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Mathematica can obtain Series[MittagLefflerE[\[Alpha], t], {t, 0, 4}]. I need to calculate

Series[MittagLefflerE[\[Alpha], -t^\[Alpha]], {t, 0, 4}].

Any suggestions?

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  • $\begingroup$ Here's a simpler problem that demonstrates a similar phenomenon: Series[Exp[t^a], {t, 0, 4}]. $\endgroup$ – J. M. will be back soon Aug 14 '17 at 20:55
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Spoiler: Mathematica behaves very well and like expected. It's a math problem.

Lets remember what the taylor-series does:

$$\hat{\mathcal{T}}_{x0}^n(f(x))\approx\sum_{m=0}^n\frac{1}{m!}\frac{\partial^m f(x_0)}{\partial x^m}(x-x_0)^m$$

Now, you take the series around $0$ which means we can simplify to

$$\sum_{m=0}^n\frac{1}{m!}\frac{\partial^m f(0)}{\partial x^m}x^m$$

Okay we can look at the first tree terms (without the factorial and constant term) with:

((D[MittagLefflerE[a, -t^a], {t, #}] & /@ Range[1, 3]) /. t -> 0)*
 t^Range[0, 2]

And you'll notice something, you get strange powers of zero. Okay take a look at those powers:

DeleteDuplicates@
  Cases[(D[MittagLefflerE[a, -t^a], {t, #}] & /@ Range[1, 3]), 
   t^(_?(! FreeQ[#, a] &)), Infinity] /. t -> 0

{0^(-1 + a), 0^a, 0^(-2 + a), 0^(-2 + 2 a), 0^(-3 + a), 0^(-3 + 2 a)}

Okay, thats bad. if only one a results in a zero-exponent, your whole series will be not defined and in the other case, all terms are equal zero. We can check that pretty fast:

n = 7;
Simplify[((D[MittagLefflerE[a, -t^a], {t, #}] & /@ Range[1, n]) /. 
    t -> 0)*t^Range[0, n - 1], a > n]

{0, 0, 0, 0, 0, 0, 0}

So Mathematica, remarkably, realizes this and refuses to evaluate to zero or a not defined term. Your left with your input.

What can you do to fix your problem?

Just set $x_0\neq 0$

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