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1

To address the issue use some Off[NumberForm::sigz] in Manipulate[...]. It is also better to use simple Epilog with Text[Row[...] (code below): Some code picture:


2

DeleteCases[list, {___, Null, ___}] DeleteCases[list, {___, , ___}] Also with Select and ContainsAny. Select[Not@*ContainsAny[{Null}]]@list


2

Observing strictly that the domain of x as the upper limit of the summation index is the integers, the limit exists, it can be calculated easily with Mathematica and it is different from zero. We need to consider this sum \[Sigma]WH[x_] := Sqrt[\[Pi]^2/12 + Simplify[Sum[(x!*x!)/(k!*(2 x - k)!)*(-1)^(x - k)/(x - k), {k, 0, x - 1}], x \[Element] ...


2

Use the third argument to IntegerDigits to specify the length of the digit sequence required, padded on the left by zero. Thus, reconstructArray = ArrayReshape[IntegerDigits[code, 2, 8*7], {8, 7}]


3

I think you will have to ask a mathematician if the limit is really 0 (or even real) since in Mathematica you can get this s[x_] := Sqrt[ Pi^2/12 + Sum[(x!*x!)/(k!*(2 x - k)!)*(-1)^(x - k)/(x - k), {k, 0, x - 1}]] s[x] // FullSimplify Limit[s[x], x -> Infinity] // FullSimplify So the limit seems to be complex, but I am not sure, I am not a ...


1

Are you looking for a shorter function to create a pattern? Or just a way to write the pattern without a function? Your function could be shorter by writing ep[x_, e_] := _?(Abs[x - #] <= e &) lst = 10.4 + RandomReal[{-.02, .02}, 10] Cases[lst, ep[10.4, 10.^-2]] Cases[lst, _?(Abs[10.4 - #] <= .01 &)] (* {10.4099, 10.4196, 10.3874, 10.3976, ...



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