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SOLVED - GFCI - should my neutral and ground have continuity?



Turning arguments into exponents


What is -(-2,3,4)?Replace rule also matching complex numbersCollecting terms of even exponentsApplying functions repeatedly on parts of complicated expressionsSplitBy with exponentsTurning table into association for ClassifyCollect with symbolic exponentscoordinate change in a list of pointsTurning a NB into a Repeatable Script?How to Append Sequence of Numbers to List?Expression replacement with variable exponents






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








4












$begingroup$


I'd like to replace every instance of L[stuff], e.g.,



L[-1,-2,-2]


in an expression with x^(-FromDigits[stuff]), e.g.



x^122


.
But I'm running into an issue-- I'd like to do this with the code



L[-1, -2, -2] /. L[a___] -> x^(FromDigits[-a])


but this produces



2/x^6 + x^4 + x^5.


In trouble-shooting, I noticed that while



L[-1, -2, -2] /. L[a___] -> a


produces the output I expected, namely -1,-2,-2, I found that



L[-1, -2, -2] /. L[a___] -> -a


just produces 4 rather than 1,2,2. I must be misunderstanding how Sequence behaves when it shows up in the right side of a Rule since when I try just inputting -Sequence[-1,-2,-2] I get the expected result 1,2,2. Does anyone see where I'm going wrong? Thanks!










share|improve this question









$endgroup$







  • 2




    $begingroup$
    funny, I asked a very similar question last week, I wonder if its the same issue. Anyway, here is a possible workaround: L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a]).
    $endgroup$
    – AccidentalFourierTransform
    8 hours ago











  • $begingroup$
    @AccidentalFourierTransform yup it would appear to be exactly the same issue and my reaction to FullForm[-b] was exactly the same as yours lol
    $endgroup$
    – Diffycue
    8 hours ago

















4












$begingroup$


I'd like to replace every instance of L[stuff], e.g.,



L[-1,-2,-2]


in an expression with x^(-FromDigits[stuff]), e.g.



x^122


.
But I'm running into an issue-- I'd like to do this with the code



L[-1, -2, -2] /. L[a___] -> x^(FromDigits[-a])


but this produces



2/x^6 + x^4 + x^5.


In trouble-shooting, I noticed that while



L[-1, -2, -2] /. L[a___] -> a


produces the output I expected, namely -1,-2,-2, I found that



L[-1, -2, -2] /. L[a___] -> -a


just produces 4 rather than 1,2,2. I must be misunderstanding how Sequence behaves when it shows up in the right side of a Rule since when I try just inputting -Sequence[-1,-2,-2] I get the expected result 1,2,2. Does anyone see where I'm going wrong? Thanks!










share|improve this question









$endgroup$







  • 2




    $begingroup$
    funny, I asked a very similar question last week, I wonder if its the same issue. Anyway, here is a possible workaround: L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a]).
    $endgroup$
    – AccidentalFourierTransform
    8 hours ago











  • $begingroup$
    @AccidentalFourierTransform yup it would appear to be exactly the same issue and my reaction to FullForm[-b] was exactly the same as yours lol
    $endgroup$
    – Diffycue
    8 hours ago













4












4








4





$begingroup$


I'd like to replace every instance of L[stuff], e.g.,



L[-1,-2,-2]


in an expression with x^(-FromDigits[stuff]), e.g.



x^122


.
But I'm running into an issue-- I'd like to do this with the code



L[-1, -2, -2] /. L[a___] -> x^(FromDigits[-a])


but this produces



2/x^6 + x^4 + x^5.


In trouble-shooting, I noticed that while



L[-1, -2, -2] /. L[a___] -> a


produces the output I expected, namely -1,-2,-2, I found that



L[-1, -2, -2] /. L[a___] -> -a


just produces 4 rather than 1,2,2. I must be misunderstanding how Sequence behaves when it shows up in the right side of a Rule since when I try just inputting -Sequence[-1,-2,-2] I get the expected result 1,2,2. Does anyone see where I'm going wrong? Thanks!










share|improve this question









$endgroup$




I'd like to replace every instance of L[stuff], e.g.,



L[-1,-2,-2]


in an expression with x^(-FromDigits[stuff]), e.g.



x^122


.
But I'm running into an issue-- I'd like to do this with the code



L[-1, -2, -2] /. L[a___] -> x^(FromDigits[-a])


but this produces



2/x^6 + x^4 + x^5.


In trouble-shooting, I noticed that while



L[-1, -2, -2] /. L[a___] -> a


produces the output I expected, namely -1,-2,-2, I found that



L[-1, -2, -2] /. L[a___] -> -a


just produces 4 rather than 1,2,2. I must be misunderstanding how Sequence behaves when it shows up in the right side of a Rule since when I try just inputting -Sequence[-1,-2,-2] I get the expected result 1,2,2. Does anyone see where I'm going wrong? Thanks!







list-manipulation replacement symbolic polynomials sequence






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 8 hours ago









DiffycueDiffycue

2521 silver badge6 bronze badges




2521 silver badge6 bronze badges







  • 2




    $begingroup$
    funny, I asked a very similar question last week, I wonder if its the same issue. Anyway, here is a possible workaround: L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a]).
    $endgroup$
    – AccidentalFourierTransform
    8 hours ago











  • $begingroup$
    @AccidentalFourierTransform yup it would appear to be exactly the same issue and my reaction to FullForm[-b] was exactly the same as yours lol
    $endgroup$
    – Diffycue
    8 hours ago












  • 2




    $begingroup$
    funny, I asked a very similar question last week, I wonder if its the same issue. Anyway, here is a possible workaround: L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a]).
    $endgroup$
    – AccidentalFourierTransform
    8 hours ago











  • $begingroup$
    @AccidentalFourierTransform yup it would appear to be exactly the same issue and my reaction to FullForm[-b] was exactly the same as yours lol
    $endgroup$
    – Diffycue
    8 hours ago







2




2




$begingroup$
funny, I asked a very similar question last week, I wonder if its the same issue. Anyway, here is a possible workaround: L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a]).
$endgroup$
– AccidentalFourierTransform
8 hours ago





$begingroup$
funny, I asked a very similar question last week, I wonder if its the same issue. Anyway, here is a possible workaround: L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a]).
$endgroup$
– AccidentalFourierTransform
8 hours ago













$begingroup$
@AccidentalFourierTransform yup it would appear to be exactly the same issue and my reaction to FullForm[-b] was exactly the same as yours lol
$endgroup$
– Diffycue
8 hours ago




$begingroup$
@AccidentalFourierTransform yup it would appear to be exactly the same issue and my reaction to FullForm[-b] was exactly the same as yours lol
$endgroup$
– Diffycue
8 hours ago










2 Answers
2






active

oldest

votes


















5












$begingroup$

Here's an explanation of why you get 4. As mentioned above, your problem is the use of Rule instead of RuleDelayed, which means the RHS of the rule evaluates before the rule is used. So, let's see what the RHS of the rule evaluates to:



-a //FullForm



List[Times[-1,a]]




So, when evaluated, the rule becomes:



L[a___] -> Times[-1, a]


Now, it should be clear what happens. When the rule is applied, a gets bound to the sequence -1, -2, -2, and so the output (before evaluation) is:



Times[-1, -1, -2, -2]


which evaluates to:




4







share|improve this answer









$endgroup$




















    5












    $begingroup$

    It's just a matter of using RuleDelayed, or rather :> instead of ->.



    L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a])



    x^122




    L[-1, -2, -2] /. L[a___] :> -a



    1, 2, 2







    share|improve this answer









    $endgroup$















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      Here's an explanation of why you get 4. As mentioned above, your problem is the use of Rule instead of RuleDelayed, which means the RHS of the rule evaluates before the rule is used. So, let's see what the RHS of the rule evaluates to:



      -a //FullForm



      List[Times[-1,a]]




      So, when evaluated, the rule becomes:



      L[a___] -> Times[-1, a]


      Now, it should be clear what happens. When the rule is applied, a gets bound to the sequence -1, -2, -2, and so the output (before evaluation) is:



      Times[-1, -1, -2, -2]


      which evaluates to:




      4







      share|improve this answer









      $endgroup$

















        5












        $begingroup$

        Here's an explanation of why you get 4. As mentioned above, your problem is the use of Rule instead of RuleDelayed, which means the RHS of the rule evaluates before the rule is used. So, let's see what the RHS of the rule evaluates to:



        -a //FullForm



        List[Times[-1,a]]




        So, when evaluated, the rule becomes:



        L[a___] -> Times[-1, a]


        Now, it should be clear what happens. When the rule is applied, a gets bound to the sequence -1, -2, -2, and so the output (before evaluation) is:



        Times[-1, -1, -2, -2]


        which evaluates to:




        4







        share|improve this answer









        $endgroup$















          5












          5








          5





          $begingroup$

          Here's an explanation of why you get 4. As mentioned above, your problem is the use of Rule instead of RuleDelayed, which means the RHS of the rule evaluates before the rule is used. So, let's see what the RHS of the rule evaluates to:



          -a //FullForm



          List[Times[-1,a]]




          So, when evaluated, the rule becomes:



          L[a___] -> Times[-1, a]


          Now, it should be clear what happens. When the rule is applied, a gets bound to the sequence -1, -2, -2, and so the output (before evaluation) is:



          Times[-1, -1, -2, -2]


          which evaluates to:




          4







          share|improve this answer









          $endgroup$



          Here's an explanation of why you get 4. As mentioned above, your problem is the use of Rule instead of RuleDelayed, which means the RHS of the rule evaluates before the rule is used. So, let's see what the RHS of the rule evaluates to:



          -a //FullForm



          List[Times[-1,a]]




          So, when evaluated, the rule becomes:



          L[a___] -> Times[-1, a]


          Now, it should be clear what happens. When the rule is applied, a gets bound to the sequence -1, -2, -2, and so the output (before evaluation) is:



          Times[-1, -1, -2, -2]


          which evaluates to:




          4








          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 7 hours ago









          Carl WollCarl Woll

          86.7k3 gold badges114 silver badges221 bronze badges




          86.7k3 gold badges114 silver badges221 bronze badges























              5












              $begingroup$

              It's just a matter of using RuleDelayed, or rather :> instead of ->.



              L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a])



              x^122




              L[-1, -2, -2] /. L[a___] :> -a



              1, 2, 2







              share|improve this answer









              $endgroup$

















                5












                $begingroup$

                It's just a matter of using RuleDelayed, or rather :> instead of ->.



                L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a])



                x^122




                L[-1, -2, -2] /. L[a___] :> -a



                1, 2, 2







                share|improve this answer









                $endgroup$















                  5












                  5








                  5





                  $begingroup$

                  It's just a matter of using RuleDelayed, or rather :> instead of ->.



                  L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a])



                  x^122




                  L[-1, -2, -2] /. L[a___] :> -a



                  1, 2, 2







                  share|improve this answer









                  $endgroup$



                  It's just a matter of using RuleDelayed, or rather :> instead of ->.



                  L[-1, -2, -2] /. L[a___] :> x^(FromDigits[-a])



                  x^122




                  L[-1, -2, -2] /. L[a___] :> -a



                  1, 2, 2








                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 8 hours ago









                  That Gravity GuyThat Gravity Guy

                  2,2311 gold badge6 silver badges15 bronze badges




                  2,2311 gold badge6 silver badges15 bronze badges



























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