Lecture Outlines
You will cover a wide variety of materials during classtime, so your constant attendance is important. To help you in organizing your study materials, the list below gives an overview of the basic concepts covered during a given lecture period.
Exam 1 content
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We spent the first part of class today discussing some logistics for the course, including taking a look at the syllabus. At the very end of class we introduced the area problem.
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We continued our discussion of the definite integral today. We began by doing some basic examples in which the region whose area we were calculating could be split up into subregions that were familiar geometric shapes; using formulas we already knew, this allowed us to calculate some definite integrals explicitly. We also spent time discussing what the notion of "net area" means. In particular, we saw that when we're integral "left to right" (meaning the lower bound of the definite integral is a smaller number than the upper bound), areas above the $y=0$ line count as positive, and areas below $y=0$ count negative.
We then began to explore some of the theoretical properties that definite integrals have, particularly when it comes to breaking the integral of interest into separate pieces. For instance, when $a< c< b$, we saw that $$\int_a^b f(x)~dx = \int_a^c f(x)~dx + \int_c^b f(x)~dx.$$ We also saw that integrating over a single point gives an area of zero, and that we can even make sense of what it means to integrate "backwards" (aka, "right to left", which occurs when the lower bound of the integral is a larger number than the upper bound). In this case, our sign conventions for net area are reversed: areas above the $y=0$ axis give a negative contribution, and areas below $y=0$ contribute positively.
Towards the end of class we started to see the limitations of the tools we had developed up to this point. The key sticking point is that the formulae we know from elementary geometry only give us areas of very special regions, namely trapezoids (in their many forms, including triangles and rectangles) and "nice" circular subregions (like semicircles or quarter circles). This makes computing areas like $\int_0^\pi \sin(x)~dx$ unapproachable, but even seemingly innocuous functions like $y=x^2$ also have areas that are impossible to compute without new ideas.
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We finished last class feeling somewhat dispirited because functions as simple as $y=x^2$ seemed to have areas that we couldn't compute. We decided instead to come up with a method for approximating the desired areas; this wouldn't give a precise area, but at least it would let us leverage what we do know (namely, computing areas for rectangles, trapezoids, etc.) to come up with (what will hopefully be) decent approximations of the quantities we want. To approximate $\int_a^b f(x)~dx$, we devised the following algorithm:
- divide the integral $[a,b]$ into $n$ subintervals of equal width $\Delta x = \frac{b-a}{n}$
- for each $1 \leq i \leq n$, select a "sample point" $x_i^*$ within the $i$th subinterval $[x_{i-1},x_i]$
- approximate $f(x)$ by assuming that $f$ is constant along $[x_{i-1},x_i]$ with value $f(x_i^*)$; this means the desired area along this subinterval is approximately the area of the rectangle with height $f(x_i^*)$ and width $\Delta x$
- add up each of these approximated areas to come up with an approximation for the entire definite integral: $$\int_a^b f(x)~dx \approx \sum_{i=1}^n f(x_i^*)\Delta x.$$
We spent some time going through this algorithm using a few options for selecting the sample points via the left-hand, right-hand, and midpoint Riemann sums. We also saw another approximation tool called the trapezoidal rule that uses trapezoids (instead of rectangles) to approximate the desired area.
Despite these good tools, we are inevitably stuck with approximations when we use them for a specific value of $n$. What can we do to make those approximations as good as they can be? For one, if we increase the value of $n$ (i.e., if we use more subintervals), then this yields a "finer grain" method for approximating, and therefore better approximations. In fact, one of the big theorems in calculus is that we can take this process to the extreme using limits! Specifically, if $f$ is sufficiently "nice" on the interval $[a,b]$, then one has that the definite integral is the limiting value of the Riemann sums: $$\int_a^b f(x)~dx = \lim_{n\to \infty} \sum_{i=1}^n f(x_i^*)\Delta x.$$ This is huge!
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Last class we saw that Riemann sums are a useful way to approximate areas, and that one can in fact take a limit of a Riemann sum to compute areas exactly: $$\int_a^b f(x)~dx = \lim_{n\to \infty} \sum_{i=1}^n f(x_i^*)\Delta x.$$ Today we tried to implement this idea so we could (finally!) determine $\int_1^4 x^2~dx$. We had to spend some time setting things up so we could do the evaluation, and this involved making a choice for how to select our "sample points" (we used the midpoint rule) and then determining what the expressions $f(x_i^*)\Delta x$ looked like. Next we tried to evaluate the given sum. This involved some fairly nasty looking terms that we needed to sum up. We did the hard work to evaluate one of these terms, and that gave us part of the answer. Unfortunately, it left us with 5 additional computations to perform that were even more brutal than the one we already slogged through! We decided that this all just seemed too horrific to continue using. Despite the fact that the limit of a Riemann sum makes good geometric sense, it didn't seem to be very efficient when it came to actual computations.
With this in mind, we needed some new idea to try to more efficiently crack the area problem. Towards that end, we introduced the notion of an antiderivative. In short, if $F$ has derivative $f$, then we say that $F$ is an antiderivative for $f$. We use the notation $\int f(x)~dx$ to denote "the most general antiderivative of $f$," and saw that this introduced a (so-called) integration constant "$+C$" into our calculations. We then looked back at some of the derivative rules that we know so we could use them to compute a corresponding antiderivative. For instance, the power rule tells us that $\frac{d}{dx}\left[x^n\right]=nx^{n-1}$, and we turned this into an antiderivative rule that we called the anti-power rule: $$\int x^m~dx = \frac{1}{m+1}x^{m+1}+C \qquad \qquad\text{(for }m\neq -1\text{).}.$$ We saw a handful of other antiderivative rules, like $\int \sec^2(x)~dx = \tan(x)+C$ and $\int \frac{1}{x}~dx = \ln|x|+C$.
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Last class we introduced antiderivatives, but we didn't talk about why we should care about them. Today we saw a pretty startling answer for why they matter: they provide the key for unlocking the area problem. Specifically, if $F$ is an antiderivative for a "nice" function $f$ on the interval $[a,b]$, then we have $$\int_a^b f(x)~dx = F(b)-F(a).$$ This means that if we can compute antiderivatives, then we can compute answers to the area problem! We used this theorem to compute areas for some functions that were previously impossible (or at least very very tedious) to calculate, like $\int_1^4x^2~dx$ and $\int_0^\pi \sin(x)~dx$.
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We have seen that computing antiderivatives is the key for computing answers to the area problem, so now we need to work on gaining proficiency in computing antiderivatives. Of course, there's a number of antiderivatives we can just "observe" since they arise from differentiation rules. For instance, since we know $\frac{d}{dx}\left[\sin(x)\right] = \cos(x)$, this means we have $\int \cos(x)~dx = \sin(x)+C$. Indeed, every differentiation rule can be turned into a corresponding antidifferentiation rule.
In today's class we focused most of our attention on the antidifferentiation tool that comes from the chain rule. We called this $u$-substitution, and saw that it's particularly effective when there is a composition in the integrand, and when the derivative of the "inner" function of that composition shows up as a factor of the integrand. We did some sample problems.
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We started class by computing a definite integral using $u$-substitution. Along the way, we saw that it was important to substitute not only functions in the integrand and the differential term, but even the bounds of integration.
After this, we then started to think about the antidifferentiation rule that comes from the produce rule. We called this integration by parts, and it tells us that $$\int u~dv = uv - \int v~du.$$ We gave an mnemonic LIATE to help us remember how to select a good choice for the "$u$" function. We practiced this technique.
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We did more problems involving integration by parts, including some that required us to "get creative" using some algebra facts. At the end of class, I handed out this work sheet (solutions) and this worksheet (solutions) which give us more practice with these skills.
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In today's class we thought about how integrands of a particular form can sometimes benefit from having clever substitutions. In particular, we studied integrals that took the form $$\int \cos^n(x)\sin^m(x)~dx,$$ and saw that for these integrals its often wise to choose a substitution of $u=\cos(x)$ is the power of sine is odd (i.e., if $m$ is odd), and its often wise to choose a substitution of $u=\sin(x)$ when the power of cosine is odd (i.e., when $n$ is odd). The magic here is that whatever power of the "other" trig function that remains will be an even power, and therefore can be substituted in the language of the "new variable" by using the Pythagorean theorem. If both powers are odd, then both substitutions have a good shot at being useful. If both powers are even, then one has to use some more complex techniques (like trigonometric reduction formulas).
Then we studied integrals of the form $$\int \tan^n(x)\sec^m(x)~dx,$$ and saw that there were also some clever substitutions to choose. If the power of secant is even, then a $u=\tan(x)$ substitution can sometimes be useful; if the power of tangent is odd, then a $u=\sec(x)$ substitution can sometimes be useful.
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In today's class we saw how the trigonometric substitutions from last class periods can make some surprising appearances. We spent a considerable amount of time computing $$\int \frac{1}{x^2\sqrt{x^2+4}}~dx,$$ and saw that an initial substitution of $x=2\tan(\theta)$ was useful for converting the initial integral into one of the form $\int \cos^n(\theta)\sin^m(\theta)~d\theta$. (We could then use tricks from last time to finish this integral off.) Resubstituting at the end of the integral created some hairy expressions, but we saw that we could use Pythagoras and right triangles to evaluate expressions like $\sin(\theta)$, even though we defined $\theta$ implicitly through the equation $x=2\tan(\theta)$.
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Today we thought about what we do when the Fundamental theorem of Calculus is not applicable for a certain kind of area we want to compute, namely when we want the area under a "nice" function over an infinite interval of the form $[a,\infty)$. In this case we defined $$\int_a^\infty f(x)~dx = \lim_{R \to \infty} \int_a^R f(x)~dx.$$ The genius of this idea is that it lets us use the Fundamental theorem to calculate the value of $\int_a^R f(x)~dx$, after which we can just study what happens to this quantity as $R$ grows to infinity. We computed a number of examples.
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Today we concluded our discussion of improper integrals. We studied two more types of "improper integrals of type 1," namely those of the form $\int_{-\infty}^a f(x)~dx$ (where $f(x)$ is continuous on $(-\infty,a]$) and those of the form $\int_{-\infty}^\infty f(x)~dx$ (where $f(x)$ is continuous on $(-\infty,\infty)$). For the former, we set $$\int_{-\infty}^a f(x)~dx = \lim_{R \to -\infty} \int_R^a f(x)~dx,$$ and we say that the original integral converges to $L$ if the limit exists and is a finite number $L$; that it diverges to $\pm \infty$ if the limit is either $+\infty$ or $-\infty$; and that it diverges (or does not exist) if the given limit does not exist. In the case of the integral $\int_{-\infty}^\infty f(x)~dx$, we define this by first choosing some number $a$ (we can choose $a$ however we like), and then compute the two improper integrals $$\int_{-\infty}^a f(x)~dx \qquad \text{ and } \qquad \int_a^\infty f(x)~dx.$$ If the first converges to $L_1$ and the second converges to $L_2$, then we say that $\int_{-\infty}^\infty f(x)~dx$ converges to $L_1+L_2$. On the other hand, if either of these two improper integrals diverges, then we say that $\int_{-\infty}^\infty f(x)~dx$ diverges. (It only takes one "piece" of the integral diverging to make the whole integral diverge!)
We then began discussing a second type of improper integral. Whereas improper integrals of type 1 are "infinitely wide," these new improper integrals can be "infintely tall." Specifically, suppose we have a function $f(x)$ which is continuous on the half-open, half-closed interval $(a,b]$. In this case, we define $$\int_a^b f(x)~dx = \lim_{t \to a^+} \int_t^b f(x)~dx.$$ As before, we say the integral converges to $L$ if the former limit exists and equals the number $L$; otherwise we say that the integral diverges. In an analogous way, if $f(x)$ is continuous on $[a,b)$, then we define $$\int_a^b f(x)~dx = \lim_{t \to b^-} f(x)~dx,$$ and we use similar "converges" or "diverges" language. These kinds of integrals are called "improper integrals of type 2." Note that improper integrals of type 2 do not announce themselves quite as explicitly as improper integrals of type 1, so be on the lookout for them!
Finally, we discussed the question of what happens if we have an integral that has some "improperness," but that "improperness" isn't lined up to sit in exactly the positions we'd like to use the techniques we've described already. The key in this case is to take the interval over which you are integrating, and split it up so that each subinterval has exactly one improperness, and with that improperness sitting in exactly the right spot (namely, either on the far left or on the far right of the subinterval). We then compute the constituent improper "subintegrals." If any one of them diverges, then the original integral diverges too. On the other hand, if they all converge, then the original integral converges to the sum of each of these "subintegral" values.
Exam 2 content
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The first part of our discussion today had us wrestling with how to handle limits which take indeterminate forms. In particular, if a limit of $\frac{f(x)}{g(x)}$ has the form "$\frac{\infty}{\infty}$" or "$\frac{0}{0}$," then a result called l'Hopital's rule tells us that if the limit of $\frac{f'(x)}{g'(x)}$ is $L$ (for some number $L$), then the corresponding limit of $\frac{f(x)}{g(x)}$ equals $L$ as well. We used this to evaluate some "tricky" limits. We also saw that l'Hopital's rule can be used to evaluate other indeterminate form limits, like $0\cdot \infty$ or $\infty - \infty$ or $1^\infty$.
For the second half of the course, we wrestled with a different problem: if we are studying some improper integral but don't know how to anti-differentiate the integrand, then is it possible for us to determine if the integral converges or not? (This is a loosening of our usual question, which would ask for the value of the integral in the case that it converges; in this setting, we don't worry as much about what precise value the integral takes.) We saw a comparison test for integrals, and began to use it to study the convergence/divergence of $\int_0^\infty \frac{\sin^2(x)}{1+x^2}~dx$.
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We started class today by completing our examples from last class, ultimately showing that $\int_0^\infty \frac{\sin^2(x)}{1+x^2}~dx$ converges by comparing it to the integral $\int_0^\infty \frac{1}{1+x^2}~dx$. We spent time thinking about the deep question of determining which function one uses when trying to run a comparison, and settled on the philosophy that we ought be be focusing on "dominating terms" to select our function for comparison. (We also need to make sure that the function we are comparing to has the property that it "compares well" to the function we're looking at --- meaning its always "bigger" or "smaller" than our function --- and also that our new function is integrable.)
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At the start of class today we mentioned the $p$-test, which is a theorem that gives us a fast way to determine when certain improper integrals converge or diverge. (This is useful because it makes running a comparison test easier.) We spent the bulk of our time focusing on our first "new" application of integrals: computing area between curves. Generally speaking, if one wants to compute the net area "below" $f$ and "above" $g$ between $x=a$ and $x=b$, we argued that this area is captured by $$\int_a^b f(x)-g(x)~dx.$$
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We thought more about how to compute areas with integrals, this time focusing on regions that are bounded by (potentially) several curves. When finding areas of these kinds of regions, it is important that one has a rough idea of what the region looks like; for this, a "cartoon sketch" of the relevant graphs usually suffices. We saw that the geometry of the region influences how we set up the integral. In the case that the region has a "uniform top boundary" and a "uniform bottom boundary," we saw that one can use an integral with respect to $x$ to calculate the area. In this case, if the top boundary is $y=f(x)$ and the bottom boundary is $y=g(x)$, and if the left-most point in the region has $x$-coordinate $a$, and the right-most point in the region has $x$-coordinate $b$, then the relevant area is $$\int_{x=a}^{x=b} f(x)-g(x)~dx.$$ These are the so-called "Type I" regions.
On the other hand, if the region has a "uniform left boundary" (described by the curve $x=h(y)$) and a "uniform right boundary" (given by the curve $x=k(y)$), and if the bottom-most point in the region has $y$-coordinate $c$, and the top-most point in the region has $y$-coordinate $d$, then the area of the given region is $$\int_{y=c}^{y=d} h(y)-k(y)~dy.$$ These types of regions are called "Type II" regions.
We also observed that not every region is Type I or Type II. If a region is neither of these things, one often has to cut the region into smaller subregions which are one of these types.
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In today's class we explored a somewhat unexpected application of integration: the computation of volumes. The idea was that one could approximate the volume of a solid by taking small "cross sectional slices" of the volume, and then approximating the volume of each "slice" as the product of the cross-sectional area of the slice times its thickness. Following this procedure through increasingly small slices, we were able to argue that the volume of the solid can be thought of as the definite integral of cross sectional area. We used this method to compute the volume of a cone. We used this as a starting point for thinking about how we compute the volume generated by revolving a $2$-dimensional region around an axis of rotation, particularly in the case where one has "sliced" the region in a direction perpendicular to the axis of rotation.
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We started today's class by formalizing some of the arguments from last class period related to computing the volume of a solid of revolution which is created by rotating a region around an axis in the case where the direction of the "slicing" of the region is perpendicular to the axis of rotation. This gave us a formula for volume that we called the washer method. This formula is $$V=\int_a^b \pi R^2-\pi r^2~du,$$ where here $R$ represents the distance from the axis of rotation to the "outer" boundary of the region, $r$ represents the distance from the axis of rotation to the "inner" boundary of the region, and $u$ stands for the variable ($x$ or $y$) that is appropriate for the direction of "slicing." For instance, if the region is Type I (meaning it is sliced vertically) and the axis of rotation is some line $y=c$, then we could use this technique with $u=x$. In any event, note that the expressions we use for $R$ and $r$ depend on the geometry of the region we're looking at, together with how that region sits relative to the axis of rotation. It is always a good idea to start these kinds of problems by sketching the region and the axis of rotation so you can parse that geometry correctly!
In the second half of the class we considered what to do in the case where the direction of the "slices" for the region are instead parallel to the axis of rotation. This gave us a technique we called the shell method, which has a formula of the form $$V=\int_a^b \ell * 2\pi r~du,$$ where this time $\ell$ represents how "long" each slice of our region is, $r$ represents how far that slice is from the axis of rotation, and $u$ represents the variable ($x$ or $y$) that is appropriate for the direction of the slicing.
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We spent today's class reviewing the techniques for volumes of revolution we've discussed in the past few days. This included working through a number of problems together. I gave the following challenge problems; I've included the solutions here too!
- Write an integral that computes the volume of the solid of revolution for the region bounded by $y=0$, $y=x^2$, and $y=-x+2$ around the axis $y=7$. Note that since the region is type II (as we saw in class) and the axis is horizontal, this means that cuts are parallel to the axis of rotation. Therefore we need to use the shell method, in which case we get $$\int_{y=0}^{y=1}((-y+2)-\sqrt{y})\cdot 2\pi(7-y)~dy.$$
- Write an integral that computes the volume of the solid of revolution generated by revolving the region bounded by $y=0$, $y=x^2$, and $y=-x+2$ around the axis $x=-3$ using both the washer and the shell method. This time if we view the region as type II, then since the axis is vertical (but slices are horizontal) we will use the washer method. The boundary $x=-y+2$ is the furthest from the axis in this case, so the "big radius" is $R=(-y+2)-(-3)$. On the other hand, the boundary $x=\sqrt{y}$ is the closest to the axis, and so the "small radius" is $r=(\sqrt{y})-(-3)$. So the washer method says the volume is $$\int_{y=0}^{y=1} \pi((-y+2)-(-3))^2 - \pi((\sqrt{y}-(-3))^2~dy.$$ If we wan to use the shell method, then since the axis is vertical we would need to try to view this region as type I. This requires us to split it into two subregions. For $0 \leq x \leq 1$, we see the top boundary is $y=x^2$ and the bottom boundary is $y=0$; for $1 \leq x \leq 2$ we see that the top boundary is $y=-x+2$ and the bottom boundary is $y=0$. Hence the shell method gives the desired volume as $$\int_0^1 ((x^2)-(0))\cdot 2\pi(x-(-3))~dx+\int_1^2 ((-x+2)-0)\cdot 2\pi(x-(-3))~dx.$$ (It turns out this equals $\frac{41\pi}{6}$.)
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Today we started our last application of integration theory: probability. To get some intuition, we gave a quick overview from some ideas in the realm of discrete probability, which you can think of as the study of probability when there are only finitely many outcomes under investigation. Our running example was rolling a die. For instance, if we roll a standard six-sided die, then the likelihood that any particular face appears after the roll is the same as the likelihood of any other face appearing. Using this, we were able to quickly compute the probability of (for example) rolling a number that is at most $2$. We observed a few guiding principles. The first is that if we can compute the probability $p(x)$ for each possible "simple" outcome $x$, then we can compute the probability of more "complex" outcomes $S$ as $P(S) = \sum_{x \in S}p(x)$. (Here "simple" outcome means a single possible result of our experiment, like rolling a $3$ on a die, or pulling a 7 of clubs from a deck of cards; a "complex" outcome is one that describes a whole range of possible results, like rolling an even number on a die, or pulling an ace from a deck of cards.) The function which keeps track of the probabilities for "simple" outcomes is called a probability mass function. Another guiding principle is that the total probability available in a given problem is $1$; this just says that the likelihood that some outcome occurs is 100 percent.
With these ideas as a guide, we asked how we could study probabilities in the situation where there are a continuum of possible outcomes; by this we mean that when we run an experiment, we get back some number $X$, and that $X$ could be any number between some prescibed values $a$ and $b$. In this scenario we argued that the probability for any individual single outcome should be $0$. (This was akin to noting that the "area above a point" is zero, since such an object has no width.) Instead, we saw that the kinds of events that we should be examining probabilities for is when the value $X$ takes on some range of values.
With this as motivation, for a "continuous random variable" $X$ we saw that there is some probability density function (i.e., pdf) $f_X$ which has the property that for any $a$ and $b$ we have $$P(a \leq X \leq b) = \int_a^b f_X(x)~dx.$$ Furthermore, we said that the pdf should satisfy two basic properties (to make sure that it's going to give us values back that are sensible in the context of probability:
- total probability should be $1$, which means that $\int_{-\infty}^\infty f_X(x)~dx=1$, and
- probability should never be negative, which means that $f_X(x) \geq 0$ for all $x$.
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For the majority of class today we worked to complete our discussion of probability. We defined the notion of expected value $\mathbb{E}(X)$ (aka, mean, which is also denoted $\mu_X$), discussing how it is intuitively the "long term average" for a random variable $X$ and can be computed as $$\mathbb{E}(X)=\int_{-\infty}^\infty xf_X(x)~dx,$$ where here $f_X$ represents the density function for $X$. We said that the standard deviation is another important statistic attached to a random variable (defined as $\sqrt{\mathbb{E}(X^2)-\left(\mathbb{E}(X)\right)^2}$) that tells us about how the density function is "spread" about the mean. We discussed normally distributed random variables, seeing their density functions and noting that they are impossible to compute precisely (with the exception of a few special values that you can compute using some tricks). This is a real disappointment, since normally distributed random variables are so important in describing real world phenomena.
Towards the end of class we motivated the next big idea in the course, which is to try to create higher degree tangent polynomials in hopes of finding better tools for estimating functions.
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We spent today finding the formula for the degree $2$ tangent polynomial to a function $y=f(x)$ at a point $(x_0,f(x_0))$. We wanted such an object to be a degree two polynomial centered at $x_0$, so we knew it should take the form $$T_2(x)=c_0+c_1(x-x_0)+c_2(x-x_0)^2$$ for some appropriately chosen values of $c_0,c_1$ and $c_2$. We also knew we wanted such a polynomial to have the following properties:
- since we want this "tangent polynomial" to actually pass through the point of tangency, we need to ensure that $(x_0,f(x_0))$ is on the graph of $T_2(x)$, and therefore we need $T_2(x_0) = f(x_0)$;
- we want this "tangent polynomial" to match the slope of the original function at $x=x_0$ (isn't this what tangency is supposed to be about, after all?), so we need to ensure that $T_2'(x_0)=f'(x_0)$; and
- we want this "tangent polynomial" to match the concavity of the original function at $x=x_0$, so we need to ensure that $T_2''(x_0)=f''(x_0)$.
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Building on the ideas from last class, we were able to create even more high degree tangent polynomials. In fact, if $n$ is any positive integer, we were able to create a degree $n$ tangent polynomial to $y=f(x)$ at the point $(x_0,f(x_0))$; we called this the degree $n$ Taylor polynomial, and we computed it as $$\begin{align*}T_n(x) &= f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2}(x-x_0)^2 + \cdots + \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n \\ &= \sum_{i=0}^n \frac{f^{(i)}(x_0)}{i!}(x-x_0)^i.\end{align*}$$ We saw how we can use a high degree Taylor polynomial to give estimates for even the most complicated functions, at least assuming the point we're estimating is close to the point of tangency. (If you want to play around with the tool we used in class, you can use this Geogebra applet that lets you input a function, a point of tangency, and a degree for the Taylor polynomial.)
We also saw that there's some hope that these Taylor polynomials might give extremely accurate approximations, even (in some cases) for points that are quite far from the point of tangency. This motivated a handful of important questions, like
- how can we know how far away from the point of tangency we can get in order for these high degree Taylor polynomials to give good approximations for the function, and
- can create something like the limit of these tangent polynomials, and (to hybridize these two questions)
- is it possible that if we took the limit of these tangent polynomials, they would actually be equal to the function itself?
To end the class, we started to introduce the first idea we need in order to answer these last few questions. To do this, we introduced the notion of a sequence. Intuitively, a sequence is just an infinite, ordered list of numbers. We saw several examples of sequences.
Exam 3 content
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In today's class we introduced the notion of the limit of a sequence, both in terms of its intuition as well as the technical definition itself. We tried both of these ideas out in practice on a few sequences. We stated the technical definition, and started to think about why this technical definition captures the intuitive idea that we have for a limit. We saw how to start verifying that the technical definition holds in the case of a particular sequence.
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For the first half of class today, we discussed some of the relevant topics for the second midterm. Afterwards, we continued our discussion of the technical definition of the limit of a sequence, and thought about how to implement the relevant ideas in specific settings.
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For the last few class periods we have been thinking about the technical definition for the limit of a sequence. While this definition is important (since it makes precise what we mean when we say "limit"), it has some obvious drawbacks. For one, the definition doesn't tell us how to find a limit; it just gives us a mechanism for verifying whether or not a given number is the limit of a sequence. Second, the definition is quite cumbersome to use.
For these reasons, we'd like to develop some theorems that give us better insight into how to determine the limit of a sequence, and which make computing limits easier. We gave two main answer to this question in the form of the "piggyback theorem" and the "kangaroo pouch" theorem. The first result of these tells us how to relate the limit of a sequence to the limit of a function related to that sequence; this is important because we already have a lot of tools for evaluating limits of functions (e.g., l'Hopital's rule). The second result gives us a way to evaluate limits of sequences that are built by taking some other sequence and plugging it into a function
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Now that we understand sequences and limits, we can use this knowledge to start making some progress towards our overall goal: creating infinite degree tangent polynomials. The first issue we need to wrestle with is that an "infinite polynomial" would require us to add an infinite number of terms together. The problem is that we only know how to add a finite number of things together; what would it mean to add up an infinite number of terms?
To answer this question, we defined the notion of a series associated with a sequence $\{a_n\}_{n=1}^\infty$. This series --- which we denote $\sum_{n=1}^\infty a_n$ --- is defined as the limit of the associated sequence of partial sums $\{s_n\}$. That sequence of partial sums is defined so that $$\begin{align*}s_1&=a_1\\s_2&=a_1+a_2\\s_3&=a_1+a_2+a_3\\&\vdots\\s_n&=a_1+a_2+\cdots+a_n\\&\vdots \end{align*}$$
A natural question to ask is how we can compute partial sums effectively, and thereby compute series. Unfortunately, for the vast majority of series it is extremely difficult (if not impossible) to find a "clean formula" for the $n$th partial sum of the series. For this reason, most series are simply impossible to compute exactly.
Despite this, there are a few kinds of series where we can compute their value exactly, and today's class focused on the two most prominent cases: telescoping series and geometric series. A telescoping series is one in which terms in the partial sums have massive cancellation, so that each partial sum "collapses" to some relatively simple expression. One can then use this simple expression to evaluate the limit of partial sums, and therefore the series itself. We saw how to implement this in practice with the series $\sum_{n=1}^\infty \frac{1}{n(n+1)}$. We also studied the series $$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots=\sum_{n=1}^\infty \frac{1}{2^n}.$$ (This involved a photo-realistic drawing of a frog jumping across a log.) We began to think about how our "frog on a log" problem fits within a larger class of series, namely ones that are associated to a geometric sequence.
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Last class we started to think about geometric series. In this class, we gave a theorem that allows us to easily determine when a geometric series converges, and (if it does) what the series converges to. We saw that if $\sum_{n=c}^\infty a_n$ is a geometric series that starts at $a$ and has rate $r$, then the series converges precisely when $|r| < 1$, in which case it converges to the value $\frac{a}{1-r}$. We saw lots of examples of this in action.
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Last time we saw that the geometric series theorem is a very powerful --- in fact, the definitive --- tool for analyzing series that happen to be geometric. But what if our series isn't geometric? Unfortunately, outside of telescoping series, the reality is that most series are enormously difficult to calculate precisely. Instead, one is often forced to be content with a much more coarse description of a series behavior. Instead of determining the value to which a series converges, we instead simply ask whether we can determine whether it converges or not. Though seemingly innocuous, this is already a fairly difficult question to resolve, and there are lots and lots of mathematical tools that have been developed to answer this question. Today we focus on two of those tools.
The first of them is the divergence test. We saw that if a series converges, then this means that the partial sums must approach some constant $L$. But since the difference of two consecutive partial sums yields the corresponding term of the sequence (or, more precisely, since $s_n-s_{n-1}=a_n$), this tells us that if a sum $\sum a_n$ converges, then $\lim_{n\to\infty}a_n = 0$. Phrased a different way, this says that if the terms of a series fail to converge to $0$, then we know that the series $\sum a_n$ itself must diverge.
The second test we discussed allows us (in certain situations) to relate the convergence of divergence of a series to the convergence or divergence of a certain improper integral. This result is called the integral test (but later we redubbed it the Linda and Heather test...at least for people who are willing to have silly names for theorems). Unfortunately, to apply this test we need to check a lot of conditions; on the plus side, if we can verify the appropriate conditions, this gives us a way to use our knowledge about integrals to answer questions about (much harder to understand) series.
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We started class by finishing an example from the previous class. Afterwards we invented our next (and last) test for series convergence: the ratio test. This test is excellent on a lot of levels, including the fact that it's relatively easy to use, has few hypotheses, and often (though not always) leads to definitive conclusions. Specifically, if $\sum a_n$ is a series, the ratio test asks us to compute $$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|.$$ If this limit exists and is less than $1$ in absolute value, then the test tells us that the series converges. If the limit exists and is greater than $1$ in absolute value, the test tells us the series diverges. Otherwise, the test is inconclusive. We will see that the ratio test is going to be our "go-to" test as we analyze "infinite degree polynomials" in our next class.
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We now have a suite of tools for analyzing series, and so we can proceed to talk about the function-theoretic analog of series: power series. A power series centered at $a$ and with coefficients $\{c_n\}_{n=0}^\infty$ is defined to be $$\sum_{n=0}^\infty c_n(x-a)^n.$$ The big questions one is interested in for a series are: what values of $x$ can we "plug into the power series" so that the resultant series is convergent, and whether or not we can find some formula that lets us compute those convergent values. We saw lots of examples of power series, and thought about their centers and sequence of coefficients. We also saw examples of some geometric power series, namely $$\begin{align*} 1+x+x^2+x^3+\cdots = \frac{1}{1-x} &\quad \text{ for }|x|< 1\\1-x^2+x^4-x^6+x^8-\cdots=\frac{1}{1+x^2} &\quad \text{ for }|x|< 1\end{align*}.$$
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Picking up on the theorems from last class period, we started today by defining the notions of interval and radius of convergence. The interval of convergence is the set of values of $x$ that we can plug into the power series so that we get a convergent series at that point; the radius of convergence measures how far we can move from the center of the series and get a convergent series. We studied these ideas in the contexts of several specific examples, including the power series we saw last class period (which were geometric) as well as some new power series (which required the ratio test). We noted that when we use the ratio test and get back a finite interval of convergence, it's typically the case that determining convergence of the series at the endpoints of the interval is a lot of effort and require other series analysis techniques (e.g., perhaps the integral test for series, or even other tests that we didn't discuss in class). For that reason, we will typically be interested in the interior of the interval of convergence (which is just a fancy way to say that we won't worry about whether a series converges at the endpoints of its interval of convergence or not).
At the end of the class we returned to the question of whether or not we can compute the value that a power series converges to. Generally speaking, this is a very hard question if one starts with a power series drawn out of the blue. On the other hand, we do have a way for manufacturing a power series that we know is associated to a particular function: we can take our old friends the Taylor polynomials associated to some function $f(x)$ at a point of tangency $a$, and turn this into a power series by letting the sum go off to infinity. The big theorem we mentioned at the end of class is that on the interval, a function is equal to its Taylor series. This is the "big idea" that we have been chasing for the last few weeks, and will allow us to really understand and make use of our ideas around "infinite degree tangent polynomials."
Additional content
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Earlier in the semester we were able to use some basic calculus skills (essentially taking derivatives) to create higher degree tangent polynomials. In particular, if $f(x)$ is some function and $x=a$ is some point in the domain of that function, then we created the degree $n$ tangent polynomial for $f$ centered at $a$: $$T_n(x)=\sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k.$$ Now that we know about power series, we are able to extend this family of polynomials to the "infinite degree tangent polynomials" we had wanted to create all along! This is what we will now call the Taylor series for $f$ centered at $a$: $$T_\infty(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k.$$ (In the case that $a=0$, we will call this series the MacLaurin series for $f$.) We saw last time that this Taylor series achieves the dream we hoped for: as long as $x$ is in the interval of convergence, then we have $f(x) = T_\infty(x)$.
With all this in mind, it's now natural for us to ask how we can actually compute some of these Taylor series, and in particular to ask where this "infinite degree tangent polynomial" really does what we hoped it might do. We did this by building series for $f(x)=e^x$ and $f(x)=\sin(x)$; for both of these functions we saw that the associated series has an infinite radius of convergence, and so each of these functions is actually equal to their associated power series for all $x$. Specifically, we found $$ \begin{align*} e^x&=\sum_{k=0}^\infty \frac{1}{k!}x^k = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots &\quad \text{(for all $x$)}\\ \sin(x)&=\sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots &\quad \text{(for all $x$)}. \end{align*} $$ We also studied the Taylor series for $f(x)=\ln(x)$ centered at $x=1$ (though we didn't have quite enough time to consider its interval of convergence).
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In today's class we thought about how we can take some "known" Taylor series for functions and use them to create "new" Taylor series via integration and differentiation. For instance, we took the series representation for $\sin(x)$ and differentiated it to created a series representation for $\cos(x)$:$$\begin{align*} \cos(x) &= \frac{d}{dx}\left[\sin(x)\right]\\ &=\frac{d}{dx}\left[x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\right]\\ &=1-\frac{3x^2}{3!}+\frac{5x^4}{5!}-\frac{7x^6}{7!}+\cdots\\ &=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots. \end{align*}$$ In fact, we even had a theorem that told us that since the radius of convergence for $\sin(x)$ is $R=\infty$, then the radius of this series representation for $\cos(x)$ is also $R=\infty$. We did a similar calculation --- but this time using integration instead of differentiation --- to use the series representation for $\frac{1}{1+x^2}$ into a series for $\arctan(x)$:$$\arctan(x)=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots \qquad \text{ for all }x \text{ in }[-1,1].$$
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Last class period we thought about how we could create "new" Taylor series by differentiating or integrating "old" Taylor series. In today's class, we thought about a similar process, but instead of "doing calculus" to a known series, we instead "do arithmetic. For instance, we computed a MacLaurin series for $e^{-x^2}$ by plugging $-x^2$ into the series representation for $e^x$, and we also computed a series expression for $\sin(x)\cos(x)$ by multiplying the Taylor series for each of these functions.
Towards the end of class we thought more about why Taylor series are so useful. The big picture answer is that Taylor series give us access to understanding complex functions by replacing them with objects that have many of the properties of polynomials. Without this, it would be incredibly difficult to understand very basic questions about these complicated functions (e.g., what their actual values are at specific inputs). However, we can also use series to tackle more specific questions. For instance, we can use series to compute integrals that would otherwise be not be computable (if, for example, we can't find an antiderivative), or to evaluate limits that would otherwise require horrific applications of l'Hopital's rule.
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At the start of class we talked about a few more applications of series, including ways they show up in probability, combinatorics, and complex analysis. We then spent the balance of the class looking through this quick introduction on Fourier series and experimenting with some online sound applications.