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I'd like to know if is there any way to get an approximation for the roots of the equation below by hand.

$$ax^{13}+bx^{12}+c=0.$$

You are allowed to use calculator to calculate powers, logarithms, roots, etc. (for example, $\text{(some number)}^{13}$, $\text{(some number)}^{1/12}$, etc.).

This problem came from the equation

$$5328.31=50000\frac{(1+i)^{13}\cdot i}{(1+i)^{13}-1}\cdot \frac{1}{1+i}$$

from where I have to calculate the value of $i$ (interest rate). If we write $x=1+i$, then the equation becomes

$$-8.3838x^{13}+9.3838x^{12}-1=0.$$

Pedro
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3 Answers3

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Inspired by this question of mine, we can approximate the solution "quite" easily using Padé approximants.

Let the equation be $$\frac{(1+i)^{13}\cdot i}{(1+i)^{13}-1}\cdot \frac{1}{1+i}-r=0$$ Building the simplest $[1,1]$ Padé approximant around $i=0$, we have $$0=\frac{\frac{2}{39} i (26 r+7)+\frac{1}{13} (1-13 r)}{1-\frac{4 i}{3}}$$ Canceling the numerator gives $$i=\frac{3 (13 r-1)}{2 (26 r+7)}$$ So, using your numbers $$i\approx 0.0591605$$ while the exact solution would be $0.0600001$ (I suppose that the true solution is $0.06$).

More difficult would be to build the $[1,2]$ Padé approximant, but it is doable. Canceling the numerator gives $$i=\frac{4 \left(338 r^2+65 r-7\right)}{3 \left(-845 r^2+546 r+35\right)}$$ and, using your numbers $$i\approx 0.0600597$$ which is much better.

Simpler (but less accurate) would be to develop the expression as a Taylor series around $i=0$; this would give $$r=\frac{1}{13}+\frac{6 i}{13}+\frac{8 i^2}{13}+O\left(i^3\right)$$ Using the first two terms would lead to $i\approx 0.0642268$ (we know that this is an overestimate of the solution by Darboux theorem since it corresponds to the first iterate of Newton method using $i_0=0$). Using the three terms implies solving a quadratic for which the positive root is $i\approx 0.0595056$.

Community
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Claude Leibovici
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Iterate:

$$x_{n+1}=\frac {-c}{a(x_n)^{12}}-\frac{b}{a} $$

$$x_0=\frac {-b}{a} $$

I will try to edit my answer and put bounds that indicate the rate of convergence

Amr
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The problem of finding an algebraic formula is closed since longtime ago: it is not possible. Therefore you need an approximation and the particular values of the coefficients are very important in each case of course.

For your equation $8.3838x^{13}-9.3838x^{12}+1=0$ you have the equivalent one $$8.3838(x^{13}-x^{12})=x^{12}-1\iff 8.3838=\frac{x^{12}-1}{x^{13}-x^{12}}=\frac{1-1/x^{12}}{x-1}$$

Consider separately the function $f(x)=\frac{1-1/x^{12}}{x-1}$.

You have f(-1)=0. $f(1.06)\approx 8.384$ and $f(-0.794)\approx 8.34$ and other approximated values. You can see the special value $x=1.06$ gives a "good" approximated root.

Here the graph of $f$

enter image description here

Ataulfo
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    Just to be clear, the theorem that you allude to is that there are no general solutions for polynomials of degree five or more--but the polynomial given here isn't one with arbitrary coefficients, so the theorem doesn't apply and there may be a general solution of an equation of the form $ax^n + bx^{n-1}+c=0$. After all there definitely is a general solution to equations of the form $ax^n + c = 0$. – Addem Nov 27 '15 at 07:31
  • I think there is an irrevocable difference between the two equations $ax^n+bx^{n-1}+c=0$ and $ax^n+c=0$. However the trinomial $x^n+ax+b$ has been nicely studied, in particular it is known its discriminant. – Ataulfo Nov 27 '15 at 09:26
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    Obviously they're different; my point was just that conditions on the coefficients could make the polynomial solvable. – Addem Nov 27 '15 at 16:10
  • My conviction is that there is no chance for that. Reason indirectly: if there is a formula for more or less general cases,it should have already been discovered by great mathematicians. What may occur is when your equation has integer roots so you can factorize – Ataulfo Nov 28 '15 at 12:18
  • With a calculator you can take the value $x=1.06$ I gave you and you can to approximate more what you want. I take it just from the graph without trying to refine the result with a calculator. You can do it if you are interested to find x. – Ataulfo Nov 28 '15 at 12:27
  • Here is a general solution to degree six polynomials of the form $a_6x^6 + ... + a_0$ with the condition on the coefficients that $a_i = 0$ if $i\not\equiv 0 \mod{3}$. Then they have the form $ax^6 + bx^3 + c$ and obviously factor using the quadratic on $ay^2 + by + c$. The cube-root of solutions to this are roots, and then by factoring the corresponding linear factors and solving the resulting quartic polynomial will give all roots. – Addem Nov 28 '15 at 16:18
  • So yeah, conditions on the coefficients might lead to general solutions. I'm just brainstorming fairly trivial conditions, but there might be more subtle conditions that no mathematician has found because they aren't all that interesting to research. – Addem Nov 28 '15 at 16:20
  • From what you say in your last comment is clear to me that you do not know the very vast world of mathematical research. Regards. – Ataulfo Nov 29 '15 at 16:27
  • Wow, you're really taking this personally. I wasn't trying to insult you, but just clarify some math. But fine, have a hurt ego rather than learn anything. – Addem Nov 29 '15 at 16:30
  • Please believe me, I had no intention of insulting. If you visualize it this way, I beg you to excuse me. (Only I wanted to meant that almost everything has already been seen in mathematics with extreme thoroughness). Excuse me please. Regards. – Ataulfo Nov 30 '15 at 18:33