## Math 100: Week 4 Saturday, Sep 28 2013

This is a post for my math 100 calculus class of fall 2013. In this post, I give the 4th week’s recitation worksheet (no solutions yet – I’m still writing them up). More pertinently, we will also go over the most recent quiz and common mistakes. Trig substitution, it turns out, is not so easy.

Before we hop into the details, I’d like to encourage you all to avail of each other, your professor, your ta, and the MRC in preparation for the first midterm (next week!).

1. The quiz

There were two versions of the quiz this week, but they were very similar. Both asked about a particular trig substitution

$\displaystyle \int_3^6 \sqrt{36 - x^2} \mathrm{d} x$

And the other was

$\displaystyle \int_{2\sqrt 2}^4 \sqrt{16 - x^2} \mathrm{d}x.$

They are very similar, so I’m only going to go over one of them. I’ll go over the first one. We know we are to use trig substitution. I see two ways to proceed: either draw a reference triangle (which I recommend), or think through the Pythagorean trig identities until you find the one that works here (which I don’t recommend).

We see a ${\sqrt{36 - x^2}}$, and this is hard to deal with. Let’s draw a right triangle that has ${\sqrt{36 - x^2}}$ as a side. I’ve drawn one below. (Not fancy, but I need a better light).

In this picture, note that ${\sin \theta = \frac{x}{6}}$, or that ${x = 6 \sin \theta}$, and that ${\sqrt{36 - x^2} = 6 \cos \theta}$. If we substitute ${x = 6 \sin \theta}$ in our integral, this means that we can replace our ${\sqrt{36 - x^2}}$ with ${6 \cos \theta}$. But this is a substitution, so we need to think about ${\mathrm{d} x}$ too. Here, ${x = 6 \sin \theta}$ means that ${\mathrm{d}x = 6 \cos \theta}$.

Some people used the wrong trig substitution, meaning they used ${x = \tan \theta}$ or ${x = \sec \theta}$, and got stuck. It’s okay to get stuck, but if you notice that something isn’t working, it’s better to try something else than to stare at the paper for 10 minutes. Other people use ${x = 6 \cos \theta}$, which is perfectly doable and parallel to what I write below.

Another common error was people forgetting about the ${\mathrm{d}x}$ term entirely. But it’s important!.

Substituting these into our integral gives

$\displaystyle \int_{?}^{??} 36 \cos^2 (\theta) \mathrm{d}\theta,$

where I have included question marks for the limits because, as after most substitutions, they are different. You have a choice: you might go on and put everything back in terms of ${x}$ before you give your numerical answer; or you might find the new limits now.

It’s not correct to continue writing down the old limits. The variable has changed, and we really don’t want ${\theta}$ to go from ${3}$ to ${6}$.

If you were to find the new limits, then you need to consider: if ${x=3}$ and ${\frac{x}{6} = \sin \theta}$, then we want a ${\theta}$ such that ${\sin \theta = \frac{3}{6}= \frac{1}{2}}$, so we might use ${\theta = \pi/6}$. Similarly, when ${x = 6}$, we want ${\theta}$ such that ${\sin \theta = 1}$, like ${\theta = \pi/2}$. Note that these were two arcsine calculations, which we would have to do even if we waited until after we put everything back in terms of ${x}$ to evaluate.

Some people left their answers in terms of these arcsines. As far as mistakes go, this isn’t a very serious one. But this is the sort of simplification that is expected of you on exams, quizzes, and homeworks. In particular, if something can be written in a much simpler way through the unit circle, then you should do it if you have the time.

So we could rewrite our integral as

$\displaystyle \int_{\pi/6}^{\pi/2} 36 \cos^2 (\theta) \mathrm{d}\theta.$

How do we integrate ${\cos^2 \theta}$? We need to make use of the identity ${\cos^2 \theta = \dfrac{1 + \cos 2\theta}{2}}$. You should know this identity for this midterm. Now we have

$\displaystyle 36 \int_{\pi/6}^{\pi/2}\left(\frac{1}{2} + \frac{\cos 2 \theta}{2}\right) \mathrm{d}\theta = 18 \int_{\pi/6}^{\pi/2}\mathrm{d}\theta + 18 \int_{\pi/6}^{\pi/2}\cos 2\theta \mathrm{d}\theta.$

The first integral is extremely simple and yields ${6\pi}$ The second integral has antiderivative ${\dfrac{\sin 2 \theta}{2}}$ (Don’t forget the ${2}$ on bottom!), and we have to evaluate ${\big[9 \sin 2 \theta \big]_{\pi/6}^{\pi/2}}$, which gives ${-\dfrac{9 \sqrt 3}{2}}$. You should know the unit circle sufficiently well to evaluate this for your midterm.

And so the final answer is ${6 \pi - \dfrac{9 \sqrt 2}{2} \approx 11.0553}$. (You don’t need to be able to do that approximation).

Let’s go back a moment and suppose you didn’t re\”{e}valuate the limits once you substituted in ${\theta}$. Then, following the same steps as above, you’d be left with

$\displaystyle 18 \int_{?}^{??}\mathrm{d}\theta + 18 \int_{?}^{??}\cos 2\theta \mathrm{d}\theta = \left[ 18 \theta \right]_?^{??} + \left[ 9 \sin 2 \theta \right]_?^{??}.$

Since ${\frac{x}{6} = \sin \theta}$, we know that ${\theta = \arcsin (x/6)}$. This is how we evaluate the left integral, and we are left with ${[18 \arcsin(x/6)]_3^6}$. This means we need to know the arcsine of ${1}$ and ${\frac 12}$. These are exactly the same two arcsine computations that I referenced above! Following them again, we get ${6\pi}$ as the answer.

We could do the same for the second part, since ${\sin ( 2 \arcsin (x/6))}$ when ${x = 3}$ is ${\sin (2 \arcsin \frac{1}{2} ) = \sin (2 \cdot \frac{\pi}{6} ) = \frac{\sqrt 3}{2}}$; and when ${x = 6}$ we get ${\sin (2 \arcsin 1) = \sin (2 \cdot \frac{\pi}{2}) = \sin (\pi) = 0}$.

Putting these together, we see that the answer is again ${6\pi - \frac{9\sqrt 3}{2}}$.

Or, throwing yet another option out there, we could do something else (a little bit wittier, maybe?). We have this ${\sin 2\theta}$ term to deal with. You might recall that ${\sin 2 \theta = 2 \sin \theta \cos \theta}$, the so-called double-angle identity.

Then ${9 \sin 2\theta = 18 \sin \theta \cos \theta}$. Going back to our reference triangle, we know that ${\cos \theta = \dfrac{\sqrt{36 - x^2}}{6}}$ and that ${\sin \theta = \dfrac{x}{6}}$. Putting these together,

$\displaystyle 9 \sin 2 \theta = \dfrac{ x\sqrt{36 - x^2} }{2}.$

When ${x=6}$, this is ${0}$. When ${x = 3}$, we have ${\dfrac{ 3\sqrt {27}}{2} = \dfrac{9\sqrt 3}{2}}$.

And fortunately, we get the same answer again at the end of the day. (phew).

2. The worksheet

Finally, here is the worksheet for the day. I’m working on their solutions, and I’ll have that up by late this evening (sorry for the delay).

Ending tidbits – when I was last a TA, I tried to see what were the good predictors of final grade. Some things weren’t very surprising – there is a large correlation between exam scores and final grade. Some things were a bit surprising – low homework scores correlated well with low final grade, but high homework scores didn’t really have a strong correlation with final grade at all; attendance also correlated weakly. But one thing that really stuck with me was the first midterm grade vs final grade in class: it was really strong. For a bit more on that, I refer you to my final post from my Math 90 posts.

## Math 100: Week 3 and pre-midterm Tuesday, Sep 24 2013

This is a post for my Math 100 class of fall 2013. In this post, I give the first three weeks’ worksheets from recitation and the set of solutions to week three’s worksheet, as well as a few administrative details.

Firstly, here is the recitation work from the first three weeks:

1. (there was no recitation the first week)
2. A worksheet focusing on review.
3. A worksheet focusing on integration by parts and u-substitution, with solutions.

In addition, I’d like to remind you that I have office hours from 2-4pm (right now) in Kassar 018. I’ve had multiple people set up appointments with me outside of these hours, which I’m tempted to interpret as suggesting that I change when my office hours are. If you have a preference, let me know, and I’ll try to incorporate it.

Finally, there will be an exam next Tuesday. I’ve been getting a lot of emails about what material will be on the exam. The answer is that everything you have learned up to now and by the end of this week is fair game for exam material. This also means there could be exam questions on material that we have not discussed in recitation. So be prepared. However, I will be setting aside a much larger portion of recitation this Thursday for questions than normal. So come prepared with your questions.

Best of luck, and I’ll see you in class on Thursday.

## Notes on the first week (SummerNT) Monday, Jul 1 2013

We’ve covered a lot of ground this first week! I wanted to provide a written summary, with partial proof, of what we have done so far.

We began by learning about proofs. We talked about direct proofs, inductive proofs, proofs by contradiction, and proofs by using the contrapositive of the statement we want to prove. A proof is a justification and argument based upon certain logical premises (which we call axioms); in contrast to other disciplines, a mathematical proof is completely logical and can be correct or incorrect.

We then established a set of axioms for the integers that would serve as the foundation of our exploration into the (often fantastic yet sometimes frustrating) realm of number theory. In short, the integers are a non-empty set with addition and multiplication [which are both associative, commutative, and have an identity, and which behave as we think they should behave; further, there are additive inverses], a total order [an integer is either bigger than, less than, or equal to any other integer, and it behaves like we think it should under addition and multiplication], and satisfying the deceptively important well ordering principle [every nonempty set of positive integers has a least element].

With this logical framework in place, we really began number theory in earnest. We talked about divisibility [we say that $a$ divides $b$, written $a \mid b$, if $b = ak$ for some integer $k$]. We showed that every number has a prime factorization. To do this, we used the well-ordering principle.

Suppose that not all integers have a prime factorization. Then there must be a smallest integer that does not have a prime factorization: call it $n$. Then we know that $n$ is either a prime or a composite. If it’s prime, then it has a prime factorization. If it’s composite, then it factors as $n = ab$ with $a,b < n$. But then we know that each of $a, b$ have prime factorizations since they are less than $n$. Multiplying them together, we see that $n$ also has a prime factorization after all. $\diamondsuit$

Our first major result is the following:

There are infinitely many primes

There are many proofs, and we saw 2 of them in class. For posterity, I’ll present three here.

First proof that there are infinitely many primes

Take a finite collection of primes, say $p_1, p_2, \ldots, p_k$. We will show that there is at least one more prime not mentioned in the collection. To see this, consider the number $p_1 p_2 \ldots p_k + 1$. We know that this number will factor into primes, but upon division by every prime in our collection, it leaves a remainder of $1$. Thus it has at least one prime factor different than every factor in our collection. $\diamondsuit$

This was a common proof used in class. A pattern also quickly emerges: $2 + 1 = 3$, a prime. $2\cdot3 + 1 = 7$, a prime. $2 \cdot 3 \cdot 5 + 1 = 31$, also a prime. It is always the case that a product of primes plus one is another prime? No, in fact. If you look at $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 + 1=30031 = 59\cdot 509$, you get a nonprime.

Second proof that there are infinitely many primes

In a similar vein to the first proof, we will show that there is always a prime larger than $n$ for any positive integer $n$. To see this, consider $n! + 1$. Upon dividing by any prime less than $n$, we get a remainder of $1$. So all of its prime factors are larger than $n$, and so there are infinitely many primes. $\diamondsuit$

I would also like to present one more, which I’ve always liked.

Third proof that there are infinitely many primes

Suppose there are only finitely many primes $p_1, \ldots, p_k$. Then consider the two numbers $n = p_1 \cdot \dots \cdot p_k$ and $n -1$. We know that $n - 1$ has a prime factor, so that it must share a factor $P$ with $n$ since $n$ is the product of all the primes. But then $P$ divides $n - (n - 1) = 1$, which is nonsense; no prime divides $1$. Thus there are infinitely many primes. $\diamondsuit$

We also looked at modular arithmetic, often called the arithmetic of a clock. When we say that $a \equiv b \mod m$, we mean to say that $m | (b - a)$, or equivalently that $a = b + km$ for some integer $m$ (can you show these are equivalent?). And we pronounce that statement as ” $a$ is congruent to $b$ mod $m$.” We played a lot with modular arithmetic: we added, subtracted, and multiplied many times, hopefully enough to build a bit of familiarity with the feel. In most ways, it feels like regular arithmetic. But in some ways, it’s different. Looking at the integers $\mod m$ partitions the integers into a set of equivalence classes, i.e. into sets of integers that are congruent to $0 \mod m, 1 \mod m, \ldots$. When we talk about adding or multiplying numbers mod $\mod m$, we’re really talking about manipulating these equivalence classes. (This isn’t super important to us – just a hint at what’s going on beneath the surface).

We expect that if $a \equiv b \mod m$, then we would also have $ac \equiv bc \mod m$ for any integer $c$, and this is true (can you prove this?). But we would also expect that if we had $ac \equiv bc \mod m$, then we would necessarily have $a \equiv b \mod m$, i.e. that we can cancel out the same number on each side. And it turns out that’s not the case. For example, $4 \cdot 2 \equiv 4 \cdot 5 \mod 6$ (both are $2 \mod 6$), but ‘cancelling the fours’ says that $2 \equiv 5 \mod 6$ – that’s simply not true. With this example in mind, we went about proving things about modular arithmetic. It’s important to know what one can and can’t do.

One very big and important observation that we noted is that it doesn’t matter what order we operate, as in it doesn’t matter if we multiply an expression out and then ‘mod it’ down, or ‘mod it down’ and then multiply, or if we intermix these operations. Knowing this allows us to simplify expressions like $11^4 \mod 12$, since we know $11 \equiv -1 \mod 12$, and we know that $(-1)^2 \equiv 1 \mod 12$, and so $11^4 \equiv (-1)^{2\cdot 2} \equiv 1 \mod 12$. If we’d wanted to, we could have multiplied it out and then reduced – the choice is ours!

Amidst our exploration of modular arithmetic, we noticed some patterns. Some numbers  are invertible in the modular sense, while others are not. For example, $5 \cdot 5 \equiv 1 \mod 6$, so in that sense, we might think of $\frac{1}{5} \equiv 5 \mod 6$. More interestingly but in the same vein, $\frac{1}{2} \equiv 6 \mod 11$ since $2 \cdot 6 \equiv 1 \mod 11$. Stated more formally, a number $a$ has a modular inverse $a^{-1} \mod m$ if there is a solution to the modular equation $ax \equiv 1 \mod m$, in which case that solution is the modular inverse. When does this happen? Are these units special?

Returning to division, we think of the greatest common divisor. I showed you the Euclidean algorithm, and you managed to prove it in class. The Euclidean algorithm produces the greatest common divisor of $a$ and $b$, and it looks like this (where I assume that $b > 1$:

$b = q_1 a + r_1$

$a = q_2 r_1 + r_2$

$r_1 = q_3 r_2 + r_3$

$\cdots$

$r_k = q_{k+2}r_{k+1} + r_{k+2}$

$r_{k+1} = q_{k+3}r_{k+2} + 0$

where in each step, we just did regular old division to guarantee a remainder $r_i$ that was less than the divisor. As the divisors become the remainders, this yields a strictly decreasing remainder at each iteration, so it will terminate (in fact, it’s very fast). Further, using the notation from above, I claimed that the gcd of $a$ and $b$ was the last nonzero remainder, in this case $r_{k+2}$. How did we prove it?

Proof of Euclidean Algorithm

Suppose that $d$ is a common divisor (such as the greatest common divisor) of $a$ and $b$. Then $d$ divides the left hand side of $b - q_1 a = r_1$, and thus must also divide the right hand side. So any divisor of $a$ and $b$ is also a divisor of $r_1$. This carries down the list, so that the gcd of $a$ and $b$ will divide each remainder term. How do we know that the last remainder is exactly the gcd, and no more? The way we proved it in class relied on the observation that $r_{k+2} \mid r_{k+1}$. But then $r_{k+2}$ divides the right hand side of $r_k = q_{k+2} r_{k+1} + r_{k+2}$, and so it also divides the left. This also carries up the chain, so that $r_{k+2}$ divides both $a$ and $b$. So it is itself a divisor, and thus cannot be larger than the greatest common divisor. $\diamondsuit$

As an aside, I really liked the way it was proved in class. Great job!

The Euclidean algorithm can be turned backwards with back-substitution (some call this the extended Euclidean algorithm,) to give a solution in $x,y$ to the equation $ax + by = \gcd(a,b)$. This has played a super important role in our class ever since. By the way, though I never said it in class, we proved Bezout’s Identity along the way (which we just called part of the Extended Euclidean Algorithm). This essentially says that the gcd of $a$ and $b$ is the smallest number expressible in the form $ax + by$. The Euclidean algorithm has shown us that the gcd is expressible in this form. How do we know it’s the smallest? Observe again that if $c$ is a common divisor of $a$ and $b$, then $c$ divides the left hand side of $ax + by = d$, and so $c \mid d$. So $d$ cannot be smaller than the gcd.

This led us to explore and solve linear Diophantine equations of the form $ax + by = c$ for general $a,b,c$. There will be solutions whenever the $\gcd(a,b) \mid c$, and in such cases there are infinitely many solutions (Do you remember how to see infinitely many other solutions?).

Linear Diophantine equations are very closely related a linear problems in modular arithmetic of the form $ax \equiv c \mod m$. In particular, this last modular equation is equivalent to $ax + my = c$ for some $y$.(Can you show that these are the same?). Using what we’ve learned about linear Diophantine equations, we know that $ax \equiv c \mod m$ has a solution iff $\gcd(a,m) \mid c$. But now, there are not infinitely many incongruent (i.e. not the same $\mod m$) solutions. This is called the ‘Linear Congruence Theorem,’ and is interestingly the first major result we’ve learned with no proof on wikipedia.

Theorem: the modular equation $ax \equiv b \mod m$ has a solution iff $\gcd(a,m) \mid b$, in which case there are exactly $\gcd(a,m)$ incongruent solutions.

Proof

We can translate a solution of $ax \equiv b \mod m$ into a solution of $ax + my = b$, and vice-versa. So we know from the Extended Euclidean algorithm that there are only solutions if $\gcd(a,m) \mid b$. Now, let’s show that there are $\gcd(a,m)$ solutions. I will do this a bit differently than how we did it in class.

First, let’s do the case when $gcd(a,m)=1$, and suppose we have a solution $(x,y)$ so that $ax + my = b$. If there is another solution, then there is some perturbation we can do by shifting $x$ by a number $x'$ and $y$ by a number $y'$ that yields another solution looking like $a(x + x') + m(y + y') = b$. As we already know that $ax + my = b$, we can remove that from the equation. Then we get simply $ax' = -my'$. Since $\gcd(m,a) = 1$, we know (see below the proof) that $m$ divides $x'$. But then the new solution $x + x' \equiv x \mod m$, so all solutions fall in the same congruence class – the same as $x$.

Now suppose that $gcd(a,m) = d$ and that there is a solution. Since there is a solution, each of $a,m,$ and $b$ are divisible by $d$, and we can write them as $a = da', b = db', m = dm'$. Then the modular equation $ax \equiv b \mod m$ is the same as $da' x \equiv d b' \mod d m'$, which is the same as $d m' \mid (d b' - d a'x)$. Note that in this last case, we can remove the $d$ from both sides, so that $m' \mid b' - a'x$, or that $a'x \equiv b \mod m'$. From the first case, we know this has exactly one solution mod $m'$, but we are interested in solutions mod $m$. Just as knowing that $x \equiv 2 \mod 4$ means that $x$ might be $2, 6, 10 \mod 12$ since $4$ goes into $12$ three times, $m'$ goes into $m$ $d$ times, and this gives us our $d$ incongruent solutions. $\diamondsuit.$

I mentioned that we used the fact that we’ve proven 3 times in class now in different forms: if $\gcd(a,b) = 1$ and $a \mid bc$, then we can conclude that $a \mid c$. Can you prove this? Can you prove this without using unique factorization? We actually used this fact to prove unique factorization (really we use the statement about primes: if $p$ is a prime and $p \mid ab$, then we must have that $p \mid a$ or $p \mid b$, or perhaps both). Do you remember how we proved that? We used the well-ordered principle to say that if there were a positive integer that couldn’t be uniquely factored, then there is a smaller one. But choosing two of its factorizations, and finding a prime on one side – we concluded that this prime divided the other side. Dividing both sides by this prime yielded a smaller (and therefore unique by assumption) factorization. This was the gist of the argument.

The last major bit of the week was the Chinese Remainder Theorem, which is awesome enough (and which I have enough to say about) that it will get its own post – which I’m working on now.

I’ll see you all in class tomorrow.

## A proof from the first sheet (SummerNT) Monday, Jun 24 2013

In class today, we were asked to explain what was wrong with the following proof:

Claim: As $x$ increases, the function

$\displaystyle f(x)=\frac{100x^2+x^2\sin(1/x)+50000}{100x^2}$

approaches (gets arbitrarily close to) 1.

Proof: Look at values of $f(x)$ as $x$ gets larger and larger.

$f(5) \approx 21.002$
$f(10)\approx 6.0010$
$f(25)\approx 1.8004$
$f(50)\approx 1.2002$
$f(100) \approx 1.0501$
$f(500) \approx 1.0020$

These values are clearly getting closer to 1. QED

Of course, this is incorrect. Choosing a couple of numbers and thinking there might be a pattern does not constitute a proof.

But on a related note, these sorts of questions (where you observe a pattern and seek to prove it) can sometimes lead to strongly suspected conjectures, which may or may not be true. Here’s an interesting one (with a good picture over at SpikedMath):

Draw $2$ points on the circumference of a circle, and connect them with a line. How many regions is the circle divided into? (two). Draw another point, and connect it to the previous points with a line. How many regions are there now? Draw another point, connecting to the previous points with lines. How many regions now? Do this once more. Do you see the pattern? You might even begin to formulate a belief as to why it’s true.

But then draw one more point and its lines, and carefully count the number of regions formed in the circle. How many circles now? (It doesn’t fit the obvious pattern).

So we know that the presented proof is incorrect. But lets say we want to know if the statement is true. How can we prove it? Further, we want to prove it without calculus – we are interested in an elementary proof. How should we proceed?

Firstly, we should say something about radians. Recall that at an angle $\theta$ (in radians) on the unit circle, the arc-length subtended by the angle $\theta$ is exactly $\theta$ (in fact, this is the defining attribute of radians). And the value $\sin \theta$ is exactly the height, or rather the $y$ value, of the part of the unit circle at angle $\theta$. It’s annoying to phrase, so we look for clarification at the hastily drawn math below:

The arc length subtended by theta has length theta. The value of sin theta is the length of the vertical line in black.

Note in particular that the arc length is longer than the value of $\sin \theta$, so that $\sin \theta < \theta$. (This relies critically on the fact that the angle is positive). Further, we see that this is always true for small, positive $\theta$. So it will be true that for large, positive $x$, we’ll have $\sin \frac{1}{x} < \frac{1}{x}$. For those of you who know a bit more calculus, you might know that in fact, $\sin(\frac{1}{x}) = \frac{1}{x} - \frac{1}{x^33!} + O(\frac{1}{t^5})$, which is a more precise statement.

What do we do with this? Well, I say that this allow us to finish the proof.

$\dfrac{100x^2 + x^2 \sin(1/x) + 50000}{100x^2} \leq \dfrac{100x^2 + x + 50000}{100x^2} = 1 + \dfrac{1}{100x} + \dfrac{50000}{100x^2}$

, and it is clear that the last two terms go to zero as $x$ increases. $\spadesuit$

Finally, I’d like to remind you about the class webpage at the left – I’ll see you tomorrow in class.

## Math 90: Concluding Remarks Sunday, Dec 30 2012

All is said and done with Math 90 for 2012, and the year is coming to a close. I wanted to take this moment to write a few things about the course, what seemed to go well and what didn’t, and certain trends in the course. that I think are interesting and illustrative.

First, we might just say some of the numbers. Math 90 is offered only as pass/fail, with the possibility of ‘passing with distinction’ if you did exceptionally well (I’ll say what that meant here, though who knows what it means in general). We had four people fail, three people ‘pass with distinction,’ and everyone else got a passing mark. Everything else will be after the fold.

## Math 90: Week 11 and Midterm Solutions Sunday, Nov 18 2012

We had a midterm this week, and did more review during recitation. The solutions are now available below the fold

## Math 90: Week 10 Wednesday, Nov 7 2012

We deviated from our regular course of action this week, so we did not have preset examples to do in classes. So instead, I will say a few things, and this can be the new posthead for questions.

## Math 90: Week 8 Quiz Wednesday, Oct 24 2012

There was a quiz this week – in this post, we consider the solutions, common mistakes, and the distribution.

## Math 90: Week 8 Wednesday, Oct 24 2012

Today, we had a set of problems as usual, and a quiz! (And I didn’t tell you about the quiz, even though others did, so I’m going to pretend that it was a pop quiz)!. Below, you’ll find the three problems, their solutions, and a worked-out quiz.

## Math 90: Week 5 Wednesday, Oct 3 2012

A few administrative notes before we review the day’s material: I will not be holding office hours this Wednesday. And there are no classes next Monday, when my usual set of office hours are. But I’ve decided to do a sort of experiment: I don’t plan on reviewing for the exam specifically next week, but a large portion of the class has said that they would come to office hours on Monday if I were to have them. So I’m going to hold them to that – I’ll be in Kassar House 105 (the MRC room) from 7-8:30 (or so, later perhaps if there are a lot of questions), and this will dually function as my office hours and a sort of review session.

But this comes with a few strings attached: firstly, I’ll be willing to answer any question, but I’m not going to prepare a review; secondly, if there is poor turnout, then this won’t happen again. Alrighty!

The rest is after the fold –

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