# help with simplifying function

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]


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

• Use 1/4 instead of 0.25 in the exponent. – J. M. will be back soon Oct 3 '18 at 10:01
• The question is not clear: the solution you have shown is already analytic. Can you show, what are you after? – Alexei Boulbitch Oct 3 '18 at 10:08
• @J.M.issomewhatokay. Thanks a lot for your comment. Please have a look at my edit. – james Oct 3 '18 at 10:09
• @AlexeiBoulbitch I want the output to look like the analytic equation from the text book.( I edited my question) – james Oct 3 '18 at 10:10
• Aha, then use please the advice of @J. M. is somewhat okay given above. – Alexei Boulbitch Oct 3 '18 at 10:19

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:

Done. Have fun!

• Thanks a lot !! – james Oct 3 '18 at 11:34

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


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


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

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