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How could I expand this equation (0.837 + x^0.5 + x)^4.870

Neither

Expand[(0.837 + x^0.5 + x)^4.870]

nor

Series[(0.837 + x^0.5 + x)^4.870, {x, 0, 6}]

work.

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    $\begingroup$ You can do Series[(0.837 + x^(1/2) + x)^4.870 , {x, 0, 6}] i.e. switch x^0.5 to x^(1/2) or simply Series[(0.837 + x^0.5 + x)^4.870 // Rationalize, {x, 0, 6}] $\endgroup$
    – Artes
    Apr 27, 2014 at 16:09
  • 2
    $\begingroup$ Usually an equation has an equal sign.. $\endgroup$ Apr 28, 2014 at 21:55

1 Answer 1

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It very strongly depends upon the answer to the question, why do you want to make the expansion and what is the range of x? For this reason in addition to the answer of Artes in the comment above you might think of the following approach for 0<x<1. This makes a fit of your function to a cubic polynomial:

    lst = Table[{x, (0.837 + x^0.5 + x)^4.870}, {x, 0, 1, 0.01}];
ft = Fit[lst, {x, x^2, x^3}, x];
Show[{
  ListPlot[lst, PlotStyle -> Blue],
  Plot[ft, {x, 0, 1}, PlotStyle -> Red]
  }]

That is how the fit looks like: enter image description here The blue points is your function and the red line is the fit. If this is a satisfactory coincidence, your expansion (or better to say, your approximation) is:

    ft

(*  26.173 x + 20.6356 x^2 + 113.437 x^3  *)
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