Question
Estimate the error that results when $\sqrt{1 + x}$ is replaced by $1 + \frac{1}{2}x$ if $|x| < 0.01$
Definition
Taylors formula is $f(x) = P_n(x) + R_n(x)$ where $P_n(x)$ is
\begin{equation}
\begin{aligned}
P_n(x) = f(a)
+ \frac{f'(a)}{1!}(x – a)
+ \frac{f''(a)}{2!}(x – a)^2
& + \ldots
+ \frac{f^{(n)}(a)}{n!}(x – a)^n \\
\end{aligned}
\end{equation}
And $R_n (x) $ is (\emph{where $\xi$ is between $a$ and $x$ })
\begin{equation}
\begin{aligned}
R_n(x) = \frac{f^{(n + 1)}(\xi)}{(n + 1)!}(x – a)^{(n + 1)}
\end{aligned}
\end{equation}
Working
I'm not sure how to go about this, would I say that this is a first order
approximation as
\begin{equation}
\begin{aligned}
P(x) & = 1 – \frac{1}{2}x \\
P'(x) &= – \frac{1}{2} \\
P''(x) &= 0
\end{aligned}
\end{equation}
Then the remainder term would be
\begin{equation}
\begin{aligned}
R_n(x) = \frac{f^{(n + 1)}(\xi)}{(n + 1)!}(x – a)^{(n + 1)}
\end{aligned}
\end{equation}
Where $n + 1 = 2$. For $f(x) = \sqrt{1 + x}$ this would be
\begin{equation}
\begin{aligned}
R_n(x) & = \frac{- \frac{1}{4} (1 + \xi)^{-3/2}}{(3)!}
\end{aligned}
\end{equation}
And $\xi $ is between $-0.01$ and $0.01$
This would give the maximum error as
\begin{equation}
\begin{aligned}
R_n(x) & = \frac{- \frac{1}{4} (1 \pm 0.01 )^{-3/2}}{(3)!} \approx -0.0423
\end{aligned}
\end{equation}
The error is greatest when $\xi = -0.01$.
Best Answer
expanding about $x=0$ $$ \begin{equation} \begin{aligned} P_n(x) = f(0) + \frac{f'(0)}{1!}(x - 0) + \frac{f''(a)}{2!}(x - 0)^2 & + \ldots + \frac{f^{(n)}(0)}{n!}(x - 0)^n \\ \end{aligned} \end{equation}\\ $$ $$ \begin{equation} \begin{aligned} P_n(x) = 1 + \frac{1}{2}x + \frac{-\frac14}{2!}x^2 & + \ldots \end{aligned} \end{equation}\\ $$so w..r.t.$|x| < 0.01$ $$\begin{equation} \begin{aligned} R_n(x) & = \frac{- \frac{1}{4(1 + \xi)^{+3/2}} }{2!} \end{aligned} \end{equation}\leq \frac{- \frac{1}{4} (1 + (-0.01))^{-3/2}}{2!}=\frac{- \frac{1}{4} (0.99)^{-3/2}}{2!}$$