Power trip (part 3)

On further thought the inequality

\displaystyle a<\frac{b-a}{\ln(b/a)}<b for a and b>a positive real numbers

can be tightened up on the left side. I used

\displaystyle \left(\frac{1}{b}\right)(b-a)<\ln(b/a)<\left(\frac{1}{a}\right)(b-a)

but in fact this can be improved to

\displaystyle \left(\frac{1}{b}\right)(b-a)<\ln(b/a)<\left(\frac{1}{b}\right)(b-a)+\frac{1}{2}\left(\frac{1}{a}-\frac{1}{b}\right)(b-a)

which leads to

\displaystyle a\left(\frac{2b}{b+a}\right) < \frac{b-a}{\ln(b/a)} < b.

For b\gg a, the left side is \approx 2a, while for b = a(1+2\epsilon) where \epsilon<<1, the left side is \approx a(1+\epsilon) = (a+b)/2.


Power trip (part 2)

That proof that e^a>a^e for any positive a\ne e relies on \ln(e)=1, so doesn’t generalize to anything nearly as simple relating a^b to b^a. But let’s see what we can do.

Here’s an inequality which, if you’re like me, looks pretty mysterious:

\displaystyle b>\frac{b-a}{\ln(b/a)}>a for a and b>a positive real numbers.

I mean, the difference between a and b doesn’t “know” anything about a or b, right? And neither does their ratio. But the difference over the log of the ratio is always between a and b? Weird!

(Okay, maybe you’re not so easily impressed, since if you know b-a and b/a you can get a and b. Fine. I think it’s mysterious anyway.)

Well, let’s consider a function f(x) with derivative df(x)/dx = f'(x) monotonically decreasing (we could do increasing too, similarly). Let a and b>a be in the range where f(x) and f'(x) exist and f'(x)>0. Then

\displaystyle \int_a^b f'(x)dx = f(b)-f(a)

But that integral is the area under the curve from a to b and so

\displaystyle f'(a)(b-a)>f(b)-f(a)>f'(b)(b-a).

Then (b-a)>[f(b)-f(a)]/f'(a) and [f(b)-f(a)]/f'(b)> (b-a) or

\displaystyle \frac{f(b)-f(a)}{f'(b)}> (b-a)>\frac{f(b)-f(a)}{f'(a)}.

But that makes sense given the definition of the derivative and the monotonic property of f'(x); with a little thought you could’ve written this down directly. As obvious as it is, it takes a decidedly less obvious form if you specify f(x) = \ln(x), f'(x) = 1/x, which gives

\displaystyle b\ln(b/a)>(b-a)>a\ln(b/a)

or the formerly mysterious

\displaystyle b>\frac{b-a}{\ln(b/a)}>a QED.

I called the original post “Power trip” and this one “Power trip (part 2)”, and there haven’t been any powers here. Sorry. Here:

\displaystyle a\ln(b/a) < b-a < b\ln(b/a)


\displaystyle (b/a)^a < e^{b-a} < (b/a)^b.

That form is… neither obvious nor interesting, though. Unless a=e in which case the left part is

\displaystyle (b^e)/(e^e) < (e^b)/(e^e)


\displaystyle b^e<e^b

which is where we came in.