8
In 13.0.0 on Windows 10
AsymptoticSum[Sqrt[-i^2/n^2 + 1]/n, {i, 1, n}, n -> Infinity]
Pi/4
It should be noticed that
AsymptoticSum[Sqrt[-i^2 + n^2]/n^2, {i, 1, n}, n -> Infinity]
returns the input.
7
Perhaps
Total@MapThread[F[#1, #2] &, {Range[0, 4], Range[-2, -10, -2]}]
(*F[0, -2] + F[1, -4] + F[2, -6] + F[3, -8] + F[4, -10]*)
7
In Mathematica 13,we can just use sum
pts = {{0, -2}, {1, -4}, {2, -6}, {3, -8}, {4, -10}};
Sum[f @@ p, {p, pts}]
f[0, -2] + f[1, -4] + f[2, -6] + f[3, -8] + f[4, -10]
6
Inner[f, 0 ~ # ~ 4, -2 # @ 5]& @ Range
f[0, -2] + f[1, -4] + f[2, -6] + f[3, -8] + f[4, -10]
Sum[f[i - 1, -2 i], {i, 5}]
f[0, -2] + f[1, -4] + f[2, -6] + f[3, -8] + f[4, -10]
Tr @ Array[f[# - 1, -2 #] &, 5]
f[0, -2] + f[1, -4] + f[2, -6] + f[3, -8] + f[4, -10]
h = Tr @* Thread @* f;
h[# - 1, -2 #] & @ Range @ 5
f[0, -2] + f[1, -4] + f[2, -...
6
You can modify the ranges as needed.
Clear[alist, blist, clist, f]
alist = Range[0, 4]
blist = Range[-2, -10, -2]
clist = Transpose[{alist, blist}]
Total[f @@@ clist]
EDIT
Another solution could be:
Total@Thread[f[alist, blist]]
f[0, -2] + f[1, -4] + f[2, -6] + f[3, -8] + f[4, -10]
5
Numerically solving,
Needs["NumericalCalculus`"]
lim = NLimit[Sum[Sqrt[-i^2 + n^2]/n^2, {i, 1, n}], n -> Infinity,
WorkingPrecision -> 15]
(* 0.785398 *)
RootApproximant[lim/Pi]*Pi
(* π/4 *)
4
This can be done with Mathematica as follows. First, on the domain of the function of \[Theta] defined by Sum[Tan[\[Theta] + k*Pi/n], {k, 0, n - 1}] + n*Cot[n*Pi/2 + n*\[Theta]] its derivative equals zero as the result of
FullSimplify[D[Sum[Tan[\[Theta] + k*Pi/n], {k, 0, n - 1}] +
n*Cot[n*Pi/2 + n*\[Theta]], \[Theta]], Assumptions -> n \[Element] ...
4
Try midpoint discretization x=1/(2n)+i/n, i,1,n-1:
dx = 1/n;
Sqrt[1 - x^2] dx /. x -> dx/2 + i dx // FullSimplify (*Sqrt[1 - (1 + 2 i)^2/(4 n^2)]/n*)
Unfortunately Mathematica can't solve the limit
Limit[Sum[Sqrt[1 - (1 + 2 i)^2/(4 n^2)]/n , {i,1, n - 1}], n -> Infinity]
but numerically evaluation gives expected result ~Pi/4
Table[{n,NSum[Sqrt[1 - (...
4
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
Verifying,
(% // Activate) == Normal[Series[f[x], {x, ...
3
Is this what you were looking for?
SeriesCoefficient[Exp[x], {x, 10, n}]
And more generally
SeriesCoefficient[Exp[x], {x, n, n}]
2
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 = ...
2
$Version
(* "13.0.0 for Mac OS X x86 (64-bit) (December 3, 2021)" *)
Clear["Global`*"]
c[k_] = SeriesCoefficient[1/(z^2 + 9), {z, 3 I, k}]
sum = Inactive[Sum][c[k]*(z - 3*I)^k, {k, -1, Infinity}] //
Simplify[#, {k >= -1, Element[k, Integers]}] &
Verifying,
sum // Activate
(* 1/(9 + z^2) *)
EDIT: Your "expected result&...
2
Using linearExpand from How to do algebra on unevaluated integrals? :
Clear[linearExpand];
linearExpand[e_, x_, head_] :=
e //. {op : head[arg_Plus, __] :> Distribute[op],
head[arg1_Times, rest__] :>
With[{dependencies = Internal`DependsOnQ[#, x] & /@ List @@ arg1},
Pick[arg1, dependencies, False] head[
Pick[arg1, ...
1
Use a replacement Rule to swap the order when appropriate.
Clear["Global`*"]
swap = Integrate[Sum[f_, iter1_, opts1___], iter2_, opts2___] :>
Sum[Integrate[f, iter2, opts2], iter1, opts1];
expr[n_Integer?Positive] = Assuming[b > 0, (Integrate[
Sum[Exp[-b t^2]*r[j], {j, 1, n}], {t, -∞, +∞}] /. swap)]
(* Sum[(Sqrt[Pi]*r[j])/Sqrt[b], ...
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