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Let $\alpha$ and $\beta$ and $\gamma$ be 3 real numbers.

Prove that there exist only one polynomial $P(x)$ of the second degree such that $$\begin{cases}P(1)=\alpha \\ P(2)=\beta \\ P(3)=\gamma\end{cases}$$

I don't even know how to start?? Perhaps this is a new type of question to me and so if I know how to solve this one I will be able to solve similar questions!

Also this is second part of the question maybe that would help: Determine $P(x)$ in every case: $$\alpha=\beta=\gamma=2500\text{ and } \alpha=3;\beta=6;\gamma=9$$

Thanks!!

VosPost
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    http://en.wikipedia.org/wiki/Lagrange_polynomial – OR. Dec 21 '13 at 17:42
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    $p(x)=ax^2+bx+c$, so we get a system of equations $a+b+c=\alpha$, $4a+2b+c=\beta$ and $9a+3b+c=\gamma$. Solving this we get $a=\gamma-\alpha$ and so on... If I understood the problem correctly. – Poppy Dec 21 '13 at 17:43
  • @VosPost: This one comes to mind (but mentioning it is rather pointless for this question). A sphere and an interior point $P$ (not the center) is given. Determine the locus of the points $Q$ with the following property: $P$ and $Q$ are diagonally opposite points of a rectangular block, of which the three vertices connected to $P$ by an edge lie on the sphere. IMO 1978 if I recall correctly. – Marc van Leeuwen Dec 21 '13 at 18:04

4 Answers4

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The statement is false. For $\alpha=1$, $\beta=2$, $\gamma=3$ there is only one polynomial of degree at most $2$ that satifies the evaluations, namely $P(x)=x$; it is not of degree$~2$. There are many similar counterexamples.

Marc van Leeuwen
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Hints:

Solve the system

$$\begin{align*}a+b+c&=\alpha\\ 4a+2b+c&=\beta\\ 9a+3b+c&=\gamma\end{align*}$$

and show the polynomial $\;p(x)=ax^2+bx+c\;$ fulfills the conditions. And about uniqueness: suppose there's another polynomial $\;g(x)\;$ that fulfills the same conditions. What can you say about the roots of $\;h(x):=f(x)-g(x)\;$ ?

DonAntonio
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  • " there's another polynomial g(x) that fulfills the same conditions. What can you say about the roots of h(x):=f(x)−g(x) ?" Euuuuuuuuuuuuuuh.... No idea.... – VosPost Dec 21 '13 at 17:49
  • And how can I show that the polynomial p(x)=ax2+bx+c fulfills the conditions. – VosPost Dec 21 '13 at 17:50
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    Why should there be any other discussion about uniqueness? Isn't a second degree polynomial uniquely specified by its coefficients? – hhsaffar Dec 21 '13 at 17:51
  • And how do you intend to show that the given conditions uniquely specify the coefficients of quadratic polynomial, @hhsaffar ? – DonAntonio Dec 21 '13 at 18:11
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assume the quadratic equation $$ ax^2+bx+c= P(x)$$ when you substitute the values of $x$ into the function. you get $3$ equations in $a,b$ and $c$ which has unique solutions.

Suraj M S
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  • This simply restates what needs to be proved, without actually proving it. – Greg Martin Dec 21 '13 at 18:44
  • rather than giving a readymade solution i prefer such hints so that the person who asks the question can put some effort himself to get it. and also i dont this is a restating of the question. – Suraj M S Dec 21 '13 at 19:23
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Assume that there are 2 polynomials $p(x)$ and $g(x)$
Then $q(x)=p(x)-g(x)$ is of degree at most $2$ and the equation $q(x)=0$ has $3$ solutions
the numbers $1,2,3$ which is a contradiction.
So, $g(x)=0$ which means $p(x)=g(x)$

Konstantinos Gaitanas
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