Perhaps I'm an extreme brainlet but this has been pissing me off ever since high school and couldn't find an answer

Perhaps I'm an extreme brainlet but this has been pissing me off ever since high school and couldn't find an answer.
I was told that when a zero's multiplicity is even, the graph touches the zero and then bounces off. If the multiplicity is odd - it goes through the x axis. I cannot understand why that happens and never had it explained - just said that it does. Help me Veeky Forums

It has something to do with local symmetry, obviously the x2-factor creates a symmetrical 0, while x3-factor is asymetrical. Make a taylor series at the local zeros and this will be clear.

A zero's multiplicity isn't even defined outside of polynomials afaik, or what kind of multiplicity can be related to sin(ax) ?

factor the polynomial p(x)=(x-z)^2n * (x-y)^k(x-w)^j....
at the even zero z, the sign of (z-y)^k(z-w)^j.... is fixed
at a hair to z such that z±h does not cross any other zero
since it doesn't cross any other zero, the sign of (z±h-y)^k(z±h-w)^j.... stays the same as it needs to cross a zero to reverse the sign of one of the linear products
now (z±h-z)^2n * (z±h-y)^k(z±h-w)^j.... = (±h)^2n * (z±h-y)^k(z±h-w)^j....
since (±h)^2n is positive regardless of the ± and the sign of the other term doesn't change, the polynomial stays on the same side of the x axis around the zero

QED

>since it doesn't cross any other zero, the sign of (z±h-y)^k(z±h-w)^j.... stays the same as it needs to cross a zero to reverse the sign of one of the linear products

This is the very statement you're trying to prove. Circular logic imo

yeah, his explanation didn't make sense to me either because of this part

Mean Value Theorem: (x-y) is continuous. If z is negative and z+h is positive (or vice versa) then there exist a 0

Let [math] a [/math] be a zero of a polynomial function [math] f [/math] with multiplicity [math] n [/math] .

Then [math] f(x)=(x-a)^n g(x) [/math] for some polynomial function [math] g(x) [/math] with [math] g(a) \neq 0 [/math] .
Because [math] g(x) [/math] is continuous (since it's polynomial) and [math] g(a) \neq 0 [/math], there exists a region around [math] a [/math] such that [math] g(x) [/math] is either only >0, or only

Let our function be rewritten as f(x) = g(x) * k(x),
with g(x) = (x-z)^2n the even zero part, and k(x) = (x-y)^k(x-w)^j.... any kind of rest.

Suppose we define z such that z±h does not cross any other zero, so g(z) = 0, g(z±h) > 0.

For this to prove, we need to show that k(z+h) - k(z-h) =/= 0, which has been the starting point of choosing h in the first place.

How does the mean value theorem help proving that such h exist, for arbitrary polynomial k?

>A zero's multiplicity isn't even defined outside of polynomials afaik
It is for holomorphic functions.

>there exists a region around a such that g(x) is either only >0, or only 0 or < 0 ?

Isn't it because a negative times a negative equals a positive so you're guaranteed to eventually end up getting the same result for a different input?

I don't think any explanation itt is correct

Explain how they're wrong then, brainlet.

In the following, "ε-region around a" means this interval (a-ε,a+ε).
Also, the region (α-ε,α+ε) I talked about in the proof, is now named (a-δ,a+δ) because it is the costumary notation.

g being continuous at α means that:
Given any ε-region around g(a) there exists a δ-region around a such that f maps it inside the ε-region.

Since g(a) is not 0 you can take an ε-region around it such that it doesn't contain 0. (if g(α) is greater than 0 then that region will contain only positive numbers, if less than 0 then only negative numbers)
By continuity, there exists a δ-region around a such that it gets mapped inside that ε-region.
This means that it all the elements of (α-δ, α+δ) get mapped to either only positive numbers or only to negative numbers.

> and g(a)≠0, there exists a region around a such that g(x) is either only >0, or only because g(x) keeps constant sign

This needs to be shown.

imagine you have a (x-a)^2 in the factors of a polynomial. (or any other even power)

the rest of the factors behave like a constant around x=a, so your polynomial is equivalent to C*(x-a)^2 around a. It behaves just like it.

And what does that look like you faggot? Like a parabola.

the first explanation is just circular logic like other user pointed
the second proof just assumes something is true without showing it
>It behaves just like it
nice proof

>This needs to be shown.
Really now? Here .
It should be freaking obvious.

essentially because g(x) is continuous? Ok I think I got it.

>nice proof

you might want to look into equivalent functions. Tired of underage american high schoolers trying to sound smart desu.

Oh and it wasn't a proof, I don't care enough to prove it so go fuck yourself with a cactus.

t. american high schooler

"it behaves just like it locally" means a very specific thing. see , which anyone who learned calculus (aka 18+) should know

>shit I lost an argument, find a good insult
>american high schooler

thanks for proving my point

you guys are so dumb it's not even funny anymore.

I think I got it, drawing it out helped. It really does suck to be a brainlet. I'm going to attempt to explain it in a little simpler way so that other brainlets like me can understand.
So let's take this polynomial as an example:
(x-3)^2 * (x-1)(x-2)
Since (x-3)^2 is always positive because it's a square, the polynomial value will only become negative when the rest is negative. The rest in this case is (x-1)(x-2) which is only negative in (1, 2)
Essentialy if the multiplicity is even it only crosses the axis when the rest is negative
This is nowhere close to a proof but definitely helped me understand it more intuitively

just listen to me mate.

around a=3, (x-1)(x-2) remains close to 2. So the polynomial you have behaves like 2*(x-3)^2 around a=3.

If you want plot 2*(x-3)^2 on top of your graph.

Imagine factorising the polynomial so you get something like (x-a)^n*(x-b)^m*(x...

Now look at the zero at x=a. Clearly all the other shit doesn't change sign as we pass x=a as the other factors are non-zero at a, so we just take them to be some constant K in the infinitesimal region around a.

So now we have K*(x-a)^n, which means just around x=a we have a function of the form x^n. Now just think of these functions, how they look at the origin. For even n the function goes in the same y-direction as x goes to positive or negative infinity, for odd n in opposite y-directions.

Now just imagine that around x=a with some prefactor K and you should be able to see why it works that way

>help proving that such h exist, for arbitrary polynomial k?

It's a polynomial so the number of roots is finite therefore the set of difference between k's zeros and z has a minimum element. Let h be smaller than that minimum.

>Clearly all the other shit doesn't change sign as we pass x=a as the other factors are non-zero at a
this is the hard part
how do we know it doesn't change sign? I imagine K could change the sign when you go from the left of the zero to the right

then the zero would not have an even order.

If each factor is x minus a constant then that factor can only change sign when x equals that constant. That means only the factor that is x minus the root we are looking at can change sign. Nothing difficult about that