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3

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.


0

This calculations is essentially Euclidean algorithm for gcd in a single variable polynomial ring. Write $\alpha=\sqrt[3]2.$ Given $a,b,c$ consider the polynomial $g(x)= a+bx+cx^2$. We want to find the inverse for $g(\alpha)$. This $g(x)$ is relatively prime to the irreducible polynomial $f(x)=x^3-2$ (the latter being the minimal polynomial of $\alpha$.) ...


2

We know that $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. Making the obvious substituions we get$$a^3+2b^3+4c^3-6abc=(a+b\sqrt[3]{2}+c\sqrt[3]{4})(a^2+b^2\sqrt[3]{4}+2c^2\sqrt[3]{2}-ab\sqrt[3]{2}-ac\sqrt[3]{4}-2bc),$$so$$\frac{1}{a+b\sqrt[3]{2}+c\sqrt[3]{4}}=\frac{(a^2-2bc)+(2c^2-ab)\sqrt[3]{2}+(b^2-ac)\sqrt[3]{4}}{a^3+2b^3+4c^3-6abc}.$$For this ...


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

In case the hints in the comments were not enough, I have written up a complete solution. Hover your mouse over to view it. I wanted to break it up into multiple paragraphs, but I couldn't get the spoiler tags to work that way, so sorry if the formatting is a bit bad. Now, as @Omnomnomnom remarked, $2015$ is not a multiple of $3$.


2

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 $. $$\therefore f(2015) = 2015^2= 4060225$$


-2

$2^3 - 3^{5/8} + 2^2 + 3^{5/8} + 2^1$ goes to $8 - 3^{5/8} + 3^{5/8} + 4 + 2$ goes to $8 + 4 + 2$ gives 14


2

Let $$\displaystyle A= a_1 +\sum_{i=1}^{2012} \frac{a_{i+1}^3}{a_i^2+a_ia_{i+1}+a_{i+1}^2} = a_1 +\sum_{i=1}^{2012} \frac{a_{i+1}^3-a_i^3+a_i^3}{a_i^2+a_ia_{i+1}+a_{i+1}^2}$$ $$\displaystyle = a_1 +\sum_{i=1}^{2012} (a_{i+1} -a_i) +\sum_{i=1}^{2012} \frac{a_i^3}{a_i^2+a_ia_{i+1}+a_{i+1}^2}$$ The first sum is solved using telescoping series property to get ...


0

This quantity is $$\left.\prod_{q=1}^n \left(1+\frac{x}{q}\right)\right|_{x=1}$$ which is $$\prod_{q=1}^n \left(1+\frac{1}{q}\right) = \prod_{q=1}^n \frac{q+1}{q} = n+1$$ by telescoping cancellation.


0

Sorry i was being silly calculating $\mathbb{E}(X)=\sum_{k} \mathbb{P}(X=k)$ rather than $\mathbb{E}(X)=\sum_{k} k\mathbb{P}(X=k)$ i blame it on a long day!! I want to just run through the rest of their solution as they miss a lot of detail, they don't explain why the expectation divided by $65$ gives the answer. Here is my extended solution which aims to ...


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 ...


0

Obviously, the solution of $y^{\prime}=4\sqrt{y-x^2}$ with condition $y(1)=2$ is : $$y(x)=2x^2$$ I let find $y(3)$. What is more interresting (but outside the scope of the question) is how to solve the ODE for the general solution: Note : the case $C=0$ corresponds to the particular solution $y=2x^2$ . The parametric form is not valid in this case ...


1

Since $$-2\lt x\lt 6\Rightarrow 0\le x^2\lt 6^2$$ and $$-4\lt y\lt -2\Rightarrow (-2)^2\lt y^2\lt (-4)^2\Rightarrow -(-4)^2\lt -y^2\lt -(-2)^2,$$ one has $$-16=0-(-4)^2\lt x^2-y^2\lt 6^2-(-2)^2=32.$$


-5

Mathematica gives the following implicit solution $$\frac{x \sqrt{y(x)-x^2}}{2 x^2-y(x)}+\frac{y(x)}{2 \left(2 x^2-y(x)\right)}+\frac{1}{2} \log \left(2 x^2-y(x)\right)-\tanh ^{-1}\left(\frac{x}{\sqrt{y(x)-x^2}}\right)=c_1$$


0

Denote $$b=a-x, \quad d=c-y. \qquad(x,y\in\mathbb{N}).\tag{1}$$ Then $$ a+b+c+d = 2a-x+2c-y = 2010,\\ a^2-b^2+c^2-d^2=(2a-x)x+(2c-y)y=2010.\tag{2} $$ Since $x,y\in\mathbb{N}$, then unique solution of $(2)$ has $$x=y=1.\tag{3}$$ So, solution has form $$ (a,\;a-1,\;c,\;c-1).\tag{4} $$ $(2),(4)\Rightarrow$ $$a+c-1=1005,$$ $$c=1006-a.$$ So, solution has ...


1

It should be $n+1$. \begin{eqnarray} \sum_{S\subseteq Q}\frac{1}{p(S)}=\frac{\sum_{S\subseteq Q}p(S)}{n!}=\frac{(1+1)(2+1)\cdots(n+1)}{n!}=n+1 \end{eqnarray}


0

Hint: Looking at $(1+a)(1+b)(1+c)=1+a+b+c+ab+ac+bc+abc$, what do yo think $(1+1)(1+\frac12)(1+\frac13)\cdots(1+\frac1n)$ expands to?


1

I'm going to do the case where $n=2$, I'll leave the generalisation to you :) First, by continuity, for all $i\in\{1,2\}$ and $r\in\mathbb{R}$, $f(r{e_i})=r^2f(e_i)$. Now, denote $f(e_i)=A_i$. Then, for every $r_1,r_2\in\mathbb{R}$, you know by the given identity that: $$f(r_1,r_2)=2f(r_1,0)+2f(0,-r_2)-f(r_1,-r_2)=2r_1^2A_1+2r_2^2A_2-f(r_1,-r_2)$$ So we ...


0

Based on how you have defined $q:=$ number of pairs of linearly independent vectors. The answer for maximum value of $q$ should be $\binom{n}{2}$. For example, if $$S=\left\{\begin{bmatrix}1\\0\end{bmatrix}, \begin{bmatrix}1\\1\end{bmatrix}, \begin{bmatrix}1\\2\end{bmatrix}, \ldots , \begin{bmatrix}1\\n-1\end{bmatrix}\right\}.$$ Then $|S|=n$, $S$ definitely ...


0

If $n \geq 2$ then $n-1 \leq q \leq \binom{n}{2}$. There are two linearly independent vectors in $S$ let us choose two $v_0, v_1$, then in the worst case all the other vectors are scalar multiples of one of these. But if $0 \neq v \in S$ is a scalar multiple of $v_0$ then $v_1, v$ is is a linearly independent pair. So the smallest possible $q $ is $n-2+1$. ...


0

Since we are unable to explore $\sin\left(\frac{\pi}{2}\sin\left(\frac{\pi}{2}x\right)\right)$ let's consider more general case: Let $f$ be concave monotonic integrable function such, that $f(0)=0,\ f(1)=1$ and $f^{-1}$ is also integrable. Consider $\displaystyle g(x)=\int\limits_0^x \left(\left(f(t)-t\right)-\left(t-f^{-1}(t)\right)\right)\,dt$ and ...


0

Thanks to the comment given by David Z, I got the answer. There can only be two possibilities, first is three cameleons and two cameleons, and second is five cameleons. The first case is already calculated in the question. Here I will only focus on second case. Assume that the five chameleons are named $1,2,3,4,5$. Suppose we pick one chameleon randomly, ...


1

This is not an answer (neither a proof of anything) but it is too long for a comment. Consider $$A_n=\tan \left( 2^{-n}\,\theta\right) \sec \left( 2^{1-n}\,\theta\right)$$ and expand it as a Taylor series built at $\theta=0$ up to order $p$ ; this let you with infinite sums of terms in geometric progression. After summations over $n$, you end with a sum ...


1

The number of permutations without fixed points in $S_5$ is given by: $$ 5!\left(\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}\right)=\color{red}{44}\tag{1} $$ by the inclusion-exclusion principle, hence the wanted probability is: $$ \frac{44}{4^5} = \color{red}{\frac{11}{256}}\approx 4,3\%.\tag{2}$$


1

hint Call $\tan{\frac{\theta}{2^n}}=t_n$ and use the identity $\cos{\frac{\theta}{2^{n-1}}}=\frac{1-t_n^2}{1+t_n^2}$ the term of the series is $$\frac{t_n(1+t_n^2)}{1-t_n^2}$$ And you decompose that into a sum of rational fractions


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} ...


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.


2

Hint: examine the conditions you are given in the question - what can you say about the value of the expression in your last equation? Can you find a condition which makes it zero?


1

every submatrix of such a matrix of size $m$ must have determinant $m!$ or $-m!$. Prove! (observe for such matrix of size $n$ the $n-1$ cofactors must be $(n-1)! $ or $-(n-1)!$ expand along a suitable row or column ) So to prove that there does not exist one such matrix with $n \geq 3$, one just have to prove there cannot be one of size 3. Note that only ...


2

This is basically the same argument as the probability approach used in the original solution, but somewhat simplified and hopefully easier to understand. To help avoid special cases for the $n_i$, let's start by appending zeroes to the list $a_1,\ldots,a_N,0,\ldots,0$ so it gets length $2^k-1\ge N$. We will make a solution to this consisting of ...


1

By using the factorization over $\mathbb{Z}[i]$, that is a Euclidean domain hence a UFD, it is not difficult to prove that: $$ r_2(n)=\#\{(x,y)\in\mathbb{Z}^2:x^2+y^2=n\} $$ is given by: $$ r_2(n) = 4(\chi_4 * 1)(n) = 4\,\sum_{d\mid n}\chi_4(d)=4\left(d_1(n)-d_3(n)\right) $$ where $d_1(n)$ is the number of divisors of $n$ of the form $4k+1$ and $d_3(n)$ is ...


2

It is known that the Gaussian integers $\mathbb{Z}[u]$ is an UFD with $4$ units: $1, i, -1, -i$ $5$ can be factorized as $(2+i)(2-i)$ where $2 \pm i$ are primes in $\mathbb{Z}[i]$. Furthermore, $2 + i$ and $2-i$ are inequivalent to each other (i.e their ratio is not an unit). The $UFD$ property of $\mathbb{Z}[i]$ tells us when one factorize $5^k$ as ...


2

Work in the ring of Gaussian integers, $\mathbf{Z}[i]$, a UFD, where we have the factorization $x^2+y^2= (x+iy)(x-iy)$. As $5=(2+i)(2-i)$, we have $5^k =(2+i)^k(2-i)^k$. Now writing $(2+i)^k =a+ib$, you get a solution $5^k=a^2+b^2$.


1

instead of solving this problem let's solve a simpler version of it : Given a,b>0 and a^2+b^2=1, prove a+b≥a^3+b^3+ab you need to use what you have been given : we know a^2 + b^2 = 1 adding 2ab to both sides of equation we have : a^2 +b^2 + 2ab = 1 + 2ab then : (a+b)^2 = 1 + 2ab (1) what we need to prove is that a+b≥a^3+b^3+ab (2) by ...


1

Hint: $ ab+cd \leq |ab| + |cd| \leq (|a| + |c| ) ( |b| + |d| ) $ Applying the above concept, we have $$ \det M = \sum_\sigma sgn(\sigma) \prod_i M_{i \sigma(i) } \leq \sum_{i=1}^n \prod_{j=1}^n | M_{i j } | \leq \prod_{j=1}^n ( \sum_{i=1}^n |M_{i j} | ) \leq \left( \frac{ \sum_{i,j} |M_{i j } | } { n} \right)^n = \frac{1}{n^n}.$$ Follow the above chain ...


1

If we call the point where the two circles meet $X$, then clearly $AXP$ and $ABP$ are similar. From this we deduce $\frac{AP}{PB}=\frac{PX}{AP}$, therefore $AP^2=PX.PB$. Similarly, $QXP$ and $QBP$ are similar and this time we obtain that $\frac{PQ}{PB}=\frac{PX}{PQ}$, and so we have that $PQ^2=PX.PB$ and so $AP=PQ$. I've left out a few details to be ...


2

$\int_0^1f=1/2\int_0^1f(x)dx+1/2\int_0^1f(y)dy=-1/2\int_{\Pi/2}^0f(\cos t)\sin tdt+1/2\int_{0}^{\pi/2}f(\sin t)\cos tdt\leq 1/2.\pi/2$


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 ...


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 ...


1

Let the center of the circles be S and T. Let r be the radius of S and R the radius of T. $PQ^2=PT^2-R^2$ (pythagoraean theorem) $=r^2+(R+r)^2-2r(R+r)cos(<TSP) -R^2$ (law of cosines) Drop a perpendicular from S to BT to see that $cos( <TSP) = \frac{R-r}{R+r}$, so $PQ^2=r^2+(R+r)^2-2r(R-r) -R^2$ $=4r^2$ so PQ=2r=AP. Sorry I don't have a ...


1

First let us assume that $c_1, c_2, c_3$ are coprime, that is $(c_1, c_2, c_3) = 1$. Suppose first that there exist integer numbers $u_1, u_2, u_3$ and $v_1, v_2, v_3$ such that $$ \begin{vmatrix} c_1 & c_2 & c_3 \\ u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \end{vmatrix} = 1 $$ In that case we can take $a = c \times u$ and $b = c \times ...


18

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 ...


0

Another answer (on top of all the already existing excellent one). Suppose $a\neq (1,0,0....)$ (called $i$) Then $a \times i$ is integer and orthogonal to $a$. If $a=i$, take $j$


1

Partial answer (The estimate below are a bit hard to make precise. Feel free to edit/criticize/downvote): Let $O = \frac{1}{5}(A_n+B_n+C_n+D_n+E_n)$ be the center of mass of the five points $A_n, B_n, C_n , D_n, E_n$. Note that $O$ is really independent of $n$ as the next five points are midpoints of the previous five. By a translation of the initial data, ...


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.


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.


0

The given integral vector v has at least one rational vector w such that w ,v are perpendicular. We take the cross product u of v and w and this is rational. Now we find a suitable common multiple of u and w to eliminate the denominators. Thus we get new integral vectors u',w' perpendicular to v.


1

In fact the result holds for any dimension. Take a primitive vector $c$ (coordinates coprime). Then there is $L\in GL(Z)$ such that $Lc$ be the unit vector of the canonical basis. You take the cross product of all the lines of $L$ not corresponding to the $0$ of the vector and you get your initial vector $c$. At the very basis of it is Euclid. You can ...


2

This is only a proof of a special case of the problem. The problem becomes simple if we have in addition the assumption that $(c_{1}, c_{2}) \mid c_{3}.$ For, taking $a_{2} := c_{1},$ $b_{3} := 1,$ $a_{3} := 0,$ $a_{1} := -c_{2}$ fulfills the first two equations and from the third we have $c_{1}b_{1} + c_{2}b_{2} + c_{3} = 0.$ Since $(c_{1}, c_{2}) \mid ...



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