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

$f(x)=\sum _{k=0}^{\mathrm{\infty }}{a}_k{x}^k=a_0+a_{1}x+a_2{x}^2+a_3{x}^3+a_4{x}^4+\cdots$

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

$f^{\prime }(x)=a_1+2a_{2}x+3a_3{x}^2+4a_4{x}^3+5a_5{x}^4+\cdots$

$f^{″}(x)=2a_2+2\cdot 3a_{3}x+3\cdot 4a_4{x}^2+4\cdot 5a_5{x}^3+\cdots$

$f^{‴}(x)=2\cdot 3a_3+2\cdot 3\cdot 4a_{4}x+3\cdot 4\cdot 5a_5{x}^2+\cdots$

$f^{⁗}(x)=2\cdot 3\cdot 4a_4+2\cdot 3\cdot 4\cdot 5a_{5}x+\cdots$

Setting $x$ equal to zero, we obtain

$f(0)=a_0,f^{\prime }(0)=a_1,f^{″}(0)=2a_2,f^{‴}(0)=2\times 3a_3,f^{⁗}(0)=2\times 3\times 4a_4.$

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 $n\le k$ for $k>1.$ Expressing the coefficients $a_k$ in terms of the derivatives of $f(x)$ at $x=0,$ we obtain

$f(x)=\sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k)}(0)}{k!}{x}^k=f(0)+f^{\prime }(0)x+f^{″}(0)\frac{x^2}{2!}+f^{‴}(0)\frac{x^3}{3!}+\cdots$

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 $x_0$:

$f(x)=\sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k)}(x_0)}{k!}(x-x_0{)}^k.$