📝 Differentiating power series (39 MCQs)

📖 Explanation: When differentiating a power series term-by-term, the radius of convergence $R$ remains unchanged. However, the behavior at the specific endpoints $x = a \pm R$ is not guaranteed to be preserved. A series might converge conditionally at an endpoint, but its derivative could diverge at that same point due to the multiplication by $k$ in the coefficients. Therefore, while the open interval $(a-R, a+R)$ is safe, endpoints require separate testing for the differentiated series.

📖 Explanation: The target series $\sum (-1)^k x^{k-1}$ is precisely the term-by-term derivative of $f(x) = \sum \frac{(-1)^k}{k} x^k$. Since power series can be differentiated within their radius of convergence, and the derivative of $\frac{(-1)^k}{k} x^k$ is $(-1)^k x^{k-1}$, this is the direct path. The radius of convergence remains $R=1$. Integration would yield powers of $x^{k+1}$, and substitution ignores the given function's structure. This tests understanding of the inverse relationship between series coefficients and calculus operations.

📖 Explanation: Power series representing physical quantities like position are infinitely differentiable within their open interval of convergence $(-R, R)$. Since $R=5$, the interval is $(-5, 5)$. The point $t=4$ is strictly inside this interval. Therefore, both velocity and acceleration can be obtained by successive term-by-term differentiation. The radius does not shrink with differentiation, and proximity to the boundary (as long as it is within) does not invalidate the analytical derivative. This applies calculus concepts to kinematic modeling.

📖 Explanation: This question mixes integration, differentiation, and convergence types. By definition, $g(x)$ is the antiderivative of the series. By the Fundamental Theorem of Calculus, g'(x) recovers the original series $\sum c_k x^k$ at any point where that series is continuous. Power series are continuous within their open interval of convergence $(-R, R)$. Absolute convergence is a sufficient condition for many operations but is not necessary for the basic FTC relationship inside the radius. The student confuses sufficient conditions for swapping limits with the fundamental property of antiderivatives.

📖 Explanation: Graphical analysis of partial sums provides visual evidence of theoretical properties. Inside the interval $(-R, R)$, $S_N(x)$ converges uniformly to $f(x)$, and S_N'(x) converges to f'(x). However, at the boundary $x=R$, uniform convergence often fails. Wild oscillations in the derivative approximations S_N'(x) near the boundary signal that the derived series does not converge nicely there. This aligns with the theorem that differentiation preserves the radius $R$ but not necessarily convergence at the endpoints. Visualizing this helps distinguish between interior smoothness and boundary pathology.

📖 Explanation: When differentiating $\sum_{k=0}^{\infty} c_k x^{2k}$, the $k=0$ term is a constant $c_0$. Its derivative is 0. However, applying the power rule mechanically yields $2(0)x^{2(0)-1} = 0 \cdot x^{-1}$. While the limit is 0, the expression $x^{-1}$ is singular at $x=0$. Rigorous manipulation requires separating the constant term or shifting the index to start at $k=1$ before writing the general term. This tests attention to detail in index manipulation and domain definitions, distinguishing formal algebraic correctness from loose heuristic computation.

📖 Explanation: First, determine the series for $h(x)$. Substituting $x^2$ into $\sum a_n x^n$ gives $\sum a_n x^{2n}$. This series converges when $|x^2| < 2$, i.e., $|x| < \sqrt{2}$. So $h(x)$ has radius $R_h = \sqrt{2}$. Differentiating $h(x)$ term-by-term yields h&#039;(x) = \sum 2n a_n x^{2n-1}. Theorem states differentiation preserves the radius of convergence. Thus, h&#039;(x) also has radius $\sqrt{2}$. Students often mistakenly apply the radius of $f$ (which is 2) or forget that substitution alters the effective radius before differentiation occurs.

📖 Explanation: If a function $f(x)$ is represented by a power series $\sum c_k x^k$ on an open interval containing 0, then by uniqueness of power series representations, $c_k$ must equal $f^{(k)}(0)/k!$. Therefore, f&#039;&#039;(0) must exactly equal $2c_2$. Any discrepancy implies either a calculation error or that the series does not actually converge to the function (e.g., the function is non-analytic). This reinforces the deep link between series coefficients and derivatives, ruling out misconceptions about asymptotic vs. exact equality within the radius.

📖 Explanation: Method B uses the standard geometric series $\sum (-1)^k x^k$, which is immediate. Method A requires starting with $\sum (-1)^{k+1} x^k / k$, differentiating to get $\sum (-1)^{k+1} k x^{k-1} / k$, simplifying to $\sum (-1)^{k+1} x^{k-1}$, and re-indexing. Method B bypasses the factorial cancellation and index shift entirely. While both yield the same result, Method B minimizes algebraic steps and potential indexing mistakes. This compares procedural fluency and strategic selection of tools, highlighting that 'differentiating power series' is a tool, not always the optimal first step.

📖 Explanation: Even within the radius of convergence, a truncated series $\sum_{n=0}^N a_n t^n$ approximates $f(t)$ with some error. For the derivative, the coefficients become $n a_n$. As $n$ increases, this factor amplifies the magnitude of higher-order terms. Consequently, for a fixed truncation $N$, the remainder term for the derivative is generally larger than for the function itself, especially as $t$ approaches $R$. This explains why the derivative approximation degrades faster. It connects the theoretical operation of differentiation to practical numerical analysis and modeling limitations.

📖 Explanation: The radius of convergence for the derived series is identical to the original, so $R=4$. The center is still 3. Thus, the open interval is $(3-4, 3+4) = (-1, 7)$. Crucially, term-by-term differentiation is guaranteed to represent the derivative only within this open interval. Behavior at $x=-1$ and $x=7$ depends on the specific coefficients and must be tested separately; it is not automatically inherited. Option B correctly identifies the guaranteed domain without making unwarranted assumptions about endpoints.

📖 Explanation: Differentiating $\sum x^k = \frac{1}{1-x}$ gives $\sum k x^{k-1} = \frac{1}{(1-x)^2}$. Multiplying by $x$ yields $\sum k x^k = \frac{x}{(1-x)^2}$. Differentiating this result gives $\sum k^2 x^{k-1}$. Multiplying by $x$ again yields $\sum k^2 x^k$. The student stopped after the first derivative and misidentified the resulting series. The key technique for generating $k^p$ factors is the operator $x \frac{d}{dx}$ applied repeatedly. This tests multi-step reasoning and operator manipulation rather than simple recall.

📖 Explanation: This addresses a subtle misconception. Convergence of $\sum a_n R^n$ does not imply convergence of $\sum n a_n R^{n-1}$. Counterexample: $\sum_{n=1}^{\infty} \frac{(-1)^n}{n} x^n$ converges at $x=1$ (alternating harmonic). Its derivative is $\sum_{n=1}^{\infty} (-1)^n x^{n-1}$, which at $x=1$ becomes $\sum (-1)^n$, a divergent oscillating series. This demonstrates that endpoint convergence is fragile under differentiation. Students must distinguish between interior convergence (always preserved) and boundary behavior (not preserved).

📖 Explanation: Multiplying two series $(\sum a_n t^n)(\sum b_n t^n)$ yields Cauchy products $c_n = \sum a_k b_{n-k}$. Differentiating this convoluted form requires handling nested sums. Alternatively, knowing (fg)&#039; = f&#039;g + fg&#039;, one can differentiate the simple series for $e^{-t}$ and $\cos(t)$ individually (which is trivial) and then multiply/add. This leverages linearity and the product rule to avoid complex coefficient algebra. It highlights strategic efficiency in series manipulation for physical modeling.

📖 Explanation: Substituting $f(x) = \sum a_n x^n$ and f&#039;(x) = \sum n a_n x^{n-1} into f&#039; = 2xf yields $\sum n a_n x^{n-1} = \sum 2 a_n x^{n+1}$. Shifting indices allows equating coefficients of like powers, producing a recurrence (e.g., $a_{n+2}$ in terms of $a_n$). This transforms calculus into algebra. While separation of variables works here, series differentiation is a universal method for ODEs without elementary closed forms. This tests understanding of series as a solution technique for differential equations.

📖 Explanation: The infinite series for $\cos(x)$ converges everywhere. Its derivative $-\sin(x)$ also converges everywhere. However, we are using a *truncated* partial sum $S_{10}(x)$. Near the edges of the approximation interval, the neglected tail of the series contributes significantly. Differentiation multiplies the $k$-th term by $k$, effectively weighting higher frequencies more heavily. Thus, the truncation error in the derivative is amplified relative to the function. This explains why derivative approximations often look worse near boundaries even when the function approximation looks good.

📖 Explanation: A fundamental theorem of power series states that term-by-term differentiation preserves the radius of convergence exactly. R&#039; = R. This holds regardless of absolute convergence, sign alternation, or boundary behavior. The limit superior formula $1/R = \limsup |c_k|^{1/k}$ yields the same value for $k c_k$ as for $c_k$ because $\lim k^{1/k} = 1$. Options suggesting inequality confuse radius (interior) with interval (endpoints). This is direct recall but foundational for all subsequent HOTS questions.

📖 Explanation: The student correctly notes that divergence at an endpoint is preserved (if $\sum a_n$ diverges, $\sum n a_n$ usually diverges more strongly). However, logically, 'Divergence implies Divergence' does not entail 'Convergence implies Convergence'. As shown in previous counterexamples, a series can converge at an endpoint while its derivative diverges. The logical structure is flawed. This tests error analysis in reasoning about implications, distinguishing between valid contrapositives and invalid converses in the context of series endpoints.

📖 Explanation: The integrand $\frac{e^x-1}{x} = \sum_{n=1}^{\infty} \frac{x^{n-1}}{n!}$ is naturally suited for term-by-term integration to find the area. Differentiation would increase the power of $x$ in the denominator or introduce negative powers, complicating the definite integral evaluation. While differentiation is a powerful tool, recognizing when *not* to use it is part of conceptual understanding. Here, integration is the direct path to the numerical value. This tests strategic tool selection.

📖 Explanation: By chain rule, g&#039;(x) = f&#039;(x^3) \cdot 3x^2. The series for f&#039;(u) = \sum n a_n u^{n-1}. Substituting $u=x^3$, f&#039;(x^3) = \sum n a_n x^{3n-3}. Multiplying by $3x^2$ gives g&#039;(x) = \sum 3n a_n x^{3n-1}. We seek the coefficient of $x^5$. Set $3n-1 = 5 \Rightarrow 3n=6 \Rightarrow n=2$. The coefficient is $3(2)a_2 = 6a_2$. Wait, let's re-evaluate option A vs C. If $n=2$, coeff is $6a_2$. Option A says 0. Option C says $6a_2$. Correct answer is C. (Self-correction: Ensure options match derivation. Derivation shows $6a_2$. Option C is correct.)

📖 Explanation: This distinguishes power series from other series. Power series $\sum c_k (x-a)^k$ can always be differentiated term-by-term within $R$. Trigonometric series $\sum c_k \sin(kx)$ do not share this automatic property. The derived series $\sum \frac{\cos(kx)}{k}$ converges conditionally (not absolutely/uniformly everywhere), and the original series' convergence does not guarantee the differentiated series represents the derivative. This tests the boundary of the 'Differentiating Power Series' topic by contrasting it with Fourier analysis, preventing overgeneralization.

📖 Explanation: Differentiating $\sum_{k=1}^{\infty} \frac{x^k}{k^2}$ gives $\sum_{k=1}^{\infty} \frac{k x^{k-1}}{k^2} = \sum_{k=1}^{\infty} \frac{x^{k-1}}{k}$. Factoring $1/x$ yields $\frac{1}{x} \sum_{k=1}^{\infty} \frac{x^k}{k}$. Recognizing the Maclaurin series for $-\ln(1-x)$, this simplifies to $-\frac{\ln(1-x)}{x}$. This confirms the CAS output. This application links series differentiation to special functions and symbolic verification, demonstrating utility beyond textbook exercises.

📖 Explanation: The series derivation via repeated differentiation of geometric series is correct: $\frac{d^2}{dx^2}(1-x)^{-1} = 2(1-x)^{-3}$, so $(1-x)^{-3} = \frac{1}{2} \sum k(k-1)x^{k-2}$, which simplifies to $\sum \binom{k+2}{2} x^k$ or similar. The student's coefficient form is plausible. However, the radius of convergence is dictated by the distance to the nearest singularity in the complex plane, which is $x=1$. Thus $R=1$. The exponent 3 affects coefficient growth, not the location of the singularity. This corrects a common dimensional misconception.

📖 Explanation: This invokes Abel's Theorem for derivatives. If $a_n \geq 0$, then f&#039;(x) = \sum n a_n x^{n-1} is a power series with non-negative coefficients. If \lim_{x \to 1^-} f&#039;(x) = L exists and is finite, then by Abel's Theorem (monotone convergence variant), the series $\sum n a_n$ converges and its sum equals $L$. Non-negativity is crucial; without it, the limit existing doesn't guarantee series convergence. This tests deep theoretical connections between limits and series sums.

📖 Explanation: The validity of $\frac{d}{dx} \lim S_n = \lim \frac{d}{dx} S_n$ requires uniform convergence of the sequence of derivatives S_n&#039; and convergence of $S_n$ at least at one point. Power series possess this property on any compact subset inside the radius of convergence due to the Weierstrass M-test. Arbitrary series may converge uniformly themselves but have derivatives that do not converge uniformly, breaking the interchange. This explains the 'why' behind the theorem.

📖 Explanation: The series is geometric. Closed form is $\frac{C}{r-g}$. Derivative is $\frac{-C}{(r-g)^2}$. But differentiating the series term-by-term gives $\sum -t \frac{C(1+g)^{t-1}}{(1+r)^{t+1}}$. This expresses sensitivity as a sum of time-weighted cash flows, which is the definition of Macaulay Duration. Series differentiation reveals the structural components (time weighting) that the closed form obscures. This demonstrates modeling insight where series representation is analytically superior for interpretation.

📖 Explanation: For $e^x$, f&#039;(x) = f(x). The series $\sum \frac{x^k}{k!}$ differentiated is $\sum \frac{k x^{k-1}}{k!} = \sum \frac{x^{k-1}}{(k-1)!}$. Ideally, these are identical. However, truncating at $N=20$ means the derivative loses the highest term and shifts. $y_2$ as described matches the form of the derivative series. The divergence illustrates that finite truncations of $e^x$ do not satisfy f&#039;=f exactly; the error grows with $x$. Graphs visualize the breakdown of functional identities under truncation.

📖 Explanation: Direct differentiation of $\arctan(x)$ becomes messy quickly. The Maclaurin series $x - x^3/3 + x^5/5 - \dots$ provides coefficients directly. By definition, coefficient of $x^n$ is $f^{(n)}(0)/n!$. For $n=5$, coeff is $1/5$ (wait, sign: $k=2 \implies (-1)^2 = +1$? No, $\arctan x = x - x^3/3 + x^5/5$. Sign is positive. So $f^{(5)}(0)/120 = 1/5 \implies f^{(5)}(0) = 24$. Note: Option A says -24. Let's recheck. $\arctan x = \sum (-1)^k x^{2k+1}/(2k+1)$. $k=0: x$. $k=1: -x^3/3$. $k=2: +x^5/5$. Coeff is $+1/5$. So answer should be 24. If Option A says -24, it's a distractor or I need to check signs. Actually, $\arctan x$ derivatives at 0: f&#039;(0)=1, f&#039;&#039;&#039;(0)=-2, f^{(5)}(0)=24. Option A text in my draft had -24. I will correct the option to 24 in final output or select the correct logic. Let's assume the option meant 24. Revised Option A: '...Coefficient is 1/5. Thus f^(5)(0) = 24.' This validates the method.

📖 Explanation: Mechanical application of power rule to $k=0$ term $x^0=1$ gives derivative 0. Writing it as $0 \cdot x^{-1}$ introduces a singularity. Properly, the constant term vanishes upon differentiation, and the new series starts at $k=1$. At $x=0$, only the $k=1$ term ($1 \cdot x^0$) survives, yielding 1. All higher terms vanish. This highlights the importance of index management and treating constant terms separately in series differentiation.

📖 Explanation: The radius is determined by $\limsup |a_n|^{1/n}$. For the modified series, we examine $\limsup |a_n n^p|^{1/n} = \limsup |a_n|^{1/n} \cdot (n^{1/n})^p$. Since $\lim_{n \to \infty} n^{1/n} = 1$, the factor $(n^{1/n})^p \to 1^p = 1$. Thus, the limsup is unchanged. Polynomial factors in $n$ do not affect the radius of convergence. This generalizes the differentiation result (where $p=1$) and strengthens conceptual grasp of root test mechanics.

📖 Explanation: The product rule on the whole function gives e^{-\lambda t} (S&#039;(t) - \lambda S(t)). Setting to zero requires solving S&#039;(t) = \lambda S(t). If $S(t)$ is a series, substituting the series forms allows matching coefficients or truncating to find approximate roots. This formulation separates the decay envelope from the polynomial dynamics. It shows how series differentiation facilitates solving transcendental equations arising in pharmacokinetics by reducing them to algebraic relations among coefficients.

📖 Explanation: This combines series differentiation with nonlinear ODEs. f&#039; = \sum n a_n x^{n-1}. $f^2 = \sum c_n x^n$ where $c_n = \sum_{k=0}^n a_k a_{n-k}$. Matching coeff of $x^{n-1}$ in f&#039; to $x^{n-1}$ in $f^2$: $n a_n = \sum_{k=0}^{n-1} a_k a_{n-1-k}$. For $n=1$: $a_1 = a_0^2 = 1$. $n=2$: $2a_2 = 2a_0 a_1 = 2 \implies a_2=1$. $n=3$: $3a_3 = 2a_0 a_2 + a_1^2 = 2(1)(1) + 1 = 3 \implies a_3=1$. This tests recursive coefficient generation via differentiation.

📖 Explanation: This is the core definition. Key elements: valid inside open interval $(-R, R)$, preserves radius $R$, represents the derivative f&#039;(x). Distractors include common misconceptions about endpoints or radius changes. Essential foundation for HOTS.

📖 Explanation: Inside $(-R, R)$, the differentiated series converges to f&#039;(x). At endpoints, even if $\sum a_n R^n$ converges to $f(R)$, $\sum n a_n R^{n-1}$ may diverge or converge to something else. Preservation of the functional relationship is guaranteed only in the interior. This nuance is critical for rigorous analysis.

📖 Explanation: Strategy B is direct: $\cos(x^2) = \sum (-1)^k x^{4k}/(2k)!$. Multiply by $x$: $\sum (-1)^k x^{4k+1}/(2k)!$. Integrate: $\sum (-1)^k x^{4k+2}/[(4k+2)(2k)!]$. Evaluate at 0.1. Strategy A works but adds substitution steps. Strategy C moves away from the integral. B leverages the simplicity of integrating monomials. Tests efficient algorithm selection.

📖 Explanation: Standard pitfall. $k=0$ term is 1. Derivative is 0. Formula $k x^{k-1}/k! = x^{k-1}/(k-1)!$ fails at $k=0$. Correct series starts at $k=1$. This ensures all terms are well-defined. Tests precision in series notation.

📖 Explanation: Wild oscillations in partial sums of derivatives near the boundary indicate lack of uniform convergence. This happens when $n a_n$ doesn't decay sufficiently fast (e.g., $a_n \sim 1/n^2 \implies n a_n \sim 1/n$, harmonic-like divergence/oscillation). Smooth $S_N$ suggests $a_n$ decays faster than $1/n$. Visual diagnostics link coefficient asymptotics to convergence quality.

📖 Explanation: Original series $\sum (-1)^{k-1} x^k/k$ converges at $x=1$ conditionally. To prove sum is $\ln 2$, relate to integral of derivative. f&#039;(x) = 1/(1+x) = \sum (-1)^k x^k. Integrate 0 to 1: $\ln 2 = \sum (-1)^k /(k+1)$. Justification requires Abel's theorem or AST at endpoint since geometric series diverges at 1. Connects differentiation, integration, and endpoint theory.

📖 Explanation: For polynomials, identity is trivial. For infinite series with $R=1$, f&#039;(1) is defined as limit of difference quotient or limit of derivative. Equality \sum k a_k = f&#039;(1) holds if the series of derivatives converges. Mere convergence of $\sum a_k$ or continuity of $f$ is insufficient (counterexamples exist). Requires convergence of $\sum k a_k$. Tests extension of finite identities to infinite domains with precise conditions.