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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.

DJBunk
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    Rationalize the expression, e.g., Series[E^x^(.5) // Rationalize[#,0]&, {x, 0, 1}] – Bob Hanlon Jul 02 '14 at 18:40
  • @BobHanion - Thanks but that doesn't seem to work for an arbitrary variable like \beta...? – DJBunk Jul 02 '14 at 18:42
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    What result do you expect for arbitrary beta? – Daniel Lichtblau Jul 02 '14 at 18:49
  • @DanielLichtblau - For arbitrary beta, not necessarily anything, but I have tried things like Assuming[\beta>0,Series[....]] and that didn't help. – DJBunk Jul 02 '14 at 18:54
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    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 Jul 02 '14 at 19:11
  • @DanielLichtblau - oh, ok. I would expect if beta is real and > 0, and x is small $e^{x^\beta} $$\approx 1 + x^\beta$ and Mathematica would give me something along these lines. – DJBunk Jul 02 '14 at 19:13
  • There are a couple of impediments, one in theory and one in practice. The first is that such a result might not make sense for β<0, say (where an expansion of the exponential at infinity would perhaps be a better idea). The practical issue is that SeriesData, the encapsulated result of a (proper) Series expansion in Mathematica, only supports explicit Puisiux series. So exponents must be explicit rational numbers. This is motivated by a number of considerations such as need to do manipulations on them, need for an explicit ordering of terms, and the like. – Daniel Lichtblau Jul 02 '14 at 19:49

1 Answers1

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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

Basheer Algohi
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