Is this a valid proof?

[YoungPhysicist]

Recently I came up with a proof of “ for a nth degree polynomial, there will be n roots”

Since the derivative of a point will only be 0 on the vertex of that function,and a nth degree function, suppose $f(x)$has n-1 vertexes, $f’(x)$ must have n-1 roots.

Is the proof valid?

 

[fresh_42]

Recently I came up with a proof of “ for a nth degree polynomial, there will be n roots”

Since the derivative of a point will only be 0 on the vertex of that function,and a nth degree function, suppose $f(x)$has n-1 vertexes, $f’(x)$ must have n-1 roots.

Is the proof valid?

What about $f(x)=x^2\,?$

 

[jedishrfu]

It seems this kind of proof would necessarily need to be an induction proof and that your version seems to be too simple.

However perhaps @fresh_42 or @Mark44 could provide a better answer.

Have you looked online for any similar proofs?

 

[jedishrfu]

I found this discussion on math stack exchange about it

https://math.stackexchange.com/ques...-a-polynomial-of-degree-n-has-at-most-n-roots

 

[Mark44]

Recently I came up with a proof of “ for a nth degree polynomial, there will be n roots”

Since the derivative of a point will only be 0 on the vertex of that function,and a nth degree function, suppose $f(x)$has n-1 vertexes, $f’(x)$ must have n-1 roots.

Is the proof valid?

No. Where is the vertex of, say $f(x) = x^3$? Does the graph of this function have 3 - 1 = 2 vertices?

Also, $f'(x) = 3x^2$ Are there 3 - 1 = 2 roots of the equation $3x^2 = 0$?

 

[YoungPhysicist]

No. Where is the vertex of, say $f(x) = x^3$? Does the graph of this function have 3 - 1 = 2 vertices?

Also, $f'(x) = 3x^2$ Are there 3 - 1 = 2 roots of the equation $3x^2 = 0$?

How about saying that they have two “identical” vertexes and roots?

 

[fresh_42]

How about saying that they have two “identical” vertexes and roots?

If you count multiplicities, then you are right, but your proof doesn't work anymore. Each polynomial can be written as a product of terms $x-z_i$ where the $z_i$ are the possibly complex roots. And they do not have to be different, so they can occur multiple times.

If you change your statement to "at most degree many roots", then your idea is close to how it is usually proven. But the geometry is tricky here: what about inflection points? What if there aren't real roots between two local extrema? One normally counts the number of sign changes to determine the maximal number of possible roots. This is the better indicator than local extrema.

 

[YoungPhysicist]

If you count multiplicities, then you are right, but your proof doesn't work anymore. Each polynomial can be written as a product of terms $x-z_i$ where the $z_i$ are the possibly complex roots. And they do not have to be different, so they can occur multiple times.

If you change your statement to "at most degree many roots", then your idea is close to how it is usually proven. But the geometry is tricky here: what about inflection points? What if there aren't real roots between two local extrema? One normally counts the number of sign changes to determine the maximal number of possible roots. This is the better indicator than local extrema.

Kind of get the point,thanks!

 

[Mark44]

No. Where is the vertexof, say $f(x) = x^3$?

How about saying that they have two “identical” vertexes and roots?

But this function doesn't have any vertices (plural of vertex) at all! I think you might be oversimplifying things, possibly limiting what you're thinking about to parabolas and power functions of even degree.

 

[fresh_42]

You can use the derivatives for the count of sign changes. It doesn't necessarily get you one, but if there is a sign change, there has to be a local extremum in between. Thus there are at most degree many roots.

 

[YoungPhysicist]

So the proof should be:

A polynomial equation of nth degree can always be factored to nth roots like

$$(x-t)$$ for nth times where t here is a complex.

Is it like that?

 

[fresh_42]

So the proof should be:

A polynomial equation of nth degree can always be factored to nth roots like

$$(x-t)$$ for nth times where t here is a complex.

Is it like that?

"Like" yes, but a bit very, very sloppy.

Let $p(x)\in \mathbb{R}[x]$ be a real polynomial of degree $n$ in one variable. Since $\mathbb{R}[x] \subseteq \mathbb{C}[x]$ we can interpret $p(x)$ as a complex polynomial, too. Then there are complex numbers $z_1,\ldots ,z_k$ such that $p(x)=(x-z_1)^{n_1} \cdot \ldots \cdot (x-z_k)^{n_k}$ and $n_1+ \ldots + n_k =n\,.$ Especially we have $k \leq n\,.$

This is one possibility. The other is to stay in $\mathbb{R}[x]$ and observe, that the limits of $p(x)$ for $x \to \pm \infty$ are also $\pm \infty$. Therefore there is a first and a last zero (or none). Every time the graph of $p(x)$ crosses the $x-$axis, we get a sign change in $p(x)$. However, between two of those sign changes, there has to be at least one point with $p'(x)=0$, maybe more. By induction we know, that at most $n-1$ such locations are possible. So the $n-1$ in betweens mean at most $n=(n-1)+1$ possible zeroes.

 

[mathwonk]

To essentially iterate what fresh has said, you can use the principle that a graph cannot change direction between two consecutive critical points (zeroes of the derivative). Hence at most one zero of the function can occur between any two zeroes of the derivative. Thus the function has at most one more zero than its derivative.

To prove the first principle stated, use the fact that a local extremum must occur on any interval where the function changes direction.

Since also a function's derivative must have a zero between any two consecutive zeroes of the function, the derivative must also have at least one fewer zeroes than the function itself. This proves again that if the function has n zeroes, the derivative has at least n-1 zeroes. This is the same proof "said backwards". Both proofs rest on the same somewhat deep fact, that a continuous function has a local extremum on any interval on which it takes the same value at both ends.

Note however the derivative can have more zeroes than this. for e.g. X^2 + 1. the derivative actually has more zeroes than the function itself.

A proof by algebra, as suggested above, uses only the basic "factor theorem", that if x=a is a root of the polynomial f(x), then (x-a) is a factor. It follows, since factoring f(x) out by (x-a) lowers the degree, that f(x) can have at most deg(f) factors, hence also at most deg(f) roots. This proof assumes only that we are working in a realm of numbers where the product of two non zero numbers is again non zero.

 

[FactChecker]

The short answer is no. This theorem is only true when complex numbers are allowed. It looks like you are working in the Reals, in which case it is false. It is difficult to prove something that is false.

 

[Stavros Kiri]

It is difficult to prove something that is false

Proving it's false is the proof in this case.

 

[FactChecker]

Proving it's false is the proof in this case.

Because the theorem is so important, I think that it is good to simply state that it is true over the complex numbers, but not over the reals. The proposed proof does not make any sense in the complex plane.

 

[Stavros Kiri]

Because the theorem is so important, I think that it is good to simply state that it is true over the complex numbers, but not over the reals. The proposed proof does not make any sense in the complex plane.

Due to multivalueness?

 

[FactChecker]

Due to multivalueness?

No. I'm not sure which part of my statement you are asking about, but multivaluedness is not the basic problem. The OP seems to say that the real derivative and geometry on the real line is enough to prove the Fundamental Theorem of Algebra, but it is not -- it is not true on the real line and the proof is more complicated over the complex plane. For correct proofs of the Fundamental Theorem of Algebra, see https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra#Proofs

 

[Stavros Kiri]

I'm not sure which part of my statement you are asking about

The proposed proof does not make any sense in the complex plane.

Due to multivalueness?

If you count multiplicities, then you are right, but your proof doesn't work anymore. Each polynomial can be written as a product of terms $x-z_i$ where the $z_i$ are the possibly complex roots. And they do not have to be different, so they can occur multiple times.

 

[FactChecker]

Multiple zeros is not the problem. As @mathwonk pointed out, $x^2+1$ has two distinct zeros, $\pm i$, in the complex plane, but no zeros on the real line. How does the proposed "proof", which is completely without detail, reach a conclusion about that polynomial based on its derivative, $2x$?

 

[Stavros Kiri]

Multiple zeros is not the problem. As @mathwonk pointed out, $x^2+1$ has two distinct zeros, $\pm i$, in the complex plane, but no zeros on the real line. How does the proposed "proof", which is completely without detail, reach a conclusion about that polynomial based on its derivative, $2x$?

I do not disagree

 

[WWGD]

What do y

Recently I came up with a proof of “ for a nth degree polynomial, there will be n roots”

Since the derivative of a point will only be 0 on the vertex of that function,and a nth degree function, suppose $f(x)$has n-1 vertexes, $f’(x)$ must have n-1 roots.

Is the proof valid?

Mean by vertex of the function?