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)

so

\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)

or

\displaystyle b^e<e^b

which is where we came in.

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Power trip

Here’s a mathematical paper of which I can smugly say I understood every word.

Ha ha. No, seriously, I like it. A visual proof that e^\pi>\pi^e.

But as Math with Bad Drawings pointed out someone else pointed out, the proof doesn’t depend on \pi but can be generalized to any number > e, and as Math with Bad Drawings pointed out, a similar proof works for any positive number < e. That is, for all a > 0 and a\ne e, e^a>a^e.

Proof here if you don’t want to click through to MwBD:

 

If a > e,

\displaystyle \frac{1}{e}(a-e) = \frac{a}{e}-1 > \int_e^a \frac{1}{x}dx = \ln(a)-\ln(e) = \ln(a)-1

so

\displaystyle \frac{a}{e} > \ln(a)

or

\displaystyle a > \ln(a^e)

or

\displaystyle e^a > a^e

and similarly, if b < e,

\displaystyle \frac{1}{e}(e-b) = 1-\frac{b}{e} < \int_b^e \frac{1}{x}dx = \ln(e)-\ln(b) = 1-\ln(b)

so

\displaystyle \frac{b}{e} > \ln(b)

or

\displaystyle b > \ln(b^e)

or

\displaystyle e^b > b^e

QED