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I would like to replace the expresion $(n1-1)$ to $z1$ in following:

2*n1*p[n1-1,n2]

2*(z1+1)*p[z1,n2]

I have tried Replace, ReplaceAll, /., //. but nothing seems to be working. Only part inside $p[n1-1, n2]$ is replaced. Can anyone help me with this?

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    $\begingroup$ (* stuff *) /. n1 -> z1 +1 doesn't do what you want? $\endgroup$
    – J. M.'s missing motivation ♦
    Commented Apr 8, 2013 at 12:58

2 Answers 2

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If you type:

2 n1 p[n1-1, n2] /. {n1 -> z1 +1}

then the answer is:

2 (1 + z1) p[z1, n2]

This is shorthand for the function:

ReplaceAll[2 n1 p[n1 - 1, n2], n1 -> z1+1]

which explicitly uses the rule (in the second argument) to change the first argumemnt.

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  • $\begingroup$ Ah strange asterixes appear in your answer. I figured I'd (propose to) edit them, but you made an edit as well in the meantime. Oh well, please fix the asterixes :). $\endgroup$
    – Jacob Akkerboom
    Commented Apr 8, 2013 at 13:24
  • $\begingroup$ thanks Jacob -- I don't know where they came from... $\endgroup$
    – bill s
    Commented Apr 8, 2013 at 13:27
  • $\begingroup$ No problem, I think they came from the original problem statement, that has now been fixed by Istvan Zachar :). Hm.. I think Maria was trying to also replace the n1 that appears outside the function p, so I think this is probably not what she wants.. $\endgroup$
    – Jacob Akkerboom
    Commented Apr 8, 2013 at 13:30
  • $\begingroup$ I just realized that and changed it. But now I realize this is exactly what JM suggested... sorry JM.. I'm happy to delete and give to you... $\endgroup$
    – bill s
    Commented Apr 8, 2013 at 13:33
  • $\begingroup$ Yesterday I posted the silliest answer only to delete it a minute later, I can relate :P $\endgroup$
    – Jacob Akkerboom
    Commented Apr 8, 2013 at 13:37
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If $n1$ is set initially, when any function that contains this variable is entered, the conversion will be automatic.

n1 = z1 + 1; 
2*n1*p[n1 - 1, n2]
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