Numerical Analysis

(1) The mean value theorem makes g a local contraction. (2) is too strong: the start $x_0=x\ast$ itself converges trivially, and preimages of x* land there in finitely many steps. (3) Taylor: $e_n$₊${}_1=$ ½$g^{\prime \prime }(\xi )e_n^2.(4)g=x-f/f^{\prime }$ has $g^{\prime }=f\cdot f^{\prime \prime }/(f^{\prime }{)}^2$, which vanishes where f does — exactly why Newton converges quadratically at simple roots.

(b) is false: it can diverge or cycle from a poor start.

(1) Standard result: if f′(r) ≠ 0 and f is twice continuously differentiable, the error satisfies $e_n$₊${}_1\approx$ Ce${}_n^2$, which is second-order convergence. (2) At a root of multiplicity m the iteration degrades to linear convergence with asymptotic factor (m − 1)/m, so m = 2 gives factor 1/2. (3) Fails — a poor starting point can cycle or diverge, as with f(x) = $x^{1/3}$ or a guess landing near a turning point. (4) Fails: Newton needs f′ and a starting point, not a bracketing interval; that requirement belongs to bisection, which is why Newton can leave the interval entirely.

$x_1=x_0-f(x_0)/f^{\prime }(x_0)=1-(-1)/2=\mathrm{1.5.}($The next iterate 17/12 ≈ 1.4167 already carries three correct digits of $\sqrt{2}$ — quadratic convergence at work.)

(1) The difference of two such polynomials has n + 1 roots but degree ≤ n. (2) Simpson's symmetric nodes buy one degree beyond quadratic: exactness degree 3. (3) is the standard error term. (4) is Runge's phenomenon: for $1/(1+25x^2)$ on [−1, 1], equispaced interpolants diverge near the endpoints as n grows — more nodes can make things worse.

The rule integrates every polynomial of degree ≤ 1 exactly (it is exact for the chord it uses) and fails on $x^2$: on [0, 1] it gives 1/2 instead of 1/3. Simpson's rule, by contrast, reaches degree 3.

$y_1=1+0.05\sqrt{6}\approx 1.1225;y_2=1.1225+0.05\sqrt{3.15+2.245+1}\approx 1.1225+0.05\cdot 2.529\approx 1.249\approx 1.25$.

(1)/(2) are the standard orders — the global order sits one below the local order because ~1/h steps accumulate. (3) restates global order 1. (4) fails: explicit Euler gives $y_n$₊${}_1=(1+h\lambda )y_n$, stable only when |$1+h\lambda$| < 1, i.e. h < 2/|$\lambda$|. That restriction is what stiffness means; implicit Euler removes it.