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MathGroup Archive 2006

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Re: Solve with dot products

  • To: mathgroup at smc.vnet.net
  • Subject: [mg70741] Re: Solve with dot products
  • From: dh <dh at metrohm.ch>
  • Date: Wed, 25 Oct 2006 01:40:08 -0400 (EDT)
  • References: <ehkch0$isb$1@smc.vnet.net>

Hi Oliver,
the manual and you are both right. The manual says about InverseFunction:
  "InverseFunction[f][y] gives the value of x for which f[x] is equal to y"
therefore, InverseFunction[Dot,1,3][d,b,c] means, that value of x, that 
makes Dot[x,b,c] equal to d and this is exactly your a. This solution is 
implicite, wheras your solution is explicite and because of the 
inversion, b and c are reversed.

Daniel

Oliver Friedrich wrote:
> Hallo,
> 
> I have some difficulties with interpretation of a solution that Solve 
> returns.
> 
> In[2]:=
> Solve[a.b.c==d,a]
> 
> From In[2]:=
> Solve::ifun: Inverse functions are being used by Solve, so some solutions 
> may \
> not be found.
> 
> Out[2]=
> {{a\[Rule]InverseFunction[Dot,1,3][d,b,c]}}
> 
> When I do that equation by hand I get something that I can't put into 
> correspondence with the solution above.
> 
> * means Inverse of a
> (a.b)*=b*.a*
> 
> a.b.c=d    	    	    	|(_)*
> 
> c*.b*.a*=d*    	    	|c.
> 
> b*.a*=c.d*    	    	|b.
> 
> a*=b.c.d*    	    	|(_)*
> 
> a=d.c*.b*
> 
> b and c are swapped compared to the Mathematica solution.
> 
> 1) Where's my error?
> 2) What's the interpretation of such an InverseFunction expression, I 
> don't get along the docu.
> 3) What's the z in InverseFunction[x,y,z][args] good for when I could 
> count the number of arguments in args?
> 
> Thanks a lot for your help
> 
> 


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