# This Quantum World/Appendix/Taylor series

#### Taylor series

A well-behaved function can be expanded into a power series. This means that for all non-negative integers ${\displaystyle k}$  there are real numbers ${\displaystyle a_{k}}$  such that

${\displaystyle f(x)=\sum _{k=0}^{\infty }a_{k}x^{k}=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots }$

Let us calculate the first four derivatives using ${\displaystyle (x^{n})'=n\,x^{n-1}}$ :

${\displaystyle f'(x)=a_{1}+2\,a_{2}x+3\,a_{3}x^{2}+4\,a_{4}x^{3}+5\,a_{5}x^{4}+\cdots }$
${\displaystyle f''(x)=2\,a_{2}+2\cdot 3\,a_{3}x+3\cdot 4\,a_{4}x^{2}+4\cdot 5\,a_{5}x^{3}+\cdots }$
${\displaystyle f'''(x)=2\cdot 3\,a_{3}+2\cdot 3\cdot 4\,a_{4}x+3\cdot 4\cdot 5\,a_{5}x^{2}+\cdots }$
${\displaystyle f''''(x)=2\cdot 3\cdot 4\,a_{4}+2\cdot 3\cdot 4\cdot 5\,a_{5}x+\cdots }$

Setting ${\displaystyle x}$  equal to zero, we obtain

${\displaystyle f(0)=a_{0},\quad f'(0)=a_{1},\quad f''(0)=2\,a_{2},\quad f'''(0)=2\times 3\,a_{3},\quad f''''(0)=2\times 3\times 4\,a_{4}.}$

Let us write ${\displaystyle f^{(n)}(x)}$  for the ${\displaystyle n}$ -th derivative of ${\displaystyle f(x).}$  We also write ${\displaystyle f^{(0)}(x)=f(x)}$  — think of ${\displaystyle f(x)}$  as the "zeroth derivative" of ${\displaystyle f(x).}$  We thus arrive at the general result ${\displaystyle f^{(k)}(0)=k!\,a_{k},}$  where the factorial ${\displaystyle k!}$  is defined as equal to 1 for ${\displaystyle k=0}$  and ${\displaystyle k=1}$  and as the product of all natural numbers ${\displaystyle n\leq k}$  for ${\displaystyle k>1.}$  Expressing the coefficients ${\displaystyle a_{k}}$  in terms of the derivatives of ${\displaystyle f(x)}$  at ${\displaystyle x=0,}$  we obtain

 ${\displaystyle f(x)=\sum _{k=0}^{\infty }{f^{(k)}(0) \over k!}x^{k}=f(0)+f'(0)x+f''(0){x^{2} \over 2!}+f'''(0){x^{3} \over 3!}+\cdots }$

This is the Taylor series for ${\displaystyle f(x).}$

A remarkable result: if you know the value of a well-behaved function ${\displaystyle f(x)}$  and the values of all of its derivatives at the single point ${\displaystyle x=0}$  then you know ${\displaystyle f(x)}$  at all points ${\displaystyle x.}$  Besides, there is nothing special about ${\displaystyle x=0,}$  so ${\displaystyle f(x)}$  is also determined by its value and the values of its derivatives at any other point ${\displaystyle x_{0}}$ :

 ${\displaystyle f(x)=\sum _{k=0}^{\infty }{f^{(k)}(x_{0}) \over k!}(x-x_{0})^{k}.}$