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I am computing the 4th order talyor approximation of this function $$\big(\frac{b}{g-x}\big)^{0.25}$$

The analytic textbook result is:

$$b^{0.25} \big(\frac{1}{g^{0.25}}+\frac{0.25 x}{g^{1.25}}+\frac{0.15625x^2}{g^{2.25}}+\frac{0.117188 x^3}{g^{3.25}}+\frac{0.0952148 x^4}{g^{4.25}}\big)$$

When I do it in Mathematica:

nonLin = (b/(g - x))^0.25
taylorLin = Normal[Series[nonLin , {x, 0, 4}] // Simplify]

I get this:

$$x^4 \left(\frac{0.126465 b^4}{g^8 \left(\frac{b}{g}\right)^{3.75}}-\frac{0.03125 \left(\frac{b}{g}\right)^{0.25}}{g^4}\right)+\frac{0.117188 b^3 x^3}{g^6 \left(\frac{b}{g}\right)^{2.75}}+\frac{b x^2 \left(\frac{0.25 g}{\left(\frac{b}{g}\right)^{0.75}}-\frac{0.09375 b}{\left(\frac{b}{g}\right)^{1.75}}\right)}{g^4}+\frac{0.25 x \left(\frac{b}{g}\right)^{0.25}}{g}+\left(\frac{b}{g}\right)^{0.25}$$

Is there a method to obtain the same nice representation as the analytic textbook solution ? ... to simplify it somehow?

EDIT: If I use 1/4 in the exponent, the reuslt looks much nicer. However, it could still be simplified: 

nonLin = (b/(g - x))^(1/4)
taylorLin = Normal[Series[nonLin , {x, 0, 4}] // Simplify]

$$\frac{195 x^4 \sqrt[4]{\frac{b}{g}}}{2048 g^4}+\frac{15 x^3 \sqrt[4]{\frac{b}{g}}}{128 g^3}+\frac{5 x^2 \sqrt[4]{\frac{b}{g}}}{32 g^2}+\frac{x \sqrt[4]{\frac{b}{g}}}{4 g}+\sqrt[4]{\frac{b}{g}}$$

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    $\begingroup$ Use 1/4 instead of 0.25 in the exponent. $\endgroup$
    – J. M.'s eventual burnout ♦
    Oct 3, 2018 at 10:01
  • $\begingroup$ The question is not clear: the solution you have shown is already analytic. Can you show, what are you after? $\endgroup$
    – Alexei Boulbitch
    Oct 3, 2018 at 10:08
  • $\begingroup$ @J.M.issomewhatokay. Thanks a lot for your comment. Please have a look at my edit. $\endgroup$
    – james
    Oct 3, 2018 at 10:09
  • $\begingroup$ @AlexeiBoulbitch I want the output to look like the analytic equation from the text book.( I edited my question) $\endgroup$
    – james
    Oct 3, 2018 at 10:10
  • $\begingroup$ Aha, then use please the advice of @J. M. is somewhat okay given above. $\endgroup$
    – Alexei Boulbitch
    Oct 3, 2018 at 10:19

4 Answers 4

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Try this:

nonLin = (b/(g - x))^0.25;
taylorLin1 = 
  Simplify[Normal[Series[nonLin, {x, 0, 4}]], {b > 0, g > 0}];
taylorLin2 = b^0.25*Expand[taylorLin1/b^0.25]

The results looks as follows:

enter image description here

Done. Have fun!

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You can take a constant factor (b/g)^.25 from the sum, then the sum is similar to what's in the book

Normal[Series[(1/(1 - x/g))^.25, {x, 0, 4}]]

Out[]= 1. + (0.25 x)/g + (0.15625 x^2)/g^2 + (0.117188 x^3)/g^3 + (
 0.0952148 x^4)/g^4
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This is using the reply found Factorization here.

It is not giving the EXACT same result, but it is a matter of just looking the output and perform some minimal algebra in the end or just simplifying terms by hand in the end. 

nonLin = (b/(g - x))^(1/4)
taylorLin = Series[nonLin, {x, 0, 4}]
sfactor[k_, p_, func_] := 
 HoldForm[StandardForm[k]]*StandardForm[func@(p*1/k)]
sfactor[(b)^(1/4), taylorLin, Apart] // Normal

Hopefully helpful.

Cheers!!!

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Instead of using Simplify afterwards, you can provide an assumption to Series:

Series[(b/(g - x))^0.25, {x, 0, 4}, Assumptions->b>0&&g>0] //TeXForm

$b^{0.25`} \left(\frac{1}{g}\right)^{0.25`}+0.25` b^{0.25`} \left(\frac{1}{g}\right)^{1.25`} x+0.15625` b^{0.25`} \left(\frac{1}{g}\right)^{2.25`} x^2+0.1171875` b^{0.25`} \left(\frac{1}{g}\right)^{3.25`} x^3+0.09521484375` b^{0.25`} \left(\frac{1}{g}\right)^{4.25`} x^4+O\left(x^5\right)$

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