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I need to generate the result given by

List = {"0", "\!\(\*FractionBox[\(\[Pi]\), \(\[Tau]\)]\)", 
  "\!\(\*FractionBox[\(2  \[Pi]\), \(\[Tau]\)]\)", 
  "\!\(\*FractionBox[\(3  \[Pi]\), \(\[Tau]\)]\)"}

but with help of Range command.

How can I do this?

Thank you!

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4
  • $\begingroup$ How about Range[0, 3] Pi/tau? Btw, don't use List as variable! It's a used symbol in Mathematica. $\endgroup$
    – halirutan
    Commented Mar 17, 2015 at 6:00
  • $\begingroup$ Thank you, but I need list of strings, where Pi and tau are fixed characters, but not variables. Any alternative? $\endgroup$
    – Andrey Kazak
    Commented Mar 17, 2015 at 6:24
  • 1
    $\begingroup$ ToString[#, StandardForm] & /@ (Pi Range[0, 3]/\[Tau])? $\endgroup$
    – Kuba
    Commented Mar 17, 2015 at 6:57
  • $\begingroup$ This is not I want, because the result depends on value of Tau. I need string with Tau literally without substitution. For example if I set: Tau = 1; the result will be different from the desired one. $\endgroup$
    – Andrey Kazak
    Commented Mar 17, 2015 at 10:02

2 Answers 2

Reset to default
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Not too pretty, but handles non-integer steps:

If[#==0,ToString[0],("\!\(\*FractionBox[\("<>If[Numerator@#===1,"", 
ToString[Numerator@#]]<>"  \[Pi]\), \("<>If[Denominator@#===1,"", 
ToString[Denominator@#]]<>"\[Tau]\)]\)")]&/@Range[0,3,1/2]

enter image description here

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3
  • $\begingroup$ Hm.... I will try it. But is there a shorter and more simple way to do it? $\endgroup$
    – Andrey Kazak
    Commented Mar 17, 2015 at 10:24
  • $\begingroup$ This statement produces the result I want. Is there a way to simplify this piece of code? Thank you! $\endgroup$
    – Andrey Kazak
    Commented Mar 17, 2015 at 13:11
  • $\begingroup$ @AndreyKazak: If your range is only integer values * pi/tau, just remove all the If stuff and keep the first ToString. Other than that, no really simpler ways come to mind. $\endgroup$
    – ciao
    Commented Mar 17, 2015 at 21:57
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If you need strings, then you can just create the symbolic list first and transform every entry into a string afterwards:

ToString[#, TraditionalForm] & /@ (Range[0, 3] Pi/τ)

If you have assigned a global value to tau and you can, as Marius already pointed out, use Block to localize the variable

τ = 4;
Block[{τ},
  ToString[#, TraditionalForm] & /@ (Range[0, 3] Pi/τ)
]
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4
  • $\begingroup$ This is not I want, because the result depends on value of Tau. I need string with Tau literally without substitution. $\endgroup$
    – Andrey Kazak
    Commented Mar 17, 2015 at 10:01
  • $\begingroup$ Put it in a Block where you localize [Tau], then the global value does not matter. $\endgroup$
    – Marius Ladegård Meyer
    Commented Mar 17, 2015 at 18:57
  • $\begingroup$ Didn't get your point. Can you provide an example please? $\endgroup$
    – Andrey Kazak
    Commented Mar 17, 2015 at 19:27
  • $\begingroup$ Block[{\[Tau]}, ToString[#, TraditionalForm] & /@ (Range[0, 3] Pi/\[Tau])] $\endgroup$
    – Marius Ladegård Meyer
    Commented Mar 17, 2015 at 20:16

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