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let $f$ and $g$ be infinitley differentiable functions and $a_k = \frac{f^{(k)}(a)}{k!}$ and $b_e = \frac{g^{(e)}(a)}{e!}$ be cofficients of Taylor Polynomial at $a$ then what would be the coefficients of $fg$.

rather than asking my specific question I asked this general question so other can benefit too

So I think we would need to multiply the two polynomials but that's just my intuition and I don't know how to justify it and I don't think it would be as simple.

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Your intuition is good.

Multiplying the series gives an n-th term coefficient of

$$c_n = a_0b_n + a_1b_{n-1} + \dots + a_{n-1}b_1 + a_nb_0= \sum_{i=0}^n a_i b_{n-i}$$

which is the same as doing the Taylor series of $fg$ the long way, since

$$c_n = \frac{(fg)^{(n)}(a)}{n!} = \frac{\sum_{i=0}^n \binom{n}{i}f^{(i)}(a)g^{(n-i)}(a)}{n!} = \sum_{i=0}^n \frac{f^{(i)}(a)}{i!} \frac{g^{(n-i)}(a)}{(n-i)!} = \sum_{i=0}^n a_i b_{n-i}$$

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  • $\begingroup$ In words: the coefficients of the product of two power series is the convolution of the coefficients of the factors. $\endgroup$ – J. M. is a poor mathematician Mar 26 at 6:11

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