## Important Stuff

- The rules of the game (and detailed course schedule)
- Worksheet 1: Rolle's Theorem and MVT
- Assignment 1 (Due: July 18)
- Worksheet 2: Integration Massacre
- Assignment 2 (Due: July 25)
- Worksheet 3: Sequences & Series Showdown
- Assignment 3 (Due: August 1)
- Worksheet 4: The Battle for the Power Series
- Survival Guide To The Midterm (or Twenty Things You Should Definitely Know)
- Assignment 4 (Due: August 15)
- Solutions to the Midterm
- Worksheet 5: No retreat, no surrender
- The Final Battle: Practice basic skills for the final exam

## Preparation material for the final

Our final exam is on Monday, August 20, from 9:00 to 12:00 in room TC 341. It covers everything we have done. If the midterm taught us something is that the assignments and worksheets are a good guide for the kind of problems that you should expect. In addition to the practice midterms, here you have one and two practice finals from the previous years. Also, in case you feel weak concerning parametric curves and polar coordinates, go over this assignment from the Winter term. On August 15 we will go over Worksheet 5. From August 16 to 17 we will have two sessions devoted to get ready (solving problems and more problems!!) for the final.**Addendum:** I just compiled a list of problems intended to provide you with basic training for the final. Probably this will not be enough but it is definitely a good place to start.

## Preparation material for the midterm

The midterm on August 7 covers sections 4.2, 7.1, 7.4, 7.8 and 11.1 to 11.10, including the suggested exercises listed in the detailed course schedule as well as problems in assignments and worksheets. Here is a very short study guide in case you need a place to start. On August 2 and 3 we (and by "we" I mean "you") will go over problems taken from these practice midterms (you can also bring your own problems if you like):A man's training is endless. Sometimes, you start to feel that you have reached perfection at something, so you stop doing what you were doing. Whereas, he who wishes to reach perfection must always remember that he is very far from it. Only he who is not satisfied with just being an achiever and always striving for new achievement can consider himself to be a great man.—Hagakure

- Midterms covering from 4.2 to 11.2: Winter 2010 and Winter 2011.
- Midterms covering from 11.2 to 11.11: Winter 2010 and Winter 2011 (of course, for your training you can skip the problems requiring 11.11 skills.)

**One more thing:**the solutions to these practice midterms are available online. They are easy to find. For your own sake, do NOT look at the solutions. Reading solutions is not studying but it will give you the feeling that you are ready for the test even if you are not. You must work hard solving problems (and even get a little frustrated from time to time) to strengthen your understanding of the material. Problems are supposed to be difficult, otherwise they would not be problems at all.

## Course Log

### August 16-17

The Final Battle: Practice basic skills for the final exam### August 15

Worksheet 5: No retreat, no surrender### August 14

A minor dose of calculus ingenuity allows us to calculate areas of regions and lengths of curves defined using polar equations.The area of a region bounded by a polar curve of the form $r=f(\theta)$ with $a\leq \theta\leq b$ can be calculated using the integral $$A=\int_a^b \frac{[f(\theta)]^2}{2}\, d\theta.$$ We deduced this integral in class.

Now, for the arc length, we can use the fast parametrization for polar equations of the form $r=f(\theta)$: $$C(\theta)=(f(\theta)\cos \theta,f(\theta)\sin \theta).$$

Using now the formula to calculate arc lengths of parametric curves, it is easy to see that the arc length of the curve $r=f(\theta)$ with $a\leq\theta\leq b$ is $$L=\int_a^b \sqrt{[f(\theta)]^2+[f'(\theta)]^2}\, d\theta.$$

It takes a while to get used to polar coordinates. If you want to practice, you can start with exercises 47, 48 and 54 from section 10.3 and 10, 11, 26 and 45 from section 10.4.

Now we've covered all the material required for this course. We'll have three final exam training sessions starting tomorrow with our final Worksheet.

### August 13

Last week of classes! This went really fast.Polar coordinates give you another way of thinking about points on the plane. Some shapes will be way easier to represent using polar coordinates than Cartesian ones.

Given a point on the plane, its Cartesian coordinates are given by orthogonal projections on the two axis. The polar coordinates are given by the distance of the point to the origin (this is the $r$ coordinate) and the angle measured from the positive $x$-axis (the $\theta$ coordinate).

A picture is better than words:

A peculiarity about polar coordinates (in oposition to Cartesian coordinates) is that they are not unique: a point on the plane can be represented by infinitely many different pairs of polar coordinates. For instance, the point $(-1,1)$ can be represented by $(\sqrt{2}, \frac{3\pi}{4}+2\pi n)$ for any $n\in \mathbb{Z}$ and also by $(-\sqrt{2}, -\frac{\pi}{4}+2\pi n)$ with $n\in\mathbb{Z}$.

Still, it is easy to go from polar coordinates to Cartesian and back. If $P=(r,\theta)$ is a point in polar coordinates, its Cartesian representation is $(r\cos\theta, r\sin\theta)$. And if $P=(x,y)$ is in Cartesian coordinates, then $r^2=x^2+y^2$ and $\tan\theta=y/x$.

A polar curve is a set of points $(r,\theta)$ that satisfy an equation of the form $F(r,\theta)=0$. Usually, we will restrict ourselves to equations of the form $r=f(\theta)$. The basic examples are $r=A$ and $\theta=B$, where $A$ and $B$ are arbitrary numbers. What curves do they describe?

More involving examples will not be visualizable without a little effort. Sometimes you will be able to transform your polar equation into a known Cartesian equation with some algebra. Sometimes there will not be a nice Cartesian equation behind. In that case, the only tool at your disposal is plotting the graph of $r=f(\theta)$ in an auxiliary (Cartesian) plane $\theta r$ and then try to pull out information from that graph about the graph of the curve on the plane $xy$. We did five examples of this process in class. Five examples done by someone else are not enough to master the technique. You should practice on your own. Try with exercises 29-46 on page 663.

Finally, if one wants to calculate the slope of points given by polar curves of the form $r=f(\theta)$, the best way to go is to note that there is a fast parametrization of the curve given by $$C(\theta)=(f(\theta)\cos\theta, f(\theta)\sin\theta).$$ Therefore, we can calculate slopes of tangent lines on polar curves using the same process we learned for parametric curves. Unfortunately, we ran out of time when we were trying to calculate the horizontal tangents of the cardioid $r=1+\sin\theta$. Tomorrow we will start with that.

### August 10

Got the graph? You could have used this site.

$$C(t)=(16 \sin^3 t, 13 \cos t - 5 \cos 2t - 2 \cos 3t - \cos 4t) \text{ with } 0\leq t\leq 2\pi$$Today we focused on the kind of information we can pull out of parametric representations of curves using calculus.

Given $C(t)=(x(t), y(t))$ with $a\leq t \leq b$, we defined the tangent vector (or velocity vector) of $C$: $$v(t)=C'(t)=(x'(t), y'(t)).$$

Seen as a vector with its tail at $C(t)$, $C'(t)$ gives us the direction of $C$ at $C(t)$. Thus, the slope of $C$ at $C(t)$ is $$m(t)=\frac{y'(t)}{x'(t)}.$$

The length (a.k.a. norm) of the velocity vector gives us the speed of a particle tracing the parametrization of the curve: $$\text{Speed}(t)=\sqrt{(x'(t))^2+(y'(t))^2}.$$

With the help $C'(t)$ it is now easy to find the points on the curve where the tangent is horizontal or vertical. You can also find the slope of the tangent line at $C(t)$ and thus write its cartesian equation.

With an argument similar to the one we used for functions, we can calculate the distance travelled by a particle tracing the parametric curve $C(t)=(x(t), y(t))$ with $a\leq t \leq b$, with the integral $$L=\int_a^b \sqrt{(x'(t))^2+(y'(t))^2}\, dt$$

Note that if you look at the integrand as the speed of the particle traveling on the curve, the formula makes even more sense.

Using this formula, we showed that one "jump" of the cycloid of a circle of radius $r$ is $8r$ long.

Crucial point: If you want to calculate the length of the curve, you must make sure your parametrization traverses the curve only once.

In order to practice our newly adquired skills, we did exercise 25 (p. 651). This is the graph of the curve:

### August 9

Parametric curves are mathematical descriptions of curves that include not only their final shape but the way (speed, direction, first point, last point, pauses, &c.) they are traced.

For a parametric curve (on the plane) to be defined, we get two functions $x(t)$ and $y(t)$ and a range for the parameter $t$. Something like $a\leq t\leq b$. The curve, then, is given by points of the form $C(t)=(x(t), y(t))$ when $t$ moves along the given range of values.

This instalation created by the artist Robert Howsare generates one parametric curve (What could be the parameter?):

Here you can play with a digital version of the same machine. Instructions and control panel in Spanish, though.

We can find several parametrizations for the same curve. For instance $C(t)=(\cos t, \sin t)$ and $C(t)=(\cos 2t, \sin 2t)$, with $0\leq t\leq 2\pi$, are two parametrizations of the circle centered at $(0,0)$ of radius $1$. The difference is that the second one traces the circle twice (if you think of $t$ as time, it goes faster.)

We found the general formula for a parametrization of a circle (traced only once) of radius $r$ and centered at $(h,k)$. You don't need to remember this formula by heart. It is easy to deduce.

For the ellipse, you just need to apply a little modification to the parametrization of the circle.

(By the way: all this stuff is way easier if you think of trig functions as describing points on the unit circle. Trig functions from triangles are just a side product of this. And most basic trig identities are immediate from the representation on the circle.)

Following the book, we deduced, by a short geometrical argument, the parametrization of the cycloid. You can learn more about the cycloid here. Used with care, Wikipedia is a wonderful thing.

The same kind of argument allowed us to find the solutions to exercises 41 and 42 (from section 10.1). This is the graph of the curve that you get in exercise 42:

As a homework, you're supposed to figure out (using any means at your disposal) the graph of the following parametric curve (with $0\leq t\leq 2\pi$): $$C(t)=(16 \sin^3 t, 13 \cos t - 5 \cos 2t - 2 \cos 3t - \cos 4t).$$

Also try exercise 45.

### August 8

How to calculate the length of curves? One way already known by the ancient Greeks is to approximate the length of a curve using line segments joining points on the curve. With this principle, we were able to calculate the perimeter of a circle as the limit of the sequence of perimeters of regular polygons inscribed in the circle.When the curve is given by a differentiable function $y=f(x)$, with $a\leq x\leq b$, we deduced, with the help of the Mean Value Theorem, that the arc length is given by the integral $$L=\int_a^b \sqrt{1+(f'(x))^2}\, dx.$$

This formula, however, is a little problematic. The square inside the square root may create unsolvable monsters really fast. Try, for instance, the arc length of $y=\sin x$ with $0\leq x\leq \pi$. (Probably you can deal with the resulting integral using power series, now that I think about it.)

Given a function $y=f(x)$ differentiable on $(a,b)$, we can define the arc length function that assigns to any $x\in(a,b)$ the length on the curve from $(a, f(a))$ to $(x,f(x))$. This function is given by the integral $$s(x)=\int_a^x \sqrt{1+(f'(t))^2}\, dt.$$

### August 7: The Midterm

**Funfact:**The standing high jump was featured in the Olympics from 1900 to 1912.### August 2 and August 3

"Several large, artificial constructions are approaching us," ZORAC announced after a short pause. "The designs are not familiar, but they are obviously the products of intelligence. Implications: we have been intercepted deliberately by a means unknown, for a purpose unknown, and transferred to a place unknown by a form of intelligence unknown. Apart from the unknowns, everything is obvious."

The midterm is this Tuesday. Intense practice is the only sensible strategy. For two days, we'll be solving problems from here, here, here and here.—James P. Hogan,*Giants Star*Also, as promised, I wrote this Survival Guide To the Midterm. It collects twenty basic things you need to know for the test but of course it will not replace the experience you gain by solving actual problems. Practice, practice and more practice: that's the only reliable way to success.

### August 1

August already. After watching the last set of a tennis game, we went back to our differential equations. Today we learned what it means for a differential equation to be separable. A differential equation is separable if it can be written in the form $$\frac{dy}{dx}=g(x)f(y).$$Separable equations are easy to solve. It's just a matter of integrating two things and then move expressions around. Sometimes you will not be able to find an actual expression for the solution but just a relation between the solution and its independent variable, but that's OK.

Since separable equations are just equations of certain form, usually you will have to manipulate your equation to be able to see it as separable. You may even need to use substitutions. For instance, after applying the substitution $u=x+y$, the differential equation $y'=x+y$ becomes separable. We did that in class.

Certain settings produce natural separable equations. Othogonal trajectories and mixing problems are two examples of this.

Want to check if you're ready for the midterm? Here you have four more practice problems:

- Suppose that $a_n>0$ for all $n\geq 1$ and $\sum_{n=1}^\infty a_n$ converges. Does $$\sum_{n=1}^\infty \frac{a_n}{1+a_n}$$ converge too?
- Calculate $$\int \ln(x^2-1)\, dx.$$ This integral requires the application of all the integration methods we know.
- Find the radius of convergence of $$\sum_{n=1}^\infty \frac{n! (x-2)^n}{1\cdot 3\cdot \cdots \cdot (2n-1)}.$$
- Express $\left(\frac{x}{x^2-1}\right)^2$ as a power series. Find its domain of convergence.

### July 31

A differential equation is an equation that contains an unknown function and one or more of its derivatives. The order of a differential equation is the highest derivative that occurs in the equation.Usually, a differential equation has several (infinitely many) solutions. Sometimes, an initial condition is added (something of the form $y(t_0)=y_0$) and this restricts the number of possible solutions.

Leibniz's notation is very useful when it comes to deal with differential equations because you can manipulate the symbols as if they were numbers. Using this, we found solutions for $$\frac{dP}{dt}=k P\text{ and }\frac{dP}{dt}=kP(1-\frac{P}{M}).$$

The calculations we did in class to solve these equations are available in section 9.4.

In the second part of the session, we continued our training for the midterm going over the following problems:

- Let $a_1=1$ and $a_{n+1}=\frac{1}{3}\left(a_n + 4 \right)$. Does it converge? If so, find $\lim_{n\to\infty} a_n$.
- Use Taylor series to calculate $$\lim_{x\to 0}\frac{\sin x - x}{x^3}.$$
- Calculate the limit $$\lim_{n\to\infty}\frac{n\sin n}{n^2+1}$$ or prove that the sequence diverges.
- Suppose that $$\sum_{n=1}^\infty a_n=\frac{e}{\pi}.$$ Evaluate $$\lim_{n\to\infty}\frac{1}{1+\frac{1}{1+a_n}}.$$
- Let $c_n\geq 0$ and suppose that $$\sum_{i=1}^N c_i < 2 - \frac{1}{N}.$$ Is $\sum_{n=1}^\infty c_n$ convergent?

### July 30

First, one announcement: from Tuesday to Thursday this week the problem clinic will be in MC106 instead of MC108.Now, let's talk math: Given $f(x)$, we defined the Taylor polynomials of $f$ at $a$ as $$T_n(x)=\sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(x-a)^i.$$

When $f(x)$ happens to be equal to its Taylor series, the $T_n(x)$ are polynomial approximations of the original function. There are at least two ways of estimating how good are these approximations:

- If the Taylor series looks like an alternating series, then you can use the Alternating Series Estimation Theorem to calculate the error.
- Otherwise, if you're trying to get a bound for $$|R_n(x)|=|f(x)-T_n(x)|$$ you can also use Lagrange's Remainder Theorem: recall that this theorem gives you $$R_n(x)=\frac{f^{(n)}(c)}{(n+1)!}(x-a)^{n+1} $$ for a certain $c$ depending of $x$. If you manage to find $M$ such that $|f^{(n)}(c)|< M$ and $N$ such that $|x-a|^{n+1} < N$, then $$|R_n(x)| < \frac{M N}{(n+1)!}.$$

In the second part of the session, we went over the following problems:

- Given $f(x)=e^{x^2}$, calculate $T_3(x)$ at $a=0$ and calculate the error for $0\leq x\leq 0.1$. (I made a
**HUGE**mistake solving this problem on the board for trying to take a shortcut. You really have to take four derivatives of $f$ in order to find a bound on the error.) - Calculate $$\lim_{n\to\infty} \frac{10^n}{n!}.$$ (Raden came up with a very nice solution for this using the ratio test for series. My ugly solution required the use of some not so elegant inequalities. I like Raden's solution better.)
- Calculate the sum $$\sum_{n=1}^\infty [\tan^{-1}(n+1)-\tan^{-1}(n)].$$ (Hint: Think about telescopes.)
- If $f(x)=e^{x^2}$, calculate $f^{(144)}(0)$. (Hint: you don't really need to take any derivatives for this one.)

### July 27

We went over Worksheet 4: The Battle for the Power Series*A monk asked Kegon, "How does an enligthtened one return to the ordinary world?" Kegon replied, "A broken mirror never reflects again; fallen flowers never go back to the old branches."*### July 26

Remember: we would like to find a condition for a function $f$ to be equal to its Taylor series. This will require a few definitions.First, let $$T_n(x)=\sum_{i=0}^n \frac{f^{(i)}(a)}{i!}(x-a)^i.$$ We call $T_n(x)$ the $n$th Taylor polynomial of $f$ at $a$.

Second, let $R_n(x)=f(x)-T_n(x)$. We call $R_n(x)$ the $n$th remainder of the Taylor series of $f$ at $a$.

By definition, if $f(x)$ equals its Taylor series (in its domain of convergence $D$), then $$\lim_{n\to\infty} R_n(x)=0$$ for any $x\in D$.

The book then uses something called Taylor's inequality to prove that $R_n(x)\to 0$ for $f(x)=e^x$ and $f(x)=\sin x$. Unfortunately, they do not provide a complete proof of the inequality. Thus, instead of relying on a black box, we decided to use the Lagrange's Remainder Theorem. This result was proven in class and it can be easily applied to $f(x)=e^x$ and $f(x)=\sin x$ in order to show that they are equal to their corresponding Maclaurin series.

In particular, we proved that $$e^x=\sum_{n=0}^\infty \frac{x^n}{n!},$$ just as we wanted.

Here is the statement of Lagrange's Remainder Theorem:

**Theorem.**Given $f$ be infinitely differentiable on $(-R,R)$ and $x\in (-R,R)$, there is $c\in (-R,R)$ satisfying $|c-a| < |x-a|$ such that $$R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}.$$A version of the proof offered in class is available in this notes (pp. 7-8) by professor Rasul Shafikov.

### July 25

Assignment 3 is out. Most of it is rather easy but problem (2)c is seriously hard. Totally doable with the tools we have learned so far, yes, but hard. Consider yourself warned.Today we met the many faces of the equation $$\sum_{n=0}^\infty x^n=\frac{1}{1-x}\text{ for } |x| < 1.$$ With very little effort, it is possible to pull out from this simple equation power series representations for a bunch of different functions by using algebraic manipulations, derivatives or integrals.

In addition, for a function $f$ and a point $a$ in its domain, we defined the Taylor series of $f$ at $a$ as the series $$\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!}(x-a)^n.$$

Key (although kinda subtle) fact: if a function $f$ has a power series representation centred at $a$, then this power series must be its Taylor series at $a$.

To appreciate the subtlety, note that the Taylor series of $f(x)=e^x$ at $0$ (also known as the Maclaurin series of $f$) is $$\sum_{n=0}^\infty \frac{x^n}{n!},$$ but this does not imply right away that $$e^x=\sum_{n=0}^\infty \frac{x^n}{n!},$$ since we have not proven yet that $f(x)=e^x$ has a power series representation centred at $0$. We need stronger tools for that. More on this tomorrow.

### July 24

A power series centred at $a\in\mathbb{R}$ is an expression of the form $$\sum_{n=0}^\infty c_n (x-a)^n,$$ where the $c_n\in\mathbb{R}$ and $x$ is a variable.The domain of convergence of a power series is the set of $x$ for which the series converges. We proved in class that there are only three possibilities (or maybe six, it depends on the actual phrasing) for the domain of convergence of a given series. The important thing is that it is always a point or a ball around $a$.

For calculating the domain of convergence of a power series we will generally use the ratio test. The ratio test does not cover all possible cases, though. You have to test by hand the values of $x$ that make the limit of the ratio test equal to $1$. Try with $$\sum_{n=0}^\infty\frac{n^2 (x+3)^2}{2^{n+1}}.$$

Finally, we mentioned that on the domain of convergence of a power series, you can integrate or differentiate the series term by term keeping the same radius of convergence (this result will not be proven. It requires machinery a bit beyond the scope of our course). Using this, we noticed that if $$f(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$$ then the domain of convergence of $f$ is $\mathbb{R}$ and $f'(x)=f(x)$. Our conjecture is that $f(x)=e^x$. On Thursday we will prove this.

### July 23

A series $\sum a_n$ is called absolutely convergent if $\sum |a_n|$ converges. If a series converges but it is not absolutely convergent then it is called conditionally convergent. Conditionally convergent series are dangerous: under rearrangement of terms they can change value. The trick with $$\sum_{n=1}^\infty (-1)^{n-1} \frac{1}{n}$$ was an example of this.The main fact about absolutely convergent series is that they are also convergent. Proving this takes less than two minutes. Sometimes it is easier to test if a series is convergent after applying absolute value to each of its terms. Try the following series (is it convergent?): $$\sum_{n=1}^\infty \frac{\cos n^n+\sin^n(\ln n)}{n^n}.$$

There are also special tests to check if a series is absolutely convergent. The ratio test and the root tests are two of them. They will be very helpful in a few days, so keep them in mind. With the ratio test it is possible to prove that the series $$\sum_{n=1}^\infty nr^n$$ is convergent for $|r|< 1$. Calculating the actual value to which the series converges is part of your mission for Assignment 2 (Problem 5).

While going over one of the applications of the ratio test we ended up facing the following limit:$$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^n.$$ We know that the value of this limit is $e$. Eric said that was

*the definition*of $e$. However, we now have all the tools to prove this. On Friday's worksheet we will attempt, among others, this rather easy exercise.### July 20

Worksheet 3: Sequences & Series Showdown! Test your skills!There is surely nothing other than the single purpose of the present moment. A man's whole life is a succession of moment after moment. There will be nothing else to do, and nothing else to pursue. Live being true to the single purpose of the moment.

—Yamamoto Tsunetomo,*Hagakure*### July 19

Two sections for the price of one. First, we went over the comparison tests. Remembering the conditions for convergence and divergence of geometric series and p-series may come handy here.After this, we saw a sketch of the proof of the alternating series test(a.k.a. Leibniz's Theorem). This is the only test we have so far for series with negative terms. It offers a simple criterion for the convergence of alternating series, i.e. series of the form $$\sum_{n=1}^\infty (-1)^{n-1}b_n$$ with $b_n>0$.

We tested the convergence of $\sum_{n=1}^\infty (-1)^{n-1}\frac{1}{n}$ and $\sum_{n=1}^\infty (-1)^{n-1}\frac{n^2}{1+n^3}$. After playing a bit with the first one, we noticed that rearranging the terms of an infinite series can be really dangerous. I mentioned but did not prove that the series $\sum_{n=1}^\infty (-1)^{n-1}\frac{1}{n}$ can be rearranged to converge to

*any*value you want. An proof of this striking fact is available in*Calculus*by Michael Spivak, my favorite Calculus book.The sketch of the proof of the alternating series test requires the following lemma: Given a sequence $\{a_n\}$, if the even subsequence $\{a_{2n}\}$ and the odd subsequence $\{a_{2n+1}\}$ converge to the same value, then the sequence $\{a_n\}$ also converges to that value. Try to prove it. It is a good practice exercise of $\epsilon-N$ arguments.

### July 18

Assignment 2 is out! (Due: July 25).Calculating the limit of series is usually hard. Deciding if a series is convergent or not is a little easier. During the next few sessions we will learn a bunch of methods to test the convergence of series. The first of them is the integral test. This test relates the convergence of a series to the existence of certain improper integral. You should know by heart the precise statement of the test and have a rough idea of how to prove it. In a way, if you understand its proof then you know exactly why it works. This gives you more control over the test.

Using the integral test we can figure out for which values of $p>0$ the series $$\sum_{n=1}^\infty \frac{1}{n^p}$$ is convergent.

Also, the integral test gives us a bound on the error when we use partial sums as estimates of the value of convergent series.

Want to practice? Use the integral test to check whether the following series converges or not: $$\sum_{n=2}^\infty \frac{1}{n (\ln n)^2}$$

### July 17

Series are a special kind of sequences. Given a sequence $\{a_n\}$ and the infinite sum of its terms $$\sum_{i=1}^\infty a_i,$$ we study the sequence of partial sums $$s_n=\sum_{i=1}^n a_i.$$ If the sequence $s_n$ converges to a number $L$, we write $$\sum_{i=1}^\infty a_i=L.$$This is the geometric series: $$\sum_{n=1}^\infty a r^{n-1}.$$ We proved that it converges if $|r| < 1$ (to what value?) and diverges otherwise. Use the geometric series to calculate, for instance, $$\sum_{n=1}^\infty 2^{2-2n} 3^n.$$ In class we use the geometric series to prove that $0.999999999999\ldots$ is actually equal to $1$.

Like challenge? Try exercise 77 in page 713.

### July 16

Second week and brand new topic! We introduce numeric sequences because our goal is to represent functions by using*infinite polynomials*(so we can integrate them easily, for instance.) In practice, these polynomials will be*limits*, whatever that means, of sequences of polynomials. Numeric sequences are the natural starting point for this project, one of the big milestones of Calculus.We say that a sequence $\{a_n\}$ converges to a number $L$ if for any $\epsilon>0$ there is $N$ so that $|a_n-L|<\epsilon$ for any $n>N$. This definition is crucial. You should understand it quite well and be capable of using it to prove the convergence of sequences like $a_n=\frac{n+1}{n}$ or $b_n=p^n$ (for $ 0 < p < 1$!!! – What happens with $p>1$?).

Sometimes, though, applying the $\epsilon-N$ definition of convergence can be quite involving. The squeeze theorem and the monotonic sequence theorem are tools that allow us to conclude that a sequence converges without the hassle of finding the right $N$ for each $\epsilon$. In class we found, using the monotonic sequence theorem, the limit of $a_1=\sqrt{2}$ and $a_n=\sqrt{2+a_{n-1}}$. We proved that the limit is $2$ and we did not have to look at the sequence at all! Just to make sure, these are the first 50 terms of that sequence (with 20 decimals of precision):

1.4142135623730950488, 1.8477590650225735123, 1.9615705608064608983, 1.9903694533443937725, 1.9975909124103447854, 1.9993976373924084402, 1.9998494036782890818, 1.9999623505652022853, 1.9999905876191523430, 1.9999976469034038199, 1.9999994117257644383, 1.9999998529314357023, 1.9999999632328585876, 1.9999999908082146258, 1.9999999977020536551, 1.9999999994255134137, 1.9999999998563783534, 1.9999999999640945883, 1.9999999999910236471, 1.9999999999977559118, 1.9999999999994389779, 1.9999999999998597445, 1.9999999999999649361, 1.9999999999999912340, 1.9999999999999978085, 1.9999999999999994521, 1.9999999999999998630, 1.9999999999999999657, 1.9999999999999999914, 1.9999999999999999978, 1.9999999999999999994, 1.9999999999999999998, 1.9999999999999999999, 2.0000000000000000000, 2.0000000000000000000, 2.0000000000000000000, 2.0000000000000000000, 2.0000000000000000000, 2.0000000000000000000, 2.0000000000000000000, 2.0000000000000000000.

The sequence gets pretty close to its limit very fast.

### July 13

Integration massacre!

We went over this worksheet.### July 12

So far, for an integral $\int_a^b f(x)\,dx$ to be defined, $f(x)$ has to be continuous and $a$ and $b$ need to be actual real numbers. Improper integrals allow us to have $\pm \infty$ as limits of integration and also deal with non-continuous integrands. The solution is simple: in order to avoid the problematic points, you replace them with variables and use limits. Want to practice? Try these out: $$\int_{-\infty}^0 e^x\sin x\,dx \text{, } \int_0^5 \frac{dx}{x^2-9} \text{ and } \int_{-\infty}^\infty \frac{dx}{x^2-2x+2}.$$ The comparison test for improper integrals may come handy when our integral calculation skills are not good enough and we only need to check if a certain improper integral converges (or not). In order to use the test, we need to come up with another (simpler) function. Finding this function will usually be the hard part of the problem. For instance, try to check whether the following integral is convergent or divergent: $$\int_0^\infty\frac{\tan^{-1} x}{2+e^x}\,dx.$$ We did an example similar to this one in class. You can actually try a similar strategy here.### July 11

Partial fractions allow you to deal with integrals of functions of the form $$f(x)=\frac{P(x)}{Q(x)},$$ where $P(x)$ and $Q(x)$ are polynomials with $\deg(P) < \deg(Q)$. It converts horrible rational functions into sums of simpler ones that we can actually handle.In class, we went over three examples. In the first one we found the integral $$\int \frac{du}{(u^2+1)^2}.$$ We now know how to deal with something like this by using the substitution $u=\tan v$ (and so $du=\sec^2 v \,dv$). We also learned that completing the square can be a very useful skill when we have to deal with quadratic denominators.

If you are curious about the theory of factorization of polynomials and want to learn more, check this notes by professor Rasul Shafikov.

If you want to test your partial fraction skills, try to calculate: $$ \int \frac{dx}{1+x^4} \text{, } \int \frac{dx}{\sqrt{1+e^x}} \text{ and }\int \sqrt{\tan x}\, dx.$$ In the first one, find the factorization of $x^4+1$ by completing the square (What square?). The second and the third require appropriate substitutions to become rational functions. The third one, for instance, will look totally doable once you have subsituted $u=\sqrt{\tan x}$. Start by finding $dx$ in terms of $u$.

On a side note, today we met Enxin and Masoud, our brave TAs for this course. They will be offering office hours every weekday from 4pm to 5pm. Javier is trying to get a room reserved for this. Until then, on MWF you should go see Enxin in MC103A and on TuTh you can visit Masoud in MC105.

### July 10

We discussed the basics of integration, the substitution method and the integration by parts method. Among other things, we tried to go over Exercise 36 of section 7.1: $$\int \cos(\sqrt{x})\, dx.$$ Javier made the mistake of applying the integration by parts method right away. That is actually a really bad idea. The integral is way easier if, as it was suggested by someone during class, you first make the substitution $u=\sqrt{x}$. Try that out and see what happens.As extra homework:

- Calculate $\int \sec x\,dx$. Hint: multiply the integrand by $$\frac{\sec x +\tan x}{\sec x +\tan x}.$$ Then try the substitution $u=\sec x +\tan x$. Isn't it beautiful?
- Calculate $$\int \frac{x \arctan x}{(1+x^2)^3}\,dx$$

### July 9

We talked briefly about the formal definition of the derivative and proved, after a few key theorems, that if a function has derivative zero (in an interval) then it is a constant.We handed out:

- The rules of the game (and detailed course schedule)
- Assignment 1 (Due date: July 18, 2012)
- Worksheet on Rolle's Theorem and Mean Value Theorem (Warning: it contains some typos.)