# Revision history [back]

Un gros MERCI Emmanuel! I got the solution I needed for the 3rd part of this piecewise funtion (it agrees with mathematica) I need the k and the x values that are real values: 50.4900997166127 0.881173769148990

Here is the code. It took a little bit of digging to get the substitutions. It would also be nice to know if there is an option in solve to just return the real values :)

fp2 = sqrt(r^2-(x-c)^2) dfp2 = derivative(fp2, x) fp3 = -k*(x-1)^3 dfp3 = derivative(fp3, x) sys = [fp2 == fp3,dfp2 == dfp3] SS = solve(sys, (x, k), algorithm="sympy") print(SS[0][k].subs(c=0.7,r=0.2)) print(SS[0][x].subs(c=0.7,r=0.2)) print(SS[1][k].subs(c=0.7,r=0.2)) print(SS[1][x].subs(c=0.7,r=0.2))

50.4900997166127 0.881173769148990 1.04977199683675*I 0.368826230851010

​Bien à vous! Pat Browne Rimouski Québec

 2 No.2 Revision Emmanuel Charpentier 6412 ●6 ●44 ●125

Un gros MERCI Emmanuel! I got the solution I needed for the 3rd part of this piecewise funtion (it agrees with mathematica) I need the k and the x values that are real values:

 50.4900997166127
*****50.4900997166127
0.8811737691489900.881173769148990*****


Here is the code. It took a little bit of digging to get the substitutions. It would also be nice to know if there is an option in solve to just return the real values :)

 fp2 = sqrt(r^2-(x-c)^2)
dfp2 = derivative(fp2, x)
fp3 = -k*(x-1)^3
dfp3 = derivative(fp3, x)
sys = [fp2 == fp3,dfp2 == dfp3]
SS = solve(sys, (x, k), algorithm="sympy")
print(SS[0][k].subs(c=0.7,r=0.2))
print(SS[0][x].subs(c=0.7,r=0.2))
print(SS[1][k].subs(c=0.7,r=0.2))
print(SS[1][x].subs(c=0.7,r=0.2))print(SS[1][x].subs(c=0.7,r=0.2))
50.4900997166127
*****50.4900997166127
0.881173769148990 0.881173769148990*****
1.04977199683675*I
0.3688262308510100.368826230851010


​Bien à vous! Pat Browne Rimouski Québec

 3 No.3 Revision Emmanuel Charpentier 6412 ●6 ●44 ●125

Un gros MERCI Emmanuel! I got the solution I needed for the 3rd part of this piecewise funtion (it agrees with mathematica) I need the k and the x values that are real values:

 *****50.4900997166127
0.881173769148990*****


Here is the code. It took a little bit of digging to get the substitutions. It would also be nice to know if there is an option in solve to just return the real values :)

  fp2 = sqrt(r^2-(x-c)^2)
dfp2 = derivative(fp2, x)
fp3 = -k*(x-1)^3
dfp3 = derivative(fp3, x)
sys = [fp2 == fp3,dfp2 == dfp3]
SS = solve(sys, (x, k), algorithm="sympy")
print(SS[0][k].subs(c=0.7,r=0.2))
print(SS[0][x].subs(c=0.7,r=0.2))
print(SS[1][k].subs(c=0.7,r=0.2))
print(SS[1][x].subs(c=0.7,r=0.2))

*****50.4900997166127
*****
50.4900997166127
0.881173769148990*****
0.881173769148990
*****
1.04977199683675*I
0.368826230851010


​Bien à vous! Pat Browne Rimouski Québec