polynomial of degree n has n real roots (1 Viewer)

[gordo]

supposing it has real coefficients,
how do u prove it?

 

[Slidey]

You can't prove it.

You actually want to prove that:

"A polynomial of degree n>=1 with complex coefficients has n complex roots."

This "proof" assumes Gauss' proof (beyond 4unit scope):

"A polynomial of degree n>=1 with complex coefficients has at least one root"

From this:

Suppose P(x) has degree n, then
P(x)=Q(x), then
P(x)=(x-a₁)Q₁(x), then
P(x)=(x-a₁)(x-a₂)Q₂(x), then
P(x)=(x-a₁)(x-a₂)(x-a₃)Q₃(x), et cetera n times, till:
P(x)=c_n(x-a₁)(x-a₂)...(x-a_n-1)(x-a_n), where c_n is the leading coefficient

EDIT: Fixed leading coefficient: meant to be c_n, but was a_n

 

[Slidey]

The fundamental theorem of algebra:

Theorem 1: "Every polynomial of degre n>=1, n a natural numbers, has at least one zero over the complex field."

Theorem 2: "Every polynomial of degre n>=1, n a natural number, has n zeros over the complex field."

Assuming Theorem 1, we can use induction to prove Theorem 2

S(n) = Theorem 2
S(1) is true: ax+b has ONE zero at x=-b/a
Assume S(k) is true, consider S(k+1):
Let P(x) be a poly of degree k+1. By Theorem 1, there is a number, c, such that P(c)=0, thus
P(x)=(x-c)Q(x), where Q(x) is a polynomial of degree k.
But if S(k) is true, Q(x) must have k zeros, and these k zeros must be zeros of P(x).
Thus if S(k) is true, then S(k+1) is true. Since S(1) is true, by induction, S(n) is true for all natural n.

EDIT: Fixed a few things.

 

[Trev]

Slide, please tell me you don't know all this stuff 'off by heart', and that you are just copying/referring to something else?

 

[Slidey]

The first post was me doing stuff off by heart.

The second post was me referring to a proof of Theorem 2 above.

Although from now on it is very likely I will remember/figure out how to prove it, yes. It's simple, really:

*Assume FTA
*Induction
*S(n): P(x) has n degree and n roots
*For S(k+1), take out a factor. Quotient has degree k, apply S(k).

Voila!

 

[KFunk]

I'm going to go do a little search for it in a sec, but assuming that it will be a bit beyond me what is the general gist of what Gauss' proof consists of?

 

[Slidey]

His actual proof? I'm not sure.

But there's many variations of the proof. All of which aren't just beyond our scope, but well beyond it, I think.

http://encyclopedia.laborlawtalk.com/Fundamental_theorem_of_algebra

http://planetmath.org/encyclopedia/FundamentalTheoremOfAlgebra.html

 

[niall045]

Um, Gauss is very much beyond the scope of the HSC, maybe even uni stuff. Im not sure how far uni goes but you wont have to prove the FTA. If they try and make you then i personally give you permission to kick the exam.

 

[martin]

Fundamental Theorem of Algebra: Every polynomial equation having complex coefficients and degree at least 1 has at least one complex root. This theorem was first proven by Gauss.

The proof isn't in the HSC but is well within the grasp of a university course. I've actually seen proofs (that I can remember) in three different courses:
- complex analysis - follows from Liouville's Thm.
- abstract algebra - pretty much the same as the Complex Analysis proof.
- topology - something tricky that I can't remember.

If you ever want a proof that doesn't use lots of complicated machinery search for an 'elementary' proof (that doesn't mean it will be easy though). Doing this I found this webpage http://www.cut-the-knot.com/fta/fta_sketch2.shtml which has lots of proofs of FTA. This one looks like it might be understandable but its really only a sketch of a proof.

 

[withoutaface]

How about something like this:

Opening remarks
1. Any polynomial can be expressed as a monic by dividing by the leading coefficient, hence no generality is lost through dealing with monics only.
2. x+y=z can always be solved even if x and z are both constants, due to y's ability to vary to become z-x.

Induction
Prove for n=1
x+c=0, clearly there is 1 root, hence holds for x=1

Assume for n=k
x^k+ d_k-1x^k-1+...+d₀=0
If it has k roots, then it can be expressed:
(x-a_k)(x-a_k-1)...(x-a₁)=0

Hence show for n=k+1
x^k+1+p_kx^k+...+p₀=0
Can be expressed
x^k+1+(m+d_k-1)x^k+...+d₀=0
(x+m)(x^k+ d_k-1x^k-1+...+d₀)=0
(x+m)(x-a_k)(x-a_k-1)...(x-a₁)=0
Clearly, this has k+1 roots, hence holds for n=k+1 if holds for n=k, and holds for n=1, therefore by mathematical induction, any polynomial of degree n (n>=1) has n solutions.

There's probably something wrong with what I wrote above, can someone point it out for me?

PS Gordo you were in my ENGG1801 lecture today.

 

[martin]

x^k+1+p_kx^k+...+p₀=0
Can be expressed
x^k+1+(m+d_k-1)x^k+...+d₀=0
(x+m)(x^k+ d_k-1x^k-1+...+d₀)=0
(x+m)(x-a_k)(x-a_k-1)...(x-a₁)=0

There's probably something wrong with what I wrote above, can someone point it out for me?

I don't really follow this whole bit. Where does m come from? Its obviously meant to be something to do with the root but you seem to have pulled it out of the air. Also note
(x+m)(x^k+d_k-1x^k-1+d_k-2x^k-2 +...+ d₁x+d₀) =
x^k+1 + (m+d_k-1)x^k + (md_k-1+d_k-2)x^k-1 +...+ (d₀+md₁)x + md₀

so I think some of those lines don't quite follow.

 

[withoutaface]

well for the expansion you get the coeffiecient for x^n-1 from within the right brackets moving forward to become part of x^n's coefficient due to being multiplied by x, and then you have the other part coming from x^n's original coeffiecient multiplied by m. Hence (md_n+d_n-1)xⁿ results in the expansion.

m's just any random number which i've included from working backwards so that the whole thing factorises nicely.

 

[gordo]

PS Gordo you were in my ENGG1801 lecture today.

ahh cool

haha that lecture was funny, i left after 15 mins go move my car and when i got back it was over lol

 

[martin]

well for the expansion you get the coeffiecient for x^n-1 from within the right brackets moving forward to become part of x^n's coefficient due to being multiplied by x, and then you have the other part coming from x^n's original coeffiecient multiplied by m. Hence (mdn+dn-1)xn results in the expansion.

I still don't understand your argument. In your original post you have a d₀ but when you expand out your result you get md₀.

Also, your proof seems to work for real numbers as well as complex, so you have proved that x²+x+1=0 has 2 real solutions, which isn't true.

 

[withoutaface]

Hmm ok fair enough, thanks

 

[withoutaface]

Actually on thinking about it further, the md₀ is nothing but a clerical error, and doing the proof for real roots would involve the restrictions that m and all roots of n=k are real, thereby getting a rather trivial result that:
x^k+1+...=(x+m)(x-a_k)...
has k+1 real roots where m and all a's are real.

 

[Slidey]

Indeed. It's easier to leave expansion out of it and leave the polynomial in factored form and simply apply S(n) assumed for n=k, S(n) being the statement that a polynomial of degree n has n roots.