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let $p(x) = a_0 + a_1x + ....+a_nx^n. If\; p(-2) = -15, \ p(1) = 9, \ p(-1) = 1,\ p(0) = 7, \ p(2) = 13\; and\; p(3) = 25$ then the smallest possible value of $n$.

what I tried is that $p(0) = 7\,, a_0 = 7$, $p(1) = a_0 + a_1 + a_2 +.....+a_n = 9$. (1)

$p(-1) = a_0 - a_1 + a_2..... = 1$. (2)

$p(2) = a_0 + 2a_1 + 4a_2 + ..... = 13$.(3)

$p(-2) = a_0 - 2a_1 + 4a_2.... = -15$. (4)

by adding (1) and (2) we get $p(1) + p(-1) = 2[a_0 + a_2 +...] = 10$.

and $p(2) + p(-2) = 2[a_0 + 4a_2 +...] = -2$

is there any way to proceed from here?

biswa prakash mishra
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  • Can't prove it, but I have $n = 3$ with the polynomial $x^3-2x^2+3x+7$ – John Lou May 02 '17 at 02:02
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    Welcome to the Mathematics SE Biswa Prakash.It is very helpful to know what trials and errors you have made. This way we can better assess your capabilities and act accordingly. Ergo "What have you tried?"If you don't do so, most probably questions get closed here, I.e. no new answers can be added. Since you are new user, I am not voting to close this question. But be sure to add your effort unless this question will be closed later. – Jaideep Khare May 02 '17 at 02:03
  • By brute force: there are $6$ conditions, so the interpolating polynomial would have degree at most $5,$. Write down the system of equations (which simplifies quickly) and solve for the coefficients. Turns out that the highest power with a non-zero coefficient is the $3^{rd},$, and the solution is the already posted cubic. – dxiv May 02 '17 at 03:18
  • P.S. After the latest edit: see my previous comment, let $n=5,$, and solve it. – dxiv May 03 '17 at 04:50

3 Answers3

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We start with $$-15,1,7,9,13,25$$ (By the way, why is your $p(1)$ out of sequence?)

Take differences once: $$16,6,2,4,12$$

Take differences twice: $$-10,-4,2,8$$

Take differences a third time: $$6,6,6$$

The final sequence is constant, i.e. a polynomial of degree $0$. Working backwards, the original sequence is a polynomial of degree $3$.

TonyK
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Using Lagrange interpolation formula,so you can get a polynomial satisfying the given conditions, as for the smallest degree,maybe the Chinese Remainder theorem is useful.

Jiabin Du
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  • Can you elaborate – Shailesh May 02 '17 at 04:28
  • On this problem,you can do it explicitly,p(x)=xg(x)+7,g(x) satisfies 5 conditions,requiring deg(g) smallest. g(x)=q(x)(x+1)+a,q(x) 4 conditions with smallest degree,do this procedures again,you can get the polynomial as desired. – Jiabin Du May 02 '17 at 05:54
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As John Lou mentions, $p(x)=x^3-2x^2+3x+7$ satisfies all the equalities. A polynomial of degree two couldn't, because the slopes cannot decrease and then increase again. Or, even easier, a parabola has constant convexity, while the set of points given imply concave on $(-2,0)$ and convex on $(1,3)$.

Martin Argerami
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