Generalizing a test from high-school.
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The converse is obviously not true because that rule only involves "a", "b", and "c" and ignores the rest coefficients. For example, (x-1)^3=0 has no complex roots but (x-1)^3-1=0 does. The two equations have the same a, b, c.
19:46 Good Place To Start the Homework
The result is a special case of the famous Newton's inequalities. In the XIX century Sylvester proved that the set of real polynomials having all of its roots real is semi-algebraic. More precisely, the set of real polynomials P of degree n that have at least one nonreal root can be characterized with at most (n-1) polynomial inequalities in the coefficients of P and n.
Maybe applying Decartes's rule of sign to p(x) and p(-x) could be a good alternative to include more cases.
I don't think you can create an iff statement if you're going to focus on just the leading 3 coefficients. In general, an nth degree polynomial's overall shape is dictated by the leading coefficient. The other terms only allow for some wiggling in some finite interval. So you can easily pick an enormous constant so that any wiggling occurs away from the x-axis, which will result in imaginary roots.
For example, take f = x(x+L)^2 where L denotes some large positive value.
Foiling gives us f = x^3 + 2Lx^2 + (L^2)x. So our condition (n-1)b^2 - 2nac = (3-1)(2L)^2 - 2(3)(1)(L^2) = 2L^2 > 0.
Assuming that an iff held, then this would imply that all the roots are real. It just so happens that the roots are real in this case (0, -L, -L).
f' = (x+L)^2 + 2x(x+L) = (x+L)(3x+L). Thus, f is decreasing when x belongs to (-L,-L/3) and increasing when x belongs to (-L/3,inf) i.e. minimum achieved when x = -L/3.
So I could create a new function g(x) = f(x) + d where d is a constant term larger than f(-L/3). This would still return a condition of 2L^2, but now 2 of the roots would be imaginary.
much easier counterexample: x^1000000 only has real roots, x^1000000+1 only has complex ones
For cubic the condition is here delta=(bc)^2+18abcd-4ac^3-27(ad)^2-4db^3
Delta>0 - real roots
Delta=0 - equal roots
Delta
The Wikipedia article for "Discriminant" has a nice write-up about how the idea of the discriminant can be generalized from quadratics to any degree. The discriminant is a polynomial in all the coefficients, so there are no shortcuts (like just the first three coefficients). It is zero if there are repeated roots, positive if all the roots are distinct and real, and negative otherwise.
This is not true. If all roots are distinct, then the discriminant is negative if and only if there is an odd number of pairs (z,\bar z) of complex roots (with conjugate).
Spot the error in the thumbnail.
Ah yes. Unless it's a polynomial of degree n+2 just written in a weird order.
Maybe it's an intentional clickbait
n-2
N+2
Not really an error.
I believe y=x^3 shows that the converse is not true. With n = 3, a = 1, and b = c = 0, we have (n - 1)b^2 - 2nac = (2)(0)^2 + 2(3)(1)(0) = 0, but not all solutions are real.
Edit: This is not an example. I was mixing up zeros and roots of unity. But y=x^3 - 1 is an example. That's what I get for doing math when I've only just woken up 😵💫
Bro, y=x^3 has a triple zero at zero which is real. I think you wanted to say y=x^3-1 is the example of the converse not being true😅
@@janeknowakowski5732 oh yeah, I'll blame it on having just woken up
@@allanjmcphersonThe inequality in the video is a strict inequality (strictly less than 0) but in your example it's equal to 0. x^3-x-1 is an example where the sum of the squared differences of the roots is equal to 3 but only 1 real root (discriminant equals -23).
@@afuyeas9914 Where in the world do you obtain discriminants equal to -23 (is 6) and 256 (for another example you gave in another comment, and it was 0)?
@@afuyeas9914 yes, but to show that statement is not an "only if" statement, it suffices to find an example where this generalized discriminant is non-negative but there are non-real roots. The real problem with my example, as has been pointed out, is that all the roots of y=x^3 are real.
To answer the final question: unfortunately, starting at n=4 there isn't a rule that is an iff statement regarding the nature of the roots. A discriminant can be positive and there aren't any real roots (x^4+1, discriminant equals 256)
The discriminant of x^4+1 equals to 256? What? It is equal to 0: b=c=0.
@@r.maelstrom4810please, google what is the discriminant of a polynomial
@@r.maelstrom4810 What Michael calculated was not the discriminant. The discriminant involves the product of the (xi-xj)^2 instead of their sum like in the video. This product will involve more than just a,b,c when the degree is more than 2. For a degree 4 polynomial with the discriminant written in terms of the coefficients a,b,c,d,e there will be 16 terms, but for the case of x^4+1 where b=c=d=0, the only nonzero term is 256a^3e^3 = 256.
It is always a pleasure to learn from Michael
Maybe newton inequalities for symmetric polynomial do if and only if . Also for homework x^n=1 for n>2
Really intriguing question that I for decades have never seen or thought of.
Seems at first glance, it would be equivalent to solving the quintic, for $n=5$.
However, the problem is merely finding a number $D$ such that $D=f(a_{i})$ (where $a_{i}$ are the polynomial coefficients) can determine if $\sum_{i=0}^{n}\left(a_{i}x^{i}
ight)$ has a (pair of) complex root(s).
For $n=2$, $D=b^{2}-4ac$.
perhaps if we consider all posible sums of the terms (xi-xj)^2 it'll be an iff. So i mean we can consider the sum of (xi-xj)^2, the sum of the double products of (xi-xj)^2 * (xl - xk)^2, ... untill we get to the whole product of all the terms (xi-xj)^2. Maybe if we suppose all of those are >=0, it follows that the numbers will all have to be real
The converse is obviously not true because the equation only captures those polynomials where the total magnitude of the imaginary parts of the complex roots exceeds the total magnitude of the real parts of those roots plus the magnitudes of all the real roots. So e.g. if i and -i are roots, any combination of real roots whose absolute values exceed 2 are not going to be detected by this test.
I am not sure about odd degree polynomials, but even degree polynomials don't have a "iff" version of the rule. For every even polynomial with only real roots, we can change the coefficient of x^0 term so that the new polynomial doesn't intersect the x-axis ie. it has an complex root.
This works olso for every non 1 odd degree polynomial by translating it to have only 1 real solution.
But this doesn't only prove that if such a rule exists, it must consider the coefficient of x^0
2 is an even number, isn't it?
I was thinking rules uses the first three coefficients and only one inequality. Should have been more specific.
Also, yeah the argument doesn't apply to degree 2.
If you want polynomial that it is guaranteed to have all real roots
then take square matrix with random but real entries then multiply it by its transposition
and calculate its characteristic polynomial
Would Sturm sequence be better ?
One does not simply discover your channel without one of your comments on a maths video.
5:55 is comparison of sum with 0 valid (i.e. the sum is real) because the expression is possible to write in terms of symmetric polynomials of 1st and 2nd order which have real values?
when I get lost in the forest of n's, j's, k's etc. (which I often do) ; I simply bask in the brilliance of Dr. Penn...😄
Can you make a video proof of rational root theorem
I found a theorem of Sturm which seems to partially characterize the existence of non-real roots for a polynomial, as follows:
Let p(x) be a polynomial with no repeated roots, and define a seuqence of polynomials {q_n} by q_0=p, q_1=p', and q_{k+2} is given by the remainder of q_k divided by q_{k+1}, which can always be done because the degree of q_k is strictly decreasing. In particular, {q_n} has at most as many nonzero terms as the degree of p. The sequence {q_n} is known as the Sturm sequence of p.
For a real number s, let V(s) be the number of sign changes in the sequence {q_n(s)}, that is, the number of indices k such that q_k(s)*q_{k+1}(s)
There is something wrong with your explanation. "Let p(x) be a polynomial with no repeated root"??? This supposition can NOT be part of a theorem for know something about roots of a polynomial. And I went to check, the Sturm Sequence doesn't demand the polynomial under analysis to have only simple roots.
About "deg(q_0)>deg(q_1) because p has no repeated roots" is NOT right. The inequality is ALWAYS true, simply because derivative decreases degree in exactly one unity, except when p is constant. So, if p is NOT constant, or deg(p)≥1,
deg(p') = deg(p)-1
Your last paragraph doesn't seem right.
"If p(x) has repeated roots ..."
Ok ...
"... the number of repeated roots of p with multiplicity k ..."
This already sound wrong, because you are not talking about what you should be talking about.
"... is the degree of gcd of p and the (k-1)-th order derivative of p"
That's simply not right. Look, every polynomial P can be written as a product
P(x) = a(x-x1)...(x-xn)
If P has a repeated root with multiplicity k≥2, this means
P(x) = (x-x0)^kQ(x)
If that's the case, look what happens with its derivative
P'(x) = k(x-x0)^{k-1}Q(x) + (x-x0)^kQ'(x)
= (x-x0)^{k-1}(k+(x-x0)Q'(x)
This means x0 is also a root for P'(x). Reciprocally, if
gcd(P(x),P'(x)) ≠ 1
then P has a repeated root. The multiplicity k will be the number of derivatives plus 1 such that x-x0 divides the derivative.
@@samueldeandrade8535 Thank you for the feedback and correction. About the repeated root, there's a theorem that states that if the discriminant is 0 if and only if the polynomial has repeated roots. Alternatively, given a polynomial p(x), we can find its derivative p'(x) and thus gcd(p,p'), and p has repeated roots if and only if gcd(p,p') is not 1. So we can determine the number of repeated roots for a polynomial from just its coefficients, counted with multiplicity.
@@19divide53 man, this is very interesting. That's the first time I am seeing such a thing. Polynomial Theory is just so rich it is ins4ne. Unfortunately, many teachers can't make kids realise that.
I was super confused by the thumbnail. I thought the question was about power series where the first n terms are 0.
I have seen a similar problem in Indian statistical institute entrance exam for bmath students.
which year?
It was a part of the 2016 entrance if I remember correctly
hello professor
Today, Michael has his Typo already in the thumbnail...
Irrelevant typo though.
@@samueldeandrade8535 That's the fun part: All his builtin mistakes are usually irrelevant (e.g. wrong formula, but not being used for further calculation; or mistake in the last formula)...
@@rainerzufall42 oh are the typos so frequent??? I heard people saying something about it, but never really realised there is already a collection of typos. Maybe someone could make a channel about it. I mean, we have the channel A Good Place To Stop ... Where is Penn's Typos channel? Or should it Michel Pencil? Michal Pen? Pichael Menn?
typo*
@@yongdetao7005 are you correcting the capital T?
No, I was waiting to see if you either spotted it yourself or just wrote it correctly in the next line... 😉
However, you didn't spot that you wrote X sub-j instead of X sub-k at 13:23...
At first glance, my answer is : the number of complex zeros is the number of zeros of the antiderivative - the degree of the original polynomial.
Nevermind I was thinking if it wrong.
x^3-1 yields 0 but has non-real roots.
Excelente!!!
Me encantó el resultado. No lo conocía
x^3+x^2+1=0
Homework? 😡
An alternative way to arrive at the same criterion as in the video is to note that by mean value theorem, if a polynomial f completely factorizes over the reals, then so does its derivative f'. Iterating, if f has degree n, then if f factorizes over the reals, so does its (n-2)nd derivative f^{(n-2)}. Applying the discriminant criterion to f^{(n-2)} then yields the conclusion.
A slightly stronger criterion can be obtained by noting that the equivalence (discriminant
I'm concerned about the entire proof method. How can we compare sum (x_i - x_j)^2 to 0 if we don't know that the x_i are all real to begin with? Perhaps the proof can be saved by looking at the absolute value squared instead?
the sum is indeed real. just look at its closed form in terms of a,b,c and the degree of the polynomial, which are all real
You are correct that his proof is incomplete because the lemma should also include the statement that the sum is always a real number. The proof was then actually completed when he found the closed form in terms of a,b,c, and n. So regardless of whether any of the x_j are real, the sum is in fact always real and if it is negative then at least one (actually two because of conjugate pairs) must be complex.