# Tag Info

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 ...

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}\\ &= ... 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)\ .$$7 While the method already posted is nice, can't resist this hint - from the first and last equation, by the inequality of power means$$\sqrt[2014]\frac{x^{2014}+y^{2014}+z^{2014}}3 \geqslant \sqrt[2012]\frac{x^{2012}+y^{2012}+z^{2012}}3$$with equality iff |x|=|y|=|z|=1. Now using the second equation, the unique answer is obvious. P.S. This obviously ... 6 This is open: http://www.openproblemgarden.org/op/quartic_rationally_derived_polynomials From looking at a few of the relevant papers, this appears to be an extremely difficult problem. For example, the full classification of cubics with this property required the theory of elliptic curves. Note that if we allow two of the roots of p to coincide, ... 6 You can write the expression as$$ \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 - ...

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 $\cos(\theta)$.

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:

3

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

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 ...

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

3

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" ...

2

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 ...

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 ...

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 ...

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 ...

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.

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.

1

Hint: $p(x)$ is in fact a cubic. The system of equations you have to solve is: $$\begin{eqnarray}p'(1) = 0&\Rightarrow&3a& + &2b& + &c& & &=& 0\\p(1) = 6&\Rightarrow&a&+&b&+&c&+&d &=& 6\\p'(3) = 0&\Rightarrow&27a&+&6b&+&c& & &=&0\\ p'(3) ... 1 From the last equation, we get a+b++c=3Z. Substituting in the first two equations, and simplifying a bit, we get$$3XZ=a+\frac{1}{2}b,\tag{1}$$and$$3YZ=b\frac{\sqrt{3}}{2}.\tag{2}$$The second equation gives b=\frac{(2)(3YZ)}{\sqrt{3}}, which simplifies to b=(2YZ)\sqrt{3}. Now from the first equation we get a=3XZ-YZ\sqrt{3}. Finally, since ... 1 Over the real numbers, every polynomial can be factored into quadratic polynomials. (This is a consequence of the mean value theorem from calculus.) But in other fields, some polynomials of high degree cannot be factored at all. For example, the polynomial x^p - x + a cannot be broken into factors of smaller degree when the coefficients are taken to lie in ... 1 This is an illustration of using DFT to find the product of two polynomials. It is based on the fact that Fourier transform (discrete or continuous) converts convolution into multiplication. The product of two polynomials amounts to convolution of their coefficients:$$ \left(\sum_{j=0}^n a_j x^j\right)\left(\sum_{k=0}^n b_k x^k\right) = \sum_{m=0}^{2n} ...

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 ...

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.

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 ...

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 ...

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.

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 ...

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