Estimate the error that results when $\sqrt{1 + x}$ is replaced by $1 + \frac{1}{2}x$ if $|x| < 0.01$ 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 $\xi = -0.01$.
 A: 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.$|x| < 0.01$
$$\begin{equation}
  \begin{aligned}
    R_n(x) & = \frac{- \frac{1}{4(1 + \xi)^{+3/2}} }{2!}
  \end{aligned}
\end{equation}\leq \frac{- \frac{1}{4} (1 + (-0.01))^{-3/2}}{2!}=\frac{- \frac{1}{4} (0.99)^{-3/2}}{2!}$$
A: You can just expand $f(x) \approx 1+\frac 12x-\frac 18x^2+\ldots$ as a Taylor series.  The first two terms of the replacement are correct, so the error for small $x$ is dominated by the $-\frac 18x^2$ term.  The error will then be negative and no less than $-\frac 18(0.01)^2=-.0000125$
A: You have that
$$f(x)=f(0)+\dfrac{f'(0)}{1!}x+R_1(x).$$ Since $f(x)=\sqrt{1+x}$ it is $f(0)=1$ and $f'(0)=\dfrac 12.$ Thus 
$$\sqrt{1+x}=1+\dfrac{1}{2}x+R_1(x).$$
Now, $$R_1(x)=\dfrac{f''(\xi)}{2!}x^2.$$ Since $$f''(\xi)=-\dfrac14 (1+\xi)^{-3/2}$$ we get
$$\sqrt{1+x}-\left(1+\dfrac{1}{2}x\right)=-\dfrac18 (1+\xi)^{-3/2}x^2.$$ Now, if $|x|\le 0.01$ we have to get a bound for the RHT. We have that 
$$\dfrac18 (1+\xi)^{-3/2}x^2\le \dfrac{0.01^2}{8},$$ where we have used that $$(1+\xi)^{-3/2}\le 1.$$
A: Usually the Laplace integral form of the remainder yields a tighter estimate of the error. At order $1$, it is
$$R_1(x)=\int_0^x\!\frac{f''(t)}{2!}(x-t)\,\mathrm d\mkern1mu t=-\frac18\int_0^x\!\!\frac 1{(1+t)^{3/2}}(x-t)\,\mathrm d\mkern1mu t$$
Let's compute this integral:
\begin{align}
\int_0^x\!\!\frac 1{(1+t)^{3/2}}(x-t)\,\mathrm d\mkern1mu t &=\int_0^x\!\!\frac {(x+1)-(1+t)}{(1+t)^{3/2}}\,\mathrm d\mkern1mu t \\
&=(x+1)\int_0^x\!\!\frac {1}{(1+t)^{3/2}}\,\mathrm d\mkern1mu t -\int_0^x\!\!\frac {1}{\sqrt{1+t}}\,\mathrm d\mkern1mu t\\
&=(x+1)\ \frac{-2}{\sqrt{1+t}}\Biggr\vert_0^x-2\sqrt{1+t}\,\biggr\vert_0^x \\
&=-2\sqrt{x+1}+2(x+1)-2\sqrt{x+1}+2\\
&=2\bigl(\sqrt{x+1}-1\bigr)^2
\end{align}
As on $(-0.01, 0.01)$, we have $-0.02<\sqrt{x+1}-1<0.02$, so that the integral is non-negative and $<0.0004$, and finally
$$-5\cdot 10^{-5}<R_1(x)\le 0. $$
