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New answers tagged summation

1

This is a known feature of Mathematica. See for example, question 199454 "Problem with extracting a constant multiplier out of sum" and the answers given there. The fundamental problem is that there is only a limited amount of simplifications that Mathematica or any other Computer Algebra System can do automatically. Some simplifications that we can see ...

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TrueQ will return True only if the input is explicitly True. You can use TrueQ to "assume" that a test fails (i.e., is False) when its outcome is not clear. It is generally not used to test if mathematical expressions are equivalent (Equal) since some form of simplification or transformation is often required to obtain a result which is explicitly True. ...

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Use KroneckerDelta[n] or DiscreteDelta[n] instead of 0^n.

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I don't see how the right-hand side of f[a_] can result from a (unless f separates each term in a rather than keeping the sum intact). To obtain the right-hand side of f[a_] it seems that one only needs to know the maximum number of terms in the products. Here's a function that will do that: f[n_] := Sum[SymmetricPolynomial[i, Table[Subscript[x, j], {j, n}...

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Update: A function that takes an expression of the specific form and an integer to construct the desired sums: ClearAll[sumF] sumF[poly_, p_] := Module[{var, indices, iteratorlist, parts = {First@#, #2} & @@ DeleteDuplicates /@ Transpose[Cases[poly /. Subscript[a_, Subscript[b_, c_]] :> Subscript[a, Superscript[b, c]], ...

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I did this: myEquation = \!$$\*UnderoverscriptBox[\(\[Sum]$$, $$i$$, $$p$$]$$Subscript[x, i]$$\) + \!$$\*UnderoverscriptBox[\(\[Sum]$$, $$i$$, $$p$$]$$\*UnderoverscriptBox[\(\[Sum]$$, $$j$$, $$i$$]Subscript[x, i] Subscript[x, j]\)\) + \!$$\*UnderoverscriptBox[\(\[Sum]$$, $$i$$, $$p$$]$$\*UnderoverscriptBox[\(\[Sum]$$, $$j$$, $$i$$]\(\*...

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use assumptions with PositiveIntegers instead of NonNegativeIntegers: Assuming[Element[n, PositiveIntegers], Sum[KroneckerDelta[a, n] f[a], {a, Infinity}]] (* f[n] *) Assuming[Element[n, PositiveIntegers], Sum[DiscreteDelta[a - n] f[a], {a, Infinity}]] (* f[n] *)

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You can also use Cases to construct a function that computes the index-sum and use it with GatherBy to group the monomials: ClearAll[f0, f1] f0 = Total @* Cases[(y | Derivative[__][y])[i_, _]^p_. :> p i]; f1 = Total /@ GatherBy[SortBy[f0]@MonomialList[#], f0]&; f1 @ Expr

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You can get the result in a list form for example as follows: Reverse[MonomialList[Expr /. y[x_, z_] -> EPS^x*y[x, z] /. Derivative[A_, B_][y][x_, z_] -> EPS^x*Derivative[A, B][y][x, z], EPS]] /. EPS -> 1 This replaces all instances of y or its derivatives by EPS to the power of first argument of the original y times itself. Then one can ...

0

Your question "Do we have any way to evaluate the equation correctly?" is yes because Mathematica allows you to define arbitrary rules to perform manipulation of expressions. In your particular use case the simplistic code 2 Sum[A[i], {i, n}] == Sum[2 A[i], {i, n}] /. Sum[k_ x_, {y_, z_}] :> k Sum[x, {y, z}] returns True as you wanted. However, the ...

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Simplify just transforms the equation into another, hopefully simpler form. Do not expect it to "solve" problems. It is not its purpose. Simplify also does not give any guarantees about success. It only guarantees that the output is equivalent to the input. Simplify does a heuristic search by applying various transformation. If it does not find a simpler ...

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Are you sure the result is false? sum = Sum[(Sin[22 n]/(7 n))^3, {n, 1, ∞}] Block[{$MaxExtraPrecision = 10000}, N[sum - 1/2 (π - 22/7)^3, 50] ] $$\left|\sum a_n-x\right|<10^{-10000.}$$ On a second thought, FullSimplify[sum] yields 1/686 (-22 + 7 π)^3 and so the sum is correct, and MMA is able to confirm it. Neat! 6 See Piecewise. It is like the cases environment in$\LaTeX$. In native math, one would use "else" instead of True. By the way, version 12 returns$$\begin{cases} \frac{\coth \left(\sqrt{a}-i b\right)+\coth \left(\sqrt{a}+i b\right)}{2 \sqrt{a}} & \arg \left(b+i \sqrt{a}\right)\geq 0 \\ \frac{\coth \left(\sqrt{a}-i b\right)}{2 \sqrt{a}}+\frac{\... 2 Not a solution but a long comment. tl;dr — I don't think this can be done for$f(x)=\tan(x)$. One of the conditions for the Euler-Maclaurin formula to work is that the function must be continuous. This is not the case here. More concretely for this case, using the pigeonhole principle it's easy to see that as$n$grows larger, some index values$x\in\{0..n\...

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