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0

Quadratic equation has real roots iff $D \ge 0$. In your case $ D = (tr(A))^2 - 4det(A) \ge 0 $. Hence matrix has real eigenvalue iff $ (tr(A))^2 \ge 4det(A) $, equivalent $ det(A) \le (tr(A)/2)^2 $


0

Hint: The discriminant has to be non-negative to have real roots of a quadratic.


2

This is an approach based on pattern recognition. Recall the Pascal's triangle $$1\\ 1 \quad 1\\ 1 \quad 2 \quad 1\\ 1 \quad 3 \quad 3 \quad 1\\ 1 \quad 4 \quad 6 \quad 4 \quad 1$$ The coefficients of the polynomial at hand are $$1 \quad 4 \quad 8 \quad 8 \quad 4$$ They look "close" to $5^{th}$ row of above triangle. This suggest us to rewrite our ...


2

The first thing I would try are degree one factors, which by the Rational Roots Theorem must have the form $n + d$ where $d$ is an integer divisor of $4$. Once these possibilities are exhausted, the only factors left to check are quadratic ones: $$ n^4 + 4n^3 + 8n^2 + 8n + 4 = (n^2 + an + b)(n^2 + cn + d) $$ where $a,b,c,d$ are integers with the ...


0

No, a very small coefficient perturbation can affect the roots dramatically, meaning lets say a root changes tens orders of magnitude more than coefficient change. This is actually for many people one of the most surprising results in mathematics. Most of the time, this will not be the case, however this still poses a difficult problem in numerical analysis. ...


2

Here's a link to the method I mentioned in the comments. First, we construct a difference table for the degree $n$ polynomial $P(k)$ $$\begin{array}{c|c|c|c|c|c|c|c} k&P(k)&D_1(k)&D_2(k)&\ldots\ldots\ldots&D_{n-2}(k)&D_{n-1}(k)&D_n(k)\\ \hline\\ 0&1&1&1&\ldots\ldots\ldots&1&1&1\\ ...


2

Following are two approachs to show $P(n+1) = 2^{n+1} - 1$. Method I is more systematic and use finite differences. Method II is more elementary and do the job with induction. Method I - finite differences Given any function $f(x)$ and positive number $h$, the finite difference $\Delta_h f(x)$ is the function defined by $$\Delta_h f(x) ...


1

Let $v$ be the vector with $1$ on each coordinate. Let $A_i$ be column vectors of $A$. Then : $$det(A)=det(A_1+tv,A_2+tv,...,A_n+tv)$$ Let us set $B_k^0:=A_k$ and $B_k^1:=v$ then expanding $det(A)$ by multilinearity : $$det(A)=\sum_{i=0}^n\sum_{E\in\mathcal{P}_i(\{1,...,n\})}t^k.det((B_k^{\chi_E(k)})_{1\leq k\leq n}) $$ But if $i\geq 2$ and $E\in ...


3

It's certainly well-known. It is an immediate corollary of the Chinese Remainder Theorem. Notice that you could also work with $R^S$ for any finite set $S$. (And you don't mean $R^R$, which doesn't exist; you mean $R^S$, where $S$ is the underlying set of the ring $R$.)


1

I figured out a way thanks to the links in the comments above. One way to bracket the roots is to find a Sturm chain as follows. Here's an example with Mathematica, assuming f1 is the polynomial. First, find f2: f1[x] = x^5 - 3 x - 1 f2[x] = D[f1[x], x] Then repeatedly execute the following and evaluate f1 at a desired point x each time, where x is not a ...


1

Yes this is correct. In fact you've noted an important result that sometimes in a field of positive characteristic, polynomials of degree >1 can vanish when you apply a formal derivative. This doesn't happen in fields of characteristic $0$. In fact, you need this idea to understand separable extensions (or more inseparable extensions) from Galois theory. ...


2

Your argument using L'Hopilat rule is correct you need just to add the condition $a>1$ For $a> 1$ it's true that $a^x$ is very larger then $x^n$ to see this you compose with a logarithm: $$\lim_{x\to \infty} \frac{a^x}{x^n}=\lim_{x\to \infty} e^{\displaystyle x\ln(a)-n\ln(x)} =e^{+\infty}=+\infty$$ because the linear functions are always larger than ...


0

This is clearly impossible if $V \cap W \neq \varnothing$. Assume otherwise. Then $V \cup W$ is a disjoint union, and $\mathcal{O}(V \cup W) = \mathcal{O}(V) \times \mathcal{O}(W)$. Since $\mathcal{O}(\mathbb{A}^n) \to \mathcal{O}(V \cup W)$ is surjective, we can therefore find $f \in \mathcal{O}(\mathbb{A}^n)$ such that $f |_V = a$ and $f |_W = b$. ...


1

One way to do this is: $$n|m\implies X^n-1|X^m-1 \tag 1$$ to prove it take $m=nt$ then : $$X^m-1=(\color{#0a0}{X^n})^t-1=(\color{#0a0}{X^n}-1)\left(\sum_{i=0}^{t-1}(\color{#0a0}{X^n})^i\right) \tag 2$$ Or you can see this as a consequence of $Y-1|Y^t-1$ for $Y=X^n$. Another result which follows from this is that if you replace $X$ by a specific number in ...


2

If $k\mid n$, then every root of $x^{p^k-1}-1$ is also a root of $x^{p^n-1}-1$, so the field $\mathbb{F}_{p^k}$ is a subfield of $\mathbb{F}_{p^n}$, by recalling that, in an algebraic closure of $\mathbb{F}_p$, the field with $p^r$ elements is uniquely determined as the set of roots of $x^{p^r}-x$. Let's see the converse. If $\mathbb{F}_{p^k}$ is a subfield ...


2

Let $y$ be the common root. We then have $$y^3 + ay+b = 0 \text{ and }y^3+cy+d = 0$$ This means we have $$ay+b = cy+d \implies y = \frac{d-b}{a-c}$$ Hence, $$\left(\frac{d-b}{a-c}\right)^3 + a \left(\frac{d-b}{a-c}\right) + b = 0$$ This gives us $$(d-b)^3 + a(d-b)(a-c)^2 + b(a-c)^3 = 0 \implies (b-d)^3 = (a-c)^2(ad-ab+ba-bc)$$ which simplifies to what you ...


0

Put the value of $p$ in any of the equations, say the first one to get, $$(b-d)^3=b(a-c)^3-(a-c)^2a(b-d)=(a-c)^2(ab-bc-ab+ad)=(a-c)^2(ad-bc)$$


11

$$9x-x^3 = x(9-x^2)=x(3-x)(3+x)$$ where in the last step I have used the very important and useful "difference of two squares" identity: $$ (A-B)(A+B)=A^2-B^2 $$


2

Notice that $9 - x^2$ is the product of $(3 - x)(3+x)$ $$9 x - x^3 =x (9 - x^2) = x(3-x)(3+x)$$


0

Hint. To find all the coefficients of the expansion, you may consider what we call elementary symmetric polynomials: $$e_k(x) = \sum_{1 \leq i_1 < i_2 \dots < i_k \leq n} x_{i_1}x_{i_2}\dots x_{i_k}.$$ Have a look here: elementary symmetric polynomials.


3

What we see here is the formal derivative. If $A$ is a ring, then we can define the formal derivative $D\colon A[X]\to A[X]$ via $$D(\sum_{k=0}^n a_kX^k):=\sum_{k=0}^{n-1}(k+1)a_{k+1}X^k$$ (where $k+1$ is not an element of $\mathbb Z$, but rather viewed as the element $\underbrace{1+1+\ldots+1}_{k+1}\in A$, which may be zero $\in A$ even if it is nonzero ...


1

The polynomial $x^4+1$ is irreducible in $\Bbb Q$ but it's reducible in every $\Bbb Z_p$ for any prime $p$, for instance in $\Bbb Z_2$ we have: $$x^4+1=x^4-2x^2+1=(x^2-1)^2 $$


3

You have $(x+1)^2=x^2+2x+3$ over $\mathbb F_2$, but $x^2+2x+3$ is irreducible over $\mathbb Q$: It has no Zero.


4

I already accepted @Rolf Hoyer s excellent answer, this is just its simplification, too large for comment, and using the same notation. $A = \sum_{i=1}^n \alpha_i =\alpha_1 + \alpha_2 + \ldots + \alpha_n$ $B = \sum_{1 \le i<j\le n} \alpha_i\alpha_j = \alpha_1\alpha_2 + \alpha_1\alpha_3 + \ldots + \alpha_{n-1}\alpha_n$ $Z = \prod_{i=1}^n \alpha_i ...


0

Factoring we get $$x^2(x^2-x+5)>3(x-2).$$ Since $P(x)=x^2-2x+5$ has a negative discriminant, it is always positive, thus the LHS is non-negative, becoming zero only when $x^2$ does, namely $x=0$. Hence the inequality is trivial for $x\le2$. For $2<x<3$ we have $$x^2(x^2-x+5)>28>9>3(x-2)$$ and finally for $x\ge3$ ...


2

Separating the even and odd terms we need to show $x^4+5x^2+6 > x^3+3x$ . That is $(x^2+2)(x^2+3)> (x^3+3x)$. As $x^2+3$ is always positive, that factor can be cancelled on both sides, and we need to show $(x^2+2)> (x+1)$. This is equivalent to showing $(x^2+1)>x$. For $x<0$, and $x>1$ this inequality is trivial. For $x\in[0,1]$ not hard.


6

We have: $$4(x^4-x^3+5x^2-3x+6)=(x^2+3)((2x-1)^2+7)$$


2

$\bf{My\; Solution::}$ Let $x,y,z$ be the roots of an Cubic equation in terms of variable $t$ So $$(t-x)\cdot (t-y)\cdot (t-z) = 0\Rightarrow t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz = 0$$ Now Given $$x+y+z=4\;\;,x^2+y^2+z^2=14\;\;, x^3+y^3+z^3=34$$ So $$(x+y+z)^2 = (x^2+y^2+z^2)+2(xy+yz+zx)\Rightarrow (xy+yz+zx) = 1$$ And $$x^3+y^3+z^3-3xyz = ...


0

If we consider that the problem is to fit data using $(x_i,y_i)$ data points for a model $$y = \dfrac{a}{x^4} + \dfrac {b}{x^2} + c$$ define two variables $u_i=\dfrac{1}{x_i^4}$ and $v_i=\dfrac{1}{x_i^2}$ which make $$y=a u+b v +c$$ and the problem is just to perform a multilinear regression for which the normal equations are very simple to establish and use ...


18

I will assume that the reader is capable of finding that there are exactly 12 such polynomials in degree $3$ or less, by brute force if necessary. Assume the degree of such a polynomial $f(x)$ is at least 4 for the sake of contradiction. Let the leading coefficients be $f(x) = x^n - Ax^{n-1} + Bx^{n-2} - C x^{n-3} + Dx^{n-4} + \ldots$. This convention is ...


0

$$\begin{align} \sum_{i=0}^k(-1)^i {k\choose i} {m-i\choose k} &=\sum_{i=0}^k(-1)^i {k\choose i} {m-i\choose m-k-i}\\ &=\sum_{i=0}^k (-1)^i{k\choose i }{-k-1\choose m-k-i}(-1)^{m-k-i}&&(1)\\ &=(-1)^{m-k}\sum_{i=0}^k {k\choose i }{-k-1\choose m-k-i}\\ &=(-1)^{m-k}{-1\choose m-k}&&(2)\\ &=(-1)^{m-k}{m-k\choose ...


3

This turns out to be a detailed explanation of Jack's answer. $$ \begin{align} \left(\frac{1-x^{13}}{1-x}\right)^5 &=(1-x^{13})^5(1-x)^{-5}\\ &=\sum_{j=0}^5(-1)^j\binom{5}{j}x^{13j}\sum_{k=0}^\infty(-1)^k\binom{-5}{k}x^k\tag{1}\\ &=\sum_{j=0}^5(-1)^j\binom{5}{j}x^{13j}\sum_{k=0}^\infty\binom{k+4}{k}x^k\tag{2}\\ ...


6

$$[x^{30}]\left(\frac{1-x^{13}}{1-x}\right)^5 = [x^{30}]\sum_{k=0}^{5}\binom{5}{k}(-1)^k x^{13k}\sum_{n\geq 0}\binom{n+4}{4}x^n \tag{1}$$ hence the LHS of $(1)$ equals: $$\binom{5}{0}\binom{34}{4}-\binom{5}{1}\binom{21}{4}+\binom{5}{2}\binom{8}{4}=\color{red}{17151.}\tag{2}$$


0

Define a map $\varphi:\mathbb{Q}[x,y]\to\mathbb{Q}[t],\ x\mapsto t^2, y\mapsto t^3$. It's clear that $\text{im}\varphi=\mathbb{Q}[t^2,t^3]$, so it suffices to show that $\ker\varphi=(x^3-y^2)$. Apparently $\ker\varphi\supseteq(x^3-y^2)$, to prove the reverse containing relation, for any $f(x,y)\in\mathbb{Q}[x,y]$, divede it by $x^3-y^2$ and we get a ...


2

The coefficient attached to $x^{30}$ will be the number of ways you can add up to $30$ by using the numbers $0$-$12$ up to five times. (Here order matters) For instance $1+1+2+10+6=30$ is one way. $10+10+10+0+0=30$ is another and so is $0+10+10+10+0 = 30$. The reason for this is more apparent for smaller polynomials. For instance let's calculate the ...


0

The equation $y-a/x^4-b/x^2-c=0$ yields $$ f(x,y)= - a - bx^2 - cx^4 + x^4y=0, $$ which is an affine algebraic curve, see here.


1

We have $$\frac{a}{x^4}+\frac{b}{x^2}+c=\frac{a}{x^4}+\frac{bx^2}{x^4}+\frac{cx^4}{x^4}=\frac{a+bx^2+cx^4}{x^4}$$ This is a rational function.


0

Proposition: Let $f$ be a polynomial, and let $a \in \mathbb{R}, n \in \mathbb{N}$. If for each $0 \leq k \leq n-1$ we have $f^{(k)}(a) = 0$, then $$f(x) = (x-a)^n g(x)$$ for some polynomial $g$. Obs: $g$ may be identically zero (which implies that $f = 0$ as well). Since we either have $f = 0$ or $\deg(g) = \deg(f) - n$, this result should be enough ...


0

Use induction : $$ \lim_{x\rightarrow \infty} \frac{A}{Bx+C} = 0$$ If $k$ is fine then $$ \lim_{x\rightarrow \infty} \frac{P_k(x)}{P_{k+1} (x)} =_{LHospital} \lim_{x\rightarrow \infty} \frac{P_{k}(x)'}{P_{k+1}(x)' } = 0$$


2

$$p_k=a_1x^k+a_2x^{k-1}+...\\p_{k+1}=b_1x^{k+1}+b_2x^{k}+...\\ \lim_{x \rightarrow \infty} \frac{p_{k}}{p_{k+1}}=\lim_{x \rightarrow \infty} \frac{a_1x^k+a_2x^{k-1}+...}{b_1x^{k+1}+b_2x^{k}+...}=\\\lim_{x \rightarrow \infty} \frac{x^K(a_1+a_2\frac{1}{x}+a_3\frac{1}{x^2}...)}{x^{k+1}(b_1+b_2\frac{1}{x}+b_3\frac{1}{x^2}..)}=\\ \lim_{x \rightarrow \infty} ...


4

Methodology Proof Divide the numerator and denominator by $x^k$. Then the numerator has a constant term and a finite number of of terms of the form $$ \frac{a_j}{x^j} \to 0 \quad \text{as } \quad x \to \infty$$ while the denominator has a linear term which goes to $\infty$.


1

So, the answer seems to be, $$\sum_{i\le r,\ j\le s,\ i+j\le\max(r,s)}p_{ij}x^iy^j$$


0

And to show that the root is not rational, use the Bezout theorem. In this case, the only rational numbers that you have to consider are divisors of ten: $\{\pm 1, \pm 2, \pm 5, \pm 10\}$. But you already know that the polynomial is increasing and $f(1) = -8$, $f(2) = 24$, so there are no rational roots.


1

To show the existence of the root, just let $f(x) = x^5 + x - 10$ and notice that $f(1) = -8$ and $f(2) = 24$, so you can apply the intermediate value theorem. If it had more than one real root, it would have a local extrema : show that the derivative of $f$ has no real roots.


0

Hint study the variations of the function: $$f(x)=x^5+x-10$$ in particular $f$ is increasing because $f'>0$


2

Probably the easiest way to tackle this is by saying: $a$ is algebraic implies that the field extension $\mathbb{Q}(a):\mathbb{Q}$ is algebraic, and thus $[\mathbb{Q}(a):\mathbb{Q}] < \infty$. Similarly, we have $[\mathbb{Q}(b):\mathbb{Q}] < \infty$. So by the tower law, $[\mathbb{Q}(a,b):\mathbb{Q}] = ...


1

The easiest way is by using these facts, which are easily proved: $a\in\mathbb C$ is algebraic iff $\mathbb Q[a]$ is finite-dimensional over $\mathbb Q$. If $a$ and $b$ are algebraic then $\mathbb Q[a,b]$ is finite-dimensional over $\mathbb Q$. The result then follows because $a\pm $ and $ab$ are in $\mathbb Q[a,b]$ and subspaces of finite-dimensional ...


0

First, it is clear that $-1$ is a root. So, $(x+1)$ is a factor of that polynomial. Thus, $4x^3-7x-3=(x+1)(ax^2+bx+c)=ax^3+bx^2+cx+ax^2+bx+c$ By equating the coefficients of powers, we get: $a=4$ $b+a=0$ so $b=-4$ $c+b=-7$, so $c=-3$ So, we have $(x+1)(4x^2-4x-3)=4(x+1)(x^2-x-\frac{3}{4})$ Now, using the "usual" method to find roots of quadratic ...


0

Yes you are almost correct, the correct factor is $(2x+1)(2x-3)(x+1)$ which is equal to your answer multiplied by 4. Another note, In Abstrat Algebra, factorization highly depends on what field are we considering. But based from the tag of the question I know that we are pertaining in the field of Real Numbers.



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