Polynomial extrapolation

Question

If $p(x)$ is a polynomial of degree $n$ and $p(k) = k/(k + 1)$ for $k = 0, 1, 2, \ldots,n$, find $p(n + 1)$.

I'm not looking for a solution, I just want help starting it.

*Edit: Typo in original post,

Answer 1

I present a method of solution, but it might not be what you're looking for because it is tedius. Using the standard way to solve these kinds of problems, let $$p(x)=a_0+a_1x+\cdots+a_nx^n$$

You can see using $p(0)=0$ that $a_0=0$. For the remaining $n$ coefficients we have a system of equations

$$a_1+a_2+\cdots+a_n = \dfrac{1}{2}$$ $$2a_1+2^2a_2+\cdots+2^na_n = \dfrac{2}{3}$$ $$\vdots$$ $$na_1+n^2a_2+\cdots+n^na_n = \dfrac{n}{n+1}$$

Or

$$\left[\begin{matrix} 1 & 1 & 1 & \cdots & 1\\ 2 & 2^2 & 2^3 & \cdots & 2^n\\ \vdots & \vdots & \vdots & \cdots & \vdots\\ n & n^2 & n^3 & \cdots & n^n\end{matrix}\right]\left[\begin{matrix} a_1\\a_2\\ \vdots\\ a_n\end{matrix}\right] = \left[\begin{matrix} 1/2 \\ 2/3 \\ \vdots\\ n/(n+1)\end{matrix}\right] $$

The matrix here is the transpose of a Vandermonde matrix, and an inverse formula is known for it (but quite messy). Using the fact that $\left(A^{-1}\right)^T=\left(A^T\right)^{-1}$, we can find what $a_1,...,a_n$ are by multiplying the inverse to both sides, and finally evaluate the polynomial at $n+1$.

Answer 2

$q(x)=(x+1)p(x)-x$ is a polynomial of degree $n+1$ and has the $n+1$ roots $x=0,1,...,n$, i.e.,

$$q(x)=(x+1)p(x)-x=cx(x-1)...(x-n)$$

Inserting $x=-1$ results in a value for $c$, which determines the solution.

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$$q(-1)=1=c(-1)(-1-1)...(-1-n)=c(-1)^{n+1}(n+1)!$$

Now setting $x=n+1$ gives

$$q(n+1)=(n+2)p(n+1)-(n+1)=c(n+1)(n+1-1)...(n+1-n)=c(n+1)!=(-1)^{n+1}$$

Answer 3

There may be a slicker way of doing this, but I'd look into Lagrange interpolation. Since you have $n+1$ points of a polynomial of degree $n$ it should be uniquely specified.