# (1+x)^1/x near zero

1. Feb 11, 2010

### trekkiee

I noticed some unexpected behavior in the real-valued f(x)=(1+x)^1/x, as a function of real numbers, when plotting it on wolfram alpha. I inputed:

plot (1+x)^1/x from x=-0.0000001 to x=0.0000001

and saw that it unexpectedly seemed to oscillate near zero. I took a closer look with:

plot (1+x)^1/x from x=-0.00000000001 to x=0.00000000001

and saw that it definitely seems to oscillate near zero. My original rough graph on paper using a hand calculator suggested the curve was smooth near zero, and even windows calculator's 32 decimal places were unable to reveal the oscillation when I manually calculated many different values near zero.

I don't think f(x) is smooth near zero, though, since

d/dx[f(x)]=f(x)d/dx[1/xln(1+x)]

so all of f(x)'s derivatives have a discontinuity at zero. Since Leonhard Euler showed limf(x)=e as x approaches zero, and this limit is the same as you approach zero from the left or from the right, then the discontinuity of f(x) at zero is removable and the function

f(x)=(1+x)^1/x, x not = 0
f(0)=0 x=0

is continous. But the discontinuities in the derivatives don't appear to be removable. Each derivative will have a term containing the factor 1/x (not the 1/x in the exponent of f(x)) which cannot be removed. E.g.

d/dx[f(x)]=f(x)[(1/x)(1/(1+x)+(-1/x^2)(ln(x+1))]

the 1/x in the 1st term is not removable.

I'm not saying it doesn't oscillate near zero, I'm just saying that this is unexpected. I need confirming opinions and confirming logic to explain it so I can believe it.
So the questions are:
1. Does f(x) oscillate near zero?
2. Why does f(x) oscillate near zero?
3. Why does this behavior not appear until |x|<~10^-7?
4. Is it a glitch in Mathimatica? [doubt it]
5. If it doesn't oscillate [which it probably does] then how is the behavior of the derivatives explained near zero?
6. If it oscillates, then why is it so hard to replicate this oscillation manually on a calculator?

Note#1: This behavior concerns the real-valued f(x). I don't think Re[f(z)] or Im[f(z)], z complex, are of interest here.
Note#2: This behavior concerns (1+x)^1/x. I dont think either (1+1/x)^x or lim(1+1/x)^x as x goes to infinty are of interest here.

Thx in advance and Kudos to the person with the explanation!

2. Feb 11, 2010

### tiny-tim

Hi trekkiee!

(try using the X2 tag just above the Reply box )
When you say "oscillate near zero", you mean that it has infinitely many zero derivatives approaching zero?

But f'(x) = 0 if (1+x)ln(1+x) - x = 0,

and the derivative of that is ln(1+x), which always has the same sign as x.

3. Feb 11, 2010

### jacksonwalter

Hm..just read it, don't really have time to work on it. Maybe try setting f(x) = 1 and seeing if it has multiple intersections on x:(0,1)?

4. Feb 11, 2010

### Hurkyl

Staff Emeritus
But what about when you combine both terms together? For a rough approximation of how things might work out, recall that log(1+x) ~ x when x ~ 0.

5. Feb 11, 2010

### Bohrok

Last edited by a moderator: Apr 24, 2017
6. Feb 11, 2010

### D H

Staff Emeritus

Synposis: Those oscillations aren't real. They are merely a result of how Wolfram alpha does real arithmetic. The singularity is removable. In the vicinity of x=0, the function f(x)=(1+x)^(1/x) is approximately a straight line with y intercept = e and slope = -e/2.

7. Feb 12, 2010

### Hurkyl

Staff Emeritus
Well, you should be clued in by the extremely small amplitude of the oscillations that you're simply looking at noise.

8. Feb 12, 2010

### raymo39

its too long to rewrite

Last edited by a moderator: Apr 24, 2017
9. Feb 12, 2010

### trekkiee

Thx all for the replies. I should have known the problem was mathematica's [wolfram alpha] behavior when doing certain calculations.

10. Feb 12, 2010

### jacksonwalter

Wolfram is probably using a Taylor series with many terms to do the ln (1+x) calculation, which would produce the oscillatory behavior.