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Suppose that $P(x)$ is a polynomial of degree $n$ such that $$P(k)=\dfrac{k}{k+1}$$ for $k=0,1,\ldots,n$. Find the value of $P(n+1)$.

I could not relate this question with this one How to find $P(n+1)$, given $P(x)$ for $x = 0,1,\ldots,n$?, so maybe it is not a duplicate

I have absolutely no idea, how the function became a polynomial and how to approach it. Please help!

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  • $\begingroup$ If $P(x) = a_{0} + a_{1}x + \cdots + a_{n}x^{n}$ then the $n + 1$ values of the $a_{i}$'s are unknowns. But now you have $n + 1$ linear equations in $n + 1$ unknowns. (You are given the values of the $x_{i}$'s so they are not unknown. $\endgroup$
    – Jay
    Commented Jan 13, 2014 at 13:36
  • $\begingroup$ Didn't get you...please elaborate! $\endgroup$
    – Hawk
    Commented Jan 13, 2014 at 13:38
  • $\begingroup$ Let $Q(x) = (x + 1)P(x) - x$. Then you can name certain roots of $Q$. If you compare this with the degree of $Q$, this will give you a lot of information. $\endgroup$
    – user120974
    Commented Jan 13, 2014 at 13:42
  • $\begingroup$ Look at $P(0) = a_{0} + a_{1}0 + a_{2}0^2 + \cdots + a_{n}0^{n} = a_{0}$. $P(1) = a_{0} +a_{1}({\frac{1}{2}})^{1} + a_{2}({\frac{1}{2}})^{2} + \cdots + a_{n}({\frac{1}{2}})^{n}$. See the pattern? $\endgroup$
    – Jay
    Commented Jan 13, 2014 at 14:11

3 Answers 3

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$Q(x) = (x+1)P(x)-x$ has degree $n+1$ and $n+1$ roots at $x=0$, $x=1$, through $x=n$, so it is of the form $Q(x) = ax(x-1)(x-2)\cdots (x-n)$. Moreover, $Q(-1) = 1$, and thus $\displaystyle a = (-1)^{n+1} \frac{1}{(n+1)!}$.

It follows that $Q(n+1) = (-1)^{n+1}$ and thus that $\displaystyle P(n+1) = \frac{n+1+(-1)^{n+1}}{n+2}$.

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Hint: Consider the polynomial $Q(x)=(x+1)P(x)$. Then $Q(k)=k$.

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Suppose that $\,p,f,g\,$ are polynomials over $\,\Bbb C\,$ (or any field) and $\,a,\,a_i\in \Bbb C\,$ are distinct points.

Suppose $\ p(x) = f(x)/g(x)\ $ for $\ x = a_i,\ i = 1,\ldots, k = \deg(gp) > \deg f,\, $ and $\, g(a) = 0.$

Then $\ q\, := g\,p-f \:$ has degree $\,k\,$ and $\,k\,$ distinct roots $\,a_i$ so $\ q = c\,(x\!-\!a_1)\cdots (x\!-\!a_{k})\,=:\,c\,h $

for some constant $\,c.\ $ But $\,\ g(a)=0\,\Rightarrow\, -f(a) = q(a) = c\,h(a),\ $ so $\ c = -f(a)/h(a).$

Therefore $\ p\, =\, \dfrac{f+q}g\, = \,\dfrac{f}g - \dfrac{f(a)h}{g\, h(a)}\, =\, \dfrac{f}g\, -\, \dfrac{f(a)}g \dfrac{(x-a_1)\cdots(x-a_{k})}{ (a-a_1)\cdots(a-a_{k})}$

In OP: $\,\ \begin{eqnarray}f = x,\ \ g = x\!+\!1\ \ \\ a_i = i-1,\ a=-1\end{eqnarray}\ $ so $\,\ p = \dfrac{x}{x+1} +\, \dfrac{1}{x+1}\, \dfrac{\,\ x\ \,\cdots\,\ (x\,-\,n)}{ -1\,\cdots(-1-n)}=\smash{\dfrac{x+(-1)^{n+1}{x\choose n+1}}{x+1}}$

Therefore, evaluating the prior at $\ x = n+1\ $ we deduce that $\ p(n+1)\, =\, \smash{\dfrac{n+1+(-1)^{n+1}}{n+2}}$

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