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type_speed
11-29-2006, 12:34 AM
Prove:
given that f(x) = Anx^n + An-1X^n-1+ ....A1X + A0, and An does not = 0
=> if f(x) has a root of the form (A+Bi), then it must have a root of the form
(A-Bi). (complex roots)

All I've figured out so far is to utilize the Conjugate roots theorem but our professor gave us a hint that we should use the remainder theorem or something. Basically I need to prove why both roots exists in a function

HallsofIvy
01-16-2007, 04:34 PM
Consider the simple quadratic case, a_2x^2+ a_1x+ a_0= 0. Suppose one root is A+ Bi and the other is C+ Di. The equation can be written a_2(x-A-Bi)(x-C-Di)= 0. Go ahead and multiply those out. Remembering that the coefficients must be real so the 0, you get two equations to solve for C and D in terms of A and B. What are they?

Talvi Watia
02-04-2007, 07:02 AM
Prove:
given that f(x) = Anx^n + An-1X^n-1+ ....A1X + A0, and An does not = 0
=> if f(x) has a root of the form (A+Bi), then it must have a root of the form
(A-Bi). (complex roots)

All I've figured out so far is to utilize the Conjugate roots theorem but our professor gave us a hint that we should use the remainder theorem or something. Basically I need to prove why both roots exists in a function

f0 : sum { dx dy dz ... dn ; su (1,1,1,1,...,n) }
f1 : matrix { 1 ... n }

are you looking for real roots or all complex number answers.

if real only n, abs ( +- ei^n )
if all real and complex, n^2

Talvi Watia
02-04-2007, 07:04 AM
Consider the simple quadratic case, a_2x^2+ a_1x+ a_0= 0. Suppose one root is A+ Bi and the other is C+ Di. The equation can be written a_2(x-A-Bi)(x-C-Di)= 0. Go ahead and multiply those out. Remembering that the coefficients must be real so the 0, you get two equations to solve for C and D in terms of A and B. What are they?

this is simple as for n=4 you would have 16 answers. some of them may be the same.

HallsofIvy
02-05-2007, 03:42 PM
this is simple as for n=4 you would have 16 answers. some of them may be the same.
??? An nth degree polynomial equation has at most n solutions. A 4th degree equation might have 4 (complex) solutions, some of which might be the same. Where did you get 16?

In any case, what type speed is asking for is a proof that if all coefficients of a polynomial are real then any complex root must also have its complex conjugate as a root. My point is that both (a+ bi)+ (a-bi)= 2a and (a+ bi)(a- bi)= a^2+ b^2 are real numbers.

Talvi Watia
02-06-2007, 12:24 AM
??? An nth degree polynomial equation has at most n solutions. A 4th degree equation might have 4 (complex) solutions, some of which might be the same. Where did you get 16?

In any case, what type speed is asking for is a proof that if all coefficients of a polynomial are real then any complex root must also have its complex conjugate as a root. My point is that both (a+ bi)+ (a-bi)= 2a and (a+ bi)(a- bi)= a^2+ b^2 are real numbers.

if you solve for 4 variables with a matrix you have this:

A B C D
A aa ab ca da
B ba bb bc db
C ca cb cc cd
D da db dc dd

when you introduce complex numbers, you have to be real careful, as i is non commutative. for example ca*i may or may not equal ac*i.

this is especially important when you look at higher orders of complex numbers, such as quaternions.

i^2=j^2=k^2=ijk=-1

complex numbers can have their own set of rules.

ijk=-1
ijkk=-k
-ij=-k
ij=k
ijj=jk
-i=jk
-ii=ijk
1=ijk

so we run into a problem.

so be careful if you are using 1 = A + Bi + Ci + Di, since it may actually represent 1 = A + Bi + Cj + Dk, especially in forms of spatial coordinates, for example 1 = X + Yi + Zj + Wk.

big problems can arise when you assume i^2 = -1 in all cases.

Talvi Watia
02-06-2007, 12:33 AM
Consider the simple quadratic case, a_2x^2+ a_1x+ a_0= 0. Suppose one root is A+ Bi and the other is C+ Di. The equation can be written a_2(x-A-Bi)(x-C-Di)= 0. Go ahead and multiply those out. Remembering that the coefficients must be real so the 0, you get two equations to solve for C and D in terms of A and B. What are they?

as far as quadratic equations, n=2. so you would have 4 complex answers.

HallsofIvy
02-06-2007, 10:32 AM
You have completely lost me. Multiplication of complex numbers IS commutative.

Multiplication of quaternions is not commutative but quaternions have nothing to do with the problem presented. (And quaternions are NOT "higher orders of complex numbers".)

Epsilon=One
02-06-2007, 01:25 PM
...quaternions have nothing to do with the problem presented. (And quaternions are NOT "higher orders of complex numbers".)This is beginning to clear, in my mind, a series of posts that has left me befuddled. (I might add, a state that is not all that difficult to accomplish.)

Talvi Watia
02-07-2007, 03:29 AM
You have completely lost me. Multiplication of complex numbers IS commutative.

Multiplication of quaternions is not commutative but quaternions have nothing to do with the problem presented. (And quaternions are NOT "higher orders of complex numbers".)

right. but what do you do if say you have:

f(x) = AnX^n+(A-1)nX^(n-1)+...+X+1
f(y) = AnY^n+(A-1)nY^(n-1)+...+Y+1
f(z) = AnZ^n+(A-1)nZ^(n-1)+...+Z+1

f(x) has a root of A+Bi
f(y) has a root of C+Dj
f(z) has a root of E+Fk

Solve all roots for f(x)*f(y)*f(z). (or at least quantatize the number of roots in terms of n.)

This is what I was talking about earlier. If you are dealing with i and only i, you get you use most of the Euclidian tricks, such as using abs(A+Bi)=sqrt(A^2+B^2)

But in doing so you are actually eliminating half of the answer.

since abs(A+Bi) can be A+Bi or -A-Bi.

So naturally, if you have one root is (A+Bi) and another is (A-Bi) it is only natural you may also have (-A-Bi) and (-A+Bi) as roots also. They will provide the same answer.

Talvi Watia
02-07-2007, 05:57 AM
Prove:
given that f(x) = Anx^n + An-1X^n-1+ ....A1X + A0, and An does not = 0
=> if f(x) has a root of the form (A+Bi), then it must have a root of the form
(A-Bi). (complex roots)

All I've figured out so far is to utilize the Conjugate roots theorem but our professor gave us a hint that we should use the remainder theorem or something. Basically I need to prove why both roots exists in a function

Since nobody is getting this, I'll just explain this. I was trying to give a hint earlier as to where this type of problem leads to...

What your professor was referring to is Bairstow's method. Basically what this involves is to use Newton's method to adjust the co-efficients u and v in the quadratic X^2+xu+v until its roots are also the roots of the polynomial being solved.

f(x) = Anx^n + An-1X^n-1+ ....A1X + A0

We notice a series can define this set: Sum { 0 - n } (AnX^n)

thus f(x) = Sum { 0 - n } (AnX^n)

now if we long divide this function by X^2+Xu+v

we get

f1(x) = (x^2+xu+v)(sum {m=0 -> m=n}(BmX^m)+(cx+d) <--- a remainder.

if you solve for each variable we find
Bn = B(n-1) = 0
Bm = A(m+2)-uB(m+1)-vB(m+2) { m = n-2,..,..,0 }
c = A(1) - uB(0) - vB(1)
d = A(0) - vB(0)

the quadratic divides even when c(u,v) = d(u,v) = 0

and following this... matrices... so I wont go further as it seems everyone is lost.

basically what happens is you are given one root. (A+Bi)

you must run that root through the set and show that it will balance with (A-Bi) no matter what.

You cannot just assume a simple quadratic set which happens to have these 2 roots as a balance will hold true for all n.

I can think of equations that have (A+Bi) that do not have (A-Bi) so cannot assume they will always balance.

for example f(x)=(A+Bi)(X+1) you only have 2 roots.

technically 1 real 2 complex, as abs(i), i can be +-i

this is where +-e^[Ci(pi)] is used as a determinate. where C is the number of real roots - complex roots.

HallsofIvy
02-09-2007, 02:37 PM
right. but what do you do if say you have:

f(x) = AnX^n+(A-1)nX^(n-1)+...+X+1
f(y) = AnY^n+(A-1)nY^(n-1)+...+Y+1
f(z) = AnZ^n+(A-1)nZ^(n-1)+...+Z+1

f(x) has a root of A+Bi
f(y) has a root of C+Dj
f(z) has a root of E+Fk

Solve all roots for f(x)*f(y)*f(z). (or at least quantatize the number of roots in terms of n.)
Well, first, functions don't HAVE "roots". Equations have roots. I assume you are talking about the ZEROES of those polynomials or the roots of f(x)= 0, etc. And the zeroes of f(x)*f(y)*f(z) are precisely those of the original f(x).

This is what I was talking about earlier. If you are dealing with i and only i, you get you use most of the Euclidian tricks, such as using abs(A+Bi)=sqrt(A^2+B^2)

But in doing so you are actually eliminating half of the answer.

since abs(A+Bi) can be A+Bi or -A-Bi.

So naturally, if you have one root is (A+Bi) and another is (A-Bi) it is only natural you may also have (-A-Bi) and (-A+Bi) as roots also. They will provide the same answer.
So you are asserting that the roots of x^2- 2x+ 2= 0 are not only 1+ i and 1- i but also -1-i and -1+ i?

I don't THINK so!
(1+i)^2- 2(1+i)+ 2= 1+2i- 1- 2- 2i+ 2= 0
(1-i)^2- 2(1-i)+ 2= 1- 2i- 1-2+ 2i+ 2= 0
but
(-1-i)^2-2(-1-i)+ 2= 1+ 2i- 1+ 2+ 2i+ 2= 4+ 4i, not 0,
and
(-1+i)^2-2(-1+i)+ 2= 1- 2i- 1+ 2- 2i+ 2= 4- 4i, not 0.

Talvi Watia
02-11-2007, 06:23 AM
Well, first, functions don't HAVE "roots". Equations have roots. I assume you are talking about the ZEROES of those polynomials or the roots of f(x)= 0, etc. And the zeroes of f(x)*f(y)*f(z) are precisely those of the original f(x).

So you are asserting that the roots of x^2- 2x+ 2= 0 are not only 1+ i and 1- i but also -1-i and -1+ i?

I don't THINK so!
(1+i)^2- 2(1+i)+ 2= 1+2i- 1- 2- 2i+ 2= 0
(1-i)^2- 2(1-i)+ 2= 1- 2i- 1-2+ 2i+ 2= 0
but
(-1-i)^2-2(-1-i)+ 2= 1+ 2i- 1+ 2+ 2i+ 2= 4+ 4i, not 0,
and
(-1+i)^2-2(-1+i)+ 2= 1- 2i- 1+ 2- 2i+ 2= 4- 4i, not 0.

So you basically found the inverse root of abs is not zero, but balanced at (4 -+ 4i) as the root in complex space. Nice to know.

But mind you
e^i(pi)=1.

So
e^4(1 -+ i)(pi) = e^4( +- 1 ) = +- e^4

I would say
-abs(f(x)) = [1]

at 4 dimensions.

Inverse of 1 = 0.

If that is your argument for, i as of itself, is commutative when ijk may not be ... I will accept that conjecture; if not, I'm sure it is proven somewhere.

Epsilon=One
02-11-2007, 11:49 AM
This thread has been a most interesting and entertaining dialogue between "Halls" and "Talvi."

You two bring much credit and respect to PMF.

e^4(1 -+ i)(pi) = e^4( +- 1 ) = +- e^4

I would say
-abs(f(x)) = [1]

at 4 dimensions.

Inverse of 1 = 0.

If that is your argument for, i as of itself, is commutative when ijk may not be ... I will accept that conjecture; if not, I'm sure it is proven somewhere.This is most interesting; as: 1, itself, = 0 is a most important concept of Pulsoid Theory.