Plotting $\log^{1/n}(1/(1 - u^n))$

Question

The function: $$f(x)=\log^{1/\alpha}\left(\frac{1}{1 - x^\alpha}\right),\quad x\in[0,1),\ \alpha\in\mathbb{R^+}$$ Looks timid enough. To my understanding, the function is well defined over the entire interval, and moreover should be almost identical to $f(x)=x$ for $\alpha\gg 1$.

Yet, when trying to numerically evaluate this seemingly straight-line function, I run into problems whenever we get too close to $x=0$. For instance, here's how Mathematica plots out our function for $\alpha =30$:

$\quad\quad\quad\quad\quad\quad$ enter image description here

What is going on here? and how can we properly evaluate this function?

score 1 · Answer 1 · answered Jan 31 '14 at 05:53

1

Here is what it looks like with $\alpha = 4$.

The problem you are running into is how precise Mathematica can store numbers. Your plot drops off around $0.29$. If you evaluate $0.29^{30}$, you get approximately $7 \times 10^{-17}$ which is basically $0$ to the computer. Hence your function immediately becomes $\log 1$ and is 0. The problem is only made worse as the numbers get closer to $0$ and as $\alpha$ gets larger.

answered Jan 31 '14 at 05:53

John Habert

4,001

Yes indeed :) Now how do we evaluate it properly? – Nathaniel Bubis Jan 31 '14 at 05:55
You'd need something that uses infinite precision. But unless your scale is super small, you'll never see this part anyway. It will be very very close to the $x$-axis. – John Habert Jan 31 '14 at 05:57
Why? the function should evaluate to $x$, not some infinitesimal. – Nathaniel Bubis Jan 31 '14 at 05:58
You're right. With infinite precision it should work correctly. I was still thinking about the incorrect graph. – John Habert Jan 31 '14 at 06:01

score 1 · Accepted Answer · answered Jan 31 '14 at 07:15

As pointed out by John Habert, the problem is related to machine accuracy and you will be in trouble as soon as $x^a<\epsilon$, $\epsilon$ being the largest number such that $1+\epsilon=1$ (typically in the order of $10^{-16}$).

Then, for the situations where $x^a<\epsilon$, you could replace your function by its Taylor expansion built around $x=0$ which is looking like $x \left(1+\frac{x^a}{2 a}\right)$

Plotting $\log^{1/n}(1/(1 - u^n))$

2 Answers2