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In the following paper in equation 104, we have the following general formula for the square of a polynomial:

https://dspace.mit.edu/bitstream/handle/1721.1/106948/Roy_Aggresive%20flight.pdf?sequence=1&isAllowed=y

enter image description here

But on this Stack Exchange answer, the square of a polynomial has a different general formula.

General square form of Taylor expansion polynomial

Which ones correct?enter image description here

Can you explain how the convolution in the paper is reached at?

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    $\begingroup$ Those formulas are answering two different questions. One is about the coefficients of the square of a polynomial and one is about the square of a sum. If you set $a_k = p_k x^k$ in the second formula and collect like terms you'll get the first formula. $\endgroup$
    – Qiaochu Yuan
    Commented Dec 11, 2020 at 3:36

1 Answer 1

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$\left(P^2\right)_n$ denotes the coefficient on $x^n$ of $P(x)^2$. Applying the linked answer, $$P^2=\left(\sum_{k=0}^Np_kx^k\right)^2 = \sum_{k=0}^N\sum_{j=0}^N p_kx^kp_jx^j$$

The $x^k \cdot x^j$ can be combined as $x^{k+j}$ and instead of summing over $k$ and $j$, you can sum over $n = k+j$ and $j$ to get $$\sum_{n=0}^{N}x^n \sum_{j=0}^Np_jp_{n-j}$$

The coefficient on $x^n$ is then $\sum_{j=0}^Np_jp_{n-j}$.

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    $\begingroup$ Should not the polynomial reach a degree $2N$ if the original one was $N$? $\endgroup$
    – yes
    Commented Aug 4, 2022 at 13:17

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