This Quantum World/Appendix/Taylor series

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[edit] 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 ak 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 ak 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\over2!}+f'''(0){x^3\over3!}+\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 x0:

f(x)=\sum_{k=0}^\infty {f^{(k)}(x_0)\over k!}(x-x_0)^k.
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