$y = \sin(\frac{\arcsin(x)}{n}), n≥1$

I know that:

$\lim \limits_{x \to 0} \frac{x}{y} = n$

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

  • $\begingroup$ You might try using double-angle or angle sum formulas $\endgroup$ – abiessu Apr 21 '16 at 1:32
  • $\begingroup$ @abiessu : I can't find anything on wiki/List_of_trigonometric_identities $\endgroup$ – reuns Apr 21 '16 at 1:36
  • $\begingroup$ if you can expand $\sin \theta/n$ in trigonometric functions of $\theta$ you would be done. However, when you look at math.stackexchange.com/questions/125774/… (specifically at $\cos (4x)$ expansion), imagine you replace $x=\theta/4$, so you have now $\cos(\theta)$ on left and an order $4$ polynomial in $\cos(\theta)$ on the RHS. Hence, to achieve your goal (stated at beginning hereby), you need to solve a degree $4$ polynomial equation. I guess for $n$, you have to solve an order $n$ polynomial, which is impossible by the famous theorem. $\endgroup$ – Chip Apr 21 '16 at 1:39

Let $\theta = \arcsin x$ and we have to assume that $|x| \lt \dfrac{1}{\sqrt 2}$.

Then $\tan \theta = \dfrac{x}{\sqrt{1-x^2}}$ and $|\tan \theta| \lt 1$.

\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}


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 $(1 + x)^z = \sum_{k=0}^\infty \binom zk x^k$

  • $\begingroup$ How can the OP calculate $\binom{1/n}{k}$? $\endgroup$ – Piquito Apr 21 '16 at 14:48

As mentioned in the comments, there is no way to simplify while maintaining the accurate relation. If approximately there could be convenient simplifications. First let me clarify the question a bit. Suppose $\arcsin(x) = \theta$, then $\sin(\theta) = x$, so what you try to achieve is if $\sin(\theta)$ is known, what's $\sin(\theta/n)$.

A straight forward attempt is do a Taylor expansion of $\arcsin(x)$, for large $n$ the sine function can be expanded around $0$ as well. If two terms are kept of the arcsine expansion: $$ \arcsin(x) = x + \frac{x^3}{6}+O(x^5) $$ Then $$ y = \sin \left(\frac{1}{n} (x + \frac{x^3}{6}) \right) \approx \frac{1}{n} (x + \frac{x^3}{6}) $$ Below is a figure (red is approximation above, blue is the accurate value), the approximation works well up to $x = 0.8$ for $n \geq 2$.




In the figure below you have an approximate representation of $\sin (x)$ and $\sin (\frac xn)$.

Your problem is to calculate the length of the segment $\overline {CD}$. Since you have $n( \frac xn)=x$, you can first find $\sin (nX)$ and then replace $X$ by $\frac xn$. The expression of $\sin (\frac xn)$ for each value of $n$, given in function of $\sin( x)$ is more and more complicated as it grows $n$. For example, in $$\sin (3x)=4\cos^2(x)\sin (x)-\sin (x)=3\sin (x)-4\sin^3(x)$$ replacing $x$ by $\frac x3$ you get $$\sin(x)=3\sin(\frac x3)-4\sin^3(\frac x3)$$ so you have the cubic equation to solve $$4t^3-3t+a=0$$ where $a=\sin (x)$.

Similarly, for $\sin(\frac x5)$, you have to solve the quintic $$16t^5-20t^3+5t-a=0$$ deduced of the formula $\sin (5x)=16\cos^4(x)\sin(x)-12\cos^2(x)\sin (x)+\sin (x)$.

In general for $\sin (\frac xn)$ you have to solve an equation of degree $n$.

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy