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21

There exists $\alpha \in\left[\dfrac{16}{3},\dfrac{17}{3}\right]$ with the required property. To see this, we will construct an interval sequence $$\left[\dfrac{16}{3},\dfrac{17}{3}\right]=[\alpha_{1},\beta_{1}]\supset [\alpha_{2},\beta_{2}]\supset\cdots\supset[\alpha_{n},\beta_{n}],$$ where $\alpha_{n}$ and $\beta_{n}$ are such that ...


18

Let $I = \displaystyle\int_{0}^{1}f(x)\,dx$. Substituting $x = \sin \theta$ yields $I = \displaystyle\int_{0}^{\pi/2}f(\sin \theta)\cos\theta\,d\theta$. Substituting $x = \cos \theta$ yields $I = \displaystyle\int_{0}^{\pi/2}f(\cos \theta)\sin\theta\,d\theta$. Hence, $I = \dfrac{1}{2}\displaystyle\int_{0}^{\pi/2}\left[f(\sin ...


8

Let $\phi=\frac{1+\sqrt{5}}{2}$. Then $\phi^2=\phi+1$. By induction, we have $F_n > \phi^{n-2}$ for all $n\ge 1$, and so $$ \sum_{n=1}^{N} \frac{n}{F_n} < \sum_{n=1}^{\infty} \frac{n}{F_n} < \sum_{n=1}^{\infty} \frac{n}{\phi^{n-2}} = \sum_{n=1}^{\infty} \frac{n\phi^2}{\phi^{n}} = \phi^2 \sum_{n=0}^{\infty} \frac{n}{\phi^{n}} = ...


7

Here's one rough approach which works for $n$ sufficiently large. The idea is that for large $n$, we have that $$ F_n \approx \phi^n/\sqrt{5} $$ since the conjugate root is less than one, and so it tends to zero. So consider now the generating function $$ G(q) = \sum_{n=0}^\infty \frac{q^n}{F_n} $$ We note that your sum approaches ...


7

\begin{align} \sum_{n=p+1}^{m}{\frac{a_n}{s_n^2}}&=\sum_{n=p+1}^{m}{\frac{s_n-s_{n-1}}{s_n^2}} \\ &=\sum_{n=p+1}^{m}\frac{s_n-s_{n-1}}{s_{n-1}s_n}\frac{s_{n-1}}{s_n} \\ &=\sum_{n=p+1}^{m}\frac{s_{n-1}}{s_n}\left(\frac{1}{s_{n-1}}-\frac{1}{s_n}\right) \\ &< \sum_{n=p+1}^{m}\left(\frac{1}{s_{n-1}}-\frac{1}{s_n}\right) \hspace{8 mm} ...


6

This seems to be a difficult problem. In the following I propose a shape that has area strictly $<{\pi\over2}$ and cannot be placed on the integer lattice without hitting a lattice point. In the figure the lattice is turned by $45^\circ$, whence $r={1\over\sqrt{2}}$. The offset $x$ is a small parameter. One computes $$a^2=r^2+x^2, \quad b^2=a^2-(r-x)^2 ...


5

\begin{eqnarray}ab+bc+ca=\frac{1}{2}((a+b+c)^2-(a^2+b^2+c^2))=47\end{eqnarray} \begin{align*}a^3+b^3+c^3-3abc&=(a+b+c)(a^2+b^2+c^2-(ab+bc+ca))\\ abc&=44\end{align*} So $a, b, c$ are roots of the polynomial $x^3-12x^2+47x-44=0$.


5

Lemma $$F_{n}\ge\dfrac{n(n+1)(n+2)}{42},n\ge 5$$ proof:use induction, since $$\dfrac{1}{42}[k(k+1)(k+2)+(k+1)(k+2)(k+3)]=\dfrac{1}{42}(k+1)(k+2)(2k+3)$$ and $$(k+1)(2k+3)\ge (k+3)(k+4)$$ so $$\dfrac{n}{F_{n}}\le \dfrac{42}{(n+1)(n+2)}$$ so $$\sum_{k=1}^{n}\dfrac{k}{F_{k}}\le 1+2+\dfrac{3}{2}+\dfrac{4}{3}+42\left(\dfrac{1}{6}-\dfrac{1}{n+2}\right)<13$$


5

Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2$ and $abc+abd+acd+bcd=4w^3$. Hence, $16u^2-12v^2=1$ and our inequality is equivalent to $3v^6-4uv^2w^3+w^6\geq0$. By Roll's theorem there are $x>0$, $y>0$ and $z>0$, for which $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$. After this substitution we need to prove that ...


5

If $\overrightarrow{a}=(a_1,a_2,a_3)$,and $a_1,a_2$ are not both zero, then take $\overrightarrow{b}=(-a_2,a_1,0)$. If $a_1=a_2=0$, take $(1,0,0)$. I will leave it to you to figure out why this works.


5

Notice that the original equation we are given tells us that $f$ and $f'$ both belong to the same class of differentiable functions. This means we can restrict ourselves to $C^\infty$ functions, and it makes sense to take derivatives. Hence $$ f'(x) - f''(\pi-x) = 0.$$ The original equation also gives a relation between $f$ and $f'$. By changing variables to ...


5

If $x=0$ and $\sin y$ is transcendental - say $\sin y = \pi/4$ and $\cos y>0$ - then $\sin(x+z)=\sin z$ being rational means $\cos z$ is algebraic, so $$\sin(y+z)=\sin y\cos z + \sqrt{1-\sin^2 y} \sin z$$ being rational means $\sin y$ is algebraic. So there is no such $z$ in this case. Indeed, for any $x$ there are at only countably many $y$ so that ...


4

Hint I will show it for two functions and then the idea can be generalized. Suppose $\not\exists \, x_0 \in [a,b]$ such that $f_1(x_0)=0=f_2(x_0)$, i.e they do not both vanish at the same point then consider the function $h(x)=(f_1(x))^2+(f_2(x))^2$ in the ideal $I$. Now corresponding to this $h$, we will have the function ...


4

$n+3=x^3$, $n=x^3-3$. $n^2+3n+3=(x^3-3)^2+3(x^3-3)+3=x^6-3x^3+3=(x^2-(1/x))^3+x^{-3}>(x^2-1)^3$, but also $x^6-3x^3+3<(x^2)^3$, so it's not a cube.


4

Here's an idea: Suppose $f$ is positive and continuous on $[1,\infty).$ The integral analogue of our problem is: If $\int_1^\infty f = \infty,$ and $F(x) = \int_1^x f,$ then $$\int_2^\infty \frac{f(x)}{(F(x))^2}\,dx < \infty.$$ This is simple to verify, since $f= F'.$ That strongly suggests $\sum (a_n/s_n^2) < \infty$ in the series case.


4

You are asking for so called Brocard's point of the triangle. One of constructions can be found on Wikipedia: http://en.m.wikipedia.org/wiki/Brocard_points


4

Assume, $a\ne 0$ because the orthogonality does not make sense for $a=0$. It is true for every $n$. Choose some two components of the vector a, of which one is not $0$. Swap them and change one of the signs. Set the other entries $0$. Then, you have found a vector $b$ orthogonal to $a$ with integer coefficients.


4

$$x+y-12=z \Rightarrow x^2+y^2-(x+y-12)^2=12$$ This leads to $$xy-12x-12y+78=0$$ or $$(x-12)(y-12)=66$$ Now check all the possible ways of writing $66$ as a product of 2 integers.


4

If $g$ is a polynomial of degree $m$ with leading coefficient $a$, i.e. $g(x) = a x^m + \ldots$ where $\ldots$ consists of terms of lower order, and $m \ge 1$ then $f(g(x)) = 2013 a x^m + \ldots$ while $g(f(x)) = 2013^m a x^m + \ldots$, so $f(g(x)) - g(f(x)) = (2013 - 2013^m) a x^m + \ldots$. Thus $f(g(x)) - g(f(x))$ can't be $0$ unless $m \le 1$. If we try ...


4

1. Complex Analysis Technique. Consider the function $$q(z) = i\sqrt{z-a\vphantom{d}}\sqrt{z-b\vphantom{d}}\sqrt{z-c\vphantom{d}}\sqrt{z-d\vphantom{d}}$$ for $z \in \Bbb{C}\setminus([a,b]\cup[c,d])$, where the square root $\sqrt{z} = \exp(\frac{1}{2}\log z)$ means the principal square root. By noting that $$ \lim_{\epsilon \downarrow 0} ...


4

Clearly: $F_n > F_{n-1}$, for $n \geq 3$ Thus, $F_n = F_{n-1} + F_{n-2} > 2*F_{n-2} > 2^2*F_{n-4}>...>2^{(n-3)/2}*F_3 = 2^{(n-1)/2}$ Now, your inequality can be written as: $\sum\limits_{i=3}^n \frac{i}{F_i} < 10$ The $LHS < \sum\limits_{i=3}^n \frac{i}{2^{(i-1)/2}} = 2^{1/2}*\sum\limits_{i=3}^n \frac{i}{2^{i/2}} < ...


4

Claim: $\det(A^{-1}A^\top+I)\geq 2^n$. Proof: Note that $\det(A^{-1}A^\top)=1$. Let $\lambda_1, \cdots, \lambda_n$ be the eigenvalues of $A^{-1}A^\top$. Then $\lambda_1\cdots\lambda_n=1$ and the eigenvalues of $A^{-1}A^\top+I$ are $\lambda_1+1, \cdots, \lambda_n+1$. So \begin{align*} \det(A^{-1}A^\top+I)&=(\lambda_1+1)\cdots(\lambda_n+1)\\ &\geq ...


4

I think there is something missing. It looks like obvious: $$\lim_{x\rightarrow\infty}\frac{f(x)}{x}=\frac{0}{\infty}=0.$$ Here i'm not using other hp.


3

Let $a_1, a_2, ..., a_{1007}$ be distinct positive real numbers. Set $a_{-i}=-a_i$ for i=1,...,1007, and let $a_0 = 0$. Then the sequence $(a_{-1007}, a_{-1006}, ..., a_{-1}, a_0, a_1, ..., a_{1006}, a_{1007})$ can be reduced to the constant sequence $(0,0,...,0)$ in $1007$ moves, where in each move we operate on a different pair $(a_{-i}, a_i) = (-a_i, ...


3

Putting $ x = y = 0 $, we get $ f(0) = 0 $. Putting $ y = 0 $, we have $ f(x^2) = x^4 $. Putting $ z = x^2 $, we have $ f(z) = z^2 $ for all $ z \geq 0 $. $$\therefore f(2015) = 2015^2= 4060225$$


3

Hint : The minimal polynomial of the matrix $A$ must be a divisor of the polynomial $$x^3-x^2-3x+2=(x-2)(x^2+x-1)$$


3

Hint: The product of conjugates elements is an integer (the norm of that element), which is equal to $2$ (see Vi├Ęte's formulae). The norm of this element, setting $\omega=\mathrm e^{\tfrac{2\mathrm i\pi}3}$, is: $$(a + b\sqrt[3]{2} + c\sqrt[3]{4})(a + b\omega\sqrt[3]{2} + c\omega^2\sqrt[3]{4})(a + b\omega^2\sqrt[3]{2} + c\omega\sqrt[3]{4})=2.$$ Thus ...


3

I would use vector notations (based on picture above). Use $\times$ as cross product. To make sure that all areas sum with correct sign - follow clockwise rotation. Then: $$S_{ADCB}*2 = |\vec{CA}\times\vec{CD} + \vec{AC}\times\vec{AB}| = |\vec{CA}\times\vec{CP*3} + \vec{AC}\times\vec{AM*3}|$$ and $$S_{ADCB} = 3 * S_{AMCP}$$ $$S_{MNPQ}*2 = ...


3

For amusement, let us solve the ODE/IVP for the case $y(1) > 2$ the hard way and then discover the solution for the case $y(1) = 2$ as some sort of a limit. Let $z = \sqrt{y-x^2}$, we have $y = x^2 + z^2$ and $$y' = 4\sqrt{y-x^2} \iff (x^2+z^2)' = 4z \iff zz' + x = 2z \iff z' = 2 - \frac{x}{z}$$ Let $u = \frac{x}{z} \iff z = \frac{z}{u}$. We will ...



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