BC Series Help

Test Preparation AP Mathematics & Computer Science
1

<p>First of all, just to make sure, these are the only things that BC covers that AB doesn't, right?</p>

<p>-parametric/polar/vector calculus (BC topic)
-sequence/series calculus (BC topic)
-indeterminate forms: L'hopital's & improper integrals (BC topic)</p>

<p>And these extra parts to AB topics:
-length of a curve & surface area of a revolution
-Euler's method
-Newton's method
-logistic equations
-antiderivation by parts & partial fractions</p>

<p>I hope I haven't missed anything.</p>

<p>Anyway, I'm taking AB as a course and self-studying BC. My AB class is done learning things already. I've been able to understand most of the stuff in BC except series. I've looked at so many websites and Youtube videos, but I'm still confused (which may be partly because I suddenly get really sleepy every time I try to learn about them...). </p>

<p>Does anyone have an easy-to-understand resource will help me? Or would someone really awesome at calculus be willing to answer my questions until I get it?</p>

2

<p>I’d be glad to help; pose your questions.</p>

3

<p>Great, thank you!</p>

<p>Is a power series a type of geometric series? Or is a geometric series a type of power series?
What is a Taylor series compared to a power series? Is a Taylor series the power series for a particular function?</p>

<p>I have more questions, I just can’t think of many at the moment.</p>

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<p>Also, what’s the relationship between L’hopital’s rule/improper integrals and series?</p>

5

<p>Improper integrals are really primarily helpful for performing the Integral Test for Series, which is really primarily used for establishing the validity of the p-Series Tests.</p>

<p>Geometric Series and power series are really different entities. A Geometric Series is a series where each term of the series is related by a common ratio, r, where the convergence or divergence of that particular series is determined by the size of |r|. Most geometric series are really comprised of just numerical values, although they don’t have to be.</p>

<p>A power series, on the other hand, is a series of terms of the form a0 + a1(x - c) + a2(x - c)^2 + a3(x - c)^3 + …, where each of the a0, a1, a2, a3, etc. are different constants. A Taylor series is one particular type of power series that represents a function f centered at x = c, where a0 = f(c), a1 = f ‘(c), a2 = f "(c) / 2!, a3 = f "’(c) / 3!, and in general, an = the nth derivative of f (written as f(n)(x) where the (n) is a subscript) evaluated at c. The idea is that the series and the function f share the same concavity, and so over a certain interval, the function and the series converge to that same location, given enough terms.</p>

<p>I can break some of those pieces down further if need be.</p>

6

<p>Thank you very much! Your explanations are very clear.</p>

<p>I’m still a bit confused about the relationship between geometric series and power series. One of the resources I’ve been using ([Power</a> Series](<a href=“http://www.sosmath.com/calculus/powser/powser01.html]Power”>Power Series)) implies that there’s a relationship between the two (to be honest, I’m not able to follow their logic entirely). Does such a relationship only exist in special cases?</p>

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<p>A lot of time when developing power series, many authors (including the author of that webpage) will look at particular series simply to illustrate examples of power series that have a common ratio. And in some cases, they use the knowledge of geometric series to actually create the power series in the first case.</p>

<p>For instance, if you’re looking at a geometric series, the sum for any convergent geometric series can be represented as FIRST/(1 - RATIO), where FIRST represents the first term of the series and RATIO represents the common ratio, r. This series could also be represented as FIRST + FIRST * RATIO + FIRST * RATIO ^ 2 + FIRST * RATIO ^ 3 + …</p>

<p>Well, some functions are already set up in this format. For instance, f(x) = 1 / (1 - x) is set up in that exact same format, where FIRST = 1 and RATIO = x. So connecting these ideas, we could represent 1 / (1 - x) as 1 + 1<em>x + 1</em>x^2 + 1*x^3 + …</p>

<p>And that’s really all the authors tend to do when connecting these ideas. They say, “Here’s what we already know about a geometric series that converges. Let’s construct some other series that same way.” Not all power series are constructed this way, but they use the background of geometric series as a way to introduce students to the concept of power series. The key is to recognize that not all power series are actually connected to geometric series.</p>

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<p>Was re-reading through my post #5, and in f(n)(x), it should read that (n) is a *superscript<a href=“like%20an%20exponent”>/I</a>, not a subscript.</p>

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<p>Ohh thank you. No worries about the f(n)(x) thing, I knew what you meant. (:</p>

<p>The number of convergence/divergence tests overwhelms me. Sometimes I don’t know when to use which one. These tests I find pretty straightforward: geometric series, nth term, integral/p-series, and alternating series. </p>

<p>However, I don’t always know when to use these tests: ratio, direct comparison, and limit comparison. Are there general guidelines for using those 3 tests? Can the ratio test be used for testing any series (though other tests may be easier)?</p>

<p>Is the ratio test the only one that tests for absolute convergence directly in the formula?</p>

<p>Also, specifically about the AP test, will I have to know trig substitution? The AP outline (<a href=“College Board - SAT, AP, College Search and Admission Tools”>College Board - SAT, AP, College Search and Admission Tools) doesn’t mention it, but it’s mentioned in the syllabus which my teacher gave me to work from.</p>

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<p>Quite honestly, you can usually ignore direct comparison. The whole premise behind it is that if you have a series whose terms are all smaller than (or equal to) a known convergent series, then the series converges. Similarly, if you have a series whose terms are all larger than (or equal to) a known divergent series, then the series diverges.</p>

<p>Limit comparison, however, does the vast majority of the series that direct comparison does, and it tends to be just as easy a test to work with. I tend to use limit comparison the most when I have series that look like p-series but that aren’t quite p-series. So, for instance, if you have Sigma(1/n^2 - 1), that series looks suspiciously like Sigma(1/n^2), which is known to converge. By Limit Comparison, you get a limit of 1, which is both positive and finite, so the original series converges also. Note that direct comparison would not apply here: since n^2 - 1 < n^2, 1/(n^2 - 1) > 1/(n^2), and we can’t conclude anything about a series with terms that are larger than a known convergent series.</p>

<p>Ratio Test will ultimately end up being your best friend when you look at intervals of convergence. You tend to use it any time you’ve got a bunch of different kinds of functions all intermingled: factorials, powers, linear terms, you name it. I would say that if you don’t know what test to use, Ratio Test is probably the test to use.</p>

<p>The Ratio Test is inconclusive whenever the Ratio you get is equal to 1, and accordingly, it doesn’t work in those cases. When you start looking at intervals of convergence, you’ll see that there are cases where the Ratio Test fails to tell you what’s going on. Other than that, I can’t think of a case offhand where you can’t use it, but the limits might not be fun.</p>

<p>The Root Test also tests for absolute convergence directly in the formula as well.</p>

<p>Trig substitution isn’t directly on the AP test, although I saw a question one year on the exam, where you were directed to perform a substitution of x = sin (theta) in a definite integral and asked what the resulting integral would be in terms of theta. Although when most people are talking about trig substitution, they’re usually referring to cases where you have to determine which trig function to substitute for x.</p>

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<p>Math Prof, you don’t know how much you’ve helped me. I actually feel kind of comfortable with series now. Thank you thank you thank you!</p>