Use Rolle's theorem to show repeated root has zero gradient

[Happiness]

Is this an abuse of Rolle's theorem?

Rolle's theorem
If a real-valued function f is continuous on a proper closed interval [a, b], differentiable on the open interval (a, b), and f(a) = f(b), then there exists at least one c in the open interval (a, b) such that f'(c) = 0.

$[x_1, x_1]$, which means $x_1\leq x\leq x_1$, is not an interval but a point. And $(x_1, x_1)$, which means $x_1<x<x_1$, doesn't make sense. So how can we apply Rolle's theorem?

 

[jambaugh]

Ok, treat the point as an interval, [ x_1,x_1] then the open interval (x_1,x_1), its interior, is the empty set\emptyset. It's perfectly sensible, just trivial. Note the term "proper" as applied to the closed interval in the hypothesis of Rolle's theorem, your example is not a proper closed interval (a<b) but rather a "degenerate" one. Failing to satisfy the premise, the theorem cannot be applied.

 

[Erland]

How does your textbook define "repeated root"?

 

[Happiness]

How does your textbook define "repeated root"?

Suppose $f(x)=A(x-\alpha_1)^{m_1}(x-\alpha_2)^{m_2}...(x-\alpha_r)^{m_r}$, where $\alpha_i$'s are complex and $m_i$'s are integers. If $m_i>1$, then $\alpha_i$ is repeated root of $f(x)$.

 

[Delta2]

Suppose $f(x)=A(x-\alpha_1)^{m_1}(x-\alpha_2)^{m_2}...(x-\alpha_r)^{m_r}$, where $\alpha_i$'s are complex and $m_i$'s are integers. If $m_i>1$, then $\alpha_i$ is repeated root of $f(x)$.

if the exercise assumes that f(x) is of the above form then you don't need Rolle's theorem to prove it, you just take the derivative of f(x) (derivative of a product) which will be $f'(x)=Am_1(x-a_1)^{m_1-1}g(x)+A(x-a_1)^{m_1}g'(x)$ where it is easy to see that for example $a_1$ will be root of $f'(x)$ if $m_1-1>=1$

$g(x)$ is the rest of the product besides the first term

 

[Happiness]

if the exercise assumes that f(x) is of the above form then you don't need Rolle's theorem to prove it, you just take the derivative of f(x) (derivative of a product) which will be $f'(x)=Am_1(x-a_1)^{m_1-1}g(x)+A(x-a_1)^m_1g'(x)$ where it is easy to see that for example $a_1$ will be root of $f'(x)$ if $m_1-1>=1$

True. The question does not restrict $f(x)$ to be just polynomials. But at the same time the book does not provide a definition of a repeated root for functions that are not polynomials.

 

[fresh_42]

Then you're lost. What chance do you have to define multiple roots if neither by differentiating nor by terms of the form $(x-a)^m$? The proposed proof is still valid even if $f$ isn't a polynomial, as long as the root term is a factor.
Read another book.

Edit: What would a multiple root $0$ in $x sinx$ be?

 

[Delta2]

I guess we should take as granted a definition like that f is of the form $f(x)=(x-a_1)^{m_1}g(x)$ where $g(x)$ is some other not necessarily polynomial function. Using that definition we ll have to prove using Rolle's theorem that $a_1$ is root of $f'(x)$ if $m_1>1$...

Well we can do it without using Rolle's theorem as I showed in my earlier post just with the extra assumption that g is differentiable at $x=a_1$.

Really can't think at the moment how to properly use Rolle's theorem for this...

 

[Erland]

I think the idea should be something like this: If we have two distinct roots $x_1$ and $x_2$, Rolle's theorem gives a point $x_3$ between $x_1$ and $x_2$ such that $f'(x_3)=0$. Now, if we let $x_1$ and $x_2$ approach each other, so that in the limit we have a repeated root $x_1=x_2$, $x_3$ is also approaching this common point, so that in the limit, $f'(x_1)=f'(x_2)=f'(x_3)=0$.

But can this argument be made rigorous? I doubt it. I agree that it probably is abuse of Rolle's theorem.

 

[Happiness]

Edit: What would a multiple root $0$ in $x sinx$ be?

The root 0 is a double root and the other (non-zero) roots $n\pi$ are single roots?

 

[fresh_42]

I agree that it probably is abuse of Rolle's theorem.

At least it reads like simply calculating the limit in differences called differentiation.

 

[fresh_42]

The root 0 is a double root and the other (non-zero) roots $n\pi$ are single roots?

Yes, but how is "double root" here defined? That is the crucial point in the whole discussion. As long as it's not clear we could debate for weeks.

 

[Happiness]

Yes, but how is "double root" here defined? That is the crucial point in the whole discussion. As long as it's not clear we could debate for weeks.

I think the generalized definition of a repeated root is as follows:

Write the function $f(x)$ in the form $f(x)=A(x-\alpha_1)^{m_1}(x-\alpha_2)^{m_2}...(x-\alpha_r)^{m_r}g(x)$, where $\alpha_i$'s are complex, $m_i$'s are integers and $g(x)$ is 1 or a power series around the center 0. If (for a particular $i$) the maximum $m_i$ that can be achieved by factoring out $(x-\alpha_i)$ from the power series is more than 1, then $\alpha_i$ is a repeated root of $f(x)$.

 

[fresh_42]

Usually one speaks of roots when polynomials are considered. Otherwise it's simply zeros. (I could be wrong and it's different in English, but to my best knowledge the term root reflects on zeros of polynomials only. So my previous question about $x sin x$ has been already misleading.) Then, in the definition above, you will get poles as well as roots if integer exponents are allowed. I agree on what Erland has said. It looks like an abuse of the theorem of Rolle to simply calculate $\lim_{x_1 → x_2} \frac{f(x_2)-f(x_1)}{x_2 - x_1},$ i.e. the derivation at $x_2$. The most general way might be to define a zero $\alpha$ as double zero, if $f(x)$ can be written $g(x) \cdot h(x)$ and $g(α) = h(α) = 0$ or by power series. And one has to assume differentiability at $α$ anyway, so why not differentiate $f = gh$?

 

[Erland]

About zeros and roots. I learned that functions (may) have zeros and that equations (may) have roots.
So, if f(a)=0, then a is zero of the function f(x) and a root of the equation f(x)=0. Some times, authors blur this distiction, which I don't like.

 

[Happiness]

Then, in the definition above, you will get poles as well as roots if integer exponents are allowed.

The definition above leads to some weird result.

$f(x)=\frac{1}{1-x}=1+x+x^2+x^3+...$
$=(1+x)+x^2(1+x)+x^4(1+x)+...$
$=(1+x)(1+x^2+x^4+...)$

Then, by the definition, $x=-1$ is a "root" of $f(x)$. But it's neither a root nor a pole.

 

[fresh_42]

Don't mess around with groupings of divergent series. This will lead to unpleasant discussions.

 

[Happiness]

Don't mess around with groupings of divergent series. This will lead to unpleasant discussions.

If we can't rearrange the terms in a divergent power series, then how do we factor out the $(x-\alpha_i)$'s to obtain the poles that you mentioned? If we can't factor them out, then doesn't it mean we will only get roots and not poles?

 

[fresh_42]

Simply stay inside the radius of convergence. Around the zeros the series will probably converge.

 

[Happiness]

Simply stay inside the radius of convergence. Around the zeros the series will probably converge.

Could you give an example of how the above definition allows us to obtain the values of poles?

It seems like it's not possible, at least for rational functions. Consider $f(x)=\frac{1}{a-x}$, which has a pole at $x=a$.

$f(x)=\frac{1}{a}\frac{1}{1-\frac{x}{a}}=\frac{1}{a}[1+\frac{x}{a}+(\frac{x}{a})^2+...]$

If the above definition give rise to roots, we must be able to factor out $(x-a)$ but when $x=a$, the power series on the RHS is divergent and we can't rearrange the terms. So it seems we can't get the poles by using the above definition.

 

[fresh_42]

Sorry, I didn't catch your "factoring out" part of your definition.