# Taylor Maclaurin series

Can someone explain to me how this equals? I'm taking a calculus III course at the moment, and I'm doing Taylor and Maclaurin series at the moment, and this is the last step of a problem, but i don't see how this equals each other (probably because I've never dealt with a sum times sum problem before or I can't recall doing one before anyways). If it at all matters, this was the original problem: f(x) = (sinx)ln(1+x). thanks.

• Try looking at finite truncations of the sums above and multiplying them - for example, compute $\left(x-\frac{x^3}{6}\right)\left(x-\frac{x^2}{2}+\frac{x^3}{3}\right)$, and see how the product matches up. As you can imagine, taking more and more terms of each sum on the left hand side will let you work out more of the coefficients on the right hand side. – πr8 Apr 30 '16 at 10:39

Open up each parentheses and carry on the product grouping together equal powers of $\;x\;$ :

$$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)!}x^{2n+1}\cdot\sum_{n=1}^\infty\frac{(-1)^{n-1}}nx^n=\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots\right)\left(x-\frac{x^2}2+\frac{x^3}3-\ldots\right)=$$

$$=x^2-\frac12x^3+\frac16x^4+\ldots$$

FYI, the first series is $\;\sin x\;$, which converges for all (real or complex) $\;x\;$ , whereas the second one is $\;\log(1+x)\;$ , which converges only for $\;|x|<1\;$

• As it stands, I'm not sure this makes things that much easier to understand - might be worth doing some expanding to explain where the $\frac{-1}{2}, \frac{1}{6}$ coefficients come from? – πr8 Apr 30 '16 at 10:30
• @πr8 Thank you. I think this is as simple as it gets for anyone asking a question of this level (Taylor series-expansions). Grouping together coefficients of same powers of $\;x\;$ is high school stuff (with polynomials), so I can't see how the asker could not understand where those coefficients in my answer come from. Anyway, if he has any problem I'll be happy to address his doubts, if any. Not only that, I also added what functions these power series represent and in what domains... – DonAntonio Apr 30 '16 at 10:35
• Right, I'm sure they understand how to multiply polynomials as well, but I still think it would be worthwhile to carry out a few explicit terms of the expansion to explain where these coefficients come from. Otherwise all you've done is basically restate the equation in the OP's post. – πr8 Apr 30 '16 at 10:37
• @πr8 I beg to differ. I explicitly wrote the few first terms of each sum to make clear where the powers of each $\;x\;$ come from in the product. Anyway, everybody keeps saying not to give completely detailed answers and, again in my opinion, this is enough for anyone doing this kind of mathematics. Maybe we need to wait until the asker says anything, shall we? – DonAntonio Apr 30 '16 at 10:43
• Actually, I still wasn't getting at first, but the discussion between both of you made it clearer, surprisingly, and I just got it now. So thank you, both of you. Not sure why I wasn't getting it, seemed straight forward, maybe just the summation concept was just throwing me off for whatever reason. – Barqs Apr 30 '16 at 11:02