# Series Expansions in Mathematica

Bug introduced in 5.0 or earlier and fixed in 11.1.1

Define variable z with the following assumptions

$Assumptions = 0 < z < 1  I put in the following two codes in Mathematica t1 = Simplify[Series[ArcTanh[(1 - Sqrt[z])^2], {z, 0, 3}]] t2 = Simplify[Series[ArcTanh[(1 - Sqrt[z])^2] // TrigToExp, {z, 0, 3}]]  The codes are identical to each other except for the extra TrigToExp. Yet, the outputs are completely different $$t_1 = -\frac{\log (z)}{4}-\frac{\sqrt{z}}{4}+\frac{z}{16}+\frac{5 z^{3/2}}{48}+\frac{z^2}{16}+\frac{z^{5/2}}{40}+O\left(z^{7/2}\right) ~~~~~~~~~~~~~~~\\ t_2 = -\frac{\log (z)}{4}-\frac{\sqrt{z}}{4}+\frac{z}{16}+\frac{5 z^{3/2}}{48}+\frac{9 z^2}{128}+\frac{9 z^{5/2}}{320}+\frac{z^3}{768}+O\left(z^{7/2}\right)$$ Why are the two results different and which result should I trust? • Just add a few more terms Series[... {z,0,7}] you will see that the two Series will diverge only for the higher order terms. The first expression without TrigToExp seems to converge faster to the correct values. – grbl Dec 2, 2016 at 23:59 • You can wrap Assuming around those series Assuming[0 < z < 1, t1 = Simplify[Series[ArcTanh[(1 - Sqrt[z])^2], {z, 0, 3}]]] instead of setting a global variable like$Assumptions. Dec 3, 2016 at 1:11
• @grbl - if you write that up i'll choose it as the answer. Thanks! Dec 3, 2016 at 17:16
• There is indeed a problem with the direct (as opposed to TrigToExp'ed) version. Looking into it. Dec 5, 2016 at 20:34

The which one can you trust, can be answered by plotting the error,

$$\left |\tanh ^{-1}\left(\left(1-\sqrt{z}\right)^2\right)-\text{t1} \right | \text{ and } \left |\tanh ^{-1}\left(\left(1-\sqrt{z}\right)^2\right)-\text{t2} \right |$$

t1 = Simplify[Series[ArcTanh[(1 - Sqrt[z])^2], {z, 0, 3}]] //Normal;

t2 = Simplify[Series[ArcTanh[(1 - Sqrt[z])^2] // TrigToExp, {z, 0, 3}]] //Normal;

Plot[{Abs[ArcTanh[(1 - Sqrt[z])^2] - t1],
Abs[ArcTanh[(1 - Sqrt[z])^2] - t2]}, {z, 0, 1},
PlotStyle -> {Blue, Green}, PlotLabels -> {"f(z)-t1", "f(z)-t2"}]


Looks like t2 would be my choice for 0 < z < 1 / 2 and t1 for the rest.

• You are welcome. Dec 3, 2016 at 17:30