Split (1)

train · 1.46M rows

text stringlengths 83 79.5k
H: Defining an ideal in the tensor algebra In the wikipedia article about exterior algebra: The exterior algebra $Λ(V)$ over a vector space $V$ over a field $K$ is defined as the Quotient algebra of the tensor algebra by the two-sided Ideal $I$ generated by all elements of the form $x \otimes x$ such that $x \in V$. I want to show that $I$ is an ideal of $T(V)$ but i was confused about the elements of $I$, i mean what is exactly $I$ as a set and more particularly as a subset of $T(V)$. AI: If $S$ is a subset of a (non commutative) ring $R$, the ideal generated by $S$ consists of all elements of the form $$ \sum_{i=1}^{n} a_i s_i b_i $$ where $n$ is an arbitrary integer, $a_i,b_i\in R$ and $s_i\in S$. This set is obviously closed under addition (by construction) and contains $0$; it's also closed by left and right multiplication by elements of $R$, so it's an ideal. Any ideal containing $S$ as a subset must contain these elements, so this set is indeed the ideal generated by $S$. In the particular case, you can think to $I$ as the set of elements in $T(V)$ of the form $$ \sum_{i=1}^n A_i(x_i\otimes x_i)B_i $$ where $x_i\in V$ and $A_i,B_i\in T(V)$. I'm afraid there's not much more that can be said, even in the finite dimensional case. What's important is that the relation $$ x\wedge x=0 $$ is satisfied in $E(V)=T(V)/I$ (where $\wedge$ denotes the induced operation on the quotient ring) for all $x\in V$ because $x\otimes x\in I$ by definition. This is just one particular construction of the exterior algebra: others are possible, but they will always give an isomorphic ring, because the exterior algebra satisfies a universal property. Therefore it's not really useful knowing what the elements of $I$ look like.
H: $f = 0 $ almost everywhere implies $\int_{\mathbb{R}} f = 0 $ my try: Suppose $\phi $ is imple function such that $ 0 \leq \phi \leq f$, then since $ f = 0 $ almost everywhere, then $\phi = 0 $ almost everywhere. Let $A = \{ x : \phi(x) = 0\}$. Therefore, $A^c = \{x : \phi(x) > 0 \}$ is a null set. Hence $$ \int_{\mathbb{R}} \phi = \int_A \phi + \int_{A^c} \phi = 0 + 0 = 0 $$ Since $\mathbb{R} = A \cup A^c $. therefore $\int_\mathbb{R} f = 0$ is this a correct solution? AI: Generally let $$f_+ := \max(f,0); f_- := \max(-f,0)$$ Then $$f = f_+ + f_-$$ Since $f_+, f_- \ge 0$, $$f = 0 \Rightarrow f_+ = f_- = 0$$ Since this holds a.e., we have $$f_+, f_- = 0 \quad\text{a.e.}$$ Proceed with your argument using a series of step functions for both, $f_+$ and $f_-$ to get $$\int_{\mathbb R} f = \int_{\mathbb R} f_+ - \int_{\mathbb R} f_- = \lim_{n\to\infty} \int_{\mathbb R} f_+^{(n)} - \int_{\mathbb R} f_-^{(n)} = 0$$ Where you use $f_+^{(n)} \nearrow f_+, f_-^{(n)} \nearrow f_-$.
H: Minimization problem with parameter $a>0,\; b>0,\; S=$parameter $>0$. $$a+b+\dfrac{S-2}{2(a+b)} \longrightarrow min$$ With condition that $a\cdot b =1$ Using inequality of arithmetic and geometric means we get: $$a+b+\dfrac{S-2}{2(a+b)} \geq 2+ \dfrac{S-2}{2(a+b)}$$ With equality when $a=b$. This is true, but the answer to this problem is that if $S\leq 10$ this will indeed be the answer, but if $S>10$, $a=b$ no longer gives the best result. I am mainly interested in relatively "simple" and short solutions and methods of this problem. AI: For $S\ge 10$, you can use AM-GM like this: $$(a+b)+\dfrac{S-2}{2(a+b)} \ge 2\sqrt{\frac{S-2}{2}}$$ with equality iff $2(a+b)^2=S-2$. From this and $ab = 1$, you can solve for $a, b$.
H: About the parity of the partition function. I am reading this Kolberg's article, where he proofs that the partition function takes both even an odd values infinitely often. http://www.mscand.dk/article.php?id=1555 Although I'm sure it's simple (the proof is very short), I can't understand how he gets the contradiction, assuming the partition funtion takes odd (or even, in the other case) values for all $n\geq a$ Can anyone help me with this? Thanks a lot. AI: By assumption, say, we have $p(n)\equiv 0 \mod 2$ for all $n\ge a$, but $p(0)=1$ is not congruent $0$ modulo $2$. This is a contradiction to Euler's identity $$ p(a(3a-1)/2)\pm \cdots \pm p(2a-1) \pm p(0)=0, $$ because all terms except the last one are congruent $0$ modulo $2$. (For $p(0)=1$ see the first line of Kolberg's paper.)
H: Proof using chain rule. For a homework, I need to demonstrate that $$ \left(\partial z \over \partial x\right)^2 - \left(\partial z \over\partial y\right)^2 = 4{\partial z \over \partial u}{\partial z \over \partial v}$$ where $z = f(x,y)$ and $u = x+y$ and $v = x-y$. I tried by chain rule, but nothing. Can anyone help me? AI: Using the notation $\partial_x$ for $\frac{\partial}{\partial x}$ etc. you have $$\partial_x u = \partial_y u = \partial_x v = -\partial_y v = 1$$ so now $$\partial_x z = \partial_u z \partial_x u + \partial_v z \partial_x v = (\partial_u z + \partial_v z)$$ and $$\partial_y z = \partial_u z \partial_y u + \partial_v z \partial_y v = (\partial_u z - \partial_v z)$$ concluding $$(\partial_xz)^2 - (\partial_yz)^2 = (\partial_xz + \partial_yz)(\partial_xz - \partial_yz) = 2\partial_uz \cdot 2\partial_vz = 4\partial_uz\partial_vz$$ q.e.d.
H: How do I find y' in this problem? $$y^3 = cos(xy) $$ This is what I have so far: $$(3y^2y'/y+xy') = -sin(xy)y $$ $$ y' = -sin(xy) (3y^2/y)$$ I'm sure I did the last part wrong, but that was the best I could come up with. AI: I do not see where your second line comes from, but the derivative of $y^3$ with respect to $x$ is by the chain rule $3y^2y'$, the right hand side gives $$ -\sin(xy) \cdot (xy)' = -\sin(xy)\cdot (y + xy') $$ also by the chainrule so we have \begin{align} 3y^2y' &= -\sin(xy)(y+xy')\\ \iff y'(3y^2 + x\sin(xy) &= -y\sin(xy)\\ \iff y' &= -\frac{y\sin(xy)}{3y^2 + x\sin(xy)} \end{align*}
H: Combination Factorial formula Is there a formula for working out combinations using the factorial of the 'choice' number? E.g.: for a group of 6, what are the total possible combinations of up to 4 selections? I have been doing this by adding the individual combinations, so (in the above example) I would add: $$ \def\c#1#2{\binom{#1}{#2}} \c64+ \c63 + \c62 + \c61 = 15+20+15+ 6 = 56 $$ to arrive at 56, but I'm interested to know if there is a formula that I can use. I've looked at the combinations questions but didn't see one similar. Many thanks AI: According to WolframAlpha the result is not very simple. The only easy cases I can think of is summing over all possibilities ($2^n$) or summing over half the possibilities. This only really works if you have an odd number of total people, but can be adapted to an even number $2k$ of people if you take care what you do with the special $k$-group case. The rest are, I believe, easiest found by getting to one of these points and add / subtract as necessary. For instance, with $6$ people there are $2^6 = 64$ ways to choose groups in total, and since you want groups with no more than $4$ people it's easier to subtract the number of $5$-groups and $6$-groups as well as the empty group to get $$ 2^6 - \binom 6 5 - \binom 6 6 - \binom 6 0 = 64 - 6 - 1 - 1 = 56 $$ which in my opinion is easier to calculate than your summing over all valid combinations. Not much easier in this case, but it might help considerably once you get to the double digits.
H: Uniqueness Proof for Division Algorithm using Contradiction Let $a, b, \mathbb \in \mathbb {Z}$ and let there exist integers $q, q_1, r, r_1$ such that the two pairs $(q,r)$ and $(q_1,r_1)$ satisfy the properties: $$\ \ \ \ a = qb+r \quad \ \ \ \ \ \ \ ; 0 \le r \lt b \\ also, \ a = q_1 b + r_1 \quad ; 0 \le r_1 \lt b$$ It is assumed that $\|r-r_1\| \lt b$. Also, assume for the sake of contradiction that $q \neq q_1$. Then come up with another inequality involving $\|r-r_1\|$ that contradicts $\|r-r_1\| \lt b$. AI: Hint: Just follow the given hint, noting that $$ 0 = a -a = (q_1 - q)b + (r_1 - r) \iff r - r_1 = (q_1 - q)b $$ So $\def\abs#1{\left\|#1\right\|}$$\abs{r_1 - r} = \abs{q_1 - q}b$, if now $q \ne q_1$, what can you say about $\abs{r_1 -r }$?
H: Solving $(3 {Q} {s^{2}})/(100h (h^{2} - 3hs + 3s^{2}))$ for $h$ $$Y = \frac{3 \times{Q} \times{s^{2}}}{100h \left(h^{2} - 3hs + 3s^{2}\right)}$$ where Q, s, and h are different variables. The question I have is how to re-arrange this equation in terms of h? So far, I have $$Y = \frac{3 \times{Q} \times{s^{2}}}{100h^{3} - 300h^{2}s + 300hs^{2}}$$ $$Y = \frac{3Qs^{2}}{100h^{3} - 300h^{2}s + 300hs^{2}}$$ $$100h^{3}Y - 300h^{2}sY + 300hs^{2}Y = 3Qs^{2} $$ $$100h^{3} - 300h^{2}s + 300hs^{2} = \frac{3Qs^{2}}{Y} $$ $$\frac{1}{3}h^{3} - h^{2}s + hs^{2} = \frac{3Qs^{2}}{300Y} $$ $$\frac{1}{3}h^{3} - h^{2}s + hs^{2} = \frac{3Qs^{2}}{300Y} $$ now I'm stuck... where do I go from here? I'm not sure what to do with s not being in the first term on the left? AI: Starting from your last result: If you multiply by 3 the left hand side is nearly the complete power $(h-s)^3$, here are the single steps to solve for $h$: $$\frac{1}{3}h^{3} - h^{2}s + hs^{2} = \frac{3Qs^{2}}{300Y} $$ $$h^{3} - 3h^{2}s + 3hs^{2} = \frac{9Qs^{2}}{300Y} $$ $$h^{3} - 3h^{2}s + 3hs^{2}-s^3 + s^3 = \frac{9Qs^{2}}{300Y} $$ $$(h-s)^3 + s^3 = \frac{9Qs^{2}}{300Y} $$ $$(h-s)^3 = \frac{9Qs^{2}}{300Y} - s^3 $$ $$h-s = \left(\frac{9Qs^{2}}{300Y} - s^3\right)^{\frac{1}{3}}$$ $$h = s +\left(\frac{9Qs^{2}}{300Y} - s^3\right)^{\frac{1}{3}}$$ Note: If would be easier to use the idea with the original equation.
H: Let $L$ be the language defined by the regular expression $(a \vee b \vee c)(a \vee b \vee c)$ 1) How does $\|L\| = 9$? 2) $\|L^\| = \aleph_0$? Thank you AI: HINTS: Either use the multiplication principle, or write out all words of the language in systematic fashion from the regular expression. If $w\in L$, then $w,ww,www,\ldots\in L^$. With a little more work you could actually say exactly what the lengths of words in $L^*$ are, and how many of each length there are.
H: Chain rule on matrix differentiation of trace in Kalman gain proof I was reading the following derivation of the Kalman filter gain http://www.robots.ox.ac.uk/~ian/Teaching/Estimation/LectureNotes2.pdf on pg 5-6. From the following $$L=\min_{K_{k+1}} trace(P_{k+1\|k+1})$$ With $$ P_{k+1\|k+1} = \boldsymbol{ (I-K_{k+1} H_{k+1}) P_{k+1\|k} (I-K_{k+1} H_{k+1})^{T} + K_{k+1}R_{k+1}K_{k+1}^T }$$ K,H, and P are all nxn matrices. It uses the following trace differentiation identity: $${\frac{\partial}{\partial \boldsymbol{A}}} (trace(\boldsymbol{ABA^T})) = 2\boldsymbol{AB}$$ But from there, there is a bit of a skip to the result of the differentiation $${\frac{\partial L}{\partial \boldsymbol{ K_{k+1}}}} = -2(\boldsymbol{I-K_{k+1}H_{k+1})P_{k+1\|k}H_{k+1}^T+2K_{k+1}R_{k+1} } = 0$$ What I am wondering about is the origin of the $H_{k+1}^T$ term in the above equation. It appears to me that chain rule is applied here. But most online sources on matrix chain rule or trace identities does not show a direct example that the chain rule can be applied to the derivative of a trace. ie. This is what I think happens in the intermediate step: $$\boldsymbol{A=I-K_{k+1} H_{k+1}; B=P_{k+1\|k}}$$ $${\frac{\partial}{\partial \boldsymbol{K_{k+1}}}} (trace(\boldsymbol{ABA^T})) = 2\boldsymbol{AB} * {\frac{\partial (trace (\boldsymbol{A}))}{\partial \boldsymbol{ K_{k+1}}}}$$ Question: Is the above chain rule application valid? If so, I would appreciate a pointer to a proof. And then using $$ {\frac{\partial trace(\boldsymbol{K_{k+1}H_{k+1}} )} {\boldsymbol{\partial K_{k+1}}}} = \boldsymbol{H_{k+1}^T} $$ To arrive at $$ = -2(\boldsymbol{I-K_{k+1} H_{k+1}) * P_{k+1\|k} * H_{k+1}^T}$$ If the intermediate steps is actually different, please let me know and link to any proofs of the identities used as I am self-learning linear algebra. Thanks. AI: Firstly $\nabla_A(ABA^T)=A(B+B^T)=2AB$ only if $B$ is symmetric. Here $L=trace((I-KH)Q(I-KH)^T+KRK^T)$ where $Q,R$ are symmetric matrices. The derivative is $DL_K:Z\rightarrow trace(-ZHQ(I-KH)^T+(I-KH)Q(-ZH)^T+ZRK^T+KRZ^T)=trace(-2ZHQ(I-KH)^T+2ZRK^T)=trace((-2HQ(I-KH)^T+2RK^T)Z)$. I wrote that follows in at least 10 posts but nobody reads my post (tears). Yet, no students know how to calculate a gradient. Incredible ! A gradient is associated to a scalar product; we take, here, $(U,V)=trace(U^TV)$. The associated gradient is defined by, for every $Z$, $(\nabla_K(L),Z)=DL_K(Z)$. Thus $\nabla_K(L)=-2(I-KH)QH^T+2KR$. EDIT (answers to frank): 1),2) $DL_K$ is a linear application and $Z$ is the variable. For instance, if $f:K\rightarrow KAK$, then $Df_K:Z\rightarrow ZAK+KAZ$, the derivative of a product ! ($(uv)'=u'v+uv'$). If $f:K\rightarrow -KHQ$, then $Df_K:Z\rightarrow -ZHQ$. $(ua)'=u'a$. 3) Here $Df_X:(h,k,l)\rightarrow [3z,4y^3,3x][h,k,l]^T$. We use the scalar product over vectors $(u,v)=trace(u^Tv)=u^Tv$. Then $\nabla_Xf=[3z,4y^3,3x]^T$. More generally, to define a gradient with the help of a scalar product has a geometric meaning. You do not need a basis. 4) The derivative uses trace and the gradient not. It is linked to the chosen scalar product. 5) Consider the application $f:K\in\mathcal{M}_n(\mathbb{R})\rightarrow KAK\in\mathcal{M}_n(\mathbb{R})$, then the derivative is the linear application $Df_K:Z\in\mathcal{M}_n(\mathbb{R})\rightarrow ZAK+KAZ\in\mathcal{M}_n(\mathbb{R})$. In the same way, if $g(K)=K^3$, then $Dg_K(Z)=ZK^2+KZK+K^2Z$. If you choose coordinates, then the matrix of the derivative is its Jacobian. For 3), $f:\mathbb{R^3}\rightarrow \mathbb{R}$ and let $X=[x,y,z]^T$. $Df_X$ is a linear application $(h,k,l)\in\mathbb{R^3}\rightarrow \mathbb{R}$ and , consequently, its matrix (the Jacobian) is $1\times 3$, that is a row. The gradient is the transpose of the previous matrix, that is a vector.
H: Finding $x$ in the domain of the function where the tangent line is horizontal $$ y = x^3 - 4x^2 +4x + 2$$ I have to find all the values of the domain of the function where the tangent line at the point (x,f(x)) is horizontal. I wasn't so sure on how to start this problem, so the first thing I did was derive the function: $$y' = 3x^2-8x + 4$$ If this isn't right, can someone correct me? AI: That's absolutely the way to start. Now, the tangent line being horizontal means that the slope--which is $y'$--should be...what? That gives you the equation that you need to solve.
H: Finding the value of $f_{x+y}$ for multivariate normal distributon Given that bivariate normal distribution is I need to find the value of $f_{X+Y}$ and the variable are standard normal. If the variables are standard normal, $$f_{X+Y}(z)=\int_{-\infty}^{+\infty}f_{X,Y}(x,z-x)\,\mathrm dx\\ =\int_{-\infty}^{+\infty}f_X(x)f_Y(z-x)\,\mathrm dx\\ =\int_{-\infty}^{+\infty}\frac1{\sqrt{2\pi}}\exp\left(-\frac12 x^2\right)\frac1{\sqrt{2\pi}}\exp\left(-\frac12(z-x)^2\right)\,\mathrm dx\\ = \frac1{2\pi} \int_{-\infty}^{+\infty}\exp\left(-\frac12\left(x^2+(z-x)^2\right)\right)\,\mathrm dx\\ = \frac1{2\pi}\exp(-1/2z^2) \int_{-\infty}^{+\infty}\exp(zx)\,\mathrm dx$$ I get stuck after this because the integral turns out to be zero. Need some guidance to solve this question. AI: A painless approach is to understand what the parameters of the density of the bivariate normal $(X,Y)$ mean. One sees that $X=\mu_X+\sigma_XU$ and $Y=\mu_Y+\sigma_YV$ where $U$ and $V$ are standard normal, and that $\rho=\mathrm{Corr}(X,Y)=\mathrm{Cov}(U,V)$. Hence, $$ V=\rho U+\tau W,\qquad\tau=\sqrt{1-\rho^2}, $$ where $(U,W)$ is i.i.d. standard normal. Thus, $X+Y=\mu_X+\mu_Y+Z$ where $$ Z=\sigma_XU+\sigma_YV=(\sigma_X+\rho\sigma_Y)U+\sigma_Y\tau W $$ is centered normal with variance $$ E[Z^2]=(\sigma_X+\rho\sigma_Y)^2+\sigma_Y^2\tau^2. $$ Finally, $X+Y$ is normal with mean $\mu$ and variance $\sigma^2$, where $$ \mu=\mu_X+\mu_Y,\qquad\sigma^2=\sigma_X^2+2\rho\sigma_X\sigma_Y+\sigma_Y^2. $$
H: Find the limit -> Infinity with radicals First guess to multiply by $x^{-1.4}$ so the radical in numerator with $x^7$ becomes 1 and other stuff becomes 0. But then denominator becomes $-\infty$. What is the right approach? $$ \lim_{x \to \infty} \frac{\sqrt[5]{x^7+3} - \sqrt[4]{2x^3 - 1}}{\sqrt[8]{x^7 + x^2 + 1} - x} $$ AI: For starters: Note, that the denominator has as highest power of $x$ the $x^1$, so we multiply by $1 = \frac {1/x}{1/x}$, giving $$\frac{\sqrt[5]{x^7+3} - \sqrt[4]{2x^3 - 1}}{\sqrt[8]{x^7 + x^2 + 1} - x} = \frac{\sqrt[5]{x^2 + 3x^{-5}} - \sqrt[4]{2x^{-1} - x^{-4}}}{\sqrt[8]{x^{-1} + x^{-6} + x^{-8}} - 1} $$ Now the denominator converges to $-1$ for $x \to \infty$, and the numerator diverges to $\infty$, hence the fraction diverges to $-\infty$.
H: semigroup ideals For the semigroup $$‎S_{3 \times 3} = \bigl\{(a_{ij}) \bigm\| a_{ij} \in \mathbb Z_2 = \{0,1\}\bigr\}$$ ‎‎(the set of all $‎3‎\times‎‎3$ matrices with entries from $\mathbb Z_2$) under multiplication. Find all the ideals of $S_{‎3‎\times‎‎3}$. AI: Hint: For any $A \in S_{3\times 3}$ there are $B,C \in {\rm GL}(3, \mathbb Z_2)$ (the invertible elements of $S_{3\times 3}$) such that $$ BAC = \begin{pmatrix} 0\mid1 & 0 & 0 \\ 0 & 0\mid1 & 0 \\ 0 & 0 & 0 \mid 1 \end{pmatrix} $$ where the number of ones is determined by the rank of $A$.
H: Almost the reverse triangle inequality I'm wondering if one can get such an inequality (for $n$ a positive integer) : $$\exists C>0\ \forall(x,y,u,v)\in \left(\mathbb{R}^n\right)^4,\quad \left\| \left\|x-y \right\|^2-\left\|u-v \right\|^2 \right\|\le C\left( \left\|x-u \right\|^2+\left\|y-v \right\|^2 \right) $$ I've tried using the fact that $\|x-y\|^2-\|u-v\|^2 = \left( \|x-y\|+\|u-v\|\right)\left( \|x-y\|-\|u-v\|\right)$ but I've got nothing... Thanks a lot. AI: Suppose $x,u,v,y$ are colinear and displayed in order, and suppose $\|u-v\|=a$ together with $\|x-u\|=\|y-v\|=b$ are both alternative. Thus $\|x-y\|=a+2b$. Now your inequality is $$4ab+4b^2=(a+2b)^2-a^2\leq 2Cb^2$$ which is obviously invalid for all $a,b$.
H: Calculate the covariance matrix of $\hat{\beta}$ $y=X\beta+u$ where $u \sim N(0,\Sigma)$ and $\Sigma$ is symmetric & idempotent. $X: nk$, $y:n1$, $\beta=k1$, $u:n1$ vector. Suppose you apply LS(least square) to the model. Calculate the covariance matrix of $\hat{\beta}$. Since $\hat{\beta}=(X'X)^{-1}X'y$, $E(\hat{\beta})=\beta$ so $\hat{\beta}$ is an unbiased estimator, $Cov(\hat{\beta})=E(\hat{\beta}-E(\hat{\beta}))(\hat{\beta}-E(\hat{\beta}))'=E[(X'X)^{-1}X'uu'X(X'X)^{-1}]=\Sigma (X'X)^{-1}$ Is it right? AI: you may use this property: $Cov\left( Ay \right)=ACov\left( y \right)A'$ where $A={{\left( {X}'X \right)}^{-1}}{X}',Cov\left( y \right)=\text{ }\!\!\Sigma\!\!\text{ }$ so ${{\left( {X}'X \right)}^{-1}}{X}'\text{ }\!\!\Sigma\!\!\text{ }X{{\left( {X}'X \right)}^{-1}}$.
H: Characterizing trees where every diametral path shares an edge In a graph $G$, a diametral path is a path of length $\text{diam}(G)$ joining two vertices that are at a distance $\text{diam}(G)$ from each other. Given a tree $T$, consider the set of all diametral paths of $T$. For example, if $T$ is say the star graph $S_4$, it is not true that the set of all diametral paths would share an edge. But when do they? How can we characterize trees where every diametral path has an edge in common? Looking at some examples, it seems always true for at least bicentered trees. The intuition would be that every such path passes through the center, and thus have the edge between the two centers in common. AI: Being bicentered is sufficient, but not necessary (consider $P_3$ with two edges attached at each end). For odd diameter, being bicentered is necessary and sufficient. For even diameter $2r$, the condition would be that the (unique) center has exactly two neighbours leading to a vertex at distance $r$ from the center. This is not much of a characterization, but I doubt if you can find anything better.
H: showing that two surface is isometric I tried to show that a parametrized surface $S$ in $\mathbb{R}^3$ given by $(u, v)$ ->$(u, v, u^2)$ is isometric to the flat plane. At first, I found their first fundamental form, but they are different. (I choose a surface patch defined by $(u, v)$ -> u$\mathbb{q}$ + v$\mathbb{p}$ where $\mathbb{p}$ and $\mathbb{q}$ are perpendicular and both unit vectors.) In presseley's text, there is a theorem following that two surface are local isometric if and only if they have the same first fundamental form. So I concluded that they are not isometric How can I show that? AI: Let $E$ be the plane defined by $z=0$. In your case you may consider $f\colon R\to R$ defined by $$f(x)=-\frac{1}{4}\left(\ln(-2x+\sqrt{1+4x^2})-2x\sqrt{1+4x^2}\right).$$ Verify that $f'(x)=\sqrt{1+4x^2}$. Define $\phi\colon S\to E$, $\phi(u,v,u^2):=\bigl(f(u),v,0\bigr)$. It's easy to verify that $\phi$ is an isometry; check the FFFs. I've constructed $f$ in an obvious way. Edit (much later): The construction of $f$ is obvious since the FFF of the given map is $$\begin{pmatrix}1+4u^2& 0\\ 0&1\end{pmatrix}.$$
H: Triangle inequality- complex I am trying to prove the triangle inequality purely algebraically. Let $z=x+iy$, $w=u+iv$. Then, $\|z+w\|^2$=$\|(x+u)+i(y+v)\|^2$=$(x+u)^2+(y+v)^2$=$x^2+2xu+u^2+y^2+2yv+v^2$ I tried the other way: $(\|z\|+\|w\|)^2$=$(\sqrt{x^2+y^2}+\sqrt{u^2+v^2})^2$=$x^2+y^2+u^2+v^2+2 \sqrt{x^2+y^2} \sqrt{u^2+v^2}$ I'm trying to do this without using the complex conjugate or anything. Any help is appreciated. AI: Hint: Use Cauchy-Schwarz inequality: $$ \sqrt{x^2+y^2} \sqrt{u^2+v^2}\geq xu+yv $$ To prove it directly: $$ \sqrt{x^2+y^2} \sqrt{u^2+v^2}\geq xu+yv\implies\\ x^2u^2+x^2v^2+y^2u^2+y^2v^2\geq x^2u^2+2xvyu+y^2v^2\implies\\ x^2v^2+y^2u^2 - 2xvyu\geq 0 \implies\\ (xv-yu)^2 \geq 0 \\ $$
H: Change of an angle in a triangle I've a triangle ABC. Where AC is the hypotenuse and the angle ABC is 90 degress. AB is $15 km$ and changes with a speed of $600 km/h$. BC is $5 km$ and changes with a speed of $0 km/h$. At what speed changes the angle CAB in terms of $rad/h?$ I call AB for $x(t)$ and BC for $y(t)$. Then I know: $$x(t) = 15 km$$ $$x'(t) = 600 km/h$$ $$y(t) = 5 km$$ $$y'(t) = 0 km/h$$ To find the angle CAB (now called $\theta$): $$ tan(\theta) = \frac{y(t)}{x(t)} \leftrightarrow \theta = arctan\left(\frac{y(t)}{x(t)}\right)$$. I derivate this to get the change of $\theta$ in $rad/h$. $$ \frac{1}{1+\left(\large\frac{y(t)}{x(t)}\right)^2} \times \left(\frac{y'(t)x(t)-y(t)x'(t)}{x(t)^2}\right) = 0.0075~rad/h$$ I know that the answer should be $12~rad/h$ so somewhere I'm wrong. But where? AI: For clarification here, $x(0)=15, y(0)=5$ is what is meant here. OP, you need to write it like this, otherwise things get real messy. As for an answer: We know that: $$ x(0)=15\\ y(0)=5\\ \frac{dx}{dt}=x'(t)=600\\\frac{dy}{dt}=y'(t)=0 $$ So from this we can get $x(t)=600t+15$ and $y(t)=5$. We're aiming for $\frac{d\theta}{dt}$, and to get this we can multiply $\frac{dx}{dt}$ and $\frac{d\theta}{dx}$, as the $dx$ cancels out. We already have $\frac{dx}{dt}$, so to get $\frac{d\theta}{dx}$: $$\begin{align} tan(\theta) & = \frac{y(t)}{x(t)} \\ & = \frac{5}{x} \\ \end{align} $$ We'll treat x as a variable by itself for the moment: $$\theta = arctan(\frac{5}{x})\\ \begin{align}\frac{d\theta}{dx} & = \frac{-5}{x^2}\cdot\frac{1}{1+(\frac{5}{x})^2}\\ & = \frac{-5}{x^2}\cdot\frac{1}{\frac{1}{x^2}(x^2+25)}\\ & = \frac{-5x^2}{x^2(x^2+25)}\\ & = \frac{-5}{x^2+25} \end{align}$$ so substituting into $\frac{dx}{dt}\cdot\frac{d\theta}{dx}=\frac{d\theta}{dt}$, we get: $$\begin{align} \frac{d\theta}{dt} & = \frac{-5\cdot600}{x^2+25}\\ & = \frac{-3000}{(600t+15)^2+25}\\ & = \frac{-3000}{360000t^2+18000t+225+25}\\ & = \frac{-3000}{360000t^2+18000t+250}\\ & = \frac{-60}{7200t^2+3600t+5} \end{align}$$ As $t=0$, we can substitute back into our equation for $\frac{d\theta}{dt}$ to get: $$\begin{align} \frac{d\theta}{dt} & = \frac{-60}{5}\\ & = -12\\ \end{align}$$ So at $t=0$, $\theta$ is decreasing at 12 rad/h.
H: optimizing prime number algorithm I am doing a function to return a list of prime number up to "n", one what to optimize the algorithm is the following: "The next most obvious improvement would probably be limiting the testing process to only checking if the potential prime can be factored by those primes less than or equal to the square root of the potential prime, since primes larger than the square root of the potential prime will be complementary factors of at least one prime less than the square root of the potential prime.(taken from http://en.wikibooks.org/wiki/Efficient_Prime_Number_Generating_Algorithms) Can someone explain this in simpler terms with an example? Thanks AI: Let's say you're trying to find primes below 150. Then, what the statement is saying is that you need to look out for the primes below sqrt(150) i.e. [2,3,5,7,11] only. Why's that you say? Well, if there were to be a number above 11, it'll have to multiply by another prime within the list [2,3,5,7,11]. In other words, if 13 were to be that next prime on the primes list, you'll have to multiply 13 by the primes already on the prime list, namely [2,3,5,7,11]. Also, 13*12>150. This automatically means that the number is not prime.
H: Ring of $p$-adic integers $\mathbb Z_p$ There are a few ways to define the $p$-adic numbers. If one defines the ring of $p$-adic integers $\mathbb Z_p$ as the inverse limit of the sequence $(A_n, \phi_n)$ with $A_n:=\mathbb Z/p^n \mathbb Z$ and $\phi_n: A_n \to A_{n-1}$ (like in Serre's book), how to prove that $\mathbb Z_p$ is the same as $$\mathbb Z_p=\left \{ \sum_{i=n}^\infty a_i p^i \ \| \ n \in \mathbb Z, \ a_i \in\left \{ 0,1,...,p-1 \right \} \right \} \ \ ?$$ I found a proof here but it's very long and technical. So maybe there are other ways to prove it? I'm looking for a shorter proof. Best regards. AI: Your displayed equation is wrong. The power series should start with non-negative powers of $p$, i.e. the sum should start at 0 instead of an arbitrary integer $n$. Intuitively, the two definitions are the same because both say that a $p$-adic integer is a compatible choice of residues modulo higher and higher powers of $p$. First, there are $p$ choices for a residue modulo $p$. Once you have made your choice (which is given by $a_0$ in your second definition), you have $p$ possible choices for the residue modulo $p^2$. There are of course $p^2$ choices for this residue altogether, but only $p$ of those will be compatible with the first choice you have made, i.e. will reduce to $a_0$ modulo $p$. This second choice is given by $a_0+pa_1$ in your second definition. You keep going in this way, fixing a residue modulo $p^3$, $p^4$, etc. Every time, there are only $p$ possible new choices that will be compatible with the ones you have already made. If this was the ordinary integers, your sequence $a_i$ would stabilise at 0 at some point, but in the $p$-adics you are allowed to keep going forever. To sum it up, the isomorphism between the two rings takes an infinite sequence $\sum_{i=0}^\infty a_ip^i$ and sends it to the inverse system $$\left(\sum_{i=0}^n a_ip^i\in \mathbb{Z}/p^n\mathbb{Z}\right)_n,$$ with $\phi_n$ given by reduction-mod-$p^n$ maps. It is completely trivial (and you should do it, not look it up!) to check that this is indeed a ring isomorphism.
H: How do I solve a derivative that has an absolute value $x \times \|x\|$ I have a function x * \|x\| To get the derivative I used the first principals: $$ f'(x) = ( (x-h) * \| x + h \| - (x * \|x\|) )/ h $$ So if x is + I got $$ x ^ 2 + xh - xh - h^2 - x^2 / h $$ $$ -h^2/h$$ $$ -h $$ $$ 0 $$ If x is negative: $$ x^2 - 2xh + h^2 - x^2$$ $$ -2xh+h^2 $$ $$ h (-2x + h)/h$$ $$ -2x + h $$ $$ -2x $$ So I checked with a derivative calculator and it says the answer is 2x.... So I'm not exactly sure why I got -2x, and why do we only used the negative part, why is the answer not -2x for negative x and 0 for positive x... why is only one chosen? AI: Since $\|x\|=\sqrt{x^2}$ we have $f(x)=x\cdot\sqrt{x^2}\Rightarrow f'(x)=\sqrt{x^2}+x\cdot\dfrac{1}{2\sqrt{x^2}}\cdot2x=\|x\|+\dfrac{x^2}{\left\|x\right\|}$. EDIT: Further simplification: $f(x)=\dfrac{\left\|x\right\|^2+x^2}{\left\|x \right\|}= \dfrac{2\left\| x\right\|^2}{\left\| x\right\|}=2\left\| x\right\|$.
H: Hausdorffness of a space We know that if $X$ and $Y$ are two topological spaces with $Y$ Hausdorff, then for any two continuous functions $f:X\to Y$ and $g:X\to Y$, $\{x\in X: f(x)=g(x)\}$ is a closed set. Is the converse true? Let us suppose that $Y$ be a topological space and for any topological space $X$ and for any continuous functions $f:X\to Y$ and $g:X\to Y$, $\{x\in X: f(x)=g(x)\}$ is a closed set in $X$. I want to prove that $Y$ is Hausdorff. Is it really possible? Any hint will be very much helpful. AI: I think we might as well spill the beans here. Let $f = \pi_1: X \times X \to X$, and let $g = \pi_2: X \times X \to X$.
H: What does the result of derivating out all of the units from a "number" tell you? I apologize for the terrible title, I'm not sure of the right way to say it. Consider an equation which outputs in Watts $ \frac{kg \cdot m^2}{s^3} $. If you derivative out all the changes in kilograms, meters, and seconds what does the final result tell you? AI: Each time you take a derivative, it can come with a unit attached. Watts is already sort of a derived quantity. If I write Watts = $\frac d{dt}$Energy, we see that Watts has one more power of time in the denominator than Energy. Going the other way, we can integrate Watts $dt$ to get energy. Formally, you could take $\frac d{dx}$ Watts and get something with units $\frac {kg \cdot m}{s^3}$ but I don't have any physical sense for what it would be.
H: Question About 3-Dimensional Integration I'm having trouble with determining the volume of this solid. $V := {(x,y,z) : 0 \le x \le 1, x^2 \le y \le \sqrt{x}, -x^{1/3} - y^{1/3} \leq z \leq x^{1/3} + y^{1/3}}$ I started off thinking $\mathrm{vol}(V) \leq \int_0^1 \int_{x^2}^{\sqrt{x}} x^{1/3} + y^{1/3} dy dx$ but I evaluated that integral to be $0$. Any help would be appreciated. Thank you. AI: That integral cannot be $0$ -- your integrand is non-negative, and so the integral being $0$ would imply that either the integrand is always 0 or the region you're integrating over has 0 area. It is also the case that you're integrand is not quite right. Close, though! This is a straight forward triple integral problem: if $V$ is the region in question, then $$ \DeclareMathOperator*{\Vol}{Vol}\Vol(V)=\iiint_V1\,dV=\int_0^1\int_{x^2}^{\sqrt{x}}\int_{-x^{1/3}-y^{1/3}}^{x^{1/3}+y^{1/3}}1\,dz\,dy\,dx=\int_0^1\int_{x^2}^{\sqrt{x}}2(x^{1/3}+y^{1/3})\,dy\,dx. $$ For the inner integral, you now get $$ \int_{x^2}^{\sqrt{x}}2(x^{1/3}+y^{1/3})\,dy=\left.2x^{1/3}y+2\cdot\frac{3}{4}y^{4/3}\right\vert_{y=x^2}^{\sqrt{x}}=2x^{1/3}(\sqrt{x}-x^2)+\frac{3}{2}(x^{2/3}-x^{8/3}). $$ Then, it is a matter of integrating this last expression for $x\in[0,1]$.
H: Transformation (coordinates) - PDE If I have $-\cos(x)u_x$ for $u\in C^2(\Omega), \Omega\subset\mathbb{R}^2$ and $$ \xi:=-x-\cos(x)+y, \eta:=x-\cos(x)+y $$ as transformation of the coordinates, what is then $-\cos(x)u_x$ after this transformation? Do not know, how to solve this. AI: We have $$ x = \frac 12(\eta - \xi) $$ and by the chain rule \begin{align} \def\pd#1#2{\frac{\partial #1}{\partial #2}} \pd ux &= \pd u\xi\pd\xi x + \pd u\eta \pd\eta x\\ &= (\sin x - 1)\pd u\xi + (1 + \sin x)\pd u\eta \end{align} so $$ -u_x\cos x = -\Biggl(\left(\sin\frac {\eta-\xi}2 - 1\right)u_\xi + \left(1+\sin\frac {\eta -\xi}2\right)u_\eta\Biggr)\cos \frac{\eta-\xi}2 $$ Addendum (answer to question in comment about expressing $y$): We have $$ \frac 12(\xi + \eta)= y - \cos x = y - \cos \frac{\eta - \xi}2 \iff y = \frac 12(\xi + \eta) + \cos \frac{\eta - \xi}2 $$ To calculate $u_{xx}$ we have $\def\x{\frac{\eta-\xi}2}$ \begin{align} u_{x\xi} &= \Biggl(\left(\sin \x -1\right)u_\xi + \left(1+\sin\x\right)u_\eta\Biggr)_\xi\\ &= -\cos\x \cdot u_\xi + \left(\sin \x - 1\right)u_{\xi\xi} - \cos\x\cdot u_{\eta} + \left(1 + \sin\x\right)u_{\eta\xi}\\ u_{x\eta} &= \Biggl(\left(\sin \x -1\right)u_\xi + \left(1+\sin\x\right)u_\eta\Biggr)_\eta\\ &= \cos\x \cdot u_\xi + \left(\sin \x - 1\right)u_{\xi\eta} + \cos\x\cdot u_{\eta} + \left(1 + \sin\x\right)u_{\eta\eta}\\ u_{xx} &= u_{x\xi}\xi_x + u_{x\eta}\eta_x\\ &= u_{x\xi}\left(\sin\x - 1\right) + u_{x\eta}\left(1 + \sin\x\right) \end{align}
H: definition of opposite catagory I have been given two definitions of what an opposite category is. The first is one which is in my textbook and is follows: If $\mathcal{C}$ is a catagory with objects $ob(\mathcal{C})$ and morphisms $C(U,V)$ for $U,V\in ob(\mathcal{C})$ we define the opposite catagory $\mathcal{C}^{op}$ with $ob(\mathcal{C})= ob(\mathcal{C})^{op}$ and we define the morphisms by simply reversing the arrows of $C(U,V)$. The second definition I have is that $\mathcal{C}^{op}$ is defined as having the same objects as $\mathcal{C}$ and morphisms $C(V,U)$ To me these do not seem the same. If I take the catagory of sets as an example would the morphisms not be different as in the first definition they need not be functions but in the second they are? Thanks for any help. AI: Hmmmm. What you want is probably : $$C^{op}(U,V)= C(V,U). $$
H: How many ways can I express $X$ as a sum unique natural numbers to the power of $N$ I also posted the problem here on Stackoverflow: http://ejj.mobi/g5i5gh but that was concerned with the programming solution. Now I am much more interested in the theory behind this problem, or rather the fact that I don't understand it. Problem Given two integers $X$ and $N$ how many ways can you express $X$ as a sum of the $Nth$ powers of unique natural numbers. My Solution Compute a range of numbers $1 .. r$ where $r$ is the largest number $< X$ that could possibly be included in its sum. (i.e. for an $X$ of 10 the range would be $1..3$ because $4^2 > 10$. After finding this range I would raise it to the power of $N$ it. From this range I computed the permutations of all subsets of $1..r$. So $1..3$ generated $[1],[4],[9],[1,4],[1,9],[4,9],[1,4,9]$ then the number of ways to express $X$ by powers of $N$ was simply the number of permutations that summed to $X$; in this case $[1,9]$ In my opinion this is a terrible solution: it's brute-force and inelegant, there has to be a more elegant way that uses a relationship that I can't [don't] see. Any ideas/suggestions would be appreciated. AI: I don't have a simple formula, but dynamic programming will make this easier. You can use the fact that the sum of the squares up to $k$ is $\frac 16k(k+1)(2k+1)$ So as you were, find the greatest $r$ such that $r^2 \le X$. If the sum of all the squares up to $r-1$ is less than $X$, you must have $r$ in the list. In your example, because $1+4 \lt 10$, you must have $3^2$ because you can't get there otherwise. Taking a larger example, let $X=1000$, so $r=31$ The sum of the smaller squares is larger than $1000$ you can get by without $31$. The number of representations is then the number of representations of $1000-31^2=39$ plus the number of representations of $1000$ using squares no larger than $30$. You need to be prepared for failure-$39$ has no such representation. The sum of the squares gets larger than $1000$ at $14$, so you have to start with a number at least that large.
H: related rates of change A balloon is rising vertically above a level, straight road at a constant rate of $1$ ft/sec. Just when the balloon is $65$ ft above the ground, a bicycle moving at a constant rate of $ 17$ ft/sec passes under it. How fast is the distance between the bicycle and the balloon is increasing $3$ seconds later? I am at a loss what to begin with? just a hint would do.. AI: Find the height of the balloon at time $t$ seconds Find the horizontal distance of the bicycle at time $t$ seconds Find the distance between the balloon and the bicycle at time $t$ seconds Differentiate with respect to $t$ Let $t = 3$.
H: Proof the Legendre polynomial $P_n$ has $n$ distinct real zeros I need a proof to show that the inequality $m < n$ leads to a contradiction and $P_n$ has $n$ distinct real roots, all of which lie in the open interval $(-1, 1)$. AI: By Rodrigues formula for Legendre polynomials, $$\displaystyle P_n(x) = \frac{1}{2^nn!}\frac{d^n}{dx^n} (x^2 - 1)^{n}\tag{1}$$ $P_n(x)$ is the $n^{th}$ derivative of a polynomial with roots at $-1$ and $1$. Repeat applying Gauss-Lucas theorem $n$ times, we know all roots of $P_n(x)$ lie on the closed line segment $[-1,1]$ in the complex plane. By direct expansion of $(1)$, one can check that $P_n(\pm 1) \ne 0$, so the roots of $P_n(x)$ are all real and belongs to $(-1,1)$. Since Legendre polynomials are solutions of the Legendre's differential equation: $$\frac{d}{dx}\left[(1-x^2)\frac{d}{dx}P_n(x)\right] + n(n+1)P_n(x) = 0$$ which is a $2^{nd}$ order ODE, $P_n(x)$ cannot have any double roots. To see this, let's say $P_n(x)$ has a double root at $\alpha \in (-1,1)$. $P_n(x)$ will then be a solution of following initial value problem: $$\frac{d}{dx}\left[(1-x^2)\frac{d}{dx}y(x)\right] + n(n+1)y(x) = 0,\quad \begin{cases}y(\alpha) = 0,\\y'(\alpha) = 0\end{cases}$$ Apply Picard-Lindelöf theorem to this $2^{nd}$ order ODE, the "uniqueness" part of the theorem tell us $P_n(x)$ vanishes over some neighbor of $\alpha$. Since $P_n(x)$ is not identically zero, this is impossible. As a consequence, all roots of $P_n(x)$ are simple.
H: compactness of the real projective plane Let $\mathbb{P}_2(\mathbb{R})$ =( $\mathbb{R}^3$-{0})/~ where x~$r$x for any nonzero point x $\in \mathbb{R}^3$ and any nonzero $r \in \mathbb{R}$ I want to show that the $\mathbb{P}_2(\mathbb{R})$ is compact.. I used quotient map to solve, but it doesn't work. Would you inform me? AI: Hint: $\mathbb{P}^2(\mathbb{R})$ is homeomorphic to the space $S^2/{\sim}$ where $x\sim -x$ for all $x\in S^2=\{x\in\mathbb{R}^3\mid \\|x\\|=1\}$. We know $S^2$ is compact by Heine-Borel.
H: Find $a$ with $a^2-a\equiv -1\pmod p$ when $p$ is a prime with $p\equiv 1\pmod 3$ I'm trying to prove that the primes with form $3k+1$ are not Eisenstein prime. This step: to find $a$ such that $a^2-a\equiv -1\pmod p$ when $p$ is a prime with $p\equiv 1\pmod 3$ is the only obstacle now, and I have thought of many methods, but none of them work. I know that when proving prime of form $p=4n+1$ aren't Gaussian primes, we need $a$ such that $a^2 \equiv -1\pmod p$. This can be obtained by Wilson's theorem, that $(4n)!\equiv -1\pmod p$, which gives $(4n)!\equiv(1\times 2\times \cdots \times(2n))((-2n)\times\cdots\times(-2)\times(-1))=(1\times 2\times \cdots \times(2n))^2\pmod p$. And this constructs the $a$ we want. But in this case, things are different. I've tried via mathematica, that the $(3k)!$ or $(6m)!$ cannot be factorized into $a(a-1)$. Such as $720=24\times30$. But when I minus some multiple of the primes, it becomes $24\times 23\equiv-1(\mod 7)$. This special case can't give me more details. So I wonder if there is some method to find $a$ or prove existence without finding it? Only hints are needed! Thanks! AI: If $p \equiv 1 \bmod 3$, then you can form the polynomial $X^{(p-1)/3} - 1$ which has the zero $1$. Therefore, in $F_p$, $$X-1 \mid X^{(p-1)/3} - 1 \; \Rightarrow \; X^3 - 1 \mid X^{p-1} - 1.$$ On the other hand, $X^{p-1}-1$ decomposes as $(X-1) \cdot (X-2) \cdot ... \cdot (X-(p-1))$ by Fermat's little theorem. Therefore $X^3 - 1$ splits into linear factors. Since $X^2 + X + 1$ divides $X^3 - 1$, it also splits into linear factors $X^2 + X + 1 = (X-a)(X-b)$. This gives you your $a$.
H: Finding coordinates of a path I am trying to find out whether to geometrical shapes overlap. This is based on an SVG-file. The following is an example: Group 1: <g transform="matrix(1.0 0.0 0.0 1.0 257.0 124.95)"> <g transform="matrix(1.7880859 0.0 0.0 0.9476929 -0.05 0.0)"> <path d="M 18.9 8.15 L -18.85 8.15 L -18.85 -8.1 L 18.9 -8.1 L 18.9 8.15 Z"/> </g> </g> Group 2: <g transform="matrix(1.0 0.0 0.0 1.0 244.05 85.95)"> <path transform="matrix(0.009765625 0.0 0.0 0.009765625 64.3499984741211 9.0)" d="M 328.0 0.0 L 328.0 -176.0 L 12.0 -176.0 L 12.0 -260.0 L 348.0 -728.0 L 420.0 -728.0 L 420.0 -260.0 L 520.0 -260.0 L 520.0 -176.0 L 420.0 -176.0 L 420.0 0.0 L 328.0 0.0 Z M 420.0 0.0 L 420.0 0.0 Z M 328.0 -260.0 L 328.0 -576.0 L 101.0 -260.0 L 328.0 -260.0 Z M 101.0 -260.0 L 101.0 -260.0 Z"/> </g> These items are in svg format. What it means is the following: The outer element (g)is a group which contains some shapes. A group can contain other groups (such as in the first example a group has a matrix transformation which consists of 6 values: 1: x scale 2: y skew 3: x skew 4: y scale 5: x translate 6: y translate So, for the first item that means that x is scaled by 1.0, y is scaled by one, x is translated by 257 and y is translated by 124.95 - a path also has a matrix transform which abides to the same rules - a path has a value (d) which defines how it's drawn. It works by drawing the line following the instructions (shown by a letter). The following instructions are provided: L: Line to M: Move to C: curve to Z: close path Now, given these two groups, how can I find out whether they overlap or not? What I have tried so far is establishing the left most point and the top most point of each group. I did this by taking the total translation and then looking for the left most point in the path and adding these together. However, I don't know how to take the curve into consideration and the results are incorrect (based on manual inspection). Also, once I have these, how would I go checking whether they overlap? I'm sorry if this question is a bit unclear (or not fit for math.stackexchange). AI: Let two groups of points plus the curves or lines connecting them be labeled $A$ and $B$. Calculate the center of each by taking the arithmetic mean of all the points (including the points of the curves/lines) and label these $A_c,B_c$. Find the largest and smallest radius of any point within each group, and label these $A_{r_{max}},A_{r_{min}},B_{r_{max}},B_{r_{min}}$. Finally, let $d$ be the distance between $A_c$ and $B_c$. Then you have some guaranteed conditions and some uncertain conditions for whether one group is contained within the other: $(1)$ If $A_{r_{max}}\lt B_{r_{min}}-d$, then $A$ is definitely contained within $B$. (Reversing the labeling of $A$ and $B$ has no effect on the truth of this.) Note that the distances here are all positive real numbers. $(2)$ If $A_{r_{max}}\gt B_{r_{max}}$ and $A_{r_{min}}\lt B_{r_{min}}-d$, then neither shape is contained within the other. $(3)$ If $d\gt A_{r_{max}}+B_{r_{max}}$ then the two shapes have no common area, i.e., they do not intersect. Other conditions of the above inequalities would require deeper analysis of the shapes and their points to determine whether there is any common ground. Note that unless $A$ or $B$ is transformed in a way that changes the shape, the arithmetic mean of the points of that shape does not change.
H: Picard group of a Affine scheme How do we define a Picard group of an Affine scheme? Is there way to define as for commutative ring? Thanks AI: The Picard group can be defined for any scheme $X$ as the group of isomorphism classes of invertible sheaves ( $\mathcal{O}_X$-modules that are locally free of rank $1$). Given such an $\mathcal{L}$, its inverse is given by $\mathcal{L}^{\vee} = \mathcal{Hom}(\mathcal{L},\mathcal{O}_X)$ where Hom there is not global Hom but rather sheaf Hom (which is a sheaf). The multiplication in the group is given by the tensor product, and to check $\text{Pic} (X)$ is a group the only non-trivial calculation is to see that $$\mathcal{L} \otimes_{\mathcal{O}_X} \mathcal{L}^\vee = \mathcal{O}_X.$$ If you want to check this isomorphism, I suggest you consider the tensor-hom adjunction.
H: Let G be a group, and let f(x)=x^(-1) for all x in G. Is f a function from G to G? If so, is it 1-1, onto? Let G be a group, and let f(x)=x^(-1) for all x in G. Is f a function from G to G? If so, is it 1-1, onto? Abstract algeba.. I know what it means to be 1-1 and onto.. 1-1: f(a)=f(b)=>a=b onto: there exists an x for every y. AI: Is $f$ a function from $G$ to $G$? Yes, note that for each $x\in G, x^{-1}$ is also an element in $G$. If the function one-to-one? If $f(g) = f(h)$ then $g^{-1} = h^{-1}$, so ... If $f$ onto? Given $g\in G$ can you find an element $h\in G$ such that $f(h) = h^{-1} = g$? How would you pick such an $h$?
H: Calculate limit using Stolz-Cesàro theorem Can someone help me calculate this limit using the Stolz-Cesàro theorem? $\lim_{n\to \infty } \frac{1+\frac12+......+\frac1n}{\ln n}$ AI: $\ln n$ is unbounded and increasing and hence we can use the theorem: $$ \displaystyle\lim_{n\to \infty } \frac{1+\frac12+......+\frac1n}{\ln n}=\displaystyle\lim_{n\to \infty } \frac{\frac1{n+1}}{\ln (n+1)-\ln n}\\ =\displaystyle\lim_{n\to \infty } \frac{\frac1{n+1}}{\ln \frac {(n+1)}n}\\ =\displaystyle\lim_{n\to \infty } \frac{\frac{-1}{(n+1)^2}}{\frac1{(n+1)}-\frac 1n}\\ =\displaystyle\lim_{n\to \infty } \frac{\frac{-1}{(n+1)^2}}{\frac{-1}{n(n+1)}}=1\\ $$ This is intuitively clear because, the harmonic series can be written as: $$ 1+\frac12+......+\frac1n=\ln n+\gamma+\epsilon_n $$ where $\gamma$ is Euler constant and $\epsilon_n$ goes to zero as $n\to\infty$.
H: if $f : \mathbb{R} \rightarrow \mathbb{R} $ is continuous and $f(x) \to 0$ as $x \rightarrow \pm \infty $. Real Analysis (Continous function) If if $f : \mathbb{R} \rightarrow \mathbb{R} $ is continuous and $f(x) \to 0$ as $x \rightarrow \pm \infty $. How to prove that $f(x)$ is bounded ? AI: Since $\lim_{x\to \pm \infty} f(x) = 0$, then there exists $R>0$, such that $$\|f(x)\|\leqslant 1,\qquad \mbox{ for } \|x\|\geqslant R.$$ Since $f$ is continuous, by compactness of $[-R,R]$ there exists $C>0$, such that $$\|f(x)\|\leqslant C, \mbox{ for } x\in [-R,R].$$ We can conclude that $$\|f(x)\|\leqslant \max\{ 1, C\},$$ for every $x\in \mathbb{R}$.
H: Limit in form of 0/0 If I multiply by I get zero/2x anyway. What manipulation needed to get 2/3? AI: Remember the identity $$a^3 - b^3 = (a-b)(a^2 + ab + b^2).$$ If you think of $a$ as $(1+x)^{1/3}$ and $b$ as $(1-x)^{1/3}$, you find that multiplying by $$\frac{(1+x)^{2/3} + (1+x)^{1/3}(1-x)^{1/3} + (1-x)^{2/3}}{(1+x)^{2/3} + (1+x)^{1/3}(1-x)^{1/3} + (1-x)^{2/3}}$$ will work.
H: A set of functions which is open in the space $C^1[0,1]$ Let $f:[0,1]\to [0,1]$ be a $C^1$ and increasing function such that $i)$ If $f(p)=p$ then $\|f'(p)\|\ne 1$ I want to prove that there exist an $\varepsilon>0$ such that if $g\in C^1$ and $\|\|f-g\|\|_{C^1}<\varepsilon$ then $g$ also have the two properties. I don't know if this definition requires some other property to the function $f$, but it's something like this (maybe it's this) The norm $\|\|.\|\|_{C^1}$ is defined on the space $C^1[0,1]$ by $\|\|g'\|\|_{C^1} = \|\|g\|\|_{\infty}+\|\|g'\|\|_{\infty}= \sup_{x\in [0,1]}{\|g(x)\|}+\sup_{x\in [0,1]}{\|g'(x)\|}$ EDIT: I considered two assumptions, but as someone said before $i) \Rightarrow ii)$ so I'll only consider this one. Thanks! AI: Let $U$ be the set of all $f$ with the property that if $f(p)=p$ then $f'(p) \ne 1$. Let $E$ be its complement, so that $f \in E$ iff there exists $p$ with $f(p)=p$ and $f'(p)=1$. Let us show that $E$ is closed in $C^1([0,1])$. Suppose $f_n \in E$ and $f_n \to f$ in $C^1$-norm; we must show $f \in E$. By definition of $E$, for each $n$ there exists $p_n$ such that $f_n(p_n) = p_n$ and $f'(p_n)=1$. Since $[0,1]$ is compact, we may pass to a subsequence and assume that $p_n$ converges to some point $p$. Now $f_n$ converges uniformly to $f$, so $f_n(p_n) \to f(p)$ and hence $f(p)=p$. Also, $f_n'$ converges uniformly to $f'$, so $f_n'(p_n) \to f'(p)$ and hence $f'(p)=1$. So $f \in E$.
H: Can this polynomial transformation produce new symmetry? I've got a polynomial transformation on $\mathbb{R}^6$, and I have a conjecture about it, but I'm having a hard time proving it. The transformation looks like this: $ u:= abcde + abc + abe + ade + cde + a + c + e\\ v:= bcdef + bcd + bcf + bef + def + b + d + f\\ w:= cdefa + cde + cda + cfa + efa + c + e + a\\ x:= defab + def + deb + dab + fab + d + f + b\\ y:= efabc + efa + efc + ebc + abc + e + a + c\\ z:= fabcd + fab + fad + fcd + bcd + f + b + d,$ so as you can see, all six formulas are the same, up to a cyclic permutation of the variables. (This arose in the study of periodic continued fractions, but I don't think that really matters right now.) Now, it's clear that if the variables $a,b,c,d,e,f$ satisfy the relations $\{a=c=e, b=d=f\}$, this would imply the corresponding relations $\{u=w=y, v=x=z\}$. Is the converse true? I think it should be, but every way that I try to prove it leads me into horrible thickets of algebra with no way out. Does anyone see a nice way to argue this from symmetry? If the claim is not true, then what's a good strategy for producing a counterexample? I'm only interested in cases where all variables are strictly positive, otherwise we could get a straightforward counterexample using $0$'s. A related question: Do the relations $\{u=x,v=y,w=z\}$ imply the relations $\{a=d,b=e,c=f\}$? AI: With $a=c=f=0$, your equations $u=w=y, v=x=z$ become $bde=0$. So take $b=0$ and $d \ne 0$. EDIT: With the additional constraint that all variables are strictly positive, the answer is yes. Using Maple, I took a "plex" Groebner basis of the ideal generated by the polynomials $u-w,w-y,v-x,x-z$, and the first element factored as $f e^2 d (d-f) (c d e f+c d+c f+e f) (c d e f+c d+c f+e f+2)$. If all variables are strictly positive, the only way this can be $0$ is $d=f$. Substituting $d=f$ into the fourth basis element and factoring gives $ce{f}^{3} \left( ef+2 \right) \left( cf+1 \right) \left( c-e \right) \left( cef+2\,c+e \right)$ so we get $c=e$. Then substituting $d=f$ and $c=e$ into the 15'th basis element gives $bf{e}^{2} \left( -f+b \right)$ so $b=f$. And finally, substituting $d=f$, $c=e$, $b=f$ into the $21$'st basis element gives $e{f}^{2} \left( -e+a \right)$ so $a=e$. ANOTHER EDIT: In principle it might be possible to attack your second question the same way, but so far (after a couple of hours) Maple 17 is still working on a Groebner basis for this one. FURTHER EDIT: It did produce a basis. The first basis element is $\left( def+d-e+f \right) \left( c-f \right) \left( cde+c-d+e \right) \left( cdef+cd+cf+ef+2 \right) \left( cdef+cd+cf+ef \right)$, so besides $c=f$ we could have $e = (d+f)/(1-df)$ with $df < 1$ or $e = (d-c)/(1+cd)$ with $c < d$. After some fiddling, I found a solution $\left\{ a=1/2,b=2,c=3/4,d=1,e=2,f=1/3 \right\}$ to $\{u=x,v=y,w=z\}$. AND MORE: $\{a=3,b=1/2,d=3,f=1,e=(3-c)/(3c+1)\}$ for $0 < c < 3$.
H: Marble ring: no two blacks are adjacent Question: How many ways are there to arrange $20$ marbles, $6$ black and $14$ white, in a ring (circular arrangement) such that no two of the black marbles are next to each other? Comments: It is easier to find the probability, as there are shortcuts one can make without having to find the number of ways. It is also much easier to do it as if the marbles were in a row. But, in a ring, finding the number of ways is harder. Any ideas? AI: To ensure the condition, glue a white marble to the right of each black marble. Now the task is to arrange 6 black(+white) marbles and 8 white marbles. In principle, there are $14\choose 6$ ways to do so. However, (most) patterns come in sets of 14 differing only by rotation. And some (namely $7\choose 3$) point-symmetric patterns come only in sets of 7 differing by rotation. We should also check for patterns with 7fold or 14fold symmetry, but they can be excluded because there are only 6 black marbles. In total there are $$ \frac17{7\choose 3}+\frac1{14}\left({14\choose 6}-{7\choose 3}\right)=217$$ patterns up to rotation (what about reflection?).
H: Disjunctive normal form expansion I do not understand this at all. Find the sum-of-products expansions of these Boolean functions. $F(x, y, z) = x + y + z$ $F(x, y, z) = (x + z)y$ $F(x, y, z) = x$ $F(x, y, z) = x y$ How is $x$ not just $x$? This makes no sense to me at all and my book wants me to memorize about 20 laws and stuff to transform these to some form that I don't understand. I don't understand the point of this, I don't understand the goal of this and I don't understand the procedure. I can't even find an explanation online that makes sense to me. What am I supposed to do? $x+y+z$ looks like a sum of products to me already. What is wrong with it? Why isn't $(x+y)z$ just $xz + zy$? My book says that is wrong. Books answers: $xyz + xy\bar z + x \bar yz + x\bar y \bar z + \bar x yz + \bar x y \bar z + \bar x\bar y z$ *$ xyz + xy \bar z + \bar xyz$ $xyz + xy \bar z + x \bar y z + x \bar y\bar z $ $ x\bar yz + x\bar y \bar z$ AI: I think the goal of the answers your book is using is to guarantee that there is exactly one term which is 1 when the whole thing is 1. One way to do this is to expand the functions into essentially glorified truth tables, and this is what appears to be happening. You can see that every term has three factors, so you're just trying to figure out how to modify your intuitive answers (like $x+y+z$) into something that will do this. A little memorization would probably be good, but you can always logic one of these out, like so: Consider $(x+y)z$. We can still use normal algebra rules, so this is the same as $xz+yz$. That looks a lot like an "intuitive SoP" so that's the first big step. The first term doesn't care about $y$ so we could write it as $xzy+xz\bar y$. The second term doesn't care about $x$ so we could write it as $yzx+yz\bar x$. So the whole thing becomes $xyz+x\bar yz + xyz + \bar xyz$. But $xyz$ is in there twice, so we can take it out. That leaves us with $xyz+x\bar yz+\bar xyz$. You can probably tease out the general algorithm at play: for any term not including $x_i$, multiply it by $(x_i+\bar{x_i})$. Do this until every term in the expanded version has all variables, and then remove duplicates. [Personally I think this is hideous and kind of ruins the point of using function notation in the first place, but… you have to play along with the rules sometimes. Plus, in any case it's probably good to at least know that such a thing can be done.]
H: Find number of ways of selection of one or more letters from the word: AAAABBCCCDEF Hello the question is above , I can't understand such type of questions, please help. the answer is 479 but how does it come? AI: As $A$ occurs $4$ times, we can choose A in $4+1$ ways So, the number of ways of selection of no or more letters from the word: AAAABBCCCDEF will be $N=(4+1)(2+1)(3+1)(1+1)(1+1)(1+1)$ Clearly, this contains one combination of choosing no letter So, the number of ways of selection of one or more letters will be $N-1$
H: Is ${\mathbb Z} \times {\mathbb Z}$ cyclic? Not sure where to go with this, but I don't think it is cyclic.. AI: Hint: supose $\;m,n\in\Bbb Z\;$ are such that $\;\Bbb Z\times\Bbb Z=\langle (m,n)\rangle\;$, then among other things there must exist $\;x\in\Bbb Z\;$ s.t. $$x(m,n)=(1,1)\implies xm=1=xn\implies\ldots ?$$ Well, now it must also be true that there exists $\;y\in\Bbb Z\;$ s.t. : $$y(m,n)=(0,1)\implies ym=0\;,\;yn=1$$ So...
H: Zariski's article on cohomology in algebraic variety Zariski is supposed to have written an article on cohomology in algebraic geometry. It is supposed to be very good and cited as such for instance here. Can anyone supply a precise reference to this? I couldn't find such an article by search for author Zariski in AMS journals. Perhaps it is part of some volume and does not appear in straight search results. AI: The search problem is likely that Zariski was only one of the authors. Here's the full reference: Martin, W. T.; Chern, S. S.; Zariski, Oscar Scientific report on the Second Summer Institute, several complex variables. Bull. Amer. Math. Soc. 62 (1956), 79–141. The full text is even available online: I don't think subscription is required, but I can't tell from here. Link: http://www.ams.org/journals/bull/1956-62-02/S0002-9904-1956-10018-9/
H: Standard inductive problem Question: Prove that $2^n \geq (n+1)^2$ for all $n \geq 6$. I have tried to prove this below and I'm interested if my method was correct and if there is a simpler answer since my answer seems unnecessarily long for such a simple claim. Inductive hypothesis $$2^n \geq (n+1)^2$$ We need to show that $2^{n+1} \geq (n+2)^2$ or alternatively $$2^n2 \geq (n+1)^2 \frac{(n+2)^2}{(n+1)^2}$$ claim $$\frac{(n+2)^2}{(n+1)^2} <2, \forall n \geq 6$$ notice that it is true for $n=6$ and $$\frac{\frac{(n+2)^2}{(n+1)^2}}{\frac{((n+1)+2)^2}{((n+1)+1)^2}}<1$$ so $\frac{(n+2)^2}{(n+1)^2}$ is decreasing as $n$ gets larger so we have proven the above claim. So we have that for all $n \geq 6$ we have $2^n \geq (n+1)^2$ and by our induction hypothesis and $\frac{(n+2)^2}{(n+1)^2} <2$ so we can conclude that $2^n2 \geq (n+1)^2 \frac{(n+2)^2}{(n+1)^2}$. AI: Except for showing a base case I think your method is good. Note that without a base case you do not have an induction, so make sure you pick e.g., $n=6$ and note that $2^6=64\ge (6+1)^2=49.$ You might consider trying to show the inductive hypothesis by showing that $n\ge 6\implies (n+1)^2\ge 2n+3$ and thus $$2\cdot 2^n\ge 2(n+1)^2 \ge (n+1)^2+2n+3=(n+2)^2$$ where $2n+3$ is the difference between $(n+1)^2$ and $(n+2)^2$.
H: Prove that $6$ divides $n(n + 1)(n + 2)$ I am stuck on this problem, and was wondering if anyone could help me out with this. The question is as follows: Let $n$ be an integer such that $n ≥ 1$. Prove that $6$ divides $n(n + 1)(n + 2)$. Note: An integer $a$ divides an integer $b$, written $a\|b$, if there exists $q ∈ Z$ such that $b = qa$. Alternatively, $a\|b$ if dividing $b$ by $a$, $b ÷ a$, results in an integer. Should I do a proof by induction? All help/input is appreciated! AI: The links provided by lab bhattacharjee solve the problem in more generality, but for the sake of completeness... Let us do an inductive proof. For $n = 0$ things are trivial. Suppose we know the claim for $n$, and we want to prove it for $n+1$. By inductive assumption, $$n(n+1)(n+2) = n^3+2n^2 + 2n = 6 q$$ for some $q$. Now, $$(n+1)(n+2)(n+3) = n(n+1)(n+2) + 3(n+1)(n+2) = 6q + 3(n+1)(n+2)$$ A simple argument (use induction again, if you like) shows that $(n+1)(n+2)$ is divisible by $2$, so you can write $(n+1)(n+2) = 2p$. Consequently, $$(n+1)(n+2)(n+3) = 6(q+p)$$ which finishes the problem.
H: Find the total work done by the force field F Find the total work done in moving a particle in a force field given by: $F=(y^2-x^2)i+(2-x+y)j$ along the curve $y=x^3$, from $(-1,-1)$ to $(1,1)$ Help is appreciated! AI: In the case of a constant force and a linear motion, work is given by finding the component of the force acting in the direction of motion and then multiplying that by the distance over which you travel. Things are more difficult when you're moving in the plane. Let ${\bf x}(t)$ give the path along which you move, then $\dot{\bf x}(t) = {\bf v}(t)$ gives the direction of (instantaneous) motion and so the scaler product $\langle {\bf F},{\bf v}\rangle$, also called the dot product and written ${\bf F} \cdot {\bf v}$, gives the component of the force in the direction of (instantaneous) motion. The instantaneous version of multiplication is integration, and so $W = F\times d$ becomes $$W=\int_{t_1}^{t_2} {\bf F} \cdot {\bf v} \, \operatorname{d}\!t$$ Your paths has the equation $y=x^3$ with $-1 \le x \le 1$. We can parametrise this as $x(t)=t$ and $y(t)=t^3$ where $-1 \le t \le 1$. In vector form: ${\bf x}(t) = t{\bf i} + t^3{\bf j}$. The force can be re-written: \begin{eqnarray} {\bf F}(x,y) &=& (y^2-x^2){\bf i} + (2-x+y){\bf j} \\ \\ {\bf F}(t) &=& (t^6-t^2){\bf i} + (2-t+t^3){\bf j} \end{eqnarray} Finally, ${\bf v}(t) = \dot{\bf x}(t) = {\bf i} + 3t^2{\bf j}$. Hence: \begin{eqnarray} W &=& \int_{t_1}^{t_2} {\bf F} \cdot {\bf v} \, \operatorname{d}\!t \\ \\ &=& \int_{-1}^1 \left((t^6-t^2){\bf i} + (2-t+t^3){\bf j}\right) \cdot \left({\bf i} + 3t^2{\bf j}\right) \, \operatorname{d}\!t \\ \\ &=& \int_{-1}^1 t^6-t^2+6t^2-3t^3+3t^5 \, \operatorname{d}\!t \\ \\ &=& \int_{-1}^1 t^6+5t^2-3t^3+3t^5 \, \operatorname{d}\!t \\ \\ &=& \left[ \tfrac{1}{7}t^7+\tfrac{5}{3}t^3-\tfrac{3}{4}t^4+\tfrac{1}{2}t^6 \right]_{-1}^1 \\ \\ &=& \tfrac{76}{21} \end{eqnarray}
H: Why output of FFT is same as input data size ? What I understand from DFT formula below I can decide the N my self. I can try to use just 16 bins to describe a function or I may even use 4 , it won't be very accurate but I can do it right? The confusing part is even most thrusted websites say next sentence taken from this link: The most general case allows for complex numbers at the input and results in a sequence of equal length . That is extremely strange to me because lets say I have 100000 data , and complexity of fourier is Nlog(N), do I have to go with 100000log(100000). Why can't I go with less accurate but more efficient N=16. Why libraries won't implement like this? Ps: As a computer science graduate I am sorry if my question is not scientific enough but I asked it to stackoverflow, DSP site etc.. but couldn't clear my mind. I hope I made my point this time . AI: You can certainly use $16$ or $4$ bins, yes. I find it strange that libraries don't implement arbitrary bin counts, but I suppose I would shrug and say it's probably because they're written for people dealing with much more data than you. (Unless you're just trying to use 16 bins for large data sets for your own amusement, but then you should probably be writing your own DFT functions :P )
H: How to find $\lim_{x\rightarrow\infty}\bigl(\frac{x+1}{x-2}\bigr)^{2x-1}$ $$\lim_{x\rightarrow\infty}\left(\frac{x+1}{x-2}\right)^{2x-1}$$ What are the steps to solve it? Probably the division should be multiplied by some expression. AI: As $\displaystyle \frac{x+1}{x-2}=1+\frac3{x-2},$ $$\lim_{x\to\infty}\left(\frac{1+x}{-2+x}\right)^{-1+2x}$$ $$=\left(\lim_{x\to\infty}\left(1+\frac3{-2+x}\right)^{\frac{x-2}3}\right)^{\lim_{x\to\infty}\frac{3(2x-1)}{x-2}}$$ $$=e^6$$ as $\lim_{x\to\infty}\frac{3(2x-1)}{x-2}=3\lim_{x\to\infty}\frac{2-\frac1x}{1-\frac2x}=6$ and $\lim_{y\to\infty}\left(1+\frac1y\right)^y=e$
H: Existence of Algebra of anticommuting idempotents Background and motivation: I'm wondering about the existence of an algebra which is in some ways similar to the exterior algebra, but is generated by idempotents rather than nilpotents. Let $V$ be a finite-dimensional real vector space and let $\{e_i\}$ be a basis of $V$. Given that the exterior algebra $\Lambda(V)$ associated to a vector space is the algebra $T(V)/I$, where $T(V)$ is the tensor algebra associated with $V$ and $I$ is the ideal generated by elements of the form $x\otimes x$, where $x\in V$, it then follows that the inclusion of $V$ in $\Lambda(V)$ is an injection. Let $n_i$ denote the image of $e_i$ under this inclusion. It then follows that the elements $n_i$ satisfy the anticommutation relations $\{n_i, n_j\} = 0$. I am curious about the existence or non-existence of an algebra $A(V)$ associated with $V$ which satisfies the following properties: first, as with the exterior algebra, it is possible to inject $V$ into $A(V)$. Second, if $p_i$ is the image of $e_i$ under this injection, then the elements $p_i$ satisfy the anticommutation relations $\{p_i, p_j\} = \delta_{ij} p_i$. If such an algebra exists, by universality of the tensor algebra $T(V)$ it should be isomorphic to $T(V)/J$ for some ideal $J$. What would be this ideal J? AI: $J$ would be the ideal of $T(V)$ generated by elements of the following set $\{p_i\otimes p_j+p_j\otimes p_i\mid i\neq j, i,j\in I\}\cup\{(p_i\otimes p_i)-p_i \mid i\in I \}$ This effectively forces the relations you seek to hold for the $p_i$. (I haven't checked the details that $p_i\mapsto p_i$ is an injection, but my gut says it is so, since none of the relations look like they can introduce linear dependence among the $p_i$.)
H: Random Variable with density and E(X) $X$ is a random variable with value $0,1,2$ and $E(X)=1$, $E(X^{2})=3/2$. Find $f(x)$=the density of $X$ and find $E(X^{7})$=? Here is what I did: $E(X) = 0f(0)+1f(1)+2f(2)$ $E(X^{2}) = 0^{2}f(0)+1^{2}f(1)+2^{2}f(2)$ Then I got: $f(1)+2*f(2)=1$ $f(1)+4f(2)=3/2$ Then I got: $2f(2)$ = ? I know this is suppose to help you get $f(2)$ My question is how do I get $2f(2)$ = ? to help me get $f(2)$=? from this. AI: Let $a=\Pr(X=0)$, $b=\Pr(X=1)$, and $c=\Pr(X=2)$. Then, as you did, we obtain $b+2c=1$ and $b+4c=\frac{3}{2}$. Solving this system of two linear equations, we obtain $c=\frac{1}{4}$ and $b=\frac{1}{2}$. To solve, use the fact that $(b+4c)-(b+2c)=\frac{3}{2}-1=\frac{1}{2}$, and therefore $2c=\frac{1}{2}$. Thus $E(X^7)=\frac{1}{2}\cdot 1^7+\frac{1}{4}\cdot 2^7$.
H: Proving logical equivalence: $P \Leftrightarrow P \vee (P \wedge Q)$ I'm a first year CS student about to write his first term test and this question is part of our practice package. I have not been successful in writing a sequence of equivalences to justify this proof. $P \Leftrightarrow P \vee (P \wedge Q)$ I've tried using the distributive property and simplifying conjunction and disjunction pairs on the right side but have only got to this step so far. $$P \vee (P \wedge Q) \\ \Leftrightarrow (P \vee P) \wedge (P \wedge P) \\ \Leftrightarrow P \wedge P \wedge Q \\ \Leftrightarrow P \wedge Q$$ Is there a "trick" for these types of proofs? A special technique? Any form of help is appreciated! AI: We prove each direction of the double implication. $$A \equiv B \Leftrightarrow (A\Rightarrow B) \land (B \Rightarrow A)$$Note that $$P \implies (P \lor (P\land Q))\equiv \lnot P \lor P \lor (P\land Q)\equiv T \lor (P\land Q) \equiv T\tag{$\Rightarrow$}$$ Note: In the preceding string of equivalencies, $T$ denotes "true", and infact, we see that the forward implication is a tautology. Now, for the $\Leftarrow$ direction: $$P\lor (P \land Q) \equiv (P\lor P) \land (P \lor Q) \equiv P \land (P \lor Q) \implies P\tag{$\Leftarrow$}$$ Therefore $$P \iff (P \lor (P \land Q))$$
H: Degree of Johnson graphs and their subgraphs What is the degree of Johnson graph $J(n,k)$ where $n>k$? http://en.wikipedia.org/wiki/Johnson_graph What are some good examples of subgraphs of Johnson graphs? The Johnson graph I am interested in is $J(n,2)$. AI: From the Wikipedia page: Johnson graphs have ${n \choose k}$ vertices and $\frac{k(n-k)}{2}{n \choose k}$ edges. So there is a natural number to try.
H: Calculate a ratio from part of a range. I'm building a smartphone javascript application but my question today is really only math related. To give you a quick idea of what I'm doing, my code checks the smartphone's gyroscope to determine how much it is rotated. Based on that rotation, it's supposed to pan a background image around. So in the image below, the red box is stationary, while the photo is supposed to pan around in the background as you rotate your device. Anyway, here's the math problem (focusing only on moving the X axis): The gyroscope gives a reading of 0 to 360 The photo in the background gets moved from the top left, so its max and min values are 0 to -100 Therefore the formula for how much to move the image based on the device rotation would be a simple ratio: (Gyroscope Reading * -100)/360 However, I don't want people to need to turn their device all the way over to achieve the full animation, so I'm only interested in the gyroscope readings of about 140 to 220. So, what formula would achieve 140 = 0 and 220 = -100? The formula should maintain the ratio for all points in between. AI: You want a line that goes through $(140,0)$ and $(220,-100)$. The two point form says it is $y-0=\frac {-100-0}{220-140}(x-140)$. $x$ is the gyro, $y$ the output. I left the constants so you could match them up with the equation in case you want to change them in the future.
H: To Prove That a Certain Set is a Manifold Definitions and Notation: Let us write $\underbrace{\mathbb R^n\times \cdots\times\mathbb R^n}_{m \text{ times}}$ as $(\mathbb R^n)^m$. A rigid motion in $\mathbb R^n$ is a function $L:\mathbb R^n\to \mathbb R^n$ such that $\|\|L(x)-L(y)\|\|=\|\|x-y\|\|$ for all $x,y\in \mathbb R^n$. Let $U$ and $V$ be open sets in $\mathbb R^n$. A function $h:U\to V$ is said to be a diffeomorphism if it is differentiable and has a differentiable inverse. A subset $M$ of $\mathbb R^n$ is said to be a $k$-dimensional manifold in $\mathbb R^n$ if for every point $x\in M$ we have: There is an open set $U$ containing $x$, an open set $V\subseteq \mathbb R^n$, and a diffeomorphism $h:U\to V$ such that $h(U\cap M)=V\cap(\mathbb R^k\times\{0\})=\{y\in V:y_{k+1}=\cdots=y_n=0\}$. (The above definition is taken from Spivak's Calculus on Manifolds) The Problem: It is known that any rigid motion in $\mathbb R^n$ can be written as a translation composed with an isometry. Let $p=(p_1,\ldots,p_m)\in(\mathbb R^n)^m$. A point $q=(q_1,\ldots,q_m)\in(\mathbb R^n)^m$ is said to be congruent to $p$ if there exists a rigid motion $L$ in $\mathbb R^n$ such that $q_i=L(p_i)$ for $1\leq i\leq m$. Let $M(p)$ be the set of all the point in $(\mathbb R^n)^m$ which are congruent to $p$. I need to show that $M(p)$ is a smooth manifold. Right now I am just trying to show that $M(p)$ is a manifold. I think the isometries and translations are themselves manifolds and that may be the key to solve this problem. I don't know anything about manifolds except for the definition I have given above and I am new to these things. Can anybody see how to do that? AI: Let $G$ denote the group of isometries of $ R^n$ and let $H$ denote the stabilizer of your point configuration $p$. Then $M(p)$ is naturally diffeomorphic to $G/H$. Note that both $G$ is a Lie group and $H$ is its closed Lie subgroup. Hence, $G/H$ is a smooth manifold.
H: Showing divergence Suppose $a_n>0$, $s_n=a_1+ \dots+ a_n$, and $\sum a_n $diverges. I need to prove that $\sum \frac{a_n}{1+a_n}$ diverges. My attempt: We have $\forall n \in \mathbb{N}^, a_n>0$ $\forall n \in \mathbb{N}^, \frac{a_n}{1+a_n}-a_n\\= \frac{a_n^2}{1+a_n} \sim {a_n}^2 $ Since $\sum a_n$ diverges, then $\sum {a_n}^2$ diverges (is this true ?) By the equivalence theorem, we can deduce that the series $\sum\big(\frac{a_n}{1+a_n}-a_n\big) $ diverges. Since $\forall n \in \mathbb{N}^*, \frac{a_n}{1+a_n}=\big(\frac{a_n}{1+a_n}-a_n\big)+a_n$ Then: $\sum\frac{a_n}{1+a_n}$ diverges Can someone verify my proof ? AI: HINT: argue for a contradiction, and assume $\sum\frac{a_n}{1 + a_n}$ converges. Then, it must be the case that $\frac{a_n}{1 + a_n} \to 0$. Use this to show that $a_n$ is eventually bounded by $1$, then use this to show that for some $N$, if $n>N$ then $$ a_n \leq C\frac{a_n}{1 + a_n} $$ and finish using the comparison test.
H: What is the connectivity between Boltzmann's entropy expression and Shannon's entropy expression? What is the connection between Boltzmann's entropy expression and Shannon's entropy expression? Shannon's entropy expression: $$ S= -K\sum_{i=1}^np_i\log (p_i) $$ AI: Mathematically they only differ in the constant multiplication. Indeed the Shannon entropy of a discrete random variable $X$, in the current literature of information theory, is defined as: $$ H(X)=\mathbb E(-\log P_X(X))=-\sum_{i=1}^{n}p_i\log p_i. $$ The base of the logarithm may vary therefore one can present Shannon entropy as follows: $$ H(X)=C \mathbb E(-\log P_X(X))=-C\sum_{i=1}^{n}p_i\ln p_i $$ Boltzmann entropy is as follows: $$ S=-K\sum_{i=1}^{n}p_i\log p_i. $$ where $K$ is Boltzmann constant. There is an interesting story behind the term of Entropy in information theory. It was indeed Von Neumann who suggested Shannon to use "Entropy" for its basic mathematical concept. Beyond the similar appearance, they are used in information theory and statistical physics for different purposes. In information theory, Shannon Entropy is intuitively measure of uncertainty. However there exists other notions such as Renyi Entropy in the literature. It is then used as a fundamental concept for treating the problem of source coding, channel coding and rate distortion etc. The entropy is a fundamental concept here. It is either derived from a set of axioms or is defined directly. In statistical mechanics, the entropy is derived from other existing concepts such as partition function or Helmholtz free energy. A deeper connection between two concepts can be explored by finding connections between statistical physics and information theory. Work of Jaynes is a valuable piece connecting these two domains. For more recent connection between two domains look at: Information, Physics, and Computation, Marc Mézard and Andrea Montanari Statistical Physics and Information Theory, Neri Mehrav here and here.
H: If the square of an integer is odd, then the integer is odd The statement is: If the square of the integer x is odd, then x is odd. My textbook says an indirect proof is applicable here but I came up with the following: State that If $x^2$ is odd, then $x$ is odd. Next, we assume that $x$ is odd. We know if $x$ is an integer, then $x = 2k+1$ given some integer $k$. This gives us $x^2 = 2k+1$ $(2k+1)^2 = 2k+1$ We know that the square of an odd integer is odd and that $2k+1$ is odd because $k$ is an integer. Therefore, QED. Have I overlooked something? Can this be performed with a direct proof somehow? AI: The direct proof has to start with you assuming $x^2$ is odd, then proving $x$ must be odd. You have sort of assumed that both $x$ and $x^2$ are odd, and then shown that 'nothing bad happens' which is not a rigorous proof. An indirect proof (via contrapositive) will probably be easier. Note that in this case, the contrapositive is "if $x$ is not odd, then the square of $x$ is not odd" i.e. "if $x$ is even, then the square of $x$ is even". You can then use the same ideas, i.e. you can say "if $x$ is even, then $x = 2k$ for some integer $k$..." then you have to show that, if you take $x^2$, there is some other integer $m$ so that $x^2 = 2m$. You should try to figure out what $m$ is on your own, but if you are still stuck after some effort, $m = 2k^2$
H: Did I answer this probability question correctly? Hello I have the following question to answer: Imagine three boxes, each of which has three slips of paper in it each with a number marked on it. The numbers for box A are 2, 4 and 9, for box B 1, 6 and 8, and for box C 3, 5 and 7. One slip is drawn, independently and with equal probability, from each box. Compute P{A Slip > B slip} (a slip taken from A is greater than a slip from B) My attempted solution: First note the possible outcomes: (highlighted are outcomes that satisfy A>B) {2,1} \|\| {2,6} \|\| {2,8} {4,1} \|\| {4,6} \|\| {4,8} {9,1} \|\| {9,6} \|\| {9,8} Since there are 5 possible outcomes where A>B and since the events are independent from one another then the probability is simply: (5/9) * P{B\|A}, where B is a slip chosen from B that is smaller than A The probability P{B\|A} is P{B\|2} + P{B\|4} + P{B\|9} = 0+1/3+1 = 4/3 So finally P{Aslip>Bslip} = (5/9)(4/3) = 20/27 or 74% AI: There are $3$ equally likely possibilities for A's number, $2$, $4$, and $9$. If A got $2$, then the probability that A beat B is $\frac{1}{3}$, for B must get $1$. If A got $4$, then again the probability that A beat B is $\frac{1}{3}$. If A got $9$, then the probability that A beat B is $1$, or more prettily $\frac{3}{3}$. Thus the probability that A beats B is $\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{3}{3}$. Remark: Your counting argument gave the same result, but then got spoiled. The problem will become interesting when you compute the probability that B beats C, and the probability that C beats A.
H: How to construct the graph from an adjacency matrix? I have the following adjacency matrix: a b c d a [0, 0, 1, 1] b [0, 0, 1, 0] c [1, 1, 0, 1] d [1, 1, 1, 0] How do I draw the graph, given its adjacency matrix above (I've added a,b,c,d to label vertices). I don't understand how the vertex $d$ (e.g., the row $d$) is adjacent to the vertex $b$, but the vertex $b$ (the row $b$) is not adjacent to the vertex $d$ (the column $d$). Is this possible? Thanks! EDIT: Maybe it's directed? If so, would that explain why d --> b, but b =/ d? AI: It is a directed graph, i.e. \begin{equation} \begin{matrix} a & & b\\ \updownarrow & X & \uparrow\\ c &\leftrightarrow & d \end{matrix} \end{equation} where I used $X$ to denote a double-sided link between $a$-$d$ and $c$-$d$. The direction of the edges is convention defined and could be reversed.
H: Norms on $\mathcal{M}_{n \times n}(\mathbb{R})$ On the space of all matrices $n \times n$ with real coefficients $\mathcal{M}_{n \times n}(\mathbb{R})$ we define two norms: $\|\|A\|\|_1 := sup_{x\neq 0; x \in \mathbb{R}^n} \frac{\|\|Ax\|\|}{\|\|x\|\|} \\ \|\|A\|\|_2 := max \|A_{ij}\|.$ It's quite evident that these are indeed norms, but I'm having trouble proving that there exists a $C_n>1$ s.t. $\frac{1}{C_n } \le \frac{\|\|A\|\|_1}{\|\|A\|\|_2} \le C_n$. Could you help me solve this? AI: $\\|A\\|_2\leq \\|A\\|_1\leq n\\|A\\|_2$ see http://en.wikipedia.org/wiki/Matrix_norm
H: Continuously differentiable curves How one can show that the unit cube $[0,1]^d$ cannot be covered by countably many continuously differentiable curves? Thanks in advance AI: By Sard's theorem, if $f:\mathbb R\to\mathbb R^d$ is continuously differentiable and $d>1$, then $f(\mathbb R)$ has measure zero. So, a countable union of continuously differentiable curves would have measure zero in $\mathbb R^d$, while the unit cube has measure $1$.
H: Can there be a real solution to the square root of -1? Basically what the title says: Can there be a real solution to the square root of $-1$ (or any negative number in fact) or is it defined to be unreal? Because of this: $$ \begin{align} \sqrt{-1} & = (-1)^\frac 12 \\[6pt] & = (-1)^\frac 24 \\[6pt] & = \sqrt[4]{(-1)^2} \\[6pt] &= \sqrt[4]{1} \\[6pt] &= 1 \end{align}$$ AI: Nope, unfortunately. Your logic seems to imply that $1^2 = -1$ which isn't the case. It also implies that root 4 of 1 is only 1. This is kinda why complex numbers were invented. Now, you could invent your own number system and call them "real numbers" or re-define the square root to mean something else.
H: hitting time of Brownian motion I'm desperately trying to prove that for a standard BM $B_t$ the stopping time $T:=\inf\{t\geq0: B_t\geq\sqrt{1+t}\}$ is a.s. finite, i.e. $\mathbb{P}[T<\infty]=1$. I actually tried to play around with $S_t=\sup_{s\in[0,t)}\frac{B_s}{\sqrt{1+s}}$ which should have the half-normal distribution, with the parameters coming from a $\mathcal{N}(0,\frac{s}{1+s})$ distributed r.v. But I cannot see how to prove that $S_t$ must be bigger than one for a certain $t$ a.s. Thanks in advance! AI: The law of the iterated logarithm says that $B_t$ hits the barrier $\sqrt{at\log\log t}$ almost surely and at unbounded times, for every $a\lt2$ (the version of the law for random walks in discrete time is enough). A fortiori, your $T$ is almost surely finite.
H: Showing rational numbers are algebraic A polynomial with integer coefficients is an expression of the form: $f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$ where $a_n$, $a_{n-1}, \ldots, a_1, a_0$ are integers and $a_n$ is not equal to $0$. a zero of the polynomial is a $c \in \mathbb{R}$ such that $f(c)=0$ A real number is said to be algebraic if it is a zero polynomial with integer coefficients 1) Show that every rational number is algebraic 2) Show that if $a$, $b$ and $k$ are positive integers, then the $k$-th root of $a/b$ is algebraic I don't even know where to start on this. What is a zero of a polynomial with integer coefficients? AI: The rational number $5/7$ is a zero of the polynomial $7x+(-5)$. We have $n=1$, $a_1=7$, $a_0=-5$. So try showing that works with every rational number.
H: Why does there seem to be so much error in the laws of sines and cosines? I've been computing the angles of a triangle with sides a = 17, b = 6 and c = 15 using the law of cosines to find the first angle and then the law of sines to find the other 2. I follow the convention of naming the angles opposite these sides A, B and C respectively. Here are my results: $ C = \arccos( \frac {6^2+17^2-15^2}{2(6)(17)}) = 60.647$ degrees to 3 d.p. $ B = \arcsin( \frac {6 \sin C}{15}) = 20.405$ degrees to 3 d.p. $ A = \arcsin( \frac {17 \sin B}{6}) = 81.051$ degrees to 3 d.p. Clearly, adding these should give $180$ degrees, but it gives 162 degrees to 3 s.f. Assuming I haven't made any mistakes, the error seems quite high and I'm just wondering if anyone knows why this is? It seems high enough to challenge the validity of the laws. AI: OK, I did the Law of Cosines 3 times and came up with 60.647 , 20.404 and 98.949 respectively for angles A, B and C. Remember, the Law of Cosines does not have an ambiguous case, unlike the Law of Sines. I suspect (without further investigating) that his may be the culprit. My advice: Always use the Law of Cosines whenever you can. In this case, when all sides are known, clearly a case for Law of Cosines
H: How to do this integral analytically? How to evaluate the following integral analytically $$f(x,y)=\int_{\Omega }d^2ke^{i(xk_x+yk_y)}$$, where the region $\Omega=\left \{ (k_x,k_y):k_x^2+k_y^2\leq K^2 \right \}$ with radius $K>0$. What's the exact form of $f(x,y)$? Thank you very much. AI: Use polar coordinates, and the fact that $x k_x+y k_y = r \rho \cos{(\theta-\phi)}$, where $r=\sqrt{x^2+y^2}$ and $\rho=\sqrt{k_x^2+k_y^2}$. We therefore get $$f(x,y) = \int_0^K d\rho \, \rho \int_0^{2 \pi} d\theta \, e^{i r \rho \cos{(\theta-\phi)}}$$ The inner integral is a Bessel function, so we get $$f(x,y) = 2 \pi \int_0^K d\rho\, \rho \, J_0(r \rho) = \frac{2 \pi}{r^2} \int_0^{K r} du \, u\, J_0(u) = 2 \pi K^2 \frac{J_1(K r)}{K r}$$
H: Solving imaginary equation $z^3 = 5i + 5$ I need help solving the equation : $z^3 = 5i + 5$. I'm basically just starting to learn imaginary numbers and how to solve them. Straight forward solution is a pain ( was trying to figure out with step-by-step solution from wolframalpha ). After that I was reading wikipedia about them and came across with the De Moivre's formula that states $z^n = \|z\|^n (\cos(n\phi) + i\sin(n\phi))$ but I have no clue how to use it with this example. I would greatly appreciate some help on this. Maybe it is possible to do it another way? Thanks in advance. AI: $$z^3=5(1+i)$$ $$=5\sqrt{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})$$ $$=5\sqrt{2}(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))$$ Do you now see some thing worthy???
H: Prof that series converges $(x_n)$, $x_1=\sqrt2$, $x_2=\sqrt{2}^\sqrt{2}$, $x_3=(\sqrt{2}^\sqrt{2})^\sqrt2$ $x_\left(n+1\right) = (\sqrt2)^\left(x_n\right)$ Does this series converge? AI: Here are the steps you need to prove. First prove that $1 < x_n < 2$ using induction. Nest prove that $x_n$ is a monotone increasing sequence using induction. Now recall that due to completeness of the real numbers a monotone increasing sequence bounded above converges. Call the limit as $L$. Make use of limit laws/rules to conclude that $L = \sqrt{2}^L$ Prove that $L = 2$
H: Given real matrices $A,B$, exists $\tilde{B}$ so that $\langle Bx,x \rangle = \langle \tilde{B}x,x \rangle$ Given $A, B \in \mathbb{R}^{n \times n}$, show there exists $\tilde{B}\in \mathbb{R}^{n \times n}$ such that $\langle Bx, x\rangle = \langle \tilde{B}x, x \rangle$ for all $x\in \mathbb{R}^n$ $\langle (A^2 + \tilde{B}^2) x,x\rangle \geq \langle (A\tilde{B} + \tilde{B}A)x, x\rangle$ for all $x\in \mathbb{R}^n$ My attempt so far: Note that $(A-\tilde{B})^2$ positive semi-definite is sufficient for (2). For this it is sufficient that $A-\tilde{B}$ be symmetric. Note that $B-\tilde{B}$ must have no eigenvalues in $\mathbb{R}\setminus \{0\}$. Of course if $A,B$ symmetric, we can take $\tilde{B} := B$. I'd like a hint for this one; please don't feed me the answer. Thanks as always, folks AI: Let $B =B_{s}+B_{a}$ where $B_{a}$ and $ B_{a}$ are symmetric and antisymmetric parts. Then $\tilde{B} =B_{s}+X_{a}$ Where $X_{a}$ is any antisymmetric matrix. (This will take care of condition (1)) similarily $A =A_{s}+A_{a}$, then condition (2) implies $(A_{s}-B_{s})^2+(A_{a}-X_{a})^2$ is positive definite. $(A_{s}-B_{s})^2$ is already positive definite while $(A_{a}-X_{a})^2$ is negative definite. Now you have to choose $X_{a}$ so that whole $(A_{s}-B_{s})^2+(A_{a}-X_{a})^2$ is positive definite.
H: Prove by mathematical induction that $\sum_{i=1}^{n}\frac{i}{2^i}\leq2$ for $n\ge 1$ I have this exercise by my professor that I have no idea how to solve. Any help would be greatly appreciated: Using the method of mathematical induction show that for all $n \geq 1$, $n \in\mathbb{N}$ $$\sum_{i=1}^{n}\frac{i}{2^i}\leq2$$ AI: HINT: Start by gathering some numerical data: $$\begin{align} &\sum_{i=1}^1\frac{i}{2^i}=\frac12=2-\frac32\\ &\sum_{i=1}^2\frac{i}{2^i}=\frac12+\frac24=\frac44=2-\frac44\\ &\sum_{i=1}^3\frac{i}{2^i}=\frac44+\frac38=\frac{11}8=2-\frac58\\ &\sum_{i=1}^4\frac{i}{2^i}=\frac{11}8+\frac4{16}=\frac{26}{16}=2-\frac6{16}\\ &\sum_{i=1}^5\frac{i}{2^i}=\frac{26}{16}+\frac5{32}=\frac{57}{32}=2-\frac7{32} \end{align}$$ This strongly suggests that $$\sum_{i=1}^n\frac{i}{2^i}=2-\frac{n+2}{2^n}\;,$$ and you might find it easier to prove this equality by induction; it certainly would imply the desired inequality.
H: How to integrate $\int\limits_{\|z\| = R} \frac{\|dz\|}{\|z-a\|^2}$ I need to integrate, $\int\limits_{\|z\| = R} \frac{\|dz\|}{\|z-a\|^2}$ where $a$ is a complex number such that $\|a\|\ne R$. So first I tried polar coordinates, which gives something I cannot continue. Then I tried to write $\|dz\| = rd\theta = dz/ie^{i\theta}$ and I have $$\int\limits_{\|z\| = R} \frac{dz}{ie^{i\theta}(z-a)(\overline{z}-\overline{a})}$$ which makes me want to use cauchy's integral formula, but I'm not sure if it has a pole at $z = a$ or not. How to I calculate this integral? AI: First sub $z=R e^{i \phi}$, $dz = i R e^{i \phi} d\phi$, $\|dz\|=R d\phi$. Then realize that $$\|z-a\|^2 = R^2 + \|a\|^2 - 2 R \|a\| \cos{\phi}$$ (I set an arbitrary phase to zero - it won't matter for the integration.) The integral then becomes $$R \int_0^{2 \pi} \frac{d\phi}{R^2 + \|a\|^2 - 2 \|a\| R \cos{\phi}}$$ Now - and this might seem weird - we go back to a complex representation so we may evaluate the integral using the residue theorem. That is, set $\zeta = e^{i \phi}$, $d\phi = -i d\zeta/\zeta$ and get that the integral is equal to $$i R \oint_{\|\zeta\|=1} \frac{d\zeta}{\|a\| R \zeta^2 - (\|a\|^2+R^2) \zeta + \|a\| R} $$ To evaluate via the residue theorem, we find the poles of the integrand, which are at $\zeta=\|a\|/R$ and $\zeta=R/\|a\|$. Clearly, the analysis depends on whether $\|a\|$ is greater than or less than $R$. For example, when $\|a\| \lt R$, the integral is, by the residue theorem, $$i 2 \pi (i R) \frac{1}{2 \|a\| R (\|a\|/R) - \|a\|^2-R^2} = \frac{2 \pi R}{R^2-\|a\|^2}$$ The analysis is similar for $R \lt \|a\|$. The end result is that $$\oint_{\|z\|=R} \frac{\|dz\|}{\|z-a\|^2} = \frac{2 \pi R}{\left\|R^2-\|a\|^2 \right\|}$$
H: Combination Question [GRE PREP] Three men (out of 7) and three women (out of 6) will be chosen to serve on a 7 member committee. In how many ways can the committee be formed? I did 7C3 to get 35 men. Then i did 6C3 to get 20 women. Then i decide to add up 20 + 35 and get 55 but it is suggested i have to multiply 35 and 20 instead. I want to know why is it that we are multiplying 35 and 20 instead of adding them up? AI: The choices of men and women are made independently. Independent $\Longleftrightarrow$ Multiply. Why? A concrete example might help. I want to choose 1 fruit from $\{apple,orange,banana\}$ and 1 drink from $\{water,tea\}$. Then I have $3 \times 2 = 6$ choices: $(apple,water)$, $(orange,water)$, $(banana,water)$, $(apple,tea)$, $(orange,tea)$, and $(banana,tea)$. For each fixed choice of the first item I can choose any of the second item. So if I have $n$ choices for item #1 and $m$ choices for item #2: I get $\underbrace{m+m+\cdots+m}_{n-\mbox{times}} = n \cdot m$ choices.
H: Distributive properties of quanitifiers What is the difference between $\forall x~(~P(x) \to Q(x)~)$ and $\forall x P(x) \to \forall x Q(x)$ To me they seem to be the same thing, what difference does it make where the quantifiers go? AI: $\forall x (P(x) \rightarrow Q(x))$ means that for every $x$ that you care to pick, if for that particular $x$ $P(x)$ holds, then $Q(x)$ also holds for that particular $x$. For $x$ that don't satisfy $P(x)$, you know nothing about $Q(x)$. $\forall x P(x) \rightarrow \forall x Q(x)$ means that if $P(x)$ holds for every $x$ at the same time, then $Q(x)$ also holds for every $x$ at the same time. If there is even a single $x$ for which $P(x)$ does not hold, it doesn't tell you anything about $Q(x)$ for any $x$. Example: With the domain ${\mathbb N}$, take $P(x) = x \geq 2$ and $Q(x) = x \geq 3$. Then $\forall x (P(x) \rightarrow Q(x))$ is false, since $x = 2$ is a counterexample. However, $\forall x P(x)$ is false, and therefore the statement $\forall x P(x) \rightarrow \forall x Q(x)$ is true. As an aside, note that the first statement $\forall x (P(x) \rightarrow Q(x))$ implies the second one $\forall x P(x) \rightarrow \forall x Q(x)$.
H: probability without replacement and unknown number in urn An urn contains N red chips, N blue chips, and N white chips all numbered 1 through N. Two chips are drawn at random and without replacement. Let A be the event that the two chips drawn are the same color; and let B be the event that they have the same number. a) Find $P(A)$ b) Find $P(B)$ C) Find the probability that the two chips drawn are either the same color or the same number. Here is what I have so far, but I am not sure as to whether or not it is correct. ANSWER: a) $P(A)= \binom{R}{2} + \binom{B}{2} \binom{W}{2}$ b) $P(B)= \frac{3}{R+B+W} \times \frac{2}{R+B+W-1}$ c) $P(\text{same color or same number})= P(A)+P(B)$ Any help that can be offered with a detailed explanation as to how I would be able to relate it to future problems would be amazing! AI: For the first problem we now that there $\binom{N}{2}$ to choose a pair of red chips, also this holds for the the blue chips and white chips. Also there are $\binom{3N}{2}$ to ways to draw two chips, because there are $3N$ chips in the urn. So the probability is: $$P(A) = \frac{3\binom{N}{2}}{\binom{3N}{2}} = \frac{\frac{3N(N-1)}{2}}{\frac{3N(3N-1)}{2}} = \frac{N-1}{3N-1}$$ For the second case there are 3 balls with same number at the beginning, while for the second draw there are 2 balls with the same number as the one we draw earlier. Note that the first draw doesn't matter, because we only need to draw the same numbered ball as the one we draw earlier. So the probability is: $$P(B) = N \times \frac{3}{N+N+N} \times \frac{2}{N+N+N-1}$$ $$P(B) = \frac{2}{3N-1}$$ And for the third case we need to check when exactly one of the conditions is fulfiled, we are using the exclusive OR. So using inclusion-exclusion we have: $$P(A \oplus B) = P(A) + P(B) - P(A \cup B)$$ But note that the event $A$ and event $B$ never intersect, because it's impossible to draw two balls from the same colour and with same number same number, so we have: $$P(A \oplus B) = P(A) + P(B) - P(A \cup B)$$ $$P(A \oplus B) = \frac{N-1}{3N-1} + \frac{2}{3N - 1}$$ $$P(A \oplus B) = \frac{N+1}{3N-1}$$
H: Rational Zeros Theorem to show irrationality Show that $\sqrt{3}+\sqrt{5}$ is irrational using the rational zeros theorem, you have to find a polynomial that has $\sqrt{3}+\sqrt{5}$ as a zero AI: Let x = √3 + √5 Then x² = 8+2√15 and (x²-8)²=60 from which you can find your polynomial. With the Rational Zero test it is easy to verify that there are no rational zeros
H: What is the meaning of $\frac{0}{0}$? I asked my teacher what is the real meaning of $\cfrac{0}{0}$, and the answer I got was "nobody knows". I can't leave this subject "as is". I need a decent explanation, at least an explanation to why "nobody knows". I'm sure you'll come up with a few good answers. AI: I really hope your calculus teacher didn't say "nobody knows": $\frac{0}{0}$ is pretty well understood. It's just that it's undefined (without further context). What I mean by "context" is: it's possible to give meaning to limits of the form $$\lim_{x \to a} \frac{f(x)}{g(x)}$$ where $\lim_{x \to a}f(x) = 0$ and $\lim_{x \to a} g(x) = 0$, but this depends on your choice of $f$ and $g$. For instance $$\lim_{x \to 0} \dfrac{\sin x}{x} = 1, \qquad \lim_{x \to 0} \dfrac{e^{kx}-1}{x} = k, \quad \lim_{x \to 0} \dfrac{x}{x^3} = \infty, \quad \cdots$$ These are all different, despite the fact they're all of the form $\frac{0}{0}$. So it's not that nobody knows, it's just that it's meaningless! But we can assign meaning if we know how the $0$s in the fraction came to be.
H: Finding Polynomials to Satisfy a Condition I need to find polynomials $x(n), y(n)$ s.t. $x(n)(n^{2}+n+1)+y(n)(n^{2}+1)=1$, $\forall n \in \mathbb{Z}$. I tried distributing it out: $x(n)n^{2} + x(n)n + x(n) + y(n)n^{2}+y(n)=1$. I understand that for this to work, one of the terms has to equal one, and the other terms have to cancel each other out. But the answer isn't coming to me yet. Can anyone give me a hint? AI: Hint: You have $(n^2+1)\left(x(n)+y(n)\right) + n x(n) = 1$. Note that $n^2+1 + n(-n)=1$.
H: Rectangle divided into three triangles with two lines. One angle is given, what are all the others? Let's suppose I have a rectangle divided into three triangles in the following way. No lengths of either the rectangle or triangles are known, only one angle is known. I would like to know how to calculate all the other angles. Here is a sample image of the problem: This looks so simple but I have no idea even where to start. Where is the trick? AI: Without further information, you cannot determine any of the other angles, other than the right angles (since you are given a rectangle). If we know one other non-right angle, we can compute all the other angles by simply knowing that the angles of a triangle sum to $180^\circ$.
H: Characteristic $n$ and local rings Prove that: a) If A is a local ring then A has characteristic zero or a power prime. Proof. Suppose M is the unique maximal ideal of A then $A/M$ is Field in particular integral domain then $Char( A/M ) = 0$ or $p$ with some $p$ prime. If $Char( A/M ) = 0$ then $\forall n \in \mathbb N, $ , $\forall a+M \in A/M $ we have that $ n(a + M )= na + M \ne M$ then $ na \notin M$ $\forall n \in \mathbb N $, so $ n 1 \notin M $, this implies $n 1 \ne m $ $ \forall m \in M $ $ \forall n \in \mathbb N $ then $ n 1 \ne 0 $ $ \forall n \in \mathbb N $ finally $Char ( A )= 0$ Suppose that $Char( A/M ) = p $ and $Char( A ) = n$ then$ n 1 \ne m$ $ \forall m \in M $ , so $ M + n 1 = M $ and since $Char( A/M ) = p $ this implies that $ p \| n$ then $n = pq$ , so we have $ n= p^l m$ where $ (p,m)=1$ but with this there are $ x,y\in \mathbb Z$ such that $ 1= px 1 + my 1$, so $ px 1 $ is unity or $my 1$ is unity. Since $p 1 \in M $ then $px 1 \in M $ this implies $px 1 \notin A^* $ then $ my 1 \in A^* $ ,so for $ a \in A$ such that $my 1 a = 1$ then $m 1$ is unity with this we have that $o(m1) = o(1)$ and since $p^l(m1)=0$ this implies that $$p^lm \| p^l$$ then $$ m=1 $$. Finally we get that $Char( A ) = p^l $ for some integer l and p prime. Is this correct? b) Let A a ring commutative with identity and with characteristic n. If $n= ab$ with $ (a,b)=1$ then A is isomorphic to the direct product of two rings,one of them is characteristic a and the other one is characteristic b. I cannot prove this can someone help me please. AI: (b) Since $a\mathbb{Z}$ and $b\mathbb{Z}$ are coprime, the same is true for $aA$ and $bA$. The Chinese Remainder Theorem implies that $A \cong A/aA \times A/bA$. Now prove that $\mathrm{char}(A/aA)=a$ (resp. for $b$). (a) follows from (b) since a local ring has only trivial idempotents.
H: How do I find $\lim_{x \to 0} \frac{\cos3x-\cos x}{\tan 2x^2}$? Can't understand how to solve limit like this: $$\lim_{x \to 0} \frac{\cos3x-\cos x}{\tan2x^2}$$ My attempt is: $$\lim_{x \to 0} \frac{\cos3x-\cos x}{\tan2x^2}=\lim_{x \to 0} \frac{\cos3x}{\tan2x^2}- \frac{\cos x}{\tan2x^2}=\lim_{x \to 0}\frac{1}{\tan2x^2}\left(\cos3x-\cos x \right) = \lim_{x \to 0} \dots$$ And could you explain me how to do next step? (I know I have to get rid of trigonometry but I don't know how to do this.) AI: We have $\cos(3x) - \cos(x) = -2\sin(2x)\sin(x)$. Hence, $$\dfrac{\cos(3x) - \cos(x)}{\tan(2x^2)} = -2 \dfrac{\sin(2x) \sin(x)}{\tan(2x^2)} = -2 \dfrac{\sin(2x)}{2x} \cdot \dfrac{\sin(x)}{x} \cdot \dfrac{2x^2}{\tan(2x^2)}$$ Now let $x \to 0$ and recall the following limits: $\lim_{y \to 0}\dfrac{\sin(y)}y = 1$ and $\lim_{y \to 0}\dfrac{\tan(y)}y = 1$.
H: Show function $f(x)=\frac{r^2 \cdot a}{a^2+x^2}$ is strictly decreasing I want to show, that $$f(x)=\frac{r^2 \cdot a}{a^2+x^2}$$ with $x,a,r \in \mathbb{R}$ is strictly decreasing for $x>0$ A function is called strictly decreasing, if $x_1<x_2 \Rightarrow f(x_1)>f(x_2)$. Perhaps someone can give me a hint how to start. Thanks a lot in advance. AI: You need $a > 0$ for this to be true. What you want to show is that if $0 < x_1 < x_2$ that you have $$\frac{r^2 a}{a^2 + x_2^2} < \frac{r^2 a}{a^2 + x_1^2} $$ Putting everything on one side this is the same as showing $$\frac{r^2 a}{a^2 + x_2^2} - \frac{r^2 a}{a^2 + x_1^2} < 0$$ Writing with a single denominator this becomes $$r^2a{x_1^2 - x_2^2 \over (a^2 + x_1^2)(a^2 + x_2^2)} < 0$$ This is true since the denominator is positive, and the numerator is negative since $0 < x_1 < x_2$. Note also that if you have derivatives, you can just take the derivative to get $$f'(x) = -r^2a{2x \over (x^2 + a^2)^2}$$ Since this is negative for $x > 0$ the function is decreasing.
H: Which sign does $\lim_{n\to\infty}(-2)^n$ have? How can I express $\lim_{n\to\infty}(-2)^n$ using $\infty$? Which sign does it have, plus or minus? AI: Neither. $\lim_{n \to \infty}(-2)^n$ does not exist.
H: All subsets of $\mathbb{N}$ are open in $\mathbb{N}$? Why precisely is this true? I've always sort of overlooked this and taken it for granted, but why exactly are all subsets of $\mathbb{N}$ open in $\mathbb{N}$? AI: All subsets of $\Bbb N$ are open in $\Bbb N$ if and only if $\Bbb N$ is given the discrete topology, and one can certainly endow $\Bbb N$ with other topologies. However, the topology that $\Bbb N$ inherits from $\Bbb R$ is the discrete topology, and that’s probably the one that you have in mind. To see why this is the discrete topology, let $n\in\Bbb N$; then $(n-1,n+1)$ is an open set in $\Bbb R$, and $$(n-1,n+1)\cap\Bbb N=\{n\}\;,$$ so $\{n\}$ is an open set in the subspace topology on $\Bbb N$. Arbitrary unions of open sets are open, and every non-empty subset of $\Bbb N$ is a union of sets of the form $\{n\}$, so every subset of $\Bbb N$ is open in the topology inherited from $\Bbb R$.
H: Bounding integral of the form $\int_0^\infty \|z+x\|^{-n}dx$ I need to bound a sum of the form $\sum_{j=1}^m \|z+j\|^{-n}$ with $\Re z\ge1$ and $n \ge 2$. I am searching for a bound of the form $$\sum_{j=1}^m \|z+j\|^{-n} \le C \|z\|^{-n+1}.$$ It is easy to see that $$\sum_{j=1}^m \|z+j\|^{-n} \le \int_{0}^{m+1} \|z+x\|^{-n}dx \le \int_{0}^{\infty} \|z+x\|^{-n}dx.$$ My problem is that I cannot compute the integral. By setting $z=r+i s$, we have $$\int_0^\infty ((r+x)^2+s^2)^{-\frac n2} dx.$$ My idea was to use integration by parts to transform it to an integrable form, but so far I cannot find how I am going to do that. Any help is welcome. AI: We have $(r+x)^2 + s^2 \geq (r+x)^2$. Hence, we have $$\int_0^{\infty} \dfrac{dx}{((r+x)^2 + s^2)^{n/2}} \leq \int_0^{\infty} \dfrac{dx}{(r+x)^n} = \dfrac{r^{-n+1}}{n-1} < r^{-n+1}$$ We also have $(r+x)^2 + s^2 \geq x^2+s^2$. Hence, we have \begin{align} \int_0^{\infty} \dfrac{dx}{((r+x)^2 + s^2)^{n/2}} & \leq \int_0^{\infty} \dfrac{dx}{(x^2+s^2)^{n/2}} = \int_0^{\pi/2} \dfrac{\vert s \vert \sec^2(t) dt}{\vert s \vert^n \sec^n(t)}\\ & = \vert s \vert^{-n+1} \int_0^{\pi/2} \cos^{n-2}(t)dt < \vert s \vert^{-n+1} \dfrac{\pi}2 \end{align} Hence, we have (make use of the fact that $\vert r \vert + \vert s \vert \leq 2 \vert z\vert$) $$\int_0^{\infty} \dfrac{dx}{((r+x)^2 + s^2)^{n/2}} \leq C \vert z \vert^{-n+1}$$
H: Prove that $ℤ[i]^= \{1,-1,i,-i\}$ Prove that $ℤ[i]^= \{1,-1,i,-i\}$. $\{1,-1,i,-i\} ⊂ ℤ[i]^$ is trivial. But I'm not sure about the other inclusion. Let $(a+bi) \in ℤ[i]^$. Then there exist $c,d \in ℤ$ such that $(a+bi)(c+di)=1$. Then $ac+bci+adi-bd=1$. Then $bc-ad=0$ and $ac=1+bd$. What can I conclude from this ? AI: Since $z\to\|z\|^2$ is multiplicative, $1=\|a+bi\|^2\|c+di\|^2=(a^2+b^2)(c^2+d^2)$ Since $(a^2+b^2)$ and $(c^2+d^2)$ are positive integers divisible by $1$ they are equal to $1$. Now, either 1) $a=0$ and ($b=1$ or $b=-1$) or 2) $b=0$ and ($a=1$ or $a=-1$) This because $a$, $b$, $c$, and $d$ are integers.
H: What is the definite integral $\int_{-1}^1 \frac{1}{x} dx$ equal to? What is the definite integral $\displaystyle\int_{-1}^1 \frac{1}{x} dx$ equal to? Is it $0$ like it would be if you just integrated and plugged in the bounds, or does the discontinuity at $x=0$ invalidate that line of reasoning and make this integral undefined. If you split up the integral $$\int_{-1}^0 \frac{1}{x} dx+\int_{0}^1 \frac{1}{x} dx.$$ You get $\infty-\infty$ which is undefined. I find it difficult to prove either of these solutions. AI: The improper integral $$ \int^{1}_{-1} \frac{1}{x} \, dx $$ is indeed, as M.Strochyk says, divergent. At least in the sense of Riemann integrability, it doesn't make any sense to try and evaluate this integral because we require our integrand to be bounded over the region we are integrating, which doesn't occur in this case. Also, for this reason, statements such as $$ \int^{1}_{-1} \frac{1}{x} \, dx = \int^{1}_{0} \frac{1}{x} \, dx + \int^{0}_{1} \frac{1}{x} \, dx$$ also make no sense, seeing as the statement is true if and only if the integrals on both sides exist. However, because the function $f : \mathbb{R} \setminus \{0\} \rightarrow \mathbb{R}$ with $f(x)=1/x$ is odd, and we often like to use that if $g(x)$ is an odd function on some interval $I$ with $0 \in I$ then $$ \int^{a}_{-a} g = 0$$ provided $a,-a \in I$, in some sense we would like to give the original integrand (I'd mark it with a star if I knew how, but alas I do not) a similar value of $0$. One way in which we can do it is by defining the Cauchy principal value, in which case after reading the definition we see that $$ \text{P.V.} \int^{1}_{-1} \frac{1}{x} \, dx = \lim_{a \rightarrow 0^{+}} \left( \int^{-a}_{-1} \frac{1}{x} \, dx + \int^{1}_{a} \frac{1}{x} \, dx \right) = 0 $$
H: What is the Y-intercept? In class, my math teacher was teaching us how to draw functions on a coordinate plane, and he mentioned something about the Y-Intercept being an important step in creating/solving a function. But, what exactly is a Y-Intercept? AI: The $y$-intercept of a function $f(x)$ is the point where the function intersects the $y$-axis (if in fact it does intersect the $y$-axis) and it is found by evaluating $f(0)$, i.e., finding the value of $f(x)$ when $x = 0$. For example, the line $f(x) = y = 3x + 2$ intersects the $x$-axis at when $x = 0$: when $\;y = 3\cdot 0 + 2 = 2$. This is the function's y-intercept. The parabola $f(x) = y = 2x^2 + 8$ intersects the $y$-axis when $x = 0:\;$ when $\;y = 2(0)^2 + 8 = 8$. This is the function's y-intercept.
H: Prove that there are no $x,y ∈ \mathbb N$ for which $x^2-y^2 = 10$ I began by factoring and got $(x+y)(x-y) = 10$ Then I tried cases and was able to prove the ones where $x$ and $y$ are equal-> because the equation will result to zero. and also where $x < y$, because the answer will be negative. How can I prove when $x > y$... or is there an easier way to do it? I think this is a proof by contrapositive? AI: Note that $x+y$ and $x-y$ must be of the same parity. If both are even, then $4 \mid (x^2-y^2)$. If both are odd, then $(x^2-y^2)$ is also odd. $10$ is neither divisible by $4$ nor odd.
H: Prove open set is not closed The question might sound ridiculous, but I am not able to prove it with rigor. I tried proving it by the following definitions ONLY. Open set: A set $U$ is open if for every $a$ belonging to $U$, there is some $r = r(a) > 0$ such that the ball $B_r(a)$ is contained in $U$. Closed set: A set $A$ is closed if it contains all of its limit points. I tried using that $A$ closed and open would render $A'$ open and closed (provable using the above definitions) but that didn't lead me anywhere. AI: I believe that the essential question here is this: Why isn't any non-empty proper subset of $\Bbb R^n$ both open and closed? This is another way of saying that $\Bbb R^n$ is connected: there do not exist nonempty, disjoint open sets $A,B\subseteq \Bbb R^n$ such that $A\cup B=\Bbb R^n$ (that is, there do not exist $A$ and $B$ that separate $\Bbb R$. It turns out that the toughest part is showing that $[0,1]$ is connected. This is done by using two facts: Every non-empty set of real numbers which is bounded above has a supremum. Between any two real numbers there is another real number. The proof starts like this: Suppose for the sake of contradiction that $A$ and $B$ separate $[0,1]$, and suppose WLOG that $0\in A$. Let $m=\inf B$. See if you can continue from there.
H: Why can't this number be written as a sum of three squares of rationals? This may be a very naive question and I apologize in advance. Suppose that $n$ is a positive integer which cannot be written as a sum of three squares $a^2+b^2+c^2$ for integers $a,b,c\in\mathbb{Z}$. Does it follow that $n$ cannot be written as a sum of three squares $a'^2+b'^2+c'^2$ for rationals $a',b',c'\in\mathbb{Q}$? Why? AI: There is a fairly easy answer; a positive integer is the sum of three squares if and only if it is not of the form $4^k (8n +7).$ Or, more useful for us, if and only if it is not $4^k m$ with $m \equiv 7 \pmod 8.$ If you have the sum of three rational squares equal to such a number, call it $t,$ then we can multiply through by the square of the least common denominator, which forces everything to be integers, so we have all integers in $x^2 + y^2 + z^2 = t L^2.$ Now, $L$ factors as some power of $2$ times an odd number, so let $L = 2^r s $ with odd $s.$ The important bit is that $s^2 \equiv 1 \pmod 8$ (CHECK THIS). So, $L^2 = 4^r s^2.$ Finally, we knew $t = 4^km$ with $m \equiv 7 \pmod 8,$ because $t$ was not integrally the sum of three squares. So $t L^2 = 4^{k+r} m s^2, $ where $m s^2 \equiv 7 \pmod 8.$ Therefore $t L^2$ is also not the sum of three integer squares and we are done. It is true that positive integers are the sum of three integer squares if and only if they are the sum of three rational squares. This result, for a number of quadratic forms, is generally referred to as the Davenport-Cassels Theorem, but was first proved by Aubry about 1912 I think. It appears in Serre's little book, A Course in Arithmetic, with three squares on pages 45-46, and Davenport-Cassels on pages 46-47. There are infinitely many quadratic forms with the same property, they represent a number over the integers if and only if they represent it over the rationals. The stronger property used in the Davenpost-Cassels hypothesis, for positive quadratic forms, occurs for just 70 forms of the type called "even lattices." The restriction is that the "covering radius" be strictly below $\sqrt 2.$ A full list is given by G. Nebe see http://www.math.rwth-aachen.de/~Gabriele.Nebe/pl.html and the pdf listed as Even lattices with covering radius stricktly smaller than sqrt{2}. Beiträge zur Algebra und Geometrie, Vol. 44, No. 1, 2003, 229-234 Note that I forgot the case R=A2A1A1 in Theorem 7. This root system yields two further lattices, one with covering radius =sqrt{2} and one with c.r. strictly smaller than sqrt{2} There are 103 positive forms in three variables that have your property, that they represent an integer over the integers if and only if they represent it over the rationals. Each is summarized by six coefficients preceded by a "discriminant" I call Delta. Any primitive, positive ternary with the property is "equivalent" to one of these 103. ADC stands for Aubry-Davenport-Cassels. Let's see, only thirteen of the forms below belong on Nebe's list. Her hypotheses are far more restrictive than the adc property. Her list has 70 entries because the number of variables gets as large as ten. $$Q(x,y,z) = a x^2 + b y^2 + c z^2 + r y z + s z x + t x y,$$ and $$\Delta = 4 a b c + r s t - a r^2 - b s^2 - c t^2$$ Delta a b c r s t --------------------------------------------------------- 2: 1 1 1 1 1 1 ADC 2 = 2 3: 1 1 1 0 0 1 ADC 3 = 3 4: 1 1 1 0 0 0 ADC 4 = 2^2 5: 1 1 2 1 1 1 ADC 5 = 5 6: 1 1 2 0 0 1 ADC 6 = 2 * 3 6: 1 1 2 1 1 0 ADC 6 = 2 * 3 7: 1 1 2 0 1 0 ADC 7 = 7 8: 1 1 2 0 0 0 ADC 8 = 2^3 9: 1 1 3 0 0 1 ADC 9 = 3^2 10: 1 1 3 1 1 0 ADC 10 = 2 * 5 10: 1 2 2 2 1 1 ADC 10 = 2 * 5 11: 1 1 3 0 1 0 ADC 11 = 11 12: 1 1 3 0 0 0 ADC 12 = 2^2 * 3 12: 1 2 2 1 1 1 ADC 12 = 2^2 * 3 12: 1 2 2 2 0 0 ADC 12 = 2^2 * 3 13: 1 2 2 1 0 1 ADC 13 = 13 14: 1 1 5 1 1 1 ADC 14 = 2 * 7 15: 1 1 4 0 1 0 ADC 15 = 3 * 5 15: 1 2 2 1 0 0 ADC 15 = 3 * 5 16: 1 2 2 0 0 0 ADC 16 = 2^4 17: 1 2 3 2 1 1 ADC 17 = 17 18: 1 2 3 2 1 0 ADC 18 = 2 * 3^2 18: 2 2 2 1 2 2 ADC 18 = 2 * 3^2 20: 1 1 5 0 0 0 ADC 20 = 2^2 * 5 20: 1 2 3 1 0 1 ADC 20 = 2^2 * 5 20: 1 2 3 2 0 0 ADC 20 = 2^2 * 5 21: 1 2 3 0 0 1 ADC 21 = 3 * 7 21: 1 2 3 1 1 0 ADC 21 = 3 * 7 22: 1 2 3 0 1 0 ADC 22 = 2 * 11 24: 1 1 6 0 0 0 ADC 24 = 2^3 * 3 24: 1 2 3 0 0 0 ADC 24 = 2^3 * 3 24: 1 2 4 2 1 1 ADC 24 = 2^3 * 3 25: 2 2 2 -1 1 1 ADC 25 = 5^2 28: 2 2 3 2 2 2 ADC 28 = 2^2 * 7 30: 1 1 10 0 0 1 ADC 30 = 2 * 3 * 5 30: 1 3 3 1 1 1 ADC 30 = 2 * 3 * 5 32: 1 2 4 0 0 0 ADC 32 = 2^5 33: 1 2 5 1 1 1 ADC 33 = 3 * 11 36: 1 2 5 2 0 0 ADC 36 = 2^2 * 3^2 36: 1 3 3 0 0 0 ADC 36 = 2^2 * 3^2 36: 1 3 4 3 1 0 ADC 36 = 2^2 * 3^2 40: 1 2 5 0 0 0 ADC 40 = 2^3 * 5 42: 1 1 11 1 1 0 ADC 42 = 2 * 3 * 7 44: 1 2 6 2 0 0 ADC 44 = 2^2 * 11 45: 2 2 3 0 0 1 ADC 45 = 3^2 * 5 46: 1 3 5 3 1 1 ADC 46 = 2 * 23 48: 1 2 6 0 0 0 ADC 48 = 2^4 * 3 49: 1 2 7 0 0 1 ADC 49 = 7^2 50: 1 4 4 3 1 1 ADC 50 = 2 * 5^2 56: 1 3 5 2 0 0 ADC 56 = 2^3 * 7 60: 2 2 5 0 0 2 ADC 60 = 2^2 * 3 * 5 60: 2 3 3 0 0 2 ADC 60 = 2^2 * 3 * 5 63: 1 3 6 3 0 0 ADC 63 = 3^2 * 7 70: 1 2 9 0 1 0 ADC 70 = 2 * 5 * 7 72: 2 2 5 1 1 1 ADC 72 = 2^3 * 3^2 72: 2 3 3 0 0 0 ADC 72 = 2^3 * 3^2 75: 1 4 5 0 0 1 ADC 75 = 3 * 5^2 78: 1 5 5 4 1 1 ADC 78 = 2 * 3 * 13 84: 1 1 21 0 0 0 ADC 84 = 2^2 * 3 * 7 90: 1 1 30 0 0 1 ADC 90 = 2 * 3^2 * 5 92: 2 3 5 2 0 2 ADC 92 = 2^2 * 23 99: 2 3 5 3 1 0 ADC 99 = 3^2 * 11 100: 2 2 7 -1 1 1 ADC 100 = 2^2 * 5^2 100: 2 3 5 0 0 2 ADC 100 = 2^2 * 5^2 112: 2 3 5 2 0 0 ADC 112 = 2^4 * 7 120: 1 3 10 0 0 0 ADC 120 = 2^3 * 3 * 5 121: 1 3 11 0 0 1 ADC 121 = 11^2 126: 3 3 5 3 3 0 ADC 126 = 2 * 3^2 * 7 140: 1 2 18 2 0 0 ADC 140 = 2^2 * 5 * 7 147: 3 3 5 -2 2 1 ADC 147 = 3 * 7^2 150: 2 5 5 5 0 0 ADC 150 = 2 * 3 * 5^2 156: 2 3 7 0 2 0 ADC 156 = 2^2 * 3 * 13 169: 2 5 5 -3 1 1 ADC 169 = 13^2 180: 2 2 15 0 0 2 ADC 180 = 2^2 * 3^2 * 5 200: 1 5 10 0 0 0 ADC 200 = 2^3 * 5^2 234: 2 3 11 3 2 0 ADC 234 = 2 * 3^2 * 13 240: 2 5 6 0 0 0 ADC 240 = 2^4 * 3 * 5 252: 3 3 7 0 0 0 ADC 252 = 2^2 * 3^2 * 7 289: 3 5 6 1 2 3 ADC 289 = 17^2 294: 5 5 5 -3 3 4 ADC 294 = 2 * 3 * 7^2 300: 1 10 10 10 0 0 ADC 300 = 2^2 * 3 * 5^2 350: 3 3 10 0 0 1 ADC 350 = 2 * 5^2 * 7 360: 1 3 30 0 0 0 ADC 360 = 2^3 * 3^2 * 5 450: 5 5 6 0 0 5 ADC 450 = 2 * 3^2 * 5^2 468: 1 6 21 6 0 0 ADC 468 = 2^2 * 3^2 * 13 490: 3 3 14 0 0 1 ADC 490 = 2 * 5 * 7^2 588: 3 7 7 0 0 0 ADC 588 = 2^2 * 3 * 7^2 600: 2 5 15 0 0 0 ADC 600 = 2^3 * 3 * 5^2 700: 5 6 6 2 0 0 ADC 700 = 2^2 * 5^2 * 7 720: 2 6 15 0 0 0 ADC 720 = 2^4 * 3^2 * 5 882: 2 11 11 1 2 2 ADC 882 = 2 * 3^2 * 7^2 900: 3 10 10 10 0 0 ADC 900 = 2^2 * 3^2 * 5^2 980: 6 6 7 0 0 2 ADC 980 = 2^2 * 5 * 7^2 1014: 1 13 23 13 1 0 ADC 1014 = 2 * 3 * 13^2 1200: 1 10 30 0 0 0 ADC 1200 = 2^4 * 3 * 5^2 1764: 1 21 21 0 0 0 ADC 1764 = 2^2 * 3^2 * 7^2 1800: 5 6 15 0 0 0 ADC 1800 = 2^3 * 3^2 * 5^2 2028: 2 7 39 0 0 2 ADC 2028 = 2^2 * 3 * 13^2 2450: 1 9 70 0 0 1 ADC 2450 = 2 * 5^2 * 7^2 3042: 3 17 17 8 3 3 ADC 3042 = 2 * 3^2 * 13^2 3600: 3 10 30 0 0 0 ADC 3600 = 2^4 * 3^2 * 5^2 4900: 2 18 35 0 0 2 ADC 4900 = 2^2 * 5^2 * 7^2 6084: 6 13 21 0 6 0 ADC 6084 = 2^2 * 3^2 * 13^2
H: Is there a fast way to compute coefficient of some term of the product of some series'? The example in wikipedia is $$A=1-3x+5x^2-7x^3+9x^4-11x^5+\cdots$$ $$B=2x+4x^3+6x^5+\cdots$$ $$AB=2x-6x^2+14x^3-26x^4+44x^5+\cdots$$ And the term $x^5$ is given by $$44x^5=(1\cdot6x^5)+(5x^2\cdot4x^3)+(9x^4\cdot2x)$$ So if you wanted the 5th term of $AB$, you would get $44$ in this example. AI: Given $A=\sum_{k=0}^\infty a_kx^k, B=\sum_{k=0} b_kx^k$, their product $AB$ is known as the series $$ C=\sum_{k=0}^\infty c_kx^k, $$ where $c_k=\sum_{j=0}^k a_jb_{k-j}$. Look up Cauchy product for more info.
H: Derivative where x=a If $f(x)=x^3+3x+2$ Find the number(s) a such that the tangent lines to the graphs of $f(x)$ and $f'(x)$ at $x=a$ are the same. So far I have come up with: $f'(x)=3(x^2+1)$ And when I graph both these functions out on my calculator the only point I am finding where $x=a$ is at $1$ (which could be incorrect as it is hard to tell exact points using only graphs). First is this correct? and second is there a better way of coming to that conclusion rather than graphing it? AI: For "the tangent lines to the graphs of $f(x)$ and $f'(x)$ at $x=a$ are the same" to be true, you need two things: The tangent lines to pass through the points $\left( a, f(a)\right)$ and $\left( a, f'(a)\right)$ which, assuming the tangents are not vertical, requires $ f(a)= f'(a) $ The tangent lines to have the same gradient which, assuming the tangents are not vertical, requires $ f'(a)= f''(a) $ The tangent lines are not vertical $-$ you can check $f'(x)$ and $f''(x)$ $-$ so these will give you two equations to solve: one a cubic and the other a quadratic. Solve one of them and see which of its solutions satisfy the other.
H: If the row-reduced form of matrix $A$ has a row of zeros, its columns do not span $\mathbb{R}^n$ Can someone explain why it is that, if the row-reduced form of an $n\times m$ matrix $A$ has a row of zeros, the columns of matrix $A$ do not span $\mathbb R^n$? I'm not seeing the bigger picture here. AI: Presuming the matrix $\mathbf{A}$ in question is $n \times n$, as a $n \times m$ matrix implies there's $m$ column vectors and if $m<n$, it's clear those columns vectors do not span $\mathbb{R}^n$. I'll leave you to think about $m>n$ after reading my response for the $n \times n$ case: A row of zeroes in the matrix $\mathbf{A}$ directly implies $\mathbf{A}$ is not invertible, or has a nontrivial solution to the homogenous matrix equation $\mathbf{A\vec{x}}=\mathbf{0}$, or there exists a $\mathbf{\vec{x}}\ne \vec{\mathbf{0}}$ such that $\mathbf{A\vec{x}}=\mathbf{0}$. And since there's a nontrivial solution, we have that the columns of the matrix $\mathbf{A}=[\mathbf{v_1}\ \mathbf{v_2}\ ...\ \mathbf{v_n}]$ are not linearly independent (since a collection of vectors $\mathbf{v}_1, \mathbf{v}_2, ...\mathbf{v}_n$ are linearly independent iff $\mathbf{v}_1x_1+\mathbf{v}_2x_2+...+\mathbf{v}_nx_n=\vec{\mathbf{0}} \implies x_1=x_2=...=x_n=0$). Since a set of $n$ vectors are not linearly independent, we directly have that the dimension of the span of these $n$ linearly dependent vectors is less than $n$, which is less than the dimension of $\mathbb{R}^n$. Or: $\text{dim}\{\text{span} \{\mathbf{v_1},\mathbf{v_2}, ..., \mathbf{v_n}\}\}<n=\text{dim}\{\mathbb{R}^n\}$. Thus the columns of $\mathbf{A}$ cannot span all of $\mathbb{R}^n$. For the original poster, muros, did this help?
H: Integral of the product of two Sine Functions Evaluated at Regular Intervals = 2 * Num. of Intervals Used? Basically the I am considering the sum of the product of two sine functions: $$\sum_{k=0}^{N-1} sin(2 \pi {1\over N} k) sin(2 \pi {1 \over N }k)$$ I am trying to make sense of the the case when you multiply these two sine functions at 8 regular intervals (say at t = 0, t = 1/8, t = 2/8 ... up to say t = 7/8) and add up all the results together. With 8 intervals, the result is 16. I am trying to make sense of this. Could someone explain please why evaluating the functions like this, and adding up the result gives two times the number of intervals used? How does the equation is used above would relate to (EDIT I mean by that I know the integral is supposed to represent the area under the curve defined by the function in this case the product of two sine functions but how shall I interpret the result of the equation above?): $$\int_{t=0}^{t=1} sin(2 \pi t) sin(2 \pi t) dt.$$ It would be great if someone could help. Thank you. AI: You have the general identity $\sin^2 x=\frac 12(1-\cos(2x))$ If you do the sum at evenly spaced points through the period, the points where the cosine is positive will cancel with other points where the cosine is negative, so the sum will be one half the number of points, or $\frac N2$. It works the same way for the integral, and for the same reason.
H: show it is a random variable Let X and Y be random variables and let A be an event. Show that the function $$Z(\omega)=\begin{cases}X(\omega) \quad \text{if} \; \omega \in A\\ Y(\omega) \quad \text{if} \; \omega \in A^c \end{cases}$$ is a random variable. I considered the case in which $Z(\omega)=I_A(\omega)$, where $$I_{A}(\omega)=\begin{cases}1 \quad \text{if} \; \omega \in A\\ 0 \quad \text{if} \; \omega \in A^c \end{cases}$$ since the indicator is a measurable function. Is my intuition correct? Many thanks AI: If $A$ is measurable, then the indicator function $1_A$ is measurable. Sums and products of measurable functions are measurable, hence $Z = X \cdot 1_A + Y \cdot 1_{A^c}$ is measurable.
H: Prove that if $A$ is a set with $m$ elements, $B$ is a set with $n$ elements, and $A \cap B = \emptyset$, then $A \cup B$ has $m+n$ elements. I'm working through a real analysis textbook, so I don't want the full answer. I'm only looking for a hint on this problem. The book starts the proof like this: let $f$ be a bijection of $\mathbb{N}_m$ onto $A$, and let $g$ be a bijection of $\mathbb{N}_n$ onto $B$. Define $h$ on $\mathbb{N}_{m+n}$ as $h(i):=f(i)$ for $i=1, ..., m$, and $h(i):=g(i)$ for $i=m+1,...,m+n$. Show that $h$ is a bijection from $\mathbb{N}_{m+n}$ onto $A \cup B$. Ok, so with that, I started out proving that $h$ is an injection into $A \cup B$. The usual method for this is to prove that if $h(i)=h(j)$, where $i,j=1,...,m+n$, then $i=j$. I think I proved this by looking at four cases. $1 \leq i,j \leq m$. In this case, $h(i)=f(i)$ and $h(j)=f(j)$, so because $f$ is a bijection, we know that $f(i)=f(j)$ implies that $i=j$. $ 1 \leq i \leq m$ and $m+1 \leq j \leq m+n$. We know that $i \neq j$ (obviously), so we want to prove that in this case, $h(i) \neq h(j)$. By definition, $h(i)=f(i)$ and $h(j)=g(j-m)$. Because $f$ is a bijection into $A$ and $g$ is a bijection into $B$, and $A \cap B = \emptyset$, we know that $f(i) \neq g(j-m)$, so $h(i) \neq h(j)$. This case is identical to case 2, except that $m+1 \leq i \leq m+n$ and $1 \leq j \leq m$, so the logic is reversed. The final case, where $m+1 \leq i,j \leq m+n$, is similar to case 1, because since $h(i)=g(i)$ and $h(j)=g(j)$, we can use the fact that $g$ is a bijection to prove that $g(i)=g(j)$ implies that $i=j$. This is where I got stuck. I know I need to prove that $h$ is surjective, but am I on the right track for proving that it's injective? I think proving that $h$ is a surjection shouldn't be too hard, but I don't know if I'm approaching this part of the proof correctly. AI: Your function $h$ isn’t well-defined: the domain of $g$ is $N_n$, not $\{m+1,\ldots,m+n\}$. You could, however, let $h(i)=g(i-m)$ for $i\in\{m+1,\ldots,m+n\}$. This is the definition that you actually use in Case 2, but in Case 4 you revert to the incorrect definition. Once you repair that error, your proof that $h$ is injective will be fine. (And surjectivity of $h$ won’t be much of a problem, since $f$ and $g$ are surjective.)

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.