What's the formula for a equation that can produce the continuum from the red to green lines in this graph below? I can easily get anywhere from the green line to the blue with $$y = B_0 + B_1x + B_2x^3$$ but getting anywhere from the blue line to the red line requires an inverse of the cube or x^(1/3) which doesn't work for negative numbers. Is there any continuous formula that can produce the red line? Alternatively, how do I use imaginary numbers in a linear regression?
enter image description here

Here's some data if anyone can fit a model to it.

  x, y 
-10,  2.154434525
 -9,  2.080083671
 -8,  1.999999861
 -7,  1.912931059
 -6,  1.817120484
 -5,  1.709975855
 -4,  1.587400979
 -3,  1.442249517
 -2,  1.259921021
 -1,  1
  0,  0
  1, -1
  2, -1.259921021
  3, -1.442249517
  4, -1.587400979
  5, -1.709975855
  6, -1.817120484
  7, -1.912931059
  8, -1.999999861
  9, -2.080083671
 10, -2.154434525
  • $\begingroup$ That's a much more specific and simpler question than the one in your earlier comment, which was about choosing a function from a smooth transition between those curves. I'll try to deal with both. $\endgroup$
    – Glen_b
    May 9, 2014 at 22:53
  • $\begingroup$ What's the distinction between $x_1$ and $x$? are they the same variables or different variables? $\endgroup$
    – Glen_b
    May 9, 2014 at 22:54
  • $\begingroup$ Sorry same variables. $\endgroup$
    – Ricky Sahu
    May 9, 2014 at 23:00

1 Answer 1


Firstly, $x^{1/3}$ does work for negative $x$. Any real number other than 0 has three cube roots, one real and two complex, and the real root for negative $x$ is negative.

You might have a problem with software refusing to compute cube roots of negative numbers, but that's easy to get around, and indeed, to generalize to cases where the root isn't an odd integer:

$$t(x) = \text{sign}(x)\cdot |x|^p$$

[An alternative form that avoids the use of the $\text{sign}$ function is $t(x) = x\cdot |x|^{p-1}$, but with that form you have to specifically set $t(0)=0$ separately. So for $p=\frac{1}{3}$, you have $t(x)=x\cdot |x|^{-\frac{2}{3}}$, except at 0, which has $t(0)=0$.]

As it is, your data is exactly $y = -x^{1/3}$. Here's how you can see it (this is in R)

> y
 [1]  2.154435  2.080084  2.000000  1.912931  1.817121  1.709976  1.587401  1.442250  1.259921  1.000000
[11]  0.000000 -1.000000 -1.259921 -1.442250 -1.587401 -1.709976 -1.817121 -1.912931 -2.000000 -2.080084
[21] -2.154435
> y^3
 [1]  10   9   8   7   6   5   4   3   2   1   0  -1  -2  -3  -4  -5  -6  -7  -8  -9 -10

... but most software won't compute cube roots for negative arguments (there's a reason why, which isn't probably worth pursuing here).

So we use my above suggestion:

f= -sign(x)*abs(x)^(1/3)

enter image description here

The circles are the data, the dots are the fitted values (f).

A couple of side notes:

The red dashed line in the plot in your question there was actually generated by using $p=1/2$ in $t(x) = \text{sign}(x)\cdot |x|^p$, rather that $1/3$, but the situation in your earlier question was slightly different. The green dashed line was actually generated using $p=2$ in the same formula (though then divided by 1000 to have the same endpoints as for $p=1/2$).

Note that if you didn't know $p$, you could actually estimate it. If there are no other predictors yo might do it via transformation, but with other predictors present you can still fit $p$ as a parameter in a nonlinear regression.

  • $\begingroup$ As @Glen_b indicates, this beast has a simple name, cube root. stata-journal.com/sjpdf.html?articlenum=st0223 has more on cube roots. Some but not all of the content is Stata-specific. $\endgroup$
    – Nick Cox
    May 9, 2014 at 23:38
  • $\begingroup$ Thanks I figured the sign(x) abs(x)^3 was the simplest way. I was hoping for some other continuous function without sign, but this gets the job done. Thanks for pointing me in the right direction through the chain of posts! $\endgroup$
    – Ricky Sahu
    May 9, 2014 at 23:42
  • $\begingroup$ You can avoid $\text{sign}$ easily enough, if you can bear a slight complication. I'll amend my answer. $\endgroup$
    – Glen_b
    May 10, 2014 at 0:39

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