13

The functionality of InputAutoReplacements has been replaced by the (much nicer in my opinion) AutoOperatorRenderings functionality: Notice how the operators are instantly replaced by their nicely-displayed variants, but the can still be easily edited. On a fresh install of 12.3, the relevant setting contains these substitutions: CurrentValue[...


9

Try ((# - Sqrt[#])/(# + Sqrt[#]) == (#^2 - #)/36) & /@ {1, 4, 8} {True, True, False} or (x |-> ((x - Sqrt[x])/(x + Sqrt[x]) == (x^2 - x)/36)) /@ {1, 4, 8} {True, True, False} ( |-> is Function, version 12.3!)


7

Perhaps Map is what you are looking for Map[(((x - Sqrt[x])/(x + Sqrt[x]) == (x^2 - x)/36)) /. x -> # &, {1,4, 8}] (*{True, True, False}*)


7

You could try any of the following: 1) Simply substitute Cos for x deqn=D[x[t],t,t]+2 x[t] deqn2=deqn/. x->Cos Yours wasn't working because x''[t] is not literally x[t]. 2) Change the definition of x in a Block and use that deqn=D[x[t],t,t]+2 x[t] deqn2=Block[{x = Cos}, deqn] Define a function for your equation which accepts a variable Head. deqnB[x_]=D[...


4

The implementation depends on how you define your variables. I have used x[i] as an expression for $x_i$. P = (1/12) (x[1]^12 + x[2]^6 + 2 x[3]^4 + 2 x[4]^3 + 2 x[6]^2 + 4 x[12]^1); P2 = P /. x[i_] -> w^i + b^i (* 1/12 ((b + w)^12 + (b^2 + w^2)^6 + 2 (b^3 + w^3)^4 + 2 (b^4 + w^4)^3 + 2 (b^6 + w^6)^2 + 4 (b^12 + w^12)) *) P2 // Expand (* b^12 + ...


4

You can also use ReplaceAll, you just have to tell it to do multiple replacements for the different values of x: ((x - Sqrt[x])/(x + Sqrt[x]) == (x^2 - x)/36) /. {{x -> 1}, {x -> 4}, {x -> 8}} (* {True, False, True} *)


3

You can get the desired function s of chii very fast with GroebnerBasis forcing it to delete the taui. eqs = {s == (\[Tau][1, 2] \[Tau][3, 4] \[Tau][5, 6])/(\[Tau][1, 6] \[Tau][2, 3] \[Tau][4, 5]), \[Chi]1 == (\[Tau][1, 2] \[Tau][3, 4])/(\[Tau][1, 3] \[Tau][2, 4]), \[Chi]2 == (\[Tau][3, 4] \[Tau][5, 6])/(\[Tau][3, 5] \[Tau][4, 6]), \[...


3

eqns = {s == (\[Tau][1, 2] \[Tau][3, 4] \[Tau][5, 6])/( \[Tau][1, 6] \[Tau][2, 3] \[Tau][4, 5]), \[Chi]1 == (\[Tau][1, 2] \[Tau][3,4])/(\[Tau][1, 3] \[Tau][2, 4]), \[Chi]2 == (\[Tau][3, 4] \[Tau][5, 6])/(\[Tau][3, 5] \[Tau][4, 6]), \[Chi]3 == (\[Tau][1,2] \[Tau][5, 6])/(\[Tau][1, 6] \[Tau][2, 5])} /. \[Tau][x_,y_] -> \[Tau][x] - \[Tau][y] Note that we ...


2

The simple way is: test /. i -> j


2

In addition to the above suggestions, if you do ever need to replace x'[t] or x''[t] or the like, it can be done via x'[t] /. Derivative[n_][x][t] -> Derivative[n][Cos][t] (* -Sin[t] *)


2

Define Pi/2 as a function, since Derivative expects functions. Sum[D[xx[[i]][\[Tau]], \[Tau]], {i, 4}] /. t -> Function[\[Tau], Pi/2] (* Derivative[1][x][\[Tau]] + Derivative[1][y][\[Tau]] + Derivative[1][z][\[Tau]] *)


1

If you are trying to do replacement in the front end, you can try something like ac = Cells[CellStyle -> {"Input"}]; content = NotebookRead /@ ac; newContent = content /. {"aa1" -> "ab1", "cc1" -> "cb1"}; Scan[NotebookWrite[ac[[#]], newContent[[#]]] &, Range@Length@newContent];


1

This is more an extended comment than an answer. The core issue is that D[t[τ], τ] /. t -> Pi/2 (* Derivative[1][Pi/2][τ] *) does not evaluate to zero. But, D[t[τ], τ] /. t -> 3/2 (* 0 *) and even D[t[τ], τ] /. t -> N[Pi]/2 (* 0 *) do. Evidently, D does not recognize Pi as numeric upon substitution, even though Mathematica does in general. ...


1

The following function works for your example, at least: fromHlog[strExpr_String] := ToExpression[StringReplace[strExpr, {"[" -> "{", "]" -> "}"}], TraditionalForm] /. {Hlog[x_, m_List] :> G[Sequence @@ m, x]}; though I'm not familiar with the Maple syntax and so it may have some pitfalls. Usage: fromHlog[...


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