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