inequality - Inequalities from Taylor expansions of $log$ functions

Consider the following Taylor expansion of the natural logarithm (denoted by $\log$ here):

$$ \log(1+x) = x - x^2/2 + x^3/3 - x^4/4 + x^5/5 - \cdots $$

It appears that from this expansion, inequalities can be generated.
$ \log(1+x) \leq x $ is well known for all $x > -1$. The Taylor expansion however
motivates further inequalities which, on numerical inspection, appear valid for all $x > -1$:

$$ \log(1+x) \leq x - x^2/2 + x^3/3 \\
\log(1+x) \leq x - x^2/2 + x^3/3 - x^4/4 + x^5/5 \\
\cdots $$

Further, for even powers also inequalities appear to hold. For $-1 < x \leq 0$:
$$ \log(1+x) \leq x - x^2/2 \\
\log(1+x) \leq x - x^2/2 + x^3/3 - x^4/4 \\
\cdots $$

and for $x \geq 0$ the opposite:
$$ \log(1+x) \geq x - x^2/2 \\
\log(1+x) \geq x - x^2/2 + x^3/3 - x^4/4 \\
\cdots $$

The very same procedure also works with the Taylor expansion of $ (1+x) \log(1+x)$. Possibly other examples can be found.

Questions:

• does it hold indeed for expansions up to all powers of $x$?


• is this a special feature of the $\log$ function?


• is there a general rule when this procedure of "generating inequalities from Taylor expansions with alternating signs" will work?

Thanks for your help!

Answer

The derivatives of $f(x) = \log (1+x)$ are
$$
f^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{(1+x)^n} \quad (n \ge 1)

$$
If we denote the $n$th Taylor polynomial with $T_n$
$$
T_n(x) = \sum_{k=1}^n \frac{(-1)^{n-1}}{n} x^n
$$
and the remainder with $R_n$
$$
\log(1+x) = T_n(x) + R_n(x)
$$
then Taylor's theorem with the mean-value form of the remainder gives

$$
R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!} x^{n+1} = \frac{(-1)^nx^{n+1}}{(n+1)(1+\xi)^{n+1}}
$$
for some $\xi$ between $0$ and $x$.

So in this case, $f^{(n)}(x)$ has alternating signs independent of $x$,
and it follows that
$$
\log(1+x) \begin{cases}
< T_n(x) & \text{ for } -1 < x < 0 \\

< T_n(x) & \text{ for } 0 < x \le 1, n \text{ odd} \\
> T_n(x) & \text{ for } 0 < x \le 1, n \text{ even}
\end{cases}
$$

The case $-1 < x < 0$ is also obvious because all terms in the Taylor expansion are negative.

For $0 < x \le 1 $ it would also follow because the Taylor series
is alternating with decreasing absolute values.

---

The same reasoning can be applied to $g(x) = (1+x)\log(1+x)$
because $g'(x) = 1 + \log(1+x)$, so that $g^{(n)}(x)$ has alternating
signs for $n \ge 2$.

---

Alternating terms in the Taylor series alone are not sufficient
to draw any conclusion about the relationship of $f(x)$ and $T_n(x)$, a simple counter example is

$$
f(x) = x-\frac{x^2}2 + \frac{x^3}3 - x^4 \, , \quad T_2(x) = x-\frac{x^2}2
$$
with
$$
f(x) \begin{cases}
< T_2(x) & \text{ for }x < \frac 13 \\
> T_2(x) & \text{ for } x > \frac 13
\end{cases}
$$