Representing Series as Power Series - AP Calculus BC

A signal model uses $f(x)=\frac{1}{1-3x}$ for $|x|<\frac{1}{3}$ ; which power series represents $f(x)$ ?
Answer: $\displaystyle \sum_{n=0}^{\infty} 3^n x^{n}$ This problem requires finding the power series representation of $f(x) = \frac{1}{1-3x}$ . The standard geometric series formula is $\frac{1}{1-r} = \sum_{n=0}^{\infty} r^n$ for $ |r| < 1 $. Here, we have $ r = 3x $, so $f(x) = \sum_{n=0}^{\infty} (3x)^n = \sum_{n=0}^{\infty} 3^n x^n$ . Choice A incorrectly shifts the power of x to n+1, which would give $x \cdot \sum 3^n x^n = \frac{x}{1-3x}$ instead. The key strategy is to identify the form $\frac{1}{1-r}$ and substitute r directly into the geometric series formula.
For $|x|<1$ , what is the power series for $f(x)=\dfrac{1}{1-3x}$ about $x=0$ ?
Answer: $\displaystyle \sum_{n=0}^{\infty}3^{n}x^{n}$ This problem requires finding the power series representation of f(x) = 1/(1-3x) about x = 0. We start with the known geometric series formula 1/(1-u) = Σ(n=0 to ∞) $u^n$ for |u| < 1. By substituting u = 3x, we get 1/(1-3x) = Σ(n=0 to ∞) $(3x)^n$ = Σ(n=0 to ∞) $3^n$ $x^n$ , which converges for |3x| < 1 or |x| < 1/3. Choice C might tempt students who think the series should start at n = 1, but the geometric series includes the n = 0 term, giving us 1 + 3x + 9x² + ... The key strategy is recognizing when a function matches the form 1/(1-u) and making the appropriate substitution.
A wave approximation uses $q(x)=\frac{1}{(1-x)^2}$ for $|x|<1$ ; which power series represents $q(x)$ ?
Answer: $\displaystyle \sum_{n=0}^{\infty} (n+1)x^{n}$ This problem requires the power series for q(x) = 1/(1-x)². We know that d/dx[1/(1-x)] = 1/(1-x)², and the derivative of Σ(n=0 to ∞) $x^n$ is Σ(n=1 to ∞) $nx^{n-1}$ . However, we can reindex this as Σ(n=0 to ∞) (n+1) $x^n$ , giving the series 1 + 2x + 3x² + 4x³ + .... Choice B would give 0 + x + 2x² + 3x³ + ..., missing the constant term 1. The strategy for 1/ $(1-x)^k$ is to differentiate the geometric series k-1 times, adjusting coefficients accordingly.
A signal is modeled by $g(x)=\ln(1+x)$ for $|x|<1$ . Which power series represents $g(x)$ ?
Answer: $\displaystyle \sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}$ Representing functions as power series is a key skill in AP Calculus BC, enabling approximation and integration of functions like logarithms. The series for ln(1+x) derives from integrating the geometric series for 1/(1+t) = sum_{n=0}^∞ $(-1)^n$ $t^n$ for |t|<1. Integrating from 0 to x gives ln(1+x) = integral sum_{n=0}^∞ $(-1)^n$ $t^n$ dt = sum_{n=0}^∞ $(-1)^n$ $x^{n+1}$ /(n+1), then reindexing to n=1 yields sum_{n=1}^∞ $(-1)^{n-1}$ $x^n$ / n. This corresponds to choice A. Choice C is a common distractor but fails as it has $(-1)^n$ instead of $(-1)^{n-1}$ , flipping the signs and representing -ln(1+x). A transferable strategy is to derive new series by integrating or differentiating known ones, adjusting indices accordingly for the desired function.
For $|x|<1$ , what is the power series for $q(x)=\dfrac{1}{(1-x)^2}$ about $x=0$ ?
Answer: $\displaystyle \sum_{n=0}^{\infty}(n+1)x^{n}$ This problem requires finding the power series representation of q(x) = 1/(1-x)² about x = 0. We recognize that q(x) is the derivative of -1/(1-x), which has the series -Σ(n=0 to ∞) $x^n$ . Taking the derivative term by term: d/dx[-Σ(n=0 to ∞) $x^n$ ] = -Σ(n=0 to ∞) $nx^{n-1}$ = -Σ(n=1 to ∞) $nx^{n-1}$ . Reindexing with m = n-1 gives -Σ(m=0 to ∞) (m+1) $x^m$ , so q(x) = Σ(n=0 to ∞) (n+1) $x^n$ . Choice C starts at n = 1 with coefficient n, missing the n = 0 term which contributes 1 to the series. The strategy of differentiating known series is powerful for finding series of related functions.
In a finance model, $g(x)=\frac{5}{1+2x}$ for $|x|<\frac{1}{2}$ ; which power series equals $g(x)$ ?
Answer: $\displaystyle 5\sum_{n=0}^{\infty} (-2)^n x^{n}$ This problem asks for the power series of $g(x) = \frac{5}{1+2x}$ . We can rewrite this as $5 \cdot \frac{1}{1 - (-2x)}$ , matching the geometric series form $\frac{1}{1-r} = \sum r^n$ . With $r = -2x$ , we get $g(x) = 5 \cdot \sum_{n=0}^{\infty} (-2x)^n = 5 \cdot \sum_{n=0}^{\infty} (-2)^n x^n = \sum_{n=0}^{\infty} 5(-2)^n x^n$ . Choice A incorrectly writes $(2x)^n$ instead of $(-2x)^n$ , missing the alternating signs from the $+2x$ in the denominator. When the denominator has $1 + ax$ , remember to use $r = -ax$ in the geometric series.
For all $x$ , what is the power series for $r(x)=e^{2x}$ about $x=0$ ?
Answer: $\displaystyle \sum_{n=0}^{\infty}\frac{2^{n}x^{n}}{n!}$ This problem asks for the power series representation of r(x) = $e^{2x}$ about x = 0. We start with the known Maclaurin series $e^u$ = Σ(n=0 to ∞) $u^n$ /n! for all u. Substituting u = 2x gives $e^{2x}$ = Σ(n=0 to ∞) $(2x)^n$ /n! = Σ(n=0 to ∞) $2^n$ $x^n$ /n!, which converges for all x since the exponential series has infinite radius of convergence. Choice B incorrectly factors out just one factor of 2 instead of $2^n$ , forgetting that $(2x)^n$ = $2^n$ $x^n$ . When substituting into the exponential series, remember to raise the entire argument to the nth power.
For all real $x$ , what is the Maclaurin series for $f(x)=e^{2x}$ ?
Answer: $\displaystyle \sum_{n=0}^{\infty} \frac{2^n x^n}{n!}$ This question tests the skill of representing functions as power series, using the exponential series. The Maclaurin series for $e^y$ is $\sum_{n=0}^{\infty} \frac{y^n}{n!}$ for all real y. Substituting y = 2x gives $\sum_{n=0}^{\infty} \frac{(2x)^n}{n!} = \sum_{n=0}^{\infty} \frac{2^n x^n}{n!}$ , which is the series for $e^{2x}$ . This holds for all real x since the exponential series converges everywhere. A tempting distractor is choice A, which has (n+1)! in the denominator, altering it to something like an integral of the exponential. A transferable strategy for representing functions as power series is to substitute scaled variables into known Taylor series like the exponential.
For $|x|<1$ , which power series equals $f(x)=\ln(1-x)$ centered at $0$ ?
Answer: $\displaystyle -\sum_{n=1}^{\infty} \frac{x^n}{n}$ This question tests the skill of representing functions as power series, specifically for the logarithm with a sign change. For f(x) = ln(1-x), note that it's -ln(1 - (-x)), but directly, it's the integral of -1/(1-x) = - integral of sum_{n=0}^∞ $x^n$ dx. Integrating term by term yields - sum_{n=0}^∞ $x^{n+1}$ / (n+1), or - sum_{n=1}^∞ $x^n$ / n. This matches choice B for |x| < 1. A tempting distractor is choice C, which includes alternating signs and represents ln(1+x) instead. A transferable strategy for representing functions as power series is to adjust signs in substitutions and use integration on geometric series derivatives.
For $|x|<1$ , what is the power series for $p(x)=\dfrac{x^2}{1+x}$ about $x=0$ ?
Answer: $\displaystyle \sum_{n=0}^{\infty}(-1)^n x^{n+2}$ This problem asks for the power series representation of p(x) = x²/(1+x) about x = 0. We start with the geometric series 1/(1+x) = 1/(1-(-x)) = Σ(n=0 to ∞) $(-x)^n$ = Σ(n=0 to ∞) $(-1)^n$ $x^n$ for |x| < 1. Multiplying both sides by x² gives x²/(1+x) = x² · Σ(n=0 to ∞) $(-1)^n$ $x^n$ = Σ(n=0 to ∞) $(-1)^n$ $x^{n+2}$ . The series still converges for |x| < 1 since multiplication by x² doesn't change the radius of convergence. Choice D incorrectly adjusts the sign pattern to $(-1)^{n+1}$ , but multiplying by x² only shifts the powers, not the alternating pattern. When multiplying a power series by $x^k$ , shift all powers by k while keeping coefficients unchanged.
For $|x|<1$ , a calibration curve is $u(x)=\sqrt{1+x}$ . Which power series represents $u(x)$ ?
Answer: $\displaystyle \sum_{n=0}^{\infty}\binom{1/2}{n}x^{n}$ Representing functions as power series is a key skill in AP Calculus BC, applied in calibrations with binomial expansions. The generalized binomial series $(1+y)^a$ = sum_{n=0}^∞ binom{a}{n} $y^n$ for |y|<1. For u(x) = $(1+x)^{1/2}$ , set a=1/2 and y=x, directly giving sum_{n=0}^∞ binom{1/2}{n} $x^n$ . This is choice A. Choice D tempts with binom{-1/2}{n}, which represents 1/sqrt(1+x) instead. A transferable strategy is to use the binomial theorem for non-integer exponents, confirming the leading term matches the function at x=0.
A wave approximation uses $r(x)=\dfrac{1}{1+x^2}$ for $|x|<1$ . Which power series represents $r(x)$ ?
Answer: $\displaystyle \sum_{n=0}^{\infty}(-1)^n x^{2n}$ Representing functions as power series is a key skill in AP Calculus BC, used in approximations like wave functions. To find the series for r(x) = 1/( $1+x^2$ ), recognize it as 1/(1 - ( $-x^2$ )). Substitute into the geometric series sum_{n=0}^∞ $r^n$ = 1/(1-r) with r = - $x^2$ , yielding sum_{n=0}^∞ $(-x^2)^n$ = sum_{n=0}^∞ $(-1)^n$ $x^{2n}$ . This is choice C. Choice A distracts by lacking the even powers, representing a different function like 1/(1+x). A transferable strategy is to identify quadratic terms in the denominator and substitute into the geometric series, ensuring only even powers result.
For $|x|<4$ , what is the power series for $g(x)=\dfrac{1}{4+x}$ about $x=0$ ?
Answer: $\displaystyle \sum_{n=0}^{\infty}(-1)^n\frac{x^{n}}{4^{n+1}}$ This problem asks for the power series representation of g(x) = 1/(4+x) about x = 0. We first factor out 4 from the denominator to get g(x) = 1/4 · 1/(1+x/4). Using the geometric series 1/(1-u) = Σ(n=0 to ∞) $u^n$ with u = -x/4, we obtain 1/(1-(-x/4)) = Σ(n=0 to ∞) $(-x/4)^n$ . Therefore, g(x) = 1/4 · Σ(n=0 to ∞) $(-1)^n$ $(x/4)^n$ = Σ(n=0 to ∞) $(-1)^n$ $x^n$ / $4^{n+1}$ , which converges for |x/4| < 1 or |x| < 4. Choice A incorrectly places the factor of 1/4 inside the sum without adjusting the exponent. When factoring constants from denominators, always factor completely before applying the geometric series formula.
A chemical model uses $t(x)=\frac{1}{2-x}$ for $|x|<2$ ; which power series equals $t(x)$ ?
Answer: $\displaystyle \sum_{n=0}^{\infty} \frac{x^{n}}{2^{n+1}}$ This problem asks for the power series of $t(x) = \frac{1}{2-x}$ . We rewrite this as $t(x) = \frac{1}{2(1 - \frac{x}{2})} = \frac{1}{2} \cdot \frac{1}{1 - \frac{x}{2}}$ . Using the geometric series with $r = \frac{x}{2}$ , we get $t(x) = \frac{1}{2} \cdot \sum_{n=0}^{\infty} \left( \frac{x}{2} \right)^n = \frac{1}{2} \cdot \sum_{n=0}^{\infty} \frac{x^n}{2^n} = \sum_{n=0}^{\infty} \frac{x^n}{2^{n+1}}$ . Choice A would give $\sum_{n=0}^{\infty} \frac{x^n}{2^n} = \frac{1}{1 - \frac{x}{2}} = \frac{2}{2-x}$ , which is twice the desired function. When the denominator has the form a-x, factor out a and use $r = \frac{x}{a}$ in the geometric series.
For $|x|<2$ , a temperature correction uses $h(x)=\dfrac{1}{2+x}$ . Which power series represents $h(x)$ ?
Answer: $\displaystyle \sum_{n=0}^{\infty}(-1)^n\frac{x^n}{2^{n+1}}$ Representing functions as power series is a key skill in AP Calculus BC, useful for modeling phenomena like temperature corrections. For h(x) = 1/(2+x), rewrite it as 1/(2(1 + x/2)) = (1/2) * 1/(1 - (-x/2)). Use the geometric series sum_{n=0}^∞ $r^n$ = 1/(1-r) with r = -x/2, giving (1/2) sum_{n=0}^∞ $(-x/2)^n$ = (1/2) sum_{n=0}^∞ $(-1)^n$ $x^n$ / $2^n$ = sum_{n=0}^∞ $(-1)^n$ $x^n$ / $2^{n+1}$ . This matches choice B. Choice A tempts by omitting the 1/2 factor, representing 1/(1 + x/2) instead, which is twice the actual function. A transferable strategy is to factor constants and adjust the geometric form to fit the denominator, ensuring the radius of convergence aligns.
A sensor response is $h(x)=\ln(1+x)$ for $|x|<1$ ; which power series represents $h(x)$ ?
Answer: $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^{n}}{n}$ This problem requires the power series for h(x) = ln(1+x). The standard Taylor series for ln(1+x) is Σ(n=1 to ∞) $(-1)^{n+1}$ $x^n$ /n, which starts at n=1 (not n=0) since ln(1) = 0. This gives x - x²/2 + x³/3 - x⁴/4 + ..., with alternating signs starting positive. Choice B has $(-1)^n$ instead of $(-1)^{n+1}$ , which would start the series with a negative term -x, giving the series for -ln(1+x). The key insight is that ln(1+x) has a known Taylor series that must be memorized, with careful attention to the sign pattern.
A probability model uses $u(x)=e^{2x}$ ; which power series represents $u(x)$ for all real $x$ ?
Answer: $\displaystyle \sum_{n=0}^{\infty} \frac{(2x)^{n}}{n!}$ This problem requires the power series for u(x) = $e^{2x}$ . The exponential series is $e^y$ = Σ(n=0 to ∞) $y^n$ /n! for all y. With y = 2x, we get u(x) = Σ(n=0 to ∞) $(2x)^n$ /n! = Σ(n=0 to ∞) $2^n$ $x^n$ /n!. Choice A incorrectly separates the 2 from $x^n$ , giving only $2x^n$ instead of $(2x)^n$ = $2^n$ $x^n$ in the numerator. The key is to substitute the entire expression 2x into the exponential series, keeping 2x grouped together when raising to the nth power.
For $|x|<2$ , what is the power series for $v(x)=\dfrac{x}{2-x}$ about $x=0$ ?
Answer: $\displaystyle \sum_{n=0}^{\infty}\frac{x^{n+1}}{2^{n+1}}$ This problem asks for the power series representation of v(x) = x/(2-x) about x = 0. We first rewrite v(x) = x · 1/(2-x) = x · (1/2) · 1/(1-x/2). Using the geometric series 1/(1-u) = Σ(n=0 to ∞) $u^n$ with u = x/2 gives 1/(1-x/2) = Σ(n=0 to ∞) $(x/2)^n$ . Therefore, v(x) = x · (1/2) · Σ(n=0 to ∞) $(x/2)^n$ = (x/2) · Σ(n=0 to ∞) $x^n$ / $2^n$ = Σ(n=0 to ∞) $x^{n+1}$ / $2^{n+1}$ , which converges for |x/2| < 1 or |x| < 2. Choice A incorrectly separates the factors of 2, placing $2^n$ in the denominator instead of $2^{n+1}$ . When multiplying a series by x/k, both the power of x and the power of k increase by 1.
A displacement model is $p(x)=\arctan(x)$ for $|x|<1$ . Which power series represents $p(x)$ ?
Answer: $\displaystyle \sum_{n=0}^{\infty}(-1)^n\frac{x^{2n+1}}{2n+1}$ Representing functions as power series is a key skill in AP Calculus BC, applied in models like displacement with inverse trig functions. The series for arctan(x) comes from integrating the geometric series for 1/( $1+t^2$ ) = sum_{n=0}^∞ $(-1)^n$ $t^{2n}$ for |t|<1. Integrating from 0 to x yields arctan(x) = sum_{n=0}^∞ $(-1)^n$ integral $t^{2n}$ dt = sum_{n=0}^∞ $(-1)^n$ $x^{2n+1}$ /(2n+1). This is choice B. Choice D fails as a distractor by using $(-1)^{n+1}$ , which negates the series and represents -arctan(x). A transferable strategy is to integrate known geometric series adaptations for inverse functions, carefully handling the powers and signs.
An engineering approximation uses $v(x)=\frac{1}{1+x^2}$ for $|x|<1$ ; which power series represents $v(x)$ ?
Answer: $\sum_{n=0}^{\infty} (-1)^n x^{2n}$ This problem asks for the power series of $v(x) = \frac{1}{1+x^2}$ . We recognize this as the geometric series form $\frac{1}{1-r}$ with $r = -x^2$ . Therefore, $v(x) = \sum_{n=0}^{\infty} (-x^2)^n = \sum_{n=0}^{\infty} (-1)^n (x^2)^n = \sum_{n=0}^{\infty} (-1)^n x^{2n}$ . This gives $1 - x^2 + x^4 - x^6 + \dots$ , with only even powers and alternating signs. Choice A would give $\sum x^{2n} = \frac{1}{1-x^2}$ , which has the wrong denominator (minus instead of plus). When you see $\frac{1}{1+x^2}$ , use $r = -x^2$ to get the alternating signs on even powers.

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