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I have 500 terms. and I want replace them with below function

rule1 := {       
 E^(Wr - (z___*Mr)/4 - (Gr*((Yr*Mr)/x___ + x___*Cr)) - 
             (Mr*((2*Mr)/y___ + (y___*Cr)/(2*Mr)))/2) :>
          
Ex[-(x+y+z)/4*Er^2]*
 (-(Mr^2/x) - Mr^2/y - Cr/(x+y+z))
       } ;

but I have some terms that don't have z (z=0) or y (y=0) for example:

E^(Wr - 1/2 Mr ((2 Mr)/y + (Cr y)/(2 Mr)) - Gr (Cr x + (Mr Yr)/x) - (
 Mr z)/4)

and rule1 couldn't contain all terms. how can I modify rule1 to contain all terms?

for example, I face this problem

term1 = E^(Wr - 1/2 Mr ((2 Mr)/y + (Cr y)/(2 Mr)) - Gr (Cr x + 
       (Mr*Yr)/x) - (Mr z)/4) 
        + E^(Wr - 1/2 Mr ((2 Mr)/y + (Cr y)/(2 Mr)) - Gr (Cr x + (Mr 
        Yr)/x))
        + E^(Wr - 1/2 Mr ((2 Mr)/y + (Cr y)/(2 Mr)) - (Mr z)/4) 
        + E^(Wr - (1/2) Mr ((2 Mr)/y + (Cr y)/(2 Mr)));

after replacing only last term has changed

term1 //. rule1
E^(Wr - 1/2 Mr ((2 Mr)/y + (Cr y)/(2 Mr))) 
+ E^(Wr - 1/2 Mr ((2 Mr)/y + (Cr y)/(2 Mr)) - Gr (Cr x + (Mr Yr)/x)) 
+ E^(Wr - 1/2 Mr ((2 Mr)/y + (Cr y)/(2 Mr)) 
- (Mr z)/4) + (-(Mr^2/x) - Mr^2/y - Cr/(x + y + z)) Ex[1/4 Er^2 (-x - y 
- z)]
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  • $\begingroup$ You can not use the same name for 2 different things:" x___ + x_*Cr". Further in "(2*Mr)/y___ + (y___*Cr" y is the same expression, is this what you want? $\endgroup$
    – Daniel Huber
    Nov 11, 2021 at 9:42
  • $\begingroup$ dear @DanielHuber thanks for your comment I edited it but still, it doesn't work. $\endgroup$
    – asal
    Nov 11, 2021 at 11:19
  • $\begingroup$ rule1 now matches your expression: E^(Wr - 1/2 Mr ((2 Mr)/y + (Cr y)/(2 Mr)) - Gr (Cr x + (Mr Yr)/x) - ( Mr z)/4) $\endgroup$
    – Daniel Huber
    Nov 11, 2021 at 11:48
  • $\begingroup$ dear@DanielHuber Again, the problem is not solved I think I did not explain my problem correctly, so I corrected the text, maybe my problem will be clearer $\endgroup$
    – asal
    Nov 12, 2021 at 6:48

1 Answer 1

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It is getting too complicated for a comment.

You may define the different cases as "Alternatives" like:

term1 = (E^(Wr - 1/2 Mr ((2 Mr)/y + (Cr y)/(2 Mr)) - 
       Gr (Cr x + (Mr*Yr)/x) - (Mr z)/4)
    + E^(Wr - 1/2 Mr ((2 Mr)/y + (Cr y)/(2 Mr)) - 
       Gr (Cr x + (Mr Yr)/x))
    + E^(Wr - 1/2 Mr ((2 Mr)/y + (Cr y)/(2 Mr)) - (Mr z)/4)
    + E^(Wr - (1/2) Mr ((2 Mr)/y + (Cr y)/(2 Mr))));

rule1 := (E^(Wr - (Mr*((2*Mr)/y___ + (y___*Cr)/(2*Mr)))/2)) | 
    (E^(Wr - (Gr*((Yr*Mr)/x___ + 
            x___*Cr)) - (Mr*((2*Mr)/y___ + (y___*Cr)/(2*Mr)))/
         2)) | (E^(Wr - (z___*Mr)/
         4 - (Mr*((2*Mr)/y___ + (y___*Cr)/(2*Mr)))/
         2)) | (E^(Wr - (z___*Mr)/
         4 - (Gr*((Yr*Mr)/x___ + 
            x___*Cr)) - (Mr*((2*Mr)/y___ + (y___*Cr)/(2*Mr)))/2)) :> 
   Ex[-(x + y + z)/4*Er^2]*(-(Mr^2/x) - Mr^2/y - Cr/(x + y + z));

And then you get:

term1 /. rule1

enter image description here

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  • $\begingroup$ Dear @DanielHuber Thank you very much for your valuable comment. it solved my problem $\endgroup$
    – asal
    Nov 13, 2021 at 8:09

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