Power series with decimal and arbitrary powers

Question

Is there a way to make Mathematica do a series of the form:

 Series[ E^{\beta}$ , {x, 0, 1}]

I have noticed that

 Series[E^x^(1/2), {x, 0, 1}]

work but

 Series[E^x^(.5), {x, 0, 1}]

does not. Thanks for the help.

EDIT: I have tried things like

  Assuming[{\bet>0},Series[ E^{\beta}$ , {x, 0, 1}]]

which hasn't helped.

Rationalize the expression, e.g., Series[E^x^(.5) // Rationalize[#,0]&, {x, 0, 1}] — Bob Hanlon, Commented Jul 2, 2014 at 18:40
@BobHanion - Thanks but that doesn't seem to work for an arbitrary variable like \beta...? — DJBunk, Commented Jul 2, 2014 at 18:42
@DanielLichtblau - For arbitrary beta, not necessarily anything, but I have tried things like Assuming[\beta>0,Series[....]] and that didn't help. — DJBunk, Commented Jul 2, 2014 at 18:54
That did not answer my question. I understand you have in mind, say, the symbol beta. What I do not know is what result you expect Series[Exp[x^beta],{x,0,1}] to deliver. I am looking for a response along the lines of "something + somethingElse*x^somepower + O[x]^someOtherPower" but with the various somethings made explicit. — Daniel Lichtblau, Commented Jul 2, 2014 at 19:11

Basheer Algohi · Accepted Answer · 2014-07-02 23:32:35Z

1

This is not an answer but a general note.

for the case of this function:

E^x^(1/2)

you have to note that the derivative of E^x^(1/2) at x=0 is ComplexInfinity.

to show you that, the general series of a function is as follows:

Clear[f]
s = Series[f[x], {x, 0, 1}] // Normal

if you set f[x] to your function it will result in ComplexInfinity.

    f[x_] := E^x^(1/2)
      s

(* ComplexInfinity*)

Many be you need to change the point that you are calculating the series about

answered Jul 2, 2014 at 23:32

Basheer Algohi

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