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Find the polynomial p of least degree that takes these values: $p(0) = 2$, $p(2) = 4$, $p(3) = −4$, $p(5) = 82.$ Use divide differences to get the correct polynomial. It is not necessary to write the polynomial in the standard form $a_0+a_1x+a_2x^ 2+· · ·.$

I am not sure how to use the divided differences method, the explanations I look up are all very confusing. I can't find an actual example of a problem similar like this to even mimick - or see what's going on.. Any help would be great

Shammy
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2 Answers2

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Taking a shot at this question...

You have four points, so we need a third-degree polynomial, $p_3(x)$.

We have $p(0)=2$, $p(2)=4$, $p(3)=-4$ and $p(5)=82$.

Calculating the divided differences...

$$p[0,2]=\frac{p(2)-p(0)}{2-0}=\frac{2}{2}=1$$ $$p[2,3]=\frac{p(3)-p(2)}{3-2}=\frac{-8}{1}=-8$$ $$p[3,5]=\frac{p(5)-p(3)}{5-3}=\frac{86}{2}=43$$ $$p[0,2,3]=\frac{p[2,3]-p[0,2]}{3-0}=\frac{-9}{3}=-3$$ $$p[2,3,5]=\frac{p[3,5]-p[2,3]}{5-2}=\frac{51}{3}=17$$ $$p[0,2,3,5]=\frac{p[2,3,5]-p[0,2,3]}{5-0}=\frac{20}{5}=4$$

The resulting polynomial should be... $$p_3(x)=2+1(x-0)-3x(x-2)+4x(x-2)(x-3)$$

Verifying... $$p_3(0)=2\,\checkmark$$ $$p_3(2)=2+2=4\,\checkmark$$ $$p_3(3)=2+3-3(3)(1)=-4\,\checkmark$$ $$p_3(5)=2+5-3(5)(3)+4(5)(3)(2)=82\,\checkmark$$

Tim Thayer
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    Shouldn't it be $p[0,2,3,5]$ ? – Shammy Oct 02 '15 at 02:42
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    Absolutely. I'm fixing it... – Tim Thayer Oct 02 '15 at 02:50
  • I finally figured out how to do it via reading my textbook. Working it out then will be back to mark your answer as a solution or not. – Shammy Oct 02 '15 at 02:54
  • Thanks so much Tim. Would you mind checking out my attempt at another problem similar to this? [Interpolating via Newton Polynomials][1]. [1]:http://math.stackexchange.com/questions/1460358/having-trouble-interpolating-a-polynomial-using-newtons-method – Shammy Oct 02 '15 at 03:29
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Let $x_0 = 0$, $x_1 = 2$, $x_3 = 3$, $x_4 = 5$ and $y_k = p(x_k)$ for $k = 0,1,\dots,5$. The Newton polynomial $p$ is $$ P(x) = [y_0] + [y_0,y_1](x-x_0) + \cdots + [y_0,\ldots,y_4](x-x_0)(x-x_1)\cdots(x-x_3) $$ where the divided differences can be put in tabular form as shown on wikipedia:

0   2 
        1
2   4      -3
       -8      4
3  -4      17
       43
5  82

Finally, we get $$ \begin{align*} p(x) &= 2 + x - 3x(x-2) + 4x(x-2)(x-3) \\ &= 4x^3 - 23x^2 + 31x + 2. \end{align*} $$

Éric Guirbal
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