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MathGroup Archive 1999

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Re: Simple edit ...

  • To: mathgroup at smc.vnet.net
  • Subject: [mg19363] Re: [mg19319] Simple edit ...
  • From: "Wolf, Hartmut" <hwolf at debis.com>
  • Date: Fri, 20 Aug 1999 23:09:39 -0400
  • Organization: debis Systemhaus
  • References: <199908150342.XAA00982@smc.vnet.net.>
  • Sender: owner-wri-mathgroup at wolfram.com

Michael L. Stokes schrieb:
> 
> Consider the following examaple;
> 
> z = (a+Sqrt[d+e])/(a Sqrt[d+e])
> 
> z //. Sqrt[d+e] ->L
> 
> The replacement function only replaces the instance of Sqrt[d+e] in the
> numerator and leaves the instance in the denominator alone.  Why does
> this not work and what is the work around?
> 

Hello Mike,

look at the FullForm of z

In[1]:= z // FullForm
Out[1]//FullForm=
Times[Power[a, -1], Power[Plus[d, e], Rational[-1, 2]], 
  Plus[a, Power[Plus[d, e], Rational[1, 2]]]]

So for replacement you have to define a pattern that matches both
Power[Plus[d, e], Rational[1, 2] and Power[Plus[d, e], Rational[-1, 2].

This simple guess doesn't do

In[2]:= (d + e)^(n_Integer/2) // FullForm
Out[2]//FullForm=
Power[Plus[d, e], Times[Rational[1, 2], Pattern[n, Blank[Integer]]]]

but working backwards

In[3]:= Power[Plus[d, e], Rational[n, 2]] // InputForm
Out[3]//InputForm=
(d + e)^(Rational[n, 2])

you'll get it. So...

In[4]:= z /. (d + e)^(Rational[n : (1 | -1), 2]) :> L^n
Out[4]= (a + L)/(a*L)

...will do, but for the following more complicated examathematica 

In[5]:= zz = Expand[(a + Sqrt[d + e])^3]/(a Sqrt[d + e]^5);

better use

In[6]:= zz /. (d + e)^(Rational[n_Integer, 2]) :> L^n
Out[6]= (a^3 + 3*a*d + 3*a*e + 3*a^2*L + d*L + e*L)/(a*L^5)

Kind regards, hw



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