Find Generalized Series with Symbolic Variable

Question

CoefficientList[Series[Exp[x], {x, a, 3}], x]

Gives the following expression, $$ \left\{-\frac{1}{6} e^a a^3+\frac{e^a a^2}{2}-e^a a+e^a,\frac{e^a a^2}{2}-e^a a+e^a,\frac{e^a}{2}-\frac{a e^a}{2},\frac{e^a}{6}\right\} $$ However, I'd like to be able to find what the generalized coefficient formula should be for any number of terms. Sadly, I cannot give Series a symbolic argument for n and SeriesCoefficient doesn't seem like what I'm looking for either.

The values should match the original expression, so if $a=0.5$ then: $${0.996102, 1.03045, 0.41218, 0.274787} $$

I want to find a closed form for the collected coefficients.

Answer 1

I'm assuming you want the "formula" for what CoefficientList[Series[f[x], {x, a, n}], x] gives. If you want the term associated with $x^j$, that is the following:

c[j_, n_] := Sum[Binomial[k, j] (-1)^(k - j) a^(k - j) (D[f[x], {x, k}] /. x -> a)/k!, {k, j, n}]

Here is a specific example:

f[x_] := Exp[x]
n = 3;

CoefficientList[Series[f[x], {x, a, 3}], x]
(* {E^a - a E^a + (a^2 E^a)/2 - (a^3 E^a)/6, E^a - a E^a + (a^2 E^a)/2, E^a/2 - (a E^a)/2, E^a/6} *)

c[j_, n_] := Sum[Binomial[k, j] a^(k - j) (D[f[x], {x, k}] /. x -> a)/k!, {k, j, n}]
Table[c[j, n], {j, 0, n}]
(* {E^a - a E^a + (a^2 E^a)/2 - (a^3 E^a)/6, E^a - a E^a + (a^2 E^a)/2, E^a/2 - (a E^a)/2, E^a/6} *)

This just uses the definition of a Taylor series and the binomial theorem.

But if you just want the "formula" for what Series does, then the following is maybe what you want:

n = 3
Series[f[x], {x, a, n}] // Normal
(* E^a + E^a (-a + x) + 1/2 E^a (-a + x)^2 + 1/6 E^a (-a + x)^3 *)
c[j_, n_] := (x - a)^j (D[f[x], {x, j}] /. x -> a)/j!
Table[c[j, n], {j, 0, n}]
(* {E^a, E^a (-a + x), 1/2 E^a (-a + x)^2, 1/6 E^a (-a + x)^3} *)

Answer 2

With SeriesCoefficient, the order and expansion point can be symbolic.

$Version

(* "13.3.1 for Mac OS X ARM (64-bit) (July 24, 2023)" *)

Clear["Global`*"]

coef[n_] = SeriesCoefficient[Exp[x], {x, a, n}]

Verification,

Sum[coef[n]*(x - a)^n, {n, 0, Infinity}]

(* E^x *)

EDIT:

Comparing CoefficientLists

m = 20;

CoefficientList[Sum[coef[n]*(x - a)^n, {n, 0, m}], x] == 
 CoefficientList[Series[Exp[x], {x, a, m}], x]

(* True *)

Answer 3

If I understand correctly what you want the following is your friend

Table[CoefficientList[Series[Exp[x], {x, a, placeholder}], x] // 
   First, {placeholder, 0, 17}] // 
 FindSequenceFunction[#, order + 1] &

Gamma[1 + order, -a]/Gamma[1 + order]