Suppose that $P(x)$ is a polynomial of degree $n$ such that $P(k)=\dfrac{k^{2}}{k^{3}+1}$ for $k=0,1,\ldots,n$. Find the value of $P(n+1)$
I tried by making a $f(x) = (x^3+1) P(x) - x^2$. But this equation will have $n+3$ roots, out of which $n+1$ will be $0,1,2,...,n$. What about the other two roots? Can we simplify $x^2 \over x^3+1$ more to help?
If $p(x)$ is a polynomial with degree $\leq n$ and $\delta$ is the forward difference operator, $(\delta p)(x)=p(x+1)-p(x)$, we have that $$ \left(\delta^{n+1}p\right)(x) = \sum_{k=0}^{n+1}\binom{n+1}{k}(-1)^k\,p(x+k) = 0\tag{1} $$ hence in our case $$ P(n+1)=(-1)^n\sum_{k=0}^{n}\binom{n+1}{k}(-1)^k \frac{k^2}{k^3+1}.\tag{2} $$
Addendum. Can we compute a closed form for the RHS of (2)? With a bit of patience, yes.
Let $\omega=\exp\left(\frac{\pi i}{3}\right)=\frac{1+i\sqrt{3}}{2}$. By partial fraction decomposition we have
$$ \frac{k^2}{k+1} = \frac{1}{3}\left(\frac{1}{k+1}+\frac{1}{k-\omega}+\frac{1}{k-\overline{\omega}}\right)\tag{3} $$
and since $\frac{1}{k+1}=\int_{0}^{1}x^k\,dx$,
$$ \sum_{k=0}^{n}\binom{n+1}{k}\frac{(-1)^k}{k+1} = \frac{1+(-1)^n}{n+2}\tag{4} $$
$$ \sum_{k=0}^{n}\binom{n+1}{k}(-1)^k\left(\frac{1}{k-\omega}+\frac{1}{k-\overline{\omega}}\right)\\ = \frac{(-1)^n(2n+1)}{n^2+n+1}+\underbrace{2\,\text{Re}\,B\left(n+2,\frac{-1+i\sqrt{3}}{2}\right)}_{\text{wildly oscillating term.}}\tag{5} $$
Using (Lagrange) polynomial interpolation: \begin{align*} P(x) &= \sum_{k=0}^n \frac{k^2}{k^3+1} \left( \prod_{j \neq k} \frac{x - j}{k-j} \right) \\ &= \sum_{k=0}^n \frac{k^2}{k^3+1} \left( \frac{x (x-1) \cdots (x - (k-1)) (x-(k+1)) \cdots (x-n) }{ k (k-1) \cdots (k-(k-1)) (k-(k+1)) \cdots (k-n)}\right) \\ &= \sum_{k=0}^n \frac{k^2}{k^3+1} \left( \frac{x (x-1) \cdots (x - (k-1)) (x-(k+1)) \cdots (x-n) }{ k! (n-k)!(-1)^{n-k} } \right) \\ \end{align*} Substituting, \begin{align*} P(n+1) &= \sum_{k=0}^n \frac{k^2}{k^3+1} \left( \frac{(n+1) (n) \cdots (n+2-k) (n-k) \cdots (1) }{ k! (n-k)!(-1)^{n-k} } \right) \\ &= \sum_{k=0}^n \frac{k^2}{k^3+1} \left( \frac{(n+1)!}{ k! (n-k)!(-1)^{n-k} (n+1-k)} \right) \\ &= \sum_{k=0}^n (-1)^{k-n} \frac{k^2}{k^3+1} {n+1 \choose k} \\ \end{align*}
Partial answer
Let $P(x)$ be
$$P(x)=\sum_{k=0}^{n}{{{x(x-1)(x-2)...(x-n)}\over{k!(n-k)!(-1)^{(n-k)}(x-k)}}P(k)}$$
It satisfies $P(x) = {k^2 \over k^3 + 1}$ for $k = 0,1,...,n$.
This is gotten from using the term $x(x-1)(x-2)...(x-n) \over x-k$ to zero all the other terms, while dividing by $k!(n-k)!(-1)^{(n-k)}$ makes the fraction $1$.
Then
$$P(n+1)=\sum_{k=0}^{n}{{(n+1)!\over {k!(n-k)!(-1)^{(n-k)}(n+1-k)}}P(k)}$$
$$=\sum_{k=0}^{n}{{(n+1)!\over {k!(n+1-k)!}}(-1)^{(n-k)}P(k)}$$
$$=\sum_{k=0}^{n}{{n+1 \choose k}(-1)^{(n-k)}P(k)}$$
$$=\sum_{k=0}^{n}{{n+1 \choose k}(-1)^{(n-k)}{k^2 \over k^3+1}}$$
Assuming to be working in complex case, we can write $f(x) = cx.(x-1)..(x-n)(x-a_1)(x-a_2) $. Now put the value of $x$ as the roots of $x^3 +1$. you shall get three equations solve them to get your three constants.
\dfracin titles please. – Did Jul 01 '17 at 19:09