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H: non-intuitive results in random sampling I play a card game (Magic The Gathering) that involves creating a deck of cards that can loosely be split into two categories: land and spells. You are permitted to play only one land per turn, and the land are used to cast your spells. Due to the one-per-turn limitation, it is advantageous to have access to exactly X land on turn X (otherwise your hand is wasted on land you can't play.) For purposes of my current experiment, I consider myself to be in a good position if on turn X, I have access to (that is, have either drawn or played) either X, X+1, or X+2 land. Having access to X-1 land means I'm not playing spells of cost X on turn X, while my opponent is, so I'm probably at a disadvantage. Having X+3 or more land on turn X means I don't have a good choice of spells to play. I desire the happy medium, with the potential buffer of 2 extras. You start the game with 7 cards, and draw an additional card every turn. There are other effects that can cause you to draw more, discard more, play additional lands, or play more powerful lands, but these are edge cases that are not included in this experiment. In order to gather data, I wrote a small java application that enables me to simulate drawing cards from my deck. The experiment consists of a 40 card deck with either 16 or 17 land. (A good rule of thumb is 40% land.) For simulation purposes, I do monte-carlo of 1 million hand draws. Here are the results for having 16 land. This table reads "On turn 4, I have access to 5 land 21.9% of the time." The "good" column is the sum of X, X+1, and X+2 for any given turn. 16 Land Turn Lands 1 2 3 4 5 6 7 8 0 1.9% 0.9% 0.5% 0.2% 0.1% 0.0% 0.0% 0.0% 1 11.5% 7.2% 4.3% 2.5% 1.4% 0.7% 0.4% 0.2% 2 27.4% 21.0% 15.2% 10.5% 6.7% 4.2% 2.5% 1.4% 3 31.9% 31.0% 27.6% 22.9% 17.8% 13.1% 9.2% 6.0% 4 19.7% 25.2% 28.3% 28.9% 27.3% 24.0% 19.8% 15.4% 5 6.4% 11.5% 17.0% 21.9% 25.4% 27.1% 26.7% 24.6% 6 1.0% 2.9% 5.9% 10.0% 14.7% 19.3% 23.0% 25.4% 7 0.1% 0.4% 1.2% 2.7% 5.3% 8.7% 12.8% 17.0% 8 0.1% 0.4% 1.1% 2.4% 4.6% 7.4% 9 0.1% 0.4% 1.0% 2.1% 10 0.1% 0.4% Good: 71% 77% 73% 61% 45.4% 30.4% 18.4% 9.9% In the same manner, here is the result for 17 land. Lands 1 2 3 4 5 6 7 8 0 1.3% 0.6% 0.3% 0.1% 0.1% 0.4% 0.0% 0.0% 1 9.2% 5.4% 3.0% 1.6% 0.8% 2.8% 0.2% 0.1% 2 24.6% 17.8% 12.2% 7.9% 4.8% 9.9% 1.5% 0.8% 3 32.3% 29.8% 25.1% 19.7% 14.4% 20.8% 6.4% 4.0% 4 22.5% 27.4% 29.4% 28.3% 25.2% 27.2% 16.2% 11.7% 5 8.4% 14.3% 20.0% 24.5% 27.1% 22.4% 25.3% 21.8% 6 1.5% 4.1% 8.0% 12.9% 18.0% 11.7% 25.2% 26.2% 7 0.1% 0.6% 1.8% 4.1% 7.5% 3.9% 16.3% 20.6% 8 0.2% 0.7% 1.9% 0.8% 6.8% 10.6% 9 0.1% 0.3% 0.1% 1.8% 3.5% 10 0.3% 0.7% 11 0.1% Good: 66.1% 75.0% 74.5% 65.7% 52.6% 16.4% 24.9% 14.8% Okay, now that's a lot of data. I've included it for your review. Here's what I don't get - let's take a look at the "good" results for 16 versus 17, paying very close attention to turn 6 for 17 land. Turn 1 Turn 2 Turn 3 Turn 4 Turn 5 Turn 6 Turn 7 Turn 8 16 Land 70.8% 77.2% 72.9% 60.8% 45.4% 30.4% 18.4% 9.9% 17 Land 66.1% 75.0% 74.5% 65.7% 52.6% 16.4% 24.9% 14.8% It would appear that on turn 7, my liklihood of having 7,8, or 9 land is higher (substantially) than having 6,7, or 8 land on turn 6. How is that value so low on turn 6? I've summed up all the columns, and they of course add up to 100% (barring a couple rounding errors). I've re-ran the simulation multiple times and it consistently comes up with the same value. Here are the questions that I feel will help me solve this mystery: What could possibly cause the "good curve" for 17-land to dip so much on turn 6? How could it be more likely that a 16 land deck could have a higher chance of having the same amount of land access than a 17 land deck after the same number of draws? Are there formulas that allow me to calculate the "land-likelihood" that might shed insight into this anomaly? AI: How sure are you that your java application is working properly? The whole $6$ lands column is out of line with the rest of the $17$ lands table. It would fit very nicely if it were dropped down one row, as would the resulting Good total of $38.0$%. It really looks as if the entries in that one column are all one row too high.
H: $P\in \operatorname{Syl}_p(S_n)$ implies that $P\in \operatorname{Syl}_p(A_n)$ and $\|N_{A_n}(P)\|=\frac{1}{2}\|N_{S_n}(P)\|$ $\newcommand{\Syl}{\operatorname{Syl}}$ This is an exercise (with hint about the second part) in my own language book in Group theory, however, maybe it is a lemma or theorem in an standard book which I don't have. It is: Let $p$ is an odd prime number and $n$ is a natural number such that $p\leq n$. Therefore if $P\in \Syl_p(S_n)$ then $P\in \Syl_p(A_n)$ and $\|N_{A_n}(P)\|=\frac{1}{2}\|N_{S_n}(P)\|$. What I have done: Suppose that $P\in \Syl_p(S_n)$ where in $S_n$ is the symmetric group on $n$ letters. For having $P\in \Syl_p(A_n)$, I clearly should prove $P\leq A_n$ first. I can see no clues here to prove that unless I assume $P\nleq A_n$. With this assumption, $P$ has an odd permutation. The rest is unclear to me. :( Maybe, I am losing some obvious facts for proving the first part? For the second part that is $\|N_{A_n}(P)\|=\frac{1}{2}\|N_{S_n}(P)\|$; there is a hint noting use the Frattini Argument. I can handle this part. :) Indeed, I see $S_n=N_{S_n}(P)A_n$ and the rest is routine. Thanks for helping me about the first part. AI: Suppose $\|P\| = p^k$ and $n! = p^k m$. Since $n!$ is divisible by 2 and $p$ is odd, $m$ is divisible by 2. Let $m' = m/2$. Then $n!/2 = p^k m'$. Since $m'$ is not divisible by $p$, there exists $Q \in Syl_p(A_n)$ such that $\|Q\| = p^k$. Since $Q \in Syl_p(S_n)$, $P$ and $Q$ are conjugate. Since $A_n$ is a normal subgroup of $S_n$, $P \in Syl_p(A_n)$.
H: Are bimodules over a commutative ring always modules? Let $R$ be a commutative ring. It is true that every module over $R$ is an $(R,R)$-bimodule. Is the converse true? In other words is it possible that there is an $R$-module where left multiplication and right multiplication do not co-incide? I thought perhaps a counterexample would be of the following form. If $S$ is an $R$-algebra, if we form the product $S \otimes_R S$ and think of it as an $S$-module where multiplication is given by $s(s' \otimes s'') = ss' \otimes s''$ and $(s' \otimes s'') s = s' \otimes s''s$. If we can pick the right element $s$ it is possible those two tensors are not equal. I cannot find any concrete counterexamples though. If you do this with $\mathbb{Q} \otimes_\mathbb{Z} \mathbb Q$ for example it doesn't work. Thanks for any help. AI: As Zhen Lin comments above, the converse is not true. For another example, suppose $S$ is a ring containing $R$ as a commutative subring such that $R$ is not in the center of $S$. Then $S$ is naturally an $(R,R)$-bimodule using the ring multiplication on $S$. But if $R$ is not in the center of $S$ then there exists $r \in R$, $s \in S$ such that $rs \neq sr$. So the left and right actions don't agree. For a simple example, take $R$ to be the subring of diagonal matrices in the matrix ring $\mathbb{M}_n(k)$ for a field $k$ and $n \geq 2$.
H: $L_p$ space convergence revised Here's what I really wanted to ask Suppose that $f_n \rightarrow f$ a.e. If we have $\int{\left\|\lvert f_n\rvert^p - \lvert f\rvert^p\right\|} \rightarrow 0$, is it true that $\\|f_n\\|_p \rightarrow \\|f\\|_p$? What I want to do is let $g_n = \lvert f_n\rvert^p, g=\lvert f\rvert^p$ and then use the fact that $\int{\lvert g_n - g\rvert} \rightarrow 0$ if and only if $\int{\lvert g_n\rvert} \rightarrow \int{\lvert g\rvert}$. Then we get $\\|f_n\\|_p^p \rightarrow \\|f\\|_p^p$ and thus $\\|f_n\\|_p \rightarrow \\|f\\|_p$. Is this valid? I'm kind of a beginner at this stuff. AI: Still it is unnecessary to require $f_n\to f$ a.e. To see this fix $\varepsilon>0$. Then exists $N_0(\varepsilon)\in\mathbb N$ such that for any $n>N_0$ $$\int \|\|f_n\|^p-\|f\|^p\| < \varepsilon$$ by your assumption. Then it is just a matter of triangular inequality. Indeed, for $n>N_0$, it follows $$\int \|f\|^p-\int \|f_n\|^p\leq \int \|\|f_n\|^p-\|f\|^p\|<\varepsilon. $$ Moreover, reversing the arguments we get $$\int\|f_n\|^p\leq \int\|f\|^p+\int\|\|f_n\|^p-\|f\|^p\|$$ which implies $$\int \|f_n\|^p-\int \|f\|^p\leq \int \|\|f_n\|^p-\|f\|^p\|<\varepsilon.$$ We conclude by the arbitrariness of $\varepsilon>0$ and the coninuity of $x\mapsto \sqrt[p] x.$
H: Orders of moves in the Rubik's cube Playing with my Rubik's cube, I was thinking of facts about it that are immediate to mathematicians but novel to others. Here's one: Given a Rubik's cube in the solved state, any sequence of moves, if repeated long enough, will eventually return the cube back to the solved state To a mathematician, that follows from "The moves of a Rubik's cube form a finite group. Therefore all the elements have finite order." Each atomic move of Rubik's cube has order 4. The simplest compound move is R U (rotate the right face 90° clockwise, then rotate the top face 90° clockwise). What's the order of R U? Secondly, what's the greatest order of an element in the Rubik's cube group? I tried repeating R U for a long while. I lost count (~50 repetitions), but it did eventually return home. AI: Take a solved cube and do $RU$; then trace the cycle structure of the permutation it realizes. The combination moves 5 corner cubies in a cycle where a cubie is twisted by a third of a turn when it gets back to its original position, so that's a factor of 15. It also twists a the FRU corner by one third of a turn; that's taken care of by the factor of 15 too. Then it permutes 7 edges cyclically, but this time every edge has the correct orientation when it gets back. So the order is the least common multiple of 7 and 15, namely 105. (For a subercube we need another factor of 4 to get the centers back into the original orientation).
H: Is independence preserved under the Girsanov transformation? Let $(B_{t},\mathcal{F}_{t})$ be a standard 1-d Brownian motion on some $(\Omega, \mathcal{F}, \mathbb{P}) \ $, and let's assume $\mathcal{F}_{t}$ is the augmentation of $\mathcal{F}_{t}^{W}$. Let $T>0$ and define $Z_{T}=\exp\{W_{T}-\frac{1}{2}T\} \ $, then $\frac{dP^{T}}{dP}=Z_{T}$ defines a probability measure $P^{T}$ on $\mathcal{F}_{T}$, which is equivalent (i.e. mutually absolutely continuous) w.r.t. $P$ on $F_{T}$. If we assume $A \in \mathcal{F}_{T}, \ B \in \mathcal{F}_{T}$ are two independent events under $P$, then do we have that $A$ and $B$ are independent under $P^{T}$? Intuitively I think the answer should be no, but I can't think of a counterexample. Could anyone help on this? Thanks a lot! AI: Try $A=[W_T\gt0]$ and $B=[\|W_T\|\lt x]$ with $x\gt0$, then $A$ and $B$ are independent under $\mathrm P$ because the distribution of $W_T$ under $\mathrm P$ is symmetric. Independence under $\mathrm P^T$ would mean that $$ \mathrm E(\mathrm e^{W_T}\,;\,0\lt W_T\lt x)\cdot\mathrm e^{T/2}=\mathrm E(\mathrm e^{W_T}\,;\,\|W_T\|\lt x)\cdot\mathrm E(\mathrm e^{W_T}\,;\,W_T\gt0). $$ If this identity holds for every $x\gt0$, differentiating it with respect to $x$ yields $$ 2\mathrm e^{x}\cdot\mathrm e^{T/2}=(\mathrm e^{x}+\mathrm e^{-x})\cdot\mathrm E(\mathrm e^{W_T}\,;\,W_T\gt0), $$ which cannot hold for every $x$.
H: Brute force method of solving the cube: How many moves would it take? Given that Rubik's cube has finitely many positions, one possible "brute force" method to solve it would be to determine once a sequence of moves which eventually reaches every possible position of the cube, and then whenever you want to solve a cube, you just mindlessly follow the sequence until you eventually reach the solved cube. This strategy is absolutely failsafe, provided you have an extraordinary memory and enough time (so, it's not a strategy for humans, but maybe for bored gods waiting for their just created universe to finally develop intelligent life ;-)). However, the question is: How many moves does the shortest possible sequence which visits every reachable position have? Of course, it's easy to give a lower bound: At least as many moves as there are reachable positions, which according to Wikipedia is $43\,252\,003\,274\,489\,856\,000 \approx 4{.}3 \cdot 10^{19}$. However, I doubt that there's a sequence which gives a not yet visited position at every move, so the actual number of moves is likely even higher. Is there maybe even a constructive method to create this "bored god's algorithm" (so that even gods with less than stellar memory can apply it)? AI: At http://bruce.cubing.net/index.html it is claimed that there is a Hamiltonian circuit for the corresponding graph. Thus there is a sequence of moves that goes through each position exactly once and then comes back to the original position.
H: Compact subspace Is the subspace $$C^k([0,T]) \subset C^{k-n}([0,T])$$ compact? I think the answer is no. But since $C^k$ is compactly embedded in $C^{k-n}$, it seems like it should be yes in some way. Can I do anything here? ($C^k$ is the space of $k$ times continuously differentiable functions) AI: What is true, I think, is that the unit ball of $C^k([0,T])$ is compact in $C^{k-n}([0,T])$. EDIT: that should be "relatively compact", not "compact".
H: Hermitian/positive definite matrices and their analogues in complex numbers I've heard a couple of times some people say that in a way, Hermitian matrices are to matrices as real numbers are to complex numbers. I know two examples where this is sort of true: Complex conjugate vs conjugate transpose We can realize complex numbers as a set of matrices, where complex number $z = a + bi$ corresponds to a matrix $$A = \begin{bmatrix}a & -b \\ b & a\end{bmatrix}$$ Then the conjugate $\bar{z} = a - bi$ corresponds to the conjugate transpose $A^$. A complex number is real if and only if $z = \bar{z}$. On the other hand, matrix $A$ is called Hermitian if and only if $A = A^$. Polar form vs polar decomposition: We can represent every complex number $z$ as $z = re^{i\varphi}$, where $r \geq 0$ and $\varphi \in [0, 2\pi[$. This representation is unique when $z \neq 0$. On the other hand, any $n \times n$ matrix $A$ with complex entries can be represented as $A = RU$ where $R$ is positive semidefinite (thus Hermitian) and $U$ is unitary. This representation is unique when $A$ is invertible. Also, if $\det A = re^{i\varphi}$, then $\det R = r$ and $\det U = e^{i\varphi}$. Taking the second example really far, you could say positive semidefinite matrices are like nonnegative real numbers and unitary matrices are like points on the unit circle. Are there more examples? I think this is interesting, but is thinking like this useful at all? It seems like oversimplifying things and I doubt if this leads to anything other than some fun facts. AI: Well, one trivial connection is that if you look at $1\times 1$ matrices (which have only a single complex entry), then you'll find that it is real iff it is Hermitian, its complex conjugate is its conjugate transpose, and its polar decomposition is the polar form. Also, just like a complex number can be uniquely decomposed into a real and an imaginary part ($z = a+\mathrm ib$ with real $a,b$), a complex matrix can be uniquely decomposed into a Hermitian and an "anti-Hermitian" part, i.e,. $M =A + \mathrm iB$ with $A$ and $B$ Hermitian. And just like $\Re(z)=\frac12(z+\bar z)$ and $\Im(z)=\frac1{2\mathrm i}(z-\bar z)$, the Hermitian part of a matrix is $\frac12(M+M^)$ and the "anti-Hermitian" part is $\frac1{2\mathrm i}(M-M^)$. Moreover, just like $\bar zz$ is a non-negative real number, $M^*M$ is a positive semidefinite matrix. Another point: Hermitian matrices have real eigenvalues, and unitary matrices have eigenvalues of the form $\mathrm e^{\mathrm i\phi}$. About the usefulness of the analogy: In classical physics, observables should be real. In quantum physics, observables are represented by Hermitian matrices. Also, the quantum analogue to probability densities, which are non-negative functions with integral $1$, are density operators, which are positive semidefinite matrices with trace $1$. So there's indeed some connection.
H: Why Does Finitely Generated Mean A Different Thing For Algebras? I've always wondered why finitely generated modules are of form $$M=Ra_1+\dots+Ra_n$$ while finitely generated algebras have form $$R=k[a_1,\dots, a_n]$$ and finite algebras have form $$R=ka_1+\dots +ka_n$$ It seems to me that this is an flagrant abuse of nomenclature, and has certainly confused me in the past. Is there some historical reason for this terminology? Any references to the genesis of these terms would be greatly appreciated! AI: The terminology is actually very appropriate and precise. Consider that "A is a finitely generated X" means "there exists a finite set G such that A is the smallest X containing G". Looking at your examples, suppose $M$ is a finitely generated module, generated by $a_1,\dots,a_n$. Then $M$ contains $a_1,\dots,a_n$. Since it is a module, it must contain all elements of the form $Ra_i$ and their sums, so it must contain the module $Ra_1+\dots+Ra_n$. However, since this latter object is in fact a module, $M$ need not contain anything else and is in fact equal to this module. If $R$ is a finitely generated algebra, we can go through the same procedure as before. However, since algebras have an additional operation (multiplication), we must allow not only sums of elements of the form $ka_n$ but also their products. This gives us that $R$ must contain all polynomial expressions in the elements $a_1,\dots,a_n$, i.e. it must contain the algebra $k[a_1,\dots,a_n]$. Again, since this latter object is in fact an algebra, $R$ need not contain anything else and is equal to this algebra. A finite algebra seems to be a name for an algebra which is finitely generated as a module. Your example is then consistent with what I wrote above. I do admit that the name seems somewhat misleading.
H: Exponent and group quotient We know tha if $1\not=H\trianglelefteq G$ which consists of all the elements of $G$ which have finite order, then in $G/H$, no elements(except element neutral) has finite order. My question is: If $1\not=H\trianglelefteq G$ ($H\not=G$) has exponent finite, then what can say on $G/H$? AI: Unless I have misinterpreted your question, nothing can be said about $G/H$. Take for example $G= H \times K$, with $H$ finite (and hence of finite exponent) and $K$ arbitrary and non-trivial. $G/H$ is isomorphic to $K$ and this group can have any property you want.
H: Period of $x^n \mod y$? Let $x$, $y$, $n$ be positive integers, where $n$ and $y$ shall be constants and $x$ a variable. Then it is trivial that the period of $x \bmod y$ in $x$ is $y$, since the function simply drops to zero and starts over when $x$ reaches $y$. Now, for me at least, it is much less obvious, what the period of $x^n \bmod y$ in $x$ should be. Visually (looking at plots of the function at different $n$ and $y$) the results suggest that the period still stays $y$. How can I see that analytically? AI: There are many cases where the period is $y$ (and, as Andre points out, many cases where it is not). If $y$ is squarefree (that is, if there is no integer $d\gt1$ such that $y$ is a multiple of $d^2$), then $x^n$ is zero if and only if $x$ is 0 (modulo $y$), so the period has to be $y$.
H: A regular expression for the words that don't contain the sequence $ab$ over $\{a,b,c\}$ The following is an exercise in a book I am reading: Let $\Sigma=\{a,b,c\}$, define $L$ to be the language of all words over $\Sigma$ that do not contain $ab$ as a sub-word. Find a regular expression for $L$. I am unable to solve this problem, I would like to know how to tackle this type of question, what is the solution and what is the thought process to get to it AI: One easy if tedious approach is to design a finite-state automaton that recognizes $L$ and then apply the algorithm that converts such an automaton to a corresponding regular expression. Going about it directly, I observe that if $w\in L$, every $a$ in $w$ must be immediately followed by another $a$ or a $c$ or be at the end of the word; otherwise there are no restrictions. Similarly, every $b$ must be immediately preceded by another $b$ or a $c$ or be at the beginning of the word. Assume for the moment that $w$ does not begin with $b$ or end with $a$. If $w$ begins with $a$, it must begin $aa^c(b\mid c)^$. This pattern may be repeated any number of times: $\big(aa^c(b\mid c)^\big)^$. To allow it to end with $a$, just tack on $a^$: $\big(aa^c(b\mid c)^\big)^a^$. To allow it to begin with $b$, prefix $b^$: $b^\big(aa^c(b\mid c)^\big)^a^$. In fact, a little thought reveals that we can instead prefix $(b\mid c)^$ to cover all cases: $$(b\mid c)^\big(aa^c(b\mid c)^\big)^a^\tag{1}$$ Added: Gerry Myerson’s approach in the comments is more elegant, though I’d carry it out a little differently. $\Sigma^=(a\mid b\mid c)^$, and we want to keep all of it that doesn’t have an $a$ followed immediately by a $b$. Thus, except at the end of a word we want to replace the selection $a$ by one of $ac,aac,aaac$, etc. If we allowed infinite disjunctions, this would give us $$(b\mid c\mid ac\mid aac\mid aaac\mid\dots)^a^\tag{2}$$ for $L$, where the last $a^$ is to allow the word to end with $a$. We don’t allow such expressions, but $ac\mid aac\mid aaac\mid\dots$ can be written legitimately as $aa^c$, and $(2)$ can then be written legitimately as $$(aa^c\mid b\mid c)^a^\;.\tag{3}$$ It’s not hard to see that the languages described by $(1)$ and $(3)$ are subsets of $L$: neither regular expression permits $ab$. To see that these languages include all of $L$, note first $\lambda$, the empty word, is in both. Now let $w=x_1^{n_1}\dots x_m^{n_m}\in L$ be non-empty, where $x_k\in\Sigma$ and $n_k\in\Bbb Z^+$ for $k=1,\dots,m$, and $x_k\ne x_{k+1}$ for $k=1,\dots,m-1$. Assume further that $w$ is minimal in length among all words of $L$ not matching one of the regular expressions $(1)$ and $(3)$. If $a$ does not occur in $w$, $w$ clearly matches both $(1)$ and $(3)$, so let $i$ be the first index such that $x_i=a$, and let $u=a^{n_i}x_{i+1}^{n_{i+1}}\dots x_m^{n_m}$. To show that $w$ matches $(1)$, it suffices to show that $u$ matches $\big(aa^c(b\mid c)^\big)^a^$; to show that $w$ matches $(3)$, it suffices to show that $u$ matches $(3)$. Both of these are clear if $u=a^{n_i}$, since in that case $u$ matches the final $a^$ of each regular expression. Otherwise, $x_{i+1}=c$, and $a^{n_i}c$ matches $aa^*c$ in both regular expressions. Now let $v$ be what’s left of $u$ after the initial $a^{n_i}c$ is removed. Then $w$ matches $(1)$ iff $v$ matches $(1)$, and $w$ matches $(3)$ iff $v$ matches $(3)$. But $v\in L$, and $\|v\|<\|w\|$, so by hypothesis $v$ matches both $(1)$ and $(3)$, and hence so does $w$. Thus, each of these regular expressions represents $L$.
H: Proof of first isomorphism theorem of group (Proof of (first) Isomorphism Theorem) Let $f : G \rightarrow H$ be a surjective group homomorphism. Let $K = \operatorname{ker} f$. Then the map $f' : G/K \rightarrow H$ by $f'(gK) = f(g)$ is well-defined and is an isomorphism. Proof: If $g'K = gK$, then $g' = gk$ with $k ∈ K$, and $f(g') = f(gk) = f(g) f(k) = f(g) e = f(g)$ so the map $f'$ is well-defined. It is surjective because $f$ is. For injectivity, if $f'(gK) = f'(g'K)$, then $f(g) = f(g')$, and $e_{H} = f(g)^{-1} · f(g') = f(g^{−1}) · f(g) = f(g^{−1}g')$ Thus, $g^{−1}g' ∈ K$, so $g' ∈ gK$, and $g'K = gK$. In summary, the normal subgroups of a group are exactly the kernels of surjective homomorphisms. First of all, I am not getting how $f(g)^{-1} \cdot f(g')$ leads to $f(g^{-1}) \cdot f(g)$. Secondly, I am unsure how the later part of the proof establishes injectivity. And lastly, at http://en.wikipedia.org/wiki/Isomorphism_theorem#Discussion, there is a diagram that shows the relationship, and can anyone explain this diagram with some insights from the above proof? AI: For your first question, it is essentially assumed by hypothesis and the definition of $f'$ that $f(g')=f(g)$. But recall that for a homomorphism $f$, $f(g)^{-1}=f(g^{-1})$. This can be seen since $f(g^{-1})f(g)=f(g^{-1}g)=f(e_G)=e_H$, so $f(g^{-1})$ is indeed a left inverse for $f(g)$, and the same argument shows it is a right inverse. So just substitute to conclude $f(g)^{-1}\cdot f(g')=f(g^{-1})\cdot f(g)$. Secondly, writing $x=gK$ and $y=g'K$, the proof shows that if $f'(x)=f'(y)$, then $x=y$. This is the standard method to show a function is injective. Another way to look at it is the equivalent statement that if $x\neq y$, then $f'(x)\neq f'(y)$. So $f'$ always takes distinct inputs to distinct outputs, that is, $f'$ is injective.
H: Summing digits of powers of 2 to get 1 2 4 8 7 5 pattern If you add the decimal digits of multiples of 9,possibly repeatedly, you get 9. For instance, with 18 we have $$18\mapsto 1+8=9$$ and with 909 we have $$ 909\mapsto 9+0+9=18\mapsto 1+8=9.$$ This is proven. If you do the same with powers of two, you develop a repeating pattern, 1,2,4,8,7,5, even when you descend negative powers. I don't know if this stops eventually, but it seems like there might be a proof for it. Can anyone prove this? \begin{align} .0625\mapsto 6+2+5=13&\mapsto 4 \\ .125&\mapsto 8 \\ .25 &\mapsto 7 \\ .5 &\mapsto 5 \\ 1&\mapsto 1 \\ 2 &\mapsto 2 \\ 4 &\mapsto 4 \\ 8&\mapsto 8 \\ 16&\mapsto 7 \\ 32&\mapsto 5 \\ 64\mapsto 10 &\mapsto 1 \\ 128 \mapsto 11 &\mapsto 2 \\ 256 \mapsto 13 &\mapsto 4 \\ 512 &\mapsto 8 \\ 1024 &\mapsto 7 \\ 2048 \mapsto 14&\mapsto 5 \\ 4096\mapsto 19 \mapsto 10 &\mapsto 1 \\ \end{align} AI: We look at $2^n$, where $n$ ranges over the non-negative integers. The key is the fact that $2^6$ has remainder $1$ on division by $9$. Using congruence notation, we have $2^6\equiv 1\pmod{9}$. Let $n$ be any non-negative integer. We can express $n$ as $6q+r$, where $0\le r\le 5$ ($q$ stands for quotient, $r$ for remainder). It follows that $$2^n=2^{6q+r}=(2^6)^q 2^r\equiv (1)^q 2^r\equiv 2^r\pmod{9}.$$ So the remainder when you divide $2^n$ by $9$ depends only on $r$. For $r=0$, $1$, $2$, $3$, $4$, and $5$, these remainders are, as you observed, $1$, $2$, $4$, $8$, $7$, and $5$. To connect this with sums of (decimal) digits, observe that a decimal number like $6852$ is just $(6)(10^3)+(8)(10^2)+(5)(10^1) +(2)(10^0)$. But for any non-negative integer $k$, we have $10^k\equiv 1\pmod{9}$. So $6852\equiv 6+8+5+2 \pmod{9}$. Thus the remainder when $6852$ is divided by $9$ is the same as the remainder when $6+8+5+2$ is divided by $9$. Asimilar remark holds for any non-negative integer expressed in decimal form. Since remainders when $2^n$ is divided by $9$ cycle with period $6$, and the "casting out nines" process gives us these remainders, the pattern you observed continues forever. You extended the pattern to negative exponents. This is an interesting observation that I do not recall seeing before. Express $2^{-n}$ as a decimal, by noting that $$2^{-n}=\frac{1}{2^n}=\frac{5^n}{10^n}.$$ Then (essentially) you looked at the digit sum (modulo $9$) of $5^n$. Modulo $9$, the numbers $5^n$ cycle with period $6$, for the same reason as with $2^n$. Now calculate $5^n$ modulo $9$, for $n=0,1, 2, 3, 4, 5$. We get that $5^n$ is congruent in turn to $1$, $5$, $7$, $8$, $4$, and $2$ modulo $9$. So the pattern does indeed continue "backwards." Remark: For a more general approach, please see Euler's Theorem.
H: Integral $\int_0^\infty x\exp(-\frac{x^2}{2}\ )\mathrm{d}x$ I was trying to solve $$\int_0^\infty x \exp \left( { - \frac{{{x^2}}}{2}} \right)\;{\text{d}}x$$ I was trying to use $$(f(g(x)))'=f'(g(x))\cdot g'(x)$$ (So $[e^{f(x)}]'=[e^{f(x)}]\cdot f'(x)$ ?), but I got the wrong answer. (I substituted $0$ and $\infty$ to $-\exp \left(-\frac{x^2}{2}\right)$ and got 1. But the test solution says $\sqrt{e}$.) Could you point out what's wrong with my solution? AI: We know that $$\int f'(g(x))g'(x)\, dx=f(g(x))$$ This is the basis of a basic integral substitution. If we let $g(x)=u \implies g'(x)\, dx=du$ we can change the above integral into $$\int f'(u)\, du=f(u)=f(g(x))$$ Using this method, we let $$f(x)=\exp(-x)$$ $$g(x)=u= \frac{x^2}{2} \implies x=\sqrt{2 u} \implies dx= \frac{1}{\sqrt{2u}}\, du$$ so the integral becomes $$\int x\exp(-x^2/2)\, dx=\int \sqrt{2u}\cdot\exp(-u)\, \cdot\frac{1}{\sqrt{2u}}\, du=-\exp(-u)=-\exp(-\frac{x^2}{2})$$
H: How do you go from $x\in\cup\{\mathcal{P}(A)\|A\in\mathcal{F}\}$ to $\exists A\in\mathcal{F}(x\in \mathcal{P}(A))$? These are statement 4 of example 2.3.6 and its solution from section 2.3 in Daniel J. Velleman's "How to Prove It - A Structured Approach" (great book), where the author asks the reader to analyze the logical form of several statements. On the solution to this particular statement, i.e.: $x\in\cup\{\mathcal{P}(A)\|A\in\mathcal{F}\}$ he argues that, according to the definition of union given earlier: $\cup\mathcal{F}=\{x\|\exists A\in\mathcal{F}(x\in A)\}=\{x\|\exists A(A\in\mathcal{F}\wedge x\in A)\}$ the statement means that "... $x$ is an element of at least one of the sets $\mathcal{P}(A)$, for $A\in\mathcal{F}$. In other words, $\exists A\in\mathcal{F}(x\in\mathcal{P}(A))$." Intuitively it makes sense, but I can't write it down formally. If I state that $x\in\cup\mathcal{F}$ is true, I know that $\exists A\in\mathcal{F}(x\in A)$. But I get lost when I try to replace $\mathcal{F}$ with $\mathcal{P}(A)$, and can't figure out the rest. AI: Define $\mathcal{G} = \{P(A) : A \in \mathcal{F}\}$. Then $x \in \bigcup \{P(A) : A \in \mathcal{F}\}$ is equivalent to $x \in \bigcup \mathcal{G}$ By definition of union, it is equivalent to $(\exists B \in \mathcal{G})(x \in B)$ Recall that $(\exists B \in \mathcal{G}$ if and only if there $(\exists A \in \mathcal{F})(B = \mathcal{P}(A))$. So replacing $B$ by $\mathcal{P}(A)$, you get $(\exists A \in \mathcal{F})(x \in \mathcal{P}(A))$
H: An exercise from a linear algebra book (hoffman and kunze) I am not sure how to solve this exercise: Find all solutions of $$\begin{align}2x_1-3x_2-7x_3+5x_4+2x_5&=-2\\ x_1-2x_2-4x_3+3x_4+x_5&=-2\\ 2x_1\qquad-4x_3+2x_4+x_5&=3\\ x_1-5x_2-7x_3+6x_4+2x_5&=-7\;. \end{align}$$ my row-reduced echelon matrix is: $$\pmatrix{1&0&-2&1&0&1\\ 0&1&1&-1&0&2\\ 0&0&0&0&1&1\\ 0&0&0&0&0&0}$$ I don't see my mistake right now, I think I did something wrong xD Thank you very much! AI: From the reduced form, you have the following equations: $$x_5 = 1, \ \ x_1-2 x_3 + x_4 = 1, \ \ x_2+x_3 - x_4 = 2.$$ Since this gives $x_1 = 1+2x_3-x_4$ and $x_2 =2-x_3+x_4$, we can choose $x_3,x_4$ arbitrarily and determine the corresponding $x_1,x_2$. Hence the solution set is given by: $$\begin{bmatrix} 1 \\ 2 \\ 0 \\ 0 \\ 1 \end{bmatrix}+ x_3 \begin{bmatrix} 2 \\ -1 \\ 1 \\ 0 \\ 0 \end{bmatrix} + x_4 \begin{bmatrix} -1 \\ 1 \\ 0 \\ 1 \\ 0 \end{bmatrix},$$ with $x_3, x_4$ arbitrary.
H: Find $w$ such that $w^8=15-15i$ Find the complex number, lying in the second quadrant, and having the smallest possible real part, which satisfies the equation $$w^8=15-15i$$ AI: $$z=15-15i\Longrightarrow \|z\|=15\,\sqrt 2\,\exp({7\pi i}/{4}+2k\pi i),\,k\in\Bbb Z$$ $$\Longrightarrow w^8=z\Longrightarrow w=z^{1/8}=15^{1/8}\,2^{1/16}\,\exp({7\pi i}/32+{k\pi i}/{4})$$ Now just observe that as $\,k\,$ runs from $\,0\,$ to $\,7\,$, we get all the possible (eight) values on the right-hand side above...
H: collinear three points in a linear equation $ax+by+z$ So, there is a plane equation $ax+by+z = 0$. Suppose there are three points $(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$, $(x_3,y_3,z_3)$ that is on the given plane. Then, $\begin{vmatrix} x & y & z & 1\\x_1 & y_1 & z_1 & 1\\x_2 & y_2 & z_2 & 1\\x_3 & y_3 & z_3 & 1 \end{vmatrix}=0$. What happens if the three points are collinear? AI: (Too long for a comment.) You can prove the identity $$\begin{vmatrix} x & y & z & 1\\x_1 & y_1 & z_1 & 1\\x_2 & y_2 & z_2 & 1\\x_3 & y_3 & z_3 & 1 \end{vmatrix}=\det\left(\begin{pmatrix}x & y & z\\x & y & z\\x & y & z\end{pmatrix}-\begin{pmatrix}x_1 & y_1 & z_1 \\x_2 & y_2 & z_2 \\x_3 & y_3 & z_3\end{pmatrix}\right)$$ If the three points given are collinear, then any row you pick can be written as a linear combination of the two other rows, making the determinant zero, and thus you wind up with the not very useful equality $0=0$.
H: Evaluate $\sum_{i=1}^n \frac{i}{n^2}$ My book asks to simplify this problem: $$\sum_{i=1}^n\frac{i}{n^2}$$ This equals: $$\sum_{i=1}^n\frac{1}{n^2}i$$ Now, shouldn't $$\sum_{i=1}^n\frac{1}{n^2}=\frac{1}{n}\text{ ?}$$ Because you are summing $1/n^2$, $n$ times, so that $n / n^2 = 1/n$. So it should be $$\frac{1}{n}\sum_1^n i = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \mathbf{\frac{n+1}{2}}$$ But the answer in my book is $\mathbf{(n+1)/2n}$. AI: Your answer would be correct if you were given $$\left(\sum_{i=1}^n \frac{1}{n^2} \right)\left(\sum_{i=1}^n i\right).$$ The problem is that $$\sum_{i=1}^n \frac{i}{n^2} \not = \left(\sum_{i=1}^n \frac{1}{n^2} \right)\left(\sum_{i=1}^n i\right).$$ Instead, you should have $$\sum_{i=1}^n \frac{i}{n^2} = \left( \frac{1}{n^2} \right)\left(\sum_{i=1}^n i\right),$$ which explains the book's answer. This is just the distributive property of multiplication over addition.
H: Function is pointwise limit of integrals This is a question from some old masters exams. Let $\phi_{n}$ be a sequence of continuous, real functions on $\mathbb{R}$ such that $\phi_{n}(x) = 0$ for all $\|x\|\geq 1/n$ and $\phi_{n}(x)\geq 0$ for all $x\in\mathbb{R}$, and $$\int\limits_{-1}^{1}{\phi_{n}(x)dx}=1$$ For each continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$, let $$f_{n}(x)=\int\limits_{-\infty}^{\infty}{\phi_{n}(x-y)f(y)dy}$$ Prove that $f_{n}(x)$ converges pointwise to $f(x)$ and prove that if $f(x)=0$ for $\|x\|\geq 10$, then $f_{n}$ converges uniformly to $f$. AI: For pointwise convergence, note that it suffices to show that $$\int_{-\infty}^\infty \phi_n(x-y)(f(y)-f(x))dy\to 0$$ for all $x$. Fix some $x$. Since $f$ is continuous, for any $\epsilon>0$ we have some $\delta>0$ such that $$\|x-y\|<\delta\implies \|f(x)-f(y)\|<\epsilon.$$ Choose $N\in\mathbb N$ such that $N> 1/\delta$. Then for $n\geq N$ we have $$\begin{align} \left\|\int_{-\infty}^\infty \phi_n(x-y)(f(y)-f(x))dy\right\| &=\left\|\int_{x-1/n}^{x+1/n} \phi_n(x-y)(f(y)-f(x))dy\right\|\\ &\leq \int_{x-1/n}^{x+1/n} \phi_n(x-y)\|f(y)-f(x)\|dy\\ &< \epsilon \int_{x-1/n}^{x+1/n} \phi_n(x-y)dy = \epsilon \end{align}$$ hence $f_n\to f$ pointwise. For uniform convergence, note that if $f(x)=0$ when $\|x\|\geq 10$ then $f$ has compact support, hence we can choose this $\epsilon$ independently of $x$ hence the convergence is uniform in $x$.
H: Given a list of axioms, can one prove that a statement is unprovable without exhibiting a model? Suppose I am given the following axioms for a vector space $V$ over the field $F$: For addition of vectors Closure Associativity Existence of an additive identity Existence of additive inverse Commutativity For multiplication by scalars Closure Associativity The two distributive laws From this I would like to prove that the statement $\forall v \: 1v = v$, where $v \in V$ and $1$ is the multiplicative identity of $F$, cannot be deduced from the above axioms. One possible method would be to exhibit a model, that is, a set of objects, a field, with operations of addition and multiplication by the elements of the field defined, where all the above axioms are satisfied but the statement $\forall v \: 1v = v$ is not. Is there some method by which I could prove the unprovability of the statement without exhibiting a model? AI: For your particular example for Vector Spaces, the only thing I can think of to prove that an axiom is unproveable from the rest is to exhibit a model for which some collection of axioms hold and some other collection does not. When your theory is sufficently strong, there exists another method to prove that a statement is unproveable. The Godel Incompleteness statements states that certain type of theories can not prove its own consistency. Peano Arithmetics, Second Order Arithmetics, Various systems of ZFC, etc are example of such theories for the incompleteness theorem holds. An actual use of this is the proof that ZFC can not prove that there exists (weakly) inaccessible cardinals. (You can look up the definition of inaccessible cardinals and other large cardinal properties.) The idea is that if ZFC proved the existence of in inaccessible cardinal, then since inaccessible cardinal are "very large" in some sence, you can use it to prove the consistency of ZFC by producing a model of $ZFC$. This would contradict the incompleteness theorem; hence, $ZFC$ can not prove the existence of inaccessible cardinals. (Also many set theorist believe that $ZFC$ can not prove that inaccessible cardinals must not exist.) Something a bit more down to earth: The fact that you can not lose the Hydra Game is unproveable in Peano Arithmetic. Basically, the Hydra Game is that you have a Tree. At each step $n$, you cut some node of the tree. You go down one node, and duplicate what left on that node $n$ times. To win the game you need to cut off all the "heads". See this website for a better description : http://math.andrej.com/2008/02/02/the-hydra-game/ You can try to prove using sufficiency strong mathematics that whatever you do, you will eventually win. However, Peano Arithemtics can not prove this result. The fact that you can not lose the Hydra Game can be used to prove the consistency of Peano Arithemtics. Again by the incompleteness theorem, Peano Arithmetics can not prove that you can never lose the Hydra Game.
H: Partitioning $\mathbb{N}$ into distinct AP’s Can $\mathbb{N}$, the set of natural numbers, be partitioned into a finite number of subsets that are in arithmetic progression with distinct steps ? AI: No. See https://mathoverflow.net/questions/25313/finitely-many-arithmetic-progressions.
H: Tensor-commutative abelian groups Say that an abelian group $A$ is tensor-commutative if the equality $x\otimes y=y\otimes x$ holds in $A\otimes_{\mathbb Z}A$ for all $x,y$ in $A$. The first question is somewhat vague: Question 1. Can one characterize the tensor-commutative abelian groups? It is easy to see that subquotients of $\mathbb Q$ are tensor-commutative, but I haven't been able to come up with other examples, whence Question 2. Are there tensor-commutative abelian groups which are not subquotients of $\mathbb Q$? [This is (I think) a natural continuation of Paul Slevin's question Are bimodules over a commutative ring always modules? More precisely, Paul asks implicitly which are the commutative rings $R$ such that any $R$-module admits only the boring $(R,R)$-bimodule structure. These are exactly the tensor-commutative rings. (In particular this condition depends only on the additive group of $R$.)] EDIT. As explained here, the answer to Question 2 is yes. So, it seems natural to ask Question 3. Is there a tensor-commutative ring which is not a subquotient of $\mathbb Q$? AI: If $A$ is a tensor-commutative abelian group, its localization $A_{\mathbb Q}=A\otimes_{\mathbb Z}\mathbb Q$ is at most one dimensional as a rational vector space. Indeed, one easily sees that $A_{\mathbb Q}$ is a «tensor-commutative $\mathbb Q$-vector space» and this is only possible if $A_{\mathbb Q}$ is at most one dimensional. This tells us that the rank of $A$ is at most $1$. Since torsion-free groups of rank $1$ are isomorphic to subgroups of $\mathbb Q$, this deals with the torsion-free case. On the other end of the spectrum, suppose $A$ is torsion and tensor-commutative. It is the direct sum of its $p$-primary components for all $p$, and each of these is tensor-commutative. Conversely, if the $p$-primary components are tensor-commutative, $A$ is (because the tensor product of the components corresponding to different primes is simply zero) We need only look at $p$-torsion tensor-commutative groups. Let $A$ be such a thing. Then $A/pA$ is a $\mathbb Z/p\mathbb Z$ vector space which is tensor-commutative, so it must be of dimension at most $1$, that is, zero or a cyclic group of order $p$. Maybe one can keep this going and show that $A$ must be a Prüfer $p$-group in this case? To do this one wants to know if $pA$ is also tensor-commutative.
H: $g(x)$ is a function such that $g(x+1)+g(x-1)=g(x)$, $x \in \mathbb{R}$. For what value of $p$, $g(x+p)=g(x)$. $g(x)$ is a function such that $g(x+1)+g(x-1)=g(x)$, $x \in \mathbb{R}$.For what value of $p$, $g(x+p)=g(x)$. $g(x+2)+g(x)=g(x+1)$ AI: This is really a question about sequences hence assume that $x_{n+2}-x_{n+1}+x_n=0$ for every $n$, for some sequence $(x_n)_{n\in\mathbb Z}$. The characteristic equation of this recursion is $r^2-r+1=0$, with roots $r=\mathrm e^{\pm\mathrm i\pi/3}$, hence $x_n=A\mathrm e^{n\mathrm i\pi/3}+B\mathrm e^{-n\mathrm i\pi/3}$ for some $A$ and $B$. Now, $\mathrm e^{2\mathrm i\pi}=1$ hence, for every $(A,B)$, $x_{n+6}=x_n$. No period smaller than $6$ is valid for every such sequence $(x_n)_{n\in\mathbb Z}$, as the example of $x_n=\cos(n\pi/3)$ shows. Edit: The link with the original question is as follows. For every $0\leqslant x\lt1$, $x_n=g(x+n)$ defines a sequence $(x_n)_{n\in\mathbb Z}$ as above, hence there exists two parameters $A(x)$ and $B(x)$ such that $$ g(x+n)=A(x)\mathrm e^{n\mathrm i\pi/3}+B(x)\mathrm e^{-n\mathrm i\pi/3}, $$ for every integer $n$. There is no relation whatsoever between the families $(g(x+n))_{n\in\mathbb Z}$ and $(g(x'+n))_{n\in\mathbb Z}$ for $x\ne x'$ in $[0,1)$. Hence, each solution $g:\mathbb R\to\mathbb C$ is described uniquely by two functions $A:[0,1)\to\mathbb C$ and $B:[0,1)\to\mathbb C$ and any two such functions $(A,B)$ define uniquely a solution $g_{A,B}$.
H: Group Theory - Proving $(ab)^{-1} = (a^{-1}) (b^{-1}) $ I'm trying to prove the below equation, where $ a,b \in G $ and $(G, ) $ is a group. $(ab)^{-1} = (a^{-1}) * (b^{-1}) $ I'm not really sure how to do it though. I tried doing something like $(ab)^{-1} (ab) = e = a a^{-1} * b^{-1} * b $ $(ab)^{-1} (ab) = e = a (a^{-1} * b^{-1}) * b $ However, I can't get the RHS to $ (a * b)* (a^{-1} * b^{-1}) $ without assuming communitivity. How would I go about doing this? Any help is appreciated. Thanks Edit: Given that correction, would this be correct? $ (ab)^{-1} (ab) = e = a(bb^{-1})a^{-1}$ Then $ a(bb^{-1})a^{-1} = ab* (b^{-1}a^{-1}) = (ab)(b^{-1}a^{-1})$ Thus $ (ab)^{-1} = (b^{-1}a^{-1})$ AI: Most of the answers above tell you how to verify what $(a \ast b)^{-1}$ is once you have found it. Of course this is all very well and good if you "already know the answer", but gives no insight in how "you find the answer". But this is group theory, and groups are in a sense "designed for solving equations". So let's write an equation and solve it. An inverse for $a \ast b$ is an element $x \in G$ such that: $(a \ast b) \ast x = e = x \ast (a \ast b)$. Therefore (using the first equation): $a^{-1} \ast ((a \ast b) \ast x) = a^{-1} \ast e$ $(a^{-1} \ast a) \ast (b \ast x) = a^{-1}$ $e \ast (b \ast x) = a^{-1}$ $b \ast x = a^{-1}$ $b^{-1} \ast (b \ast x) = b^{-1} \ast a^{-1}$ $(b^{-1} \ast b) \ast x = b^{-1} \ast a^{-1}$ $e \ast x = b^{-1} \ast a^{-1}$ $x = b^{-1} \ast a^{-1}$ Having found what any "right-inverse" of $a \ast b$ must be, and since $G$ is a group, it must be that $x$ is also a left-inverse, although it may now be verified directly that $b^{-1} \ast a^{-1}$ is also a left inverse.
H: Evaluate $\lim_{s\to0}\frac{s\sqrt{e^s}}{e^s-1}$ without using L'Hospital rule. Evaluate $\lim_{s\to0}\frac{s\sqrt{e^s}}{e^s-1}$ without using L'Hospital rule. I've done this: Since $\sqrt{e^s}\to1$as $s\to1$, so $\lim_{s\to0}\frac{s\sqrt{e^s}}{e^s-1}=\lim_{s\to0}\frac{s}{e^s-1}=1$ Are my steps correct? Is step 1 to step 2 available? Thank you. AI: Well, indeed you are right: you can prove it just based on the limit $\lim\limits_{x\to 0}(1+x)^{1/x} = \mathrm e$. By taking the logarithm of both parts, you obtain $\lim\limits_{x\to 0}\frac{\log(1+x)}{x} = 1$. If you put $s =\log(1+x)$ then you get $$ \lim\limits_{s\to 0}\frac{s}{\mathrm e^s-1} = 1 $$ as you mentioned. Now, if you have $\lim\limits_{s\to 0}f(s) = F$ and $\lim\limits_{s\to 0}g(s) = G$ then $$ \lim\limits_{s\to 0}f(s)g(s) = FG $$ and the limit in LHS exists - that's why you can go from step 1 to step 2 in your proof.
H: How does the supremum of the cardinalities of a set of ordinals relate to the supremum of the ordinals? Suppose we have a set of ordinals such that their supremum is a cardinal (not in the original set). $\sup\{\alpha\}=\kappa$ I'm interested in the supremum of the cardinalities of those ordinals: $\sup\{\|\alpha\|\}$. I've seen many proofs that seem to assume that $\sup\{\|\alpha\|\}=\kappa$ also. I can see how this would be the case if the set $\{\alpha\}$ contained an infinite subset of cardinals whose supremum was also $\kappa$, but I can think of a simple counterexample that would seem to indicate that's not always the case. Take our set to be $\{\alpha\mid\omega\le\alpha\lt\omega_1\}$. The supremum of this set is clearly $\omega_1$ (with cardinality $\aleph_1$ if you prefer). But each ordinal $\alpha$ is countably infinite, so $\forall\alpha,\|\alpha\|=\omega$ (or $\aleph_0$ if you insist). But then $$\sup\{\|\alpha\|\}=\sup\{\omega\}=\omega\ne\omega_1=\sup\{\alpha\}.$$ What am I missing? Motivation: I'm trying to show that if $\kappa$ is an infinite cardinal, these definitions for cofinality are equivalent: $$cof(\kappa)=\inf\{\beta\mid\{\alpha_\xi\}_{\xi\lt\beta}\text{ is cofinal in }\kappa\}\equiv\inf\{\delta\mid\sum_{\xi<\delta}\kappa_\xi=\kappa\}.$$ To establish the RHS, (after taking $\{\kappa_\xi\}=\{\|\alpha_\xi\|\}$) it's easy to show the summation is bounded above by $\kappa$. To show that it is also bounded below by $\kappa$, an example proof I found claims that since each $$\kappa_\xi\le\sum_{\xi<\delta}\kappa_\xi$$ then the summation, as an upper bound of all of the $\kappa_\xi$'s, must be larger than the smallest upper bound (supremum) of the $\kappa_\xi$'s, so $$\kappa\le\sup\{\alpha_\xi\}=\sup\{\|\alpha_\xi\|\}=\sup\{\kappa_\xi\}\le\sum_{\xi<\delta}\kappa_\xi$$ but I'm not so sure about the first equality. Any tips would be appreciated. AI: I'll first make a comment about the concern in the title of the question, and it seems to break down into successor and limit cases. If $\kappa = \lambda^+$ is a successor cardinal, then $\| \alpha \| \leq \lambda$ for all ordinals $\alpha < \kappa$, and it easily follows that $\sup \{ \| \alpha \| : \alpha < \kappa \} = \lambda$. (This is what you noticed for $\kappa = \aleph_1 = \aleph_0^+$.) If $\kappa$ is a limit cardinal, then $\mu^+ < \kappa$ for each $\mu < \kappa$. From here it easily follows that $\sup \{ \| \alpha \| : \alpha < \kappa \} = \kappa$. The above dichotomy also holds whenever $\{ \alpha_\xi \}_{\xi < \nu}$ is an arbitrary family of ordinals cofinal in $\kappa$. Of course, it is relatively easy to show that successor cardinals are regular: If $\delta < \kappa = \lambda^+$ and $\{ \alpha_\xi \}_{\xi < \delta}$ is an (increasing) sequence of ordinals $< \kappa$, then let $\beta = \sup_{\xi < \delta} \alpha_\xi = \bigcup_{\xi < \delta} \alpha_\xi$. Note that $$\| \beta \| = \left\| \bigcup_{\xi < \delta} \alpha_\xi \right\| \leq \sum_{\xi < \delta} \| \alpha_\xi \| \leq \sum_{\xi < \delta} \lambda = \| \delta \| \cdot \lambda = \lambda < \kappa.$$ Thus $\{ \alpha_\xi \}_{\xi < \delta}$ is not cofinal in $\kappa$. Note that this also contains the idea for proving the summation characterisation of cofinality in this case: if $\{ \kappa_\xi \}_{\xi < \delta}$ is a family of cardinals $< \kappa = \lambda^+$, then $\sum_{\xi < \delta} \kappa_\xi \leq \sum_{\xi < \delta} \lambda = \| \delta \| \cdot \lambda = \max \{ \|\delta\| , \lambda \}$, so if this sum equals $\kappa = \lambda^+$, it must be that $\| \delta \| = \kappa$.) For a limit cardinal $\kappa$, letting $\mu = \mathrm{cof} ( \kappa )$, note that there is an (increasing) sequence $\{ \kappa_\xi \}_{\xi < \mu}$ of cardinals $< \kappa$ which is cofinal in $\kappa$. It is relatively easy to show that $\sum_{\xi < \mu} \kappa_\xi = \kappa$. Clearly $\kappa_\eta \leq \sum_{\xi < \mu} \kappa_\xi$ for all $\eta < \mu$, and therefore $\kappa \leq \sum_{\xi < \mu} \kappa_\xi$. Given $\delta < \mu$ note that $\kappa_\xi < \kappa_\delta$ for all $\xi < \delta$ and so $\sum_{\xi < \delta} \kappa_\xi \leq \sum_{\xi < \delta} \kappa_\delta = \| \delta \| \cdot \kappa_\delta = \max \{ \| \delta \| , \kappa_\delta \}$. As $\kappa_\delta < \kappa$, and $\| \delta \| < \kappa$, it follows that $\sum_{\xi < \delta} \kappa_\xi < \kappa$. Thus, $\sum_{\xi < \mu} \kappa_\xi \leq \kappa$. But suppose $\delta < \mu$ and $\{ \kappa_\xi \}_{\xi < \delta}$ is any sequence of cardinals $< \kappa$. Then $\{ \kappa_\xi \}_{\xi < \delta}$ is not cofinal in $\kappa$, and so there is a cardinal $\nu < \kappa$ such that $\kappa_\xi \leq \nu$ for all $\xi < \delta$. Then $$\sum_{\xi < \delta} \kappa_\xi \leq \sum_{\xi < \delta} \nu = \| \delta \| \cdot \nu = \max \{ \| \delta \| , \nu \}.$$ As $\nu , \| \delta \| < \kappa$, we have that $\sum_{\xi < \delta} \kappa_\xi < \kappa$.
H: How to calculate blood viscosity formula using $cm^3$ (lt) and $cm^2$ (area) units I'm reading a blood viscosity related theory and I have an example formula which I don't understand. It turns lt (fluid liters) or $cm^3$ into $cm^2$ and comes out with a number. I don't understand how the conversion is made from area to volume and vice-versa. My formula is: $Q = S * v$ $S = 2,5 cm$ $Q = 5Lmin^{-1}$ $v = ?$ So my simlpe equation is: ${\bf (5 cm^3min^{-1}) = 2,5 cm * v}$ In my (friend's) notes I see the following fraction: $\frac{\frac{510^{-3}}{60}}{2,510^{-4}}$ NowI don't understand how the $5cm^3$ got turned into $\frac{5*10^{-3}}{60}$ and why the 2,5 must be multiplied with $10^{-4}$. It's like converting the two units at once to match each other, but I'm missing something... the math between :-) (if there's any). Thanks AI: Well $1L$ is not $1\ \rm{cm}^3$ but $1000\ \rm{cm}^3$ and I think that $S$ is a section so that it should be in $\rm{cm}^2$. If you work in $\rm{cm}$ and in $\rm{s}$ you'll get $$v=\frac{5000\,\rm{cm}^3/60\,\rm{s}}{2.5\,\rm{cm}^2}=\frac {100}3\frac{\rm{cm}}{\rm{s}}$$ You could too use the meter as a unit and in this case you'll get : $$v=\frac{5\cdot 10^{-3}\,\rm{m}^3/60\,\rm{s}}{2.5\cdot 10^{-4}\,\rm{m}^2}=\frac {1}3\frac{\rm{m}}{\rm{s}}\approx 0.3333 \frac{\rm{m}}{\rm{s}}$$ (so that the $10^{-4}$ is right ! :-)) For this kind of problem remember to always write the units (as I did here) and handle them as variables (FrenzY DT proposed you the interesting link to 'dimensional analysis'). Here you had : $$\rm{cm}=10^{-2}\,\rm{m}$$ so that you may use : $$2.5(\rm{cm})^2=2.5\left(10^{-2}\,\rm{m}\right)^2=2.5\cdot 10^{-4}\,\rm{m}^2$$ $$\frac {(\rm{cm})^3}{(\rm{cm})^2}=\rm{cm}$$ the same way you had : $$\rm{m}=60\,\rm{s}\quad \text{i.e.}\ \ 1\ \text{minute}=60\ \text{seconds}$$
H: Modular Multiplicative Inverse & Modular Exponentiation Equation I was solving a problem containing that equation. $$key=(\sum_{K=0}^n\frac{1}{a^K})\mod m$$ Given: $1 \le a \le 2,000,000,000$ $0 \le n \le 2,000,000,000$ $2 \le m \le 2,000,000,000$ $a$ and $m$ are coprime My Approach: To Solve the SECOND part Using Geometric Sequence sum I transfered the equation to this form I Used Euclid's Extended Algorithm to get the Modular Multiplicative Inverse of "$a$" then I used Modular Exponentiation to calculate its modulus to "$m$" Question: What about the first part? how would I calculate it given the problem limits? AI: If $\gcd(a-1,m)=1$, then there will be no problems using the sum formula for a geometric series. Otherwise I suggest the following that is kind of like the square-and-multiply approach to modular exponentiation. Write $q=1/a$ (= the modular inverse of $a$) and denote the geometric sum by $$ S(k)=\sum_{i=0}^{k-1} q^i.\qquad\text{<- Edit: A typo here. The upper bound was off by one.} $$ Then we have the length-doubling recurrence relation (all arithmetic done modulo $m$, so if you are so minded, imagine $\%m$ at the end of each addition or product - I prefer to think of this as doing arithmetic in the ring $\mathbb{Z}/m\mathbb{Z}$): $$ S(2k)=S(k)(1+q^k), $$ as well as the add-one recurrence: $$ S(k+1)=q S(k) +1. $$ With the aid of these you can follow the usual square-and-multiply (or rather, double-and-add) logic. So for example to calculate $S(113)$ you need to first calculate $S(112)$ and then apply the add-one -formula. To calculate $S(112)$ you first calculate $S(56)$, and then apply the length-doubling -formula. Apply these ideas recursively. The complexity looks like to be of the order $O\left((\log_2n)^2\right)$ modular multiplications given that you need to compute all those $q^k$:s for the length doubling formula. It does look like, there is a lot of redundancy there (in that the $q^k$ you will need in the next recursive call has probably already been computed). I'm sure you can build a more efficient recursive procedure from these elements, as it is possible to predetermine exactly which powers $q^k$ are needed! May be what you end up with has linear complexity (in terms of the number of bits in $n$ using the cost of a modular multiplication as a unit)? Some further shaving may be possible by reorganizing the calculation further.
H: Which symbol should be used for an empty set? Currently, a discussion started on the German Wikipedia article for Empty Set (the German discussion), whether $\emptyset$ or $\varnothing$ should be used or is more common as a symbol for an empty set. Do you know any sources for this question? Do you know a source for the statement, that Nicolas Bourbaki was the first who used this symbol? AI: I would say the following should be not too controversial: $\emptyset$ and $\varnothing$ are typographical variants of the same mathematical symbol designating the empty set The symbol was introduced by Bourbaki, was inspired by the Norwegian character Ø, but is a distinct character from it The intention was most probably to create a symbol related to $0$ (zero), not to O (Oh), distinguished from it by striking it through. After all the empty set has all kinds of relations with the number $0$, but none with the letter O. (By contrast big-Oh and little-o symbols derive from the word "order".) The symbol has absolutely no relation (apart from appearance) with the lower-case Greek letter phi, with typographical variants $\phi$ and $\varphi$.
H: What is linearity of Expectations? In reading about the average case analysis of randomized quick sort I came across linearity of expectations of indicator random variable I know indicator random variable and expectation. What does linearity of Expectation mean ? AI: Let $\xi_1,\xi_2:\Omega\to\mathbb R$ be two random variables on the same probability space $(\Omega,\mathscr F,\mathsf P)$ . The expectation of either is defined by $$ \mathsf E\xi_i:= \int_\Omega \xi_i(\omega)\mathsf P(\mathrm d\omega). $$ The linearity of the expectation means that for any constants $\alpha_1,\alpha_2\in\Bbb R$ it holds that $$ \mathsf E[\alpha_1\xi_1+\alpha_2\xi_2] = \alpha_1\mathsf E\xi_1+\alpha_2 \mathsf E\xi_2 $$ which follows directly from the linearity of the Lebesgue integral in the definition of the expectation. Hence, the functional $\mathsf E$ defined over the space of random variables on the probability space $(\Omega,\mathscr F,\mathsf P)$ is linear. For the independence over the product, yet again if $\xi_1,\xi_2,\dots,\xi_n$ are random variables on the same probability space as above, and they are mutually independent then $$ \mathsf E\left[ \prod_{i=1}^n\xi_i\right] = \prod_{i=1}^n\mathsf E\xi_i $$
H: Prove that: $\frac{a}{a+1}+\frac{b}{(a+b+1)^2}+\frac{1}{a+b+1}\le1 $ If $a\geq0$, $b\geq 0$ then the following inequality holds: $$\frac{a}{a+1}+\frac{b}{(a+b+1)^2}+\frac{1}{a+b+1}\le1 $$ There are at least three things to try here: a). Use AM-GM for the denominator of the second fraction, $(a+b)\le \frac{(a+b+1)^2}{4}$ b). Use the fact that $\frac{b}{(a+b+1)^2}\le\frac{b}{(a+b+1)(a+b)}$ c). Consider that $a\geq b$ or $b\geq a$ simply or combined with a). and\or b). No one of these attempts has brought me to a nice, simple solution and I'm trying to see if there is something around that missed me. This is an inequality given at a high school competition math. I'd appreciate to receive your valuable feedback in terms of this question. Thanks. AI: Alternatively you can use b') $\frac{b}{(a+b+1)^2}\le\frac{b}{(a+b+1)(a+1)}$. If you insert it, the LHS is $1$.
H: Is $f$ is computable by a finite automaton, the dual of $f$ is thus computable also? et $A$ be a finite alphabet. Let $A^$ denote the language of all words in $A$, and $\epsilon$ the empty word. Let $\rho : A^ \to A^$ denote the "reverting" map, that transforms $a_1a_2\ldots a_n$ into $a_na_{n-1} \ldots a_1$. I consider here a simplified version of finite automata on $A$. Thus I define an Ewan automaton on $A$ as a triple $(S,tr,s_0)$ where $S$ is a finite set of states, tr is a “transition” map $S \times A \to S$ and $s_0$ is an element of $S$. For such an automaton, there is a unique map $F : A^{} \to S$ satisfying for any $w\in A^$ and $a\in A$, $$ () F(\epsilon)=s_0, \ F(wa)=tr(F(w),a) $$ Now let $f$ be a map $A^{} \to X$ where $X$ is a finite set. If “f can be computed using the automaton”, i.e. if we can find $S,tr,s_0,F$ satisfying () as above, and a map $\phi : S \to X$ such that $f=\phi \circ F$, I say that $f$ is automatic. Given such an $f$, one may consider the dual map $f^{d}=f \circ \rho $. My question is : is $f$ is automatic, must $f^{d}$ be automatic also ? On the examples I’ve seen the answer is yes, although the automaton computing $f^{d}$ can be much larger than the one computing $f$. AI: Yes. It suffices to show that $f: A^* \to X$ is automatic iff $f^{-1}(x)$ is regular for all $x \in X$: If $f^{-1}(x)$ is regular then so is ${(f^d)}^{-1}(x) = \rho(f^{-1}(x))$ since the regular languages are closed under the mirror operation. If $f: A^* \to X$ is computed by an automaton $(S,\mathrm{tr}, s_0)$ via $\phi: S \to X$ then you get an automaton recognizing $f^{-1}(x)$ by marking $\phi^{-1}(x)$ as accepting states. If for each of the (finitely many) $x \in X$ there is an automaton $M_x = (S_x, tr_x, s_{0,x}, F_x)$ recognizing $f^{-1}(x)$ ($F_x$ is the set of accepting states) you can construct an automaton computing $f$ by running all the $M_x$ in parallel and setting $\phi(s) := x$ if $s\in F_x$.
H: Polynomial of degree $-\infty$? I'm reading E.J Barbeau Polynomials. I'm in a page where he asks a polynomial of degree $-\infty$. Then I thought about $77x^{-\infty}+1$, but when I went for the answers, the answer to this question was zero. Then I thought about making $n^{-\infty}$ on Mathematica and it outputed $Indeterminate$ as a result. I thought the problem was in my understanding of exponantiation, then I tried to "algebrize" it. (I guess that's the name of the procedure) Then I thought: $2^3=\overbrace{2\cdot 2\cdot 2}^{\text{3 times}}$ That would lead me to: $a^b=\overbrace{a\cdot a\cdot a\cdot ...}^{\text{b times}}$ And in this case: $a^{-\infty}=\overbrace{a\cdot a\cdot a\cdot ...}^{{-\infty}\text{ times}}$ But this gave me no insight of what could be done to better understand this. I can't see why $n^{-\infty}=0$ so clearly. With the last example, I'm thinking that there will be no $a$'s to multiply, can you help me? Addendum: I thought about some other thing: $$2^{-8}=\frac{1}{256}=\frac{1}{2^8}$$ Then considering this example, I would get: $$a^{-\infty}=\frac{1}{\infty}=0$$ Right? AI: IMO it comes down to conventions. We say the zero polynomial has degree $-\infty$. Let's see why this is a good convention: Usually the degree is the highest power with a non-vanishing coefficient. Following this logic it is not really clear what the degree of the zero-polynomial should be. We could just say it has no degree, or we could say it is just a special case of a degree $0$ polynomial (i.e. a constant polynomial), or maybe it's something different? What properties does the degree have? More specifically what happens if I add or multiply two polynomials $P$ and $Q$ of degree, say, $n$ and $m$? You can check that the degree of the sum of $P$ and $Q$ will be smaller or equal to the maximum of the degrees of $P$ and $Q$, while the product will have degree $m+n$. In particular if we multiply any polynomial $P$ with the zero polynomial we want: $$\deg 0=\deg P\cdot 0=\deg P+ \deg 0$$ To make sense of this equation $\deg 0$ has to be $\pm \infty$ but $+\infty$ doesn't agree with the property for sums. So $-\infty$ remains as the only sensible choice.
H: Dual of a 2-category Let $C$ be a $2$-category. There are two ways of dualizing $C$: The first one is well-known and also generalizes to arbitrary $(\infty,1)$-categories: We dualize "at each stage". The second one only dualizes the $2$-morphisms: We keep the underlying $1$-category, however we define $2$-morphisms $f \to g$ between $1$-morphisms $f,g : a \to b$ to be $2$-morphisms $g \to f$ in our given $C$. Thus we get a $2$-category $C'$. More abstractly, we may view $C$ as a category enriched over $\mathsf{Cat}$ and apply the usual dualization functor $\mathsf{Cat} \to \mathsf{Cat}$ to optain another category enriched over $\mathsf{Cat}$, namely $C'$. Question. Is this kind of dualization well-known? Does $C'$ have a name? Is there a common notation? How do we call $2$-functors $D \to C'$ (which are roughly functors $C \to D$ which are covariant on $1$-morphisms and contravariant on $2$-morphisms)? AI: There are three ways of dualising a 2-category $\mathfrak{K}$. $\mathfrak{K}^\textrm{op}$ is the 2-category with the same 0-cells and 2-cells, but with 1-cells reversed. This is basically the enriched version of taking opposite categories. $\mathfrak{K}^\textrm{co}$ is the 2-category with the same 0-cells and 1-cells, but with 2-cells reversed. An important example of this is the operation of forming the opposite of a 1-category: that is a 2-functor $\mathfrak{Cat}^\textrm{co} \to \mathfrak{Cat}$. Of course, we can also do both: $\mathfrak{K}^\textrm{coop} = (\mathfrak{K}^\textrm{co})^\textrm{op}$.
H: Savings when driving the inside edge of a curve. I was wondering how many kilometers/meters I would save if I always drove on the inside edge of the curves between my home and workplace, compared to driving in the middle of the lane. If the lanes are approximately $3 m$ wide and I estimate that over $100 km$ I drive an average of: 60x 90-degree curves with a length of $10 m$. 20x traffic circles with a driving length of $45 m$ (half circle). 30x 45-degree curves at $1000 m$. 30x 30-degree curves at $250 m$. How many meters would I save on a distance of $100 km$? AI: This is a cute problem because the answer turns out not to depend on how far you drive, only on how much you turn. Suppose the road follows a curve whose turning radius in the middle of the lane is $r$ meters, and it turns for a total of $\theta$ degrees. If you drove in the middle of the lane, you would have travelled a distance $2\pi r\cdot\theta/360$ meters, because the whole circle is $2\pi r$ meters long but you've only covered a fraction $\theta/360$ of it. On the inside of the lane, you travel $2\pi(r-1.5)\theta/360$ meters, assuming the distance from the middle of the lane to the inside is $1.5$ meters. The difference is $2\pi\cdot1.5\cdot\theta/360 = \pi\theta/120$ meters, independent of $r$. (This is the same as the distance you would travel if you went $\theta$ degrees around a circle of radius $1.5$ meters. This fact may or may not seem obvious once you think about it more.) Now that you have multiple curves in the road, your shortest path depends on how they're arranged, but the simplest assumption is that there are enough straight stretches in the road between curves that you can position yourself on the inner side of each turn without significantly affecting your distance travelled. In that case, all we have to do is add up the sixty $90^\circ$ curves, twenty $180^\circ$ turns, thirty $45^\circ$ curves, and thirty $30^\circ$ curves, to get a total turning of $11250^\circ$. This corresponds to a saving of $11250\pi/120 \approx 294$ meters.
H: Sequence that maps to arbitrary positive real number How do you construct a sequence of functions $f_n(x)$ such that $$s = \limsup_{n\rightarrow \infty} \sqrt[n]{f_n(x)}$$ for all $s > 0$? I know it's possible to this with a different sequence $$s = \limsup_{n\rightarrow \infty} (1 + \frac{x}{n})^n$$ where $x = \log(s)$. The motivation is from proofs on radius of convergence which rely on the definition of the radius $$r = \frac{1}{\limsup_{n\rightarrow \infty} \sqrt[n]{f_n(x)}}$$ and out of curiosity I tried to construct a function similar to that for $e^x$ that could map to any $s = 1/r$ but couldn't. AI: If $f_n(x)=x^n$, then $\sqrt[n]{f_n(x)}=x$ and you are there.
H: Is there a sign for "x's negation equals y"? For example, let M and N be two real numbers. M is smaller than N. Now I negate the inequality, such that now M is greater or equal to N. $M < N ≟ M \geq N$ Is there a sign to replace '≟' in the expression above? AI: It looks like you're talking about equivalence of a proposition to the negation of another. In which case, the "exclusive or" should do: $$ (M < N) \;\veebar\; (M \geqslant N) $$ asserts that exactly one of the two propositions $M < N$ and $M \geqslant N$ is true. As suggested in the comments (and in Ilya's answer), though, explicitly writing one of $$\begin{gather} (M < N) \;\equiv\; \neg (M \geqslant N) \\ (M < N) \;\iff\; \neg (M \geqslant N) \end{gather}$$ would probably be clearer.
H: Equations - Solving for x I have this problem: $$9x^3 - 18x^2 - 4x + 8 = 0$$ However, I'm not sure how to find the values of $x$. I moved the 8 over and factor out an $x$, but the trinomial it created can't be factored. Could someone enlighten me? AI: Factor the function. $9x^3-18x^2-4x+8=9x^2(x-2)-4(x-2)=(x-2)(9x^2-4)=(x-2)(3x-2)(3x+2)$ $(x-2)(3x-2)(3x+2)=0$ $x=2$ or $2/3$ or $-2/3$
H: Proving that a bounded, continuous function has a supremum Theorem: Let $f$ be a continuous real-valued function on a closed rectangle $R$ in $\Bbb R^2$. Then, (a) $\quad f$ is bounded on $R$ (b) $\quad $There exist points $c$ and $d$ in $R$ so that $f(c) = \sup_{p\in R} f(p), \quad f(d) = \inf_{p\in R} f(p)$. (c) $\quad f$ is uniformly continuous on $R$. My Work (a) $\quad$ To prove that $f$ is bounded on $R$, we need to show that there is an $U$ so that $\|f(p)\| \le U$ for all $p \in R$. Suppose this is false. Then for each large integer $n$, there is a $p_n \in R$ such that $\|f(p_n)\| > n$. By Bolzano-Weierstrass, the bounded sequence $\{p_n\}$ has a convergent subsequence $\{p_{n_k}\}$ that converges to $p_0$ as $k \to \infty$. Since $f$ continuous, $f(p_{n_k}) \to f(p_0)$ as $k \to \infty$. But this is impossible since $f(p_{n_k})$ is unbounded by virtue of $f(p_n)$ being unbounded. So $f$ is bounded on $R$. (b) $\quad$ Since $S = \{f(x) : x \in R \}$ is nonempty, $S$ has a supremum. Call this supremum $M$. Consider a sequence $\{c_n\}$ and the interval $M - \frac{1}{n} \le f(c_n) \le M$. Since this interval is closed, $f$ must have a convergent subsequence $\{c_{n_k}\}$ that converges as $k \to \infty$. Taking $N \ge 1/\epsilon$, we have $M - (M - 1/n) = 1/n \le \epsilon$ for $n \ge N$. So since $f$ is continuous we must have $f(c_n) \to f(c) = M$. My Question: I feel pretty sure of my proof of (a), but (b) seems like it lacks a key component. And I am not sure how to prove (c) for any arbitrary $f$, given that the definition of uniform continuity for a continuous $f$ on a set $E \subseteq \Bbb R ^2$ is given $\epsilon > 0$, there exists a $\delta > 0$ such that $$ \\|q - p \\| \le \delta \ \Rightarrow \ \|f(q) - f(p) \| \le \epsilon \quad \forall \ q, p \in E $$ Edit Revised part (c): If $f$ is not uniformly continuous on $R$, then, from the definition of uniform continuity, there are two sequences $(x_n)$ and $(y_n)$ in $R$ and an $\epsilon_0>0$ so that for each $n$ we have $$\tag{1}\Vert x_n-y_n\Vert\le 1/n\ \ \ \text{ and }\ \ \ \Vert f(x_n)-f(y_n)\Vert\ge\epsilon_0.$$ Consider a convergent subsequence $(x_{n_k})$ of $(x_n)$. This converges to some $x\in R$. So for some large integer $K \ge 1/\epsilon$, and $k \ge K$, we have $\Vert y_{n_k} - x \Vert \le \epsilon$ so $(y_{n_k})$ converges to $x$ as well. From this we have that $(x_{n_k}), (y_{n_k}) \to x$, but $\Vert f(x_{n_k}) - f(y_{n_k}) \Vert \ge \epsilon_0$ which implies that $f$ is not continuous at $x$ which contradicts the hypothesis. AI: For part b), you cannot start by saying "$f$ achieves a maximum...". This is what you are trying to prove. Instead, start by observing that the set $\{f(x)\|x\in R\}$ is nonempty and bounded above. It thus has a supremum, $U$. Now (carefully) define your $c_n$ so that $U-1/n\le f(c_n)\le U$ and continue with your argument. For part c), you could argue as follows: If $f$ is not uniformly continuous on $R$, then, from the definition of uniform continuity, there are two sequences $(x_n)$ and $(y_n)$ in $R$ and an $\epsilon_0>0$ so that for each $n$ we have $$\tag{1}\Vert x_n-y_n\Vert\le 1/n\ \ \ \text{ and }\ \ \ \Vert f(x_n)-f(y_n)\Vert\ge\epsilon_0.$$ Consider a convergent subsequence $(x_{n_k})$ of $(x_n)$. This converges to some $x\in R$. Show that the corresponding subsequence $(y_{n_k})$ of $(y_n)$ also converges to $x$. Now, show that this and $(1)$ imply that $f$ is not continuous at $x$.
H: Multiplying Reciprocal Exponents This is the problem I have: $$y^{\frac{3}{2}} = 5y$$ What I tried so far was raising $y^{\frac{3}{2}}$ to the $\frac{3}{2}$, making it equal $1$, but I had trouble raising $5y$ to that power. AI: You have the right idea, you are trying to get rid of the fractional power. However, the best way to do this is to raise each side to the power $2$, so we have: $$\left(y^{\frac{3}{2}}\right)^{2}=\left(5y\right)^{2}\implies y^{3}=25y^{2},$$ Which is then much easier to manipulate. Hope this helps! If you are trying to find all solutions to the equation, then you must solve a cubic equation: $$y^{3}-25y^{2}=0$$ This can be factorized as: $$y^{2}\left(y-25\right)=0,$$ Which gives us the following solutions: $$y=\{0,25\}$$ Where $0$ is a duplicated root.
H: Triangular grid with 4 edges per vertex I am trying to create a triangular grid/mesh for a rectangular domain in $\mathbb{R}^2$ with the property that each vertex is shared by (at most) four edges. Is this possible to accomplish? AI: Only if you have few enough triangles. You can split the domain with a diagonal into two triangles and each vertex has at most 3. It helps to think about the total angle around the vertices. You have four vertices at the corners that contribute a total of $2\pi$ in angle. Each vertex on the edge of the rectangle contributes $\pi$ and each vertex inside the rectangle contributes $2\pi$. If you have no added points, you need two triangles to supply the $2\pi$. Each point on the edge adds $\pi$ to the total angle, so adds one triangle. Each point in the interior adds $2\pi$ to the total angle, so you need two more triangles. You can only have twice as many edges as vertices, so letting $V_a$ be the number of vertices on the edge (including the corners), $V_b$ the number of vertices inside, $E$ the number of edges, and $F$ the number of faces (one more than the number of triangles to count the exterior region) we have $V_a+V_b-E+F=2, F-1=2+(V_a-4)+2V_b, E \le 2(V_a+V_b)$, so $2V_a+3V_b-E=3$. Clearly $V_b$ has to stay small. It works with $1,2$, but not with $3$ The above assumes that it is not acceptable to have a vertex in the middle of one side of a triangle. If that restriction is removed, you can have more points in the interior, as shown below. You can also cut up the rectangle in other ways.
H: What is the rationale behind the evaluation of the Expectation operator? We know that the expectation operator is defined for a random variable $x$, as such: $$ \mathbb{E} \left\{x\right\} = \int_{-\infty}^{\infty} x \: p_x(x) \; \mathrm{d}x $$ Where $p_x{x}$ is the PDF of the random variable $x$. If there is an arbitrary(?) function $f$ acting on the random variable $x$, then the expected value of this function can also be written as: $$ \mathbb{E}\left\{f(x) \right\} = \int_{-\infty}^{\infty} f(x) \: p_x(x) \: \mathrm{d}x $$ My questions are: On many algorithms that I study, (statistical in nature), one often finds themselves taking the expected value of some entity, that is a function of the random variable $x$. In the reverse case, one can also find themselves poking around and manipulating the probability distribution function of $x$, and then we can 'take it back' into an expression using the expectation operator. Upon evaluating the expected value of $x$ however, ($\mathbb{E[x]})$, I often come across this estimation formula: $$ \mathbb{E}\left\{x\right\} \approx \frac{1}{N}\sum_{n=1}^{N} x[n] $$ and similarly, $$ \mathbb{E}\left\{f(x)\right\} \approx \frac{1}{N}\sum_{n=1}^{N} f(x[n]) $$ Where each $x[n]$ is an individual realization of the random variable $x$. My question is, why is this formula true, and how did it come about? Every book I read seems to just include it as if it fell from the sky one day and no explanation is given as to why it is true. Could someone please give an intuitive and mathematical explanation for why - and more importantly, how this happens to be true? What is the history/rationale behind it? Many thanks. AI: As you are aware, the expected value of a random variable $X$ is computed as: $\mathbb{E}\{X\} = \int_{-\infty}^{\infty}z f_{X}(z)dz$, where $f_X$ is the probability density function (PDF) of $X$. Note that the above integral may not converge (e.g. if $X$ is a Cauchy random variable), and the PDF may not exist as well. But let's assume we are not getting into such problems, which is indeed the case in most practical settings. Often, we will not know the PDF, or even its functional form. In such a case, one resorts to estimating the PDF from realizations of the random variable. That is, assume we have $N$ realizations of $X$ - $\{x_i\}_{i=1}^N$. Let us consider the following estimate: $\hat{f}_X(z) = \frac{1}{N}\sum_{i=1}^N \delta(z-x_i)$, where $\delta(.)$ is the Dirac delta function. So essentially, we are treating the random variable as a uniform discrete random variable, and putting a mass of $1/N$ over each of the observed values. The estimate of the expectation of $X$ becomes: $\hat{\mathbb{E}}_N\{X\} = \int_{-\infty}^{\infty}z \frac{1}{N}\sum_{i=1}^N \delta(z-x_i)dz$ $= \frac{1}{N}\sum_{i=1}^N \int_{-\infty}^{\infty}z \delta(z-x_i)dz$ By the sifting property of the delta function, you will note that the integral becomes $x_i$. Hence, we arrive at the following expression for the sample expectation: $\hat{\mathbb{E}}_N\{X\} = \frac{1}{N}\sum_{i=1}^N x_i$ Note that $x_i$'s are themselves random, since another set of draws from $X$ is likely to give me a different set of realizations. Thus the above estimate of the expected value is itself a random variable - dependent on the draws $\{x_i\}_{1}^N$ and the number of draws $N$. Since all $x_i$s are distributed identically (with PDF $f_X$), and are independent draws, the Laws of Large Numbers tell us that $\mathbb{E}_N\{X\} \rightarrow \mathbb{E}\{X\}$ with probability $1$ (almost surely), and in probability, as $N \rightarrow \infty$.
H: Derivative of function defined by integral Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be continuous and let $a$ be a nonzero real number. Show that the function $F(x)=\frac{1}{2a}\int_{-a}^{a}{f(x+t)dt}$ is differentiable and has continuous derivative. My thoughts were that $\frac{d}{dx}(\frac{1}{2a}\int_{-a}^{a}{f(x+t)dt})= \frac{1}{2a}\frac{d}{dx}(\int_{-a}^{a}{f(x+t)dt}) = \frac{1}{2a}(f(x+a)-f(x-a))$, which is continuous. Is this correct? It seemed a little too easy. AI: Set $x+t=u$, so that $t=a$ means $u=x+a$ and $t=-a$ means $u=x-a$. Then $$ F(x)=\frac{1}{2a}\int_{x-a}^{x+a} f(u)\, du. $$ You can now apply the fundamental theorem of calculus.
H: Function to represent curve between two 2D points I need a (simplest) function that interpolates values in range from predefined point $A$ to $B$ with rules: it must be smooth curve direction near $B$ must be the same as predefined $D$ vector I have a variant to build circle arc but is too complex. Maybe some sort of splines, but I don't know how to represent it with my $A$, $B$ points and $D$ vector. AI: Given a starting point $A$, ending point $B$ and final direction $D$, you can define a quadratic Bezier curve. Let $M=B-D$, $M$ being the middle point used to define the curve. The curve $f$ is given by $$f(t)=(1-t)[(1-t)A+tM]+t[(1-t)M+tB],\ 0\leq t\leq 1$$ You can adjust the smoothness of the curve by changing the placement of $M$. You can do this by choosing a scaling factor $x>0$ and setting $M=B-xD$.
H: A list from each element of another? Sorry to edit this so much at this late stage, but the question and answers are confused so much by my incorrect use of terminology and the such, I feel that I should clear this up. Where $a$ and $b$ are lists. $$a = \{\mathbf{true, false, true, false, false}\}$$ $$b = f(a) = \{0.8, 0, 0.8, 0, 0\}$$ In this case each $\bf true$ in $a$ has become a $0.8$ in $b$, as with $\bf false$ and $0$. What is $f$? AI: First, note that you are working with multisets rather than sets, since you are allowing repeated elements. What you have in the example is a function $f: a \rightarrow b$ defined by $$ f(x) = \begin{cases} 0.8 & \text{if } x = \text{true}\\ 0 & \text{if } x = \text{false}. \end{cases} $$ You can replace the domain $a$ with any multiset of $0.8$'s and $0$'s. If you want to introduce new elements, you will need to specify how $f$ acts on them.
H: What is the essential difference between the terminologies "associative" and "semi-group". Can only a map $$:S\times S\rightarrow S$$ be associative? If I look at $$(a b)* c=a* (b* c),$$ then it seems I have to rule out the more general case $$*:A\times B\rightarrow C.$$ But $A=B=C$ is really the only restriction for a binary operation to be a magma. Hence, if my assumption is true, for a relation to be associative is the exact same thing as being a semi-group. This redundancy in notation (one could just state that a relation together with its set is a semigroup) just surprised me. (Here is a previous thread about a related point.) AI: Let me make a general philosophical point about definitions. A mathematical definition is much more than just its bare semantic content. A definition also carries an implicit intent which must be understood from the way it is used in practice. For example, the mathematical definitions of directed (multi)graph and quiver are equivalent. So why do some people say directed (multi)graph and others say quiver? It is the intent behind the term: people who say directed (multi)graph want to do graph theory of some kind, whereas people who say quiver specifically want to study quiver representations, quiver varieties, etc. There are also various examples in category theory of constructions which, on the level of category theory, are formally equivalent (e.g. pullback and fiber product) but which are named after special cases which occur in different contexts. Choosing to use one name over the other evokes a particular context and activates certain intuitions related to whatever you're going to use the construction for. So, back to your question: strictly speaking the word "associative" is (in my experience) only used to describe a property of a map $S \times S \to S$, so saying "$f$ is associative" is formally equivalent to saying "$f$ defines a semigroup." But these two words occur in different contexts. When you want to talk about a given operation being associative, it is usually in the context of potentially several other operations (e.g. addition in the context of multiplication), whereas when you want to talk about a semigroup, you are usually only talking about the one operation. Failing to recognize this difference in intent will make it very difficult for people to understand you in practice even if you are, on a formal semantic level, saying the correct thing.
H: Sum independent of the rearrangements of its terms? Is it true, that in every concept of an infinite sum in a Banach space, encountered in an introductory functional analysis course, convergence is independent of the rearrangement of its terms ? AI: $\mathbb R$ is a Banach space, and its concept of infinite sums is certainly not independent of rearrangements. The standard example is the alternating harmonic series $\sum_n \frac{(-1)^n}{n}$, which can be rearranged to have any limit you like, or none at all.
H: Evaluating $\int e^{ax} x^b (1-x)^c \mathrm{dx}$ Edit: clarify question The integrand looks kind of like a gamma density function, and kind of like a beta density function, so maybe it has a somewhat nice solution? $$\int e^{ax} x^b (1-x)^c \mathrm{dx}$$ Wolfram alpha does not want to do it. AI: You can expand out the $(1-x)^c$ to get terms of the form $\int e^{ax}x^n dx$. Wolfram Alpha then gives a solution in terms of the incomplete Gamma function. This is a form that can be integrated by parts-set $dv=e^{ax}dx, u=x^n$ and step down the exponents, giving $\int e^{ax}x^n dx=\frac {x^n e^{ax}}a -\frac na \int x^{n-1}e^{ax}dx$
H: Name a ring of 2 by 2 matrices where $a^3 = a$ and a belonging to this ring? I need an example of a ring consisting of 2 by 2 matrices where $a^3=a$ with $a$ belonging to this ring. If someone can list the elements I would be satisfied. What I'm trying to get at it is conceptualize why a ring $R$ is always commmuative when $a^3=a$. I know of one such example and that is the factor ring $\mathbb{Z}/3\mathbb{Z}.$ Does anyone know how to prove this statement mathematically as well as giving me an example of a ring of 2 by 2 matrices? AI: There is a theorem due to Jacobson that says if for every $a\in R$ there exists an $n\in\mathbb{N}$ such that $a^n=a$, then $R$ is commutative. (See this, or this for example). Obviously the identity matrix cubed is itself... is this the sort of thing you're looking for?! In general matrix rings are going to have a lot of idempotent elements $e$ such that $e^2=e$, and for all of those $e^3=e$ as well. For an example where $a^2\neq a$, you could use $\begin{bmatrix}0&1\\1&0\end{bmatrix}$.
H: Derivative of vector field I have just started learning differential geometry and I am trying to answer the following exercise (it's not for homework): Let $M$ be an $n$-dimensional manifold with chart $(U, \varphi_U)$ and let $(x^1, \dots, x^n)$ be the corresponding local coordinates. Define a vector field on $U$ by $X = X^i\frac{\partial}{\partial x^i}$. We will try to define a "derivative of vector fields" by $$d_Y X = d_Y X^i \frac{\partial}{\partial x^i}$$ By changing coordinates to $(\bar{x}^1, \dots, \bar{x}^n)$, show that unlike $df$, this operator is not independent of the choice of local coordinates. I think I understand the method, but I am not certain of the details. I believe what I have to do is first apply $d_Y$ to $X$ and then change coordinates, and then change coordinates and apply $d_Y$ to $X$, and show that the two results are not then same. Here is what I have attempted: Let $Y = Y^i\frac{\partial}{\partial x^i}$ (the question doesn't define $Y$ but presumably it's just another vector field). Then $$d_Y X = d_Y X^i \frac{\partial}{\partial x^i}$$ $$= dX^i(Y) \frac{\partial}{\partial x^i}$$ $$= \frac{\partial X^i}{\partial x^j}Y^j \frac{\partial}{\partial x^i}$$ This is just another vector field, so transforming it according to the contravariant transformation we get the following messy expression: $$\frac{\partial \bar{x}^i}{\partial x^j} \frac{\partial X^j}{\partial x^k} Y^k \frac{\partial}{\partial \bar{x}^i},$$ which is $d_Y X$ in $\bar{x}^i$ coordinates (if I'm doing this correctly). Now I am trying to first change coordinates and apply $d_Y$ to $X$: first, in the new coordinates the vector field $X$ will be $\bar{X} = \frac{\partial \bar{x}^i}{\partial x^j}X^j \frac{\partial}{\partial \bar{x}^i}$ (contravariant transformation). So applying $d_Y$ to this we have $$d_Y \bar{X} = d_Y \left(\frac{\partial \bar{x}^i}{\partial x^j}X^j \right) \frac{\partial}{\partial \bar{x}^i}$$ Now I'm not sure what to do because it looks like $d_Y \left(\frac{\partial \bar{x}^i}{\partial x^j}X^j \right)$ will be terribly messy, and I don't really know how to evaluate it. Also, I don't know if I'm supposed to transform $Y$ to new coordinates as well (this would make the calculation even uglier) or if it's just $X$ that is transformed. Could anyone help? AI: You are basically on track. It is not $d_Y \bar{X}$, though you are supposed to calculate, rather still $d_Y X$, since $X$, as a vector field, is known to be independent of the representation. If you want to calculate $d_y($ugly expression$)$ in the new coordinate system then you will, in fact, have to express $Y = \bar{Y}^k\frac{\partial }{\bar{x}^k }$ And yes, you are supposed to then calculate $$d_{Y}\left(\frac{\partial \bar{x}^k}{\partial x^j}X^j\right) = \frac{\partial}{\partial \bar{x}^l}\left(\frac{\partial \bar{x}^k}{\partial x^j}X^j\right)\bar{Y}^l$$ and compare it to the other term you got earlier. It's just the product rule, though, nothing to be afraid of. If you want to dig deeper into differential geometry this will become daily business, so the earlier you start the further you may get ;-) (Writing this kind of thing, esp if it get's even more complicated, with $\TeX$, can, admittedly, be kind of awkward).
H: calculator errors and rational root theorem In this question we were asked for roots of $9x^3 - 18x^2 - 4x + 8 = 0$ and I remarked on the rational root theorem. In fact, the roots are $2, \frac 23, \frac {-2}3$. When checking them it would be nice to use an exact form of the fractions, but in fact many will use a calculator. When checking $\frac 23$, mine makes an error of about $10^{-11}$. It is not hard to estimate the error if the value is really a root, using the machine epsilon times the size of the terms in the polynomial, so one will not reject a true root, but what about the other direction? For a polynomial $f(x)$ with integer coefficients of reasonable size, how small can $f(x_0)$ be where $x_0$ is a root candidate from the rational root theorem but not a root? If the real roots are rational we are limited by the fact that the rationals can't be too close together. One of my first efforts was $(x-2+i)(x-2-i)(3x-8)=3x^3-20x^2+47x-40$, which has $f(2)=-2$, pretty small compared to the terms but not small enough to fool anybody. Can we do better? AI: I propose $f(x) = A x^n -A x +1$. Then $f(1/A)=1/A^{n-1}$. This is basically best possible, although one could make the trap less obvious. If $f(x) = Ax^n + \cdots$, then $f(p/A)$ will always be a rational with denominator $A^{n-1}$, so we can't get smaller than $1/A^{n-1}$.
H: Lifting a homomorphism In a set of lecture notes, I recently came across the remark that any continuous homomorphism from $\mathbb{R} \to S^1$ can be lifted to continuous homomorphism from $\mathbb{R} \to \mathbb{R}$. How do I conclude that the lift is a homomorphism too? I tried many ways to solve this but failed. I could conclude, for example, that if the lift is $\bar{f}$, then for any $x_1, x_2 \in \mathbb{R}$, $\bar{f}(x_1+x_2)=\bar{f}(x_1)+\bar{f}(x_2)+2n\pi$ for some $n\in \mathbb{Z}$. If this $n$ were fixed, I might be able to say something, however this may not be true. I feel that this must be a very trivial problem, but I have spent a lot of time on it, without a result. AI: Fix $x_1$ and vary $x_2$ a little, then $\bar f(x_1+x_2)$ and $\bar f(x_2)$ change a little, while $\bar f(x_1)$ stays fixed. Hence $2n\pi$ also can only change a little. Since $\mathbb Z$ is discrete $n$ has in fact to be constant. Thus $n$ is constant on connected components of $\mathbb R$. Since $\mathbb R$ is connected you are done.
H: Can this type of series retain the same value? Let $H$ be a Hilbert space and $\sum_k x_k$ a countable infinite sum in it. Lets say we partition the sequence $(x_k)_k$ in a sequence of blocks of finite length and change the order of summation only in those blocks, like this (for brevity illustrated only for the first two blocks $$(x_1,\ldots,x_k,x_{k+1},\ldots,x_{k+l},\ldots )$$ becomes $$(x_{\pi(1)},\ldots,x_{\pi(k)},x_{\gamma(k+1)},\ldots,x_{\gamma(k+l)},\ldots ),$$ where $\pi$ and $\gamma$ are permutations. If we denote the elements of the second sequence with $x'$, does anyone know, what will happen to the series $\sum _k x'_k$ in this case ? Can it stay the same ? Does staying the same requires additional assumptions ? AI: If both series converge, it doesn't change anything. This can be easily seen by considering partial sums. Put $k_j$ as the cumulative length of the first $j$ blocks. Then clearly $\sum_{j=1}^{k_n} x_j=\sum_{j=1}^{k_n} x_j'$ for any $n$, so assuming both series converge, we have that $$\sum_j x_j=\lim_{n\to \infty}\sum_{j=1}^{k_n} x_j=\lim_{n\to \infty}\sum_{j=1}^{k_n} x'_j=\sum_j x_j'$$ On the other hand, we can change a (conditionally) convergent series into a divergent one using this method – if we take the alternating harmonic sequence $x_n=(-1)^n/n$, sorting large enough blocks so that all the positive elements come before all the negative elements, we can get infinitely many arbitrarily large “jumps”, preventing convergence. On another note, if the length of permuted blocks is bounded, then I think this kind of thing could not happen (again, by considering partial sums).
H: Comparison of two infinity Possible Duplicate: Different kinds of infinities? Today I got to know that two infinity can be compared, But I want to know how is this possible? infinity will be infinity. If it doesn't have any particular value, how can we say that this infinity is small and other one is greater. Can anyone help me? AI: The question about several types of infinity has come up several times before. As linked to in the comments about, you can read through the question and answer given in What Does it Really Mean to Have Different Kinds of Infinities? or Are all infinities equal?. You ask about if you can compare infinities. Well, you might say that you can. IF you take for example the natural numbers $1,2,3,...$, then there are an infinite number of them. We say that the set is infinite. But, you can also count them. If we look at the real numbers, then the fact is that you cannot count these. So in a way, the infinite number of real number is "greater" than the infinite number of natural numbers. But all this comes down to the question about how you measure the size of something. If someone says that something is bigger than something else, then they should always be able to define exactly what that means. We don't (I don't) like when questions become philosophical, then it has (In my opinion) left the realm of mathematics. So if someone tells you that one infinity is greater than another infinity, ask them exactly what they mean. How do you measure sizes of infinities? If they are a mathematician, they will be able to give you a precise definition (study Andre's answer). But, what we usually think about when we compare numbers (or elements in a set) is a set with some kind of ordering on. Without going into any detail, there are different types or orderings, but you can think about how we can order the set consisting of the real numbers in the usual way (ex $7 > 3$). But in this example we are just talking about the real numbers. And infinity is not a number. One more thing to keep in mind is that we will some times write that a limit is equal to infinity. Like $$\lim_{x \to a} f(x) = \infty. $$ However, when we write this, we don't think (I don't) of $\infty$ as an element in the set of real numbers (it isn't). All we mean by writing that the limit is infinity is that the values of $f(x)$ become arbitrarily large as $x$ "gets" close to $a$. Just a few things.
H: An explicit calculation of Galois group This is a question requires to compute the Galois group of $X^4+1$ over $\mathbb{Q}$, $\mathbb{Q}(i)$, $\mathbb{F}_3$ and $\mathbb{F}_5$. Here is a brief of what I can think of. For the first two, the splitting field is $\mathbb{Q}(\sqrt{2},i)$. Since $X^2+1$ and $X^2-2$ are irreducible, the Galois group should be $C_2 \times C_2$ and $C_2$. I am not sure what happens when the field is finite. ================================= I thought about the first case again last night but I messed up myself a little bit. Take $\zeta_8 = \frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$ to be the eighth partition of unity. The splitting field of $X^4+1$ over $\mathbb{Q}$ should be $\mathbb{Q}(\zeta_8)$, since $\zeta_8$ generates other roots which are $\zeta_8^3,\zeta_8^5, \zeta_8^7$. It follows $Gal(\mathbb{Q}(\zeta_8)/\mathbb{Q})=Aut(C_8)=C_4$. On the other hand, $\zeta_8$ and $\zeta_8^3$ generates $\sqrt{2}$ and $i$, which leads to the answer I tried to gave last time. In this case, the Galois group seems to be $C_2 \times C_2$. For sure the degree of extension is $4$, but I cannot figure out which approach is correct. AI: Over $F_3$ you have $\sqrt{2} = \sqrt{-1} = i$, so the extension is quadratic. Over $F_5$, you have $i = \sqrt{-1} = \sqrt{4} = \pm 2$, so again the extension is quadratic. Since extensions of finite fields are always cyclic, this was to be expected.
H: Number of Zeros of a Section vs Integral First Chern Class I've often read that the first chern class can be seen as "the number of zeroes a section must have". How precise can this statement be made? I'm only interested in Line bundles. I actually know how a generic section looks like on my manifold. From this I can even tell what the "lowest" number of zeros is. Is it safe to say that minimum number of zeros $> 0$ $\to$ $c_1 \neq 0$? AI: If a line bundle has a non-zero section, it is a trivial line bundle and that tells you what the $c_1$ is. Conversely, if every section has a zero, then the bundle is not trivial and since complex line bundles are actually classified by their first Chern class, then $c_1$ cannot be zero.
H: How to find the eigen vector for the given matrix? Sorry, I made error here. Please check Edit!! Given matrix is $$ A = \begin{bmatrix} 2 & 5\\ 4 & -2 \end{bmatrix}$$ How to find the eigen-vector? I'm stuck here $$\begin{bmatrix} 2 - 4i & 5\\ 4 & -2 -4i \end{bmatrix} \times \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix}$$ Problem is here. Problem Set Part I Problems EDIT:: $$ A = \begin{bmatrix} 2 & -5\\ 4 & -2 \end{bmatrix}$$ AI: "the eigen-vector" is vague. You're looking for an eigenvector associated with an eigenvalue. The eigenvalues of the given matrix are: $\pm 4i.$ So solve for $v_1$ and $v_2$ in $$ \pmatrix{ 2 & -5\\ 4 & -2} v_1 = 4i v_1 \\ \pmatrix{ 2 & -5\\ 4 & -2} v_2 = -4i v_2$$ Edit (from the comments): Let the eigenvalues be $x_1, x_2.$ It's known (see your textbook) that $$ x_1 + x_2 = \text{Trace}(A) = 0. \\ x_1 x_2 = \text{Det}(A) = 16.$$ Solve for $x_1, x_2$ to get the eigenvalues $\pm 4i.$ Edit (from the comments): Do you know about elimination? You will need $1$ step to solve $$ \begin{bmatrix} 2 - 4i & -5\\ 4 & -2 -4i \end{bmatrix} \times \begin{bmatrix} x\\ y \end{bmatrix} = \begin{bmatrix} 0\\ 0 \end{bmatrix} $$ Multiply the first row by $\dfrac{−(2+4i)}{5}$ and add it the second row. Solve for $y$ then substitute back in to solve for $x.$
H: Find the domain of $f(x)=\ln(3x+2)$ Find the domain of $f(x)=\ln(3x+2)$ I can find domain, but is it the same for a $\log$ function? And also, do I have to rid the equation of the $\ln$ before I can find the domain? I'm really confused as to how to start out this problem. Please help. Edit: I got the answer and now I ask a question (irrelevant to my problem but I am curious), why does the real portion of $f(x)=\log x$ resemble the graph of a square root function. In a sense of, starting at a single point and going in two directions: up/down & left/right. AI: Hint: $\log t$ is defined precisely if $t$ is positive. The base is irrelevant. Why: The explanation depends on how you define the logarithm. Let $b\gt 0$, with $b\ne 1$. Then $\log_b t$ can be defined as the number $w$ such that $b^w=t$. But $b^w$ is always positive, so $t$ must be positive. Thus $\log_b t$ is undefined if $t \le 0$. To show that $\log_b t$ is defined whenever $t\gt 0$, we need some information about the function $b^w$. Assume that $b\gt 1$. Then $b^w$ is steadily increasing, and is close to $0$ when $w$ is large negative, and is huge when $w$ is large positive. Thus the function $b^w$ takes on every positive value once and only once. A similar argument settles the (much less common) situation where $0\lt b\lt 1$.
H: Direct Sum and Sum We know if $R$ is semi simple ring and $M$ is $R$-module then for any submoldule $N$ of $M$ there exists submodule $N'$ such that $N \oplus N'=M$. Now I have trouble with sum and its supposed answer. let $R$ be ring (Not semi simple) and $M$ is $R$-module and $N$ is submodule of $M$. Does exist submodule $N'$ such that $N+N'=M$? AI: No. Take for example $R={\bf Z}=N$, $M=\bf Q$. Assume by contradiction that we have $N'$ a proper subgroup of $\bf Q$ such that $N'+{\bf Z}=\bf Q$. Any subgroup of $\bf Q$ is of the form $\langle \frac{p}{q_n}\vert n\in{\bf N}\rangle$ with $p,q_n\in \bf N$, $q_{n}\vert q_{n+1}$, and $(p,q_n)=1$ for all $n$, so suppose $N'$ is of that form. If $p=1$, then ${\bf Z}\subseteq N'$, so $N'=N'+\bf Z$, so $p>1$. $N'+\mathbf Z=\mathbf Q$, so in particular, there are $m,n,k\in \bf N$ such that $m\frac{p}{q_n}-k=\frac{1}{p}$. But then $p\vert q_n$, which is a contradiction, so there is no such $N'$.
H: Calculate a whitening matrix without using inverses? Consider a random column vector $\mathbf{x}$, of dimension $m$. That is, it is a random vector, composed of $m$ random variables. The PDF of the random vector $\mathbf{x}$ is thus the joint-PDF of its $m$ random variable components. Let us assume that one wishes to whiten the random variables of this vector. That is, you want to de-correlate them, and then scale their variances to be unity. One way to do this is to compute a whitening matrix $\mathbf{V_{\mathrm{mxm}}}$, and one way to compute the whitening matrix is by: $$ \mathbf{V_{\mathrm{mxm}}} = \mathbf{E} \mathbf{D^{\mathrm{-\frac{1}{2}}}}\mathbf{E^{T}} $$ where $\mathbf{E}$ is the column wise eigen-vector matrix, and $\mathbf{D}$ its corresponding eigen-value matirx, coming from the eigenvector decomposition of x's covariance matrix, $\mathbf{R_{\mathrm{xx}}} = \mathbf{E}\mathbf{D}\mathbf{E^T}$. The $\mathbf{D^{-\frac{1}{2}}}$ implies the " inverse matrix square root". In fact there are many other types of whitening matricies that can be constructed. My question is, to my knowledge, any construction of a whitening matrix needs to have an inverse operation within it. For me, this is fine if my dimensionality is 'small', but I hesitate to use this method for larger dimensions of $m$. What other methods of computing a whitening matrix might exist that do not involve the computation of inverses? Much obliged. AI: Let $X$ be the original random with covariance matrix $\Sigma$. Since $\Sigma$ is Hermitian symmetric, we can do an eigen-value decomposition with an orthonormal basis: $\Sigma = E \Lambda E^T$ Now, the whitening matrix is $W = (E \Lambda^{1/2})^{-1}$ since covariance matrix of $WX$ is $I$. The good thing about having an orthonormal basis is that $E^{-1} = E^T$. Hence, $W = \Lambda^{-1/2} E^{T}$. Hence, you only need to take inverse of the diagonal matrix, which is a matrix with inverse of the original diagonal entries. This is not computationally expensive at all.
H: Graph $f(x)=e^x$ Graph $f(x)=e^x$ I have no idea how to graph this. I looked on wolframalpha and it is just a curve. But how would I come up with this curve without the use of other resources (i.e. on a test). AI: You'd like to capture some of the important behavior of the graph, yes? For example, end-behavior (what happens as $x\to-\infty$ or $x\to\infty$), intercepts, where it's increasing/decreasing, any peaks or troughs. In this case, you'll want to convey the idea that $f(x)$ is always positive, continuous, and increasing (should be evident from your sketch, and the fact that you'll not draw any jumps, peaks or troughs) You'll want to show that $f(0)=1$. You'll want to show that as $x\to-\infty$, $f(x)$ levels off and approaches $0$ asymptotically (again, should look that way from your sketch, but you may want to explicitly indicate the horizontal asymptote). You'll want to give the idea that as $x\to\infty$, $f(x)\to\infty$ (should look that way). You'll want to plot a few other points, too, like $f(1)=e\approx 2.8$ and $f(-1)=1/e\approx 0.35$. That should cover it pretty well.
H: What is the isomorphism between $\wedge^n(V)$ and $\mathbb{R}$? Let $V$ denote an $n$-dimensional real vector space, and $\wedge^n$ denote the $n$-fold exterior product. What is the isomorphism between $\wedge^n(V)$ and $\mathbb{R}$? In the book Introduction to Manifolds, by Loring W. Tu, 3rd edition, Chapter: Orientation of Manifolds, it says: A linear isomorphism $\wedge^n(V)\cong\mathbb{R}$, identifies the set of non-zero $n$-covectors on $V$ with $\mathbb{R} \setminus \{0\}$". AI: Note that you assume that $V$ is of dimension $n$. The $n$-fold outer product of $V$ can now be described as follows: Choose a basis $v_1,\dots,v_n$ of $V$. The outer product is generated (as a vector space) by expressions of the form $v^1\wedge\dots\wedge v^n$ where $v^1,\dots,v^n\in\{v_1,\dots,v_n\}$, where one such expression $v^1\wedge\dots\wedge v^n$ is the negative of $w^1\wedge\dots\wedge w^n$ if $w^1\wedge\dots\wedge w^n$ is obtained from $v^1\wedge\dots\wedge v^n$ by exchanging one component $v^i$ and another component $v^j$. This implies that $v^1\wedge\dots\wedge v^n=0$ if there are distinct $i$ and $j$ with $v^i=v^j$. It follows that $v_1\wedge\dots\wedge v_n$ and $-v_1\wedge\dots\wedge v_n$ are the only non-zero generators, and they are linearly dependent (obviously). Hence the outer product is generated by $v_1\wedge\dots\wedge v_n$. Mapping this generator to $1$ extends to your desired isomorphism. As pointed out in the comments, there are many isomorphisms, and the one pointed out here depends on the choice of the basis $v_1,\dots,v_n$.
H: Graph $f(x)=\ln x+2$ Graph $f(x)=\ln x+2$ And find all intercepts and asymptotes. I know exactly how the graph looks and I have a sketch in front of me. Now, for the intercepts, $x$-int=$-1$ and $y$-int$=-\infty+2$ which I'm guessing is just $-\infty$ because adding $2$ wouldn't make much of a difference on that level. So, I have my intercepts (correct me if they are wrong please) now for the asymptotes. For vertical: $\log(x)+2 \rightarrow \infty$ as $x\rightarrow 0$ thus making the vertical asymptote $0$. And for horizontal, it has to do with when $y \rightarrow\pm\infty$. But how do I find this. (The inly reason I know that the vertical asymptote is $0$ is because it's a $\log x$ function moved up $2$ units). AI: It's not correct to say that the x-intercept is at $x=-1$. The x-intercept occurs when $f(x) = 0$, so we have to solve the following: $$\begin{align} \ln x + 2 &= 0 \\ \ln x &= -2 \\ x &= e^{-2} \approx 0.13\end{align} $$ While your y-intercept is sort of correct, it doesn't make a lot of sense to say the the regular $\ln x $ funcion has a y-intercept at $y=-\infty$. Rather, you should say that the y-intercept is the value of $f(x)$ at $x=0$. But $f(x) = \ln x + 2$ is not defined at $x=0$, so this funcion doesn't have a y-intercept. You mention at the end that you know this funcion is like $\ln x$ moved up two units. That should be all you need to know: It will still have the vertical asymptote at $x=0$, it will still go to infinity as $x$ goes to infinity, it will still have only one x-intercept. The shape is the same, but moved vertically.
H: Galilean transformations How do you prove that every galilean transformation of the space $\mathbb R \times \mathbb R^3$ can be written in a unique way as the composition of a rotation, a translation and uniform motion? Thanks! AI: As mentioned, this problem appears in V.I Arnold's "Mathematical Methods of Classical Mechanics". A simple proof can be found on page 7 of this nice lecture series on the topic. In essence, you use the affine property of the transformation to show that the time cannot be dilated, the space component must be multiplied by a orthogonal matrix plus a constant vector. The uniqueness follows by considering to different parameterizations and proving equality.
H: A natural question about convergence Let $u,u_k \in C^{0}(K)$ where $K \subset \mathbb{R}^{n}$ is a compact set. Assume that $u_k \rightarrow u$ uniformly. Is these hypotheses sufficient to guarantee that \begin{equation} \mbox{med}(\{u_k>0\}) \rightarrow \mbox{med}(\{u>0\}) \end{equation} or \begin{equation} \mbox{med}(\{u_k \ge0\}) \rightarrow \mbox{med}(\{u \ge 0\})? \end{equation} AI: This is not true. Let $K=[-1,1]$ and $$u_k(x)=\begin{cases} \frac{1}{k} &\text{if } x\geq \frac12\\ \frac{2x}{k} &\text{if } \frac{-1}{2}<x<\frac12\\ \frac{-1}{k} &\text{if } x\leq \frac{-1}{2}\end{cases}$$ and note that $u_k\to 0$. We have $\mathrm{med}(\{u_k>0\})=1$ for all $k$ yet $\mathrm{med}(\{u>0\})=0$, and $\mathrm{med}(\{u_k\geq 0\})=1$ for all $k$ yet $\mathrm{med}(\{u\geq 0\})=2$.
H: $(p,\frac{n}{p^\alpha})=1$ then $p\nmid\binom{n}{p^\alpha}$ Let $n=p^\alpha m$ wherein $(p,m)=1$. Then we have $$p\nmid\binom{n}{p^\alpha}$$ What I have done is just playing with $\binom{n}{p^\alpha}$ like:$$\binom{n}{p^\alpha}=\frac{n!}{(n-p^\alpha)!p^\alpha!}=m\binom{p^\alpha m-1}{p^\alpha-1}$$ Honestly, the rest is unclear to me. Thank you for your help. EDIT: $p$ is prime. AI: I assume that $p$ is prime. Kummer's Theorem says that the number of factors of $p$ in $\binom{n}{k}$ is the sum of the base-$p$ digits of $n$ minus the sum of the base-$p$ digits of $k$ and $n-k$, all divided by $p-1$. In the case above, the base-$p$ representation of $n$ looks like $$ d_md_{m-1}d_{m-2}\dots d_{\alpha}\underbrace{000\dots000}_{\alpha\text{ zeros}} $$ where $d_\alpha\not=0$. Furthermore, $p^\alpha$ looks like $$ 1\underbrace{000\dots000}_{\alpha\text{ zeros}} $$ and $n-p^\alpha$ looks like $$ d_md_{m-1}d_{m-2}\dots (d_{\alpha}-1)\underbrace{000\dots000}_{\alpha\text{ zeros}} $$ The sum of the base-$p$ digits in $n-p^\alpha$ and $p^\alpha$ is $$ \sum_{j=\alpha}^md_j $$ which is the same as the sum of the base-$p$ digits in $n$. Thus, there are no factors of $p$ in $\binom{n}{p^\alpha}$.
H: Statistical Significance Dice Probability I think I just need to be pushed to the right formula or algorithm... Imagine you've got a "real" dice which is not an optimal one. And you want to know the confidence intervals for each result. So you rolled the dice a couple of times and get the following absolute probabilities as result: #eyes #occurrences ------------------ 1 10 2 11 3 24 4 13 5 14 6 11 You actually want to know weather e.g. this 24 times 3 eyes is just a random result or weather it's really more probable. If so, how much more probable is it (for sure)? So I would like to calculate a 99%-confidence interval for the probabilities. How to calculate this? I probably know this from statistics in university, but just forgot it... so you don't need to go to much into detail. Just need the right formula/algorithm to look for... Thanks for your help. --- edit --- Just to make clear, why I do not just lookup "Confidence Interval" at wikipedia. I would know how to calculate everything if there would be only two cases (e.g. like a coin... 0 and 1). Then I would be able to apply the formula, but I just didn't use such statistics for some years now and just don't see the solution how to reduce the problem. I just think about taking the result in question (e.g. 3 eyes) as "p" and all other results as "\not p"; does that work? AI: The outcome can be dichotomized to the events A={you roll a 3} and B={ you roll something other than a 3}. Let p be the probability of rolling a three. Then the number of threes rolled is binomial with n=83 (in the example) and p unknown. Using the binomial distribution you can construct your 99% confidence interval for p. If the interval doesn't contain 1/6 and the lower bound is above 1/6, you can conclude at the alpha=0.01 level that the die tends to roll more 3s than you would get by chance.
H: Obtaining a monotone subsequence from a sequence of functions It is true that given any sequence of numbers, there is a monotone subsequence within. I thought obtaining a monotone subsequence out of a sequence of functions might be too much to ask. But given a sequence of $L^1$ functions, could we say that it admits a monotone subsequence? EDIT: My motive for asking this was to decide the following problem. Given a sequence of non-negative measurable functions $\{f_n\}$, it is known that for all measurable sets $E$, $\int_Ef_n\to 0$. Could we conclude from here that $f_n\to 0$? AI: Certainly not, since functions are not totally ordered. For example, if we let $f_n$ be the indicator function of $(n,n+1)$ then each $f_n\in L^1$ but none of the functions are comparable. What you are trying to prove is however true, at least in the sense of $L^1$ convergence. Let $E$ be the whole set, and note that $$\\|f_n\\|_1=\left\|\int_E f_nd\mu\right\|=\int_Ef_nd\mu\to0$$ so $f_n\to 0$ in $L^1$, since $\\|\cdot\\|_1$ is a norm on the $L^1$ space, which consists of equivalence classes of functions which differ on sets of measure $0$.
H: An $(R,S)$-bimodule is a left $R \otimes_k S^{\text{op}}$-module Let $k$ be a commutative ring, and let $R,S$ be $k$-algebras. To me "$R$ is a $k$-algebra" means that $R$ is a $k$-module such that $a(rs)=(ar)s=r(as)$ for all $a\in k$ and $r,s \in R$. Let $M$ be a $(R,S)$-bimodule. I am trying to show that $M$ is a left $R \otimes_k S^{\text{op}}$-module. Define a map $\phi \colon R \times S^{\text{op}} \to M$ given by $\phi(r,s) = (rm)s$. I believe that for me to progress with the proof this map has to have the property that $\phi (ar,s) = \phi (r, as)$ for $a \in k$. However $\phi(ar,s) = [(ar)m]s$ and $\phi(r, as) = (rm)(as)$. However I have no idea how to get that $a$ to pass to the other side of the $m$. How is this achieved? Is there implicitly a $k$-module structure on $M$? AI: We assume $R$ and $S$ are unital. Hence there exists the canonical homomorphism $k \rightarrow R$. Hence by restricting the actions of $R$ on $M$ to $k$, $M$ can be a $k$-module. Similarle $M$ can be a $k$-module through $S$. We assume the both $k$-module structures on $M$ coincide. Let $T = End_k(M)$ be the $k$-algebra of $k$-endomorphisms on $M$. For $r \in R$, we denote by $\lambda(r) \in T$ the endomorphism $x \rightarrow rx$. For $s \in S$, we denote by $\rho(s) \in T$ the endomorphism $x \rightarrow xs$. $\lambda\colon R \rightarrow T$ and $\rho\colon S^{\text{op}} \rightarrow T$ are $k$-algebra homomorphisms. Define a map $\phi \colon R \times S^{\text{op}} \to T$ by $\phi(r, s) = \lambda(r)\rho(s)$. For $a \in k$, $a\lambda(r)\rho(s) = \lambda(ar)\rho(s) = \lambda(r)\rho(as)$. Hence $a\phi (r,s) = \phi (ar, s) = \phi (r, as)$. Hence there exists a $k$-linear map $f\colon R \otimes_k S^{\text{op}} \rightarrow T$ such that $f(r\otimes s) = \lambda(r)\rho(s)$. Since $\lambda(r)\rho(s) = \rho(s)\lambda(r)$, $f$ is a $k$-algebra homomorphism. Hence $f$ makes $M$ into a left $R \otimes_k S^{\text{op}}$-module.
H: Expansion of $x^4\over1+x^2$ into a power series I calculated the generating function $A(x)$ of the recurrence $a_n = a_{n-2} - 2a_{n-3}$, $(n \ge 0, a_0 = a_1 = 0, a_2 = 2)$ and I have no clue on how to expand it into a power series in order to read the coefficients. $$A(x) = {-2x^4 \over 1+x^2}.$$ AI: Your generating function is incorrect. You have the recurrence $$a_n=a_{n-2}-2a_{n-3}+2[n=2]\;,$$ where the last term is an Iverson bracket, and I assume that $a_n=0$ for $n<0$. Then $$\begin{align} A(x)&=\sum_{n\ge 0}a_nx^n\\ &=\sum_{n\ge 0}a_{n-2}x^n-2\sum_{n\ge 0}a_{n-3}x^n+2x^2\\ &=x^2A(x)-2x^3A(x)+2x^2\;, \end{align}$$ so $$A(x)=\frac{2x^2}{1-x^2+2x^3}\;.$$ Your function is $$\begin{align} \frac{-2x^4}{1+x^2}&=-2x^4\cdot\frac1{1-(-x^2)}\\ &=-2x^4\sum_{n\ge 0}(-x^2)^n\\ &=-2x^4\sum_{n\ge 0}(-1)^nx^{2n}\;, \end{align}$$ in which the coefficient of $x^2$ and of every odd power of $x$ is $0$. However, $a_2=2\ne 0$, and $a_3=a_1-2a_0=0$, so $a_5=a_3-2a_2=-4\ne 0$.
H: Proving that $\frac{\csc\theta}{\cot\theta}-\frac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$ Prove $\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}=\tan\theta\sin\theta$ So, LS= $$\dfrac{\csc\theta}{\cot\theta}-\dfrac{\cot\theta}{\csc\theta}$$ $$\left(\dfrac{1}{\sin\theta}\cdot \dfrac{\tan\theta}{1}\right)-\left(\dfrac{1}{\tan\theta}\cdot \dfrac{\sin\theta}{1}\right)$$ $$\dfrac{\tan\theta}{\sin\theta}-\dfrac{\sin\theta}{\tan\theta}$$ Now, considering the fact that I must have a common denominator to subtract, would this be correct: $$\dfrac{\tan^2\theta}{\tan\theta\sin\theta}-\dfrac{\sin^2\theta}{\tan\theta\sin\theta}\Rightarrow \dfrac{\tan^2\theta-\sin^2\theta}{\tan\theta\sin\theta}$$ I feel like I'm close to the answer because the denominator is the RS of the OP. Please help. Do not give me the answer. AI: $$\frac{\tan\theta}{\sin\theta}-\frac{\sin\theta}{\tan\theta}=\frac1{\cos\theta}-\cos\theta=\frac{1-\cos^2\theta}{\cos\theta}=\dots$$
H: Lebesgue measure: $\mu(f-g(D))=0\Rightarrow\mu(f(D))=\mu(g(D))$ Let $f,g:\mathbb{R}\rightarrow\mathbb{R}$ be Lebesgue measurable functions, and let $D$ be a subset of $\mathbb{R}$. If $f(D)$ and $g(D)$ are measurable subsets of $\mathbb{R}$, and $(f-g)(D):=\{f(x)-g(x):x\in D\}$ is a measurable subset of $\mathbb{R}$ with measure $0$, is it necessarily true that $\mu(f(D))=\mu(g(D))$ (where $\mu$ denotes Lebesgue measure)? This question was originally posed to me without the measure being specified as Lebesgue, and I was able to find a counter-example (using "unit mass" measure). However, for Lebesgue measure it appears to be true... it is clearly true if $f$ and $g$ are simple functions... although I don't see how a more general proof would go. Thanks in advance for any help. AI: No. Consider $f(x) = x$ and $g(x) = x - \lfloor x \rfloor$, the fractional part.
H: Euler's theorem on a polynomial The following theorem is said to be due to Euler. However, it seems not so well known. How do you prove this? Theorem Let $K$ be a field. Let $n \geq 1$ be an integer. Let $\alpha_1,\dots,\alpha_n$ be distinct elements of $K$. Let $f(X) = (X - \alpha_1)\cdots(X - \alpha_n)$. Then $\sum_i \alpha_i^{k}/f'(\alpha_i) = 0$ for $0 \leq k \leq n - 2$. $\sum_i \alpha_i^{n-1}/f'(\alpha_i) = 1$. AI: We first work over $\mathbb{C}$. Lemma: Let $\frac{P(z)}{Q(z)}$ be a rational function such that $Q(z)$ has distinct roots. Then the residue of $\frac{P(z)}{Q(z)}$ at $z = a$ (where $Q(a) = 0$) is $\frac{P(a)}{Q'(a)}$. Proof. The residue is $\lim_{z \to a} \frac{P(z) (z - a)}{Q(z)} = \frac{P(a)}{Q'(a)}$ by l'Hopital's rule. $\Box$ It follows that $$\frac{1}{f(z)} = \sum_i \frac{1}{f'(\alpha_i) (z - \alpha_i)}$$ hence $$\frac{z}{f(z)} = \sum_i \frac{z}{f'(\alpha_i) (z - \alpha_i)} = \sum_i \frac{1}{f'(\alpha_i) \left( 1 - \frac{\alpha_i}{z} \right)}.$$ Rewriting the above identity in terms of $t = \frac{1}{z}$ gives $$\frac{t^{n-1}}{(1 - t \alpha_1)...(1 - t \alpha_n)} = \sum_i \frac{1}{f'(\alpha_i) (1 - t \alpha_i)}$$ and comparing the first $n-1$ coefficients of the Taylor series in $t$ of both sides gives the desired result. The final identity we prove is a sequence of polynomial identities among various symmetric polynomials in the $\alpha_i$. Consequently, if it holds over $\mathbb{C}$, it holds identically, so in particular over any field.
H: What do finite, infinite, countable, not countable, countably infinite mean? Possible Duplicate: Different kinds of infinities? Those terms keep confusing me. Could somebody give a clear and unambiguous definition? How do we prove a set is countable, finite, ...? I know that there are more real numbers than there are integers. How do we prove this? What about other sets like $\mathbb{Z, Q, C}$? AI: The definitions are: We say that a set $A$ is finite if and only if there exists some $k\in\mathbb N$ such that there exists $f\colon A\to\{n\in\mathbb N\mid n<k\}$ such that $f$ is a bijection. Namely, $A$ is finite if we can write it as $\{a_0,\ldots,a_{k-1}\}$. We say that a set $A$ is countable if and only if there exists $f\colon A\to\mathbb N$ which is injective. Clearly every finite set is countable, but also some infinite sets are countable. Note that some places define countable as infinite and the above definition. In such cases we say that finite sets are "at most countable". We say that a set $A$ is uncountable if and only if it is not countable. Now to show that $\mathbb{N,Z,Q}$ are countable we do the following: Note that $\mathbb N$ is countable by definition (the identity function witnesses that); The set $\mathbb Z$ is countable since we can consider the following function: $$f(m)=\begin{cases}2m& m\geq 0\\ -2m+1& m<0\end{cases}$$ We can show that this is an injective function (in fact a bijection) by considering the two cases. To show that $\mathbb Q$ is countable we use the following facts: If $A$ is countable then $A\times A$ is countable. If $A$ is infinite and countable and $B$ is a set such that there is $f\colon A\to B$ which is onto $B$, then $B$ is countable, and vice versa: if $B$ is countable then such surjective function exists. Now we have that since $\mathbb Z$ is countable so is $\mathbb Z\times\mathbb Z$; we consider the function $f\colon\mathbb Z\times\mathbb Z\to\mathbb Q$ defined as: $$f(m,k)=\begin{cases}\frac{m}{k}&k\neq 0\\ 0 &k=0\end{cases}$$ It is not hard to see that this is a surjective function, since every rational number can be written as a fraction $\frac{m}{k}$. Therefore $\mathbb Q$ is countable too. Lastly, $\mathbb R$ is uncountable. For this we use Cantor's theorem which states that if $A$ is a set then there is no surjective function from $A$ onto $\mathcal P(A)$ (the set of all subsets of $A$). We can use the above theorem to show that $\mathbb R$ is in fact with bijection with $\mathcal P(\mathbb N)$, and therefore $\mathbb R$ is not countable. Since the above shows that $\mathbb R$ is uncountable, and $\mathbb R\subseteq\mathbb C$ we have that the complex numbers are also uncountable.
H: Geometric interpretation of linear transformation I have a linear transformation, given by the following matrix $$ \begin{pmatrix} x_1\\ x_2 \end{pmatrix} \mapsto \begin{pmatrix} 2 & 2\\ -1 & -1\\ \end{pmatrix} \begin{pmatrix} x_1\\ x_2 \end{pmatrix} $$ How can I determine what this corresponds to geometrically, when I apply it to the $x_1,x_2$-plane. I have tried to visualise the transformation by hand, and using a fieldplot in Maple, to get an idea of what is happening. My idea was then to decompose it into scaling, rotation, reflection or some other simple transformations. My question is: Which geometric transformation does the above linear map correspond to, and in general, what is a good strategy for solving this kind of problem? AI: The better approach is to see what happens with the elements of the canonical base. So, you have two vectors $e_1=(1,0)$ and $e_2=(0,1)$. You have the unity square generated by them and its image could be a rectangle, or parallelogram, or a line, etc.... In our case, $e_1$ is mapped to the vector $(2,-1)$. The same for $e_2$. So the image is a line, namely, generated by $v=(2,-1)$.
H: Localizations of the Boolean Ring P(X) Given a set $X$, we can construct the Boolean ring whose elements are the power set of $X$. The multiplication therein is intersection, and the addition is symmetric difference. I am interested in finding information on the study of such rings, for instance, information about the spectrum of such a ring. In that case, I am specifically interested in trying to find a systematic description of what the localization of such a ring at an object (i.e. a subset of $X$) would look like. Doing some basic calculations, strange things seem to happen. For instance, if $X$ is the natural numbers, and we localize at (i.e. invert all powers of) the set $\{1,2\}$, we get constructions like $\{1,2\}+1/\{1,2\}$, which is an object that when multiplied by (intersected with) the set $\{1,2\}$, yields $X-\{1,2\}$, which is pretty wild. This is a pretty algebraically simple subject, but I don't recall ever seeing anything written down about it. Thanks for any help! AI: The spectrum is the space $\beta X$ of ultrafilters on $X$. More generally, the spectrum of any Boolean ring is a Stone space, and this actually gives a contravariant equivalence of categories. For more details, see this blog post. For way more details, see Johnstone's Stone Spaces. Algebraically, the Boolean ring you are interested in is $\mathbb{F}_2^X$. Inverting the indicator function $\chi_S$ of a subset $S \subseteq X$ just gets you $\mathbb{F}_2^S$ (it kills of all the copies of $\mathbb{F}_2$ at which $\chi_S$ is equal to $0$). If $X = \mathbb{N}$ and $S = \{ 1, 2 \}$ then the localization is just $\mathbb{F}_2^{ \{ 1, 2 \} }$ and $\chi_S + \frac{1}{\chi_S} = 2 \chi_S = 0$ in the localization.
H: Strategies for arranging coins in a circle Suppose you have $n$ identical circular coins and you would like to arrange them on the table so that their edges touch and their centers lie on a circle. Mathematically, there is no trouble. "Just" put the center of each coin at $re^{2ik\pi/n}$ for $k$ in $\{0, 1\ldots, n-1\}$ and some suitable $r$. But in practice, one can't easily calculate $e^{2ik\pi/n}$ and one wouldn't be able to position the coins even if the coordinates of their centers were given. What I want are heuristics that allow one to position the coins approximately correctly, which can be executed by someone with an ordinarily good eye and ordinarily good hands, without any special measuring devices. Good solutions for $n\le 3$ are trivial. There is also a good solution for the special case of $n=6$, which is to arrange the six coins around a seventh. In practice it does not seem too hard to arrange four coins into a square, by first estimating the right angles and then looking to see if the resulting quadrilateral is visibly rhombic. But I would be glad to see a more methodical approach. This is a soft question. I expect the solution to be informed by mathematics, but not purely mathematical. AI: For a large number of coins, I would apply an iterative scheme inspired by mean curvature flow. In other words, just keep nudging in coins that are jutting out, until the curvature of the whole arrangement is uniform. We can consider the arrangement of coins as a polygon whose vertices are the centers of the coins, and whose edges connect adjacent coins and are all of equal length. The polygon is regular if and only if all its internal angles are equal, which holds if and only if each angle is equal to (i) the average of all the angles, and/or (ii) the average of the angles at adjacent vertices. If you can compare adjacent angles by eye, or tell which vertices have the largest and smallest angles of all, then you can nudge them in the right direction. In practice, I found that this has a couple of limitations. First, it's easy to push coins inwards because the rest of them will automatically move out of the way, but if you push a coin outward, you have to carefully reconnect the arrangement. Second, the procedure converges quite slowly. You quickly arrive at a nice smooth oval, after which moving coins one at a time is a frustratingly constrained way to round out the arrangement.
H: Span of an empty set is the zero vector I am reading Nering's book on Linear Algebra and in the section on vector spaces he makes the comment, "We also agree that the empty set spans the set consisting of the zero vector alone". Is Nering defining the span of the empty set to be the set containing the zero vector or is this something you can prove from the definition of span? I sense it is the latter, but the proof seems a bit tricky since you would be saying that {0} = Span of the indexed set of vectors in the empty set. But since the empty set has no vectors, it is not clear to me what its span would be. AI: Depending on how you define the span, this is either a definition or it follows from the definition of span (and judging by the wording it is probably the former). What's Nering's definition of span? (One definition of span is the following: the span of a collection of vectors is the intersection of all subspaces containing them. The span of no vectors is therefore the intersection of all subspaces, which is $\{ 0 \}$.)
H: Strong Law of Larger Numbers with Coin flips I am reviewing some probability for a class I am about to take, and I came up with this example to try to help my intuition. Suppose I take the the space of all sequences of the two symbols $H,T$. Let $X_i$ be the function on this space that is 1 if the $i$th entry is $H$ and $0$ otherwise. Then I think that these can be considered as independent identically distributed random variables, so that the Strong Law of Large Numbers should say that $$ \lim \frac1n\sum_{i=1}^n X_i = \frac 12 a.e. $$ This implies that the set of sequences of $H,T$ which in the limit are not balanced have measure $0$. The measure here is the one so that the set of all sequences starting with $H$ has measure $\frac12$, the set of all sequences starting with $HH$ has measure $\frac14$, etc. So my question is: Is this correct? Is there such a measure on this space of sequences? If so, is there a nice way to see the conclusion at the beginning of the paragraph above without quoting the strong law of large numbers? AI: The space you're talking about, the space of all one-way infinite binary sequences, is known as Cantor Space (especially within logic). I think of it as the set of all paths through the full infinite binary branching tree. As Qiaochu said, the 'fair coin measure' can be thought of as a product measure. But I prefer to just recall that by Caratheodory's extension theorem, it suffices to define any set function $\mu$ on the algebra of basic clopens (that is, sets of the form $$[\sigma]:=\{A \in \{0,1\}^{\mathbb{N}} : A \text{ extends } \sigma \},$$ where $\sigma$ is some finite binary string) such that $\mu[\sigma] = \mu[\sigma 1] + \mu[\sigma 0]$ and $\mu[\varnothing] = 1$. ("$\varnothing$" denotes the empty finite binary string; all elements of $\{0,1\}^{\mathbb{N}}$ extend it.) So to get the measure you want, you just define your $\mu$ such that $\mu[\sigma 0] = \mu[\sigma 1] = \mu[\sigma]/2$. Notice that in terms of the tree picture, you're just saying that no matter where you are in the tree, it costs the same amount of measure to go the the left as to the right (so my tree is growing upwards). Notice that this measure (often called the Lebesgue measure because of what @Michael said in the comments) is just a special case of a Bernoulli measure, which corresponds to a coin with probability $p$ of heads and $1-p$ of tails (or if you prefer $p$ to go left in the tree and $1-p$ to go right). There are other ways to see that almost every number must have a 'balanced' frequency of 0's and 1's, but none of them are much nicer than using the law of large numbers. For example, one could apply, as @Qiaochu suggested, the Borel-Cantelli lemma. To do so, you want to exploit something that happens infinitely often (because of the $\limsup$). If some $A \in \{0,1\}^{\mathbb{N}}$ does not have a balanced number of 1's and 0's, then frequency of 0's compared to the frequency of 1's is outside of the interval from $[1- 1/n, 1+1/n]$ for some $n$ infinitely often. I haven't checked the whether the resulting sets have measures whose total sum is finite (as is required by the B-C), but something like this should work. One could also apply Birkhoff's Ergodic Theorem (which is just a generalization of the law of large numbers anyway) or the theory of Algorithmic Randomness to conclude that almost every normal has a balanced frequncy of 0's and 1'. In fact, you can conclude much more, that almost every real number is Normal in every base.
H: Is the number of primes congruent to 1 mod 6 equal to the number of primes congruent to 5 mod 6? (I know that there's an infinity of primes congruent to 5 mod 6, but I don't know if there is an infinity of primes congruent to 1 mod 6.) But what I'd really like to know is whether or not the number of primes ≡1 mod 6 less than a given n can be said to be asymptotically equal to the number of primes ≡5 mod 6 less than that n. So the "total" number of primes ≡1 mod 6 would be equal to the number of primes ≡5 mod 6. (I hope this is understandable. I'd also like to know how to phrase this inquiry in a more standard fashion, if someone would tell me how to fix it.) AI: Yes, this is true. A more general statement follows from a suitably strong form of Dirichlet's theorem on arithmetic progressions, namely that asymptotically the proportion of primes which are congruent to $a \bmod n$ (for $\gcd(a, n) = 1$) is $\frac{1}{\varphi(n)}$. In the particular case that $n = 6$ it is possible to give a more elementary proof of a weaker result. Define the Dirichlet L-function $$L(s, \chi_6) = \sum_{n=1}^{\infty} \frac{\chi_6(n)}{n^s}$$ where $\chi_6$ is the unique nontrivial Dirichlet character $\bmod 6$. This takes the form $\chi_6(n) = 1$ if $n \equiv 1 \bmod 6$, $\chi_6(n) = -1$ if $n \equiv 5 \bmod 6$, and $\chi_6(n) = 0$ otherwise. The Euler product of this L-function is $$L(s, \chi_6) = \prod_p \left( \frac{1}{1 - \chi_6(p) p^{-s}} \right) = \prod_{p \equiv 1 \bmod 6} \left( \frac{1}{1 - p^{-s}} \right) \prod_{p \equiv 5 \bmod 6} \left( \frac{1}{1 + p^{-s}} \right).$$ It is possible to explicitly evaluate $L(1, \chi_6)$ and in particular to show that it is not zero; in fact, $$L(1, \chi_6) = \int_0^1 \frac{1 - x^5}{1 - x^6} \, dx$$ and this can be evaluated using partial fractions (but note that the integrand is always positive so this number is definitely positive). So we conclude that $$-\log L(s, \chi_6) = \sum_{p \equiv 1 \bmod 6} \log (1 - p^{-s}) + \sum_{p \equiv 5 \bmod 6} \log (1 + p^{-s})$$ approaches a nonzero constant as $s \to 1$ (if summed in the appropriate order) even though the first and second terms separately approach $\mp \infty$. So the contributions coming from primes in each residue class cancel out asymptotically. This is not quite as strong as the desired statement, though; if you fill in all the details in what I've said you'll show that the Dirichlet density of the primes congruent to $\pm 1 \bmod 6$ are the same but this should still be true for the natural density and this requires a further argument (I am not sure how much further, though). For more details see any book on analytic number theory, e.g. Apostol.
H: Definition of operation that creates matrix from two vectors I'm looking for something similar to Kronecker's product that creates matrix based on two vectors I'm thinking about either: $$ (1,2,3) @ (4,5,6) = \begin{bmatrix} (1, 4) & (2, 4) & (3, 4) \\[0.3em] (1, 5) & (2, 5) & (3, 5) \\[0.3em] (1, 6) & (2, 6) & (3, 6) \end{bmatrix} $$ or $$ (1,2,3) @ (4,5,6) = \begin{bmatrix} 1 \cdot 4 & 2 \cdot 4 & 3 \cdot 4 \\[0.3em] 1 \cdot 5 & 2 \cdot 5 & 3 \cdot 5 \\[0.3em] 1 \cdot 6 & 2 \cdot 6 & 3 \cdot 6 \end{bmatrix} $$ where obviously @ is my imaginary operation. AI: The second one is sometimes called the outer product.
H: L'Hôpital's Rule for $ \frac{\infty}{\infty-\infty} $? I'm currently trying to find this improper integral: $$ \int^{\infty}_{-\infty}\frac{1}{\sqrt{x^{2}+1}} dx $$ I started off by splitting it into a proper integral, and then into the sum of two integrals: $$ = \lim_{a\rightarrow\infty} \int^{a}_{-a}\frac{1}{\sqrt{x^{2}+1}}dx = \lim_{a\rightarrow\infty}(\int^{0}_{-a}\frac{1}{\sqrt{x^{2}+1}}dx + \int^{a}_{0}\frac{1}{\sqrt{x^{2}+1}}dx) $$ To calculate the integrals I used the trig. substitution $ x=b\tan\theta $ with $ b=1 $, which would give the differential $ dx=sec^{2}\theta d\theta $. The new limits of integration would then be $ [-\frac{\pi}{2},0] $ and $ [0,\frac{\pi}{2}] $ because as $ x\rightarrow\pm\infty $, $ \theta\rightarrow\pm\frac{\pi}{2} $, so the integrals and limit can be rewritten as: $$ = \lim_{a\rightarrow\pi/2}(\int^{0}_{-a}\frac{\sec^{2}\theta}{\sqrt{\tan^{2}\theta+1}}d\theta + \int^{a}_{0}\frac{\sec^{2}\theta}{\sqrt{\tan^{2}\theta+1}}d\theta) $$ ...which can then simplify to: $$ = \lim_{a\rightarrow\pi/2}(\int^{0}_{-a}\frac{\sec^{2}\theta}{\sqrt{\sec^{2}\theta}}d\theta +\int^{a}_{0}\frac{\sec^{2}\theta}{\sqrt{\sec^{2}\theta}}d\theta) = \lim_{a\rightarrow\pi/2}(\int^{0}_{-a}\frac{\sec^{2}\theta}{\|\sec\theta\|}d\theta+\int^{a}_{0}\frac{\sec^{2}\theta}{\|\sec\theta\|}d\theta) $$ The absolute values on the secants can be removed because on the interval $ [-\frac{\pi}{2},\frac{\pi}{2}] $, the secant function is positive. $$ = \lim_{a\rightarrow\pi/2}(\int^{0}_{-a}\frac{\sec^{2}\theta}{\sec\theta}d\theta+\int^{a}_{0}\frac{\sec^{2}\theta}{\sec\theta}d\theta) = \lim_{a\rightarrow\pi/2}(\int^{0}_{-a}\sec\theta d\theta+\int^{a}_{0}\sec\theta d\theta) $$ The antiderivative of $ \sec\theta = \ln\|\sec\theta+\tan\theta\|+C $, so the integrals become: $$ = \lim_{a\rightarrow\pi/2}(\ln\|\sec\theta+\tan\theta\|\bigg\|^{0}_{-a} + \ln\|\sec\theta+\tan\theta\|\bigg\|^{a}_{0}) $$ $$ = \lim_{a\rightarrow\pi/2}((\ln\|\sec(0)+\tan(0)\|-\ln\|\sec(-a)+\tan(-a)\|)+(\ln\|\sec(a)+tan(a)\|-\ln\|\sec(0)+tan(0)\|)) $$ Since $ \sec(0) = 1 $ and $ \tan(0) = 0 $, the value of $ \ln\|\sec(0)+tan(0)\| = \ln(1) = 0 $. The limit can be rewritten as: $$ = \lim_{a\rightarrow\pi/2}((0-\ln\|\sec(-a)+\tan(-a)\|)+(\ln\|\sec(a)+tan(a)\|-0)) $$ $$ = \lim_{a\rightarrow\pi/2}(-\ln\|\sec(-a)+\tan(-a)\|+\ln\|\sec(a)+tan(a)\|) $$ The tangent function has been shown to be odd, and the secant function even, so $ \sec(-a) = \sec(a) $ and $ \tan(-a) = -\tan(a) $. Therefore, applying and then commuting the addition, we have: $$ = \lim_{a\rightarrow\pi/2}(\ln\|\sec(a)+tan(a)\|-\ln\|\sec(a)-\tan(a)\|) $$ Subtraction of logarithms become division, so $ \ln\|\sec(a)+tan(a)\|-\ln\|\sec(a)-\tan(a)\| $ $ = \ln\left\|\frac{\sec(a)+\tan(a)}{\sec(a)-\tan(a)}\right\| $, which becomes: $$ = \lim_{a\rightarrow\pi/2}\left(\ln\left\|\frac{\sec(a)+\tan(a)}{\sec(a)-\tan(a)}\right\|\right)$$ Here's where I'm confused: can you take the natural log of the limit of the fraction (i.e., $$ \ln\left\|\lim_{a\rightarrow\pi/2}\left(\frac{\sec(a)+\tan(a)}{\sec(a)-\tan(a)}\right)\right\| $$ ), or does the limit not exist? And, if you can take the natural log of the limit, how would you go about evaluating the limit of the fraction? Since $ \sec(\frac{\pi}{2}) "=" \infty $ and $ \tan(\frac{\pi}{2}) "=" \infty $, would there be some form of L'Hôpital's Rule you'd have to use, since $ \frac{\infty}{\infty-\infty} $ is indeterminate? AI: $${\sec x+\tan x\over\sec x-\tan x}={1+\sin x\over1-\sin x}$$
H: The logic behind partial fraction decomposition In the general case of any function would be interesting but my question is concerning the general case of polynomials with integer powers. I can use the method of partial fractions in the simple case required for an introductory course on integration, but I'm not sure I really understand it or could describe WHY it works. For an arbitrary example $\frac{P(x)}{Q(x)}$ where $P$ and $Q$ are some string of polynomials, I know to check that this is indeed a proper fraction before continuing. Next I know that the number of constants I need to find is determined by the degree of $Q$. But why is this? I don't really get how this is known in advance. Further, in the case of repeated roots, what is the mechanism at play behind how a fraction like this is expanded? [i.e. $(x+1)^3$ in the denominator would be expanded with $\frac{A}{(x+1)} + \frac{B}{(x+1)^2} + \frac{C}{(x+1)^3}]$. I'm not really that advanced in mathematics but I know that in the case of a repeated root it would be necessary to include all the powers of the root to "account for" all the possible cases of what you may have started with. From the theoretical aspect though, why does this work? Why do we know that the degree of $Q$ determines the number of constants and that we can find all the constants by incrementing every power of the root in the case of a repeated root? I want lift the rug up and see what mechanisms are at play in order to understand this better. AI: Great question! First, let's tackle just the case where the roots of $Q(x)$ are all distinct. One way to conceptualize what's going on is the following: if $r$ is a root of $Q(x)$, then as $x \to r$, the function $f(x) = \frac{P(x)}{Q(x)}$ (we always assume $P, Q$ have no common roots) goes to infinity. How quickly does it go to infinity? Well, write $Q(x) = (x - r) R(x)$. Then $$\frac{P(x)}{Q(x)} = \frac{1}{x - r} \left( \frac{P(x)}{R(x)} \right)$$ and $R(r) \neq 0$. So we see that as $x \to r$, this expression goes to infinity like $\frac{1}{x - r}$; more precisely, it goes to infinity like $\frac{c_r}{x - r}$ where $c_r = \frac{P(r)}{R(r)}$. This number is referred to in complex analysis as the residue of the pole at $x = r$. So the upshot of all of this is that we can subtract this pole away and write $$f(x) - \frac{c_r}{x - r} = \frac{1}{x - r} \left( \frac{P(x)}{R(x)} - \frac{P(r)}{R(r)} \right).$$ The expression in parentheses approaches $0$ as $x \to r$, and in fact it is a rational function whose numerator is divisible by $x - r$, so we can actually divide by $x - r$. The result is a new rational function which no longer has a pole at $r$. We can repeat this process for every root of $Q$ until we get a rational function with no poles whatsoever. But this must be a polynomial. So now we've written $f$ as a sum of fractions of the form $\frac{c_r}{x - r}$ plus a polynomial. (Note that in general we need to consider the complex roots of $Q$.) Okay, so what if $Q$ has repeated roots? Then $f$ might go to infinity more quickly as $x \to r$. If $r$ is a root with multiplicity $n$, then writing $Q(x) = (x - r)^n R(x)$ we now have $$\frac{P(x)}{Q(x)} = \frac{1}{(x - r)^n} \left( \frac{P(x)}{R(x)} \right)$$ where $R(r) \neq 0$. So we see that as $x \to r$, this expression goes to infinity like $\frac{1}{(x - r)^n}$; more precisely, it goes to infinity like $\frac{c_{r,n}}{(x - r)^n}$ where $c_{r, n} = \frac{P(r)}{R(r)}$. So we can do the same thing as before and just subtract this off, getting $$f(x) - \frac{c_{r,n}}{(x - r)^n} = \frac{1}{(x - r)^n} \left( \frac{P(x)}{R(x)} - \frac{P(r)}{R(r)} \right).$$ The expression in parentheses approaches $0$ as $x \to r$, so again it is divisible by $x - r$, but this time we're not done! We still have to subtract off terms that look like $\frac{1}{(x - r)^k}$ where $k < n$ until the resulting rational function no longer goes to infinity as $x \to r$. The above is nice as far as it goes, but let me mention that algebraically partial fraction decomposition is more general than rational functions over $\mathbb{C}$. It also generalizes to, for example, rational numbers! Like rational functions, rational numbers also have partial fraction decompositions, like $$\frac{5}{12} = \frac{2}{3} - \frac{1}{4}.$$ Explaining all of this in a unified framework requires the language of abstract algebra, in particular the notion of a group, of a field, and of a principal ideal domain. Partial fraction decomposition in this setting describes the additive group of the field of fractions of a principal ideal domain using essentially the Chinese remainder theorem, but that's a story for another day... (It might not seem like uniqueness of partial fraction decomposition holds for rational numbers because we can also write $\frac{5}{12} = \frac{3}{4} - \frac{1}{3}$, but the correct notion of uniqueness here is subtle; it is uniqueness "mod $1$.")
H: Probability of hitting a bullseye in five shots The probability of a target shooter hitting the bullseye on any one shot is 0.2. a If the shooter takes five shots at the target, find the probability of: i) missing the bullseye every time ii) hitting the bullseye at least once b) What is the smallest number of shots the shooter should make to ensure a probability of more than 0.95 of hitting the bullseye at least once? i) $n=5,p=0.2$ $(5)$($(0.2)$=1 $1(1-0.2)=0.8$ $(0.8)^5= 0.327$ I don't is this the right way to do but I got the right answer. I don't know how to do ii) and B) , appreciate your help! AI: Hints: For ii, you have the probability of missing every time from i, so all the rest is the chance of hitting at least once. For b, the same logic of deriving ii from i applies, but the number of shots is a variable. If you want 0.95 to hit at least once, what is the chance you miss them all? Then find a number of shots that gets the chance below that.
H: Hilbert's Original Proof of the Nullstellensatz Does anyone have a link to Hilbert's Original Proof of the Nullstellensatz, or know a book where it's printed? I'd be interested to see what it was like. I only really know the Noether normalisation and Zariski proofs. While these are both good, it would be nice to have it 'from the horse's mouth'! Many thanks in advance. AI: According to Eisenbud, Hilbert's Nullstellensatz can be found in: Hilbert, D. (1893). Über die vollen Invariantensysteme. Math. Ann. 42, pp. 313–373. Also freely available on the GDZ. The statement of the theorem appears at the bottom of p. 320, if I'm not mistaken. (I don't read German, however!)
H: 3D Projection (Perspective). Not what I would expect... I have some code based on the 3d perspective projection formula I found on wiki. http://en.wikipedia.org/wiki/3D_projection It works great... mostly :) I recently started making the code do z ordering/culling... but I was getting weird results. After a lot of debugging I came to realise my formula for working out the zorder value is wrong. So... I created this in sketchup. The params I use to do the projection are CoordinateToProject, CameraPosition, CameraRotation, ViewerPosition. 1) What exactly is the formula I need calculate the zorder value/distance, please? 2) Why does my code not support -Z? Is it a most likely a bug in my code or a limitation of the formula? THANKS :) Edit: I am now happy with the answer to number 2. I just need to know how to calculate the z order.... any links/help would be appreciated!!!! AI: Could you give more information on which coordinates you use(left handed, right handed)? The exact matrix will relate on the handedness. If you are using an OpenGL style projection, you would better look at this: glFrustum Notice that both nearZ and farZ are positive. finally a hint: you need to do more than just projection since you may want linear z value(1/z) EDIT: I find this may help you: projection matrix EDIT for question #1: this is a perspective projection matrix $$ \begin{bmatrix} \frac{2n}{r-l} & 0 & -\frac{r+l}{r-l} & 0\\ 0 & \frac{2n}{t-b} & -\frac{t+b}{t-b} & 0\\ 0 & 0 & \frac{f}{f-n} & -\frac{fn}{f-n}\\ 0 & 0 & 1 & 0 \end{bmatrix} $$ ATTENTION please: this is a projection transformation for left handed coordinate system and it is a column-major matrix(every column forms a frame bases), and it maps Z to [0,1] in NDC(if you do not know what it is, search "Normalized Device Coordinates"), and notice that the Z is NOT linear(that is, a median point in Z direction will not be mapped to 0.5)
H: A simple depreciation problem Question 2: The share price of a public company fell 4% on each day of a week, from Monday to Friday. What is the price of a share at the end of the trading day on Friday, if the share was valued at $20 on the Sunday of the same week? I could do this the long way by calculating the depreciation for each day as: monday: 20 -4% of 20 tuesday:monday-4% of monday wed: tuesday-4% of tuesday and so on to friday This method takes a lot of time. What is the shortest way of finding the answer? could depreciation formula could be used for it? AI: The closing price on Friday is $20(0.96)^5$. Similarly, if the price increases by say $6\%$ each day, then in $5$ days it gets multiplied by $(1.06)^5$.
H: Determing the order when moving to the other side of the equation. Sorry if the title is vague. But I don't really know the terms I'm looking for (Nor do I know any tags that go with it). I usually end up being confused with the little(and perhaps trivial things) in math. The following for example:$$\mathbf{a} \cdot \mathbf{b} = \\| \mathbf{a} \\| \; \\| \mathbf{b} \\| \cos \theta$$to$$\theta = \arccos \left ( \frac{ \mathbf{a} \cdot \mathbf{b}}{\\| \mathbf{a} \\| \; \\| \mathbf{b} \\|}\right )$$ What are the rules that demand it's $\frac{ \mathbf{a} \cdot \mathbf{b}}{\\| \mathbf{a} \\| \; \\| \mathbf{b} \\|}$ and not $\frac{\\| \mathbf{a} \\| \; \\| \mathbf{b} \\|}{\mathbf{a} \cdot \mathbf{b}}$? Perhaps this is a silly question. But, What are the rules when moving from one side of the equation to the other? AI: The general rule is, you can do anything to one side of an equation, so long as you do the same thing to the other side. So, if you have the equation, $$r=st$$ you can multiply both sides by $1/s$ (assuming that $s\ne0$) to get $$(r)(1/s)=(1/s)st$$ which becomes $${r\over s}=t$$ There's no way you can go from $r=st$ to $s/r=t$ by doing the same thing to both sides of the equation.
H: Reconciling the classical formulation of a surface integral with a general integral over a manifold So I was just brushing up on some calculus when I came across a problem. I was trying to perform a surface integral I found online through the more general formulation of a differential form on a manifold. This led to some trouble. I'm considering the integral over the vector field $F(x,y,z) = (2x,2y,2z)$ along a cylinder of radius $1$ and height $5$ paramtrized by $S(\theta, t) = (\cos(\theta),\sin(\theta),t).$ Calc 2 would tell me to just take the dot product of the vector field and the normal vector to the surface, but I wanted to do this from a manifolds perspective, according to http://en.wikipedia.org/wiki/Differential_form's integration section. Following its lead, I identify $F(x,y,z) = 2x dx + 2y dy + 2z dz$ as my differential form. But now I immediately run into trouble as I'm trying to integrate a 1-form on a 2-manifold. Is there something intrinsically different about a surface integral versus an integral along a manifold? It seems like a surface integral assigns to each 1-form a 2-form that measures flow through a manifold rather than actually integrating along its surface. AI: Vector fields correspond to either one-forms or two-forms in three dimensions. I usually use the notation that follows: if $\vec{F} = \langle a,b,c \rangle$ then the corresponding work form is $$ \omega_{\vec{F}} = adx +bdy+cdz $$ whereas the corresponding flux form is $$ \Phi_{\vec{F}} = ady \wedge dz +bdz \wedge dx+cdx \wedge dy $$ surface integrals measure flux so you should use the flux form.
H: Is the fundamental theorem of calculus independent of ZF? By the fundamental theorem of calculus I mean the following. Theorem: Let $B$ be a Banach space and $f : [a, b] \to B$ be a continuously differentiable function (this means that we can write $f(x + h) = f(x) + h f'(x) + o(\|h\|)$ for some continuous function $f' : [a, b] \to B$). Then $$\int_a^b f'(t) \, dt = f(b) - f(a).$$ (This integral can be defined in any reasonable way, e.g. one can use the Bochner integral or a Riemann sum.) This theorem can be proven from Hahn-Banach, which allows you to reduce to the case $B = \mathbb{R}$. However, Hahn-Banach is independent of ZF. Recently I tried to prove this theorem without Hahn-Banach and found that I couldn't do it. The standard proof in the case $B = \mathbb{R}$ relies on the mean value theorem, which is not applicable here. I can only prove it (I think) under stronger hypotheses, e.g. $f'$ continuously differentiable or Lipschitz. So I am curious whether this theorem is even true in the absence of Hahn-Banach. It is likely that I am just missing some nice argument involving uniform continuity, but if I'm not, that would be good to know. AI: I believe that one of the standard proofs works. Let $F(x) := \intop_a^x f^\prime (t) dt$. Then $F$ is differentiable and its derivative is $f^\prime$ due to a standard estimate that has nothing to do with AC. $(F-f)^\prime = 0$, hence it is constant. This boils down to the one-dimensional case: just consider $g := \Vert F-f-F(a)+f(a) \Vert$. It is a real-valued function with zero derivative, and $g(a)=0$, so we can use the usual "one-dimensional" mean value theorem.
H: How to prove that if $\det(A)=0$ then $\det(\operatorname{adj}(A))=0$? How to prove that if $\det(A)=0$ then $\det(\operatorname{adj}(A))=0$? I have been trying to solve this but I can't use $$\det(A^{-1})=\det \Big(\frac{1}{\det(A)} \operatorname{adj}(A) \Big)$$ because $\det(A)=0$ and $\frac{1}{0}$ is not allowed. AI: Let's write $A'$ for the adjoint of $A$. $AA'=(\det A)I$, so $\det A\det A'=(\det A)^n$ (where $A$ is an $n\times n$ matrix). If $\det A\ne0$, this yields $\det A'=(\det A)^{n-1}$. By continuity, this last equation is true even when $\det A=0$, and you're done.
H: about a good book - Vector Calculus[by Jerold E. Marsden, Anthony J. Tromba ] I start reading Vector Calculus by Jerold E. Marsden, Anthony J. Tromba and I want to know if there is a book with the answers of the exercises. I like a lot this book, it seems to be made for a beginner but also for those who want to assimilate new knowledge. Thanks :) AI: Marsden and Tromba-from which I learned vector analysis under the eminent differential geometer Josef Dodziuk-indeed has a full solutions manual that's available from the publishers. It's actually a pretty good book,if overpriced. You can get a better book for fraction of the price from Dover: Vector Calculus by by Peter Baxandall and Hans Liebeck. It's a little more difficult then Marsden/Tromba and requires some linear algebra as prerequisite, but's beautifully written with many, many graphical examples,good exercises and distinguishes many of the finer points of calculus of several variables in 2 and 3 space. It's certainly worth the effort and you should get a copy as collateral reading.
H: Axioms of set theory and logic Zermelo–Fraenkel set theory is the most common foundation of mathematics with eight axioms and axiom of choice (ZFC): http://plato.stanford.edu/entries/set-theory/ZF.html But one can see that the statement of ZFC depends on first-order logic symbols and definitions, for example $=, \in, \forall, \exists$, predicate, formula, etc... And in many books, they didn't concept clearly about axioms of first-order logic, even through use set theory definition. My question is how set theory and first-order logic can combine in a unique system of axioms as the foundation of Mathematics. For example, There are objects called sets, denote by $X,Y,Z,..., A,B,C$ There are symbols $=, \in, \forall, \exists$ Symbols $=, \in$ can be put between two any sets $X,Y$ to be (primary) propositionals: $X\in Y$, $X=Y$. There are two logic values: $T$ (true), $F$ (false). Each propositional has a unique value, $T$ or $F$. There are logic operators $\neg, \vee, \wedge, \rightarrow, \leftrightarrow$ Logic operators can combine propositionals to create new propositionals, their logic values follow the standard logic rules. For a set $A$ and a propositional $P$, then $\forall A, P,\ \exists A, P$ are propositionals. etc... There must have a system of axioms for propositionals and sets (include ZFC) and how can we make them be minimal? AI: The usual approach is to define a concept of a first order language $\mathcal{L}$. They are usually specified by the nonlogical symbols. Well-formed formulas in the language $\mathfrak{L}$ are strings of symbols of $\mathfrak{L}$ along with the logical symbols such as $($, $)$, $\wedge$, $\neg$, variables etc. You can look up in a logic textbook the inductive definition of well-formed formulas, but something like $x \wedge y$ is a well-formed formula, but $(()\neg\wedge xy \neg$ is not a well-formed formula. A first order theory $T$ in the language $\mathfrak{L}$ is then a collection of well-formed sentences (no free variable) in the language $\mathfrak{L}$. You would then define the deduce relation $T \vdash \varphi$ to mean that there exists a proof of $\varphi$ using $T$. A proof is just a string of of sentences $\phi_1, ..., \phi_n$ such that $\phi_n = \varphi$, each $\phi_i$ is in $T$, a logical axiom of first order logic, follows from modus ponen or generalization using previous $\phi_j$, where $j < i$. So the above is the definition of a arbitrary first order theory in an arbitrary first order language $\mathfrak{L}$. Now let $\mathfrak{L} = \{\in\}$ a first order language consisting a single binary relation. $ZFC$ is then the first order theory in the language $\mathfrak{L}$ consisting of the "eight axioms" you mentioned above. (Note that ZFC has infinitely may axioms. For example, the axiom schema of specification is actually one axiom for each formula.) The benefit of this approach where the general definition of first order logic is developed first is that you apply this to study first order logic in general and other first order theories such that the theory of groups, rings, vector space, random graphs, etc. Also first order logic is developed in the metatheory. That is for example, a theorem of ZFC (even if it is about infinite cardinals greater than $\aleph_1$) has a finite proof in the metatheory. However, within ZFC you can formalize first order logic. Then you can consider question about whether $ZFC$ can prove it own consistency. By taking the approach of developing first order theories in general, you also gain a certain perspective. Some people think that ZFC is something special since it can serve as a foundation for much of mathematics. Through this approach, $ZFC$ is really just another first order theory in a very simple language consisting of a single non-logical symbol. People often have a hard time grasping the idea that $ZFC$ can have different models, for instance one where the continuum hypothesis holds and one where it does not. However, almost everyone would agree that that there exists more than one model of group theory (i.e. more than one group). Sometimes it is helpful to know that results about arbitrary first order theory still apply when one is working in ZFC set theory.
H: Exponential of the differential operator I am not sure whether this question is even well-posed. But today I learnt that $e^Df(x) = f(x+1)$ where $D$ is differential operator and $$e^D \triangleq \sum_{i=0}^{\infty} \frac{D^i}{i!}.$$ (ref. Dan Piponi's answer) So I was curious as to whether the differential equation $$\frac{df(t)}{dt} = e^Df(t) = f(t+1)$$ has any solutions apart from $f = 0$? AI: We first make a heuristic calculation. The equation in question is $Df = e^{D}f$. Thus it suffices to solve that $D = e^D$. Treating $D$ as a complex number, it is not hard to see that this equation has a solution in $\Bbb{C}$. For example, we may exploit Lambert-W function to specify such a number. Now let $r$ satisfy $r = e^r$. Then the function $f(t) = e^{r t}$ satisfies the desired property. Indeed, we prove that this is the case. We have $$Df(t) = r e^{r t} = e^{r}e^{r t} = e^{r(t+1)} = f(t+1) = e^{D}f(t).$$ Therefore $f(t)$ satisfies the desired property. If we confine ourselves to real-valued functions, taking real part or imaginary part will also give such non-trivial examples.
H: Extending the length of a curcumference by 1 meter I once heard the following statement: Take a piece of string and measure the length around a ball. Now add 1 meter to the string and stretch it out evenly around the ball. Obviously the diameter/radius will increase. Not sure by how much though. Do the same thing for a ball the size of the earth. Add 1 meter to the string and (amazingly, at least to me) the increased radius/diameter is the same for a ball the size of the earth as for a small ball held in your hand. Is this true? If yes, why is this true and by how much is the radius extended? AI: How about a proof without words? (Of course, one should imagine the limit as the number of segments tends to infinity.)
H: Reference: Finite Groups and Geometry The examples $C_n, D_n, A_4, S_4, A_5$ are the first nice examples of groups to relate with 2-D and 3-D Euclidean geometry. These groups can be investigated by studying action on sphere as described in M. Artin's book on Algebra. Next, Fano plane is another nice object to relate the simple group of order 168 with (finite) geometry. Then, by considering the double cover $SU(2)$of $SO(3)$, we get few more finite groups (called double/binary groups). The good references to study these groups and geometry (as far as I saw) are Geometries and Groups (Viacheslav, Shafarevich) Groups: A path to Geometry (R. P. Burn) Groups and Geometry (P.M. Neumann, etc.) On Quaternions and Octonions (Conway, Smith). Can one suggest more references (books, notes, articles) to see more finite groups with geometry? AI: Non-euclidean tessellations and their groups by William Magnus. For example, the author shows that $T^(\ell,m,n)=\langle a,b,c \| (ab)^{\ell}=(bc)^m=(ac)^n=1 \rangle$ is the group of symmetries of a triangular tessellation on the plan if $\displaystyle \frac{1}{\ell}+ \frac{1}{m} + \frac{1}{n} =1$, on the hyperbolic plane if $\displaystyle \frac{1}{\ell}+ \frac{1}{m} + \frac{1}{n} <1$ or on the sphere if $\displaystyle \frac{1}{\ell}+ \frac{1}{m}+ \frac{1}{n}>1$. Thus, $T^(\ell,m,n)$ is finite iff $\displaystyle \frac{1}{\ell}+ \frac{1}{m}+ \frac{1}{n}>1$!
H: Definition of $\operatorname{span}S$ and equivalence to $S$ I'm getting a little confused with the notation for spanning sets. In our notes we have the following definition. Let $S$ be a vector space over the field $F$. Then: $$\operatorname{span}S = \left\{\sum\limits_{i=1}^n a_iv_i:a_i \in F, v_i \in S\right\}$$ We say that $S$ spans $V$ if $V = \operatorname{span}S$ From this definition it seems that a spanning set is the set of all vectors in a vector space. Later in the notes we then say that the set $ \{(1,0), (0,1) \} $ spans $\mathbb{R}^2$. I understand why that set spans $\mathbb{R}^2$ but am getting a bit confused by the different ways span is used. Is $\operatorname{span}S$ the set of all possible vectors in the vector space $S$, hence equivalent to $S$? Also, I'm asked to determine if $\{1+x, x^2 \}$ spans $P_2(\mathbb{R})$. I'm thinking no because there is no way to get a polynomial such that $x$ and $x^0$ have different coefficients. Is this correct? AI: The problem here comes right at the beginning: the first sentence of the definition should read Let $S$ be a subset of a vector space $V$ over the field $F$. Since $S$ can be any subset of $V$, $\operatorname{span}S$ clearly need not be all of $V$, and if it’s not, then $S$ does not span $V$. For instance, if $S=\{(1,1)\}\subseteq\Bbb R^2$, then $$\operatorname{span}S=\{a(1,1):a\in\Bbb R\}=\{(a,a):a\in\Bbb R\}\;;$$ pictorially, $S$ is the graph of $y=x$, which is certainly not all of $\Bbb R^2$. Your reasoning in your last paragraph is correct. Added: Here’s another way of looking at it that may be helpful. Suppose that $S$ is any old set of vectors in some vector space $V$ over a field $F$. $S$ certainly need not be a subspace of $V$, because it needn’t be closed under vector addition and scalar multiplication. We might ask, therefore, what is the bare minimum that needs to be added to $S$ to get a subspace of $V$. If $v\in S$ and $a\in F$, we’ll have to have $av$ in order to get closure under scalar multiplication. That means that if $v_1,\dots,v_n\in S$ and $a_1,\dots,a_n\in F$, we’ll have to have $a_1v_1,\dots,a_nv_n$, and then to get closure under vector addition we’ll have to have $a_1v_1+\ldots+a_nv_n$. In other words, we’ll have to have every vector in the set $$\left\{\sum_{k=1}^na_kv_k:n\ge 0\text{ and }v_1,\dots,v_n\in S\text{ and }a_1,\dots,a_n\in F\right\}\;.$$ It turns out that this is it: once we have all of these vectors, we actually have a subspace of $V$. (Proving this is the exercise that I mentioned in the comments.) This subspace of $V$ is the smallest subspace of $V$ that contains the set $S$, so we call it the span of $S$, written $\operatorname{span}S$, and say that $S$ spans it.

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