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8

Suppose $a$ is a root of $x^2+x+1=0$, then we have both $$a+1=-a^2$$ and $$a^3=1$$ Let $f(x)=(x+1)^{2n+1}+x^{n+2}$ then $$f(a)=(-a^2)^{2n+1}+a^{n+2}=-a^{4n+2}+a^{n+2}=-a^{n+2}+a^{n+2}=0$$ Since the two distinct roots of the quadratic are also roots of $f(x)$ we can use the remainder theorem to conclude that the remainder is zero.

7

If $n=0$, then it is trivial. For $n>0$, we have $$(x+1)^{2n+1}+x^{n+2}\\=(x^2+2x+1)(x+1)^{2n-1}+x^{n+2}\\=(x^2+x+1)(x+1)^{2n-1}+x(x+1)^{2n-1}+x^{n+2}\\=(x^2+x+1)(x+1)^{2n-1}+x((x+1)^{2n-1}+x^{n+1})$$. By using induction, suppose $(x+1)^{2n-1}+x^{n+1}$ can be divided by $x^2+x+1$, then, $(x+1)^{2n+1}+x^{n+2}$ also can be divided by $x^2+x+1$.

5

Put $e^{2\pi i/11}=:\omega$ and $e^{i\theta}\omega^n=:z_n$. Then $$\sin\left(\theta+{2n\pi\over 11}\right)={\rm Im}(z_n)={1\over 2i}\bigl(e^{i\theta}\omega^n-e^{-i\theta}\omega^{-n}\bigr)$$ and $$\sin^{14}\left(\theta+{2n\pi\over 11}\right)={-1\over 2^{14}}\sum_{k=0}^{14}(-1)^k{14\choose k}e^{i(14-2k)\theta}\>\omega^{(14-2k)n}\ .$$ Now ...

4

Here's another way of looking at this through a reverse lens: Let's solve $u^3+3u-4=0$ by Cardano's method, putting $u=x+y$. Then $(x+y)^3-3xy(x+y)-(x^3+y^3)=0$ and we require: $$x^3+y^3=4$$ and $$-3xy = 3 \text { so that }xy=x^3y^3=-1$$ Then we note that $x^3$ and $y^3$ are roots of the quadratic $$z^2-4z-1=0$$So that $$z=\frac{4\pm\sqrt{16+4}}{2}=2\pm ... 4 Use this formula (found here, and mentioned recently on MSE here):$$\prod _{k=1}^{n-1}\,\sin \left({\frac {k\pi }{n}} \right)=\frac{n}{2^{n-1}} .$$Let n=13, which gives$$\left(\sin{\frac{\pi}{13}} \cdot \sin{\frac{2\pi}{13}} \cdot \sin{\frac{3\pi}{13}} \cdots \sin{\frac{6\pi}{13}}\right)\left(\sin{\frac{7\pi}{13}} \cdot \sin{\frac{8\pi}{13}} \cdot ...

4

Here is how I would explain the three "bold statements". (i) If $z\in C_r$ then $\vert z^n-f(z)\vert =\vert z^n\vert\, \left\vert 1-\frac{f(z)}{z^n}\right\vert= \left\vert 1-\frac{f(z)}{z^n}\right\vert\, r^n$. Since $f(z)/z^n\to 1$ as $\vert z\vert\to \infty$, you have $\left\vert 1-\frac{f(z)}{z^n}\right\vert\leq\frac 12$ if $z\in C_r$ and $r$ is large ...

4

Assuming $a,b,c>0$, then as you noted the equalities are just cosine laws for $3$ triangles which form a larger triangle with sides $2,\sqrt3,\sqrt7$, because the angles add up to $2\pi$. That triangle is right, because $4+3=7$, so we can find the lengths analytically if we draw it like this: Here we set $A=(0,0)$, $B=\left(0,\sqrt3\right)$ and ...

3

Here is an algebraic solution that yields all the real answers. Let us introduce the complex numbers: $$x=\frac{-b+i a}{\sqrt{3}},\quad y=\frac{2a+c}{4}-i\frac{\sqrt{3}}{4}c.$$ The first two equations are equivalent to the statement: $\vert x\vert=\vert y\vert=1$, and the third equation tells us that $$\vert \sqrt{3} x-2i y\vert^2=\left\vert ... 3 Assume p(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots +a_1x+a_0 with a_n\ne 0. Show that p(x+1)-p(x)=na_nx^{n-1}+(\text{lower terms}). This tells you that p must be of degree n=2 and that a_2=1, so p(x)=x^2+a_1x+a_0. Now you can comfortably compute p(x+1)-p(x) explicitly and compare with 2x+1. 3 Hint \  Evaluated at \,x = -1\, each odd term \,x^{2n+1}\, has value -1 and each even term \,x^{2n}\, has value 1, and there are an equal number of odd and even terms, so they sum to 0\,\, i.e. \,f(-1) = 0. Your cases are sums of terms \,x^{2n}+x^{2n+1} = (1+x)x^{2n}\, so you can explicitly factor out \,x+1\, which shows that \,x = ... 3 Here is a method you can try. Let s_1=a+b+c, s_2=a^2+b^2+c^2, p_2=ab+bc+ac, p_3=abc a, b, c are the roots of the cubic$$0=(x-a)(x-b)(x-c)=x^3-s_1x^2+p_2x-p_3$$We don't know p_2 but can calculate it using s_1^2=s_2+2p_2, and then solve the cubic to find a,b,c. 3 HINT: We have$$\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega+1=0$$Dividing either sides by \omega^3,$$\omega^3+\frac1{\omega^3}+\omega^2+\frac1{\omega^2}+\omega+\frac1\omega+1=0$$... 2 For positive integer n If \sin(2n+1)x=0, (2n+1)x=m\pi\iff x=\frac{m\pi}{2n+1}  where m is any integer From (3) of this, \displaystyle \sin(2n+1)x=2^{2n}s^{2n+1}+\cdots+(2n+1)s=0 where s=\sin\frac{m\pi}{2n+1} So the roots of \displaystyle 2^{2n}s^{2n+1}+\cdots+(2n+1)s=0  are \sin\frac{m\pi}{2n+1}; 0\le m\le2n So the roots of ... 2 Hint: try to check if complex roots of x^2+x+1 are also roots for (x+1)^{2n+1}+x^{n+2}. Also there is another way to prove it. Let see that ((x+1)^2-x)\cdot (x+1)^{2n-1} also divided by x^2+x+1. So we just need to prove that x\cdot (x+1)^{2n-1}+x^{n+2} is divided by x^2+x+1. In this way we can come to prove that (x^k\cdot (x+1)^{2n+1-2\cdot ... 2 If you're not aware of it, you should look into arithmetic circuit complexity [1]; in this framework, your function \Phi(f) is (essentially) the minimum size of an arithmetic formula computing f. Unfortunately, your statement is bound to be false, simply because there are just not enough `small' expressions compared to the number of possible functions ... 2 You can invent your own notation: say r(a,b,c,d,e;n) for the n^\mathrm{th} smallest root of the quintic with coefficients a, b, c, d, e. You may object that this is kind of a cop-out, and you'd be right, but it's also basically what we do with quadratics. We get excited when we find out that we can "solve" x^2 = 2 by writing x = \sqrt 2, but all ... 2 It depends on the function f and I'll assume that the interpolation points are distinct. The closest result I know is the following: Theorem: If f:[0,1]\rightarrow\mathbb{C} is analytic and analytically continuable to a function that is analytic in a region containing the closed "stadium" of radius 1 (consisting of all the points in the complex-plane ... 2 Proof without words. This one shows that$$ax^2+bx+c=a\left(x+\dfrac b{2a}\right)^2+c-\dfrac{b^2}{4a}$$from which the quadratic formula can be easily derived. Credits to LucasVB. I hope this helps. Best wishes, \mathcal Hakim. 2$$\frac{2x^n+x^{n-1}+x^{n-2}+...+x^2+x+5}{x-\frac{1}{2}}=\frac{(x-\frac{1}{2})(2x^{n-1}+2x^{n-2}+\dots+2x^2+2x+2)+6}{x-\frac{1}{2}}=2x^{n-1}+2x^{n-2}+\dots+2x^2+2x+2+\frac{6}{x-\frac{1}{2}}$$So the quotient is: 2x^{n-1}+2x^{n-2}+\dots+2x^2+2x+2 The remainder is: 6 And the sum of coefficients in the quotient is: ... 1 Hint P(x)=Q(x)q(x)+r(x) where r(x)=ax^2+bx+c is a quadratic, and Q(0)=Q(1)=Q'(1)=0 Set x=0 to find r(0)=1 Set x=1 to find r(1)=n+(n-1)+\dots + 1+1 Differentiate with respect to x and set x=1 to obtain 2a+b=n^2+(n-1)^2+\dots +1 This should give you three equations in three unknowns (the coefficients of r(x)). Since you didn't show ... 1 I believe that you are asking the question does LOGSPACE = P, which remains an open question similar to the P vs NP problem. It is known that the use of constant space \subset LOGSPACE \subset PSPACE, where PSPACE uses polynomial space, and it known that LOGSPACE is contained within P. It is generally believed that LOGSPACE \subset P \subset NP ... 1 Hint: If the complex roots of f are z_1 through z_4, then f(x)=a(x-z_1)(x-z_2)(x-z_3)(x-z_4). But since f has no real roots, its complex roots come in conjugate pairs, so the linear factors combine together two and two to form upwards-pointing parabolas.$$ f(x) = a((x-a_1)^2+s)(x-a_2)^2+t) $$for some a_1,a_2\in \mathbb R and s,t> 0. 1 Note that (x^2-1)^2=x^4-2x^2+1 is non-negative. You will find some information about general results here, and this looks as though it may be of interest. 1 If p(0) and p(1) are both odd, then p(x) cannot have any integer roots. If it did, then there would be a solution to p(x)\equiv0 mod 2. But p(0)\equiv p(1)\equiv1\not\equiv0 mod 2. If you're not comfortable with modular arithmetic here, suppose p(x)=(x-r)q(x) with an integer root r. Then -rq(0)=p(0) implies r is odd, but ... 1 You are correct. Just distribute everything as you normally would, then mod out all the coefficients by 13 when you are finished. After just distributing, you should get 12x^3 + 53x^2 + 59x + 11. Then, you should get 12x^3 + x^2 + 7x + 11 after modding out. Of course, there are some tricks to make the distribution a lot easier. For example, ... 1 Any function f from \Bbb Z_p to \Bbb Z_p can be written as a polynomial: by Fermat's little theorem,$$ f(i) = \sum_{t=0}^{p-1} f(t)\big( 1-(i-t)^{p-1} \big). $$In your case, you want the answer to be k when 0\le i\le j and (a+b)c when j<i\le p+1; so the appropriate polynomial is$$ f(i) = k \sum_{t=0}^{j} \big( 1-(i-t)^{p-1} \big) + (a+b)c ...

1

Since your map is polynomial, it cannot be bijective from $\mathbb{R}^2$ to $\mathbb{R}^3$ as it goes from dimension $2$ to dimension $3$. In fact the image needs to be in a variety of dimension $\le 2$, which corresponds to say that the points of the image satisfy some polynomial equation. In your case, you can easily check that $(0,0,1)$ is not in the ...

1

Those polynomial functions can be compressed using geometric sums, $$x^{2n-1}+x^{2n-2}+...+x+1=\frac{x^{2n}-1}{x-1}$$ The numerator allows another factorization $$x^{2n}-1=(x^2-1)(x^{2(n-1)}+x^{2(n-2)}+...+x^2+1)$$ so that the original polynomial is equal to $$(x+1)(x^{2(n-1)}+x^{2(n-2)}+...+x^2+1)$$ This could of course also be seen directly by ...

1

Just a different interpretation: For fixed $x\in [0,1)$ the equation gives a recursion or difference equation for $n\in\Bbb Z$: $$p(x+n+1)-p(x+n)=2n+2x+1.$$ This has the constant sequences as homogeneous solutions and $An^2+Bn$ as type of a particular inhomogeneous solution. Inserting an comparing gives $$A(2n+1)+B=2n+2x+1\iff A=1\land A+B=2x+1\iff ... 1 Because when plugging -1, all the terms with even degrees will sum up to n/2+1, and all the terms with odd degrees will sum up to -n/2-1, adding 1 you get 0, and thus -1 will be a root. (n being the number of terms of those polynomials)$$\underbrace{x^n+x^{n-2}+\cdots+x^2+1}_{\displaystyle \color{white}{\overset{}{\color{black}{\dfrac ...

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