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wildrabbitt 2019-10-25 09:00

roots of quadratics
 
Hi,


in the A-level maths curriculum, there's a topic to do with [B]roots of quadratics[/B].


A typical question would be for example,


The quadratic \(2x^2+3x-6\) has roots \(\alpha\) and \(\beta\).


(i) Write down \(\alpha+\beta\) and \(\alpha\beta\).


(ii) Hence show that \(\alpha^3+\beta^3=\frac{-135}{8}\).


(iii) Find a quadratic whose roots are \(\alpha+\frac{\alpha}{\beta^2}\) and \(\beta+\frac{\beta}{\alpha^2}\).




Does anyone know if the reason quadratics are studied in this way is anything more than just because they can be?


Does getting to understand how to solve such questions give a student any skills for things that could be studied later on in maths?


I'm wondering if this has a connection with group theory but I don't know what the connection is, if there is one.

Nick 2019-10-25 09:57

[QUOTE=wildrabbitt;528881]
Does anyone know if the reason quadratics are studied in this way is anything more than just because they can be?

Does getting to understand how to solve such questions give a student any skills for things that could be studied later on in maths?

I'm wondering if this has a connection with group theory but I don't know what the connection is, if there is one.[/QUOTE]
In this case, the coefficients of the quadratic equation are all integers but the roots α and β are irrational numbers.
However, since \((X-\alpha)(X-\beta)=X^2-(\alpha+\beta)X+\alpha\beta\), the values α+β and αβ are integers too.
What's more, it follows that [B]any symmetric expression[/B] in α and β (i.e. which stays the same when you permute them)
can be written in terms of α+β and αβ and therefore is also an integer.
So this is a way of staying inside the integers instead of having to calculate approximately with irrational numbers, which makes it of great practical importance.

You are right that there is also a connection with group theory.
Permuting the roots of a polynomial equation gives us important symmetries
(known as automorphisms) n the fields containing them, which help us understand their structure.
This was first worked out by Évariste Galois, a French mathematician who died aged 21 in a duel, and the theory is still known today as Galois Theory.
It's quite a showpiece of mathematics!

Dr Sardonicus 2019-10-25 12:14

Some basic facts about the "theory of equations" are at work here.

One (the "factor theorem") is that if f(x) is a polynomial with coefficients in a field (here, the field of rational numbers) and f(t) = 0, then (x - t) is a factor of f(x). This is pertinent to your question (iii).

Also WRT to your question (iii), note that each of the two quantities which are supposed to be roots can be obtained from the other by permuting α and β.

One reason quadratic polynomials are studied (especially as an introduction to the "theory of equations") is that the results are theoretically interesting, and the requisite calculations can actually be done by hand without too much effort. With higher-degree polynomials, the calculations can become too laborious to do by hand, but, armed with a grasp of the theory obtained from the tractable quadratic case, you can at least understand the general [i]form[/i] of the results.

One result for which your your question (ii) is a "jumping off point" is known as "Newton's identities."

MattcAnderson 2019-10-29 11:17

Not an expert on group theory. But there is the trivial group of just one element. This could be labeled identity or something else. There is no group with count nill.

In my humble opinion the easiest group to understand is cyclic group. These cyclic group must have prime order. To be clear, the count of number of elements is a prime number.

In the end, don't know of a connection between quadratic surds, and mathematical groups.

Surely we can learn from the question of Original Poster.

R.D. Silverman 2019-10-29 12:56

[QUOTE=MattcAnderson;529177]Not an expert on group theory. But there is the trivial group of just one element. This could be labeled identity or something else. There is no group with count nill.

In my humble opinion the easiest group to understand is cyclic group. These cyclic group must have prime order. .[/QUOTE]

You acknowledge that you are "Not an expert on group theory".

So why do you feel compelled to post?

Hint: Your claim that a cyclic group must have prime order is wrong.
Further hint: There is a cyclic group of order n for all n \in N.

Dr Sardonicus 2019-10-29 14:50

Yes, there are cyclic groups of all positive integer orders. The cyclic groups of prime order are the only (nontrivial) [i]simple[/i] cyclic groups. Which reminds me...

[url=https://www.math.upenn.edu/~pemantle/songs/groups]"Simple Groups", to the tune of "Sweet Betsy from Pike"; words by Saunders Mac Lane[/url]

Be it noted, the [url=https://plus.maths.org/content/enormous-theorem-classification-finite-simple-groups]classification of finite simple groups[/url] was achieved in 1981.

R.D. Silverman 2019-10-29 16:42

[QUOTE=Dr Sardonicus;529197]Yes, there are cyclic groups of all positive integer orders. The cyclic groups of prime order are the only (nontrivial) [i]simple[/i] cyclic groups. Which reminds me...

[url=https://www.math.upenn.edu/~pemantle/songs/groups]"Simple Groups", to the tune of "Sweet Betsy from Pike"; words by Saunders Mac Lane[/url]

Be it noted, the [url=https://plus.maths.org/content/enormous-theorem-classification-finite-simple-groups]classification of finite simple groups[/url] was achieved in 1981.[/QUOTE]

This reminds me of a joke that I heard a long time ago.

I was taking an algebra course from Birkhoff. He was introducing the notion of
simple groups. He stated (more or less)

A simple group is one with no self conjugate subgroups... Simple, isn't it???

Nick 2019-10-30 08:29

[QUOTE=Dr Sardonicus;529197]The cyclic groups of prime order are the only (nontrivial) [I]simple[/I] cyclic groups.[/QUOTE]
Better: the cyclic groups of prime order are the only simple abelian groups.


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