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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