1

enter image description here

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?

user130306
  • 1,890

2 Answers2

1

Let $\Delta_k$ be the $k^{\text{th}}$ finite diffetence. then

$$P(x)=P(x_0)+$$

$$\sum_{k=1}^n\frac{\Delta_k}{k!}(x-x_0)(x-x_1)...(x-x_{k-1})=$$ $$3+1.(x-x_0)+\frac 32(x-x_0)(x-x_1)=$$

$$3+(x-0)+\frac 32(x-0)(x-1)=$$

$$3+x+\frac 32x^2-\frac 32x=$$ $$\frac 32x^2-\frac 12x+3.$$

hamam_Abdallah
  • 62,951
1

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.

Théophile
  • 24,627
  • ah I understand. thank you. so what would be a better formulation of the question? – user130306 Nov 12 '18 at 19:47
  • A better way would be to look up "Lagrange interpolation" now, from the comments. – Dietrich Burde Nov 12 '18 at 19:59
  • @user130306 Glad to help. A better formulation would be: "Let $f$ be a polynomial of minimum degree satisfying [etc.]". In that case, you can use finite differences, or whatever other technique, to find a unique solution. In the simplified example I gave above, the unique solution would be $f(x) = 0$ for all $x$, i.e., the zero polynomial. In general, if the table has $n$ points, then you can find a unique polynomial of degree at most $n-1$. (Typically it would be degree $n-1$, but it's also possible that the points line up in a nice way so that a polynomial of small degree works.) – Théophile Nov 12 '18 at 19:59