Zeros of order $m$

Question

Suppose that $f$ has a root $\alpha$ of order $m$. Prove that $\alpha$ is a root of order $2m-1$ for $$ \theta(x)=f(x+f(x))-f(x). $$ My attempt: We can take $f(x)=(x-\alpha)^{m}q(x)$ where $q(\alpha)\neq0$. However, I can not see the result from this.

Answer 1

Using the linear Taylor polynomial with the quadratic remainder term one gets \begin{align} f(x+f(x))-f(x)&=f'(x)f(x)+\frac12f''(x+θf(x))f(x)^2 \end{align} where the first term has root multiplicity $2m-1$ and the second $2m$, so that the sum has multiplicity $2m-1$.

Answer 2

\begin{align*} f(x) &= (x-\alpha)^{m}g(x) \\ f'(x) &= m(x-\alpha)^{m-1}g(x)+(x-\alpha)^{m}g'(x) \\ f(x+f(x)) &= f(x+(x-\alpha)^{m}g(x)) \\ &= f(x)+(x-\alpha)^{m}g(x)f'(x)+\ldots \\ f(x+f(x))-f(x) &= (x-\alpha)^{m} g(x) f'(x)+\ldots \\ &= m(x-\alpha)^{2m-1} g(x)+\ldots \end{align*}