Higher-Degree Approximations and Taylor Polynomials

At $a=0$ the derivatives $\cos,-\sin,-\cos,\sin,\cos$ give $1,0,-1,0,1$, so

$T_4(x)=1+0\cdot x-\frac{x^{2}}{2!}+0\cdot x^{3}+\frac{x^{4}}{4!}=1-\frac{x^{2}}{2}+\frac{x^{4}}{24}$

$T_4(0.2)=1-0.02+\frac{0.0016}{24}=0.98006667$

Since $\cos$ is even the $x^{5}$ coefficient vanishes, so $T_4=T_5$ and $|R_5(0.2)|\le\frac{1\cdot(0.2)^{6}}{6!}=8.9\times10^{-8}$ — good to seven decimals.

Matching gives $c_0=2$, $c_1=-5$, $c_2=4$, $c_3=-1$, and $f^{(k)}(1)=k!\,c_k$:

$f(1)=2,\quad f'(1)=-5,\quad f''(1)=2!\cdot 4=8,\quad f'''(1)=3!\cdot(-1)=-6$

At $x=1.2$ the increment is $x-1=0.2$:

$T_3(1.2)=2-5(0.2)+4(0.2)^{2}-(0.2)^{3}=2-1+0.16-0.008=1.152$

The trap is reporting $f''(1)=4$; the coefficient is $f''(1)/2!$, not $f''(1)$.

Start from $e^{u}\approx 1+u+\frac{u^{2}}{2}+\frac{u^{3}}{6}$ and put $u=-x^{2}$, legal for a Maclaurin polynomial because $u=0$ when $x=0$. Each $u^{j}$ becomes degree $2j$, so three terms past the constant reach degree $6$:

$T_6(x)=1-x^{2}+\frac{x^{4}}{2}-\frac{x^{6}}{6}$

This has degree at most $6$ and differs from $f$ by less than any multiple of $x^{6}$, so by uniqueness it is $T_6$. Then $f^{(k)}(0)=k!\,c_k$:

$f^{(4)}(0)=4!\cdot\tfrac12=12,\qquad f^{(6)}(0)=6!\cdot\left(-\tfrac16\right)=-120$

Also $c_3=c_5=0$, so $f'''(0)=f^{(5)}(0)=0$ — as an even function requires.

$f'=\frac12(1+x)^{-1/2}$, $f''=-\frac14(1+x)^{-3/2}$, $f'''=\frac38(1+x)^{-5/2}$, so at $0$ the values are $1,\frac12,-\frac14,\frac38$ and

$T_3(x)=1+\frac{x}{2}-\frac{x^{2}}{8}+\frac{x^{3}}{16},\qquad T_3(0.1)=1.0488125$

Now $f^{(4)}(t)=-\frac{15}{16}(1+t)^{-7/2}$, whose size decreases on $[0,0.1]$, so $M=\frac{15}{16}$ at $t=0$:

$|R_3(0.1)|\le\frac{\frac{15}{16}(0.1)^{4}}{4!}=3.91\times10^{-6}$

So $1.0488085<\sqrt{1.1}<1.0488165$.

$f(8)=2$; $f'=\frac13x^{-2/3}$ gives $f'(8)=\frac{1}{12}$; $f''=-\frac29x^{-5/3}$ gives $f''(8)=-\frac{1}{144}$, so $c_2=-\frac{1}{288}$:

$T_2(x)=2+\frac{x-8}{12}-\frac{(x-8)^{2}}{288},\qquad T_2(8.5)=2+\frac{1}{24}-\frac{1}{1152}=2.04079861$

$f'''(t)=\frac{10}{27}t^{-8/3}$ is positive and decreasing, so on $[8,8.5]$ its maximum is at $t=8$: $M=\frac{10}{27\cdot 256}=1.447\times10^{-3}$. With $n=2$ and $|x-a|=0.5$,

$|R_2(8.5)|\le\frac{M(0.5)^{3}}{3!}=3.01\times10^{-5}$

Every derivative of $e^{x}$ is $e^{x}$, and on $[-1,1]$ it is increasing, so $|f^{(n+1)}(t)|\le e<3=M$; also $|x-0|\le 1$. Hence

$|R_n(x)|\le\frac{3\cdot 1^{n+1}}{(n+1)!}=\frac{3}{(n+1)!}$

Force this under $10^{-5}$: $(n+1)!>3\times10^{5}$. Since $8!=40\,320$ and $9!=362\,880$, the smallest choice is $n+1=9$, so $n=8$ and the bound is $3/9!=8.27\times10^{-6}$.

Check $n=7$ genuinely fails: at $x=1$, $T_7(1)=2.7182540$ against $e=2.7182818$, an error of $2.8\times10^{-5}$.