Is there a rule to find the divisor of a cubic polynomial? Wikipedia says that for a cubic polynomial ax^3 + bx^2 + cx + d that one of the products of factorization is a one degree polynomial of the form qx - p and that q must be a divisor of a and p must be a divisor of d. Does this mean that a must always be exactly divisible by q and d must always be exactly divisible by p in order for a divisor to divide a polynomial into a quotient and remainder?
There's a neat theorem that says, a polynomial is reducible over the integers iff it's reducible over the rationals
In other words, if you have a polynomial with integer coefficients, if you can factor it into smaller polynomials with rational coefficients, you can factor it into smaller polynomials with integer coefficients
@AkivaWeinberger thanks for confirming but why does the divisor x - 3 divide the cubic polynomial x^3 - 2x^2 + 0x - 4 into the quotient x^2 + x + 3 and remainder -13 when d in this cubic polynomial is 4 which is not divisible by the divisor's second term p which is 3?
@AkivaWeinberger thanks that clears it up, that's the piece of information I was missing from the Wikipedia article.
I was trying to find p and q for any cubic polynomial based on thinking that for any cubic polynomial ax^3+bx^2+cx+d a must always be exactly divisible by q and d must always be exactly divisible by p. Now I know that only holds when qx - p divides ax^3+bx^2+cx+d without a remainder, how can I find p and q of a divisor (i.e. finding qx - p) when given only the dividend (i.e. cubic polynomial)?
This is equivalent to finding a root p/q of the polynomial
You can do trial and error, but note that there's not always a solution (the polynomial might not have a rational root)
There's also an exact formula for the root, Cardano's formula (the cubic equivalent of the quadratic formula), and numerical methods like Newton's method
Let $K$ be a convex set in a vector space such that $0$ is an internal point of $K$, let $p_K$ be the associated Minkowski functional, and let $z \notin K$. Why must $p_K(z) \ge 1$ hold?
@AkivaWeinberger I planned to do trial and error if p and q satisfied some criteria such as always being a factor of a and d respectively but now I know that only holds when qx - p divides ax^3+bx^2+cx+d without a remainder. Do you know of any condition/criteria for p and q when qx -p divides ax^3+bx^2+cx+d with a remainder, or in other words do you know of any condition/criteria for p and q regardless of whether qx -p divides ax^3+bx^2+cx+d with or without a remainder?
If it were true that $z \in p_K(z) \cdot K$, then if $p_K(z) < 1$ were the case, I could write $z = p_K(z) \cdot \frac{z}{p_K(z)} + (1 - p_K(z)) \cdot 0 \in K$, which would be a contradiction...
However, I don't know if it's true that $z \in p_K(z) \cdot K$.
I am also not completely familiar with the terminology here so please clarify that when you say "there's not always a solution" you mean that there's not always a root when you divide a polynomial by its divisor and that this happens when the polynomial might not have a root that is an integer (i.e. what you mean by a rational root)?
Here is the 'tangent-chord' construction of infinitely many rational points from a single such point.
Suppose we have a point $p_0=(x_0,y_0)$ on the cubic curve $f(x,y)=x^3+y^3=9$. The tangent vector at this point is given by $$\left(\frac{\partial f}{\partial y},-\frac{\partial f}{\partial x}\r...
Is the product of factorizing a reducible polynomial always in the form (qx - p) (ax^2 + bx + c)? that is, is the coefficient of the first term of the lower degree polynomial always positive and the coefficient of the second term always negative?
For some reason I thought for a cubic polynomial ax^3+bx^2+cx+d only one specific divisor can be used to obtain all the roots of the cubic polynomial but since any polynomial less than degree 3 can divide a cubic polynomial into a quotient does that mean that the roots of a cubic equation can be infinite?
Can someone clarify this. May be I should start thinking of a root as simply a solution to a polynomial therefore any first degree or second degree polynomial should have a solution (i.e. should be able to be solved by either factorization or quadratic formula)
So I'm just realizing that a cubic polynomial cannot just be divided by a single polynomial of the form qx -p, any polynomial of the form qx - p can be the divisor
To your first question, that coefficient is only a matter of convention; after all, qx-p=qx+(-p). To your second question, it is not true that any polynomial can divide a cubic (if that's what you mean) and the number of roots of a cubic is at most 3. To your third statement, a root is by definition a solution to the equation "polynomial = 0". Whether polynomials have a solution depends on the field you're working on (x^2+1 has no real solution). The fourth statement is wrong, like the second.
For clarity, a polnomial P(x) divides a polynomial Q(x) if there is another polynomial R(x) such that Q(x)=P(x)R(x).
@Thorgott I want to know: are you an academic student? I’m asking it beacuse your replies to me are very proficient (just like sir @Semiclassical and I know he is not a student) . If you don’t mind much then please just tell me only at what stage of learning are you in ?
@Thorgott Why don't you write out a sketch of what you think the proof is? I can't criticize a vague handwaving proof. But I can say for sure that a proof that merely claims some ring to be a field cannot be correct. I don't know what you mean by "plain induction". It should be clear from the proof I gave that all you need is a well-ordering of the initial field. If it is countable then obviously you don't need any transfinite recursion.
@Thorgott thanks a lot, that helped. However in my second question I meant that a polynomial can reduce (i.e. divide) another polynomial into a quotient and no remainder or a quotient and a remainder. For example for the polynomial x^3 - 2x^2 + 0x - 4 the polynomial x - 3 can divide it into the quotient x^2 + x + 3 and the remainder 13 and also the polynomial x - 5 can divide the same cubic polynomial into the quotient x^2 + 3x + 15 and the remainder 71.
@Thorgott From these two quotients you can obtain four solutions (i.e. 4 roots). I think that your statement "For clarity, a polnomial P(x) divides a polynomial Q(x) if there is another polynomial R(x) such that Q(x)=P(x)R(x)." is true only when polynomial P(x) divides polynomial Q(x) without a remainder.
@MyWrathAcademia What I gave you is the definition of what it means for one polynomial to divide the other. Of course, this means there is no remainder. But saying "a polynomial divides another polynomial with a remainder" is vapid, because that's always true (except for the zero polynomial). I don't see how dividing by a polynomial that leaves a remainder gives you a root.
@Thorgott This is how dividing by a polynomial that leaves a remainder gives you a root: Dividing the polynomial x^3 - 2x^2 + 0x - 4 by the polynomial x - 5 gives the quadratic x^2 + 3x + 15 and the remainder 71. The quadratic x^2 + 3x + 15 via the quadratic formula has the solutions x = 2.653311931459 or x = -5.653311931459, and the root of the polynomial x - 5 is of course 5. So the cubic ploynomial x^3 - 2x^2 + 0x - 4 has these three solutions as its three roots. Feel free to correct me.
@Thorgott My idea of what it means to divide a polynomial and what a remainder is differs from yours, I believe that divide does not mean to exactly divide something leaving no remainder, and I didn't consider 0 to be a remainder so may be once you consider 0 to be a remainder then may be you are right that it is always true that a polynomial divides another polynomial with a remainder.
btw I allowed SF to play for a while on the position before that dubious queen move, and in 4 moves it moved the king to b8, rook to g8, and queen to e8
I’ll note that, if one number isn’t divisible by another, then long division can either be used to find the remainder or to find the decimal expansion for the ratio
@Leaky: I think you need a separate chess chatroom.
@StupidQuestionsInc Oh, my apologies. I was too obsessed with breaking it into two pieces. Interestingly, following my suggested idea leads to an example where one of the pieces need not converge, but presumably the other piece (which would be a mess) cancels out the bad stuff. Think about what they're doing with a picture (like setting up a double integral).
One of the sums is getting the $(k,\ell)$ with $k\le \ell$ and the other is getting the terms with $k\ge\ell$. (Change the order of summation on one.) I guess the diagonal is appearing twice. Not sure if I'm missing a small detail or not.
@TedShifrin @Thorgott thanks a lot, I think I'm understanding how to know whether a divisor of a cubic polynomial can get distinct roots for that ploynomial. So a factors result must always be a root. Does that mean that only a polynomial qx - p that is a factor (x is a root) can be used to find the other roots of the cubic polynomial of the form ax^3 + bx^2 + cx + d where these other roots are the solutions of the quadratic formula?
@Thorgott since you confirmed that one of the factors of the cubic polynomial ax^3 + bx^2 +cx + d is always in the form qx - p (or in other words, 1x + (-p)) I think I'm finally understanding how to find which values for p and q in px - q to use to divide the polynomial ax^3 + bx^2 + cx + d in order to find three roots.
@StupidQuestionsInc: I guess I didn't pay enough attention to the indices in your original question. It looks like they did make sure not to include the diagonal in the second sum.
@Lukas: I can do that, but I honestly don't know what the European standards are. If you're applying to graduate school, I think you're worrying way too much about it.
@Lukas: In this country they care about your courses, your grades, and most of all letters of recommendation. All the seminar talks and conferences you've attended should be a bonus. You should certainly mention those in whatever little essay you write.
If $qx-p$ is a factor, the corresponding root is $p/q$. Dividing the cubic by this linear factor gives a quadratic whose roots can be calculated using the quadratic formula, yes. However, it isn't at all easy to find such a linear factor. The conditions that q divides d and p divides a discussed earlier are necessary, but not sufficient.
@Lukas: As I said, in this country I don't think we even ask for CVs. You can list all relevant academic experiences somewhere, I guess, but that essay gives you an opportunity to impress people.
@Balarka: Learn for math or learn for actual grammar and literature?
Yes, French literature is lovely (à mon avis). For that you probably should just take a few courses. For reading math, get a French math dictionary and just start reading. If you read math you sort of know a bit, it'll be pretty easy.
I can send you my little paper in French (if I didn't already).
Well, you won't find passé simple in mathematics, but you will find it in older literature for sure.
Talk about some specific mathematics you've studied or worked on and why it is so interesting to you. I mean specifics. They hear general bullshit from everyone.
So when you say "student seminar" you should make it clear that it's a "graduate student seminar."
hi there. Trivial question. What is the notation used to express the inverse of a multivariate function? Do you specify the variable with which the inverse is being computed? Say y = f(x,w). x=f^{-1}(y,w) is one inverse, and the other is w=f^{-1}(y,x). The f is the same. Does it matter, or is it evident by the variables in parenthesis which is the one being inverted?
In my particular case, I am evaluating the inverse. So, actually I have f^{-1}(3,4). I thought if I use f_x^{-1}(3,4) it would be clear, but subscript usually means derivative.
Here's what I can offer (if I understand correctly). You have a function $f\colon A\times B\rightarrow C$; then you can define functions $f_a\colon B\rightarrow C, b\mapsto f(a,b)$ for $a\in A$ and $f^b\colon A\rightarrow C,a\mapsto f(a,b)$ for $b\in B$ and in case these are bijective, you can properly talk about their inverses.
let me rephrase. Assume $f\colon A\times B\rightarrow C$. How would you, notationwise, differentiate between $f^{-1}\colon C\times B\rightarrow A$ and $f^{-1}\colon A\times C\rightarrow B$?
perhaps using $f_C^{-1}$ and $f_B^{-1}$ respectively?
These notations are not standard, they are not even permissible. $f^{-1}$ already has a specific meaning different from what you are talking about. I'm afraid I don't know a standard notation for what you're looking for.
Consider the function
$$ y = f(x,w) $$
I am looking for "proper" notation for the "partial inverse" of $f(\cdot)$ with respect to each variable. Something like:
$$ x = f^{-1}(y,w) $$
$$ w = f^{-1}(x,y) $$
But this does not differentiate them. Perhaps
$$ x = f_x^{-1}(y,w) $$
$$ w = f_w^{-1...
A question, based on one I just saw on main: Suppose $f(x)/x\to 1$ and $g(x)/x\to 1$ as $x\to 0$. Then it is certainly true that $f(x)/g(x)\to 1$ as $x\to 0$. Must $f^{n}(x)/g^{n}(x)\to 1$ as $x\to 0$?
Oh wait, the example I linked is for $x\rightarrow\infty$, but it should work with some modification.
And yes, you do get $f^{\prime}(0)=1$, so any counterexample would need some discontinuity in the derivative (in particular, if you're assuming $C^1$, then the statement should ohld).
So, by L'Hospital, we get the statement is true for higher $n$ if $f$ is sufficiently differentiable and its higher derivatives zeros don't accumulate in $0$, though the second condition seems rather awkward.
I was trying to throw something together with x^2sin(1/x), but I didn't get it to work. Now I'm trying to think of other examples of functions with discontinuous derivatives, but I actually don't know many.
It suffices to find a differentiable function $f$ with $f(x)/x\rightarrow1$ as $x\rightarrow0$ and $f^{\prime}$ discontinuous at $0$. Then we can take $g(x)=x$ and it will be a counterexample.
@Thorgott I now understand exactly that a root of a polynomial is simply the zero of the corresponding polynomial function. Apologies for my previous questions, they were coming from a lack of understanding of polynomials and roots of polynomials. Your example of plugging 5 into that polynomial and the Wikipedia article got me to perfectly understand what a root of a polynomial is. Thanks a lot for your clear and concise explanations.
Consider the function
$$ y = f(x,w) $$
I am looking for "proper" notation for the "partial inverse" of $f(\cdot)$ with respect to each variable. Something like:
$$ x = f^{-1}(y,w) $$
$$ w = f^{-1}(x,y) $$
But this does not differentiate them. Perhaps
$$ x = f_x^{-1}(y,w) $$
$$ w = f_w^{-1...
