📝 Newton's Method in calculus (26 MCQs)

📖 Explanation: When f'(x_n)=0, the denominator in the Newton step \frac{f(x_n)}{f'(x_n)} becomes undefined, causing a division‑by‑zero situation. Consequently the algorithm cannot produce a new iterate and stops, which is precisely the behavior described in option B.

📖 Explanation: Starting at $x_{0}=1$ gives $f(1)=-1$ and a derivative of 3, producing $x_{1}=1+\frac{1}{3}\approx1.333$. Subsequent iterates overshoot and grow larger, moving away from the root, so the behavior described in option C—divergence—is accurate.

📖 Explanation: Quadratic convergence means the error at step $n+1$ is proportional to the square of the previous error. This is expressed as $e_{n+1}=C e_n^{2}+o(e_n^{2})$, where $C$ depends on f''(r) and f'(r). Option A captures this defining property.

📖 Explanation: Newton's method's order of convergence depends on the first non‑zero derivative after the first. When f''(r)=0 but the root is simple, the higher‑order term does not affect the leading error term, so the method retains its usual quadratic rate, as stated in option A.

📖 Explanation: Newton's method requires a well‑defined derivative at each iterate. For $\cos x$, the root near $x\approx1.5708$ has derivative $-\sin x$ which is $-1$, not zero; however, the iteration function $g(x)=x+\tan x$ can produce large jumps if the derivative of the iteration map exceeds one in magnitude, leading to divergence. The most direct cause among the options is the large step size caused by a near‑zero denominator, which corresponds to option D.

📖 Explanation: For a root of multiplicity $m$, the standard Newton step loses its quadratic speed because the derivative is proportionally small. Multiplying the correction term by $m$ compensates for this effect, restoring the quadratic convergence property. Hence the adjusted formula in option A is the correct one.

📖 Explanation: Oscillation indicates that the iteration map g(x)=x-\frac{f(x)}{f'(x)} has a repelling fixed point, meaning the magnitude of its derivative at that point exceeds one. When |g'(r)|>1, nearby points are pushed away, often causing alternating behavior. This explanation matches option B.

📖 Explanation: Applying Newton's formula gives $x_{n+1}=x_n-\frac{x_n^{2}-4}{2x_n}= \frac{2x_n^{2}-x_n^{2}+4}{2x_n}= \frac{x_n^{2}+4}{2x_n}= \frac{x_n}{2}+\frac{2}{x_n}$. This matches option C.

📖 Explanation: Any positive starting value that is not zero will converge to the root $x=2$; however, larger values such as $x_{0}=5$ keep the iterates away from the singularity at zero and avoid division by a small derivative, ensuring stable convergence. Option D therefore satisfies the guarantee.

📖 Explanation: Newton's method exhibits quadratic convergence, so the error roughly squares each step, leading to a dramatic reduction in the number of iterations—typically under 5—to reach $10^{-6}$. In contrast, the bisection method halves the interval each step, needing about 20 iterations. Hence option A correctly describes the comparison.

📖 Explanation: For a simple root, Newton's method reduces the error roughly by the square of the previous error, i.e., $e_{n+1}\approx C e_n^{2}$. Starting with an error of order $10^{-1}$, the next error becomes about $10^{-2}$, corresponding to a reduction factor near 0.1. However, the constant $C$ for this function yields a slightly larger factor, making 0.25 a reasonable approximation.

📖 Explanation: Near $x=0$, the derivative of $f$ is $e^{0}=1$ while the derivative of $g$ is $\cos 0=1$. However, the curvature (second derivative) of $f$ is also 1, whereas for $g$ it is 0, making the error constant $C$ for $f$ smaller. Consequently, Newton's method converges slightly faster for the exponential function, as indicated in option A.

📖 Explanation: Newton's method is derived from the fixed‑point iteration $x_{n+1}=g(x_n)$. At a simple root $r$, the derivative of the iteration function satisfies g&#039;(r)=0, guaranteeing a contraction and thus local convergence. While |g&#039;(r)|<1 is sufficient for general fixed‑point methods, the specific property for Newton's method is the vanishing derivative, option C.

📖 Explanation: Expanding $f$ in a Taylor series about the simple root $r$ and substituting into the Newton step yields e_{n+1}= \frac{f&#039;&#039;(r)}{2f&#039;(r)}e_n^{2}+O(e_n^{3}). The coefficient of $e_n^{2}$ is the error constant $C$, which is \frac{f&#039;&#039;(r)}{2f&#039;(r)}. This matches option A.

📖 Explanation: The root at $x=1$ is simple because f&#039;(1)=5\neq0. Newton's method applied to a simple root always exhibits quadratic convergence, regardless of the polynomial degree. Hence the correct description is quadratic convergence, as given in option A.

📖 Explanation: Newton's step replaces the function locally by its tangent line, which is defined only when the derivative exists. Without a derivative, the tangent cannot be constructed, and the formula x_{n+1}=x_n-\frac{f(x_n)}{f&#039;(x_n)} becomes meaningless. This necessity is captured in option D.

📖 Explanation: Each Newton iteration computes the x‑intercept of the tangent line drawn at the current approximation. This point becomes the next approximation, embodying the geometric picture of moving from the current point to where its tangent crosses the horizontal axis. Option A accurately describes this process.

📖 Explanation: Newton's method for a cubic equation produces three distinct basins of attraction, each leading to one of the three complex roots. The boundaries between these basins are highly intricate, often exhibiting fractal structure. Consequently, the initial guess determines which root the iteration converges to, as expressed in option C.

📖 Explanation: A line search modifies the Newton update by multiplying the correction term by a factor $\alpha\in(0,1]$ chosen to reduce the residual $|f(x_{n+1})|$. This safeguards against large, potentially divergent steps, especially when the initial guess is far from the root, thereby enhancing global convergence. Option B captures this strategy.

📖 Explanation: For $f(x)=\sqrt{x}$, the derivative f&#039;(x)=\frac{1}{2\sqrt{x}} blows up as $x\to0^{+}$. The Newton step $x_{n+1}=x_n-\frac{\sqrt{x_n}}{1/(2\sqrt{x_n})}=0$ forces the next iterate to zero in a single step, producing quadratic convergence to the root. Hence option A correctly describes the behavior.

📖 Explanation: Newton's method reduces the error quadratically, but once the error magnitude approaches machine epsilon, the subtraction in the iteration amplifies round‑off. At that stage the error no longer squares effectively, and the observed convergence degrades to linear or stalls. Option C accurately reflects this limitation.

📖 Explanation: Replacing the exact derivative with a finite‑difference estimate introduces an additional error term of order $h$ (the step size). This perturbation typically reduces the overall convergence order from quadratic to linear, because the error constant no longer squares each iteration. Option B describes this common outcome.

📖 Explanation: At a saddle point the derivative vanishes, making the Newton step undefined or leading to division by a small number. Consequently, the iteration cannot progress and typically stalls, remaining near the saddle point without converging to a root. Option C correctly characterizes this behavior.

📖 Explanation: For a simple root where f&#039;(r)\neq0, Newton's method reduces the error roughly as the square of the previous error, a property known as quadratic convergence. This is a fundamental characteristic taught in introductory numerical analysis courses. Hence the correct answer is quadratic.

📖 Explanation: The Banach Fixed Point Theorem (also called the Contraction Mapping Theorem) provides sufficient conditions—namely, that the iteration function be a contraction on a closed interval—to ensure a unique fixed point and local convergence. Newton's method satisfies these conditions when the function is sufficiently smooth near a simple root, making option C the appropriate choice.

📖 Explanation: The Newton‑Kantorovich theorem asserts that if the inverse Jacobian at the initial point exists and the product \|f&#039;(x_0)^{-1}f(x_0)\| is bounded by $2/L$, where $L$ is a Lipschitz constant for the derivative, then Newton's method converges. This condition is expressed in option A.