📝 Quadratic approximation formula (35 MCQs)

📖 Explanation: The definition of the local quadratic approximation requires that the polynomial and its first two derivatives match the function at the expansion point. Since the second derivative of $c_2x^2$ is $2c_2$, setting this equal to f''(0) necessitates that c_2 = f''(0)/2!. Omitting the division by 2 results in a polynomial whose curvature is double what it should be, leading to immediate divergence from the true function behavior near the expansion point.

📖 Explanation: Since $p(x)$ matches $f(x)$ up to the second derivative, the Taylor expansion of the difference $E(x)$ begins with the cubic term (f'''(0)/3!)x^3. Therefore, when dividing $E(x)$ by $x^2$, the result approaches (f'''(0)/3!)x, which goes to 0. However, if the question implies the *coefficient* governing the error's growth relative to the next power, it relates to the third derivative. Strictly speaking, $\lim_{x\to 0} E(x)/x^2 = 0$. If the option meant $E(x)/x^3$, it would be f'''(0)/6. Given standard analysis contexts, recognizing that the error is of order $x^3$ is key; thus, the ratio to $x^2$ vanishes, but the *nature* of the error is dictated by the unmatched third derivative.

📖 Explanation: At a stable equilibrium, U'(0)=0, so the local linear approximation is simply the constant $U(0)$. This predicts zero net force everywhere locally, implying no dynamics. The local quadratic approximation includes the U''(0)x^2/2 term. Since stability implies U''(0) > 0, this term provides the parabolic potential well necessary to model a restoring force F = -U'(x) \approx -U''(0)x. Without this second-order term, the mathematical model fails to capture the fundamental physics of oscillation, regardless of how close $x$ is to zero.

📖 Explanation: For $f(x)=\ln(1+x)$, f'(x)=1/(1+x) and f''(x)=-1/(1+x)^2. At $x=0$, f''(0)=-1. The linear approximation $L(x)=x$ lies above the curve for $x>0$ due to concavity. The quadratic approximation $Q(x)=x-x^2/2$ incorporates this negative curvature. For $x=0.1$, $L(0.1)=0.1$ and $Q(0.1)=0.095$. The true value is approximately $0.0953$. The quadratic approximation is significantly more accurate because it accounts for the rate of change of the slope. The colleague’s reasoning ignores the geometric significance of the second derivative in correcting systematic bias.

📖 Explanation: From the graph description: $f(0)=1$ implies the constant term is 1. A horizontal tangent means f'(0)=0, eliminating any linear $x$ term. Concave up implies f''(0) > 0. Examining the options: $1+x$ has non-zero slope; $1-x^2$ is concave down; $x^2$ has $p(0)=0$. Only $1+2x^2$ satisfies $p(0)=1$, p'(0)=0, and p''(0)=4 > 0. This demonstrates translating visual geometric features directly into polynomial coefficients without explicit formula derivation.

📖 Explanation: If $f(x)$ is odd, then $f(-x) = -f(x)$. Differentiating twice gives f''(-x) = -f''(x), meaning f'' is also an odd function. Assuming continuity/differentiability, f''(0) must be 0. Consequently, the coefficient c_2 = f''(0)/2! = 0. Thus, the quadratic term vanishes entirely. The local quadratic approximation reduces to f(0) + f'(0)x. Since $f(0)=0$ for odd functions, p_2(x) = f'(0)x, which is exactly the local linear approximation. Students often mistakenly believe 'quadratic' implies a non-zero squared term exists.

📖 Explanation: When expanding around $x=a$, writing $p(x) = c_0 + c_1(x-a) + c_2(x-a)^2$ simplifies differentiation significantly. Evaluating $p(a)$ immediately isolates $c_0$; evaluating p'(a) isolates $c_1$; and evaluating p''(a) isolates $c_2$. If we used powers of $x$ (i.e., $Ax^2+Bx+C$), solving for $A, B, C$ would require solving a coupled system of linear equations involving $a$. The shifted basis diagonalizes the mapping between coefficients and derivative values, making the derivation trivial and computationally stable.

📖 Explanation: The formal definition of the nth-degree Taylor/Maclaurin polynomial requires the existence of derivatives up to order n at the expansion point. While $x^2$ might be a good regression fit or share symmetry, $f(x)=|x|$ is not differentiable at $x=0$. Therefore, f'(0) is undefined, and strictly speaking, the local quadratic approximation does not exist. Confusing 'best fit' with 'Taylor polynomial' is a common misconception. The Taylor polynomial is a specific analytic construct dependent on pointwise differentiability, not just geometric proximity or integral error minimization.

📖 Explanation: The local quadratic approximation for $\cos(x)$ at 0 is $1 - x^2/2$. We examine $g(x) = \cos(x) - (1 - x^2/2)$. g(0)=0, g'(0)=0, g''(0)=0. The fourth derivative test or series expansion shows $g(x) \approx x^4/24 > 0$ near zero, so $\cos(x) > 1-x^2/2$ locally. However, as $|x| \to \infty$, $1-x^2/2 \to -\infty$ while $\cos(x) \geq -1$, so the inequality holds globally? Wait. Actually, $1-x^2/2$ drops below -1 when $x^2 > 4$. Cosine stays above -1. So $\cos(x) \geq 1-x^2/2$ actually holds for ALL real x. Let's re-evaluate. Is there a crossover? No. The parabola opens downward and goes to negative infinity. Cosine oscillates between -1 and 1. Once the parabola drops below -1, cosine is always greater. Near zero, Taylor remainder analysis confirms cosine is greater. Thus, the inequality actually holds for all real x. This challenges intuition about polynomial vs trigonometric bounds.

📖 Explanation: The true function $\sqrt{1+x}$ has f''(x) = -1/(4(1+x)^{3/2}), which is negative (concave down). The correct quadratic bends downward. If the student uses a positive quadratic coefficient, their parabola bends upward. For $x>0$, the true function increases slower than linear, while the incorrect approximation increases faster than linear. This creates a compounding error where the gap widens quadratically rather than being minimized. Recognizing the sign of curvature is crucial for determining whether an approximation acts as an upper or lower bound.

📖 Explanation: The local quadratic approximation is determined exclusively by f(a), f'(a), f''(a). It ignores function behavior anywhere else. In contrast, least squares regression minimizes $\int_{-h}^h (f(x)-p(x))^2 dx$, incorporating information from the whole interval. As $h \to 0$, the least squares fit converges to the Taylor polynomial, but for any finite $h$, they differ. The Taylor polynomial optimizes local contact (tangency of order 2), whereas regression optimizes average fidelity. Understanding this distinction prevents misapplying local tools to global fitting problems.

📖 Explanation: By definition, if $p(x) = 3 - 2x + x^2$ is the local quadratic approximation, then $f(0)=3$, f'(0)=-2, and f''(0)/2! = 1 \implies f''(0)=2. The limit expression is essentially asking for the coefficient of the $x^2$ term in the Taylor expansion, or equivalently $\lim_{x\to 0} \frac{f(x) - p(x) + x^2}{x^2}$. Since $f(x) - p(x) = o(x^2)$, the limit simplifies to $\lim \frac{o(x^2)}{x^2} + 1 = 0 + 1 = 1$. This tests understanding that the approximation defines the limiting ratio of the function's deviation from lower-order terms.

📖 Explanation: Newton's method for solving f'(x)=0 applies the root-finding algorithm to the derivative: x_{n+1} = x_n - f'(x_n)/f''(x_n). Alternatively, constructing the local quadratic model q(x) = f(x_n) + f'(x_n)(x-x_n) + \frac{1}{2}f''(x_n)(x-x_n)^2 and finding its minimum involves setting q'(x)=0, which yields f'(x_n) + f''(x_n)(x-x_n) = 0. Solving for $x$ gives the exact same update rule. Thus, Newton's optimization is precisely the iterative minimization of successive local quadratic approximations. This unifies the geometric and algebraic perspectives.

📖 Explanation: The Maclaurin series for $\sin(\theta)$ is $\theta - \theta^3/3! + \dots$. The second derivative at 0 is $-\sin(0) = 0$. Therefore, the coefficient of $\theta^2$ is zero. The local quadratic approximation is $p_2(\theta) = \theta + 0\theta^2 = \theta$. The expression $\theta - \theta^3/6$ is the local *cubic* (and quartic) approximation. Calling it quadratic is terminologically incorrect, even though it contains a lower-degree term. Precision in terminology matters: 'quadratic' refers to the degree of the polynomial used, not just 'any improvement over linear'.

📖 Explanation: This classic counterexample has $f^{(n)}(0) = 0$ for all $n$. Thus, every Taylor polynomial, including the quadratic, is identically zero. Despite $f(x) > 0$ for all $x \neq 0$, the local quadratic approximation provides absolutely no information about the function's positive values. This highlights a critical limitation: local polynomial approximations rely on derivatives encoding local behavior. For 'flat' or non-analytic smooth functions, derivatives fail to capture the function's structure, rendering Taylor-based approximations useless despite infinite differentiability.

📖 Explanation: Tangent is an odd function with an inflection point at the origin. Geometrically, the curve crosses its tangent line at $x=0$. Analytically, f''(x) = 2\sec^2(x)\tan(x), so f''(0)=0. The local quadratic approximation is indeed $x$. The second student confuses local behavior with global behavior. While $\tan(x) > x$ for $x>0$, this deviation is cubic ($x^3/3$), not quadratic. Locally, the quadratic term vanishes due to symmetry. Super-linear growth does not necessitate a non-zero quadratic coefficient if the dominant nonlinear term is of odd degree.

📖 Explanation: Physical smoothness corresponds to mathematical differentiability. Position continuity is $C^0$, slope continuity is $C^1$, and acceleration continuity relates to $C^2$ (since acceleration involves the second derivative of position with respect to time, linked to curvature). The local quadratic approximation explicitly matches $f(0)$, f'(0), and f''(0). Matching these three quantities ensures that the transition from the straight segment (where y'=0, y''=0) to the curve is seamless up to the second order. Linear approximation would leave a discontinuity in acceleration (jerk), causing discomfort.

📖 Explanation: For a local maximum at $x=a$: 1) f&#039;(a)=0 \implies B=0. 2) Second derivative test requires f&#039;&#039;(a) < 0. Since C = f&#039;&#039;(a)/2, this implies $C < 0$. The value $A = f(a)$ is simply the function value at the maximum, which can be any real number depending on the vertical shift of the function. Option D is incorrect because $C=0$ makes the second derivative test inconclusive (could be inflection or higher-order extremum). Option B describes a minimum. This links calculus extrema tests directly to polynomial coefficient signs.

📖 Explanation: The true approximation uses f&#039;&#039;(0)/2. The computed version uses (f&#039;&#039;(0)+\epsilon)/2. The difference between the computed and true polynomial is $\frac{\epsilon}{2}x^2$. This demonstrates that errors in higher-order derivative estimates become increasingly significant further from the expansion point. Near zero, the $x^2$ factor suppresses the error, but as $x$ grows, this quadratic amplification dominates. This explains why high-order Taylor approximations can be numerically unstable far from the center unless derivatives are computed with extreme precision.

📖 Explanation: Note that $f(x) = \ln(x) = \ln(1 + (x-1)) = g(x-1)$. The local quadratic approximation respects this translation. If $Q_g(u) = u - u^2/2$ approximates $g(u)$ near $u=0$, then approximating $f(x)$ near $x=1$ corresponds to approximating $g(u)$ near $u=x-1=0$. Thus, $Q_f(x) = Q_g(x-1) = (x-1) - (x-1)^2/2$. Direct calculation confirms f&#039;(1)=1, f&#039;&#039;(1)=-1, yielding the same result. This illustrates the translational invariance of Taylor expansions: shifting the input variable shifts the expansion center without altering the functional form of the polynomial coefficients relative to that center.

📖 Explanation: Differentiability is a prerequisite for Taylor polynomials. For $f(x)=\sqrt{x}$, f&#039;(x) = 1/(2\sqrt{x}). As $x \to 0^+$, f&#039;(x) \to \infty. Since the derivative is unbounded/non-existent at the expansion point, neither the linear nor quadratic coefficients can be defined. Geometrically, this corresponds to a vertical tangent. While the curve is smooth visually, analytically it lacks the finite slope required for polynomial approximation at that specific point. One must expand around $x=a>0$ instead.

📖 Explanation: Let $x = v^2/c^2$. Then $\gamma = (1-x)^{-1/2}$. The local quadratic (in $x$) is $1 + x/2 + 3x^2/8$. Substituting back: $K \approx mc^2(1 + v^2/2c^2 + 3v^4/8c^4 - 1) = \frac{1}{2}mv^2 + \frac{3}{8}m\frac{v^4}{c^2}$. The first term is classical KE. The second term is the leading-order relativistic correction. This demonstrates how local quadratic approximation bridges modern and classical physics, with higher-order terms quantifying deviations from classical theory at higher velocities.

📖 Explanation: Since $p(x)$ matches $f$ up to the second derivative, the difference $D(x) = f(x) - p(x)$ behaves like \frac{f&#039;&#039;&#039;(a)}{6}(x-a)^3 near $a$. If $p(x) > f(x)$ for $x > a$, then $D(x) < 0$ for $x > a$. Since $(x-a)^3 > 0$ for $x > a$, the coefficient f&#039;&#039;&#039;(a)/6 must be negative. Therefore, f&#039;&#039;&#039;(a) < 0. This connects the visual crossing behavior of the approximation and the function directly to the sign of the first unmatched derivative.

📖 Explanation: By FTC, f&#039;(x) = e^{-x^2}. At $x=0$, $f(0)=0$, f&#039;(0)=e^0=1. For the second derivative: f&#039;&#039;(x) = -2xe^{-x^2}, so f&#039;&#039;(0)=0. Thus, the local quadratic approximation is $0 + 1\cdot x + 0\cdot x^2 = x$. Even though the integrand is complex, the local behavior of the accumulation function at the lower limit is simply linear. The quadratic term vanishes because the integrand $e^{-t^2}$ has a critical point (maximum) at $t=0$, making its rate of change zero initially.

📖 Explanation: A single quadratic polynomial has exactly one extremum and no inflection points. Real-world data often exhibits complex curvature changes. A global quadratic forces a simplistic parabolic shape onto potentially complex data, leading to large systematic errors. Piecewise quadratics (splines) allow different curvature parameters in different regions, adapting to local features while maintaining smoothness at knots. This flexibility makes them vastly superior for modeling non-monotonic or multi-modal phenomena where a single local approximation at one point cannot represent the entire domain.

📖 Explanation: Mathematical definitions are precise. The local quadratic approximation is uniquely determined by derivatives up to order 2. For $x^4$, these are all zero at the origin. Thus, $p(x)=0$ is the *only* correct local quadratic approximation. The fact that it poorly represents the function's positivity reflects that $x^4$ is 'flatter' than any quadratic at the origin. The remedy is not to redefine the quadratic approximation, but to recognize that a higher-degree polynomial is needed. This reinforces that 'local quadratic' is a specific operator, not a generic 'simple curve fit'.

📖 Explanation: The given formula is precisely the Taylor expansion of $V(S+\Delta S)$ truncated at the second order. $\Delta = \partial V/\partial S$ corresponds to the first derivative coefficient. $\Gamma = \partial^2 V/\partial S^2$ corresponds to the second derivative. The factor of $1/2$ comes from the Taylor series denominator $2!$. Financial practitioners use this local quadratic model to estimate risk exposure for small price moves. Understanding this as a Taylor polynomial clarifies why it works well for small $\Delta S$ but requires higher-order Greeks (Vanna, Volga) for larger moves.

📖 Explanation: Approximation and nonlinear transformation do not commute. The local quadratic of $g(u)=u$ is exact. Substituting $u=\ln(x)$ gives $\ln(x)$, which is exact but not a polynomial in $x$. Conversely, taking the quadratic in $x$ and substituting $x=e^u$ yields $(e^u-1) - (e^u-1)^2/2$, which expands to $u + u^2/2 + ... - (u^2 + ...)/2 = u + O(u^3)$. The quadratic terms cancel! This shows that the 'best' quadratic in one coordinate system differs from the transformed 'best' quadratic in another. Truncation errors interact with the Jacobian of the transformation.

📖 Explanation: This is a classic coefficient error. The Taylor polynomial is P(0) + P&#039;(0)t + \frac{P&#039;&#039;(0)}{2!}t^2. Given P&#039;&#039;(0)=2, the term is $\frac{2}{2}t^2 = t^2$. The researcher used the raw second derivative value as the coefficient, forgetting the factorial divisor. While the objection about long-term growth (A) is biologically valid, the immediate *mathematical* rejection based on the provided derivatives is the incorrect coefficient. In applied math exams, distinguishing between model validity and calculation errors is crucial. Here, the calculation is objectively wrong regardless of biological realism.

📖 Explanation: In multivariable calculus, the local quadratic approximation at a critical point is given by the quadratic form associated with the Hessian matrix. For $f=x^2+xy+y^2$, first derivatives are zero at origin. Second partials: $f_{xx}=2, f_{yy}=2, f_{xy}=1$. The Hessian is [[2,1],[1,2]]. The quadratic approximation is $\frac{1}{2}[x,y]H[x;y] = x^2 + xy + y^2$. Since the original function is already quadratic, the approximation is exact. This connects single-variable f&#039;&#039;(0)/2 to the multivariable Hessian quadratic form, showing the unified structure of second-order approximations.

📖 Explanation: $f(x)=\arcsin(x)$ is odd: $\arcsin(-x) = -\arcsin(x)$. Odd functions have Taylor series with only odd powers. Therefore, the coefficient of $x^2$ must be zero. Calculus check: f&#039;(x)=(1-x^2)^{-1/2}, f&#039;&#039;(x)=x(1-x^2)^{-3/2}. At $x=0$, f&#039;&#039;(0)=0. Thus, the local quadratic approximation is indeed just $x$. The curvature appears at the cubic level ($x^3/6$). Trusting the symmetry argument allows quick verification of CAS outputs without full recalculation.

📖 Explanation: Taylor approximation requires analytical derivatives at a point, which are rarely available in discrete signal processing. Quadratic interpolation fits a parabola through three adjacent data points $(t_{i-1}, s_{i-1}), (t_i, s_i), (t_{i+1}, s_{i+1})$. This estimates local behavior using finite differences rather than infinitesimal derivatives. As sampling density increases, the interpolating quadratic converges to the Taylor quadratic, but for finite data, they differ. Interpolation is data-driven; Taylor is model-driven. Confusing them leads to errors when analytical forms are unknown.

📖 Explanation: At a pure inflection point ($x^3$), the quadratic term vanishes. Adding an arbitrarily small $\epsilon x^2$ breaks the odd symmetry and introduces a non-zero quadratic component. Topologically, the zero polynomial and $\epsilon x^2$ are distinct (one is flat, one is curved). This discontinuity in the space of approximations indicates structural instability. In practical computation, noise can mimic this $\epsilon$, causing the estimated quadratic approximation to fluctuate wildly between flat and curved near inflection points. This warns against over-interpreting small quadratic coefficients near suspected inflections.

📖 Explanation: Algebraic operations on Taylor polynomials are valid provided one truncates to the appropriate degree afterward. $\cos(x) = 1 - x^2/2 + O(x^4)$. Squaring: $(1 - x^2/2 + O(x^4))^2 = 1 - x^2 + x^4/4 + O(x^4)$. Discarding terms above degree 2 gives $1 - x^2$. Direct verification: $d^2/dx^2 (\cos^2 x)$ at 0 is $-2$. Coefficient is $-2/2 = -1$. Matches. This method is often faster than repeated product-rule differentiation. It demonstrates the power of treating series as algebraic objects, a key skill in advanced calculus.