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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!

1 Answers 1

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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 $$ f(x) \begin{cases} < T_2(x) & \text{ for }x < \frac 13 \\ > T_2(x) & \text{ for } x > \frac 13 \end{cases} $$

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    Great answer. Since you just edited it: can we come up with sufficient conditions? – 2017-02-12
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    @Andreas: If the series has terms with alternating signs and decreasing absolute values then the sum lies always between two subsequent partial sums, i.e. the Taylor polynomials are alternately larger and smaller than the function value (compare https://en.wikipedia.org/wiki/Alternating_series_test). As I mentioned above, that can be applied in the case $0< x \le 1$. – 2017-02-12
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    In your counterexample, the argument using the remainder doesn't work though. We have $R_2(x) = \frac{f^{(3)}(\xi)}{3!} x^{3} = (\frac{1}{3} - 4 \xi) x^3$, so $R_2(1) = \frac{1}{3} - 4 \xi$ for some $\xi$ between 0 and 1. Hence we don't know about the sign of $R_2(1)$. – 2017-02-12
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    @Andreas: That counterexample was meant only as an example that alternating signs in the Taylor series alone are not sufficient for any conclusion ($f(x) > T_2(x)$ for $0 < x < 1/3$ and $f(x) < T_2(x)$ for $x > 1/3$). The remainder formula does not help either in that example and $x=1$ – 2017-02-12
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    As to my question "is this a special feature of the log function" - do we find other functions $f(x)$ which have derivatives $f^{(n)}(x)$ with no sign change for all $x > 0$, and with signs alternating with $n$ for all $n > 0$? Of course, other than the trivial modification $f(x) = a + b \log (x)$. – 2017-02-12
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    @Andreas: What about $\exp(-x)?$. Or $\sin(x), \cos(x)$ for $0 \le x \le \pi/2$. – 2017-02-12
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    Yes, that holds as well. – 2017-02-14