New answers tagged

1

In most elementary linear algebra courses you're looking for the characteristic polynomial as means to study the eigenvalues. Given an $n \times n$ matrix $A$, $A$ has eigenvalue $\lambda \in \mathbb{R}$ if there exists $v \in \mathbb{R}^n$ such that $$Av = \lambda v$$ But then we have $(A - \lambda I)v = 0$. By definition, $A-\lambda I$ is then not ...


0

As mentioned in the comments, you just find $\det(A-\lambda I)$ (or $\det(\lambda I-A)$ if you want the leading term positive). Alternatively, if you find all of the (complex) eigenvalues $\lambda_1, \lambda_2, \lambda_3$, counted with multiplicity, then the characteristic polynomial will be $(\lambda-\lambda_1)(\lambda-\lambda_2)(\lambda-\lambda_3)$. In ...


0

We may factor as follows: $$x^4+4x^3-2x^2+4x+1=(x+a)(x+b)(x+c)(x+d)$$ $$a=1+\sqrt2+r$$ $$b=1+\sqrt2-r$$ $$c=1-\sqrt2+ir'$$ $$d=1-\sqrt2-ir'$$ where $r=\sqrt{2+\sqrt8}$ and $r'=\sqrt{-2+\sqrt8}$ By Vieta's formulas, $$b^{(m)}_{2m}=\sum_{1\le i_1<i_2<\dots<i_{2m}\le4m}x_{i_1}x_{i_2}x_{i_3}\dots x_{i_{2m}}$$ where $x_{i_{4n+1}}=a$, $x_{i_{4n+2}...


0

Here is some information pertaining to P(j,s): I get the impression that if this function is used at all for anything the values that matter $s\in(0,1)$. Now for the actual functions: $P(1,s)=-1$ $P(2,s)=-s^6+s^5+2 s^4-2 s^2-2 s+1$ $P(3,s)=-s^{30}+s^{29}+2 s^{28}-3 s^{25}-3 s^{24}-4 s^{23}+s^{22}+6 s^{21}+7 s^{20}+7 s^{19}+2 s^{18}-4 s^{17}-13 s^{16}-7 ...


1

Hint $$I=(1 + 4 x - 2 x^2 + 4 x^3 + x^4)^m=((x+1)^4-8x^2)^m=\sum_{i=0}^{m}(-8)^i(x+1)^{4m-4i}x^{2i}$$ $$I=\sum_{i=0}^{m}\sum_{j=0}^{4m-4i}(-8)^i x^{4m-2i-j}$$


2

I don't quite understand the work you showed without any formatting (not sure where one statement begins and the next ends). I'll write out how I approached the problem. We know that for some polynomials $f, g$ $$P(x) = (x-1)f(x) + 5$$ $$P(x) = (x-2)g(x) + 3$$ Multiplying $(x-2)$ to the top equation and $(x-1)$ to the bottom gives $$(x-2)P(x) = (x-1)(x-2)...


2

It is the remainder theorem that if you divide $P(x)$ by the degree two $x^2-3x+2$, the remainder will be a degree one polynomial. Hence, $$P(x) = (x^2-3x+2)Q(x) + (ax+b)$$ Now, this is an identity for all values of $x$. Thus if in particular you plug in $x=1$ on both sides, you'll get an equation. But notice that $x=1 \implies x^2-3x+2=0$, so you're left ...


0

In $ax^2+bxy+cy^2+5x−2y+3 = (x−y+1)(dx+ey+f)$, put $x=0, y=1$ to get $c+1 = 0$ and hence $c=-1$. Put $y=0, x=-1$ to get $a-2=0$ and hence $a=2$. Putting $x=1, y = 2$, we get $a+2b+4c+5-4+3 = 0$ and hence $b=-1$.


1

Since there is no remainder the original equation will be the product of $x-y+1$ with another unknown equation. In order to create the $x^2$, $y^2$, and $xy$ terms we expect the form to be $dx+ey+f$. $$ax^2+bxy+cy^2+5x−2y+3 = (x−y+1)(dx+ey+f)$$ $$ax^2+bxy+cy^2+5x−2y+3 = dx^2+(e-d)xy-ey^2+(d+f)x+(e-f)y+f$$ working right to left $$f=3$$ $$e-f=-2, e=1$$ $...


2

In my opinion this is a bit simpler to prove, if we interpret the matrix $P_n$ as a linear transformation. Consider the space $V_n$ of polynomials of degree $\le n+1$ (over, say $\Bbb{Q}$, but you are welcome to use reals or complex numbers, or any other field actually). The mapping $T: f(x)\mapsto f(x+1)$ for all $f(x)\in V_n$ is obviously linear. My key ...


1

Let $f \in K[x^q - x]$ and $\alpha \in K$. Then there is a $\bar{f} \in K[x]$, with $f(x) = \bar{f}(x^q-x)$. Since $(K^\times,\cdot)$ is a group with a finite number of elements, we have that $ 1 = \alpha^{\vert K^\times \vert} = \alpha^{q-1}$. Hence we get \begin{align*} f(x + \alpha) &= \bar{f}( (x + \alpha)^q - x - \alpha ) \\ &= \...


0

Newton's interpolation formula is $$ f(n) = d_0 \binom{n}{0} + d_1 \binom{n}{1} + d_2 \binom{n}{2} + d_3 \binom{n}{3} +\cdots $$ where $d_i$ are the numbers in the first column of the repeated differences array. This proves that a polynomial takes integral values at integers iff it is an integer linear combination of the binomial polynomials. See Integer-...


2

A well known result is that any integer-valued polynomial is the sum of integer multiples of $\binom{x}{k} =\dfrac{x(x-1)...(x-k+1)}{k!} $. https://en.wikipedia.org/wiki/Integer-valued_polynomial By multiplying by the largest $k!$, we get your statement.


0

If $a=0$ then $x=0$. If $a\ne0$ then we can divide both sides by $a$ to get \begin{equation} \left( 1-\dfrac{n}{a}\right)x^{n-1}+x^{n-2}+x^{n-3}+\cdots+x+1=0 \end{equation} If $a=1$ then the equation becomes \begin{equation} \left( 1-n\right)x^{n-1}+x^{n-2}+x^{n-3}+\cdots+x+1=0 \end{equation} for which $x=1$ is a solution. If $a\ne1$ then the $n$ in ...


1

If $x^2 + kx + 1$ is a factor, we can write: $ax^4+bx^3+c = (x^2+kx+1)(\gamma_2x^2+\gamma_1x+\gamma_0)$ and solve for $\gamma_0, \gamma_1, \gamma_2$ Multiplying out the expressions: $ax^4+bx^3+c = \gamma_2x^4 + (k\gamma_2+\gamma_1)x^3 + (\gamma_2+k\gamma_1+\gamma_0)x^2 + (\gamma_1+\gamma_0k)x+\gamma_0$. So we must have $\gamma_2 = a$ (equate $x^4$ terms) ...


2

This is a polynomial of degree $n-1$ in $x$ (unless $a=n$). If, for example, $a$ is an integer and $p$ is a prime such that $p$ but not $p^2$ divides $a$ while $p$ does not divide $n$, then Eisenstein's criterion says it is irreducible over the rationals. In general, for $n \ge 6$ we would expect it not to have roots expressible in radicals. EDIT: For ...


0

If we do a bit of reorganizing we can get $$nx^{n-1}=a\left(\sum_{k=0}^{n-1}x^k\right)$$ Using the partial sum formula for geometric series we get $$nx^{n-1} =a\frac{1-x^{n}}{1-x}$$ Doing some work with it... $$(1-x)nx^{n-1}=a(1-x^n)$$ $$a(1-x^{n})-(1-x)nx^{n-1}=0$$ $$(n-a)x^n-nx^{n-1}+a=0$$ At this point there doesn't seem to be any known general ...


5

An efficient algorithm to factor a polynomial into irreducible polynomials is given in this article. The lattice basis reduction algorithm they developed for this purpose is the famous LLL algorithm which has many applications besides its use in polynomial factorization problems.


6

It can't be done. There are formulas for the roots of a quadratic, cubic or quartic in terms of radicals, but not (in general) for the roots of a polynomial of degree $5$ or higher. For example, the roots of $x^5 + 2 x + 1$ can't be written in terms of radicals. See e.g. Abel-Ruffini theorem


1

Given any two fields $K$ and $L$ with $L$ properly containing $K$, the homogeneous polynomial $x+y$ has zeros in $L$ that it doesn't have in $K$, so there is no such thing as a field over which it has all of its roots. [I suspect OP has something else in mind, but I can't for the life of me figure out what.]


3

Since $x_1x_2$ is irrational, there is an automorphism $\sigma$ of $\overline {\Bbb Q}$ that changes $x_1x_2$ into something else. Since $\sigma$ acts on a permutation on the roots, we must have $\sigma(x_1)=x_i$ and $\sigma(x_2) = x_j$ where $i,j \in \{1;2;3;4\}$ and $i \neq j$, and importantly, $x_1x_2 \neq x_ix_j$ Since $x_1+x_2$ is rational it is fixed ...


0

$\begin{array}\\ (x-4i+2)(x-4i+1)-(x-4i+3)(x-4i) &=((x-4i)^2+3(x-4i)+2)-((x-4i)^2+3(x-4i))\\ &=2\\ \end{array} $ so each individual term is greater, so their product is greater.


-1

The following proof is wrong (see quid's comment): From $$2>0$$ follows $$x^2+3x+2>x^2+3x$$ and further $$(x+2)(x+1)>x(x+3)$$ for all $x$. substitute $x$ by $x-4k$ and you get $$(x-4k+2)(x-4k+1)>(x-4k)(x-4k+3)$$ for all $x$ and all $k$. and so we have $$\prod_{k=1}^{n}(x-4k+2)(x-4k+1)>\prod_{k=1}^{n}(x-4k+3)(x-4k)$$


11

Suppose that all the roots are non zero. Put $x_1+x_2=u$, $a=x_1x_2$, $x_3+x_4=v$, $b=x_3x_4$, we suppose that $u\in\mathbb{Q}$, then it is also the case for $x_3+x_4$, as the sum of all the roots is rational, and that $a,b$ are irrationals (as the product $ab$ is rational, if $a$ is irrational, then it is also the case for $b$). The polynomial with roots $...


1

To put (a very minor improvement on) Camilo Arosemena's comment into a full answer : if we write $T=\lbrace t_1<t_2<\ldots<t_n\rbrace$, then we can take $$ \begin{array}{lcl} \alpha &=& -(t_1\sqrt{p_1}+t_2\sqrt{p_2}+\ldots+t_n\sqrt{p_n}),\\ \beta &=& \sqrt{p_1}+\sqrt{p_2}+\ldots+\sqrt{p_n}+\sqrt{p_{n+1}} \end{array} $$ where the $...


3

Let $f(x)$ be the left side. Note that $f(x) \to 0$ as $x \to \pm \infty$, $+\infty$ as $x \to a_i-$ and $-\infty$ as $x \to a_i+$. So there will be at least one root in each interval $(-\infty, a_1)$, $(a_1, a_2)$, ..., $(a_{n-1}, a_n)$. Since the equation is equivalent to a polynomial equation with degree $n$, there are at most $n$ roots. Therefore ...


0

The equation is nothing but a polynomial of degree $n$ in disguise (except at $x=a_i$). Hence, it will have $n$ roots. You can show that the roots are interlaced by consecutive $a_i$'s and hence will have $n$ real roots. We have $$p(x) = 2016 \prod_{k=1}^n (a_k-x) - \sum_{i=1}^n \left(a_i \prod_{\overset{k=1}{k \neq i}}^n (a_k-x) \right)$$ We have $$p(a_1) &...


3

Don't know if this is the quickest/easiest/best/etc. way but here's one way.. $P$ has two inflection points, which means $P''(x)$ has two distinct real zeros. Therefore $P''(x)$ has degree at least 2, which means $P(x)$ has degree at least 4. Since $P(x)$ is a polynomial of least degree, we can conclude that $P(x)$ must have degree 4. Therefore we can ...


-1

$$P(j,s) = \frac{1}{1 - s^{\ell_0(j)} }\sum_{\ell =1}^{\ell_0(j)}\left((-1)^\ell \left( \frac{s^{\ell(\ell+3)/2 - (j+1)\ell - j(j-1)/2}}{1-s^\ell} \right) \prod_{k=\ell}^{\ell_0(j)}(1-s^k)^2\right) $$ $$= \sum_{\ell =1}^{\ell_0(j)}((-1)^\ell s^{\ell(\ell+3)/2 - (j+1)\ell - j(j-1)/2 } \quad \frac{\prod_{k=\ell}^{\ell_0(j)}(1-s^k)^2 }{( 1-s^\ell)( 1 - s^{\...


2

Solution credits to Rui Yao: According to a property of finite differences, if we set $c_n\in \mathbb{Z}$ to be the highest coefficient of $f(x)$, which has degree $n$, then $$n!c_n=\Delta ^n [f](x)=\sum_{i=0}^{n}\binom{n}{i} f(i)(-1)^{n-i}$$ Thus $$|n!c_n|=|\sum_{i=0}^{n}\binom{n}{i} f(i)(-1)^{n-i}|\le \sum_{i=0}^{n}|\binom{n}{i} f(i)(-1)^{n-i}|\...


2

$ x^{14} - x^{12} + \ldots + x^2 -1 = \frac{7}{x}, $ I am considering $ x \neq 0 $ multiply both side by $ x^2 $ and add you get $ x^{16} = 1+ 7x + \frac{7}{x}. $ Take minimum of R.H.S.


7

Note that your polynomial is a geometric series with first term $-x$, common ratio $-x^2$ so that it can be written as $$\frac{x(x^{16} - 1)}{x^2 + 1} = 7$$ So $$x^{16} = \frac{7(x^2 + 1)}{x} + 1$$ And now, all you need to do is show that $\frac{x^2 + 1}{x} = x + \frac{1}{x} > 2$, which is not hard.


2

Notice that the left hand is equal to $(x^3-x)(x^{16}-1)/(x^4-1)$; hence the equation will be transfer into: $(x^{16}-1)=7(x^4-1)/(x^3-x)=7(x^2+1)/x$; notice that $(x^2+1)/x$ is graeter than 2, hence the right hand is greater than 7*2, therefor $x^{16}>15$.


1

$$a^3 +b^3 +(-9)^3-3ab(-9)=(a+b-9)(a^2+b^2-ab+9a+9b+81)=0$$ therefore \begin{cases} a+b-9=0\\ \qquad\operatorname{or}\\ a=b=-9 \end{cases} since $a>b>0$ thus $$a+b-9=0$$ Set $f(x)=ax^2+bx-9$. We have $f(1)=a+b-9=0$, thus $Q=1$ and $P=\frac{-9}{a}$ finally $$4Q-aP=4+9=13$$


1

The solution with the smallest maximum is $56;$ 56 54 41 37 61 56 54 29 62 57 42 34 63 42 42 42 69 56 51 29 69 66 63 24 72 57 42 33 75 75 72 21 81 63 36 36 84 42 42 42 84 76 39 31 84 84 63 21 90 69 42 30 91 62 54 26 98 84 42 28 Next, if we fix two of the numbers, call them $W = 56, Z = 54,$ we ...


2

If I understand correctly, the goal is to write the given quartic polynomial as the difference of a quartic polynomial and a cubic polynomial. Hence, $$\begin{array}{rl} x^4 + x^3 + x^2 + x + 1 &= (x^4 + (1 + t_3) x^3 + (1 + t_2) x^2 + (1 + t_1) x + 1) - (t_3 x^3 + t_2 x^2 + t_1 x)\\ &= (x^4 + (1 + t_3) x^3 + (1 + t_2) x^2 + (1 + t_1) x + 1) - x (...


0

$1+x+x^2+x^3+x^4$ is the same as ($x^5-1$)/($x-1$), in general $1+x+x^2+x^3...x^n$ is the same of ($x^n-1$)/($x-1$). There is no factorization to that above.


1

Notice that this is a geometric series, first term $1$, common ratio $x$ and $5$ terms. So using the sum-formula for a finite geometric series gives us $$1 + x + x^2 + x^3 + x^4 = \frac{1 - x^5}{1-x}$$ If, instead of a simplification, you merely want to prove the identity given in your question, then observe that $$(x^2 + 3x + 1)^2 = x^4 + 6x^3 + 11x^2 + ...


4

Your polynomial is just $$\Phi_5(x)=\frac{x^5-1}{x-1} = \prod_{k=1}^{4}\left(x-e^{\frac{2\pi i k}{5}}\right)=\left(x^2-2\cos\frac{2\pi}{5}x+1\right)\left(x^2-2\cos\frac{4\pi}{5}x+1\right)$$ and we just have to recall that: $$ 2\cos\frac{2\pi}{5}=\frac{-1+\sqrt{5}}{2},\qquad 2\cos\frac{4\pi}{5}=\frac{-1-\sqrt{5}}{2},$$ We may reach the same decomposition by ...


0

Attempt a factorization of the form $(x^2+ax+1)(x^2+bx+1)$ and see what conditions you get for $a,b$.


1

The first equation does not make much sense to me because on the left hand side there is a velocity (in units of length over time) and at the right hand side we first see a variable $s$ considered a position, which has units of meter. So there there is something fundamentally unphysical in that initial equation, unless you declare s to be the velocity at ...


0

If the polynomial in the denominator of the generating functions has only simple roots (i.e. roots with multiplicity 1), a full partial fraction decomposition of the generating function with only have terms $\sim 1/(x-r)$ with roots $r$, and replacing these terms by their geometric series shows that the original series has the required shape (keyword: Binet ...


1

$$\text{Velocity}=\text{s}-\frac{jx^2}{200000}\Longleftrightarrow x=\pm\frac{200\sqrt{5}\cdot\sqrt{\text{s}-\text{Velocity}}}{\sqrt{j}}$$ Assuming $j\ne0$. So, we get: $$x=\pm\frac{200\sqrt{5}\cdot\sqrt{100-0}}{\sqrt{1}}=\pm2000\sqrt{5}$$ But time has to be positive so it is $x=2000\sqrt{5}\approx4472.14$ $$\text{Velocity}=\text{s}-\frac{jx^2}{200000}\...


0

Any strictly concave function has the property you want. You can use any polynomial of the form $$f(x)=-a(x-b)^n+c$$ where $n$ is an even natural number, $a>0$ and $c\in\mathbb{R}$. The maximum of $f$ is at $(b,c)$.


1

Theorem: $F[x]$ is an Euclidean domain for every field $F$. the kernel of the map namely K is an ideal of the ring $Q[x]$ .As $Q[x]$ is a PID , $K=(p(x))$ , for any monic polynomial $p(x)$ of lowest degree contained in $K$. By rational root theorem $f(x)$ is irreducible in $Q[x]$. Also it is of lowesr degree monic irreducible polynomial in $k$. Because if $...


1

Note that$$-2y^4+4y^3-2y^2-1=-2y^2(y-1)^2-1<0$$ for real $y$, so the remaining roots are not real. We need $$ y^2(y-1)^2=-\frac{1}{2}, $$ or $$ y(y-1)=\pm\frac{i\sqrt{2}}{2}. $$ For each choice of sign, this is a quadratic that you can solve with the usual formula. Specifically, $$ y^2-y(\pm)_1\frac{i\sqrt{2}}{2}=0\implies y=\frac{+1(\pm)_2\sqrt{1(\mp)...


3

$2y^4 - 4y^3 + 2y^2 + 1 =0$ is nothing but $2y^2\cdot (y-1)^2 + 1 = 0$ which is sum of square of two non negative terms.


0

I assume your definition of a variety is the following: Definition. Let $k$ be a field, $d$ be an integer greater than $1$ and $V$ be a subset of $k^d$. $V$ is a variety if and only if there exists $f_1,\cdots,f_n$ in $k[X_1,\cdots,X_d]$ such that: $$V=\{x\in k^d\textrm{ s.t. }\forall i\in\{1,\cdots, n\},f_i(x)=0.\}.$$ Let $I$ be an ideal of $K[X_1,\...


18

This is a statement of the Fundamental Theorem of Algebra that is completely correct, but in a way kind of obscures what is actually going on if someone doesn't think about what this actually means... Consider a complex polynomial $p(z) = a_0 + a_1z + ... + a_nz^n$. By the fundamental theorem of algebra, this has at least 1 zero. It is a well-known result ...


26

The theorem may as well state that every polynomial equation of degree $n$ has exactly $n$ roots (counted with their multiplicity). The statements are equivalent, for, if your polynomial $p(z)$ of degree $n$ has one root $\lambda$, then you can factor it as $$ p(z) = (z-\lambda)q(z), $$ where the degree of $q$ is $n-1$ Then you can recursively apply the ...



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