📝 Taylor polynomials formula (35 MCQs)

📖 Explanation: The third-degree Taylor polynomial includes the term \frac{f'''(0)}{3!}x^3. If f'''(0) is incorrect, the error introduced is proportional to $x^3$. For $x=0.1$, $x^3 = 0.001$, making the error negligible. However, for $x=2.0$, $x^3 = 8$, amplifying the coefficient error by a factor of 8000 relative to the smaller input. This demonstrates that errors in higher-order derivatives have a disproportionately catastrophic effect on approximation accuracy as the distance from the center of expansion increases, highlighting the local nature of Taylor approximations.

📖 Explanation: Since the difference between the functions is $x^4$, and the second-degree Taylor polynomial only utilizes derivatives up to order two, the term $x^4$ vanishes during the calculation of f(0), f'(0), and f''(0). Thus, $p_2(x)$ is identical for both. Furthermore, since the first non-zero neglected term in the Taylor series determines the leading order of the error, and both functions share the same lower-order behavior, the local approximation quality defined by the second-degree polynomial is effectively equivalent regarding the matched derivatives. The distinction only appears in higher-order remainders.

📖 Explanation: For $\cos(x)$, the Maclaurin series is alternating. The error bound for an alternating series satisfying the Alternating Series Test is simply the magnitude of the first omitted term. For $n=2$, the next term is $x^4/4!$. At $x=0.5$, this is $0.5^4/24 \approx 0.0026$, which is too large. For $n=4$, the polynomial includes the $x^4$ term (since the $x^3$ term is zero). The first omitted non-zero term is $-x^6/6!$. Evaluating $0.5^6/720 \approx 2.17 \times 10^{-7}$, which satisfies the $10^{-6}$ requirement. Note that $p_3(x) = p_4(x)$ for cosine, so technically degree 3 suffices algebraically, but standard error analysis often targets the even degree for cosine bounds; however, strictly speaking, the polynomial of degree 4 guarantees the bound via the next term test.

📖 Explanation: By definition, the zeroth-degree Taylor polynomial $p_0(x) = f(a)$ matches only the value, corresponding to Curve A. The first-degree polynomial $p_1(x)$ matches the value and the first derivative (slope), representing the tangent line; however, the prompt states B has the wrong concavity, implying it matches slope but fails curvature, consistent with a linear approximation or a mismatched quadratic. Wait, re-reading: Curve B is tangent (matches slope) but wrong concavity. This describes $p_1$ perfectly as it has no intrinsic concavity (linear). Curve C matches slope and concavity, defining $p_2$. Therefore, A corresponds to $p_0$, B to $p_1$, and C to $p_2$.

📖 Explanation: While the conclusion that $R=1$ is correct, the student's reasoning based solely on the real-valued domain restriction is incomplete. In 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 $\ln(1+z)$, the branch point is indeed at $z=-1$, which is distance 1 from the origin. However, relying only on 'undefined at -1' ignores cases where functions are smooth on the real line but have limited radii due to complex singularities (e.g., $1/(1+x^2)$). Thus, while the answer is right, the justification lacks the necessary theoretical depth regarding analytic continuation.

📖 Explanation: According to the Lagrange Error Bound, $|R_n(x)| \leq \frac{M}{(n+1)!}|x-x_0|^{n+1}$, where $M$ is the maximum of $|f^{(n+1)}(c)|$ on the interval between $x_0$ and $x$. For $f(x)=e^x$, every derivative is $e^x$, which is strictly increasing. Therefore, for any $x \in [0,1]$, the maximum value of the derivative on $[0,x]$ occurs at $c=x$. Consequently, the error bound function $\frac{e^x}{(n+1)!}x^{n+1}$ is the product of two positive increasing functions on $[0,1]$, guaranteeing that the worst-case error bound increases monotonically with $x$, peaking at $x=1$.

📖 Explanation: In physics, a stable equilibrium point $x_0$ is defined as a local minimum of the potential energy function $U(x)$. By calculus, a local extremum requires U&#039;(x_0) = 0, eliminating the linear term U&#039;(x_0)(x-x_0) from the Taylor expansion. The behavior near equilibrium is dominated by the quadratic term \frac{U&#039;&#039;(x_0)}{2}(x-x_0)^2. For the point to be a minimum (stable), we must have U&#039;&#039;(x_0) > 0. If U&#039;&#039;(x_0) < 0, it would be a maximum (unstable equilibrium). Thus, the absence of the linear term is a consequence of equilibrium, and the positive quadratic coefficient confirms stability, linking mathematical Taylor coefficients directly to physical system properties.

📖 Explanation: The Maclaurin series for $\sin(x)$ converges rapidly when $x$ is close to 0. For $x=0.1$, very few terms are needed for high precision. However, $89^\circ \approx 1.55$ radians is far from 0. Using the Maclaurin series here would require many more terms to achieve the same accuracy due to the slower decay of $x^n/n!$ for larger $x$. Instead, expanding $\sin(x)$ about $x_0 = \pi/2$ (where $\sin(\pi/2)=1$) transforms the problem into approximating $\cos(x-\pi/2)$ near zero, restoring rapid convergence. This highlights the critical importance of selecting an expansion center close to the evaluation point for numerical efficiency.

📖 Explanation: Both methods are mathematically valid and must yield the same result due to the uniqueness of Taylor series. However, direct differentiation of composite functions like $\sin(x^2)$ becomes exponentially complex due to repeated application of the chain and product rules. Substituting $x^2$ into the known series $u - u^3/6 + \dots$ yields $x^2 - x^6/6 + \dots$ almost instantly. Note that for the *fourth-degree* polynomial, the substitution method immediately reveals that the term involving $x^4$ has a coefficient of zero (since the series contains only even powers of $x^2$, i.e., $x^2, x^6, \dots$), whereas direct differentiation would require calculating up to the fourth derivative only to find it vanishes. Substitution is overwhelmingly superior.

📖 Explanation: The Lagrange Error Bound states $|R_n(x)| \leq \frac{\max|f^{(n+1)}(c)|}{(n+1)!}|x|^{n+1}$. Since $|x|^{n+1}$ is symmetric for $\pm x$, any asymmetry in the error growth must arise from the derivative term $\max|f^{(n+1)}(c)|$. If the error grows faster for $x>0$, it implies that the magnitude of the relevant higher-order derivative is larger on the positive interval $[0, x]$ than on the corresponding negative interval $[x, 0]$. For example, $e^x$ has derivatives that grow for $x>0$ and decay for $x<0$, causing precisely this asymmetric error profile. Radius of convergence is a property of the series limit, not the finite remainder asymmetry.

📖 Explanation: The function $e^{-x^2}$ is famous for lacking an elementary antiderivative. Therefore, one cannot symbolically integrate it to get a closed-form function $F(x)$ and then expand $F(x)$. The standard procedure is to expand $e^{-x^2}$ as a power series $\sum (-1)^n x^{2n}/n!$, which converges everywhere, and then integrate term-by-term to obtain a new series representation for the definite integral. This transforms an impossible symbolic integration problem into a manageable summation problem. While integrating term-by-term does increase the power (improving convergence slightly), the primary driver is the non-elementary nature of the integral itself.

📖 Explanation: The third-degree Taylor polynomial depends only on f(a), f&#039;(a), f&#039;&#039;(a), and f&#039;&#039;&#039;(a). If all these are zero, $p_3(x) \equiv 0$. However, this provides no information about higher-order derivatives. The function $f(x) = (x-a)^4$ has f(a)=f&#039;(a)=f&#039;&#039;(a)=f&#039;&#039;&#039;(a)=0, so its third-degree Taylor polynomial is indeed zero. Yet, $f(x)$ is clearly not the zero function for $x \neq a$. Even stronger counterexamples exist, such as $f(x) = e^{-1/(x-a)^2}$ (with $f(a)=0$), which has all derivatives zero at $a$ but is non-zero elsewhere. Thus, a vanishing Taylor polynomial of finite degree does not imply the function vanishes.

📖 Explanation: The Taylor series for $1/(1-x)$ is the geometric series $\sum x^n$, which diverges for $|x| \geq 1$. A Padé approximant represents the function as a ratio of polynomials $P(x)/Q(x)$. For $1/(1-x)$, the Padé approximant of order [0/1] is exactly $1/(1-x)$, capturing the pole at $x=1$ perfectly. Even for functions where the exact form is unknown, Padé approximants often provide accurate approximations beyond the radius of convergence of the Taylor series and can model singularities that polynomials cannot. Taylor polynomials, being polynomials, cannot exhibit vertical asymptotes and thus fail qualitatively near poles, whereas rational approximants succeed.

📖 Explanation: An even function satisfies $f(-x) = f(x)$. Differentiating this identity shows that f&#039;(-x) = -f&#039;(x) (odd), f&#039;&#039;(-x) = f&#039;&#039;(x) (even), and generally $f^{(k)}(0) = 0$ for all odd $k$ because an odd function evaluated at zero must be zero (assuming continuity/differentiability). Since the Maclaurin polynomial coefficients are $f^{(k)}(0)/k!$, all coefficients for odd powers of $x$ vanish. Therefore, the Maclaurin polynomial of an even function consists exclusively of even-powered terms. This symmetry property saves significant computation time and serves as a useful check for calculation errors.

📖 Explanation: The discrepancy is $1.005 - 1.0049875 \approx 0.0000125$. This is a classic example of truncation error, which arises when an infinite series is approximated by a finite partial sum. The next term in the binomial expansion of $(1+x)^{1/2}$ is $\frac{(1/2)(-1/2)}{2!}x^2 = -x^2/8$. For $x=0.01$, this term is $-0.0001/8 = -0.0000125$, which exactly accounts for the observed difference. Round-off error relates to floating-point precision, not series termination. The binomial series is valid for $|x|<1$, so validity is not the issue. The linear coefficient $1/2$ is correct.

📖 Explanation: The absolute value function $|x|$ has a sharp corner (cusp) at $x=0$, meaning the left-hand derivative (-1) and right-hand derivative (+1) do not agree. Since the first derivative does not exist at the expansion center, the Taylor polynomial cannot be formed. In physical modeling, this signifies that smooth polynomial approximations fail to capture non-smooth phenomena like friction transitions or idealized hard-wall potentials at the singularity. One must either use piecewise definitions, generalized functions, or alternative basis sets capable of representing non-differentiable features. This limitation underscores that Taylor series are inherently tools for smooth (analytic or at least $C^\infty$) local approximation.

📖 Explanation: Uniqueness of Taylor series applies to the *infinite* series representation of an analytic function. However, a *finite* Taylor polynomial only encodes information about derivatives up to a specific order. Any function of the form $\sin(x) + x^6 g(x)$ (where $g$ is sufficiently differentiable) will have the exact same 5th-degree Taylor polynomial as $\sin(x)$. For instance, $f(x) = \sin(x) + x^6$ or $f(x) = \sin(x) + e^{-1/x^2}$ both match the given polynomial. Therefore, observing a finite polynomial does not uniquely identify the generating function; it only constrains its local jet at the expansion point. Additional global or structural information is needed to confirm identity.

📖 Explanation: Picard iteration or successive substitution for y&#039; = y, y(0)=1 starts with $y_0(x)=1$. Then $y_{n+1}(x) = 1 + \int_0^x y_n(t) dt$. This yields $y_1 = 1+x$, $y_2 = 1+x+x^2/2$, etc. Each iteration adds the next term of the Maclaurin series for $e^x$. This constructive proof demonstrates that the Taylor series is not merely a post-hoc approximation but emerges naturally from the differential equation's structure. It validates the series solution method and shows that the Maclaurin series for $e^x$ is the unique fixed point of the integral operator associated with the ODE, linking analytical series methods directly to existence-uniqueness theory.

📖 Explanation: The Lagrange Remainder Theorem requires that $f$ be $(n+1)$-times differentiable on an open interval containing both the center $x_0$ and the evaluation point $x$. For $\ln(x)$ centered at 1, the domain is $(0, \infty)$. The point $x=0$ lies on the boundary of the domain where $\ln(x)$ and all its derivatives blow up to negative infinity. Since $f$ is not differentiable at 0, the remainder formula cannot be applied on any interval including 0. This reflects the fundamental fact that the Taylor series for $\ln(x)$ at 1 has radius of convergence 1, and the singularity at 0 represents the barrier beyond which the polynomial approximation ceases to be valid or bounded.

📖 Explanation: This is a classic illustration of the distinction between real smoothness and complex analyticity. On the real line, $1/(1+x^2)$ is $C^\infty$ everywhere. However, the radius of convergence of a Taylor series centered at 0 is determined by the distance to the nearest singularity in the complex plane. The denominator vanishes at $x = \pm i$, which are distance 1 from the origin. Thus, $R=1$. Beyond this disk, the power series diverges regardless of the function's real-valued behavior. This demonstrates that real-variable intuition about 'smoothness' is insufficient for predicting series convergence; one must consider the function's analytic structure in the complex domain.

📖 Explanation: Taylor polynomials are designed to match derivatives at a single point, resulting in excellent accuracy near that point but potentially large errors at the interval boundaries. For applications requiring uniform accuracy over a fixed range (like graphics or embedded systems), minimax (Chebyshev) approximation is superior because it distributes the error evenly across the interval, minimizing the worst-case deviation. A degree-n minimax polynomial typically achieves significantly lower maximum error over $[a,b]$ than the degree-n Taylor polynomial. While Taylor series are theoretically fundamental, practical numerical implementation favors equioscillating error distributions for robustness and guaranteed precision bounds across the entire operational domain.

📖 Explanation: A fundamental theorem of power series states that differentiation and integration do not change the radius of convergence. If $\sum a_n x^n$ has radius $R$, then $\sum n a_n x^{n-1}$ also has radius $R$. The behavior at the endpoints may change (convergence might be lost or gained at the boundary), but the radius itself remains invariant. This is because the limit superior defining $R$ depends on the asymptotic growth rate of coefficients, and multiplying by $n$ does not alter this exponential growth rate. Thus, knowing $R=3$ for $f$ immediately tells us $R=3$ for f&#039;, f&#039;&#039;, and $\int f$.

📖 Explanation: For $F(v) = kv^2$, we have $F(0)=0$ and F&#039;(v)=2kv \Rightarrow F&#039;(0)=0. The first-order Taylor polynomial is p_1(v) = F(0) + F&#039;(0)v = 0. This approximation suggests there is no drag force at all for small velocities, which is physically useless and mathematically degenerate for this specific function. To capture the quadratic drag, one must use at least the second-order polynomial $p_2(v) = kv^2$. This illustrates a critical modeling pitfall: blindly applying linearization to functions with vanishing first derivatives at the operating point yields trivial results. One must inspect the derivative structure before choosing approximation order.

📖 Explanation: The Maclaurin series for $\cos(x)$ contains only even powers of $x$: $1 - x^2/2! + x^4/4! - \dots$. The coefficient of $x^3$ is f&#039;&#039;&#039;(0)/3!. Since f&#039;&#039;&#039;(x) = \sin(x) and $\sin(0)=0$, the cubic term vanishes. More fundamentally, because $\cos(x)$ is even ($\cos(-x)=\cos(x)$), its Taylor expansion cannot contain odd powers, as odd powers would break the symmetry. Therefore, $p_3(x) = p_2(x) + 0 \cdot x^3 = p_2(x)$. This means that for cosine, using a cubic polynomial offers no additional accuracy over the quadratic one; one gains nothing until the quartic term.

📖 Explanation: Leveraging known Maclaurin series through substitution is a powerful verification and computation tool. The standard series for $e^u$ is $\sum u^n/n!$. Replacing $u$ with $2x$ gives $1 + (2x) + (2x)^2/2! + \dots = 1 + 2x + 4x^2/2 + \dots = 1 + 2x + 2x^2 + \dots$. This matches the colleague's result instantly. This method is not only faster than computing derivatives via chain rule but also less error-prone. It relies on the uniqueness of power series representations: if a valid series manipulation produces a power series for $f$, it must be the Taylor series. This exemplifies efficient mathematical practice over brute-force computation.

📖 Explanation: The binomial expansion of $(1-u)^{-1/2}$ converges only for $|u|<1$. Here $u=(v/c)^2$, so convergence requires $v<c$. At $v=c$, the series diverges, mirroring the physical reality that $\gamma \to \infty$ as $v \to c$, requiring infinite energy to accelerate a massive particle to light speed. The mathematical singularity of the Taylor series at the radius boundary is not a defect but a feature: it encodes the fundamental causal structure of spacetime. This deep correspondence between analytic properties of series expansions and physical limits demonstrates how mathematical formalism can reflect ontological constraints in theoretical physics.

📖 Explanation: Taylor polynomials represent functions in the shifted monomial basis $\{(x-a)^k\}$. For a polynomial of degree $n$, its Taylor polynomial of degree $\geq n$ centered at any point is exactly equal to the original function, just rewritten. Expanding $1 + 3(x-1) + 3(x-1)^2 + (x-1)^3$ using binomial theorem recovers $x^3$ perfectly. The student's confusion stems from expecting the standard monomial form $x^3$ rather than recognizing that $x^3$ and its Taylor expansion at 1 are the same mathematical object in different coordinate representations. This reinforces that Taylor series for polynomials are exact identities, not approximations, when carried to sufficient degree.

📖 Explanation: While Taylor series converge analytically within their radius, finite high-degree Taylor polynomials can suffer from severe numerical issues. Near the edges of the interval, cancellation errors between large alternating terms can destroy precision. Additionally, the conditioning of polynomial evaluation worsens with degree. Unlike Chebyshev or minimax approximations which are optimized for uniform error, Taylor polynomials concentrate accuracy at the center, leaving the periphery vulnerable to oscillatory artifacts and amplified round-off. For wide intervals, orthogonal polynomial expansions or piecewise low-degree approximations are numerically superior. This distinction between theoretical convergence and practical numerical stability is crucial in scientific computing.

📖 Explanation: Directly computing Taylor coefficients via $f^{(n)}(0)/n!$ would require differentiating $\int_0^x \sin(t^2)dt$ repeatedly. The first derivative is $\sin(x^2)$, the second is $2x\cos(x^2)$, the third involves products and chains that grow combinatorially complex. Conversely, $\sin(u) = \sum (-1)^n u^{2n+1}/(2n+1)!$. Substituting $u=t^2$ gives $\sin(t^2) = \sum (-1)^n t^{4n+2}/(2n+1)!$. Integrating term-by-term yields $f(x) = \sum (-1)^n x^{4n+3}/[(4n+3)(2n+1)!]$ effortlessly. This exemplifies a key strategy: when a function is defined as an integral of a known series-expansible function, integrate the series rather than differentiate the integral. It transforms a hard calculus problem into simple algebra.

📖 Explanation: Many students incorrectly assume $\tan(x)$ shares sine's series because both are odd and start with $x$. However, $\tan(x) = \sin(x)/\cos(x)$. Performing polynomial division: $(x - x^3/6 + \dots) / (1 - x^2/2 + \dots) \approx (x - x^3/6)(1 + x^2/2) \approx x + x^3/2 - x^3/6 = x + x^3/3$. Alternatively, f&#039;(x)=\sec^2 x \Rightarrow f&#039;(0)=1; f&#039;&#039;&#039;(x) = 2\sec^4 x + 4\sec^2 x \tan^2 x \Rightarrow f&#039;&#039;&#039;(0)=2. So coefficient is $2/3! = 1/3$. The distractor $x-x^3/6$ tests the common misconception that tangent and sine have identical low-order behavior. Correct derivation requires either division or careful differentiation.

📖 Explanation: The power of the Lagrange Error Bound lies in converting an unknown quantity into a computable bound. While $c$ is unknown, if we can find $M = \max_{z \in I} |f^{(n+1)}(z)|$ on the interval $I$ between $x_0$ and $x$, then $|R_n(x)| \leq M|x-x_0|^{n+1}/(n+1)!$ holds for all possible $c$. We don't need to know $c$; we only need to bound the derivative. For functions like sine, cosine, or exponentials on bounded intervals, finding $M$ is straightforward. This transforms the existential quantifier ('there exists c') into a universal guarantee ('error is at most...'), making the theorem practically useful for rigorous numerical certification.

📖 Explanation: This is the canonical counterexample in analysis. All derivatives of $e^{-1/x^2}$ at 0 exist and equal 0, so the Taylor series is $0 + 0x + 0x^2 + \dots \equiv 0$. However, $f(x) > 0$ for all $x \neq 0$. Thus, the Taylor series does not converge to the function in any neighborhood of 0. By definition, a function is analytic at a point if its Taylor series converges to the function in some open interval around that point. Since this fails, $f$ is $C^\infty$ but not analytic at 0. This demonstrates that infinite differentiability is strictly weaker than analyticity, a subtle but crucial distinction in advanced calculus.

📖 Explanation: For $\ln(1.1)$, direct series uses $u=0.1$, converging as $0.1^n$. The transformed series $\ln((1+x)/(1-x)) = 2(x + x^3/3 + x^5/5 + \dots)$ requires solving $(1+x)/(1-x) = 1.1 \Rightarrow x = 1/21 \approx 0.0476$. This series converges as $x^{2n+1} \approx (0.0476)^{2n+1}$. Since $0.0476^3 \approx 10^{-4}$ vs $0.1^3 = 10^{-3}$, the transformed series achieves each additional digit of accuracy in roughly half the terms. Historically, this was crucial for hand-computation of logarithm tables. It exemplifies how algebraic manipulation of series arguments can dramatically accelerate convergence, a technique still relevant in algorithm design.

📖 Explanation: Matching derivatives at a point ensures tangency of high order but does not prevent crossings elsewhere. For $f(x)=\sin(x)$ and $p_3(x)=x-x^3/6$, they intersect whenever $\sin(x) = x-x^3/6$. Besides $x=0$, numerical solution shows intersections near $x \approx \pm 1.89$ (within the region where the cubic approximation is reasonable). In fact, for oscillatory functions, the polynomial approximation will typically cross the function multiple times near the expansion center as it attempts to match curvature. The condition $f^{(k)}(x_0)=p_n^{(k)}(x_0)$ constrains local behavior infinitesimally but imposes no global restriction on intersection count.

📖 Explanation: Applying the ratio test to $\binom{m}{n}x^n$ gives $|\frac{a_{n+1}}{a_n}| = |\frac{m-n}{n+1}x| \to |x|$ as $n \to \infty$. Thus, absolute convergence requires $|x|<1$. At $|x|=1$, the ratio test is inconclusive, and convergence depends delicately on $m$: for $m>0$, it converges absolutely at both endpoints; for $-1<m<0$, conditionally at $x=1$ and diverges at $x=-1$; for $m \leq -1$, diverges at both. The restriction $|x|<1$ is therefore the universal safe zone guaranteed by the ratio test, while endpoint analysis requires separate treatment. This nuanced convergence behavior distinguishes general binomial series from polynomial cases.