# Power series with decimal and arbitrary powers

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.

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Rationalize the expression, e.g., Series[E^x^(.5) // Rationalize[#,0]&, {x, 0, 1}] –  Bob Hanlon Jul 2 at 18:40
@BobHanion - Thanks but that doesn't seem to work for an arbitrary variable like \beta...? –  DJBunk Jul 2 at 18:42
What result do you expect for arbitrary beta? –  Daniel Lichtblau Jul 2 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 2 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 Jul 2 at 19:11

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

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