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Define $P_n$ as follows: $P_n(a,b,c) = \frac {a^n}{(a-b)(a-c)}+\frac {b^n}{(b-a)(b-c)}+\frac {c^n}{(c-a)(c-b)}$ with $n \in N$.

I know that $P_2(a,b,c)=1,$ and $ P_3(a,b,c)=a+b+c$.

How can I find $P_3, P_4, P_5$?

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    I did a major edit in accordance with the discussion following my tentative answer - comments already made refer to the unedited version.2012-10-28

1 Answers 1

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This may be barking up the wrong tree, but is offered as a potential hint because I found it interesting. Take $P_n(a,b,c) = \frac {a^n}{(a-b)(a-c)}+\frac {b^n}{(b-a)(b-c)}+\frac {c^n}{(c-a)(c-b)}$ We must clearly have $a,b,c$ distinct.

Now make some observations.

$P_n$ is symmetric, so is expressible as a rational function of elementary symmetric polynomials.

If we do the natural thing, and choose a common denominator of $\pm (a-b)(b-c)(c-a)$ this is antisymmetric, and so each factor divides the numerator (eg swapping $a$ and $b$ must change the sign of the numerator too, if the whole fraction is to be symmetric, which makes $(a-b)$ a factor of the numerator).

Because we have multiplied each term through by a linear factor, and then divided through by three linear factors, we find that our final symmetric polynomial is homogeneous of degree (n-2).

Observe that in multiplying through to get a common denominator we get$\frac{a^n(c-b)+b^n(a-c)+c^n(b-a)}{(a-b)(b-c)(c-a)}$ and all the terms in the numerator involve just two of $a,b,c$, so there will be no terms in the final polynomial which involve all of $a,b,c$.

It remains to organise the divisions and to identify which terms do emerge in the specific cases in the question. So, for example, gather terms as if you have a polynomial in $a$ and divide through by $(b-c)$. Then take out a quadratic factor $(a-b)(c-a)$ with care over signs.

Now let's look at $P_4$ as requested in the comments and see how to do that one. We can write the numerator as $a^4(c-b)+a(b^4-c^4)+bc(c^3-b^3)=$$(c-b)\left[a^4-a(b^3+b^2c+bc^2+c^3)+bc(c^2+bc+b^2)\right]$

Now we know we can take a factor $(a-b)(a-c)=a^2-(b+c)a+bc$ out of the second factor, and considering the signs we can just cancel this with the denominator. The factorisation is easy, because we get another quadratic factor, and we know the sum and product of the roots. So we get $P_4(a,b,c) = a^2+(b+c)a+(b^2+bc+c^2) = a^2+b^2+c^2+ab+bc+ca$

Note also that $P_0=0$

$P_5$ is more tricky computationally using this method, but there is no additional twist.

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    may be $P_4=a^2+b^2+c^2$ $P_5=a^3+b^3+c^3$, just a guess2012-10-28