The Taylor series expansion is what the function is expanded in the form of the sequence of numbers, and it would be very convenient. The following equation which is called as ''Maclaurin expansion'' is more commonly used.
\begin{eqnarray}
f(x)&=&f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+\frac{1}{3}f'''(0)x^3+\cdots+\frac{f^n(0)}{n!}x^n+\cdots \nonumber \\
&=&\sum_{n=0}^\infty \frac{f^n(0)}{n!}x^n \nonumber
\end{eqnarray}
The below equations show the Taylor series expansion (Maclaurin series expansion) of the major functions.
In terms of the all of ''$x$'':
\begin{eqnarray}
e^x&=&1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots+\frac{x^n}{n!}+\cdots \nonumber \\
e^{-x}&=&1-x+\frac{x^2}{2}-\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots+(-1)^n\frac{x^n}{n!}+\cdots \nonumber \\
sin\;x&=&x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots+\frac{(-1)^n}{(2n+1)!}x^{2n+1}+\cdots \nonumber \\
cos\;x&=&1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots+\frac{(-1)^n}{(2n)!}x^{2n}+\cdots \nonumber
\end{eqnarray}
In terms of $|x| < 1$:
\begin{eqnarray}
(1-x)^a&=&1-ax+\frac{a(a-1)}{2}x^2-\frac{a(a-1)(a-2)}{3!}x^3+\cdots \nonumber \\
&\;&\;\;\;\;\;\;\;\;\;\;+(-1)^n\frac{a(a-1)(a-2)\cdot\cdot(a-n+1)}{n!}x^n+\cdots \nonumber \\
(1+x)^a&=&1+ax+\frac{a(a-1)}{2}x^2+\frac{a(a-1)(a-2)}{3!}x^3+\cdots \nonumber \\
&\;&\;\;\;\;\;\;\;\;\;\;+\frac{a(a-1)(a-2)\cdots(a-n+1)}{n!}x^n+\cdots \nonumber \\
(1-x^2)^a&=&1-ax^2+\frac{a(a-1)}{2}x^4-\frac{a(a-1)(a-2)}{3!}x^6+\cdots \nonumber \\
&\;&\;\;\;\;\;\;\;\;\;\;+(-1)^n\frac{a(a-1)(a-2)\cdot\cdot(a-n+1)}{n!}x^{2n}+\cdots \nonumber \\
(1+x^2)^a&=&1+ax^2+\frac{a(a-1)}{2}x^4+\frac{a(a-1)(a-2)}{3!}x^6+\cdots \nonumber \\
&\;&\;\;\;\;\;\;\;\;\;\;+\frac{a(a-1)(a-2)\cdots(a-n+1)}{n!}x^{2n}+\cdots \nonumber \\
\sqrt{(1-x)}&=&1-\frac{1}{2}x-\frac{1}{8}x^2-\frac{3}{48}x^3-\frac{15}{384}x^4-\cdots-(\frac{1}{2})^n\frac{(2n-3)!!}{n!}x^n+\cdots \nonumber \\
\sqrt{(1+x)}&=&1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{3}{48}x^3-\frac{15}{384}x^4+\cdots+(-1)^{n-1}(\frac{1}{2})^n\frac{(2n-3)!!}{n!}x^n+\cdots \nonumber \\
\sqrt{1-x^2}&=&1-\frac{1}{2}x^2-\frac{1}{8}x^4-\frac{3}{48}x^6-\frac{15}{384}x^8-\cdots-(\frac{1}{2})^n\frac{(2n-3)!!}{n!}x^{2n}-\cdots \nonumber \\
\sqrt{1+x^2}&=&1+\frac{1}{2}x^2-\frac{1}{8}x^4+\frac{3}{48}x^6-\frac{15}{384}x^8+\cdots+(-1)^{n-1}(\frac{1}{2})^n\frac{(2n-3)!!}{n!}x^{2n}+\cdots \nonumber \\
\frac{1}{1-x}&=&1+x+x^2+x^3+x^4+x^5+\cdots+x^n+\cdots \nonumber \\
\frac{1}{1+x}&=&1-x+x^2-x^3+x^4-x^5+\cdots+(-1)^nx^n+\cdots \nonumber \\
\frac{1}{1-x^2}&=&1+x^2+x^4+x^6+x^8+x^{10}+\cdots+x^{2n}\dots \nonumber \\
\frac{1}{1+x^2}&=&1-x^2+x^4-x^6+x^8-x^{10}+\cdots+(-1)^nx^{2n}+\cdots \nonumber \\
\frac{1}{\sqrt{1-x}}&=&1+\frac{1}{2}x+\frac{3}{8}x^2+\frac{15}{48}x^3+\cdots+(\frac{1}{2})^n(2n-1)!!\frac{x^{n}}{n!}+\cdots \nonumber \\
\frac{1}{\sqrt{1+x}}&=&1-\frac{1}{2}x+\frac{3}{8}x^2-\frac{15}{48}x^3+\cdots+(-\frac{1}{2})^n(2n-1)!!\frac{x^{n}}{n!}+\cdots \nonumber \\
\frac{1}{\sqrt{1-x^2}}&=&1+\frac{1}{2}x^2+\frac{3}{8}x^4+\frac{15}{48}x^6+\cdots+(\frac{1}{2})^n(2n-1)!!\frac{x^{2n}}{n!}+\cdots \nonumber \\
\frac{1}{\sqrt{1+x^2}}&=&1-\frac{1}{2}x^2+\frac{3}{8}x^4-\frac{15}{48}x^6+\cdots+(-\frac{1}{2})^n(2n-1)!!\frac{x^{2n}}{n!}+\cdots \nonumber \\
tan^{-1}x&=&x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots+(-1)^n(\frac{1}{2n+1)}x^{2n+1}+\cdots \nonumber
\end{eqnarray}
The circular constant, ''$\pi$'' represents ''$tan^{-1}x$'' , and assuming that ''$x=1$'', it becomes
$tan^{-1}\{1\}=\pi/4$. Accordingly, the following equation can be indicated.
\begin{eqnarray}
\pi&=&4-\frac{4}{3}+\frac{4}{5}-\frac{4}{7}+\frac{4}{9}-\frac{4}{11}+\cdots+(-1)^n\frac{4}{(2n+1)}+\cdots \nonumber
\end{eqnarray}
This is called as the formula of Leibniz. In the meantime, Mr.Kikuo Takano improved the convergence of this formula, and
when Mr.Yasumasa Kaneda calculated the circular constant, ''$\pi$'' in the year 2002 up to a trillion two hundred forty one billion and a hundred millionth place after the decimal point, the following formula of Mr.Kikuo Takano was used.
\begin{eqnarray}
\frac{\pi}{4}=12tan^{-1}\frac{1}{49}+32tan^{-1}\frac{1}{57}-5tan^{-1}\frac{1}{239}+12tan^{-1}\frac{1}{110443} \nonumber
\end{eqnarray}