Numerical Methods: Difficult Problems with Solutions

Solution:
For [tex]f\in C^2[a, b][/tex] the remainder of the composite trapezoidal rule with [tex]m[/tex] equal subintervals of length [tex]h=\frac{b-a}{m}[/tex] is
[tex]R=-\frac{(b-a)h^2}{12}f''(\xi)=-\frac{(b-a)^3}{12m^2}f''(\xi),\qquad \xi\in(a, b)[/tex]
The function [tex]f(x)=\frac{1}{x+2}[/tex] is smooth on [tex][1, 11][/tex], since [tex]x+2\gt 0[/tex] there, and
[tex]f'(x)=-\frac{1}{(x+2)^2},\qquad f''(x)=\frac{2}{(x+2)^3}[/tex]
The second derivative is positive and decreasing on [tex][1, 11][/tex], so
[tex]M_2=\max_{[1, 11]}|f''(x)|=f''(1)=\frac{2}{27}[/tex]
With [tex]b-a=10[/tex] and [tex]m=5[/tex] (step [tex]h=2[/tex]):
[tex]|R|\le\frac{10^3}{12\times 25}\times \frac{2}{27}=\frac{10}{3}\times \frac{2}{27}=\frac{20}{81}\approx 0.247[/tex]
Check: the trapezoidal value is [tex]\frac{1}{3}+2\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\frac{1}{11}\right)+\frac{1}{13}\approx 1.5000[/tex], while the exact value is [tex]\ln\frac{13}{3}\approx 1.4663[/tex]; the actual error [tex]\approx 0.034[/tex] is indeed within the bound.

Solution:
By Taylor's formula with the Lagrange remainder, for every [tex]x[/tex] there is a point [tex]\xi[/tex] between [tex]0[/tex] and [tex]x[/tex] such that
[tex]e^x=T_6(x)+\frac{f^{(7)}(\xi)}{7!}x^7[/tex]
Since [tex]f^{(7)}(x)=e^x[/tex], the error is
[tex]|e^x-T_6(x)|=\frac{e^{\xi}|x|^7}{7!}[/tex]
For [tex]x\in[-1, 1][/tex] the point [tex]\xi[/tex] also lies in [tex][-1, 1][/tex], and since [tex]e^x[/tex] is increasing, [tex]e^{\xi}\le e[/tex]; moreover [tex]|x|^7\le 1[/tex]. Hence
[tex]|e^x-T_6(x)|\le\frac{e}{7!}=\frac{e}{5040}\approx 5.4\times 10^{-4}[/tex]
For example, [tex]T_6(1)=\frac{1957}{720}\approx 2.71806[/tex] approximates [tex]e\approx 2.71828[/tex] with an actual error [tex]\approx 2.3\times 10^{-4}[/tex], within the bound. The factor [tex]e^{\xi}[/tex] cannot be dropped: the bound [tex]\frac{1}{7!}\approx 1.98\times 10^{-4}[/tex] is false at [tex]x=1[/tex].

Solution:
Let [tex]\varphi(x)=1+0.5\sin x[/tex], so that [tex]x_{k+1}=\varphi(x_k)[/tex]. By the mean value theorem, for any [tex]a,b\in\mathbb{R}[/tex] there is [tex]c[/tex] between them with [tex]\varphi(a)-\varphi(b)=0.5\cos c\,(a-b)[/tex]. Since [tex]|\cos c|\le 1[/tex],
[tex]|\varphi(a)-\varphi(b)|\le\frac{1}{2}|a-b|[/tex]
i.e. [tex]\varphi[/tex] is a contraction with constant [tex]q=\frac{1}{2}\lt 1[/tex]. Consequently
[tex]|x_{k+1}-x_k|=|\varphi(x_k)-\varphi(x_{k-1})|\le\frac{1}{2}|x_k-x_{k-1}|\le\ldots\le\frac{1}{2^k}|x_1-x_0|[/tex]
For [tex]p\gt k[/tex] the triangle inequality gives
[tex]|x_p-x_k|\le\sum_{j=k}^{p-1}|x_{j+1}-x_j|\le|x_1-x_0|\sum_{j=k}^{\infty}\frac{1}{2^j}=\frac{|x_1-x_0|}{2^{k-1}}[/tex]
which tends to [tex]0[/tex] as [tex]k\to\infty[/tex]. So [tex](x_k)[/tex] is a Cauchy sequence and, by the completeness of [tex]\mathbb{R}[/tex], it converges to some [tex]\xi[/tex]. Since [tex]\varphi[/tex] is continuous, letting [tex]k\to\infty[/tex] in [tex]x_{k+1}=\varphi(x_k)[/tex] gives
[tex]\xi=\varphi(\xi)=1+0.5\sin \xi[/tex]
so the limit is a root of the equation. Moreover, the root is unique: if also [tex]\eta=\varphi(\eta)[/tex], then [tex]|\xi-\eta|=|\varphi(\xi)-\varphi(\eta)|\le\frac{1}{2}|\xi-\eta|[/tex], which forces [tex]\xi=\eta[/tex]. Hence every starting point leads to the same root [tex]\xi\approx 1.4987[/tex].