The first difference is 1, 4, 7, 10, 13, 16
The second difference is 3, 3, 3, 3, 3, 3
Since the second difference is constant this would be quadratic and I would have
$\frac{3}{2}n^{2}$
So now I will take the differences between the original sequence and the values of $\frac{3}{2}n^{2}$ for n = 1, 2, ...
3 - 3/2, 4 - 6, 8 - 27/2, 15 - 24, 25 - 75/2
Which is 3/2, -2, -11/2, -9, -25/2
so the difference between each of those terms are
$-3 \frac{1}{2}, -3 \frac{1}{2}, -3 \frac{1}{2}, -3 \frac{1}{2}, -3 \frac{1}{2}$
So then I have $\frac{3}{2} n^{2} -3.5n+5$ but when I plug in zero i get 5, and I should get 3, so what did I do wrong?
$\endgroup$ 32 Answers
$\begingroup$The question is not well formed, since the wording implies that there is a unique answer. In fact, there are infinitely many polynomials passing through those points, and the answer can be anything you want it to be. If you're able to find a polynomial through a given set of points, then just pick your favourite number $y$, add $(2018, y)$ to the table, and use your technique to find the polynomial.
To see it more clearly, let's consider a simpler version of the question:
Let $f$ be a polynomial that agrees with the table:
$$\begin{array}{c | c} \text{Input} & \text{Output} \\ \hline 0 & 0\end{array}$$ Find $f(2018)$.
Well, (infinitely) many polynomials pass through $(0,0)$: they would have the form$$f(x) = a_1x + a_2x^2 + a_3x^4 + \ldots + a_mx^m$$for some $m$ and coefficients $a_i$. We can force $f(2018)$ to be whatever we choose. If you wanted $f$ to pass through, say, $(2018, \pi)$, then you could let $f(x) = \frac{\pi}{2018}x$, to give one among the infinitely many possibilities.
$\endgroup$ 3 $\begingroup$Let $\Delta_k$ be the $k^{\text{th}}$ finite diffetence. then
$$P(x)=P(x_0)+$$
$$\sum_{k=1}^n\frac{\Delta_k}{k!}(x-x_0)(x-x_1)...(x-x_{k-1})=$$$$3+1.(x-x_0)+\frac 32(x-x_0)(x-x_1)=$$
$$3+(x-0)+\frac 32(x-0)(x-1)=$$
$$3+x+\frac 32x^2-\frac 32x=$$$$\frac 32x^2-\frac 12x+3.$$
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