Is is possible to manually solve $10x = 1 - (1+x)^{-12}$ with a simple calculator only?

Question

Context: although I'm not a mathematician, I'm a curious software engineer which remember high school Math and had my time with calculus at college.

I'm trying to manually calculate the amortization rate of a loan (so really none of the fields i now something about). given that I've already deduced the amortization formula, i was given a problem like this:

principal = U\$$1000$, monthly payment = U\$$100$, $12$ monthly payments in total. What is the monthly rate of this loan?

I've ended at this: $$10x = 1 - (1+x)^{-12}$$

from all possible roots, there's a single positive real root ~0.0292 that matches the solution (the loan monthly rate should be 2.92% per month, so this is right towards the correct answer).

my problem is: if i had to solve this with pen and paper without any computer to aid me (like wolfram alpha or any online solver), how should i manipulate this to get that $0.0292$ root? is this possible?

Answer 1

It is quite straightforward to set up a simple recursion relation to obtain the unique positive real root. Such a recursion can be efficiently implemented in certain kinds of hand calculators; in particular, the calculator needs to have an Ans ("answer") key, and the ability to input algebraic expressions that are functions of the previous output.

In your case, consider the polynomial $$f(x) = (10x-1)(1+x)^{12} + 1. \tag{1}$$ Its derivative is $$\begin{align} f'(x) &= \frac{d}{dx}[10x-1](1+x)^{12} + (10x-1) \frac{d}{dx}[(1+x)^{12}] \\ &= 10(1+x)^{12} + (10x-1)(12)(1+x)^{11} \\ &= (1+x)^{11}(10+10x+120x-12) \\ &= 2(65x-1)(1+x)^{11}. \tag{2} \end{align}$$ So the Newton's method recursion is $$\begin{align} x_{n+1} &= x_n - \frac{f(x_n)}{f'(x_n)} = x_n - \frac{(10x_n-1)(1+x_n)^{12} + 1}{2(65x_n-1)(1+x_n)^{11}} \\ &= \frac{(1-11x_n+120x_n^2) - (1+x_n)^{-11}}{2(65x_n-1)}. \tag{3}\end{align}$$ Now all that is needed is a suitable initial guess $x_0$. The original equation suggests that if there is no interest, i.e. $x = 0$, then the total paid is $1200$, which exceeds the present value of the loan by $20\%$. So if we assume that interest is not compounded (i.e., simple interest), the monthly rate would be $0.20/12 \approx 0.016667$. This will serve as our choice of $x_0$.

Then to implement this recursion on a calculator, we type in

0.2/12

which outputs $0.016666667$ or something like that. Then we input

((1 - 11 * Ans + 120 * Ans^2) - (1 + Ans)^(-11))/(2 * (65 * Ans - 1))

which will return $0.0975032$. Then you recall the last input (like the Ans button, the name of this button might vary, but it is typically just an upward arrow button), and press = or Ans again, which causes the calculator to apply the previous output as the argument to the previous function you typed. You will get $0.0664072$. If you do this repeatedly--recall the previous input, then compute its value based on the answer you obtained in the previous iteration--then you should get a sequence of numbers like this:

$$\begin{array}{c|c} n & x_n \\ \hline 0 & 0.0166667 \\ 1 & 0.0975032 \\ 2 & 0.0664072 \\ 3 & 0.0460899 \\ 4 & 0.03476 \\ 5 & 0.0301476 \\ 6 & 0.0292615 \\ 7 & 0.0292286 \\ 8 & 0.0292285 \\ 9 & 0.0292285 \\ \end{array}$$ and we stop when the output no longer changes. This represents, to within the display precision of the calculator, the fixed point of the recursion relation, and therefore the desired root of $(1)$.

Answer 2

$$ 10x=1−(1+x)^{−12}\\ 10x=1-\frac{1}{(1+x)^{12}}\\ 10x\cdot(1+x)^{12}=(1+x)^{12}-1\\ (10x-1)\cdot(1+x)^{12}=-1 $$

This eventually simplifies into a polynomial with degree $13$, which is impossible to solve by hand.

Answer 3

Another way to look at the problem that might make the following answer make a little more sense is to instead try to find roots to the polynomial

$$(x+1)^{12}-10x(x+1)^{12}-1.$$

We will need to be careful to omit the value $-1$ from consideration as the function you have given is not defined at $x=-1$.

In general, polynomials of degree $d\ge 5$ are not solvable, i.e. there is no clean way to write out all their roots in terms of addition, subtraction, multiplication, or taking $n^{th}-$roots of rational numbers. While there are specific cases where an algorithm exists, since your given polynomial has degree $13$, I'd be very surprised to see an algorithm to explicitly determine all the roots.

HOWEVER, if you are willing to put in a little bit of elbow grease, Newton's Method for finding roots of polynomials often offers a relatively quick way of finding values very close to being a root. The algorithm is as follows:

1. Let $x_0$ be any value, preferably one that you suspect is close to being a root.

2. Let $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$, and repeat until satisfied.

The general idea of Newton's Method is that, looking at a very small range of inputs, the tangent line of a function at a point is a pretty good estimate of the function. Thus, for a very small range the line $$ y-f(x_0) = f'(x_0)(x-x_0) $$ is a good estimate for our polynomial $f$. A root will be where $y=0$, so we find the root of our tangent line is (after moving terms around) $$ x =x_0 -\frac{f(x_0)}{f'(x_0)} = x_1. $$ As we iterate further through the algorithm, our estimates get more and more refined.

In your particular example, using $$ f=(x+1)^{12}-10x(x+1)^{12}-1 \qquad x_0 = \frac{1}{10} $$ the first five estimates for Newtons Method are $$ x_1 = 0.0681369 \qquad x_2 = 0.0471467 \qquad x_3 = 0.0352813 \qquad x_4 = 0.0303015 \qquad x_5=0.0292729 $$

Hopefully, this helps! Thanks for the awesome question :D !!

Answer 4

If you're OK with a "good enough" estimate and taking a non-standard approach, you can consider the following. First, rearrange the original equation to:

$$(1+x)^{12} = (1-10x)^{-1}$$

and we can take the log of both sides:

$$12 \log(1+x) = - \log (1-10x)$$

Now it's well known that:

$$ \begin{align} \log (1+x) &= x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \\ \log (1-y) &= - y - \frac{y^2}{2} - \frac{y^3}{3} - \frac{y^4}{4} - \cdots \end{align} $$

Substituting $y = 10x$ and taking just the first three terms of each expansion gives us:

$$12 \left(x - \frac{x^2}{2} + \frac{x^3}{3} \right) \approx 10x + \frac{100x^2}{2} + \frac{1000x^3}{3}$$

Dividing through by $x$ (we know $x \ne 0$) and rearranging gets us to:

$$2 \approx \frac{112x}{2} + \frac{988x^2}{3} \implies 494x^2+84x-3 \approx 0$$

which has a positive root at $x \approx 0.03031$, about $4\%$ off the mark.

If you use only the first two terms rather than three, you get the cruder approximation $x=\frac1{28} \approx 0.03571$, which, given it's a linear approximation, at $22\%$ off is still not bad.

When you have an equation that involves exponential growth or compound interest of any sort, there's almost always a trick involving logarithms. Also, don't ever take out a loan at $2.9\%$ monthly interest, even the Mob has better rates.

Answer 5

With only a simple scientific calculator fixed point iteration can be your best friend. We can rewrite the relationship as

$$x = \frac{1}{10}\left(1-(1+x)^{-12}\right) \equiv f(x)$$

which can be solved by taking $f(f(\cdots f(x)\cdots))$ repeatedly until a solution stabilizes for a suitable starting point. For example, one can prove that

$$0<x<1 \implies 0 < f(x) < 0.1$$ so choosing a starting point that isn't $0$ should get you to the solution very quickly. For a good inital starting seed, consider the nonzero solution to the approximate equation

$$10x \approx 12x + 66x^2 - 144x^2 \implies x = \frac{1}{39} \approx 0.0256$$

Then

$$f(0.0256) = 0.0262$$

$$f(0.0262) = 0.0266$$

Etc.