Multiplying power series (2): Cauchy product

Given two power series:

$A= a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+...$

$B= b_0+b_1x+b_2x^2+b_3x^3+b_4x^4+...$

The Cauchy product of two power series are defined as below:

$A\cdot B= a_0b_0+(a_0b_1+a_1b_0)x+(a_0b_2+a_1b_1+a_2b_0)x^2+...$

$(a_0b_3+a_1b_2+a_2b_1+a_3b_0)x^3+(a_0b_4+a_1b_3+a_2b_2+a_3b_1+a_4b_0)x^4$

From here, we can construct the "Cauchy triangle" to exhaustively multiply the coefficients of two power series. We simply collect and group terms that is multiplied with the same degree of $x$. The result is the triangle below:



$[x^0]a_0b_0$

$[x^1]a_0b_1+a_1b_0$

$[x^2]a_0b_2+a_1b_1+a_2b_0$

$[x^3]a_0b_3+a_1b_2+a_2b_1+a_3b_0$

$[x^4]a_0b_4+a_1b_3+a_2b_2+a_3b_1+a_4b_0$

$[x^5]a_0b_5+a_1b_4+a_2b_3+a_3b_2+a_4b_1+a_5b_0$

$[x^6]a_0b_6+a_1b_5+a_2b_4+a_3b_3+a_4b_2+a_5b_1+a_6b_0$

$[x^7]a_0b_7+a_1b_6+a_2b_5+a_3b_4+a_4b_3+a_5b_2+a_6b_1+a_7b_0$

$[x^8]a_0b_8+a_1b_7+a_2b_6+a_3b_5+a_4b_4+a_5b_3+a_6b_2+a_7b_1+a_8b_0$

$[x^9]a_0b_9+a_1b_8+a_2b_7+a_3b_6+a_4b_5+a_5b_4+a_6b_3+a_7b_2+a_8b_1+a_9b_0$

$[x^{10}]a_0b_10+a_1b_9+a_2b_8+a_3b_7+a_4b_6+a_5b_5+a_6b_4+a_7b_3+a_8b_2+a_9b_1+a_{10}b_0$

This method of multiplying series avoid the issue of misalignment of the long multiplication technique. It can be generalized to form the method of indeterminate coefficients, which I will introduce in another post.

Series representations for $\sin^2(x)$

To obtain the series for this function, rewrite $\sin^2(x)$ as $\sin(x)\cdot\sin(x)$. Let $a_n$ denotes the coefficients of the first $\sin(x)$ and $b_n$ denotes the coefficieints of $\sin(x)$. We notice that $\sin(x)$ is an odd function, therefore, all even terms are going to be $0$. The coeffcients are:

$a_0=0; b_0=0$

$a_1=1; b_1=1$

$a_3=-\dfrac{1}{3!}; b_3=-\dfrac{1}{3!}$

$a_5=\dfrac{1}{5!}; b_5=\dfrac{1}{5!}$

$a_7=-\dfrac{1}{7!}; b_7=-\dfrac{1}{7!}$

$a_9=\dfrac{1}{9!}; b_9=\dfrac{1}{9!}$

$a_{11}=-\dfrac{1}{11!}; b_{11}=-\dfrac{1}{11!}$

Notice that $sin^2(x)$ is an even function. Therefore, all odd terms are going to be $0$. If we apply the Cauchy triangle, we will have:

$[x^0]a_0b_0$

$[x^0]=0\cdot0=0$

$[x^2]a_0b_2+a_1b_1+a_2b_0$

$[x^2]=0\cdot\left(-\dfrac{1}{3!}\right)+1\cdot1 \left(-\dfrac{1}{3!}\right)\cdot 0 =1$

$[x^4]a_0b_4+a_1b_3+a_2b_2+a_3b_1+a_4b_0$

$[x^4]=0+\left(-\dfrac{1}{3!}\right)+0+\left(-\dfrac{1}{3!}\right)+0=-\dfrac{1}{3}$

$[x^6]a_0b_6+a_1b_5+a_2b_4+a_3b_3+a_4b_2+a_5b_1+a_6b_0$

$[x^6]=0+\left(\dfrac{1}{5!}\right)+0+\left(-\dfrac{1}{3!}\right)\left(-\dfrac{1}{3!}\right)+0+\dfrac{1}{5!}=\dfrac{2}{45}$

$[x^8]a_0b_8+a_1b_7+a_2b_6+a_3b_5+a_4b_4+a_5b_3+a_6b_2+a_7b_1+a_8b_0$

$[x^8]=0+\left(-\dfrac{1}{7!}\right)+0+\left(-\dfrac{1}{3!}\right)\left(\dfrac{1}{5!}\right)+0+\left(\dfrac{1}{5!}\right)\left(-\dfrac{1}{3!}\right)+0+\left(-\dfrac{1}{7!}\right)=-\dfrac{1}{315}$

$[x^{10}]a_0b_10+a_1b_9+a_2b_8+a_3b_7+a_4b_6+a_5b_5+a_6b_4+a_7b_3+a_8b_2+a_9b_1+a_{10}b_0$

$[x^{10}]= 0+\left(\dfrac{1}{9!}\right)+0+\left(-\dfrac{1}{3!}\right)\left(-\dfrac{1}{7!}\right)+0+\left(\dfrac{1}{5!}\right)\left(\dfrac{1}{5!}\right)+0+\left(-\dfrac{1}{7!}\right)\left(-\dfrac{1}{3!}\right)+0$
$+\dfrac{1}{9!}=\dfrac{2}{14175}$

$[x^{12}]a_0b_{12}+a_1b_{11}+a_2b_{10}+a_3b_9+a_4b_8+a_5b_7+a_6b_6+a_7b_5+a_8b_4+$

$a_9b_3+a_{10}b_2+a_{11}b_1+a_{12}b_0$

$[x^{12}]= 0+\left(-\dfrac{1}{11!}\right)+0+\left(-\dfrac{1}{3!}\right)\left(\dfrac{1}{9!}\right)+0+\left(\dfrac{1}{5!}\right)\left(-\dfrac{1}{7!}\right)+0+\left(\dfrac{1}{7!}\right)\left(-\dfrac{1}{3!}\right)+0+$

$\left(\dfrac{1}{9!}\right)\left(-\dfrac{1}{3!}\right)-\dfrac{1}{11!}=-\dfrac{2}{467775}$

So we are finally able to derive the series representation for $\sin^2(x)$:

$\sin^2(x)=x^2-\dfrac{1}{3}x^4+\dfrac{2}{45}x^6-\dfrac{1}{315}x^8+\dfrac{2}{14715}x^{10}-\dfrac{2}{467775}x^{12}+O(x^{14})$

Series representations for $\cos^2(x)$

Since $\cos(x)$ is an even function, all odd terms are zero. The coefficients are:

$a_0=1; b_0=1$

$a_2=-\dfrac{1}{2!}; b_2=-\dfrac{1}{2!}$

$a_4=\dfrac{1}{4!}; b_4=\dfrac{1}{4!}$

$a_6=-\dfrac{1}{6!}; b_6=-\dfrac{1}{6!}$

$a_8=\dfrac{1}{8!}; b_8=\dfrac{1}{8!}$

$a_{12}=-\dfrac{1}{12!}; b_{12}=-\dfrac{1}{12!}$

Notice that $\cos^2(x)$ is an even function. Therefore, all odd terms are going to be $0$. If we apply the Cauchy triangle, we will have:

$[x^0]a_0b_0$

$[x_0]=1\cdot1=1$

$[x^2]a_0b_2+a_1b_1+a_2b_0$

$[x^2]=-\dfrac{1}{2!}+0-\dfrac{1}{2!}\cdot 0 =-1$

$[x^4]a_0b_4+a_1b_3+a_2b_2+a_3b_1+a_4b_0$

$[x^4]=\dfrac{1}{4!}+0+\left(-\dfrac{1}{2!}\right)\left(-\dfrac{1}{2!}\right)+0+\dfrac{1}{4!}=\dfrac{1}{3}$

$[x^6]a_0b_6+a_1b_5+a_2b_4+a_3b_3+a_4b_2+a_5b_1+a_6b_0$

$[x^6]=-\dfrac{1}{6!}+0+\left(-\dfrac{1}{2!}\right)\left(-\dfrac{1}{6!}\right)+0+\left(\dfrac{1}{4!}\right)\left(-\dfrac{1}{2!}\right)+0+\left(-\dfrac{1}{6!}\right)=-\dfrac{2}{45}$

$[x^8]a_0b_8+a_1b_7+a_2b_6+a_3b_5+a_4b_4+a_5b_3+a_6b_2+a_7b_1+a_8b_0$

$[x^8]=\dfrac{1}{8!}+0+\left(-\dfrac{1}{2!}\right)\left(-\dfrac{1}{6!}\right)+0+\left(\dfrac{1}{4!}\right)\left(\dfrac{1}{4!}\right)+0+\left(-\dfrac{1}{6!}\right)\left(-\dfrac{1}{2!}\right)+0+\dfrac{1}{8!}$
$=\dfrac{1}{315}$

$[x^{10}]a_0b_10+a_1b_9+a_2b_8+a_3b_7+a_4b_6+a_5b_5+a_6b_4+a_7b_3+a_8b_2+a_9b_1+a_{10}b_0$

$[x^{10}]= -\dfrac{1}{10!}+0+\left(-\dfrac{1}{2!}\right)\left(\dfrac{1}{8!}\right)+0+\left(\dfrac{1}{4!}\right)\left(-\dfrac{1}{6!}\right)+0+\left(-\dfrac{1}{6!}\right)\left(\dfrac{1}{4!}\right)+0$
$+\left(\dfrac{1}{8!}\right)\left(-\dfrac{1}{2!}\right)+0-\dfrac{1}{10}=-\dfrac{2}{14175}$

$[x^{12}]a_0b_{12}+a_1b_{11}+a_2b_{10}+a_3b_9+a_4b_8+a_5b_7+a_6b_6+a_7b_5+a_8b_4+$
$a_9b_3+a_{10}b_2+a_{11}b_1+a_{12}b_0$

$[x^{12}]= \dfrac{1}{12!}+0+\left(-\dfrac{1}{2!}\right)\left(-\dfrac{1}{10!}\right)+0+\left(\dfrac{1}{4!}\right)\left(\dfrac{1}{8!}\right)+0+\left(-\dfrac{1}{6!}\right)\left(-\dfrac{1}{6!}\right)+0$
$+\left(\dfrac{1}{8!}\right)\left(\dfrac{1}{4!}\right)+0+\left(-\dfrac{1}{10!}\right)\left(-\dfrac{1}{2!}\right)+\dfrac{1}{12!}=\dfrac{2}{467775}$

So we are finally able to derive the series representation for $\cos^2(x)$:


$\cos^2(x)=1-x^2+\dfrac{1}{3}x^4-\dfrac{2}{45}x^6+\dfrac{1}{315}x^8-\dfrac{2}{14715}x^{10}+\dfrac{2}{467775}x^{12}+O(x^{14})$

If we add $\sin^2(x)+\cos^2(x)$, we will have:

$\color{red}{x^2-\dfrac{1}{3}x^4+\dfrac{2}{45}x^6-\dfrac{1}{315}x^8+\dfrac{2}{14715}x^{10}-\dfrac{2}{467775}x^{12}+}...$
$+1\color{red}{-x^2+\dfrac{1}{3}x^4-\dfrac{2}{45}x^6+\dfrac{1}{315}x^8-\dfrac{2}{14715}x^{10}+\dfrac{2}{467775}x^{12}+}=1$

Red terms denote elimination

This gives us the familiar identity $\color{green}{\sin^2(x)+\cos^2(x)=1}$ 

Series representation for $\sinh^2(x)$ and $\cosh^2(x)$

The Maclaurin series for $\sinh^2(x)$ and $\cosh^2(x)$ are going to be very similar to $\sin^2(x)$ and $\cos^2(x)$, except that the signs are all positive. We can derive the series using the same method that we did for $\sin^2(x)$. This is left to the readers as an excercise.

$\sinh^2(x)=x^2+\dfrac{1}{3}x^4+\dfrac{2}{45}x^6+\dfrac{1}{315}x^8+\dfrac{2}{14715}x^{10}+\dfrac{2}{467775}x^{12}+O(x^{14})$

$\cosh^2(x)=1+x^2+\dfrac{1}{3}x^4+\dfrac{2}{45}x^6+\dfrac{1}{315}x^8+\dfrac{2}{14715}x^{10}+\dfrac{2}{467775}x^{12}+O(x^{14})$

Subtracting $\cosh^2(x)$ from $\sinh^2(x)$, we will have:

$1+\color{red}{x^2+\dfrac{1}{3}x^4+\dfrac{2}{45}x^6+\dfrac{1}{315}x^8+\dfrac{2}{14715}x^{10}+\dfrac{2}{467775}x^{12}+}...-$
$\color{red}{x^2-\dfrac{1}{3}x^4-\dfrac{2}{45}x^6-\dfrac{1}{315}x^8-\dfrac{2}{14715}x^{10}-\dfrac{2}{467775}x^{12}}...=1$


So we obtain the familiar identity: $\color{green}{\cosh^2(x)-\sinh^2(x)=1}$