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1

Yes. More generally: Let $A = \bigoplus_{i\in\mathbb{Z}} A_i$ be a graded domain und $f\in A\setminus \{0\}$ homogeneous. If $f$ factors in $A$ as $f=gh$ then $g,h$ are homogeneous. Proof: Since we are in a domain and $f\neq 0$, the factors $g,h$ are non-zero as well. Let $g$ have non-zero component of lowest degree $d_{min}$ and of highest degree ...


2

Hint: This is based on @columbus8myhw's hint. Consider the polynomial $$kf(k)-1$$ which has degree 99 is zero for all $k=1,2,3, \cdots, 99$. So you know the polynomial $kf(k)-1$ has the form $$C(k-1)(k-2)(k-3)\cdots(k-99)$$ and you can find $C$ when you put $k=0$. In general, its a good idea to find a polynomial that has as many zeros as its degree ...


0

Take your second diagram and replace $10$ cm and $2$ cm along the top with $5x$ and $2$, respectively, and similarly replace $10$ cm and $4$ cm on the side with $3x$ and $-1$. Now, this is a little strange since $-1$ is negative, but the principle remains the same. The total area of the rectangle is $(5x+2)(3x-1)$, which you can see as the area of four ...


0

Binomial multiplication is really anything that can be written in the form $(a+b)\times(c+d)$. We already know that the area of the rectangle is given by $12\times14$, so you just need to split $12$ into the right $a$ and $b$ and split $14$ into the right $c$ and $d$.


3

Yes. Let $f\in\mathbb R[X_1,\ldots,X_n]$ be a polynomial that vanishes on an open subset of $\mathbb R^n$, wlog. an open neighbourhood of $0$. Then for any $(a_1,\ldots,a_n)\in\mathbb R^n$ the polynomial $g(T)=f(a_1T,\ldots, a_nT)\in\mathbb R][T]$ vanishes in a neighbourhood of $0$, hence in infinitely many points, hence is the zero polynomial. Now assume ...


2

The polynomial $x^{16}-x\in\mathbb{F}_2[x]$ has roots precisely equal to the elements of $\mathbb{F}_{16}$, and the subfields of $\mathbb{F}_{16}$ are $\mathbb{F}_{16}$, $\mathbb{F}_{4}$, and $\mathbb{F}_{2}$, which have degrees $$[\mathbb{F}_{16}:\mathbb{F}_2]=4\qquad [\mathbb{F}_{4}:\mathbb{F}_2]=2\qquad [\mathbb{F}_{2}:\mathbb{F}_2]=1$$ Therefore ...


0

HINT: Start with this... $10 \times 10 + 10 \times 2 = 10(10 + 2)$ Do the same for the other two terms $10 \times 4 + 2 \times 4$ (pull out common factor). You should then have a common binomial to factor out leading to 2 binomials multiplied together. You may be expecting to see a variable like $x$ (based on your last sentence) in your binomials but you ...


0

There is no mistake. 1 is a factor of every integer. So +-1/a0 and +-1/an are possible factors for the polynomial function...


1

You are getting bitten by the (1-x)^(m-r) term when x=1 and m=r. The sum and the add commands handle that differently. Your m is a fixed integer, and for finite summation you should be using the add command and not the sum command. The sum command is for symbolic summation. m := 20: sum( 0^(m-r), r=0..m ); 0 add( 0^(m-r), ...


5

As Lucian pointed out this follows immediately from the properties of cyclotomic polynomials. We have the factorization (into polynomials irreducible over $\Bbb{Q}$) $$ x^n-1=\prod_{d\mid n}\Phi_d(x). $$ Your observation follows from this as the factorization $$ a^n-b^n=b^n[\left(\frac ab\right)^n-1]=b^n\prod_{d\mid n}\Phi_d(\frac ab)= \prod_{d\mid n}b^{\deg ...


1

In your example, you can choose $p=\pm1$ and $q=3$ to find that $\pm\frac 1 3$ is a possible zero of your polynomial. Indeed, $\pm\frac 1 9$ cannot be a rational zero.


0

The sum of the roots of the first equation is $$a+c=-a$$ so $c=-2a$. And the product of the roots of the second equation is $$bd=d$$ so $d=0$ or $b=1$. If $d=0$ the second equation is $x^2+cx$ whose roots are $0$ and $-c$. That is, $b=0$ or $b=-c$. If $b=0$ the roots of the first equation are $0$ and $-a$, so $a=0$ and $c=0$. This gives $$a=b=c=d=0$$ If ...


0

By Vieta's formulas we have $$ a + c = -a\\ ac = -b\\ b + d = -c\\ bd = d $$ From last equation, $b = 1$ or $d = 0$. I) $b=1$. So, $$ a + c = -a\\ ac = -1\\ 1 + d = -c $$ From second eq. $c = -1/a$ and $$ a - \frac1a = -a \Longrightarrow 2a^2 = 1\\ d = \frac1a - 1 $$ So, $a = 1/\sqrt2, b = 1, c =- \sqrt2, d = \sqrt2-1$ and $a = -1/\sqrt2, b = 1, c = \sqrt2, ...


0

Use the coefficients relation, Look here For example, you want to split: $8+10x+9x^2+3a^3+x^4 = (x^2+px+q)(x^2+rx+s)$ The polynomial coefficients must be equal, so: $3 = p+q$ $9 = q+s+pr$ $10 = ps+qr$ $8 = qs$ this is a system of equtions on $\mathbb{Z}$...


0

The abel ruffini theorem in galois theory is not a tautology its just an implication which leads us to a contradiction. In simple terms if a polynomial is solvable by radicals then a normal extension of fields must exist and since there are some quintic equations which cannot contain a ladder of normal extensions hence we arrive at the conclusion that not ...


1

By Newton's identities you can find the values of the elementary symmetric functions of the unknowns $X_i$. In other words when we consider the equation $$ (X-X_1)(X-X_2)\cdots(X-X_N)=X^N+a_1X^{N-1}+\cdots +a_N=0 $$ that has the unknown numbers $X_i$ as its solutions, we can calculate coefficients $a_i, i=1,2,\ldots,N,$ in terms of the power sums $S_i, ...


1

Denote the above polynomial by $f(a,b,c)$. Then $f(a,b,c)=0$ for $a=b$ or $a=c$ or $b=c$. Hence we can write $$ f(a,b,c)=g(a,b,c)(a-b)(a-c)(b-c). $$ For $n=2$ we have $g(a,b,c)=1$. For $n\ge 3$ the homogeneous polynomial $g$ is given by $$ g(a,b,c)=a^{n-2}b^{n-2}+a^{n-2}b^{n-3}c+a^{n-2}b^{n-4}c^2+\ldots +b^{n-2}c^{n-2}, $$ where all monomials $a^ib^jc^k$ ...


1

$10x³-(2y+5)x²+(y-4)x+76 = 0$ Solving for y, we get $y = \dfrac{10x^3 - 5x^2 - 4x + 76}{2x^2 - x} = 5x + \dfrac{148}{2x-1} - \dfrac{76}{x}$ The only values of x that are divisors of 76 and make $\dfrac{148}{2x-1}$ an integer are $x \in \{1, 19\}$ So {$(x,y) = \{(1,77),(19,95)\}$} It's still possible that $\dfrac{148}{2x-1}$ and $\dfrac{76}{x}$ could ...


0

since $68590-70395+1729+76=0$ hence $(x-19)$ is definitely a factor of this polynomial so we must frame this equation in this particular manner: $$10x^3-195x^2+91x+76=10x^3-190x^2-5x^2+95x-4x^2+76=(x-19)(10x^2-5x-4)$$


-1

You can start by knowing that $x<1+max(a_0,a_1,a_2,a_3,a_4)$, so letting $x=1/y$, we have $1/x<1+(max(a_4,a_3,a_2,a_1,1))/a_0.$ Depending the specific calues of your coefficients,a linear substitution for x may improve this.


0

From Hamza's post one solution is -1.5,others you can find dividing polynomial to the $(x+1.5)$,then we get: $$2x^{ 3 }-5x^{ 2 }+18x+45=0\\ \left( x+1.5 \right) \left( 2x^{ 2 }-8x+30 \right) =0\\ \left( x+1.5 \right) \left( x^{ 2 }-4x+15 \right) =0\\ { x }_{ 1 }=-1.5,x_{ 2 }=2+i\sqrt { 11 } ,{ x }_{ 3 }=2-i\sqrt { 11 } $$


1

You can remark that $\frac{-3}{2}=-1.5$ is a solution.


1

Perhaps for Maple: $$0^0\neq 1\text{ ?}$$ I was able to replicate the problem It is odd but $\mathrm{subs}(x=1,B(x))=4$ in Maple. Also $ Z\mathrm{ := unapply (B(x),x)}$ gives $Z(1)=4$. Maple Primes is a better site for Maple questions


0

Since $P_k(s|t)$ is a polynomial in $s$ for a given $t$, $$ P_k(s|t)=\sum_{i=1}^k f_i(t)s^i $$ If $$ P_k(s+u|t+u)=P_k(s|t) +P_k(s|u) $$ Then $$ \sum_{i=1}^k f_i(t+u)\sum_{j=0}^i \binom{i}{j} s^ju^{i-j}=\sum_{i=1}^k f_i(t)s^i +\sum_{i=1}^k f_i(u)s^i $$ Equating the coefficient on $s^k$ we get $$ f_k(t+u)= f_k(t) + f_k(u) $$ Hence $f_k$ is linear and the ...


1

Let we set: $$q_m(x)=(1-x)^m\sum_{k=0}^{m}\binom{2m+1}{2k+1}\left(\frac{x}{x-1}\right)^k .$$ Since: $$\sum_{k=0}^{m}\binom{2m+1}{2k+1}w^{2k+1} = \frac{1}{2}\left(\left(1+w\right)^{2m+1}-\left(1-w\right)^{2m+1}\right)$$ we have: $$ q_m(x)=(1-x)^m ...


1

Notice that $a_0=q(0)$, $a_1=q'(0)$, $a_2={1\over2}q''(0)$ and so on. It should be easy to evaluate the $a_k$ this way.


1

(Too long for a comment.) Hm, there may be a typo in the paper or an unqualified statement. When you have an equation in $x$ of form, $$P_1(x) = P_2(x)\sqrt{P_3(x)}$$ like your $(2)$ above, the straightforward way to get its degree is to square both sides, then equate it to zero, $$\big(P_1(x)\big)^2-\Big(P_2(x)\sqrt{P_3(x)}\Big)^2 = 0\tag3$$ As you ...


2

Posting this as an answer because it is too long for comments. Below is the SAGE sessions I used to compute. M = matrix([[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1],[1,-1,0,0],[-1,1,0,0],[-1,0,0,0],[0,-1,0,0],[1,-1,-1,0],[-1,1,0,-1]]) P = Polyhedron(M); P L = LatticePolytope(M.rows()); L print L.poly_x("") print P.f_vector() ​ A 4-dimensional polyhedron in ...


1

Ok. Unless I made a mistake the good news are that this always holds. The bad news is that it holds even without the assumption that $Q$ is not a permutation. But we can also make the observation that if the restriction of $Q$ is not injective then the total imbalance increases. See below what exactly I mean by this. Observe the following: the numbers ...


1

First off, as Svetoslav comments, the graph is not a parabola since the expression for the function is cubic. You are correct that $f(5) = 0$ and $f(6)=0$. But there is another zero. Dividing $x^3-18x^2 + 107x-210$ by $x-5$ or $x-6$ will yield a quadratic that you can factor to find the third root. That gives one more root of the polynomial, or ...


2

The way to prove this depends a bit on what one knows. The comment suggests the in the context of OP the following is known: Each (non-zero) ideal of $F[X]$ is generated by a unique monic polynomial (which is a bit stronger than $F[X]$ is a PID). If this is known one can chose $d$ as the monic generator of the ideal generated by the $p_i$, so $(d)= ...


2

Can you, please, describe your question in more details? You can use any iterative procedure to solve it... You could use Newton-Raphson, for example. You need an algorithm, is that right? You can define your functions as: $$ g(r) = -b + \sum _{i = 1} ^N a_{i}(1 + r)^{-c_{i}} $$ $$ g'(r) = \frac {dg}{dr} = \sum_{i=1}^N -c_ia_i(1+r)^{-(c_i+1)} $$ The ...


1

W.r.t. "Do I need to know at least the interval I want to search in?": Yes, typically you do. At least for subdivision-based solvers such as Sturm's method or pretty much anything based on Descartes' rule of sign or its variations. For other algorithms like Durand-Kerner, Aberth-Ehrlich or homotopy continuation methods, you generally don't need bounds. It ...


1

This is not a polynomial, it is a rational function of polynomials. You want to solve the equation $\frac{x^3-4x}{4x^2-4x+1}=-10$ You can clear the denominator and get a cubic, but if the solution is not rational (as here) it is a mess. Alpha reports three real roots, around $-41.077, 0.32271, 0.75437$


0

In the most general sense, a Taylor series is not guaranteed to converge everywhere - try plotting the first few terms of a Taylor expansion of $\frac{1}x$ around $x=1$. If you only look in the interval $0<x<2$, you'll see that it keeps getting better and better, but if we look at $x>2$ we'll see that when we have an odd number of terms, the Taylor ...


1

The Taylor series of an analytic function $f(z)$ about $z=a$ has radius of convergence $r$ ($0 < r < \infty$) if and only if $r$ is the largest radius such that $f$ can be defined to be analytic in the open disk $\{z \in \mathbb C: |z - a| < r\}$. For example, since $1/(1 + x^2)$ has singularities at $x = \pm i$, the radius of convergence of its ...


0

Well, in the case of $M_1(t)=1$ the proof doesn't work, as you pointed out correctly. I would guess the conditions on the theorem are sloppy, as the polynomials probably weren't meant to be constant. The proof works when both polynomials $M_1(t)$ and $M_2(t)$ are nonconstant, as then $\deg(M_1(t))+\deg(M_2(t))=\deg(M(t))=\dim(V)$ and ...


2

Following is a more systematic approach uses Newton's identities. Let $e_1, e_2, e_3$ be the elementary symmetric polynomials associated with $a, b, c,$ i.e. $$\begin{cases} e_1 &= a + b + c\\ e_2 &= ab + bc + ca\\ e_3 &= abc \end{cases} \quad\iff\quad (x-a)(x-b)(x-c) = x^3 - e_1 x^2 + e_2 x - e_3 $$ and let $p_k = a^k + b^k + c^k, k \in ...


2

The Bernstein representation is great to give an intuition how and why the algorithms based on Descartes' rule of signs work: essentially, the signs of the coefficients of the "localized" polynomial which are computed in the classical VCA method show whether the corresponding vertices of the Bézier control polygon are located above or below the $x$-axis. If ...


1

One has $$0 = (a+b+c)^2 = a^2 + b^2 +c^2 + 2(ab+bc+ca).$$ Then, $$(\frac{a^2 +b^2 +c^2}{2})^2 = (ab+bc+ca)^2 = a^2b^2 + b^2c^2 +c^2a^2 + 2abc(a+b+c) = a^2b^2 + b^2c^2 +c^2a^2.$$ Finally, one has $$(a^2+b^2+c^2)^2 = a^4+b^4+c^4 + 2(a^2b^2 + b^2c^2 +c^2a^2) = a^4+b^4+c^4 + 2(\frac{a^2 +b^2 +c^2}{2})^2$$ Thus, $$\frac{1}{2}(a^4+b^4+c^4) = (\frac{a^2 +b^2 ...


0

Here is a link to a short note establishing that the roots of a polynomial are $C^\infty$ functions of the coefficients using the implicit function theorem.


0

$$\dfrac{\sum a^4}2-\left(\dfrac{\sum a^2}2\right)^2=\dfrac{\sum a^4-\sum2b^2c^2}4$$ $$\sum a^4-\sum2b^2c^2=(a^2+b^2-c^2)^2-(2ab)^2$$ $$(a^2+b^2-c^2)^2-(2ab)^2=(a^2+b^2-c^2-2ab)(a^2+b^2-c^2+2ab)$$ $$a^2+b^2-c^2+2ab=(a+b)^2-c^2=(a+b+c)(a+b-c)$$


1

Unfortunately the answer is: subdivide and go on. The rule-of-signs-predicate is not able to tell whether there are any roots, and if you have no additional means to do so, you have no other option. However, there are theorems which tell you that this won't happen too often. Short summary of the "easy" cases: If you count the sign variations $v$ of the ...


4

One of the reasons why the methods you mentioned do not work for multiple roots is that, in this situation, the root-finding problem is numerically ill-posed. That is, an arbitrarily small perturbation of the input will change the structure of your solution (just as you recognized, multiple roots will split into clusters). The same holds for the computation ...


2

Note that $\frac{1+\alpha}{1-\alpha}<0$ is equivalent to $1+\alpha$ and $1-\alpha$ having opposite signs, which happens when $\alpha>1$ or $\alpha <-1$. So all we have to do is count the number of roots of $3x^2-2x+5$ in $[-1,1]$. Let $p(x) = 3x^3-2x+5$. Now $p(-1)=4$ and $p(1)=6$. Moreover, $p'(x)=9x^2-2$, which has roots at $x=\pm\sqrt{2}/3$. ...


0

Following the method outlined in this answer we can write the original equation in the form $$\frac{\sigma p}{p} = \frac{\sigma^3 r}{r}$$ where $\sigma p(x) = p(2x)$ and $r(x)=8-x$. Using the "additive notation" (see the referenced post) we obtain $$p=\frac{\sigma^3-1}{\sigma-1}r=(\sigma^2+\sigma+1)r=(4x-8)(2x-8)(x-8)$$ unique up to a constant factor.


2

The following is essentially @drhab's solution, but uses only one idea repeatedly. From $$ (x-8)p(2x) = 8(x-1)p(x) $$ we see $x-8$ divides $p(x)$. Let $p(x) = (x-8)p_1(x)$ and substitute, yielding $$ 2(x-8)(x-4)p_1(2x) = 8(x-1)(x-8)p_1(x) $$ From this we see $x-4$ divides $p_1(x)$. Let $p_1(x) = (x-4)p_2(x)$ and substitute, yielding $$ ...


5

Usage of the multinomial coefficient $(k_1, k_2, \cdots, k_n)$!: $$ \big( 1 + x^5 + x^7\big)^{20} = \sum_{k_1=1}^{20} \sum_{k_2=1}^{20-k_1} (k_1, k_2, 20 - k_1 - k_2)! x^{5k_1} x^{7k_2}, $$ where $$ (k_1, k_2, \cdots, k_n)! = \frac{ (k_1 + k_2 + \cdots + k_n )! } { k_1! k_2! \cdots k_n!}. $$ So we get $k_1=2$ and $k_2=1$, thus $$ (2,1,17)! = ...


0

This is the base t expansion of a number, which is of course unique. Now if the polynomial factors it must factor as P*1. Since 1 has a unique base t expansion it must be that one of the polynomials is 1, hence the other is our given polynomial. So our polynomial does not factor. -M


5

$17$ can only be obtained by using two $5$s and one $7$ . These two $5$s can be obtained in $\binom{20}2$ ways which is $190$ and the $7$ can be got in from one of the remaining 18 brackets. So $190$ x $18$ = $3420$ is the answer.



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