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-3

f[x_, n_] := (1/2 + ArcTan[n x]/Pi)^n; Limit[f[x, n], n -> Infinity, Assumptions -> x > 0] E^(-(1/(Pi x)))

3

Consider the following. $Assumptions (* default value at session start *) True$Assumptions = 0 < a < 1; $Assumptions = 0 < b < 1;$Assumptions 0 < b < 1 $Assumptions = 0 < a < 1 && 0 < b < 1;$Assumptions 0 < a < 1 && 0 < b < 1 Refine[If[a*b > 1, Print[Yes], Print[No]]] No ...

5

What you need to use here is Refine: Refine[If[t - u < 0, Print[Yes], Print[No]], t < u] Yes Ad if you need to work with $Assumptions as well, then you could do this:$Assumptions = t < u; Refine[If[t - u < 0, Print[Yes], Print[No]], $Assumptions] Yes The last line is also the way it would look if you were to wrap your If statement in ... 1 An alternative to prevent the expression from evaluating before it is passed on to Assuming: MyAssumptions := Assuming[α > 0 && ϵ > 0 && t > 0, #] &; MyAssumptions[ Unevaluated@ FullSimplify@ Integrate[(z^2 Exp[-α t (z^2 + ϵ)])/(z^2 + 1), {z, 0, ∞}]] 5 It can be zero but not negative. To show this we first rewrite the assumptions so they are actually usable, removing those that are redundant and getting the Thread to work by getting rid of inner lists (things in curly braces).$Assumptions = Flatten[{Thread[{c0, c1, p, w, e} > 0], Thread[0 <= {r, i, x, alpha, beta, gamma, 1 - alpha - ...

2

Your assigning a pattern to $Assumptions won't work because the Mathematica assumption mechanism is simply not geared to accept patterns. It is not based on pattern matching. To get$Assumptions to behave as you say you want it to, you would have to use the system hook $NewSymbol and do something like what is discussed in this answer. But even that won't ... 8 df2 = D[df1, μ];$Assumptions = Flatten[{Thread[{c1, c2, λ, μ} > 0], Element[{c1, c2, λ, μ}, Reals], μ > λ}]; FullSimplify@Positive[df2] (* True *) FullSimplify@Sign[df2] (* 1 *) Or, you can use your assumptions directly as the rhs of the Assumptions option, or as the first argument of Assuming, without setting the value of the global variable ...

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