How can I find the error bound when trying to approximate $\sqrt3$ with the expansion up to and including $x^3$ for $\sqrt{4-9x}$?
I thought about using : $$|R_n(x)|\leq\left|\frac{M(x-c)^{n+1}}{(n+1)!}\right|$$
But I struggled to find the upper bound $M$, otherwise $c=0, x=\frac{1}{9},n=3$ right?
(Edit, n=3 and not 4)
On
Binomial expansion gives the Maclaurin series with remainder: $$(4-9x)^{1/2}={1/2\choose 0}4^{1/2}+{1/2\choose 1}4^{1/2-1}(-9x)+{1/2\choose 2}4^{1/2-2}(-9x)^2+{1/2\choose 3}4^{1/2-3}(-9x)^3+R_{3}(x).$$
Hence, by using the bound $\left|{1/2\choose n}\right|<\left|{1/2\choose 4}\right|<0.04$ for all $n>4$, $$\left|R_{3}(\frac19)\right|=\left|\sum_{n=4}^\infty (-1)^n{1/2\choose n}4^{1/2-n}\right|<\sum_{n=4}^\infty \left|{1/2\choose n}\right|4^{1/2-n}<\sum_{n=4}^\infty\frac{0.08}{4^n}\approx 0.000416667.$$
On
Another way to express the error term in Taylor approximation is the integral form.
This states that if $f^k(x)$ is absolutely continuous on $[a, x]$ one has
$$R_k(x) := f(x) - P_k(x) = \int_a^x{\frac{f^{(k+1)}(t)}{k!}(x-t)^k\ dt}$$
where $P_k(x)$ is the Taylor series of $f(x)$ up to the $k$th term.
Letting $k=3$, $\ a = 0$, $\ x = 1/9$ and $f(x) = \sqrt{4-9x}$, $\ $ you have
$$|R_3(1/9)| \ = \left| \int_0^{1/9}{\frac{f^{(4)}(t)}{3!}\left(\frac{1}{9}-t\right)^3\ dt}\ \right| $$
$$= \left|\ \frac{1}{6} \int_0^{1/9}{\left(-9^4\frac{15}{16}(4-9t)^{-7/2}\right)\left(\frac{1}{9}-t\right)^3\ dt}\ \right|$$
$$= \ \frac{45}{32} \int_0^{1/9}{\frac{(1 - 9t)^3}{(4-9t)^3\sqrt{4-9t}}\ dt}.$$
Substituting $u = 1 - 9t$, it follows that
$$|R_3(1/9)| = \ \frac{5}{32} \int_0^{1}{\frac{u^3}{(u+3)^3\sqrt{u+3}}\ du} \ \lt\ \frac{5}{32\sqrt{3}} \int_0^{1}{\frac{u^3}{(u+3)^3}\ du}$$
Note: $\ $ Observing that the integrand is bounded above by $u^3/27$ one can achieve the exact same bound as @cs89. Also, evaluating the integral will give a smaller error term, however, it contains logs which I assume aren't allowed in this case.
Instead, you can use the fact that $h(u) := \frac{u^3}{(u+3)^3}$ is convex when $u\in(0, 1]$ to apply the trapezoid rule as follows
$$\int_0^{1}{\frac{u^3}{(u+3)^3}\ du} \ =\ \int_0^{1/2}{h(u)\ du} \ +\ \int_{1/2}^{1}{h(u)\ du}$$
$$\lt\ \left(\frac{1}{2}\cdot\frac{1}{2}\cdot\left(h(1/2)-h(0)\right)\right) +\ \left(\frac{1}{2}\cdot h(1/2) + \frac{1}{2}\cdot\frac{1}{2}\cdot\left(h(1) - h(1/2) \right)\right)$$
$$=\ \frac{1}{4}\left(\frac{1}{343} - 0\right) + \frac{1}{2}\cdot\frac{1}{343} + \frac{1}{4}\left(\frac{1}{64} - \frac{1}{343}\right)\ =\ \frac{1}{256} + \frac{1}{686}$$
Thus, using the standard bound $\sqrt{3} > 3/2$, $\ $ one obtains
$$|R_3(1/9)|\ \lt\ \frac{5}{32}\cdot\frac{2}{3}\left(\frac{1}{256} + \frac{1}{686}\right)$$
$$=\ \frac{5}{96}\left(\frac{1}{128} + \frac{1}{343}\right)\ \approx\ 0.0005587475.$$
As I understand it, you are considering the function $f : x \mapsto \sqrt{4-9x}$, performing a Taylor expansion of order $n=3$ at $0$, to obtain an approximation of its value at $x = \frac 1 9$, which is $\sqrt{3}$.
In order to quantify the remainder, one possible way to go is to use the mean-value form of the remainder in Taylor's theorem, which states that: $$ f(x) = f(0) + x f'(0) + \frac{x^2}{2!} f''(0) + \frac{x^3}{3!} f^{(3)}(0) + R_3(x) $$ where $$ R_3(x) = \frac{x^4}{4!} f^{(4)}(\xi_x) \quad \text{where} \quad \xi_x \in [0,x]. $$ Hence, to obtain an explicit bound, you just have to bound the fourth derivative on $f$ on $[0,\frac 19]$.
The fourth derivative of the function $s \mapsto \sqrt{s}$ is $-\frac{15}{16} x^{-\frac72}$. So, by composition, you have: $$ f^{(4)}(x) = - 9^4 \frac{15}{16} (4-9x)^{-\frac 72}. $$ Since this is an increasing function of $x$ and $\xi_x \in [0,x]$, you have, for $x = \frac 19$, $$ |f^{(4)}(\xi_x)| \leq |f^{(4)}(x)| = 9^4 \frac{15}{16} 3^{-\frac 72}. $$ Putting everything together, you end up with $$ |R_3(x)| \leq \frac{15}{16} \frac{1}{4!} 3^{-\frac 72} \approx 0.000835. $$