Question

Verify the product rule for Formal Power Series. Very specifically: Let f(x) be the generating function for a sequence san) and g(x) be the generating function for a sequence sbn1. Using the definitions of multiplication and differentiation of FPS, write down a formula for the derivative of [f(x)g(x)]. Then write down a formula for f(x)g(x) + f(x)g'(x), where f(x) denotes the derivative of f(x). Then show that the two formulas describe the same formal power series. (i.e. both series have the same terms.)

Answer 1

$f(x)=a_0+a_1x+a_2x^2+\cdots ,g(x)=b_0+b_1x+b_2x^2+\cdots$

Then $f(x)g(x)=(a_0+a_1x+a_2x^2+\cdots)(b_0+b_1x+b_2x^2+\cdots)$

Equals $f(x)g(x)=a_0a_1+(a_0b_1+a_1b_0)x+(a_0b_2+a_1b_1+a_2b_1)x^2+\cdots$

Its derivative is $(f(x)g(x))'=(a_0b_1+a_1b_0)+2(a_0b_2+a_1b_1+a_2b_1)x+\cdots$

And $f'(x)=a_1+2a_2x+\cdots ,g'(x)=b_1+2b_2x+\cdots$

So that $f'(x)g(x)=(a_1+2a_2x+\cdots)(b_0+b_1x+b_2x^2+\cdots)$

$f'(x)g(x)=b_0a_1+(2a_2b_0+a_1b_1)x+\cdots$ and

$f(x)g'(x)=a_0b_1+(2a_0b_2+a_1b_1)x^2+\cdots$

And so

aibi)2.

That is, $f'(x)g(x)+f(x)g'(x)=(a_0b_1+a_1b_0)+(2a_2b_0+2a_1b_1+2a_0b_2)x+\cdots$

Which we note is same as $(f(x)g(x))'=(a_0b_1+a_1b_0)+2(a_0b_2+a_1b_1+a_2b_1)x+\cdots$

Thus, we have $f'(x)g(x)+f(x)g'(x)=(f(x)g(x))'$ for formal power series

$\blacksquare$

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