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As the title says, I want to know how to solve this equation. Just cannot find a way

I want to know how much % profit someone has to make EVERY MONTH in average from a stock within $4$ months, to double the money. In other words: I have to make r % profit on every single month of the 4 month to double (r=100%) the money.

Johnny
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    What is $r^5 -1 $ ? –  Oct 11 '17 at 15:08
  • Anyway you need to solve a polynomial that has degree more than $2$. –  Oct 11 '17 at 15:11
  • Clearly, $2<r<3$. So $r\simeq 2.84922$. – Dietrich Burde Oct 11 '17 at 15:12
  • thanks guys. for me as a math noob, can you show me the exact way? I have also changed the description of my question to what my goal is with this equation. – Johnny Oct 11 '17 at 15:15
  • You can factor, but then to find the actual solutions you must use newtons method – K Split X Oct 11 '17 at 15:17
  • You need to reformulate the whole question, including the title and the tag. Are we discussing the profit, or are we discussing the equation? This may not be the same. – Dietrich Burde Oct 11 '17 at 15:18
  • @Johnny Your equation has a very ugly real root. If you want I can explain how to solve a fourth degree polynomial equation in the general case with another example of polynomial, which has human's roots. – Michael Rozenberg Oct 11 '17 at 15:37
  • Add 1 to both side, left side is a geometric progression , so we have: (r^5 - 1)/(r - 1) = 101 ; so we have r(101 - r^4) = 101, r = 2 gives 170 for lef side, r=3 gives 60 for left side, so 2 < r < 3. By trial and error you can find r approximately. – sirous Oct 11 '17 at 15:47
  • I don't think that equation does what you want. – Thomas Andrews Oct 11 '17 at 15:49
  • If you started with one dollar, then after a month you'd have $1+r$ dollars, and then $(1+r)^2$ dollars. So you want $(1+r)^4=2$, ro $r=\sqrt[4]{2}-1$. – Thomas Andrews Oct 11 '17 at 15:50

2 Answers2

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Why would you need to know something like this: $$ -1/4-1/12\,\sqrt {3}\sqrt {{\frac {4\, \left( 2260+6\,\sqrt {48382278} \right) ^{2/3}-5\,\sqrt [3]{2260+6\,\sqrt {48382278}}-4808}{\sqrt [3] {2260+6\,\sqrt {48382278}}}}}+1/12\,\sqrt {6}\sqrt {{\frac {-2\,\sqrt {{\frac {4\, \left( 2260+6\,\sqrt {48382278} \right) ^{2/3}-5\,\sqrt [ 3]{2260+6\,\sqrt {48382278}}-4808}{\sqrt [3]{2260+6\,\sqrt {48382278}} }}} \left( 2260+6\,\sqrt {48382278} \right) ^{2/3}+15\,\sqrt {3}\sqrt [3]{2260+6\,\sqrt {48382278}}-5\,\sqrt [3]{2260+6\,\sqrt {48382278}} \sqrt {{\frac {4\, \left( 2260+6\,\sqrt {48382278} \right) ^{2/3}-5\, \sqrt [3]{2260+6\,\sqrt {48382278}}-4808}{\sqrt [3]{2260+6\,\sqrt { 48382278}}}}}+2404\,\sqrt {{\frac {4\, \left( 2260+6\,\sqrt {48382278} \right) ^{2/3}-5\,\sqrt [3]{2260+6\,\sqrt {48382278}}-4808}{\sqrt [3] {2260+6\,\sqrt {48382278}}}}}}{\sqrt [3]{2260+6\,\sqrt {48382278}} \sqrt {{\frac {4\, \left( 2260+6\,\sqrt {48382278} \right) ^{2/3}-5\, \sqrt [3]{2260+6\,\sqrt {48382278}}-4808}{\sqrt [3]{2260+6\,\sqrt { 48382278}}}}}}}} $$

when numerical solution will give you $2.849217207$ approximately.

Edit:

But in all seriousness you must factor and then use alternative methods such a Newton Raphson to solve for the zeroes. Otherwise its virtually impossible by just hand

K Split X
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GEdgar
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    I lol'd so hard – K Split X Oct 11 '17 at 15:17
  • Thanks a lot for your reply. So generally, is the way I want to go the right way to solve the problem as I've written in the description? "I want to know how much % profit someone has to make EVERY MONTH in average from a stock within 44 months, to double the money." In other words: I have to make r % profit on every single month of the 4 month to double (r=100%) the money. – Johnny Oct 11 '17 at 15:22
  • No, you are not going the right way. Because of compounding, the profit after four months is $(1+r)^4-1$, not $r+r^2+r^3+r^4$. To find the value of r that creates 100% profit after 4 months you need to solve $(1+r)^4-1=1$, which gives a value for r of about 19%. – gandalf61 Oct 11 '17 at 15:47
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The underlying problem is that you've got the wrong equation.

If you make a profit of $r$% after each month, and started with $n$, then after one month, you'd have $n\left(1+\frac{r}{100}\right)$, after two months, $n\left(1+\frac{r}{100}\right)^2$, and in general, after $m$ months:

$$n\left(1+\frac{r}{100}\right)^m$$

When $m=4,$ you want to have $2n$, so you want:

$$\left(1+\frac{r}{100}\right)^4=2$$

which yields:

$$r=100(\sqrt[4]2-1)\approx 18.92$$

Thomas Andrews
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