The following result is based on comments above exchanged between Michael E2 and me. Because
SeriesCoefficient[Sqrt[(x - t)^2 + c^2], {x, Infinity, n}]
returns unevaluated, and
SeriesCoefficient[Sqrt[u^2 + c^2], {u, Infinity, n}]
(* Piecewise[{{(c^2)^((1 + n)/2)*Binomial[1/2, -n/2], Mod[n, -2] == 0 && n <= 0}}, 0] *)
is manifestly incorrect (i.e., a bug), begin instead with
SeriesCoefficient[Sqrt[1 + z], {z, 0, n}];
s1 = %[[1, 1, 1]] (x - t) (c/(x - t))^(2 n)
(* (c/(-t + x))^(2 n) (-t + x) Binomial[1/2, n] *)
To test the correctness of this result, perform
Sum[s1, {n, 0, Infinity}] // Simplify
(* Sqrt[1 + c^2/(t - x)^2] (-t + x) *)
which is correct. However, it is not the desired result, because the terms are functions of x - t
instead of x
. So, expand s1
in t
.
SeriesCoefficient[(t - x)^(1 - 2 n), {t, 0, m}, Assumptions -> n >= 0];
s2 = Simplify[%[[1, 1, 1]] s1 (t - x)^-(1 - 2 n), x - t > 0 && n ∈ Integers] t^m
(* c^(2 n) t^m (-x)^(-m - 2 n) x Binomial[1/2, n] Binomial[1 - 2 n, m] *)
Again, a test produces the correct result.
Sum[s2, {n, 0, Infinity}, {m, 0, Infinity}] // Simplify
(* Sqrt[1 + c^2/(t - x)^2] (-t + x) *)
Finally, all terms of a given order in x^-j
must be summed to obtain the desired result. Doing so yields
f[j_] = Piecewise[{{x - t, j == 0}, {Sum[If[Mod[j + 1 - m, 2] == 0,
s2 /. x -> 1 /. n -> (j + 1 - m)/2, 0], {m, 0, j}] x^-j, j > 0}}];
To test its accuracy, compare it with the series expansion provided in the question.
Series[Sqrt[c^2 + (t - x)^2], {x, Infinity, 5}] // Normal
(* -t + (c^6 - 12 c^4 t^2 + 8 c^2 t^4)/(16 x^5) + (-3 c^4 t + 4 c^2 t^3)/(8 x^4) +
(-c^4 + 4 c^2 t^2)/(8 x^3) + (c^2 t)/(2 x^2) + c^2/(2 x) + x *)
FullSimplify[Sum[f[j], {j, 0, 5}] == %]
(* True *)
Thus, the sum of f[j]
is the desired result.
All calculations were performed with
$Version
(* 11.1.0 for Microsoft Windows (64-bit) (March 13, 2017) *)
DifferenceRoot
answer? I can obtain that for expansion about0
but not for expansion aboutInfinity
. $\endgroup$Indeterminate
...it fails to do the Limit of a general DifferenceRoot answer obtained by the new MMA11.1 feature to make parameterized multiple derivatives. $\endgroup$SeriesCoefficient[Sqrt[c^2 + (u)^2], {u, Infinity, n}]
returns a formula (binomial series). Not equivalent, but perhaps useful. $\endgroup$SeriesCoefficient[Sqrt[1 + y], {y, 0, n}]
, and you can get theu
series withy == c^2/u^2
and multiplying they
series byu
. $\endgroup$