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1

I usually use Simplify with assumptions: Simplify[ Reduce[1/(E^0)^x + 1/(E^1)^x + 1/(E^3)^x == 0, x], C[1] ∈ Integers] (* x == I π (1 + 2 C[1]) + Log[-Root[1 + #1^2 + #1^3 &, 1]] || x == 2 I π C[1] + Log[Root[1 + #1^2 + #1^3 &, 2]] || x == 2 I π C[1] + Log[Root[1 + #1^2 + #1^3 &, 3]] *) That way I'm fairly confident the ...


5

As far as I know, there is no easy, general way to handle this kind of algebra with Sum expressions. What follows is an attempt to use replacement rules to handle a wider range of cases than chris's example. I don't consider it to be the canonical answer that is required, but perhaps someone might be able to use it as a starting point. I use Inactive on ...


3

stripF = ToExpression[ToString[#, StandardForm]] &; stripF /@ {Style[1201/100000, FontFamily -> "Charter", FontSize -> 20], Style[NumberForm[10.01, {10, 2}], FontFamily -> "Academy Engraved LET", FontSize -> 50, FontColor -> RGBColor[0, 1, 0]]} (* {1201/100000, 10.01} *) Also: ...


2

You may never know how many wrappers there are but those functions have this in common that first argument is what we only care about. f[x_?NumericQ] := N @ x; f[x_] := f @ First[x]


4

Yes we can ! MapAt[Integrate[#, {x, -Infinity, Infinity}] &, f[x], 1] // PowerExpand (* n *) tt = f[x]^2 /. Power[Sum[a__, b__], 2] :> sum[a (a /. i -> j) // Release, b, b /. {i -> j}] MapAt[Integrate[#, {x, -Infinity, Infinity}] &, tt, 1] /. sum -> Sum // PowerExpand


2

Update: This matches the evaluated form of your desired output for each example: fn = Collect[#, x^y_ /; FreeQ[y, _Integer], Apart] &; Test: expr1 = c0 x^(a + b) + c1 x^(a + b + 1) + c2 x^(a + b + 2) + c3 x^(a + b + 2) + c4 x^(a + b + 2) + c5 x^(a) + c6 x^(a + 1) + c7 x^(a + 1); target1 = x^(a) (c5 + x (c6 + c7)) + x^(a + b) (c0 + x c1 + x^2 (c2 ...



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