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H: Let $A$ be $2\times 2$ Orthogonal matrix such that det$A= -1$. Show that $A$ represents reflection about the line in $R^2$. Every matrix is corresponding to a linear transformation. Reflection is a linear transformation called $T$. So $T^2=I$. Also modulus of eigen values of $A$ is $1$.Since it has determinant $-1$ so odd no eigen values are $-1$. I am stucking here. I can't understand how to proceed further AI: The general form for such a matrix looks like this $$M = \begin{bmatrix}\cos \theta &\sin \theta \\ \sin \theta &-\cos \theta\end{bmatrix}$$ This matrix is orthogonal, has determinant -1. Now what does it do to a point $(x,y)$? $$M(x,y) = (x\cos\theta + y\sin\theta, x\sin\theta-y\cos\theta)$$ This point is actually the reflection of point$(x,y)$ about the line making angle $\theta/2$ with x axis. I got to this conclusion by plugging in values of $\theta$, but you can try proving it by proving them equivalent. This is just a hint on how to proceed
H: Why does this sequence generate all numbers in 2^n? Here's a sequence, for fixed $n \in \mathbb{N}$: $$ j_{i+1} = (5 j_i + 1) \pmod{ 2^n} $$ This update rule can be simplified to: $$ j_{i} = \frac{5^i - 1}{4} \pmod{ 2^n}$$ One can easily verify the above via proof by induction. Now when you start with $j_1 = 1$, you will eventually generate all numbers in the range $[2, 2^i]$ and then come back to 0 (in which case the term after is 1 again). I wrote a quick python program to verify this: ctr = 1 j = 1 exp = 20 # This is the n in 2^n while j: j = ((5j) + 1) % 2exp ctr += 1 print("{} elements found.".format(ctr)) And up to exp = 25 this works. Running the above gives us 1048576 elements found. I have a bit of knowledge in group theory, and from what this looks like it's similar to the intuition of cyclic groups. However, I can't come up with a plausible explanation for why this is true. Any help would be greatly appreciated! EDIT: I also wanted to verify that indeed there were no duplicates, so I used a hashset for checking up to $2^{16}$ elements. ctr = 1 j = 1 exp = 16 seen = set([1]) while j: j = ((5j) + 1) % 2**exp ctr += 1 if j in seen: print("DUPLICATE FOUND") seen.add(j) print("{} elements found.".format(ctr)) And it appears that there were no duplicates either. AI: EDIT: This answer is wrong. See KCd's comment on the main question for a correct solution.
H: The series $\sum_{n=1}^\infty nc_n (x-a)^{n-1}$ converges uniformly to $f'$ on the interval $[a-r, a+r]$ (Tao Vol.2, P.77) Let $\sum_{n=0}^\infty c_n(x-a)^n$ be a power series, and let $R$ be the radius of convergence. Suppose that $R>0$ (the series converges at least one other point than $x= a$). Let $f : (a- R, a+ R) \to \mathbb{R}$ be the function $f(x) : = \sum_{n=0}^\infty c_n (x-a)^n$. I need to prove the following statement: (Differentiation of power series) The function $f$ is differentiable on $(a- R, a+R)$, and for any $0<r <R$, the series $\sum_{n=1}^\infty nc_n (x-a)^{n-1}$ converges uniformly to $f'$ on the interval $[a-r, a+r]$. The author suggests to use the following theorem: Theorem 3.7.1. Let $[a,b]$ be an interval, and for every integer $n \ge 1$, let $f_n : [a,b] \to \mathbb{R}$ be a differentiable function whose derivative $f'_n:[a,b] \to \mathbb{R}$ is continuous. Suppose that the derivatives $f'_n$ converge uniformly to a function $g: [a,b] \to \mathbb{R}$. Suppose also that there exists a point $x_0$ such that the limit $\lim_{n \to \infty} f_n(x_0)$ exits. Then the function $f_n$ converge uniformly to a differentiable function $f$, and the derivative of $f$ equals $g$. Let $F_N(x) = \sum_{n=0}^N c_n(x-a)^n$ for $x\in[a-r, a+r]$. $F'_N(x) = \sum_{n=1}^{N} nc_n(x-a)^{n-1}$ is continuous. I need to show that $(F'_N (x))_N$ converges uniformly to $g$. Then, by Theorem 3.7.1, $(F_N(x))_N$ converges uniformly to $f$, and $f$ is differentiable and $\sum_{n=1}^{\infty} nc_n(x-a)^{n-1} =f'(x) = g(x)$. To show uniform convergence, it suffices to show that $\sum_{n=1}^\infty \|\|nc_n(x-a)^{n-1}\|\|_{\infty}$ converges. $\|\|nc_n(x-a)^{n-1}\|\| _\infty = \sup\{\|nc_n\|\|x-a\|^{n-1} :x \in [a-r, a+r]\} = \|nc_n\|\|r\|^{n-1}.$ I am trying to prove the convergence of $\sum_{n=1}^\infty\|nc_n\|\|r\|^{n-1}$ using the root test. Thus, to finish the proof, I have to show that $$\lim \sup_{n\to \infty}\|n\|^{1/n}\|c_n\|^{1/n} \|r\|^{(n-1)/n} <1.$$ I don't know how to prove this inequality. I know that $\|n\|^{1/n} \to 1$ and $\lim \sup \|c_n\|^{1/n} = 1/R$. Am I using the right approach? How can I finish the proof? I would appreciate if you give some help. AI: Let $r<r'<R$. Then $\lim \sup \|c_n\|^{n} =\frac 1 R<\frac 1 {r'}$ so there exist $n_0$ such that $\|c_n\|^{n} <\frac 1 {r'}$ for all $n \geq n_0$. Hence $\lim \sup n^{1/n}\|c_n\|^{1/n}r^{1-\frac 1 n} =\lim \sup \|c_n\|^{1/n}r^{1-\frac 1 n} \leq \frac 1 {r'} r <1$.
H: Evaluating $\int\frac {dx}{1+x^2}$ Evaluate $$\int\frac{dx}{1+x^2}$$ Please help me find my mistake. I have integrated $\frac {1}{1+x^2}$ and gotten the correct result by making a mistake in the substitution. I imagined a triangle, with $1 = \cos\theta$ and $x = \sin\theta$ I then integrated $d\theta$ and got $\arctan$ a numerical result as this was a definite integral. While the result was correct, I realized that I should have substituted $dx$ for $\cos \theta$ but if I do this I get the wrong result. I think I am making a mistake in the substitution thinking. Thank you! AI: The folly: When you put $1=\cos\theta$, you are fixing the values of $\theta$ to be equal to $2n\pi+\frac{\pi}{2}$, where $n$ is an integer. Fixing it: If you do know the following, well and good, else another method follows: $$\frac{d}{dx}\arctan x=\frac{1}{1+x^2}$$ The other method: Substitute $x=\tan\theta$. Or, $dx=\sec^2\theta\cdot d\theta$. So the integral becomes: $$I=\int\frac{dx}{1+x^2}=\int\frac{\sec^2\theta\cdot d\theta}{1+\tan^2\theta}$$ Hence, $$I=\int\frac{\sec^2\theta}{\sec^2\theta}d\theta=\int d\theta$$ Can you finish?
H: Is the series $\frac{\sin (an)}{\ln(n+1)}$ convergent? Is the series $\displaystyle \sum\frac{\sin (an)}{\ln{(n+1)}}$ convergent? ( $a\in \mathbb{R}$ ) I know that $\displaystyle \sum\frac{1}{\ln{n}}$ diverges. I am trying to use the comparison test but fails to come up with a good result. AI: $\sum a_n \sin (nx)$ converges for every $x$ whenever $a_n$ decreases to $0$. This is proved in books on Fourier series using the idea of 'summation by parts'. Ref.: See p. 112 of Fourier Series' by Edwards R E.
H: Why is the unique mapping from $\emptyset$ to $Y$ inclusion? I am sorry, but I edited my question several times. I am reading "Set Theory and General Topology" by Takeshi Saito(in Japanese). The author wrote as follows: When $X$ is a subset of $Y$, a mapping from $X$ to $Y$which maps $x \in X$ to $x \in Y$ is called inclusion. When $X$ is the empty set, the inclusion $\emptyset \to Y$ is the unique mapping from $\emptyset$ to $Y$. $\forall x(x \in \emptyset \implies \text{every element } x \in \emptyset \text{ is mapped to } x \in Y \text{ by the mapping } \emptyset \to Y)$. This is vacuously true. But I think the following is also vacuously true: $\forall x(x \in \emptyset \implies \text{every element } x \in \emptyset \text{ is mapped to } y \in Y \text{ such that } y \neq x \text{ by the mapping } \emptyset \to Y).$ So, the mapping $\emptyset \to Y$ is both inclusion and non-inclusion. Then, the author could write as follows: When $X$ is a subset of $Y$, a mapping from $X$ to $Y$which maps $x \in X$ to $x \in Y$ is called inclusion. When $X$ is the empty set, the non-inclusion $\emptyset \to Y$ is the unique mapping from $\emptyset$ to $Y$. But the author wrote: the inclusion $\emptyset \to Y$ is the unique mapping from $\emptyset$ to $Y$. Why? AI: Because $\emptyset$ is a subset of $Y$ and the unique application maps every element of the $\emptyset$ in itself. Clearly this requirement is void since the empty set has no element, so it is automatically satisfied. I don't know if it's useful but a way you can define a function is this: A function from $X$ to $Y$ is a subset $F$ of the cartesian product $X \times Y$ such that $$\forall x \in X \exists ! y \in Y \| (x,y) \in F$$ (To be a bit more precise here I'm defining the graphic of a function, but if you fix domain and codomain, it's the same thing) With this formalism you can define an inclusion mapping from $X$ to $Y$ as: $$\Delta_X=\{(x,x)\|x \in X\}$$ Where $ \Delta_X$ is called the diagonal of $X$. If $X=\emptyset$: clearly: $$\Delta_{\emptyset}=\emptyset$$ And this defines the immersion of $\emptyset$ in $Y$ Regarding the edit, yes they are all vacuosly true. It's even true that $x \in \emptyset$ is mapped into a tomato or a cow. The fact that the second one is true doesn't imply that tue first one is not true. I'll say you more, if you consider $f: \emptyset \to \mathbb{R}$ , this function is technically increasing and decreasing at the same time. The empty sets element satisfy any property(except existence) because every property becomes null. On the second edit. The definition of inclusion is formally: $\forall x \in X, f(x)=x$ Now clearly: $$\forall x \in \emptyset, f(x)=x$$ So $f$ is an inclusion. Now you are saying tha since: $$\forall x \in \emptyset, f(x)=tomato$$ Then $f$ is a non-inclusion. But this is technically wrong. The latter proposition isn't the negation of the former. The negation of the former is: $$\exists x \in \emptyset \| f(x)≠x$$ The "tomato" proposition doesn't imply this one if the set is void.
H: Solve the integral $\int_1^3\!\sqrt{x-\sqrt{x+\sqrt{x-...}}}\,\mathrm{d}x$ As an extension to my discussion in one of the answers to my previous question on simplifying the integrand, I'd like to evaluate the following integral: $$\int_1^3\!\sqrt{x-\sqrt{x+\sqrt{x-...}}}\,\mathrm{d}x$$ The above radical, when solved, yields 4 possible solutions:$$1) y=\frac{1}{2}(-\sqrt{4x-3}-1)\\2)y=\frac{1}{2}(\sqrt{4x-3}-1)\\3)y=\frac{1}{2}(1-\sqrt{4x+1})\\4)y=\frac{1}{2}(\sqrt{4x+1}+1)$$ Definitely, only one of these solutions has to be considered as an integrand. Since the limits of integration are positive(and square roots are involved), I suspect that the integrand must be positive as a whole. Thus, solutions $(1$) and $(3)$ are ruled out. However, I cannot decide which expression amongst $(2)$ and $(4)$ is legitimate. It was brought to my attention that this involves the notion of convergence, a concept I'm not yet completely familiar with(I have a naive understanding of convergence in infinite geometric series). Thus, I'd like to know: Which of the above 4 solutions to the radical is legitimate for solving this integral, and why? AI: I found the problem. If we define the sequence $$f_n(x)=\sqrt{x−\sqrt{x+f_{n−1}(x)}}$$ with $f_0(x)=0$ (which is the only subsequence such that $x=1$ is not complex valued) then values like $1.1$ immediately become imaginary after two iterations. The integral can be done, but this function would probably be very complicated to get an analytical form because it deviates eggregiously from the square root shape. As a practical matter, this integral probably cannot be done by hand, and definitely not as a real valued function at all, which means the original sequence was ill defined. If the integral were from $2$ to $3$, instead, then it would neatly converge to option $2$.
H: How does WolframAlpha solve this recursion? I have the following recursion: $$x_n=\frac{n-1}{n}x_{n-1}+\frac{1}{n}\left\lfloor\frac{n}{2}\right\rfloor.$$ WolframAlpha gives a solution to this recursion as $$x_n=\frac{C_1+\left\lceil\frac{1-n}{2}\right\rceil^2-\left\lceil\frac{1-n}{2}\right\rceil+\left\lfloor\frac{n}{2}\right\rfloor^2+\left\lfloor\frac{n}{2}\right\rfloor}{2n},$$ for some arbitrary parameter $C_1$. How is this solution found? Moreover, is this solution unique? AI: If you set $y_n = n x_n$ then the recursion is $y_n = y_{n-1} + \left\lfloor\frac n2\right\rfloor$ which may in turn be simplified by considering even and odd $n$ separately. All you need to know then is that $\sum_{k=1}^n k = \frac{n^2+n}2$. The parameter $C_1$ appears because no initial value for the recursion is given. As for the question of uniqueness, once you are given, e.g. $x_0$ or $x_1$, the whole sequence is fixed, so the correctness of the solution implies it is the unique solution.
H: Definition of vertices of a polytop in $\mathbb{R}_{+}^d$ Having the following set: $V=\left\{v\in{}\mathbb{R_{+}^{d}}\hspace{2pt}:\hspace{2pt}\sum_{i=1}^dv_i=K\hspace{4pt}\text{and}\hspace{4pt}\forall{}v_i\hspace{2pt},v_i\leq{}1\right\}$. That is, all the vectors with entries between 0 and 1 that sum up to $K\leq{}d$. I'm trying to show that $V$ is a polytop whose vertices are of the form $(1,1,...,1,0,...,0)$,$(0,1,0,0,1,...,0)$. i.e. all the vectors with $K$ ones and $N-K$ zeros. I would like to show that this is exactly the group of vertices, that is, all the vertices are of that form and all the vectors from that form are vertices of $V$. I'm struggling with that because I'm trying to figure out the definition of a vectice and not sure how those satisfy it. AI: For convex polytopes like the set $V$ you have, the definition of a vertex can be stated in an interesting linear-algebraic way: A point $v \in V$ is a vertex if and only if, for all $x, y \in V$ and $t \in (0,1)$, if $tx + (1-t)y = v$ then $x = y = v$. In general, this is the definition of an extreme point of a convex set $V$, and when $V$ is a convex polytope, the extreme points are called vertices. Edit: also, the set of extreme points of a compact polytope in $\mathbb{R}^n$ is the minimal set (by inclusion) whose convex hull is the polytope – this is a consequence of the Krein-Milman Theorem.
H: How to prove that $\|f\|\leqslant\\|f\\|_\infty$ almost everywhere? Let $f\in\mathcal L^\infty(X,\Sigma,\mu)$, where $(X,\Sigma,\mu)$ is a measure space, and $\mathcal L^\infty(X,\Sigma,\mu)$ is the set of essentially bounded functions. That is, $\\|f\\|_\infty:=\inf\{c>0:\mu\{x\in X:\|f(x)\|>c\}=0\}<\infty$. How can we prove that $\|f(x)\|\leqslant\\|f\\|_\infty$ almost everywhere? It almost seems obvious, since this condition is equivalent to $\mu\{x\in X:\|f(x)\|>\\|f\\|_\infty\}=0,$ which appears in the definition of $\\|\cdot\\|_\infty$ with $c\to\\|f\\|_\infty$. However, $\\|f\\|_\infty$ is the infimum of the set satisfying this condition. And there is in general no need for an infimum of a set $A$ to satisfy the condition defining $A$. For example, the infimum of $\{x\in\mathbb R:x>0\}$ certainly does not satisfy $x>0.$ AI: There exist $c_n$ decreasing to $\\|f\\|_{\infty}$ such that $\mu \{x: \|f(x)\| >c_n\}=0$ for all $n$. Let $E$ be the union of the sets $\{x: \|f(x)\| >c_n\}$. Then $\mu (E)=0$ and $x \notin E$ implies $\|f(x)\| \leq c_n$ for all $n$. Letting $n \to \infty$ we see that $\|f(x)\| \leq \\|f\\|_{\infty}$ whenever $x \notin E$.
H: Solving : $ \left(1+x^{2}\right) \frac{d y}{d x}+2 x y=4 x^{2} $ Solve the following ODE : $$ \left(1+x^{2}\right) \frac{d y}{d x}+2 x y=4 x^{2} $$ After rearrangement I get : $$\frac{dy}{dx}=\frac{4x^{2}-2xy}{1+x^{2}}$$ please help me after this step. AI: $\newcommand{\d}{\mathrm{d}}$ This is a linear DE. Rearranging this we have that \begin{align} &\frac{\d y}{\d x} + \frac{2x}{1 + x^2}\cdot y = \frac{4x^2}{1 + x^2}\\ \implies&\frac{\d y}{\d x} + P(x)\cdot y = Q(x) \end{align} Now the integrating factor is $$ I = e^{\int{{2x}/{1 + x^2}}dx} = e^{{{\ln(1 + x^2)}}} = 1 + x^2$$ On multiplication by the integrating factor and on integrating we obtain this $$y \cdot I = \int Q(x) \cdot I dx + C $$ And on substitution of $I$, we have that \begin{align} &~y (1 + x^2) = \int 4x^2 + C\\ \implies &~ y \cdot (1 + x^2) = \frac{4}{3} \cdot x^3 + C \end{align}
H: Probability of guessing everything wrong I was doing some AP Biology practice, and noticed that instead of Bio, I was learning more about probability. Tests have A B C and D, say you could guess one, get an answer whether it was right or wrong, guess again, and again and again, until eventually you got it right. For guessing the first one right, the probability is just $\frac{1}{4}$, guessing the first wrong and second right is $\frac{3}{4} \cdot \frac{1}{3} = \frac{1}{4}$, the first two wrong and third right, $\frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} = \frac{1}{4}$, and in the case of all everything wrong, the same value, as it is the same equation. Its just interesting, that all these values are the same, even though intuitively it seems like getting the first one right or getting them all wrong would be the more likely one, just wondering if there is a good explanation for why this happens, as it really surprised me to actually look at the numbers. AI: Let us say that even after guessing the correct answer, we keep on guessing until we have guessed all possible answers. Obviously, this does not alter the probability. The question become “what is the probability that the correct answer is the first, second, third, fourth guess?”. The answer is of course $\frac{1}{4}$
H: $2310 $ points are evenly marked on circle.How many regular polygons can be drawn by joining some of these points. Question:- $2310$ points are evenly marked on circle.How many regular polygons can be drawn by joining some of these points($N$ -sided polygon should be counted only once). I came across this question in my assignment.I wasn't able to solve this question since I don't know method to tackle such kind of problems involving polygons.Can somebody tell me what is the method to solve this problem! Thank you for your help!! $ $ $ $ AI: Hint: What is a number of all divisors of 2310? If $m = p_1^{n_1} p_2^{n_2}... p_k^{n_k}$ then the number of all divisors is: $$(n_1+1)(n_2+1)...(n_k+1)$$
H: Covering spaces and fibre bundles, using the of $SU(2)$ and $SO(3)$ I understand there are many questions on here that show give an explicit map to show that $SU(2)$ is a double cover of $SO(3)$ (seevia quaternions). I am trying and use the fact that $SU(2)$ is a double cover of $SO(3)$ to write $SU(2)$ as some fibre bundle. But I seem to encounter some contradictions. I want to know what is incorrect. Here are my ideas: (1) Since the Lie algebras of $SO(3)$ and $SU(2)$ are isomorphic. They have isomorphic connected components. (2) Covering spaces can always be thought of as discrete fibre bundles over the base space. (3) But $SU(2)$ is simply connected. I think (2) is wrong. (4) If two is wrong what is the correct way to think about covering spaces. As far as I understand $SU(2)$ has one connected component, and $SO(3)$ has two connected components. So it seems like maybe $SO(3)$ should be a double cover of $SU(2)$? AI: (1) Is wrong. The kernel of $SU(2)\to SO(3)$ is $\{\pm I\}$. The element $-I$ is central in $SU(2)$, but $SO(3)$ has trivial center, so they cannot be isomorphic. (Both $SU(2)$ and $SO(3)$ are connected.) (2) True. (3) True. (4) It's wrong to say $SO(3)$ has two connected component. It's true to say $SO(3)$ is the identity component of $O(3)$, the unique other component being comprised of reflections (and "improper reflections").
H: Expected Utility Maximization This is from Markowitz's Risk-Return Analysis: The Theory and Practice of Rational Investing (Volume One) Chapter 1. Suppose, for example, that a decision maker can choose any probabilities $p_0$, $p_1$, $p_2$ that he or she wants for specified dollar outcomes $D_0$ < $D_1$ < $D_2$ and that they have a given expected value $p_0$$D_0$ + $p_1$$D_1$ + $p_2$$D_2$ = $k$ For example, if $D_0$ < 0 were the price of a lottery ticket with possible prizes $D_1$ and $D_2$, then $k$ = 0 would define a “fair” lottery, while $k$ < 0 would afford the lottery organizer a profit. We may arbitrarily let the utilities of $D_0$ and $D_2$ be $u_0$ = 0 and $u_2$ = 1; then the utility of $D_1$ is $u_1$ ∈ (0,1). For a typical lottery, \|$D_0$\| is quite small as compared to $D_1$ and $D_2$. With $k$ ≤ 0, this implies that feasible $p_1$ and $p_2$ are small, with $p_1$ + $p_2$ well under 0.5, and therefore with $p_0$ well over 0.5. My Questions: If $D_0$ is the price of a lottery ticket, how could it possibly be less than zero? Why include the price of a lottery ticket in an EV calculation? The prizes $D_1$ and $D_2$ have a probability associated with them, that makes sense when calculating expected value. But the price of a lottery ticket? What does it mean for a ticket price to have a probability "well over 0.5" For $k$ ≤ 0, it only makes sense that $D_0$ must be negative, but again, how could the price of a lottery ticket be negative? What am I not understanding here? If $D_0$ were a positive return with a given probability, then it wouldn't be possible for $k$ to be less than zero, in which case, how would one define a "fair" or a profitable lottery? I'm so confused. Surely I'm reading this wrong. AI: What you're not understanding is the fact that we're calculating your expected winnings. So what your author means is that if you buy a lottery ticket at a price $\|D_0\|$ (which is positive), then if you don't win the lottery you'll have earned $D_0$ (which is negative), your overall wealth decreased by $\|D_0\|$. So it seems like we're modelling the scenario where a lottery ticket costs $\|D_0\|$ and with probability $p_1$, I gain some amount money $M_1>0$. In this case, I've paid for the ticket and won $M_1$, so my gains have been $D_1:=M_1+D_0$ (note again that $D_0$ is negative). Similarly, with probability $p_2$, I gain some greater amount of money $M_2>0$, in which case I'll have gained $D_2:=M_2+D_0$ - I've still paid for my ticket. Thus, with probability $p_0:=1-(p_1+p_2)$, I'll have entered the lottery and not won, and thus lost the money that I paid for the ticket. Therefore, my expected winnings are as written above. The important part is to keep in mind which quantity we're actually keeping track of - in this case, it's your wealth after the lottery is over.
H: $b^* a^* ab \leq \Vert a\Vert^2 b^* b$ in a $C^$-algebra. Let $A$ be a $C^$-algebra and $a,b \in A$. In a proof I'm reading the following is claimed: $b^* a^* ab \leq \Vert a\Vert^2 b^* b$. I want to understand this: Here is my reasoning: we view $A \subseteq \tilde{A}$ with $\tilde{A}$ the unitisation of $A$. Then we know that $a^* a \leq \Vert a^* a \Vert 1$ since this holds in every unital $C^$-algebra (by a Gelfand-representation argument). Then $$b^ a^a b \leq b^ \Vert a^* a \Vert 1 b = \Vert a \Vert ^2 b^* b$$ Is the above correct? I find arguments with unitisations always a bit tricky. AI: Yes, your argument is correct. Here is another version, using the GNS representation theorem: If $a,b$ are bounded linear operators on the Hilbert space $H$ and $\xi\in H$, then $$ \langle b^\ast a^\ast a b \xi,\xi\rangle=\lVert ab \xi\rVert^2\leq \lVert a\rVert^2\lVert b\xi\rVert^2=\lVert a\rVert^2\langle b^\ast b\xi,\xi\rangle. $$ Thus $b^\ast a^\ast a b\leq \lVert a\rVert^2 b^\ast b$.
H: Jacobian of a system of equations including an ODE for Newton-Raphson I want to use the Newton-Raphson method to solve a system of equations and in order to do so, I need to calculate the Jacobian. The system given describes the stages (called $k$) of an implicit numerical method, which I want to solve for using Newton-Raphson: $0 = - \binom{k_1}{k_2} + f\left(\binom{y_{n1}}{y_{n2}} + h \left( B_{11} \binom{k_1}{k_2} + B_{12} \binom{k_3}{k_4} \right) \right) \\ 0 = - \binom{k_3}{k_4} + f\left(\binom{y_{n1}}{y_{n2}} + h \left( B_{22} \binom{k_3}{k_4} + B_{21} \binom{k_1}{k_2} \right) \right)$ where the ODE $f$ is the Van der Pol oscillator $ \ddot{x}(t) - \mu (1-x^2(t)) \dot{x}(t)+x(t)=0$, $y_{n}$ are approximations computed by the numerical method (always known), $h$ is the stepsize (scalar) and $B_{ij}$ are also constant scalars (entries of a 2x2 matrix). To solve for the four entries of the k vector numerically, the system's Jacobian is required and I could not figure it out correctly. I expect it to look something like this: $ J = -I_4 + h B J_f$ where $I_4$ denotes the identity matrix of dimension 4, $h$ and some matrix form of $B$ might be needed due to the chain rule (really not sure) and $J_f$ is the Jacobian of the vdP oscialltor: $$ J_f = \left[ \begin{matrix} 0 & 1 \\ -2 \mu x \dot{x}-1 & \mu(1-x^2) \\ \end{matrix} \right] $$ I am mostly unsure concerning four things: Do I apply the chain rule? What is the argument of $J_f$? How can I get the correct dimensions ($J$ should be 4x4 I think). Maybe by some multiplication of $B$ and $J_f$? I need to plug in my guess somewhere in the Jacobian, so the four entries of the vector $k$ must appear somewhere. In case it helps: I use MATLAB to run the whole numerical method and also to compute the Newton-Raphson inside the method. AI: The general idea of Newton-like methods is that you approximate the equation by some linearization. This linearization does not need to be very exact to still get a reasonably fast convergence. Here just iterating the stage equations as a fixed-point iteration gives linear convergence with a factor $O(h)$. If the linearization is also $O(h)$ exact, then the resulting Newton-like iteration will still be linear, but now with a factor $O(h^2)$. Depending on how the initial guess, the predictor, is computed, this can result in a drastic reduction of required corrector steps in the Newton-like iteration to get to an error level that is below the method step truncation error. $\newcommand{\D}{\mathit{\Delta}}$ So if $J$ is the Jacobian at $y_n$, then you can decompose $f(y+\D y)=f(y)+J\D y+R(\D y)$. For the given 2-stage method this gives the system $$ \vec k_1-hJ(B_{11}\vec k_1+B_{12}\vec k_2)=f(\vec y_n)+R(h(B_{11}\vec k_1+B_{12}\vec k_2)) \\ \vec k_2-hJ(B_{21}\vec k_1+B_{22}\vec k_2)=f(\vec y_n)+R(h(B_{21}\vec k_1+B_{22}\vec k_2)) $$ In an iteration step the left side is a linear system for the new values, while the right side is computed from the old values. This form demonstrates that the right side is constants plus small superlinear corrections, in the practical computation the right side will be computed as $f(y)+R(\D y^{old})=f(y+\D y^{old})-J\D y^{old}$. \begin{align} \pmatrix{I_2-hB_{11}J&-hB_{12}J\\-hB_{21}J&I_2-hB_{22}J} \pmatrix{\vec k_1\\\vec k_2} = \pmatrix{f(y+\D_1 y^{old})-J\D_1 y^{old}\\f(y+\D_2 y^{old})-J\D_2 y^{old}} \end{align} Using a Kronecker product, the matrix on the left could also be written as $I_4-hB\otimes J$.
H: $N\subseteq G_x$ then is $N$ in the kernel of the group action Let $G$ be a group that acts transitive on $X$. Show that if $N$ is a normal subgroup in $G$ and for a $x\in X$ holds $N\subseteq G_x$. Then lies $N$ in the kernel of the group action. [This is one part of a task in a textbook, so that $G$ acts transitive might not be needed.] My proof goes like this: A group action $G\times X\to X$ can be translated into a group homomorphism $\varphi: G\to\operatorname{Sym}(X), g\mapsto \tau_g$ (left translation, so $\tau_g(x)=gx$). We want to show $N\subseteq\ker\varphi=\{g\in G: \tau_g(x)=x\quad\text{for every}\,\,x\in X\}=\{g\in G: gx=x\quad\text{for every}\,\, x\in X\}$. Here I am unsure if it is actually correct to add 'for every $x$ in $X$". I thought so, because $g\in\ker\varphi$ if $\tau_g$ is the identity. So it has to hold $gx=x$ for every $x\in X$. We have that for some $y\in X$ it is $N\subseteq G_y$. Let $n\in N$. Since $n\in G_y$ we have $ny=y$. Also $G$ acts transitive. So $Gx=\{gx: x\in X\}=X$. Which means for every $x\in X$ we have a $g_x\in G$ with $g_xx=x$. A proof of the statment should now go something like this: $nx=ng_xx\stackrel{\text{N normal}}{=}g_xnx\stackrel{?}{=}g_xx=x$ Where I do not know how to justify this last equality from the assumption (if possible) or neither do I know if I really need that $G$ acts transitive here. Do you mind checking on my thoughts, and correct them if necessary? Thanks in advance. AI: Transitivity is crucial here. For example, consider action of $G$ on $G\cup\{p\}$ which is the translation on $G$ but fixes $p$. Then $G_p=G$ contains all its normal subgroup,s but obviously a nontrivial normal subgroup cannot fix anything else. I'll leave you to prove $G_{gx}=gG_xg^{-1}$. Then $N=gNg^{-1}\subseteq G_{gx}$ for all $g\in G$, so $N\subseteq\bigcap_{g\in G}G_{gx}=\ker(G\to S_X)$ by transitivity.
H: How to evaluate $ \:\sum _{n=3}^{\infty \:}\frac{4n^2-1}{n!}\:\: $? I am trying to evaluate: $$ \:\sum _{n=3}^{\infty \:}\:\:\frac{4n^2-1}{n!}\:\: \quad (1)$$ My attempt: $$ \:\sum _{n=3}^{\infty \:}\:\:\frac{4n^2-1}{n!}\:\: \quad = 4\sum _{n=0}^{\infty \:} \frac{(n+3)^2}{(n+3)!} + \sum _{n=0}^{\infty \:} \frac{1}{(n+3)!}$$ The last form is very similar to the expotential series, but I can't get it from here. Any ideas? AI: Hint: $$ \frac{k^2}{k!} = \frac{k-1+1}{(k-1)!} = \frac{1}{(k-2)!} + \frac{1}{(k-1)!} $$
H: If $p(x)$ be the polynomial left as remainder when $x^{2019}-1$ is divided by $x^6+1$. What is the remainder left when $p(x)$ is divided by $x-3$? If $p(x)$ be the polynomial left as remainder when $x^{2019}-1$ is divided by $x^6+1$. What is the remainder left when $p(x)$ is divided by $x-3$? It is obvious that remainder is a constant. How do I solve it? AI: The other answers give cooler results, but keeping in mind the algebra-precalculus tag: $$x^{2019}-1=(x^6+1)(x^{2013}-x^{2007}+\cdots+(-1)^kx^{2013-6k}+\cdots+x^9-x^3)+(x^3-1)$$ $$\therefore x^3-1=(x-3)(x^2+3x+9)+26$$
H: How to interpret $P_{V^{\perp}}m$, where $P_{V^{\perp}}$ is an orthogonal projection and $m$ is a matrix? How would you interpret $P_{V^{\perp}}m$ in the following extract? Given is a vector $v\in \mathbb{R}^n$ and a symmetric tracefree $n \times n$-matrix $m$. Let $V^{\perp}$ be the orthogonal complement of the span of $v$ in $\mathbb{R}^n$ and $P_{V^{\perp}}$ be the orthogonal projection from $\mathbb{R}^n$ onto $V^{\perp}$. Since $m$ is self-adjoint, then so is the restriction of $P_{V^{\perp}}m$ to $V^{\perp}$. Hence there exists at least one eigenvector of the operator. I try to understand the last sentence. I read that every selfadjoint transformation $T:X \rightarrow X$ ($X$ is an Euclidean space) has a real eigenvextor. But I am not sure whether this is here applicable because I thought $P_{V^{\perp}}m$ is the composition $P_{V^{\perp}} \circ f: \mathbb{R}^n \rightarrow \mathbb{R}^n \rightarrow V^{\perp}$ for a linear map $f(w) = mw$. AI: The restriction of $P_{V^{\perp}} \circ f$ to $V^{\perp}$ is well defined as for $x \in V^{\perp}$, $m(x) \in \mathbb R^n$ and $(P_{V^{\perp}} \circ f)(x) \in V^{\perp}$. Then this restriction is self-adjoint as both $P_{V^\perp}$ and $f$ are self-adjoint. Therefore the restriction of $P_{V^{\perp}} \circ f$ to $V^\perp$ has an eigenvector. By the spectral theorem, it is even diagonalizable.
H: How to find the number of terms in the expansion of $(a+b+c)^2$? How to find the number of terms in the expansion of $(a+b+c)^2$? Can anyone help me in finding its formula with proof? AI: We basically have to divide the power of $2$ amongst the $ a , b, c$. Let the powers be $ x, y , z $. $ x+y+z = 2$ where x , y,z are integers. So the answer is $6$. Using the formula $C_{ r-1}^{n+r-1}$
H: Simplifying $\sum_{k=0}^{24}\binom{100}{4k}.\binom{100}{4k+2}$ How to evaluate the following series: $$\sum_{k=0}^{24}\binom{100}{4k}\binom{100}{4k+2}$$ What I have tried : Considering expansion of $\displaystyle (1+x)^n= \binom{n}0+ \binom n1 x + \binom n2 x^2+\cdots$ By this I can get easily the result : $$\sum_{k=0}^{100}\binom{100}{k}\binom{100}{100-k}=\binom{200}{100}$$ Which seems to be almost similar to $\displaystyle \sum_{k=0}^{24}\binom{100}{4k}\binom{100}{98-4k}$. I think what I should do next is substitute $i$ and $-i$ and add equations. What I am not sure is which ones to add. Any suggestion or other approaches? Much Appreciated. AI: Call the sum $S$. Using the symmetry of Pascal's triangle, $S$ equals $$\sum_{k=0}^{24}\binom{100}{4k}\binom{100}{98-4k}.$$ Putting $j=24-k$ gives $$S=\sum_{k=0}^{24}\binom{100}{96-4k}\binom{100}{2k+2}.$$ Therefore $$2S=\sum_{r=0}^{49}\binom{100}{2r}\binom{100}{98-2r}.$$ This is the $X^{98}$ coefficient of $F(X)^2$ where $$F(X)=\sum_{r=0}^{50}\binom{100}{2r}X^{2r}=\frac{(1+X)^{100}+(1-X)^{100}}{2}.$$ Then $$F(X)^2=\frac{(1+X)^{200}+2(1-X^2)^{100}+(1-X)^{200}}4.$$ I get $$8S=\binom{200}{98}-2\binom{100}{49}+\binom{200}{98}$$ etc.
H: Inequality like $AB \leq \frac{\epsilon}{2n}A^2 + \frac{n}{2\epsilon}B^2$ In this paper on page 18 there is an inequality that I cannot prove. I will only look at the term with $K$, since the term with $b$ follows analogously I think. So we have, using $L_{\sigma}M>0$ to ease the notation: $$ L_\sigma M \sum_{i=1}^n \Big(1 + \frac{L_\sigma M}{n}\Big)^{n-i} \int_{\frac{i-1}{n}}^{\frac{i}{n}} \|\|K(t) - K_{i-1}^{(n)}\|\|\;dt \leq \frac{\epsilon L_\sigma M}{2n} \sum_{i=1}^n \Big(1 + \frac{L_\sigma M}{n}\Big)^{2(n-1)} \\+ \frac{n L_\sigma M}{2\epsilon} \sum_{i=1}^n \Big( \int_{\frac{i-1}{n}}^{\frac{i}{n}} \|\|K(t) - K_{i-1}^{(n)}\|\|\;dt\Big)^2, $$ for any $\epsilon>0$ and for $n\geq 1$. Now I think that this does not make use of any of the properties of $K$ or anything else, so I think I need to prove the identiy (for all $1\leq i \leq n$) $$ A_iB_i \leq \frac{\epsilon}{2n} A_i^2 + \frac{n}{2\epsilon} B_i^2, $$ where $A_i=\Big(1 + \frac{L_\sigma M}{n}\Big)^{n-i}$ and $B_i = \int_{\frac{i-1}{n}}^{\frac{i}{n}}\|\|K(t) - K_{i-1}^{(n)}\|\|\;dt$. Now from above I know that $A_i = \Big(1 + \frac{L_\sigma M}{n}\Big)^{n-i} \geq 1$, I also know that $K^{(n)}\to K$, but I don't know how to use that. Something I noticed which might be useful: $$ A_iB_i \leq A_i^2 + B_i^2 $$ AI: If $A_i, B_i \in \mathbb{R}$, then this is literally Young inequality for products: $$ab \leq \frac{a^2}{2 \varepsilon} + \frac{\varepsilon b^2}{2} $$ https://en.wikipedia.org/wiki/Young%27s_inequality_for_products
H: How to find the $n$-th term of the series 1, 22, 333, 4444, 55555..........? How to find the $n$-th term of the series $1, 22, 333, 4444, 55555..........?$ Here the $i^{\text{th}}$ term is the concatenation of $i$ for $i$ times. for ef $13$ the term = $13$ concatenated $13$ times = $13131313131313131313131313$ AI: For a number of $n$ of $m$ digits, that quantity is $$n\cdot\sum_{k=0}^{n-1}10^{km}=n\cdot\frac {10^{mn}-1}{10^m-1}$$ The number of digits of $n$ is $m=\left\lfloor \log_{10}n\right\rfloor+1$, therefore $$a_n=n\cdot\frac{10^{\left(\left\lfloor \log_{10}n\right\rfloor+1\right)n}-1}{10^{\left\lfloor \log_{10}n\right\rfloor+1}-1}$$
H: Proof of the convergence of infinite series of positive terms. I would like to prove the fact that the series of positive terms converges. Precondition; $\sum_{n=0}^{\infty} a_n$ is the series of positive terms, and $\sum_{n=0}^{\infty} a_n$ converges. Problem; $\sum_{n=0}^{\infty} (a_n)^2$ converges. I tried to prove that $\sum_{k=0}^n a_k$ is Cauchy sequence $\implies \sum_{k=0}^n (a_k)^2$ is Cauchy sequence. However, I couldn't prove this. How can I prove that $\sum_{n=0}^{\infty} (a_n)^2$ converges? AI: Since $\sum_{n=0}^∞ a_n $ is convergent, we have $a_n \to 0$ , hence there is $N$ such that $a_n <1$ for $n >N$. Then we have $$0<a_n^2 < a_n$$ for $n>N.$ Can you proceed ?
H: About modified rock-paper-scissors game (The game that I am about to ask is not my original creation; it is from a Korean game program 'RunningMan'. Also, I am also Korean, and please understand my poor English.) The game is about a modified rock-paper-scissors game. 8 people are playing this game, and each of them have 3 cards written rock, paper or scissor. At each round, the referee(not one of the 8 people) shows a card written either rock, or paper,or scissor. 8 people then play one of rock, paper, scissor after seeing the referee's card. The least number of used cards(excluding the referee's card) get the points. If the least number of cards wins the referee's card, the person(or the people with the least number of cards) get 5 points. If they are the same as the referee's card, they get 3 points. If they lose by the referee's card, they get 1 point. Game is continued until there is a person who reaches 11 points. The number of people playing the game, the points given to the winning people, the ending point could be changed. (For example, when the referee shows the rock card, and 4 people show rock, 3 people show paper, 1 person shows scissor, the least used card is scissor, and since rock wins scissor, the person with scissor gets 1 point.) My question is, Is there a winning strategy to this game, or is there at least a partial winning strategy? I am not familiar with the game theory, and to me, the game seems pretty much complicated to have a winning strategy. Any help would be very nice. Thanks a lot in advance. AI: In a Nash equilibrium, suppose the other seven all play the same mixed strategy; Win $x\%$ of the time, Draw $y\%$ of the time and Lose $z\%$ of the time. Note $x+y+z=100\%$. Suppose, when player 8 plays Win, he is in the smallest group $p\%$ of the time; when he plays Draw, he is in the smallest group $q\%$ of the time; and when he plays Lose he is in the smallest group $r\%$ of the time. Note $p+q+r\gt100\%$ because there are sometimes two smallest groups. You can calculate $p,q,r$ as functions of $x,y,z$. It's going to be messy. For example, the probability the other seven play one Win and six Draws is $7xy^6$. That contributes to both $p$ and $r$ because Player 8 scores in that case if he plays either Win or Lose. Player 8's expected winnings are $5p$, $3q$ or $r$ depending on his choice of card. It is a Nash equilibrium if these three values are equal. Finding the right $x,y,z$ is a tricky numerical problem.
H: How to use Bars and Stars method for equations with more than 1 non unity coefficients? I know we can find the non negative integral solutions of the equation $x+y+z=24$ using Bars and Stars method. The same can be extended to provide the solutions for equations like $2x+y+z = 24$. But is there any way to find the non negative integral solutions of $3x + 2y + z = 24$ and subsequently a generalised formula of non negative integral solutions of $nx + py + qz = k$ where $n,p,q>1$? AI: I don't have an exact solution, but since the $d$-simplex $\sum_{i=1}^da_ix_i\le k$ has volume $k^d/(d!\prod_ia_i)$ for $a_i,\,k\ge0$, $\sum_{i=1}^da_ix_i=k$ should have $\sim k^{d-1}/((d-1)!\prod_ia_i)$ solutions.
H: For which odd number $k$ does $\ \varphi(n) \mid n-k \ $ has infinitely many solutions? The Lehmer-totient problem: For a prime number $\ n\ $ we have $\ \varphi(n)=n-1\ $. In particular, we have $\ \varphi(n) \mid n-1\ $. Is there a composite number $\ n\ $ with $\ \varphi(n)\mid n-1\ $ ? It is not known if there is any such composite $n$ but may be Lehmer-totient problem is the tip of the iceberg. I was running computations on variation of this problem and I observed that for every small odd $k$ such as $k = 3,5,7,\ldots 25$ there were only a couple of solutions of $\ \varphi(n) \mid n-k\ $ for $n < 10^{10}$. E.g. for $k = 3$, the only solutions so far are $n = 9, 195$ and $5187$. The data suggests that at most there are finitely many solutions for a given odd $k$. Question: Is there any fixed odd $k > 1$ such that $\ \varphi(n) \mid n-k\ $ has infinitely many solutions? AI: This is an open conjecture by R. L. Graham (see R. K. Guy, Unsolved problems in Number theory, Second ed., Springer-Verlag, p. 93): Conjecture: For all positive integers $k$ there are infinitely many $n$ such that $$ ϕ(n)\|(n − k) $$ The conjecture has been proved for $k = 0$, $k = 2^a$ for $a> 0$ and for $k = 2^a\cdot 3^b$ with $a,b>0$. For odd $k\ge 3$ the conjecture is still open.
H: Given $f$ is a Lebesgue measurable function and $\int_0^1 x^{2n}f = 0 ~~~ \forall n$ , then show that $f = 0$ a.e. Given $f$ is a Lebesgue measurable function and $\int_0^1 x^{2n}f\,d\mu = 0 \quad \forall n$, then show that $f = 0$ a.e. Of course, if it was given that $f \geq 0$ then this was pretty trivial. My attempt : By contradiction. Suppose $\exists A \subset [0,1]$ s.t. $\mu(A) >0$ and $f(x) \neq 0~~~~ \forall x \in A.$ Without loss of generality we may assume that $f(x) > 0~~~~\forall x \in A .$ And $\forall \epsilon>0 \exists $ a sequence of intervals such that $\mu(A \setminus I_n) <\epsilon$ $$\int_{[0,1]} x^{2n}f(x)\,d\mu = \int_Ax^{2n}f(x)\,d\mu + \int_{[0,1]\setminus A} x^{2n}f(x)\,d\mu$$ Now the first integral is strictly greater than 0. I was hoping I could use $x^{2n}$ to reduce $x^{2n}f$ on $I_n$, and so get the required contradiction. Can someone help further my attempt, or give any other hint to solve this question. Deepest appreciation in advance. AI: Here is a sketch of the proof. I believe you can fill in the details to complete the proof: Assumption. $f$ is integrable on $[0,1]$ and satisfies $\int_{0}^{1}x^{2n}f(x)\,\mathrm{d}x=0$ for all $n\geq 0$. Then by noting that the space of even polynomials on $[0,1]$ is dense in $C([0,1])$ w.r.t. the supremum norm, we have $\int_{0}^{1}\varphi(x)f(x)\,\mathrm{d}x=0$ for any continuous function $\varphi$ on $[0, 1]$. Now observe: Lemma. Let $E\subseteq[0,1]$ be measurable. Then there exists a sequence of $\varphi_n:[0,1]\to[0,1]$ of continuous functions such that $\varphi_n \to \mathbf{1}_E$ almost everywhere. Using this, we can find a sequence $\varphi_n : [0, 1] \to [-1, 1]$ of continuous functions such that $\varphi_n \to \operatorname{sign}(f)$ a.e. Then by the dominated convergence theorem, $$ 0 = \lim_{n\to\infty} \int_{0}^{1} \varphi_n(x) f(x) \, \mathrm{d}x = \int_{0}^{1} \lim_{n\to\infty} \varphi_n(x) f(x) \, \mathrm{d}x = \int_{0}^{1} \|f(x)\| \, \mathrm{d}x, $$ which is sufficient to conclude the desired claim. Addendum - Proof of Lemma. 1st Proof. Since $C([0,1])$ is dense in $L^1([0,1])$, there exists $(\psi_n)_{n\geq 1} \subseteq C([0, 1])$ such that $\psi_n \to \mathbf{1}_E$ in $L^1$. By passing to a subsequence if necessary, we can also assume that this convergence is pointwise a.e. Then the claim follows setting $\varphi_n = \max\{0, \min\{\psi_n, 1\}\}$. $\square$ In case OP is not familiar to $L^p$-theory, here is a more elemtary proof: 2nd Proof. Pick a closed set $F_n$ such that $F_n \subset E$ and $$ \mu(E \setminus F_n) < 2^{-n}.$$ Then there exists a continuous function $\varphi_n : [0, 1] \to [0, 1]$ such that $\varphi_n \equiv 1$ on $F_n$ and $$\mu(\operatorname{supp}(\varphi_n) \setminus F_n) < 2^{-n}. $$ For instance, set $\varphi_n(x) = \max\{0,1-k \operatorname{dist}(x, F_n)\}$ for a sufficiently large $k$. Now we claim that $\varphi_n \to \mathbf{1}_E$ almost everywhere. We will establish this by invoking the idea of Borel-Cantelli Lemma. Indeed, define the function $N : [0, 1] \to \mathbb{N}_0 \cup\{\infty\}$ by $$ N(x) = \sum_{n=1}^{\infty} \mathbf{1}_{\{ \varphi_n(x) \neq \mathbf{1}_E(x) \}}. $$ Then we note that, for each $x \in [0, 1]$ and $n \geq 1$, we have \begin{align} [\varphi_n(x) \neq \mathbf{1}_E(x)] &\quad\Rightarrow\quad [x \in E \text{ and } \varphi_n(x) \neq 1] \text{ or } [x \notin E \text{ and } \varphi_n(x) \neq 0] \\ &\quad\Rightarrow\quad [x \in E \setminus F_n] \text{ or } [x \in \operatorname{supp}(\varphi_n)\setminus E]. \end{align} From this, we get $$ \int_{[0,1]} N(x) \, \mathrm{d}x = \sum_{n=1}^{\infty} \mu(\{ x : \varphi_n(x) \neq \mathbf{1}_E(x) \}) \leq \sum_{n=1}^{\infty} 2^{-(n-1)} < \infty. $$ In particular, this implies that $N$ is finite a.e., which in turn tells that $\varphi_n(x) = \mathbf{1}_E(x)$ eventually holds a.e. Therefore $\varphi_n \to \mathbf{1}_E$ pointwise a.e. as desired. $\square$
H: Zorn's Lemma - partial order, or preorder? According to Wikipedia Zorn's Lemma says; every partially ordered set containing upper bounds for every chain necessarily contains at least one maximal element. According to Nlab Zorn's Lemma says something slightly different; Every preorder in which every sub-total order has an upper bound has a maximal element. Are these different? Is the nlab version stronger than the other? I can see how most of it corresponds; sub-total order $\mapsto$ chain. at least one maximal $\mapsto$ a maximal but I have to conclude with its weaker preorder condition, the nlab version is a stronger theorem. Is this correct? Am I right in thinking that when there are multiple maximal elements, they might look like a cyclic set that fails antisymmtry such $a\preceq b\preceq c\preceq a$, and in this case every distinct element $a,b,$ and $c$ are maximal elements? I'm unclear whether it can be the case that there are separate distinct maximal chains e.g. $a\preceq a$ and $b\preceq b$ which do not satisfy $a\preceq b$. AI: There is no essential difference in the sense that one version is stronger than the other. It is more a question of: one version of the lemma is translated into another. Double translations are neutral. If $(P,\preceq)$ denotes a preorder then it induces on a natural way a partial order $(P',\leq)$. Elements of $P'$ are equivalence classes wrt to relation:$$x\sim y\iff x\preceq y\wedge y\preceq x$$ Relation $[x]\leq[y]$ on $P'$ defined by $x\preceq y$ appears to be well defined. A maximal element $[m]$ in $P'$ is actually a set of maximal elements in $P$ and for every $x,y\in[m]$ we have $x\preceq y\wedge y\preceq x$ (so yes, there is cycling there). On the partial order we have the lemma of Zorn and "translating" it to the original preorder gives a formulation of the lemma for preorders (and vice versa). IMV it is a good habit to connect any preorder that you meet at once with the partial order that is induced by it. Edit: Usually an element $m$ is defined to be maximal if: $$m\leq a\implies m=a\text{ for every }a$$ This works fine in partial orders but not in preorders. A definition that works for both is: $$m\text{ is maximal if }m\leq a\implies a\leq m\text{ for every }a$$
H: Which matrices in $\operatorname{SL}_2(\mathbb R)$ have orthogonal squares? This is a self answered question, which I post here since, it turned out to be quite nice, and wasn't trivial for me. Of course, I would be happy to see other approaches. Question: Characterize all the matrices $X \in \operatorname{SL}_2(\mathbb R)$, with distinct singular values whose square $X^2 \in \operatorname{SO}(2)$. (If the singular values are equal, then $X$ itself already lies in $ \operatorname{SO}(2)$). I guess that a natural way to do that is through the SVD form of $X$. Edit: It turns out that $X^2 \in \operatorname{SO}(2) \iff X^2=-Id$. I find this quite surprising. I would be happy to see other proofs which start with the assumption $X^2 \in \operatorname{SO}(2)$ and deduce $X^2=-Id$. AI: Well, perhaps surprisingly, the answer does not depend on the singular values of $X$ at all (except on them being different from each other). Claim: Let $X=U\Sigma V^T \in \operatorname{SL}_2,U,V \in \operatorname{SO}(2)$, $\Sigma$ positive-definite. Suppose that the singular values of $X$ are distinct. Then the following are equivalent: $X^2 \in \operatorname{SO}(2)$ $V^T U \in \{R_1,R_2\}$, where $R_1:=\begin{pmatrix} 0 & -1 \\\ 1 & 0\end{pmatrix}$ and $R_2:=\begin{pmatrix} 0 & 1 \\\ -1 & 0\end{pmatrix}$. $X^2=-Id$. This means that we can choose $V \in \operatorname{SO}(2)$ as we want to, and then given $V$, $U$ has only two options, $VR_1$ or $VR_2$. (so we essentially have two distinct copies of $\operatorname{SO}(2)$). Proof: $X^2=U\Sigma V^T U\Sigma V^T=U\Sigma Q\Sigma V^T$ where $Q=V^T U \in \operatorname{SO}(2)$. Thus $$ X^2 \in \operatorname{SO}(2) \iff A:=\Sigma Q\Sigma \in \operatorname{SO}(2) \iff $$ $$A^TA=\Sigma Q^T \Sigma^2 Q\Sigma = Id \iff Q^T \Sigma^2 Q=\Sigma^{-2} \iff (Q^T \Sigma Q)^2=(\Sigma^{-1})^2 \iff Q^T \Sigma Q=\Sigma^{-1},$$ where the last equivalence followed from the uniqueness of the symmetric positive-definite square root. Thus we showed that $X^2 \in \operatorname{SO}(2) \iff Q^T \Sigma Q=\Sigma^{-1}$. Note that if $\Sigma=\operatorname{diag}(\sigma_1,\sigma_2)$, then $\Sigma^{-1}=\operatorname{diag}(\sigma_1^{-1},\sigma_2^{-1})=\operatorname{diag}(\sigma_2,\sigma_1)$, so $\Sigma^{-1}$ is obtained from $\Sigma$ by switching the diagonal entries. (here we used the fact $\sigma_1 \sigma_2=1$ i.e. $X \in \operatorname{SL}_2$). Define $P= \begin{pmatrix} 0 & 1 \\\ 1 & 0\end{pmatrix}. $ Direct calculation shows that $\Sigma^{-1}=P\Sigma P=P^T\Sigma P$. Combining this with $Q^T \Sigma Q=\Sigma^{-1}$, we obtain $$ X^2 \in \operatorname{SO}(2) \iff P^T\Sigma P=Q^T \Sigma Q \iff (QP^T) \Sigma =\Sigma (QP^T) \iff QP^T \in \operatorname{O}^{-}(2) \, \, \text{ is diagonal.} $$ (Since a matrix which commutes with a diagonal matrix with distinct entries must be diagonal. Here we use the fact that the singular values of $X$ are distinct). Since the only diagonal orthogonal matrices in $\operatorname{O}^{-}(2) $ are $\begin{pmatrix} 1 & 0 \\\ 0 & -1\end{pmatrix}, \begin{pmatrix} -1 & 0 \\\ 0 & 1\end{pmatrix}$, we deduce that $QP^T$ must be equal to one of those matrices, which implies that $Q=R_1:=\begin{pmatrix} 0 & -1 \\\ 1 & 0\end{pmatrix}$ or $Q=R_2:=\begin{pmatrix} 0 & 1 \\\ -1 & 0\end{pmatrix}$. So, we proved that $X^2 \in \operatorname{SO}(2) \iff V^TU=R_i$ for $i \in \{1,2\}$. Edit: We finally prove that $X^2 \in \operatorname{SO}(2) \iff X^2=-Id$. We proved that $X^2 \in \operatorname{SO}(2) \iff \iff Q^T \Sigma Q=\Sigma^{-1} \iff \Sigma Q \Sigma=Q \iff X^2=UQV^T$ where in the last assertion we have used the fact that $X^2=U\Sigma Q\Sigma V^T$. Using $Q=V^TU$ we got $X^2 \in \operatorname{SO}(2) \iff X^2=UV^TUV^T$. Since $Q=R_i$, we have $U=VR_i$, so $X^2=UV^TUV^T$ becomes $$ X^2=(VR_i)V^T(VR_i)V^T=VR_i^2V^T=-Id, $$ since $R_i^2=-Id$.
H: Universal coefficients for an integral domain Let $R$ be an integral domain with unit. In Burde-Zieschang's Knots it is claimed (page 219) that for a pair of spaces $(X,Y)$ $$H_i(X,Y;R) \cong H_i(X,Y;\mathbb{Z}) \otimes_{\mathbb{Z}} R.$$ Is it the case that $R$, as a $\mathbb{Z}$-module, is projective or free (so that $\mathrm{Tor}^{\mathbb{Z}}_1(-, R)$ always vanishes)? Or otherwise why does the $\mathrm{Tor}$ term never appear? AI: It's not true in general, for instance with $R= \mathbb Z/p, p$ prime it fails badly. e.g. $p=2, X=\mathbb RP^2, Y=\emptyset$, then $H_2(\mathbb RP^2;\mathbb Z) = 0$ but $H_2(\mathbb RP^2;\mathbb Z/2) \cong \mathbb Z/2$ Maybe the author meant characteristic $0$ integral domain ? (in which case the underlying abelian group is torsion-free and hence, since $\mathbb Z$ is principal, flat)
H: What do the double bars and the 2 2 mean in this equation? I was trying to understand this equation which is used to find the Rotation and translation between a set of 3D points x, and y that brings them closest together. It says you look for the solution where it is minimized. R I think is the 3x3 rotation matrix and t is a 3x1 translation matrix. I think it starts by saying its taking the average or 1/n of the sum of each matched point from the x and y list. And because it is y -x the minimum value would be when the position of x and y were made equal. I don't follow the bars or the 2 over 2 at the end. Bars mean absolute value to me and the 22 I don't know what that means. AI: The bars stand for norm. The subscript $2$ indicates that this is the 2-norm, or euclidean norm, which is the one defining the usual euclidean distance between vectors or euclidean length of a vector. Mathematically, given a vectors $x=(x_1,\dots, x_n)$ the euclidean norm of $x$ is defined to be $\|\|x\|\|_2=\sqrt{x_1^2+\cdots+x_n^2}.$ The superscript 2 is just a power, meaning that the norm is squared, so $\|\|x\|\|^2_2=x_1^2+\cdots+x_n^2.$ This is done because square roots are problematic in optimizations problems due to non-differentiability at the origin and because squaring is a monotone function for non-negative values, so the minimum is attained at the same point as the original function.
H: How can one prove that $f(x) = 4x - \ln(x^2 + 1)$ is injective? Let $f : \mathbb{R} \rightarrow \mathbb{R}$ be defined by $f(x) = 4x - \ln(x^2 + 1)$. Prove that $f$ is injective. How I thought to solve this problem was to use the derivative of the function and show that it is strictly positive and strictly increasing, thus demonstrating that $f(x)$ is in fact strictly increasing and this in turn proving that the function is injective, but I got stuck on showing that the derivative is strictly increasing. $$f'(x) = 2\frac{2x^2 - x + 2}{x^2 + 1}$$ Let $a, b \in \mathbb{R}$, so that $a > b$. $$f'(a) > f'(b) \Longleftrightarrow 2\frac{2a^2 - a + 2}{a^2 + 1} > 2\frac{2b^2 - b + 2}{b^2 + 1}$$ And here is where I got stuck. AI: Notice that $2x^2-x+2$ and $x^2+1$ are always positive and hence $f'(x) >0\ \forall\ x \in\mathbb R$. So, $f(x) $ is always increasing.
H: Are Turing machines that move before assigning Turing complete? A state in a Turing machine is an assignment and a move direction both depending on the current input. But if we define Turing machines to be a move first and then an assignment both depending on the current input, would this also be Turing complete? e.g. instead of: State X: if 1 then change to 0 and move Left and goto state Y if 0 then change to 0 and move right and goto state Z We had rules such as: State X: if 1 then move left and change to 0 and goto state Y if 0 then move right and change to 1 and goto state Z Would this also represent a Turing complete system? Also what about if we had a system like the following where each state always moved in one direction?: State X: move Left if 1 then change to 0 and goto state Y if 0 then change to 1 and goto state Z AI: The first example is, according to the comments, a finite state machine, and thus not Turing complete. Its only inputs are the head programming and the first tape symbol, and it cannot read anything it has written itself. The second example, however, is Turing complete. One simple way to encode a standard Turing machine is to use only every second cell on the tape for actual information, and use the remaining ones to "turn around", if needed. So instead of the standard X: If 0, write 1, move left, change to Y If 1, write 0, move right, change to Z you would have something like XR: If 0, write 1, move right, change to YLLLL If 1, write 0, move right, change to ZRR and a corresponding code for XL. The state LLLL says, regardless of writing, move left, change to LLL, and corresponding for states LLL and LL, and for states RRRR, RRR and *RR.
H: Finding the joint distribution of two independent uniform random variables $X,Y$ given event $E$ If $E = \{\text{either} \ X < 1/3 \ \text{or} \ Y<1/3\}$ for $X\sim Unif(0,1)$ and $Y\sim Unif(0,2)$, does the joint distribution of $X,Y$ given event $E$ exist? I am assuming that $X$ and $Y$ are independent. AI: Let $(\Omega,\mathcal F, \mathbb P)$ be a probability space, on which you have two independent random variables $X \sim \mathcal U((0,1)), Y \sim \mathcal U((0,2))$. Let $E := \{ \omega \in \Omega : X(\omega) < \frac{1}{3} $ or $Y(\omega) < \frac{1}{3} \}$. You want to find the distribution of $(X,Y)$ but treated as a random variable from $(\Omega,\mathcal F , \mathbb P_E)$ where $\mathbb P_E$ is a probability measure on $(\Omega,\mathcal F)$ such that $\mathbb P_E(A) = \frac{1}{\mathbb P(E)}\mathbb P(E \cap A)$ for any $A \in \mathcal F$. Note that $\mathbb P(E) = 1 - \mathbb P(X > \frac{1}{3})\mathbb P(Y > \frac{1}{3}) = 1 - \frac{2}{3} \frac{5}{6} = \frac{4}{9}$ due to independence We'll find the CDF, that is $F(t,s) = \mathbb P_E(X \le t, Y \le s)$. For $s$ or $t$ less than $0$ it is obviously $0$. If both of $s,t$ are greater than $\frac{1}{3}$ it is obviously $1$. Take any $s \in (0, \frac{1}{3}), t>0$ $$ F(t,s) = \frac{9}{4}\mathbb P(X \le s, Y \le t \cap \{ X < \frac{1}{3} \ or \ Y < \frac{1}{3} \}) = \frac{9}{4} \mathbb P(X \le s, Y \le t) = \frac{9}{8}st1_{(0,2)}(t) + \frac{9}{4}s1_{(2,\infty)}(t)$$ since in that case $\mathbb P(X \le s) = s, \mathbb P(Y \le t) = \frac{t}{2}1_{(0, 2)}(t) + 1_{(2,\infty)}(t)$ and we have independence. Moreover, for $s >0, t \in (0,\frac{1}{3})$ we get: $$ F(t,s) = \frac{9}{4} \mathbb P(X \le s)\mathbb P(Y \le t)= \frac{9}{8}ts1_{(0,1)}(s) + \frac{9}{8}t1_{(1,\infty)}(s)$$
H: A particular problem on series Problem: Show that there exist $c>0$ such that for all $N\in \mathbb N$ we have $$ \sum_{n=N+1}^{\infty}\left(\sqrt{n+\frac{1}{n}}-\sqrt{n}\right)\le \frac{c}{\sqrt{N}} $$ I have no clue how to solve this. All I know is this fact $$\int_0^1\left(\sum_{n\in \mathbb N}\frac{1}{ \sqrt{n^3+nx}}\right)\mathrm dx=\sum_{n\in \mathbb N}\left(\int_0^1\frac{1}{ \sqrt{n^3+nx}}\mathrm dx \right)$$ $$=2\sum_{n=1}^{\infty}\left(\sqrt{n+\frac{1}{n}}-\sqrt{n}\right)$$ as the series of function is uniformly convergent . Note: I am quoting this fact regarding the series of function because the above problem was meant to be solved as a consequence of this fact . But other methods are welcome as well. AI: A possible approach is to use Bernoulli's inequality: $$ \sqrt{n+\frac{1}{n}} = \sqrt{n} \sqrt{1+\frac{1}{n^2}} \le \sqrt{n} \left( 1 + \frac{1}{2n^2}\right) $$ That leaves you with $$ \sum_{n=N+1}^\infty \frac{1}{2n^{3/2}}\le \int_N^\infty \frac{1}{2x^{3/2}}\mathrm dx $$ which can be estimated above by the corresponding integral.
H: Confused about linear independence of two vectors in $\mathbb R^3$ If I put the two vectors 1,0,0 and 0,1,0 next to each other $$\begin{pmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{pmatrix}$$ I can see that they are independent since I cannot write $(1,0,0)^t$ as $(0,1,0)^t$. But I've heard that "If you have a row of $0$'s in your matrix, it's linearly dependent" But the matrix above has a row of zeroes but is still linearly independent. Which means I'm understanding something wrong. What am I missing? AI: That statement (“If you have a row of $0$'s in your matrix, it's linearly dependent”) holds if you have $n$ vectors in $\Bbb R^n$. That is not the case here: you have two vectors in $\Bbb R^{\bf 3}$.
H: Determining convergence or Divergence of $\sum\limits_{n=1}^{\infty} (\frac{3n}{3n+1})^n$ $$\sum\limits_{n=1}^{\infty} \left(\frac{3n}{3n+1}\right)^n$$ A cauchy root test will not work since $\lim_{n \to \infty}\sqrt[n]{(\frac{3n}{3n+1})^n} = 1$. However, by inspection, as $n \to \infty$ the sum reduces to: $$\sim^{\infty} \sum\limits_{n=1}^{\infty} 1$$ which diverges. Hence, can I conclude from this that the original series diverges? AI: $\lim\limits_{n\to \infty} a_n=\lim\limits_{n\to \infty}\left(\frac{3n}{3n+1}\right)^n = \mathrm{e}^{\lim\limits_{n\to \infty}n\ln\left(\dfrac{3n}{3n+1}\right)} =\mathrm{e}^T.$ Where $T = \lim\limits_{n\to \infty}\dfrac{\ln\dfrac{3n}{3n+1}}{\dfrac{1}{n}}= \lim\limits_{n\to \infty}\dfrac{-3n^2(3n+1)}{3n(3n+1)^2}=-\dfrac{1}{3}.$ Thus $\lim\limits_{n\to \infty} a_n = \mathrm{e}^{-\dfrac{1}{3}} \ne 0.$ So we can conclude that the series $\sum\limits_{n=1}^{\infty} a_n = \sum\limits_{n=1}^{\infty} (\frac{3n}{3n+1})^n $ is divergence.
H: Given a real world dataset of film budgets and their actual gross profit, find the optimal budget So I am currently working on a project for a Data Science Bootcamp that I started about a month ago and we are examining the film industry to gleam certain insights that we determine. What I am looking to do is take this dataset that I have compiled that involves the budget of the film and the domestic/international gross of the film and return what the optimal budget to make the most profit is. I have been googling and I am not certain what to google to get an answer. I have found several bits of information that gives me a feeling I am heading in the right direction, but I can't seem to find exactly what I am looking for. This is probably something simple that I am just overlooking, but I am having a difficult time finding it. Though I have a sneaking suspicion that this may be a linear regression problem. A huge plus if you can also relate it to a way I could graph it to determine this. AI: In mathematical terms, you have an independent variable $X_i$, the cost of producing film $i$; and a dependent variable $Y_i$, the percentage profit realised by film $i$ (percentage profit is probably more relevant than absolute profit - it is better to spend ten million dollars on ten $1$ million dollar films that make a profit of half a million dollars each than on one ten million dollar film that makes a profit of two million dollars). You now need a statistical model which connects the dependent and independent variables: $Y_i = f(X_i, \beta_0, \beta_1 \dots ) + \epsilon_i$ where $f$ is your predictor function, $\beta_0, \beta_1$ etc. are parameters and $\epsilon_i$ is an error term. For example, in a linear model you would have a linear predictor function $f= \beta_0 + \beta_1 X_i$ You can then use your data to estimate the values of your model parameters that best fit your model to the data, and you can get an objective measure of how good that fit is. If you are using a linear model then the right tool for the job is linear regression analysis, but a non-linear model will need some other type of regression analysis.
H: simplified formula of expression having adjoint of a matrix Well, the question is : let A be a square matrix of order $n_n$ and adj(A) is the adjoint of matrix A then is it possible to get a simplified formula for the expression $$adj(adj(adj(...adj(A))))$$ here adj() has been written 'r' times. The reason I am asking this question is that there exists a simplified formula for the determinant of the given expression if $\|A\|$ represents determinant of a matrix A then $$\|adj(adj(adj(...adj(A))))\| = \|A\|^{(n-1)^r}$$ This can be easily proved using the identity $$Aadj(A) = \|A\|I_n$$ for $r=2$ I know that $$adj(adj(A)) = \|A\|^{n-2}*A$$ but I can't generalize this result. So, is it possible to get a general form of this expression? All help is greatly appreciated. AI: Let $A$ be $n\times n$. When $n=1$, $\operatorname{adj}(A)$ is equal to $1$ by convention. In turn, the $m$-fold adjugate matrix of $A$ is always equal to $1$. Note that the identity $\operatorname{adj}\left(\operatorname{adj}(A)\right)=\det(A)^{n-2}A$ only applies to the case where $n\ge2$. When $n=2$, $\operatorname{adj}\left(\operatorname{adj}(A)\right)=A$. Hence the $m$-fold adjugate of $A$ is $\operatorname{adj}(A)$ when $m$ is odd, or $A$ when $m$ is even. Now suppose $n>2$. If $A$ is singular, $\operatorname{adj}\left(\operatorname{adj}(A)\right)=0$. Therefore the $m$-fold adjugate of $A$ is $\operatorname{adj}(A)$ when $m=1$, or zero when $m>1$. If $n>2$ and $A$ is nonsingular, we have, for every integer $k$, \begin{aligned} \operatorname{adj}(\det(A)^kA) &=\operatorname{det}\left(\det(A)^kA\right)\left(\det(A)^kA\right)^{-1} =\det(A)^{(n-1)k+1}A^{-1},\\ \operatorname{adj}(\det(A)^kA^{-1}) &=\operatorname{det}\left(\det(A)^kA^{-1}\right)\left(\det(A)^kA^{-1}\right)^{-1} =\det(A)^{(n-1)k-1}A. \end{aligned} It follows that the $m$-fold adjugate matrix of $A$ is $$ \operatorname{adj}^m(A)=\begin{cases} \det(A)^{a_m}A^{-1}&\text{when $m\ge1$ is odd},\\ \det(A)^{a_m}A&\text{when $m\ge1$ is even}.\\ \end{cases} $$ where $a_m$ is given by the recurrence relation \begin{aligned} a_1&=1,\\ a_2&=n-2,\\ a_m &=\begin{cases} (n-1)a_{m-1}+1&\text{when $m\ge3$ is odd},\\ (n-1)a_{m-1}-1&\text{when $m\ge3$ is even},\\ \end{cases}\\ &=\begin{cases} (n-1)\left[(n-1)a_{m-2}-1\right]+1&\text{when $m\ge3$ is odd},\\ (n-1)\left[(n-1)a_{m-2}+1\right]-1&\text{when $m\ge3$ is even},\\ \end{cases}\\ &=\begin{cases} (n-1)^2a_{m-2}-(n-2)&\text{when $m\ge3$ is odd},\\ (n-1)^2a_{m-2}+(n-2)&\text{when $m\ge3$ is even},\\ \end{cases}\\ &=\begin{cases} (n-1)^{m-1}-\frac{(n-1)^{m-1}-1}{(n-1)^2-1}(n-2)&\text{when $m$ is odd},\\ \frac{(n-1)^m-1}{(n-1)^2-1}(n-2)&\text{when $m$ is even}.\\ \end{cases}\\ \end{aligned}
H: Big $O$ notation and limit superior I'm reading the Wikipedia's link for big $O$ https://en.wikipedia.org/wiki/Big_O_notation and it says [...] both of these definitions can be unified using the limit superior: $$f(x) = O(g(x)) \; \mbox{as} \; x \rightarrow a$$ if $$\limsup_{x \rightarrow a} \left\|\frac{f(x)}{g(x)}\right\|< \infty.$$ But, why is this so? I mean, why is the limit superior used and not the normal limit? AI: The definition can be applied in situations where the limit $\lim_{x\to a} \|f(x)/g(x)\|$ may not exist. For example, let $a=0$. Suppose $f(x)=\sin (1/x)$ and $g(x)$ is the constant function $1$. We have $\|f(x)\|\leq g(x)$ for all $x$ and so $f(x)=O(g(x))$ as $x\to 0$. But $$ \lim_{x\to 0} \|f(x)/g(x)\|=\lim_{x\to 0}\|\sin (1/x)\| $$ does not exist. Examples with $a=\infty$ might also be useful since that situation is more common. Something similar to the previous example works: $\sin(x)=O(1)$ as $x\to\infty$, but $\lim_{x\to\infty}\|\sin(x)\|$ does not exist. Or define $f:\mathbb{R}^+\to \mathbb{R}^+$ such that $f(x)=x^2$ if $x$ is an integer, and $f(x)=0$ otherwise. Then $f(x)=O(x^2)$ (as $x\to\infty$), but $f(x)/x^2$ is the indicator function of the positive integers, and so the limit as $x\to\infty$ does not exist.
H: Quotient of ring by radical ideal. Let $R$ be a commutative ring with identity, and let $I$ be an ideal of $R$. Let $J$ be the radical of $I.$ What can we say about the quotient ring $R/J?$ Does it have any special properties? In particular, if $J$ is the nilradical of $R,$ what can we say about $R/J?$ I couldn't find any special properties of this. Please help me with this. AI: Consider a commutative unital ring $R$ with a proper ideal $I.$ Consider the ideal $$J = \operatorname{rad}(I) = \{r \in R \,\|\, r^n \in I \text{ for some integer } n \geq 1 \}.$$ Observe that in $R / J,$ the nilpotent elements are precisely the cosets whose representatives belong to $J.$ Explicitly, we have that $r + J$ is nilpotent if and only if $r^n + J = (r + J)^n = 0 + J$ for some integer $n \geq 1$ if and only if $r^n$ is in $J$ for some integer $n \geq 1$ if and only if $r \in J.$ Consequently, the nilradical of $R / J$ is precisely the set of elements in $J,$ i.e., the only nilpotent element of $R / J$ is the coset $0 + J.$ Consequently, the ring $R / J$ is reduced, i.e., there are no nonzero nilpotent elements in $R / J.$ On the other hand, the nilradical of a commutative unital ring $S$ is the intersection of all prime ideals in $S.$ Considering that every proper ideal (and hence every prime ideal) is contained in a minimal prime ideal, and every prime ideal contains a minimal prime ideal, it follows that the nilradical of $S$ is the intersection of all minimal prime ideals of $S,$ i.e., $\operatorname{nil}(S) = \bigcap_{P \in \operatorname{MinSpec}(S)} P,$ where we define$$\operatorname{MinSpec}(S) = \{P \,\|\, P \text{ is a minimal prime ideal of } S \}.$$ By the previous paragraph, we have that $J = \bigcap_{P \in \operatorname{MinSpec}(R) \,\|\, P \supseteq J} P = \bigcap_{P \in \operatorname{MinSpec}(R) \,\|\, P \supseteq I} P.$
H: Real sequences and convergence almost everywhere. Let $(X,\mu,\mathcal{A})$ be a finite measure space and $f_n$ measurable functions such that $f_n \to 0 $ almost everywhere. Show that exists a sequence $a_n \to +\infty$ auch that $a_nf_n \to 0$ a.e. I managed (by using the Borel-Cantelli lemma) to find a subsequence $a_{n_m}$ such that $a_{n_m}f_{n_m} \to 0 $ a.e using the convergence in measure(since we have convergence a.e),but i could not solve it. Can someone give me a hint? I do not seek a full solution. Thank in advance. AI: Assuming that $\mu(X)<\infty$, $f_n\to 0$ a.e. iff for every $\epsilon>0$, $\mu(\sup_{k\ge n}\|f_k\|>\epsilon)\to 0$ as $n\to\infty$. In this case $\{a_n\}$ can be constructed as follows. Pick the sequence $n_j$ s.t. $\mu(\sup_{k\ge n_j}\|f_k\|>\epsilon/j)\le j^{-1}$ and for each $j\ge 1$, set $a_{n_j}=\cdots=a_{n_{j+1}-1}=j$.
H: What is a good algorithm to find the clique number of a vertex of a graph? In this question, "graph" means a non-oriented, simple graph with no loop and no label on the edges or vertices. A clique in a graph $G$ is a complete subgraph of $G$. The clique number $\omega_v(G)$ of a vertex $v$ of $G$ is the maximum of the order (=number of vertices) of all cliques of $G$ that contains $v$. Is there a good algorithm that computes the clique number of a vector $v$ in a graph? The graphs I consider are represented by their adjacency matrix, but an algorithm that works on the list of edges of a graph would be fine too. I know that the clique number $\omega$ of $G$ is the maximum of the order over all cliques in $G$. Therefore, $\omega_v(G)=\omega(N(v))$, where $N(v)$ is the neighborhood of $v$ ($v$ included). Also, the clique number of a graph is the independence number of its complement. But I am not sure whether these information are useful or not. AI: You've probably already figured out that this problem is NP-hard: if you could solve this problem, then you could find the clique number of a graph $G$ by just adding a vertex $v$ to $G$ and connecting it to all previously existing vertices of $G$, then finally querying what $\omega_v(G \cup v)$ is. The clique number of $G$ would then be $\omega(G \cup v) - 1$. This gives rise to a nice heuristic (though it can still be very slow). You can first consider the induced subgraph (call it $H$) of $G$ on the vertices $v \cup N(v)$ where $N(v)$ is the neighbors of $v$. Then you could find a maximum clique $C$ for $H \setminus v$ in $O\left(3^{\|V(H)\| / 3}\right)$ time. Since all vertices in $H \setminus v$ are guaranteed to be connected to $V$, it follows that $C \cup v$ would be a maximum clique in $H$ (and subsequently) a maximum clique in $G$ containing $v$. Given that the fastest maximum clique-finding algorithms we know run in $O\left(3^{V / 3}\right)$ time, the equivalence of your problem to the maximum clique problem would suggest that the best algorithm known for your problem also runs in $O\left(3^{V / 3}\right)$ time. Otherwise, we would have have just found a faster algorithm for solving the maximum clique problem, via the reduction I outlined in the first paragraph.
H: Is this Hankel matrix positive definite? It's an exercise given in my book which says that Question: Consider a matrix $A=(a_{ij})_{5×5}$, $1\leq i,j \leq 5$ such that $a_{ij}=\frac{1}{n_i+n_j+1}$, where $n_i,n_j\in\mathbb{N}$. Then in which of the following is $A$ positive definite? (a) $n_i=i$, for all $i=1,2,3,4,5$ (b) $n_1<n_2<n_3<n_4<n_5$. (c) $n_1=n_2=n_3=n_4=n_5$ (d) $n_1>n_2>n_3>n_4>n_5$ I am unable to see an easy method to tell whether this matrix is positive definite. I found this matrix on Google search to be a special matrix known as a Hankel matrix. Can anybody help me solve this? It seems to me that any Hankel matrix with the terms in the rows being monotonic is positive definite. If it is not, is there a counterexample? My last question: Is there a sufficient condition for a Hankel matrix to be positive definite? Thanks in advance. AI: Note that $$\int_0^1x^{m+n}\,dx=\frac1{m+n+1}.$$ Therefore $(1/(n_i+n_j+1))$ is the Gram matrix for the $x^{n_i}$ with respect to the inner product $$\left<f,g\right>=\int_0^1f(x)g(x)\,dx$$ on the space of continuous functions $[0,1]\to\Bbb R$ So this Gram matrix will be positive definite iff the $x^{n_i}$ are linearly independent on $[0,1]$.
H: Normalizers of Sylow $p$-subgroups have restricted order I saw an old post. It gives the following result. Assume that $H$ is a subgroup of a finite group $G$, and that $P$ is a Sylow $p$-group of $H$. If $N_G(P) \subset H$ then $P$ is a Sylow p-subgroup of $G$. My first question: I wonder if there should be “$m=k$” at the end of the proof. “$\alpha=s$” is the desired result and in fact we can only get $\alpha=s$ from $p^{\alpha-s}m\equiv k({\rm mod}~p)$. “$m=k$” gives $$\|N_G(P)\|=p^sk=p^{\alpha}m=\|G\|$$ and hence $N_G(P)=G$, which we don’t necessarily have. My second question: If what I said in the first question is correct, then we still have $m\equiv k ({\rm mod}~p)$. If that is so, then the order of the normalizer of a Sylow $p$-subgroups is rather restricted. Is that right? Is there anything that I should note? Any help is appreciated. Thank you! AI: Your first question: Yes, $m\equiv k\bmod p$, not $m=k$. Second question: The order of a Sylow$p$-normalizer can be anything you want, in for example the group $P\times G$. If you mean inside a given group then yes, it has to be such that the index is congruent to $1$ modulo $p$. I'm not aware of any other serious restrictions, except things like Burnside's normal $p$-complement theorem. This has the corollary that if the Sylow $p$-subgroup is abelian and $N_G(P)=P$ then there is a normal $p'$-subgroup $K$ such that $G=KP$.
H: Let $X=\{f:\Bbb{R}\to\Bbb{R}\}$. Define $f\oplus g=h$ by $h(x)=f(x)+g(x)$. (i) Show $(X, \oplus)$ is a group. (ii) Show $\{f\in X\mid f(0)=0\}\le X$. Let $X= \{f:\mathbb{R}\to\mathbb{R} \}$. Define an operation $\oplus$ such that $f \oplus g = h$, where $h(x) = f(x) + g(x)$ for all $x \in \mathbb{R}$ and $+$ is the regular addition from real numbers. (i) Show that $(X, \oplus)$ is a group. (ii) Show that $Y = \{f \in X \mid f(0)=0\} \subset X$ is a subgroup. So I know the definitions for (i) we need to show the associative property, existence of the identity and the inverse element. For (ii) we need to show that $Y \ne \emptyset$, that it's closed under the group operation and closed under inverse. I haven't done these problems with functions and I'm not sure where to start here. How can I even use $f\oplus g = h$ for the associative property? $g \oplus f=h?$ This would lead me to $f(x)+g(x)=g(x)+f(x)$? AI: For closure, we have, for any $f,g\in X$, $$(f\oplus g)(x)=f(x)+g(x)$$ for all $x\in\Bbb R$. Hence $f\oplus g\in X$, since $f(x), g(x)\in \Bbb R$. Associativity of $\oplus$ follows from that of $+$: for any $f,g,h\in X, x\in \Bbb R$, we have $$\begin{align} (f\oplus (g\oplus h))(x)&=f(x)+(g\oplus h)(x)\\ &=f(x)+(g(x)+h(x))\\ &=(f(x)+g(x))+h(x)\\ &=(f\oplus g)(x)+h(x)\\ &=((f\oplus g)\oplus h)(x), \end{align}$$ so $f\oplus (g\oplus h)=(f\oplus g)\oplus h$. The identity is the zero function $0_X:X\to X$ given by $x\mapsto 0$. Indeed, for any $f\in X$, $x\in \Bbb R$, we have $$\begin{align} (f\oplus 0_X)(x)&=f(x)+0_X(x)\\ &=f(x)+0\\ &=f(x), \end{align}$$ so $f\oplus 0_X=f$, and, similarly, $0_X\oplus f=f$. The inverse of an arbitrary $f\in X$ is $(-f):\Bbb R\to\Bbb R$ given by $x\mapsto -f(x)$. To verify: $$\begin{align} (f\oplus (-f))(x)&=f(x)+(-f)(x)\\ &=f(x)-f(x)\\ &=0_{\Bbb R}\\ &=0_X(x), \end{align}$$ so $f\oplus(-f)=0_X$, and, dually, $(-f)\oplus f=0_X$. Hence $(X,\oplus)$ is a group. To show $(Y,\oplus)\le (X, \oplus)$, I will use the one-step subgroup test. We are given that $Y\subset X$. (It is, nonetheless, easy to show; just examine the definition of $Y$.) Since $0_X(x)=0$ for all $x\in\Bbb R$, we have, in particular, that $0_X(0)=0$. Thus $0_X\in Y\neq \varnothing$. Let $f,g\in Y$. Then $f,g\in X$ such that $f(0)=g(0)=0$, so $$\begin{align} (f\oplus (-g))(0)&=f(0)+(-g)(0)\\ &=0-g(0)\\ &=0-0\\ &=0, \end{align}$$ so $f\oplus (-g)\in Y$ as closure implies $f\oplus (-g)\in X$. Hence $(Y,\oplus)\le(X,\oplus).$
H: Only Two Isomorphism Classes of Groups of Order Four So my textbook says that there are two classifications of groups of order four. Those two are: $\mathbb{Z}_4\cong\{0,1,2,3\}$ under $+_4$, and $\mathbb{K}_4\cong$ symmetry group of a (non-square) rectangle. It also says if $P\cong Q$, and $P$ has $k$ elements of order $n$, then $Q$ has $k$ elements of order $n$. And that groups of order 8 or less can be classified entirely by the orders of their elements. So, let $(O,N)$ describe a group, $A$, such that $O$ represents an order (where $O$ divides $\|A\|$) and $N$ represents the number of elements (in the underlying set of $A$) with that respective order $O$. Let $G$ be a group such that $\|G\|=4$. $G$ can only be broken down in one of the following ways the following ways: $G_1=(1,1),(2,3)$ $G_2=(1,1),(2,2),(4,1)$ $G_3=(1,1),(2,1),(4,2)$ $G_4=(1,1),(4,3)$ Since one and only one element can have an order of one, and the other three elements can either have an order of two or four (ignoring order). Given this, $\mathbb{Z}_4\cong G_3$, and $\mathbb{K}_4\cong G_1$, leaving both $G_2$ and $G_4$ without a group to isomorphise with. I've also seen online that a cyclic group has exactly one generating element, whereas my textbook says that a cyclic group has at least one generating element. I feel like a clarification there might clear this up. Or is it the case that $G_2$ and $G_4$ aren't possible? If so, how would I prove that (for larger orders)? AI: $G_2$ is impossible. If $\|G\|=4$ and $x$ has order $4$, then $x^{-1}$ has order $4$ and is distinct from $x$. So we cannot have $(4,1)$. $G_4$ is impossible. If $\|G\|=4$ and $x$ has order $4$, then $x^2$ has order $2$. And if there is no element of order $4$ then every non-identity element in $G$ has order $2$ by Lagrange's Theorem. So we cannot have $(2,0)$. Finally, it is certainly not the case that a cyclic group has a unique generating element. Indeed, in a cyclic group of prime order, every non-identity element is a generator. (Perhaps you can provide a source where you saw this?)
H: Gap between the Uniform and Uniform convergence on compacts topologies Let $X$ and $Y$ be the set $C_0(\mathbb{R})$ (of functions vanishing at infinity) equipped with the topologies of compact-convergence and uniform convergence respectively. The second is clearly stronger than the first (as the closure of the compactly-supported functions differ in these cases) but what is an example of a function $f \in C_0(\mathbb{R})$ and a sequence $\{f_n\}$ there such that $\lim\limits_{n\to \infty} f_n(x) =f(x)$ in $X$ $\lim\limits_{n\to \infty} f_n(x) \neq f(x)$ in $Y$? AI: I think you can start with any non-constant function $g$ in $C_0(\mathbb{R})$ and build such a sequence by defining $f_n(x) = g(x-n)$. Now in the topology of the compact convergence this sequence converges to the function $f \equiv 0$, since for any compact set $K\subset \mathbb{R}$ you can find some integer $n$ such that $f_n(x)$ is pushed so far to the right that its value on $K$ gets arbitrarily small. However in the topology of the uniform convergence this sequence never converges to $f\equiv 0$ (nor to any other function) since its supremum and infimum remain constant and by hypothesis at least one of them is nonzero.
H: How can I express $(1+2+\dots+(k+1))^2$ using a $\sum$ instead? Good morning from México, I am in my first semester of Mathematics and I started proving by induction that: $$\sum_{i=0}^n i^3 = \left(\sum_{i=0}^n i\right)^2$$ This question has been answered before, three times actually, but not with the approach I am looking for so I am stuck here (I know I can do it with the other approaches like $((n(n+1))/2)^2$, but I want to know if it is possible doing this): $$\sum_{i=0}^k i^3 +(k+1)^3= \left(\sum_{i=0}^k i\right)^2+...$$ Next to the $+$ I tried $(k+1)^2$ (which obviously does not equalize my equation) but after plugging in values to check, I knew I was wrong. So my question after all: Is it possible to express: $(1+2+3+...+(k+1))^2$ with sigma? From that point I can finish proving it. Thank you. AI: $$ \left(\sum_{i=0}^{k+1} i\right)^2=\left(\left(\sum_{i=0}^{k} i\right)+k+1\right)^2=\left(\sum_{i=0}^{k} i\right)^2+2\left(\sum_{i=0}^{k} i\right)(k+1)+(k+1)^2$$ $$=\left(\sum_{i=0}^{k} i\right)^2+k(k+1)^2+(k+1)^2=\left(\sum_{i=0}^{k} i\right)^2+(k+1)^3$$ $$=\left(\sum_{i=0}^{k} i^3 \right)+(k+1)^3= \sum_{i=0}^{k+1} i^3 . $$ Is that what you wanted?
H: Computing the pdf after clipping Given a random variable X with uniform distribution on [-b,b], I want to compute the probability density function of Y = g(X) with $$g(x) = \begin{cases} 0, ~~~x\in[-c,c]\\ x, ~~~\text{else} \end{cases}$$, and $b>c$. There is a discontinuity at $x = c$ and $x=-c$. Therefore, the distribution of Y has both an absolutely continuous component $f_Y(y)=\frac{1}{2b}$ on $y\in (-b,-c)$ and $y\in (c,b)$ and a discrete component. This is the point, where I am not sure...my idea was the following: The discrete component consists of two mass points at $-c$ and $c$. This results in $$P_Y = \frac{b-c}{b} P_Y^{AC} + \frac{c}{b}P_Y^{D}$$ with $AC$...absolutely continuous $D$...discrete Could this be right? AI: The density function of Y, as you already observed, is a mixed density with a discrete component. This component has probability $P(X \in[-c;c])=[c-(-c)]\cdot\frac{1}{2b}=\frac{c}{b}$ This because when $X \in [-c;c]$ Y concentrate all this mass of probability in one single value: $Y=0$ Thus this mixed density (not absolute continuous) is the following $$ f_Y(y) = \begin{cases} \frac{c}{b}, & \text{if $y=0$} \\ \frac{1}{2b}, & \text{if $y \in[-b;-c] \cup [c;b]$} \\ 0, & \text{elsewhere} \end{cases}$$
H: What does it mean for a complex function to be real-differentiable? There is a proposition I read which claims Let $U$ be an open set. If $f(z)$ is real-differentiable on $U$ and satisfies Cauchy Riemann Equation then it is complex differentiable on $U$. I am now confused as to what does it mean for a complex function to be real-differentiable? Does it simply mean if we write $f(z)=u(x,y)+iv(x,y)$ where $u(x,y)$ and $v(x,y)$ are real-valued functions, then $u,v$ are both differentiable in $\mathbb{R}^2$? However, that does not seem to justify the name real-differentiable at all. Many thanks in advance! AI: You can see $f$ as a map $$\begin{array}{l\|rcl} f : & \mathbb R^2 & \longrightarrow & \mathbb R^2 \\ & (x,y) & \longmapsto & (u(x,y),v(x,y)) \end{array}$$ And $f$ is supposed to be differentiable. The term real-differentiable is used as $f$ is from $\mathbb R^2$ to $\mathbb R^2$.
H: Base locus under a pull-back Let $\pi: Y \to X$ be a surjective morphism of smooth projective varieties. Question: Suppose the linear system $\|D\|$ of a divisor $D$ on $X$ is base-point free. Is the linear system $\|\pi^D\|$ also base-point free? Here $\pi^$ is the pull-back. I have seen there are statements about pulling back ample or very ample line bundles, but these do not seem quite the same. AI: The assumption that $\|D\|$ is base-point free means that there is a surjective morphism $$ V \otimes \mathcal{O}_X \to \mathcal{O}_X(D). $$ The pullback functor is right-exact, hence the pullback of the above morphism $$ V \otimes \mathcal{O}_Y \to \pi^\mathcal{O}_X(D) = \mathcal{O}_Y(\pi^D) $$ is surjective, which means that $\pi^*D$ is base-point free.
H: Limit of $L^p$ norm is $L^\infty$ norm variation I'm familiar with the result that $$\lim_{p \to \infty} \|\|f\|\|_p=\|\|f\|\|_\infty$$ when $f \in L^p([0,1])$, but I've come a cross a variation of this fact that I'm having trouble showing. The assertion is that given $f \in L^\infty(\mathbb{R})$ $$\lim_{n \to \infty}\left(\int \frac{\|f(x)\|^n}{1+x^2} \, dx\right)^\frac{1}{n}=\|\|f\|\|_\infty$$ The function $\frac{1}{1+x^2}$ inside the integrand is what is tripping me up. I'm not sure how to deal with it in order to run the typical argument. AI: Apply the result you are familiar with to the measure $\mu:=\frac 1{1+x^2}dx$: the quantity $\left(\int \frac{\|f(x)\|^n}{1+x^2} \, dx\right)^\frac{1}{n}$ is the $L^n$ norm with respect to this measure. Moreover, $\mu(A)=0$ if and only if the Lebesgue measure of $A$ is zero hence the $L^\infty$-norm with respect to $\mu$ is the same as the $L^\infty$-norm with respect to the Lebesgue measure.
H: Matrix whose square is in Jordan normal form Let $$A = \begin{bmatrix}J_0^2 \\ & J_0^2 & \\ && J_{1/4}^3\end{bmatrix}\in M_7(\mathbb{Q})$$ Find, with proof, a matrix $B$ so that $B^2 = A$. I'm not sure how to find this matrix. Clearly, $A$ is not invertible as it is contains rows solely consisting of zeros. I know how to convert a matrix into Jordan form, but I'm not sure how this can be applied here. Also, I know that for any invertible matrix $A$ in $M_n(\mathbb{C})$, there exists a matrix $B$ so that $B^2 = A$. I can find a matrix $B$ so that $B^2 = J_{1/4}^3$, where $$B = \begin{bmatrix}1/2 & 1 & -1 \\ & 1/2 & 1\\ & & 1/2\end{bmatrix} = \frac{1}2 \left(I + \frac{1}2N + {1/2\choose 2}N^2 \right)$$ where $N^k$ is the matrix where $N_{i, i+k} = 4^k$ and $N_{i,j} = 0$ otherwise. I think I might need to solve some system of equations to find the matrix $B$. Clarification: $A := J_k^m$ is the $m\times m$ matrix with entries given by $A_{m,m}= k$ and for $1\leq i\leq m-1, A_{i,i} = k$ and $A_{i,i+1} = 1$ and $A_{i,j}=0$ otherwise. AI: Think of $A$ as a block diagonal matrix with blocks $\begin{bmatrix} J_0^2 & 0 \\ 0 & J_0^2 \end{bmatrix}$ and $\begin{bmatrix} J_{1/4}^3 \end{bmatrix}$. If you find matrices that square to each of these blocks then form a block diagonal matrix out of them and the result will square to $A$. You've already figured out a matrix for the second block, for the first block use $\begin{bmatrix} 0 & J_0^2 \\ I & 0 \end{bmatrix}$.
H: Is there a single continuous function satisfy all these properties? I'm finding one $C^2$, non-decreasing function $f$: $[0, 1] \to \mathbb{R}$ that \begin{align} f'(0) &= 0, \\ f'(0.5) &= \max f'(x) \text{ } \forall \text{ } x \in [0, 1] \\ f'(1) &= 0, \text{ and} \\ f''(0.5) &= 0. \\ \end{align} A Gaussian function with $b = 0.5$ would be my 2nd choice of $f'(x)$ but I wish it reaches, not only approaches $0$ at both ends. Any ideas? AI: What about $f(x) = \cos \pi \left( x-1\right)$?
H: Write the polynomial of degree $4$ with $x$ intercepts of $(\frac{1}{2},0), (6,0)$ and $(-2,0)$ and $y$ intercept of $(0,18)$. Write the polynomial of degree $4$ with $x$ intercepts of $(\frac{1}{2},0), (6,0) $ and $ (-2,0)$ and $y$ intercept of $(0,18)$. The root ($\frac{1}{2},0)$ has multiplicity $2$. I am to write the factored form of the polynomial with the above information. I get: $f(x)=-6\big(x-\frac{1}{2}\big)^2(x+2)(x-6)$ Whereas the provided solution is: $f(x)=-\frac{3}{2}(2x-1)^2(x+2)(x-6)$ Here's my working: Write out in factored form: $f(x) = a\big(x-\frac{1}{2}\big)^2(x+2)(x-6)$ I know that $f(0)=18$ so: $$18 = a\big(-\frac{1}{2}\big)^2(2)(-6)$$ $$18 = a\big(\frac{1}{4}\big)(2)(-6)$$ $$18 = -3a$$ $$a = -6$$ Thus my answer: $f(x)=-6\big(x-\frac{1}{2}\big)^2(x+2)(x-6)$ Where did I go wrong and how can I arrive at: $f(x)=-\frac{3}{2}(2x-1)^2(x+2)(x-6)$ ? AI: You didn't go wrong anywhere. $-6(x-\frac{1}{2})^2(x+2)(x-6) = -\frac{3}{2}(2x-1)^2(x+2)(x-6)$ That is because $-6(x- \frac 1{2})^2 = -6(\frac 12[2x-1])^2= -6\cdot (\frac 12)^2(2x-1)^2 = -6\cdot \frac 14(2x-1) = -\frac 32(2x-1)$. Is there some rule that that says fractions in the $(ax + b)$ parts are frowned upon? If so, if you get $(x + \frac ab)^k$ and can replace it with $(\frac 1b)^k(bx + a)^k$ but I don't see why you should have to. (In fact, I much prefer your notation as it directly indicates the roots and solutions... and indicates what the leading coeficient will actually be when expanded out. And what the heck kind of sense does removing a horrifying offensive fraction from one area make if you are just going to have to put an equally offensive fraction somewhere else?) But.... weren't you supposed to expand this out? So far as I can tell neither answer has done that. If you expand it out, you will see both answers are exactly the same.
H: Showing $P\left(\bigcap_{n=1}^{\infty} B_n\right)=1$ if $P(B_n)=1$ for every $n$ Resnick - Probability path 2.11: Let $\{B_n, n\geq 1\}$ be events with $P(B_n)=1$ for every $n$. Show that $$P\left(\bigcap_{n=1}^{\infty} B_n\right)=1$$ I was thinking to use a sequence such that it is equal to $\bigcap\limits_{n=1}^{\infty} B_n$, for $n$ going to infinity, and then apply the continuity property of a probability measure. Although, since it is not mentioned about the limit of $B_n$, neither if it is non increasing, I don't know how to proceed. Any help would be appreciated. Thanks! AI: Taking complements one gets $$ P\Big[\Omega\setminus\bigcap_n B_n\Big]=P\big[\bigcup_n(\Omega\setminus B_n)\Big]\leq\sum_nP[\Omega\setminus B_n]=0$$ Hence $P\Big[\bigcap_nB_n\Big]=1$.
H: What am I allowed to assume in questions like this? Suppose $T \in \mathcal{L}(V, W)$, and $(w_1, \ldots, w_m)$ is a basis of $\operatorname{range}(T)$. Prove that there exists $(\varphi_1, \ldots, \varphi_m) \in \mathcal{L}(V, \mathbf{F})$ such that $$ T(v) = \varphi_1(v)w_1 + \cdots + \varphi_m(v)w_m $$ for every $v \in V$. $V$ and $W$ are both vector spaces over a field $\mathbf{F}$. I always find questions like this hard to understand, and I never know what exactly I am 'allowed' to do. I see two ways of interpreting it: We define $(\varphi_1, \ldots, \varphi_m) \in \mathcal{L}(V, \mathbf{F})$. These linear maps cannot change, and must be valid for every $v \in V$ such that $T(v) = \varphi_1(v)w_1 + \cdots + \varphi_m(v)w_m$. For every $v \in V$, we must be able to define some $(\varphi_1, \ldots, \varphi_m) \in \mathcal{L}(V, \mathbf{F})$ such that $ T(v) = \varphi_1(v)w_1 + \cdots + \varphi_m(v)w_m $ holds. Note that key difference here being we can redefine $(\varphi_1, \ldots, \varphi_m)$ for each $v \in V$. Clearly there is only one correct interpretation. Which one is it? And if the other interpration was intended, how would the question have been worded differently? AI: The first interpretation is the correct one. One way you can tell this is the order of the quantifiers. The statement is: There exist $\phi_1,\ldots,\phi_m \in \mathcal L(V,\mathbf F)$ such that for all $v \in V$ the equation $T(v) = \phi_1(v) w_1 + ... + \phi_m(v) w_m$ holds. In particular, the quantifier on $v$ coming after the quantifiers on the $\phi$'s tells you that the $\phi$'s cannot depend on $v$. And then, like any existence proof, what you actually have to do, as you say, is to define $\phi_1,\ldots,\phi_m \in \mathcal L(V,\mathbf F)$ and then use their definitions to verify that the required equation holds (for all $v \in V$).
H: How do I evaluate $\sum_{m,n\geq 1}\frac{1}{m^2n+n^2m+2mn}$ I saw a problem here which state to evalute $$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^2n+n^2m+2mn}$$ My attempt Let $$f(m,n)=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^2n+n^2m+2mn}$$ and interchanging $m,n$ as $n,m$ we have $$f'(n,m) = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{mn^2+nm^2+2mn}$$ then I add $f(m,n)+f'(n,m)$ which gives the same which gives me the twice the original series. I failed to evaluate the series with the process I have applied. How do I evaluate the series? Any help will be appreciated. AI: Using integration trick Note that $$\dfrac{1}{nm^2+n^2m+2mn} =\dfrac{1}{mn(m+n+2)}=\dfrac{1}{mn}\int_0^1 x^{m+n+1}\,dx$$ then we have $$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{1}{mn} \int_0^1 x^{m+n+1}\,dx=\int_0^1x\left(\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \dfrac{x^{m}\cdot x^n}{mn}\right)\,dx \\= \int_0^1 x\ln^2(1-x)\,dx =\int_0^1 (1-x)\ln^2 x\,dx $$ hence integrating (by parts) we have the result $\dfrac{7}{4}$. Without integration trick Expanding the inner summation gives us $$\sum_{n\geq 1} \left(\frac{1}{n(n+3)}+\frac{1}{2n(n+4)}+\frac{1}{3n(n+15)}+\cdots\right)$$ making partial fraction of each summand we have $$\sum_{n\geq 1}\left(\frac{1}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)+\frac{1}{8}\left(\frac{1}{n}-\frac{1}{n+4}\right)+\cdots\right)=\sum_{n\geq 3} \frac{H_n}{n(n-2)}\cdots(1)$$ from here we can carry on our work by using the generation function of harmonic number, ie $$\sum_{n\geq 1} x^nH_n=\frac{\ln(1-x)}{x-1}$$ If we dont wish to make repeated integration and multiplying process then let's continue our work from $(1)$ we have above $$\sum_{n\geq 3}\frac{H_n}{2}\left(\frac{1}{n-2}-\frac{1}{n}\right)=\frac{1}{2}\left(H_3+\frac{H_4}{2}\right)+\frac{1}{2}\sum_{n\geq 3}\left(\frac{H_{n+2}}{n}-\frac{H_n}{n}\right)=\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}\right)+\frac{1}{4}\left(1+\frac{1}{2}+\cdots +\frac{1}{4}\right)+\sum_{n\geq 3}\frac{1}{2}\left(\frac{H_n}{n}-\frac{H_n}{n}+\frac{1}{n}\left(\frac{1}{n+1}+\frac{1}{n+2}\right)\right)\\=\frac{1}{2}\left(\frac{11}{6}+\frac{25}{24}\right)+\sum_{n\geq 3}\left(\frac{2}{(n+2)(n+3)}+\frac{3}{n(n+1)(n+2)}\right)=\frac{1}{2}\left(\frac{69}{24}+\frac{1}{2}+\sum_{n\geq 1}\frac{1}{n(n+1)(n+2)}-\frac{3}{6}-\frac{3}{24}\right)=\frac{1}{2}\left(\frac{69}{24}+\frac{1}{2}+\frac{3}{2\cdot 2!}-\frac{5}{8}\right)=\frac{7}{4}$$ Recall that for all $ p>0$ $$\sum_{n\geq 1}\frac{1}{n(n+1)(n+2)\cdots(n+p)}=\frac{1}{p\cdot p!}$$
H: Subgroup of Plane Isometries Isomorphic to $O_2(\mathbb{R})$ Let $\mathcal{Isom}(\mathbf{E})$ be the group of the isometries of the euclidean plane $\mathbf{E}$, and for every point $P \in \mathbf{E}$, let $\mathcal{Isom}_P$ be the subgroup of all isometries of $\mathbf{E}$ that fix $P$. It is well known that $\mathcal{Isom}_P \simeq O_2(\mathbb{R})$. Is the converse true? That is, if $H$ is a subgroup of $\mathcal{Isom}(\mathbf{E})$, and $H \simeq O_2(\mathbb{R})$, is is true that there exists some point $P \in \mathbf{E}$ such that $H = \mathcal{Isom}_P$? Even though it seems quite a reasonable statement, every attempt that I made to prove it has failed up to now, so any help is welcome in answering this question. Thank you very much in dvance for your kind attention. AI: $H$ has a subgroup $K$ of index $2$ isomorphic to $SO_2$ and so commutative. Then $K$ has an element $g$ of order $3$. The isometries of the plane of order $3$ are rotations through angle $\pm2\pi/3$. So $g$ is a rotation centred at a some point $P$. The centralizer of $g$ in the isometry group of the plane is the set of all rotations with centre $P$. So $K\subseteq\text{Rotations}_P$ The elements of $H-K$ don't centralise $g$ but normalise the subgroup it generates. The isometries which do this are the reflections fixing $P$. Therefore $H$ is a subgroup of the stabiliser of $P$. But there are proper subgroups of $O_2$ which are isomorphic to $O_2$ so we cannot say that $H$ is the stabiliser of $P$, but if say $H$ is a closed subgroup of the isometry group, then it is.
H: Proving that an open subgroup of $\mathbb{R}$ must be all of $\mathbb{R}$ Let $G$ be an open subset of $\mathbb{R}$ which is also a subgroup of the group $(\mathbb{R}, +)$. Show that $G = \mathbb{R}$. (Hint: $0$ belongs to $G$ and hence $(-\epsilon,\epsilon)$ is a subset of $G$ for some $\epsilon > 0$. Use the fact that $G$ is closed under addition.) AI: The hint pretty much tells you what to do, but I’ll expand it a bit. Start with the fact that $0\in G$, since it’s the identity element. Since $G$ is open, there must be an open interval $(-E,E)$ around $0$ that is contained in $G$. For any $g\in G$ we have $g+g\in G$, and $g+g+g\in G$, and in general $ng\in G$ for each positive integer $n$. Let $I=(-E,E)$, and for $n\ge 1$ let $I_n=\{ng:g\in I\}$. Each $I_n\subseteq G$. What is $\bigcup_\limits{n\ge 1}I_n$?
H: Proving Segal's Category $\Gamma \simeq \mathbf{FinSet}_^{op}$ I'm trying to show that Segal's category $\Gamma$ is equivalent to $\mathbf{FinSet}_^{op}$, the opposite of the category of finite pointed sets with basepoint preserving morphisms. I'm intuitively gathering that $\Gamma$ can be shown to be the skeleton of $\mathbf{FinSet}_^{op}$, or equivalently that $\Gamma^{op}$ is the skeleton of $\mathbf{FinSet}_$, but I'm having trouble just understanding what the opposites of these categories are. For reference $\Gamma$ is the category consisting of finite sets, and morphisms from $S$ to $T$ are maps $\theta:S \to P(T)$ where $P(T)$ is the power set of $T$, such that distinct points in $S$ get mapped to disjoint subsets of $T$. If $\theta:S\to P(T)$ and $\varphi:T \to P(U)$ are two such morphisms, then their composite is given by a morphism $\psi: S \to P(U)$ such that $$ \psi(\alpha) \;\; =\;\; \bigcup_{\beta \in \theta(\alpha)} \varphi(\beta). $$ My biggest issue here is that I can't even conceive of the opposite morphisms. How can these be found? If I wanted to demonstrate $\Gamma^{op} \simeq \mathbf{FinSet}_$ and then use duality, I'd have to find the opposite morphisms $\theta^{op}$, but would these be mappings from $P(T)$ to $S$, or from $T$ to $P(S)$? What's even the form of the identity morphisms in $\Gamma$? I thought possibly these would be the mappings $1_S:S\to P(S)$ such that $1_S(\alpha) = \{\alpha\}$. In what sense can $1_S$ be seen as an isomorphism in $\Gamma$? In what sense could we naturally find a notion of "opposite" here? For instance, for $A \in P(T)$ I'm doubtful we could define $\theta^{op}(A) = \theta^{-1}(A)$ since $\theta$ was a set-valued function to begin with. Either way, there's no guarantee that $A\in P(T)$ would be in the image of $\theta$. AI: Your typical morphism from $S$ to $T$ is a map $S\to\wp(T)$ satisfying the condition you list. This induces a map $\phi$ from $T_$ to $S_$ as follows, where $S_$ is the disjoint union of $S$ and a basepoint $$. This map $\phi$ is defined by $\phi(t)=s$ whenever $t\in\theta(s)$ for some $s$ (necessarily unique as the $\theta(s)$ are disjoint, and $\phi(t)=$ if $t=$ or $t$ is not in any of the $\theta(s)$. Then $S\mapsto S_$ and $\theta\mapsto\phi$ are the object and morphism maps of the required contravariant functor.
H: Define a linear operator which has as Kernel the line $y=-x$ and as image the line $y=$x So the question asks basically to determine a linear operator $F: \mathcal{R}^2 \rightarrow \mathcal{R}^2$ which has as Kernel the line $y=-x$ and as image the line $y=x$. Here is what i tried: i supposed i could write the operator in the form $F(x,y)=(ax+by,cx+dy)$. Thus $Ker(F)$={$(x,y) \in \mathcal{R}^2 : (ax+by,cx+dy)=(0,0)$} and i defined a=b=1 and c=d=0, so that i would have $Ker(F)$={$(x,y) \in \mathcal{R}^2 : (x+y)=(0,0)$} and the solution is x=-y if i'm not wrong. But for the image with these same guesses i got $Im(F)$={$(x+y,0) : (x,y) \in \mathcal{R}^2$} ={$(x+y)(1,0) : (x,y) \in \mathcal{R}^2$} But idk if i can conclude that the image is y=x. I don't have arguments to support it or i'm just wrong in my trial. Any help would be great, thank you AI: Let $f: \mathbb{R}^{2} \to \mathbb{R}^{2}$ such that if $(x,y) \in \mathbb{R}^{2}$ so $\ker(f)=\text{ span}\{(-1,1)\}$ and $\text{im}(f)=\text{ span}\{(1,1)\}$. Then we need a linear transformation such that $$f(-1,1)=(0,0)$$ and $$f(0,1)=(1,1)$$ Note that a basis for $\mathbb{R}^{2}$ is $\beta_{\mathbb{R}^{2}}=\{(-1,1),(0,1)\}$. Also by rank-nullity theorem:$$\dim(\text{im}(f))+\dim(\ker(f))=1+1=\dim(\mathbb{R}^{2})=2$$ Now, let $$(x,y)=\alpha_{1}(-1,1)+\alpha_{2}(0,1) \quad \implies \alpha_{1}=-x, \quad \wedge \quad \alpha_{2}=x+y$$ so $$f(x,y)=-xf(-1,1)+(x+y)f(0,1) \quad \implies \quad f(x,y)=(x+y,x+y)$$ Finally, you can see that $\ker(f)=\text{ span}\{(-1,1)\}$ and $\text{im}(f)=\text{ span}\{(1,1)\}$.
H: Image of an open set under an injective immersion map is Borel measurable? Let $M$ and $N$ be two smooth manifolds and $f:M\to N$ be an injective immersion. Let $U\subset M$ be an open subset, then is $f(U)$ necessarily a Borel measurable subset of $N$? AI: You can just cover U with open subsets where f is an imbedding. Then f(U) is a countable union of borel measurable sets Edit: Daniel’s answer in the comments is obviously better
H: Simple exercise on a linear operator $T$ I'm given the following linear operator $$T(a,b)=(-2a+3b,-10a+9b)$$ on the vector space $V=\mathbb{R}^2$. I have to find the eigenvalues of $T$ and an ordered basis $\beta$ for $V$ such that $[T]_\beta$ is a diagonal matrix. I've tried using the standard basis $\beta=\{(1,0), (0,1)\}$ and getting $$[T]_\beta= \begin{bmatrix} -2 & 3\\ -10 & 9 \end{bmatrix} $$ which is not a diagonal matrix. When I deal with this type of exercises, if I find a basis such that $[T]_\beta$ is diagonal, the diagonal elements are the eigenvalues and I don't need to write the characteristic polynomial anymore, right? In the case $[T]_\beta$ is not a diagonal matrix, can I still use it to write the characteristic polynomial $p(\lambda)=\det([T]_\beta-\lambda I_2)$ and then find the eigenvalues and write the corresponding diagonal matrix $D$? Is this procedure valid for any chosen basis $\beta$ and the corresponding $[T]_\beta$? AI: Yes, it is valid for any choosen basis $\beta$ and the corresponding $[T]_{\beta} $. And this is the general rule for finding characteristics polynomial and for finding the eigenvalue,$\det([T]_\beta-\lambda I_2) = 0 $ And here one thing, you can't write a $[T]_{\beta} $ as a diagonal matrix, if it is not diagonal with respect to $\beta $. In stead, you can find diagonal matrix similar to $[T]_{\beta} $ , whenever $[T]_{\beta} $ is diagonalizable.
H: Prove that $c_{m} \in[a, b],$ for all $m \geq 1, \lim _{m \rightarrow \infty} c_{m}$ exists and find its value. Let $f:[0,1] \rightarrow[0, \infty)$ be a continuous function. Let $$ a = \inf_{0 \leq x \leq 1} f(x) ~\text{ and }~ b = \sup_{0 \leq x \leq 1} f(x) . $$ For every positive integer $m$, define $$ c_{m}=\left[\int_{0}^{1}(f(x))^{m} d x\right]^{1 / m} $$ Prove that $c_{m} \in[a, b],$ for all $m \geq 1, \lim\limits_{m \rightarrow \infty} c_{m}$ exists and find its value. My work: To show limiting value is definite I have to show the value of integration is a finite one! For doing the I am thinking to apply Leibniz rule.But I am not sure how to go step by step.Any hint or suggestion would be highly appreciated! AI: The estimates $a \leq c_n \leq b$ follow from $$ a^m = \int_0^1 a^m\, dx \leq \int_0^1 f(x)^m\, dx \leq \int_0^1 b^m\, dx = b^m. $$ Let us prove that $(c_m)$ converges to $b$. If $b=0$ then also $a=0$, so that the claim follows from the first part. Assume that $b>0$, let $\varepsilon \in (0, b)$, and let $I\subset [0,1]$ be a set where $f \geq b-\varepsilon$ (since we are assuming $f$ continuous, we can take a suitably small interval containing the maximum point of $f$). Denoting by $L>0$ the length of the interval $I$, we have that $$ \int_0^1 f(x)^m\, dx \geq \int_I (b-\varepsilon)^m\, dx = (b-\varepsilon)^m L, $$ so that $$ c_m \geq (b-\varepsilon) L^{1/m}. $$ Since $L^{1/m} \to 1$ as $m\to +\infty$, we have that $$ \liminf_m c_m \geq b-\varepsilon. $$ Since $c_m\leq b$ for every $m$, the claim follows by the arbitrariness of $\varepsilon$:
H: Decreasing Sequence of Open Sets in Tychonoff Pseudocompact Spaces has Nonempty Intersection of Closures Let $(U_n)$ be a decreasing sequence of non-empty open sets in a Tychonoff pseudocompact space $X$. Then, show that $\cap \overline U_n \neq \phi$ This was part of a problem in Willard. I was able to do the rest of the parts, but this one has still eluded me. Any help would be appreciated! AI: $\newcommand{\cl}{\operatorname{cl}}$From what you said, I’m assuming that you’ve proved the equivalence of $2$(a) and $2$(c) in Exercise $17$J. For $n\in\Bbb N$ let $V_n=X\setminus\cl U_n$, and let $\mathscr{V}=\{V_n:n\in\Bbb N\}$. If $\bigcap_n\cl U_n=\varnothing$, then $\mathscr{V}$ is an open cover of $X$, so there is a finite $\mathscr{V}_0\subseteq\mathscr{V}$ such that $\bigcup\{\cl V_n:V_n\in\mathscr{V}_0\}=X$. Moreover, since the sets $U_n$ are decreasing, the sets $V_n$ are increasing, so in fact there is an $n\in\Bbb N$ such that $\cl V_n=X$. But then $U_n\subseteq X\setminus\cl V_n=\varnothing$, which is impossible. Added: Here’s one way to prove (b) directly from (a), which will let you complete your circle of implications. If there is an $n_0\in\Bbb N$ such that $U_n=U_{n_0}$ for all $n\ge n_0$, then clearly $\bigcap_n\cl U_n\ne\varnothing$, so we may as well assume that $U_n\supsetneqq U_{n+1}$ for each $n\in\Bbb N$, so that for each $n\in\Bbb N$ there is an $x_n\in U_n\setminus U_{n+1}$. $X$ is completely regular, so for each $n\in\Bbb N$ there is a continuous $f_n:X\to[0,n]$ such that $f(x_n)=n$, and $f_n[X\setminus U_n]=\{0\}$. Suppose that for each $x\in X$ there are an open set $V_x$ and a $k(x)\in\Bbb N$ such that $x\in V_x$, and $V_x\cap U_k=\varnothing$ whenever $k>k(x)$. Then we can define a function $$f:X\to\Bbb R:x\mapsto\sum_{n\in\Bbb N}f_n(x)\;,$$ since $f_n(x)=0$ for all $k\ge k(x)$. I’ll leave it to you to check that the function $f$ is continuous; use the fact that every $x\in X$ has an open nbhd $V_x$ that intersects only finitely many of the sets $U_n$. This is impossible, since $X$ is pseudocompact, and $f$ is clearly unbounded. Thus, there must be some $x\in X$ such that every open nbhd of $x$ intersects infinitely many of the sets $U_n$. And since the sets $U_n$ are nested, this means that every open nbhd of $x$ intersects every $U_n$ and hence that $x\in\cl U_n$ for each $n\in\Bbb N$, so that $x\in\bigcap_{n\in\Bbb N}\cl U_n$.
H: One-sided limit of q norm on $[0,1]$ I'm trying to show that for $f \in L^q([0,1])$ for all $q$ with $1≤q≤p<\infty$, we have $$\lim_{q \to p^-}\|\|f\|\|_q=\|\|f\|\|_p$$ It's easy to show that $$\lim_{q \to p^-}\|\|f\|\|_q≤\|\|f\|\|_p$$ But the other direction is proving difficult. I've tried to use an epsilon argument similar to what you use to show $$\lim_{p \to \infty}\|\|f\|\|_p=\|\|f\|\|_\infty$$but it hasn't worked out. AI: Suppose first that $f\in L^p$. Let $(q_n)$ be a sequence of numbers smaller than $p$ converging to $p$. Then using the dominated convergence theorem, $\lVert f\rVert_{q_n}^{q_n}\to\lVert f\rVert_{p}^{p}$ (bound $\lvert f\rvert^{q_n}$ by $\lvert f\rvert^p\mathbf{1}_{\{\lvert f\rvert\geqslant 1\}}+\mathbf{1}_{\{\lvert f\rvert\lt 1\}}$). If $f$ does not belong to $L^p$, let $f_N:=\lvert f\rvert\mathbf{1}_{\{\lvert f\rvert\leqslant N\}}$. Then for all $N\geqslant 1$, $$ \liminf_{q\to p^-}\lVert f\rVert_q\geqslant \liminf_{q\to p^-}\lVert f_N\rVert_q =\lVert f_N\rVert_p,$$ where we used what we established in the first paragraph. By monotone convergence, $\lVert f_N\rVert_p\to +\infty$ as $N$ goes to infinity.
H: Understanding why the integral test is applicable in $\sum_{n=1}^{\infty} \frac{\ln n}{n^2}$? Here is the statement of the integral test and question (e) that I want to solve: My questions are: 1- I know that here $f(x) = \frac{\ln x}{x^2} $ as this is a question of MATH subject GRE test and $\log n$ means $\ln n.$ But I do not know why my function is positive? as at $x=1$ I know that my function is equal 0. 2- why my function is monotonically decreasing? I know that I should calculate $\frac{a_{n}}{a_{n+1}}$ and it should be greater than 1 or checking the difference between $a_{n}$ and $a_{n+1}$ but the formula I get does not give me any inference. Could anyone help me in answering those questions, please? I calculated the improper integral $\int_{1}^{\infty} \frac{\ln x}{x^2} $ and I got that it converges to 1. I have no problem in that step. AI: $f'(n)=2n \frac{\frac{1}{2} - \ln(n)}{n^4} < 0 $ for $n\ge 2 $ So, $f(n)$ monotone decreasing for $n\ge 2 $ . And, take $\sum_{n=2}^{\infty} \frac{\ln(n)}{n^2} $, take out the first term.
H: Bilinear form and quotient space Let $U,V$ be two finite dimensional vector spaces of a field $K$ and let $f:U \times V \to K$a bilinear form. The set $U_0 = \{u \in U: f(u,v) = 0,\forall v \in V \}$ is called the left kernel of $f$ and $V_0 = \{v \in V: f(u,v) =0, \forall u \in U\}$ is called the right kernel of $f$. Show that $\dim U/U_0 = \dim V/V_0$ and the kernels of $g: U/U_0 \times V/V_0 \to K$ given by $g(u+U_0,v+V_0) = f(u,v)$ is the null space. The second statement is obvious, but how can I show the first?? that is, $\dim U/U_0 = \dim V/V_0$. AI: Using the bilinear form, you get a linear map $U/U_0 \to (V/V_0)^*$ and this is injective. Now compare dimensions and you get an inequality (the spaces are finite-dimensional, so dual spaces do not change the dimension).
H: If $\sum a_n^p$ converges, then $\sum a_n^q$ also converges? for $1 \leq p \leq q$ and $a_n > 0$, for all $n \in \mathbb{N}$. I was able to give a counterexample to the reciprocal, but I'm having trouble proving this one (I even tried to look for a counterexample, but I couldn't). Any suggestions on how to do it, or any counterexample? AI: Notice that $a_n^p$->0 so $a_n$->0 and you can find a $n_o \in \mathbb N$ that $a_n<1$ for every $n>=n_o$. So for $n>=n_o:0<a_n^q$<$a_n^p$ and your $\sum_{n=n_0}^\infty a_n^q$ converges.
H: Dual of a $\mathbb{C}G$-module, can we get $g^{-1}$ outside $\theta(\cdot)$? I am a bit lost with dual modules. Let $G$ be a group, $\mathbb{C}$ be the complex numbers and let $V$ be a $\mathbb{C}G$-module. Then as I understand it, the dual is $V^=\operatorname{Hom}_{\mathbb{C}G}(C,\mathbb{C}G)$. So if we have $\theta\in V^$ the action is given by $(g\cdot \theta)(v)=\theta(g^{-1}\cdot v)=g^{-1}\theta(v)$ where after the last equality we get the product inside $\mathbb{C}G$. Is this correct? AI: There is a difference between the dual of a $kG$-module, and the dual of an $R$-module, where $R=kG$. If $R$ is a ring and $M$ is a right $R$-module, then the dual of $M$, $\mathrm{Hom}_R(M,R)$, is a left $R$-module. This is not the same as the dual of a right (or left) $kG$-module, which is the dual of $M$ as a vector space $\mathrm{Hom}_k(M,k)$, with the group action as you have described, notably on the same side as the original action. There are two different concepts, one used in representation theory of groups and one used in module theory for arbitrary rings.
H: A problem involving sum of digits of integers Question: The sum of the digits of a natural number $n$ is denoted by $S(n)$. Prove that $S(8n) \ge \frac{1}{8}S(n)$ for each $n \in \mathbb N$. [source:Latvia 1995] At first I thought this problem can be solved using induction on the number of digits. Say without the leftmost digit (say $a_n$) of $n$ the number becomes $n'$. Then I tried to find some relation between $S(n)$ and $S(n')$ also $S(8n)$ and $S(8n')$. But that doesn't seem to be working in general. I don't have much experience dealing with sums of digits, it seems somewhat random. Apart from any hint or solution in this problem, I am also interested in general strategies or any references in order to deal with such digit-sum problems. Thanks in advance. AI: For all natural numbers $a,b$ we have: $S(10a)=S(a)$, and $S(a+b)\le S(a)+S(b)$, with equality only when there is no carry when doing the grade school addition $a+b$. A consequence of these is that for all natural numbers $n$ $$ \begin{aligned} S(125n)&=S(100n+20n+5n)\\ &\le S(100n)+S(20n)+S(5n)\\ &=S(n)+S(2n)+S(5n)\\ &=S(n)+S(n+n)+S(n+n+n+n+n)\\ &\le 8S(n). \end{aligned} $$ In other words, $$S(n)\ge\frac18 S(125n).$$ Plugging in $n=8m$ gives, for all natural numbers $m$, the inequality $$ S(8m)\ge\frac18 S(1000m)=\frac18 S(m). $$
H: is there a type of number disjoint complex numbers is there a type of number disjoint complex numbers? Has anyone found a type of number or is there exists such number s.t. $\omega^$ (or something else) that $\omega^\not\in\mathbb{C}$ but grouped in a new type of number with its meanings such as $\omega^*\in\mathbb{G_z}$ AI: Check out hypercomplex numbers in Wikipedia. https://en.wikipedia.org/wiki/Hypercomplex_number
H: Find limits of $\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$ without l'Hopital's rule or Taylor Expansion. Find the limit $\displaystyle \lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$ without l'Hopital's rule or Taylor expansion. My Try $\displaystyle =\lim _{x \to 0} \frac {\cos x - \frac{\sin x}{x}} {x \sin x}$ $=\frac {\displaystyle\lim _{x \to 0}\cos x - \lim _{x \to 0}\frac{\sin x}{x}} {\displaystyle\lim _{x \to 0}x \sin x}$ $=\frac{1-1}{0}$ But still I end up with $\frac00$ Any hint for me to proceed would be highly appreciated. P.S: I did some background check on this question on mathstack and found they have solved this with l'Hopital's rule and the answer seems to be $\frac{-1}{3}$. What is $\lim _{x \to 0} \frac {x \cos x - \sin x} {x^2 \sin x}$? AI: Famously $\lim_{x\to0}\frac{\sin x}{x}=1$ can be proved without such techniques, and implies $\lim_{x\to0}\frac{1-\cos x}{x^2}=\frac12$. With a little more effort (e.g. by approximating a circular arc as a parabola), you can also show $\lim_{x\to0}\frac{x-\sin x}{x^3}=\tfrac16$. So $\lim_{x\to0}\frac{x-\sin x}{x^2\sin x}=\tfrac16$ and$$\lim_{x\to0}\left(\frac{\cos x-1}{x\sin x}+\frac{x-\sin x}{x^2\sin x}\right)=-\frac12+\frac16=-\frac13.$$
H: Find a positive-definite matrix $B$ such that $\langle A,B\rangle_F=0$ Let $A$ denotes an $n\times n$ Hermitian matrix. Is there any positive-definite matrix, $B$, such that $\langle A,B\rangle_F=0$? AI: Such a positive definite $B$ exists if and only if $A$ is zero or indefinite. If $B$ is positive definite and $A$ is semidefinite but nonzero, then $B^{1/2}AB^{1/2}$ is semidefinite but nonzero. Hence $\langle A,B\rangle_F=\operatorname{tr}(AB^\ast)=\operatorname{tr}(B^{1/2}AB^{1/2})\ne0$. If $A$ is zero or indefinite, let $A=U\operatorname{diag}(x)U^\ast$ be a unitary diagonalisation. Then either $x$ is zero or $x$ has a pair of entries of different signs. Hence there exists a positive vector $y$ that is orthogonal to $x$. Let $B=U\operatorname{diag}(y)U^\ast$ and we are done.
H: How to find the series $\sum_{n=0}^{\infty} \frac{(-1)^n x^{4n}}{(4n!)}$ It is known that, $$\sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!}=\cos(x)$$ and $$\sum_{n=0}^{\infty} \frac{x^{2n}}{(2n)!}=\cosh(x)$$ From this follows that $$\sum_{n=0}^{\infty} \frac{x^{4n}}{(4n)!}=\frac{\cos(x)+\cosh(x)}{2}$$ How to find $$\sum_{n=0}^{\infty} \frac{(-1)^n x^{4n}}{(4n)!}$$ AI: To expand a little on @Angina Seng's point, essentially the thing to notice is that if we can find the fourth root of ${-1}$, and replace $x$ with $x$ multiplied by this fourth root you get $${\frac{((-1)^{\frac{1}{4}}x)^{4n}}{(4n)!}=\frac{((-1)^{\frac{1}{4}})^{4n}x^{4n}}{(4n)!}=\frac{(-1)^nx^{4n}}{(4n)!}}$$ There are actually four different fourth roots of ${-1}$: $${e^{\frac{i\pi}{4}},e^{\frac{-i\pi}{4}},e^{\frac{3i\pi}{4}},e^{\frac{-3i\pi}{4}}}$$ So any one of these will suffice. If we take the principal root (${e^{\frac{i\pi}{4}}}$) though, this would make your answer $${\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n}}{(4n)!}=\frac{\cos\left(e^{\frac{i\pi}{4}}x\right)+\cosh\left(e^{\frac{i\pi}{4}}x\right)}{2}}$$ Just as others have said. But we don't have to pick that root, $${\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n}}{(4n)!}=\frac{\cos\left(e^{\frac{3i\pi}{4}}x\right)+\cosh\left(e^{\frac{3i\pi}{4}}x\right)}{2}}$$ is also just as valid Edit: I won't write the full working (I will leave it as an exercise to you) - but we can actually further then simplify the result to $${\sum_{n=0}^{\infty}(-1)^n\frac{x^{4n}}{(4n)!}=\cos\left(\frac{x}{\sqrt{2}}\right)\cosh\left(\frac{x}{\sqrt{2}}\right)}$$
H: Evaluate the sum $\sum_{m,n\geq 1}\frac{1}{m^2n+n^2m+kmn} $ Motivated by the double summation posted here and recently here too. I came up with the general closed form for one parameter $k>0$ where $k$ being a positive integer. $$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{m^2n+n^2m+kmn}=\frac{H^2_{k}-\psi_1(k+1)}{k}+\frac{\zeta(2)}{k}$$ where $H_n$ is nth Harmonic number and $\psi_1(x)$ is trigamma function. How do I prove the above equality is true? AI: Note that $$\frac{1}{m^2n+n^2m+kmn}=\frac{1}{mn}\int_0^1 x^{m+n+k-1}dx $$ and thus $$\sum_{m,n\geq 1}\frac{1}{mn}\int_0^1 x^{k-1}\cdot x^{m+n} dx=\int_0^1 x^{k-1} \sum_{m,n\geq 1}\frac{x^{m+n}}{mn}dx=\int_0^1x^{k-1} \ln^2(1-x)dx \cdots(1) $$ where we exploit the series of $\ln(1-x)$. Now by definition of beta function we have $$B(k,y)=\int_0^1 x^{k-1} (1-x)^{y-1}dx\cdots(2)$$ having second derivatives of $(2)$ at $y=1$ we obtained the right hand expression of $(1)$. That $$\lim_{y\to 1^{+}}B(k,y)=\int_0^1 x^{k-1} \ln^2(1-x)dx$$ We evaluate left hand side of $(2)$ $$\lim_{y\to 1^+}\frac{\partial^2}{\partial y^2} B(k,y)= \lim_{y\to 1^+}\frac{\partial }{\partial y}B(k,y)\left(\psi_0(y)-\psi_0(k+y)\right)$$ on further differentiation and setting $y=1$ we have $$ B(k,1)\left((-\psi_0(k+1)-\gamma)^2-\psi_1(k+1)+\frac{\pi^2}{6}\right)=\frac{H_k^2-\psi_1(k+1)}{k}+\frac{\zeta(2)}{k}$$ now we have desired closed form as $$\frac{H_{k+1}^2-\psi_1(k+2)}{k+1}+\frac{\zeta(2)}{k+1}=\frac{H_k^2+H_{k}^{(2)}}{k}$$ Alternatively We show that $$ \int_0^1 x^{k-1} \ln^2(1-x)dx =\frac{{H_k^2}+H_k^{(2)}}{k}$$ Using the integration trick used in the book, (Almost) Impossible integrals, sums and series by Sir Cornel loan Vălean, page no 59-60. Consider the integral of the form for all $n\geq 1$.$$I(n) = \displaystyle \int_0^1 x^{n-1} \ln(1-x) dx =\frac{1}{n}\displaystyle \int_0^1 \frac{d}{dx}(x^{n}-1)\ln(1-x) dx$$ and by integration by parts we yield $$I(n)=\frac{1}{n}\left[\underbrace{(x^{n}-1)\ln(1-x)}_{0}\right]_0^1-\frac{1}{n}\int_0^1\frac{1-x^{n}}{1-x}dx $$ $$=-\frac{1}{n}\int_0^1\sum_{j=1}^{n} x^{j-1} =-\frac{1}{n}\sum_{j=1}^{n}\int_0^1 x^{j-1}dx=-\frac{1}{n} \sum_{j=1}^n\frac{1}{j} =-\frac{H_n}{n}\cdots (3)$$ Further, consider the integral $ I(k)=\displaystyle \int_0^1 x^{k}\ln^2(1-x) dx$ for $ k\geq 0 $ which further can be written as $\displaystyle \frac{1}{k+1}\int_0^1 \frac{d}{dx}(x^{k+1}-1)\ln^2(1-x) dx$ and hence on Integration by parts we see that $$I(k)=\frac{1}{k+1}\left[\underbrace{(x^{k+1}-1)\ln^2(1-x)}_{0}\right]_0^1-\frac{2}{k+1}\int_0^1\frac{1-x^{k+1}}{1-x}\ln(1-x)dx $$ $$=-\frac{2}{k+1}\int_0^1\sum_{n=1}^{k+1} x^{n-1}\ln(1-x)=-\frac{2}{k+1}\sum_{n=1}^{k+1}\int_0^1 x^{n-1}\ln(1-x)dx$$ Plugging the result from $(3)$ to last integral we have $$\frac{2}{k+1}\sum_{n=1}^{k+1}\frac{H_n}{n}=\frac{2}{k+1}\left(\frac{H_{k+1}^2+H_{k+1}^{(2)}}{2}\right)=\frac{H_{k+1}^2 +H_{k+1}^{(2)}}{k+1}$$ Further note that nth partial sum of $\displaystyle H_{k+1}^{(2)} = \zeta(2) -\psi^1(k+2)$ giving us required result $$\frac{H_{k+1}^2-\psi^1(k+2)}{k+1}+\frac{\pi^2}{6(k+1)} $$ to get the desired integral we let $k\to k-1$ which finally result us $$\frac{H_{k}^2+H_{k}^{(2)}}{k}=\frac{H_{k}^2-\psi_{1}(k+1)}{k}+\frac{\zeta(2)}{k}$$
H: If $\sum_n \sqrt{a_n a_{n+1}}$ converges, then $\sum_n a_n$ might not converge. This is the other direction of this question that reads Let $a_n>0$ (!) for each $n\in\mathbb{N}$. Then $\sum_{n=1}^\infty a_n<\infty$ implies $\sum_{n=1}^\infty \sqrt{a_n a_{n+1}}<\infty$. I need to prove that "the converse of the last statement is false". I don't want the answer, but to ask if I can use the following argument to prove the statement. To start, I'm confused with the statement "the converse of the last statement is false". Is it enough if I prove, if $\sum_n \sqrt{a_n a_{n+1}}$ diverges, then $\sum_n a_n$ must diverge? If this is true, then the proof is easy since I can choose $a_n=1/n$ and prove that $\sum_n a_n$ converges. AI: Addressing the logic of the question rather than the mathematical content: The original implication is: Let $a_n>0$ for each $n\in\mathbb{N}$. Then $\sum_{n=1}^\infty a_n<\infty$ implies $\sum_{n=1}^\infty \sqrt{a_n a_{n+1}}<\infty$. This is an assertion about every sequence in the world, so proving it true requires a general proof that works for all sequences. The converse of the original implication is: Let $a_n>0$ for each $n\in\mathbb{N}$. Then $\sum_{n=1}^\infty \sqrt{a_n a_{n+1}}<\infty$ implies $\sum_{n=1}^\infty a_n<\infty$. If you wanted to prove this true, you would again need an argument that worked for all sequences. But to prove it false, you just need a single counterexample—that is, provide a single sequence $\{a_n\}$ such that $\sum_{n=1}^\infty \sqrt{a_n a_{n+1}}<\infty$ is true but $\sum_{n=1}^\infty a_n<\infty$ is false.
H: Can't solve quadratic function I'm trying to get a simple quadratic function of the form $y=ax^2+bx+c$ which goes through the following points: $$(0;0) \;(\frac d2;2)\; (d;0)$$ The solution I've calculated $3$ times on paper has always been $y=\frac8d- \frac{8x^2}{d^2}$, but this can't be true, for an arbitrary $d$, putting $x$ as $0$ should result in $0$, but $\frac8{10}-8\times{\frac0{100}}$ is $\frac 8{10}$, and not $0$. Neither does input $d$ result in $0 \to y=\frac 8d$ Am I just approaching this problem incorrectly or am I daft? AI: The key bit to notice is that you've been given the roots of such a quadratic. Namely the roots are ${x=0}$ and ${x=d}$. This must mean your quadratic can be written of the form $${y=Ax(x-d)}$$ (since any polynomial can be factorised over it's roots) where ${A}$ is a constant to be determined. If we plug in ${x=\frac{d}{2}}$, want the result to be ${2}$. As an equation, that is $${A\frac{d}{2}\left(\frac{d}{2}-d\right)=2}$$ We can then rearrange for ${A}$ to get $${A=\frac{2}{\frac{d}{2}\left(\frac{d}{2}-d\right)}=\frac{-8}{d^2}}$$ Hence $${y=\frac{-8}{d^2}x(x-d)=-\frac{8}{d^2}x^2 + \frac{8}{d}}$$
H: Show that a finite D-dimensional Hilbert Space has $D^2$ operators I'm studying Group Theory and i've just arrived at this chapter about Hilbert Spaces and the author states that a this finite Hilbert Space has $D^2$ operators, one being the trivial one, by completeness: \begin{equation} \sum_{j=0}^D\,\|\,j\,\rangle\,\langle\,j\,\|=1 \quad,\quad\langle\,j\,\|\,k\,\rangle=\delta_{jk} \end{equation} So, it leaves $(D^2-1)$ non-trivial linear operators. My questions are: How do I prove that for this space, it has $D^2$ operators? The author also states that it has $(D-1)$ diagonal combinations such that: \begin{equation} (\|1\rangle\langle1\|-\|2\rangle\langle2\|),...,(\|1\rangle\langle1\|+\|2\rangle\langle2\|+...+\|D-1\rangle\langle D-1\|-(D-1)\|D\rangle\langle D\|) \end{equation} How can I show this? Thanks in advance for any help, and by the way, it's not a homework. It's a study of my own to my undergraduate thesis. AI: Hopefully you are misquoting: what the author means is that there are $D^2$ linearly independent linear operators, forming a basis of the linear operators on this $D$-dimensional Hilbert space. For example, consider the operators $\| i \rangle \langle j\|$ for $i, j = 1 \ldots D$.
H: Why isn't the Disjoint Union in Set a product in addition to being a coproduct? So I'm understanding that in $Set$ that the cartesian product is a categorical product, and further get why the disjoint union is a categorical coproduct, but why is it also not a product? I want to understand where I'm going wrong in my reasoning for why it seems like it could be a product. If I define disjoint union to be $$ A \sqcup B := \{ a \in A\ \| \ (0, a) \} \cup \{ b \in B\ \| \ (1, b) \} $$ then I can define a projection $p: A \sqcup B \to A$, as $$ p := \{ (i, x) \in A \sqcup B, i =0\ \|\ x \}$$ and respectively $q: A \sqcup B \to B$ by setting $i = 1$ above. Now with that, pick an arbitrary object $V$ with morphisms $f: V \to A, g: V \to B$, then there exists a morphism $V \to A \sqcup B$, defined as $$ h := \{ v \in V \ \| \ (0, f(v) \} \cup \{ v \in V \ \| \ (1, g(v) \}$$ Can you explain where my reasoning falls apart? If I were to take a guess, it's that my definition of disjoint union is improper, because the indexes can be arbitrary, hence there cannot be cannonical projection functions from the disjoint union to its parts, because the indexes separating the parts are neither known, nor do they tell us if they correspond to $A$ or $B$. Does that hunch sound correct? Even in that, my instinct is to ask why we couldn't construct a definition of disjoint union where you always know how to access elements of any of its objects in a canonical way. That is, where the index always starts at zero, and for each successive object between a disjoint union of two or more sets, the next index is defined to be the successor to the largest existing index in the disjoint union. AI: Your problem is with the definition of $p$, which doesn't really make sense. You could define it as a partial function sending $a\in A$ to itself and undefined on $B$, but by definition it needs to be defined on all of $A\sqcup B$. You could define it arbitrarily on $B$, but that just wouldn't satisfy the requirements for a product. Say you have a set $V$ and two functions $f:V\to A,g:V\to B$, there doesn't necessarily exist a function $H :V\to A\sqcup B$ such that its composition with $p$ and $q$ are $f$ and $g$. EDIT. Here's an example. $A=\{1,2,3\},B=\{4,5,6\}$, $V=\{0\}$. There are $3$ maps from $V$ to $A$ and $3$ from $V$ to $B$, so, by definition of the product, there are $9$ maps from $V$ to the product of $A$ and $B$ (since they are precisely characterized by specifying one map from $V$ to $A$ and one map from $V$ to $B$). But there are only $6$ maps from $V$ to $A\sqcup B$, so this is impossible. I am sure someone who is more into categories than me would feel that a different contradiction ought to be presented, but it seems to me this is the easiest thing to write down. Perhaps you should read the proof of the fact the product is unique up to (unique) isomorphism, and see where it fails when you try to define $A\sqcup B$ as a product.
H: Calculus of Variations: Looking for theorem that ensures that a given variational problem has maxima and minima Is there a theorem that garuantees that a variational problem $I[y] = \int_a^bF(x,y,y')dx$ has local/ global maxima and minima? Perhaps similar to the extreme value theorem for continuous functions on compact sets?! Specifically once I obtain a solution to the Euler- Lagrange equations how can I show that such a solution is an extemum/ minimum / maximum? Are there similar theorems for constrained problems? Many thanks! AI: If $F$ is a continuous function and your constraint set is compact, the Weierstrass theorem still applies. The problem is that compactness in function spaces is much more complicated than in finite-dimensional vector spaces. What it sounds like you really want is second-order sufficient conditions: "Specifically once I obtain a solution to the Euler- Lagrange equations how can I show that such a solution is an extemum/ minimum / maximum?" The corresponding versions of the SOSC's for optimal control come in two flavors. Arrow's sufficient conditions are on the fundamentals of the problem, while Mangasarian's are on the optimized Hamiltonian. This is a pretty good reference, if you are at a university: https://www.jstor.org/stable/2525753?seq=1 Otherwise, googling around for "Arrow-Mangasarian Sufficient Conditions, Optimal Control" will turn up thousands of references.
H: Is it true that $\mathbb{E}[X^n] > \mathbb{E}[X^{n-1}]\mathbb{E}[X]$ for all $n \geq 2$? Let $X$ denote a random variable with a smooth distribution over $[0, 1]$. Is it true that $$\mathbb{E}[X^n] > \mathbb{E}[X^{n-1}]\mathbb{E}[X]$$ for all integers $n = 2, 3, ...$? This seems to be a simple application of Jensen's inequality: however, the exact proof eludes me! My thoughts so far: In the case of $n = 2$, we know that $$\mathbb{E}[X^2] > \mathbb{E}[X]\mathbb{E}[X] = \mathbb{E}[X]^2$$ by Jensen's inequality (since the function $f(x) = x^2$ is strictly convex on $[0, 1]$). Similarly, Jensen's inequality tells us that $$\mathbb{E}[X^n] > \mathbb{E}[X]^{n} = \mathbb{E}[X]^{n-1}\mathbb{E}[X]$$ but that is, unfortunately, not quite the inequality I am after. Finally, in the case where $X$ is uniformly distributed, one can calculate that $$ \mathbb{E}[X^n] = \frac{1}{n+1}$$ and similarly $$ \mathbb{E}[X^{n-1}] = \frac{1}{n}$$ $$ \mathbb{E}[X] = \frac{1}{2}$$ which allows one to verify that $$\mathbb{E}[X^n] = \frac{1}{n+1} > \mathbb{E}[X^{n-1}]\mathbb{E}[X] = \frac{1}{2n}$$ for all $n > 1$. So this inequality holds in the uniform case, at least. AI: By Hölder's inequality, for any random variable $X$ for which $\mathbb E[\|X\|^n]$ exists we have $$ \mathbb E[\|X\|^{n-1}] \le (\mathbb E[\|X\|^n])^{(n-1)/n} $$ and $$ \mathbb E[\|X\|] \le (\mathbb E[\|X\|^n])^{1/n} $$ The product of these is $$ \mathbb E[\|X\|^{n-1}]\; \mathbb E[\|X\|] \le \mathbb E[\|X\|^n]$$ Moreover, equality holds iff $\|X\|$ is almost surely equal to a constant.
H: What's the minimal positive integer solution to $13n^2+1 = m^2$ What's the minimal positive integer solution to $13n^2+1 = m^2$? My first intuition was that $m$ has to be $13k+1$ or $13k-1$. But how do I proceed to find the minimum solutions? The minimum I found through computer program is $(m,n) = (649,180)$ AI: Rearrange $$1 = m^2 - 13 n^2$$ Continued fraction of $$\sqrt{13} = [3; \overline{1, 1, 1, 1, 6}].$$ Convergents: are m/n = 4, 7/2, 11/3, 18/5, 119/33, 137/38, 256/71, 393/109, 649/180. checking each one we find the smallest solution is $$649^2 - 13\cdot 180^2 = 1.$$
H: Expected value and Lindeberg condition The Lindeberg condition (classic) states that for a random variable $X$ with finite mean and variance, $\mu$ and $\sigma^2$, and for every $\varepsilon>0$ $$E(X^2 \boldsymbol{I}_{\{ X>\varepsilon\sqrt{n}\sigma\}}) \rightarrow 0, \quad \mbox{as} \quad n\rightarrow \infty,$$ where $\boldsymbol{I}$ is the indicator function. Now I'd like to prove this. Let's say $X$ is a continuous random variable with distribution $P_X$, then $$E(X^2 \boldsymbol{I}_{\{ X>\varepsilon\sqrt{n}\sigma\}}) = \int_{X>\varepsilon \sqrt{n}\sigma}x^2 dP_X$$ I think I get the intuition of this integral: for a fixed $\varepsilon$ when $n$ gets larger, the set $X>\varepsilon\sqrt{n}\sigma$ gets smaller. So I'm integrating on a set that "tends" to the empty set. I thought the proof may be something like $$\sigma^2 + \mu^2 = \int_{\mathbb{R}}x^2 dP_X = \int_{X\leq\varepsilon \sqrt{n}\sigma}x^2 dP_X + \int_{X>\varepsilon \sqrt{n}\sigma}x^2 dP_X,$$ but I'm not sure. Thanks AI: Note that since the mean and variance are finite, then $$ \sigma^2 + (EX)^2 = EX^2 = M < \infty $$ Now, you know that $EX^2 = \int x^2 dP_x$, so as you did, you break down the integral $$ EX^2= \int_{X\leq \epsilon \sqrt n \sigma} x^2 dP_x + \int_{X>\epsilon \sqrt n \sigma} x^2 dP_x $$ Then, taking the limit $$ \lim_{n \to \infty } EX^2 = \int_{X\leq \infty} x^2 dP_x +\lim_{n\to \infty} \int_{X>\epsilon \sqrt n \sigma} x^2 dP_x = M + \lim_{n \to \infty} \int_{X>\epsilon \sqrt n \sigma} x^2 dP_x $$ Which proves what you wanted to show.
H: Convergence of a series with divergent odd and even sub-term series If $\sum_{n=0}^{\infty} a_{2 n}$ and $\sum_{n=0}^{\infty} a_{2 n+1}$ both diverge, then prove/disprove $\sum_{n=0}^{\infty} a_{n}$ diverges as well. Can we take the counters as $$\sum_{n=1}^\infty a_n$$ Where $a_n=\left\{ \begin {array}{ll} \frac{1}{n}& n \textit{ is odd}\\ \frac{-1}{n}& n \textit{ is even} \end{array} \right.$ AI: That works perfectly. You will want to be careful in proving that $\sum_{n=1}^\infty a_n$ actually does converge - using the limit of the partial sums.
H: How can I prove that $\sum_{n=0}^{\infty}\frac{\sin(7n)}{13^n}=\frac{13 \sin(7)}{170 - 26 \cos(7)}$? $$\sum_{n=0}^{\infty}\frac{\sin(7n)}{13^n}=\frac{13 \sin(7)}{170 - 26 \cos(7)}$$ Have no clue how to prove it. Possibly rewrite $\sin(7n)$ as $\frac{1}{2\sin(7)}\left(\cos(7n-7)-\cos(7n+7)\right)$. But what next? AI: To follow up on my comment: using the geometric series we have $$ \sum_{n=0}^{\infty}\frac{\sin(7n)}{13^n} = \Im\left(\sum_{n=0}^{\infty}\left(\frac{e^{7i}}{13}\right)^n\right) $$ $$ =\Im\left(\frac{13}{13-e^{7i}}\right) $$Multiply by the conjugate and take the imaginary part. $$ =\Im\left(\frac{13(13-e^{-7i})}{13^2+1-26\cos(7)}\right) $$ $$ =\frac{13\sin(7)}{170-26\cos(7)} $$
H: Does $-6(x-\frac{1}{2})^2$ = $-\frac{3}{2}(2x-1)^2$? I'm confused about a question I posted this morning. I am trying to understand if $-6(x-\frac{1}{2})^2$ can be rewritten as $-\frac{3}{2}(2x-1)^2$? I tried multiplying out the expression $-6(x-\frac{1}{2})^2$ to a polynomial form $36x^2-36x+9$ but that didn't take me closer to understanding my goal. I noticed that I can remove the fraction inside $-6(x-\frac{1}{2})^2$ by doubling the contents: $-6(x-\frac{1}{2})^2$ <> $-6(2x-1)^2$ # used <> for does not equal I don't think I can simply half the factor -6 to get $-3(2x-1)^2$ As you can no doubt see, I am confused. Does $-6(x-\frac{1}{2})^2$ = $-\frac{3}{2}(2x-1)^2$ ? If it does could someone show me how to transform from $-6(x-\frac{1}{2})^2$ to $-\frac{3}{2}(2x-1)^2$ in granular baby steps? AI: Look at my answer, it's quite simple to prove $$-6\left(x-\frac{1}{2}\right)^2=-6\left(\frac{2x-1}{2}\right)^2=-6\frac{\left(2x-1\right)^2}{2^2}=-6\frac{\left(2x-1\right)^2}{4}=-\frac 32(2x-1)^2$$
H: What is the normalised arc length measure on the unit circle? Consider the following fragment from Murphy's '$C^*$-algebras and operator theory': Can someone explain what the normalised arc length measure on $\Bbb{T}$ is? Is this some translation of Lebesgue measure on the circle? AI: Are you familiar with Haar measure? If not, take the map $[0,1)\mapsto \mathbb{T}$ given by $t\mapsto e^{2\pi it}$. This is a bijection, and you can transfer the Lebesgue measure from $[0,1)$ onto the unit circle. Note that translation invariance on the unit interval translates (no pun intended) into invariance under multiplication on the unit circle. Haar measure is a generalization of Lebesgue measure to other (sufficiently well-behaved) groups. The structure $(\mathbb{R},+)$ is a locally compact Hausdorff group, and the Lebesgue measure on $\mathbb{R}$ is translation invariant, i.e. $$\lambda(x+A)=\lambda(A)$$ for any real $x$ and any measurable set $A$. The construction I outlined above yields a measure $m$ on the unit circle that satisfies $$ m(z·A)=m(A) $$ for every $z$ in the unit circle and every measurable set $A$. Note that $\mathbb{T}$ is also a locally compact Haussdorff group. Also, the measure $m$ is regular since the Lebesgue measure is. These properties characterize general Haar measures, see here for more details https://en.wikipedia.org/wiki/Haar_measure If you know about Fourier analysis, looking at Fourier transforms on $\mathbb{T}$ and Fourier series of periodic functions can be unified under the same framework: you're studying the generalized Fourier transform with respect to different Haar measures. See Rudin's Fourier Analysis on Groups for more information.
H: Sum of i.i.d random variables equals to infinity Let $\{X_i\} $ be i.i.d random variables in $\mathbb{R}+$. When $\sum_{i=1}^{\infty} X_i=\infty$? AI: There exists $\epsilon > 0$ such that $\mathbb{P}(X_1 \geq \epsilon) > 0$. Then by the 2nd Borel-Cantelli Lemma, $$\mathbb{P}(X_n \geq \epsilon \text{ i.o.}) = 1.$$ Now on the event $\{X_n \geq \epsilon \text{ i.o.}\}$, $$ \sum_{n=1}^{\infty} X_n \geq \sum_{n=1}^{\infty} \epsilon \mathbf{1}_{\{ X_n \geq \epsilon\}} = \infty. $$
H: How to calculate line integral using Green's theorem I had this specific task in my math exam and didn't solve it correctly. Also, I, unfortunately, don't have any correct result. So I am asking you, if anyone can solve and explain it to me. I would be super grateful. Sorry for my bad English, it's my second language. Calculate the line integral where l is upper section of circle $$x^2+y^2=16x$$ form point A(16,0) to point B(0,0) $$\int_l (e^xsiny-7y) dx + (e^xcosy-7)dy$$ I tried doing this : First, I wrote circle like this: $$ x^2−16x+y^2=0,(x−2\sqrt{2})^2+y^2=2\sqrt{2} $$ Then, I wrote that $$ P=e^xsiny−7y $$ and $$ Q=e^xcosy−7 $$ After that I calculated derivatives $$ \frac{dP}{dy} $$ and $$ \frac{dQ}{dx} $$ , put them back in double integral, using Green's theorem. Meaning I did the following: $\iint_D (\frac{dQ}{dx} - \frac{dP}{dy})\,dx\,dy$ I used the points A and B to define the boundaries, but failed when I got zero as an result. AI: You have the right idéa! Greens theorem states the if $l$ is a closed curve oriented counter clockwise and $D$ is the area which $l$ encloses then $$\int_{l}P\,dx+Q\,dy=\iint_{D}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\,dx\,dy.$$ In your case the simple closed line $l$ is the $x$-axis and the part of the circle $x^{2}+y^{2}=16x$ which lies above the $x$-axis. We also have $P=e^{x}\sin(y)-7y$ and $Q=e^{x}\cos(y)-7$. We then calculate the partial derivatives: \begin{align} \frac{\partial Q}{\partial x} &= e^{x}\cos(y) \\ &\text{and}\\ \frac{\partial P}{\partial y} &= e^{x}\cos(y)-7\end{align} and therefore $$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=e^{x}\cos(y)-(e^{x}\cos(y)-7)=7.$$ This is really nice since \begin{align}\int_{l}(e^{x}\sin(y)-7y)\,dx+(e^{x}\cos(y)-7)\,dy &=\iint_{D}7\,dx\,dy\\ &= 7\times \text{Area}(D).\end{align} We can re-write the circles equation as $$(x-8)^{2}+y^{2}=8^{2}$$ which is a circle with middle point $(8,0)$ and radius $8$. This means that $D$ is precisely the half disc that is part of this circle and lies above the $x$-axis. Therefore the area of $D$ is $$\frac{1}{2}(8^{2}\pi)$$ so the answer is $$7\times 32\times\pi= 224\pi$$
H: Probability of exactly $2$ sixes in $3$ dice rolls where $2$ dice have $6$ on $2$ faces? Three dice are rolled. One is fair and the other two have 6 on two faces. Find the probability of rolling exactly 2 sixes. My textbook gives an answer of $\frac{20}{147}$ but I get an answer of: $$\frac{1}{6}\frac{2}{6}\frac{4}{6}+\frac{1}{6}\frac{4}{6}\frac{2}{6}+\frac{5}{6}\frac{2}{6}\frac{2}{6}=\frac{8}{216}+\frac{8}{216}+\frac{20}{216}=\frac{36}{216}=\frac{1}{6}$$ I just want to know where I am going wrong or could the textbook be mistaken ? AI: Note that the official answer is correct if you make the (unnatural) assumption that the two non-standard dice have seven sides (two of which show $6$). In that case the answer is $$\frac 27\times \frac 27\times \frac 56+2\times \frac 27\times \frac 57\times \frac 16=\frac {20}{147}$$ To be sure, this was arrived at by reverse engineering, not by any sensible reading of the problem.
H: Show that if $x \in \partial (A \cap B)$ and $x \not\in (A \cap \partial B)\cup (B \cap \partial A)$, then $x \in \partial A \cap \partial B$. This question is for my exam prep. I want to solve the following example: Show that if $x \in \partial (A \cap B)$ and $x \not\in (A \cap \partial B)\cup (B \cap \partial A)$, then $x \in \partial A \cap \partial B$, where $A, B \subseteq \mathbb{R^n}$. AI: Fix a neighbourhood $U$ of $x$. Since $x\in \partial (A\cap B)$, there is some $y\in A\cap B$ which belongs to $U$. In particular, $y\in A$. This shows that $U\cap A\neq\emptyset$. We'll be done if we can argue that $U\setminus A\neq\emptyset$. Since $x\not \in A\cap \partial B$, we have that either $x\not \in A$ or $x\not\in\partial B$. If $x\not \in A$, then we're done. Suppose thus that $x\not\in \partial B$. That means there is some neighbourhood $V$ of $x$ which is either disjoint from $B$ or disjoint from $B^c$. By replacing $V$ with $U\cap V$, we can assume wlog that $V\subset U$. Again, $x\in\partial (A\cap B)$, so there is some $z\in A\cap B$ with $z\in V$. It follows that $V\cap B^c=\emptyset$. On the other hand, we can use that $x\in\partial (A\cap B)$ to infer the existence of $w\not\in A\cap B$ and $w\in V$. This means that $w\not \in A$ or $w\not\in B$. As $V\cap B^c=\emptyset$, it follows that $w\in B$, so $w\not\in A$. But then $w\in V\setminus A\subset U\setminus A$. In particular, $U\setminus A\neq \emptyset$. We've shown that $x\in\partial A$. By symmetry, $x\in\partial B$, and we're done.
H: Verify that $\mathcal{A}$ is a $\pi$-system and also determine $\lambda\langle\mathcal{A}\rangle$, the $\lambda$-system generated by $\mathcal{A}$. Let $\Omega$ be a nonempty set and $\{A_{i}\}_{i\in\mathbb{N}}$ be a sequence of subsets of $\Omega$ such that $A_{i+1}\subset A_{i}$ for all $i\in\mathbb{N}$. Verify that $\mathcal{A} = \{A_{i}:i\in\mathbb{N}\}$ is a $\pi$-system and also determine $\lambda\langle\mathcal{A}\rangle$, the $\lambda$-system generated by $\mathcal{A}$. MY ATTEMPT Let $A\in\mathcal{A}$ and $B\in\mathcal{A}$. Then either $A\subseteq B$ or $B\subseteq A$. Without loss of generality, let us assume that $A\subseteq B$. Then we have that $A\cap B = A\in\mathcal{A}$. Thus $\mathcal{A}$ is a $\pi$-system. Since $\mathcal{A}$ is a $\pi$-system, one has that $\lambda\langle\mathcal{A}\rangle = \sigma\langle\mathcal{A}\rangle$, the $\sigma$-algebra generated by $\mathcal{A}$. But I do not know to proceed from here. Can someone help me with this? AI: As you noted, $\lambda(\mathcal A)=\sigma(\mathcal A)$. First define $\mathcal B=\{A_1,A_2\cap A_1^c,A_3\cap A_2^c,...\}$ and (exercise!) $\sigma(\mathcal A)=\sigma(\mathcal B)$ so that $\lambda(\mathcal A)=\sigma(\mathcal B)$. Now $\sigma(\mathcal B)$ is just going to be the set of all possible countable unions of sets of the type $(\bigcup A_n)^c\cup A_i\cap A_{i-1}^c$ or $A_i\cap A_{i-1}^c$ since $\mathcal B$ is a set of partitioning sets for $\Omega$.
H: Prove that montonicity and continuity imply bijectivity. Here's the question: Let $f: [a,b] \to [f(a),f(b)]$ be monotonically increasing and continuous. Prove that $f$ is bijective. Proof Attempt: Let $f: [a,b] \to [f(a),f(b)]$ be monotonically increasing and continuous. Since it is monotonically increasing, it is injective. So, all we have to do is to prove surjectivity. Let $K \in (f(a),f(b))$. We define the following sets: $$E = \{x \in [a,b]: f(x) < K\}$$ Since $E$ is nonempty and bounded above, it follows that it has a least upper bound. We denote this by $c = \sup(E)$. Now, I claim that: $$\lim_{x \to c} f(x) = K$$ This, in fact, follows from the monotonicity of $f$. If we want $\|f(x)-K\| < \epsilon$, then we can always choose a $\delta_1 > 0$ such that $0 < c-x < \delta_1 \implies \|f(x)-K\| < \epsilon$. Similarly, we can always choose a $\delta_2 > 0$ such that $0 < x-c < \delta_2 \implies \|f(x)-K\| < \epsilon$. In other words, the left & right hand limits are $K$ so the limit above is $K$. Since $f$ is continuous, we conclude that $\lim_{x \to c} f(x) = K = f(c)$. However, this just shows that $f$ is surjective. Hence, $f$ is injective and surjective so it must be bijective. Does the proof above work? If it doesn't, then why? How can I fix it? The specific thing that I'm a little unsure of is if my second last paragraph is justified or not. Like, I'm pretty sure the argument works for the left-hand limit but I'm pretty ehh about the right-hand limit. AI: Hint : Do you understand intuitively (I suggest a drawing) why you can get arbitrarily close to $K$ on both sides of $c$ ?
H: How can we show that each component of a 2-regular graph is a cycle? I have done quite a few examples and this statement turns out to be true in each case. However I'm not sure how to formally prove it. Does there exist a theorem that can help with such type of proof? AI: You need to show that a connected 2-regular graph is a cycle. Take a vertex $x$ of the graph, and consider the longest simple path $p$ (meaning no edge appears twice in $p$)starting with $x$. Let $y$ be the last vertex of $p$. Two cases a) $y=x$ and b) $y\ne x$. In the first case $p$ is a cycle, and no vertex in the graph outside $p$ is connected to a vertex in $p$, so the graph is disconnected (a contradiction) or is equal to $p$. b) There is a vertex $z$ in the graph which is connected to $y$ by an edge which is not in $p$ (because $y$ has degree $2$). Since $p$ was a maximal simple path either $z=x$ or $z$ is equal to another vertex of $p$. The second option is impossible because $p$ was the largest simple path starting at $x$. If $x=z$ then we get a cycle and the rest is as in case a).

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