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$$ P(x+n+1) = \sum_{i=0}^n (-1)^{n-1} \binom {n+1} i P(x+i)$$

What does the above formula do ? I came across this formula in a set of lecture notes, so the explanation following was a tad short.

The following problems are to be solved using this formula, but I'm stuck:

  1. A polynomial $P$ of degree $n$ satisfies $P(i)= \binom {n+1} {i}^{-1} $ for $i=0,1,2,\ldots,n$. Find $P(n+1)$.

  2. A polynomial $P$ of degree $n$ which satisfies $P(i)= {1 \over i}$ for $i=1,2,\ldots,n+1$. Find $P(n+2)$.

The problems are routine exercise problems, but I just can't get around the above formula .

Elmar Zander
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donvoldy666
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  • Well, what it does is compute the value of $P$ at some point $y$, if you only know the values at the points $y-1, y-2,\ldots,y-n-1$, given that the polynomial has degree at most $n$. And that's exactly what the two exercises want you to do $\ldots$ – Elmar Zander Mar 23 '13 at 09:55

2 Answers2

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Try to use different values of $x$ to solve your exercices (for example $x=0$ for the first question and $x=1$ for the second one)

JJ Fleck
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I assume you're having trouble understanding what the equation itself is. Perhaps I can help.

If you take the right side over to the left (correcting a minor typo - "n-1" should be "n-i" in the power of -1), the equation can also be written like this:

$$ \sum_{i=0}^{n+1} (-1)^{n+1-i} \binom {n+1} i P(x+i)=0 $$ What this is, is actually the iterated difference. That is, if $\Delta f(0) = f(1)-f(0)$, and $\Delta^2 f(0) = \Delta (f(1)-f(0)) = f(2)-2f(1)+f(0)$, and so on, then our equation says

$$ \Delta^{n+1} P(x) = 0 $$ And if our polynomial is of order $n$ or lower, then this must be true. To see this, consider that, for $P(x)=x^n$, we have

$$ \Delta x^n = (x+1)^n-x^n = \sum_{k=0}^{n-1}\binom{n}{k}x^k $$

And $$ \Delta C = 0 $$ for any constant $C$. So, iterating the $\Delta$ operation $n+1$ times, you end up with $\Delta^{n+1} x^n = 0$.

Glen O
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