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 $x_{0}$:

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