Thursday, 2 January 2020

trigonometry - How to simplify y=sin(fracarcsin(x)n),n1?




y=sin(arcsin(x)n),n1



I know that:



limx0xy=n



But I can't figure out what the curve of x/y practically represents. Is there an obvious simple solution?


Answer



Let θ=arcsinx and we have to assume that |x|<12.




Then tanθ=x1x2 and |tanθ|<1.



\begin{align} \cos \dfrac{\theta}{n} + i \sin \dfrac{\theta}{n} &= (\cos \theta + i \sin \theta)^{\frac 1n} \\ &= (\cos \theta)^{\frac 1n}(1 + i \tan \theta)^{\frac 1n} \\ &= x^{\frac 1n}\left(1 + i\dfrac{x}{\sqrt{1-x^2}}\right)^\frac 1n\\ &= x^{\frac 1n}\sum_{k=0}^\infty \binom{1/n}{k} i^k\left(\dfrac{x}{\sqrt{1-x^2}}\right)^k \\ \cos \dfrac{\theta}{n} &= x^{\frac 1n}\sum_{k=0}^\infty (-1)^k\binom{1/n}{2k}\left(\dfrac{x^2}{1-x^2}\right)^k \\ \sin\dfrac{\theta}{n} &= x^{\frac 1n} \dfrac{x}{\sqrt{1-x^2}} \sum_{k=0}^\infty (-1)^k\binom{1/n}{2k+1}\left(\dfrac{x^2}{1-x^2}\right)^k \\ \end{align}



NOTES







We define \binom zn where z \in \mathbb R and 0 \le n \in \mathbb Z as follows



(z)_n = \begin{cases} 1 & \text{If $n = 0$.}\\ z(z-1)(z-2)\cdots(z-n+1) &\text{If $n \ge 1$.} \end{cases}




then \binom zn = \dfrac{(z)_n}{n!}



It can be shown that, if |x| < 1, then (1 + x)^z = \sum_{k=0}^\infty \binom zk x^k


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