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How can an equation having ColorDistance function be solved?



The 2019 Stack Overflow Developer Survey Results Are InCan't solve equation having complex coefficientsRecursive functions taking list argumentsThe system can not be solvedHow to interpolate colors on scattered points for a smooth visualizationHow to add a common color-bar to a set of 2d maps produced from evaluating a function along three projections in Mathematica 8.0?How to imitate the opacity plot in EditColorFunctionHandling oscilloscope bitmap dataHow can I combine the ChromaticityPlot3D with a Sphere?Can this be solved even faster?










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$begingroup$


I wish to extract the colors on the L axis in the CIELab color space such that the CIE2000 distance between a corresponding color and LABColor[0., 0., 0.] will be gradually increasing by a factor 0.01. For example, LABColor[0, 0, 0], LABColor[0.01733965, 0, 0], LABColor[0.0344219, 0, 0], LABColor[0.0512525, 0, 0], LABColor[0.0678371, 0, 0], LABColor[0.0841806, 0, 0], LABColor[0.100288, 0, 0] are the first few colors in this series. I have got these values by manually checking. However, to automate this, when I tried the following (to get the second element in the series)



Solve[ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], 
DistanceFunction -> "CIE2000"] == 0.01, l]


I got an error message.



How can I automate this process to complete the series.










share|improve this question









$endgroup$
















    6












    $begingroup$


    I wish to extract the colors on the L axis in the CIELab color space such that the CIE2000 distance between a corresponding color and LABColor[0., 0., 0.] will be gradually increasing by a factor 0.01. For example, LABColor[0, 0, 0], LABColor[0.01733965, 0, 0], LABColor[0.0344219, 0, 0], LABColor[0.0512525, 0, 0], LABColor[0.0678371, 0, 0], LABColor[0.0841806, 0, 0], LABColor[0.100288, 0, 0] are the first few colors in this series. I have got these values by manually checking. However, to automate this, when I tried the following (to get the second element in the series)



    Solve[ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], 
    DistanceFunction -> "CIE2000"] == 0.01, l]


    I got an error message.



    How can I automate this process to complete the series.










    share|improve this question









    $endgroup$














      6












      6








      6


      1



      $begingroup$


      I wish to extract the colors on the L axis in the CIELab color space such that the CIE2000 distance between a corresponding color and LABColor[0., 0., 0.] will be gradually increasing by a factor 0.01. For example, LABColor[0, 0, 0], LABColor[0.01733965, 0, 0], LABColor[0.0344219, 0, 0], LABColor[0.0512525, 0, 0], LABColor[0.0678371, 0, 0], LABColor[0.0841806, 0, 0], LABColor[0.100288, 0, 0] are the first few colors in this series. I have got these values by manually checking. However, to automate this, when I tried the following (to get the second element in the series)



      Solve[ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], 
      DistanceFunction -> "CIE2000"] == 0.01, l]


      I got an error message.



      How can I automate this process to complete the series.










      share|improve this question









      $endgroup$




      I wish to extract the colors on the L axis in the CIELab color space such that the CIE2000 distance between a corresponding color and LABColor[0., 0., 0.] will be gradually increasing by a factor 0.01. For example, LABColor[0, 0, 0], LABColor[0.01733965, 0, 0], LABColor[0.0344219, 0, 0], LABColor[0.0512525, 0, 0], LABColor[0.0678371, 0, 0], LABColor[0.0841806, 0, 0], LABColor[0.100288, 0, 0] are the first few colors in this series. I have got these values by manually checking. However, to automate this, when I tried the following (to get the second element in the series)



      Solve[ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], 
      DistanceFunction -> "CIE2000"] == 0.01, l]


      I got an error message.



      How can I automate this process to complete the series.







      list-manipulation equation-solving color






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Apr 5 at 8:13









      MajisMajis

      1,477415




      1,477415




















          1 Answer
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          8












          $begingroup$

          f[l_?NumericQ] := 
          ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], DistanceFunction -> "CIE2000"]

          FindRoot[f[l] == 0.01, l, .5]
          (* l -> 0.0173396 *)


          ColorDistance does not work with symbolic inputs. Write a function that evaluates with numeric inputs only.



          Solve is meant for exact symbolic computations. Use FindRoot instead.






          share|improve this answer









          $endgroup$












          • $begingroup$
            Perfect. Now I can get the complete list. Thanks.
            $endgroup$
            – Majis
            Apr 5 at 8:45











          Your Answer





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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

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          8












          $begingroup$

          f[l_?NumericQ] := 
          ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], DistanceFunction -> "CIE2000"]

          FindRoot[f[l] == 0.01, l, .5]
          (* l -> 0.0173396 *)


          ColorDistance does not work with symbolic inputs. Write a function that evaluates with numeric inputs only.



          Solve is meant for exact symbolic computations. Use FindRoot instead.






          share|improve this answer









          $endgroup$












          • $begingroup$
            Perfect. Now I can get the complete list. Thanks.
            $endgroup$
            – Majis
            Apr 5 at 8:45















          8












          $begingroup$

          f[l_?NumericQ] := 
          ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], DistanceFunction -> "CIE2000"]

          FindRoot[f[l] == 0.01, l, .5]
          (* l -> 0.0173396 *)


          ColorDistance does not work with symbolic inputs. Write a function that evaluates with numeric inputs only.



          Solve is meant for exact symbolic computations. Use FindRoot instead.






          share|improve this answer









          $endgroup$












          • $begingroup$
            Perfect. Now I can get the complete list. Thanks.
            $endgroup$
            – Majis
            Apr 5 at 8:45













          8












          8








          8





          $begingroup$

          f[l_?NumericQ] := 
          ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], DistanceFunction -> "CIE2000"]

          FindRoot[f[l] == 0.01, l, .5]
          (* l -> 0.0173396 *)


          ColorDistance does not work with symbolic inputs. Write a function that evaluates with numeric inputs only.



          Solve is meant for exact symbolic computations. Use FindRoot instead.






          share|improve this answer









          $endgroup$



          f[l_?NumericQ] := 
          ColorDistance[LABColor[0, 0, 0], LABColor[l, 0, 0], DistanceFunction -> "CIE2000"]

          FindRoot[f[l] == 0.01, l, .5]
          (* l -> 0.0173396 *)


          ColorDistance does not work with symbolic inputs. Write a function that evaluates with numeric inputs only.



          Solve is meant for exact symbolic computations. Use FindRoot instead.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Apr 5 at 8:26









          SzabolcsSzabolcs

          164k14448946




          164k14448946











          • $begingroup$
            Perfect. Now I can get the complete list. Thanks.
            $endgroup$
            – Majis
            Apr 5 at 8:45
















          • $begingroup$
            Perfect. Now I can get the complete list. Thanks.
            $endgroup$
            – Majis
            Apr 5 at 8:45















          $begingroup$
          Perfect. Now I can get the complete list. Thanks.
          $endgroup$
          – Majis
          Apr 5 at 8:45




          $begingroup$
          Perfect. Now I can get the complete list. Thanks.
          $endgroup$
          – Majis
          Apr 5 at 8:45

















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