Suppose I have a polynomial $$p(x)=\sum_{l=0}^{k} (-1)^la_l x^l$$
where $a_k$ are some positive constants. I don't mind to consider $b_l=(-1)^la_l $.
I am trying to compute $[p(x)]^2$ and I want to express the result as:
$$[p(x)]^2=\sum_{i=0}^{2k} c_i x^{i}$$ in which the $c_i$s are functions of the $b_l$ (or $a_l$).
Any help is appreciated.
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$\begingroup$ Your second displayed formula maybe should have $[p(x)]^2$ on its left side. $\endgroup$– coffeemathSep 17, 2015 at 15:25
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$\begingroup$ Well have you tried expanding the product to see what the first few terms look like? $(b_0 + b_1 x + b_2 x^2 + \dots) \cdot (b_0 + b_1 x + b_2 x^2 + \dots) = \dots ?$ $\endgroup$– Najib IdrissiSep 17, 2015 at 15:39
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$\begingroup$ @NajibIdrissi Yes.. it yields: $b_0^2+(2b_0b_1)x+(2b_0b_2+b_1^2)x^2+(2b_1b_2)x^3+b_2^2x^4$. But I am not able to generalize it.. $\endgroup$– tamSep 17, 2015 at 15:51
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1 Answer
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For any two polynomials $f(x)=\sum_{i=0}^n\alpha_ix^i$ and $g(x)=\sum_{i=0}^m\beta_ix^i$ you have $$ f(x)g(x)=\sum_{i=0}^{m+n}\gamma_ix^i $$ where $\gamma_i=\sum_{j=0}^i\alpha_j\beta_{i-j}$ (interpret $\alpha_s=0$ if $s>n$ and $\beta_s=0$ if $s>m$).
Therefore, (again interpreting $a_s=b_s=0$ if $s>k$) $$ c_i=\sum_{j=0}^ib_jb_{k-j}=(-1)^i\sum_{j=0}^ia_ja_{i-j}. $$ If $i=2n+1$ is odd, this becomes $$ c_i=2\sum_{j=0}^{n+1}b_jb_{2n+1-j}=-2\sum_{j=0}^{n+1}a_ja_{2n+1-j} $$ and if $i=2n$ is even, you get $$ c_i=b_n^2+2\sum_{j=0}^{n-1}b_jb_{2n-j}=a_n^2+2\sum_{j=0}^{n-1}a_ja_{2n-j} $$
answered Sep 17, 2015 at 16:05