Taylor and Maclaurin Series - AP Calculus BC

1. For $h(x)=\cos(x)$, what is the Maclaurin series representation of $h(x)$?
   Answer: $\sum_{n=0}^{\infty}(-1)^n\dfrac{x^{2n}}{(2n)!}$ Finding the Maclaurin series for $h(x) = \cos(x)$ is a fundamental Taylor/Maclaurin series problem. The cosine function has derivatives that cycle through $\cos(x)$, $-\sin(x)$, $-\cos(x)$, $\sin(x)$, and evaluating at $x=0$ gives the pattern $1, 0, -1, 0, 1, ...$. This produces the series $\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}$, containing only even powers of $x$ with alternating signs. Option B incorrectly includes odd powers, which would appear in the sine series but not cosine. Remember that cosine is an even function, so its Maclaurin series contains only even powers of $x$.
2. For $F(x)=\ln x$, what is the Taylor series centered at $x=1$ for $F(x)$?
   Answer: $\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{(x-1)^n}{n}$ This problem asks for the Taylor series of $F(x) = \ln x$ centered at $x = 1$, a classic Taylor series application. We compute derivatives: $F'(x) = \frac{1}{x}$, $F''(x) = -\frac{1}{x^2}$, $F'''(x) = \frac{2}{x^3}$, and evaluate at $x = 1$ to get $F^{(n)}(1) = (-1)^{n-1}(n-1)!$ for $n \geq 1$. The Taylor series formula gives $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{(x-1)^n}{n}$, noting that $F(1) = \ln(1) = 0$ so the series starts at $n = 1$. Option C has the wrong sign pattern with $(-1)^n$ instead of $(-1)^{n-1}$, which would make the first term negative. For Taylor series centered at $a$, always use powers of $(x-a)$ and evaluate derivatives at $x = a$.
3. A spring's displacement is $s(x)=\sin(3x)$. What is the Maclaurin series for $s(x)$?
   Answer: $\sum_{n=0}^{\infty}(-1)^n\dfrac{(3x)^{2n+1}}{(2n+1)!}$ To find the Maclaurin series for $s(x) = \sin(3x)$, we apply the Taylor/Maclaurin series expansion method. The Maclaurin series for $\sin(u)$ is $\sum_{n=0}^{\infty}(-1)^n\frac{u^{2n+1}}{(2n+1)!}$, so substituting $u = 3x$ yields $\sum_{n=0}^{\infty}(-1)^n\frac{(3x)^{2n+1}}{(2n+1)!}$. This expansion captures the alternating signs and odd powers characteristic of the sine function. Option E temptingly omits the $(-1)^n$ factor, which would incorrectly make all terms positive and fail to represent the oscillating nature of sine. Remember that sine series always have odd powers with alternating signs, while cosine series have even powers with alternating signs.
4. A spring's displacement is $h(x)=\sin x$; what is its Maclaurin series through the $x^7$ term?
   Answer: $x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}$ This problem involves finding the Maclaurin series expansion for the function h(x) = sin x. To construct the series, use the Taylor series formula centered at 0, computing derivatives: sin(0)=0, cos(0)=1, -sin(0)=0, -cos(0)=-1, sin(0)=0, cos(0)=1, and so on, showing odd powers with alternating signs. Dividing by n! gives terms x/1! - $x^3$/3! + $x^5$/5! - $x^7$/7!, but since 1!=1, it's x - $x^3$/6 + $x^5$/120 - $x^7$/5040. The even derivatives at 0 are zero, so only odd terms appear. A tempting distractor like choice A fails because it uses positive signs for all terms, ignoring the alternating pattern from the sine function's derivatives. A transferable series-construction strategy is to recall the patterns for trigonometric functions, where sine has odd powers and alternating signs starting positive.
5. What is the Maclaurin series for $f(x)=\sin(3x)$?
   Answer: $\displaystyle \sum_{n=0}^{\infty} (-1)^n\frac{(3x)^{2n+1}}{(2n+1)!}$ This question tests proficiency in Maclaurin series, expansions centered at zero. The sine function has a known Maclaurin series: $\sin(u) = \sum_{n=0}^{\infty} (-1)^n \frac{u^{2n+1}}{(2n+1)!}$. Substitute u = 3x to obtain $\sin(3x) = \sum_{n=0}^{\infty} (-1)^n \frac{(3x)^{2n+1}}{(2n+1)!}$. This expansion is valid for all x since the sine series converges everywhere. A tempting distractor is choice C, which includes an extra (-1) factor as $(-1)^{n+1}$, but this would negate the series to $-\sin(3x)$, failing to match the positive leading term. A transferable strategy is to use composition by substituting into known series for elementary functions like sine or cosine.
6. Suppose $r(x)=\sqrt{1+x}$. What is the Maclaurin series for $r(x)$?
   Answer: $\sum_{n=0}^{\infty}\binom{\tfrac12}{n}x^n$ The Maclaurin series for $r(x) = \sqrt{1+x} = (1+x)^{1/2}$ uses the generalized binomial series, a key Taylor/Maclaurin series technique. The binomial series states that $(1+x)^\alpha = \sum_{n=0}^{\infty}\binom{\alpha}{n}x^n$ where $\binom{\alpha}{n} = \frac{\alpha(\alpha-1)...(\alpha-n+1)}{n!}$. For $\alpha = \frac{1}{2}$, we get $\sum_{n=0}^{\infty}\binom{1/2}{n}x^n$. The first few terms are $1 + \frac{1}{2}x - \frac{1}{8}x^2 + ...$. Option D incorrectly uses $\binom{-1/2}{n}$, which would give the series for $\frac{1}{\sqrt{1+x}}$ instead. Remember that the binomial coefficient's upper number matches the exponent in $(1+x)^\alpha$.
7. Let $f(x)=\ln(1+x)$. What is the Maclaurin series for $f(x)$?
   Answer: $\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{x^n}{n}$ This question requires finding the Maclaurin series for $f(x) = \ln(1+x)$ using Taylor/Maclaurin series techniques. To derive this series, we compute derivatives: $f'(x) = \frac{1}{1+x}$, $f''(x) = -\frac{1}{(1+x)^2}$, $f'''(x) = \frac{2}{(1+x)^3}$, and evaluate at $x=0$ to get the pattern $f^{(n)}(0) = (-1)^{n-1}(n-1)!$ for $n \geq 1$. This gives the series $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}$, noting that the series starts at $n=1$ since $\ln(1) = 0$. Option B incorrectly starts at $n=0$, which would include a $\frac{1}{0}$ term that is undefined. When finding Maclaurin series, always check the starting index and the sign pattern carefully.
8. What is the Maclaurin series for $f(x)=\frac{x}{1+x^2}$, valid for $|x|<1$?
   Answer: $\displaystyle \sum_{n=0}^{\infty} (-1)^n x^{2n+1}$ This question tests Maclaurin series knowledge, expansions at x=0. For x/($1+x^2$), note it as x * 1/($1+x^2$), and 1/($1+x^2$) = ∑ $(-1)^n$ $(x^2)^n$ from n=0 to ∞ for |x|<1. Multiply by x: ∑ $(-1)^n$ $x^{2n+1}$ from n=0 to ∞. This matches the form directly. A tempting distractor is choice B, with even powers $x^{2n}$, but this lacks the initial x multiplier, failing to produce the odd-powered series. A versatile strategy is to factor or multiply known geometric series to build compound functions.
9. For $f(x)=\ln(1+2x)$, what is the Maclaurin series for $f(x)$ through the $x^4$ term?​​
   Answer: $2x-2x^2+\frac{8}{3}x^3-4x^4+\cdots$ This problem requires finding the Maclaurin series for f(x) = ln(1 + 2x) using Taylor/Maclaurin series techniques. The Maclaurin series for ln(1 + u) is u - u²/2 + u³/3 - u⁴/4 + ..., so substituting u = 2x gives ln(1 + 2x) = 2x - (2x)²/2 + (2x)³/3 - (2x)⁴/4 + ... = 2x - 4x²/2 + 8x³/3 - 16x⁴/4 + ... = 2x - 2x² + 8x³/3 - 4x⁴ + .... Choice C might tempt students who forget to divide by the denominators 2, 3, 4, etc., leading to coefficients like -4x² instead of -2x². The key strategy is to substitute the inner function (2x) into the known series for ln(1 + u) and simplify each term carefully.
10. What is the Taylor series for $f(x)=\ln x$ centered at $x=1$, in powers of $(x-1)$?
    Answer: $\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1}\frac{(x-1)^n}{n}$ This question involves Taylor series centered at x=1, not zero. For ln(x), express as ln(1 + (x-1)). Use the known series ln(1+u) = ∑ $(-1)^{n-1}$ $u^n$ / n from n=1 to ∞ for |u|<1. Substitute u = x-1 to get ∑ $(-1)^{n-1}$ $(x-1)^n$ / n from n=1 to ∞. It converges for |x-1|<1. A tempting distractor is choice D, with $(-1)^n$ instead of $(-1)^{n-1}$, but this negates the series to -ln(x), failing for x>1 near 1. A transferable method is to shift the variable to center at the desired point and apply known series expansions.
11. Let $p(x)=\dfrac{1}{1-x}$ represent a geometric response; what is its Maclaurin series through $x^5$?
    Answer: $1+x+x^2+x^3+x^4+x^5$ This problem involves finding the Maclaurin series expansion for the function p(x) = 1/(1 - x). To construct the series, recognize it as the geometric series sum_{n=0}^∞ $x^n$, which converges for |x| < 1. Through $x^5$, this is explicitly 1 + x + $x^2$ + $x^3$ + $x^4$ + $x^5$. This can be derived by long division or noting the infinite sum formula. A tempting distractor like choice A fails because it alternates signs, which would be correct for 1/(1 + x) instead of 1/(1 - x). A transferable series-construction strategy is to identify geometric forms and adjust the sign in the denominator to determine if signs alternate.
12. Find the Taylor series for $f(x)=\cos x$ centered at $x=\pi$, written in powers of $(x-\pi)$.​​
    Answer: $-\sum_{n=0}^{\infty}(-1)^n\dfrac{(x-\pi)^{2n}}{(2n)!}$ This problem asks for the Taylor series of f(x) = cos(x) centered at x = π, demonstrating Taylor/Maclaurin series at non-zero centers. At x = π, we have cos(π) = -1, -sin(π) = 0, -cos(π) = 1, sin(π) = 0, and this pattern repeats. The Taylor series is f(x) = -1 + 0·(x-π) + 1·(x-π)²/2! + 0·(x-π)³/3! - 1·(x-π)⁴/4! + ... = -[1 - (x-π)²/2! + (x-π)⁴/4! - ...] = -Σ[$(-1)^n$$(x-π)^{2n}$/(2n)!]. Choice C incorrectly uses odd powers (2n+1) instead of even powers (2n), missing that sine terms vanish at x = π. The strategy is to evaluate derivatives at the center point and identify the pattern in the coefficients.
13. Find the Maclaurin series for $f(x)=\dfrac{1}{1+3x}$ through the $x^4$ term.​​
    Answer: $1-3x+9x^2-27x^3+81x^4+\cdots$ This problem requires finding the Maclaurin series for f(x) = 1/(1 + 3x) using Taylor/Maclaurin series methods. The geometric series 1/(1 - u) = 1 + u + u² + u³ + ... can be adapted by writing 1/(1 + 3x) = 1/(1 - (-3x)). Substituting u = -3x gives 1/(1 + 3x) = 1 + (-3x) + (-3x)² + (-3x)³ + (-3x)⁴ + ... = 1 - 3x + 9x² - 27x³ + 81x⁴ + .... Choice A might tempt students who forget the alternating signs that arise from powers of (-3x). The key insight is recognizing when to use the geometric series formula with appropriate substitutions.
14. Compute the Maclaurin series for $f(x)=\arctan x$ through the $x^7$ term.​​
    Answer: $x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots$ This problem asks for the Maclaurin series of f(x) = arctan(x), demonstrating Taylor/Maclaurin series for inverse trigonometric functions. The derivative of arctan(x) is 1/(1 + x²) = 1 - x² + x⁴ - x⁶ + ... (geometric series with u = -x²). Integrating term by term gives arctan(x) = ∫[1 - x² + x⁴ - x⁶ + ...]dx = x - x³/3 + x⁵/5 - x⁷/7 + .... Choice D incorrectly includes factorials in the denominators (like x³/3!), confusing this with the series for sin(x) or cos(x). The strategy for inverse functions is often to find the series for the derivative first, then integrate term by term.
15. Find the Taylor series for $f(x)=\ln x$ centered at $x=1$.​
    Answer: $\displaystyle \sum_{n=1}^{\infty} (-1)^{n+1}\frac{(x-1)^n}{n}$ To find the Taylor series for ln(x) centered at x=1, we use the fact that d/dx[ln(x)] = 1/x. From the previous problem, we know 1/x = Σ(-1)ⁿ(x-1)ⁿ for |x-1|<1. Integrating term by term: ∫[1 - (x-1) + (x-1)² - ...]dx = (x-1) - (x-1)²/2 + (x-1)³/3 - ... = Σ(-1)ⁿ⁺¹(x-1)ⁿ/n, starting from n=1. Choice B has incorrect signs (-1)ⁿ instead of (-1)ⁿ⁺¹, which would give -ln(x) instead of ln(x). The strategy is to differentiate or integrate known series to find related series.
16. What is the Maclaurin series for $f(x)=\ln(1+x)$, valid for $|x|<1$?​
    Answer: $\displaystyle \sum_{n=1}^{\infty} (-1)^{n+1}\frac{x^n}{n}$ This question tests your ability to find the Maclaurin series for ln(1+x). To construct this series, we start by recognizing that the derivative of ln(1+x) is 1/(1+x), which has the geometric series representation 1-x+x²-x³+... = Σ(-1)ⁿxⁿ for |x|<1. Integrating term by term gives us ∫(1-x+x²-x³+...)dx = x-x²/2+x³/3-x⁴/4+... = Σ(-1)ⁿ⁺¹xⁿ/n, starting from n=1. Choice A incorrectly omits the alternating signs that come from the geometric series of 1/(1+x). The key strategy is to recognize when you can use the derivative or integral of a known series to find a new series.
17. What is the Maclaurin series for $f(x)=\frac{1}{1+2x}$, valid for $|x|<\tfrac{1}{2}$?
    Answer: $\displaystyle \sum_{n=0}^{\infty} (-2x)^n$ This question evaluates understanding of Maclaurin series, which are power series centered at zero. The function $1/(1+2x)$ resembles the geometric series form $1/(1-r) = \sum_{n=0}^{\infty} r^n$ for $|r|<1$. Set $r = -2x$, so $1/(1+2x) = \sum_{n=0}^{\infty} (-2x)^n$, valid for $|2x|<1$ or $|x|<1/2$. This directly matches the expanded form without further manipulation. A tempting distractor is choice B, which uses $(2x)^n$ without the negative sign, but this would represent $1/(1-2x)$, failing for the given function due to the incorrect sign in the denominator. A general approach to constructing series is to identify a base series like the geometric one and substitute appropriately to fit the target function.
18. For $r(x)=\arctan x$, what is the Maclaurin series through the $x^7$ term?
    Answer: $x-\dfrac{x^3}{3}+\dfrac{x^5}{5}-\dfrac{x^7}{7}$ This problem involves finding the Maclaurin series expansion for the function r(x) = arctan x. To construct the series, integrate the geometric series for 1/(1 + $x^2$) = sum $(-1)^n$ $x^{2n}$, yielding arctan x = sum $(-1)^n$ $x^{2n+1}$ / (2n+1). Through $x^7$, this is x - $x^3$/3 + $x^5$/5 - $x^7$/7. Derivatives confirm: r'(x)=1/($1+x^2$), and higher terms follow the pattern. A tempting distractor like choice D fails because it uses factorial denominators, confusing it with trigonometric series like sin x. A transferable series-construction strategy is to derive inverse trig series by integrating known geometric or binomial expansions.
19. What is the Maclaurin series for $f(x)=\ln(1+x)$, valid for $|x|<1$?
    Answer: $\displaystyle \sum_{n=1}^{\infty} (-1)^{n-1}\frac{x^n}{n}$ This question assesses knowledge of Maclaurin series, which are Taylor series expansions centered at zero. To construct the Maclaurin series for $\ln(1+x)$, begin with the geometric series for $1/(1-u) = \sum_{n=0}^{\infty} u^n$ for $|u|<1$. Substitute $u = -x$ to get $1/(1+x) = \sum_{n=0}^{\infty} (-1)^n x^n$. Integrate term by term from 0 to $x$, yielding $\ln(1+x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}$, and reindex by letting $k=n+1$ to get $\sum_{k=1}^{\infty} (-1)^{k-1} \frac{x^k}{k}$. A tempting distractor is choice A, which lacks the alternating sign, but this would represent $-\ln(1-x)$ instead, failing to incorporate the negative substitution correctly. A transferable strategy for series construction is to start with a known series like the geometric one and apply operations such as substitution, integration, or differentiation to derive the desired expansion.
20. Find the Maclaurin series for $f(x)=e^{2x}$ in powers of $x$.​
    Answer: $\displaystyle \sum_{n=0}^{\infty} \frac{(2x)^n}{n!}$ To find the Maclaurin series for $e^{2x}$, we use the known series for $e^u$ = Σuⁿ/n! and substitute u = 2x. This gives us $e^{2x}$ = 1 + 2x + (2x)²/2! + (2x)³/3! + ... = Σ(2x)ⁿ/n! starting from n=0. Each term contains (2x)ⁿ = 2ⁿxⁿ, so we must keep the 2 inside the parentheses raised to the nth power. Choice B incorrectly factors out the 2, writing 2xⁿ instead of 2ⁿxⁿ, which would give $2e^x$ rather than $e^{2x}$. When substituting into a known series, always substitute the entire expression and simplify carefully.

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