<p>Hey guys,
I keep doing pretty well on the free-response section of the Calc BC except when I get to the series part. I have no idea how to tackle those.
Does anybody know which review book will help me prepare for that part?
I tried PR, and it didn't help much.
Well, I get the first two parts of the series questions, but the last two parts are hard in almost every exam. I want to get those last two parts of the questions.</p>
<p>Any suggestions?</p>
<p>Try Cliffs. It has a decent review of that concept. Also, just search for it on the internet.</p>
<p>Try Peterson's. It was really helpful.</p>
<p>do u have any particular questions we mite help with?</p>
<p>I have the most trouble with series on the free-response as well... I probably should've paid attention in class during that chapter.</p>
<p>Barron's has been pretty helpful so far, though I haven't finished reading the sequences/series chapter.</p>
<p>Hey, thanks a lot guys.
I will try all, Peterson's, Cliff's, and Barrons.
Yes, it's in the 2000 Free Response Section, but they have the explanations for it, so it shouldn't be a problem to understand the problem.
Thank you, nikolayh</p>
<p>Yeah, I thought Barron's was pretty good. Watch out for PR's though, I heard that they tell you to just use the first omitted term (which I think only works for alternating series) for estimating error, rather than using Lagrange. I'd think that this could potentially be a big problem, since all the steps of the FRP for such a question would be wrong.</p>
<p>I remember seeing a Lagrange Error Bound question on one of the FRQs that I did a few days ago, so you definitely need to know that.</p>
<p>Edit: question 6C on the 2004 FRQ</p>
<p>wow, thats weird, i really didn't think that would be on there. But, once you figure it out, its pretty easy.</p>
<p>anyone care to explain because i don't really understand error bounds.</p>
<p>When you use a Maclaurin or Taylor series to approximate a function, you have a level of error based on the number of terms you use (a series taken to 8 terms will have less of an error than a series only taken to 3). The error bound is the maximum possible error that a series has at a given point. For alternating series, this error bound is equal to the absolute value of first omitted term (1 - 1/2 + 1/4 - 1/8 will have an error bound of 1/16, if it were representing a function).</p>
<p>However, if the series doesn't alternate, you need to use Lagrange Error Bound to figure out the maximum possible error. The formula isn't too hard, but the formatting here would make it hard to read here (try <a href="http://mathworld.wolfram.com/LagrangeRemainder.html)%5B/url%5D">http://mathworld.wolfram.com/LagrangeRemainder.html)</a>. The idea is to take the first omitted term, but take the (n+1)th derivative of that term at C, with C being a number between a and x (a is where the Taylor series is centered, and x is the value for which you're trying to get an approximate value with the Taylor series).</p>
<p>So, let's say I want to find the error bound of (1 + x + x^2/2 + x^3/3!) for the value of e (I'll use the above series - The Maclaurin series for e^x). the first omitted terms is (e^c)*(x^4/4!) - [Remember that the nth derivative for e^x is just e^x. We don't write this for each term in the Maclaurin series because e^0 is 1] Now, since we're using a Maclaurin series (a=0) to represent e (x=1), c has to be between 0 and 1 inclusive. Also, since we're finding the maximum possible value for that term, we'll use the maximum value for e^c: e (c=1). So to get our error, we'll evaluate (e/4!), which is .11326 or so.</p>
<p>
[quote]
For alternating series, this error bound is equal to the absolute value of first omitted term (1 - 1/2 + 1/4 - 1/8 will have an error bound of 1/16, if it were representing a function).
[/quote]
How did you get 1/16 from that? I am sorry, but this error is something I haven't worked on yet.</p>
<p>
[quote]
the first omitted terms is (e^c)*(x^4/4!) - [Remember that the nth derivative for e^x is just e^x. We don't write this for each term in the Maclaurin series because e^0 is 1]
[/quote]
Can do another example that is a bit harder? Thanks for your time.</p>
<p>
[quote]
How did you get 1/16 from that? I am sorry, but this error is something I haven't worked on yet.
[/quote]
For convergent alternating series, the error is less than the absolute value of the first omitted term.</p>
<p>The series he used was 1 - 1/2 + 1/4 - 1/8 + ...
therefore, the first omitted term is -1/16. The absolute value of that is 1/16.</p>