# 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 $k$  there are real numbers $a_{k}$  such that

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

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

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

Setting $x$  equal to zero, we obtain

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

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

 $f(x)=\sum _{k=0}^{\infty }{f^{(k)}(0) \over k!}x^{k}=f(0)+f'(0)x+f''(0){x^{2} \over 2!}+f'''(0){x^{3} \over 3!}+\cdots$ This is the Taylor series for $f(x).$

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

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