error of a Taylor polynomial

wahaj

the error of a taylor series of order(I think that's the right word) n is given by
[tex] \frac{f^{n+1} (s)}{n!} (x-a)^n [/tex]

I think this is right. The error in a linear approximation would simply be

[tex] \frac{f''(s)}{2} (x-a)^2 [/tex]

My question is what is s and how do I find it. Use linear approximation of √(47) if you need to use an example because I just did that question so it might be easier to explain.

jbunniii

Actually, it should be
$$\frac{f^{n+1} (s)}{(n+1)!} (x-a)^{n+1}$$

I think this is right. The error in a linear approximation would simply be

[tex] \frac{f''(s)}{2} (x-a)^2 [/tex]

All that you know for sure is that ##s## is a number between ##a## and ##x##, and it depends on ##x##.

OK, you didn't specify a value for ##a##, so I'll pick ##a = 49##. Then, if I still remember how to take derivatives, we should have
$$f(x) = x^{1/2}$$
$$f'(x) = \frac{1}{2}x^{-1/2}$$
$$f''(x) = \frac{-1}{4}x^{-3/2}$$
so
$$\begin{align}f(a) + f'(a)(x-a) + \frac{1}{2}f''(s)(x-a)^2 &= (49)^{1/2} + \frac{1}{2}(49)^{-1/2}(x-49) + \frac{-1}{8}s^{-3/2}(x-49)^2\\
&= 7 + \frac{1}{14}(x-49) - \frac{1}{8}s^{-3/2}(x-49)^2\\
\end{align}$$
where ##s## is some number between ##a## and ##x##. Setting ##x=47##, we get
$$\sqrt{47} = 7 - \frac{1}{7} - \frac{1}{2}s^{-3/2}$$
where ##s## is some number satisfying ##47 \leq s \leq 49##. Putting it another way, if we use only the linear approximation ##7 - 1/7##, then the approximation error is
##-\frac{1}{2}s^{-3/2}## for some ##47 \leq s \leq 49##. We can bound this error by taking the worst case ##s## (the value that maximizes the absolute approximation error), which in this case is obtained by choosing ##s=47##. Then the approximation error is guaranteed to satisfy
$$|\textrm{approximation error}| \leq \frac{1}{2}(47)^{-3/2} = 0.00155176...$$
Actually, we can also obtain a lower bound for the approximation error by choosing the best-case value of ##s##, namely ##s = 49##. The error will be at least ##\frac{1}{2}(49)^{-3/2} = 0.00145772...##. Thus we in fact have
$$0.00145772... \leq |\textrm{approximation error}| \leq 0.00155176...$$
We can check this: the actual value of ##\sqrt{47}## is
$$\sqrt{47} = 6.8556546...$$
and our approximation is
$$7 - 1/7 = 6.85714285...$$
and the error is therefore ##6.85714285... - 6.8556546... = 0.00148825...## which is indeed within the computed range.

wahaj

Thanks that makes more sense than my book