0
$\begingroup$

I was trying to solve the set of equations $a+b+c=1, a^2+b^2+c^2=2, a^3+b^3+c^3=3$ for $a^n+b^n+c^n=x$ I took $x=f(n)$ and found that $f(n)=f(n-1)+f(n-2)/2+f(n-3)/6$ when $f(1)=1, f(2)=2, f(3)=3$ I then went on to try to find an explicit formula but I don’t know how to get started, can anyone please help?

$\endgroup$
3
  • 1
    $\begingroup$ It might helps you. math.stackexchange.com/questions/3521010/… $\endgroup$ – dust05 Jun 3 '20 at 6:04
  • $\begingroup$ Actually I understood that but I quite couldn’t relate it much with my question, I would be glad if you can explain me further $\endgroup$ – Asv Jun 3 '20 at 7:14
  • $\begingroup$ I think I misunderstood your point. $\endgroup$ – dust05 Jun 3 '20 at 9:48
1
$\begingroup$

What do you want to find? the values of $a, b,c $ or the general equation for $f_n$? I'm not sure your requirement, but two questions are related.

You have explicit formula, $f(n) = a^n + b^n + c^n$. So it is enough to find each values for $a, b, c$.

Let $a+b+c = A$, $ab + bc + ca = B$, $abc = C$. Then $a, b, c$ are three roots of $x^3 - A x^2 +Bx - C = 0$.

We have $A = f_1 = 1$. Note that $f_1^2 = f_2 + 2B$, i.e. $B = -1/2$. Also one can deduce that \begin{align*} f_1^3 &= a^3 + 3 a^2 b + 3 a^2 c + 3 a b^2 + 6 a b c + 3 a c^2 + b^3 + 3 b^2 c + 3 b c^2 + c^3\\ & = f_3 + 3ab(a+b) + 3bc(b+c) + 3ca(c+a) + 6C \\ & = f_3 + 3ab(A - c) + 3bc(A-a) + 3ca(A-b) + 6C \\ &= f_3 + 3AB - 9 C + 6C\\ 1& = 3 -3/2 - 3C \\ C& = 1/6 \end{align*} so $a, b, c$ satisfies $x^3 - x^2 - x/2 -1/6 =0$. Actually, this is exactly the characteristic polynomial of the recurrence formula you found. It dose not have rational solutions, so let three solutions be $\alpha, \beta, \gamma$, then we have $$ f(n) = \alpha^n+ \beta^n + \gamma^n$$

Numerically, $\alpha \approx 1.4308$, $\beta \approx -0.21542 - 0.26471 i$, $\gamma =\overline{\beta}$.


I'm not sure if i understood your question. Are you interested in the following general situation? :

$x_n = a_1^n + \cdots + a_k^n$, $x_1 = 1, x_2 = 2, \cdots, x_k = k$

In this case, if you proceed as above, this would be helpful.

$\endgroup$
1
  • $\begingroup$ Thanks a lot, actually I wasn’t able to see that a, b and c would be the roots of a cubic $\endgroup$ – Asv Jun 3 '20 at 10:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.