How to solve in terms of k the following:

\begin{equation} 200=20 \times \frac{1-(1+k)^{-4}}{k}+\frac{228.59}{(1+k)^4} \end{equation}

Any help highly appreciated

  • $\begingroup$ I doubt there is a great closed formula. Numerical methods work, of course. There's a solution near $k\sim .1295$ for instance. $\endgroup$
    – lulu
    Jun 22, 2020 at 10:45
  • $\begingroup$ What have you tried? Avoid no-clue question: How to ask a good question. $\endgroup$
    – Ѕааԁ
    Jun 22, 2020 at 10:47
  • $\begingroup$ @Saad I don't know how to start... $\endgroup$
    – Gina
    Jun 22, 2020 at 10:49
  • $\begingroup$ Does $228,59$ stand for $228.59$? $\endgroup$
    – bjcolby15
    Jun 22, 2020 at 10:52
  • $\begingroup$ Yes I edited the original post $\endgroup$
    – Gina
    Jun 22, 2020 at 10:54

2 Answers 2


Consider $$\begin{equation} y=20 \times \frac{1-(1+k)^{-4}}{k}+\frac{22859}{100(1+k)^4} \end{equation}$$

Use Taylor series around $k=0$ gives $$y=\frac{30859}{100}-\frac{27859 }{25}k+\frac{26859 }{10}k^2-\frac{26359 }{5}k^3+\frac{182413}{20}k^4+O\left(k^5\right)$$ Using series reversion leads to $$k=-t+\frac{134295}{55718}t^2-\frac{10691793215 }{1552247762}t^3+\frac{3664144204520905 }{172976281606232}t^4+O\left(t^5\right)$$ where $t=\frac{25 }{27859}\left(y-\frac{30859}{100}\right)$.

Making $y=200$ leads to $$k=\frac{3430730696218578011844685765476873}{26674028799972940581873746705242112}\approx 0.1286$$ while the exact solution, obtained using Ferrari's method for quartics, is $0.1295$

Another thing which could be done is to approximate $y$ by a $[2,2]$ Padé approximant which would be $$y\sim\frac {\frac{30859}{100}-\frac{12973524100179}{66968788100}k+\frac{16134307928979 }{133937576200} k^2} {1+\frac{1997920043 }{669687881}k+\frac{3294721405 }{1339375762}k^2}$$ Solving the quadratic for $y=200$ would give $k\approx 0.129504$ to be compared to $0.129497$.

  • $\begingroup$ Whoa. Those are some big numbers :D $\endgroup$
    – K.defaoite
    Jun 22, 2020 at 13:19
  • 2
    $\begingroup$ @K.defaoite. What do you want to do if numerical methods are not used ? This looks like a finance problem where you need to find the interest rate. Cheers :-) $\endgroup$ Jun 22, 2020 at 13:25

Transform $1+k \rightarrow z$, which gives

$$a=c\frac{1-z^{-4}}{z-1}+\frac{d}{z^4}$$ where $a=200,c=20,d=228.59$, since these constants do not have any apparent bearing to the problem.

Then, multiplying $z^4$ throughout, we have $$az^4 = c\dfrac{z^4-1}{z-1} + d \implies az^4-cz^3-cz^2-cz-(c+d)=0$$

which being a quartic equation, we actually know how to find it's roots by Ferrari's method.

The roots you find will be the values of $z=1+k$, so remember to subtract $1$ to get your answer.

  • 2
    $\begingroup$ That is not a biquadratic equation as it has terms of degree $\;1.\.3\;$ . That's just a quartic equation. $\endgroup$
    – DonAntonio
    Jun 22, 2020 at 11:53
  • 1
    $\begingroup$ @DonAntonio I apologize, my bad. I meant a degree 4 equation only, I wasn't aware of the difference between the terminologies of quartic and biquadratic. Thank you for the correction, I learnt this today. :D $\endgroup$ Jun 22, 2020 at 13:39

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