8
$\begingroup$

From both sides, this approaches infinity, but when evaluated exactly at $n = -1$, yields $\ln (x)$.

This seems similar to the behaviour of solutions to linear ODEs with characteristic polynomials (as the determinant approaches 0, a factor of $x$ appears next to the repeated root).

How can I explain this phenomenon to a layman? Obviously, I could show a proof of both, but that is probably not enough to be satisfactory. How does an infinity turn into a logarithm at the drop of a hat?

As an aside, is there a name for this behaviour?

$\endgroup$
5
  • $\begingroup$ when you say "from both sides" are we talking about using fractional calculus and taking the limit as $n$ approaches $0$ or integer intervals? $\endgroup$ Oct 4, 2015 at 6:15
  • $\begingroup$ Yes, fractionally, with $n = -0.\overline{9}$ from one side and $n = -1.\overline{0}1$ from the other. $\endgroup$
    – SKK
    Oct 4, 2015 at 6:19
  • 2
    $\begingroup$ @apt-get I think you mean $\lim_{n\to1}n$, as $-0.\overline9 = -1$ and $-1.\overline01$ isn't a real number. $\endgroup$
    – Frank Vel
    Oct 4, 2015 at 7:02
  • 3
    $\begingroup$ This limiting process isn't well-defined, since the indefinite integral is only defined up to a constant of integration. This has been discussed on this site several times before. Here's one related question: math.stackexchange.com/questions/498339/…. I'm pretty sure there has been at least one other question that's even more relevant, but I can't find it at the moment... $\endgroup$ Oct 4, 2015 at 9:13
  • $\begingroup$ That makes sense, combined with GEdgar's explanation below. Could I explain the situation with the DE as well by taking the initial conditions into account and the constants $c_0$ and $c_1$ that arise as a result of them? $\endgroup$
    – SKK
    Oct 4, 2015 at 22:35

2 Answers 2

3
$\begingroup$

Another thought. If $0<a<b$, $n\ne -1$, then $$ \int_a^b x^n \;dx = \frac{b^{n+1}-a^{n+1}}{n+1} $$ and the limit of that as $n \to -1$ is $\ln b - \ln a$. Not infinity.

So, in a certain sense, your confusion comes from ignoring the constant of integration.

$\endgroup$
0
$\begingroup$

My answer is a very "soft" answer, but it's a soft question, and the point is to explain to laymen, so I'll go ahead.

As you mentioned, you could of course show them the proofs, which might convince them that it is true but wouldn't explain why.

I would simply skip the calculus involved and point out that $e$ itself can be arrived at by the formula:

$$\lim_{a\to\,0}{(1+n)^{\frac{1}{n}}}$$

And point out that by the "intuitive" expectancy, $1+0 = 1$, and $1$ raised to any power (even infinity) is still $1$, so you would think the answer to the above limit would be $1$. Instead, it's $e$.

So I'd give them the "soft" (but nonetheless true) conclusion that in limits where you are dealing with infinity and zero in combination, $e$ is likely to show up. And since $ln$ just means "log base $e$", this is just another form of $e$ showing up.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .