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Comment: How to go about this question has been a Herculean task for me.

I would be glad if an explanation is provided alongside with the calculation as regards making $ r $ the subject of formula in the equation: $$ pv = \frac{1}{r} - \frac{1}{r(r + 1)^t} $$

How far I have gone solving the problem: At first, I did take the LCM of the denominators at the RHS of the equation. I solved the problem to the point where I got stuck at

$$ (r + 1)^t (1 - pvr) = 1 $$

Nzewi Ernest Kenechukwu
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  • What have you tried? – Varun Vejalla Jul 27 '19 at 14:54
  • At first, I did take the LCM of the denominators at the RHS of the equation. I solved the problem to the point where I got stuck at $$ (r + 1)^t (1 - pvr) = 1 $$ – Nzewi Ernest Kenechukwu Jul 27 '19 at 15:06
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    You should show your work in your question rather than in the comments, where it might be missed. To do so, click on the [edit] button at the bottom of your post. – N. F. Taussig Jul 27 '19 at 15:12
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    For general $t$, there is no way to solve this equation analytically. You must resort to numerical methods. –  Jul 27 '19 at 15:31
  • @YvesDaoust - Thanks! Although, I am not pretty good in numerical analysis, but that hint alone proves why I couldn't solve it analytically like before. Still, thanks. – Nzewi Ernest Kenechukwu Jul 27 '19 at 15:34
  • Is $r$ supposed to be small (such as interest rate) ? – Claude Leibovici Jul 27 '19 at 16:00
  • I suppose the subject of the formula $ \frac{1}{r} - \frac{1}{r(r + 1)^t} $ is r. As you already know what the misnomerd subject is suppose to be, the problem degenerates into translating that neojargon into coherence. What is the solution for r in the equation $ pv = \frac{1}{r} - \frac{1}{r(r + 1)^t} $? – William Elliot Jul 28 '19 at 00:41

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As told in comments, you cannot make $r$ the subject and some numerical method is required for finding the zero of function (for simplicity I changed notations : $p$ stands for $pv$) $$f(r)=(r + 1)^t (1 - p\,r) - 1$$ The simplest to use is Newton method which, starting from an initial guess $r_0$ will update it according to $$r_{n+1}=r_n-\frac{f(r_n)}{f'(r_n)}$$

However, since $r$ is small $(r \ll 1)$, we can make approximations using series expansions around $r=\frac 1p$. This would give $$f(r)=-1-p \left(r-\frac{1}{p}\right) \left(\frac{1}{p}+1\right)^t-p t \left(r-\frac{1}{p}\right)^2 \left(\frac{1}{p}+1\right)^{t-1}+O\left(\left(r-\frac{1}{p}\right)^3\right)\tag1$$ Ignoring the $\left(r-\frac{1}{p}\right)^2$ term and higher powers of $r$, this would give an estimate $$r_0=\frac1p-\frac{\left(\frac{1}{p}+1\right)^{-t}}{p}$$ For illustation puposes, let us try using $r=0.005$ and $t=180$; this would give $p=118.504$; reusing these numbers would lead to $r_0=0.00657919$ which is much too high. However, Newton iterations will be $$\left( \begin{array}{cc} n & r_n \\ 0 & 0.006579187739 \\ 1 & 0.005481598716 \\ 2 & 0.005061970447 \\ 3 & 0.005001209158 \\ 4 & 0.005000000473 \\ 5 & 0.005000000000 \end{array} \right)$$

Claude Leibovici
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