Multiplying power series (1): Long multiplication

We can set out to multiply power series the same way we do with polynomials, numbers or fractions. We will first introduce the scheme of long multiplication and its difficulty, then we will move on to define the Cauchy products of two power series. The latter method has the advantage of correcting the misalignment of the former.

Series representation for $e^x\sin(x)$

We know that

$e^x= 1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{x^6}{6!}+\dfrac{x^7}{7!}+\dfrac{x^{8}}{8!}+\dfrac{x^{9}}{9!}+\dfrac{x^{10}}{10!}+O(x^{11})$

$\sin(x)=  x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\dfrac{x^9}{9!}-\dfrac{x^{11}}{11!}+\dfrac{x^{13}}{13!}-\dfrac{x^{15}}{15!}+\dfrac{x^{17}}{17!}+O(x^{19})$

$$\begin{equation} \frac{     \frac{         \begin{array}[b]{r}                 1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{x^6}{6!}+\dfrac{x^7}{7!}+\dfrac{x^{8}}{8!}+\dfrac{x^{9}}{9!}+\dfrac{x^{10}}{10!}...    \\             \times~~~  x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\dfrac{x^9}{9!}-\dfrac{x^{11}}{11!}+\dfrac{x^{13}}{13!}-\dfrac{x^{15}}{15!}+\dfrac{x^{17}}{17!}...         \end{array}         }         {     \begin{array}[b]{r}         x+x^2+\dfrac{x^{3}}{2!}\color{red}{+\dfrac{x^{4}}{3!}}+\dfrac{x^{5}}{4!}+\dfrac{x^{6}}{5!}+\dfrac{x^{7}}{6!}\color{red}{+\dfrac{x^{8}}{7!}}+\dfrac{x^{9}}{8!}~~~~~~~~~~~~~~~   \\         ~~-\dfrac{x^{3}}{3!}\color{red}{-\dfrac{x^{4}}{3!}}-\dfrac{x^{5}}{3!2!}-\dfrac{x^{6}}{3!3!}-\dfrac{x^{7}}{3!4!}\color{red}{-\dfrac{x^{8}}{3!5!}}-\dfrac{x^{9}}{3!6}~~    \\         ~~~    \\          \dfrac{x^{5}}{5!}+\dfrac{x^{6}}{5!}+\dfrac{x^{7}}{2!5!}\color{red}{+\dfrac{x^{8}}{3!5!}}+\dfrac{x^{9}}{2!7!} \\           -\dfrac{x^{7}}{7!}\color{red}{-\dfrac{x^{8}}{7!}}-\dfrac{x^{9}}{2!7!}\\     \end{array}     } } { x+x^2+\dfrac{x^3}{3}-0-\dfrac{x^5}{30}-\dfrac{x^6}{90}-\dfrac{x^7}{630}-0+\dfrac{x^9}{22680} } \end{equation}$$

Terms in red are those that are totally eliminated

$e^x\sin(x)=x+x^2+\dfrac{x^3}{3}-\dfrac{x^5}{30}-\dfrac{x^6}{90}-\dfrac{x^7}{630}+\dfrac{x^9}{22680}...$

Series representation for $e^x\cos(x)$

Similarly, we know that

$e^x= 1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+\dfrac{x^6}{6!}+\dfrac{x^{7}}{7!}+\dfrac{x^{8}}{8!}+\dfrac{x^{9}}{9!}++\dfrac{x^{10}}{10!}+O(x^{11})$

$\cos(x)=  1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\dfrac{x^8}{8!}-\dfrac{x^{10}}{10!}+\dfrac{x^{12}}{12!}-\dfrac{x^{14}}{14!}+\dfrac{x^{16}}{16!}+O(x^{18})$


$$\begin{equation} \frac{     \frac{         \begin{array}[b]{r}                 1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+\dfrac{x^6}{6!}+\dfrac{x^{7}}{7!}+\dfrac{x^{8}}{8!}+\dfrac{x^{9}}{9!}+\dfrac{x^{10}}{10!}...    \\             \times~~~ 1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\dfrac{x^8}{8!}-\dfrac{x^{10}}{10!}+\dfrac{x^{12}}{12!}-\dfrac{x^{14}}{14!}+\dfrac{x^{16}}{16!}...         \end{array}         }         {     \begin{array}[b]{r}        1+x\color{red}{+\dfrac{x^2}{2!}}+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}\color{red}{+\dfrac{x^6}{6!}}+\dfrac{x^7}{7!}+\dfrac{x^{8}}{8!}+\dfrac{x^{9}}{9!}+\dfrac{x^{10}}{10!}...   \\         +~~\color{red}{-\dfrac{x^{2}}{2!}}-\dfrac{x^{3}}{2!}-\dfrac{x^{4}}{2!2!}-\dfrac{x^{5}}{2!3!}\color{red}{-\dfrac{x^{6}}{2!4!}}-\dfrac{x^{7}}{2!5!}-\dfrac{x^{8}}{2!6!}~~    \\         ~~~    \\          \dfrac{x^{4}}{4!}+\dfrac{x^{5}}{4!}\color{red}{+\dfrac{x^{6}}{2!4!}}+\dfrac{x^{7}}{3!4!}+\dfrac{x^{8}}{4!4!} \\           \color{red}{-\dfrac{x^{6}}{6!}}-\dfrac{x^{7}}{6!}-\dfrac{x^{8}}{2!6!}\\            \dfrac{x^{8}}{8!} ~\\     \end{array}     } } { 1+x+0-\dfrac{x^3}{3}-\dfrac{x^4}{6}-\dfrac{x^5}{30}-0+\dfrac{x^7}{630}+\dfrac{x^8}{2520} ...} \end{equation}$$

$e^x\cos(x)=1+x-\dfrac{x^3}{3}-\dfrac{x^4}{6}-\dfrac{x^5}{30}+\dfrac{x^7}{630}+\dfrac{x^8}{2520}+...$

Series representation for $\sin(x)\cos(x)$


$$\begin{equation} \frac{     \frac{         \begin{array}[b]{r}                 1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\dfrac{x^8}{8!}-\dfrac{x^{10}}{10!}+\dfrac{x^{12}}{12!}-\dfrac{x^{14}}{14!}+\dfrac{x^{16}}{16!}...    \\             \times~~~ x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\dfrac{x^9}{9!}-\dfrac{x^{11}}{11!}+\dfrac{x^{13}}{13!}-\dfrac{x^{15}}{15!}+\dfrac{x^{17}}{17!}...         \end{array}         }         {     \begin{array}[b]{r}        x-\dfrac{x^3}{2!}+\dfrac{x^5}{3!}-\dfrac{x^7}{7!}+\dfrac{x^9}{8!}-\dfrac{x^{11}}{10!}+\dfrac{x^{13}}{12!}-\dfrac{x^{15}}{14!}+\dfrac{x^{17}}{16!}...   \\         +~~-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{2!3!}-\dfrac{x^{7}}{3!4!}+\dfrac{x^{9}}{3!6!}-\dfrac{x^{11}}{3!8!}+\dfrac{x^{13}}{3!10!}-\dfrac{x^{15}}{3!12!}+\dfrac{x^{17}}{3!14!}~~    \\         ~~~    \\          \dfrac{x^{5}}{5!}-\dfrac{x^{7}}{2!5!}+\dfrac{x^{9}}{4!5!}-\dfrac{x^{11}}{5!6!}+\dfrac{x^{13}}{5!8!}-\dfrac{x^{15}}{5!10!}+\dfrac{x^{17}}{5!12!} \\           -\dfrac{x^{7}}{7!}+\dfrac{x^{9}}{2!7!}-\dfrac{x^{11}}{4!7!}+\dfrac{x^{13}}{6!7!}-\dfrac{x^{15}}{7!8!}+\dfrac{x^{17}}{7!10!}\\            \dfrac{x^{9}}{9!}-\dfrac{x^{11}}{2!9!}+\dfrac{x^{13}}{4!9!}-\dfrac{x^{15}}{6!9!}+\dfrac{x^{17}}{8!9!}~~\\            -\dfrac{x^{11}}{11!}+\dfrac{x^{13}}{2!11!}-\dfrac{x^{15}}{4!11!}+\dfrac{x^{17}}{6!11!}~~\\              \dfrac{x^{13}}{13!}-\dfrac{x^{15}}{2!13!}+\dfrac{x^{17}}{4!13!}~~\\              -\dfrac{x^{15}}{15!}+\dfrac{x^{17}}{2!15!}~~\\              \dfrac{x^{17!}}{17!} \end{array}     } } { x-\dfrac{2x^3}{3}+\dfrac{2x^5}{15}-\dfrac{4x^7}{315}+\dfrac{2x^9}{2835}-\dfrac{4x^{11}}{155925}+\dfrac{4x^{13}}{6081075}-\dfrac{8x^{15}}{638512875}+\dfrac{2x^{17}}{10854718875} } \end{equation}$$

$\sin(x)\cos(x)=x-\dfrac{2x^3}{3}+\dfrac{2x^5}{15}-\dfrac{4x^7}{315}+\dfrac{2x^9}{2835}-\dfrac{4x^{11}}{155925}+\dfrac{4x^{13}}{6081075}-\dfrac{8x^{15}}{638512875}+....$

From here we can calculate the series for $2\sin(x)\cos(x)=\sin(2x)$. Admittedly this can be achieved easier by using the method of substitution, but we derive the series just for convenience.

$2\sin(x)\cos(x)=2\left(x-\dfrac{2x^3}{3}+\dfrac{2x^5}{15}-\dfrac{4x^7}{315}+\dfrac{2x^9}{2835}-\dfrac{4x^{11}}{155925}+\dfrac{4x^{13}}{6081075}-\dfrac{8x^{15}}{638512875}+\right)....$

$\sin(2x)=2x-\dfrac{4x^3}{3}+\dfrac{4x^5}{15}-\dfrac{8x^7}{315}+\dfrac{4x^9}{2835}-\dfrac{8x^{11}}{155925}+\dfrac{8x^{13}}{6081075}-\dfrac{16x^{15}}{638512875}+....$

Weakness of long multiplication method:

The weakness of long multiplication is that when we multply two power series, their terms may not be well aligned, which make computation harder and messier.

Consider the following example. We simply switch the arrangement of $e^x\cos(x)$:


$$\begin{equation} \frac{     \frac{         \begin{array}[b]{r}                  1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\dfrac{x^8}{8!}-\dfrac{x^{10}}{10!}+\dfrac{x^{12}}{12!}-\dfrac{x^{14}}{14!}+\dfrac{x^{16}}{16!}....    \\             \times~~~ 1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{x^6}{6!}+\dfrac{x^7}{7!}+\dfrac{x^{8}}{8!}+\dfrac{x^{9}}{9!}+\dfrac{x^{10}}{10!} ...         \end{array}         }         {     \begin{array}[b]{r}        1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\dfrac{x^8}{8!}-\dfrac{x^10}{10!}+\dfrac{x^{12}}{12!}-\dfrac{x^{14}}{14!}+\dfrac{x^{16}}{16!}...   \\         +~~\color{green}{x-\dfrac{x^{3}}{2!}+\dfrac{x^{5}}{4!}-\dfrac{x^{7}}{6!}+\dfrac{x^{9}}{8!}-\dfrac{x^{11}}{10!}+\dfrac{x^{13}}{12!}-\dfrac{x^{15}}{15!}+\dfrac{x^{17}}{16!}~~}    \\         ~~~    \\          \dfrac{x^{2}}{2!}+\dfrac{x^{4}}{2!2!}+\dfrac{x^{6}}{2!4!}+\dfrac{x^{8}}{2!6!}+\dfrac{x^{10}}{2!8!}-\dfrac{x^{12}}{2!10!}+\dfrac{x^{14}}{2!12!} \\ \end{array}     } } { } \end{equation}$$

The line that is colored in green gets in the way of our computation. The terms are not well-aligned, making calculation harder than it should be. For this reason, in the next article, I will introduce a method of multiplying two power series without setting out the scheme of long multiplication. This method is called the Cauchy product.