Ap Calculus BC 2002 Form B Question 6??

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Ap Calculus BC 2002 Form B Question 6??

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Answer 1 · 2018-04-28T03:41:58

a) We have:

$ln (\frac{1}{1 + 3 x}) = ln (\frac{1}{1 - (- 3 x)})$ $ln(1/(1 + 3x)) = ln(1/(1 - (-3x)))$

If we let $x = 3 x$ $x = 3x$ in the second series, we get the maclaurin series is $\infty \sum n = 1 \frac{{(3 x)}^{n}}{n}$ $sum_(n = 1)^oo (3x)^n/n$

b) This is simply a case of $ln (\frac{1}{1 - x})$ $ln(1/(1 - x))$ with $x = - 1$ $x = -1$ , so we have:

$\infty \sum n = 1 \frac{{(- 1)}^{n}}{n} = ln (\frac{1}{1 - (- 1)}) = ln (\frac{1}{2})$ $sum_(n = 1)^oo (-1)^n/n = ln(1/(1 - (-1))) = ln(1/2)$

c) We start by noticing that the series on the left is an alternating series, so as long as the absolute value of the terms are getting smaller and smaller and the series is converging to $0$ $0$ it will converge.

This will hold true whenever $p > 0$ $p > 0$ . However, by the p-series, whenever $c$ $c$ in $\frac{1}{x^{c}}$ $1/x^c$ is larger than $1$ $1$ , the series converges. We want the p-serices with $\frac{1}{n^{2 p}}$ $1/n^(2p)$ to diverge, therefore, we will require $2 p \leq 1$ $2p ≤ 1$ , or $p \leq \frac{1}{2}$ $p ≤ 1/2$ . Thus $p$ $p$ must satisfy $0 < p \leq \frac{1}{2}$ $0 < p ≤ 1/2$ . Many individual values of $p$ $p$ are possible, as long as they're within that interval (e.g. $\frac{1}{3}$ $1/3$ , $\frac{1}{2}$ $1/2$ , $\frac{2}{5}$ $2/5$ , etc).

d) This is another application of the p-series test. We immediately note that $\frac{1}{n^{2 p}}$ $1/n^(2p)$ converges if $p > \frac{1}{2}$ $p > 1/2$ , and $\frac{1}{n^{p}}$ $1/n^p$ diverges if $p \leq 1$ $p ≤ 1$ . Thus any value of $p$ $p$ that lies on $\frac{1}{2} < p \leq 1$ $1/2 < p ≤ 1$ will satisfy.

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Ap Calculus BC 2002 Form B Question 6??

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