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Prove that $h(u)=\frac{2}{u^2}(u-\log(1+u))$ is decreasing on $(-1,\infty)- \{0\}$

This function appears on the proof of Stirling's formula in Principles of Mathematical Analysis, by Walter Rudin.

The classic way to approach the "show a function is decreasing" is to show that its derivative is negative on the given range.

$$h'(u)= -\frac{2}{u^2}+\frac{4}{u^3}\log(1+u)-\frac{2}{u^2(1+u)}$$

And after we factor out the $-\frac{2}{u^2}$ we are left with showing that

$$g(u):= 1-\frac{2}{u}\log(1+u)+\frac{1}{1+u}$$ is positive. We could try showing that $g$ has positive derivative and show that the limit of $g$ as $u$ approaches $-1$ is positive, but this seems like rhe derivative of $g$ will be something much more complicated.

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To show $g(u)\ge 0$ it suffices to show $$k(u)=u(1+u)-2\log (1+u)\cdot (1+u)+u\le 0\tag 1$$ for all $u\in(-1,0]$

and $$k(u)=u(1+u)-2\log (1+u)\cdot (1+u)+u\ge 0\tag 2$$ for all $u\in[0,\infty)$

Now $k'(u)=2(u-\log(1+u))\ge 0$

therfore $\forall u\in(-1,0] k(u)\le k(0)=0$ and $\forall u\ge 0,k(u)\ge k(0)=0$ So we are done!

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Another approach is to notice that \begin{align*} \log (1 + u) = \int_0^u {\frac{{dt}}{{1 + t}}} & = u - \int_0^u {\frac{t}{{1 + t}}dt} = u - u^2 \int_0^1 {\frac{s}{{1 + su}}ds} \\ & =u - u^2 \int_0^1 \int_u^{ + \infty } {\frac{{s^2 }}{{(1 + st)^2 }}dt} ds \end{align*} for $u>-1$.

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    $\begingroup$ Most elegant, the integral representation for $h(u)$. +1 & thanks for this insight! $\endgroup$
    – Hanno
    Feb 26 at 13:25

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