📝 Power series uniqueness Taylor (36 MCQs)

📖 Explanation: The fundamental uniqueness theorem for power series states that if a function can be represented by a power series centered at $x_0$ on some open interval, that representation must be the Taylor series. Therefore, regardless of whether one uses substitution, integration, or differentiation to find the series, as long as it converges to the function, it is identical to the Taylor series derived via derivative formulas. This eliminates the need to verify coefficients through direct differentiation.

📖 Explanation: When a known Maclaurin series is manipulated via valid algebraic substitution (like $x \to -x^2$) within its interval of convergence, the resulting series converges to the composite function. By the uniqueness of power series representations, this new series must be the Maclaurin series for the composite function. Consequently, the coefficients obtained via substitution are exactly equal to $g^{(n)}(0)/n!$, rendering direct computation of high-order derivatives at zero redundant and inefficient compared to the substitution method.

📖 Explanation: A necessary condition for a function to have a Maclaurin series representation is that it must possess derivatives of all orders at $x=0$. The function $f(x)=|x|$ fails this criterion because its first derivative is undefined at the origin. While algebraic manipulations might formally produce a series expression, that series cannot represent $|x|$ as a Maclaurin series because the underlying function lacks the required smoothness. Power series representations imply infinite differentiability within their interval of convergence.

📖 Explanation: The uniqueness property of power series is a cornerstone of analysis. It asserts that there is only one power series centered at a specific point that converges to a given function on an open interval. Therefore, any valid method—be it direct differentiation, substitution, multiplication, or integration—that yields a convergent power series for the function must yield the exact same series. Discrepancies between methods indicate a calculation error in one of the approaches, not a difference in the mathematical objects themselves.

📖 Explanation: Both integration of a related series and direct substitution are valid techniques for generating power series. Since the target function $\frac{1}{1+x^2}$ is analytic at $x=0$, its Maclaurin series is unique. Integrating the derivative's series recovers the original function's series (up to a constant, which is fixed by the function value at the center). Thus, both pathways must converge to the exact same polynomial coefficients. This illustrates that the 'Taylor series' is an intrinsic property of the function, independent of the derivation path.

📖 Explanation: The existence of a Maclaurin series depends on the function being infinitely differentiable at the center, not on having an elementary closed-form antiderivative. Since $e^{-t^2}$ is analytic everywhere, its Maclaurin series converges for all real numbers. Term-by-term integration of this series yields a new power series that converges to $\int_0^x e^{-t^2} dt$. By the uniqueness theorem, this integrated series IS the Maclaurin series for the integral function, demonstrating that power series allow us to handle non-elementary functions rigorously.

📖 Explanation: Maclaurin series coefficients encode local geometric properties: $c_n = f^{(n)}(0)/n!$. If a graph exhibits even symmetry about the y-axis, the function is even, implying all odd derivatives at zero vanish. Consequently, the Maclaurin series can contain only even powers of x. The presence of an $x^3$ term in the proposed polynomial contradicts this geometric constraint. This highlights that graphical analysis provides immediate necessary conditions for Taylor coefficients, serving as a sanity check before algebraic derivation.

📖 Explanation: If a power series solution to a differential equation is found and subsequently identified as representing a known function $f(x)$, then by the uniqueness theorem, that series must be the Taylor (Maclaurin) series for $f(x)$. The method of discovery (recursion vs. differentiation) is irrelevant to the identity of the series. However, caution is needed: $\cos(\sqrt{x})$ is actually analytic at 0 (it equals $\sum (-1)^n x^n / (2n)!$), unlike $\cos(1/x)$. Assuming convergence to the function validates the identification.

📖 Explanation: When finding a Maclaurin series via integration, the constant of integration is determined by evaluating both the function and the series at the center $x=0$. Since \int_0^x f'(t)dt = f(x) - f(0), and the definite integral form naturally yields 0 at the lower limit, the constant is automatically handled if using definite integrals. If using indefinite integrals, one must explicitly set $C = f(0) - (\text{series value at } 0)$. In this case, since both are 0, $C=0$. Forgetting this step is a common error that works accidentally here but fails for functions like $\ln(2+x)$.

📖 Explanation: Directly computing high-order derivatives for products or compositions is often computationally prohibitive. However, if we can construct the power series via algebraic operations (multiplication, substitution) on known series, the uniqueness theorem guarantees that the coefficient of $x^n$ in the result is exactly $f^{(n)}(0)/n!$. Thus, we can extract derivative information purely from algebraic manipulation of series coefficients. This transforms a calculus problem into an algebra problem, leveraging the fact that the constructed series MUST be the Taylor series.

📖 Explanation: The theorem stating 'power series representations are Taylor series' includes a crucial domain restriction: the representation is valid only on the open interval where the series converges to the function. Outside the radius of convergence, the series diverges and ceases to represent the function, even though the function itself may be well-defined (as with $1/(1-x)$ at $x=2$). Confusing the analytic continuation of a function with its power series representation is a fundamental misconception. The series is the Taylor series only where it converges.

📖 Explanation: This classic counterexample distinguishes between 'having a Taylor series' and 'being representable by a power series.' The principle states that IF a function equals a power series on an open interval, THEN that series is the Taylor series. For $e^{-1/x^2}$, the Taylor series exists (all zeros) but does NOT equal the function on any neighborhood of 0. Therefore, the hypothesis 'represented by a power series' is false, and the principle is not violated. This highlights that infinite differentiability is necessary but not sufficient for power series representability (analyticity).

📖 Explanation: Term-by-term differentiation of a power series yields a new power series with the same radius of convergence that converges to the derivative of the original function. Since this new series is a valid power series representation for f'(x) on the interval, the uniqueness theorem dictates it must be the Taylor series for f'(x). This justifies the operational rule: differentiating a known series gives the correct Taylor series for the derivative without needing to recompute derivatives of f' from scratch.

📖 Explanation: Algebraic operations (addition, subtraction, multiplication, division) on power series that converge to functions $f$ and $g$ yield power series that converge to $f \pm g$, $fg$, or $f/g$ (provided $g(0) \neq 0$). Since the result is a valid power series representation for the target function, the uniqueness theorem guarantees it is the Taylor series. Long division of Maclaurin series is thus a legitimate and efficient method for finding Taylor series of quotients, avoiding messy quotient-rule derivatives.

📖 Explanation: While substitution is powerful, it respects domains. If $\sum c_n u^n$ converges for $|u|<R$, substituting $u=g(x)$ yields a valid representation for $f(g(x))$ only when $|g(x)|<R$. If one needs the series on a domain where $|g(x)| \geq R$, simple substitution fails even if the composite function is analytic there. Additionally, substituting expressions with constant terms (e.g., $x+1$) shifts the center, producing a Taylor series about a different point, not necessarily the Maclaurin series. Recognizing these domain and center constraints prevents misapplication.

📖 Explanation: A Maclaurin series is specifically a power series of the form $\sum a_n x^n$ with $n \geq 0$. Substituting $1/x$ into a series generates a Laurent-type series with negative powers, which is fundamentally different from a Taylor/Maclaurin series. Moreover, $\sin(1/x)$ is not analytic at 0; its behavior near 0 is oscillatory, not polynomial. The resulting expression, while perhaps asymptotically relevant for large x, cannot represent the function near 0 as a Maclaurin series. This underscores that valid substitutions must preserve the power series structure (non-negative exponents).

📖 Explanation: The identity theorem for power series states that if two power series centered at the same point agree on any open interval containing that point, their coefficients must be identical term-by-term. This is the algebraic foundation of the statement 'Power Series Representations Must Be Taylor Series.' It means the sequence of coefficients uniquely determines the function locally, and conversely, the function uniquely determines the coefficients. This allows us to equate coefficients when solving differential equations or proving identities via series.

📖 Explanation: Let $F(x) = \int_0^x e^{-t^2} dt$. The Maclaurin series for $F(x)$ is obtained by integrating the series for $e^{-t^2}$. Truncating this integrated series at degree n yields the nth Maclaurin polynomial for $F(x)$. Thus, approximating the definite integral via series truncation is equivalent to evaluating the Taylor polynomial of the accumulation function. This connects numerical approximation directly to Taylor theory, validating error bounds via the Remainder Estimation Theorem applied to $F(x)$, not just the integrand.

📖 Explanation: Repeated differentiation of $\arctan(x)$ quickly becomes algebraically complex. However, $\frac{d}{dx}\arctan(x) = \frac{1}{1+x^2}$, whose Maclaurin series is a simple geometric series. Integrating this series term-by-term produces a power series for $\arctan(x)$. By the uniqueness theorem, this integrated series IS the Maclaurin series. This exemplifies the strategic advantage of the 'representations must be Taylor series' principle: it permits choosing the easiest path to generate the unique correct series, transforming difficult calculus into manageable algebra.

📖 Explanation: Substituting $(x-1)$ for $x$ in a Maclaurin series shifts the center of expansion from 0 to 1. The resulting series $\sum (x-1)^n$ is indeed a valid power series representation for $\frac{1}{2-x}$, but it is centered at $x=1$. Therefore, it is the Taylor series about $x=1$, not the Maclaurin series (which is about 0). Understanding this distinction is vital: 'Power Series Representations Must Be Taylor Series' refers to the Taylor series ABOUT THE CENTER OF THE SERIES. Misidentifying the center leads to incorrect labeling and application.

📖 Explanation: Due to the uniqueness of power series representations, if a derived series purports to represent $\sinh(x)$ on an open interval, it must match the true Maclaurin series coefficient-for-coefficient for ALL n. A mismatch at any term proves the derived series does not represent the function (or contains an error). There is no 'approximate equality' for power series identities; they are either identical or distinct. This binary nature makes series comparison a rigorous verification tool: partial agreement is insufficient for proving representational equality.

📖 Explanation: A Maclaurin series requires the function to be defined and infinitely differentiable at $x=0$. Since $\cot(x) = \cos(x)/\sin(x)$ has a singularity at 0 (division by zero), it possesses no Maclaurin series. Attempting series division fails because the denominator series has zero constant term, preventing standard power series division (which requires invertible constant term). This reinforces that 'Power Series Representations Must Be Taylor Series' presupposes analyticity at the center. Functions with poles at the center have Laurent series, not Taylor/Maclaurin series.

📖 Explanation: While the full Maclaurin series for $e^{rt}$ converges for all t, any practical model uses a truncated polynomial. The validity of this approximation depends on the magnitude of the remainder $R_n(t)$. Even though the series representation is theoretically exact in the limit, the truncated model is only useful when higher-order terms are negligible. This connects the abstract 'representation must be Taylor series' concept to applied modeling: the Taylor polynomial is the optimal local approximant, but its utility is governed by error bounds, not just convergence.

📖 Explanation: Symmetry imposes strict constraints on Taylor coefficients. An even function (symmetric about y-axis) must have $f^{(n)}(0)=0$ for all odd n, so its Maclaurin series contains only even powers. If a proposed series includes odd terms for an evidently even function, it violates the uniqueness theorem—it cannot be the correct Maclaurin series. Graphical inspection thus serves as a powerful preliminary filter, catching errors before lengthy calculations. This leverages the deep link between geometric properties and analytic series structure.

📖 Explanation: While indefinite integration introduces an arbitrary constant, a Maclaurin series representation is uniquely determined by the function values. The series must satisfy $S(0) = f(0)$. When integrating f&#039;(x)'s series to recover $f(x)$'s series, the constant is fixed by this initial condition. This reflects the broader principle: the power series representation is unique. There is no family of Maclaurin series differing by constants; there is exactly one series that represents the function, and finding it requires anchoring via function evaluation at the center.

📖 Explanation: The Cauchy product formula for multiplying series is designed precisely so that the resulting series converges to the product of the functions (within the common interval of convergence). Since this product series is a valid power series representation for $h(x) = f(x)g(x)$, the uniqueness theorem ensures it is THE Maclaurin series for h. This validates algebraic multiplication as a legitimate series-generation technique. It also explains why the Cauchy product coefficients match those obtained via Leibniz's rule for nth derivatives of products—the series representation is unique regardless of derivation method.

📖 Explanation: Computing the 100th derivative of a composite product directly is infeasible. However, substituting into known series gives $e^{x^3} = \sum x^{3k}/k!$ and $\sin(x^2) = \sum (-1)^m x^{4m+2}/(2m+1)!$. Their product's $x^{100}$ coefficient arises from pairs where $3k + 4m + 2 = 100$. By uniqueness, this algebraically extracted coefficient IS $f^{(100)}(0)/100!$. This showcases the immense power of the principle: it converts impossible calculus into solvable combinatorics, enabling extraction of extreme-order derivative information through finite algebraic steps.

📖 Explanation: The uniqueness of power series is absolute: there is only one sequence of coefficients $\{a_n\}$ such that $\sum a_n x^n = \sqrt{1+x}$ near 0. Any valid method producing a convergent power series for this function must yield this exact sequence. Disagreement implies error in at least one derivation. This principle underpins all series manipulation: we trust algebraic results because they must coincide with calculus-based results. It transforms series from mere approximations into exact analytic representations, where method independence guarantees correctness.

📖 Explanation: Two errors compound here: First, $\sum (-1)^n x^n$ is the geometric series for $\frac{1}{1+x}$, not $\frac{1}{x}$. Second, and more fundamentally, $f(x)=1/x$ has a singularity at $x=0$, violating the prerequisite for Maclaurin series existence (infinite differentiability at center). The 'must be Taylor series' principle applies only when a power series representation EXISTS. For singular functions, no such representation exists at the singularity. Recognizing domain restrictions prevents misattributing series to incompatible functions.

📖 Explanation: Physical laws modeled by smooth functions are assumed analytic near equilibrium. When empirical data is fit to a power series near a point, and the model is valid, the uniqueness theorem ensures the fitted coefficients correspond to the true Taylor coefficients (derivatives) of the underlying physical function. This bridges experiment and theory: measured series coefficients ARE the derivatives. This application of 'representations must be Taylor series' allows extracting fundamental material properties (virial coefficients) from macroscopic measurements, grounding abstract calculus in physical reality.

📖 Explanation: Setting up a differential equation that the function satisfies and solving via series is a valid technique. If the resulting series converges to $f(x)$, uniqueness guarantees it is the Maclaurin series. For $f(x)=x^2$, the recursion would force $a_2=1$ and all other $a_n=0$, recovering the trivial series. This demonstrates that even for simple functions, the series method is consistent. The principle validates indirect characterization: any series satisfying the function's defining relations and converging to it is the unique Taylor series.

📖 Explanation: The principle has a critical hypothesis: the power series must CONVERGE TO THE FUNCTION on an open interval. Merely writing down a formal series (e.g., via Taylor formula) does not guarantee it represents the function. For non-analytic smooth functions like $e^{-1/x^2}$, the Taylor series exists but does not represent the function. Assuming representability without verifying convergence to the function misapplies the theorem. The principle links representation to Taylor series CONDITIONALLY on convergence, not unconditionally on formal construction.

📖 Explanation: Verification via inverse operations leverages uniqueness. If $S(x)$ is claimed to be the Maclaurin series for $\ln(\cos x)$, then $e^{S(x)}$ should equal the Maclaurin series for $\cos x$. Computing $e^{S(x)}$ via series composition and comparing to the known cosine series tests equality. If they match, uniqueness confirms $S(x)$ is correct. This cross-validation technique avoids redundant derivative calculations and exploits the bijective correspondence between analytic functions and their power series, turning verification into an algebraic consistency check.

📖 Explanation: Although $\sec(x)$ has singularities at $\pm \pi/2$, it is analytic at $x=0$. Thus, a Maclaurin series exists with radius of convergence $\pi/2$. Within this disk, $\cos(x) \neq 0$, so series division is valid and yields the unique Maclaurin series for sec(x). The presence of distant singularities limits the radius but doesn't prevent local representation. This clarifies that 'must be Taylor series' applies locally around the center; global behavior affects convergence radius, not local representability.

📖 Explanation: At $x=0$, $f(x)$ appears indeterminate, but the removable singularity makes it analytic. Direct derivatives require tedious limit evaluations. Instead, write $f(x) \cdot \frac{e^x - 1}{x} = 1$. Substitute known series for $\frac{e^x-1}{x}$ and unknown series $\sum B_n x^n/n!$ for f(x). Their Cauchy product must equal 1. Uniqueness forces coefficient equations that recursively determine Bernoulli numbers. This elegant approach bypasses calculus entirely, showcasing how 'must be Taylor series' enables defining important constants through pure series algebra.

📖 Explanation: Function parity directly constrains Taylor coefficients. For an odd function, $f(-x) = -f(x)$. Substituting into the series: $\sum a_n (-x)^n = -\sum a_n x^n$. Equating coefficients via uniqueness gives $a_n (-1)^n = -a_n$. For even n, this implies $a_n = -a_n$, so $a_n = 0$. Thus, recognizing symmetry allows immediate deduction of coefficient structure without computation. This is a direct consequence of the uniqueness of power series representations and provides a quick validation tool for derived series.