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I have some difficulty of determining the canonical form of a polynomial. Here is the problem:

Suppose $P$ is a polynomial with degree $5$, and value of $P$ at $-2, 0, 1, 4, 5, -3$ are $2, 4, 5, -1, -2, 1$, correspondingly. Determine the canonical form of $P$.

I know that using Lagrange interpolation, we can determine some form of P, but from that form, it's very complicated to determine the canonical form of $P$. I heard some people says that there's a way using Horner table to make the calculation less complicated and make the computation possible in the 60-minutes exam. So can anyone please suggest some way to do that, and clarify how it can apply to this problem? I really appreciate all your help. Thanks so much.

le duc quang
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  • What do you mean by the “canonical form of a polynomial”? – Steve Kass May 10 '15 at 04:29
  • Ok, I assume you mean the “regular” form $a_0+a_1x+a_2x^2+\cdots+a_5x^5$. – Steve Kass May 10 '15 at 05:07
  • @SteveKass: Yes, that's what I mean. Sorry if I use the wrong name for the form. Do you have any suggestion for me? – le duc quang May 10 '15 at 05:16
  • That name is fine. I didn’t immediately remember that the Lagrange interpolation formula gave a result that could require a lot of work to put into canonical form. – Steve Kass May 10 '15 at 05:19
  • Yeah, I think the same way as you. But when coming to this exercise, we will have 6 polynomials degree 5 and terrible coefficients. You can try and you'll see it! – le duc quang May 10 '15 at 05:21

2 Answers2

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Use Newton's interpolation, as suggested. Just sort the $x$ $$ \begin{array}{ccccccccccccccccccccccc} x & -3 & & -2 & & 0 & & 1 & & 4 & & 5\\ f & 1 && 2 && 4 && 5 && -1 && -2\\ f^{(1)} && 1 && 1 && 1 && -2 && -1\\ f^{(2)} &&& 0 && 0 && -\frac{3}{4} && \frac{1}{4}\\ f^{(3)} &&&& 0 && -\frac{1}{8} && \frac{1}{5}\\ f^{(4)} &&&&& -\frac{1}{56} && \frac{13}{280}\\ f^{(5)} &&&&&& \frac{9}{1120}\\ \end{array} $$ Now, $$P = 1 + (x + 3) + 0 (x+3) (x+2) + 0 (x+3) (x+2) x - \frac{1}{56} (x+3) (x+2) x (x-1) - \frac{9}{1120} (x+3) (x+2) x (x-1) (x-4).$$ Or in form $$ P = 1 + (x + 3) \left( 1 + (x + 2) \left( 0 + x \left( 0 + (x-1)\left( -\frac{1}{56} + (x-4)\left( \frac{9}{1120} \right)\right)\right)\right)\right) $$ Now we need expand all of that. Let $P(x) \equiv P_1(x) = a_1 + (x - x_1) P_1(x)$ and so on $$P_k(x) = a_k + (x - x_k) P_{k+1}(x).$$ $$ \begin{array}{ccccccccccccccccccccccc} x_k & -3 && -2 && 0 && 1 && 4 && 5\\ a_k & 1 && 1 && 0 && 0 && -\frac{1}{56} && \frac{9}{1120}\\ \hline \hline P_6(x) &&&&&&\frac{9}{1120}\\ P_5(x) &&&&&\frac{9}{1120} && -\frac{1}{20}\\ P_4(x) &&&&\frac{9}{1120} && -\frac{13}{224} && \frac{1}{20}\\ P_3(x) &&&\frac{9}{1120} && -\frac{13}{224} && \frac{1}{20} && 0\\ P_2(x) &&\frac{9}{1120} && -\frac{47}{1120} && -\frac{37}{560} && \frac{1}{10} && 1\\ P_1(x) &\frac{9}{1120} && -\frac{1}{56} && -\frac{43}{224} && -\frac{11}{112} && \frac{13}{10} && 4\\ \end{array} $$

Finally $$ P = 4 + \frac{13}{10} x - \frac{11}{112} x^2 - \frac{43}{224}x^3 -\frac{1}{56}x^4 +\frac{9}{1120}x^5. $$

PS I've made a couple mistakes during computations and was constantly checking my intermediate results with Mathematica. Not sure if it is possible to do all of that in exam conditions.

uranix
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They might refer to Newton interpolation, see this (link).

mvw
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  • Newton interpolation still need too much calculation, especially in calculating $c_i$. Do you have any idea? I have some thought to use Maclaurin series for Lagrange interpolation, but seem like it doesn't help much. – le duc quang May 10 '15 at 05:19