1
$\begingroup$

Let $f(x)$ be a polynomial of degree $n$ such that $$f(k) =\dfrac1k$$ for $k = 1,2,3,\ldots,(n+1)$. Determine $f(n + 2)$.

This question is quite similar to this one. How do I solve such type of problems in general?

$\endgroup$
  • 2
    $\begingroup$ $P(k)=k/(k+1)$ for an interesting range of numbers if and only if $1-P(k)=1/(k+1)$ for that same range. Looks like you may be looking for $f(x)=1-P(x-1)$ where $P(x)$ the polynomial that popped out in the linked question. Adjust $n$ by one. $\endgroup$ – Jyrki Lahtonen Jul 12 '17 at 10:58
3
$\begingroup$

Hint: Note that we have $kf(k) = 1$ for $k = 1,\ldots,n+1$, and clearly $0f(0) = 0$. So now we know $n+2$ values of the $(n+1)$-degree polynomial $xf(x)$.

As for the general bit, it's difficult to say. You should try to make everything into polynomials as nicely as possible, because polynomials are a lot easier to work with than rational functions. Case in point: I can't tell much about $f(x)$ just from knowing the solutions of $f(x) = \frac1x$, but I can tell a lot about $xf(x)$ from knowing the solutions to $xf(x) = 1$.

It won't always work, though, and the most important thing, as with problem solving in general, is to just try things and not give up when your first attempt fails. Remember that when you read a solution that someone else has written to a problem like this, you only see the attempts that succeeded. Behind each successful solution are either a lot of failed attempts at solving the problem, or the experience needed to write a solution more or less on the first attempt, built up through a lot of trial and error on other, similar problems.

$\endgroup$
1
$\begingroup$

Let $g(x):=x\,f(x)$ like Arthur suggests. Then, $g$ is of degree $n+1$. Hence, from this link, we have $$\sum_{r=0}^{n+2}\,(-1)^r\,\binom{n+2}{r}\,g(x+r)\equiv 0\,.$$ In particular, $$\sum_{r=0}^{n+2}\,(-1)^r\,\binom{n+2}{r}\,g(r)=0\,.$$ As $g(0)=0$ and $g(1)=g(2)=\ldots=g(n+1)=1$, we conclude that $$\begin{align}g(n+2)&=(-1)^{n+1}\,\sum_{r=1}^{n+1}\,(-1)^{r}\,\binom{n+2}{r} \\&=(-1)^{n+1}\,\left((1-1)^{n+2}-1-(-1)^{n+2}\right)=(-1)^{n}+1\,.\end{align}$$ Consequently, $$f(n+2)=\frac{1+(-1)^n}{n+2}\,.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.