I bought the Home Edition of Mathematica online 0.5 G of memory and costs \$150 per year. When I calculated the median of outputs from $[0,1]$ with an interval of $.000001$ I ended up with the follwoing


Then I tried a smaller interval of $.0000001$ and got.

Cloud::memlimit :  *This computation has exceeded the memory limit for your plan*

I had limited knowledge of mathematica and am taking a college course on its next year. I tried the documentations but found little. Is there a way of calculating interval up to and smaller than seven digits without switching to the standard edition that costs \$950 per year?

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    $\begingroup$ Most calculations can be done with limited memory (writing intermediate results to disk) if you are prepared to put enough effort into writing the software. $\endgroup$ – mikado Jun 18 '16 at 15:48
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    $\begingroup$ @Feyre the function 2^x-x is not monotonic on [0 1] $\endgroup$ – mikado Jun 18 '16 at 15:50
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    $\begingroup$ You may be able to compute the median of a large set by reexpressing as a minimization problem e.g. Minimize[Sum[Abs[(2^x - x) - y], {x, 0, 1, 0.00001}], y] $\endgroup$ – mikado Jun 18 '16 at 15:54
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    $\begingroup$ I see that the code in the question has been changed so my comments regarding memory are now invalid. $\endgroup$ – Mr.Wizard Jun 18 '16 at 15:59
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    $\begingroup$ FWIW, the answer is Root[{-(Log[2]/2) + ProductLog[-2^-#1 Log[2]] - ProductLog[-1, -2^-#1 Log[2]] &, 0.935553478022935551757231992253}]. $\endgroup$ – Chip Hurst Jun 18 '16 at 16:07

As mentioned in the comments, here's the answer to how I got the median.

I found the value $y$ that minimized

$$ f(y) = \int_0^1 \left| 2^x-x-y \right| dx. $$

First, to break up the absolute value in the integrand, I solved for when $2^x = y$:

Solve[(2^x - x) == y && 9/10 < y < 1, x, Reals]

enter image description here

Then I integrated and found where $f'(y) = 0$:

int = Integrate[2^x - x - y, {x, 0, b1}] 
      - Integrate[2^x - x - y, {x, b1, b2}] 
      + Integrate[2^x - x - y, {x, b2, 1}];

Solve[D[int, y] == 0 && 9/10 < y < 1, y, Reals]
{{y -> Root[{-(Log[2]/2) + ProductLog[-2^-#1 Log[2]] - 
   ProductLog[-1, -2^-#1 Log[2]] &, 0.935553478022935551757231992253}]}}

I decided to keep things symbolic in hopes of a nice looking answer. A numeric adaptation would probably be much easier.

  • $\begingroup$ That is amazing! Is there a mathematical textbook on this or is it simply logic? $\endgroup$ – Arbuja Jun 18 '16 at 16:26
  • $\begingroup$ Well I essentially just did what mikado did, but in integral form. It seems like it would be the median, but I have no source... $\endgroup$ – Chip Hurst Jun 18 '16 at 16:27

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