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Re: Re: Replacement equivalence?


carlos at colorado.edu wrote:

>Hmm... cant get it to work as stated.  Can you try the idea in this
>benchmark test
>
> ClearAll[f,a,b,c,i];  f[b_,c_]:=1/(c+b^2);
> v=Table[{i,f[b,i]/.b^2->16},{i,-6,6}];  ListPlot[v,PlotJoined->True];
>
>and see if the plot gap goes away? Thanks.
>
>  
>
This seems to work
Clear[b, x, c]
f[b_,c_]:=1/(c+b^2);
ListPlot[Table[{i, Replace[f[b, i], b -> (b =
4 â?¨ b = -4 )]}, {i, -6, 6}], PlotJoined -> True]

Hope this helps

Pratik


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