Sine
The sine function $\sin \colon \mathbb{R} \rightarrow \mathbb{R}$ is the unique solution of the second order initial value problem $y=-y;y(0)=0,y'(0)=1$.
- Sine.png
Graph of $\sin$ on $\mathbb{R}$.
- Complex sin.jpg
Domain coloring of analytic continuation of $\sin$.
Contents
Properties
Theorem
The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \sin(z) = \cos(z),$$ where $\sin$ denotes the sine function and $\cos$ denotes the cosine function.
Proof
From the definition, $$\sin(z) = \dfrac{e^{iz}-e^{-iz}}{2i},$$ and so using the derivative of the exponential function, the linear property of the derivative, the chain rule, and the definition of the cosine function, $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}z} \sin(z) &= \dfrac{1}{2i} \left[ \dfrac{\mathrm{d}}{\mathrm{d}z} [e^{iz}] - \dfrac{\mathrm{d}}{\mathrm{d}z}[e^{-iz}] \right] \\ &= \dfrac{1}{2i} \left[ ie^{iz} + ie^{-iz} \right] \\ &= \dfrac{e^{iz}+e^{-iz}}{2} \\ &= \cos(z), \end{array}$$ as was to be shown. █
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $4.3.105$
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
Proposition: $\sin(x) = x \displaystyle\prod_{k=1}^{\infty} \left( 1 - \dfrac{x^2}{k^2\pi^2} \right)$
Proof: proof goes here █
- REDIRECT Gamma(z)Gamma(1-z)=pi/sin(pi z)
