Spivak: How do we know $\arctan{x}$ and $\log{(1+x)}$ for all $x$ if we know these functions for $|x|<1$? In Ch. 20 of Spivak's Calculus, he shows that the remainder terms for $\arctan$ and $\log{(1+x)}$ become large with the order of the Taylor polynomial used to approximate these functions. Thus these approximations

are of no use whatsoever in computing $\arctan{x}$ and $\log{(1+x)}$.
This is no tragedy, because the values of these functions can be
found for any $x$ once they are known for all $x$ with $|x|<1$.

How do we find the values of these functions for any $x$ if we know the functions for $|x|<1$?
 A: With base ten logarithms, we have $\log_{10}(1234)=4+\log_{10}(0.1234)$, and in general,
$$\log_{10} x=n+\log_{10} (x/10^n),$$
where $n$ is the number of digits of $x$ before the decimal point. Natural logarithms are just this multiplied by a constant: $\log_e(x)=\log_{10}(x) / \log_{10}(e)$.
For tangents,
$$
\begin{align}
\tan(90^\circ-x)&={\sin (90^\circ-x)\over\cos (90^\circ-x)}\\
&={\cos x\over \sin x}\\
&={1\over \tan x}
\end{align}
$$
All the values of $\tan$ from $0$ to $90^\circ$ are known once the values are known up to $45^\circ$, so $$\arctan (1/x)=90^\circ-\arctan(x).$$ Thus you only need to compute values from $0$ to $1$.
A: It is easy to show, using differentiation the formula
$arctanx+arctan(\dfrac{1}{x})=\dfrac{\pi}{2}$. Hence, knowing the value
for $0<x<1$ we obtain the value of $arctanx$ for $x\geq1$. For negative $x$
we have $arctanx+arctan(\dfrac{1}{x})=-\dfrac{\pi}{2}$ and we follow the same method.
Now, for $log(1+x)$. We assume that $x\in\,[1,2)$ then $x=1+y$ where $0\leq\,y\,<1$.
So $log(1+x)=log(2+y)=log2(1+\dfrac{y}{2})$=$log2+log(1+\dfrac{y}{2})$ where $\dfrac{y}{2}<1$.
If $x\in[2,3)$ then $x=1+z$ where $1\leq z<2$, hence
$log(1+x)=log(2+z)=log2(1+\dfrac{z}{2})=log2+log(1+\dfrac{z}{2})$ where
$\dfrac{z}{2}<1$.
Continuing this process on $[3,4)$ e.t.c. we get the value of $log(1+x)$ for all $x>1$.
