Deriving the power series for $\cot(x)$ and $\coth(x)$

 In the previous post, we have successfully derived the power series for $\dfrac{x}{e^x-1}=\displaystyle\sum_{n=0}^{+\infty}\dfrac{B_nx^{n}}{n!}$. We know that the coefficients for this power series are the Bernoulli numbers. This result is important because it allows us to arrive at the power series of $\cot(x)$ and $\coth(x)$. In this post, we will see how the Bernoulli numbers are related to the expansion of $\cot(x)$ and $\coth(x)$.

(added in Feb, 11, 2021):

The derivation of both functions are particularly important to me, for it helps me to derive the summing formula of zeta function at even integers. At first, I knew that there was a connection between the power series of $\dfrac{x}{e^x-1}$, then I tried to rewrite $\cot(x)$ and $\coth(x)$ in exponential forms. After tinkering for a while, I was unable to make a connection between them and $\dfrac{x}{e^x-1}$, so I looked up on wikiproof. Their approach is exactly what I was doing, so I continued, only looking at wikiproof when I was unable to go on. Eventually the formula arrived. I was able to derive them. I believe that my symbolic manipulation ability is still very weak and deplorable, but doing these kinds of derivations certainly improve it. I fumbled at $(1)$ (see below) and re-indexing the summation formula. After seeing wikiproof, I totally understand it.

Deriving power series for $\cot(x)$:

We have proved here that $\sin(x)=\dfrac{e^{ix}-e^{-ix}}{2i}$ and $\cos(x)=\dfrac{e^{ix}+e^{-ix}}{2}$. By definition:

$$\cot(x)=\dfrac{\cos(x)}{\sin(x)}=\dfrac{e^{ix}+e^{-ix}}{2}\cdot\dfrac{2i}{e^{ix}-e^{-ix}}=i\left(\dfrac{e^{ix}+e^{-ix}}{e^{ix}-e^{-ix}}\right)\tag{1}$$

Multiply the numerator and denominator for $e^{ix}$, we have:

$$i\left(\dfrac{e^{2ix}+1}{e^{2ix}-1}\right)\tag{2}$$

This is the tricky part, we need to make the $\dfrac{2ix}{e^{2ix}-1}$ appear to use substitution to obtain the power series for $\left(\dfrac{u}{e^{u}-1}\right)$ with $u=2ix$

$$i\left(\dfrac{e^{2ix}+1}{e^{2ix}-1}\right)=i\left(\dfrac{e^{2ix}-1+2}{e^{2ix}-1}\right)=i\left(1+\dfrac{2}{e^{2ix}-1}\right)=i+\dfrac{2i}{e^{2ix}-1}\tag{3}$$

We factor out $\dfrac{1}{x}$ from $\dfrac{2i}{e^{2ix}-1}$, and we obtain:

$$i+\dfrac{1}{x}\left(\dfrac{2ix}{e^{2ix}-1}\right)=i+\dfrac{1}{x}\left(\sum_{n=0}^{+\infty}\dfrac{B_n(2ix)^{n}}{n!}\right)$$

Next, we need to "release" two terms from the sigma notation chunk above. Why do we need to release the terms and readjust the index $n=2$? Because we want to take out the first odd Bernoulli number $B_1=-\dfrac{1}{2}$. Then we can adjust the index to include all even (non-zero) Bernoulli numbers. The first 2 terms to be released is $1$ and $-ix$. We need to remember that the second term is eliminated, but the first term remained. So when we shift the index of the summation notation to $n=0$, we need to include the first term to avoid duplicating the terms.

$$i+\dfrac{1}{x}\left(\color{red}{1}-ix+\sum_{\color{blue}{n=2}}^{+\infty}\dfrac{B_n(2ix)^{n}}{n!}\right)=i+\color{red}{\dfrac{1}{x}}-i+\dfrac{1}{x}\left(\sum_{\color{blue}{n=2}}^{+\infty}\dfrac{B_n(2ix)^{n}}{n!}\right)=\dfrac{1}{x}+\dfrac{1}{x}\left(\sum_{\color{blue}{n=2}}^{+\infty}\dfrac{B_n(2ix)^{n}}{n!}\right)$$

Convert the summation index to $n=1$, and $B_n=B_{2n}$ and change all index of $n$ to $2n$ to include all even Bernoulli numbers, we have:

$$\dfrac{1}{x}+\dfrac{1}{x}\left(\sum_{\color{blue}{n=1}}^{+\infty}\dfrac{B_{2n}(2ix)^{2n}}{(2n)!}\right)$$

Shift the summation index to $n=0$, we need to eliminate $\dfrac{1}{x}$ to avoid duplicating terms. In other words $\dfrac{1}{x}$ is "absorbed" into the sum. We have:

$$\dfrac{1}{x}\left(\sum_{\color{blue}{n=0}}^{+\infty}\dfrac{B_{2n}(2ix)^{2n}}{(2n)!}\right)=\sum_{\color{blue}{n=0}}^{+\infty}\dfrac{B_{2n}2^{2n}(i^{2})^nx^{2n-1}}{(2n)!}$$

Finally, $(i^2)^n=(-1)^n$, so we arrived at the final conclusion of $\cot(x)$

$$\boxed{\sum_{n=0}^{+\infty}\dfrac{(-1)^{n}B_{2n}2^{2n}x^{2n-1}}{(2n)!}}$$

Deriving power series for $\coth(x)$

We have proved here that $\sinh(x)=\dfrac{e^{x}-e^{-x}}{2}$ and $\cosh(x)=\dfrac{e^{x}+e^{-x}}{2}$. By definition:

$$\coth(x)=\dfrac{\cosh(x)}{\sinh(x)}=\dfrac{e^{x}+e^{-x}}{2}\cdot\dfrac{2}{e^{x}-e^{-ix}}=\left(\dfrac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\right)\tag{1}$$

Multiply the numerator and denominator for $e^{x}$, we have:

$$\dfrac{e^{2x}+1}{e^{2x}-1}\tag{2}$$

This is the tricky part, we need to make the $\dfrac{2x}{e^{2x}-1}$ appear to use substitution to obtain the power series for $\left(\dfrac{u}{e^{u}-1}\right)$ with $u=2x$

$$\dfrac{e^{2x}+1}{e^{2ix}-1}=\dfrac{e^{2x}-1+2}{e^{2x}-1}=1+\dfrac{2}{e^{2x}-1}\tag{3}$$

We factor out $\dfrac{1}{x}$ from $\dfrac{2}{e^{2x}-1}$, and we obtain:

$$1+\dfrac{1}{x}\left(\dfrac{2x}{e^{2x}-1}\right)=1+\dfrac{1}{x}\left(\sum_{n=0}^{+\infty}\dfrac{B_n(2x)^{n}}{n!}\right)$$

Similarly, as above, we have

$$1+\dfrac{1}{x}\left(\color{red}{1}-x+\sum_{\color{blue}{n=2}}^{+\infty}\dfrac{B_n(2x)^{n}}{n!}\right)=1+\color{red}{\dfrac{1}{x}}-1+\dfrac{1}{x}\left(\sum_{\color{blue}{n=2}}^{+\infty}\dfrac{B_n(2x)^{n}}{n!}\right)=\dfrac{1}{x}+\dfrac{1}{x}\left(\sum_{\color{blue}{n=2}}^{+\infty}\dfrac{B_n(2x)^{n}}{n!}\right)$$

Convert the summation index to $n=1$, and $B_n=B_{2n}$ and change all index of $n$ to $2n$ to include all even Bernoulli numbers, we have:

$$\dfrac{1}{x}+\dfrac{1}{x}\left(\sum_{\color{blue}{n=1}}^{+\infty}\dfrac{B_{2n}(2x)^{2n}}{(2n)!}\right)$$

Shift the summation index to $n=0$, we need to eliminate $\dfrac{1}{x}$ to avoid duplicating terms. In other words $\dfrac{1}{x}$ is "absorbed" into the sum. We have:

$$\dfrac{1}{x}\left(\sum_{\color{blue}{n=0}}^{+\infty}\dfrac{B_{2n}(2x)^{2n}}{(2n)!}\right)=\sum_{\color{blue}{n=0}}^{+\infty}\dfrac{B_{2n}2^{2n}x^{2n-1}}{(2n)!}$$

We arrived at the final conclusion of $\coth(x)$

$$\boxed{\sum_{n=0}^{+\infty}\dfrac{B_{2n}2^{2n}x^{2n-1}}{(2n)!}}$$

Deriving power series for $\dfrac{x}{2}\cot\left(\dfrac{x}{2}\right)$

Why are we interested in the series expansion of this function? Because it is the generating function of even Bernoulli numbers. Since we have derived the series representation for $\cot(x)$ and $\coth(x)$, we can go on to easily produce the series for these functions.

$\cot(x)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^nB_{2n}2^{2n}x^{2n-1}}{(2n)!}\implies\cot\left(\dfrac{x}{2}\right)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^nB_{2n}2^{2n}\left(\dfrac{x}{2}\right)^{2n-1}}{(2n)!}$

$$=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^nB_{2n}2^{2n}x^{2n-1}}{2^{2n-1}(2n)!}=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^nB_{2n}2x^{2n-1}}{(2n)!}$$

$$\implies\dfrac{x}{2}\cot\left(\dfrac{x}{2}\right)=\boxed{\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^nB_{2n}x^{2n}}{(2n)!}}$$

Deriving power series for $\dfrac{x}{2}\coth\left(\dfrac{x}{2}\right)$

We can repeat the same process as above, utilizing the series of $\dfrac{x}{2}\cot\left(\dfrac{x}{2}\right)$ and obtain the following result:

$$\boxed{\displaystyle\sum_{n=0}^{+\infty}\dfrac{B_{2n}x^{2n}}{(2n)!}}$$

Deriving the power series of the generating function of the Bernoulli numbers.

In this thread, we are going to derive the series of $\dfrac{x}{e^x-1}$. This series is especially important because it is the generating function of the Bernoulli numbers. I have tried many times in vain to derive the series, but with the precious help of members of Mathstack Exchange, I could finally derive it.

The series is especially difficult because the $n$ order derivative is very complex as we progress. For example:

$$f(x)=\dfrac{x}{e^x-1} \implies f'(x)=\dfrac{e^x(x-1)+1}{(e^x-1)^2}\implies f'(0)=\dfrac{1(0-1)+1}{1-1}=\dfrac{1}{0}$$

We can of course rely on the L'Hopital rule to calculate the limit of $x$ approaches $0$ in all $n$ order derivatives but it is a hugely complicated calculation.

Another way must be found to derive the series for this function. The method is to rely on the equality: $$\left(\dfrac{x}{e^x-1}\right)\left(\dfrac{e^x-1}{x}\right)=1\tag{1}$$

We can easily expand $\dfrac{e^x-1}{x}$ into its power series by using the series $e^x=\sum_{n=0}^{+\infty}\dfrac{x^n}{n!}$

$$\dfrac{e^x-1}{x}=\dfrac{\color{red}{1}+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+...\color{red}{-1}}{x}=1+\dfrac{x}{2!}+\dfrac{x^2}{3!}+\dfrac{x^3}{4!}+...=\sum_{n=1}^{+\infty} \dfrac{x^{n-1}}{n!}=\sum_{n=0}^{+\infty} \dfrac{x^{n}}{(n+1)!}\tag{2}$$

Now, denote the series of $\dfrac{x}{e^x-1}=\sum_{n=0}^{+\infty}b_nx^n$ and the series of $\dfrac{e^x-1}{x}=\sum_{n=0}^{+\infty} a_nx^n$.

The coefficient of the $\sum_{n=0}^{+\infty} a_nx^n$ is $a_n=\dfrac{1}{(n+1)!}\tag{3}$

We can rewrite the equation in $(1)$ as followed:

$$\left(1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+...\right)\sum_{n=0}^{+\infty}b_nx^n=1+0x+0x^2+0x^3+...\tag{4}$$

$$=\left(\sum_{n=0}^{+\infty}b_nx^n\right)\left(\sum_{n=0}^{+\infty}a_nx^n\right)=1\tag{5}$$

We can compare the product of the coefficients $b_n$ and $a_n$ on LHS to the coefficients of the "series" on the RHS in $(4)$ and form this system of equations:

$[x^0]a_0b_0=1$

$[x^1]a_0b_1+a_1b_0=0$

$[x^2]a_0b_2+a_1b_1+a_2b_0=0$

$[x^3]a_0b_3+a_1b_2+a_2b_1+a_3b_0=0$

$[x^4]a_0b_4+a_1b_3+a_2b_2+a_3b_1+a_4b_0=0$

$[x^5]a_0b_5+a_1b_4+a_2b_3+a_3b_2+a_4b_1+a_5b_0=0$

$[x^6]a_0b_6+a_1b_5+a_2b_4+a_3b_3+a_4b_2+a_5b_1+a_6b_0=0$

Refer to this thread to review what is a Cauchy triangle when multiplying 2 power series:

From the system of equations above, we can easily deduce the value for $b_n$:

$a_0=1$

$b_0=1/a_0=\dfrac{1}{1}=1$

$b_1=-a_1b_0/a_0=-\left(\dfrac{1}{2}\cdot1\right)=-\dfrac{1}{2}$

$b_2=-(a_2b_0+a_1b_1)/a_0=-\left(\dfrac{1}{3!}\cdot1+\dfrac{1}{2!}\cdot\left(-\dfrac{1}{2}\right)\right)=\dfrac{1}{12}$

$b_3=-(a_3b_0+a_2b_1+a_1b_2)/a_0=-\left(\dfrac{1}{4!}\cdot1+\dfrac{1}{3!}\left(-\dfrac{1}{2}\right)+\dfrac{1}{2}\cdot\left(\dfrac{1}{12}\right)\right)=0$

$b_4=-(a_4b_0+a_3b_1+a_2b_2+a_1b_3)/a_0=-\left(\dfrac{1}{5!}\cdot1+\dfrac{1}{4!}\left(-\dfrac{1}{2}\right)+\dfrac{1}{3!}\\\cdot\dfrac{1}{12}+\left(\dfrac{1}{2}\cdot0\right)\right)=-\dfrac{1}{720}$

If we continue on, we will find that $b_n$ when $n$ is odd are all $0$, except that $b_1=-\dfrac{1}{2}$

We only need to compute for $b_n$ when n is even, and these are even values of $b_n$

$b_0=1$

$b_2=-\dfrac{1}{2}$

$b_4=-\dfrac{1}{720}$

$b_6=\dfrac{1}{30240}$

$b_8=-\dfrac{1}{1209600}$

$b_{10}=\dfrac{1}{47900160}$

So the coefficients for the series of $\dfrac{x}{e^x-1}$ will be:

$$\dfrac{x}{e^x-1}=1-\dfrac{1}{2}x+\dfrac{1}{12}x^2-\dfrac{1}{720}x^4+\dfrac{1}{30240}x^6-\dfrac{1}{1209600}x^8+\dfrac{1}{47900160}x^{10}+...$$

To make the Bernoulli numbers appear, we divide the denominator by $n!$. For example, for the fourth term , we divide $720$ by $4!$ and we obtain $\dfrac{1}{30}$, which the fourth Bernoulli number. We can rewrite:

$$=\color{green}{1}-\dfrac{\color{green}{1}}{\color{green}{2}\cdot1!}x+\dfrac{\color{green}{1}}{\color{green}{6}\cdot2!}x^2-\dfrac{\color{green}{1}}{\color{green}{30}\cdot4!}x^4+\dfrac{\color{green}{1}}{\color{green}{42}\cdot6!}x^6-\dfrac{\color{green}{1}}{\color{green}{30}\cdot8!}x^8+\dfrac{\color{green}{5}}{\color{green}{66}\cdot10!}x^{10}+...$$

The numbers colored in green are Bernoulli numbers.

Thus the general formula of the series is: 

$$\dfrac{x}{e^x-1}=\sum_{n=0}^{+\infty}\dfrac{B_nx^n}{n!}$$

Proving two formulae involving two trigonometric series of Euler using Euler identities

I have recently read the book "Тригонометрические ряды от Эйлера до Лебега" (Trigonometric series from Euler to Lebesgue) and on page 12, there are 2 intriguing formlae derived by Euler in 1748. The book doesn't specify from which publication these two series come. Further up on that page, there is a mention of the "введения в анализ бесконечных" (Introduction to the analysis of the infinite), which was published also in 1748. Thus I suspect that these two formulae may come from there.

The above two formulae are as followed:

$$\dfrac{1-r\cos(x)}{1-2r\cos(x)+r^2}=1+r\cos(x)+r^2\cos(2x)...=1+\sum_{n=1}^{+\infty}r^{n}\cos\left(nx\right)\tag{1}$$

$$\dfrac{r\sin(x)}{1-2r\cos(x)+r^2}=r\sin(x)+r^2\sin(2x)...=1+\sum_{n=1}^{+\infty}r^{n}\sin\left(nx\right)\tag{2}$$

In the above book, the author states that Euler didn't provide the radius of convergence for these series, he wrote: "никаких указаний на области применимости этих и подобных им разложений у Эйлера нет" (No indications of range of applications are provided by Euler for these and similar expressions) (page 13)

The radius of convergence should be $|r|<1$ 

At $r=1$ and $r=-1$, Euler obtained:

$$-\dfrac{1}{2}=\cos(x)+\cos(2x)+\cos(3x)+...\tag{3}$$

$$\dfrac{1}{2}=\cos(x)-\cos(2x)+\cos(3x)+...\tag{4}$$

Both of these trigonometric series diverge $\forall x \in \mathbb{R}$

I wish to provide a derivation of these two formulae. This is derived from an answer of Student A Level, a member of Mathstackexchange, the site is here

We proceed by working with the RHS of the equation $(1)$ and $(2)$ and try to turn it into the LHS.

We denote the series in $(1)$ as $C$, and the series in $(2)$ as B

$$C=1+r\cos x+r^2\cos2x+r^3\cos3x+\cdots\\$$

$$S=r\sin x+r^2\sin2x+r^3\sin3x+\cdots\\$$

We will use the Euler identity $\color{green}{e^{ix}=\cos(x)+i\sin(x)}$ and $\color{green}{e^{inx}=\cos(nx)+i\sin(nx)}$

We multiply $S$ with imaginary number $i=\sqrt{-1}$, and obtain the new series:

$$iS=ri\sin(x)+r^2i\sin(2x)+r^3i\sin(3x)...$$

Summing these two series, we have

$$C+iS=1+r[\cos(x)+i\sin(x)]+r^2[(\cos(2x)+i\sin(2x)]+r^3[\cos(3x)+i\sin(3x)]...$$

Using Euler identity, we have:

$$C+iS=1+re^{ix}+(re^{ix})^2+(re^{ix})^{3}+(re^{ix})^4+...$$

This is, in fact, a geometric series, and the general formula is:

$$\dfrac{1}{1-kt}=1+kt+(kt)^2+(kt)^3+(kt)^4+...$$

Let $r=k$ and $t=e^{ix}$

We have $$\dfrac{1}{1-re^{ix}}$$

Let $r=k$ and $t=e^{ix}$

We have $$\dfrac{1}{1-re^{ix}}=1+re^{ix}+(re^{ix})^2+(re^{ix})^{3}+(re^{ix})^4+...$$

On the LHS: we mutiply the fraction with $1-re^{-ix}$, so we will have

$$\dfrac{1-re^{-ix}}{(1-re^{ix})(1-re^{-ix})}=\dfrac{1-r(\cos(x)-ir\sin(x))}{1-re^{-ix}-re^{ix}+r^2}$$

This is the tricky part of the proof, we wish to turn

$$1-re^{-ix}-re^{ix}+r^2=1-r(e^{-ix}+e^{ix})+r^2=1-2r\cos(x)+r^2$$

We can work backward and realize that $\cos(x)=\dfrac{e^{ix}+e^{-ix}}{2}$, so we need to multiply $re^{-ix}-re^{ix}$ with $2r\left(\dfrac{e^{-ix}+e^{ix}}{2}\right)$. So we will have:

$$1-re^{-ix}-re^{ix}+r^2=1-2r\left(\dfrac{e^{-ix}+e^{ix}}{2}\right)+r^2=1-2r\cos(x)+r^2$$

Thus we have:

$$\dfrac{1-r(\cos(x)-i\sin(x)}{1-2r\cos(x)+r^2}=\dfrac{1-r\cos(x)+ir\sin(x)}{1-2r\cos(x)+r^2}=\dfrac{1-r\cos(x)}{1-2r\cos(x)+r^2}+\dfrac{ir\sin(x)}{1-2r\cos(x)+r^2}$$

Equating the imaginary and the real part, we thus have:

$$\boxed{\color{green}{A=\dfrac{1-r\cos(x)}{1-2r\cos(x)+r^2}=\sum_{n=1}^{+\infty}r^{n}\cos\left(nx\right)}}$$

$$\boxed{\color{green}{B=\dfrac{r\sin(x)}{1-2r\cos(x)+r^2}=\sum_{n=1}^{+\infty}r^{n}\sin\left(nx\right)}}$$

Derivation of Euler formula, Euler Identity and de Moivre formula

Euler formula

The proof for Euler formula and identity using power series is ubiquitous on the internet. However, I will include it here to amplify the power of substitution. The derivation of Euler formula is useful in obtaining the exponential form of $\sin(x)$ and $\cos(x)$.

We have:

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

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

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

We are looking for the series representation of $e^{ix}$, so we can substitute $ix$ into $x$ in the series of $e^x$, we have:

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

where $i$ is the imaginary number $\sqrt{-1}$

We have $i^1=i$; $i^2=-1$; $i^3=i^2\cdot i=-i; i^4=1$; $i^5=i$; $i^6=-1$; $i^7=-i$

Thus:

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

Rearrange the terms and factor $i$, we will have

$e^{ix}=\color{blue}{\left(1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}...\right)}+\color{green}{i\left(x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}...\right)}$

$e^{ix}=\color{blue}{\cos(x)}+\color{green}{i\sin(x)}$ (QED)

Euler identity

From this, we can easily obtain the Euler's identity:

$e^{ix}=\cos(x)+i\sin(x)$

$\implies e^{i\pi}=\cos(\pi)+i\sin(\pi)=-1+0$

$\implies \boxed{\color{green}{e^{i\pi}=-1}}$

Euler formula for hyperbolic function

Similar to the Euler formula, we can also derive

$\color{green}{\sinh(x)=x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{x^7}{7!}+\dfrac{x^9}{9!}+\dfrac{x^{11}}{11!}+\dfrac{x^{13}}{13!}+O(x^{15})}$

$\color{blue}{\cosh(x)=  1+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+\dfrac{x^{10}}{10!}+\dfrac{x^{12}}{12!}+O(x^{14})}$

$e^x=\color{blue}{1}+\color{green}{x}+\color{blue}{\dfrac{x^2}{2!}}+\color{green}{\dfrac{x^3}{3!}}+\color{blue}{\dfrac{x^4}{4!}}+\color{green}{\dfrac{x^5}{5!}}+\color{blue}{\dfrac{x^6}{6!}}+O(x^7)$

$e^x=\color{blue}{\cosh(x)}+\color{green}{\sinh(x)}$

Exponential form of $\sin(x)$

We can also derive two principal identities for $\sin(x)$ and $\cos(x)$, and then find the relationship between them and $\sinh(x)$ and $\cosh(x)$

$e^{ix}-e^{-ix}=\cos(x)+i\sin(x)-[\cos(-x)+i\sin(-x)]$

$=\cos(x)+i\sin(x)-\cos(x)-(-i\sin(x))$ (Since $\sin(x)$ is odd)

$=\color{red}{\cos(x)}+i\sin(x)-\color{red}{\cos(x)}+i\sin(x)$ (Red terms denote elimination)

$=2i\sin(x)\implies \boxed{\color{green}{\dfrac{e^{ix}-e^{-ix}}{2i}=\sin(x)}}$

Exponential form of $\sinh(x)$

$e^{x}-e^{-x}=\cosh(x)+\sinh(x)-[\cosh(-x)+\sinh(-x)]$

$=\cosh(x)+\sinh(x)-\cosh(x)-(\sinh(x)$ (Since $\sinh(x)$ is odd)

$=\color{red}{\cosh(x)}+\sinh(x)-\color{red}{\cosh(x)}+\sinh(x)$ (Red terms denote elimination)

$=2\sinh(x)\implies \boxed{\color{green}{\dfrac{e^{x}-e^{-x}}{2}=\sinh(x)}}$

Proof that $\sin(ix)=i\sinh(x)$

The relationship between $\sin(x)$ and $\sinh(x)$ can be expressed as followed:

Recall that $\sinh(x)=\dfrac{e^x-e^{-x}}{2}\implies -\sinh(x)=\dfrac{e^{-x}-e^{x}}{2}$

$\sin(ix)=\dfrac{e^{i^2x}-e^{-i^2x}}{2i}=\dfrac{e^{-x}-e^{x}}{2i}=-\dfrac{\sinh(x)}{i}=i^2\dfrac{\sinh(x)}{i}=i\sinh(x)$

Exponential form of $\cos(x)$

$e^{ix}+e^{-ix}=\cos(x)+i\sin(x)+[\cos(-x)+i\sin(-x)]$

$=\cos(x)+\color{red}{i\sin(x)}+\cos(x)-\color{red}{i\sin(x)}$ 

$=2\cos(x)\implies \boxed{\color{green}{\dfrac{e^{ix}+e^{-ix}}{2}=\cos(x)}}$

Exponential form of $\cosh(x)$

$e^{x}+e^{-x}=\cosh(x)+\sinh(x)+[\cosh(-x)+\sinh(-x)]$

$=\cosh(x)+\color{red}{\sinh(x)}+\cosh(x)-\color{red}{\sinh(x)}$ 

$=2\cosh(x)\implies \boxed{\color{green}{\dfrac{e^{x}+e^{-x}}{2}=\cosh(x)}}$

Proof that $\cos(ix)=\cosh(x)$

The relationship between $\cos(x)$ and $\cosh(x)$ is that:

$\cos(ix)=\dfrac{e^{i^2x}+e^{(-i)^2x}}{2}=\boxed{\color{green}{\dfrac{e^{-x}+e^x}{2}=\cosh(x)}}$

De Moivre Formula

This formula can be easily proven once the proof of Euler formula is established:

$[\cos(x)+i\sin(x)]^n=(e^{ix})^n=e^{inx}=\color{green}{\cos(nx)+i\sin(nx)}$

Prove de Moivre Formula by induction

To be more rigorous, we can prove De Moivre formula by using induction:

What we need to prove is 

$$[\cos(x)+i\sin(x)]^n=\cos(nx)+i\sin(nx)$$

We prove for the base case $n=1$, we have

We have 

$\cos(x)+i\sin(x)=\cos(x)+i\sin(x)$ (Proved)

Let $n=k$, assume that the equality holds for $k$

$$\color{green}{[\cos(x)+i\sin(x)]^k=\cos(kx)+i\sin(kx)}$$

We prove that this is true for $k+1$, we have for the $LHS$:

$$=[\cos(x)+i\sin(x)]^{k+1}$$

$$=\color{green}{[\cos(x)+i\sin(x)]^{k}}[\cos(x)+i\sin(x)]$$

$$=[\color{green}{\cos(kx)+i\sin(kx)]}[\cos(x)+i\sin(x)]$$

$$=[\cos(kx)\cos(x)+i\cos(kx)\sin(x)+i\sin(kx)\cos(x)+i^{2}\sin(kx)\sin(x)]$$

$$=[\cos(kx)\cos(x)+i\cos(kx)\sin(x)+i\sin(kx)\cos(x)-\sin(kx)\sin(x)]$$

$$=\color{blue}{[\cos(kx)\cos(x)-\sin(kx)\sin(x)}+i\color{orange}{[\cos(kx)\sin(x)+\sin(kx)\cos(x)]}$$

Apply the trigonometric formula:

$$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$$

$$\sin(x+y)=\sin(x)\cos(y)+\sin(y)\sin(x)$$

We have

$$\color{blue}{cos(kx+x)}+i[\color{orange}{sin(kx+x)}]=\cos[(k+1)x]+i\sin[(k+1)x]$$

$\color{blue}{cos[(kx+1)x]}+i[\color{orange}{sin[(kx+1)x}]]=\cos[(k+1)x]+i\sin[(k+1)x]$ (Q.E.D)

Method of substitution

The method of substitution is the among the most basic technique that can be applied to infinite power series. Its use is simple to grasp and easy to master. We will first start with a couple of simple series and build up from there.

Series representation of $e^{-x}$

We know that:

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

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

$=1-x+\dfrac{x^2}{2!}-\dfrac{x^3}{3!}+\dfrac{x^4}{4!}-\dfrac{x^5}{5!}+\dfrac{x^6}{6!}+O(x^8)$

Series representation of $e^{-x^2}$

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

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

$=1-x^2+\dfrac{x^4}{2!}-\dfrac{x^6}{3!}+\dfrac{x^8}{4!}-\dfrac{x^{10}}{5!}+\dfrac{x^{12}}{6!}+O(x^{14})$

Series representation of $\dfrac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}$ (Normal distribution function)

$e^{-x^2}=1-x^2+\dfrac{x^4}{2!}-\dfrac{x^6}{3!}+\dfrac{x^8}{4!}-\dfrac{x^{10}}{5!}+\dfrac{x^{12}}{6!}+O(x^{14})$

$\dfrac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}=\dfrac{1}{\sqrt{2\pi}}\left(1-\dfrac{x^2}{2}+\dfrac{x^4}{4\cdot 2!}-\dfrac{x^6}{8\cdot 3!}+\dfrac{x^8}{16\cdot 4!}-\dfrac{x^{10}}{32\cdot 5!}+\dfrac{x^{12}}{64\cdot 6!}+O(x^{14})\right)$

Series representation of $e^{2x}$

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

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

$=1+2x+\dfrac{4x^2}{2!}+\dfrac{8x^3}{3!}+\dfrac{16x^4}{4!}+\dfrac{32x^5}{5!}+\dfrac{64x^6}{6!}+O(x^8)$

$=1+2x+2x^2+\dfrac{4x^3}{3}+\dfrac{2x^4}{3}+\dfrac{4x^5}{15}+\dfrac{4x^6}{45}+O(x^8)$

Series representation of $\sin(2x)$

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

$\sin(2x)=2x-\dfrac{(2x)^3}{3!}+\dfrac{(2x)^5}{5!}-\dfrac{(2x)^7}{7!}+\dfrac{(2x)^9}{9!}-\dfrac{(2x)^{11}}{11!}+\dfrac{(2x)^{13}}{13!}+O(x^{15})$

$=2x-\dfrac{8x^3}{3!}+\dfrac{32x^5}{5!}-\dfrac{128x^7}{7!}+\dfrac{512x^9}{9!}-\dfrac{2048x^{11}}{11!}+\dfrac{8192x^{13}}{13!}+O(x^{15})$

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

$$ \Leftrightarrow  \sum_{k=0}^{\infty}\dfrac{(-1)(2x)^{2n+1}}{(2n+1)!}$$

Series representation of $\cos(2x)$

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

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

$=1-\dfrac{4x^2}{2!}+\dfrac{16x^4}{4!}-\dfrac{64x^6}{6!}+\dfrac{256x^8}{8!}-\dfrac{(1024x)^{10}}{10!}+\dfrac{(4096x)^{12}}{12!}+...$

$=1-2x^2+\dfrac{2x^4}{3}-\dfrac{4x^6}{45}+\dfrac{2x^8}{315}-\dfrac{4x^{10}}{14175}+\dfrac{4x^{12}}{467775}+...$

$$\Leftrightarrow  \sum_{k=0}^{\infty}\dfrac{(-1)(2x)^{2n}}{(2n)!}$$

Series representation of $\tan(2x)$

$\tan(x)=x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+\dfrac{17x^7}{315}+\dfrac{62x^9}{2835}+\dfrac{1382x^{11}}{155925}...$

$\tan(2x)=2x+\dfrac{(2x)^3}{3}+\dfrac{2(2x)^5}{15}+\dfrac{17(2x)^7}{315}+\dfrac{62(2x)^9}{2835}+\dfrac{1382(2x)^{11}}{155925}...$

$=2x+\dfrac{8x^3}{3}+\dfrac{64x^5}{15}+\dfrac{2176x^7}{315}+\dfrac{31744x^9}{2835}+\dfrac{2830336x^{11}}{155925}...$

Dividing power series (2): More complicated examples

This thread will include more intricate long division problems. We start by deriving the power series for $\tan(x)$ and $\cot(x)$. It is easy to note that $\tan(x)$ and $\cot(x)$ can be obtained by using their very definition. $\tan(x)=\dfrac{\sin(x)}{\cos(x)}$ and $\cot(x)=\dfrac{\cos(x)}{\sin(x)}$. We first expand in series and then divide the two power series using the scheme of long division. This method, while simple, do not reveal the relationship between these two functions' coefficients and the Bernoulli numbers. We will illuminate this point in another thread where we seek to obtain the power series for the generating function of Bernoulli numbers. For now, we contend to just obtain the power series for $\tan(x)$ and $\cot(x)$.

Series representation of $\tan(x)$

Thus $\tan(x)=x+\dfrac{x^3}{3}+\dfrac{2x^5}{15}+\dfrac{17x^7}{315}+\dfrac{62x^9}{2835}+\dfrac{1382x^{11}}{155925}...$

Series representation of $\cot(x)$

The function is not defined at $x=0$, since $sin(0)=0$, division by zero is undefined. The function doesn't have a Maclaurin series centered at $0$. Nevertheless, we can still expand it into what is called the Laurent series. The Laurent series is a power series that includes negative degree.

Thus $\cot(x)=\dfrac{1}{x}-\dfrac{x}{3}-\dfrac{x^3}{45}-\dfrac{2x^5}{945}-\dfrac{x^7}{4725}-\dfrac{2x^9}{93555}...$

Series representation of $\sec(x)$

We use the identity $\sec(x)=\dfrac{1}{\cos(x)}$ to derive the Maclaurin series for $\sec(x)$.

Thus $\sec(x)=\color{green}{1}+\dfrac{\color{green}{1}x^2}{2!}+\dfrac{\color{green}{5}x^4}{24}+\dfrac{\color{green}{61}x^6}{720}+\dfrac{277x^8}{8064}+\dfrac{50521x^{10}}{3628800}...$

Long division may give us the series, but it doesn't hint at the fact that $\sec(x)$ is one of the two generating functions of Euler numbers. In fact, since we reduce fractions to their simplest form, this obscure the general formula for the coefficients of $\sec(x)$. The series of $\sec(x)$ is as followed:

$$\sum_{n=0}^{\infty}\dfrac{(-1)^{n}E_{2n}x^{2n}}{(2n)!}$$

The first few Euler numbers are:
$E_0=1$, $E_2=-1$, $E_4=5$, $E_6=-61$, $E_8=1385$, $E_{10}=-50521$, $E_{12}=2702765...$

Thus we can rewrite the series as followed:

$\sec(x)=\color{green}{1}+\dfrac{\color{green}{1}x^2}{2!}+\dfrac{\color{green}{5}x^4}{4!}+\dfrac{\color{green}{61}x^6}{6!}+\dfrac{\color{green}{1385}x^8}{8!}+\dfrac{\color{green}{50521}x^{10}}{10!}...$

Series representation of $\tanh(x)$ and sech(x)

Using the same method above, we can derive the series for $\tanh(x)$ and $\operatorname{sech(x)}$. The difference is that instead of being a positive term series, they become alternating series. The series for $\tanh(x)$ is:

$\tanh(x)=x-\dfrac{x^3}{3}+\dfrac{2x^5}{15}-\dfrac{17x^7}{315}+\dfrac{62x^9}{2835}-\dfrac{1382x^{11}}{155925}...$

The series for $sech$ is:

$\operatorname{sech}(x)=1-\dfrac{x^2}{2!}+\dfrac{5x^4}{24}-\dfrac{61x^6}{720}+\dfrac{277x^8}{8064}-\dfrac{50521x^{10}}{3628800}...$

$$\sum_{n=1}^{\infty}\dfrac{(-1)^nE_{2n}x^{2n}}{(2n)!}$$

Series representation of $\dfrac{x}{\sin(x)}$

At $x=0$, the function has the form $\dfrac{0}{0}$. Using L'Hopital rule to find the limit $$\lim_{x\to 0} \dfrac{x}{\sin(x)}$$, we have $$\lim_{x\to 0} \dfrac{1}{\cos(x)}$$, which is $1$. So the function is defined when $x=0$. By long division, we have:

Thus $\dfrac{x}{\sin(x)}=1+\dfrac{x^2}{3!}+\dfrac{7x^4}{360}+\dfrac{31x^6}{15120}+\dfrac{127x^8}{604800}+\dfrac{73x^{10}}{3421440}...$

Series representation of $\dfrac{1}{\sin(x)}=\csc(x)$

It is now easy to obtain the series representation for $\csc(x)$, we only have to divide it by $x$, the series is:

Thus $\csc(x)=\dfrac{1}{x}+\dfrac{x}{3!}+\dfrac{7x^3}{360}+\dfrac{31x^5}{15120}+\dfrac{127x^7}{604800}+\dfrac{73x^9}{3421440}...$

Again, the function is not defined at $x=0$, so there is no Maclaurin series for $\csc(x)$. What we obtain is the Laurent series.

Dividing power series (1): Long Division

Dividing power series is a a powerful method to obtain new series from known ones. In this thread, I will show some simple examples of dividing power series. We will start from the simplest example and move on the most intricate ones. The first example is that of a geometric series. The series for $\dfrac{1}{1-x}$ is worth learning by heart because it is ubiquitous in obtaining new power series. The scheme for dividing the series is represented below. This scheme is what we call long division of power series.

Series representation of $\dfrac{1}{1-x}$

Thus $\dfrac{1}{1-x}=1+x+x^2+x^3+x^4+x^5+x^6+...$

Series representation of $\dfrac{1}{1+x}$

Thus $\dfrac{1}{1+x}=1-x+x^2-x^3+x^4-x^5+x^6+...$

Series representation of $\dfrac{1}{1+2x}$

Thus $\dfrac{1}{1+2x}=1-2x+4x^2-8x^3+16x^4-32x^5+64x^6+...$

$=1-2x+2^2x^2-2^3x^3+2^4x^4-2^5x^5+2^6x^6+...$

Series representation of $\dfrac{1}{1+3x}$

Thus $\dfrac{1}{1+3x}=1-3x+9x^2-27x^3+81x^4-243x^5+729x^6+...$

$=1-3x+3^2x^2-3^3x^3+3^4x^4-3^5x^5+3^6x^6+...$

Series representation of $\dfrac{1}{1+4x}$

Thus $\dfrac{1}{1+4x}=1-4x+16x^2-64x^3+256x^4-1024x^5+4096x^6+...$

$=1-4x+4^2x^2-4^3x^3+4^4x^4-4^5x^5+4^6x^6+...$

Series of the form of $\dfrac{1}{1+kx}$

Here I will list the first 10 series in this form:

$\dfrac{1}{1+x}=1-x+x^2-x^3+x^4-x^5+x^6+...$

$\dfrac{1}{1+2x}=1-2x+2^2x^2-2^3x^3+2^4x^4-2^5x^5+2^6x^6+$

$\dfrac{1}{1+3x}=1-3x+3^2x^2-3^3x^3+3^4x^4-3^5x^5+3^6x^6+$

$\dfrac{1}{1+4x}=1-4x+4^2x^2-4^3x^3+4^4x^4-4^5x^5+4^6x^6+...$

$\dfrac{1}{1+5x}=1-5x+5^2x^2-5^3x^3+5^4x^4-5^5x^5+5^6x^6+...$

In general:

$\dfrac{1}{1+kx}=1-kx+k^2x^2-k^3x^3+k^4x^4-k^5x^5+k^6x^6+...$

We can insert fractional values to $k$ to obtain the inverse of these series:

$\dfrac{1}{1+\dfrac{1}{2}x}=\dfrac{2}{x+2}=1-\dfrac{x}{2}+\dfrac{x^2}{2^2}-\dfrac{x^3}{2^3}+\dfrac{x^4}{2^4}-\dfrac{x^5}{2^5}+\dfrac{x^6}{2^6}...$

$\dfrac{1}{1+\dfrac{1}{3}x}=\dfrac{3}{x+3}=1-\dfrac{x}{3}+\dfrac{x^2}{3^2}-\dfrac{x^3}{3^3}+\dfrac{x^4}{3^4}-\dfrac{x^5}{3^5}+\dfrac{x^6}{3^6}...$

$\dfrac{1}{1+\dfrac{1}{4}x}=\dfrac{4}{x+4}=1-\dfrac{x}{4}+\dfrac{x^2}{4^2}-\dfrac{x^3}{4^3}+\dfrac{x^4}{4^4}-\dfrac{x^5}{4^5}+\dfrac{x^6}{4^6}...$

$\dfrac{1}{1+\dfrac{1}{5}x}=\dfrac{5}{x+5}=1-\dfrac{x}{5}+\dfrac{x^2}{5^2}-\dfrac{x^3}{5^3}+\dfrac{x^4}{5^4}-\dfrac{x^5}{5^5}+\dfrac{x^6}{5^6}...$

Series of the form of $\dfrac{1}{1-kx}$

$\dfrac{1}{1-x}=1+x+x^2+x^3+x^4+x^5+x^6+...$

$\dfrac{1}{1-2x}=1+2x+2^2x^2+2^3x^3+2^4x^4+2^5x^5+2^6x^6+$

$\dfrac{1}{1-3x}=1+3x+3^2x^2+3^3x^3+3^4x^4+3^5x^5+3^6x^6+$

$\dfrac{1}{1-4x}=1+4x+4^2x^2+4^3x^3+4^4x^4+4^5x^5+4^6x^6+...$

$\dfrac{1}{1-5x}=1+5x+5^2x^2+5^3x^3+5^4x^4+5^5x^5+5^6x^6+...$

In general

$\dfrac{1}{1-kx}=1+kx+k^2x^2+k^3x^3+k^4x^4+k^5x^5+k^6x^6+...$

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: $\color{green}{\cosh^2(x)-\sinh^2(x)=1}$

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.

Adding and subtracting power series

We can add and subtract power series the same way we do with numbers, fractions or polynomials. I wish to re-derive the Maclaurin series for $\sinh(x)$ and $\cosh(x)$ to illustrate this point.

We can use the definition of $\sinh(x)$ and $\cosh(x)$ as followed

Series representation of $\sinh(x)$:

$\sinh(x)=\dfrac{e^x-e^{-x}}{2}$

We know the series representation for $e^x$:

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

Replace $x$ by $-x$, we have:

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

$\sinh(x)=\dfrac{1}{2}\left[1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+...-\left(1-\dfrac{x}{1!}+\dfrac{x^2}{2!}-\dfrac{x^3}{3!}+\dfrac{x^4}{4!}-\dfrac{x^5}{5!}+...\right)\right]$

$=\dfrac{1}{2}\left(1+\dfrac{x}{1!}\color{red}{+\dfrac{x^2}{2!}}+\dfrac{x^3}{3!}\color{red}{+\dfrac{x^4}{4!}}+\dfrac{x^5}{5!}+...-1+\dfrac{x}{1!}\color{red}{-\dfrac{x^2}{2!}}+\dfrac{x^3}{3!}\color{red}{-\dfrac{x^4}{4!}}+\dfrac{x^5}{5!}-...\right)$

Red terms denote elimination

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

$=x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{x^7}{7!}+\dfrac{x^9}{9!}+\dfrac{x^{11}}{11!}+\dfrac{x^{13}}{13!}+O(x^{11})$

Series representation of $\cosh(x)$:

$\cosh(x)=\dfrac{e^x+e^{-x}}{2}$

$\cosh(x)=\dfrac{1}{2}\left[1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+...+\left(1-\dfrac{x}{1!}+\dfrac{x^2}{2!}-\dfrac{x^3}{3!}+\dfrac{x^4}{4!}-\dfrac{x^5}{5!}+...\right)\right]$

$=\dfrac{1}{2}\left(1\color{red}{+\dfrac{x}{1!}}+\dfrac{x^2}{2!}\color{red}{+\dfrac{x^3}{3!}}+\dfrac{x^4}{4!}\color{red}{+\dfrac{x^5}{5!}}+...+1\color{red}{-\dfrac{x}{1!}}+\dfrac{x^2}{2!}\color{red}{-\dfrac{x^3}{3!}}+\dfrac{x^4}{4!}\color{red}{-\dfrac{x^5}{5!}}+...\right)$

$=\dfrac{2}{2}+\dfrac{2x^2}{2\cdot 2!}+\dfrac{2x^4}{2\cdot 4!}+\dfrac{2x^6}{2\cdot 6!}+\dfrac{2x^8}{2\cdot 8!}+\dfrac{2x^{10}}{2\cdot 10!}+O(x^{12})$

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

We can try a few more examples to get the gist of these methods:

Series representation of $e^x+\sin(x)$:

$e^x+\sin(x)=\left(1+\dfrac{x}{1!}+\dfrac{x^2}{2!}\color{red}{+\dfrac{x^3}{3!}}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+...\right)+\left(x\color{red}{-\dfrac{x^3}{3!}}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\dfrac{x^9}{9!}...\right)$

$=1+2x+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{2x^5}{5!}+\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+\dfrac{2x^9}{9!}+\dfrac{x^{10}}{10!}+\dfrac{x^{12}}{12!}...$

Series representation of $e^x+\cos(x)$:

$e^x+\cos(x)=\left(1+\dfrac{x}{1!}\color{red}{+\dfrac{x^2}{2!}}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+...\right)+\left(1\color{red}{-\dfrac{x^2}{2!}}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\dfrac{x^8}{8!}...\right)$

$=2+\dfrac{x^3}{3!}+\dfrac{2x^4}{4!}+\dfrac{x^5}{5!}+\dfrac{x^7}{7!}+\dfrac{2x^8}{8!}+\dfrac{x^{9}}{9!}+\dfrac{x^{11}}{11!}...$

Series representation of $e^x-\sin(x)$:

$e^x-\sin(x)=\left(1+\dfrac{x}{1!}+\dfrac{x^2}{2!}\color{red}{+\dfrac{x^3}{3!}}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+...\right)-\left(x\color{red}{-\dfrac{x^3}{3!}}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\dfrac{x^9}{9!}...\right)$

$=\left(1\color{red}{+\dfrac{x}{1!}}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}\color{red}{+\dfrac{x^5}{5!}}+...\right)\color{red}{-x}+\dfrac{x^3}{3!}\color{red}{-\dfrac{x^5}{5!}}+\dfrac{x^7}{7!}-\dfrac{x^9}{9!}...$

$=1+\dfrac{x^2}{2!}+\dfrac{2x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^6}{6!}+\dfrac{2x^7}{7!}+\dfrac{x^8}{8!}...$

Series representation of $e^x-\cos(x)$:

$e^x-\cos(x)=\left(1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+\dfrac{x^5}{5!}+...\right)-\left(1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\dfrac{x^8}{8!}...\right)$

$=\left(\color{red}{1}+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}\color{red}{+\dfrac{x^4}{4!}}+\dfrac{x^5}{5!}+...\right)\color{red}{-1}+\dfrac{x^2}{2!}\color{red}{-\dfrac{x^4}{4!}}+\dfrac{x^6}{6!}-\dfrac{x^8}{8!}...$

$=x+\dfrac{2x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^5}{5!}+\dfrac{2x^6}{6!}+\dfrac{x^7}{7!}+\dfrac{x^{9}}{9!}+\dfrac{2x^{10}}{10!}...$


The last example involved two functions that we haven't yet derived since the last post. We will do this later. The series is beautiful in its own right and deserve a place on our blog:

Series representation of $\ln(1-x)+\dfrac{1}{1-x}$:

$$\ln(1-x)=-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}-\dfrac{x^4}{4}-\dfrac{x^5}{5}-O(x^6)=-\sum_{n=1}^{+\infty}\frac{x^n}{n}$$

$$\dfrac{1}{1-x}=1+x+x^2+x^3+x^4+x^5+O(x^6)=1+\sum_{n=1}^{+\infty}x^n$$

$\ln(1-x)+\dfrac{1}{1-x}=1+(-x+x)+(-\dfrac{x^2}{2}+x^2)+(-\dfrac{x^3}{3}+x^3)+(-\dfrac{x^4}{4}+x^4)...$

$$=1+\dfrac{1}{2}x^2+\dfrac{2}{3}x^3+\dfrac{3}{4}x^4+\dfrac{4}{5}x^5+\dfrac{5}{6}x^6+\dfrac{6}{7}x^7+O(x^8) = 1+\sum_{n=0}^\infty\frac{n}{n+1}x^{n+1}$$

or $$-\sum_{n=1}^{+\infty}\frac{x^n}{n}+1+\sum_{n=1}^{+\infty}x^n=1+\sum_{n=1}^{+\infty}\left(x^n-\frac{x^n}{n}\right)=1+\sum_{n=1}^{+\infty}(1-\frac 1n)x^n$$