# How to solve in terms of k the following equation? [closed]

How to solve in terms of k the following:

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

Any help highly appreciated

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

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

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 $$0.129497$$.

• Whoa. Those are some big numbers :D Jun 22, 2020 at 13:19
• @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 :-) 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.

• That is not a biquadratic equation as it has terms of degree $\;1.\.3\;$ . That's just a quartic equation. Jun 22, 2020 at 11:53
• @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 Jun 22, 2020 at 13:39