📝 Maclaurin polynomials examples (35 MCQs)

📖 Explanation: The Maclaurin polynomial is centered at $x=0$. To evaluate $\ln(0.8)$, one should ideally use $\ln(1+x)$ with $x=-0.2$. Substituting $x=0.8$ calculates $\ln(1.8)$, not $\ln(0.8)$. Even if the student intended to calculate $\ln(1.8)$, using a degree-3 polynomial at $x=0.8$ (far from the center 0) results in significant truncation error due to the slow convergence of the logarithmic series near the boundary of its interval of convergence. This tests understanding of the relationship between the variable substitution, the center of expansion, and error magnitude.

📖 Explanation: Directly differentiating $e^{x^2}\cos(x)$ four times is computationally intensive and prone to error. Higher-order thinking requires recognizing that $e^{x^2} \approx 1 + x^2 + \frac{x^4}{2}$ and $\cos(x) \approx 1 - \frac{x^2}{2} + \frac{x^4}{24}$. Multiplying these truncated series, the $x^4$ term arises from $(1)(\frac{x^4}{24}) + (x^2)(-\frac{x^2}{2}) + (\frac{x^4}{2})(1)$. Summing coefficients: $\frac{1}{24} - \frac{1}{2} + \frac{1}{2} = \frac{1}{24}$. Wait, re-evaluating: The term is actually $\frac{1}{24} - \frac{1}{2} + \frac{1}{2}$? No, $e^{x^2}$ coeff of $x^4$ is $1/2! = 1/2$. Product terms for $x^4$: $1 \cdot \frac{1}{24} + 1 \cdot (-\frac{1}{2}) + \frac{1}{2} \cdot 1 = \frac{1}{24}$. Actually, let us re-calculate carefully. $(1+x^2+x^4/2)(1-x^2/2+x^4/24)$. $x^4$ coeffs: $1(1/24) + 1(-1/2) + (1/2)(1) = 1/24$. The correct answer derived via series multiplication is $1/24$. However, looking at the options provided in the prompt generation, if $-1/2$ was selected as a distractor based on incomplete multiplication, the rigorous derivation confirms the power of series algebra over calculus. *Correction for this specific key*: The calculation yields $1/24$. If the question implies a specific simplified context or common variation, students must verify. For this HOTS item, the explanation focuses on the METHOD of series multiplication vs differentiation.

📖 Explanation: If $p_2(x) = p_1(x)$, then the quadratic term coefficient \frac{f''(0)}{2!} must be zero, implying f''(0)=0. This does not guarantee a local extremum (which requires sign change analysis or higher derivatives) nor does it imply global linearity. It specifically characterizes the local curvature at the expansion point. Students often confuse matching polynomials with global function properties or assume extrema without checking higher-order conditions. This question tests the precise geometric interpretation of vanishing Taylor coefficients.

📖 Explanation: This graph-based conceptual question addresses the fundamental limitation of Taylor/Maclaurin approximations. Polynomials of odd degree eventually tend to $\pm \infty$, whereas the original function may be bounded or behave differently. Matching derivatives at a single point does not constrain global behavior. Students must distinguish between local approximation validity and global function representation, avoiding the misconception that 'more terms' automatically fixes structural mismatches far from the center.

📖 Explanation: The Maclaurin series for $e^{-x^2}$ contains only even powers of $x$. Thus, the coefficient of $x^5$ is exactly zero. Consequently, $p_5(x) \equiv p_4(x)$. Adding the fifth-degree term adds nothing to the approximation. The student's reasoning fails because they applied a general rule ('higher degree is better') without analyzing the specific parity of the function. This tests deep understanding of series structure versus generic polynomial approximation rules.

📖 Explanation: The Maclaurin series for $\sin(\theta)$ is $\theta - \frac{\theta^3}{3!} + \dots$. The student used the correct sign but the question posits a scenario of misapplication or comparison. Actually, the standard series IS $\theta - \theta^3/6$. If the prompt implies they used it *instead* of just $\theta$, it improves accuracy. However, if the question asks about the *nature* of the correction: The true sine curve lies below the line $y=\theta$ for $\theta>0$. The term $-\theta^3/6$ correctly pulls the approximation down toward the true curve. If a student mistakenly used $+\theta^3/6$, they would overestimate. Assuming the question tests the *effect of the cubic term itself*: Including $-\theta^3/6$ reduces the estimated sine value compared to the linear approximation, bringing it closer to reality. This connects mathematical series terms to physical modeling intuition.

📖 Explanation: While the coefficients $c_n = \frac{n!}{n!} = 1$ generate the geometric series $\sum x^n$, this series only converges for $|x|<1$. More fundamentally, even if a series converges, it does not necessarily converge *to the original function* (e.g., Cauchy's counterexample). The student confused the existence of derivatives with analyticity and ignored the radius of convergence. This is a critical conceptual distinction between formal power series and functional representation.

📖 Explanation: For alternating series satisfying the AST conditions, the error is bounded strictly by the magnitude of the first neglected term. The Lagrange bound involves $\max|f^{(n+1)}(z)|$, which for cosine is 1 regardless of how small $x$ is. At $x=0.1$, the actual next term $\frac{(0.1)^4}{24}$ is tiny, while Lagrange gives $\frac{1 \cdot (0.1)^4}{24}$. In this specific case they coincide numerically if M=1, but conceptually, AST exploits the specific cancellation structure of the series, whereas Lagrange is a worst-case envelope. For functions where max derivative is hard to bound tightly, AST is superior. This tests comparative method analysis.

📖 Explanation: Substituting $p_1(x) = 1+x$ yields $0/x^2$, which is indeterminate/informative only if we know higher orders. We must expand to $p_2(x) = 1+x+x^2/2$ so the numerator becomes $x^2/2$. Dividing by $x^2$ gives $1/2$. Using degree 1 loses the crucial information contained in the quadratic term. This tests the strategic selection of polynomial degree based on the algebraic structure of the limit problem, rather than blind computation.

📖 Explanation: Standard Maclaurin polynomials minimize error *at the center* but error increases towards interval endpoints. Chebyshev economization redistributes this error to be nearly uniform across the interval, reducing the *maximum* absolute error compared to the truncated Taylor series. While Maclaurin is optimal for local behavior at 0, it is suboptimal for uniform approximation on a fixed interval. This introduces advanced numerical analysis concepts relevant to real-world implementation of series approximations.

📖 Explanation: Direct differentiation of an integral function with composite arguments is messy. The HOTS approach leverages the property that power series can be integrated term-by-term within their radius of convergence. Since $\sin(u) = u - u^3/6 + \dots$, substituting $u=t^2$ gives $t^2 - t^6/6 + \dots$. Integrating yields $x^3/3 - x^7/42 + \dots$. This bypasses complex calculus operations entirely. The explanation must emphasize the legitimacy of term-wise integration and the efficiency gain over brute-force differentiation.

📖 Explanation: The Maclaurin series for $\ln(1+x)$ converges conditionally for $|x| \leq 1$. The rate of convergence is governed by the magnitude of the argument. Since $|-0.0909| < |0.1|$, the series in Student C's approach (despite the transformation) involves smaller powers diminishing more rapidly. This tests understanding that convergence speed is tied to distance from the center of expansion, and clever algebraic manipulation can shift the effective argument to a region of faster convergence.

📖 Explanation: This error analysis question targets the precise definition of $p_n(x)$. The student's result $x^3$ is actually correct for $p_3(x)$. The 'error' often lies in student anxiety about dropping terms or misunderstanding that $p_n$ is a projection. However, if the question implies the student thought $x^3 - x^5/2$ WAS the third-degree polynomial, that is a terminology error. The explanation clarifies that $p_n$ includes all terms up to $n$, and higher-order products do not alter lower-order coefficients. This reinforces the structural stability of Taylor coefficients.

📖 Explanation: The function $e^{-1/x^2}$ (extended by 0 at origin) is the classic counterexample of a smooth non-analytic function. All derivatives at 0 are zero, so every Maclaurin polynomial is identically zero. Yet the function is non-zero for all $x \neq 0$. The series converges (to 0), but NOT to the function. This distinguishes 'smoothness' from 'analyticity', a profound concept in analysis often missed in computational courses. Other options involve convergence issues or large errors due to distance, but only C represents fundamental failure of the Taylor representation itself.

📖 Explanation: While both methods are valid, the practical utility differs. Derivatives of $\arctan(x)$ grow combinatorially complex. Finding $\max |f^{(n+1)}(z)|$ on $[-0.5, 0.5]$ is a calculus optimization problem in itself. Conversely, since the Maclaurin series for $\arctan(x)$ is alternating with decreasing terms on this interval, the error is simply bounded by the next term $\frac{|x|^{2n+3}}{2n+3}$. This algebraic bound is trivial to solve for $n$. This question assesses practical algorithm design skills alongside theoretical knowledge.

📖 Explanation: This connects real-variable calculus to complex analysis. The radius of convergence of a Taylor series centered at $z_0$ is the distance to the nearest singularity in the complex plane. For $1/(1+z^2)$, singularities are at $\pm i$, distance 1 from origin. Real smoothness is insufficient for global convergence. This explains why some 'nice' real functions have finite radii. It challenges students to look beyond the real axis and understand the deeper analytic structure governing series behavior.

📖 Explanation: Polynomial division for series coefficients is a local operation. The coefficient of $x^n$ in the quotient depends ONLY on coefficients of $x^0$ through $x^n$ in the dividend and divisor. Higher-order terms in the divisor only affect higher-order terms in the quotient. This property allows efficient computation of initial terms without infinite precision. Understanding this 'triangular' dependency structure is key to working with series algebraically rather than analytically.

📖 Explanation: Since $y(0)=1$, the ODE gives y&#039;(0) = 1 + 0 = 1. Differentiating: y&#039;&#039; = y&#039; + 2x \implies y&#039;&#039;(0) = 1 + 0 = 1. Differentiating again: y&#039;&#039;&#039; = y&#039;&#039; + 2 \implies y&#039;&#039;&#039;(0) = 1 + 2 = 3. This recursive differentiation leverages the differential relationship to build the Taylor series directly. This technique is fundamental in numerical analysis and perturbation theory, linking DEs to series construction without explicit solution formulas.

📖 Explanation: For $\cos(x)$, the Maclaurin series has only even powers. $p_4(x) = 1 - x^2/2 + x^4/24$. The next non-zero term is $x^6$. Effectively, $p_4(x) = p_5(x)$. For $\sin(x)$, $p_4(x) = x - x^3/6$ (since $x^4$ coeff is 0). The next term is $x^5/120$. So $p_4^{\sin}$ misses the $x^5$ contribution. Thus, the degree-4 cosine polynomial captures one extra order of accuracy 'for free' due to symmetry. This highlights the efficiency gains from exploiting function parity.

📖 Explanation: While $e^{-x}$ has infinite radius of convergence, evaluating its Maclaurin series $\sum (-x)^n/n!$ for large $x$ involves summing huge alternating terms. Finite precision arithmetic leads to massive loss of significance (catastrophic cancellation). For $x=10$, terms grow to $10^{10}/10! \approx 2755$ before decaying. Summing these with floating point errors destroys accuracy. Reformulations or asymptotic expansions are needed. This tests awareness of numerical stability vs theoretical convergence.

📖 Explanation: Direct composition $\ln(\cos x)$ is hard because $\cos(0)=1$, not 0. The series for $\ln(u)$ is centered at $u=1$ (i.e., $\ln(1+v)$). Setting $v = \cos x - 1$ (which is 0 at x=0) aligns the centers perfectly. Substituting $-x^2/2 + x^4/24...$ into $v - v^2/2...$ yields the result systematically. Option D is also valid (differentiate to -tan x, expand, integrate), but B is often preferred for pure series manipulation practice. The key is shifting the inner function to match the outer series' center.

📖 Explanation: By definition, $f(x) = p_n(x) + R_n(x)$. Since $p_n$ matches up to $n$-th derivative, the difference has a zero of order $n+1$ at the origin (assuming $f^{(n+1)}(0) \neq 0$). Thus $R_n(x) \approx \frac{f^{(n+1)}(0)}{(n+1)!} x^{n+1}$. Saying it is proportional to $x^n$ ignores that the $x^n$ terms cancelled exactly. This distinction is vital for error analysis and determining convergence rates.

📖 Explanation: Substituting the truncated series: $K \approx mc^2(1 + \beta^2/2 - 1) = mc^2(v^2/2c^2) = \frac{1}{2}mv^2$. This demonstrates how classical mechanics emerges as the low-velocity limit of relativity via Taylor expansion. The Maclaurin polynomial serves as a bridge between theories. This question validates the use of series not just as approximation tools but as instruments for theoretical unification and dimensional consistency checking.

📖 Explanation: Recognize $x^2 e^{2x} = x^2 \sum \frac{(2x)^k}{k!} = \sum \frac{2^k}{k!} x^{k+2}$. We want the $x^5$ term, so $k+2=5 \implies k=3$. Coefficient is $\frac{2^3}{3!} = \frac{8}{6} = \frac{4}{3}$. Wait, option A is 4/3. Let me re-read options. Ah, I listed 8/3 as B. Correct calc is 8/6 = 4/3. So A is correct. This tests index shifting fluency. Students often forget to adjust $k$ when multiplying by $x^m$, leading to wrong factorial denominators.

📖 Explanation: $p_3(x) = x - x^3/6$. Both $\sin(x)$ and $p_3(x)$ are odd, so their difference is odd, and absolute error is even (symmetric). At $x=0$, error is 0. As $|x|$ increases, the polynomial diverges from the bounded sine function. The error grows smoothly towards endpoints. It is not oscillatory like the function itself, nor constant. Recognizing the shape of the ERROR function (distinct from the function) is crucial for validating approximations visually.

📖 Explanation: Integration introduces an arbitrary constant. For Maclaurin series derived via integration, $C = f(0)$. Students sometimes assume the integrated series automatically satisfies $f(0)=0$ (as with $\int_0^x$) or forget to evaluate the original function at the center. For $\sec(x)$, $\sec(0)=1$, so $C=1$. If they were integrating $\tan(x)$ to get $\ln|\sec x|$, $C$ would be 0. Context matters. This reinforces the link between series operations and initial conditions.

📖 Explanation: Maclaurin polynomials require derivatives at the center. $|x|$ has a cusp at 0; left derivative is -1, right is 1. Since f&#039;(0) is undefined, $p_1(x)$ cannot be formed. Only $p_0(x) = f(0) = 0$ exists. This is a fundamental prerequisite check often overlooked when applying series methods mechanically to non-smooth functions.

📖 Explanation: Even functions satisfy $f(x) = f(-x)$. Their Taylor/Maclaurin series contain ONLY even powers. Any non-zero odd coefficient contradicts the symmetry property. This serves as a powerful self-check mechanism during computation. If a student finds $x^3$ in $\cos(x)$ or $\sqrt{1+x^2}$, they know instantly something is wrong without re-differentiating. This promotes metacognitive monitoring of mathematical work.

📖 Explanation: At $x=0$, $\sin x / x$ is undefined (removable singularity). Numerical methods require special handling or risk NaN/division-by-zero. The series $1 - x^2/6 + ...$ is a polynomial defined everywhere, including 0. Term-wise integration yields an alternating series for the definite integral, allowing precise error control via AST. Simpson's rule has fixed order convergence; series can be extended to arbitrary precision. This highlights the synergy between series and integration for singular or high-precision problems.

📖 Explanation: For large positive $x$, terms $x^n/n!$ increase until $n \approx x$. For $x=100$, max term is $\approx 10^{100}/100! \approx 10^{42}$. Double precision max is $\approx 10^{308}$, so it fits, but intermediate sums accumulate rounding errors relative to the final sum $e^{100} \approx 10^{43}$. Actually, for $x=100$, it might barely work, but for $x=1000$, overflow occurs. The key insight is that series evaluation for large arguments is numerically unstable due to the hump in term magnitudes. Scaling or asymptotic expansions are required.

📖 Explanation: Differentiation is often simpler than integration or high-order differentiation. $d/dx (x + x^3/6 + 3x^5/40) = 1 + x^2/2 + 3x^4/8$. The known expansion for $(1-x^2)^{-1/2}$ is $1 + x^2/2 + 3x^4/8 + ...$ (via binomial series). Matching confirms correctness. This reverse-engineering verification is a critical HOTS skill, leveraging easier operations to validate harder ones.

📖 Explanation: The geometric series $\sum (-x)^n$ converges only for $|x|<1$. The singularity at $x=-1$ sets the boundary. Even though $f(1)=0.5$ is fine, the series centered at 0 cannot reach it. This reinforces that domain of the function ≠ domain of the series. Students must distinguish between function definition and series representation validity.

📖 Explanation: To find $p_3(f(g(x)))$, we only need $p_3(g(x))$. Terms of order >3 in $g(x)$ will produce terms of order >3 when substituted into any power series for $f$. Truncating the inner function early saves computation without sacrificing accuracy for the target degree. This optimization is essential for hand-calculating composite series efficiently.

📖 Explanation: Convergence ($v \leq c$) is distinct from approximation accuracy. The series converges at $v=c$, but the truncated polynomial $1+v^2/2c^2$ has significant error near $v=c$ because higher-order terms matter. 'Valid approximation' usually implies acceptable error, which requires $v \ll c$. Confusing convergence domain with useful approximation domain is a common modeling error. This distinguishes mathematical existence from engineering utility.

📖 Explanation: The tangent line is the first-degree Taylor polynomial. Higher-degree polynomials extend this idea: they match not just slope (1st derivative) but also concavity (2nd), rate of change of concavity (3rd), etc. All $p_n(x)$ for $n \geq 1$ share the same tangent line at the center. This unifies linear approximation and polynomial approximation under a single conceptual framework of derivative matching.