# Two equations identical but opposite signs have the same Maclaurin series in Mathematica: Is that possible? [closed]

I have two equations

Sqrt[1 - 2 y] y^(5/2) Sqrt[y - 4 Sqrt[1 - 2 y] y^(3/2) + 2 y^2]


and

y^(5/2) (Sqrt[y] - 2 Sqrt[1 - 2 y] y - 2 y^(3/2))


When plotted over a set of values of interest the results are of opposite sign:

Plot[{Sqrt[1 - 2 y] y^(5/2) Sqrt[y - 4 Sqrt[1 - 2 y]  y^(3/2) + 2 y^2],
y^(5/2) (Sqrt[y] - 2 Sqrt[1 - 2 y] y - 2 y^(3/2))},
{y, 1/3, (2 + Sqrt[2])/8},
PlotLegends -> {"Sqrt[1-2y] \!$$\*SuperscriptBox[\(y$$, $$5/2$$]\) \
Sqrt[y-4 Sqrt[1-2y]  y^(3/2)+2 \!$$\*SuperscriptBox[\(y$$, $$2$$]\)]",
"y^(5/2)(Sqrt[y]-2Sqrt[1-2y]y-2y^(3/2))"}]


But when I take a Maclaurin series, the series look identical rather than having opposite signs:

Series[Sqrt[1 - 2 y] y^(5/2) Sqrt[y - 4 Sqrt[1 - 2 y]  y^(3/2) + 2 y^2], {y, 0, 10}] // Normal


Series[y^(5/2) (Sqrt[y] - 2 Sqrt[1 - 2 y] y - 2 y^(3/2)), {y, 0, 10}] // Normal


Am I just not understanding some basic mathematics?

• Let $f$ and $g$ be your 1st and 2nd expressions. The problem may have to do with a discontinuity in $\frac{df}{dx}$ at y=1/6 . The two expressions may be equal on {0, 1/6} where the MacLaurin is valid and have opposite signs in the region {1/6, 1/2} . Use Plot[{f, g}, {y, .1, .2}, PlotStyle -> {Automatic, Dashed}] and FunctionDomain[D[f, y], y, Reals] to see the discontinuity. – LouisB Oct 2 at 5:25
• @LouisB Thanks! That makes sense. – JimB Oct 2 at 5:32
• Plot them from 0 instead of 1/3. Point being, they agree in a neighborhood of the origin. – Daniel Lichtblau Oct 3 at 22:11
• Thank you for your comments that solved my issue but I’m voting to close this question because it turns out to be a mathematics question (i.e. me not remembering my real analysis decades ago) and not an issue with Mathematica. – JimB Oct 4 at 3:05