Manually solving the $\frac{1-(1+x)^{-10}}{x} = 10$?

Question

I am studying financial mathematics and I see the following expression.

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

is there any way of evaluating the value of $x$ manually? without using calculator

Answer 1

From $$ \frac{1-\frac1{(x+1)^{10}}}{x}=10 \tag1 $$ by setting $X=\dfrac1{x+1}$ one gets the equation $$ X \cdot\frac{1-X^{10}}{1-X}=10 $$

or$$ X^{10}+X^9+X^8+\cdots+X-10=0 \tag2 $$ an equation of tenth degree which doesn't seem to admit an elementary solution in terms of radicals of rational numbers.

A search for a numerical solution using any CAS is a powerful tool, here one finds $$ x=-1.74705407486515\cdots $$ as the only real solution of $(1)$

Answer 2

There's no nice way. If you don't mind scratching out calculations until your fingers bleed, you can use Newton's method (which is calculus based) or secant line method (not calculus based.) For secant line method, you make two guesses $x_1$ and $x_2$ of the solution (and they can be pretty bad guesses) and plug them into the function $$f(x) = \frac{1-(1+x)^{-10}}{x} - 10 $$

to get two $y$ values. Now you have two points $(x_1,y_1)$ and $(x_2,y_2)$. Find the equation of the straight line through those two points. Call the $x$-intercept of that line $x_3$. Throw away $x_1$ and repeat the process with $x_2$ and $x_3$ to produce $x_4$. The sequence $x_1, x_2, x_3,\ldots$ should get ever closer to the real solution.

If I were doing this, I would rewrite in the form given in the comments by El Bazzi, for something more stable (no pesky vertical asymptotes.)