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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, $\mathcal H$akim.


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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