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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 ...
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 computesa^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*} Soa, 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)$$3 For a =[a_1 , a_2 ,...,a_n ] take b=[a_n ,a_n,...,a_n, -(a_1 +a_1 +...+a_{n-1} )]. 3 Hint: The product of conjugates elements is an integer (the norm of that element), which is equal to 2 (see Viète's formulae). The norm of this element, setting \omega=\mathrm e^{\tfrac{2\mathrm i\pi}3}, is:$$(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 casey(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 ...