New answers tagged

0

The opening is a standard way but is not easy to simplify. May try in symmetrical manner, $$af(a)=bf(b)=cf(c)=abc$$ Now, $$xf(x)-abc=k(x-a)(x-b)(x-c)$$ For $f(x)$ being a polynomial, $k=1$ which is what @Raffaele commented.


1

One source of inspiration may have been looking at the binomial expansion of $(u+v)^3$ and mapping it to the cubic $t^3 + pt +q = 0$. $$\begin{align*}(u+v)^3 &= u^3 +3 u^2v +3uv^2+v^3 \\ \\ &= u^3+v^3+3uv(u+v) \\ \\ (u+v)^3 - 3uv(u+v)-(u^3+v^3)&= 0\\ \end{align*}$$ Substituting $t =u+v$ yields $$t^3-3uvt-(u^3+v^3) =0$$ Which forces the equalities ...


8

Jose already picked out an answer, but it’d also suffice to observe that the zero polynomial isn’t in $W$.


2

Hint: Use 1. Vieta formula: $$x_1+x_2+x_3+x_4+x_5= -{b\over a}$$ where polynomial $$ax^5+bx^4+....$$


0

$$(x-7)(x-1) (x+3)(x+8) ( x - A) = x^5 + (3-A)x^4 - (3A + 57)x^3 + (57A - 115)x^2 + (115A + 168)x - 168A $$


0

You mention $10$ axioms of a complex vector space, but you only need one for the vast majority of cases. Assuming scalar multiplication and usual addition we have that $$W \text{ is VS } \iff p, q\in W \implies \alpha p+\beta q \in W$$ for every $\alpha, \beta \in \Bbb C$. You simply need to find two polynomials with real zeros whose combination has a non-...


7

Both $p(x)=x^2$ and $q(x)=1$ belong to $W$. However, $p(x)+q(x)=x^2+1\notin W$, since its roots are $\pm i$, none of which is real.


0

Your question is completely unclear. Hensel lemma is saying that from $F\in \Bbb{Q}[x]$, that we transform in $f\in \Bbb{Z}[x]$ monic squarefree, there is some $N$ (easily computable but don't ask me its value) such that $f$ has a root in $\Bbb{Z}_p$ iff $f\bmod p^N$ has a root in $\Bbb{Z}/p^N\Bbb{Z}$. The correspondence is $a_N\to \lim_{n\to \infty } a_n$ ...


0

You are done. $f^5(x) = 240$ and that is a constant. The derivative of a constant function is $0$ so $f^6(x) = 0$. And $0$ is a constant function so it's derivative is $0$ and so by induction. Base case: $f^6(x) = 0$. Induction step: If $f^n(x) = 0$, then $f^{n+1}(x) = (0)' = 0$. Conclusion: $f^{n} = 0$ for all $n \ge 6$.


1

For the same reason as, for any commutative ring $R$, $R[X] $ is a free $R$-module, but not a finitely generated one.


3

Hint: Let $p_1,\cdots, p_n\in k[x]$. Can you find an upper bound on $\deg_y(p_1b_1+\cdots+p_nb_n)$ which is independent on $p_1,\cdots, p_n$?


2

With $P(x)=x(x-a)(x-b)(x-c)$, you can use a decomposition of $1/P(x)$ in simple fractions : $$\frac{1}{P(x)}=\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}+\frac{D}{x}$$ where $A,B,C$ are the terms in your expression (for example, to find $A$, you multiply by $x-a$, simplify and take $x=a$), and $D=-1/abc$. Then multiply by $x$ and let $x\to \infty$ and we are ...


0

Substitute $u = x^\frac{1}{2}$ to reexpress the equation $x^2 + bx^\frac{1}{2} + c=0$ as $$u^4 + bu + c=0 $$ whose unique depressed structure allows it to be factorized as $$\left( u^2+2s u +\frac{8s^3-b}{4s} \right) \left( u^2-2s u +\frac{8s^3+b}{4s} \right)=0 $$ where $s$ satisfies $s^6-\frac c4s^2-\left(\frac b8\right)^2=0$, a cubic equation in $s^2$. ...


4

Hint Denote $$\begin{cases} a &= \sqrt[100]{\sqrt 3 + \sqrt 2}\\ b &= \sqrt[100]{\sqrt 3 - \sqrt 2} \end{cases}$$ and notice that $ab=1$. We have $a^{100} + \frac{1}{a^{100}} = 2\sqrt 3$. Now you can write $$a^{100} + \frac{1}{a^{100}} = \left(a + \frac{1}{a}\right)^{100} - (p(a)+p(1/a))$$ where $p(x)$ is a polynomial of degree at most equal to $99$ ...


3

If you translate the graph so that the single point of inflection is the origin then the resulting function is odd.


0

I think the answer is yes, and hopefully this is a valid proof. If there's a polynomial $P(t)\in K(f(x))[t]$ that has $x$ as a root with degree less than $\deg f$, then we can write $$P(t)=\sum_{i=0}^n\frac{p_i(f(x))}{q_i(f(x))}t^i$$ where $p_i$, $q_i$ are polynomials with coefficient in $K$ and $n<\deg f$. Then plugging in $x$ we can write $$\sum_{i=0}^...


0

By the assumptions, there exist polynomials $q_1(x),q_2(x)$ such that $$ \begin{cases} p(x)=(x+3)q_1(x)+10\\ p(x)=(x-1)q_2(x)+2 \end{cases} $$ Hence $p(1)=2$ and $p(-3)=10$. Now, by the divisiob algorithm, there exist a polynomials $q(x),r(x)$ such that $$ (\ast)\quad p(x)=(x^2+2x-3)q(x)+r(x) $$ and $\deg r(x)<\deg(x^2+2x-3)=2$. Thus, $r(x)$ is of the ...


0

You should be able to produce the following equations using the familiar division algorithm. $$p(x) = (x+3)q_1(x) + 10$$ $$p(x) = (x-1)q_2(x) + 2$$ $$p(x) = (x-1)(x+3)q_3(x) + (ax+b)$$ We don't really care what $q_i(x)$ for $i=1,2,3$ are (they are the quotients obtained on division, though). Also, since the last division is by a quadratic term, the remainder ...


1

You can rearrange x^3 + x = 1 as x^3 = 1 – x, and let x = u + v. Notice that by the binomial theorem, (u + v)^3 = u^3 + 3vu^2 + 3uv^2 + v^3 = u^3 + v^3 + 3uv(u + v), suggesting that u^3 + v^3 = 1, while -1 = 3uv. -1 = 3uv implies v = -1/(3u), so u^3 – 1/(27u^3) = 1, implying 27u^6 – 1 = 27u^3. You can rearrange this as 27(u^3)^2 – 27u^3 – 1 = 0. This is a ...


2

Alternatively $$x=\frac2{\sqrt3}\sinh\left( \frac13\sinh^{-1}\frac{3\sqrt3}2\right)$$ which may be less monstrous.


1

If one re-arranges $x^{3}+x= 1$ as $x^{2} + 1 = \frac1x$ then the accompanying figure shows an alternative approach to obtaining the (real) solution iteratively, starting with some root approximation (e.g. $x_0=0.5$ in the figure), calculating $x_0^{2} + 1$ to obtain the first iteration $x_1=\frac{1}{x_0^{2}+1}$ and so on. In comparison to the unavoidable ...


4

$$ \left( \frac{1}{2} \left( 1 + \sqrt{\frac{31}{27}} \right) \right)^{1/3} + \left( \frac{1}{2} \left( 1 - \sqrt{\frac{31}{27}} \right) \right)^{1/3} $$


2

I think this is more of a confusion of language and nothing else. If $f(x) \in K[x] $ is a specific polynomial then the coefficients of $f$ are nothing but specific members of $K$. And then if you have a formula for roots of $f$ which involves a combination of some members of $K$ along with operations like $+, -, \times, /, \sqrt[n] {. } $ then the ...


0

Hint: Use $$ (a+b)^n=\sum_{k=0}^n\binom{n}{k}a^{n-k}b^k, i^{2k}=(-1)^k, i^{2k-1}=(-1)^{k+1}i $$ and you will get the answer.


0

Attacking discriminants is the natural try, but is the wrong approach for this particular problem. The problem can be solved via vieta's formulas. Since $3x^2 - 2px + q = 0$ has two roots, $r_1, r_2$, then these roots also satisfy: $$x^2 - \frac{2}{3}px + \frac{1}{3}q = 0.$$ Therefore, since $r_1$ and $r_2$ are integers, where $r_1 + r_2 = \frac{2}{3}p,$ ...


2

Using Vieta, as the sum of the roots of $g(x)$ is $\dfrac{2p}3$, for integer roots, we must have $3\mid p$. Similarly, as the product of the roots is $\dfrac{q}3$, we must have $3\mid q$. Further, if $f(x)=x\,h(x)$, we need $h(x) = x^2-px+q$ to have two integer roots, and as $3\mid q$ the product of roots, one of the roots is divisible by $3$, and as $3\mid ...


0

If you prove that $(1+i)^{2n+1}=a_n+ib_n$, for real $a_n,b_n$, then taking the conjugates shows that $$ (1-i)^{2n+1}=a_n-ib_n $$ There's no other computation to do. Let's try to see what $a_n$ and $b_n$ should be. For $n=0$, we have $a_n=b_n=1$. Then $$ a_{n+1}+ib_{n+1}=(1+i)^{2n+3}=(a_n+ib_n)(1+i)^2=2i(a_n+ib_n)=-2b_n+2ia_n $$ Hence you have $b_{n+1}=2a_n$ ...


1

Any polynomial that is solvable with radicals but with no roots in $K$ would work. The four arithmetic operations cannot take you outside of $K$. Take $K=\Bbb Q$ and $f(x)=x^2-2$, for instance.


0

Assume that $p\in\Bbb R[x]\space\land\space\deg (p)\ge 2$. May assume that $\lim\limits_{x\to \infty} p(x) = \infty$. Then there exists an interval $[c,\infty)$ such that $p\colon [c, \infty)\to [d, \infty)$ is bijective. Consider the values $n\pi$ ($n\ge n_0$) in the interval $[d, \infty)$ and their preimages $x_n \in [c, \infty)$. Now $p(\sin x_n) = \sin(...


0

Hint: $(2x+1)^4 = (8 x^2 + 12 x + 2)(2x^2 + x + 1) -6 x - 1$


0

For $|x|<1,$ $$(1+x+x^2+\cdots+x^{n-1})^2=(1-x^n)^2(1-x)^{-2}=(1-2x^n+x^{2n})(1-x)^{-2}$$ If $k<n,$ use binomial series to find the coefficients of $x^k$ in $(1-x)^{-2}$ If $n\le k<2n$ we need the coefficients of $x^k$ in $(1-x)^{-2} +(-2)\cdot$ the coefficients of $x^{k-n}$ in $(1-x)^{-2}$ What if $k\ge2n?$


0

It's $0$ unless $k\in\Bbb Z$ with $0\le k\le 2n-2$, in which case each integer $j$ with $0\le j\le n-1$ and $0\le k-j\le n-1$ contributes $1$ to the coefficient, making it $\min\{k+1,\,2n-1-k\}$.


1

$${s-\alpha}=\alpha\left(\frac{s}{\alpha}-1\right)=(-\alpha)\left(1-\frac{s}{\alpha}\right),$$ provided $\alpha\ne0$. So \begin{eqnarray} \frac{(s-a_1)(s - a_2) \; ... \; (s-a_m)}{(s-b_1)(s-b_2) \; ... \; (s-b_n)}& =&\frac{(-a_1)(-a_2)\cdots(-a_m)(1-s/a_1)(1-s/a_2)\cdots(1-s/a_n)}{(-b_1)(-b_2)\cdots(-b_n)(1-s/b_1)(1-s/b_2)\cdots(1-s/b_m)}\\ &=&...


0

Hint: If $t$ is real, what is the complex conjugate of $t+i$. What does that tell you about $Q$? Hint 2: This is a heuristic hint. If you know, or assume, that a polynomial has only real roots, then you can assume the polynomial can be put in the for of a product of monomial factors, each of the form $(x-r_i)$, where r_i is a real root of the polynomial. $$...


0

Let's look at things backwards: Suppose $Q(x)$ has integer coefficients and two integer roots; we need to show that $Q(x)-5^m$ cannot have three integer roots. By translation, we may assume that one of the roots of $Q$ is positive and the other is at $x=0$. Let's let the positive root be at $x=n$, so that $Q(x)=x(x-n)R(x)$. Now if $Q(r)=5^m$, then both $|r|$ ...


1

We have $$n^3-n-6=0\iff$$ $$(n-1)n(n+1)=1.2.3 \iff n=2$$ thus $$(m-1)m(m+1)\ne 6 \iff m\ne 2$$


2

Suppose $r_1$, $r_2$, $r_3$ are three distinct integer roots of $P(x)$, and $y_1$ and $y_2$ are integer roots of $P(x) + 5^m$. Thus $P(x) = Q(x)(x-r_1)(x-r_2)(x-r_3)$ where $Q(x)$ also has integer coefficients. Now $P(y_i) = -5^m = Q(y_i) (y_i - r_1)(y_i-r_2)(y_i-r_3)$, so $y_i - r_j$ are integers that divide $5^m$, in particular they are $\pm$ powers of $...


2

It's correct, but you're not actually using contradiction and it's indeed hard to write a real proof by contradiction of this particular statement. What you proved is that if $n^3-n-6=0$, then $n=2$. Hence for no $m\ne2$ one can have $m^3-m-6=0$.


1

It looks like your proof would be assuming $n^{3}-n-6=0, m\neq n$ (where $m,n \in \mathbb{Z}$) showing a contradiction if we had $m^{3}-m-6=0$ as well. You have shown by factoring and the fact that $m,n \in \mathbb{Z}$, that $m, n = 2$. Hence $m=n$, a contradiction.


4

The statement is actually false. Consider the polynomial $f=\Phi_{12}=X^4-X^2+1$. It's straightforward to see that $\Phi_{12}$ divides $X^{12}-1$. The latter polynomial is coprime with its derivative in $\mathbb{F}_p$ for $p > 3$. So we need to check that $\Phi_{12}$ has no double root in $\mathbb{F}_p$ for $p=2$ and $p=3$. For $p=2$, $\overline{f}(x) = (...


2

I don't quite know what you mean by "find" in your question. Do you mean how to list or name them all? The general term of a polynomial in $k$ variables $x_1, \dots x_k$ has the form $$c_{i_1, i_2, \dots i_k} x_1^{i_1} x_2^{i_2} \dots x_k^{i_k}$$ with total degree $\sum_{j=1}^k i_j$. If you want to bound the total degree then you just enforce a ...


0

Actually, since you have a quadratic polynomial, you need to have an expression where $(x,y)$ occurs twice, not once. Something like this: $$ \begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} $$ Can you multiply this out and compare the coefficients to your expression?


0

$$2x^2 + 3xy + 9y^2 = (x,y)\begin{pmatrix}2&\frac32\\\frac32&9\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$


1

There are an infinite number of solution pairs. That's a general result for quadratic Diophantine equations: they either have no solution (except perhaps for trivial solutions involving zero), or they have an infinite number of solutions, although those solutions will fall into a finite number of families. IIRC, Lagrange gave the first proof of that result (...


3

You're almost there, though your last bullet point is not quite correct; it should be $$a\mid b\quad\implies\quad b=0\ \text{ or }\ |a|\leq|b|.$$ Of course $3n-2=0$ is impossible, so indeed you get $|2n^3+5|\leq|3n-2|$. The left hand side of this inequality clearly grows faster than the right hand side, so there can only be finitely many solutions. Now it ...


3

... therefore $I[X]$ is never maximal.


1

Too long for the comment. Taking in account the brilliant condition $$f(f(1-x)) = 1 - f(f(x))$$ by JimmyK4542, one can write $$I=\int\limits_{\large^1/_8}^{\large^1/_2}f(f(x))\text dx +\int\limits_{\large^1/_2}^{\large^7/_8}f(f(x))\text dx =\int\limits_{\large^1/_8}^{\large^1/_2} \big(f(f(x))+f(f(1-x))\big)\text dx =\int\limits_{\large^1/_8}^{\large^1/_2}\...


0

$f(x)=x(x^2-px+q)=0 \implies x = 0 \text{ or } x^2-px+q=0 \implies D_1 \text{ must be a square.} \implies $ $$ p^2-4q \text{ must be a square}$$ In addition $g(x) = 3x^2-2px+q = 0 \implies D_2 = 4(p^2-3q) \text{ must be a square.} \implies$ $$ p^2-3q \text{ must also be a square}$$ Now, take $p=2$ and you will get $4-4q=4(1-q)$ and $4-3q$. From the first ...


1

As mentioned, subtraction decreases the degree if the leading coefficients are the same (which we can ensure with $cf(x)$), so all we need is polynomials of the same degree. This is the crux. See if you can work it out before reading on. We currently do not have 2 polynomials of the same degree, so let's force them out and take their difference. I will work ...


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