Ekman's compilation question (2 Viewers)

[marxman]

Q3 from ekman's polynomial set? It is stickied to the ext 2 thread - neither I nor my teacher understands how to crack it.

 

[Drsoccerball]

(Via binomial expansion).







Also if you have a question about ekman's questions I think it would be better if you posted it on the actual thread so other people can also benefit in the future .

 

[leehuan]

(Via binomial expansion).







Also if you have a question about ekman's questions I think it would be better if you posted it on the actual thread so other people can also benefit in the future .

Haven't actually come across this type of question being asked from memory though... is it really that common? The method didn't look so obvious to me without glancing at the answers in advance

 

[Drsoccerball]

Haven't actually come across this type of question being asked from memory though... is it really that common? The method didn't look so obvious to me without glancing at the answers in advance

I probably did about 7 of these before HSC.

 

[leehuan]

I probably did about 7 of these before HSC.

From where?

 

[Drsoccerball]

From where?

BOS and other sources Sy123 introduced me to this type of question.

 

[Paradoxica]

It's easy if you know about depressed cubics.....

 

[seanieg89]

(Via binomial expansion).







Also if you have a question about ekman's questions I think it would be better if you posted it on the actual thread so other people can also benefit in the future .

By itself, this working does not show that your constructed P has the lowest degree possible.

 

[Drsoccerball]

By itself, this working does not show that your constructed P has the lowest degree possible.

So we see that x = f(x) is irrational and also that x^2 =(f(x))^2 is also irrational and thus P(x) is the lowest degree polynomial with rational coefficients. How would we do this if the polynomial happened to have a leading coefficient of 100 or larger?

 

[seanieg89]

So we see that x = f(x) is irrational and also that x^2 =(f(x))^2 is also irrational and thus P(x) is the lowest degree polynomial with rational coefficients. How would we do this if the polynomial happened to have a leading coefficient of 100 or larger?

x=f(x) is irrational? what is f(x) and what does it mean for an equation to be irrational?

You just need to show that a quadratic with rational coefficients cannot have a root at this value. I have posted a proof for a similar (perhaps identical) question in the past on here, but will do so again if no-one else does.

As for higher degree analogues, this starts to get into an aspect of something called Galois theory. The theory is very rich but you can only tell quite a limited part of the story if you want to remain within the scope of MX2.

If you study this question (and rational cubics in general) more carefully, you will note that other than the root a^(1/3)+b^(1/3) (I changed the sign because why not), the other two roots of the cubic are wa^(1/3)+w^2b(1/3) and w^2a^(1/3)+wb^(1/3) where w is the primitive cube root of unity.

You can think of this fact as being a specific instance of a third degree version of the "conjugate root theorem" which deals with surdic roots of the form a+sqrt(b). Any rational polynomial that has as a root a^(1/3)+b^(1/3) (where a and b denote the quantities in this question) must have the other two quantities as roots as well.

It is an fun exercise to state and prove a precise version of this cubic conjugate root theorem. I will perhaps write a guided question on this later. (Late tonight at the earliest.)

 

[Paradoxica]

It is sufficient to demonstrate that there cannot exist a linear factor that multiples with the one considered to obtain a Quadratic Polynomial with rational coefficients. This can be done via contradiction, although there may be a more elegant method.

 

[Paradoxica]

x=f(x) is irrational? what is f(x) and what does it mean for an equation to be irrational?

You just need to show that a quadratic with rational coefficients cannot have a root at this value. I have posted a proof for a similar (perhaps identical) question in the past on here, but will do so again if no-one else does.

As for higher degree analogues, this starts to get into an aspect of something called Galois theory. The theory is very rich but you can only tell quite a limited part of the story if you want to remain within the scope of MX2.

If you study this question (and rational cubics in general) more carefully, you will note that other than the root a^(1/3)+b^(1/3) (I changed the sign because why not), the other two roots of the cubic are wa^(1/3)+w^2b(1/3) and w^2a^(1/3)+wb^(1/3) where w is the primitive cube root of unity.

You can think of this fact as being a specific instance of a third degree version of the "conjugate root theorem" which deals with surdic roots of the form a+sqrt(b). Any rational polynomial that has as a root a^(1/3)+b^(1/3) (where a and b denote the quantities in this question) must have the other two quantities as roots as well.

It is an fun exercise to state and prove a precise version of this cubic conjugate root theorem. I will perhaps write a guided question on this later. (Late tonight at the earliest.)

And all we need to finish off is the Quartic analogue of the Conjugate Root Theorem.

Then prove no such theorem exists for all higher polynomials ???



Actually wait how does one disprove the existence of a theorem. That doesn't make any sense.

 

[seanieg89]

And all we need to finish off is the Quartic analogue of the Conjugate Root Theorem.

Then prove no such theorem exists for all higher polynomials ???



Actually wait how does one disprove the existence of a theorem. That doesn't make any sense.

The general principle of roots of polynomials having Galois conjugates does not depend on degree.

Given an algebraic number s, it has a minimal degree monic rational polynomial that it is the root of. The other roots of this polynomial are said to be its conjugates. If we assume a certain form of algebraic number, then we will usually get a nice expression for its conjugates.

Eg if a rational polynomial has as a root, it must also have as a root for k=1,...,n-1 where is a primitive n-th root of unity.

The only nice thing about low degree polynomials is that all algebraic numbers of degree < 5 will be able to be expressed in a known form in terms of radicals.

Galois theory still lets us talk about symmetries between roots of higher degree polynomials.