Showing posts with label tips. Show all posts
Showing posts with label tips. Show all posts

Monday, May 20, 2024

Getting a copy of Stroll

A Stroll through Calculus covers many of the concepts of calculus in an informal way with little more than intermediate algebra (and later, some trig). It's an attempt to help people understand what's going on without intimidating proofs or formalities, because the basic notions of calculus are both straightforward and elegant.

Anyone who wants to obtain a copy of A Stroll through Calculus can use this direct link to the publisher:

https://store.cognella.com/25177

If you need any help with ordering from Cognella, feel free to email orders@cognella.com or call 800.200.3908 ext. 503.

Stroll is listed as a required textbook on ARC's online Follett bookstore pages for Math 402 and Math 499, but copies will be on reserve in the library so you don't have to purchase a copy. If you choose to buy a copy, you will get a better price from Cognella than from Follett (or Amazon), so use the provided Cognella link.

Wednesday, January 20, 2021

Monday, November 23, 2020

Tuesday, March 10, 2020

A primer on polling

Pollsters usually examine a sample that is much smaller than the actual population in question. For a statewide election in California, for example, a pollster is likely to interview a few hundred voters to discover the opinions of an electorate that comprises millions. How can this possibly work?

A sample example

A small sample can give you a surprisingly robust measure of what is going on with a large population. As long as the pollster takes some practical measures to ensure that a sample is not unduly skewed (don't find all your polling subjects in the waiting rooms of Lexus dealerships!), the sample will be representative of the whole population. That's why you'll hear people talk about picking people at random in a polling survey. It's a way to avoid biasing a sample.

Suppose, for the sake of illustration, that a voting population is evenly divided between candidates A and B. Suppose that you're going to pick two voters at random and ask them who they prefer. What could happen?

There are actually four possible outcomes: Both subjects prefer A, both subjects prefer B, the first subject prefers A while the second prefers B, and the first subject prefers B while the second prefers A. These four outcomes are equally likely, leading us to an interesting conclusion: Even a sample of size 2 gives you a correct measure of voter preference half the time!


No doubt this result should improve if we choose a larger sample. After all, while it's true that half of the possible outcomes correctly reflect the opinions of the electorate, the other half is way off, telling us there is unanimous sentiment in favor of one candidate.

Let's poll four voters this time. There are actually sixteen (24) equally likely results:


This time, six of the possible sample results (that's three-eighths) are exactly right in mirroring the fifty-fifty split of the electorate. What's more, only two of the possible samples (one-eighth of them) tell us to expect a unanimous vote for one of the candidates. The other samples (three-eighths of them) give skewed results—giving one candidate a three-to-one edge over the other—but not as badly skewed as in the previous two-person sample.

You know what's going to happen as we continue to increase the sample size: It's going to get more and more difficult to obtain really unrepresentative results. Let's look at what occurs when we increase the sample size to just eight randomly selected voters. There are 28 = 256 ways the samples can come up, varying from all for A to all for B. Okay, that's too many cases to write out individually. We'll have to group them. The following table summarizes the possibilities. For example, there are 28 cases in which we end up with 6 votes for A versus 2 for B. (In case you're curious, these numbers come from binomial coefficients, made famous in Pascal's triangle.) Check it out:

As you can see, in each case I've given the percentage of supporters for A found in the sample. There are 56 samples in which A has 62.5% support, 70 in which A has 50% support, and 56 in which A has only 37.5%. In this little experiment, therefore, we have 56 + 70 + 56 = 182 cases out of 256 in which A has support between 37.5% and 62.5%. Since 182/256 ≈ 71%, that is how often our random sample will indicate that A's support is between 37.5% and 62.5%. Observe that 37.5% = 50% − 12.5% and 62.5% = 50% + 12.5%. Since we set this up under the assumption that A's true support is 50%, our poll will be within ±12.5% of the true result about 71% of the time. Mind you, we have no idea in advance which type of sample we'll actually get when polling the electorate. We're playing the percentages, which is how it all works.

These results are pretty crude, since professional polls do much better than ±12.5% only 71% of the time, but we did this by asking only eight voters! A real poll would ask a few hundred voters, which suffices to get a result within ±3% about 95% of the time. That's why pollsters don't have to ask a majority of the voters their preferences in order to get results that are quite accurate. A relatively small sample can produce a solid estimate.

Bigger may be only slightly better

It's unfortunate that more people don't take a decent course in probability and statistics. That's why most folks are mystified by polls and can't understand why they work. They do work, as I've just shown you, within the limits of their accuracy. Pollsters can measure that accuracy and publish the limitations of their polls alongside their vote estimates. Every responsible pollster does this. (Naturally, everything I say is irrelevant when it comes to biased polls that are commissioned for the express purpose of misleading people. One should always treat skeptically any poll that comes directly from a candidate's own campaign staff.)

The controversy over sample size is constantly hyped by the statistically ignorant. More than fifty years ago, Phyllis Schlafly was harping on the same point:

The unscientific nature of the polls was revealed by Marvin [sic] D. Field, formerly with the Gallup poll and now head of one of the polls which picked Rockefeller to beat Goldwater in the California primary, who admitted to the press that he polled only 256 out of the 3,002,038 registered Republicans in California. He thus based his prediction on .000085 of Republican voters.

While her math is okay, Schlafly doesn't know what she's talking about. A sample size of 256 is quite good and should have produced a reliable snapshot of voter sentiment at the time the poll was conducted. In addition to getting the pollster's name wrong (it's Mervin), Schlafly neglected to mention that Rockefeller's wife had a baby just before the California primary, sharply reminding everyone about his controversial divorce from his first wife. You can't blame a poll for not anticipating a development like that. Otherwise, Schlafly's complaint about the poll is based on her ignorance about the sufficiency of sample sizes.

By the way, did you notice that Mervin Field's sample size was a power of 2? It would have occurred in the natural progression of samples that I modeled for you in our polling experiment. In my three different sampling examples, I doubled the sample size each time, going from 2 to 4 to 8, each time getting a significant increase in reliability. If you keep up the pattern, you get 16, 32, 64, 128, and 256. As you can see, Field went way beyond my little experiment, doubling my final sample of 8 an additional five times before he was satisfied he would be sampling enough voters for a good result.

Two caveats

There are a couple of things I should stress about the polling game we just played. First, of course, in real life we would not know the exact division of the voters beforehand. That's what we're trying to find out. It won't usually be something as nice and neat as fifty-fifty. However, as long as there's a real division between voters (in other words, not some 90% versus 10% rout), it won't be too difficult to poll enough voters to get an accurate profile.

Second, even a poll that is supposed to be within its estimated margin of error 95% of the time will be wrong and fall outside those bounds 5% of the time. That's one time in twenty. Therefore, whenever you see a political poll whose results seem way out of whack, it could be one of those flukes. Remember, polling is based on probability and statistics: it's accurate in the long run rather than in every specific instance. In a hot contest where lots of polls are taken, a candidate's campaign is likely to release only those polls that show the candidate in good shape. The 5% fluke factor may be just enough to keep hope alive among those people who believe everything they read.

Pollsters take their results with a grain of salt, so you should, too. But it's not because of sample size.

Friday, January 17, 2020

Sunday, October 29, 2017

Friday, November 11, 2016

Math 420 (fall 2016): Laplace Transforms

Every student of differential equations should know all of these Laplace transforms.


Wednesday, April 20, 2016

Wallis's formula for integrals of powers of sine and cosine

Please note that Wallis's formula is for definite integrals from 0 to π/2. You'll need to adjust the results for other intervals of integration (and for odd powers, for some intervals, you'll get zero because results in different quadrants cancel each other).


Friday, December 18, 2015

Frequently asked questions about Math 129

Q. What is Math 129?

A. Math 129 is an accelerated one-semester course that covers the entire algebra curriculum through the level of intermediate algebra. It contains the entire course content of Math 100 (Elementary Algebra) and Math 120 (Intermediate Algebra) and satisfies the prerequisite for any course with an intermediate algebra prerequisite.

Q. Who should take Math 129?

A. Math 129 is designed for exceptionally well-prepared students who can commit to two hours of math in class, five days a week, for an entire sixteen-week semester (plus many additional hours of work outside of class). Successful completion of Math 129 moves you rapidly through the math curriculum and accelerates your access to all of the courses that require algebra. Math 129 is also a good choice for students who previously did well in algebra but are re-entry students who have been away from school for several years and need a comprehensive algebra review.

Q. What is the format of Math 129?

A. Math 129 is a lecture course that meets Monday through Friday from 9:00 till 10:50 (with a 10-minute break). Many in-class quizzes are given; most of them are individual but a few are done as group work. Exams occur approximately every two weeks. Math 129 is not offered in an on-line or hybrid format.

Q. What is the most common problem that Math 129 students run into?

A. The time commitment. The class time (ten hours every week) is just the tip of the iceberg. Math 129 is really two courses in one semester, so expect to double your usual out-of-class study time and homework time. A common mistake is to treat Math 129 like an ordinary single course. It’s not.

Q. Is on-line homework or testing required?

A. Depending on the instructor, on-line homework could be required. However, the instructor assigned to teach Math 129 during the 2015-2016 school year does not require on-line homework. Homework is assigned from the textbook and collected on exam days. All exams are handwritten exams (no multiple choice tests and no Scantrons) and grading is based on mathematical accuracy, completeness, and the correct use of notation.

Q. What textbook is used for Math 129? Is it required?

A. The current textbook is Elayn Martin-Gay’s Beginning and Intermediate Algebra, 5th edition. The book is required and should be brought to class every day. The course will cover the entire book.

Q. What is the grading policy for Math 129?

A. Most of your grade is determined by the exams. Seven chapter tests are given (with the lowest score dropped), as is a comprehensive final exam (which is never dropped). The chapter tests are 70% of your grade, the final is 15% of your grade, and the homework and quizzes together constitute the remaining 15%.

Q. Is attendance mandatory?

A. Yes. Skipping class is the best way to fall behind and earn a low (or even failing) grade. Reasonable accommodation will be made for emergencies, but students are expected to make their best effort to attend every day. In keeping with Los Rios policy, students will be dropped for excessive absenteeism.

Q. Can students with learning disabilities take Math 129?

A. A good-faith effort is always made to provide reasonable accommodation. However, students who need to work slowly to maintain their accuracy should carefully take into account the challenges of an accelerated course. Time-and-a-half on exams is not difficult to provide, but Math 129 also includes dozens of short in-class quizzes that are immediately followed by demonstrations of their solutions; there is no satisfactory way to provide extra time on such quizzes.

Q. Is extra credit available?

A. Short answer: no. Longer answer: A few (but not many) exams or quizzes will offer optional problems that earn extra points for students who successfully solve them, but Math 129 is not a class where students can compensate for bad exam scores with projects or other alternative assignments.

Q. Is Math 129 offered during summer session?

A. Please tell me you’re kidding!

Friday, August 07, 2015

Getting help at ARC

Every student has had the experience of getting stuck. Not every student enjoys the experience of getting unstuck. In many cases, the solution is as simple as finding the right source of assistance.

1 Ask your teacher. Both in class and in office hours, your instructor will be available to answer questions and will expect questions. Although class time is devoted to learning new material and reviewing old material, it's not just a matter of your teacher telling you things. It should be a discussion, where you feel free to ask questions. Contact the instructor by e-mail when you need to.

2 Visit the Learning Resource Center. The college's LRC offers drop-in tutoring at many different times during the week and is usually fully operational by the second week of class. Don't hesitate to drop in, sign up, and take advantage of the LRC's tutors and assistants.

3 If you're a science, engineering, or math student, go to MESA, the Mathematics, Engineering, and Science Achievement program, which is housed in Room 131.

4 Form your own study group. Your instructor will be passing around a sign-up list for people who want to swap contact information to organize study groups.

5 Talk to a counselor about your education plan and get reliable information about the right classes to take.

6 Visit the Student Services building and get direct assistance from Financial Aid or Disabled Student Programs & Services or many other support programs.

Here is a comprehensive link to ARC Support Services, with individual links to many different programs. If you have not browsed this before, it will be worth a visit: Help

Monday, January 21, 2013

Know your support services

Get a helping hand
American River College has an extensive array of support services that you should not hesitate to take advantage of. In addition to the direct support you get from your instructor during class time and office hours, ARC offers drop-in tutoring in the Learning Resource Center, academic counseling in the administration building's Counseling Center, and a broad collection of services in the Student Services building. Most of these support services can be quickly found on the college's webpage titled ARC Support Services. There are currently 48 separate entries on that site.

Some of the programs, like the aformentioned Learning Resource Center, have their own website. The hours for drop-in tutoring are posted there. It is also the location for an open computer lab, the Reading Center, and the Writing Across the Curriculum program. Check out ARC's student services and learn about them now rather than later.

Saturday, August 25, 2012

Math Student Survival Guide

Although there is no magical secret for succeeding in a math class, there are some things that are universally true. Here's one:

You will not pass your math class
unless you spend enough time on it.

What does “enough time” mean? There is no one-size-fits-all answer. Some students breeze through math with a minimum of effort and others have to struggle. If you are taking prealgebra or algebra in college, you have probably struggled with math on occasion. If you plan to succeed, you must plan to find the necessary time for reading math, studying math, and doing your math homework. One important tip:

Do some math every day.

There will be days when you just don't have a lot of time. This happens. But do some math anyway, even if it's just 15 minutes of review or solving a homework problem. Doing some math every day is a recipe for success and maintains forward progress. “Saving up” your math for weekends is almost always a mistake.

Here's another important truth:

Students who get stuck need to get help.

It's amazing how many students quietly fail their classes instead of getting help. Community colleges are all about helping students. Even when state funding is tight and budgets are cut, we scramble to provide assistance opportunities to our students. Here's a list:
I apologize for all of the exclamation points. Teachers get a little excited when repeating advice for the millionth time, but be aware of this: We have a lot of experience seeing what works and what doesn't. Your mileage may vary, but some (maybe a lot) of what your teacher says may apply to you.

Be an active learner and take charge of your learning. The teacher can't teach you without your consent and your active participation. Take responsibility for your education.

Monday, November 22, 2010

The intersecting sphere and cylinder

The bonus task I gave you last week involved looking at the curve created by the intersection of a sphere and cylinder. For purposes of this demonstration, I will assume that the sphere has radius 4 and the cylinder correspondingly has diameter 4. Thus the sphere is x2 + y2 + z2 = 16 and the cylinder is (x − 2)2 + y2 = 4.

We will now show that the self-intersection curve crosses itself at right angles at the point (4, 0, 0).

Here is one way to do it:

We parameterize the path by r = xi + yj + zk, where we need to find appropriate functions for x, y, and z. Let's rewrite everything in cylindrical coordinates, because that form is more useful in this situation. The sphere's equation is r2 + z2 = 16, the cylinder is r = 4 cos θ, and the intersection curve we are looking for is r = (r cos θ)i + (r sin θ)j + zk.

Since the cylinder requires that r = 4 cos θ, we can substitute for r in the expression for the curve r. We obtain r = (4 cos θ cos θ)i + (4 cos θ sin θ)j + zk.

What can we do about that z? According to the sphere, we must have z = (16 − r2)1/2. Substituting r = 4 cos θ into the equation for z, we get z = (16 − (4 cos θ)2)1/2 = (16 − 16 cos2 θ)1/2 = 4 sin θ. We can now write our curve with θ as our parameter:

r(θ) = (4 cos θ cos θ)i + (4 cos θ sin θ)j + (4 sin θ)k.

Observe that r(0) = 4i and r(π) = 4i, indicating that the curve intersects itself for θ = 0 and θ = π. If we find the tangent vectors at these two points, the angle between them is supposed to be a right angle. Of course, the easiest way to find tangent vectors is to use r′(θ), the velocity vector:

r′(θ) = (−8 cos θ sin θ)i + 4( cos2 θ − sin2 θ)j + (4 cos θ)k.

Plug in θ = 0 and θ = π to find the pertinent tangent vectors:

r′(0) = 4j + 4k
r′(π) = 4j − 4k.

If we dot these vectors we get zero. The vectors are perpendicular, so the intersection curve crosses itself at ninety degrees. QED.

Sunday, August 22, 2010

Computing a weighted average

It's how your grades are computed

It's not too difficult to turn the points you earn on an assignment into a percentage. You just divide them by the total number of points possible and move the decimal point two places. For example, if you get 15 points on a 20-point quiz, your percentage score is 15/20 = 0.75 = 75%. No problem.

However, I do not use just a point system to figure grades for a class. I use a weighted average of the scores you earn in three categories: HWQ (homework & quizzes, lumped together), Exams (the chapter tests), and Final (the comprehensive final exam at the end of the semester). For example suppose that HWQ is 15% of your grade, Exams are 70% of your grade, and the Final is 15%. (If you check, you'll see that the percentages add up to 100%, as they should. I may vary these weights depending on the number of exams I give.)

Suppose that you earn 72% of all the points possible on the homework and quizzes, you average 84% on the exams, and you get 79% on the final. Did you earn a B? Let's find out. The weighted average is computed by multiplying each score by its percentage weight and then adding everything together. Recall that the percentage weight of HWQ is 15% of your grade, which we can write as the decimal 0.15. For Exams we have 70% or 0.70, and the Final is 15% or 0.15, just like HWQ. The formula for grades is

Course Grade = HWQ*0.15 + Exams*0.70 + Final*0.15.

Using the scores we pretended you got during the semester, your course grade would be given by

Course Grade = 72%(0.15) + 84%(0.70) + 79%(0.15) = 81.45%.

Congratulations! You earned a B for the class, with just a little bit to spare.

It's not too surprising. Most of the weight of the grade is on Exams. Homework and quizzes get only 15% because I treat them more as a learning experience. The final gets only 15% because it's only one exam. While other instructors might choose to give the final more weight, I prefer to limit the importance of any single exam.

In general

If you ever want to compute a weighted average for any reason, the main thing to keep in mind that the weights have to add up to 100%. If, for example, you wish to compute the weighted average of four numbers, X1, X2, X3, and X4, you can do it as long as the weights, w1, w2, w3, and w4, add up to 100%. Then you compute the sum of their respective products:

Weighted average = X1w1 + X2w2 +X3w3 +X4w4.

That's all there is to it. Of course, if you choose to use equal weights, so that w1 = w2 = w3 = w4 = 25% = 0.25, your result will just be the usual arithmetic average. (It would be simpler to add up the numbers and divide by 4.)

I'll leave it to you to figure out what to do if there are more than four numbers for which you wish to compute a weighted average.

Wednesday, June 14, 2006

Area under a curve

Math 400: Problem 1.1.2

How far does an object travel in 15 seconds if its velocity is given by the function v(t) = 20 + 7 cos t ft/sec? It helps to recall that we can represent the distance traveled by the area under a velocity curve. Let's look at a graph where t is the horizontal axis and y = v(t) is the vertical axis. The time units are in seconds and the velocity units are feet per second (ft/sec).

In the following graph, the red curve represents the velocity function. As we can see, the traveling object alternately slows down and speeds up (the effect of the cosine term), never going more slowly than 13 ft/sec and never going faster than 27 ft/sec. The area under the curve for time values between t = 0 and t = 15 (in blue) represents the distance traveled, if only we could figure it out. Learning how to do that is one of the goals of calculus. However, it's not too difficult to get a rough estimate.
In the second figure, I've drawn in the lines y = 13 and y = 27 to mark off the low and high points of the velocity curve. If the object were traveling only 13 ft/sec for 15 sec, then the distance it travels would be just (13 ft/sec)(15 sec) = 195 ft. That's the area of the blue rectangle that represents a lower bound on the area under the red curve. On the other hand, if the object always traveled 27 ft/sec (the upper bound on velocity), then the distance traveled would be (27 ft/sec)(15 sec) = 405 ft.

We have now figured out that the distance traveled is somewhere between 195 ft and 405 ft. That's a pretty broad range, to be sure, but we could make it better by using less extreme bounds. For example, we could try drawing triangles inside the humps of the cosine curve to estimate the area we left off while computing the lower bound. That should help. At some point, though, we would probably want something better than computing lots of different estimates. When we have the right calculus tools available, we'll be able to show that the exact distance traveled is 300 + 7 sin 15 ≈ 304.55 ft.

Tuesday, February 21, 2006

What the dot product means

In elementary physics we learn the work equals force times distance:

W = F × d

But it's not quite that simple when we start considering vector quantities instead of scalar. When direction enters into it, we have to account for the fact that work can be performed by a force that is not entirely aligned with the direction of the distance traveled. The cosine of the angle of divergence comes to the rescue:

W = (F cos θ) × d

Up to this point we have stuck with scalar computations. Vector notation is more powerful and useful. Let's redraw the previous figure in terms of nonparallel vectors F and d. Now we can restate our results.

If we use standard vector notation and consider what work actually equals, we end up looking at the following scalar product, which is the parallel component of F times the length of d:

W = (||F|| cos θ) × ||d||

This is something we recognize as the dot product of the vectors F and d. In fact, this is probably how the dot product came to be defined as a useful quantity in mathematics. Since we have learned to compute dot products algebraically by multiplying corresponding components and adding the results, we can compute quantities like work without having to compute angles and cosines, making the dot product into something that is easy to compute as well as useful.