Supplemental AP Calculus BC Final Exam

Let \(f\) be the function that has derivatives of all orders (i.e. can be differentiated any number of times). Let \(3-6(x-1)^2 + 5(x-3)^3 - 2(x-1)^4\) be the fourth-degree Taylor polynomial for \(f\) about \(x=1\). Does \(f\) have a local minimum, local maximum, or neither at \(x=1\)? Explain how you know.

From the equation, the following can be determined with the Taylor Series:

\[ \frac{f(1)}{0!} = 3, \frac{f'(1)}{1!} = 0, \frac{f''(1)}{2!} = -6 \]

The function has a critical point at \(x=1\) since \(f'(1) = 0\). Since \(f''(1) = -12 < 0\), the function has a local maximum at \(x=1\).
Evaluate the following integrals:
1. \(\int_{-1}^1 \frac{1}{x^2} dx\)
  
  \[ \begin{aligned} &= \int_{-1}^0 \frac{1}{x^2} dx + \int_{0}^1 \frac{1}{x^2} dx\\ &= -\frac{1}{x} \Biggr|_{-1}^{0} + \frac{1}{x} \Biggr|_{0}^{1}\\ &= \text{Divergent} \end{aligned} \]
2. \(\int_{-\infty}^{\infty} \frac{1}{1+x^2} dx\)
  
  \[ \begin{aligned} &= \tan^{-1}(x) \Biggr|_{-\infty}^{\infty}\\ &= \lim_{x \to \infty} \tan^{-1}(x) - \lim_{x \to -\infty} \tan^{-1}(x)\\ &= \frac{\pi}{2} - \frac{-\pi}{2}\\ &= \pi \end{aligned} \]
3. \(\int_{-3}^3 \frac{x}{1+|x|} dx\)
  
  \[ \begin{aligned} f(x) &= \frac{x}{1+|x|}\\ f(-x) &= \frac{-x}{1+|-x|}\\ &= \frac{-x}{1+|x|}\\ f(x) &= f(-x) \text{ (Odd function)} \\ \therefore \int_{-3}^3 \frac{x}{1+|x|} dx &= 0 \end{aligned} \]
4. \(\int \frac{dx}{e^x \sqrt{4-e^{-2x}}}\)
  
  \[ \begin{aligned} u = e^{-x}&, du = -e^{-x} dx\\ \int \frac{dx}{e^x \sqrt{4-e^{-2x}}}dx &= \int \frac{-du}{\sqrt{4-u^2}}\\ &= -\sin^{-1}(\frac{u}{2}) + C\\ &= -\sin^{-1}(\frac{e^{-x}}{2}) + C \end{aligned} \]
Find the sum of the following series:
1. \(2 - \sqrt{2} + 1 - \frac{\sqrt{2}}{2} + \frac{1}{2} - \dots\)
  
  \[ \begin{aligned} &= \sum_{n=0}^{\infty} 2(\frac{1}{2})^n - \sum_{n=0}^{\infty} \sqrt{2}(\frac{1}{2})^n\\ &= \frac{2}{1-\frac{1}{2}} - \frac{\sqrt{2}}{1-\frac{1}{2}}\\ &= 4 - 2\sqrt{2} \end{aligned} \]
2. \(\sum_{n=0}^{\infty} \frac{2^{n-1}}{n!}\)
  
  \[ \begin{aligned} &= \sum_{n=0}^{\infty} \frac{2^n}{2n!}\\ &= \frac{1}{2} \sum_{n=0}^{\infty} \frac{2^n}{n!}\\ \end{aligned} \]
  
  Using the Maclaurin Series for \(e^x\):
  
  \[ \begin{aligned} e^x &= \sum_{n=0}^{\infty} \frac{x^n}{n!}\\ \frac{1}{2} \sum_{n=0}^{\infty} \frac{2^n}{n!} &= \frac{e^2}{2} \end{aligned} \]
3. \(3 \sum_{n=0}^{\infty} \frac{(-1)^n (\sqrt{3})^{2n+1}}{2n+1}\)
  
  Given the Maclaurin Series for \(\tan^-1(x)\):
  
  \[ \begin{aligned} \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} &= \tan^{-1}(x)\\ 3\sum_{n=0}^{\infty} \frac{(-1)^n (\sqrt{3})^{2n+1}}{2n+1} &= 3\tan^{-1}(\sqrt{3}) \end{aligned} \]
4. \(\sum_{k=1}^{\infty} \ln(\frac{k}{k+1})\)
  
  \[ \begin{aligned} &= \sum_{k=1}^{\infty} \ln(k) - \ln(k+1)\\ &= \ln(1) - \ln(2) + \ln(2) - \ln(3) + \ln(3) - \ln(4) + \dots\\ &= \lim_{k \to \infty} \ln(k) - \ln(k+1)\\ &= -ln(\infty)\\ &= -\infty \text{ (Divergent)}\\ \end{aligned} \]
5. \(\sum_{n=1}^{\infty} \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}\)
  
  \[ \begin{aligned} &= \sum_{n=1}^{\infty} \frac{\sqrt{n+1}}{\sqrt{n(n+1)}} - \frac{\sqrt{n}}{\sqrt{n(n+1)}}\\ &= \frac{\sqrt{2}}{\sqrt{1 \cdot 2}} - \frac{\sqrt{1}}{\sqrt{1 \cdot 2}} + \frac{\sqrt{3}}{\sqrt{2 \cdot 3}} - \frac{\sqrt{2}}{\sqrt{2 \cdot 3}} + \frac{\sqrt{4}}{\sqrt{3 \cdot 4}} - \frac{\sqrt{3}}{\sqrt{3 \cdot 4}} + \dots\\ &= 1 \text{ (Telescoping series)} \end{aligned} \]
For what values of \(r\) is the integral \(\int_{0}^{\infty} \frac{1}{x^r} dx\) convergent?

Between \(0\) and \(1\), the integral is convergent for \(r > 1\). For \(r \leq 1\), the integral is divergent.

Between \(1\) and \(\infty\), the integral is convergent for \(r < 1\). For \(r \geq 1\), the integral is divergent.

Therefore, no combination of \(r\) values will make the integral convergent. One section of the integral will always be divergent, so the entire integral is divergent since the integral \(\int_{0}^{\infty} \frac{1}{x^r} dx\) is made up from \(\int_{0}^{1} \frac{1}{x^r} dx + \int_{1}^{\infty} \frac{1}{x^r} dx\).
The sum of an infinite geometric series with common ratio \(r\) is 15, and the sum of the squares of the terms of this series 45. What is the first term of the series?

\[ \begin{aligned} \begin{cases} \frac{a}{1-r} = 15\\ \frac{a^2}{1-r^2} = 45 \end{cases}\\ a &= 15(1-r)\\ 45 &= \frac{(15(1-r))^2}{1-r^2}\\ 45(1-r^2) &= 15^2(1-r)^2\\ 270r^2 - 450r + 180 &= 0\\ 3r^2 - 5r + 2 &= 0\\ (3r - 2)(r - 1) &= 0\\ r &= \frac{2}{3}, 1 \text{ (Cannot have } r = 1 \text{)}\\ a &= 15(1-\frac{2}{3})\\ &= 5 \end{aligned} \]
Determine if the following series are absolutely convergent, conditionally convergent, or divergent.
1. \(\sum_{n=2}^{\infty} \frac{(-1)^n}{n(\ln{n})^3}\)
  
  Using the integral test:
  
  \[ \begin{aligned} \int_{2}^{\infty} \frac{1}{x(\ln{x})^3} dx &= \frac{1}{2\ln(2)^2} \text{ (Convergent)}\\ \end{aligned} \]
  
  Passes the alternating series test, therefore it is absolutely convergent.
2. \(\sum_{n=1}^{\infty} \frac{(n!)^2}{(2n)!}\)
  
  Using the ratio test:
  
  \[ \begin{aligned} \lim_{n \to \infty} \frac{(n+1)!^2}{(2n+2)!} \cdot \frac{(2n)!}{(n!)^2} &= \lim_{n \to \infty} \frac{(n+1)^2}{(2n+2)(2n+1)}\\ &= \frac{1}{4} \text{ (Absolutely Convergent)}\\ \end{aligned} \]
3. \(\sum_{n=1}^{\infty} a_n \text{ where } a_1=2, a_{n+1} = \frac{2n+1}{5n-4}\)
  
  Using the ratio test:
  
  \[ \begin{aligned} \lim_{n \to \infty} \frac{a_{n+1}}{a_n} &= \lim_{n \to \infty} \frac{2n+1}{5n-4} \cdot \frac{5n-9}{2n-1}\\ &= \frac{2}{5} \text{ (Absolutely Convergent)}\\ \end{aligned} \]
Let \(f(x) = 1 + \frac{x}{2} + \frac{x^2}{4} + \frac{x^3}{8} + \dots = \sum_{n=0}^{\infty} \frac{x^n}{2^n}\) for \(-1 \leq x \leq 1\). Determine \(\sqrt{e^{\int_{0}^{1} f(x) dx}}\).

\[ \begin{aligned} \int_{0}^{1} f(x) dx &= \sum_{n=0}^{\infty} \frac{x^n}{2^n} dx\\ &= \sum_{n=0}^{\infty} (\frac{x}{2})^n \\ &= \frac{2}{2-x}\\ &\int_{0}^{1} \frac{2}{2-x}\\ u = 2-x&, du = -dx\\ &= \int_{1}^{2} \frac{2}{u} du\\ &= 2\ln(u) \Biggr|_{1}^{2}\\ &= 2\ln(2) + 2\ln(1)\\ &= 2\ln(2)\\ \sqrt{e^{2\ln2}} &= e^{\ln2}\\ &= 2 \end{aligned} \]
Consider the following series

\[\frac{1}{2} - \frac{1}{3} + \frac{1}{2^2} - \frac{1}{3^2} + \frac{1}{2^3} - \frac{1}{3^3} + \dots\]
1. Explain why it is not possible to use the alternating series test on this series.
  
  The series is not monotonically decreasing since the terms alternate between \(\frac{1}{2^n}\) and \(\frac{1}{3^n}\).
2. This series actually does converge! Find the sum of the series. (Hint: Split the series up)
  
  \[ \begin{aligned} &= \sum_{n=1}^{\infty} \frac{1}{2^n} - \sum_{n=1}^{\infty} \frac{1}{3^n}\\ &= \frac{\frac{1}{2}}{1-\frac{1}{2}} - \frac{\frac{1}{3}}{1-\frac{1}{3}}\\ &= 1 - \frac{1}{2}\\ &= \frac{1}{2} \end{aligned} \]
Find the values of \(p\) which make the following series converge:
1. \(\sum_{n=1}^{\infty} (-1)^n \frac{1}{n + p}\)
  
  For any value of \(p\), the series will converge since it is an alternating series.
2. \(\sum_{n=1}^{\infty} n(1+n^2)^p\)
  
  The largest dominating term is \(n^{2p+1}\), therefore the series will converge for \(2p + 1 < -1\). Therefore, the series will converge for \(p < -1\).
3. \(\sum_{n=1}^{\infty} \frac{(n!)^2}{(pn)!}\)
  
  The ratio test can be used to determine the values of \(p\) that make the series converge:
  
  \[ \begin{aligned} \lim_{n \to \infty} \frac{(n+1)!^2}{(p(n+1))!} \cdot \frac{(pn)!}{(n!)^2} &= \lim_{n \to \infty} \frac{(n+1)^2}{(pn+1)(pn+2)(pn+3)\dots(pn+p)}\\ &= \frac{1}{p^p} \\ p &\geq 2\\ \end{aligned} \]
If \(f(x) = \sin(x^3)\), then find \(f^{(15)}(0)\). (Hint: This means 15th derivative) (Hint 2: Don't try to take 15 derivatives since it would get unbearably messy)

With the Maclaurin Series:

\[ \begin{aligned} \sin(x) &= \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!}\\ &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\\ \sin(x^3) &= x^3 - \frac{x^9}{3!} + \frac{x^{15}}{5!} - \frac{x^{21}}{7!} + \cdots\\ \end{aligned} \]

By definition of the terms of the Maclaurin Series:

\[ \begin{aligned} \frac{f^{(15)}(0)}{15!}x^15 &= \frac{x^{15}}{5!}\\ \frac{f^{(15)}(0)}{15!} &= \frac{1}{5!}\\ f^{(15)}(0) &= \frac{15!}{5!}\\ &= 10,897,286,400 \end{aligned} \]
Find \(F'(0)\) if \(F(x) = \begin{cases} \frac{g(x)\sin^2x}{x} & x \neq 0 \\ 0 & x = 0 \end{cases}\) (Hint: Your answer should be in terms of \(g\))

\[ \begin{aligned} F'(0) &= \lim_{x \to 0} \frac{F(x) - F(0)}{x - 0}\\ &= \lim_{x \to 0} \frac{\frac{g(x)\sin^2x}{x} - 0}{x}\\ &= \lim_{x \to 0} \frac{g(x)\sin^2x}{x^2} \Rightarrow \frac{0}{0}\\ &= \lim_{x \to 0} \frac{2g(x)\sin x\cos x + \sin^2xg'(x)}{2x} \text{ LH} \Rightarrow \frac{0}{0}\\ &= \lim_{x \to 0} \frac{2g(x)(\cos^2x - \sin^2x) + 4g'(x)\sin x\cos x + g''(x)\sin^2x}{2} \text{ LH}\\ &= \frac{2g(0)(1 - 0) + 4g'(0)(0)(1) + g''(0)(0)}{2}\\ &= \frac{2g(0)}{2}\\ &= g(0) \end{aligned} \]