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#1 h Apr 20, 2025, 9:46 AM

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If $p(x)$ is polynomial with minimum degree such that $p(x)=\frac{x}{x^2+3x+2}$ for $x=0,1,2,...,10$ , find $p(-1)$

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#2 h Apr 20, 2025, 11:46 AM

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Pi-rate_91 wrote:

If $p(x)$ is polynomial with minimum degree such that $p(x)=\frac{x}{x^2+3x+2}$ for $x=0,1,2,...,10$ , find $p(-1)$

So $(x+1)(x+2)P(x)-x=Q(x)\prod_{k=0}^{10}(x-k)$ for some $Q(x)$ with minimal degree

Setting there $x=-1$ and $x=-2$ , we get $Q(-1)=\frac{-1}{11!}$ and $Q(-2)=\frac{-2}{12!}$

Which easily gives minimal degree $Q(x)=-\frac{10x+22}{12!}$

And so $P(x)=\frac{12!x-(10x+22)\prod_{k=0}^{10}(x-k)}{(x+1)(x+2)}$ $\forall x\notin\{-2,-1\}$ , extended with continuity at $-2,-1$

And we are looking for $\lim_{x\to-1}\frac{12!x-(10x+22)\prod_{k=0}^{10}(x-k)}{(x+1)(x+2)}$

Which is easily got with L'Hôpital's rule :
Derivative of denominator is $12!(2x+3)$ and its value at $-1$ is $12!$
Derivative of numerator is $12!-10\prod_{k=0}^{10}(x-k)-(10x+22)\prod_{k=0}^{10}(x-k)\sum_{k=0}^{10}\frac 1{x-k}$
And its value at $-1$ is $12!+10.11!-12!\sum_{k=1}^{11}\frac 1k$

And so $\boxed{P(-1)=1+\frac{10}{12}-\sum_{k=1}^{11}\frac 1k=-\frac{32891}{27720}}$

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