Taylor & Maclaurin Expansions - Calculus 2

1. Let $p(x) = 1 + 2x - x^2 + \frac{x^3}{2}$ be the third-degree Taylor polynomial for $f(x)$ centered at $x = 0$. If $g(x) = f(2x)$, what is the coefficient of $x^2$ in the third-degree Taylor polynomial for $g(x)$ centered at $x = 0$?
   Answer: $-4$ This question tests your understanding of Taylor polynomials and function composition. When you see a function transformed by composition (like $g(x) = f(2x)$), you need to find how the Taylor polynomial transforms accordingly. Given that $p(x) = 1 + 2x - x^2 + \frac{x^3}{2}$ is the third-degree Taylor polynomial for $f(x)$ at $x = 0$, we can find the Taylor polynomial for $g(x) = f(2x)$ by substituting $2x$ for $x$ in $p(x)$. The Taylor polynomial for $g(x)$ becomes: $p(2x) = 1 + 2(2x) - (2x)^2 + \frac{(2x)^3}{2}$ $= 1 + 4x - 4x^2 + \frac{8x^3}{2}$ $= 1 + 4x - 4x^2 + 4x^3$ The coefficient of $x^2$ is $-4$, making C the correct answer. Answer A ($-1$) comes from mistakenly using the original coefficient without applying the transformation. Answer B ($-2$) results from incorrectly multiplying the original coefficient by 2 instead of $2^2$. Answer D ($4$) makes the sign error of forgetting the negative sign in the original polynomial. Strategy tip: When dealing with composition $f(ax)$ in Taylor polynomials, substitute $ax$ directly into the given polynomial. Remember that the coefficient of $x^n$ gets multiplied by $a^n$ due to the power rule. Practice recognizing this pattern to avoid coefficient errors on exams.
2. Suppose $f(x)$ has the Taylor series $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(2)}{n!}(x-2)^n$ that converges for $|x-2| < 3$. What is the radius of convergence for the Taylor series of $h(x) = f(x^2 + 2)$ centered at $x = 0$?
   Answer: $R = \sqrt{3}$ When you encounter a problem involving the Taylor series of a composite function, you need to think about how the radius of convergence changes when you substitute one function into another. Given that $f(x)$ has a Taylor series centered at $x = 2$ with radius of convergence 3, this means the series converges when $|x - 2| < 3$. Now you need to find the radius of convergence for $h(x) = f(x^2 + 2)$ centered at $x = 0$. The key insight is that $h(x) = f(x^2 + 2)$ will converge when the input to $f$ satisfies the original convergence condition. Since $f$ converges when its input is within distance 3 of the center 2, you need $|(x^2 + 2) - 2| < 3$, which simplifies to $|x^2| < 3$ or $|x| < \sqrt{3}$. Therefore, the radius of convergence is $\sqrt{3}$, making D correct. Let's examine why the other options are wrong: A) $R = 3$ incorrectly assumes the radius stays the same under composition. B) $R = \sqrt{5}$ might come from incorrectly solving $x^2 + 2 < 5$ instead of using the proper convergence condition. C) $R = 9$ likely results from squaring the original radius instead of taking the square root. Remember this pattern: when you have $f(g(x))$ and need the radius of convergence, set up the inequality for when $g(x)$ falls within the convergence region of $f$, then solve for the values of $x$ that satisfy this condition.
3. The Maclaurin series for $\frac{1}{1-x}$ is $\sum_{n=0}^{\infty} x^n = 1 + x + x^2 + x^3 + ...$ for $|x| < 1$. Find the coefficient of $x^5$ in the Maclaurin series for $\frac{x^2}{(1-x)^2}$.
   Answer: $4$ First, find the series for $\frac{1}{(1-x)^2}$. Since $\frac{d}{dx}[\frac{1}{1-x}] = \frac{1}{(1-x)^2}$ and $\frac{1}{1-x} = 1 + x + x^2 + x^3 + ...$, we get $\frac{1}{(1-x)^2} = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + 6x^5 + ...$ Therefore, $\frac{x^2}{(1-x)^2} = x^2(1 + 2x + 3x^2 + 4x^3 + 5x^4 + 6x^5 + ...) = x^2 + 2x^3 + 3x^4 + 4x^5 + 5x^6 + ...$ The coefficient of $x^5$ is $4$. Choice A uses the coefficient of $x^4$. Choice C uses the coefficient of $x^6$. Choice D uses the coefficient from the original series without the $x^2$ shift.
4. The Maclaurin series for $\ln(1+x)$ is $x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...$ for $|x| < 1$. What is the radius of convergence for the Maclaurin series of $\ln(1+x^3)$?
   Answer: $R = 1$ When working with power series transformations, the key principle is understanding how substitutions affect the radius of convergence. If you have a series for $f(x)$ with radius $R$, then the series for $f(g(x))$ converges when $|g(x)| < R$. Starting with the given Maclaurin series for $\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...$ which converges for $|x| < 1$, we can find the series for $\ln(1+x^3)$ by substituting $x^3$ for every $x$: $\ln(1+x^3) = x^3 - \frac{(x^3)^2}{2} + \frac{(x^3)^3}{3} - \frac{(x^3)^4}{4} + ... = x^3 - \frac{x^6}{2} + \frac{x^9}{3} - \frac{x^{12}}{4} + ...$ For this new series to converge, we need $|x^3| < 1$, which means $|x| < 1$. Therefore, the radius of convergence is $R = 1$. Answer A ($R = \frac{1}{3}$) incorrectly assumes you divide the original radius by the exponent. Answer B ($R = 3$) mistakenly multiplies the original radius by the exponent. Answer D, while mathematically equivalent to C, suggests unnecessary complexity with the cube root notation when the answer is simply 1. Remember this pattern: when substituting $x^n$ into a power series, the new convergence condition becomes $|x^n| < R_{original}$, so you solve $|x| < R_{original}^{1/n}$ to find the new radius. Here, $|x| < 1^{1/3} = 1$.
5. The Taylor series for $f(x)$ centered at $a = 2$ is $\sum_{n=0}^{\infty} \frac{(-1)^n}{3^n}(x-2)^n$. What is $f'''(2)$?
   Answer: $-\frac{2}{9}$ From the Taylor series $f(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{3^n}(x-2)^n$, the coefficient of $(x-2)^3$ is $\frac{(-1)^3}{3^3} = -\frac{1}{27}$. Since the coefficient of $(x-2)^3$ in a Taylor series equals $\frac{f'''(2)}{3!}$, we have $\frac{f'''(2)}{6} = -\frac{1}{27}$, so $f'''(2) = -\frac{6}{27} = -\frac{2}{9}$. Choice B forgets to multiply by $3!$. Choice C has a sign error. Choice D incorrectly uses $2 \cdot 3!$ instead of $3!$.
6. If $g(x) = \int_0^x \sin(t^2) dt$, what is the coefficient of $x^7$ in the Maclaurin series for $g(x)$?
   Answer: $-\frac{1}{42}$ First, find the Maclaurin series for $\sin(t^2)$. Since $\sin(u) = u - \frac{u^3}{6} + \frac{u^5}{120} - ...$ and $u = t^2$, we get $\sin(t^2) = t^2 - \frac{t^6}{6} + \frac{t^{10}}{120} - ...$ Integrating term by term: $g(x) = \int_0^x (t^2 - \frac{t^6}{6} + ...) dt = \frac{x^3}{3} - \frac{x^7}{6 \cdot 7} + ... = \frac{x^3}{3} - \frac{x^7}{42} + ...$ The coefficient of $x^7$ is $-\frac{1}{42}$. Choice A has the wrong sign. Choices C and D incorrectly use $7!$ instead of $6 \cdot 7$.
7. The function $h(x) = \frac{1}{1+x^2}$ has the Maclaurin series $\sum_{n=0}^{\infty} (-1)^n x^{2n}$. Using this series, find an approximation for $\int_0^{0.5} \frac{1}{1+x^2} dx$ using the first three non-zero terms.
   Answer: $0.5 - \frac{1}{24} + \frac{1}{160} \approx 0.464$ The Maclaurin series is $\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 + ...$ Integrating from 0 to 0.5: $\int_0^{0.5} (1 - x^2 + x^4) dx = [x - \frac{x^3}{3} + \frac{x^5}{5}]_0^{0.5} = 0.5 - \frac{(0.5)^3}{3} + \frac{(0.5)^5}{5} = 0.5 - \frac{0.125}{3} + \frac{0.03125}{5} = 0.5 - \frac{1}{24} + \frac{1}{160}$. Choice B incorrectly uses $\frac{x^2}{2}$ instead of $\frac{x^3}{3}$. Choice C uses $\frac{x^4}{4}$ instead of $\frac{x^5}{5}$. Choice D has both errors.

Start this quiz