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}))
Please show each step!
Sorry for the typo! Added = sign after A
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}))
Please show each step!
Sorry for the typo! Added = sign after A
Now that you have the equal sign, your equation is
$\displaystyle A = P\frac{(1 + i)^n - 1}{(1 - i) + (1 + i)^n}$
The first, obvious step is to divide both sides by P:
$\displaystyle \frac{(1+ i)^n- 1}{(1- i)+ (1+ i)^n}= \frac{A}{P}$
Now multiply both sides by that denominator, $\displaystyle (1- i)+ (1+ i)^n$:
$\displaystyle (1+ i)^n- 1= \frac{A}{P}((1- i)+ (1+ i)^n)$.
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 $\displaystyle x^n- 1= \frac{A}{P}(x^n+ 2- x)$ and then
$\displaystyle \left(\frac{A}{P}- 1\right)x^n - x+ 3= 0$.
That is an nth degree polynomial equation for x. Unfortunately, there is no general method for solving nth degree polynomial equations.