# Composition of Taylor Series

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 $$[f_1\circ f_2 \circ .... \circ f_n]' = \prod_{i=1}^n(f'_{i}\circ f_{i+1}\circ ...\circ f_n)$$

• Is $y_1$ (and $y_0$) a function, as you've written at the end of your second sentence, or is it a variable? Apr 26, 2019 at 13:37
• No they are functions like $y_0(x)$ and $y_1(x)$. But the Taylor series is in the parameter $\epsilon$, so I am treating them to be evaluated at some fixed point. So $f'(y_1)$ should be interpreted as $f'|_{(y_1(x))}$ Apr 26, 2019 at 13:39
• In "the first few terms are", you seem to have replaced $g(y_0)$ with a Taylor series for $g$ at $y_1$, using $\epsilon$ as the variation in the Taylor series, i.e. treating $\epsilon$ as $y_1 - y_0$. But according to your first equation, $\epsilon = -\frac{y_1 - y_0}{g(y_0)}$, so the sign is wrong, and there's a missing factor of $g(y_0)$, unless I'm misreading something. I think perhaps your question needs more thinking through. Apr 26, 2019 at 13:44
• I am replacing $y_0$ in $g(y_0)$ just as it was replaced in $f(y_0)$ so $$g(y_0) = g(y_1 + \epsilon g(y_0)) = g(y_1) + \epsilon g'(y_1)g(y_0) + ...$$ which gives the second equality. So I think this is correct... Apr 26, 2019 at 14:02
• OK. Well, best of luck. Apr 26, 2019 at 14:45