Find x,y for
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1.5 = [1/(1+x)^3 + 1/(1+y)^3]/2.1
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0.18 = .5*ln(x/y)
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If so, please send help to Mary.J.Nordstrom@chi.frb.org
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Hello Mary,
here is one way to solve your problem using the Newton-Raphson method.
First you can write your first equation as a function of y by
substituting
x = ye^0.36 from equation 2. Hence equation 1 can be written as
f(y) = 1/(1+4.33329y)^3 + 1/(1+y)^3 - 3.15
the derivative of f(y) = -4.299987/(1+4.33329y)^4 - 3/(1+y)^4
You can now implement the Newton-Raphson method in S-plus as follows:
f1<-function(y){(1/(1+1.433329*y)^3 + 1/(1+y)^3)-3.15}
f2<-function(y){-4.299987/(1+1.433329*y)^4 - 3/(1+y)^4}
lasty<-1000
y<--.001
while(abs(y-lasty)>0.00000001){
lasty <- y
y<-y-f1(y)/f2(y)
}
> y
[1] -0.1143195
> x<-y*1.433329414
> x
[1] -0.1638576
Hope this helps.
Albert Essiam