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]
$$\frac{195 x^4 \sqrt[4]{\frac{b}{g}}}{2048 g^4}+\frac{15 x^3 \sqrt[4]{\frac{b}{g}}}{128 g^3}+\frac{5 x^2 \sqrt[4]{\frac{b}{g}}}{32 g^2}+\frac{x \sqrt[4]{\frac{b}{g}}}{4 g}+\sqrt[4]{\frac{b}{g}}$$
1/4
instead of0.25
in the exponent. $\endgroup$