I'm looking for an approximation of $\sqrt 8$ with an approximation of $10^{-4}$. It was given that $\sqrt 8 =3\sqrt\frac{8}{9}$ so I set up a general series for $ \sqrt{1+x}^\frac{1}{2} $ around zero to the fourth term:
$3-(\frac{3}{2})(\frac{1}{9})-(\frac{3}{8})(\frac{1}{9^{2}})-(\frac{3}{16})(\frac{1}{9^{3}})-(\frac{15}{128})(\frac{1}{9^{4}} ) $
Now, for Lagrange's remainder term:
$\frac{f^{(n+1)}(c)}{(n+1)!}(\frac{1}{9})^{(n+1)} $
Now, seeing as I have to solve this question under a time constraint (final exam question), how do I find the approximating without summing up every term until I find the right approximation? I'm not exactly sure as to how to use the above formula.
Best Answer
The remainder term in a power series is an upper bound for the error. So your first step is to determine which is the smallest natural number $n$ such that $\left|\frac{f^{(n+1)}(0)}{(n+1)!\hspace{1ex} 9^{n+1}}\right|<10^{-4}$. Say you guess $n=3$. Then you must find $\left|\frac{f^{(4)}(0)}{4!\hspace{1ex} 9^4}\right|$. Now $f(x)=3\sqrt{1+x}$ so taking some derivatives you can find that $f^{(4)}(x)=-\frac{45}{16}(1+x)^{-7/2}$. Thus, $f^{(4)}(0)=-\frac{45}{16}.$ Then just plug everything in. $$\left|\frac{f^{(4)}(0)}{(4)\cdot 9^{4}}\right|=\left|\frac{-45}{16\cdot4!\cdot 9^{4}}\right|\approx 1.786\cdot10^{-5}$$
Since this is smaller than $10^{-4}$, taking terms up to and including $n=3$ would be sufficient to obtain the desired level of accuracy. At this point, you could just find those terms, add them up and be done. It is possible at this point that that many terms are not necessary. To determine the fewest number of terms needed to obtain that level of accuracy, you would have to check the error term with $n=2$, and so on until you are not guaranteed to be within $10^{-4}$.