Find the explicit formula for a polynomial

460 Views Asked by At

I have this collection of ten polynomials of the same kind. It is possible, starting from these, to formulate an explicit expression for the n-th polynomial $ P_ {n} $?

$P_{0}=1$

$P_{1}=1+X_{1}$

$P_{2}=1+X_{1}+X_{2}$

$P_{3}=1+X_{1}+X^{2}_{1}+X_{2}+X_{3}$

$P_{4}=1+X_{1}+X^{2}_{1}+X_{2}+X_{1}X_{2}+X_{3}+X_{4}$

$P_{5}=1+X_{1}+X^{2}_{1}+X^{3}_{1}+X_{2}+X_{1}X_{2}+X_{2}^{2}+X_{3}+X_{1}X_{3}+X_{4}+X_{5}$

$P_{6}=1+X_{1}+X^{2}_{1}+X^{3}_{1}+X_{2}+X_{1}X_{2}+X_{1}^{2}X_{2}+X_{2}^{2}+X_{3}+X_{1}X_{3}+X_{2}X_{3}+X_{4}+X_{1}X_{4}+X_{5}+X_{6}$

$P_{7}=1 + X_{1} + X_{1}^2 + X_{1}^3 + X_{1}^4 + X_{2} + X_{1} X_{2}+ X_{1}^2 X_{2}+ X_{2}^2 +X_{1} X_{2}^2 + X_{3} +X_{1} X_{3}+ X_{1}^2 X_{3} + X_{2} X_{3} + X_{3}^2 + X_{4}+ X_{1}X_{4} + X_{2}X_{4} + X_{5} +X_{1}X_{5} +X_{6} + X_{7} $

$P_{8}=1 + X_{1} + X_{1}^2 + X_{1}^3 + X_{1}^4+X_{2} + X_{1} X_{2}+ X_{1}^2 X_{2}+ X_{1}^3 X_{2} + X_{2}^2 + X_{1}X_{2}^2 + X_{2}^3 + X_{3} + X_{1}X_{3} + X_{1}^2 X_{3} + X_{2} X_{3} +X_{1}X_{2} X_{3} + X_{3}^2 +X_{4} + X_{1} X_{4} + X_{1}^2 X_{4} + X_{2}X_{4} + X_{3} X_{4} + X_{5} + X_{1} X_{5} + X_{2} X_{5} +X_{6} + X_{1}X_{6} +X_{7} + X_{8}$

$P_{9}=1 + X_{1} + X_{1}^2 + X_{1}^3 + X_{1}^4 + X_{1}^5 + X_{2} + X_{1} X_{2} + X_{1}^2 X_{2} + X_{1}^3 X_{2} +X_{2}^2 X_{1} X_{2}^2 + X_{1}^2 X_{2}^2 + X_{2}^3 + X_{3}+ X_{1}X_{3} + X_{1}^2 X_{3}+ X_{1}^3 X_{3}+ X_{2} X_{3} + X_{1} X_{2} X_{3} + X_{2}^2 X_{3} + X_{3}^2 + X_{1}X_{3}^2 +X_{4} + X_{1}X_{4} +X_{1}^2 X_{4} + X_{2}X_{4} +X_{1}X_{2}X_{4} +X_{3}X_{4} + X_{4}^2 + X_{5} + X_{1}X_{5} +X_{1}^2 X_{5} + X_{2}X_{5} + X_{3}X_{5} + X_{6} +X_{1} X_{6}+X_{2} X_{6} + X_{7} +X_{1}X_{7} + X_{8} + X_{9}$

1

There are 1 best solutions below

0
On BEST ANSWER

It looks like some combination of the following:

  • The $X_1$ powers increase every other step, so you have something like $$ P_n=\sum_{i=0}^{\left\lfloor\frac{n+1}{2}\right\rfloor}X_1^i+\cdots. $$

  • Something similar happens for $X_2$ (and also for higher powers). This gives $$ P_n=\sum_{j=0}^n\sum_{i=0}^{\left\lfloor\frac{n+1}{j+1}\right\rfloor}X_j^i+\cdots. $$

  • There's a similar structure for $X_1X_j$ starting at $n=4$ and similar for $X_2X_j$. You can come up with similar double sums.

Also, there might be a typo in $P_9$ with the $X_2^2X_1X_2^2$ term as $X_2^2$ appears twice in the expansion.