Isolate Variable in Fraction

Question

I can approximate u with a calculator by guessing or using excel but I want to isolate it.

$100 = \dfrac{1 + \dfrac{1}{(1+u)^6}}{u}$

Can not seem to do it by hand myself. Is it possible using only simple algebra?

Answer 1

The solutions that you are looking for are the roots of a degree 7. Polynomial. $$100u(1+u)^6-(1+u)^6-1 = 0$$ There is no shorthand formula for finding the roots of a high degree polynomial. Numerical approximation is probably the way to go, or use a graphing calculator and read the roots off.

Answer 2

This looks very much as a finance problem.

Let us rewrite it as $$\frac{u}{1+\frac{1}{(1+u)^6}}=a$$ Develop the lhs as a Taylor series to get $$\text{lhs}=\frac{1}{2}u+\frac{3 }{2}u^2-\frac{3 }{4}u^3-4 u^4+\frac{51 }{8}u^5+O\left(u^6\right)$$ and use series reversion to get $$u=2 a-12 a^2+156 a^3-2392 a^4+40560 a^5+O\left(a^6\right)$$ Making $a=\frac 1 {100}$ would give $$u=\frac{2367017}{125000000}=0.0189361$$ while the exact solution is $\approx 0.0189355$

Edit

We could make the problem more general considering $$\frac{u}{1+\frac{1}{(1+u)^n}}=a$$ $$\text{lhs}=\frac{1}{2}u+\frac{n }{4}u^2-\frac{n }{8}u^3-\frac{n \left(n^2-4\right)}{48} u^4+\frac{n \left(n^2-2\right)}{32} u^5+O\left(u^6\right)$$ from which $$u=2 a-2 na^2+2 n (2 n+1)a^3-\frac{2}{3} n (2 n+1) (7 n+4) a^4+$$ $$2 n (2 n+1)^2 (3 n+2)a^5+O\left(a^6\right)$$