# Thread: Transpose This Equation To Solve For i

1. ## Transpose This Equation To Solve For i

Can anyone transpose this equation to solve for i as a function of A, P, n ie, i = .....

A = P(((1 + i)n - 1)/( (1 - I) + (1 + i)n))

Sorry for the typo! Added = sign after A

2. ## Re: Transpose This Equation To Solve For i

Originally Posted by SGS
Can anyone transpose this equation to solve for i as a function of A, P, n ie, i = .....
A - P(((1 + i)n - 1)/( (1 - I) + (1 + i)n))
There is no = sign posted. So no equation to solve.

3. ## Re: Transpose This Equation To Solve For i

Now that you have the equal sign, your equation is
$A = P\frac{(1 + i)^n - 1}{(1 - i) + (1 + i)^n}$

The first, obvious step is to divide both sides by P:
$\frac{(1+ i)^n- 1}{(1- i)+ (1+ i)^n}= \frac{A}{P}$

Now multiply both sides by that denominator, $(1- i)+ (1+ i)^n$:
$(1+ i)^n- 1= \frac{A}{P}((1- i)+ (1+ i)^n)$.

Now, rather than expand that "1+ i" to the nth power, I would let x= 1+ i. Then i= x- 1 so 1- i= 1- x+ 1= 2- x so that can be written as $x^n- 1= \frac{A}{P}(x^n+ 2- x)$ and then
$\left(\frac{A}{P}- 1\right)x^n - x+ 3= 0$.

That is an nth degree polynomial equation for x. Unfortunately, there is no general method for solving nth degree polynomial equations.