Re: Replace not spotting a replacement
- To: mathgroup at smc.vnet.net
- Subject: [mg103426] Re: Replace not spotting a replacement
- From: Alexei Boulbitch <Alexei.Boulbitch at iee.lu>
- Date: Mon, 21 Sep 2009 05:51:40 -0400 (EDT)
Hi, David, try this: In[12]:= Sqrt[2 \[Pi]] \[Tau]^(3/2) /. \[Tau] -> \[Mu]^2/2 Pi Out[12]= 1/2 \[Pi]^2 (\[Mu]^2)^(3/2) I do not understand the internal mechanism of it, but I observe that the rule holds: in replacements one should resolve expression to be substituted with respect to one of the variables. In other words, if you need that in your expression expr=expr[x,y,z] a relation f[x,y,z]=0 holds, you should resolve first the relation f[x,y,z]=0 to get say, x=g[y,z], and then you may use the replacement expr[x,y,z]/.x->g[y,z]. Regards, Alexei Hi Folks, Seems the replace function is being a bit dim here. This fails: Sqrt[2 \[Pi]] \[Tau]^(3/2) /. Sqrt[ 2 \[Pi] \[Tau]] -> \[Mu] Apologies if this is an FAQ - I couldn't see anything about it. What do I need to do here to coerce Mathematica into making this replacement? I tried using the Collect function on tau first, but that didn't seem to work. I also tried re-writing the replacement rule as Sqrt[2 \[Pi]] \[Tau]^(3/2) /. Sqrt[ 2 \[Pi] ] t^(1/2) -> \[Mu] But still no joy. Anyone know? Many thanks, David. -- Alexei Boulbitch, Dr., habil. Senior Scientist IEE S.A. ZAE Weiergewan 11, rue Edmond Reuter L-5326 Contern Luxembourg Phone: +352 2454 2566 Fax: +352 2454 3566 Website: www.iee.lu This e-mail may contain trade secrets or privileged, undisclosed or otherwise confidential information. If you are not the intended recipient and have received this e-mail in error, you are hereby notified that any review, copying or distribution of it is strictly prohibited. Please inform us immediately and destroy the original transmittal from your system. Thank you for your co-operation.