Root multiplicity for non-polynomial function

Question

I know that if $f$ is a polynomial and $f(a)=0,f'(a)=0...,f^{(k)}(a)\neq 0$, then $a$ is a root of multiplicity $k$. Does this work for a differentiable function that is not a polynomial? I have seen my textbook saying that, for instance, $1$ is a double root of the equation $x^2-2\ln x-1=0$ and I don't really know why and I also don't know what double root means here. They claim that it is a double root by using Rolle's Theorem's consequences (the actual method is called Rolle's sequence, I don't know how widespread this is, I hope it is not something used only in my country) .

Answer 1

You say that $f(x)$ has a double root on $x=x_0$ if $f(x)=(x-x_0)^2 \varphi(x)$, with $\varphi(x)\ne 0$ in a neighbourhood of $x_0$. When $f$ has a root of multiplicity $k$ (and is smooth enough) you have, like in the case of polynomials, that $$ f(x_0)= \cdots = f^{(k)}(x_0) = 0, \quad f^{(k+1)}(x_0) \ne 0 $$

And this is how you determine the multiplicity... Take a root $x_0$, compute the successive derivatives at $x_0$ and observe when does it stop from being zero.

Answer 2

Locally, a smooth (differentiable) function behaves like a polynomial, as shown by its Taylor's development. The multiplicity is given by the number of leading zero coefficients.

On the plot, the given function and it's second order Taylor approximation:

Answer 3

Statements like this should be interpreted as: $f(x)$ has a double root (or triple root, etc.) at $x_0$ if $f(x)=(x-x_0)^2g(x)$ where $g(x_0)\neq0$ and $g(x)$ has the same differentiable properties as $f(x)$.

In your example of $f(x)=x^2-2\ln x-1$, we have $$f(x)=(x-1)^2\cdot\left(\frac{x^2-2\ln x-1}{(x-1)^2}\right).$$ For $0<x<2$, we can expand $\ln(x)$ in a Taylor series centered at $a=1$, which shows that the function on the right is defined and nonzero at $x=1$. In particular, we have for $0<x<2$, $$x^2-2\ln x-1=2(x-1)^2+\sum_{n=3}^\infty\frac{-1}{n}(x-1)^n.$$ Dividing by $(x-1)^2$ gives $$\frac{x^2-2\ln x-1}{(x-1)^2}=2+\sum_{n=3}^\infty\frac{-1}{n}(x-1)^{n-2}.$$ Since the sum vanishes at $1$, the left-hand side must be in a neighborhood of $2$ when $x\approx1$.