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Let $P( x )$ be a polynomial of degree $5$.

If $P(1)=0$ , $ \ $ $P(3)=1$, $ \ $ $P(9)=2$, $ \ $ $P(27)=3$, $ \ $$P(81)=4$, $ \ $$P(243)=5$, what is the numerical coefficient of $x$ in $P( x )$?

I tried Lagrange interpolation as method and got $121/162$ but somehow in looking for the its high school level solution... any idea how to solve this is greatly appreciated. Tnx in advance...

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    you must solve an equation system – 2017-02-20
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    It lacks of sufficient details to be used. – 2017-02-20
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    You could try applying the methods used here: http://math.stackexchange.com/questions/2032335/constructing-a-cubic-given-four-points/ – 2017-02-20
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    You are going to get solutions, but I think you should emphasize that you are looking for a very elementary high-school method and that you do know how to solve it yourself. – 2017-02-20

3 Answers 3

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use the ansatz $$a_5x^5+a_4x^4+a_3x^3+a_2x^2+a_1x+a_0=P(x)$$ and plug the given values in the equation,e.g. $$P(1)=a_5+a_4+a_3+a_2+a_1+a_0=0$$ etc. can you finish this?

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Hint:

Observe that for $n=0,1,\cdots5$, you are given $P(3^n)=n$.

The resolution by Cramer yields, for the first degree coefficient

$$\frac{\left|\begin{matrix} 3^0&0&3^0&3^0&3^0&3^0\\ 3^0&1&3^2&3^3&3^4&3^5\\ 3^0&2&3^4&3^6&3^8&3^{10}\\ 3^0&3&3^6&3^9&3^{12}&3^{15}\\ 3^0&4&3^8&3^{12}&3^{16}&3^{20}\\ 3^0&5&3^{10}&3^{15}&3^{20}&3^{25} \end{matrix}\right|} {\left|\begin{matrix} 3^0&3^0&3^0&3^0&3^0&3^0\\ 3^0&3^1&3^2&3^3&3^4&3^5\\ 3^0&3^2&3^4&3^6&3^8&3^{10}\\ 3^0&3^3&3^6&3^9&3^{12}&3^{15}\\ 3^0&3^4&3^8&3^{12}&3^{16}&3^{20}\\ 3^0&3^5&3^{10}&3^{15}&3^{20}&3^{25} \end{matrix}\right|}.$$

The denominator is of the Vandermonde type and equals $(3^5-3^4)(3^5-3^3)(3^5-3^2)(3^5-3^1)(3^5-3^0)(3^4-3^3)(3^4-3^2)(3^4-3^1)(3^4-3^0)(3^3-3^2)(3^3-3^1)(3^3-3^0)(3^2-3^1)(3^2-3^0)(3^1-3^0)$.

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    Unfortunately, I don't see an easy way to compute the numerator. – 2017-02-20
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    Maybe a system of equations can be easier to solve if we transform it to the form: $P(3)-P(1)= P(3^2)-P(3)= ...=P(3^5)-P(3^4)=1$ ? – 2017-02-20
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    @Widawensen: that makes some of the coefficients vanish, but the others get a more complicated expression. After two rounds of Gaussian elimination, you end up with a $4\times4$ determinant. – 2017-02-20
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    @Widawensen You need to go to the end. – 2017-02-20
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You let $g(x)=f(3x)-f(x)-1$. Then $g(1)=g(3)=g(9)=g(27)=g(81)=0$, and I think you can now proceed from here. Yes, we have the same answer, which is $121/162$.