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I was thinking for a while what we can achieve by creating new functions with infinite expressions. The FIRST expression I couldn't handle is: $$ y= \sin(x+y)$$ I thought about deriving this expression many times and trying to do a Taylor Series on $x=t-\sin(t), t \ \in [0,\pi]$, and I thought about Newton's Method, since we can solve for $x$ and solve for an inverse function: $$\sin^{-1}(y)=x+y$$ $$x=-y+\sin^{-1}(y)$$ $$y=f(x)$$ Does anyone have a method to solve this problem? For which function is the condition real? I'd be grateful if anyone could help me.

MY TRY:

$$\frac{dy}{dx}=\cos(x+y)(1+\frac{dy}{dx})=\frac{\cos(x+y)}{1-\cos(x+y)}$$ $$\frac{d^2 y}{dx^2}=-y(1+\frac{dy}{dx})^2+\cos(x+y)\frac{d^2y}{dx^2}=\frac{-y(1+\frac{dy}{dx})^2}{1-\cos(x+y)}$$ Set $y=\sum_{n=0}^\infty a_n(x-\pi)^n$, then we have that $\frac{d^ny}{dx^n}|_{x=\pi}=n!a_n$. So: $$y|_{x=\pi}=a_0=\sin(a_0+\pi) \Rightarrow a_0=0, \cos(a_0+\pi)=-1$$ $$ \frac{dy}{dx}|_{x=\pi}=a_1=-1-a_1\Rightarrow a_1=-\frac{1}{2}$$ $$\frac{d^2y}{dx^2}|_{x=\pi}=2a_2=-a_0(1+a_1)^2-a_2=-a_2\Rightarrow a_2=0$$ So: $$y \approx -\frac{1}{2}(x-\pi) \forall x \approx \pi$$

MY 2nd TRY:

So, deriving the function many times, I saw that in the denominator of the derivatives was the expression $1-\cos(x+y)$. Then, I tried to derive $y'(1-\cos(x+y))=\cos(x+y)$ to reach at the expression: $$y^{(n)}(1-\cos(x+y))=\frac{d^{n-1}}{dx^{n-1}}(\cos(x+y))-\sum_{m=0}^{n-1} \frac{d^m}{dx^m}(y*y^{n-1-m} (1+y'))$$ and, if $y=\sum_{n=0}^\infty a_n(x-x_0)^n$, so: $$\frac{n!a_n}{1+a_1}=[\frac{d^{n-1}}{dx^{n-1}}(\frac{y'}{1+y'})-\sum_{m=0}^{n-1} \frac{d^m}{dx^m}(y*y^{n-1-m} (1+y')]_{x=x_0}$$ since $y^{(n)}|_{x=x_0}=n!a_n$ and $\cos(x+y)|_{x=x_0}=\frac{y'}{y'+1}|_{x=x_0}=\frac{a_1}{1+a_1}$.

So, for example, pick $x_0=\pi$. So:

$$a_0=\sin(\pi+a_0)=0$$ $$a_1=\frac{\cos(\pi+a_0)}{1-\cos(\pi+a_0)}=-\frac{1}{2}$$ $$a_2=0$$ $$a_3=\frac{1}{96}$$

So: $$y=-\frac{1}{2}(x-\pi)+\frac{1}{96}(x-\pi)^2+...$$

Green graph is y=sin(x+y) and blue graph is the approximation

Tio Zuca
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    You can calculate $f(x)$ in a very simple way: Take $a_0 = 00$ and $a_{n+1} = \sin(x + a_n)$.Then $f(x) = \lim_{n\to\infty} a_n$. In other words, begin at $0$ (or in any other real number) and iterate the function $u \mapsto \sin(x+u)$. In practice you would get an approximation to $f$ after a few iterations, and you can keep iterating until you have the desired precision. – jjagmath Oct 14 '23 at 02:44
  • Ok, but we can make a series for the function? – Tio Zuca Oct 14 '23 at 02:53

1 Answers1

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Use the Taylor series $$\sin ^{-1}(y)-y=\sum_{n=1}^\infty\frac{ (2 n)!}{4^{n}(2 n+1) (n!)^2}y^{2n+1}$$ and then power series reversion.

You should end with $$y=\sum_{n=0}^\infty a_n\, t^{2n+1} \qquad \text{where} \qquad \color{red}{t=\sqrt[3]{6x}}$$ the first coefficients being $$\left\{1,-\frac{3}{20},\frac{1}{1400},\frac{1}{25200},\frac{43}{1724 8000},\frac{1213}{7207200000},\frac{151439}{12713500800000},\cdots\right\}$$

You can generate as many as you wish using the explicit formula for the $n^{\text{th}}$ term given by Morse and Feshbach.

If you use the above terms and plug them in the original equation, you would find $$x=x-\frac{8074161\ 3^{2/3}}{1191190000000 \sqrt[3]{2}}x^{17/3}+\cdots$$

Use it for $x=\frac 14$; this would give $y=\color{red}{0.9212296}45$

Claude Leibovici
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