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this problem is from my Gelfand's Algebra book.

Problem 164. Prove that a polynomial of degree not exceeding 2 is defined uniquely by three of its values.

This means that if $P(x)$ and $Q(x)$ are polynomials of degree not exceeding 2 and $P(x_1)=Q(x_1), P(x_2)=Q(x_2), P(x_3)=Q(x_3)$ for three different numbers $x_1, x_2,$ and $x_3$, then the polynomials $P(x)$ and $Q(x)$ are equal.

I honestly don't know where to start.

Paul Dirac
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  • Hint: what are some roots of the polynomial $P-Q$? – mdp Nov 06 '12 at 16:30
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    If there's $V(x)=P(x)-Q(x)$, and if $P(x_1)=Q(x_1), P(x_2)=Q(x_2), P(x_3)=Q(x_3)$, then $V(x_1)=V(x_2)=V(x_3)=0$, therefore $x_1,x_2,x_3$ are the roots of $V(x)$, but how can it be? if $V(x)$ is degree 2 polynomial. – Paul Dirac Nov 06 '12 at 16:38
  • The answer to the question "how can it be?" is the rest of the proof - see Hagen's answer. – mdp Nov 06 '12 at 16:44
  • This fails if the coefficient ring is not a domain. e.g. the polynomials $\rm:x^2:$ and $1$ have the same values at the $4$ points $\pm1,, \pm3$ over $,\Bbb Z/8 =$ integers mod $8.:$ So you need to specify the coefficient ring. – Bill Dubuque Nov 06 '12 at 16:45
  • I suspect the precalculus tag restricts the coefficient set... – copper.hat Nov 06 '12 at 16:59
  • @copper.hat Probably, but some precalculus algebra does include modular arithmetic and polynomials, esp. nowadays with applications to CS (cryptography, etc). So the question needs clarification. – Bill Dubuque Nov 06 '12 at 17:05

3 Answers3

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A polynomial of degree two is defined by three of it's values. Why? If we want to obtain a $2^{\rm nd}$ degree polynomial, we must specify three constants, $a_1,a_2,a_3$. But what are this constants? Say

$$p(x)=a_1x^2+a_2 x+a_3$$

Then, knowing three values, $b_1,b_2,b_3$ of points $x_1,x_2,x_3$, we have that

$$\begin{cases} a_1x_1^2+a_2 x_1+a_3=b_1\\ {}\\a_1x_2^2+a_2 x_2+a_3=b_2\\{}\\a_1x_3^2+a_2 x_3+a_3=b_3\end{cases}$$

and this is a system of three linear equations in three unknowns $a_1,a_2,a_3$, which we can always solve (meaning we can always determine its solutions, which might be none at all). This generalizes: an $n$ degree polynomial is uniquely determined by $n+1$ of its values.

Pedro
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  • +1 The matrix formed from these equations is known as the Vandermonde matrix and the determinant has the form $(x_2-x_1)(x_3-x_1)(x_3-x_2)$ which is non-zero iff the $x_i$ are distinct. – copper.hat Nov 06 '12 at 16:55
  • @copper.hat Didn't know that, but seeing the book is titled "Algebra", it seemed the most natural approach. I'll take a look at that! – Pedro Nov 06 '12 at 17:03
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    It is very ill-conditioned from a numerical perspective, but shows 'invertibility' clearly. – copper.hat Nov 06 '12 at 17:05
  • Thank you Peter Tamaroff, but I don't understand how that proves that polynomial with degree not exceeding 2 is uniquely defined by 3 values. – Paul Dirac Nov 06 '12 at 17:24
  • @PaulDirac If you solve the linear system, you'll obtain what the coefficients of the polynomial are. Do you follow? – Pedro Nov 06 '12 at 17:29
  • @PeterTamaroff Yes, if you know $b_1,b_2,b_3$, I understand that. – Paul Dirac Nov 06 '12 at 17:32
  • And those $b_1,b_2,b_3$ you know. I mean $p(x_1)=b_1$, $p(x_2)=b_2$, $p(x_3)=b_3$. – Pedro Nov 06 '12 at 17:33
  • @PeterTamaroff Thank you, I think I understand now. – Paul Dirac Nov 06 '12 at 17:55
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Hint $\rm\:f(x) = ax^2\!+\!bx\!+\!c,\ a\ne 0\:$ and $\rm\:f(x) = f(y) = f(z)\:$ implies following determinant $= 0$, having proportional first and last columns

$$\rm 0\, =\, \left | \begin{array}{ccc} \rm f(x) &\!\! \rm x &\!\! 1 \\ \rm f(y) &\!\! \rm y &\!\! 1 \\ \rm f(z) &\!\! \rm z &\!\! 1 \end{array} \right | \, =\, \left | \begin{array}{ccc} \rm ax^2&\!\!\! \rm x &\!\!\! 1 \\ \rm ay^2 &\!\!\rm y &\!\!\! 1 \\ \rm az^2 &\!\! \rm z &\!\!\! 1 \end{array} \right | \, =\, aV(x,y,z)\, =\, a\,(x\!-\!y)(y\! -\! z)(z\!-\!x)\qquad $$

where $\rm\,V\,$ denotes that Vandermonde determinant, e.g. see this answer or Wikipedia.

Thus, if the coefficient ring is a field (or domain), one of the RHS factors must be $0$. Since $\rm\:a\ne 0\:$ one of the other factors $= 0,\:$ i.e. the roots are not distinct.

Alternatively we can iterated the Factor Theorem as here.

Remark $ $ That the second $\det$ equals the first follows from the fact that the first columns are congruent modulo the others. More precisely, the second $\det$ arises simply by applying the elementary col operation $\rm\: col_1 \leftarrow col_1 - b\cdot col_2 - c\cdot col_3,\,$ to get $\rm\,f(x) - b\cdot x - c\cdot 1 = ax^2,\:$ etc. The same argument works for higher degree polynomials. For the general case, it helps to know that the rhs matrix, after removing the factor of $\rm\:a,\:$ is the ubiquitous Vandermonde matrix,

Bill Dubuque
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Hint: $P-Q$ is a polynomial of degree not exceeding $2$ and has at least three roots $x_1, x_2, x_3$. How many roots would a nonzero polynomial of such degree have?

Pedro
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Hagen von Eitzen
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  • Thank you Hagen for the hint, I understood that results, they contradict each other right? so does that mean that $P-Q$ is zero polynomial? Should I assume that? since polynomial of degree not exceeding 2 has 3 roots? – Paul Dirac Nov 06 '12 at 16:59
  • Thanks for help, I will add rep point once I reach 15 reputations. – Paul Dirac Nov 06 '12 at 17:56