How to use finite differences to determine an equation of a polynomial given consecutive integer $x$ and corresponding $y$ coordinates of the graph?
This chart is given:
for $x=-3$, $y=-9$
for $x=-2$, $y=3$
for $x=-1$, $y=3$
for $x=0$, $y=-3$
for $x=1$, $y=-9$
for $x=2$, $y=-9$
for $x=3$, $y=3$
I found the finite differences to be 6 and degree of the polynomial with the given points to be 3 ($n=3$). But how do I find the polynomial function rule out of this information? Given no factors? Please help with a short solution. Thanks a lot in advance. I also did $y=kx^3$, $-9=k(-3)^3$, $k=1/3$. But of course the rule isn't $y=1/3x^3$.
$\endgroup$2 Answers
$\begingroup$Once you discover that the third difference is constant you know that the polynomial is third degree. Denote it by $P(x)$.
We can see that the horizontal lines $y=3$ and $y=-9$ cross the graph of $y=P(x)$ three times each. Therefore the graph of the polynomial $Q(x)=P(x)-3$ has $x$-intercepts $(-2,0),\,(-1,0)$ and $(3,0)$. Therefore for some leading coefficient $a$
\begin{eqnarray} P(x)-3&=&a(x+2)(x+1)(x-3)\\ P(x)&=&a(x+2)(x+1)(x-3)+3\\ &=&ax^3-7ax-6a+3 \end{eqnarray}
We know that when $x=0$, $y=-3$ so $a=1$. Therefore
$$P(x)=x^3-7x-3$$
The same result can be had using the three values of $-9$ since the graph of $y=P(x)+9$ has $x$-intercepts $(-3,0),\,(1,0)$ and $(2,0)$. Therefore
\begin{eqnarray} P(x)+9&=&a(x+3)(x-1)(x-2)\\ P(x)&=&a(x+3)(x-1)(x-2)-9\\ &=&ax^3-7ax+6a-9 \end{eqnarray}
In this case, $6a-9=-3$ so $a=1$. And we obtain the same result.
Note: This approach works only when an $n$th degree polynomial sequence $\{y_k\}$ has $n$ values of $x_k$ for which the $y_k$ values are equal. In general you need the method used by Ahmed S. Attaalla in his answer.
Suppose $P$ is a degree $n$ polynomial with $P(x_k)=y_k$ and that for $n$ different values of $k$ we have $y_k=c$. Then for those same $n$ values of $k$ it will be true that the polynomial $Q(x_k)=P(x_k)-c$ will have $y_k=0$ since for those values of $k$, $Q(x_k)=P(x_k)-c=c-c=0$. Knowing the $n$ $x$-intercepts of $Q(x)$ allows us to factor $Q(x)$. Adding $c$ to $Q(x)$ gives us $P(x)$. We can then use one of the given values of $(x,P(x))$ to find the value of $a$.
$\endgroup$ 1 $\begingroup$Treat your numbers as a sequence with $g(0)$ being the first term, corresponding to $f(-3)$,
$$f(x-3):=g(x) : \color{green}{-9},3,3,-3,-9,-9,3$$
$$\Delta g=g(x+1)-g(x) : \color{green}{12},0,-6,-6,0,12$$
$$\Delta^2 g : \color{green}{-12},-6,0,6,12$$
$$\Delta^3 g : \color{green}{6},6,6$$
Now assume all else is $0$. To get an umbral Taylor series representation,
$$f(x-3)=\color{green}{-9}+\color{green}{12}{x \choose 1}\color{green}{-12}{x \choose 2}+\color{green}{6}{x \choose 3}$$
So that with $x \mapsto x+3$,
$$f(x)=-9+12{{x+3} \choose 1}-12{{x+3} \choose 2}+6{{x+3} \choose 3}$$
$$=x^3-7x-3$$
$\endgroup$
