Weird Power Series Question o.O - Calc BC

Test Preparation AP Test Preparation
1

<p><a href="http://img585.imageshack.us/img585/8494/calcbcar.jpg%5B/url%5D"&gt;http://img585.imageshack.us/img585/8494/calcbcar.jpg&lt;/a&gt;&lt;/p>

<p>Uhh any ideas? I got the answer, need an explanation please.</p>

2

<p>i’d like to kno also b/c i have no effing clue haha</p>

3

<p>Just use the ratio test. It’s fairly straightforward.</p>

4

<p>I did that and got D though. :&lt;/p>

5

<p>i dont get the a0 and (2n+1)/(3n-1) part</p>

6

<p>What is the correct answer?</p>

7

<p>The correct answer is C. Could someone elaborate?</p>

8

<p>C. Here is my hastily written solution <a href=“http://imgur.com/NfOqg.jpg[/url]”>http://imgur.com/NfOqg.jpg&lt;/a&gt;
Note that at one point you end up dividing a<em>n by a</em>n, which is obviously 1. I didn’t make that step very clear in my picture, sorry. (2nd to 3rd line)</p>

9

<p>That’s what I got.
I don’t think my methodology is correct though. But I just used the ratio test. The 2n+1 cancel and 3n-1 cancel leaving you with 3/2. Idk if the first part of the problem is necessary.</p>

10

<p>I got C, too, but didn’t use the a0 term at all. </p>

<p>you should get to a point where you have to find the limit as n approaches infinity of (2n+1)(x-2)/(3n-1), where you then just foil the numerator and divide by n to get lim as n->infinity of (2x-4)/3. From -1<(2x-4)/3<1, -3<(2x-4)<3, 1<2x<7, 1/2<x<7/2. Radius of convergence is half the total, 3/2.</p>

11

<p>The correct Answer is C) 3/2</p>

<p>Work:</p>

<p>By Ratio test, we can find the radius:</p>

<p>lim n–> infinity | (an+1) / an | < 1</p>

<p>lim n–> infinity | [[[(2(n+1) + 1) / (3(n+1) - 1)] * an] * (x-2)^(n+1)] / [an * (x-2)^n]| < 1
We basically took the definition of an and added so it becomes the n+1th term</p>

<p>an cancels out, and continue to simplify, the </p>

<p>|x-2| lim n–> infinity |(2n+3)/(3n+2)| < 1</p>

<p>limit is just the constants of the ratio</p>

<p>=</p>

<p>2/3 |x-2| < 1</p>

<p>radius = 3/2</p>

<p>Sorry for the messy explanation guys, hope this helps!</p>

12

<p>Because a<em>n = (2n+1)a</em>(n-1) / (3n-1) and a<em>o = 5, a</em>1 = 5(2n+1)/(3n-1).</p>

<p>Thus, a<em>2 = 5[(2n+1) / (3n-1)]^2, a</em>3 = 5[(2n+1) / (3n-1)]^3, etc.</p>

<p>Thus, a_n = 5[(2n+1) / (3n-1)]^n</p>

<p>Thus, the series = 5[(2n+1) / (3n-1)]^n * (x-2)^n</p>

<p>Root test: lim n—>∞ nth_rt( | 5[(2n+1) / (3n-1)]^n * (x-2)^n | )</p>

<p>lim n—>∞ | [(2n+1) / (3n-1)] * (x-2) |</p>

<p>lim n—>∞ = 2(x-2)/3</p>

<p>2(x-2)/3 = 1</p>

<p>(x-2) = 3/2</p>

13

<p>Thanks a bunch guys! I was totally thrown off, especially by the a sub 0 = 5, which apparently you don’t need haha. Thanks again.</p>

14

<p><a href=“http://img90.imageshack.us/img90/889/calclimit.jpg[/url]”>http://img90.imageshack.us/img90/889/calclimit.jpg&lt;/a&gt;&lt;/p&gt;

<p>Just one more haha, could someone explain how number III is not true? There is a x^2 in the denominator…</p>

15

<p>Vertical asymptotes correspond to the zeroes of the denominator of a rational function. Sin is not a rational function, ergo no asymptote. Also, you can think about sin graphically and realize that sin of anything has to be between 1 and -1. A function would approach +/- ∞ at a vertical asymptote.</p>

16

<p>so what is the answer to that question? Just I…or I and II. How could there be a horizontal asymptote at both?</p>

17

<p>I would think just I. Horizontal asymptotes have to do with end behavior. As X approaches infinity or negative infinity, the argument of the sin in effect becomes zero, and we all know sin(0) is 0.</p>

<p>EDIT: yeah, I was right</p>

<p><a href=“sin((x+1)/x^2) from -20 to 20[/url] - Wolfram|Alpha”>sin((x+1)/x^2) from -20 to 20 - Wolfram|Alpha;

18

<p>thank you! that’s what i thought, but i wasn’t sure</p>

19

<p>Thanks again guys! Now, how about this logistic growth one? :&lt;/p>

<p><a href=“http://img856.imageshack.us/img856/9207/calclogistic.jpg[/url]”>http://img856.imageshack.us/img856/9207/calclogistic.jpg&lt;/a&gt;&lt;/p&gt;

<p>I got it right, but kind of guessed… any clear explanation from someone please?</p>

20

<p>eobaggs whats the answer to that one?</p>