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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))"}]

Plot of two functions with opposite signs

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 1

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

Series 2

Am I just not understanding some basic mathematics?

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    $\begingroup$ 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. $\endgroup$ – LouisB Oct 2 at 5:25
  • $\begingroup$ @LouisB Thanks! That makes sense. $\endgroup$ – JimB Oct 2 at 5:32
  • $\begingroup$ Plot them from 0 instead of 1/3. Point being, they agree in a neighborhood of the origin. $\endgroup$ – Daniel Lichtblau Oct 3 at 22:11
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    $\begingroup$ 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. $\endgroup$ – JimB Oct 4 at 3:05

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