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Bilinear forms (nondegenerate) on a vector space over a finite field arise from nonsingular matrices; the important ones are the "reflexive" ones for which $B(x,y) = B(y,x)$ for all $x$ and $y$. These come from nonsingular symmetric or alternating matrices (and give rise to symmetric or symplectic bilinear forms, respectively). The symmetric forms can also ...

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In characteristic $0$, an irreducible polynomial has distinct roots (in a suitable extension field). Indeed, $\gcd(f,f')$ is a divisor of $f$, so it is either $1$ or $f$ (up to multiplication by nonzero constants). Suppose it is $f$; then $f'=0$, which means $f$ has degree $1$: impossible. Now, let the characteristic be $p$. If $f$ is inseparable and ...

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You can learn this by expanding the factored form of the cubic. If we suppose $a$, $b$ and $c$ are the roots, then: $$(x-a)(x-b)(x-c)=x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc$$ It's a good thought to do this out by hand if you've not seen it before. Notice that when we multiply monomials like $(x-a)$, we just "take" one term from each monomial and multiply them ...

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I can get rid of the explicit appeal to probability by showing combinatorially that $f\upharpoonright\big(\Bbb Q\cap(0,1]\big)$ is monotone non-decreasing and then extending the result to $f\upharpoonright[0,1]$ by continuity. Note, though, that at bottom it’s really the same basic argument. Let $c$ be an integer greater than $1$, and let $g\le c$ be a ...

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Cross-posted and answered on Mathoverflow. The formula is $(2n+1)(2n+3) n!^4 / (2n)!$. See my answer to MO_210823 for the proof.

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The article you refer to goes too fast. It should not be able to reach the conclusion $p(0)=\frac{1}{6}(p_{i-2}+4p_{i-1}+p_i)$ simply from positional continuity and symmetry. This article http://www2.cs.uregina.ca/~anima/408/Notes/Interpolation/UniformBSpline.htm provides a better explanation about how to derive the basis functions for uniform cubic ...

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The $40321$ term does play a role in determining the first column. To properly apply the Routh Hurwitz criterion, you need to compute all $8+1 = 9$ rows of the Routh array, not just the first $4$ rows. Here is the completed array (computed in Excel): $\begin{bmatrix} 1& 546 & 22449 & 118124 & 40321\\ ... 0 Here's Wolfy's plot of the roots: So the roots are pretty close to 1, 2, ..., 8. 3 Let$P=(a,b)$a point and$y=f(x)$a derivable function$\mathbb{R} \rightarrow \mathbb{R}$. Obviously if$b=f(a)$we have a tangent from$P$. Otherwise a tangent to the graph of the function form$P$exists if there is a point$X=(x,f(x))$such that. $$\dfrac{f(x)-b}{x-a}=f'(x)$$ So there exists a tangent to$f(x)$from any point$P$if this equation ... 5 Answer: For all odd degree polynomials of degree$\ge 3$. If$p$is of degree$1$, then the graph of$p$is itself the tangent line to all its points. Hence for any other point there is no tangent passing through. Let$p(x)$be a polynomial of odd degree$\ge 3$. By switching to$-p(x)$if necessary, we may assume that the leading term is positive. We ... 2 Hint: for any$p\in P$, the degree of$f(p)$is even. 0 From that link, you see that the terms$c^{k,m}_s$are defined as the coefficients of a polynomial: $$\sum^{k(m-1)}_{r = 0}z^rc^{k, m}_{r - k(m - 1)/2}=(1 + z + ... + z^{m-1})^k$$ which means that to compute them you need to expand the power$(1 + z + \dotsb + z^{m-1})^k$for appropriate values of$k$and$m$. For a simple example, consider$k = 3$,$m = ...

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It appears that the following conditions are sufficient: $P_1$, $P_2$ are of the same degree $d$ (even) strictly positive at all points and, moreover, the homogenous components of top degree $d$ are also $>0$ at all points $\ne 0$ (the fact that the top components are $\ge 0$ follows from the fact that the polynomials are $\ge 0$; one wants in fact ...

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No. Let's show that $x = -1 + i$ can be a root of such a polynomial for a certain choice of $d_j$'s and $k$. The problem amounts to showing that there are some $2n$ numbers of the form $-1 - d_j + i$ whose product is a negative real number $-k^2$. In the complex plane, this amounts to picking $2n$ numbers on the line $y = 1$ with arguments strictly between ...

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The answer is zero. Define 6 variables $$\begin{cases} A_1 &= x_1 + x_2 + x_3\\ A_2 &= y_1 + y_2 + y_3\\ A_3 &= z_1 + z_2 + z_3 \end{cases} \quad\text{ and }\quad \begin{cases} B_1 &= x_1 + y_1 + z_1\\ B_2 &= x_2 + y_2 + z_2\\ B_3 &= x_3 + y_3 + z_3 \end{cases}$$ The product at hand can be rewritten as $$f = ... 0 If my computations are correct, the coefficient is 0. 0 Using the same approach as user3002473 and continuing the simplifications, the calculations simplify to$$R=\frac{z}{3} (2 x +1)G=\frac{z}{3} \left(\sqrt{3} y-x+1\right)B=-\frac{z}{3} \left(x+\sqrt{3} y-1\right)$$0 There is a simple and fast way to locate the root at u = 6 using the Rational Roots theorem. In bburGsamohT's answer the root candidates are given. If we try the first positive one, we see that it is not a root because the value of the polynomial is 55 which is not equal to 0. Computing this value looks like a waste of effort, but it turns out that we can ... 0 Lucky that there is a general solution for every cubic: http://www.sosmath.com/algebra/factor/fac11/fac11.html Though there is no formulas for EDIT quintics or higher. 5 By the rational root theorem, if u^3-54u+108 has an integer root r, then r must be an integer factor of 108. So in the worst case, you could try the possibilities 1,2,3,4,6,9,12,18,27,36,54,108 as well as their negatives and then use polynomial long division when and if you find a root. 0 It means that over \mathbf C, any non-constant polynomial splits as a product of linear factors, and this factorisation is unique (up to order), hence the roots of f are the roots of the linear factors. They observe that some factors may be repeated, which corresponds to multiple roots. The number of factors equal to x-z_i is the order of ... 0 We have$$ (R + G + B)x = R-\frac12(G + B)\quad \implies\quad (x - 1)R + \left(x + \frac12\right)G + \left(x + \frac12\right)B = 0 \\ (R + G + B)y = \frac{\sqrt{3}}{2}(G - B)\quad \implies\quad yR + \left(y - \frac{\sqrt{3}}{2}\right)G + \left(y + \frac{\sqrt{3}}{2}\right)B = 0 \\ R + G + B = z $$which is a set of linear equations. We can solve this by ... 1 We can simplify GFR's approach and go further: let X be the matrix$$ \left( \begin{matrix} 0&0&0&0\\1&0&0&0\\0&1&0&0\\0&0&1&0 \end{matrix} \right) $$Then$$M = I + X + X^2 + X^3 = \sum_{k=0}^{\infty} X^k = (I - X)^{-1}$$Consequently,$$M^n = (I-X)^{-n} = \sum_{k=0}^{\infty} \binom{-n}{k} (-X)^k = ...

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We have $$(\lambda I + N)^k = \sum_{j=0}^k \binom kj \lambda^{k-j}N^j$$ Now, taking $f(t) = \sum_{k=0}^m a_k t^k$, we have $$\sum_{k=0}^m a_k(\lambda I + N)^k = \sum_{k=0}^m a_k\sum_{j=0}^k \binom kj \lambda^{k-j}N^j =\\ \sum_{j=0}^k \left(\sum_{k=0}^m a_k \binom kj \lambda^{k-j}\right) N^j =\\ \sum_{j=0}^k \left(\frac 1{j!}f^{(j)}(\lambda)\right) N^j ... 0 Call them f(x) and g(x). You can do a polynomial version of Euclid's algorithm. Note that f(x)-g(x) is a cubic polynomial. So x*(f(x)-g(x)) is another quartic. Then 4f(x)-x(f(x)-g(x)) is another cubic. The difference between two cubics with equal leading term is a quadratic. That should be enough to find the answer. 3 HINT: Let \alpha be a common root of both the equations. Hence, it would satisfy both of them. Subtract both the equations, you get$$4\alpha^3-2\alpha^2-66\alpha+108=0$$It gives the roots \frac{-9}{2},2,3. Check that \frac{-9}{2} will not be a common root. I hope you can take it from here. The question reduces to a quadratic polynomial. 0 First of all , here is the meaning of the notation (a mathematician (even a student) never uses a notation which does not understand) :$$ f \equiv h \mod \{n,g\} \iff f(x)-h(x)\equiv q(x). g(x) \mod n$$This notation can be seen also as the definition of congruence in the ring F_n[X]. So we will her eliminate n and say only that : Let f,g,h_1\in ... 8 Maybe not exactly what you are asking, but notice that your matrix is lower triangular, and can be written as M=I+N, with I the identity and $$N=\begin{pmatrix} 0&0&0&0\\ 1&0&0&0\\1&1&0&0\\1&1&1&0 \end{pmatrix}.$$ N is nihilpotent, N^4=0 and (obviously) commutes with I, hence ... 4 These are diagonals of Pascal's Triangle. 3 Consider P(x)=x-1, which has the root x=1. Now consider Q(x)=x+3 (k=2), which has the root x=-3. 1 No, that is not necessarily true. Consider the polynomial$$P(x)=(x-1)(x-2)(x-3)$$If we select k=4, then the we get$$Q(x)=(x-1)(x-2)(x-3)+16$$which has a negative root and the others are now non-real. 0 The name comes from the notion of discriminant of a quadratic form of two variables and Taylor's formula at order 2: \begin{multline}f(x_0+h,y_0+k=f(x_0,y_0)+f'_x(x_0,y_0)h+f'_y(x_0,y_0)k\\+\frac12\bigl(f''_{xx}(x_0,y_0)h^2+2f''_{xy}(x_0,y_0)hk+f''_{yy}(x_0,y_0)k^2\bigr)+o\bigl(\lVert(h,k)\rVert^2\bigr)\end{multline} This discriminant is the reduced ... 3 Discriminants have their roots in abstract algebra and are used to determine the nature of zeroes of polynomials. From Wolfram Mathworld, if p(z) is an nth degree polynomial with leading coefficient a_n and roots r_i, the discriminant \Delta is$$\Delta(p) = a_n^{2n-2} \prod_{i > j} (r_i - r_j)^2.$$Discriminants are defined up to a constant ... 0 It's not true for cubics in general. Try$$ F(X) = X^3 + c_2 X^2 + c_1 X + c_0$$The roots of F(X) - z are f_1, f_2, f_3 with$$ \eqalign{f_1 + f_2 + f_3 &= -c_2\cr f_1 f_2 + f_1 f_3 + f_2 f_3 &= c_1\cr f_1 f_2 f_3 &= -c_0 - z\cr} $$Note that the first two equations don't involve z. We can eliminate f_3 from ... 1 If you make your conditions a bit different, it gets mathematically a bit more uniform. Instead of requiring M^n having x,y components in it, you can instead require that if \vec{r} is on the curve, then M\vec{r} is also on the curve (and therefore M^n\vec{r} too). A further simplification is to consider infinitesimal moves. For a circle, that's ... 3 This is a nontrivial task, and the best approach may depend on how the polynomial is specified (i.e., what kind of structure it has). In principle, the Routh–Hurwitz theorem can answer the question, as it gives the number of roots in the positive half plane \{z:\operatorname{Re}z>0\}. Your problem amounts to determining whether this number is the same ... 2 We can actually invent infinitely many! From what i gather you are interested in orthogonal polynomials, meaning polynomials P_i such that there exists a,b such that$$ \int_a^bP_i(x)P_j(x) dx = 0 \forall i \ne j$$To generate such polynomial sequences we can start with some initial Polynomial P_0 it can be any polynomial of your choice (An ... 5 HINT :$$x^4-(a+b)x^3+(ab+2)x^2-(a+b)x+1=(x^2-ax+1)(x^2-bx+1)$$4 Hint: You have a palindromic polynomial, hence P(x)=0 is equivalent to:$$ \left(x^2+\frac{1}{x^2}\right)-(a+b)\left(x+\frac{1}{x}\right)+(ab+2) = 0\tag{1} $$or, by setting z=x+\frac{1}{x},$$ z^2-(a+b)z+ab = (z-a)(z-b) = 0.\tag{2} $$Consider now that the range of f(x)=x+\frac{1}{x} is \mathbb{R}\setminus(-2,2). 0 I will restrict to commutative rings. If R has dimension d, then R[x] has dimension d+1 \leq \dim(R[x]) \leq 2d+1. So one property is "being of finite dimension". Further properties listed here are "integral domain", "noetherian ring", "normal ring", "UFD". Other properties are "reduced", "connected", "smooth" (over some base ring), "finitely ... 0 Here is one specialized solution. Let h=g. The the original equation becomes:$$(3n-3)(f^2+2g^2)=(9n+1)(2fg+g^2)\tag{1}$$We solve (1) for f and obtain (assuming g>0):$$f=A(n)g\tag{2}A(n)=\frac{9n+1\pm\sqrt{10(9n^2+3n-2)}}{3(n-1)}\tag{3}$$(a) For n=2, We have A(2)=-1/3,13. If g=3 j where j is arbitrary positive polynomial with ... 0 I am really not sure of this, but here's what I could do! Let \tau be in the Galois group. \tau is uniquely determined by what it does to \zeta. And \tau_{k}(\zeta)=\zeta^{k}. Now if m_{1} and m_{2} lie in the same coset of S, then m_{1}m_{2}^{-1} \in S, So m_{1}m_{2}^{-1}=t for some t in S. But we know that we can find an elememt ... 0 The expression factors as$$z^3(1+z)^3(1+z^2)^3=z^3(1+3z+3z^2+z^3)(1+3z^2+3z^4+z^6).$$Then I see no better way than to perform the multiply (though there is a symmetry)$$\begin{align} &1+3z+&3z^2+&z^3\\ &&3z^2+&9z^3+&9z^4+&3z^5\\ &&&&3z^4+&9z^5+&9z^6+&3z^7\\ ...

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You can have x = z1 and y = z 2 + z 3 + z 4 Then the problem is (x+y) 3. Later resubstitute y in the expansion and choose a = z 2 and b = z 3 + z 4. And expand those powers by pascal's triangle and finally resubstitute for a and b and expand those powers by pascal's triangle.

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$$(z+z^2+z^3+z^4)^3 = z^3\cdot\left(\frac{1-z^4}{1-z}\right)^3=z^3(1-3z^4+3z^8-z^{12})\sum_{n\geq 0}\binom{n+2}{2}z^n$$ hence: $$\begin{eqnarray*}(z+z^2+z^3+z^4)^3&=&(z^3-3z^7+3z^{11}-z^{15})\sum_{n\geq 0}\binom{n+2}{2}z^n\\&=&\sum_{n\geq 3}\binom{n-1}{2}z^n-3\sum_{n\geq 7}\binom{n-5}{2}+3\sum_{n\geq 11}\binom{n-9}{2}z^n-\sum_{n\geq ... 1 An example of such matrix is the companion matrix of the polynomial, which in this case is$$\begin{pmatrix} 0 & 0 & 0 & -D\\ 1 & 0 & 0 & -C\\ 0 & 1 & 0 & -B\\ 0 & 0 & 1 & -A \end{pmatrix}.$$1 Perhaps these steps make it clearer - as you specified giving only key steps, let me know if something needs more detail. Let f(x) = \sum_{k=0}^n a_kx^k. Immediately, f(0)=1 \implies a_0 = 1. Next, from the functional equation, consider the coefficient of the highest power term on either side. Show a_n=1. From Vieta, now we have that the ... 1 You can use Descartes' Rule of Signs which states that An equation f(x) = 0 cannot have more positive roots than there are changes of sign in f (x), and cannot have more negative roots than there are changes of sign in f (— x). Since 15a+6b+4c+8d=0 this implies that there is atleast one negative number and one positive number in the set {a,b,c,d}. ... 0 I think that I finally understood. You're requested to work only locally around the origin. Not to make an approximation on the full "ball". If that is the case, the Taylor expansion of g at the origin (0,0) up to order n will answer the question. For this you can have a look at Taylor series. You'll need to compute$$\frac{\partial^{h+k} g}{\partial ...

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$x^3−8 = (x-2)(x^2+4+2x)$ LHS=$x^3-8-7x+14$ $=(x-2)(x^2+2x+4-7)$ $(x-2)(x-1)(x+3)$

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