Revision as of 06:22, 25 March 2016

A Taylor series is a way to express a function as an infinite series under suitable differentiability conditions. The Taylor series is typically given by $$f(x) = \displaystyle\sum_{k=0}^{\infty} f^{(k)}(x_0) (x-x_0)^k,$$ where $(k)$ denotes differentiation.

Examples of Taylor series

Theorem

The following Taylor series holds for all $z \in \mathbb{C}$: $$e^{z} = \displaystyle\sum_{k=0}^{\infty} \dfrac{z^k}{k!},$$ where $e^z$ is the exponential function.

Proof

References

Theorem

Let $z_0 \in \mathbb{C}$. The following Taylor series holds: $$\sin(z)=\displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k z^{2k+1}}{(2k+1)!},$$ where $\sin$ denotes the sine function.

Proof

Using the Taylor series of the exponential function and the definition of $\sin$, $$\begin{array}{ll} \sin(z) &= \dfrac{e^{iz}-e^{-iz}}{2i} \\ &= \dfrac{1}{2i} \left[ \displaystyle\sum_{k=0}^{\infty} \dfrac{i^k (z-z_0)^k}{k!} - \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k i^k (z-z_0)^k}{k!} \right] \\ &= \dfrac{1}{2i} \displaystyle\sum_{k=0}^{\infty} \dfrac{(z-z_0)^k}{k!}i^k (1-(-1)^k). \end{array}$$ Note that if $k=2n$ is a positive even integer, then $$i^k(1-(-1)^k)=i^{2n}(1-(-1)^{2n})=0,$$ and if $k=2n+1$ is a positive odd integer, then $$i^k(1-(-1)^k)=i^{2n+1}(1-(-1)^{2n+1})=2i(-1)^n.$$ Hence we have derived $$\begin{array}{ll} \sin(z)&=\dfrac{1}{2i} \displaystyle\sum_{k=0}^{\infty} \dfrac{(z-z_0)^k}{k!}i^k (1-(-1)^k) \\ &=\displaystyle\sum_{k \mathrm{\hspace{2pt} odd},k>0}^{\infty} \dfrac{(z-z_0)^k}{k!}i^k (1-(-1)^k) \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k (z-z_0)^{2k+1}}{(2k+1)!}, \end{array}$$ as was to be shown. █

References

Theorem

Let $z_0 \in \mathbb{C}$. The following Taylor series holds for all $z \in \mathbb{C}$: $$\cos(z)= \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k z^{2k}}{(2k)!},$$ where $\cos$ denotes the cosine function.

Proof

Using the Taylor series of the exponential function and the definition of $\cos$, $$\begin{array}{ll} \cos(z) &= \dfrac{e^{iz}+e^{-iz}}{2} \\ &= \dfrac{1}{2} \left[ \displaystyle\sum_{n=0}^{\infty} \dfrac{i^n (z-z_0)^n}{n!} + \displaystyle\sum_{n=0}^{\infty} \dfrac{(-1)^n i^n (z-z_0)^n}{n!} \right] \\ &= \dfrac{1}{2} \displaystyle\sum_{n=0}^{\infty} \dfrac{(z-z_0)^n}{n!}i^n (1+(-1)^n). \end{array}$$ Note that if $n=2k$ is a positive even integer, then $$i^n(1+(-1)^n)=i^{2k}(1+(-1)^{2k})=2(-1)^k,$$ and if $n=2k+1$ is a positive odd integer, then $$i^n(1+(-1)^n)=i^{2k+1}(1+(-1)^{2k+1})=0.$$ Hence we have derived $$\begin{array}{ll} \cos(z)&=\dfrac{1}{2} \displaystyle\sum_{n=0}^{\infty} \dfrac{(z-z_0)^n}{n!}i^n (1+(-1)^n) \\ &=\dfrac{1}{2} \displaystyle\sum_{n \mathrm{\hspace{2pt} even},n>0}^{\infty} \dfrac{(z-z_0)^n}{n!}i^n (1-(-1)^n) \\ &= \displaystyle\sum_{k=0}^{\infty} \dfrac{(-1)^k (z-z_0)^{2k}}{(2k)!}, \end{array}$$ as was to be shown. █

References