Question: How to find "x" by substituting values in the equation

January 23 2014 Syeda 25
0

I have two matrices.How to find matrix "x". Equation of x is given. I can evaluate the value of 'x' at a point. 

 

z[1] = [0.3042939e-2, .135433537, .263300507, .376372832, .474749796, .556813639, .593373102, 0.3250283e-2, .134445703, .260238867]:

z[2] = [.997953273, .988583777, .96233161, .923889683, .877336956, .827642812, .801825939, .994477871, .978659033, .947179739]:
 

eq1 := x = 1.782005573*z[1]^2+2.483310992*z[1]+1.360919422-3.123062564*10^(-9)*sqrt(3.174428983*10^17*z[1]^4+8.847440780*10^17*z[1]^3+1.461867487*10^18*z[1]^2+1.178106706*10^18*z[1]+1.600992583*10^18*z[2]-1.484747647*10^18)

x = 1.782005573*z[1]^2+2.483310992*z[1]+1.360919422-0.3123062564e-8*(0.3174428983e18*z[1]^4+0.8847440780e18*z[1]^3+0.1461867487e19*z[1]^2+0.1178106706e19*z[1]+0.1600992583e19*z[2]-0.1484747647e19)^(1/2)

(1)

 


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