# Help required in a formula

• Dec 20th 2007, 02:41 AM
tpriyan
Help required in a formula
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.

Priyan
• Dec 20th 2007, 01:25 PM
topsquark
Quote:

Originally Posted by tpriyan
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.

Priyan

What, exactly, are you asking for? You need to solve the above equation for, say, y in terms of the remaining variables? (Then a, then b, then i, then n?)

-Dan
• Dec 20th 2007, 05:05 PM
a and b are straightforward. $

a = x(\frac {1-(1+i)^{n-y}}{i}+y)+b(1+i)^{-n}$

$
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 $(1+i)^n$:
$
x(\frac {1-(1+i)^{-y}(1+i)^n}{i}+y)-a+b(1+i)^{-n} = 0$

$
\frac {x}{i(1+i)^y}(1+i)^n+(\frac {x}{i}+xy-a)+b(1+i)^{-n} = 0$

$
(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}}$

$
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)}$
• Dec 21st 2007, 12:38 AM
tpriyan
Quote:

Originally Posted by topsquark
What, exactly, are you asking for? You need to solve the above equation for, say, y in terms of the remaining variables? (Then a, then b, then i, then n?)

-Dan

Yes that is what i want