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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 $|x|<1$?

xoux
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    Well-known formulas can be used: $\log(ab)$, $\arctan(1/x)$... – Jean-Armand Moroni Aug 21 '22 at 21:28
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    Alternatively, we can expand the convergence region for the natural logarithm series to include all positive real arguments or all complex arguments with positive real parts; see here. – Oscar Lanzi Aug 21 '22 at 22:12
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    $\log (x+1) = K\iff e^k = x+1\iff e^{k-1}=\frac {x+1}e$. If $|x|< 1$ then $|x+1| < 2$ and $|\frac{x+1}e| < 1$ and so $\log \frac {x+1}e$ is known. So $k-1=\log \frac {x+1}e$ and $\log(x+1) = k = 1+\log\frac {x+1}e$. – fleablood Aug 22 '22 at 00:20

2 Answers2

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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 $1$.

Benjamin Wang
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Suzu Hirose
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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 $x>1$.