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_kx^k=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+\cdots$$

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



f'(x)=a_1+2\,a_2x+3\,a_3x^2+4\,a_4x^3+5\,a_5x^4+\cdots$$



f''(x)=2\,a_2+2\cdot3\,a_3x+3\cdot4\,a_4x^2+4\cdot5\,a_5x^3+\cdots $$



f'''(x)=2\cdot3\,a_3+2\cdot3\cdot4\,a_4x+3\cdot4\cdot5\,a_5x^2+\cdots $$



f(x)=2\cdot3\cdot4\,a_4+2\cdot3\cdot4\cdot5\,a_5x+\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\times3\,a_3,\quad f'(0)=2\times3\times4\,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

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 $$x_0$$: