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New answers tagged machine-precision

-4

f[x_] = Cos[x^x]; g[x_] = Sin[x^x]; Plot[{1 - f[x]^2, 1 + g[x]^2}, {x, 142, 144}, PlotStyle -> Thin, PlotLegends -> "Expressions", WorkingPrecision -> 450]

25

Normally Plot uses machine precision numbers; your $x^x$ expression is hitting the limit of the numbers that can be represented in machine precision right about $x>143$. Note: Solve[\$MaxMachineNumber == x^x, x] (* Out: {{x -> 143.016}} *) You can increase the WorkingPrecision setting for Plot adequately, and the plot will be complete: f[x_] = ...

2

Mathematica knows how to simplify when a is exactly 2*Pi: (-b*Cos[a*b] Sin[a/2] + Sin[a*b] Cos[a/2])/(b^2 - 1) /. a -> 2*Pi // InputForm -(Sin[2*b*Pi]/(-1 + b^2)) It then applies the numerical limit for b=1. For the approximate number 2.*Pi, Mathematica can't make this simplification, and it turns out the limit is +Infinity for a<2*Pi and ...

1

This is a feature of ListPlot. It tries to avoid generating graphics where all the ticks would look the same ("1", in this case.)

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