Consider the following equation: \(\displaystyle{x}={\frac{{{\left({p}+{x}{\left({1}-{p}\right)}\right)}\cdot{c}}}{{{1}-{\left({p}+{x}{\left({1}-{p}\right)}\right)}{\left({1}-{c}\right)}}}}\)

Addison Fuller

Addison Fuller

Answered question

2022-03-27

Consider the following equation:
x=(p+x(1p))c1(p+x(1p))(1c)

Answer & Explanation

Raiden Griffin

Raiden Griffin

Beginner2022-03-28Added 13 answers

Step 1
x2(c+pcp1)x(c+p2cp1)cp=0
Discriminant is
D=(c+p2cp1)24cp(c+pcp1)
=c2+p2+1+2cp2c2p=(c+p1)2
so
x=(c+p2cp1)±(c+p1)c+pcp1;;c1landp1
denominator
pcp+c1=p(1c)1(1c)=(1c)(p1)
x1=c+p2cp1cp+12(1c)(p1)
=cp(1c)(p1)=cp(1c)(1p)
and
x2=c+p2cp1+c+p1}{2(1c)(p1)}
=2(c1)(p1)2(1c)(p1)=1
if p=1lorc=1 then x=1

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