Suppose I have smooth functions $f,g,y_0$ and $y_1$ from $\mathbb{R}$ to $\mathbb{R}$, such that $$y_1(x) = y_0(x) - \epsilon g(y_0(x))$$ Then I consider $$f(y_0(x)) = f(y_1(x) + \epsilon g(y_0(x)))$$ Is there a closed form expression for the Taylor series in the small parameter $\epsilon$ in terms of derivatives of $f$ and $g$ and only the function $y_1$?
The first few terms are
$$f(y_0) = f(y_1) + \epsilon f'(y_1) g(y_0) + \frac{1}{2}\epsilon^2 f''(y_1)g^2(y_0) +..$$ Where we interpret $f(y_0)$ as $f|_{y_0(x)}$ and treat $x$ fixed. Then we can again replace the $y_0$ in $g(y_0)$ with $$g(y_0) = g(y_1)+ \epsilon g'(y_1)g(y_0) +...$$ giving $$= f(y_1) + \epsilon f'(y_1) [g(y_1) + \epsilon g'(y_1)g(y_0) + ... ]$$ $$+ \frac{1}{2}\epsilon^2 f''(y_1)[g(y_1) + \epsilon g'(y_1)g(y_0) + ... ]^2 +...$$ Continuing to replace the $y_0$ with $g(y_0)$ like this and grouping terms gives $$f(y_0) = f(y_1)+ \epsilon [f'g](y_1) + \epsilon^2[f'g'g + \frac{1}{2}f''g^2](y_1) + \epsilon^3[f'g'^2g + \frac{1}{2}f'g''g + \frac{1}{2}f''g'g + \frac{1}{6}f'''g^3](y_1) + O(\epsilon^4)$$ But is there some way to write this as a more compact sum like $$f(y_0) \sim f(y_1) + g(y_1)\sum_{n=1}^\infty\sum_{m=0}^n \epsilon^n \alpha(n,m)f^{(n)}g^{(n-m)}(y_1)$$ I am having trouble identifying the pattern. I know there will be some product involved as well.
Edit:
Thinking about it some more it may suffice to just set $f=id$ and consider $$y_0 = y_1 + \epsilon g(y_0)$$ $$y_0 = y_1 + \epsilon g(y_1 + \epsilon g(y_0))$$ $$y_0 = y_1 + \epsilon g(y_1 + \epsilon g(y_1 + \epsilon g(y_0)))$$ $$y_0 = y_1 + \epsilon g(y_1 + \epsilon g(y_1 + \epsilon g(y_1 + ...)))$$ and somehow use the chain rule $$[f_1\circ f_2 \circ .... \circ f_n]' = \prod_{i=1}^n(f'_{i}\circ f_{i+1}\circ ...\circ f_n)$$
from Hot Weekly Questions - Mathematics Stack Exchange
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