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12

According to the documentation: $\sum _i^{i_{\max }} f$ is by default interpreted as Sum[f {$i$, $i_{\max }$}] so we can abuse this: You can use Sequence to provide di too because: $\sum _{i=i_{\min }}^{i_{\max }} f$ is by default interpreted as Sum[f {$i$, $i_{\min }$, $i_{\max }$}] ssch has found undocumented but useful pattern that is ...

0

Revised version It seems you don't want to use Plot, because this creates a continuous plot in the region. You want to draw the sum for integer values of x. It takes a good amount of time, but your sum can be evaluated analytically s = Sum[Binomial[x, l]*l^50*(-1)^(x - l), {l, 0, x}] (* (-(-1)^x)*x* HypergeometricPFQ[{2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ...

4

Plot has attributes, HoldAll, so the Sum isn't evaluated, try this: Plot[Evaluate[Sum[a^10*(-1)^(x - a), {a, 1, x}]], {x, 1, 10}] Where the sum evaluates to: before plotting: related questions/answers: Difference in Plot when using Evaluate vs when not using Evaluate How and when to use Evaluate? Behavior of expression evaluation in Plot I ...

2

I believe you are trying to limit evaluation in the sum. You can use Defer to accomplish this. I Apply (@@) it to a list for partial evaluation. Subscript[λ, n_] := Defer @@ {(2 n - 1) Pi/2 L}; Sum[Subscript[λ, n], {n, 1, 3}] (L π)/2 + (3 L π)/2 + (5 L π)/2 Or perhaps more to your liking: Subscript[λ, n_] := Defer @@ {(2 n - 1) Pi/Defer[2 L]}; ...

2

Artes's solution is the most general one that works for numerical black box functions as well. In this special case however, we have a different option: make sure that Plot sees the expression of the function. It will recognize HeavisideTheta as a function with a discontinuity, and it will plot the discontinuities as their precise location (not at a ...

1

As Artes noted, adding to Plot e.g. these options resolves the problem: PlotPoints -> 200, MaxRecursion -> 4

0

I've solved this by expanding out all of the summands and playing around with the ranges over the various summations. I had to analyse which values of k, n, s and t actually contributed to the sum, and change the summation ranges accordingly for each individual summand in the expanded expression for summ.

1

Here is one way to do it. I sure there are many more. a[k_, n_] := Sum[Re[n^#] + Im[n^#] &[ZetaZero[i] // N], {i, k}] Plot[With[{a = a[100, n]}, Sign[a] a^2/n], {n, 0, 30}]

1

EDIT In the following the role of f and g have been inadvertently exchanged,i.e. $f(n)=f(n-1)^2+2 g(n-1)^2$ and $g(n)=2 f(n-1)g(n-1)$ Therefore, just exchange. The values can be obtained by defining the recursive functions with suitable starting values for f and g. I present alternatives. It is relatively straightforward to uncouple the relations: ...

8

Number theory questions are always a huge accumulator for up votes. :) From my experience I can say that the builtin MangoldtLambda function is pretty slow. So let's define a Mangoldt function on our own. The Mangoldt function is defined by: $\Lambda(n) \equiv \left\{ \begin{array}{1 1} ln\ p & \quad \text{if n =$p^k\$ for p a prime}\\ 0 & \quad ...

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