Unexpected Behavior of (1+x)^1/x Near Zero

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In summary, the unexpected oscillations near zero in the real-valued function f(x)=(1+x)^1/x are not real and are only a result of how Wolfram alpha does real arithmetic. The singularity at x=0 is removable and in the vicinity of x=0, the function is approximately a straight line with a y-intercept of e and a slope of -e/2. The oscillations can be explained by Wolfram using a Taylor series with many terms to calculate the natural logarithm of (1+x).
  • #1
trekkiee
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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 don't 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!
 
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  • #2
Hi trekkiee! :smile:

(try using the X2 tag just above the Reply box :wink:)
trekkiee said:
f(x)=(1+x)^1/x

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

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
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
trekkiee said:
the 1/x in the 1st term is not removable
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.
 
  • #6
The same original post was made here: http://www.scienceforums.net/forum/showthread.php?t=48648.

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
Bohrok said:
However, when I graph their difference, wolfram gives a graph with a lot of oscillations that shouldn't exist, yet says the result is 0 as it should be.
Well, you should be clued in by the extremely small amplitude of the oscillations that you're simply looking at noise.
 
  • #8
i left a reply for this at www.advancedphysics.org[/URL], that should give you the right answer :)
its too long to rewrite
 
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  • #9
Thx all for the replies. I should have known the problem was mathematica's [wolfram alpha] behavior when doing certain calculations.
 
  • #10
Wolfram is probably using a Taylor series with many terms to do the ln (1+x) calculation, which would produce the oscillatory behavior.
 

What is the (1+x)^1/x function?

The (1+x)^1/x function is an exponential function where the exponent is a fraction. It can also be written as (1+x)^(1/x).

What is the behavior of (1+x)^1/x near zero?

The behavior of (1+x)^1/x near zero is unexpected because as x approaches zero, the function approaches a value of 1. This is in contrast to other exponential functions where the value approaches infinity as the input approaches zero.

Why is the behavior of (1+x)^1/x near zero unexpected?

The behavior is unexpected because it goes against our intuition and the behavior of other exponential functions. Generally, we expect an exponential function to increase or decrease without bound near zero, but (1+x)^1/x approaches a finite value of 1.

What are the applications of (1+x)^1/x near zero?

The behavior of (1+x)^1/x near zero has applications in calculus, where it can be used to find the limit of a function as x approaches 0. It is also used in finance to calculate compound interest when the interest rate is continuously compounded.

How can we graph (1+x)^1/x near zero?

To graph (1+x)^1/x near zero, we can use a graphing calculator or software to plot the points for various values of x approaching zero. We can also use the limit definition to find the value of the function at x=0 and plot that point on the graph. The resulting graph will be a curved line approaching the point (0,1).

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