AP Calculus BC Practice Quiz: Finding Taylor or Maclaurin Series for a Function

Correct Answer: A

The Maclaurin series for $e^u$ is $1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots$. To find the series for $f(x) = e^{2x}$, substitute $u=2x$ into the series for $e^u$. This gives $1 + (2x) + \frac{(2x)^2}{2!} + \frac{(2x)^3}{3!} + \dots = 1 + 2x + \frac{4x^2}{2} + \frac{8x^3}{6} + \dots = 1 + 2x + 2x^2 + \frac{4}{3}x^3 + \dots$. [cite: 3245]

Correct Answer: C

The formula for the $n$-th degree Taylor polynomial for a function $f$ about $x=a$ is $P_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \dots + \frac{f^{(n)}(a)}{n!}(x-a)^n$. For this problem, $a=1$ and $n=3$. Plugging in the given values: $P_3(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3 = 3 + (-2)(x-1) + \frac{6}{2}(x-1)^2 + \frac{-12}{6}(x-1)^3 = 3 - 2(x-1) + 3(x-1)^2 - 2(x-1)^3$. [cite: 3241, 3242]

Correct Answer: C

The general form of a Maclaurin series term is $\frac{f^{(k)}(0)}{k!}x^k$. We need to find the term in the given series that contains $x^4$. This occurs when the exponent $n+1=4$, which means $n=3$. The term is $\frac{(-1)^3 (2x)^{3+1}}{3!} = \frac{-1 \cdot (2^4 x^4)}{6} = \frac{-16x^4}{6} = -\frac{8}{3}x^4$. The coefficient of $x^4$ is $-\frac{8}{3}$. We set this equal to the general coefficient for $k=4$: $\frac{f^{(4)}(0)}{4!} = -\frac{8}{3}$. Solving for $f^{(4)}(0)$ gives $f^{(4)}(0) = -\frac{8}{3} \cdot 4! = -\frac{8}{3} \cdot 24 = -8 \cdot 8 = -64$. [cite: 3243]

Correct Answer: C

The Maclaurin series for $\frac{1}{1-u}$ is the geometric series $\sum_{n=0}^{\infty} u^n$. To find the series for $\frac{1}{1+x^2}$, we can write it as $\frac{1}{1-(-x^2)}$ and substitute $u=-x^2$. This gives $\sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty} (-1)^n x^{2n}$. To find the series for $g(x) = \frac{x}{1+x^2}$, we multiply this series by $x$: $x \cdot \sum_{n=0}^{\infty} (-1)^n x^{2n} = \sum_{n=0}^{\infty} (-1)^n x^{2n+1}$. [cite: 3244]

Correct Answer: B

To find the second-degree Taylor polynomial, we need the function value and the first two derivatives evaluated at the center $x=1$. $f(x) = \ln(x) \implies f(1) = \ln(1) = 0$. $f'(x) = \frac{1}{x} \implies f'(1) = 1$. $f''(x) = -\frac{1}{x^2} \implies f''(1) = -1$. The Taylor polynomial is $P_2(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2!}(x-1)^2 = 0 + 1(x-1) + \frac{-1}{2}(x-1)^2 = (x-1) - \frac{1}{2}(x-1)^2$. [cite: 3241, 3242]

Correct Answer: A

The Maclaurin series for $\sin(u)$ is $u - \frac{u^3}{3!} + \frac{u^5}{5!} - \dots$. To find the series for $f(x) = \sin(x^2)$, we substitute $u=x^2$ into the series for $\sin(u)$. This yields $(x^2) - \frac{(x^2)^3}{3!} + \frac{(x^2)^5}{5!} - \dots = x^2 - \frac{x^6}{3!} + \frac{x^{10}}{5!} - \dots$. [cite: 3245]

Correct Answer: C

The Maclaurin series coefficients relate to the derivatives at $x=0$. The coefficient of $x$ is $f'(0)$, and the coefficient of $x^2$ is $\frac{f''(0)}{2!}$. From the series, the coefficient of $x$ is $-\frac{1}{2}$, so $f'(0) = -\frac{1}{2}$. Since $f'(0) < 0$, the function is decreasing at $x=0$. The coefficient of $x^2$ is $\frac{1}{3}$, so $\frac{f''(0)}{2!} = \frac{1}{3}$, which means $f''(0) = \frac{2}{3}$. Since $f''(0) > 0$, the function is concave up at $x=0$. Therefore, $f$ is decreasing and concave up at $x=0$. [cite: 3243]

Correct Answer: A

The Maclaurin series for $\frac{1}{1+x} = \frac{1}{1-(-x)}$ is the geometric series $\sum_{n=0}^{\infty} (-x)^n = \sum_{n=0}^{\infty} (-1)^n x^n = 1 - x + x^2 - x^3 + \dots$. Integrating term by term gives $\int (\sum_{n=0}^{\infty} (-1)^n x^n) dx = C + \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}$. Since $\ln(1+0) = \ln(1) = 0$, the constant of integration $C$ must be 0. Thus, the series is $\sum_{n=0}^{\infty} \frac{(-1)^n x^{n+1}}{n+1} = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots$. This is equivalent to $\sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n}$. Option A is the most direct representation from the integration. [cite: 3244, 3241]

Correct Answer: A

A Taylor polynomial is a partial sum of the corresponding Taylor series. The Maclaurin series for $\cos(x)$ is $1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$. The fourth-degree Taylor polynomial, $P_4(x)$, includes all terms up to degree 4. This gives $P_4(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} = 1 - \frac{x^2}{2} + \frac{x^4}{24}$. [cite: 3242, 3245]

Correct Answer: A

The general form for the coefficient of the $(x-a)^n$ term in a Taylor series is $\frac{f^{(n)}(a)}{n!}$. In this problem, $a=2$ and we are interested in $n=10$. The coefficient of the $(x-2)^{10}$ term is given by the formula with $n=10$, which is $\frac{10+1}{3^{10} \cdot 10!} = \frac{11}{3^{10} \cdot 10!}$. We set the general form equal to this specific value: $\frac{f^{(10)}(2)}{10!} = \frac{11}{3^{10} \cdot 10!}$. Multiplying both sides by $10!$ gives $f^{(10)}(2) = \frac{11}{3^{10}}$. [cite: 3243]

Correct Answer: B

We multiply the Maclaurin series for $e^x$ and $\cos(x)$. $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots$ and $\cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots$. Multiplying these as polynomials: $e^x \cos(x) = (1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots)(1 - \frac{x^2}{2} + \dots)$. The constant term is $1 \cdot 1 = 1$. The $x$ term is $x \cdot 1 = x$. The $x^2$ term is $(1)(-\frac{x^2}{2}) + (\frac{x^2}{2})(1) = 0$. The $x^3$ term is $(x)(-\frac{x^2}{2}) + (\frac{x^3}{6})(1) = -\frac{x^3}{2} + \frac{x^3}{6} = -\frac{3x^3}{6} + \frac{x^3}{6} = -\frac{2x^3}{6} = -\frac{x^3}{3}$. The first three non-zero terms are $1 + x - \frac{x^3}{3}$. [cite: 3245]

Correct Answer: C

This series is a known Maclaurin series. It is similar in form to the series for $\sin(x)$, which is $\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}$, but it lacks the factorial in the denominator. This series is the Maclaurin series for $\arctan(x)$. It can be derived by integrating the geometric series for $\frac{1}{1+x^2}$. [cite: 3241, 3243]

Correct Answer: D

The series $1 + x + x^2 + x^3 + \dots = \sum_{n=0}^{\infty} x^n$ is a geometric series with first term $a=1$ and common ratio $r=x$. This is a foundational series for generating many other Maclaurin series. [cite: 3244]

Correct Answer: A

Since $\frac{d}{dx}(\sin x) = \cos x$, the Maclaurin series for $\cos(x)$ can be found by differentiating the series for $\sin(x)$ term by term. The series for $\sin(x)$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$. Differentiating gives $1 - \frac{3x^2}{3!} + \frac{5x^4}{5!} - \dots = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots$. This corresponds to the series $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}$. [cite: 3245]