for teaching duties; that includes office hours, tutoring, meeting with the course supervisor, and of course actually teaching.  Not factored into those 10 hours is lesson planning.  We mathematics GTAs have the luxury of designing our very own lectures, so long as those lectures cover the material in the syllabus and prepare students for the exams.  Cool, but it does take time.  

As an out-of-state student I'm only enrolled in six credits (two classes), and so I spend six hours each week absorbing lectures.  As an undergraduate the saying goes, "For each hour of lecture, you should spend two hours studying."  I can tell you right now that graduate school could care less that there are only 24 hours in a day, because for each hour of lecture you should spend 10 hours studying.  (At least for Real Analysis, which is destroying my life!)

Dating: What is that?  

Running: I spent an entire week indoors, hunched over my books.

Drinking: We are being driven to it.

Cooking: Better order in, there's no time!

Sleeping: Three hours is enough, isn't it?

Despite all of this, there is no other place I would rather be than in graduate school at Montana State University.

Moving = Annoying Tetris

Today will involve a lot of strategizing and determination in order to get the U Haul packed.  So today I will be playing Tetris with boxes, only it will be an annoying 3D version. 

Then tomorrow, the western road trip will begin!

Redefining Mathematics Education

Watch out future students of mine!

Ah... summer

For those of you in college, you can understand the mess which has taken over my desk.  My once organized, clean, and productive workstation is now a pile of various notebooks, folders, and textbooks.  Digging deeper you can see that even my notes from the beginning of the semester are thorough and easy-to-read, and as the dates near the end of the semester they are anything but thorough or easy-to-read.

The busy schedule of a college student leaves time for working, exercising, and sleeping, but at the cost of properly absorbing the course material.  When I say "properly absorbing" I mean that I have a recollection of the concepts and main ideas of the courses I've survived, but I certainly couldn't explain an elaborate proof without re-reading at least a chapter on the matter.  The last thing I want is to forget everything I have learned, and although I have a degree in mathematics and not a foreign language, the saying goes "If you don't use it, you'll lose it."

Although I'm excited to begin my graduate school experience, it is not August yet.  So, I will be giving myself a refresher course of abstract algebra.  I will actually have the time to read the book, and several times if necessary.  I will complete the problems at the end of each lesson with out the pressures of an unreasonable deadline.  Finally, I will have the time to share this review with you.

The book is Abstract Algebra, Third Edition, by I.N. Herstein.

Chapter One: Things Familiar and Less Familiar
     Section One: A Few Preliminary Remarks

There wasn't anything I took note of and only three problems in this section.  I'd say this was a nice section to get me thinking in the realm of abstract algebra.

Problems: 1-3

Stewart's Hoard: Magic Circle

Arrange the numbers 1, 2, 3, 4, 5, and 6 in the small circles so that the sum of the numbers on each large circle is fourteen.

First, notice that each small circle is intersected by two large circles.  If I give each circle its own color, we have the following.

For each small circle we have three options,
Red/Green
Green/Blue
Blue/Red
and each of these appears twice in the diagram





In order to solve this problem I first began with C(6;4) to find all the four-number combinations of the numbers 1-6.  For those of you who don't know, C(6;4) = $\dfrac{6!}{4!(6-4)!} = 15$. 
Here are the fifteen combinations:

1234     1235     1236     1245     1246    
1256     1345     1346     1356     1456
2345     2346     2356     2456     3456

Now find those that have a sum of fourteen:

1234     1235     1236     1245     1246    
1256     1345     1346     1356     1456
2345     2346     2356     2456     3456

There are three combinations and three circles and now our task is complete.  As you can see, each of the combinations shares two digits with another combination, and the other two digits with the remaining combination.  This allows us to fulfill the intersections of the circles.

Source of Problem: Professor Stewart's Hoard of Mathematical Treasures by Ian Stewart

Great News

Do you know that feeling?  When you push yourself to jump through hoops and it actually pays off?  I totally have that feeling.

For those of you who have read my past posts about the GRE, you know I've had my sights on Graduate School.  Let me tell you, the GRE is the first hoop (and the second if you are also required to take the subject test).  Then once you've decided which schools to apply to, the application process is just as brutal.  Each University wants a different version of a Statement of Purpose, and some want a separate essay concerning who you are as a person.

If you're looking for a better option than student loans to cover the high cost of Graduate School, especially if applying out-of-state, then you'll need to be aware of the extremely early deadlines for the applications.  If you are seeking a Fellowship or an Assistantship, it may not be enough to simply meet the early deadline; you may be required to submit a special form stating that you are seeking one or both of the above.  Considering schools in any state except Michigan, I opted for this route, and once each school received my application I felt relief, but I knew it still wasn't over.

Most schools require three letters of recommendation, usually from professors or instructors you've had the pleasure of working with during your undergraduate career.  Although I had previously asked three professors in the mathematics department, who all gladly agreed, to compose letters on my behalf, the applications I submitted were not complete until their letters were in the hands of those who would decide my fate.  I sent many e-mails, and even made some phone calls, checking in to ensure my applications would make the deadline.

Some were late and some on-time, but it was out of my hands.  I can say these last few months of waiting have been difficult.  I do not know the process my applications underwent, but they must have been read by many eyes.  Out of the five schools I applied to, one school replied with a, "Thanks, but no thanks," three schools said, "Congratulations!  We'd love to have you attend," and one school has yet to say anything.

Focusing on the three possibilities, this week the decision has been made.  If you knew me, then you'd know that I've been spending much of my free-time researching the three schools more closely.  Looking into the cities in which they reside, including the community, climate, and future opportunities.  Yet somehow none of these were the deciding factor.  In fact, all that time I spent researching was almost a waste.  You see, I didn't choose a school, a school chose me.

I was officially offered a Graduate Teaching Assistantship at Montana State University.  Who could pass up an offer like that?  I'm extremely happy I pushed myself and made it to this moment, knowing that it was all worth it.  Now it's just a matter of working out the details that come along with moving to a place I've never been.  I think I'll put that off for a few days, and simply bask in the glory.

Chapter 4 #3

Use the definition to find the derivative of $f(x)=\sqrt {x}$, for $x>0$.  Is $f$ differentiable at zero?  Explain.

$\lim_{t\rightarrow 0} \dfrac {\sqrt {x+t}-\sqrt {x}}{t}$

     $=\lim_{t\rightarrow 0} \dfrac {\sqrt {x+t}-\sqrt {x}}{t} * \dfrac {\sqrt {x+t}+\sqrt {x}}{\sqrt {x+t}+\sqrt {x}}$

     $=\lim_{t\rightarrow 0} \dfrac {x+t-x}{t[\sqrt {x+t}+\sqrt {x}]}$

     $=\lim_{t\rightarrow 0} \dfrac {1}{\sqrt {x+t}+\sqrt {x}}$

     $=\dfrac {1}{2\sqrt {x}}$

$f(x)=\sqrt {x}$ is not differentiable at $x=0$ because the limit of $T(x)=\dfrac {f(x)-f(x_0)}{x-x_0}$ as $x$ approaches zero does not exist.

$\lim_{x\rightarrow 0} \dfrac {\sqrt {x}-\sqrt {0}}{x-0}$

     $=\lim_{x\rightarrow 0} \dfrac {1}{\sqrt {x}}$

     $=\dfrac {1}{0}$ and is undefined.

If continuous, then Riemann integrable

Theorem 5.4 
If $f:[a,b]\rightarrow R$ is continuous, then $f\in R(x)$ on $[a,b]$.

*The domain of the function is closed and bounded, making it compact.  Having a compact domain implies the function is uniformly continuous, and with this we can say that $f$ is bounded (by previous theorems).


By the definition of uniform continuity, 

          choose an $\epsilon >0$ then there is a $\delta >0$ 

          such that $\left| x-y\right| <\delta$ with $x,y\in [a,b]$ 

          implies $|f(x)-f(y)|<\dfrac {\epsilon}{b-a}$.


Let $P=\{x_0, ..., x_n\}$ be a partition 

          such that $x_i-x_{i-1}<\delta$ for $i=1, ...,n$.

          Then there are $t_i,s_i\in [x_{i-1}, x_i]$ 

          such that $M_i(f)=f(t_i)$ and $m_i(f)=f(s_i)$.

Since $x_i-x_{i-1}<\delta$ then certainly $|t_i-s_i|<\delta$ 

          which implies$|f(t_i)-f(s_i)|<\dfrac {\epsilon}{b-a}$.  

          By a previous theorem, we know $f(t_i)\geq f(s_i)$, so we have a stronger inequality:

          $0\leq f(t_i)-f(s_i)<\dfrac {\epsilon}{b-a}$.

We now have the following:

$U(P,f)-L(P,f)=\sum _{i=1}^{n}[f(t_i)-f(s_i)](x_i-x_{i-1})$

                                $<\sum _{i=1}^{n}\dfrac {\epsilon}{b-a}(x_i-x_{i-1})$

                                $=\dfrac {\epsilon}{b-a}(b-a)=\epsilon$

Therefore, $f\in R(x)$ on $[a,b]$.

Notes

4.1 The Derivative of a Function


"The basic idea in calculating the slope of the tangent line at a point $P_0$ on the curve is to find the slope of the line through $P_0$ and another point $P$ on the curve, and then to take the limit of that slope as $P$ approaches $P_0$.  If that limit exists, it is defined to be the slope of the curve at $P_0$."

If the coordinates of point $P_0$ are $(x_0, y_0)$ and the coordinates of point $P$ are $(x, y)$, then the slope of the line passing through those points is

          $\dfrac {y-y_0}{x-x_0}$    or    $\dfrac {f(x)-f(x_0)}{x-x_0}$.

Let $f:D \rightarrow R$ with $x_0 \in D$ an accumulation point of $D$ and $x_0 \in D$ with $x \neq x_0$, define $T(x)= \dfrac {f(x)-f(x_0)}{x-x_0}$.

Let $f:D \rightarrow R$ with $x_0$ an accumulation point of $D$ and $x_0 \in D$.  For each $t \in R$ such that $x_0 + t \in D$ and $t \neq 0$, define $Q(t)= \dfrac {f(x_0 + t)-f(x_0)}{t}$. 

Prove that the two definitions above are equivalent.

Limit of a Function

 How would one prove that the limit of this function, 
$f(x)=x-2$ where $x\neq-2$,
is $L=-4$ at the excluded point?







Lets define $f:(-2,0) \rightarrow R$ by $f(x) = \dfrac {x^2 - 4} {x+2}$. 
    
By definition of a limit, if $x_0$ is an accumulation point of the domain, then $f$ has a limit $L$ at $x_0$ iff for each $\epsilon >0$, there is a $\delta >0$ such that if $0<|x-x_0|<\delta$ then $|f(x) - L|<\epsilon$.


Choose $\epsilon = \delta$
     If $0<|x+2|<\delta$ 
          Then $|f(x) - L| = |\dfrac {x^2 - 4} {x+2} +4| = |(x-2)+4| = |x+2|<\epsilon = \delta$ 

Therefore, $f(x)$ has a limit at $x=-2$, and that limit is $L=-4$.

Uniformly continuous, continued

Aha!  Class on Friday revealed some important information in how this proof works out.   

Since $D$ is a bounded set, there are $x_1, x_2, ... , x_n \in D$ such that
$D\subset \cup^n _{i=1} (x_i - \delta, x_i + \delta)$

Because D is bounded, we can find an open cover: a collection of open sets that contains the original set.

For all $x \in D$, then $x_i - \delta \leq x \leq x_i + \delta$
Then $|x - x_i| < \delta$
Which implies $|f(x) - f(x_i)| < 1$
     But that just means $f(x_i) - 1 \leq f(x) \leq f(x_i) + 1$

Therefore, $f(D)$ is bounded.

Uniformly continuous

I will begin with a definition of uniformly continuous directly from the book

A function $f: D \to R$ is uniformly continuous on $E\subset D$ iff for every $\epsilon >0$, there is \delta >0 such that if $x,y\in D$ with $|x-y|<\delta$, then $|f(x) - f(y)|<\epsilon$. If $f$ is uniformly continuous on $D$, we say $f$ is uniformly continuous.

That definition is helpful in proving this theorem

3.9 If $f: D\to R$ is uniformly continuous and $D$ is a bounded set, then $f(D)$ is a bounded set.

Choose \epsilon =1. 
     Then \exists \delta >0 such that
          if $|x-y| <\delta$
          then $|f(x) - f(y)| <1$

Since $D$ is a bounded set, there are $x_1, x_2, ..., x_n \in D$ such that
     $D\subset \cup^n _{i=1} (x_i-\delta, x_i+\delta)$
          Then $f(D)\subset \cup^n _{i=1} (f(x_i)-1, f(x_i)+1)$

Now we have $|x-x_i| <\delta$ where $x \in D$
     Then $|f(x) - f(x_i)| <1$

Thus, $f(D)$ is bounded!

Well, I find this proof easy to follow, except for the union of the neighborhoods.  How does that come from D being bounded?

It's the end of the road

Well all, today was my last day of classes as an undergraduate student. That journey has ended and a new will begin soon!

Final project

Ever hear of Matching Theory?
Although the theorems are pretty straight-forward and lacking any elegant proofs, I have found some interesting applications. For example, The Kidney Exchange by Alvin Roth of Harvard, which you can view by clicking HERE.

He utilizes concepts such as directed graphs, cycles, paths, and I'm sure some other important ideas. Unfortunately I haven't made it through the entire paper yet, but I assure you I will. My final project is dependent upon this paper for a lecture that will emphasize the applications of Matching Theory.

I'm Back!

Wow, I have been an absent blogger for too long! Perhaps I should start by shedding some light on the tasks that have been consuming my time, leaving none for blogging.

First, I am still an undergraduate student. It is my last semester as such and I'm working my tail-end off to walk across that stage. If you were to look at my class schedule you'd think it was easy, but when you throw in the study time each class requires, three part-time jobs, and then any necessary activities like eating, sleeping, or socializing, then you'd see just how crazy this semester has been.

Oh yeah, don't let me forget to tell you about finding the time to study for and to take the GRE general and subject tests. Also, researching schools to send those scores to, which also involves completing applications and writing several versions of my Statement of Purpose. All of you who may be thinking about graduate school someday, here's where I give you some advice: DO ALL THIS CRAP DURING SUMMER VACATION!

Despite all that, I'm still curious about the Fibonacci, Padovan, and other sequences. I'm anxious to reunite myself with the stack of papers lying dormant on my desk. Any suggestions on where to start?

Combinatorics

The following problem is taken from Introductory Combinatorics, third edition, by Kenneth P. Bogart.

Some rows of Pascal's triangle consist of even numbers with the exception of the two entries at the ends. Show that row 8 has this property. Do any rows beyond row 8 (that is, with n>8) have this property? Explain why not or give an example.

I like this problem, and I think you should give it a try!

:)

What a great day to have a birthday!
9-10-11

An elegant proof

I'm taking this proof from Introductory Combinatorics, Third Edition by Kenneth P. Bogart. In the first week of classes I have discovered an elegant proof that I would very much like to share with you all.

The powers of two are all even and the powers of five all end in five. Show that for any other prime p, among the first six powers of p, one of these numbers must end in one.

If a prime is not two, then no power of it is even, so a power can only end in one of the five odd digits. Then the function given by

f(i) = the last digit of pⁱ

from {1, 2, 3, 4, 5, 6} to the set of digits {1, 3, 5, 7, 9} cannot be one-to-one. Therefore pⁱ and p^j have the same last digit for some i and j between 1 and 6 with i < j. Thus p^{j -} pⁱ ends in a zero, so it is a multiple of 10. Thus pⁱ⁽p^j-i - 1) is a multiple of 10, and since neither 2 nor 5 is a factor of p, (p^j-i - 1) must be a multiple of 10 (and therefore end in zero), so adding 1 to each side tells us that p^j-i ends in a 1.