Taylor & Maclaurin Series - Calculus 2

If $h(x) = x \cos(x^3)$ , what is the coefficient of $x^7$ in the Maclaurin series for $h(x)$ ?
Answer: $-\frac{1}{2}$ When finding coefficients in Maclaurin series for products of functions, you need to systematically expand each factor and identify which terms contribute to your target power. Start with the Maclaurin series for $\cos(u) = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \frac{u^6}{6!} + \cdots$ . Substituting $u = x^3$ : $\cos(x^3) = 1 - \frac{(x^3)^2}{2!} + \frac{(x^3)^4}{4!} - \frac{(x^3)^6}{6!} + \cdots = 1 - \frac{x^6}{2} + \frac{x^{12}}{24} - \frac{x^{18}}{720} + \cdots$ Now multiply by $x$ : $h(x) = x \cos(x^3) = x\left(1 - \frac{x^6}{2} + \frac{x^{12}}{24} - \cdots\right) = x - \frac{x^7}{2} + \frac{x^{13}}{24} - \cdots$ The coefficient of $x^7$ is $-\frac{1}{2}$ , confirming answer D. Let's examine the wrong answers: A) $-\frac{1}{6}$ likely comes from incorrectly using $\frac{1}{3!}$ instead of $\frac{1}{2!}$ when expanding $\cos(x^3)$ . B) $-\frac{1}{120}$ suggests confusion with $\frac{1}{5!}$ , perhaps from misidentifying which factorial applies. C) $\frac{1}{6}$ combines the factorial error from option A with a sign error—forgetting that the $x^6$ term in $\cos(x^3)$ comes from the negative $-\frac{u^2}{2!}$ term. Study tip: When working with Maclaurin series of composite functions, write out the first several terms explicitly rather than trying to work with general formulas. This prevents factorial and sign errors that are common in these problems.
Find the Taylor series representation for $f(x) = \frac{1}{x^2}$ centered at $a = -1$ .
Answer: $\sum_{n=0}^{\infty} (n+1) (x+1)^n$ We need to find the Taylor series for $f(x) = x^{-2}$ centered at $a = -1$ . We'll compute the derivatives and evaluate them at $x = -1$ . $f(x) = x^{-2} \implies f(-1) = 1$ $f'(x) = -2x^{-3} \implies f'(-1) = 2$ $f''(x) = 6x^{-4} \implies f''(-1) = 6$ $f'''(x) = -24x^{-5} \implies f'''(-1) = 24$ In general, $f^{(n)}(x) = (-1)^n (n+1)! x^{-(n+2)}$ . Evaluating at $x = -1$ : $f^{(n)}(-1) = (-1)^n (n+1)! (-1)^{-(n+2)} = (n+1)!$ . The Taylor series coefficient is $c_n = \frac{f^{(n)}(-1)}{n!} = \frac{(n+1)!}{n!} = n+1$ . Therefore, the Taylor series is $\sum_{n=0}^{\infty} (n+1)(x+1)^n$ .
What is the coefficient of $x^4$ in the Maclaurin series expansion for the function $f(x) = \frac{\sin(x)}{1-x}$ ?
Answer: $\frac{5}{6}$ To find the coefficient, multiply the Maclaurin series for $\sin(x)$ and $\frac{1}{1-x}$ . $\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots$ $\frac{1}{1-x} = 1 + x + x^2 + x^3 + x^4 + \dots$ Multiplying these two series, we look for the terms that produce $x^4$ : $(x)(x^3) + \left(-\frac{x^3}{6}\right)(x) = x^4 - \frac{1}{6}x^4 = \left(1 - \frac{1}{6}\right)x^4 = \frac{5}{6}x^4$ The coefficient is $\frac{5}{6}$ .
The function $f(x) = \frac{1}{2-x}$ can be written as a Maclaurin series. What is the radius of convergence of this series?
Answer: $R = 2$ When you encounter a rational function like $f(x) = \frac{1}{2-x}$ and need to find the radius of convergence of its Maclaurin series, look for where the function has singularities (points where it's undefined or discontinuous). To find the Maclaurin series, you can rewrite this function in a form that matches a known series. Factor out the constant: $f(x) = \frac{1}{2-x} = \frac{1}{2} \cdot \frac{1}{1-\frac{x}{2}}$ . This matches the geometric series formula $\frac{1}{1-u} = \sum_{n=0}^{\infty} u^n$ where $u = \frac{x}{2}$ , which converges when $|u| < 1$ . For convergence, you need $\left|\frac{x}{2}\right| < 1$ , which means $|x| < 2$ . Therefore, the radius of convergence is $R = 2$ , making (D) correct. The wrong answers represent common misconceptions: (A) $R = 1$ might come from forgetting the factor of 2 in the denominator and treating this like $\frac{1}{1-x}$ . (C) $R = \frac{1}{2}$ could result from incorrectly thinking the radius equals the coefficient $\frac{1}{2}$ or making an algebraic error. (B) $R = \infty$ would suggest the series converges everywhere, but rational functions with finite poles never have infinite radius of convergence. Study tip: For rational functions, the radius of convergence equals the distance from the center of expansion (here, $x = 0$ ) to the nearest singularity. Since $f(x)$ has a pole at $x = 2$ , the radius is $|2 - 0| = 2$ .
If the Maclaurin series for $f(x)$ is $\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n^2}$ , what is the Maclaurin series for $f'(x)$ ?
Answer: $\sum_{k=0}^{\infty} \frac{(-1)^k x^k}{k+1}$ We differentiate the series for $f(x)$ term-by-term to find the series for $f'(x)$ . $f'(x) = \frac{d}{dx} \left( \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n^2} \right) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} (n x^{n-1})}{n^2} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{n-1}}{n}$ To match the format of the answer choices, we re-index the series. Let $k = n-1$ , so $n = k+1$ . When $n=1$ , $k=0$ . The sum becomes: $\sum_{k=0}^{\infty} \frac{(-1)^{(k+1)+1} x^k}{k+1} = \sum_{k=0}^{\infty} \frac{(-1)^{k+2} x^k}{k+1} = \sum_{k=0}^{\infty} \frac{(-1)^k x^k}{k+1}$ This is the Maclaurin series for $\frac{\ln(1+x)}{x}$ .
What is the Maclaurin series for $f(x) = \cosh(x) = \frac{e^x + e^{-x}}{2}$ ?
Answer: $\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$ We use the Maclaurin series for $e^x$ and $e^{-x}$ . $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \dots$ $e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \frac{x^4}{4!} - \dots$ Adding them together, the odd-powered terms cancel: $e^x + e^{-x} = 2 + 2\frac{x^2}{2!} + 2\frac{x^4}{4!} + \dots = 2 \sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$ Dividing by 2 gives the series for $\cosh(x)$ : $\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}$ .
Which of the following is the Taylor series for $f(x) = \sin(x)$ centered at $a = \pi/2$ ?
Answer: $\sum_{n=0}^{\infty} \frac{(-1)^n (x-\pi/2)^{2n}}{(2n)!}$ One way is to use a substitution and a trigonometric identity. Let $u = x - \pi/2$ , so $x = u + \pi/2$ . Then $f(x) = \sin(u + \pi/2)$ . Using the angle addition formula, $\sin(u + \pi/2) = \sin(u)\cos(\pi/2) + \cos(u)\sin(\pi/2) = \cos(u)$ . The Maclaurin series for $\cos(u)$ is $\sum_{n=0}^{\infty} \frac{(-1)^n u^{2n}}{(2n)!}$ . Substituting $u = x - \pi/2$ back gives the Taylor series for $\sin(x)$ at $a=\pi/2$ : $\sum_{n=0}^{\infty} \frac{(-1)^n (x-\pi/2)^{2n}}{(2n)!}$ .
Find the Maclaurin series for the function $f(x) = \int_0^x \frac{\sin(t)}{t} dt$ .
Answer: $\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!(2n+1)}$ First, find the Maclaurin series for $\sin(t)$ : $\sin(t) = \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n+1}}{(2n+1)!}$ . Next, divide by $t$ : $\frac{\sin(t)}{t} = \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n}}{(2n+1)!}$ . Finally, integrate term-by-term from 0 to $x$ : $\int_0^x \sum_{n=0}^{\infty} \frac{(-1)^n t^{2n}}{(2n+1)!} dt = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \int_0^x t^{2n} dt = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \left[ \frac{t^{2n+1}}{2n+1} \right]_0^x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!(2n+1)}$
Find the Taylor series for $f(x) = e^{3x}$ centered at $a=-1$ .
Answer: $\sum_{n=0}^{\infty} \frac{3^n e^{-3} (x+1)^n}{n!}$ When finding a Taylor series centered at a point other than zero, you need to use the general Taylor series formula and carefully evaluate all derivatives at the given center point. For $f(x) = e^{3x}$ centered at $a = -1$ , start by finding the pattern of derivatives. The first few derivatives are: $f'(x) = 3e^{3x}$ , $f''(x) = 9e^{3x}$ , $f'''(x) = 27e^{3x}$ . In general, $f^{(n)}(x) = 3^n e^{3x}$ . Next, evaluate each derivative at the center point $x = -1$ : $f^{(n)}(-1) = 3^n e^{3(-1)} = 3^n e^{-3}$ . The Taylor series formula gives us: $f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(-1)}{n!}(x-(-1))^n = \sum_{n=0}^{\infty} \frac{3^n e^{-3}}{n!}(x+1)^n$ This matches choice C exactly. Looking at the wrong answers: Choice A is missing the $3^n$ factor that comes from the chain rule when differentiating $e^{3x}$ . Choice B incorrectly uses $e^3$ instead of $e^{-3}$ - this happens when you forget to substitute $x = -1$ and mistakenly use $x = 1$ . Choice D has the wrong center point, using $(x-1)$ instead of $(x+1)$ , and is missing the exponential factor entirely. Study tip: Always substitute the center point into your derivatives carefully, and remember that centering at $a = -1$ means your series involves $(x+1)$ terms, not $(x-1)$ .
The Maclaurin series for $f(x) = (1-x)^{-2}$ is $\sum_{n=0}^{\infty} (n+1)x^n$ . What is $f^{(10)}(0)$ ?
Answer: $11!$ The coefficient of $x^n$ in the Maclaurin series is $c_n = \frac{f^{(n)}(0)}{n!}$ . From the given series, the coefficient of $x^n$ is $c_n = n+1$ . We want to find $f^{(10)}(0)$ , so we look at the case where $n=10$ . The coefficient is $c_{10} = 10+1 = 11$ . Using the formula, $c_{10} = \frac{f^{(10)}(0)}{10!}$ . So, $11 = \frac{f^{(10)}(0)}{10!}$ . Solving for the derivative gives $f^{(10)}(0) = 11 \cdot 10! = 11!$ .
Use the provided information to find the Maclaurin series for $g(x) = \frac{1}{1+x^2}$ .
Answer: $\sum_{n=0}^{\infty} (-1)^n x^{2n}$ We know that the derivative of $\arctan(x)$ is $\frac{1}{1+x^2}$ . Therefore, we can find the series for $g(x)$ by differentiating the series for $\arctan(x)$ term-by-term. $g(x) = \frac{d}{dx} \left( \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{2n+1} \right) = \sum_{n=0}^{\infty} \frac{d}{dx} \left( \frac{(-1)^n x^{2n+1}}{2n+1} \right)$ $= \sum_{n=0}^{\infty} \frac{(-1)^n (2n+1)x^{2n}}{2n+1} = \sum_{n=0}^{\infty} (-1)^n x^{2n}$ This is also recognizable as the geometric series for $\frac{1}{1-(-x^2)}$ .
Find the first four terms of the Taylor series for $f(x) = \frac{1}{2-x}$ centered at $a=1$ .
Answer: $1 + (x-1) + (x-1)^2 + (x-1)^3$ The most efficient method is to manipulate the function into the form of a geometric series, $\frac{1}{1-r}$ . $f(x) = \frac{1}{2-x} = \frac{1}{2 - (1 + (x-1))} = \frac{1}{1 - (x-1)}$ This is a geometric series with first term $a=1$ and common ratio $r = x-1$ . The series is $\sum_{n=0}^{\infty} (x-1)^n$ . The first four terms (for $n=0, 1, 2, 3$ ) are $1 + (x-1) + (x-1)^2 + (x-1)^3$ .
Find the first three non-zero terms of the Taylor series for $g(x) = \ln(1 + x^2)$ centered at $x = 0$ .
Answer: $x^2 - \frac{x^4}{2} + \frac{x^6}{3}$ Using the known series $\ln(1 + u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} + \cdots$ and substituting $u = x^2$ : $\ln(1 + x^2) = x^2 - \frac{(x^2)^2}{2} + \frac{(x^2)^3}{3} - \cdots = x^2 - \frac{x^4}{2} + \frac{x^6}{3} - \cdots$ . Choice B incorrectly has $x^8$ instead of $x^6$ in the third term. Choice C includes the constant term 1, which doesn't appear since $\ln(1) = 0$ . Choice D uses the original $\ln(1+u)$ series without substituting $u = x^2$ .
Consider the function $f(x) = \cos(x) \sin(x)$ . The Maclaurin series for $f(x)$ has the form $\sum_{n=0}^{\infty} a_n x^n$ . Which of the following statements about the coefficients is true?
Answer: $a_1 = 1$ and $a_3 = -\frac{2}{3}$ , with all even coefficients zero Using the identity $\cos(x)\sin(x) = \frac{1}{2}\sin(2x)$ , we can find the Maclaurin series: $f(x) = \frac{1}{2}\sin(2x) = \frac{1}{2}[2x - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - \cdots] = x - \frac{8x^3}{12} + \cdots = x - \frac{2x^3}{3} + \cdots$ . So $a_1 = 1$ and $a_3 = -\frac{2}{3}$ . Since $\sin(2x)$ only has odd powers, all even coefficients are zero. Choice A incorrectly gives a non-zero even coefficient. Choice B incorrectly claims odd coefficients are zero for $k \geq 1$ . Choice D incorrectly has $a_2 = 1$ .
Let $f(x)$ be a function such that $f(1) = 2$ , $f'(1) = -1$ , $f''(1) = 6$ , and $f'''(1) = -12$ . What is the third-degree Taylor polynomial, $T_3(x)$ , for $f(x)$ centered at $a = 1$ ?
Answer: $2 - (x-1) + 3(x-1)^2 - 2(x-1)^3$ The formula for the third-degree Taylor polynomial centered at $a=1$ is: $T_3(x) = f(1) + \frac{f'(1)}{1!}(x-1) + \frac{f''(1)}{2!}(x-1)^2 + \frac{f'''(1)}{3!}(x-1)^3$ Substituting the given values: $T_3(x) = 2 + \frac{-1}{1}(x-1) + \frac{6}{2}(x-1)^2 + \frac{-12}{6}(x-1)^3$ $T_3(x) = 2 - (x-1) + 3(x-1)^2 - 2(x-1)^3$ Distractor (A) forgets to divide by the factorials. Distractor (C) incorrectly centers the polynomial at 0. Distractor (D) makes an error in calculating $3!$ .
Which function is represented by the Maclaurin series $\sum_{n=0}^{\infty} \frac{(-1)^n (2x)^{2n+1}}{(2n+1)!}$ ?
Answer: $\sin(2x)$ The standard Maclaurin series for $\sin(u)$ is $\sum_{n=0}^{\infty} \frac{(-1)^n u^{2n+1}}{(2n+1)!}$ . The given series has the same form, $\sum_{n=0}^{\infty} \frac{(-1)^n (2x)^{2n+1}}{(2n+1)!}$ . By comparing the two series, we can see that $u = 2x$ . Therefore, the function represented by the series is $\sin(2x)$ .
What is the Maclaurin series for the function $f(x) = x \arctan(x^2)$ ?
Answer: $\sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+3}}{2n+1}$ When you encounter a Maclaurin series problem involving composite functions like $x \arctan(x^2)$ , start with known series and use substitution rather than computing derivatives from scratch. The key insight is to recall the Maclaurin series for $\arctan(u)$ : $\arctan(u) = \sum_{n=0}^{\infty} \frac{(-1)^n u^{2n+1}}{2n+1}$ . This series is valid for $|u| \leq 1$ . To find the series for $\arctan(x^2)$ , substitute $u = x^2$ : $\arctan(x^2) = \sum_{n=0}^{\infty} \frac{(-1)^n (x^2)^{2n+1}}{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{2n+1}$ Now multiply by $x$ to get $f(x) = x \arctan(x^2)$ : $x \arctan(x^2) = x \cdot \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+2}}{2n+1} = \sum_{n=0}^{\infty} \frac{(-1)^n x^{4n+3}}{2n+1}$ This confirms answer C. Looking at the wrong answers: A gives you $\arctan(x^2)$ but forgets the multiplication by $x$ . B has the wrong exponent pattern—it would come from $x \arctan(x)$ instead of $x \arctan(x^2)$ . D is missing the alternating signs $(-1)^n$ , which are essential in the arctangent series. Study tip: For composite function series, always start with a known series and substitute carefully. Keep track of every transformation—substitution, then multiplication—and verify your exponent patterns match the operations you performed.
Let $T_n(x)$ denote the $n$ -th degree Taylor polynomial of $f(x) = \ln(1+x)$ at $x=0$ . Which of the following is $T_3(x)$ ?
Answer: $x - \frac{x^2}{2} + \frac{x^3}{3}$ The Maclaurin series for $\ln(1+x)$ is a standard series: $\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}x^n}{n} = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$ The third-degree Taylor polynomial, $T_3(x)$ , consists of all terms up to and including the degree 3 term. Thus, $T_3(x) = x - \frac{x^2}{2} + \frac{x^3}{3}$ .
The Maclaurin series for a function $f(x)$ is given by $\sum_{n=0}^{\infty} (n+1)2^n x^n$ . What is the value of $f'''(0)$ ?
Answer: $192$ The coefficient of $x^n$ in a Maclaurin series is given by $c_n = \frac{f^{(n)}(0)}{n!}$ . We want $f'''(0)$ , so we need the coefficient of the $n=3$ term, $c_3$ . From the given series, $c_n = (n+1)2^n$ . For $n=3$ , $c_3 = (3+1)2^3 = 4 \cdot 8 = 32$ . Now, we solve for the derivative: $f'''(0) = c_3 \cdot 3! = 32 \cdot 6 = 192$ .
Which of the following is the Taylor series for $f(x) = \ln(x)$ centered at $a = 2$ ?
Answer: $\ln(2) + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(x-2)^n}{n \cdot 2^n}$ To find the Taylor series, we compute the derivatives of $f(x) = \ln(x)$ at $a=2$ . $f(2) = \ln(2)$ . $f'(x) = x^{-1}$ , $f''(x) = -x^{-2}$ , $f'''(x) = 2x^{-3}$ , and in general, $f^{(n)}(x) = (-1)^{n-1}(n-1)!x^{-n}$ for $n \ge 1$ . At $a=2$ , $f^{(n)}(2) = \frac{(-1)^{n-1}(n-1)!}{2^n}$ . The Taylor series coefficient is $c_n = \frac{f^{(n)}(2)}{n!} = \frac{(-1)^{n-1}(n-1)!}{n! \cdot 2^n} = \frac{(-1)^{n-1}}{n \cdot 2^n}$ for $n \ge 1$ . The series is $f(2) + \sum_{n=1}^{\infty} c_n (x-2)^n$ , which is $\ln(2) + \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(x-2)^n}{n \cdot 2^n}$ .

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