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Re: Problem with Series

Am 17.11.2010 11:29, schrieb Themis Matsoukas:
> I was using Series to linearize a function at x=0 but I run into a problem with the specific function below:
> f = Log[Coth[1/2 ArcCosh[Sqrt[1 + x^2]/x]]];
> fLinearized = Simplify[Normal[Series[f, {x, 0, 1}]]]
> Plot[{f, fLinearized, x}, {x, -10, 10}]
> Mathematica produces output without complaining but the result does not contain an explicit linear term. Specifically, fLinearized is not a linear function of x, though it does seem to have the same derivative at x=0 as the original function f. The correct answer should be f=x +O[x^2].

Try to avoid complex results in Plot functions if you are using inverse 
functions with logarithmic or terms or fractional powers. Check the 
output using Table instead of Plot.

For all complex analytical functions  g Not FreeQ of fractional powers 
or of functions f with simple poles of 1/f'(f^-1)  (the derivative of 
the inverse f^-1) the zeros of the powers and the zeros of f'@f-1 are 
branch points. Mathematica as a rule gives expressions with cuts along 
the negative real line. But if the many valued function is hidden 
somewhere in the functions tree its often impossible to trace the branch 
lines in the original plane, where it may cut the real line by chance.

for f= exp, exp(Log) has a zero at x=0

for f= ArcCosh = ArcCos(i),  Sinh[ArcCosh] = Sqrt[x^2-1] has zeros for 
x=+-1 and the branch cut goes from -1 to 1 because Cosh[x] >1 for all 
real x:

ArcCosh[0.5 +- 10^-16 I] -> 0.+- 1.0472 I

In your case you get what you want by

Clear[f, fLinearized]
f[x_] = Log[Coth[1/2 ArcCosh[Sqrt[1 + x^2]/x]]];
fLinearized[x_] = Normal[Series[f[x], {x, 0, 1}]]

Plot[Evaluate[{Re@f[x], Re@fLinearized[x], x}], {x, -10, 10}]
Plot[Evaluate[{Im@f[x], Im@fLinearized[x], x}], {x, -10, 10}]

A good way to get an impression of branch cuts and possibly a multi 
valuedness with one or more branch points is to inspect the real 
conformal coordinate map of the domain in question generates by f

ParametricPlot[{Re[#],Im[#]}&[f[x+I y]], xdomain, ydomain]


Roland Franzius

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