Given: sin⁡(x)=usin⁡(x+a),{u&lt;1} How do I expand x in powers of u?

Here is what I did. I started with x0=O(u). Define by recursion xn+1=sin−1(usin⁡(xn+a)). The first few values are x1=sin⁡(a),u+O(u2), ,x2=sin⁡(a) ,u+sin⁡(2a) ,u22+O(u3). Using the pattern and taking it to the limit, I found that x=log⁡(1−uexp⁡(−ia)1−uexp⁡(ia))2i .In the limit x=∑n=1∞sin⁡(na),unn. Another method uses exponentials. Let X=exp⁡(ix), ,A=exp⁡(ia) and substitute them in equation sin⁡(x)=usin⁡(x+a) to get (X−1X)=u(XA−1XA) and solving for X gives X2=1−uA1−Au. Using log⁡(1−x)=−∑n&gt;0xnn now gives the power series in u.

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Isis Solis

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2022-01-27

Given:

\sin (x) = u \sin (x + a), {u < 1}

How do I expand x in powers of u?

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hmotans

Beginner2022-01-28Added 8 answers

Here is what I did. I started with

x_{0} = O (u)

. Define by recursion

x_{n + 1} = \sin^{- 1} (u \sin (x_{n} + a))

. The first few values are

x_{1} = \sin (a), u + O (u^{2}),, x_{2} = \sin (a), u + \sin (2 a), \frac{u^{2}}{2} + O (u^{3})

. Using the pattern and taking it to the limit, I found that

x = \frac{\log (\frac{1 - u \exp (- i a)}{1 - u \exp (i a)})}{2 i}

.In the limit

x = \sum_{n = 1}^{\infty} \sin (n a), \frac{u^{n}}{n}

.
Another method uses exponentials. Let

X = \exp (i x),, A = \exp (i a)

and substitute them in equation

\sin (x) = u \sin (x + a)

to get

(X - \frac{1}{X}) = u (X A - \frac{1}{X A})

and solving for X gives

X^{2} = \frac{1 - \frac{u}{A}}{1 - A u}

. Using

\log (1 - x) = - \sum_{n > 0} \frac{x^{n}}{n}

now gives the power series in u.

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