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1

Hint $\,\ (x\!-\!a)^2\mid P(x)\,\color{#C00}\Rightarrow\, \color{#0a0}{x\!-\!a}\mid P'(x)\,\Rightarrow\, 0 = P'(a) = \frac{a^2}2+a+1,\,$ contra $\,a\in R\ \ $ QED ${\rm\color{#c00}{Indeed}}\,\ P = (x\!-\!a)^2 Q\,\Rightarrow\, P'\! = (\color{#0a0}{x\!-\!a})^2 Q' + 2(\color{#0a0}{x\!-\!a}) Q\,\Rightarrow\,P'(a) = 0$ Remark $\ $ It generalizes as follows. Let ...


6

A nicer trick to determine such $a$ (if it exists) is to note that $a$ is root both of $P$ and of $P'(x)=(2(x-a)Q(x)+(x-a)^2Q'(x)$. Now $P'(x)= \frac{x^2}2+x+1$, so $a$ must also be a root of $P(x)-P'(x)=\frac {x^3}6$, in other words, we must have $a=0$. But obviously $0$ isn't even a root of $P$. In other words: No such $a$ exists (and that is also the ...


0

John McGee's comment is good. Another way to do this is to define a set $S = \{f(x_1, x_2, ..., x_n) \in F[x_1, x_2, ..., x_n]: \deg(f) \leq D\}$. Or invent your own notation so long as you have clearly defined it - this happens all the time.


1

f(x)=$\frac{1}{x}$ on [1,2] has maximum at x=1, but f(x) isn't a polynomial function.


1

This is an illustration of using DFT to find the product of two polynomials. It is based on the fact that Fourier transform (discrete or continuous) converts convolution into multiplication. The product of two polynomials amounts to convolution of their coefficients: $$ \left(\sum_{j=0}^n a_j x^j\right)\left(\sum_{k=0}^n b_k x^k\right) = \sum_{m=0}^{2n} ...


0

Say the answers to this question have already convinced you, and you're thinking, "Okay, real numbers aren't enough for me". Your first instinct will probably be to look at the following fields: $\mathbb{Q}$, $\mathbb{C}$, and $\mathbb{F}_{p^n}$. But even that won't tell you the full story. Here's a phenomenon that can't occur in these fields: For any of ...


0

In $\mathbb Z_7$ - the field with three elements $0,1,2,3,4,5,6$ and addition and multiplication defined modulo $7$ we have $$x^7-x=x(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)$$which is a factorisation you don't get in the real numbers, $[x(x+1)(x-1)(x^2+x+1)(x^2-x+1)]$. You get a similar factorisation for any prime instead of $7$. One of the reasons for exploring ...


1

Over the real numbers, every polynomial can be factored into quadratic polynomials. (This is a consequence of the mean value theorem from calculus.) But in other fields, some polynomials of high degree cannot be factored at all. For example, the polynomial $x^p - x + a$ cannot be broken into factors of smaller degree when the coefficients are taken to lie in ...


0

The first misconception you get if you always think of real numbers is that there is always infinitely many polynomial of degree $k$ (for some $k\in \mathbb N$). If the coefficients are from a finite field (like $\mathbb Z/2\mathbb Z$), you have finitely many polynomials of degree $k$. Such polynomials on finite fields are used in many areas, especially ...


1

Your proof seems to work, but it looks incomplete because you need to use more explicitly the fact that the ring of polynomials is an UFD. The condition that the resultant $\mathrm{Res}(f,g)=0$ means that there are some polynomials $p,q$ such that $\deg p<\deg f$, $\deg q<\deg q$ and $fq=gp$ (so $p/q=f/g$). In a UFD this is equivalent to having a ...


2

I like user157227's basic line of attack which uses de Moivre's formula $\cos n\theta + i\sin n\theta = (\cos \theta + i\sin \theta)^n, \tag{1}$ which itself may be seen as a consequence of Euler's formula $e^{i\phi} = \cos \phi + i\sin \phi \tag{2}$ by choosing $\phi = n\theta$ and remembereing that $e^{in\theta} = (e^{i\theta})^n$; the only improvement ...


7

Induction: if $\cos n\theta=P_n(\cos\theta)$ then $$\cos(n+1)\theta+\cos(n-1)\theta=2\cos n\theta\cos\theta$$ gives $$P_{n+1}(x)=2xP_n(x)-P_{n-1}(x)\ .$$


4

$$\cos(n\theta) = \Re e^{ni \theta} = \Re(\cos(\theta)+i\sin(\theta))^n = Q(\cos(\theta),\sin^2(\theta)) = P(\cos(\theta))$$ $Q$ is a polynomial in $\cos(\theta)$ and $\sin^2(\theta)$ and $P$ is just a polynomial in $\cos(\theta)$.


0

The same idea as the linked solution appears to work: We claim that the solutions are exactly those in the sequence $x, x^n+1,(x^n+1)^n+1,\ldots$ First, let $\omega$ be an $n$-th root of unity. Substituting $x\mapsto \omega x$, we find that $P(\omega x) = \omega^k P(x)$ for some $k<n$, so $P$ is $x^k$ times a polynomial in $x^n$. In other words, there ...


7

While the method already posted is nice, can't resist this hint - from the first and last equation, by the inequality of power means $$\sqrt[2014]\frac{x^{2014}+y^{2014}+z^{2014}}3 \geqslant \sqrt[2012]\frac{x^{2012}+y^{2012}+z^{2012}}3$$ with equality iff $|x|=|y|=|z|=1$. Now using the second equation, the unique answer is obvious. P.S. This obviously ...


23

Let $e_1$, $e_2$ and $e_3$ be LHS of the the first, second and third equations respectivelly, then: $$e_3 -2e_2 + e_1 = \\ x^{2012}(x^2 - 2x + 1) + y^{2012}(y^2-2y+1) + z^{2012}(z^2-2z + 1) = 0 \Longrightarrow \\\Longrightarrow x^{2012}(x-1)^2 + y^{2012}(y-1)^2 + z^{2012}(z-1)^2 = 0 $$ Since all the sumands are product of even powers, they cannot be ...


3

Have you considered using the properties of the Laplace transform to simplify calculation? The one's I'm thinking of are: $$\mathcal{L}(e^{at}f(t))=F(s-a)$$ and, $$\mathcal{L}\left(\frac{d^n}{dt^n}f(t)\right)=s^nF(s)-\sum_{i=1}^ns^{i-1}f^{(n-i)}(0)$$ where $F(s)=\mathcal{L}(f(t))$ and $f^{(n)}$ is the n-th derivative of $f$. The $\frac{1}{n!}$ is a ...


1

Looking at the equation modulo 2 shows that $y$ is odd. Then looking at it modulo 4 shows that $x$ is odd as well. (The square of an odd number is always $1\bmod4$.) A brute force search shows that the solutions $(x,y)$ are (335,1343), (593,1151), (965,407) and (1007,1). I fail to see enough structure in the answer to be able to explain it.


0

Suppose that $a^k \in (fg)$, so $a^k \in (f)$ and $a^k \in (g)$. Since $(f)$ and $(g)$ are prime ideals, we have $a\in (f)\cap (g) = (fg)$. In general, showing that an ideal is radical is not an easy problem. One should expect to approach a general problem using Gröbner bases.


4

Using complex exponential relations, $$e^{i\theta} = \cos\theta + i \sin\theta \qquad \sin\theta = \frac{1}{2i}(e^{i\theta}-e^{-i\theta}) \qquad \cos\theta = \frac{1}{2}(e^{i\theta}+e^{-i\theta})$$ we see that sine-cosine polynomial of degree $p$ can be written as a linear combination of complex exponentials "as large as" $e^{ip\theta}$ and "as small as" ...


3

To show this is true for any polynomial of degree $n+m$, it's sufficient to show that this is the case for an arbitrary term $c^ns^m$ - since every term must be of this form for some $n,m$ - and to check that adding such terms together won't affect the derivative. Note that $$\frac{d}{dx}(c^n s^m)=mc^{n+1}s^{m-1}-nc^{n-1}s^{m+1}$$ This gives another degree ...


0

As you say if $I$ was maximal then $f$ must be irreducible. Conversely notice that $\mathbb{Z}[x]/(p,f(x))\cong\mathbb{Z}_p[x]/(f(x))$. At this point if you know about finite fields you will be done. Else since $f$ is irreducible every non-zero element of the quotient comes from an element $g\in\mathbb{Z}_p[x]$ that is coprime to $f$. Thus by the euclidean ...


0

The general sextic polynomial cannot be solved in radicals, and I don't see any reason why this family should be special in this way. So I suspect the vast majority of these polynomials do not have a closed-form solution in the sense of being able to write them down with radicals and arithmetic operations. Of course there are plenty of numerical methods for ...


1

Hint: $p(x)$ is in fact a cubic. The system of equations you have to solve is: $$\begin{eqnarray}p'(1) = 0&\Rightarrow&3a& + &2b& + &c& & &=& 0\\p(1) = 6&\Rightarrow&a&+&b&+&c&+&d &=& 6\\p'(3) = 0&\Rightarrow&27a&+&6b&+&c& & &=&0\\ p'(3) ...


2

As an alternative to finding the coefficients of $p$ explicitly, start with the "standard" zig-zagging cubic $$ f(x) = 2x^3 - 3x^2 $$ which has a maximum at $(0,0)$ and a minimum at $(1,-1)$. Now just scale and translate to get $p(x) = 6 + 4f(\frac{x-1}2)$. Then you don't even need to simplify to find $p'(0)$, just the chain rule.


2

$p'(1)=0, p'(3)=0, p(1)=6, p(3)=2$. We must have $\partial p' \ge 2$, so try $p'(x) = c(x-1)(x-3)$. This gives $p(x) = p(1)+c\int_1^x p'(t)dt = 6+{c \over 3} (x-4)(x-1)^2$. Setting $p(3) = 2$ gives $c=3$ and so we have $p(x) = 6+(x-4)(x-1)^2$. Computing $p'(0)$ gives 9.


3

Basically you need to setup system of equations and solve the coefficients of $p(x)$ Since $p(x)$ has two turning points, the minimum possible degree is $3$ Say $p(x) = ax^3+bx^2+cx+d$ $\implies p'(x) = 3ax^2+2bx+c$ from the problem we have : $p(1) = 6$ $p(3) = 2$ $p'(1) = 0$ $p'(3)=0$ four equations and four unknowns - can be easily solved


2

Let $G$ be a function define as follow, $$ \begin{array}{ccccc} f & : & \Bbb{C} & \to & \Bbb{R} \\ & & z & \mapsto & \lvert \log(\max(1,\lvert P(z)\rvert))-d\log(\max(1,\lvert z)\rvert ))\rvert \\ \end{array}$$ As $z\mapsto \max(1,\lvert P(z)\rvert)$ and $z\mapsto \max(1,\lvert z\rvert)$ are continuous and ...


0

I've implemented an $O\left( n^2\right)$ algorithm at http://jsfiddle.net/xcrqoxk5/ I believe it's the recursive solution Shooting Squirrel is talking about. Check to see if you understand how it works. Suppose you want to multiply a polynomial $b_4 x^4 + b_3 x^3 + b_2 x^2 + b_1 x + b_0$ with the polynomial $ax + 1$. By a standard multiplication, we ...


0

Exponential solution Let $b_{n}x^n+...b_{1}x+b_{0}$ be the polynomial in discussion. You first find $b_{n}$(what is it?), then using Vieta's formulas you can easily derive all the coefficients. You can use it since you get all the zeroes to the polynomial(make sure to make the additions modulo $100003$). Hint(for solution in $O(n^2)$): Think recursively. ...


0

All the answers are good and provide the method or algorithm by which the quadratic formula is derived. But i think the question has one more part in it. Why for quadratics and not for sth else? So by methods of Galois Theory (mention it just for reference), the thing that makes the quadratic solvable by such formulas (called radical formulas) is that due ...


4

Since it seems the PDF linked by Macavity is not accessible to everyone, I reproduce it here :


0

If you are using a point-value representation of polynomials, then if you have something that isn't a polynomial (e.g. because it is a rational function with nonconstant denominator), then it can't possibly be represented correctly. However, to go beyond the point your book is trying to make, you can do point-value representation of rational functions ...


1

The answer is yes. I've not worked through the technical difficulties in making a direct proof, so I'll cheat and suggest that you construct the companion matrix, and use the fact that you can write the eigenvalues as continuous functions. (I don't have a better reference off hand) All that remains is to show that if you have $n$ continuous functions ...


1

After substituting $u=x^e$ you can apply the quartic formula, or you can use the reduction techniques that are used to derive it. The result is horrible, though.


0

I solved the question (for those who wonder how this comes out : ) We have: $$f_1\left(x\right)=a_1x^2+b_1x+c_1=0$$ $$f_2\left(x\right)=a_2x^2+b_2x+c_2=0$$ Multiply $f_1\left(x\right)$ by $a_2$ and $f_2\left(x\right)$ $a_1$ then use elimination method to obtain one part of the answer. Then mulitply $f_1\left(x\right)$ by $b_2$ and $f_2\left(x\right)$ ...


1

$F(x)$ is just the original piecewise linear function you are given. The expression for $F(x)$ comes from adjusting the middle part of the original function so that it becomes equal to $F$ throughout its range. So we start with $3x+2$. When $x\lt -1$ we need this to become $-1$, and $-1-(3x+2)=-3x-3$. If $x\ge -1$ we want to leave the function unchanged, by ...


2

Note that your matrix is the sum of the matrix $T$ with all entries equal to $t$ and the matrix $-(1+t)I$. Therefore the determinant you are asking about is the value at $X=-(1+t)$ of the characteristic polynomial $\chi_{-T}$ of $-T$. Since $T$ has rank (at most) $1$, its eigenspace for eigenvalue $0$ has dimension $n-1$, so the characteristic polynomial of ...


0

The vector $v = (\alpha^2, \alpha, 1)$ is orthogonal to both $w_1=(a_1,b_1,c_1)$ and $w_2=(a_2,b_2,c_2)$ and assuming that $f_1$ and $f_2$ are independent this implies that $v$ is a multiple of the cross product $w_1 \times w_2$. That's precisely what these equalities express.


0

Here is another characterization of the result. For convenience, I will take the matrix size as $n+1$ rather than $n$. First, note that we may factor $t$ from each of the $n+1$ rows, and so the determinant may be written as $$ \det{[t(M-t^{-1} I_{n+1})]} =t^{n+1} \det(M-t^{-1} I_{n+1}) $$ where $(M)_{ij}=1-\delta_{ij}$ for $1\leq i,j\leq n+1$. Next, ...


10

Using elementary operations instead of induction is key. $$\begin{align} &\begin{vmatrix} -1 & t & t & \dots & t\\ t & -1 & t & \dots & t\\ t & t & -1 & \dots & t\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ t & t & t & \dots& -1 \end{vmatrix}\\ &= ...


6

You can write the expression as $$ \det(t C - (t+1)I)$$ where $C = \mathbf{1}\mathbf{1}^T$ is the matrix of all $1$'s, formed by the column of ones times its transpose. Using the identity $\det(I+cr) = 1+rc$, you can first factor out $(t+1)$: $$ \det(t C - (t+1)I) = (-1)^n(t+1)^n \det\left(I - \frac{t}{t+1} \mathbf{1}\right) = (-1)^n(t+1)^n \left(1 - ...


0

Here is the solution to the equations $a = Z \left(3 X-\sqrt{3} Y\right)$ $b = 2 \sqrt{3} Y Z$ $c = Z \left(-\left(3 X+\sqrt{3} Y-3\right)\right)$


1

From the last equation, we get $a+b++c=3Z$. Substituting in the first two equations, and simplifying a bit, we get $$3XZ=a+\frac{1}{2}b,\tag{1}$$ and $$3YZ=b\frac{\sqrt{3}}{2}.\tag{2}$$ The second equation gives $b=\frac{(2)(3YZ)}{\sqrt{3}}$, which simplifies to $b=(2YZ)\sqrt{3}$. Now from the first equation we get $a=3XZ-YZ\sqrt{3}$. Finally, since ...


4

Hint: The expression is equal to zero when any two of the parameters $a,b,c,d$ are equal. Can you use this, the remainder theorem and the degrees of the expression to get a result? Remainder Theorem: If $f(x)$ is a polynomial and $f(a) = 0$, then $x-a \mid f$. Bigger Hint:


1

It's an old result of Adelman and Manders "NP-complete decision problems for quadratic polynomials" (1976) that solving the equation $ax^2 + by = c$ for $x,y$ in the natural numbers is NP-complete. The key word here is natural, because this requires not only that $x^2 \equiv c \pmod {b}$ but also that $0 < x \le \sqrt{c/a}$. It's easy to solve the ...


6

This is open: http://www.openproblemgarden.org/op/quartic_rationally_derived_polynomials From looking at a few of the relevant papers, this appears to be an extremely difficult problem. For example, the full classification of cubics with this property required the theory of elliptic curves. Note that if we allow two of the roots of $p$ to coincide, ...


0

The polynomial is of degree $2^{n-1}-n$ and the value of the leading coefficient is $(n-1)!$. This is problem S311 of the current issue of Mathematical Reflections: https://www.awesomemath.org/assets/PDFs/MR4_2014_update.pdf


0

We have $C_{A}(x) = (x-1)^{5}(x-2)^{2}$ and $m_{A}(x) = (x-1)^{3}(x-2)^{2}$. Write $B = A^{2} + A + I$. Then $Av = \lambda v \implies Bv = (\lambda^{2} + \lambda + 1)v$. This equation tells us what the eigenvalues of $B$ look like, but also more: $ \ker(A-\lambda I) \subset \ker \left (B - (\lambda^{2} + \lambda + 1)I\right)$ and the reverse inclusion also ...


2

Here is a purely symbolic approach via the Snake Oil method. (See generatingfunctionology for more details). To make things simpler to read, note that $n=2m+d$ for some integer $m$ since $n$ and $d$ have the same parity. Hence the identity to be proven is equivalent to $$T_n(d)=\sum\limits_{k}{k\choose m}{2m+d\choose 2k}=2^d \frac{2m+d}{2m+2d}{m+d\choose d} ...



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