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7

The roots of this polynomial are $$e^{2k\pi i/5}$$ for $k\in\{1,2,3,4\}$. The point is that the roots for $k=1$ and $k=4$ are conjugate complex numbers, as well as the roots for $k=2$ and $k=3$. Furthermore, $$e^{2\pi i/5}+e^{8\pi i/5}=2\cos\left(\frac{2\pi}5\right)$$ and $$e^{4\pi i/5}+e^{6\pi i/5}=2\cos\left(\frac{4\pi}5\right)$$ Then ...


6

$$ (x+1)^n=((x-1)+2)^n=2^n+n 2^{n-1}(x-1)+\binom{n}{2}2^{n-2}(x-1)^2+(x-1)^3 \cdot F(x) $$ So the remainder is $$2^n+n 2^{n-1}(x-1)+\binom{n}{2}2^{n-2}(x-1)^2.$$


6

A monic polynomial is any polynomial $f(x)=a_n x^n + a_{n-1} + \cdots + a_1 x^1 + a_0 x^0$ such that $a_n=1$. Therefore, a monic polynomial of degree zero is of the form $f(x) = a_0$ where $a_n = a_0 = 1$ as $n=0$ so they may only take the form $f(x) = 1$.


5

Let $r=s/t$ where $s\in\mathbb{Z}$ and $r\in\mathbb{N}$. Assume $e^r=p/q$ where $p,q\in\mathbb{N}$. Then $$ pqt^nJ_n(r)= p^2 t^n A_n\left(\frac{s}{t}\right)+ q^2 t^n B_n\left(\frac{s}{t}\right)\in\mathbb{Z} \tag{1} $$ Note that $$ 0<J_n(x)\leq\frac{x^{2n}}{n!}\int_{-x}^xe^tdt=\frac{2 x^{2n}\sinh x}{n!} $$ So $$ 0<pqt^nJ_n(r)\leq \frac{2 pq t^n ...


5

Write $2014 = 2N$, where $N = 1007$ is odd. I claim that $$f(x) = N x^2 + 2 x = 1007 x^2 + 2x$$ has the required property. Note that $f(x)$ is odd if $x$ is odd and even if $x$ is even. Hence, if $f(i)$ and $f(j)$ have the same remainder when divided by $2014 = 2N$, then $i$ and $j$ have the same parity, and so $i-j$ is divisible by $2$. Suppose that ...


4

This is actually a very sloppy simplification of an important and complicated proof of the fundamental theorem of algebra, which secretly proves nontriviality of $\pi_1(S^1)$ behind the curtains (which involves some nontrivial bit of algebraic topology). I'll try to give a more explicit description of what's really going on in the proof by answering your ...


4

This might not be true. See for instance $f(X)=X^3-2$ (irreducible by Eisenstein) then : $$\dim_{\mathbb{Q}}(\mathbb{Q}[x]/(f(x))=\deg(f)=3 $$ However the roots of $f$ are $\sqrt[3]{2}$, $\sqrt[3]{2}j$ and $\sqrt[3]{2}j^2$. It is easy to see that $F:=\mathbb{Q}[{\sqrt[3]{2}},{\sqrt[3]{2}}j,{\sqrt[3]{2}}j^2]$ cannot be of dimension $3$ over $\mathbb{Q}$ ...


3

I don't think the following could be caracterized as "nasty determinant calculation". I don't know how one can prove the equality without indulging in some computation. Let $r=\operatorname{rank}(A)$ From a well-known theorem, derive that there exists $P,Q$ invertible $m\times m$ and $n \times n$ matrices such that $$A=P\begin{bmatrix}I_r& 0\\ 0 ...


3

Well ordered $\implies $ every chain has a minimum, but $$1>m \implies 1>m>m^2>\dots$$


3

Hint : The desired polynomial $p(x)$ must have $3$ as a root, that means $p(3)=0$. It is not difficult to see that $1$ must be added using this fact.


3

Let's decompose $P$ into distinct monomials, say, $P=a_1+ ... + a_k + b_1 + ... + b_l$ where $a_1+ ... + a_k$ is the lacunary part of $P$. Let $\Phi_\sigma$ be the automorphism on $R[X_1,...,X_n]$ corresponding to a permutation $\sigma\in S_n$. Then, $\Phi_\sigma( a_1 ) + ... + \Phi_\sigma( a_k ) + \Phi_\sigma( b_1 ) + ... \Phi_\sigma( b_l ) = ...


3

You can write the equality (E): $$(x+1)^n=P(x)(x-1)^3+a(x-1)^2+b(x-1)+c$$ where $P(x)$ is the quotient and $a(x-1)^2+b(x-1)+c$ the reminder of the euclidean division of $(x+1)^n$ by $(x-1)^3$. And your problem is to find $a,b,c$. (E) can be seen as an equality between functions. So can you differentiate both side to get further equalities. Make $x=1$ in ...


3

Let $A$ and $B$ be domains with $A\subseteq B$. We say that the ring extension $B/A$ is good if the following implication holds: $$\text{For every}\ f\in A\setminus\{0\}\ \text{and}\ h\in B,\ \text{if}\ fh\in A\ \text{then}\ h\in A\,.$$ We claim that if $B/A$ is good then $B[x]/A[x]$ is good as well. In fact, let $f\in A[x]\setminus\{0\}$ and $h\in ...


3

As you said, if $a \neq 0$ we have $f(a^{-1})=0$ which means that $f$ has a root and thus is reducible [$x-a^{-1}$ is a factor]. If $a=0$ you need to factor $x^6+5$. To do this note that $5 \equiv -100 \pmod{7}$.


2

Note that $f(a^{-1})=0$ as you noted before when $a$ is not $0$. Then we know that $(x-a^{-1})$ is a factor. We can do a "complete the sextic" approach on the function as follows: $f(x) = (x-a^{-1})(x^5) = x^6-a^{-1}x^5$ $f(x) = (x-a^{-1})(x^5+a^{-1}x^4) = x^6 + a^{-2}x^4$ Verify that $f(x) = (x-a^{-1})(x^5+a^{-1}x^4+a^{-2}x^3+a^{-3}x^2 + a^{-4}x + 2a) = ...


2

(1) Consider $f(x)=x^n-2$ then we'll show that it have no rational root. suppose $x=\frac{p}{q} ~with~gcd(p,q)=1$ be a root of $f(x)$. $\implies (\frac{p}{q})^n-2=0$ $\implies p^n-2q^n=0$ $\implies p^n=2q^n$ $\implies q\mid p^n$ but $gcd(p,q)=1$ and therefore it shows that $q=1$ So we have $p^n=2~for~ all~ n\geq2$. Which is a absurd since 2 is prime. ...


2

Let $f\mid g$ in $L[x]$. Polynomial long division yields a polynomial $q\in L[x]$ with $f = gq$. A closer look at the polynomial long division algorithm shows that the coefficients of $q$ are computed by repeatedly applying field operations to the coefficients of $f$ and $g$. So if the coefficients of $f$ and $g$ are in $K$, then the computed coefficients of ...


2

When you are doing the simplification: $\frac{2(x-2)}{(x-2)}$ = 2, you are making the claim that $\frac{(x-2)}{(x-2)}$ =1. This is true for almost any value of x, for example: x= 5 $\frac{(5-2)}{(5-2)}$ = $\frac{3}{3}$ = 1 However, in the case $\frac{(2-2)}{(2-2)}$ = $\frac{0}{0}$, which is undefined. So, that step you do is value for all values of x ...


2

We look only at $P(x)=x^{10}-x^7+x^4-x^2+1$. It is clear at a glance that $P(x)\gt 0$ if $|x|\ge 1$. Grouping as $(x^{10}-x^7)+(x^4-x^2)+1$ does it. So we look at $|x|\lt 1$. Negative $x$ in this range are easy to deal with, so we concentrate on $0\lt x\lt 1$. Since $x^4-x^7\gt 0$, we have $P(x)\gt 1-x^2\gt 0$.


2

Just another way for $(a)$ is using the AM-GMs: $$\frac12x^{10}+\frac12x^4 \ge x^7, \quad \frac12x^4+\frac12 \ge x^2$$ $$\implies x^{10}-x^7+x^4-x^2+1 \ge \frac12+\frac12x^{10}>0$$ and similarly for $(b)$: $$\frac12x^4+\frac12 \ge x^2, \quad \frac12x^4+\frac32+\frac32+\frac32 \ge 2\times 3^{3/4}x> 3x$$ Not always applicable, of course. Also you ...


2

Hint: $$f(x)=(x-3)(x+1)(x-a)$$ $$f(4)=5(4-a)=30$$


2

You don't need to solve the third degree equation.$$\begin{align}(2-x)^2\color{red}{(-2-x)}-\color{red}{(-2-x)}&=\color{red}{(-2-x)}((2-x)^2-1)\\&=-(2+x)((x-2)^2-1^2)\\&=-(x+2)(x-2-1)(x-2+1)\\&=-(x+2)(x-3)(x-1)\end{align}$$


2

Hint: $$a^2-b^2=(a-b)(a+b)$$ Apply this to $$ (2-x)^2 \cdot (-2-x) - (-2-x) = - (x+2)((2-x)^2-1^2) $$


2

There is a trick. Note that $(-2-x) = -(x+2)$ is a factor of both summands so $$\begin{align*} (2-x)^2 \cdot (-2-x) - (-2-x) & = -(x+2)\cdot((2-x)^2 - 1) \\ & = -(x+2)(x^2 - 4x + 4 - 1) \\ & = -(x+2)(x^2-4x+3) \\ & = -(x+2)(x-3)(x-1) \end{align*}$$ Where we only had to factor the quadratic $x^2-4x+3$


2

Let the original polynomial be $p(x)$. By Remainder Theorem, the remainder when dividing $p(x)$ by $(x-a)$ is $p(a)$. Applying that here, the remainder when dividing $p(x)$ by $(x-3)$ is $p(3) = -1$. To get a remainder of $0$, we simply add one to the polynomial ($-1 + 1 = 0$). It's that simple.


2

Any function of the form $$ y = ( a_1x-b_1)^2(a_2 x -b_2)^2 ( a_3 x - b_3)^2 $$ will have your triple well form with centres $b_i/a_i$, this fixes 3 free variables. To determine the heights, you want the derivative at the peaks to be zero. Solve the resulting equations and you'll have your conditions. In your example we have $$ y = x^2 (x-3)^2(x+3)^2 $$ You ...


2

Let me suppose that you want $$A(x)=\sum_{n=0}^\infty a_i x^i$$ So, let us mutliply both sides by the denominator $$8+14x-50x^2=(1-7x^2+6x^3)\sum_{n=0}^\infty a_i x^i$$ Now, decompose the product $$8+14x-50x^2=\sum_{n=0}^\infty a_i x^i-7\sum_{n=0}^\infty a_i x^{i+2}+6\sum_{n=0}^\infty a_i x^{i+3}$$ So, for the constant term $$8=a_0$$ For the first power of ...


1

I'm not sure what you are allowed to assume. Do you know that $e$ is transcendental and can you use that? If so consider $e^{\frac{p}{q}}=l$ where $l\in \mathbb{Q}$. What happens when you look at the polynomial $x^p-l^q$ ? Looking at the first part of your question I'm assuming it's unlikely you are allowed to use the fact that $e $ is transcendental ...


1

As noted elsewhere in the thread, if $f$ is irreducible with roots $\{a_1, a_2, ..., a_n\}$, it is definitely not the case that $\mathbb{Q}[x]/\langle f(x) \rangle \cong \mathbb{Q}[a_1, a_2, ..., a_n]$. It is the case, however, that $\mathbb{Q}[x]/ \langle f(x) \rangle \cong \mathbb{Q}[a_j]$ for any $1 \leq j \leq n$. More generally, this result holds for ...


1

If $a = 0$, then we have $\lvert f(w)\rvert = \lvert b\rvert$ for all $w$, so the strict inequality cannot be achieved. If $a \neq 0$, we can divide by $a$ and assume $a = 1$. If $n \geqslant 3$, then $$\{ w^n : \lvert w\rvert = r, \lvert w-1\rvert < 1\} = \{ z : \lvert z\rvert = r^n\},$$ since for every $r < 1$ the angle of the arc $\{ w : \lvert ...



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