I'm trying to approximate $\sqrt{101}$ using the Taylor series for the function $f(x)=\sqrt{x}$ centered at the point $x=100$. I need to obtain an approximation that is within $0.01$ of the correct answer. The Taylor series is given by
$$
f(x) = \sum_{k=0}^{n-1} \frac{f^{(k)}(100)}{k!}(x-100)^k + R_n(x)
$$
where $R_n(x)$ is the remainder term. For the case $n=2$, we have
$$
R_2(x) = \frac{f''(c)}{2!}(101-100)^2 = \frac{f''(c)}{2}.
$$
where $c \in [100, \, 101]$. Since $f''(x) = -1/4x^{3/2}$, we have
$$
\left|R_2(101)\right| = \left| \frac{1}{2} \cdot \frac{-1}{4c^{3/2}} \right| = \left| \frac{1}{8c^{3/2}} \right|
$$
Since we are considering the interval $[100, \, 101]$, we can bound this by
$$
\left|R_2(101)\right| \leq \frac{1}{8\cdot 100^{3/2}} = \frac{1}{8000} = 0.000125.
$$
Thus, the approximation using the first two terms of the Taylor expansion should be sufficiently accurate. Those two terms are given by
$$
f(101) = f(100) + f'(100)/2 = 10 + 1/40 = 401/40.
$$
However, using numerical software to confirm the approximation, we see that
$$
\left| (401/40) – \sqrt{101} \right| \approx 0.024875621,
$$
which is not sufficiently accurate. Can anyone tell me why I am not obtaining a sufficiently accurate approximation? Thanks in advance!
Best Answer
I also think that most of the problem is related to the mistake @Macavity noticed and, accordingly, $$\left| (201/20) - \sqrt{101} \right| \approx 0.000124379$$ But what I would like to suggest for this kind of problem is to write $$\sqrt x=\sqrt {x_0+(x-x_0)}=\sqrt {x_0} \sqrt {1+\frac{x-x_0}{x_0}}=\sqrt {x_0} \sqrt {1+y}$$ and consider the Taylor expansion of the last term around $y=0$. I suppose that this could make life slightly easier.
This would make $$\sqrt x=x_0\left(1+\frac{k}{2 x_0^2}-\frac{k^2}{8 x_0^4}+O\left(\frac{1}{x^8}\right)\right)$$ So, in your case $$\sqrt{101}=10\left(1+\frac{1}{200}-\frac{1}{80000}+\cdots\right)\approx \frac{80399}{8000} =10.04987500$$ while the "exact" value would be $\approx 10.04987562$