How to solve $c_1x+c_2x^2+...+c_nx^n = K $ type equation (internal rate of return)

Question

I'm trying to calculate internal rate of return and wonder how you would solve this equation

$50x + 20 x^2+75x^3 = 135.6$

The answer is $x = 0.9694$

But does anyone know how to do this step-by-step? Also is there any simple method for calculating $x$ when the equation does not have high exponentials? I.e.

Are the methods for solving

$c_1x+c_2x^2+...+c_nx^n = K $

The same as

$c_1x+c_2x^2+c_3x^3 = K$

Where c and K are constants.

Answer 1

In my opinion, the question in interesting with respect to many items :

• for this kind of problems related to financial mathematics, there is rarely more than one real solution
• typically, the solution corresponds to a rather small number (if $x$ is an interest rate, it is "close" to $0$; if $x$ is a return on investment, it is "close" to $1$ and so $r=1-x$ is close to $0$)
• high accuracy is never required
• analytical solutions are impossible to obtain if the degree of the polynomial is higher than $4$ and, already in the case of degree $3$ the expressions are quite ugly.

So, as LutzL already answered, the more realistic solution is a numerical method and probably Newton would be the simplest (and all calculations can be done with a simple pocket calculator). The only problem is to get easily a reasonable estimate to start the iterations.

To have very good approximations, starting from $0$, we can use either Newton, Halley or Householder methods (these are the simplest) which require respectively first, first and second, first, second and third derivatives. I shall not bother you with the formulas (they are in the indicated Wikipedia pages).

Consider your problem which is related to a return on investment. So, settings $x=1+r$ and using the binomial theorem, a cubic polynomial such as $$c_1x+c_2 x^2+c_3 x^3$$ write $$(c_1+c_2+c_3)+(c_1+2 c_2+3 c_3) r+(c_2+3 c_3) r^2+c_3 r^3$$ Now, consider solving for the root of $$F(r)=(c_1+2 c_2+3 c_3) r+(c_2+3 c_3) r^2+c_3 r^3+(c_1+c_2+c_3-a)=0$$ What we need is the value of the function and its derivatives at $r=0$. So, we have $$F(0)=c_1+c_2+c_3-a$$ $$F'(0)=c_1+2 c_2+3 c_3$$ $$F''(0)=2(c_2+3 c_3)$$ $$F'''(0)=6c_3$$ Applied to the equation you gave, this means $$F(0)=\frac{47}{5} \quad F'(0)=315 \quad F''(0)=490 \quad F'''(0)=450$$ Applying the different methods, we then get $$r_{(2)}=-\frac{47}{1575}\approx -0.0298413$$ $$r_{(3)}=-\frac{13912}{455625}\approx -0.0305339$$ $$r_{(4)}=-\frac{67697698273}{2215260140625}\approx-0.0305597$$ while the exact solution would be $-0.0305609$; this means that, using basic arithmetic operations and no algebra, we have been abale to get the result for five correct figures.

Nevertheless, I suggest you use the first estimate (it was already given by LutzL) and start Newton procedure. The successive iterates would then be $$r_0=-0.0298413$$ $$r_1=-0.0305605$$ $$r_2=-0.0305609$$

The same kind of approach could be used with the same efficiency for any degree of the expression.

Edit

In the case of a problem with interest, suppose the equation to be $$50x + 20 x^2+75x^3 = 6$$ the same method would have given $$x_{(1)}=\frac{3}{25}=0.12$$ $$x_{(2)}=\frac{357}{3125}=0.11424$$ $$x_{(3)}=\frac{87657}{781250}=0.11220096$$ while the solution would be $\approx 0.112763$. And this is not a small interest.

Answer 2

Your function is monotonically increasing and convex (as long as the $c_k$ stay non-negative). Thus you can use any of Newton's method, regula-falsi with Illinois anti-stalling, secant method, etc. with guaranteed and fast convergence.

Additionally, since the interest rate, outside of usury, is a number $r$ close to $0$ you get for $x=1+r$ the equation $$ 0=75r^3 + 245r^2 + 315r + 9.4 $$ which gives a first (and good) approximation $r=-\frac{9.4}{315}=-0.02984…$ with $x=1+r=0.9701587…$