Tag Archives: Calculus

Another proof of Taylor’s Theorem

In this note, we produce a proof of Taylor’s Theorem. As in many proofs of Taylor’s Theorem, we begin with a curious start and then follow our noses forward.

Is this a new proof? I think so. But I wouldn’t bet a lot of money on it. It’s certainly new to me.

Is this a groundbreaking proof? No, not at all. But it’s cute, and I like it.1

We begin with the following simple observation. Suppose that $f$ is two times continuously differentiable. Then for any $t \neq 0$, we see that $$f'(t) – f'(0) = \frac{f'(t) – f'(0)}{t} t.$$ Integrating each side from $0$ to $x$, we find that $$f(x) – f(0) – f'(0) x = \int_0^x \frac{f'(t) – f'(0)}{t} t dt.$$ To interpret the integral on the right in a different way, we will use the mean value theorem for integrals.

Mean Value Theorem for Integrals

Suppose that $g$ and $h$ are continuous functions, and that $h$ doesn’t change sign in $[0, x]$. Then there is a $c \in [0, x]$ such that $$\int_0^x g(t) h(t) dt = g(c) \int_0^x h(t) dt.$$

Suppose without loss of generality that $h(t)$ is nonnegative. Since $g$ is continuous on $[0, x]$, it attains its minimum $m$ and maximum $M$ on this interval. Thus $$m \int_0^x h(t) dt \leq \int_0^x g(t)h(t)dt \leq M \int_0^x h(t) dt.$$ Let $I = \int_0^x h(t) dt$. If $I = 0$ (or equivalently, if $h(t) \equiv 0$), then the theorem is trivially true, so suppose instead that $I \neq 0$. Then $$m \leq \frac{1}{I} \int_0^x g(t) h(t) dt \leq M.$$ By the intermediate value theorem, $g(t)$ attains every value between $m$ and $M$, and thus there exists some $c$ such that $$g(c) = \frac{1}{I} \int_0^x g(t) h(t) dt.$$ Rearranging proves the theorem.

For this application, let $g(t) = (f'(t) – f'(0))/t$ for $t \neq 0$, and $g(0) =f'{}'(0)$. The continuity of $g$ at $0$ is exactly the condition that $f'{}'(0)$exists. We also let $h(t) = t$.

For $x > 0$, it follows from the mean value theorem for integrals that there exists a $c \in [0, x]$ such that $$\int_0^x \frac{f'(t) – f'(0)}{t} t dt = \frac{f'(c) – f'(0)}{c} \int_0^x t dt = \frac{f'(c) – f'(0)}{c} \frac{x^2}{2}.$$ (Very similar reasoning applies for $x < 0$). Finally, by the mean value theorem (applied to $f’$), there exists a point $\xi \in (0, c)$ such that $$f'{}'(\xi) = \frac{f'(c) – f'(0)}{c}.$$ Putting this together, we have proved that there is a $\xi \in (0, x)$ such that $$f(x) – f(0) – f'(0) x = f'{}'(\xi) \frac{x^2}{2},$$ which is one version of Taylor’s Theorem with a linear approximating polynomial.

This approach generalizes. Suppose $f$ is a $(k+1)$ times continuously differentiable function, and begin with the trivial observation that $$f^{(k)}(t) – f^{(k)}(0) = \frac{f^{(k)}(t) – f^{(k)}(0)}{t} t.$$ Iteratively integrate $k$ times: first from $0$ to $t_1$, then from $0$ to $t_2$, and so on, with the $k$th interval being from $0$ to $t_k = x$.

Then the left hand side becomes $$f(x) – \sum_{n = 0}^k f^{(n)}(0)\frac{x^n}{n!},$$ the difference between $f$ and its degree $k$ Taylor polynomial. The right hand side is
$$\label{eq:only}\underbrace{\int _0^{t_k = x} \cdots \int _0^{t _1}} _{k \text{ times}} \frac{f^{(k)}(t) – f^{(k)}(0)}{t} t \, dt \, dt _1 \cdots dt _{k-1}.$$

To handle this, we note the following variant of the mean value theorem for integrals.

Mean value theorem for iterated integrals

Suppose that $g$ and $h$ are continuous functions, and that $h$ doesn’t change sign in $[0, x]$. Then there is a $c \in [0, x]$ such that $$\underbrace{\int_0^{t _k=x} \cdots \int _0^{t _1}} _{k \; \text{times}} g(t) h(t) dt =g(c) \underbrace{\int _0^{t _k=x} \cdots \int _0^{t _1}} _{k \; \text{times}} h(t) dt.$$

In fact, this can be proved in almost exactly the same way as in the single-integral version, so we do not repeat the proof.

With this theorem, there is a $c \in [0, x]$ such that we see that \eqref{eq:only} can be written as $$\frac{f^{(k)}(c) – f^{(k)}(0)}{c} \underbrace{\int _0^{t _k = x} \cdots \int _0^{t _1}} _{k \; \text{times}} t \, dt \, dt _1 \cdots dt _{k-1}.$$ By the mean value theorem, the factor in front of the integrals can be written as $f^{(k+1)}(\xi)$ for some $\xi \in (0, x)$. The integrals can be directly evaluated to be $x^{k+1}/(k+1)!$.

Thus overall, we find that $$f(x) = \sum_{n = 0}^n f^{(n)}(0) \frac{x^n}{n!} + f^{(k+1)}(\xi) \frac{x^{k+1}}{(k+1)!}$$ for some $\xi \in (0, x)$. Thus we have proved Taylor’s Theorem (with Lagrange’s error bound).

When are there continuous choices for the Mean Value Abscissa?

Miles Wheeler and I have recently uploaded a paper to the arXiv called “When are there continuous choices for the mean value abscissa?”, which we have submitted to an expository journal. The underlying question is simple but nontrivial.

The mean value theorem of calculus states that, given a differentiable function $f$ on an interval $[a, b]$, then there exists a $c \in (a, b)$ such that
$$\frac{f(b) – f(a)}{b – a} = f'(c).$$
We call $c$ the mean value abscissa.
Our question concerns potential behavior of this abscissa when we fix the left endpoint $a$ of the interval and vary $b$. For each $b$, there is at least one abscissa $c_b$ such that the mean value theorem holds with that abscissa. But generically there may be more than one choice of abscissa for each interval. When can we choose $c_b$ as a continuous function of $b$? That is, when can we write $c = c(b)$ such that
$$\frac{f(b) – f(a)}{b – a} = f'(c(b))$$
for all $b$ in some interval?
We think of this as a continuous choice for the mean value abscissa.

This is a great question. It’s widely understandable — even to students with only one semester of calculus. Further it encourages a proper understanding of what a function is, as thinking of $c$ as potentially a function of $b$ is atypical and interesting.

But I also like this question because the answer is not as simple as you might think, and there are a few nice ideas that get to the answer.

Should you find yourself reading this without knowing the answer, I encourage you to consider it right now. Should continuous choices of abscissas exist? What if the function is really well-behaved? What if it’s smooth? Or analytic?

Let’s focus on the smooth question. Suppose that $f$ is smooth — that it is infinitely differentiable. These are a distinguished class of functions. But it turns out that being smooth is not sufficient: here is a counterexample.

In this figure, there are points $b$ arbitrarily near $b_0$ such that the secant line from $a_0$ to $b$ have positive slope, and points arbitrarily near such that the secant lines have negative slope. There are infinitely many mean value abscissae with $f'(c_0) = 0$, but all of them are either far from a point $c$ where $f'(c) > 0$ or far from a point $c$ where $f'(c) < 0$. And thus there is no continuous choice. From a theorem oriented point of view, our main theorem is that if $f$ is analytic, then there is always a locally continuous choice. That is, for every interval $[a_0, b_0]$, there exists a mean value abscissa $c$ such that $c = c(b)$ for some interval $B$ containing $b_0$. But the purpose of this article isn’t simply to prove this theorem. The purpose is to exposit how the ideas that are used to study this problem and to prove these results are fundamentally based only on a couple of central ideas covered in introductory single and multivariable calculus. All of this paper is completely accessible to a student having studied only single variable calculus (and who is willing to believe that partial derivatives exist are a reasonable object). We prove and use simple-but-nontrivial versions of the contraction mapping theorem, the implicit function theorem, and Morse’s lemma. The implicit function theorem is enough to say that any abscissa $c_0$ such that $f”(c_0) \neq 0$ has a unique continuous extension. Thus immediately for “most” intervals on “most” reasonable functions, we answer in the affirmative. Morse’s lemma allows us to say a bit more about the case when $f”(c_0) = 0$ but $f'{}'{}'(c_0) \neq 0$. In this case there are either multiple continuous extensions or none. And a few small ingredients and the idea behind Morse’s lemma, combined with the implicit function theorem again, is enough to prove the main result. ## Student projects A calculus student looking for a project to dive into and sharpen their calculus skills could find ideas here to sink their teeth into. Beginning by understanding this paper is a great start. A good motivating question would be to carry on one additional step, and to study explicitly the behavior of a function near a point where $f”(c_0) = f'{}'{}'(c_0) = 0$, but $f^{(4)}(c_0) \neq 0$. A slightly more open question that we lightly touch on (but leave largely implicit) is ther inverse question: when can one find a mean value abscissa $c$ such that the right endpoint $b$ can be written as a continuous function $b(c)$ for some neighborhood $C$ containing the initial point $c_0$? Much of the analysis is the same, but figuring it out would require some attention. A much deeper question is to consider the abscissa as a function of both the left endpoint $a$ and the right endpoint $b$. The guiding question here could be to decide when one can write the abscissa as a continuous function $c(a, b)$ in a neighborhood of $(a_0, b_0)$. I would be interested to see a graphical description of the possible shapes of these functions — I’m not quite sure what they might look like. There is also a nice computational problem. In the paper, we include several plots of solution curves in $(b, c)$ space. But we did this with a meshed implicit function theorem solver. A computationally inclined student could devise an explicit way of constructing solutions. On the one hand, this is guaranteed to work since one can apply contraction mappings explicitly to make the resulting function from the implicit function theorem explicit. But on the other hand, many (most?) applications of the implicit function theorem are in more complicated high dimensional spaces, whereas the situation in this paper is the smallest nontrivial example. ## Producing the graphs We made 13 graphs in 5 figures for this article. These pictures were created using matplotlib. The data was created using numpy, scipy, and sympy from within the scipy/numpy python stack, and the actual creation was done interactively within a jupyter notebook. The actual notebook is available here, (along with other relatively raw jupyter notebooks). The most complicated graph is this one.

This figure has graphs of three functions along the top. In each graph, the interval $[0, 3]$ is considered in the mean value theorem, and the point $c_0 = 1$ is a mean value abscissa. In each, we also have $f”(c_0) = 0$, and the point is that the behavior of $f”(b_0)$ has a large impact on the nature of the implicit functions. The three graphs along the bottom are in $(b, c)$ space and present all mean value abscissa for each $b$. This is not a function, but the local structure of the graphs are interesting and visually distinct.

The process of making these examples and making these figures is interesting in itself. We did not make these figures explicitly, but instead chose certain points and certain values of derivatives at those points, and used Hermite interpolation find polynomials with those points.1

In the future I plan on writing a note on the creation of these figures.

Posted in Expository, Math.CA, Mathematics | | 1 Comment

Series Convergence Tests with Prototypical Examples

This is a note written for my Fall 2016 Math 100 class at Brown University. We are currently learning about various tests for determining whether series converge or diverge. In this note, we collect these tests together in a single document. We give a brief description of each test, some indicators of when each test would be good to use, and give a prototypical example for each. Note that we do justify any of these tests here — we’ve discussed that extensively in class. [But if something is unclear, send me an email or head to my office hours]. This is here to remind us of the variety of the various tests of convergence.

A copy of just the statements of the tests, put together, can be found here. A pdf copy of this whole post can be found here.

In order, we discuss the following tests:

1. The $n$th term test, also called the basic divergence test
2. Recognizing an alternating series
3. Recognizing a geometric series
4. Recognizing a telescoping series
5. The Integral Test
6. P-series
7. Direct (or basic) comparison
8. Limit comparison
9. The ratio test
10. The root test

The $n$th term test

Statement

Suppose we are looking at $\sum_{n = 1}^\infty a_n$ and

\lim_{n \to \infty} a_n \neq 0. \notag

Then $\sum_{n = 1}^\infty a_n$ does not converge.

When to use it

When applicable, the $n$th term test for divergence is usually the easiest and quickest way to confirm that a series diverges. When first considering a series, it’s a good idea to think about whether the terms go to zero or not. But remember that if the limit of the individual terms is zero, then it is necessary to think harder about whether the series converges or diverges.

Example

Each of the series

\sum_{n = 1}^\infty \frac{n+1}{2n + 4}, \quad \sum_{n = 1}^\infty \cos n, \quad \sum_{n = 1}^\infty \sqrt{n} \notag

diverges since their limits are not $0$.

Recognizing alternating series

Statement

Suppose $\sum_{n = 1}^\infty (-1)^n a_n$ is a series where

1. $a_n \geq 0$,
2. $a_n$ is decreasing, and
3. $\lim_{n \to \infty} a_n = 0$.

Then $\sum_{n = 1}^\infty (-1)^n a_n$ converges.

Stated differently, if the terms are alternating sign, decreasing in absolute size, and converging to zero, then the series converges.

When to use it

The key is in the name — if the series is alternating, then this is the goto idea of analysis. Note that if the terms of a series are alternating and decreasing, but the terms do not go to zero, then the series diverges by the $n$th term test.

Example

Suppose we are looking at the series

\sum_{n = 1}^\infty \frac{(-1)^n}{\log(n+1)} = \frac{-1}{\log 2} + \frac{1}{\log 3} + \frac{-1}{\log 4} + \cdots \notag

The terms are alternating.
The sizes of the terms are $\frac{1}{\log (n+1)}$, and these are decreasing.
Finally,

\lim_{n \to \infty} \frac{1}{\log(n+1)} = 0. \notag

Thus the alternating series test applies and shows that this series converges.

Continuity of the Mean Value

1. Introduction

When I first learned the mean value theorem as a high schooler, I was thoroughly unimpressed. Part of this was because it’s just like Rolle’s Theorem, which feels obvious. But I think the greater part is because I thought it was useless. And I continued to think it was useless until I began my first proof-oriented treatment of calculus as a second year at Georgia Tech. Somehow, in the interceding years, I learned to value intuition and simple statements.

I have since completely changed my view on the mean value theorem. I now consider essentially all of one variable calculus to be the Mean Value Theorem, perhaps in various forms or disguises. In my earlier note An Intuitive Introduction to Calculus, we state and prove the Mean Value Theorem, and then show that we can prove the Fundamental Theorem of Calculus with the Mean Value Theorem and the Intermediate Value Theorem (which also felt silly to me as a high schooler, but which is not silly).

In this brief note, I want to consider one small aspect of the Mean Value Theorem: can the “mean value” be chosen continuously as a function of the endpoints? To state this more clearly, first recall the theorem:

Suppose ${f}$ is a differentiable real-valued function on an interval ${[a,b]}$. Then there exists a point ${c}$ between ${a}$ and ${b}$ such that $$\frac{f(b) – f(a)}{b – a} = f'(c), \tag{1}$$
which is to say that there is a point where the slope of ${f}$ is the same as the average slope from ${a}$ to ${b}$.

What if we allow the interval to vary? Suppose we are interested in a differentiable function ${f}$ on intervals of the form ${[0,b]}$, and we let ${b}$ vary. Then for each choice of ${b}$, the mean value theorem tells us that there exists ${c_b}$ such that $$\frac{f(b) – f(0)}{b} = f'(c_b).$$
Then the question we consider today is, as a function of ${b}$, can ${c_b}$ be chosen continuously? We will see that we cannot, and we’ll see explicit counterexamples. This, after the fold.

Trigonometric and related substitutions in integrals

$\DeclareMathOperator{\csch}{csch}$
$\DeclareMathOperator{\sech}{sech}$
$\DeclareMathOperator{\arsinh}{arsinh}$

1. Introduction

In many ways, a first semester of calculus is a big ideas course. Students learn the basics of differentiation and integration, and some of the big-hitting theorems like the Fundamental Theorems of Calculus. Even in a big ideas course, students learn how to differentiate any reasonable combination of polynomials, trig, exponentials, and logarithms (elementary functions).

But integration skills are not pushed nearly as far. Do you ever wonder why? Even at the end of the first semester of calculus, there are many elementary functions that students cannot integrate. But the reason isn’t that there wasn’t enough time, but instead that integration is hard. And when I say hard, I mean often impossible. And when I say impossible, I don’t mean unsolved, but instead provably impossible (and when I say impossible, I mean that we can’t always integrate and get a nice function out, unlike our ability to differentiate any nice function and get a nice function back). An easy example is the sine integral $$\int \frac{\sin x}{x} \mathrm d x,$$
which cannot be expressed in terms of elementary functions. In short, even though the derivative of an elementary function is always an elementary function, the antiderivative of elementary functions don’t need to be elementary.

Worse, even when antidifferentiation is possible, it might still be really hard. This is the first problem that a second semester in calculus might try to address, meaning that students learn a veritable bag of tricks of integration techniques. These might include ${u}$-substitution and integration by parts (which are like inverses of the chain rule and product rule, respectively), and then the relatively more complicated techniques like partial fraction decomposition and trig substitution.

In this note, we are going to take a closer look at problems related to trig substitution, and some related ideas. We will assume familiarity with ${u}$-substitution and integration by parts, and we might even use them here from time to time. This, after the fold.

An Intuitive Overview of Taylor Series

This is a note written for my fall 2013 Math 100 class, but it was not written “for the exam,” nor does anything on here subtly hint at anything on any exam. But I hope that this will be helpful for anyone who wants to get a basic understanding of Taylor series. What I want to do is try to get some sort of intuitive grasp on Taylor series as approximations of functions. By intuitive, I mean intuitive to those with a good grasp of functions, the basics of a first semester of calculus (derivatives, integrals, the mean value theorem, and the fundamental theorem of calculus) – so it’s a mathematical intuition. In this way, this post is a sort of follow-up of my earlier note, An Intuitive Introduction to Calculus.

PLEASE NOTE that my math compiler and my markdown compiler sometimes compete, and sometimes repeated derivatives are too high or too low by one pixel.

We care about Taylor series because they allow us to approximate other functions in predictable ways. Sometimes, these approximations can be made to be very, very, very accurate without requiring too much computing power. You might have heard that computers/calculators routinely use Taylor series to calculate things like ${e^x}$ (which is more or less often true). But up to this point in most students’ mathematical development, most mathematics has been clean and perfect; everything has been exact algorithms yielding exact answers for years and years. This is simply not the way of the world.

Here’s a fundamental fact to both mathematics and life: almost anything worth doing is probably pretty hard and pretty messy.

For a very recognizable example, let’s think about finding zeroes of polynomials. Finding roots of linear polynomials is very easy. If we see ${5 + x = 0}$, we see that ${-5}$ is the zero. Similarly, finding roots of quadratic polynomials is very easy, and many of us have memorized the quadratic formula to this end. Thus ${ax^2 + bx + c = 0}$ has solutions ${x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}}$. These are both nice, algorithmic, and exact. But I will guess that the vast majority of those who read this have never seen a “cubic polynomial formula” for finding roots of cubic polynomials (although it does exist, it is horrendously messy – look up Cardano’s formula). There is even an algorithmic way of finding the roots of quartic polynomials. But here’s something amazing: there is no general method for finding the exact roots of 5th degree polynomials (or higher degree).

I don’t mean We haven’t found it yet, but there may be one, or even You’ll have to use one of these myriad ways – I mean it has been shown that there is no general method of finding exact roots of degree 5 or higher polynomials. But we certainly can approximate them arbitrarily well. So even something as simple as finding roots of polynomials, which we’ve been doing since we were in middle school, gets incredibly and unbelievably complicated.

So before we hop into Taylor series directly, I want to get into the mindset of approximating functions with other functions.

1. Approximating functions with other functions

We like working with polynomials because they’re so easy to calculate and manipulate. So sometimes we try to approximate complicated functions with polynomials, a problem sometimes called
polynomial interpolation”.

Suppose we wanted to approximate ${\sin(x)}$. The most naive approximation that we might do is see that ${\sin(0) = 0}$, so we might approximate ${\sin(x)}$ by ${p_0(x) = 0}$. We know that it’s right at least once, and since ${\sin(x)}$ is periodic, it’s going to be right many times. I write ${p_0}$ to indicate that this is a degree ${0}$ polynomial, that is, a constant polynomial. Clearly though, this is a terrible approximation, and we can do better.

Math 100: Before second midterm

You have a midterm next week, and it’s not going to be a cakewalk.

As requested, I’m uploading the last five weeks’ worth of worksheets, with (my) solutions. A comment on the solutions: not everything is presented in full detail, but most things are presented with most detail (except for the occasional one that is far far beyond what we actually expect you to be able to do). If you have any questions about anything, let me know. Even better, ask it here – maybe others have the same questions too.

And since we were unable to go over the quiz in my afternoon recitation today, I’m attaching a worked solution to the quiz as well.

Again, let me know if you have any questions. I will still have my office hours on Tuesday from 2:30-4:30pm in my office (I’m aware that this happens to be immediately before the exam – status not by design). And I’ll be more or less responsive by email.

Study study study!

Math 100: Week 4

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.

An intuitive introduction to calculus

This is a post written for my fall 2013 Math 100 class but largely intended for anyone with knowledge of what a function is and a desire to know what calculus is all about. Calculus is made out to be the pinnacle of the high school math curriculum, and correspondingly is thought to be very hard. But the difficulty is bloated, blown out of proportion. In fact, the ideas behind calculus are approachable and even intuitive if thought about in the right way.

Many people managed to stumble across the page before I’d finished all the graphics. I’m sorry, but they’re all done now! I was having trouble interpreting how WordPress was going to handle my gif files – it turns out that they automagically resize them if you don’t make them of the correct size, which makes them not display. It took me a bit to realize this. I’d like to mention that this actually started as a 90 minute talk I had with my wife over coffee, so perhaps an alternate title would be “Learning calculus in 2 hours over a cup of coffee.”

So read on if you would like to understand what calculus is, or if you’re looking for a refresher of the concepts from a first semester in calculus (like for Math 100 students at Brown), or if you’re looking for a bird’s eye view of AP Calc AB subject material.

1. An intuitive and semicomplete introduction to calculus

We will think of a function ${f(\cdot)}$ as something that takes an input ${x}$ and gives out another number, which we’ll denote by ${f(x)}$. We know functions like ${f(x) = x^2 + 1}$, which means that if I give in a number ${x}$ then the function returns the number ${f(x) = x^2 + 1}$. So I put in ${1}$, I get ${1^2 + 1 = 2}$, i.e. ${f(1) = 2}$. Primary and secondary school overly conditions students to think of functions in terms of a formula or equation. The important thing to remember is that a function is really just something that gives an output when given an input, and if the same input is given later then the function spits the same output out. As an aside, I should mention that the most common problem I’ve seen in my teaching and tutoring is a fundamental misunderstanding of functions and their graphs

For a function that takes in and spits out numbers, we can associate a graph. A graph is a two-dimensional representation of our function, where by convention the input is put on the horizontal axis and the output is put on the vertical axis. Each axis is numbered, and in this way we can identify any point in the graph by its coordinates, i.e. its horizontal and vertical position. A graph of a function ${f(x)}$ includes a point ${(x,y)}$ if ${y = f(x)}$.

Thus each point on the graph is really of the form ${(x, f(x))}$. A large portion of algebra I and II is devoted to being able to draw graphs for a variety of functions. And if you think about it, graphs contain a huge amount of information. Graphing ${f(x)= x^2 + 1}$ involves drawing an upwards-facing parabola, which really represents an infinite number of points. That’s pretty intense, but it’s not what I want to focus on here.

1.1. Generalizing slope – introducing the derivative

You might recall the idea of the ‘slope’ of a line. A line has a constant ratio of how much the ${y}$ value changes for a specific change in ${x}$, which we call the slope (people always seem to remember rise over run). In particular, if a line passes through the points ${(x_1, y_1)}$ and ${(x_2, y_2)}$, then its slope will be the vertical change ${y_2 – y_1}$ divided by the horizontal change ${x_2 – x_1}$, or ${\dfrac{y_2 – y_1}{x_2 – x_1}}$.

So if the line is given by an equation ${f(x) = \text{something}}$, then the slope from two inputs ${x_1}$ and ${x_2}$ is ${\dfrac{f(x_2) – f(x_1)}{x_2 – x_1}}$. As an aside, for those that remember things like the ‘standard equation’ ${y = mx + b}$ or ‘point-slope’ ${(y – y_0) = m(x – x_0)}$ but who have never thought or been taught where these come from: the claim that lines are the curves of constant slope is saying that for any choice of ${(x_1, y_1)}$ on the line, we expect ${\dfrac{y_2 – y_1}{x_2 – x_1} = m}$ a constant, which I denote by ${m}$ for no particularly good reason other than the fact that some textbook author long ago did such a thing. Since we’re allowing ourselves to choose any ${(x_1, y_1)}$, we might drop the subscripts – since they usually mean a constant – and rearrange our equation to give ${y_2 – y = m(x_2 – x)}$, which is what has been so unkindly drilled into students’ heads as the ‘point-slope form.’ This is why lines have a point-slope form, and a reason that it comes up so much is that it comes so naturally from the defining characteristic of a line, i.e. constant slope.

But one cannot speak of the ‘slope’ of a parabola.

Intuitively, we look at our parabola ${x^2 + 1}$ and see that the ‘slope,’ or an estimate of how much the function ${f(x)}$ changes with a change in ${x}$, seems to be changing depending on what ${x}$ values we choose. (This should make sense – if it didn’t change, and had constant slope, then it would be a line). The first major goal of calculus is to come up with an idea of a ‘slope’ for non-linear functions. I should add that we already know a sort of ‘instantaneous rate of change’ of a nonlinear function. When we’re in a car and we’re driving somewhere, we’re usually speeding up or slowing down, and our pace isn’t usually linear. Yet our speedometer still manages to say how fast we’re going, which is an immediate rate of change. So if we had a function ${p(t)}$ that gave us our position at a time ${t}$, then the slope would give us our velocity (change in position per change in time) at a moment. So without knowing it, we’re familiar with a generalized slope already. Now in our parabola, we don’t expect a constant slope, so we want to associate a ‘slope’ to each input ${x}$. In other words, we want to be able to understand how rapidly the function ${f(x)}$ is changing at each ${x}$, analogous to how the slope ${m}$ of a line ${g(x) = mx + b}$ tells us that if we change our input by an amount ${h}$ then our output value will change by ${mh}$.

How does calculus do that? The idea is to get closer and closer approximations. Suppose we want to find the ‘slope’ of our parabola at the point ${x = 1}$. Let’s get an approximate answer. The slope of the line coming from inputs ${x = 1}$ and ${x = 2}$ is a (poor) approximation. In particular, since we’re working with ${f(x) = x^2 + 1}$, we have that ${f(2) = 5}$ and ${f(1) = 2}$, so that the ‘approximate slope’ from ${x = 1}$ and ${x = 2}$ is ${\frac{5 – 2}{2 – 1} = 3}$. But looking at the graph,

we see that it feels like this slope is too large. So let’s get closer. Suppose we use inputs ${x = 1}$ and ${x = 1.5}$. We get that the approximate slope is ${\frac{3.25 – 2}{1.5 – 1} = 2.5}$. If we were to graph it, this would also feel too large. So we can keep choosing smaller and smaller changes, like using ${x = 1}$ and ${x = 1.1}$, or ${x = 1}$ and ${x = 1.01}$, and so on. This next graphic contains these approximations, with chosen points getting closer and closer to ${1}$.

Let’s look a little closer at the values we’re getting for our slopes when we use ${1}$ and ${2, 1.5, 1.1, 1.01, 1.001}$ as our inputs. We get

$\displaystyle \begin{array}{c|c} \text{second input} & \text{approx. slope} \\ \hline 2 & 3 \\ 1.5 & 2.5 \\ 1.1 & 2.1 \\ 1.01 & 2.01 \\ 1.001 & 2.001 \end{array}$

It looks like the approximate slopes are approaching ${2}$. What if we plot the graph with a line of slope ${2}$ going through the point ${(1,2)}$?

It looks great! Let’s zoom in a whole lot.

That looks really close! In fact, what I’ve been allowing as the natural feeling slope, or local rate of change, is really the line tangent to the graph of our function at the point ${(1, f(1))}$. In a calculus class, you’ll spend a bit of time making sense of what it means for the approximate slopes to ‘approach’ ${2}$. This is called a ‘limit,’ and the details are not important to us right now. The important thing is that this let us get an idea of a ‘slope’ at a point on a parabola. It’s not really a slope, because a parabola isn’t a line. So we’ve given it a different name – we call this ‘the derivative.’ So the derivative of ${f(x) = x^2 + 1}$ at ${x = 1}$ is ${2}$, i.e. right around ${x = 1}$ we expect a rate of change of ${2}$, so that we expect ${f(1 + h) – f(1) \approx 2h}$. If you think about it, we’re saying that we can approximate ${f(x) = x^2 + 1}$ near the point ${(1, 2)}$ by the line shown in the graph above: this line passes through ${(1,2)}$ and it’s slope is ${2}$, what we’re calling the slope of ${f(x) = x^2 + 1}$ at ${x = 1}$.

Let’s generalize. We were able to speak of the derivative at one point, but how about other points? The rest of this post is below the ‘more’ tag below.

Posted in Brown University, Expository, Math 100, Mathematics | | 9 Comments

A proof from the first sheet (SummerNT)

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

Claim: As $x$ increases, the function

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

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:

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.