y=sin(arcsin(x)n),n≥1
I know that:
limx→0xy=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|<1√2.
Then tanθ=x√1−x2 and |tanθ|<1.
cosθn+isinθn=(cosθ+isinθ)1n=(cosθ)1n(1+itanθ)1n=x1n(1+ix√1−x2)1n=x1n∞∑k=0(1/nk)ik(x√1−x2)kcosθn=x1n∞∑k=0(−1)k(1/n2k)(x21−x2)ksinθn=x1nx√1−x2∞∑k=0(−1)k(1/n2k+1)(x21−x2)k
NOTES
We define (zn) where z∈R and 0≤n∈Z as follows
(z)n={1If n=0.z(z−1)(z−2)⋯(z−n+1)If n≥1.
then (zn)=(z)nn!
It can be shown that, if |x|<1, then (1+x)z=∑∞k=0(zk)xk
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