Numerical Methods: Problems with Solutions

Solution:
Newton's method uses the iteration below, which requires [tex]f'(x_k)\ne 0[/tex]:
[tex]x_{k+1}=x_k-\frac{f(x_k)}{f'(x_k)}[/tex]
Here [tex]f'(x)=3x^2-1[/tex], so
[tex]f(2)=8-2=6,\qquad f'(2)=12-1=11\ne 0[/tex]
Therefore
[tex]x_1=2-\frac{6}{11}=\frac{16}{11}\approx 1.4545[/tex]
The step moves from [tex]x_0=2[/tex] towards the root [tex]x=1[/tex] of the equation, as expected.

Solution:
Let [tex]f(x)=x^3+x-4[/tex]. It is continuous, [tex]f(1)=-2\lt 0[/tex] and [tex]f(4)=64\gt 0[/tex], so by Bolzano's theorem there is a root in [tex][1, 4][/tex]; since [tex]f'(x)=3x^2+1\gt 0[/tex], the function is strictly increasing and the root is unique.
Each bisection step halves the current interval and keeps the half at whose endpoints [tex]f[/tex] has opposite signs. So after [tex]n[/tex] steps the root lies in an interval of length
[tex]\frac{4-1}{2^n}=\frac{3}{2^n}[/tex]
We need
[tex]\frac{3}{2^n}\lt 0.01\iff 2^n\gt 300[/tex]
Since [tex]2^8=256\lt 300[/tex] and [tex]2^9=512\gt 300[/tex], the least such [tex]n[/tex] is [tex]9[/tex].

Solution:
The algebraic degree of precision is the largest [tex]m[/tex] such that the formula is exact for every polynomial of degree [tex]\le m[/tex]. Both sides are linear in [tex]f[/tex], so it suffices to test the monomials [tex]1,\ x,\ x^2,\ \ldots[/tex] until the first failure:
[tex]f=1:\quad \int_0^1 1\,dx=1,\qquad \frac{3}{4}+\frac{1}{4}=1[/tex]
[tex]f=x:\quad \int_0^1 x\,dx=\frac{1}{2},\qquad \frac{3}{4}\times \frac{1}{3}+\frac{1}{4}=\frac{1}{2}[/tex]
[tex]f=x^2:\quad \int_0^1 x^2\,dx=\frac{1}{3},\qquad \frac{3}{4}\times \frac{1}{9}+\frac{1}{4}=\frac{1}{12}+\frac{3}{12}=\frac{1}{3}[/tex]
[tex]f=x^3:\quad \int_0^1 x^3\,dx=\frac{1}{4},\qquad \frac{3}{4}\times \frac{1}{27}+\frac{1}{4}=\frac{1}{36}+\frac{9}{36}=\frac{5}{18}\ne\frac{1}{4}[/tex]
Hence the formula is exact for all polynomials of degree [tex]\le 2[/tex] but not for [tex]x^3[/tex], so its algebraic degree of precision is [tex]2[/tex].