# Tag Info

## New answers tagged polynomials

1

Subtract off the principal part of each pole, including the one at $\infty$, and you have a bounded entire function, which by Liouville is constant. What you subtracted was a rational function...

4

$f_n(x) = x^n$ is a standard example where the sequence does not converge uniformly on $[0,1]$. So trying $$g_n(x) = \frac{1}{n+1} x^{n+1} = \int_0^x t^n dt$$ and this actually satisfies your conditions.

1

As $f(\frac1z)$ has no zeroes near $z=0$, the set of zeroes of $f$ is bounded, hence finite, say (with multiplicity) they are $w_1,\ldots,w_n$. Then $$h(z)=\frac{(z-z_1)\cdots (z-z_m)}{(z-w_1)\cdots (z-w_n)} f(z)$$ is entire, has no zeroes and $h(\frac1z)$ has a pole of order $m+1-n$ (if $m+1>n$) or a zero of order $n-1-m$ (if $m+1<n$) or a removable ...

5

$$\frac{|x^2+5x+6|}{|x|-3} = 1 \implies |x^2+5x+6|=|x|-3$$ Solving Algebraically: Case$1$: When, $x <-3$, $$x^2+5x+6=-x-3 \implies x^2+6x+9=0 \implies x=-3 \text{ [No solution]. }$$ Case$2$: When, $-3 < x \leq -2$, $$-x^2-5x-6=-x-3 \implies x^2+4x+3=0 \implies (x+3)(x+1)=0 \implies x=-3,-1 \text{ [No solution from this case]. }$$ Case$3$: ...

2

Since you want $\frac{|x^2+5x+6|}{|x|-3}=1$, that implies $|x^2+5x+6|=|x|-3$ as long as $|x|-3\ne 0$. Graph both equations simultaneously, and you will see they intersect at exactly $1$ spot: $(-3,0)$. Unfortunately we must reject the solution since it makes $|x|-3=0$. Thus there is no solution. By graphing them, you will also be able to see how to chop up ...

1

Hint: You can rewrite the numerator as $|(x+2)(x+3)|$. From there you can figure out the cases for which the expression inside the absolute value is negative and for which it is positive (and rewrite the numerator appropriately, i.e. as either $x^2 +5x+6$ or $-(x^2 +5x+6)$), which will also help you decide whether to rewrite $|x|$ as either $x$ or $-x$.

1

First, we note that for any two complex numbers $x, y$, we have from first principles, $$\overline{x+y} = \overline x + \overline y, \quad \overline{x\cdot y} = \overline x \cdot \overline y$$ Using these, we may show (say by induction) that for any polynomial $p(z)$ with all coefficients real, $\overline {p(z)} = p(\overline z)$. Thus in the problem we ...

2

The polynomial $f(x)$ in general need not have three distinct real roots, and two non-real roots, in order to be not solvable by radicals. Consider, say, $f(x)=x^5+3x+3$, which is irreducible by Eisenstein. Here $f(x)$ has only one real root, because its derivative is positive everywhere, so that $f$ is an increasing function. Nevertheless the Galois group ...

0

To answer this and make it answered, here is a statement. Fact A polynomial $p(x)=\sum_{0=1}^ma_ix^i$ has a nonzero root, i.e. there exists a nonzero complex root $y\neq0:p(y)=0$, iff there exist $i,j\in\{0,\dotsc,m\}$ with $i\neq j$ such that $a_i$ and $a_j$ are both nonzero. Proof. Suppose the condition holds. Suppose wlog $i<j$, and that $i$ is the ...

3

I believe that there is a mistake in your table. Brill's theorem states that the sign of the discriminant of an algebraic number field is $(-1)^{r_2}$ where $r_2$ is the number of complex places. When we have a power basis for our number field, the minimal polynomial of the generator will have $2r_2$ complex roots. Thus the column for $D>0$ should only ...

1

I don't think the table is correct. $f = (x-1)(x-2)(x-3)(x-4)(x-5)(x-6)$ has discriminant $1194393600 > 0$ and 0 complex roots. It doesn't have a direct citation from a reputable source, but the Wikipedia page you linked says that in general for $D > 0$ there is an integer $0 \leq k \leq \frac{n}{4}$ such that there are $4 k$ complex roots, and for $D ... 0 The algorithmic way uses successive Euclidean divisions, the same as the change of basis algorithm from base$10$to base$b$for numbers. In detail, you divide$3x^2+4ˆ$by$x+2$, getting a first remainder. Then you divide again the quotient by$x+2$, getting a second remainder, until the quotient is a constant. You obtain this constant as a lats remainder ... 0 Expand both sides:$3x^2+4 ≅ Ax^2+4Ax+4A+Bx+2B+C$Compare coefficients of$x$.$x^2$gives:$3=Ax$gives:$0=4A+B$constant gives:$4=4A+2B+C$Solve simultaneously for the 3 unknowns.$A=30=4\times3+B\toB=-124=4\times3+2\times-12+C\toC=4+24-12=16$Alternatively substitute in$y=x-2$to get:$3(y-2)^2+4≅Ay^2+By+C$... 0 Expand and compare the respective power of X and X^2 and the constant term you will surely get your answer.... 1 In fact, it is an integral domain: the polynomial$f^p-a$is irreducible in$k[x,y]$. In order to show this we use the Eisenstein's criterion. Note that$f^p-a=\alpha^px^p+\beta^py^p-a$. Multiplying by$\alpha^{-p}$doesn't change anything, so we may instead consider the polynomial$x^p+b^py^p-c$,$c\notin k^p$. Now note that the polynomial$b^py^p-c$is ... 4 Yes, if$f$is irreducible, then$5\mid [K:\mathbb{Q}]$where$K$is the splitting field of$f$. Hence, the Galois group$G$of$f$contains a 5-cycle. Furthermore, if it has only two non-real roots, then$G$also contains a transposition. The result now follows from the fact that if$G<S_p$(where$p$is prime) is a subgroup that contains a$p$-cycle ... 5 Since$i$is a root of$x$, and it has real coefficients, thus$-i$is a root as well, meaning that$x^2+1$divides this polynomial: $$5x^5+x^4-5x^3-x^2-10x-2=$$ $$=5(x^2+1)x^3+x^2(x^2+1)-10x(x^2+1)-2(x^2+1)=$$ $$=(x^2+1)(5x^3-10x-2)=(x^2+1)(5x+1)(x^2-2)$$ In the last step we tried to find a rational root, and since ... 0 Suppose we seek to verify that $$1 = (-1)^n \sum_{k=0}^n \frac{x_k^n}{\prod_{l=0\atop l\ne k}^n (x_l - x_k)}$$ with the$x_l$distinct. This is the same as $$1 = \sum_{k=0}^n \frac{x_k^n}{\prod_{l=0\atop l\ne k}^n (x_k - x_l)}.$$ Introduce $$f(z) = \frac{z^n}{\prod_{l=0}^n (z-x_l)}$$ which yields for the sum $$\sum_{k=0}^n \mathrm{Res}_{z=x_k} f(z).$$ ... 0 By inspection factor out$x$.$ 1\, x^2 - 2 x - 3 = 0  1+ (-3) = -2 $so$(x+1) $is a factor. Factor this also out, assuming this rule is known. Else find factor by inspection/trial & error/quadratic formula methods.. You are now left with$(x-3)$Roots are$ x= 0, -1,3.

0

$$x^3-2x^2-3x=0\Longleftrightarrow$$ $$x(x-3)(x+1)=0\Longleftrightarrow$$ $$x=0\Longleftrightarrow\space\space\vee\space\space x-3=0\Longleftrightarrow\space\space\vee\space\space x+1=0\Longleftrightarrow$$ $$x=0\space\space\vee\space\space x=3\space\space\vee\space\space x=-1$$

0

We have $$x^3-2x^2-3x = x(x^2-2x-3)=x(x-3)(x+1),$$ and hence the roots are $x=0$, $x=3$, and $x=-1$.

4

Factor out the $x$, getting $x(x^2-2x-3)=0$ Now one of the factors must be zero. You can use the quadratic formula on the second, or maybe you can factor it by inspection.

0

You can assume that $n = \deg f \ge \deg g > 0$. Then $h = (f-g)f'$ is a polynomial of degree $\le 2n-1$. Furthermore, any zero of multiplicity $m \ge 1$ of $f$ is a zero of multiplicity $m-1$ of $f'$, and it is at least a simple zero of $f-g$, so it is a zero of multiplicity $\ge m$ of $h$. This shows that $f$ divides $h$. The same argument also shows ...

1

Write $dq = pr+s$ where $s,r \in \Bbb Z[X]$, $\deg s < \deg p$ and $d$ is a (nonzero) positive integer. (this is always possible by picking for $d$ the dominant coefficient of $p$ to the $(\deg q - \deg p+1)$th power) Your hypothesis implies that $p(n)$ divides $s(n)$ for infinitely many $n$. For $n$ large enough, $|p(n)| > |s(n)|$ so you must have ...

2

Apparently, the problem was indeed too complicated, as stated in the errata here: In page 66, problem 44, there is a $=2$ missing at the end of the string of equations, thus $x^p+y^q=y^r+z^p=z^q+x^r=2$. We still have the solution $x=y=z=1$ for arbitrary $p,q,r$. Say we have a solution $x',y',z'$ where WLOG $x'>1$. That would force $y'<1$ from the ...

3

This article may be to your liking. It still uses the concept of Galois groups (it's hard to see how one could avoid that), but does not use the Galois correspondence.

1

Let $~P_n(x)~=~\dfrac{2^n}{n!}~\Big(\sqrt{1-x^2}\Big)^{2n+1}~\bigg(\dfrac1{\sqrt{1-x^2}}\bigg)^{(n)}.~$ Then its coefficients form the sequence described here.

2

$k$ has to be $O(\log n)$ for $3^k$ (or any $c^k$) to be $O(n^m)$ for some $m$. Proof: If $k = O(\log n)$, then $k < a \log n$, so $c^k < c^{a \log n} = e^{a \log c \log n} = (e^{\log n})^{a \log c } = n^{a \log c}$ which is polynomial in $n$. If $\frac{k}{\log n} \to \infty$, we can similarly show that $c^k$ grows faster than any $n^m$ for fixed ...

1

A Google search for "distribution of number of real roots in a polynomial" came up with this as the first hit: https://web.williams.edu/Mathematics/sjmiller/public_html/ntprob12/handouts/polyzeros/Fairley_NumbRealRootsRandPolySmallDeg.pdf Here is the title and abstract: THE NUMBER OF REAL ROOTS OF RANDOM POLYNOMIALS OF SMALL DEGREE* By WILLIAM B. PAIRLEY ...

0

Let $F$ be the field, we are working over. We have $g(A)^{-1} \in F[g(A)] \subset F[A]$. All elements of $F[A]$ fulfill the desired property. I will explain this answer a little bit: If $M$ is any square matrix, you can think of $F[M]$ as the ring of all polynomial expressions in $M$. Note that, if $M$ has a minimal polynomial of degree $d$, every such ...

0

FTT transforms for multiplications are only useful for very big numbers. The complexity hides a big constant factor. If you have to write algorithms for cryptography involving huge numbers, use it. But for small integers, you will waste time

2

Note that you have found the three elementary symmetric polynomials in the variables $a,b,c$, which determine a cubic polynomial whose roots are $a,b,c$. So $a,b,c$ are the solutions to \begin{equation*} \begin{aligned} &\mathrel{\phantom{=}} x^3 - (a+b+c)x^2 + (ab+bc+ac)x - abc \\ &= x^3 - 12x^2 + 47x - 60 \\ &= (x - 3)(x - 4)(x - 5). ...

2

You can, however, write any parabola $ax^2 + bx + c$ in the form $$a\left(x + \frac{b}{2a}\right)^2 + c - \frac{b^2}{4a},$$ called the vertex form of the parabola, since $$\left(-\frac{b}{2a}, c - \frac{b^2}{4a}\right)$$ is the vertex of the parabola, i.e. the maximum (if $a < 0$) or minimum (if $a > 0$) point of the parabola

1

I don’t see how to do it without any listing at all. I would do it this way, but I’d be using a fairly primitive symbolic-computation package to help me. First, I’d work over $k=\Bbb F_9$, and find an element $z\in S=\Bbb F_{81}\setminus\Bbb F_9$. Then I would do a listing of the elements of $S$, namely all $a+bz$ with $a,b\in k$ but $a\ne0$. Then I’d find ...

1

You may suppose that $P$ and $Q$ have no commun divisors in $K[x]$. Your equality $P(x)Q(-x)=P(-x)Q(x)$ show then that as $P(x)$ divide $P(-x)Q(x)$ and is prime to $Q(x)$, $P(x)$ must divide $P(-x)$. As they have the same degree, there exists $c$ such that $P(-x)=cP(x)$. Replacing $x$ by $-x$, we get $c^2=1$, hence $c=1$ or $c=-1$ and we are done.

1

Take $Q = x + x^2$, $P = xQ$. Then $F = x$, and $F$ is odd, but $Q$ isn't odd or even.

2

If your polynomial has only one real root $r$ and no imaginary roots, then that means that the multiplicity of $r$ is $n$. Therefore, we can write : $$c_n x^n + c_{n-1} x^{n-1} + \cdots + c_1 x + c_0 = c_n (x - r)^{n}$$ By expanding the RHS (using the binomial theorem), we find that the coefficient of $x^{n-1}$ is $- c_n \cdot n \cdot r$. Then, by ...

0

We need a minimal polynomial with degree equal to the dimension of $V$ to apply cyclic vector theorem. So if you take $E=V$ you need a polynomal of degree $[E:F]$ and the only good choice for this polynomial I see is the minimal polynomial of $\alpha$. What you can do is to view the cyclic vector theorem as special case of the primitive element theorem. ...

2

Hint: As $p(\alpha)=0 \implies p(p(\alpha)+\alpha)=0$, we always have $p(x)$ as a factor.

2

By the structure theorem of finitely generated modules over a principal ideal domain, a submodule of $M^2$ must be free of rank at most $2$.

2

I don't think there's any way to avoid doing a fairly large amount of listing (given that the final answer you are looking for is a list of $18$ degree $4$ polynomials...). However, you at least can avoid going through all $81$ monic polynomials one by one and trying to factor them all. The following is one strategy you might follow. If a polynomial ...

3

Inspired by this question of mine, we can approximate the solution "quite" easily using Padé approximants. Let the equation be $$\frac{(1+i)^{13}\cdot i}{(1+i)^{13}-1}\cdot \frac{1}{1+i}-r=0$$ Building the simplest $[1,1]$ Padé approximant around $i=0$, we have $$0=\frac{\frac{2}{39} i (26 r+7)+\frac{1}{13} (1-13 r)}{1-\frac{4 i}{3}}$$ Canceling the ...

1

(Nothing original here) $\sum_{k=a}^{n+a}(-1)^{k-a}k^{b}{n \choose k - a} =\sum_{k=0}^{n}(-1)^{k}(k+a)^{b}{n \choose k}$. This is the $n$-th difference of $f(x) =(x+a)^b$, which is a polynomial of degree $b$. Since the first difference ($f(x)-f(x-1)$) of a polynomial of degree $b$ is a polynomial of degree $b-1$, the $n$-th difference, up to $n = b$, is ...

1

Proof by induction over the degree of the function $n$ is probably the easiest way. Base case when $n=0$, $f(x)=c$ so $f(x)-f(x-1)=c-c=0$ takes only 1 step. Now assumes the result holds for all $f(x)$ of degree $n$. We look at the $n+1$ degree case. Let $f(x)=a_{n+1}x^{n+1}+a_nx^n+...+a_1x+a_0$. Then $$g(x)=f(x)-f(x-1)=a_{n+1}(x^{n+1}-(x-1)^{n+1})+h(x)$$ ...

0

Hint: $f_n(x)=\frac{x}{1+nx^2}\le \frac{x}{nx^2}=\frac{1}{nx}$, $x\not= 0$

1

Iterate: $$x_{n+1}=\frac {-c}{a(x_n)^{12}}-\frac{b}{a}$$ $$x_0=\frac {-b}{a}$$ I will try to edit my answer and put bounds that indicate the rate of convergence

0

The problem of finding an algebraic formula is closed since longtime ago: it is not possible. Therefore you need an approximation and the particular values of the coefficients are very important in each case of course. For your equation $8.3838x^{13}-9.3838x^{12}+1=0$ you have the equivalent one $$8.3838(x^{13}-x^{12})=x^{12}-1\iff ... 3 I think I ought to start by pointing out that we find all primitive solutions to x^2 + y^2 - z^2 = 0 (Pythagorean triples) using a parametrization in two variables, x=r^2 - s^2, y = 2 r s, z = r^2 + s^2, with \gcd(r,s) = 1 and one odd, one even. This type of 2 variable outcome is evident when, in the original question, d=e=0, so that we have a x^2 ... -4 HINT: for the term \frac{(1+i)^{13}\cdot i}{(1+i)^{13}-1}\cdot \frac{1}{1+i} we get$${\frac {4096}{8321}}+{\frac {4160\,i}{8321}}$$with i=x we get after some calculations$$44671.69000\, \left( {x}^{2}+ 0.5609606087\,x+ 0.3183225370 \right) \left( {x}^{2}+ 1.262369067\,x+ 0.9572725355 \right) \times\\ \left( {x}^{2}+ 2.082793011\,x+ 1.744814484 ...

3

$5$ is prime, thus by Fermat's theorem $a^4 \equiv 1 (\text{mod } 5)$ for all $a \in \mathbb{Z}_5$, such that $a \neq 0$. Therefore you can reduce any power of $x$ to its remainder after the division by $4$. For the first coefficient we have then $$2x^{219} = 2\cdot (x^4)^{54}\cdot x^3 = 2\cdot 1 \cdot x^3.$$ And as pointed out by André Nicolas, solving ...

Top 50 recent answers are included