Hello all,
x = [(a-b(1+i)^-n)*y] / [(1-(1+i)^(n-y))/i)+y].
i need to get the reverse formulae for a,b,i,n and y. i need this information urgently. please help.
Thanks in Advance
Priyan
a and b are straightforward.$\displaystyle
a = x(\frac {1-(1+i)^{n-y}}{i}+y)+b(1+i)^{-n}$
$\displaystyle
b = (a - x(\frac{1-(1+i)^{n-y}}{i}+y))(1+i)^n$
It does not appear to be possible to solve explicitly for i and y. I have not proved this, but I cannot see any way to do it.
n can be solved for by first forming a quadratic in $\displaystyle (1+i)^n$:
$\displaystyle
x(\frac {1-(1+i)^{-y}(1+i)^n}{i}+y)-a+b(1+i)^{-n} = 0$
$\displaystyle
\frac {x}{i(1+i)^y}(1+i)^n+(\frac {x}{i}+xy-a)+b(1+i)^{-n} = 0$
$\displaystyle
(1+i)^n = \frac{a-xy-\frac{x}{i}\pm \sqrt{(\frac {x}{i} +xy-a)^2+4\frac {x}{i(1+i)^y}b}}{2\frac{x}{i(1+i)^y}}$
$\displaystyle
n = \frac {\log (\frac{a-xy-\frac{x}{i}\pm \sqrt{(\frac {x}{i} +xy-a)^2+4\frac {x}{i(1+i)^y}b}}{2\frac{x}{i(1+i)^y}})}{\log (1+i)}$