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9

What is wrong is this: there is a difference between evaluating to zero at all points (as does $X^7-X$), and being zero. It is only over infinite fields that evaluating to zero at all points implies that the polynomial itself is zero. Your polynomial is not equal to zero, because $X$ and $X^7$ are different in $\mathbb Z_7[X]$. In fact, one of the ...


7

If $a$ is one root, then we get $$ \frac{x^3+x^2-2x+1}{x-a}=x^2+(a+1)x+(a^2+a-2)\tag{1} $$ Using the quadratic equation to solve this yields $$ \frac{-1-a\pm\sqrt{9-2a-3a^2}}2\tag{2} $$ for the other two roots. After the Question Change Changing the constant term only changes $(1)$ slightly: $$ \frac{x^3+x^2-2x-1}{x-a}=x^2+(a+1)x+(a^2+a-2)\tag{3} $$ and ...


7

HINT: Let $p(x)$ be the polynomial. Note that $$(a+b+c)(a+b+d)(a+c+d)(b+c+d)=(7-d)(7-c)(7-b)(7-a)=p(?)\;.$$


6

$xg(x)$ has no constant term, so the constant term of $2f(x)+xg(x)$ is twice the constant term of $f$. But it must be equal to $1$, which is a contradiction because $1$ is not divisible by $2$.


5

$$s^2+s\frac{R}{L}+\frac{2}{LC}=\left(s-\frac{-\frac RL+\sqrt{\frac{R^2}{L^2}-\frac{8}{LC}}}{2}\right)\left(s-\frac{-\frac RL-\sqrt{\frac{R^2}{L^2}-\frac{8}{LC}}}{2}\right)$$


5

According to marty cohen, we have \begin{align*} r^4+r^3+r^2+r+1&=\left(r^2+\frac{1}{2}r+1\right)^2-\frac{5}{4}r^2\\ &=\left(r^2+\frac{1+\sqrt{5}}{2}r+1\right)\left(r^2+\frac{1-\sqrt{5}}{2}r+1\right)\\ \end{align*} Then $$x^4+x^3y+x^2y^2+xy^3+y^4=\left(x^2+\frac{1+\sqrt{5}}{2}xy+y^2\right)\left(x^2+\frac{1-\sqrt{5}}{2}xy+y^2\right)$$


5

Observe that $$ P_n(z)=(z-\alpha_1)(z-\alpha_2)...(z-\alpha_n)$$ giving $$(\alpha_1+1)(\alpha_2+1)...(\alpha_n+1)=(-1)^nP_n(-1) $$ that is $$(\alpha_1+1)(\alpha_2+1)...(\alpha_n+1)=(-1)^n\left((-1)^n+a_{n-1}(-1)^{n-1}+...-a_1+a_0\right). $$ Hoping this helps.


4

Call $f=|F|$ the number of elements of $F$. Then the polynomial $$X^fY-XY^f \in F[X,Y]$$ annihilates all points of $F^2$.


4

Here's an efficient route, in the spirit of the remark in André Nicolas' answer: On one hand, the product of $f(x, y)$ with $(x - y)$ telescopes: $$(x - y) (x^4 + x^3 y + x^2 y^2 + x y^3 + y^4) = x^5 - y^5.$$ On the other, $$x^5 - y^5 = \prod_{k = 0}^4 (x - \zeta^k y),$$ where $\zeta := e^{2 \pi i / 5},$ and so over $\Bbb C$ (in fact, over $\Bbb Q[\zeta]$), ...


4

One way of showing irreducibility over the rationals is to reduce the problem to a more familiar one. If $f(x,y)$ were reducible over the rationals, then the polynomial $g(x)=f(x,1)=x^4+x^3+x^2+x+1$ would be reducible over the rationals. But that is not the case. One way to show this is to use the Eisenstein Criterion on $g(x+1)$. Another way is to note ...


4

The important thing to notice is that $f(x, y)$ is homogeneous; every term of the form $x^ay^b$ has the same value of $a+b$. Therefore, dividing by $y^n$, where $n$ is this sum, $\begin{array}\\ f(x,y) &= x^4+x^3y+x^2y^2+xy^3+y^4\\ &= y^4 \left((x/y)^4+(x/y)^3+(x/y)^2+(x/y)+1 \right)\\ &= y^4 \left(r^4+r^3+r^2+r+1 \right) \qquad\text{where } ...


4

The discriminant of the polynomial $p(x)=x^3+x^2-2x-1$ is $49$, which is a perfect square. It has no rational roots, so it is irreducible in $\Bbb{Q}[x]$. Together these facts imply that the Galois group of the polynomial is cyclic of order three. If $a$ is one of its zeros, we thus see that $\Bbb{Q}[a]$ is its splitting field. This means that the other ...


4

Try it like this. Let $p(x)= a_0 + a_1x + a_2x^2 + \dots + a_nx^n$ (with each $a_k \in \mathbb{Z}$) be any element that is in $J$ but not in $I$. Now note that $a_1x + a_2x^2 + \dots + a_nx^n$ is already in $I$ (because it is a multiple of $x$), hence it must be true that $a_0 \in J$ but $a_0 \notin I$. Now $a_0$ can't be an even integer -- if it were, it ...


3

We know the following to be true: $$ \tag1x_1+x_2+x_3+x_4 = -4 $$ $$ \tag2x_1x_2+x_2x_3+x_3x_4+x_4x_1+x_1x_3+x_2x_4 = 6 $$ $$ \tag3x_1 x_2x_3+x_2x_3x_4+x_3x_4x_1+x_4x_1x_2 = -a $$ $$ \tag4x_1x_2x_3x_4 = b $$ Now we can square $(1)$ to get $$ \begin{align} (x_1+x_2+x_3+x_4)^2 = &x^2_1+x^2_2+x^2_3+x^2_4\\ ...


3

$$\sum_{r=0}^n x^r=\dfrac{1-x^{n+1}}{1-x}$$ $$\implies(x^0 + x^1 + \dots + x^n)^n=(1-x^{n+1})^n(1-x)^{-n}$$ So, the coefficient of $x^n$ in $(x^0 + x^1 + \dots + x^n)^n$ $=$ the coefficient of $x^n$ in $(1-x)^{-n}$ Now the $r$th term of $(1-x)^{-n}$ is $\dfrac{(-n)(-n-1)\cdots\{-n-(r-1)\}}{r!}\left(-x\right)^r$ ...


2

Take the reciprocals of each and multiply by $abcdef$ to obtain the equations $$a^2 = \frac14 abcdef \\ b^2 = \frac19 abcdef \\ \vdots$$ Multiply all original equations to get $$(abcdef)^4 = 4\cdot 9\cdot 16 \cdots = 1$$ Now you should be able to tell the solution without using a calculator.


2

Here is a solution to question 1. I am still thinking about question 2. Solution to Question 1 First, we want to replace the polynomial $f$ with its remainder when divided by $g$. So we could write $f(x) = h(x) g(x) + r(x)$, where $\deg r < \deg g$. Unfortunately, $h(x)$ and $r(x)$ could have rational coefficients rather than integer coefficients. So ...


2

The probability is $\frac{1}{p}$. One way to see this is to imagine that instead of $(x_n,y_n')$ you take $(\beta,0)$ as an interpolation point. Then you get a unique polynomial $\tilde{P}$ of degree at most $n-1$ satisfying $\tilde{P}(x_i)=y_i'$ for $i<n$ and $\tilde{P}(\beta)=0$. Let $\tilde{y_n}=\tilde{P}(x_n)$. You know that $P'$ has degree at most ...


2

Since $P_n$ is a monic polynomial, $$ P_n(z) = \prod_{i=1}^{n}(z-\alpha_i)\tag{1} $$ hence by evaluating at $z=-1$, $$ (-1)^n P_n(-1) = \prod_{i=1}^{n}(1+\alpha_i).\tag{2}$$


2

Write $\xi_j=2\cos(2\pi\frac{j}{n})$ for $j=1,\cdots,n$. (Order will turn out not to matter.) We seek $${\rm Tr}\left[\prod_{j=1}^n (\xi_j{\rm e}_{11}+a{\rm e}_{12}+b{\rm e}_{21})\right].$$ We're denoting by ${\rm e}_{ij}$ the elementary matrices. They satisfy ${\rm e}_{ij}{\rm e}_{kl}=\delta_{jk}{\rm e}_{il}$. The noncommutative matrix monomials that ...


2

Let $$ S = (1 + x)^{2n} + x(1 + x)^{2n - 1} + x^2(1 + x)^{2n - 2} + ... + x^n(1 + x)^n.........................(1)$$ Now Multiply both side by $\displaystyle \frac{x}{(1+x)}\;,$ We get $$\displaystyle S\frac{x}{(1+x)} = x\cdot (1+x)^{2n-1}+x^2\cdot (1+x)^{2n-2}+.......+x^{n+1}\cdot (1+x)^{n-1}...............(2)$$ So $$\displaystyle ...


2

For $n=1,2,3,4$ there are general formulas. For all other $n$, there is no general method for solving such an equation.


2

One might argue in this fashion, instead: Suppose that $J$ is an ideal such that $I\subsetneq J\subseteq\Bbb Z[x],$ and take any $y\in J\setminus I.$ By definition of $I,$ we know that $x$ doesn't divide $y,$ and so if $y=px+n$ for some $p\in\Bbb Z[x]$ and some $n\in\Bbb Z,$ then $n\ne 0.$ In fact, $n$ cannot be any even integer. (Why?) Thus, if $y=px+n$ for ...


2

Do a simple example, and you will see that a linear combination of $A^{\circ(-1)}$ and $B^{\circ(-1)}$ does not suffice. The first one that comes to my mind: $\alpha=\frac12$ and $$ A=\pmatrix{1&1\\1&1}, \quad B=\pmatrix{1&1/2\\1/3&1/4}; $$ now $$ \Gamma=\pmatrix{1&3/4\\2/3&5/8}=\frac1{24}\pmatrix{24&18\\16&15} $$ and as ...


2

HINT: We can complete the square by writing $$s^2+\frac RL s+\frac2{LC}=\left(s+\frac{R}{2L}\right)^2-\left(\frac{R^2}{4L^2}-\frac{2}{LC}\right)$$ Can you factor this?


2

A positive real root is the point say $x=a$ ($\forall \ \ a>0$) where the curve intersects the x-axis. In general, to find one real root of the polynomial equation $$a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots +a_1x+a_0=0$$ Apply Newton Raphson's iteration formula given as follows $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Where, ...


2

As $x>0$ it is a positive root. There is no really nice way to get to that solution, one possibilty is to use Cardano's method which gives you a formula for these equations: $$ax^3+bx^2+cx+d=0.$$ If you don't need the exact solution I'd suggest something like Newton's method.


2

Yes, you assume correctly that the update can also be written as $$ w_k=-\frac{f(z_k)}{g'(z_k)}. $$ And correct, the derivative of $g$ can be written as sum of products according to its construction as product of linear factors $$ g(x)=(x-z_1)…(x-z_n) $$ (note that the leading coefficient of $g$ is $1$, so that this must also hold for $f$) $$ ...


2

Since $P$ has degree $n$ and $P'$ has degree $n-1$ at most, $P-P'$ has degree $n$, so it has $n$ distincts roots at most. Since $P(x_i)=y_i=P'(x_i)$ for $i\in\{1,\dots,n\}$, $P'(\beta) \neq P(\beta)=0$, so $P'$ cannot have $\beta$ has a root.


2

If $F$ is a field, then, in $F[x]$, if $f$ divides $g$, then so does $\alpha f$ for any non-zero $\alpha \in F$. Requiring the gcd to be monic makes it unique. In your example with $F = \mathbb{R}$, $6x$ or $72x$ would be just as good values for $\gcd(5x^2, 25x)$ if we didn't adopt this convention (not all authors do). If the ring of coefficients is not a ...



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