📝 Functions defined by power series (35 MCQs)

📖 Explanation: Differentiation of a power series preserves the radius of convergence $R$, meaning $g(x)$ definitely converges on $(-3, 3)$. However, behavior at the specific endpoints $x = \pm R$ is not guaranteed to remain the same. A series might converge conditionally at an endpoint while its derivative diverges there due to the multiplication by $n$ in the coefficient term. Therefore, one must re-test endpoints separately after differentiation.

📖 Explanation: While recognizing the series as $\cos x$ yields the answer quickly, the question requires using the properties of power series directly. Differentiating term-by-term gives f'(x) = \sum \frac{(-1)^k (2k) x^{2k-1}}{(2k)!} and f''(x) = \sum \frac{(-1)^k (2k)(2k-1) x^{2k-2}}{(2k)!}. Evaluating at $x=0$, only the constant terms survive. For $f(0)$, the $k=0$ term is 1. For f''(0), the $k=1$ term is -1. Summing them gives 0, demonstrating that operations on series are valid independent of closed-form identification.

📖 Explanation: The student confused the antiderivative of $\frac{1}{1+x^2}$ with $\ln(1+x^2)$; it is actually $\arctan(x)$. To correctly find the series for $\ln(1+x^2)$, one should differentiate to get $\frac{2x}{1+x^2} = 2x \sum (-1)^k x^{2k}$. Integrating this correct expansion yields $\sum (-1)^k \frac{x^{2k+2}}{k+1}$. This highlights the importance of verifying calculus operations before applying series manipulations. Simply matching forms without checking derivatives leads to plausible-looking but incorrect series representations.

📖 Explanation: Power series converge uniformly on compact subsets strictly inside the radius of convergence. Wild oscillation or divergence of partial sums near the boundary $x=R$ typically indicates the presence of a singularity (pole, branch point, or essential singularity) on the circle of convergence in the complex plane. Even if the function appears well-behaved on the real line at $x=R$, complex singularities dictate the radius. The graph visually confirms the theoretical limit of the series representation's validity.

📖 Explanation: When the function is unknown, we cannot compute $f^{(n)}(0)$ directly. Instead, we substitute $\sum c_n t^n$ and $\sum n c_n t^{n-1}$ into the ODE. Matching coefficients of like powers of $t$ generates a recurrence relation (e.g., $c_{n+1} = \frac{-k}{n+1}c_n$). This constructs the series from the governing law itself. This distinction is crucial in physics where equations define functions (like Bessel functions) that have no elementary closed form, making the recursive series definition primary.

📖 Explanation: Using the Root Test on $g(x)$, we examine $\lim |a_n x^{2n}|^{1/n} = \lim |a_n|^{1/n} |x|^2 = L|x|^2$. For convergence, we require $L|x|^2 < 1$, which implies $|x|^2 < 1/L$ or $|x| < \sqrt{1/L}$. This demonstrates how variable substitution within a power series transforms the radius of convergence. Students often mistakenly apply the original radius $1/L$ without accounting for the exponent change. This problem tests deep understanding of the Root Test mechanism versus memorized formulas.

📖 Explanation: This is a classic counterexample in analysis. Although $f(x)$ is smooth ($C^\infty$), $f^{(n)}(0) = 0$ for all $n$. Thus, the Taylor series is $\sum 0 \cdot x^n = 0$. Since $f(x) > 0$ for $x \neq 0$, the series fails to represent the function anywhere except the center. This proves that infinite differentiability is necessary but not sufficient for a function to be analytic (representable by a power series). It distinguishes between smooth functions and analytic functions.

📖 Explanation: Computing $f^{(n)}(0)$ for $\tan(x)$ directly involves messy higher-order derivatives of secants and tangents. Conversely, knowing $\sin x = x - x^3/6 + \dots$ and $\cos x = 1 - x^2/2 + \dots$, one can perform polynomial long division or equate coefficients in $(\sum a_n x^n)(\cos x) = \sin x$. This reduces calculus problems to algebra, minimizing computational error and providing a scalable algorithm for generating terms. This efficiency makes series manipulation superior for constructing expansions of composite or quotient functions.

📖 Explanation: Since the integrated series for $\int e^{-t^2} dt$ is alternating with decreasing magnitude terms for small $x$, the Alternating Series Estimation Theorem applies. The error in approximating by the $n$-th partial sum is bounded strictly by the magnitude of the first neglected term $|a_{n+1}|$. This is far simpler and sharper than the Lagrange remainder for this specific series type. Identifying the series structure (alternating vs. general) is a critical step in selecting the most efficient error estimation strategy.

📖 Explanation: Term-by-term integration preserves the radius of convergence $R$. Within $(a-R, a+R)$, the new series represents a continuous, differentiable antiderivative. However, endpoint behavior can change dramatically. A series might diverge at an endpoint while its integral converges there (e.g., geometric series vs. log series). Integration tends to 'improve' convergence at boundaries because coefficients are divided by $n$. Therefore, one cannot assume the interval of convergence remains identical; endpoints must always be re-evaluated after integration.

📖 Explanation: For non-integer $m$, $(1+x)^m$ is multi-valued or undefined for $x \le -1$ (depending on branch cuts). In complex analysis, the radius of convergence of a Taylor series centered at 0 is exactly the distance to the nearest singularity. Here, the singularity is at $x=-1$, so $R=1$. In contrast, $e^x$ is entire (no singularities in the finite complex plane), giving $R=\infty$. This connects real-variable series behavior to complex analytic structure, explaining why algebraic functions typically have finite radii.

📖 Explanation: The identity $f(x) = \sum c_n x^n$ is conditional on $x$ being within the interval of convergence. At $x=2$, the series $\sum 2^n$ diverges to infinity; it does not possess a sum in the standard sense. While analytic continuation extends the *function* $1/(1-x)$ to $x=2$, it does not extend the *series representation*. Confusing the function with its specific series representation outside the convergence disk is a fundamental misconception. The series ceases to represent the function beyond $R=1$.

📖 Explanation: The product of two power series converges absolutely at least on the intersection of their individual intervals of convergence. Since $A(x)$ converges for $|x|<2$ and $B(x)$ for $|x|<3$, both converge absolutely for $|x|<2$. Thus, their Cauchy product converges for $|x|<2$, implying $R \ge 2$. It could be larger if singularities cancel out (e.g., $(1-x) \times \frac{1}{1-x} = 1$), but it cannot be smaller than the minimum of the two radii. This tests understanding of lower bounds versus exact values.

📖 Explanation: Multiplying by $x$ gives $\sum \frac{(-1)^k x^{2k+1}}{2^{2k}(k!)^2}$. Integrating term-by-term from 0 to 1 yields $\sum \frac{(-1)^k}{(2k+2) 2^{2k} (k!)^2}$. This transforms a difficult special-function integral into a simple alternating series of rational numbers that can be summed to arbitrary precision. This exemplifies the utility of power series: they convert transcendental operations into arithmetic ones. Direct lookup or numerical quadrature misses the opportunity to leverage the series structure for exact, controllable computation.

📖 Explanation: If $\sec x = \sum s_n x^n$ and $\cos x = \sum c_n x^n$, then $(\sum s_n x^n)(\sum c_n x^n) = 1$. This requires the Cauchy product of the coefficients to satisfy convolution identities (e.g., $s_0 c_0 = 1, s_0 c_2 + s_2 c_0 = 0$), not simple element-wise inversion $s_n = 1/c_n$. Assuming $1/\sum a_n x^n = \sum (1/a_n) x^n$ is a common algebraic fallacy. Correctly finding reciprocal series requires solving a system of linear equations derived from the product condition.

📖 Explanation: A power series with $R=\infty$ defines an entire function. A fundamental theorem states that functions represented by power series are infinitely differentiable within their interval of convergence. Since the interval is $(-\infty, \infty)$, $f(x)$ has derivatives of all orders everywhere. It need not be a polynomial (e.g., $e^x$), bounded (e.g., $e^x$), or periodic (e.g., $e^x$). This question tests the basic regularity property implied by power series representation.

📖 Explanation: For $\ln(1.1)$, standard series uses $x=0.1$. Using $\ln(\frac{1+y}{1-y}) = 2(y + y^3/3 + \dots)$, we set $\frac{1+y}{1-y} = 1.1 \Rightarrow y = 1/21 \approx 0.0476$. Since convergence rate depends on the magnitude of the argument, $y \approx 0.05$ yields accuracy much faster than $x=0.1$. This demonstrates practical numerical analysis: transforming the input to accelerate series convergence is often more efficient than simply adding more terms to a slowly converging expansion.

📖 Explanation: Substituting series into f&#039; = 2xf: $\sum n a_n x^{n-1} = \sum 2 a_n x^{n+1}$. Shifting indices to match powers of $x^k$: $(k+1)a_{k+1} = 2a_{k-1}$ for $k \ge 1$. With $a_0=1$, we get $a_1=0$ (since f&#039; has no constant term). Then $2a_2 = 2a_0 \Rightarrow a_2=1$. $3a_3 = 2a_1 = 0$. $4a_4 = 2a_2 = 2 \Rightarrow a_4 = 1/2$. This recursive extraction of coefficients bypasses closed-form solutions and tests ability to manipulate series indices within differential equations.

📖 Explanation: The radius of convergence of a Maclaurin series is determined by the distance from the center (0) to the nearest singularity in the complex plane. On a real graph, a vertical asymptote, cusp, or point of non-differentiability at $x=a$ serves as a visible barrier. The series cannot converge beyond this point. Horizontal asymptotes or symmetry do not restrict $R$ (e.g., $\frac{1}{1+x^2}$ has horizontal asymptote but finite $R$ due to complex poles; $e^{-x^2}$ has asymptote and infinite $R$). Visible real singularities are definitive indicators.

📖 Explanation: Matching derivatives at a point generates the Taylor series, but does not guarantee it converges *to the function*. One must prove $\lim_{n\to\infty} R_n(x) = 0$. For $e^x$, the Lagrange remainder involves $e^c \frac{x^{n+1}}{(n+1)!}$. Since factorial growth dominates exponential growth for fixed $x$, this limit is zero. Without this step, one has only a formal series, not a functional representation. This distinguishes between 'finding a series' and 'proving a representation,' a critical theoretical nuance.

📖 Explanation: The Identity Theorem states that if $\sum a_n (x-c)^n = \sum b_n (x-c)^n$ for all $x$ in some neighborhood of $c$, then $a_n = b_n$ for all $n$. This justifies methods like undetermined coefficients and series substitution. If we derive a series for $f(x)$ via any valid method (integration, multiplication, etc.), it *must* be the Taylor series. This uniqueness underpins the consistency of all series manipulation techniques; there is no ambiguity in the representation.

📖 Explanation: Maclaurin series are power series centered at 0. If $y(x) \sim 1/x$ at infinity, it suggests algebraic decay. More critically, functions with Maclaurin series are analytic at 0. If the function actually behaves like $1/x$ globally or has a pole at 0, no Maclaurin series exists. Even if regular at 0, asymptotic behavior at infinity often correlates with finite radius of convergence (singularities in complex plane). A single power series centered at 0 cannot capture behavior across an infinite domain if singularities exist. Global representation requires other tools (asymptotic expansions, Padé approximants).

📖 Explanation: When expanding $(\sum a_k x^k)(\sum b_m x^m)$, the term $x^n$ arises from all pairs $(k,m)$ such that $k+m=n$. Thus, the coefficient is $\sum_{k=0}^n a_k b_{n-k}$. Simple term-wise multiplication $a_n b_n x^n$ ignores cross-terms like $a_1 x \cdot b_{n-1} x^{n-1}$. Understanding this combinatorial structure is essential for correctly computing products like $e^x \sin x$ or solving ODEs with variable coefficients. It reflects the algebraic structure of the ring of formal power series.

📖 Explanation: This is the canonical example illustrating that real smoothness does not imply global analyticity. The function $1/(1+x^2)$ has poles at $x = \pm i$ in the complex plane. The radius of convergence of a Taylor series centered at 0 is the distance to the nearest complex singularity, which is $|i-0|=1$. Despite perfect behavior on $\mathbb{R}$, the complex structure constrains the real series. This emphasizes that power series convergence is inherently a complex-analytic property, not merely a real-variable one.

📖 Explanation: While Lagrange remainder works generally, for alternating series with decreasing terms, the error is strictly bounded by the first omitted term $|a_{n+1}|$. For $\sin(0.1)$, this is $(0.1)^5/120 \approx 8.3 \times 10^{-8}$. Lagrange would require bounding $|\cos(c)|$ or $|\sin(c)|$, yielding a similar but less direct bound. Recognizing the alternating structure simplifies error analysis significantly. This question tests strategic selection of error estimation tools based on series characteristics rather than defaulting to the most general theorem.

📖 Explanation: Although the derivative series $\sum (-1)^n x^{2n}$ diverges at $x=1$, the integrated series $\sum (-1)^n/(2n+1)$ converges conditionally by AST. Abel's Theorem guarantees that if a power series converges at an endpoint and the function is continuous there, the series sum equals the function value. Thus, $\sum (-1)^n/(2n+1) = \arctan(1) = \pi/4$. This subtle interplay between differentiation, integration, and endpoint continuity is central to rigorous series analysis.

📖 Explanation: The recurrence $a_{n+2} = f(n) a_n$ links terms with step 2. This decouples the sequence into even terms (determined by $a_0$) and odd terms (determined by $a_1$). Since $a_0 = y(0)$ and a_1 = y&#039;(0) are arbitrary initial conditions, the general solution is $y(x) = a_0 y_{even}(x) + a_1 y_{odd}(x)$. This structural splitting naturally generates the two-dimensional solution space required for second-order linear ODEs. Understanding this link between recurrence parity and solution independence is key to the series method.

📖 Explanation: Start with $\sum x^n = (1-x)^{-1}$. Differentiate: $\sum n x^{n-1} = (1-x)^{-2}$. Multiply by $x$: $\sum n x^n = x(1-x)^{-2}$. Differentiate again: $\sum n^2 x^{n-1} = \frac{d}{dx}[x(1-x)^{-2}] = (1-x)^{-2} + 2x(1-x)^{-3}$. This path gets complicated. Better: Differentiate $(1-x)^{-1}$ twice to get $2(1-x)^{-3} = \sum n(n-1)x^{n-2}$. Multiply by $x^2/2$: $x^2(1-x)^{-3} = \sum \frac{n(n-1)}{2} x^n$. Adjusting indices/algebra yields the target. Systematic operator application (differentiate/multiply) generates rational function series efficiently.

📖 Explanation: Substituting $x^2$ for $x$ means the new series converges when $|x^2| < R$, i.e., $|x| < \sqrt{R}$. This is a direct consequence of the definition of radius of convergence. Students sometimes confuse this with differentiation/integration (which preserve $R$) or coefficient scaling. Variable transformation fundamentally alters the domain of convergence. This basic fact is essential for composing functions and understanding how series domains map under nonlinear transformations.

📖 Explanation: Theoretically, the series converges for all $x$. Practically, for $x=-20$, terms grow to $\approx 20^{20}/20! \approx 4 \times 10^7$ before decaying. Summing huge alternating terms in finite precision causes massive loss of significance (catastrophic cancellation). The true value $e^{-20} \approx 2 \times 10^{-9}$ is lost in rounding noise. This illustrates that theoretical convergence does not guarantee numerical stability. For large negative arguments, one should compute $e^{20}$ and take the reciprocal, avoiding alternating large terms.

📖 Explanation: A fundamental property of functions defined by power series is that they are infinitely differentiable ($C^\infty$) within their interval of convergence. Since $|x|$ lacks a first derivative at $x=0$, it cannot equal any power series centered at 0. This provides a quick diagnostic: any lack of smoothness at the center immediately precludes power series representation. This contrasts with Fourier series, which can represent non-smooth functions, highlighting the restrictive regularity of power series.

📖 Explanation: Since $f(x) = \sum a_n x^n$, we know $a_n = f^{(n)}(0)/n!$. Expanding $\sin(t^2) = t^2 - t^6/6 + \dots$, integrating gives $f(x) = x^3/3 - x^7/42 + \dots$. The coefficient of $x^5$ is zero, so $f^{(5)}(0) = 0$. Direct differentiation of an integral with composite argument is tedious. Series conversion turns differentiation into coefficient inspection. This showcases the power of series as a computational tool for extracting local derivative information without symbolic calculus.

📖 Explanation: While special functions are exact, they are often opaque 'black boxes'. A truncated power series $\sum_{n=0}^N a_n x^n$ is a simple polynomial: easy to differentiate, integrate, optimize, and implement in hardware. For small arguments, it captures dominant physics clearly (e.g., linear vs. quadratic regimes). It trades global exactness for local transparency and computational simplicity. This pragmatic choice drives much of applied mathematics, where interpretability and efficiency often outweigh formal closed-form elegance.

📖 Explanation: Differentiation preserves radius $R$ but not necessarily endpoint convergence. Example: $\sum (-1)^n x^n / n^2$ converges absolutely at $x=1$. Its derivative $\sum (-1)^n x^{n-1} / n$ converges conditionally at $x=1$. But consider $\sum (-1)^n x^n / n$; converges conditionally at $x=1$. Derivative $\sum (-1)^n x^{n-1}$ diverges at $x=1$. Multiplication by $n$ amplifies terms, potentially breaking conditional convergence. Endpoint behavior must always be re-tested after differentiation.

📖 Explanation: First, $e^{x^2} \neq (e^x)^2 = e^{2x}$. These are different functions. Second, even if squaring $e^x$, $(\sum a_n)^2 \neq \sum a_n^2$; one needs Cauchy product. Third, the correct series for $e^{x^2}$ comes from substituting $x^2$ into $e^u$: $\sum (x^2)^n/n! = \sum x^{2n}/n!$. The student committed multiple conceptual errors: confusing function composition with algebraic operations and misunderstanding series arithmetic. Identifying all layers of error demonstrates comprehensive mastery.