## OneKey Token Out of Battery — What to do?

It seems like the OneKey Token (used for 2 factor authentication) runs out of battery quite fast. I have barely used mine (usually use SMS), yet its battery has died surprisingly soon, compared to my other bank tokens.

What should one do in this case? Please comment below if you have other options.

Once the battery goes to zero, it appears that there are only two options:

Option 1)

Go to their office PSA Building (Alexandra Road) or International Plaza (Anson Road), to get free** replacement (**provided still within the warranty period of 1 year).

Waiting time is estimated 40-50 minutes. (and “free” may not be guaranteed, depends on whether you meet their requirement of warranty period, etc.)

Disclaimer: I have not tried this method myself. This is based on the Hardwarezone post linked below.

Option 2)

Go to this OneKey Assurity site https://portal.assurity.sg/naf-web/public/index.do to purchase a new token at $15. (Quite expensive 😦 ) I have no idea why the battery runs out so fast. Even my cheap Casio watch’s battery (which is running 24 hours a day) lasts longer than this token’s (which I have pressed less than 10 times). Advertisements Posted in math | Tagged , , , | Leave a comment ## Unknown German Retiree Proved The “Gaussian Correlation Inequality” Conjecture https://www.wired.com/2017/04/elusive-math-proof-found-almost-lost$latex boxed {P (a + b) geq P (a) times P (b)}&fg=aa0000&s=3$Case “=” : if (a, b) independent Case “>” : if (a, b) dependent Thomas Royen used only high-school math (function, derivative) in his proof in 2014. He then published it in arxiv.org website – likePerelman did with the “Poincaré Conjecture”. View original post Posted in math | Leave a comment ## Tough Math : Diophantine Equation Posted in math | Leave a comment ## Fun Math Answer : (scroll below) . . . . . . . . . . . . . . . . . . . . Answer = 3 mins See clearer if change person to taxi car, bun to passenger. 9 taxi cars send 9 passengers will take the SAME timing as 3 taxi cars send 3 passengers. View original post Posted in math | Leave a comment ## Category Adjunctions 伴随函子 Adjunction is the “weakeningof Equality” in Category Theory. Equivalence of 2 Categories if: 5.2 Adjunction definition:$latex (L, R, epsilon, eta )$such that the 2 triangle identities below ( red and blue) exist. 6.1Prove: Let C any category, D a Set.$latex boxed {text {C(L 1, -)} simeq text {R}}&fg=aa0000&s=3latex {text {Right Adjoint R in Set category is }}&fg=aa0000&s=3latex {text {ALWAYS Representable}}&fg=aa0000&s=3$1 = Singleton in Set D From an element in the singleton Set always ‘picks’ (via the function) an image element from the other Set, hence :$latex boxed {text{Set (1, R c) } simeq text{Rc }}&fg=0000aa&s=3$Examples : Product & Exponential are Right Adjoints Note: Adjoint is a more powerful concept to understand than the universal construction of Product and Exponential. View original post Posted in math | Leave a comment ## Category Theory (Stanford Encyclopedia of Philosophy) Posted in math | Leave a comment ## Beautiful Math Olympiad Problem [INMO 1993] View original post Posted in math | Leave a comment ## The genius who rejected Fields Medal & Clay Prize: Grigori Perelman Russian mathematician Grigori Perelman proved the Poincaré Conjecture in 7 years of solitude research in his Russian apartment - same 7 years of solitude forAndrew Wiles (The Fermat’s Last Theorem)in the Cambridge attic house andZhang Yitang 张益唐 (7-Million-Gap Twin Primes) in the “Subway” sandwich kitchen. 7 is the Perfect Number. 1 week has 7 days, according to the “Book of Genesis”, God created the universe in 6 days and rested “Sabbath” on the beautiful 7th day. People involved in his journey: • His mother who supported his early education in Math, all the way to his international fame and adulthood —Erdös Paulalso had a mother supporting her single son in the entire Mathematical life. • Prof Hamilton who “unintentionally”gave him the inspiration of the “Ricci-flow” tool – the first key to the door of The Poincaré Conjecture. • His closed Chinese friend the MIT… • View original post 76 more words Posted in math | Leave a comment ## The Yoneda Lemma Representable Functor F of C ( a, -):$latex boxed {(-)^{a} = text {F} iff a = text {log F}}&fg=aa0000&s=3$4.2Yoneda Lemma Prove : Yoneda Lemma:$latex text {F :: C} to text {Set}latex boxed {alpha text { :: [C, Set] (C (a, -),F) } simeq text {F a}}&fg=0000aa&s=3latex alpha : text {Natural Transformation}latex simeq : text {(Natural) Isomorphism}$Proof: By “Diagram chasing” below, shows that Left-side :$latex alpha text { :: [C, Set] (C (a, -),F) } $is indeed a (co-variant) Functor. Right-side: Functor “F a“. Note: When talking about the natural transformations, always mention their component “x”:$latex alpha_{x}, beta_{x}Yoneda Embedding (Lagatta) View original post Posted in math | Leave a comment ## SO(3) diffeomorphic to RP^3 $SO(3)\cong\mathbb{R}P^3$} Proof: We consider $SO(3)$ as the group of all rotations about the origin of $\mathbb{R}^3$ under the operation of composition. Every non-trivial rotation is determined by its axis of rotation (a line through the origin) and its angle of rotation. We consider $\mathbb{R}P^3$ as the unit 3-sphere $S^3$ with antipodal points identified. Consider the map from the unit ball $D$ in $\mathbb{R}^3$ to $SO(3)$, which maps $(x,y,z)$ to the rotation about $(x,y,z)$ through the angle $\pi\sqrt{x^2+y^2+z^2}$ (and maps $(0,0,0)$ to the identity rotation). This mapping is clearly smooth and surjective. Its restriction to the interior of $D$ is injective since on the interior $\pi\sqrt{x^2+y^2+z^2}<\pi$. On the boundary of $D$, two rotations through $\pi$ and through $-\pi$ are the same. Hence the mapping induces a smooth bijective map from $D/\sim$, with antipodal points on the boundary identified, to $SO(3)$. The inverse of this map, $\displaystyle ((x,y,z),\pi\sqrt{x^2+y^2+z^2})\mapsto(x,y,z)$ is also smooth. (To see that the inverse is smooth, write $\theta=\pi\sqrt{x^2+y^2+z^2}$. Then $x=\sqrt{\frac{\theta^2}{\pi^2}-y^2-z^2}$, and so $\frac{\partial^k x}{\partial\theta^k}$ exists and is continuous for all orders $k$. Similar results hold for the variables $y$ and $z$, and also mixed partials. By multivariable chain rule, one can see that all component functions are indeed smooth, so the inverse is smooth as claimed.) Hence $SO(3)\cong D/\sim$, the unit ball $D$ in $\mathbb{R}^3$ with antipodal points on the boundary identified. Next, the mapping $\displaystyle (x,y,z)\mapsto (x,y,z,\sqrt{1-x^2-y^2-z^2})$ is a diffeomorphism between $D/\sim$ and the upper unit hemisphere of $S^3$ with antipodal points on the equator identified. The latter space is clearly diffeomorphic to $\mathbb{R}P^3$. Hence, we have shown $\displaystyle SO(3)\cong D/\sim\cong\mathbb{R}P^3.$ Posted in math | Tagged , | Leave a comment ## SU(2) diffeomorphic to S^3 (3-sphere) $SU(2)\cong S^3$ (diffeomorphic) Proof: We have that $\displaystyle SU(2)=\left\{\begin{pmatrix}\alpha &-\overline{\beta}\\ \beta &\overline{\alpha}\end{pmatrix}: \alpha, \beta\in\mathbb{C}, |\alpha|^2+|\beta|^2=1\right\}.$ Since $\mathbb{R}^4\cong\mathbb{C}^2$, we may view $S^3$ as $\displaystyle S^3=\{(\alpha,\beta)\in\mathbb{C}^2: |\alpha|^2+|\beta|^2=1\}.$ Consider the map \begin{aligned}f: S^3&\to SU(2)\\ f(\alpha,\beta)&=\begin{pmatrix}\alpha &-\overline{\beta}\\ \beta &\overline{\alpha}\end{pmatrix}. \end{aligned} It is clear that $f$ is well-defined since if $(\alpha,\beta)\in S^3$, then $f(\alpha,\beta)\in SU(2)$. If $f(\alpha_1,\beta_1)=f(\alpha_2,\beta_2)$, it is clear that $(\alpha_1, \beta_1)=(\alpha_2,\beta_2)$. So $f$ is injective. It is also clear that $f$ is surjective. Note that $SU(2)\subseteq M(2,\mathbb{C})\cong\mathbb{R}^8$, where $M(2,\mathbb{C})$ denotes the set of 2 by 2 complex matrices. When $f$ is viewed as a function $\widetilde{f}: \mathbb{R}^4\to\mathbb{R}^8$, it is clear that $\widetilde{f}$ and $\widetilde{f}^{-1}$ are smooth maps since their component functions are of class $C^\infty$. Since $SU(2)$ and $S^3$ are submanifolds, the restrictions to these submanifolds (i.e.\ $f$ and $f^{-1}$) are also smooth. Hence $f$ is a diffeomorphism. Posted in math | Tagged , | Leave a comment ## Don’t fear the Monad Brian Beckman: You can understand Monad without too much Category Theory. Functional Programming = using functions to compose from small functions to very complex software (eg. Nuclear system, driverless car software…). Advantages of Functional Programming: • Strong Types Safety: detect bugs at compile time. • Data Protection thru Immutability: Share data safely in Concurrent / Parallel processing. • Software ‘Componentisation’ ie Modularity: Each function always returns the same result, ease of software reliability testing. Each “small” function is a Monoid. f : a -> a (from input of type ‘a‘ , returns type ‘a’) g: a -> a compose h from f & g : (strong TYPING !!) h = f。g : a -> a [Note]: Object in Category, usually called Type in Haskell, eg.’a’ = Integer) You already know a Monoid (or Category in general) : eg Clock 1. Objects: 1 2 3 …12 (hours) 2. Arrow View original post 185 more words Posted in math | Leave a comment ## Programming and Math Category Theory (CT) is like Design Pattern, only difference is CT is a better mathematical pattern which you can prove, also it has no “SIDE-EFFECT” and with strongTyping. The examples use Haskell to explain the basic category theory : product, sum, isomorphism, fusion, cancellation, functor… View original post Posted in math | Leave a comment ## Echelon Form Lemma (Column Echelon vs Smith Normal Form) The pivots in column-echelon form are the same as the diagonal elements in (Smith) normal form. Moreover, the degree of the basis elements on pivot rows is the same in both forms. Proof: Due to the initial sort, the degree of row basis elements $\hat{e}_i$ is monotonically decreasing from the top row down. For each fixed column $j$, $\deg e_j$ is a constant. We have, $\deg M_k(i,j)=\deg e_j-\deg \hat{e}_i$. Hence, the degree of the elements in each column is monotonically increasing with row. That is, for fixed $j$, $\deg M_k(i,j)$ is monotonically increasing as $i$ increases. We may then eliminate non-zero elements below pivots using row operations that do not change the pivot elements or the degrees of the row basis elements. Finally, we place the matrix in (Smith) normal form with row and column swaps. Posted in math | Tagged , | Leave a comment ## Persistent Homology Algorithm Algorithm for Fields In this section we describe an algorithm for computing persistent homology over a field. We use the small filtration as an example and compute over $\mathbb{Z}_2$, although the algorithm works for any field. A filtered simplicial complex with new simplices added at each stage. The integers on the bottom row corresponds to the degrees of the simplices of the filtration as homogenous elements of the persistence module. The persistence module corresponds to a $\mathbb{Z}_2[t]$-module by the correspondence in previous Theorem. In this section we use $\{e_j\}$ and $\{\hat{e}_i\}$ to denote homogeneous bases for $C_k$ and $C_{k-1}$ respectively. We have $\partial_1(ab)=-t\cdot a+t\cdot b=t\cdot a+t\cdot b$ since we are computing over $\mathbb{Z}_2$. Then the representation matrix for $\partial_1$ is $\displaystyle M_1=\begin{bmatrix}[c|ccccc] &ab &bc &cd &ad &ac\\ \hline d & 0 & 0 & t & t & 0\\ c & 0 & 1 & t & 0 & t^2\\ b & t & t & 0 & 0 & 0\\ a &t &0 &0 &t^2 &t^3 \end{bmatrix}.$ In general, any representation $M_k$ of $\partial_k$ has the following basic property: $\displaystyle \deg\hat{e}_i+\deg M_k(i,j)=\deg e_j$ provided $M_k(i,j)\neq 0$. We need to represent $\partial_k: C_k\to C_{k-1}$ relative to the standard basis for $C_k$ and a homogenous basis for $Z_{k-1}=\ker\partial_{k-1}$. We then reduce the matrix according to the reduction algorithm described previously. We compute the representations inductively in dimension. Since $\partial_0\equiv 0$, $Z_0=C_0$ hence the standard basis may be used to represent $\partial_1$. Now, suppose we have a matrix representation $M_k$ of $\partial_k$ relative to the standard basis $\{e_j\}$ for $C_k$ and a homogeneous basis $\{\hat{e}_i\}$ for $Z_{k-1}$. For the inductive step, we need to compute a homogeneous basis for $Z_k$ and represent $\partial_{k+1}$ relative to $C_{k+1}$ and the homogeneous basis for $Z_k$. We first sort the basis $\hat{e}_i$ in reverse degree order. Next, we make $M_k$ into the column-echelon form $\tilde{M}_k$ by Gaussian elimination on the columns, using elementary column operations. From linear algebra, we know that $rank M_k=rank B_{k-1}$ is the number of pivots in the echelon form. The basis elements corresponding to non-pivot columns form the desired basis for $Z_k$. Source: “Computing Persistent Homology” by Zomorodian & Carlsson Posted in math | Tagged , , | Leave a comment ## BM Category Theory 10.1: Monads 10.1Monadslatex begin{array}{|l|l|l|}
hline
Analogy & Compose & Identity
hline
Function & : : : : “.” & : : : : Id
hline
Monad & ” >> = ” (bind) & return :: “eta”
hline
!}}&fg=aa0000&s=3$Our brain works by: 1) Abstraction 2) Composition 3) Identity (to identify) What is a Category ? 1) Abstraction: • Objects • Morphism (Arrow) 2) Composition: Associative 3) Identity Notes: • Small Category with “Set” as object. • Large Category without Set as object. • Morphism is a Set : “Hom” Set. Example in Programming : • Object : Types Set • Morphism : Function “Sin” converts degree to R:$latex sin frac {pi}{2} = 1Note: We just look at the Category “Types Set” from external Macroview, “forget ” what it contains, we only know the “composition” (Arrows) between the Category “Type Set”, also “forget”… View original post 115 more words Posted in math | Leave a comment ## A Category Language : Haskell Haskell is the purest Functional Language which is based on Category Theory. eBook: http://learnyouahaskell.com/chapters View original post Posted in math | Leave a comment ## Online Study Guide : Abstract Algebra Posted in math | Leave a comment ## How to Take Great Notes – Study Tips – How to be a Great Student – Cornell Notes Posted in math | Leave a comment ## What is a Field, a Vector Space ? (Abstract Algebra) Suitable for Upper Secondary School and Junior College Math Students. Abstract Algebra is scary because it is abstract, and its Math Profs are mostly fierce – but not with this pretty Math lady… WHAT IS A FIELD (域) ? WHAT IS A VECTOR SPACE (向量空间) ? See all 20+ videos here: View original post Posted in math | Leave a comment ## 陈省身：数学之美 SS Chern : The Math Beauty There are 5 great Geometry Masters in history: 欧高黎嘉陈 Euclid (300 BCE, Greece), Gauss (18CE, Germany), Riemann (19CE, Germany), Cartan (20CE, France), Chern (21CE, China). Jim Simons (Hedge Fund Billionaire, Chern’s PhD Student) quoted Chern had said to him: “If you do One Thing that is reallygood, that’s all you could really expect in a life time.” 一生作好一件事, 此生无悔矣! Highlights: 1. Video below @82:00 mins, SS Chern criticised on Hardy’s famous statement: “Great Math is only discovered by young mathematicians before 30.” Chern’s response: “Don’t believe it ! 不要相信它”. 2. Chern’s Conjecture :“21世纪中国将是数学大国。 ” China will be a Math Kingdom in 21st century. http://www.bilibili.com/mobile/video/av1836134.html View original post Posted in math | Leave a comment ## CP1 and S^2 are smooth manifolds and diffeomorphic (proof) Proposition: $\mathbb{C}P^1$ is a smooth manifold. Proof: Define $U_1=\{[z^1, z^2]\mid z^1\neq 0\}$ and $U_2=\{[z^1, z^2]\mid z^2\neq 0\}$. Also define $g_i: U_i\to\mathbb{C}$ by $g_1([z^1, z^2])=\frac{z^2}{z^1}$ and $g_2([z^1, z^2])=\frac{\overline{z^1}}{\overline{z^2}}$. Let $f:\mathbb{C}\to\mathbb{R}^2$ be the homeomorphism from $\mathbb{C}$ to $\mathbb{R}^2$ defined by $f(x+iy)=(x,y)$ and define $\phi_i: U_i\to\mathbb{R}^2$ by $\phi_i=f\circ g_i$. Note that $\{U_1, U_2\}$ is an open cover of $\mathbb{C}P^1$, and $\phi_i$ are well-defined homeomorphisms (from $U_i$ onto an open set in $\mathbb{R}^2$). Then $\{(U_1,\phi_1), (U_2,\phi_2)\}$ is an atlas of $\mathbb{C}P^1$. The transition function $\displaystyle \phi_2\circ\phi_1^{-1}: \phi_1(U_1\cap U_2)\to\mathbb{R}^2,$ \begin{aligned} \phi_2\phi_1^{-1}(x,y)&=\phi_2 g_1^{-1}(x+iy)\\ &=\phi_2([1,x+iy])\\ &=f(\frac{1}{x-iy})\\ &=f(\frac{x+iy}{x^2+y^2})\\ &=(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}) \end{aligned} is differentiable of class $C^\infty$. Similarly, $\phi_1\circ\phi_2^{-1}:\phi_2(U_1\cap U_2)\to\mathbb{R}^2$ is of class $C^\infty$. Hence $\mathbb{C}P^1$ is a smooth manifold. Proposition: $S^2$ is a smooth manifold. Proof: Define $V_1=S^2\setminus\{(0,0,1)\}$ and $V_2=S^2\setminus\{(0,0,-1)\}$. Then $\{V_1, V_2\}$ is an open cover of $S^2$. Define $\psi_1: V_1\to\mathbb{R}^2$ by $\psi_1(x,y,z)=(\frac{x}{1-z},\frac{y}{1-z})$ and $\psi_2: V_2\to\mathbb{R}^2$ by $\psi_2(x,y,z)=(\frac{x}{1+z},\frac{y}{1+z})$. We can check that $\psi_1^{-1}(x,y)=(\frac{2x}{1+x^2+y^2},\frac{2y}{1+x^2+y^2},\frac{-1+x^2+y^2}{1+x^2+y^2})$. Hence $\psi_1$ is a homeomorphism from $V_1$ onto an open set in $\mathbb{R}^2$. Similarly, $\psi_2$ is a homeomorphism from $V_2$ onto an open set in $\mathbb{R}^2$. Thus $\{V_1,V_2\}$ is an atlas for $S^2$. The composite $\psi_2\circ\psi_1^{-1}: \psi_1(V_1\cap V_2)\to\mathbb{R}^2$ is differentiable of class $C^\infty$ since both $\psi_2$, $\psi_1^{-1}$ are of class $C^\infty$. Similarly, $\psi_1\circ\psi_2^{-1}$ is of class $C^\infty$. Thus $S^2$ is a smooth manifold. We can also compute the transition function explicitly: $\displaystyle \psi_2\psi_1^{-1}(x,y)=(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}).$ Note that $\psi_2\psi_1^{-1}=\phi_2\phi_1^{-1}$. Define $h:\mathbb{C}P^1\to S^2$ by $h(\phi_1^{-1}(x,y))=\psi_1^{-1}(x,y)$ and $h(\phi_2^{-1}(x,y))=\psi_2^{-1}(x,y)$. We see that $h$ is well-defined since if $\phi_1^{-1}(x,y)=\phi_2^{-1}(u,v)$ then $\displaystyle (u,v)=\phi_2\phi_1^{-1}(x,y)=\psi_2\psi_1^{-1}(x,y)$ so that $\psi_2^{-1}(u,v)=\psi_1^{-1}(x,y)$. Similarly, we have a well-defined inverse $h^{-1}: S^2\to\mathbb{C}P^1$ defined by $h^{-1}(\psi_1^{-1}(x,y))=\phi_1^{-1}(x,y)$ and $h^{-1}(\psi_2^{-1}(x,y))=\phi_2^{-1}(x,y)$. We check that (from our previous workings) \begin{aligned} \psi_1 h\phi_1^{-1}(x,y)&=(x,y)\\ \psi_2 h\phi_1^{-1}(x,y)&=\psi_2\psi_1^{-1}(x,y)\\ \psi_1 h\phi_2^{-1}(x,y)&=\psi_1\psi_2^{-1}(x,y)\\ \psi_2 h\phi_2^{-1}(x,y)&=(x,y) \end{aligned} are of class $C^\infty$. So $h$ is a smooth map. Similarly, $h^{-1}$ is smooth. Hence $h$ is a diffeomorphism. Posted in math | Tagged | Leave a comment ## Tangent space (Derivation definition) Let $M$ be a smooth manifold, and let $p\in M$. A linear map $v: C^\infty(M)\to\mathbb{R}$ is called a derivation at $p$ if it satisfies $\displaystyle v(fg)=f(p)vg+g(p)vf\qquad\text{for all}\ f,g\in C^\infty(M).$ The tangent space to $M$ at $p$, denoted by $T_pM$, is defined as the set of all derivations of $C^\infty(M)$ at $p$. Posted in math | Tagged , | Leave a comment ## Looking for Home Tutors? If you are looking for home tutors (any subject, e.g. Mathematics, Chinese, English, Science, etc.) contact us at: SMS/Whatsapp: 98348087 Email: mathtuition88@gmail.com We are able to recommend you highly qualified tutors, free of charge, no obligations. Note that usually tutors’ slots will start to fill up as the year progresses, so by mid year May/June it is going to be very hard to find a good tutor. Posted in math | Tagged , | Leave a comment ## 转 载: 矩阵的真正含义 The Connotations of Matrix and Its Determinant Excellent reading for Upper Secondary / High School (JC, IB) Math students. Posted in math | Leave a comment ## 矽谷预测AI後的10年大未來 In 15 years, AI driven driverless car will change the transport/work/environment landscape… it is true not futuristic… behind AI is advanced math which teaches computer to learn without a fixed algorithm but by analysing BIG DATA patterns using Algebraic Topology ! 世界趨勢，可作參考 矽谷预测AI後的10年大未來 現在因為人工智能(AI)的發展，配合更高速度的積體電路，科技正在加快速度的進展。據悉，在很短的5 -10年後，医療健保、自駕汽車、教育、服務業都將面臨被淘汰的危機。 1. Uber 是一家軟體公司，它沒有擁用汽車，卻能夠讓你「隨叫隨到」有汽車坐，現在，它已是全球最大的Taxi公司了。 2. Airbnb 也是一家軟體公司，它沒有擁有任何旅館，但它的軟體讓你能夠住進世界各地願出租的房間，現在，它已是全球最大的旅館業了。 3. 今年5月，Google的電腦打敗全球最厲害的南韓圍棋高手，因為它開發出有人工智能(AI)的電腦，使用能夠「自己學習」的軟體，所以它的AI能夠加速度的進步，達到比專家原先預期的、提前10年的成就。 4. 在美國，使用IBM 的Watson電腦軟體，你能夠在幾秒內，就有90%的準確性的法律顧問，比較起只有70% 準確性的人為律師，既便捷又便宜。 所以，你如果還有家人親友在讀大學的法律系，建議他們停學省錢，因為市場已大幅的縮減了，未來的世界，只需要現在10%的專業律師就夠了。 5. Watson 也已經能夠幫病人檢驗癌症，而且比醫生正確4 倍。 6. 臉書也有一套AI的軟體可以比人類更準確的鑒察(辨識)人臉，而且無所不在。 7. 到了2030年，AI的電腦會比世界上任何的專家學者還要聰明。 8. 2017年起，會自動駕駛的汽車就可以在公眾場所使用。 約在2020年，整個汽車工業就會遭遇到全面性的改變，你再不需要擁用汽車。 你可以用手機叫自動駕駛的車，來帶你到你想去的目的地。 9. 未來的世界，你再也不必擁有車，或花時間加油、停車、排隊去考駕照、交保險費，尤其是城市，將會很安靜，走路很安全，因為90%的汽車都不見了，以前的停車場，將會變成公園。 10. 現在，平均每10萬公里就有一次車禍，造成每年全球有約120萬人的死亡。 以後有AI電腦控制的自動駕駛汽車，平均每1000萬公里才有一次車禍，約減少一百萬人死亡。 因為保險費和需要保險的人極少，保險公司會面臨更多的倒閉風潮。 11. 大部份的傳統汽車公司會面臨倒閉。Tesla、 Apple、及 Google 的革命性軟體，將會用在每一部汽車上。 據悉，Volkswagen 和 Audi 的工程師非常擔心Tesla革命性的電池和人工智能軟體技術。 12. 房地產公司會遭遇極大的變化。 因為你可以在車程中工作，距離將不是選住房屋的主要條件之一。市民會選擇住在較遠、但是較空曠且環境優美的鄉村。 13. 電動汽車很安靜，會在2020變成主流。所以城市會很變成安靜，而且空氣乾淨。 14. 太陽能在過去30年也有快速的進展。 去年，全球太陽能的增產超過石油的增產。 預計，到2025年時，太陽能的價格(低廉)會使煤礦業大量的破產。 因為電費非常的便宜，淨化水及海水淡化的費用大減，人類將能解決人口增加的需水問題。 15. 健保：今年醫療設備商會供應如同「星球大戰」電影中的 Tricorder，讓你的手機做眼睛的掃瞄，呼吸氣體及血液的化學檢驗：用54個「生物指標」，就可檢驗出你是否有任何疾病的徵兆。 因為費用低，幾年後，全球人類都可以有世界級的疾病預防服務。 16. 立體列印(3D printing)：預計10 年內，3D列印設備會由近20000美元減到400美元，而速度增加100倍快。 所有的「個人化」設計鞋子，將開始用這種設備生產，其他如大型的機場，其零件也能使用這種設備供應，至於人類太空船，也會使用這種設備。 17. 今年底，你的手機就會有3D掃瞄的功能，你可以測量你的腳送去做「個人化」鞋子。據悉，在中國，他們已經用這種設備製造了一棟6 層樓辦公室，預計到2027年時， 10% 的產品會用3D的列印設備製造。 18. 產業機會： a. 工作：20年內，70-80% 的工作會消失，即使有很多新的工作機會，但是不足以彌補被智能機械所取代的原有工作。 b. 農業：將有100 機械人耕作，不必吃飯、不用住宅、及支付薪水，只要便宜的電池即可。在開發國家的農夫，將變成機械人的經理。溫室建築物可以有少量的水。

到2018年，肉可以從實驗室生產，不必養豬、雞或牛。30%用在畜牧的土地，會變成其他用途的土地。很多初創公司會供給高蛋白質的昆蟲當成食品。
c. 到2020年時 ，你的手機會從你的表情看出，與你說話的人是不是說「假話」？ 是否騙人的？ 政治人物(如總統候選人)若說假話，馬上會被當場揭發。
d. 數位時代的錢，將是Bitcoin ，是在智能電腦中的「數據」。
e. 教育：最便宜的智能手機在非州是$10美元一隻。 f. 到2020年時，全球70%的人類會有自己的手機，所以能夠上網接受世界級的教育，但大部份的老師會被智能電腦取代。所有的「小學生」都要會寫 Code，你如果不會，你就是像住在Amazon森林中的原住民，無法在社會上做什麼。你的國家，你的孩子準備好了嗎？ 參考一下；這也是矽谷 VC, Innovators,Entrepreneurs … 談的資料。 View original post Posted in math | Leave a comment ## Secondary Chinese Tuition (IP / O Level) Ms Gao specializes in tutoring Secondary Level Chinese. Can teach composition, comprehension, etc, according to student’s weaknesses. Has taught students from RI (IP Programme), MGS, and more. Familiar with IP and O Level (HCL/CL) Chinese syllabus. Website: https://chinesetuition88.com/ Contact: 98348087 Email: chinesetuition88@gmail.com Posted in math | Tagged , | Leave a comment ## Homology Group of some Common Spaces Homology of Circle $\displaystyle H_n(S^1)=\begin{cases}\mathbb{Z}&\text{for}\ n=0,1\\ 0&\text{for}\ n\geq 2. \end{cases}$ Homology of Torus $\displaystyle H_n(T)=\begin{cases}\mathbb{Z}\oplus\mathbb{Z}&\text{for}\ n=1\\ \mathbb{Z}&\text{for}\ n=0, 2\\ 0&\text{for}\ n\geq 3. \end{cases}$ Homology of Real Projective Plane $\displaystyle H_n(\mathbb{R}P^2)=\begin{cases} \mathbb{Z}&\text{for}\ n=0\\ \mathbb{Z}_2&\text{for}\ n=1\\ 0&\text{for}\ n\geq 2. \end{cases}$ Homology of Klein Bottle $\displaystyle H_n(K)=\begin{cases} \mathbb{Z}&\text{for}\ n=0\\ \mathbb{Z}_2\oplus\mathbb{Z}&\text{for}\ n=1\\ 0&\text{for}\ n\geq 2. \end{cases}$ Posted in math | Tagged | Leave a comment ## 北大 高等代数 Beijing University Advanced Algebra 辛弃疾的《青玉案·元夕》：“…众里寻他千百度；蓦然回首，那人却在灯火阑珊处。” –表达出了我的一种 (网上)意外相逢的喜悦，又表现出对心中(名师)的追求。 2011 年 北京大学教授丘维声教授被邀给清华大学 物理系(大学一年级) 讲一学期课 : (Advanced Algebra) 高等代数, aka 抽象代数 (Abstract Algebra)。 丘维声（1945年2月－）生于福建省龙岩市[1]，中国数学家、教育家。16岁时以全国高考状元的成绩考入北京大学，1978年3月至今担任北京大学数学科学学院教授，多年坚持讲授数学专业基础课程[2]。截至2013年，共著有包括《高等代数（上册、下册）》、《简明线性代数》两本国家级规划教材在内的40部著述[3]。于1993-97年的一系列文章中逐步解决了n=3pr情形的乘子猜想，并取得了一系列进展[2]。 ——————— 72岁的丘教授学问渊博, 善于启发, 尤其有别于欧美的”因抽象而抽象”教法, 他独特地提倡用”直觉” (Intuition) – 几何概念, 日常生活例子 (数学本来就是源于生活)- 来吸收高深数学的概念 (见:数学思维法), 谆谆教导, 像古代无私倾囊相授的名师。 全部 151 (小时) 讲课。如果没时间, 建议看第1&第2课 Overview 。 http://www.bilibili.com/mobile/video/av7336544.html?from=groupmessage 第一课: 导言 : n 维 方程组 – 矩阵 (Matrix)-n 维向量空间 (Vector Space) – 线性空间 (Linear Space) 第二课: 上表 (左右对称): 双线性函数 (Bi-linear functions) / 线性映射 (Linear Map) 线性空间 + 度量 norm => • Euclidean Space (R) => (正交 orthogonal , 对称 symmetric)变换 • 酉空间 Unitary Space (C)… =>变换, Hermite变换 近代代数 (Modern Math since 19CE Galois): 从 研究 结构 (环域群) 开始: Polynomial Ring, Algebraic structures (Ring, Field, Group). 第三课: 简化行阶梯形矩阵 Reduced Row Echelon Matrix 第四课: 例子 (无解) 第五课: 证明 无解/唯一解/无穷解 [几何直觉]: 任何2线 1) 向交(唯一解) ; 2) 平行 (无解) ; 3) 重叠 (无穷解) n次方程組的解也只有3个情况: 无解 : View original post 16 more words Posted in math | Leave a comment ## 中国数学考研 Graduate Math Exams 中国”考研”究生: 考题难, 重视理论基础, 不是技巧。计算量大, 时间(3小时)不够。 国家 “及格” 底线 : 58~ 90分 (总分 : 150 分) – 根据 理工 / 经管系 , 不同重点大学, 底线各异。 http://www.bilibili.com/mobile/video/av2261356.html [例子]$latex p (x) = a + bx+cx^{2}+dx^{3}latex p(x) – tan x sim x^{3}, text { when } x to 0$Find a, b, c, d ? [Solution] : 1. Don’t use l’Hôpital Rule for$latex displaystyle lim frac {f}{g}$2. Apply Taylor expansion :$latex tan x = x + frac {1}{3}x^{3} + o (x^{3})latex p (x) – tan x = a + (b -1)x + (c – frac {1}{3})x^{3} + o (x^{3})latex p(x) – tan x sim x^{3}, text { when } x to 0 latex iff boxed {a=0, b=1, c=frac {4}{3}}&fg=aa0000&s=2$View original post Posted in math | Leave a comment ## Barry Mazur – Harvard Lecture on Primes and the Riemann Hypothesis for High School Students Prelude: Harvard Lecture: The Key toopen this secret View original post Posted in math | Leave a comment ## Summary of Persistent Homology We summarize the work so far and relate it to previous results. Our input is a filtered complex $K$ and we wish to find its $k$th homology $H_k$. In each dimension the homology of complex $K^i$ becomes a vector space over a field, described fully by its rank $\beta_k^i$. (Over a field $F$, $H_k$ is a $F$-module which is a vector space.) We need to choose compatible bases across the filtration (compatible bases for $H_k^i$ and $H_k^{i+p}$) in order to compute persistent homology for the entire filtration. Hence, we form the persistence module $\mathscr{M}$ corresponding to $K$, which is a direct sum of these vector spaces ($\alpha(\mathscr{M})=\bigoplus M^i$). By the structure theorem, a basis exists for this module that provides compatible bases for all the vector spaces. Specifically, each $\mathcal{P}$-interval $(i,j)$ describes a basis element for the homology vector spaces starting at time $i$ until time $j-1$. This element is a $k$-cycle $e$ that is completed at time $i$, forming a new homology class. It also remains non-bounding until time $j$, at which time it joins the boundary group $B_k^j$. A natural question is to ask when $e+B_k^l$ is a basis element for the persistent groups $H_k^{l,p}$. Recall the equation $\displaystyle H_k^{i,p}=Z_k^i/(B_k^{i+p}\cap Z_k^i).$ Since $e\notin B_k^l$ for all $l, hence $e\notin B_k^{l+p}$ for $l+p. The three inequalities $\displaystyle l+p define a triangular region in the index-persistence plane, as shown in Figure below. The triangular region gives us the values for which the $k$-cycle $e$ is a basis element for $H_k^{l,p}$. This is known as the $k$-triangle Lemma: Let $\mathcal{T}$ be the set of triangles defined by $\mathcal{P}$-intervals for the $k$-dimensional persistence module. The rank $\beta_k^{l,p}$ of $H_k^{l,p}$ is the number of triangles in $\mathcal{T}$ containing the point $(l,p)$. Hence, computing persistent homology over a field is equivalent to finding the corresponding set of $\mathcal{P}$-intervals. Source: “Computing Persistent Homology” by Zomorodian and Carlsson Posted in math | Tagged | 1 Comment ## Part 4 群的线性表示的结构 不变子空间: Invariant Sub-space 第一课: Direct Sum 直和$latex oplus$of Representations 直和 =$latex {oplus}&fg=aa0000&s=3$第二课: 群表示可约 Reducible Representation Analogy : Prime number decomposition Irreducible Polynomial 外直和 :$latex { dot{ +} }&fg=aa0000&s=3latex boxed { displaystyle phi_{1} dot {+} phi_{2} = tilde {phi_{1}} oplus tilde {phi_{2}}}&fg=aa0000&s=3$* 第三课: 完全可约表示 Completely Reducible Representation 完全表示是可 完全分解为 不可约表示 的一种表示。 完全可约表示 => 其子表示 也 完全可约 不可约 一定是完全可约的! 一次表示一定是不可约的! [Analogy: Polynomial degree 1 (x + 1) is irreducible. ] 註: (*) 深奥课, 可以越过直接跳到结果。(证明 待以后 复习)。 集合证明: 交(和)和(交) 如果 也是⊆ , 则 交(和) =和(交) Ref 2 《高代》 Pg 250 命题 1$latex boxed {U cap (U_{1} oplus W) supseteq (U cap U_{1} ) oplus (U cap W)}&fg=aa0000&s=3$Also,$latex U cap (U_{1} oplus W) subseteq (U cap U_{1} ) oplus (U cap W)$Then,$latex boxed {U…

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## Part 3 (b) 群的线性表示和例

$latex x_{i} in Omega = big{0.x_{1},0.x_{2},…, 0.x_{i-1},1.x_{i}, 0.x_{i+1}, ….0.x_{n} big}$

第11课:Cyclic Group (循环群) Representation , Dihedron 二面体

$latex begin{pmatrix} 0 & 0 & 1 1 & 0 &… 0 & 1 &… end{pmatrix} = P (a)$3 阶 Cyclic Group (循环群) Representation

$latex boxed{ Bigr|D_{n} Bigr| = 2n }&fg=aa0000&s=3$

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