Let $\exp(x)=\sum_{n=1}^\infty \frac {x^k}{k!}$ be the exponential function defined on real numbers. It is trivial that $\exp(x)$ is monotone increasing. However, let $h_1(x)= \frac {e^x-1}{x}$, $h_2(x)=\frac {e^x - 1 - x}{x^2}$, so on and so forth. In general we let $h_k(x)=\frac {e^x - \sum_{i=0}^{k-1}\frac {x^i}{i!}}{x^k}$, where we subtract the first $k$ terms in the series, and divide by $x^k$. Notice that $h_k(x)$ is well-defined over all real numbers. Now the question is, are $h_k(x)$ all monotone increasing for $k=1,2,\ldots$ ? It is easy to solve some special cases by brutal force when $k$ is small, or you can plot the graph of $h_k(x)$ and see that they do increase monotonically. But I can't find a general pattern to prove for all $k$.
Asked
Active
Viewed 27 times
1 Answers
1
From the integral form of Taylor remainder, we can write
$$ h_k(x) = \int_{0}^{1} \frac{(1-t)^{k-1}}{(k-1)!} e^{xt} \, \mathrm{d}t. $$
Since $x \mapsto e^{xt}$ is increasing and $\frac{(1-t)^{k-1}}{(k-1)!} \geq 0$ for each $t \in [0, 1]$, the desired claim follows.
Sangchul Lee
- 167,468