Linear algebra matrix coding to solve specific formula

Question

I have a formula which cannot be expressed in terms of y.

fv = (p (1 + r) (-1 + (1 + r)^m) ((1 + r)^(m y) - (1 + x)^y))/
      (r (-1 + (1 + r)^m - x)) + c (1 + r)^(m y)

I'd like to solve it numerically with some kind of matrix operation.

For example

with

r = 0.05/12
m = 12
p = 1000
x = 0.03
c = 2000
fv = 92988.93

Can this be done with linear algebra, or is an iterative approach necessary?

Wolfram Alpha is not giving any clues for a matrix solution. Maybe some gradient descent program is required.

Answer 1

Linear algebra and matrices are not relevant here. Roughly speaking, they are useful when you have a system of linear equations. Here, you have a single equation, not a system of equations; and it is nonlinear.

If I understand correctly, given $p,m,\dots$, you are trying to find a value of $y$ that satisfies the equation. This equation can be written in the form

$$f(y) = 0$$

where $f$ is a function of $y$ (it also depends on $p,m,\dots$ but those are considered known constants). Here $f(y) = (p(1+r)\cdots) - fv$.

You can find an approximate solution to such an equation using an iterative algorithm, such as Newton's method. See https://en.wikipedia.org/wiki/Root-finding_algorithms.

Answer 2

From what I see, your equation has the form

$$A(1+a)^y+B(1+b)^y+C=0$$ where $a,b$ are small. Such a nonlinear equation cannot be solved analytically.

To find an approximate solution, you can write

$$0=A(1+a)^y+B(1+b)^y+C\approx A(1+ay)+B(1+by)+C$$ and solve for $y$. An even better approximation comes from

$$0=A(1+a)^y+B(1+b)^y+C\\\approx A\left(1+ay+a^2\frac{y(y-1)}2\right)+B\left(1+by+b^2\frac{y(y-1)}2\right)+C$$

which is quadratic in $y$ (by the generalized binomial formula).

Then you can refine with Newton's method. Few iterations will be needed.

$$y\leftarrow y-\frac{A(1+a)^y+B(1+b)^y+C}{A\log(1+a)(1+a)^y+B\log(1+b)(1+b)^y}.$$

Answer 3

Newton-Raphson solution for n

Need to find a root for this formula with one unknown, n

(p ((1 + r)^n - (1 + x)^n))/(r - x) + c (1 + r)^n - fv = 0

Using the algorithm here: https://en.wikipedia.org/wiki/Newton%27s_method#Code

This method requires the derivative of the formula.