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As the title suggests, I'm trying to represent a series through a simple summation. For example, the function

Series[Exp[x], {x, 0, 10}]

obviously gives me the series expansion of the exponential function. I would now like to obtain from this representation in the form of a summation. Is it possible to do this with mathematica?

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  • $\begingroup$ Please give an example of what output you expect in this case. It is not clear what output you want to get. $\endgroup$
    – Somos
    Jan 17, 2022 at 20:24

4 Answers 4

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In v13.2, This can be easily achiveved with Asymptotic:

Asymptotic[Exp[x], {x, 0, ∞}]

enter image description here

If you insist on summing to 10, replace the :

Asymptotic[Exp[x], {x, 0, ∞}] /. ∞ -> 10
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Clear["Global`*"]

sumRule = 
  Inactive[Series][f_, {x_, x0_, n_}] :> 
   Inactive[Sum][Assuming[{Element[k, Integers], k >= 0},
     SeriesCoefficient[f, {x, x0, k}] (x-x0)^k //
      FullSimplify],
    {k, 0, n}];

n = 10;

f[x_] = Exp[x];

Inactive[Series][f[x], {x, 0, n}] /. sumRule

enter image description here

Verifying,

(% // Activate) == Normal[Series[f[x], {x, 0, n}]]

(* True *)

f[x_] = Sin[x];

Inactive[Series][f[x], {x, 0, n}] /. sumRule

enter image description here

Verifying,

(% // Activate) == Normal[Series[f[x], {x, 0, n}]]

(* True *)
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  • $\begingroup$ As always, that's very impressive. $\endgroup$
    – user49048
    Jan 17, 2022 at 5:56
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Is this what you were looking for?

SeriesCoefficient[Exp[x], {x, 10, n}]

And more generally

SeriesCoefficient[Exp[x], {x, n, n}]
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This is one of those times when FindSequenceFunction works.

Make a list of the terms. Then let Mathematica find a function, $f$, that generates the coefficients. That is, $f(k) = a_k$, the Taylor series coefficients. We know that $f(k)$ should be $1/k!$, but we want Mathematica to figure that out.

terms = List @@ Normal[Series[Exp[x], {x, 0, 10}]];
f = FindSequenceFunction[terms /. x -> 1]

(*  1/Pochhammer[1, -1 + #1] &  *)

Well, that was a surprise. We can get a simpler expression for the coefficients in our summation like this

a[k] = FullSimplify[f[k], k ∈ PositiveIntegers]  (* 1/Gamma[k] *)

s[n_] = Inactive[Sum][a[k] x^(k - 1), {k, 1, n}]

$$\underset{k=1}{\overset{n}{\sum }}\frac{x^{k-1}}{\Gamma (k)}$$

We accept the gamma function in lieu of the factorial. We recover the exponential in the limit as $n\rightarrow\infty$

Limit[s[n], n -> ∞] // Activate   (*  E^x  *) 
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