mippes

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16 years, 288 days

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Thanks for the answer! However do you know what the 'signum(equ)infinity' means. As far as I understood it only mentions that the solution is possitive. Right? So this is not a solution. Miriam
Thanks for the answer! However do you know what the 'signum(equ)infinity' means. As far as I understood it only mentions that the solution is possitive. Right? So this is not a solution. Miriam
You are right R0 is not a function of P_m. Want I meant was the function which I integrate is a function of only P_m. Sorry for this. By re-reading my post I see that I also made another mistake. q,r,u,v and Di_d are indeed constants, but kappa is not a constant. This is a function of also P_m. The values the constants are: > q:=0.020; q := 0.020 > r:=0.55; r := 0.55 > u:=0.20; u := 0.20 > v:=0.24; v := 0.24 > Di_d:=5; Di_d := 5 kappa is defined by kappa:=P_m/5; Thanks, Miriam
You are right R0 is not a function of P_m. Want I meant was the function which I integrate is a function of only P_m. Sorry for this. By re-reading my post I see that I also made another mistake. q,r,u,v and Di_d are indeed constants, but kappa is not a constant. This is a function of also P_m. The values the constants are: > q:=0.020; q := 0.020 > r:=0.55; r := 0.55 > u:=0.20; u := 0.20 > v:=0.24; v := 0.24 > Di_d:=5; Di_d := 5 kappa is defined by kappa:=P_m/5; Thanks, Miriam
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