I want to have a series expansion of $e^{x^a}$ or $e^{c_1x^a+c_2x^b}$ at $x=0$, but Series cannot give any useful result even if the assumption $a>0$ is specified.

Series[Exp[x^a], {x, 0, 2}]
Series[Exp[x^a], {x, 0, 2}, Assumptions -> {x > 0, a > 0}]
(* -> E^x^a *)

However, if $a$ is a number, we can get a result.

Series[Exp[x^(2/3)], {x, 0, 2}]
(* -> 1+x^(2/3)+x^(4/3)/2+x^2/6+O[x]^(8/3) *)
  • $\begingroup$ Even more strangely Series[Exp[f[x]], {x, 0, 2}] works too. $\endgroup$ – b.gates.you.know.what Mar 17 '14 at 21:32
  • $\begingroup$ Perhaps something like Series[Exp[x^a], {x, 0, 2}, {a, 0, 2}]? $\endgroup$ – ciao Mar 18 '14 at 0:30
  • $\begingroup$ As long as you are only looking at real x and positive exponents why not use the exponential series and truncate it? E.g. cutoff = 10; Assuming[{x \[Element] Reals, a \[Element] Reals, b \[Element] Reals}, Sum[1/k!*(c1*x^a + c2*x^b)^k, {k, 1, cutoff}]] for 10 addends. One reasone might be that mathematica simply does not compute because it sees the result as the "most simple form for the expression" given the conditions. On the other hand as already mentioned if you consider your problem in the whole complex plain such an expansion is not unique due to the properties of complex powers. $\endgroup$ – gst Mar 18 '14 at 4:42
  • $\begingroup$ Sorry k should obviously start at k=0. $\endgroup$ – gst Mar 18 '14 at 15:40
  • $\begingroup$ note to get a result a must be rational..if you set about manuallycomputing a Taylor series you'll soon see why. This should probably go to math.stackexchange.com for further discussion. $\endgroup$ – george2079 Mar 18 '14 at 20:04

With some functions, a series expansion with explicit powers in the polynomial is not possible. For example, the expansion of $x^a$ around $0$ is just $x^a$ where the power depends on (is equal to) $a$. (If $a$ is not a nonnegative integer then it's not analytic.) If we were to expand it not around $0$ but around $1$ it would be possible to get an explicit power series.

Similarly, the powers in the result you request would depend on $a$. We can however do something like

Normal@Series[Exp[x], {x, 0, 6}] /. x -> x^a

(* ==> 1 + x^a + x^(2 a)/2 + x^(3 a)/6 + x^(4 a)/24 + x^(5 a)/120 + x^(6 a)/720 *)

You'll generally run into the same problem whenever the expression contains a symbolic power such as $x^a$, e.g. $f(x^a)$.


I don't have enough reputation to comment, but I wanted to point out why @b.gatessucks's observation jibes with the point that a proper result is returned if the exponents on x are definite numbers.

Series[Exp[f[x]], {x, 0, 2}]

is likely expanding f in a Series first, so that you wind up with a polynomial where x has definite powers,

Series[Exp[ f[0] + f'[0] x + 1/2 f''[0] x^2 + O[x]^3]]

Then Exp[that series] is expanded as expected, just like the case of x^(2/3).

As to why Series doesn't work on x^generic, I have no idea.


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