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H: How Many Ways to get N by adding k integers ? I know the number of ways getting $ n $ by adding k integers is $ \binom{N-1}{k-1} $ . Now I have applied this theory to the following problem but could not get the correct answer . Why ? I have N coins having values 1,2,3, … N. I have to select a subset of exactly K coins from those such that the selected coins sum to N. How many ways can I do it ? According to my theory for N = 8 and k=3 the answer of the above problem is $ \binom{8-1}{3-1} $ . But the answer is 2 .For N=8, K=3 I can select coins in 2 ways: {1,2,5}, {1,3,4}. Where is the problem ? AI: There indeed $\binom72=21$ ways to get $8$ as the sum of three positive integers: $$\begin{array}{} 6+1+1&1+6+1&1+1+6\\ 5+2+1&5+1+2&2+5+1&2+1+5&1+5+2&1+2+5\\ 4+3+1&4+1+3&3+4+1&3+1+4&1+4+3&1+3+4\\ 4+2+2&2+4+2&2+2+4\\ 3+3+1&3+1+3&1+3+3 \end{array}$$ Notice first that we’re keeping track of the order of the terms, so that $4+3+1$ and $1+3+4$ are counted separately; in your problem the order of the coins is irrelevant, and that’s just the single solution $\{1,3,4\}$. Moreover, we’re allowing the same number to be used more than once; in the coin problem that’s not possible, because you have only one of each coin. In other words, you’re applying a formula for one kind of problem to a very different kind of problem.
H: $\int_{-\infty}^{\infty} \frac{1}{2\pi} \exp\{ -\frac{1}{2} ((y-x)^2 + x^2) \} dx$ $$\int_{-\infty}^{\infty} \frac{1}{2\pi} \exp\{ -\frac{1}{2} ((y-x)^2 + x^2) \} dx$$ What I thought of doing was expand the inside $$\int_{-\infty}^{\infty} \frac{1}{2\pi} \exp\{ -\frac{1}{2}y^2 + xy -x^2 \} dx$$ then I can take out the $ -\frac{1}{2}y^2$ part ... but how do I proceed? I got line 2 of the image below, but how did they goto line 3? AI: Hope the following answer helps .
H: showing that $n$th cyclotomic polynomial $\Phi_n(x)$ is irreducible over $\mathbb{Q}$ I studied the cyclotomic extension using Fraleigh's text. To prove that Galois group of the $n$th cyclotomic extension has order $\phi(n)$( $\phi$ is the Euler's phi function.), the writer assumed, without proof, that $n$th cyclotomic polynomial $\Phi_n(x)$ is irreducible over $\mathbb{Q}$. I know that for n=p, p is prime, $\Phi_n(x)$ is irreducible over $\mathbb{Q}$ by Eisenstein's criterion. But I don't know how $\Phi_n(x)$ is irreducible over $\mathbb{Q}$ when n is not prime. AI: The proof which follows is the one provided by Gauss and it uses modular arithmetic in very ingenious way. We will summarize the results needed as follows: 1) For a given prime $ p$, the numbers $ 0, 1, 2, \ldots, (p - 1)$ form a finite field under the operations of addition and multiplication modulo $ p$. 2) Since these numbers form a field, say $ F_{p}$, we can talk about polynomials $ f(z)$ whose coefficients are in $ F_{p}$. The set of all such polynomials, say $ F_{p}[z]$, has the unique factorization property i.e. any such polynomial can be factored as a product of irreducible polynomials in $ F_{p}[z]$ in a unique way apart from the order of the factors. The proof is same as that used for normal polynomials with rational coefficients. 3) If $ f(z)$ is a polynomial in $ F_{p}[z]$ then $ \{f(z)\}^{p} = f(z^{p})$. This is true for constant polynomials by Fermat's theorem which says that $ a^{p} \equiv a\,\,\text{mod} (p)$. For higher degree polynomials this is achieved by induction by writing $ f(z) = az^{n} + g(z)$ and using binomial theorem to raise both sides to power $ p$. In so doing we only need to note that the binomial coefficients involved are divisible by $ p$. The proof of irreducibility of $ \Phi_{n}(z)$ is done in two stages: Stage 1: Let $ \zeta$ be a primitive $ n^{th}$ root of unity and let $ f(z)$ be its minimal polynomial i.e. $ f(z)$ is monic (leading coefficient $ 1$), has rational coefficients and irreducible and $ f(\zeta) = 0$. Since $ \zeta$ is also a root of $ z^{n} - 1 = 0$, it follows that $ f(z)$ divides $ (z^{n} - 1)$ and by Gauss Lemma $ f(z)$ has integer coefficients as well. We now establish the following: If $ p$ is any prime which does not divide $ n$ then $ \zeta^{p}$ is a root of $ f(z) = 0$. Proof: Since $ \zeta$ is also a root of $ \Phi_{n}(z) = 0$ it follows that $ f(z)$ divides $ \Phi_{n}(z)$. Thus we have $ \Phi_{n}(z) = f(z)g(z)$ where $ g(z)$ is also monic and has integer coefficients (by Gauss Lemma). Since $ p$ is coprime to $ n$ it follows that $ \zeta^{p}$ is also a primitive $ n^{th}$ root. And therefore $ \Phi_{n}(\zeta^{p}) = 0$. Assuming that $ \zeta^{p}$ is not a root of $ f(z) = 0$ (otherwise there is nothing to prove), we see that it must be a root of $ g(z) = 0$. Therefore $ \zeta$ is a root of $ g(z^{p}) = 0$. Since $ f(z)$ is the minimal polynomial of $ \zeta$, it follows that $ f(z)$ divides $ g(z^{p})$ so that $ g(z^{p}) = f(z)h(z)$ where $ h(z)$ is monic with integer coefficients. Also since $ \Phi_{n}(z)$ is a factor of $ (z^n - 1)$ so that we have $ z^{n} - 1 = \Phi_{n}(z)d(z)$ where $ d(z)$ is again monic with integer coefficients. We thus have the following equations: $$z^{n} - 1 = f(z)g(z)d(z)\tag{1}$$ $$g(z^{p}) = f(z)h(z)\tag{2}$$ We now apply the modulo $ p$ operation to each of the equations above, i.e. we replace each coefficient in the polynomials involved with its remainder when it is divided by $ p$. The resulting polynomials are all in $ F_{p}[z]$ and we will use the same letters to denote them. So the above equations are now to be interpreted as relations between some polynomials in $ F_{p}[z]$. The equation $(2)$ can now be equivalently written as $$\{g(z)\}^{p} = f(z)h(z)\tag{3}$$ Let $ k(z)$ in $ F_{p}[z]$ be an irreducible factor of $ f(z)$. Then from the above equation $(3)$, $ k(z)$ divides $ \{g(z)\}^{p}$ and so divides $ g(z)$. Thus from equation $(1)$ the polynomial $ \{k(z)\}^{2}$ divides $ (z^{n} - 1)$. Thus $ (z^n - 1)$ has repeated factors and therefore $ (z^{n} - 1)$ and its derivative $ nz^{n - 1}$ must have a common factor. Since $ n$ is coprime to $ p$, therefore the derivative $ nz^{n - 1}$ is non-zero polynomial and it clearly does not have any common factor with $ (z^{n} - 1)$. We have reached a contradiction and therefore the initial assumption that $ \zeta^{p}$ is not the root of $ f(z) = 0$ is wrong. The result is now proved. Stage 2: We have thus established that if $ f(z)$ is the minimal polynomial for any primitive $ n^{th}$ root of unity then for any prime $ p$ not dividing $ n$, $ \zeta^{p}$ (which is again a primitive $ n^{th}$ root) is also a root of $ f(z) = 0$. And since $ f(z)$ is irreducible and monic, it will act as minimal polynomial for the primitive root $ \zeta^{p}$. The same logic can be applied repeatedly and we will get the result that $ f(z)$ is the minimal polynomial for $ \zeta^{p_{1}p_{2}\ldots p_{m}}$ where $ p_{1}, p_{2}, \ldots, p_{m}$ are any primes not dividing $ n$. It follows that $ \zeta^{k}$ where $ k$ is coprime to $ n$ is also a root of $ f(z)$. Thus all the primitive $ n^{th}$ roots of unity are roots of $ f(z) = 0$. Hence $ \Phi_{n}(z)$ divides $ f(z)$. Since $ f(z)$ is irreducible it follows that $ f(z) = \Phi_{n}(z)$ (both $ f(z)$ and $ \Phi_{n}(z)$ are monic). We thus have established that $ \Phi_{n}(z)$ is irreducible. Update: User Vik78 points out in comments that Gauss had proved the irreducibility of $\Phi_{n}(z)$ for prime values of $n$ only and it was Dedekind who established it for non-prime values of $n$. Note that the problem is far simpler when $n$ is prime (one easy proof is via Eisenstein's criterion of irreducibility) and the crux of the argument presented above is essentially due to Gauss. I came to know of this proof from the wonderful book Galois Theory of Algebraic Equations by Jean Pierre Tignol. Tignol mentions in his book that Gauss proved the irreducibility of $\Phi_{n}(z)$ over $\mathbb{Q}$ for non-prime values of $n$ in 1808 (see section 12.6 titled Irreducibility of cyclotomic polynomials on page 196 in the book mentioned in last paragraph). Dedekind on the other hand proved an even more powerful theorem: Theorem: Let $\zeta_{m}$ denote a primitive $m^{\text{th}}$ root of unity. If $m, n$ are relatively prime then $\Phi_{n}(z)$ is irreducible over the field $\mathbb{Q}(\zeta_{m})$. And Gauss used this result implicitly (thinking that it could be proved in similar manner as irreducibility of $\Phi_{n}(z)$ over $\mathbb{Q}$) in obtaining solution to $\Phi_{n}(z) = 0$ via radicals.
H: Inverting a quartic equation of state I have the following equation (which is an adaptation of the Beattie-Bridgeman Equation of State): $$ P = \frac{RT}{V} + \frac{B}{V^2} + \frac{C}{V^3} + \frac{D}{V^4} $$ This is a function of the form $P = f(V)$ as R, T, B, C and D are all constant with respect to P and V. From this, my ultimate goal is to derive an equation for $dV/dP$. My first step is therefore to write the equation explicit in V, however I get stuck at: $$ PV^4 - RTV^3 - BV^2 - CV - D = 0 $$ What steps am I missing in order to be able to write this in the form $V = f(P)$? AI: You have: $$ P = {RT}{V^{-1}} + {B}{V^{-2}} + {C}{V^{-3}} + {D}{V^{-4}} $$ Now diffrentiate wrt $P$: $$ 1 = {R{d{(TV^{-1}})\over dP}} + {B}{dV^{-2}\over dP} + {C}{dV^{-3}\over dP} + {D}{dV^{-4}\over dP} $$ On the RHS the last three terms are fine and you can take $dV\over dP$ common but the first term will also introduce a $dT\over dP$ So the approach to get $dV\over dP$ will be: Take the original equation $$ P = {RT}{V^{-1}} + {B}{V^{-2}} + {C}{V^{-3}} + {D}{V^{-4}} $$ and diffrentiate it wrt $P,V,T$ to get three equations. These three equations will have 3 variables $dT\over dP$or $dP\over dT$ , $dP\over dV$ or $dV\over dP$ , $dT\over dV$or $dV\over dT$ Solve the three equations to get $dV\over dP$
H: Deduce that $\sum \frac {a_n}{r_n}$ diverges. Suppose $a_n>0$ and the series $\sum a_n$ converges. Put $r_n=\sum_{n=m}^\infty a_n$. Prove that $$\frac {a_m}{r_m}+\dots+\frac {a_n}{r_n}>1- \frac {r_n}{r_m},$$ if $m<n$, and deduce that $\sum \frac {a_n}{r_n}$ diverges. My problem is to deduce that $\sum \frac {a_n}{r_n}$ diverges with $\frac {a_m}{r_m}+\dots+\frac {a_n}{r_n}>1- \frac {r_n}{r_m}$. Thank you. AI: So you know how to prove the inequality $$\frac {a_m}{r_m}+\frac{a_{m+1}}{r_{m+1}}+\ldots+\frac {a_n}{r_n} > 1-\frac{r_{n}}{r_m}.$$ Now for any fixed $m$ the right-hand side of this inequality tends to $1$, which contradicts to Cauchy's criterion.
H: Can this situation arise for a subset of a group? Let $(G, )$ be a group, let $e_G$ be the identity in $G$, and let $H$ be a non-empty subset of $G$. Can it happen that $(H, )$ be a group in its own right (with the same binary operation) but with an identity element $e_H$ that is different from $e_G$? Of course, our requirement is that $e_G$ is not in $H$, for otherwise $e_G$ will have to be the same as $e_H$. AI: No: For if we choose any $h \in H$, we see that $$e_H h = h$$ Now by cancellation in $G$, we find that $e_H = e_G$.
H: Simple group with order $\geq n!$ cannot have subgroup of index $n$. My problem is as seen in the title: For positive integer $n>1$, prove that a simple group with order $\geq n!$ cannot have subgroup of index $n$. Could anyone give me some hints on how to approach this? AI: If $[G:H]=n$ then $G$ acts on $n$ cosets of $H$. Hence there is a homomorphism $G$ into the symmetric group $S_n$. Since $\|S_n\|=n!$ then either this homomorphism has the non-trivial kernel or $G=S_n$. But in the last case $G$ also is not simple.
H: Finding a trigonometric function for a problem The question: At Dolphin Bay the depth of the water at the end of the jetty is 6 metres at high and 4 metres at low tide. High tide occurs at 11am and low tide occurs at 5pm. a). Using the information given find a trigonometric function which models the depth of the water at the end of the jetty. b. Find the depth of the water at 12:30pm. c). A boat moored at the end of the jetty needs to leave the Bay before the depth of the water falls to 4 metres. At what time after the 11am high tide will the depth of the water be 4 metres? d). Find the next time after low tide when the depth of the water will again be 4 metres. AI: A way of representing the trigonometric function which models the depth of the water is: $$5-Asin(2\pi {t\over T})$$ Where: $t=0$ is considered to be 2 pm $A=1$ -The Amplitude $T=12$-The time period Hence $$D(t)=5-sin({{\pi t}\over 6})$$
H: Closure of globe Is it true that $\overline{K}(x,r) = \mbox{cl}K(x,r)$ ? I suppose that it is true. In euclidean space I can see that it is true. But maybe in other metric spacies we can find counterexample? If not, how can I prove that fact? AI: There are numerous counterexamples. Take the metric $d$ on $\Bbb R$ defined by $$d(x,y)=\min\{\|x-y\|,1\}\;;$$ this is a metric, and it even generates the usual Euclidean topology. However, $\overline{B}(0,1)=\Bbb R$, while $\operatorname{cl}B(0,1)=[-1,1]$.
H: The ring $\ \mathbb{Z}[\zeta]\ $ where $\ \zeta \in \mathbb{C}\setminus \{ 1 \} \ $ such that $ \ \zeta^3 = 1 \ $ Some exercise I have to make states that this is a ring. $$ \mathbb{Z}[\zeta] \ := \ \{a+b\zeta: a,b \in \mathbb{Z} \} \qquad \text{where $\zeta^3=1$ and $\zeta\neq 1$}$$ I see that this is a ring with respect to "$+$", but I can't figure out why it's closed with respect to "$*$". My problem is that it appears to me that $\zeta \cdot \zeta= \zeta^2$ can't be written in the form $a+b \zeta$. I see that $\zeta^2$ is a cube root as well, but $ 1<\zeta -\zeta^2 < 2$ or else $ 1<\zeta^2 -\zeta < 2$. I convinced myself about that by making a picture of the complex unit circle. This means that the recquired expression $a+ b \zeta = \zeta^2$ does not exist. I hope that some of you are familiar with this "ring" so that you can tell me what I did wrong. AI: If $x^3 = 1$, then $$x^3 - 1 = 0 \implies (x - 1)(x^2 + x + 1) = 0$$ So $\zeta^2 + \zeta + 1 = 0$ gives a way to write $\zeta$ in the desired form. On the complex plane, $\zeta$ is either $e^{i\pi/3}$ or $e^{-i\pi/3}$; in either case, it can be confirmed by direct computation that $\zeta^2 = -1-\zeta$.
H: from $1-\sin x $ to $2 \sin^2 \left(\frac{\pi}{4} - \frac{x}{2} \right)$ How can you go from $1-\sin x $ to $2 \sin^2 \left(\frac{\pi}{4} - \frac{x}{2} \right)$? I mean how to prove that $1-\sin x = 2 \sin^2 \left(\frac{\pi}{4} - \frac{x}{2} \right)$? AI: Use $\displaystyle \sin x=\cos\left(\frac\pi2-x\right)$ and $\displaystyle\cos2y=1-2\sin^2y$
H: Proving $\sin 6\theta \cos4\theta + \cos4\theta\sin2\theta = {(\cos 2\theta \tan 8\theta)\over \sec 8\theta}$ I need to prove the following: $$\sin 6\theta \cos4\theta + \cos4\theta\sin2\theta = {(\cos 2\theta \tan 8\theta)\over \sec 8\theta}$$ How would I do this? AI: Use $$2\sin A\cos B=\sin(A+B)+\sin(A-B),$$ $$2\sin B\cos A=\sin(A+B)-\sin(A-B),$$ and then $$\sin C+\sin D=2\sin\frac{C+D}2\cos\frac{C-D}2$$ finally $$\frac{\tan y}{\sec y}=\frac{\frac{\sin y}{\cos y}}{\frac1{\cos y}}=\sin y$$
H: Effect of Moving within the Feasible Region $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is a concave function with local maximum at $\mathbf{x}^$ in a convex, closed feasible set $\mathcal{F}\subset\mathbb{R}^n$. Now consider a suboptimal point $\mathbf{x}\in\mathcal{F}$. If we move from $\mathbf{x}$ in such a direction (within the feasible region) that 1) $\mathcal{L}_1$ norm of $\mathbf{x}-\mathbf{x}^$ increases monotonically. 2) $\mathcal{L}_2$ norm of $\mathbf{x}-\mathbf{x}^*$ increases monotonically. Does it mean $f\left(\mathbf{x}\right)$ will decrease in both of the above cases (considered independently)? AI: Edit: No. Consider $f(x,y) = -(x^2+9y^2)$ defined on $\mathbb{R}^2$. The origin is the sole local and global maximum point of $f$. If we move from $(1,1)$ to $(3,0)$ along the line $x+2y-3=0$, the $L_1$ norm increases from $2$ to $3$ and the $L_2$ norms increases from $\sqrt{2}$ to $2$, yet $f(1,1)=-10<f(3,0)=-9$. If you plot a graph, you will see that $f$ increases monotonically at first, then attains the maximum value $-6.23$ at $(x,y)\approx(2.08,0.46)$ in the course of movement and then decreases monotonically to $-9$.
H: $(\frac{1}{\kappa})^2+(\frac{\dot{\kappa}}{\kappa^2\tau})^2=r^2$ Show that for a curve lying on a sphere of radius r with nowhere vanishing torsion, the following equation is satisfied: $$\left(\frac{1}{\kappa}\right)^2+\left(\frac{\dot{\kappa}}{\kappa^2\tau}\right)^2=r^2$$ Please help me doing this. Honestly, I could not do anything. So I cannot show my efforts. Thank you. AI: This is to illustrate the relationship between two question: write $\rho=1/\kappa$ and $\sigma=1/\tau$. We have $$\rho^\prime=\left(\frac{1}{\kappa}\right)^\prime=-\frac{k^\prime}{k^2}.$$ Your equation thus becomes $$\rho^2+(\rho^\prime\sigma)^2=r^2.$$ Take derivative on both side, we get $$2\rho\rho^\prime+2(\rho^\prime\sigma)(\rho^{\prime\prime}\sigma+\rho^\prime\sigma^\prime)=0,$$ manipulate the terms few time we have $$\frac{\rho}{\sigma}+(\rho^{\prime\prime}\sigma+\rho^\prime\sigma^\prime)=\frac{\rho}{\sigma}+(\rho^\prime\sigma)^\prime=0,$$ which is equivalent to the equation in the link. Remark: Judging by your question you actually want to go from a spherical curve to its condition. Here is a hint on how to start: consider a unit-speed curve $\alpha$ lies on a sphere with center $\mathbf{c}$ and radius $r$. (since an arbitrary-speed curve involves only reparameterization, the result here can be extended.) We can thus write $$(\alpha-\mathbf{c})\cdot(\alpha-\mathbf{c})=r^2.$$ Take derivative on both side, $$2(\alpha-\mathbf{c})^\prime\cdot(\alpha-\mathbf{c})=0.$$ Recall that $T=(\alpha-\mathbf{c})^\prime$, so $$T\cdot(\alpha-\mathbf{c})=0.$$ Keep on this track and utilize different Frenet formula, you will eventually get to the desired result.
H: Integral of $\sqrt{1-x^2}$ using integration by parts I was asked to solve this indefinite integral using Integration by parts. $$\int \sqrt{1-x^2} dx$$ I know how to solve if use the substitution $x=\sin(t)$ but I'm looking for the Integration by parts way. any help would be very appreciated. AI: $$I=\int 1\cdot\sqrt{1-x^2}dx=\int dx \sqrt{1-x^2}-\int \left(\frac{\sqrt{1-x^2}}{dx}\int dx\right)dx$$ $$=x\sqrt{1-x^2}-\int\frac{-2x}{2\sqrt{1-x^2}}xdx$$ $$=x\sqrt{1-x^2}+\int\frac{1-(1-x^2)}{\sqrt{1-x^2}}dx$$ $$=x\sqrt{1-x^2}+\int\frac1{\sqrt{1-x^2}}dx-I$$ Now, $\int\frac1{\sqrt{1-x^2}}dx=\arcsin x+C$
H: Frechet derivative of compact operator is compact ... this seems to be a well known fact as mentioned in this and in this manuscript. However, I was not able to find a proof or to prove it by myself. So my question is: How to prove this? Any hint or reference is appreciated. AI: Theorem. Let $\Omega$ be an open set in a Banach space $X$ and let $F \in C(\Omega,X)$. If the Fréchet derivative $F'(x_0)$ exists for some $x_0\in\Omega$, then $F'(x_0)$ is a (linear) compact operator. Proof. Assume that $F'(x_0)$ is not compact. Then one can find $\epsilon_0>0$ and a sequence $\{y_n\}_n$ such that $\\|y_n\\|\leq 1$ and $$ \\|F'(x_0)y_k - F'(x_0)y_l\\| \geq \epsilon_0 $$ for $k \neq l$. By definition of the Fréchet derivative, there is $\delta>0$ such that $$ \\|F(x_0+h)-F(x_0)-F'(x_0)h\\| \leq \frac{\epsilon_0}{4}\\|h\\| $$ provided that $\\|h\\| < \delta$. Choose $\tau$ such that $\\|\tau y_k\\|<\delta$ and $x_0+\tau y_k \in \Omega$ for all $k \in \mathbb{N}$. Then $$ \begin{split} \\|F(x_0+\tau y_k)-F(x_0+\tau y_l)\\| & \geq \\|F'(x_0) (\tau y_k-\tau y_l)\\|-\\|F(x_0+\tau y_k)-F(x_0)-F'(x_0)\tau y_k\\| \\ &{}-\\|F(x_0+\tau y_l)-F(x_0)-F'(x_0)\tau y_l\\| \geq \frac{\epsilon_0}{2}\tau. \end{split} $$ But this means that $F$ is not compact on $\Omega$, a contradiction. QED. Another proof appears in Deimling's book on nonlinear functional analysis. However, it is based on the measure of non-compactness, and it is therefore less direct.
H: Finding a minimum value of an expression I have this question to solve: A city's temperature of y degrees Celsius on a day in February is given by $y = 16 + 8 \sin \left(\dfrac{\pi t}{12}\right)$ where t is the time in hours after 9am. a). What is the minimum temperature? b). At what time is the temperature a minimum? c). What is the temperature at 5pm? Any help would be appreciated. Thanks. AI: (b) $0=y'=\frac{2\pi}{3}\cos\left(\frac{\pi t}{12}\right) \Rightarrow t=6 \Rightarrow$ The time at which the temperature is minimal is $3$pm. (a) In $t=6$, the temperature is minimal. So the minimum temperature is 24 degrees Celsius. (c) The temperature at 5pm is calculated in $t=8$, ie, $y(8)=16+4\sqrt{3}$.
H: Center of Heisenberg group- Dummit and Foote, pg 54, 2.2 Let $H(F)$ be the Heisenberg group over the field $F$ introduced in Exercise 11 of Section 1.4. Determine which matrices lie in the center of $H(F)$ and prove that $Z(H(F))$ is isomorphic to the additive group $F$. Section 1.4 defines the Heisenberg group $H(F)$ over the field $F$ to be the group of all upper-unitriangular $3\times 3$-matrices $\begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}$ with entries in $F$. When I worked out the elements in the center, I got its elements such that the diagonal was multiplied by a scalar and the element in the first row,third column possibly non-zero. So, you see that my problem was compounded as I just did not see how the center was isomorphic to only F. Thanks for any help that I might receive. AI: Let $$ Y = \begin{pmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{pmatrix} \in Z \big( H(F) \big) $$ and $$ A = \begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix} \in H(F). $$ We need to determine conditions on $Y$ so that the commutator vanishes for arbitrary $A$. $$ 0 = AY - YA = \begin{pmatrix} 0 & 0 & az - cx \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} $$ CLAIM: $x = z = 0$. Why? If $x \ne 0$, then with $a = 0$ and $c \ne 0$, $AY - YA \ne 0$. Similarly for $z$. Notice that there are no restrictions on $y$ since it doesn't appear in the commutator. Thus, the center consists of matrices of the form $$ Y = \begin{pmatrix} 1 & 0 & y \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}. $$ You can check that they form a subgroup isomorphic to the additive group underlying $F$: $$ \begin{pmatrix} 1 & 0 & y_1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & y_2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & y_1 + y_2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}. $$
H: How to get value of x and y here? $x^2=16$ $y =\sqrt{16}$ here I know that when we solve value of $x$ then we get two values $+4$ and $-4$ But why we don't' get two values of $y$. Can you please explain this. Thanks for help. AI: This is not an inequality but a system of equations, there are two numbers that squared will give 16, -4 and 4. By definition the square root function gives only one value ( by definition ), that is why $y=4$. PS : repeating by definition is on purpose.
H: Dividing n balls into n buckets so 2 are empty. We know that the balls and buckets are distinguishable, and n>2. I was searching the site for some clues but didn't really find any of them helpful. Any thoughts? AI: I'm going to assume the question is, how many ways are there to distribute $n$ distnguishable balls among $n$ distinguishable buckets in such a way as to leave exactly two buckets empty. First, choose the two buckets to be empty --- $n\choose2$ ways to do this. Now you could have $3$ balls in one bucket, one ball in every other bucket. $n-2$ ways to choose the 3-ball bucket, $n\choose3$ ways to choose the balls to go in it, $(n-3)!$ ways to distribute the other balls. Or, you could have $2$ balls in each of two buckets, one ball in each other bucket. $n-2\choose2$ ways to choose the two buckets, $n\choose4$ ways to choose the four balls that will go into those two buckets, $4\choose2$ ways to distribute the four balls to the two buckets, $(n-4)!$ ways to dispose of the other balls. Can you combine all these numbers to get the answer?
H: $(p\lor \lnot q) \land (q \lor ¬r) \land (r \lor ¬p)$ is true $\iff$ $(p, q, r$ all have the same truth-values$)$ Explain why $(p\lor \lnot q) \land (q \lor ¬r) \land (r \lor ¬p)$ is True when p,q,and r have the same truth value and it is false otherwise. (Without using a truth table ) Please help me solve this AI: When you're completely lost, sometimes the best thing to do is to draw a truth-table, even if you can't hand that in for your assignment. Doing so can be illuminating, and may help you grasp what's going on, intuitively, so that you can confirm its truth and better understand why it's true. Note that you can also rewrite your expression: $$(p\lor \lnot q) \equiv q \rightarrow p$$ $$(q \lor \lnot r) \equiv r \rightarrow q$$ $$(r \lor \lnot p) \equiv p \rightarrow r$$ "Anding" them in reverse order gives $$(p\rightarrow r) \land (r \rightarrow q) \land (q\rightarrow p)$$ Maybe seeing this equivalent expression will help you understand intuitively why the statement is true if and only if $p, q, r$ all share the exact same truth-value.
H: Dantzig's unsolved homework problems From Wikipedia: An event in George Dantzig's life became the origin of a famous story in 1939 while he was a graduate student at UC Berkeley. Near the beginning of a class for which Dantzig was late, professor Jerzy Neyman wrote two examples of famously unsolved statistics problems on the blackboard. When Dantzig arrived, he assumed that the two problems were a homework assignment and wrote them down. According to Dantzig, the problems "seemed to be a little harder than usual", but a few days later he handed in completed solutions for the two problems, still believing that they were an assignment that was overdue. What were the two unsolved problems which Dantzig had solved? AI: I think the two problems appear in these papers: Dantzig, George B. "On the Non-Existence of Tests of 'Student's' Hypothesis Having Power Functions Independent of Sigma." Annals of Mathematical Statistics. No. 11; 1940 (pp. 186-192). Dantzig, George B. and Abraham Wald. "On the Fundamental Lemma of Neyman and Pearson." Annals of Mathematical Statistics. No. 22; 1951 (pp. 87-93). Read more at http://www.snopes.com/college/homework/unsolvable.asp#6oJOtz9WKFQUHhbw.99 EDIT: In case snopes ever goes belly up, the story can be found in Albers, Reid, and Dantzig, An Interview with George B. Dantzig: The Father of Linear Programming, College Math J 17 (1986) 292-314. The interview has also been reprinted in Albers, Alexanderson, and Reid, More Mathematical People, page 67.
H: Positive definiteness of matrix? Suppose $A$ is positive definite of order $k$. Let $X$ be of order $k \times n$. Is $W := X'AX$ necessarily positive definite? Maybe it is something like a theorem of Sylvester, but I have been unable to find a theorem. Thanks AI: $W$ will be positive semidefinite, with (strict) positive definiteness achieved if and only if $X$ is of a full column rank. This is easy to see. By definition, $w^T A w \ge 0$ for all $w$, with $w^T A w = 0$ if and only if $w = 0$. So, for any $v$ we can define $w := Xv$ and get: $$v^T (X^T A X) v = (Xv)^T A (Xv) = w^T A w \ge 0.$$ As I've said before, we get zero if and only if $w = 0$, i.e., if and only if $Xv = 0$. Note that $$Xv = 0 \iff v = 0$$ if and only if $X$ is of a full column rank. In other words, $v^T W v = 0$ only for $v = 0$ (i.e., $W$ is strictly positive definite) if and only if $X$ is of a full column rank. A "theorem of Sylvester" you mention is probably Sylvester's law of inertia, which is closely related, but not exactly what you ask.
H: why is $(2) \subseteq \mathbb{Z}[\zeta]$ a prime ideal? Previously I asked something about the ring $\mathbb{Z}[\zeta]$ where $\zeta$ is a cube root of $1$ in $\mathbb{C} \setminus \{1 \} $. Could you provide me hints for showing that $(2)$ is prime Research effort I tried to find some isomorphim with $\mathbb{Z}[X] / (X^3-1)$ with the substitution of $\zeta$, but this failed because $\ X-1 \mapsto 0 \ $, but $\ X-1 \notin (X^3+1) \ $. I don't know if there is another useful isomorphism. I also tried something straight forward. Assume that $z \cdot z'=(a+b\zeta)(c+d\zeta) = 2m +2n\zeta$ for some $n,m \in \mathbb{Z}$. I calculated that $$z z' = ac+bd\zeta^2 + (ad+bc)\zeta = ac+bd (-1-\zeta) + (ad+bc)\zeta = (ac-bd)+(ad+bc-bd)\zeta$$. I took two elements $z, z' \notin (2)$ to show that $z \cdot z' \notin (2)$. This leads to a lot of case-checking though, 9 cases if i'm not mistaking. I hope that you can give me an approach that takes less time. If you think I should do the case-checking, I'm fine with it. Thank you for your answer. AI: The said ring is isomorphic to $\mathbb Z[X]/(X^2+X+1)$. The ideal $(2)$ in it is prime iff the quotient by this ideal is an integral domain. That quotient would be $\mathbb F_2[X]/(X^2+X+1)$ where $\mathbb F_2$ is the field with two elements. Now it remains to prove that $X^2+X+1$ is an irreducible polynomial over $\mathbb F_2$. This should be doable for you, no?
H: Updating eigen decomposition for a matrix after some row changes Let us say we have a matrix $A$ which has eigen decomposition $$A=UDU^{-1}$$ If some of the rows of A are changed by multiplying a constant positive value, is there a simple way to update the eigen decomposition from existing $D$ and $U$? Many thanks. AI: There is none that I know of. What you ask about is $XA$, where $X$ is a diagonal matrix. Note that by premultiplying $A$ with some diagonal matrix $D$, you generally lose Hermitianity/similarity/normality/... and any other nice structure that $A$ may have had. So, while $A$ may have only real eigenvalues (for $A$ Hermitian/symmetric) or on a unit circle (for $A$ unitary/orthogonal), $XA$ can have random complex eigenvalues. Further, when you lose normality ($A^A = AA^$, but $(XA)^(XA) \ne (XA)(XA)^$), you also lose diagonality of $D$.
H: My conjecture on almost integers. Here when I was studying almost integers , I made the following conjecture - Let $x$ be a natural number then For sufficiently large $n$ (Natural number) Let $$\Omega=(\sqrt x+\lfloor \sqrt x \rfloor)^n$$ then $\Omega$ is an almost integer . The value of $n$ depends upon the difference between the number $x$ and its nearest perfect square which is smaller than it.Can anyone prove this conjecture. Moreover, I can provide examples like $(\sqrt 5+2)^{25}=4721424167835364.0000000000000002$ $(\sqrt 27+5 )^{15}=1338273218579200.000000000024486$ AI: Well, we have that $$(\sqrt{x}+\lfloor\sqrt{x}\rfloor)^{n} + (-\sqrt{x}+\lfloor\sqrt{x}\rfloor)^{n} $$ is an integer (by the Binomial Theorem), but $(-\sqrt{x}+\lfloor\sqrt{x}\rfloor)^{n}\to 0$ if $\sqrt{x}$ was not already an integer.
H: Translation matrix without homogeneous coordinates I am looking to translate a vector, and I found this article on wikipedia, which states that I need to use homogeneous coordinates to do so. I don't get the reason why, and I don't really see how the result of the translation matrix with homogeneous coordinates is different from a translation matrix $$ \matrix{T} = \begin{bmatrix} v_x & 0 & 0 \\ 0 & v_y & 0 \\ 0 & 0 & v_z \\ \end{bmatrix} $$ so that with my vector $\vec{p} = [p_x \, p_y \, p_z]^T$ I'd get the translation as $$ \vec{r}=\matrix{T}\vec{p}= \begin{bmatrix} v_x+p_x & 0 & 0 \\ 0 & v_y+p_y & 0 \\ 0 & 0 & v_z+p_z \\ \end{bmatrix} $$ What is the reason for using the homogeneous coordinates instead of the above? AI: As you say, you don't really need to use homogeneous coordinates to represent translations. You can translate a point $\mathbf{p}$ by a vector $\mathbf{v}$ just by adding coordinates: $T(\mathbf{p}) = \mathbf{p} + \mathbf{v} = (p_x+v_x, p_y+v_y, p_z+v_z)$. The benefit of using homogeneous coordinates and $4 \times 4$ matrices is that they give you a way to represent many different kinds of geometric transforms in a uniform way. Specifically, translations, rotations, scaling, shearing, and even central (perspective) projections can all be represented this way. There are a couple of reasons this is valuable: Firstly, if you need to compose transformations (perform several in succession, one after another), you can do this just by multiplying matrices. No special-case thinking is required. Life is tidier. Secondly, if you're building graphics hardware, then all you need to do is create devices that can multiply $4 \times 4$ matrices, and these devices will then be able to perform a wide variety of different transformations, including all the ones that are needed in computer graphics. Hardware devices like uniformity.
H: Proof of Sobolev Inequality Theroem I have a short question about the proof of Theorem 2 below. I have included Theorem 1's statement since it is used in the proof of Theorem 2. Definition: If $1 \leq p < n$, the Sobolev Conjugate of $p$ is $p* := \frac{np}{n-p}$, note that p* > p. Consider the following two Theorems related to Sobolev Inequalites: Theorem 1(Gagliardo-Nirenberg-Sobolev inequality) Assume $1\leq p<n$. There exists a constant $C$, depeding only on $p$ and $n$, such that $\|\|u\|\|_{L^{p}(\mathbb{R}^{n})} \leq C\|\|Du\|\|_{L^{p}(\mathbb{R}^{n})}$ for all $u \in C_{c}^{1}(\mathbb{R}^{n})$. Where in the proof of Theorem 1 it is determined that $C = \frac{p(n-1)}{n-p} = p - \frac{p}{n-p}$. Theorem 2 Assume $U$ is a bounded open subset of $\mathbb{R}^{n}$. Suppose $u \in W_{0}^{1,p}(U)$ for some $1 \leq p < n$. Then we have the estimate $\|\|u\|\|_{L^{q}(U)} \leq C\|\|Du\|\|_{L^{P}(U)}$ for each $q \in [1,p]$, the constant $C$ depending only on $p,q,n$ and $U$. Proof: Since $u \in W_{0}^{1,p}(U)$, there exist functions $u_{m} \in C_{c}^{\infty}(U)$ ($m=1,2...$) converging to $u$ in $W^{1,p}(U)$. We extend each function $u_{m}$ to be 0 on $\mathbb{R}^{n}-\bar{U}$ and apply Theorem 1 to discover $\|\|u\|\|_{L^{p}(U)} \leq C\|\|Du\|\|_{L^{p}(U)}$. As $\|U\| < \infty$, we furthermore have $\|\|u\|\|_{L^{q}(U)} \leq C\|\|u\|\|_{L^{p}(U)}$ if $1 \leq q \leq p$. $\square$ How does the final part of the proof follow? The part which states "...As $\|U\| < \infty$, we furthermore have $\|\|u\|\|_{L^{q}(U)} \leq C\|\|u\|\|_{L^{p}(U)}$ if $1 \leq q \leq p$." Thanks for any assistance. This proof is from book 'Partial Differential Equations' by Lawrence Evans. AI: Generalized Holder will do the trick, taking the $L^r$ norm of the indicator on $U$ and the $L^{p\ast}$ norm of $u$. Note the inequality needs $1/q=1/p\ast + 1/r$ which shows what range of $q$ this applies to.
H: Find the value of $x$ such that $2^x=10$ Given that $\log 5 = 0.7$ (to one decimal place), find the value of $x$ such that $2^x = 10$ (again to one decimal place) I don't know what to do with the information that $10^{0.7} = 5$. Why is this information useful? AI: We know,$$\log_{10}10=1$$ But $$\log_{10}10=\log_{10}(2\cdot5)=\log_{10}2+\log_{10}5$$ $$\implies \log_{10}2=1-\log_{10}5=1-0.7=0.3$$ Now taking logarithm on the given equation $$x\log_{10}2=\log_{10}10=1$$ as $\log_a(b^m)=m\log_ab$
H: Does this hold in general for inverse function? Let $z=G^{-1}(y)$ be any number such that $G(z)=y$. Now, does $z>G^{-1}(y)$ imply $G(z)>y$ in general ? does $z<G^{-1}(y)$ imply $G(z)<y$ in general ? AI: It depends if $G(x)$ is an increasing function, decreasing function, or neither. ($\sin(x)$ is neither because it is increasing sometimes, and decreasing sometimes.) For example, $G(x)=x^3$ is increasing, and your conclusions are true for that function.
H: How does this set look like? I have trouble to understand the following set: $M_2 = \{ 1 \le \\|x\\|_2 \lt 2 \} \cup \{ \vec{O} \} \subset \mathbb{R}^2 $ On the left side there are scalars and on the right side there is the null vector. Can we have sets with elements of different dimension? AI: As Prahlad mentioned in the comments, the left set is really shorthand for $\{x\in\mathbb{R}^2\mid 1\leq\\| x\\|_2<2\}$ which can be read as the set of vectors in the plane such that their $2$-norm is less than $2$ and great than or equal to $1$, and so it really is a set of vectors (which is of course just a set of points in the plane). In terms of the geometry of the set $M_2$, you can view it as the union of an annulus determined by the requisite radii and only containing its 'inner boundary circle' and not the outer, together with the point at the center of the annulus.
H: If $f$ is continuous on $[a,b],$ $f(x)\geq 0 $ $(\forall x\in [a,b])$ and $\int^{b}_{a} f = 0,$ then $f = 0$ on $[a,b].$ Here is my proof: Given $\epsilon>0, \exists$ partition $P = \{x_0 =a,...,x_n =b\}$ of $[a,b]$ such that $\sum^{n}_{i=1} (M_i-m_i)\Delta x_i < \epsilon,$ where $M_i = \sup\{f(x): x\in \Delta x_i\}$ and $m_i =\inf\{f(x): x\in \Delta x_i\}.$ Since $m_i = 0,$ we have $\sum^{n}_{i=1} M_i\Delta x_i < \epsilon$ and hence $M_i < \epsilon.$ Since $\epsilon>0$ is arbitrary, $f = 0 $ on $[a,b]. $ May I know why we need $f$ to be continuous on $[a,b]?$ Why couldn't we just have $f(x)\geq 0 $ $(\forall x\in [a,b])$ and $\int^{b}_{a} f = 0 ?$ Could there be something wrong with my proof? Please advise,thank you. AI: You need $f$ to be continuous, else you can just take $f:[0,1]\to \mathbb{R}$ which is zero everywhere except one point. Now for the proof, something is wrong - why is $m_i = 0$? The usual proof goes something like this : If $f \neq 0$, then $\exists p \in (a,b)$ such that $\epsilon := f(p) > 0$ (because if not, then continuity would force it to be zero on the boundary as well). Now by continuity, there is an open interval $(p-\delta, p+\delta)$ such that $$ f(x) > \epsilon/2 \quad\forall x\in (p-\delta,p+\delta) $$ Now let $I = [p-\delta/2,p+\delta/2]$, then $$ \int_a^b f(x)dx \geq \int_I f(x)dx \geq \frac{\epsilon\delta}{2} > 0 $$ where the first inequality follows from the fact that $f(x) \geq 0$ for all $x\in [a,b]$.
H: If a function is Frechet differentiable, does the Frechet derivative equal the Gateaux derivative? If a function is Frechet differentiable, does the Frechet derivative equal the Gateaux derivative? AI: A function $f \colon U \to Y$, where $U \subset X$ is open, is Fréchet differentiable in $x \in U$ with (Fréchet) derivative $L$, iff we have $$f(x+h) - f(x) - L(h) \in o(\lVert h\rVert).$$ So for the Gâteaux derivative, we have, for any $v \in X$, $$\frac{f(x+tv) - f(x)}{t} - L(v) = \frac{f(x+tv) - f(x) - L(tv)}{t} = \pm\frac{f(x+tv) - f(x) - L(tv)}{\lVert tv\rVert}\lVert v\rVert,$$ and the right hand side tends to $0$ for $t\to 0$ since $t\to 0$ implies $\lVert tv\rVert \to 0$. Thus Fréchet differentiability implies Gâteaux differentiability, and the Gâteaux derivative then coincides with the Fréchet derivative.
H: Two sets given : solve $A \cup X = B$ Do you know how to solve this problem? I have two sets and need to solve $A \cup X = B. $ Thanks a lot for your help. AI: Note that if in fact there exists a set $B = A \cup X$, then $A\subseteq B$ AND $X\subseteq B$. If $A$ and $X$ are disjoint, so that $A \cap X = \varnothing,$ then $$A\cup X = B \iff (A\cup X)\setminus A = B\setminus A \iff X = B\setminus A$$ If they are not disjoint, then you we need to know what $A \cap X$ is, since we'd have to look at $$X = (B\setminus A) \cup(A \cap X)$$ where $$B\setminus A, \;\text{ means }\;(B \;\text{ set-minus }\; A).$$
H: Homeomorphism of closed intervals One can prove that if $f: [a,b] \to [f(a), f(b)]$ is continuous and monotone increasing that then it is a homeomorphism. The only part one might have to think about at all is that $f$ is open but that can be shown easily by showing that $f$ is closed: If $C=[x,y]$ is closed in $[a,b]$ then because $f$ is monotone, $z < x$ implies $f(z) < f(x)$ and similarly $z > y$ implies $f(z) > f(y)$ so that $f(C) = [f(x), f(y)]$ is closed. Can this be generalised by requiring $f$ to only be increasing that is, $x \le y$ implies $f(x) \le f(y)$? AI: This isn't a valid generalisation, as mentioned in the comments, because if there are two points $x,y$ with $x<y$ but $f(x)=f(y)$, then $f$ is no longer injective and hence cannot be a homeomorphism because it is not a bijection. Aymen gave an example of such a function with this property which is any constant function because an interval is not homeomorphic to a single point (by cardinality arguments).
H: How to get the roots of a quartic function when given a quadratic factor We have the function $$x^4 + 4x^3 - 17x^2 -24x + 36 = 0.$$ $x^2 -x - 6$ is a factor of this function. Find all the roots of the polynomial. So we have $(x-3)(x+2)$, and since it is a quartic we need 2 more solutions. Intuitively I'd say you should divide our quartic by our factors to get another quadratic, but I'm sure there exists a quicker/simpler method. My question is what is the quickest method to do this? AI: Your intuition is spot on. Polynomial long division is the quickest, at least the standard, route to go here, barring any immediate "sighting" of another root. Divide your quartic by the quadratic $x^2 - x - 6$ to obtain a quadratic quotient, then factor the quotient (or use the quadratic formula) to find any remaining roots, if they exist. Your quartic divided by the quadratic $x^2 - x - 6$ gives the quadratic quotient $$x^2 + 5x -6 = (x+6)(x - 1)$$ $$x^4 + 4x^3 -17x^2 -24x + 36 = (x-3)(x + 2)(x + 6)(x - 1)$$ So there exist four roots. You can a find brief tutorial on polynomial long division at the Khan Academy, to better familiarize yourself with it.
H: Integral of absolute value of Brownian motion I know it is a really stupid question and it should be quite easy, but how can I show that $\int_0^{\infty}\|B_t\|\mathbb{d}t=\infty$ a.s. with $B_t$ being a standard brownian motion? I just don't get it. What I tried so far: I know that $\limsup_{t\rightarrow\infty}\|B_t\|=\infty$ and the liminf being 0. Hence I know that for a given $K$ there is a sequence $t_n$ s.t. $B_{t_n}>K$. But by continuity I only know that for every $t_n$ there is a $\delta_n>0$ s.t. $B_t>K$ for $t\in t_n\pm\delta$. But I don't know that these $\delta_n$ won't converge to $0$... AI: Hint: For every $t$ there exists $s\geqslant t$ such that $\|B_s\|\geqslant2$. Then, the first hitting time of $1$ after time $s$ has a constant distribution. Then sum these contributions. Another argument is to note that by scaling $(B_{2t})$ is distributed like $(\sqrt2B_t)$ hence the integral $K=\int_0^\infty\|B_t\|\mathrm dt$ is distributed like $2\sqrt2K$. Since $K\gt0$ almost surely, this implies that $K=+\infty$ almost surely. Edit: About the last implication, consider some random variable $Z$ such that $P[Z\gt0]=1$ and $Z$ equals $aZ$ in distribution, with $a\gt1$. Then for every $z\gt0$, $P[Z\gt z]=P[aZ\gt z]$ hence $P[Z\gt z]=P[Z\gt z/a]=\cdots=P[Z\gt z/a^n]\to1$ when $n\to\infty$. This implies that $Z\gt z$ almost surely, for every $z\gt0$. Thus, ...
H: Does almost sure convergence implies convergence of the mean? I asked a slightly similar question here: Does Convergence in probability implies convergence of the mean?, but now I wish to examine a stricter scenario: Let $\{X_n\}_{n=1}^\infty$ be a sequence of random variables converging a.s to a const $c$. Is it required for the sequence to be uniformly integrable in order to imply $\lim_{n\to \infty} EX_n = c$? And what about $\lim_{n\to \infty} EYh(X_n) = E[Y]h(c)$ for some random variable $Y$ and a continuous function $h$? Under which regularity conditions does the last equality holds? AI: For the first question, try $P[X_n=0]=1-1/n^2$, $P[X_n=2^n]=1/n^2$, then $X_n\to0$ almost surely but $E[X_n]$ does not converge to $0$. (The second question is unclear.)
H: Borel-measurable functions $R\rightarrow R$ For $f: R\rightarrow R$ this two proposition are identical ($B$ is Borel set). $\forall A \in B\ f^{-1}(A) \in B$ if $A$ is open subset then $f^{-1}(A) \in B$ Is it true? And if it's then how can I prove it? Thanks. AI: Consider $$ \Sigma:= \left\{A \in \mathcal{B}(\mathbb{R}): f^{-1}(A) \in \mathcal{B}(\mathbb{R})\right\}. $$ Can you show it is a $\sigma$-algebra containing the open sets? Hint 2: You are trying to prove that 2. implies 1. Assume $f^{-1}(A) \in B$ for all $A$ open. This tells you that the open sets are in $\Sigma$. Now can you show that $\Sigma$ is a $\sigma$-algebra? This should follow immediately from the fact that preimages behave nicely under any set operations.
H: How to place bets to get > 50% chance of winning? Assuming we have a simple game of fair-coin throwing, there's a 50% chance to win and a 50% chance to lose. Let's assume that we have a large amount of tokens and there is no cap to the amount placed in the bet, I have this strategy to sure-win: Bet x amount. If win, pocket x and go to 1. If lose, bet 2x. If win, pocket x and go to 1. If lose, bet 4x. If win, pocket x and go to 1. If lose, bet 8x. If win, pocket x and go to 1. If lose, bet 16x. If win, pocket x and go to 1. If lose, bet 32x. If win, pocket x and go to 1. If lose, bet 64x. If win, pocket x and go to 1. If lose, bet 128x. If win, pocket x and go to 1. If lose, bet 256x. etc... Since there is no cap to the bet amount, using this strategy would guarantee us > 50% chance of winning. Now if there is a cap at 50000x, I believe this strategy doesn't work any more (chance of winning no longer > 50%). When there is a cap at 50000x, How should we modify the strategy to guarantee us > 50% chance of winning? AI: This betting strategy is called a martingale. In fact, providing you stop playing once you have won, the probability of winning is more than a half, both with and without a cap. In reality, there is always a cap, as most people will not let you bet an amount higher than you could possibly repay. The key point is that if you lose (i.e. hit the cap), then you lose a much larger amount than when you win. This can be disastrous, which is why most casinos ban it.
H: Problem with approximation of semicontinuous function with continuous functions Assume that a function $f: \mathbb R \rightarrow \mathbb R$ is a lower semicontinuous. The Baire's theorem says that there is an increasing sequence of continuous functions $(f_n)$ which is pointwise convergent to $f$. I known a proof in case when $f$ is additionally bounded. But I don't know how to proof this theorem for unbounded $f$. I try taking $h(x)=\frac{2}{\pi} \arctan f$ and next taking an increasing sequence of continuous functions $(g_n)$ pointwise convergent to $h$. But it may happens that $g_n(x)<-1$ for some $x$ and I cannot take $f_n(x)=\tan (\frac{\pi}{2}g_n(x))$. How to prove the Baire theorem for unbounded $f$? AI: I suppose the problem is with a lower semicontinuous function that is not bounded below. If the function is bounded below (without loss of generality $f \geqslant 0$) but not above, one can approximate the bounded function $\tilde{f}(x) = \dfrac{f(x)}{1+f(x)}$ by an increasing sequence $g_n$, cap each $g_n$ at $1-\frac1n$, $h_n(x) = \min \left\{ g_n(x),\, \frac{n-1}{n}\right\}$, and use $$k_n(x) = \frac{h_n(x)}{1-h_n(x)}$$ for the approximating monotonic sequence of continuous functions. For $f$ not bounded below, if we find a continuous $g \leqslant f$, we can reduce the approximation to the above, $h = f-g \geqslant 0$ is lower semicontinuous, and if $k_n \uparrow f-g$, then $k_n+g \uparrow f$. So it remains to find a (finitely valued) continuous $g \leqslant f$. Since a lower semicontinuous function attains its minimum on any compact subset of $\mathbb{R}$, we have no problem finding a continuous function $g_{a,b}$ on $[a,b]$ that is a lower bound of $f$ there (for example a constant function). Then, using a partition of unity, we can glue those lower bounds together to obtain a global continuous $g\leqslant f$. For example, let $$\psi(x) = \begin{cases}\qquad 0 &, \lvert x\rvert \geqslant \frac34\\ 2\left(x+\frac34\right) &, -\frac34 \leqslant x \leqslant -\frac14\\ \qquad 1 &, \lvert x\rvert \leqslant \frac14\\ 2\left(\frac34-x\right) &, \frac14 \leqslant x \leqslant \frac34 \end{cases}$$ Then the translates of $\psi$ by integers form a continuous partition of unity, $$\sum_{k\in\mathbb{Z}} \psi(x-k) \equiv 1,$$ and letting $$c_k := \min \{ f(x) : x \in [k-1,\,k+1]\}$$ we obtain a continuous function $g \leqslant f$ by setting $$g(x) = \sum_{k\in\mathbb{Z}} c_k\cdot \psi(x-k).$$
H: How to find the expectation of $\sqrt{x}$ and others like it Assume a Bernoulli distribution with probability of success $p$. I understand how to find expectations for random variables in general, but how do I go about finding something like $E(\sqrt{x})$ or $E(x^6+1)$? All the examples I have seen refer to finding expectations $E(x)$. How does the intuition even work for $E(\sqrt{x})$? In a Bernoulli distribution, we assume a single trial, so how can we have an expectation for this value? AI: The definition of the expected value of a discrete random variable is $E(X) = \sum_x x\;P(X = x)$. Similarly, for functions $g(X)$ of $X$, we have $E(g(X)) = \sum_x g(x)\;P(X = x)$. Since for a Bernoulli distribution, there are only two possible outcomes $x$, the sum has only two terms: $E(g(X)) = g(0) (1-p) + g(1) p$.
H: Need help in understanding state transition diagram of a convolutional coder. How are the output bits calculated? Have a look at the above figure. I am confused in how the output bits are calculated. e.g. according to my understanding a state transition from 00 to 10 (with input bit 1) should produce output 10 instead of 11 as given in the figure. What am I doing wrong here? Note: D is the delay element. + operation is modulo-2 addition (xor). AI: The output pair (Ck1, Ck2) is given by (Bk + Bk-2, Bk + Bk-1 + Bk-2). The state is given by (Bk-1, Bk-2). So, when you are at state (0,0), Bk-1 = Bk-2 = 0. Given that Bk is 1, (Ck1, Ck2) is equal to (1+0, 1+0+0) = (1,1). The new state is (1,0) because Bk-1 will become what Bk was and Bk-2 will become what Bk-1 was. I hope it helps!
H: Question on Cauchy Sequence definition? Kaplansky defines a Cauchy sequence if for any $\epsilon > 0$ there exists sufficiently large $i, j$ such that $D(x_i, x_j) < \epsilon$ for some sequence $\{x_n\}$ in a metric space. The sequence need not converge to some specified limit point, just that $x_i$ and $x_j$ become arbitrarily close. However, if $D(x_i, x_j) < \epsilon$, doesn't that mean $x_i \to x_j$? AI: What you know is that $\lim\limits_{i,j\to\infty}D(x_i,x_j)=0$. But you know nothing about the limits $\lim\limits_{i\to\infty}D(x_i,x_j)$, $\lim\limits_{j\to\infty}D(x_i,x_j)$. As an example, take $x_i=i^{-1}$. Then $\lim\limits_{i,j\to\infty}\|i^{-1}-j^{-1}\|=0$ yet $\lim\limits_{j\to\infty}D(x_i,x_j)=i^{-1}$ and $\lim\limits_{j\to\infty}D(x_i,x_j)=j^{-1}$.
H: Compactness and sequential compactness Let $X$ be a nonempty set. If the cocountable topology $\tau$ is considered on $X$, then I want to find all compact and sequential compact subsets of $X$. Definitely only finite subsets of $X$ are compact. But I am in confusion to find all sequentially compact sets. Is there any infinite set which is sequentially compact? Plese help. AI: Again, the only sequentially compact sets are finite. In fact, you can show that the only convergent sequences are those that are eventually constant : If $(x_n) \subset X$ such that $x_n \to l$, then set $A = \{x_n : x_n \neq l\}$, then $A$ is countable, and $U = X\setminus A$ is an open neighbourhood of $l$. So $\exists N \in \mathbb{N}$ such that $x_n \in U$ for all $n\geq N$. But that implies that $x_n \notin A$, which means $x_n = l$ for all $n\geq N$ Now, let $Y \subset X$ be an infinite set, so choose a sequence $(x_n) \subset Y$ with distinct terms. Every possible subsequence of $(x_n)$ has distinct terms, and so does not converge. So, $Y$ is not sequentially compact.
H: Limit as x approaches ∞ using epsilon-delta Let $f$ and $g$ be defined on $(a,\infty)$ and suppose $\displaystyle\lim_{x\rightarrow\infty}(f)=L$ and $\displaystyle\lim_{x\rightarrow\infty}(g)=\infty$. Prove that $\displaystyle\lim_{x\rightarrow\infty}(f\circ g)=L$. I am having trouble proving this using $\epsilon-\delta$ definition. Any alternatives? Or how can it be proved using the epsilon-delta definition? AI: We know that $\lim_{x \to \infty}f(x) = L$ by definition means that for any $\varepsilon > 0$ there exists an $N$, such that $$x > N\implies \|f(x) -L\|<\varepsilon, \tag{1}$$ and also that since $\lim_{x \to \infty}g(x) = \infty$, by definition for any $O$ there exists an $M$ such that $$x > M \implies g(x)>O. \tag{2}$$ Now we need to show using these definitions that $\lim_{x \to \infty}f(g(x)) = L$. This just means showing that for any $\varepsilon>0$ there exists a $P$ such that $$x>P\implies\|f(g(x))-L\|<\varepsilon. \tag{3}$$ Since we have the liberty to choose any $O$ we want (the definition guarantees that for any $O$ an $M$ exists), just choose $O = N$. Then $(2)$ tells us that there exists some $M$ such that $x > M \implies g(x) > N$. But then our definition of the limit as $x$ approaches infinity of $f$ tells us that $g(x) > N \implies \|f(g(x)) - L\| < \varepsilon.$ And now we're done! For any $\varepsilon > 0$ there is an $M$ such that $$x>M\implies g(x) > N \implies \|f(g(x))-L\|<\varepsilon.$$ I've added the middle step $(g(x) > N)$ for clarity, but if you take it out then you can clearly see that this chain of statements is just a restatement of $(3)$, so our proof is complete. $\square$
H: How can I prove a function is odd? I have a problem of calculus and I can compute it. The problem is $\iiint_V x^5 y^7 z^9 dxdydz$, where $V = \{ (x, y, z) : x^2 + y^2 + z^2 \leq 1 \}$. And the solution is 0. this solution is exactly right. If you want I can show whole my solution. I solved it using by polar coordinates. But I heard if prove $\iiint_V x^5 y^7 z^9 dxdydz$ = -$\iiint_V x^5 y^7 z^9 dxdydz$ (because $x^5y^7z^9$ is odd ft), I don't need to compute it. If someone know How can prove it please help me. I have no idea, how can start it.. AI: A function $f(t)$ is odd when $f(-t)=-f(t)$. In this case $t=(x,y,z)$. So you just need to show that $$-f(t)=-\int\int\int_{V}x^{5}y^{7}z^{9}=\int\int\int_{V}(-x)^{5}(-y)^{7}(-z)^{-9}=f(-t)$$ Which should immediately follow when you begin.
H: Convergent subsequence 1) Let (x_n) be a sequence and let L ∈ R. Suppose that for each ϵ > 0, {k ∈ N : x_k ∈ B(L; ϵ)} is infinite. Show that (x_n) has a subsequence converging to L. 2) Suppose that S is not a compact subset of R. Explain why there is a sequence (y_n) in S so that no subsequence of (y_n) converges to any L in S. My idea is to let ϵ = 1, and let a subsequence m ∈ N, such that m_1 < m_2 < m_3 < ... and x_(m_k) ∈ B( l ; 1/k) for all k ∈ N. AI: For each $k \in \mathbb{N}$, take $\epsilon = 1/k$, then $$ B_k := \{n \in \mathbb{N} : x_n \in B(l,\epsilon)\} $$ is infinite (and thus non-empty). Now choose the $x_{n_k}$ inductively, so that $$ x_{n_k} \in B_k\setminus\{x_{n_j} : 1\leq j < k\} $$ Then $x_{n_k} \to l$ Choose an open cover $\mathcal{U}$ of $S$ which has no finite subcover. Pick $U_1 \in \mathcal{U}$, and $y_1 \in U_1\cap S$. Having chosen $y_1, y_2, \ldots, y_{k-1}$, pick an open set $U_k \in \mathcal{U}$ such that $$ S\cap U_k\setminus \left (\bigcup_{j=1}^{k-1} U_j\right ) \neq \emptyset $$ Choose $y_k$ in that set. This inductively builds a sequence $(y_n)$ with no convergent subsequence (why?)
H: Joint Probability Distribution of a Gaussian Random Variable Correlated with a Gamma Random Variable I want to know if the joint PDF of a Gaussian RV correlated with a Gamma RV can be found in closed form. The correlation is known. AI: There are lots of ways to construct bivariate distributions from given marginals. One such way is with copulae. Let the continuous random variable $X$ have pdf $f(x)$ and cdf $F(x)$; similarly, let the continuous random variable $Y$ have pdf $g(y)$ and cdf $G(y)$. We wish to create a bivariate distribution $H(x,y)$ from these marginals. The joint distribution function $H(x,y)$ is given by $$H(x,y) = C(F,G)$$ where $C$ denotes the copula function (to be defined). Then, the joint pdf $h(x,y)$ is given by: $$h(x,y)=\frac{\partial ^2H(x,y)}{\partial x\partial y}$$ Examples of copula functions are the Morgenstern copula: $$C = F G ( 1 + \alpha (1-F)(1-G))$$ and the Ali–Mikhail–Haq copula: $$ C = \frac{F G}{1-\alpha (1-F) (1-G))} $$ etc. (where $\alpha$ is a parameter such that $-1 < \alpha < 1$). The copula scheme you select will determine the correlation. However, please note that there will not be a unique solution … potentially multiple (indeed, infinitely multiple) solutions may exist (whether by copula methodology or other methods).
H: Numerical integration problem How can I solve integral equations of the form $$\int_{-3}^x e^{e^t}dt=3?$$ Is there for example Sage code for that kind of equations? Is there better method that evaluating numerically $$\int_{-3}^{-1} e^{e^t}dt<2.4,$$ $$\int_{-3}^{0} e^{e^t}dt>4.2,$$ $$\int_{-3}^{-0.5} e^{e^t}dt>3.1,$$ and so on? AI: Using the nice hint by @HaraldHanche-Olsen, we can cast this into an ODE problem and use a numerical solver. So, we have: $$y' = e^{e^x},~~ y(-3) = 0$$ Using a fourth order Runge-Kutta, with a step-size of $0.045$, we get the following values: $\ldots$ $x = -0.615, y = 2.95805$ $x = -0.57, y = 3.03628$ $x = -0.525, y = 3.11653$ If we numerically evaluate the integral at the point we found above, we get an exact match. Now that you have that range, you can use various methods to reduce the error to whatever you need.
H: Maclaurin series for $e^z /\cos z$. I want to find the Maclaurin series for the function $$f(z)=\frac{e^z}{\cos z}.$$ Right away I can tell that the radius of convergence will be $\pi/2$, since it's the distance to the nearest singularity (not sure if this explanation is rigorous enough, but I can't think of anything else). Calculating the coefficients as n'th derivatives at $0$ strikes me as fruitless here, so I'm guessing I have to resort to tricks. Of course I could just divide the series formally but that doesn't get me an explicit formula for the coefficient $a_n$. I tried rewriting the cosine as an exponential, but that too produces nothing of interest. I'm looking for a hint to get me started. AI: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $$ \expo{z} = \sum_{k = 0}^{\infty}{z^{k} \over k!} \quad \left\vert% \begin{array}{rcl} \qquad\sec\pars{z} & = & \sum_{k = 0}^{\infty}{\verts{E_{2k}} \over \pars{2k}!}z^{2k} \\ \verts{z} & < & {\pi \over 2} \\[2mm] &&E_{\nu}\ \mbox{is an}\ {\it\mbox{Euler Number}}. \end{array}\right. $$ \begin{align} &\vphantom{\Huge A^{A}} \\ {\expo{z} \over \cos\pars{z}} &= \expo{z}\sec\pars{z} = \sum_{k = 0}^{\infty}\sum_{k' = 0}^{\infty}{z^{k} \over k!}\, {\verts{E_{2k'}} \over \pars{2k'}!}z^{2k'} = \sum_{k = 0}^{\infty}\sum_{k' = 0}^{\infty}{z^{k} \over k!}\, {\verts{E_{2k'}} \over \pars{2k'}!} \sum_{n = 0}^{\infty}z^{n - k}\,\delta_{n,k + 2k'} \\[3mm]&= \sum_{n = 0}^{\infty}z^{n} \sum_{k' = 0 \atop {\vphantom{\LARGE A}n - 2k' \geq 0}}^{\infty} {1 \over \pars{n - 2k'}!}\, {\verts{E_{2k'}} \over \pars{2k'}!} \end{align} $${\large% {\expo{z} \over \cos\pars{z}} = \sum_{n = 0}^{\infty}A_{n}\,z^{n}\,, \qquad\qquad A_{n} \equiv \sum_{k = 0}^{\floor{n/2} \atop } {\verts{E_{2k}} \over \pars{2k}!\pars{n - 2k}!}} $$
H: Interested in a "more fundamental" proof for basic properties of the logical connectives Starting with the classical propositional logic, is there a rather canonical way to prove that $$p\wedge q=q\wedge p$$ for the commutativity of the conjunction and analogously for the other properties and connectives, other than using truth tables, visualizing with Venn diagrams akin Wikipedia's approach, or verbal philosophical reasoning? Put it other way, can we well-define the connectives from a deeper foundation than that? For example, in set theory, we define a union of the two sets $A$, $B$ as $$A\cap B:=\{x\,\|\,x\in A\wedge x\in B\}$$ to then move on and prove that $\cap$ is commutative. By doing so we simply delegate the proof to the very propositional (or whichever) logic we defined the operator with $$A\cap B\overset{\mathrm{def}}{=}\{x\,\|\,x\in A\wedge x\in B\}\overset{\mathrm{com}}{=}\{x\,\|\,x\in B\wedge x\in A\}\overset{\mathrm{def}}{=}B\cap A.\square$$ AI: I'm not sure this will satisfy you, but a categorically-minded way to characterize meets $a \wedge b$ and joins $a \vee b$ is via universal properties: $$x \leq a \wedge b \;\;\; \text{iff}\;\;\; x \leq a,\; x \leq b$$ $$a \vee b \leq x\;\;\; \text{iff}\;\;\; a \leq x,\; b \leq x$$ for any $x$. These are general definitions in the theory of posets or preorders, but for propositions, we can think of $\leq$ as denoting the entailment relation. The pair of entailments on the right (for each of $\wedge, \vee$) simply means both are asserted. In that case, one can prove $a \wedge b = b \wedge a$. For, we have $$x \leq a \wedge b\;\;\; \text{iff}\;\;\; x \leq a, x \leq b\;\;\; \text{iff}\;\;\; x \leq b \wedge a.$$ Now, since $a \wedge b \leq a \wedge b$, we can put $x = a \wedge b$ and reason forward to conclude $a \wedge b \leq b \wedge a$. Similarly, putting $x = b \wedge a$ and reasoning backward, we conclude $b \wedge a \leq a \wedge b$. Thus, if we take propositions to be equal if they entail one another (i.e., if we assume the antisymmetry axiom for posets), we derive $a \wedge b = b \wedge a$. Similarly we can prove $a \vee b = b \vee a$. A similar "universality argument" can be used to prove that $\wedge, \vee$ are associative, idempotent, etc. Once we have universal characterizations for $\wedge, \vee$, we can add a third that characterizes negation $$a \wedge b \leq c\;\;\; \text{iff}\;\;\; a \leq (\neg b) \vee c$$ and in this way we get classical propositional logic (more exactly, we'd add in two more to characterize the top element $\top$ ("true") and $\bot$ ("false")).
H: Determine whether $\langle f, g\rangle = \int_1^e {1 \over x} f(x)g(x)\,dx $ is a inner product. Let $C[1,e]$ be the set of continuous real-valued functions with domain $W:=[1,e]$. Let $$\langle f, g\rangle = \int_1^e {1 \over x} f(x)g(x)\,dx $$ be a function. Determine whether $\langle \cdot, \cdot\rangle$ is an inner product. (i.e. $(a,b): W \times W \to C$ (C is complex set) In order that it be a inner product it has to meet 3 conditions: $\forall f \in W $ , an inner product of $(f,f) \ge 0 $, $f = 0 \iff (f,f) = 0$ So I tried to solve this integral: $$\int {1 \over x}f(x)f(x)\,dx = \int {1 \over x}f(x)^2\,dx = {1 \over x}f(x)^2 - \int{1 \over x}2f(x) = {f(x)^2 \over x} - 2 \int {f(x) \over x} $$ Then I got stuck. can you please help? thanks in advance. AI: First, this is a good question! You said what step you're stuck at, and didn't try to get help solving the entire problem before you get through that step. I wish everybody here did that. Kudos. It's not the right approach to try to calculate the integral. You can't—$f(x)$ is arbitrary, and could be something quite terrible and complicated (and check your steps!—they don't work). But let's look again at what we'd like to prove: $$ \int_1^e \frac{1}{x} f(x)^2 dx \geq 0$$ Again, don't start manipulating the equation until you have a plan. How would we show this? How do you tell, just from looking at it, that an integral can't be negative? Pay attention to the domain of integration—if you change $1$ and $e$ to something else, the problem might be false. Try plugging in some really simple functions if you need to (like $f(x)=x$). Maybe draw a picture. But there aren't any complex mathematics needed here. Just one simple realization. Think about what an integral is. Can integrals ever be negative? How? Think about the reverse question: What conditions should $g(x)$ meet to guarantee that $\int_a^b g(x)dx$ is always non-negative?
H: How to integrate $\int_{0}^{\sqrt{3\pi}}\int_{y}^{\sqrt{3\pi}}\cos 7x^2\;dxdy$? Evaluate the integral by reversing the order of integration: $$\int_{0}^{\sqrt{3\pi}}\int_{y}^{\sqrt{3\pi}}\cos 7x^2\;dxdy.$$ I have tried a lot of times, but I did'n get something reasonable. AI: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ \begin{align} &\int_{0}^{\sqrt{\vphantom{\large A}3\pi}} \int_{y}^{\sqrt{\vphantom{\large A}3\pi}}\cos\pars{7x^{2}}\,\dd x\,\dd y = \left.\int_{0}^{\sqrt{\vphantom{\large A}3\pi}} \int_{0}^{\sqrt{\vphantom{\large A}3\pi}}\cos\pars{7x^{2}}\,\dd x\,\dd y \right\vert_{x\ >\ y} \\[3mm]&= \left.\int_{0}^{\sqrt{\vphantom{\large A}3\pi}} \int_{0}^{\sqrt{\vphantom{\large A}3\pi}}\cos\pars{7x^{2}}\,\dd y\,\dd x \right\vert_{y\ <\ x} = \int_{0}^{\sqrt{\vphantom{\large A}3\pi}}\cos\pars{7x^{2}}\, \int_{0}^{x}\dd y\,\dd x = \int_{0}^{\sqrt{\vphantom{\large A}3\pi}}\cos\pars{7x^{2}}x\,\dd x \end{align} Now, OP can take it from here.
H: Radius of Convergence of a Series How would I find the radius of convergence of the following series? $$ \sum_{n=1}^{\infty}\frac{(-1)^n}{n}z^{n(n+1)} $$ The ratio test and root test are inconclusive, so I think I have to use the definition of the radius of convergence $$\frac{1}{R}=\limsup\|a_n\|^{\frac{1}{n}}.$$ I have been told that the $n$-th coefficient of this series is not $$\frac{(-1)^n}{n}.$$ I am not sure what exactly I should equate to $\|a_n\|$. Any help is appreciated. AI: If $\|z\|\gt 1$, then $\frac{\|z\|^{n(n+1)}}n$ does not converge to $0$, and when $\|z\|\lt 1$, $\sum_n \|z\|^{n(n+1)} $ is convergent, so the radius of convergence is $1$. Indeed, the $n$-th coefficient is not $\frac{(-1)^n}n$ as if we write the series into the form $\sum_k b_kx^k$, we have $b_{k(k+1)}=\frac{(-1)^k}k$ and $b_n=0$ if $n$ is not of the form $k(k+1)$ for some integer $k$.
H: Find derivative of [(4-pi)/(4pi) x^2 - 10x + 100]? How do I find derivative of this equation? my attempt: A(x) = $$\frac{4x^2+\pi x^2}{4\pi}-10x+100.$$ so how do I find the derivative of the first part? Like do I use quotient rule for the beginning of the equation?? please show full solutions :) thanks Sincerely, Math should solve itself, I have my own problems!! AI: $$\left(\frac{4x^2+\pi x^2}{4\pi}-10x+100\right)'=\left(\frac{4+\pi}{4\pi}x^2-10x+100\right)'=\left(\frac{4+\pi}{4\pi}(x^2)'-10x'+100'\right)$$ $$=\frac{4+\pi}{4\pi}2x-10=\frac{4+\pi}{2\pi}x-10$$
H: Three series of Kolmogorov Let $X_n\geqslant 0$ be a sequence of independent random variables. The following are equivalent: $i) \sum_{n=1}^{\infty}{ X_n} <\infty$ a.s $ii)$ $\sum_{n=1}^{\infty}{ \mathbb P(X_n>1)} +\sum_{n=1}^{\infty}{ \mathbb E X_n1_{X_n\le 1}} <\infty$ $iii)$ $\sum_{n=1}^{\infty}{\mathbb E\left(\frac{X_n}{1+X_n}\right)} <\infty$ Maybe I have to use the Three series of Kolmogorov but I don't know how :/ AI: That (i) is equivalent to (ii) is exactly contained in the statement of Three series theorem of Kolmogorov up to the convergence of the series of variances. This can be tackled using $\mathbb E(X_n^2\chi_{\{X_n\lt 1\}})\leqslant \mathbb E(X_n\chi_{\{X_n\lt 1\}})$ and the square of the expectations is not problematic. Assume (ii) holds. Then $$0\leqslant \mathbb E\left(\frac{X_n}{1+X_n}\right)\leqslant \mathbb P\{X_n\gt 1\}+\mathbb E[X_n\chi_{\{X_n\leqslant 1\}}].$$ Conversely, assume (iii) holds. Then $$\mathbb E[X_n\chi_{\{X_n\leqslant 1\}}]\leqslant 2\mathbb E\left(\frac{X_n}{1+X_n}\chi_{\{X_n\leqslant 1\}}\right)\leqslant 2\mathbb E\left(\frac{X_n}{1+X_n}\right),$$ and $\mathbb P\{X_n\gt 1\}=\mathbb P\left\{\frac{X_n}{1+X_n}\gt \frac 12\right\}$. An application of Markov's inequality shows the convergence of $\sum_n\mathbb P\{X_n\gt 1\}$.
H: Isomorphism between the real plane and a set of functions I have been told that $\mathbb{R}^2$ is isomorphic to the collection of functions $2\to\mathbb{R}$. Does this statement make sense at all? And if yes, how does this isomorphism work? Thanks AI: Indeed, it is sensible. Here, $\textbf{2}:=\{0,1\}.$ Every element of $\Bbb R^2$ is an ordered pair whose first and second components are both members of $\Bbb R$. Every element of ${}^{\textbf{2}}\Bbb R$ (one notation for the set of functions $\textbf{2}\to\Bbb R$) is a function taking the first element and second element of $\textbf{2}$ to elements of $\Bbb R$. Does this suggest to you how the natural bijection can be constructed? As for isomorphic, you should be able to show that ${}^{\textbf{2}}\Bbb R$ is a group under the operation $\oplus$ given by $$(f\oplus g)(x):=f(x)+g(x).$$ The natural bijection will be operation-preserving, and so will be the desired isomorphism.
H: In a topological vector space, show if $A$ and $B$ are bounded, then $A + B$ is bounded? I get as far as this before I am stuck: Pick any neighbourhood of $0$ and call it $U$. Then there exists $a, b$ such that $A \subseteq aU$ and $B \subseteq bU$. So hence $ A + B \subseteq aU+bU$. This last part should probably be easy, but I'm having trouble showing that the latter is contained in another scaling of U. Help finishing this proof? AI: By continuity of multiplication by a scalar, we can take $V\subset U$ balanced, that is, $\lambda x\in V$ if $x\in V$ and $\|\lambda\|\leqslant 1$ and such that $V+V\subset U$ (continuity of addition). Then take $a,b\gt 0$ such that $A\subset aV$ and $B\subset bV$: then $$A+B\subset aV+bV=(a+b)\left(\frac a{a+b}V+\frac b{a+b}V\right)\subset (a+b)(V+V)\subset (a+b)U.$$
H: Error term for a cubic interpolation I have a question on one interpolation problem. The problem is below. For the given points, $x_0 = -1, x_1 = 0, x_2 = 3$ and $x_3 = 4,$ find the error term $e_3(\bar{x}) = f(\bar{x}) - p_3(\bar{x})$ for cubic interpolation of $f(x)=x^5 -5x^4.$ Give an upper bound on the absolute value of interpolating error $\|e_3(\bar{x})\|$ at $\bar{x} = 2.$ I found the interpolant $p_3(x) = -6+6(x+1)-15x(x+1)+x(x+1)(x-3).$ For the error term, do I have to subtract the interpolant from the original function?? How do I find the upper bound?? Thanks. AI: The error term is related to the article in wiki: http://en.wikipedia.org/wiki/Polynomial_interpolation#Interpolation_error at the section "Interpolation error" Since you are given the original function and the closed interval containing the points that you interpolate the function with, then you can apply the formula in wiki directly to calculate an upperbound for the error at x=2. :)
H: Inverse Laplace Transform for $F(s) = (9s-24)/(s^2-6s+13)$ Find the inverse Laplace transform of $\displaystyle F(s) = \frac{9s-24}{s^2-6s+13}$. I have tried factoring out a $3$ from the top and putting it into the form of $\displaystyle\frac{b}{(s-a)^2+b^2}$ but I can't seem to do that with this equation. Any help is appreciated, thank you in advance! AI: Use $$L\left(e^{at}\cos bt\right)=\frac{s-a}{(s-a)^2+b^2}$$ and $$L\left(e^{at}\sin bt\right)=\frac{b}{(s-a)^2+b^2}$$ Using Partial fraction $$\frac{9s-24}{(s-3)^2+2^2}=\frac{A(s-3)}{(s-3)^2+2^2}+\frac{B\cdot 2}{(s-3)^2+2^2}$$ where $A,B$ are arbitrary constants Can you take it from here?
H: Product of affine varieties vs. product of (quasi-)projective varieties Suppose that $F_1,\ldots,F_r\in k[T_1,\ldots, T_n]$ and $G_1,\ldots,G_s\in k[S_1,\ldots,S_m]$ where $k$ is an algebraically closed field. Clearly $X=V(F_1,\ldots,F_r)\subseteq\mathbb A^n_k$ and $Y=V(G_1,\ldots,G_s)\subseteq\mathbb A^m_k$ are two affine varieties and $X\times Y\subseteq\mathbb A^{n+m}_k$ has a natural structure of affine variety as follows: $$X\times Y=V(F_1,\ldots,F_r,G_1,\ldots,G_s)$$ For the product of (quasi-)projective variety we need the Segre embedding to define a structure of variety, and I don't understand the motivation. Why a straightforward argument as the above doesn't work for (quasi-)projective varieties? AI: Note that you write $X \times Y \subset \mathbb A^{m+n}$, and so you seem to be using almost without thinking about the isomorphism $\mathbb A^m\times \mathbb A^n \cong \mathbb A^{m+n}$. Okay, now for your question: Forget about the equations, and just consider $\mathbb P^m \times \mathbb P^n$. What straightforward way might you suggest to think of this as a variety? Once you can do this case, you can do any quasi-projective case. So you should think about the role of the Segre embedding in this case.
H: Application of Minkowski inequality for integrals I have a question regarding Minkowski inequality for integrals: Suppose $f:(0,\infty)\rightarrow\mathbb{R}$ is a function in $L^p$ with respect to Lebesgue measure on $(0,\infty),\ p\in(1,\infty)$. Define $F(y)=\int_{(0,1)}f(xy)d\lambda(x),\ y>0$. Show that $F\in L^p$ with respect to the Lebesgue measure on $(0,\infty)$ and $\|\|F\|\|_p={p\over{p-1}}\|\|f\|\|_p$. I am advised to use Minkowski inequality for integrals: $\|\|F\|\|_p=[\int_{(0,\infty)}\|\int_{(0,1)}f(xy)d\lambda(x)\|^{p}d\lambda(y)]^{1\over p}\leq\int_{(0,1)}[\int_{(0,\infty)}\|f(xy)\|^{p}d\lambda(y)]^{1\over p}d\lambda(x)$. However, I am stucked after reaching this step. Kindly advise to proceed on the working. Thank you very much. AI: First note that the correct inequality is (see here) $$\\|F\\|_p=\left(\int_{(0,\infty)}\left\|\int_{(0,1)}f(xy)d\lambda(x)\right\|^{p}d\lambda(y)\right)^{1/ p}\leq\int_{(0,1)}\left(\int_{(0,\infty)}\|f(xy)\|^{p}d\lambda(y)\right)^{1/ p}d\lambda(x)$$ Now let $xy=z$ and note that $$\int_{(0,1)}\left(\int_{(0,\infty)}\|f(xy)\|^{p}d\lambda(y)\right)^{1/ p}d\lambda(x)=\int_0^1\left(\int_0^\infty \frac{\|f(z)\|^p}{x}d\lambda(z)\right)^{1/p}d\lambda (x)$$ Can you proceed from here?
H: Group isomorphism between rationals with sum and rationals with multiplication It is well known that the exponential function induces an isomorphism between the additive group of real numbers and the multiplicative group $\mathbb{R}_{>0}$. I was wondering if there exists an isomorphism between the additive group of rationals and the multiplicative group $\mathbb{Q}_{>0}$. AI: There can't be. In the additive group, you can divide by $n$ for all $n \in \mathbb{Z}^+$, i.e. for all $x$ there is a $y$ with $x = n\cdot y \;(= \underbrace{y + y + \dotsb + y}_{n\text{ times}})$. In the multiplicative group of positive rational numbers, that would correspond to the existence of $n$-th roots for all $n > 0$.
H: Permutations with prior positions In how many ways $P,Q,R,S,T,U$ can be arranged such that $ P, Q$ should come before $T,U$ ? Do we have to find the ways that $P$ and $Q$ can be placed in the first four position? AI: We interpret the question as asking for the number of ways to arrange the letters so that $P$ is before $T$ and $U$, and also $Q$ is before $T$ and $U$. Consider just the $4$ letters $P,Q,T,U$. There are $(2)(2)$ ways to arrange them so that our condition is fulfilled. And there are $4!$ ways to arrange them with no restriction. Thus the probability that our arrangement of $6$ letters satisfies the condition is $\frac{4}{4!}$. There are $6!$ equally likely arrangements, so the required number of arrangements is $6!\cdot\frac{4}{4!}$. Remark: We used a probabilistic argument. However, we can use pure counting. The positions to be occupied by our $4$ important letters can be chosen in $\binom{6}{4}$ ways. By our argument above, the actual locations of $P,Q,T,U$ can be chosen in $(2)(2)$ ways. Then the remaining $2$ letters can be put in the two open solots in $2!$ ways, for a total of $\binom{6}{4}(2)(2)(2!)$.
H: Properties of the interior of a set Consider $ \tau_1 , \tau_2$ two topologies defined on X. I want to prove that these are equivalent: (i) $ \tau_2 \subseteq \tau_1 $ (ii) $A\in X$, $(A^2)^\circ \subseteq (A^1)^\circ $ I have started by proving that (i) implies (ii) by definition but I don't know if this is correct. $x\in (A^2)^\circ$ if and only if $(A^2)\in N_x^2$ but since $ \tau_2 \subseteq \tau_1$ then $(A^2)\in N_x^2$ and this implies $x\in (A^1)^\circ$?? AI: Start with $(i) \to (ii)$. For a fixed $A$, we need to show that $A^{2\circ} \subset A^{1 \circ}$. Let us take any $x \in A^{2\circ}$. By the definition, $x \in N$, where $N \in \tau_2$ and $x \in N \subset A$. But since $\tau_2 \subset \tau_1$, we also have $N \in \tau_2$, and (because the other condition $x \in N \subset A$ does not depend on the topology) we have by definition $x \in A^{1\circ}$. Next, we do $(ii) \to (i)$. Let $A \in \tau_2$; we need to show that $A \in \tau_1$. By the definition, we have $A = A^{2 \circ} $. By $(ii)$ we have $A^{2\circ} \subset A^{1 \circ}$. By the definition of interior, we have $A^{1\circ} \subset A$. Hence, $A = A^{2 \circ} \subset A^{1 \circ}\subset A$. This is only possible when all the inclusions in the chain are in fact equalities; in particular $ A^{1 \circ} = A$, and this $A \in \tau_1$.
H: Sketching graphs : Most importaint points I'm currently studying for a test which places a lot of emphasis on sketching graphs of certain functions, without anything but a ruler and a pencil. I mean tricky functions, for example: $y = \sin^2(x)$ $y=\dfrac{1-x-x^2}{x^2}$ I always try to sketch these functions by rewriting them to a form I can immediately jot down, or simplifying them in some way or another. For example, for $y=\sin^2(x)$ I thought I'd just take the absolute of $y=\sin(x)$. But it's still tough for me to actually accurately sketch functions. An example of a problem I have is that for functions such as the second, I have trouble figuring out whether a sketch are 2 non-continuous lines or 1 continuous one.. So my question is: What is a good approach to these sketching questions? Is there a certain sequence which I can follow in order to make it easier (for example, always look at $x=0$ first, then find asymptotes, then...)? AI: Some key points: Zeroes Local maxima/minima Boundary/ies of the domain, if the domain is bounded from at least one direction Boundary/ies of the support (where the function is non-zero), if the support is bounded in at least one direction Asymptotes, if such exist Points of inflection Application: $$y=\frac{1-x-x^2}{x^2}.$$ First, let's look for zeroes: $1-x-x^2=0$ when $$x=\frac{-1\pm \sqrt 5}2.$$ So estimate $\sqrt 5$ and use that, perhaps noting the exact values. Next, we can rewrite this as $$y=\frac{1-x}{x^2}-1.$$ As $x$ increases or decreases without bound, this whole thing approaches $y=-1$, so draw that line. $$y'=\frac{-x^2+2x(x-1)}{x^4}=\frac{x-2}{x^3},$$ which is $0$ exactly at $2$. $$y''=\frac{x^3-3x^2(x-2)}{x^6}=\frac{-2x+6}{x^4},$$ which is positive at $2$, so $x=2$ is a local minimum. Draw that. The function is undefined at $x=0$ and in fact has no limit. Figure out which ways it goes.
H: Balls complete in metric space I'm struggling with the following task, i'd be thankful if anybody could help me out with a solution. Note $ (X,d)= (\mathbb{R}^n,d) $, $ \ R= S^{n-1}(0)= \{ x \in \mathbb{R}^n $: $ \ \|x\|=1 $} and $z \in $ R. Then: $$ B_e(z)= \{x \in R: \ d_R(x,z)< \varepsilon \ \} \ \text{is complete in} \ (R,d_R) \leftrightarrow \varepsilon > 2$$ AI: Hint: $R$ is compact, hence complete. A subset of a complete metric space is complete if and only if it is closed. Assembling the gist of the argument from the comments to have a self-contained complete answer after it has been solved: The boundary of the open ball $B_\varepsilon(z)$ is the set $S_\varepsilon(z) = \{ x \in R: d_R(x,y) = \varepsilon\}$. Since $R$ is complete, the open ball is complete if and only if it is closed, and that is the case if and only if $S_\varepsilon(z) = \varnothing$. Since the diameter of $R$ is $2$, we have $B_\varepsilon(z) = R$ for $\varepsilon >2$, hence $B_\varepsilon(z)$ is complete for $\varepsilon > 2$. For $0 < \varepsilon \leqslant 2$, there are points in $R$ with $d_R(x,z) = \varepsilon$, hence then $S_\varepsilon(z) \neq \varnothing \iff \overline{B_\varepsilon(z)} \neq B_\varepsilon(z)$, so the ball is not closed, hence not complete for $0 < \varepsilon \leqslant 2$.
H: Probability density function with conditional distribution and correlation I am unsure with question c/d/e/f but I will give my answers for all questions I have attempted. The random variables $X$ and $Y$ have joint probability density function $f_{X,Y}(x,y) = ke^{-(x+y)},\ x>0,\ y>0,\ 0<x<y.$ (a) Sketch the region over which the joint probability density function is non-zero. (b) Show that $k = 2.$ (c) Find the marginal Distribution of $X$; name the distribution and state its mean and variance. (d) Find the conditional distribution of $Y\|X = x$ for $x>0$. Pay particular attention to the range of $Y$ conditional on $x$. (e) Calculate $Pr(X + Y<1)$. (f) Given that $E(Y) = 1.5$ and $Var(Y) = 1.25$, calculate the correlation between $X$ and $Y$. ANSWERS: (a) Don't know how to draw on here but it's the area above $y=x$ and $y>0$. (b) Here I used double integration $\int_0^\infty \int_0^y ke^{-(x+y)}dxdy = 1$ $\int_0^\infty [-ke^{-(x+y)}]_0^y dy = \int_0^\infty[-ke^{-y}+ke^{-2y}]dy$ $[ke^{-y}-\frac{k}{2}e^{-2y}]_0^\infty = [ke^{-0} - \frac{k}{2}e^{-2\cdot0}]-[ke^{-\infty}-\frac{k}{2}e^{-2\cdot \infty}] = [k - \frac{k}{2}]-[0] = \frac{k}2$ Hence $\frac{k}{2}=1$ and $k=2$ (c) $P_x(x)=\int_yP_{X,Y}(x,y)dy=\int_0^\infty2e^{-(x+y)}dy$ $[-2e^{-(x+y)}]_0^\infty= -2e^{-x}$ Presuming that the integration I have done is correct the marginal distribution of $X$ is $-2e^{-x}$ and hence this is a gamma distribution as the limits are $(0,\infty)$ and it just doesn't match the binomial distribution either. Gamma distribution: $\frac{1}{\Gamma(\alpha)\cdot \beta^\alpha}\cdot x^{\alpha-1}\cdot e^{-\frac{x}{\beta}}$ Hence $\frac{1}{\Gamma(\alpha)\cdot \beta^\alpha}\cdot x^{\alpha-1}\cdot e^{-\frac{x}{\beta}} = -2e^{-x}$ This is where I have gone wrong as using $\beta = 1$ to leave $e^{-x}$ as that i then need to find: $\frac{1}{\Gamma(\alpha)\cdot 1^\alpha}\cdot x^{\alpha-1} = -2$ and using $\alpha =1$ to make $x^0$ and getting rid of $x$ but this makes $1=-2$. (d) unsure how to do this question also. (e) Adding the line $y=1-x$ to the graph from (a)we get a new region to find $\int_0^1 \int_x^{1-x}2e^{-(x+y)}dydx$ $\int_0^1[-2e^{-(x+y)}]_x^{1-x}dx$ $\int_0^1[(-2e^{-2x})-(-2e^{-1})]dx$ $[e^{-2x}+2xe^{-1}]_0^1$ $e^0 + 0 - e^{-2} + 2e^{-1} = 1-\frac{1}{e^2}+\frac{2}{e}$ Hence $Pr(x+y<1) = 1-\frac{1}{e^2}+\frac{2}{e}$ (f) Unsure how to do. AI: On (c), we want to "integrate out" $y$. So the density function of $X$ is $$\int_{y=x}^\infty 2e^{-x}e^{-y}\,dy.$$ Integration yields $2e^{-x}e^{-x}$. So $X$ has exponential distribution, parameter $2$. On (d), recall that the conditional density of $Y$ given that $X=x$ is $$\frac{f_{X,Y}(x,y)}{f_X(x)}.$$ Note that the conditional density is $0$ for $y\lt x$. On (e), the setup is good. There is a small sign error in the calculation. The number obtained is impossible, it is greater than $1$. On (f), we will first need the covariance, that is, $E(XY)-E(X)E(Y)$. For that, we need to find $$\iint_D (xy) 2e^{-(x+y)}\,dydx,$$ where $D$ is the region where our joint density is positive.
H: Composition of polynomials - is it a simple group? I wouldn't be surprised if this can be found maybe even on Wikipedia but I'm not a native English speaker and unfortunately couldn't find this myself. So for a set of polynomials $F = \left\{ \, f(x) \mid f(x) = ax+b,\ a,b \in \mathbb{R},\ a \neq 0 \, \right\}$ is the composition group $\langle \, F; \circ \, \rangle$ simple? I really tried to contemplate possible normal subgroups but don't see any (besides consisting of whole $F$ and $\{x\}$). How to be sure? AI: The group $F$ you are considering is also called the affine group of the real line. Affine groups of affine spaces of dimension $\ge 1$ always have a non-trivial normal subgroup, namely the subgroup of translations. In particular, they are never simple. In your case, consider the map $F \to \mathbb{R}^{\times}$ defined by $a x + b \mapsto a$. This is clearly a surjective group homomorphism, and its kernel is the normal subgroup consisting of those polynomials of the form $x + b$, i.e. translations.
H: In the solution below how is (cosC+3√sinC)= 2sin(π6+C)? You can use the Law of cosine and the area formula of a triangle to solve this problem. Suppose the three angles are A,B,C opposite to the sides a,b,c, respectively. Then c2=a2+b2−2abcosC,S=12absinC and hence a2+b2+c2−4S3√==≥=≥2[a2+b2−ab(cosC+3√sinC)]2[a2+b2−2absin(π6+C)]2(a2+b2−2ab)2(a−b)20 since sin(π6+C)≤1. The equal sign holds iff a=b and sin(π6+C)=1, which implies that a=b,C=π3 or a=b=c. AI: HINT: Use $$\sin(A+B)=\sin A\cos B+\cos A\sin B$$ Hope you know or can derive the values of $\displaystyle\sin\frac\pi6,\cos\frac\pi6$
H: x raised by the power of y equals infinity. How many zeros are there in this equation? 10000^999 In my calculator it says infinity but that doesn't seem right. AI: The answer is: $3996$. $10000^1 - 4$ zeroes $(4 \cdot 1)$ $10000^2 - 8$ zeroes $(4 \cdot 2)$ ... $10000^{999} - 3996$ zeroes $(4 \cdot 999)$
H: Proof of a equality-topology Consider he topological space $ (X,\tau)$ and $A \subseteq X$. I want to prove that these are equivalent: (i)$A\in \tau$ ($A$ is open)~ (ii) For any $B\subseteq X$, where $A\cap B=\emptyset$ then, $A\cap \overline B =\emptyset$ I have started in this way: (i) $\to$ (ii) $B\subseteq X$, where $A\cap B=\emptyset$ if and only if (As $A$ is open) $A^\circ\cap B = \emptyset $ if and only if $A^\circ\cap (B^\circ)^\circ = \emptyset$ if and only if $(A\cap B^\circ)^\circ = \emptyset$ if and only if $X-(A\cap B^\circ)^\circ =X$ if and only if $\overline{X-(A\cap B^\circ)} =X$..... But I don't get to any point Thank you for your time AI: If $A$ is open, then $X -A $ is closed. If $A \cap B = \varnothing$, then $B \subset X-A$. Then $\overline{B} \subset \overline{X-A} = X-A$, so $A \cap \overline{B} = \varnothing$. Conversely, if all sets $B$ disjoint from $A$ are such that $A \cap \overline{B} = \varnothing$. Then $A \cap \overline{X-A} = \varnothing \implies \overline{X-A} \subset X-A$, or $X-A$ is closed, so $A$ is open.
H: If an element of a basis is a scalar multiple of an element of another basis, are the two bases considered distinct? Suppose that we have the bases {(1,0) (0,1)} and {(1,0) (0,-1)} We see that (0,-1) is a scalar multiple of (0,1). Are the two bases considered distinct? Thank you AI: The vectors in the two bases are distinct and will have different linear combinations to represent the same vectors. So yes, the 2 bases are distinct. However we only use one basis for a vector space and {(1,0),(0,1)} is more conventional.
H: What is that sign in the context of vectors? Suppose $v = (0, -5, 5, -6, -7)$ a vector. I need to find $$\\|v\\|_1, \\|v\\|_2, \\|v\\|_9, \\|v\\|_\infty.$$ can you please explain me what does $\\|v\\|_i$ mean? AI: It's probably the p-norm: $$\\|v\\|_p = \left(\|v_1\|^p + \| v_2\|^p + \cdots \right)^{1/p}$$ Where for $p\to\infty$, it's equal to the maximum norm: $$\\|v\\|_\infty = \max\{\|v_1\|,\|v_2\|, \ldots \}$$
H: Convolution of functions with compact support I have a question regarding convolution with compact support: Suppose $f \in L^1(\mathbb{R})$ and $g \in L^p(\mathbb{R})$, and both of them have compact support. Show that $f*g$ (convolution integral of $f$ and $g$) has compact support. Kindly advise in proceeding the working. Thank you. AI: We define $f \ast g (x)$ to be $$ f \ast g(x) = \int f(y)g(x-y) dy$$ If this integral is to be nonzero, there must be some overlap between the support of $f$ and the support of $g(x-y)$. What does changing $x$ do to the support of $g(x-y)$? If both $f$ and $g$ are compact support, can you see why taking $x$ large enough will force the supports of the functions in the integral to be disjoint?
H: Fixed Point Iteration is Not Converging to the Desired Root What is the fixed point for the following function? $f(x) = 2sin(\pi x) + x = 0$ between $[1,2]$ I expressed $f(x)$ as $x=g(x)$ such that $g(x)$ can be $- 2sin(\pi x)$ or $\frac{1}{\pi}sin^{-1}(\frac{-x}{2})$. However, in both cases, the method is not coverging to a root lying between in the interval $[1, 2]$. What to do? Please help me. AI: Note that if $1 \le x \le 2$, you don't want the principal branch of arcsin. You could try $x = 2 + \dfrac{\arcsin(-x/2)}{\pi}$ (for the solution in $[1.5,2]$, or $x = 1 + \dfrac{\arcsin(x/2)}{\pi}$ (for the solution in $[1,1.5]$).
H: Exponent Upon Exponent? I can't understand this. Can you please make a clearer explanation? AI: Write $y=x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}$. Then $$y = x^{x^{x^{\cdot^{\cdot^{\cdot}}}}} = x^{(x^{x^{x^{\cdot^{\cdot^{\cdot}}}}})} = x^y$$ The book is saying that if $x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}=2$ then we can stick $y=2$ into the above, giving $2=x^2$, and hence $x=\sqrt{2}$. And likewise, if $y=4$ then $4=x^4$, so again $x=\sqrt{2}$. (In fact, as I'm sure the book goes on to say, the value of $x^{x^{x^{\cdot^{\cdot^{\cdot}}}}}$ when $x=\sqrt{2}$ is in fact $2$ and not $4$.)
H: Determining which Probability Distrubitive Function to use I have two questions that I am to solve as practice. I am having difficulties with determining which probability function to use. The firstquestion is: Todd decides to keep buying a lottery ticket each week until he has 4 winners (of some prize). Suppose 30% of the tickets win some prize. Find the probability he will have to buy 10 tickets. For this question, I should apply the Binomial distribution function. let X be the number of lottery tickets bought X ~ $Binomial(10, 0.3)$ $P(x) = ((10) nCr (x))(0.3)^x(0.7)^(10-x)$ $P(4) = (10 nCr 4) * (0.3)^4*(0.7)^6=0.200$ However, the solution is 0.08. The second question is: An oil company runs a contest in which there are 500,000 tickets; a motorist receives one ticket with each fill-up of gasoline, and 500 of the tickets are winners. If a motorist has ten fill-ups during the contest, what is the probability that he or she wins at least one prize? Here my random variable would the number of tickets received. I would use the Hypergeometeric distributive function because we have a finite set. I attempted to solve the above question, but the solution is not what I am getting. Here is my attempted solution: let X be the number of tickets received through gas fill-ups X ~ $Hypergometeric(500 000, 500, 10)$ $P(x) = (500 nCr x)(450000 nCr 10-x)/(500000 nCr 10)$, Where x is an element of [0,10] Since the question explicitly asks for at least 1 ticket, so: $P(X >= 1) = 1 - P(0) = 1 - [(500 nCr 0)(450000 nCr 10)/(500000 nCr 10)]= 0.651$ However, the answer is 0.010. I know I am doing something terribly wrong, but I am not too sure what it is. I am also sorry for not properly using Latex. AI: For the first question, we want exactly $3$ successes in the first $9$ trials, and success on the $10$-th. The probability is $$\binom{9}{3}p^3(1-p)^6 p,$$ where $p=0.3$. For the second problem, it is easier to find first the probability of no prizes. The probability the first results in no prize is $\frac{499500}{500000}$. Given that the first rsults in no prize, the probability the second results in no prize is $\frac{499500}{499999}$. And so on. For the probability of no prize, you will get a product of $10$ terms. Remark: We get an excellent numerical approximation for the probability of no prize using $\left(\frac{499500}{500000}\right)^{10}$. Even more simply, the probability we win on any buy is about $\frac{500}{500000}$. Since multiple wins are highly unlikely, and $1$ in $1000$ tickets is a winner, the probability of a win is about $10\cdot\frac{1}{1000}$.
H: Open cover with no finite subcover Let (x_n) be a sequence, let $L$ ∈ R, and for each ϵ>0, {k ∈ N : x_k ∈ B($L$; ϵ)} Suppose S is not a compact subset of R. There is some ϵ_L > 0, such that {k ∈ N : x_k ∈ B($L$; ϵ_L)} is finite. Question: Let O:={B($L$,ϵ_L): $L$ ∈S}. Explain why $O$ is an open cover for S that has no finite subcover. Note: I understand for something like [0,1], and finding some open covers. Also that it has finitely many subcovers. But I am confused with the open cover of something more general. I see how it is possible, but I dont know how to show it. I know that the intersection of infinitely many open sets may fail to be open. AI: Think of the pigeonhole principle. We have an infinite collection of numbers $x_n$, and infinitely many boxes to put them in: the collection of open sets in $O$. If a find subcover existed, we would be stuffing our infinite collection of numbers into finitely many open sets. One of these open sets in $O$ would have to have infinitely many elements of $x_n$. This gives a contradiction.
H: To show that $\Lambda^pL(V\rightarrow W)$ and $L(\Lambda^pV\rightarrow W)$ are not necessarily isomorphic Let $V$ and $W$ be two vector spaces. Use $L(V\rightarrow W)$ to represent the vector space of linear map from $V$ to $W$. It is proved that $\Lambda^p(V^)\cong (\Lambda^pV)^$, where $\Lambda$ is the exteror power, which is the case $W=\mathbb{R}$. The general case seems to be false. But I failed to provide an counterexample. Can anyone help? AI: Just compute the dimensions. Suppose $\dim V = n$ and $\dim W = m$. On the one hand, $\Lambda^p L (V \to W)$ has dimension $\frac{(n m)!}{p! (n m - p)!}$; on the other hand, $L (\Lambda^p V \to W)$ has dimension $\frac{n! m}{p! (n - p)!}$. Of course, these two quantities are equal when $m = 1$, but already for $m = 2$ we have $\frac{(2 n)!}{p! (2 n - p)!}$ vs $\frac{2 \cdot n!}{p! (n - p)!}$, and the former is in general bigger than the latter. (For instance, take $p = n = 2$.)
H: Why does this whole term go away with the $dx$? I'm learning how to take indefinite integrals with U-substitutions on khanacademy.org, and in one of the videos he says that: $$\int e^{x^3+x^2}(3x^2+2x) \, dx = e^{x^3+x^2} + \text{constant}$$ I understand that the differential goes away, but not how the whole $(3x^2+2x)$ term go away together with the $dx$. AI: The $u$ substitution here is $u = x^3 + x^2$. Then $du = (3x^2 + 2x) \, dx$. This allows us to rewrite $$\int e^{x^3 + x^2}(3x^2 + 2x) \, dx = \int e^u \, du$$ At which point we integrate and substitute back. You can see that you cannot do a $u$-substitution without accounting for the change from $dx$ to $du$, and it is this that takes care of the $(3x^2 + 2x)$.
H: Convergence of the arithmetic mean Let $(a_n)_{n \in \mathbb{N}}$ be a convergent sequence with limit $a \in \mathbb{R}$. Show that the arithmetic mean given by: $$s_n:= \frac{1}{n}\sum_{i=1}^n a_i \tag{A.M.} $$ also converges to $a$. I have read: arithmetic mean of a sequence converges but unfortunately the answers there don't help me much because I don't understand their substitutions and most of it all, why their substitutions seem to work. What I know from the problem is that since $a_n$ is convergent and $\epsilon >0$ is given, I can say that: $$\exists N_1 \in \mathbb{N}, \forall n \geq N_1: \|a_n-a\|<\epsilon_1 $$ I also know that since $a_n$ is convergent, it is bound, so $(a_n) < M, \ \forall n \in \mathbb{N}$ I need to show that: $$\exists N_2 \in \mathbb{N}, \forall n \geq N_2: \|s_n-a\|< \epsilon_2$$ I started as follows: $$ \left\|s_n-a \right\| = \left\|\frac{1}{n}\sum_{i=1}^na_i -a\right\|= \left\|\frac{1}{n}\sum_{i=1}^m(a_i-a)+\frac{1}{n}\sum_{i=m+1}^n(a_i-a) \right\| \\ \leq \frac{1}{n} \sum_{i=1}^m\|(a_i-a)\|+\frac{1}{n}\sum_{i=m+1}^n\|a_i-a\|$$ I believe to understand that the left sum after the $\leq$ is finite, bound and doesn't depend on the values that $n$ takes on. However, I don't understand where all the substitutions come from and make this proof so seemingly easy to complete. Is there a general idea I can follow to complete such proofs? Because I know that the last step is to show that the given sum is smaller than $\epsilon$. I also know that I should bring the condition $a_n < M$ into place somewhere, but I don't know where. If I choose an $n \geq N: \|a_n-a\|< \epsilon'$ what does that tell me about $\|a_i-a\|$? I know that they are seemingly the same just with a different index. AI: Here is the intuition behind the proof. As you have done, we split the sum into two pieces. Each piece is controlled differently. To control the first piece of the sum, we note that our sequence is bounded by some constant $M$, so that this sum is at most $$M\frac{m}{n}$$ That is, $m$ terms of value at most $M$, multiplied by the $1/n$ in front. To control the second part of the sum, we note that our sequence $a_n$ is getting close to $a$. Thus, if we choose $m$ big enough, the $\|a_n - a\|$ term is less than $\epsilon /2$ for $n>m$. This means that the second sum is at most $$\frac{\epsilon}{2}\frac{n-m}{n} \leq \frac{\epsilon}{2}$$ The last observation is that if $n$ is chosen very large, we can also have $$M\frac{m}{n} \leq \frac{\epsilon}{2}$$ Putting these together gives us our bound. To recap: We want to split our sum into a piece with a bounded number of terms and another piece with a growing number of terms. On the piece with a bounded number of terms, we use the bound on the sequence. On the piece with a growing number of terms, we just make sure to start far out enough that the terms are small. Finally, we shrink $\frac{1}{n}$ to make both terms as small as we please.
H: Find the inverse Laplace transform $f(t)=L^{-1}\left\{F(s)\right\}$ of the function $F(s)=\dfrac{7s−22}{s^2−6s+13}. $ Find the inverse Laplace transform $f(t)=L^{-1}\left\{F(s)\right\}$ of the function $F(s)=\dfrac{7s−22}{s^2−6s+13}. $ $f(t)=L^{-1}\left\{\frac{7s-22}{s^2-6s+13}\right\}$. I was trying to break $F(s)$ into simpler rational fractions by partial fraction, but I could not factor $s^2-6s+13$. Can anyone give me some hints? AI: Hint: complete the square.....
H: Basic number theory question, proving something is an integer. Let $a$ and $n$ be two non-zero natural numbers that are relatively prime. Show that there exists $b \in \mathbb Z$ such that $ab \equiv 1\pmod n$. So $(a,n)=1$ and we know there exists $\alpha, b \in \mathbb Z$ such that $$\alpha a + b n =1$$ and $$b=\frac{1-\alpha a}{n}$$ Where did I go wrong this problem? AI: The puzzlement is due to a non-optimal choice of letters. There exist integers $x$ and $y$ such that $ax+ny=1$. Taking $b=x$, we get that $ab\equiv 1\pmod{n}$.
H: how to prove this is a metric given the following conditions I need help wrapping my head around the concepts of metrics and how to prove that something is a metric. For example, prove that if $p_1$ and $p_2$ are metrics in $X$, then $p_1 + p_2$ and $\max\{p_1, p_2\}$ are also metrics. Are the functions $\min \{p_1, p_2 \}$ and $p_1*p_2$ metrics and why? AI: All you have to do is to check the definition: You know that a metric $d$ is a function $d \colon X \to [0,\infty]$ such that the following three properties are satisfied: $1)$ $d(x,y) = 0$ if and only if $x = y$ $2)$ $d(x,y) = d(y,x)$ for any $x,y \in X$ &$3)$ $d(x,y) \le d(x,z) + d(z,y)$ for any $x,y,z \in X$. I will show that if $d_1$ and $d_2$ are metric on the same set $X$ then $d_1 + d_2$ is a metric. You can play with all the others :D By definition $(d_1 + d_2)(x,y) = 0$ if and only if $d_1(x,y) + d_2(x,y) = 0$ since they are positive this happen if and only if $d_1(x,y) = 0$ and $d_2(x,y) = 0$. But $d_1$ is a metric then this can happen of and only if $x = y$. Symmetry is even simpler if possible: $(d_1 + d_2)(x,y) = d_1(x,y) + d_2(x,y) = d_1(y,x) + d_2(y,x) = (d_1 + d_2)(y,x)$. Triangle inequality: $(d_1 + d_2)(x,y) = d_1(x,y) + d_2(x,y) \le d_1(x,z) + d_1(z,y) + d_2(x,z) + d_2(z,y) = (d_1 + d_2)(x,z) + (d_1 + d_2)(z,y)$. And we are done: $d_1 + d_2$ is a metric on $X$. Clearly it is not always that simple: there are metrics for which the triangle inequality is not just a one line proof (for example, if $d$ is a metric, $d_1 := \frac{d}{1 + d}$ turns out to be a metric, but you need to work to prove the triangle inequality, or at least you need to work more that we needed to prove that $d_1 + d_2$ is a metric :D )
H: Is $\mathbb{Q}[α]=\{a+bα+cα^2 :a,b,c ∈ \mathbb{Q}\}$ with $α=\sqrt[3]{2}$ a field? I'm making some exercises to prepare for my ring theory exam: Is $\mathbb{Q}[α]=\{a+bα+cα^2 :a,b,c ∈ \mathbb{Q}\}$ with $α=\sqrt[3]{2}$ a field ? If $(a+bα+cα^2)(a'+b'α+c'α^2)=1$, then (after quite some calculation and noticing that $α^3=2$ and $α^4=2α$): \begin{align} aa'+2bc'+2cb'&=1 \\ ab'+ ba'+2cc'&=0 \\ ca'+bb'+ac' &= 0 \end{align} I'm not sure how to proceed, and if I'm heading in the right direction. Any help would be appreciated. Something else I was thinking about, this ring I have seems to be isomorphic to: $$\mathbb{Q}[X]/(X^3-2)$$ But this is not a maximal ideal, as it is contained in the ideal $(X^3,2)$. Would this be correct reasoning ? AI: One approach would be to consider $\mathbb{Q}[\alpha]$ as a $\mathbb{Q}$-vector space of dimension $3$. Let $\beta = a + b\alpha + c\alpha^2$. Now consider the numbers $1$, $\beta$, $\beta^2$, and $\beta^3$. Because $\mathbb{Q}[\alpha]$ has dimension $3$, these are linearly dependent, so there are $u,v,w,x\in \mathbb{Q}$ with $u+v\beta + w\beta^2 + x\beta^3 =0$. We can assume $u\neq 0$ (why?), so $\beta(v+w\beta+x\beta^2)=-u$, and we're almost done...
H: Write a Mathematica program to find the four known factorions so i have come up with code that will find the four factorion numbers. I have a bunch of ideas of how it needs to be done theoretically but its still not quite clicking for me not too mention i am new to mathematica so the syntax is a bit confusing. So can anyone show me the proper code with brief explanations? AI: Actually, the numbers are so small we can do this very quickly by just testing for each combination below a certain threshold: Reap[If[Total@Factorial[IntegerDigits[#]] == #, Sow@#] & /@ Range[50000]][[2]] Which returns {{1, 2, 145, 40585}}
H: Order of Double Coset I am working on a homework problem (so don't just give me the answer) from Herstein's Topics in Algebra, which goes as follows: If $G$ is a finite group, show that the number of elements in the double coset $AxB$ is $$\dfrac{o(A)o(B)}{o(A\cap xBx^{-1})}$$ It makes sense to me, but my attempts at a proof seem to fall short of undeniability. I have been trying to show that $o(A \cap x B x^{-1})$ equals the number of duplicate terms in the list of all possible products of an element from $A$ and an element from $xB$. Suppose $a_0 \in A \cap xBx^{-1}$. Then $\exists b_0 \in B : a_0 = x b_0 x^{-1}$. In the list of products of the form $axb, a \in A, b \in B$, any term involving $a_0$ will be of the form $a_0 x b = x b_0 x^{-1} x b = x b_0 b$ for some $b \in B$. But then this list is just the left coset $xB$, which is already accounted for in the product list of $AxB$ by setting $a=e$. This is where I run into trouble. I can't seem to crystallize this argument. Am I going about this the right way? And if you see how I should extend this argument, can you nudge me in the right direction? Thanks Math.SE. AI: Edit: I didn't notice that "don't just give me the answer" in the first version of my answer. The following are some hints. Firstly, the formula $$o(HK)=\frac{o(H)o(K)}{o(H\cap K)}$$ holds for subgroups $H$ and $K$ of $G$, but may not hold for subsets $H$ and $K$ of $G$. Secondly, note that $$o(AxB)=o(AxBx^{-1}).$$
H: Showing the parametrically representation of hyperbolic paraboloid. And how to find the curves $u$ and $v$ be constant. Show that the hyperbolic paraboloid can be represented parametrically as $$r(u,v)=\langle a(u+v), b(u-v), uv\rangle$$ Find the curves $u$ is constant and $v$ is constant. I guess I need to use the hyperbolic paraboloid equation. But I cannot solve this. Please help me doing it. Thank you very much AI: You've written $$r(u,v)=\langle a(u+v), b(u-v), uv\rangle$$ and that's the same thing as saying $\langle x,y,z\rangle = \langle a(u+v), b(u-v), uv\rangle$, so that $$ \begin{align} x & = a(u+v), \\ y & = b(u-v), \\ z & = uv. \end{align} $$ So $$ \frac{y^2}{b^2} - \frac{x^2}{a^2} = \frac{(b(u-v))^2}{b^2} - \frac{(a(u+v))^2}{a^2}. $$ You can cancel $a$ and $b$ and then do routine simplifications and see if it turns into $-4uv$, which then turns into $-4z$, so that $\frac{y^2}{b^2} - \frac{x^2}{a^2} = \frac{z}{c}$ with $c = -\frac14$. That shows that the parametrized surface you've got is a subset of the surface defined by the equations involving $x$, $y$, and $z$. Then you need to show that it's the whole set. One way to do that would be by solving for $u$ and $v$ in terms of $x$ and $y$, thereby showing that all $(x,y,z)$ points on the surface actually appear within the surface parametrized by $u$ and $v$.
H: Expected Number of Cars in a Parking Space Another expectation problem: There is a parking space of length 4. Cars come and randomly choose any position to park over the interval [0, 4]. Each car occupies a space of length 1. Calculate expected number of cars that can park. I am trying to solve it using indicator random variables, but I am not able to form any (probably because the parking lane is not discrete....??). I am confused here. Can I have some hint over here please ? AI: Hint: draw a square, with $x$ being the center of the first car and $y$ the center of the second car. We must have $0.5 \le x,y \le 3.5, \|x - y\| \ge 1$, so shade the part of the square that is allowable. Now what fraction of that area can accommodate another car? As you will always fit two cars and never fit four cars, all you care about is whether the third car fits.
H: A set of formulas that classifies two-element structures Give a set of formulas $\Gamma$ such that for any structure $\mathcal{A}=\langle A;-;-\rangle$ it holds that $\mathcal{A} \models \Gamma$ if $A$ has exactly two elements. AI: The formula you mention the comments works, but should be parenthesised as follows: $$\exists x \exists y \forall z ((z=x \vee z=y) \wedge x \ne y)$$ It says there are two elements which are not equal such that any element is one or the other... which is precisely the statement that there are exactly two elements. In general it's not true that $(A \vee B) \wedge C \equiv A \vee (B \wedge C)$, and without including parentheses it's unclear which of these $A \vee B \wedge C$ is taken to mean.
H: Twin prime "test" via congruence I decided to try getting a test for a "twinness" of a prime via Wilson's theorem. Wilson's theorem says that integer $n > 1$ is a prime iff $$(n-1)! \ \equiv -1 \pmod n $$ Now, if both $n$ and $n+2$ are prime, we get two equations: \begin{cases} (n-1)! \ \equiv -1 \pmod n & (1)\\ (n+1)! \ \equiv -1 \pmod{n+2} & (2) \\ \end{cases} The equation (1) means that there exists integer $k$ such that $$(n-1)! = k n-1$$ By writing the equation (2) such that there is integer $v$ that $$(n+1) n (n-1)! = -1+ v(n+2)$$ and then replacing $(n-1)!$ in (2) from (1), we get: $$(n+1) n (k n-1) = -1+ v(n+2).$$ This can be written as $$k(n^3+n^2)+v(-2-n)=n^2+n-1 \ \ \ \ (3)$$ So, I would think that $n$ is the first of twin of primes iff there exists integers $k$ and $v$ that (3) is true. But, take $n=7.$ Now we get $$392 k = 55 + 9 v$$ which has solution $\{k = 2, v = 81\}.$ This shouldn't be possible. Where is the error? AI: Your logic is not reversible, so the congruence you derive is a necessary condition, not an if-and-only-if. For instance, there is no hope of deducing (1) from (3), because you have eliminated any information there was about $(n-1)!$ from (3).
H: Express $\sin\frac{\pi}{8}$ and $\cos\frac{\pi}{8}$ with $\cos\frac{\pi}{4}$ I've been trying with no success expressing this functions. a) $\sin\frac{\pi}{8}$ with $\cos\frac{\pi}{4}$ b) $\cos\frac{\pi}{8}$ with $\cos\frac{\pi}{4}$ I've tried formula of the double angle ($\sin$ and $\cos$) and the ecuation $\cos^2X+\sin^2X=1$. Can anyone point what to use or where to start? I appreciate it, thanks. UPDATE I think this is achived with the half formula as njguliyev and nbubis said. $$\sin\frac{\pi}{8} = \sin(\frac{1}{2} * \frac{\pi}{4}) = \pm\sqrt\frac{1-\cos\frac{\pi}{4}}{2}$$ $$\cos\frac{\pi}{8} = \cos(\frac{1}{2} * \frac{\pi}{4}) = \pm\sqrt\frac{1+\cos\frac{\pi}{4}}{2}$$ AI: Hint: You my want to take a look at the half angle formulas.
H: Poisson Process Notation: P(N(h) = 2) =P(N1(h) = 1; N2(h) = 1) This is from the book: Stochastic Processes by Sheldon Ross Does the ";" sign mean conditional probability in the second part of the equation? AI: $$P(A,B)=P(A;B)=P(A\cap B){}$$
H: Determine the value of h such that the matrix is the augmented matrix of a consistent linear system. Determine the value of h such that the matrix is the augmented matrix of a consistent linear system. $$ \begin{bmatrix} 1 && h && 4 \\ 3 && 6 && 8 \end{bmatrix} $$ I'm entirely unsure how to go about solving this. This how far I got: $$ \sim\begin{bmatrix} 1 && h && 4 \\ 0 && (6-3h) && -4 \end{bmatrix} R_2' = R_2-3R_1 $$ $$ 6-3h=-4 \\ 6+4=3h \\ 10=3h \\ h=3/10 \text{ ?} $$ New progress $$ \sim\begin{bmatrix} 3 && 6 && 8 \\ 1 && h && 4 \end{bmatrix} R_1 \leftarrow\rightarrow R_2 $$ $$ \sim\begin{bmatrix} 3 && 6 && 8 \\ 0 && (h-2) && 4/3 \end{bmatrix} R_2' = - \frac13 R_1 $$ $$ \sim\begin{bmatrix} 3 && 6 && 8 \\ 0 && 3(h-2) && 4 \end{bmatrix} R_2' = 3R_2 $$ $$ \sim\begin{bmatrix} 3 && 6 && 8 \\ \\ 0 && 1 && \frac{4}{3(h-2)} \end{bmatrix} R_2' = R_2 \div 3(h-2) $$ $$ \sim\begin{bmatrix} -3 && 0 && z \\ \\ 0 && 1 && \frac{4}{3(h-2)} \end{bmatrix} R_1' = 6R_2 - R_1 \\ z = -8 + \frac{24}{3(h-2)} = \frac{-32(h-2) + 24}{3(h-2)} $$ $$ \sim\begin{bmatrix} 1 && 0 && z \\ \\ 0 && 1 && \frac{4}{3(h-2)} \end{bmatrix} R_1' = -\frac13 R_1 \\ z = - \frac{-32(h-2) + 24}{(h-2)} $$ AI: Do these row operations: Swap $R_1$ and $R_2$ $R_2 \rightarrow -R_1/3 + R_2$ $3R_2$ Divide $R_2/(3(h-2))$ $R_1 \rightarrow 6R_2 - R_1$ Divide $R_1/3$ This should provide the value of $h$ that makes the system inconsistent. Spoiler - Do Not Peek $\begin{bmatrix} 1 && 0 && \dfrac{8(h-3)}{3(h-2)} \\ 0 && 1 && \dfrac{4}{3(h-2)}\end{bmatrix}$ Notes: Sometimes you can also look at the determinant of the system, in this case it is $6-3h$ (what value of $h$ would make this inconsistent). You could have also reduced and solved the system as another approach. You get $y = \dfrac{4}{3(2-h)}$ and can substitute back. There are other approaches.
H: I need help figuring out how this sequence converges. $A_n = ( 1 + \frac 2n ) ^ n$; I know the end result is convergent at $e^2$, but how do I figure that out? I've started setting it up as $An = (1 + (2/n))^{(n/2)2}$ as someone has suggested but i don't understand how that results in $e^2$. AI: The suggestion that you received relies on your knowing that $$\lim_{x\to 0^+}(1+x)^{1/x}=e\;.$$ Given that knowledge, you can substitute $x=\frac2n$ and compute $$\begin{align} \lim_{n\to\infty}\left(1+\frac2n\right)^n&\lim_{x\to 0^+}(1+x)^{2/x}\\\\ &=\lim_{x\to 0^+}\left((1+x)^{1/x}\right)^2\\\\ &=\left(\lim_{x\to 0^+}(1+x)^{1/x}\right)^2\\\\ &=e^2\;. \end{align}$$ With a little practice one needn’t actually make the substitution: $$\begin{align} \lim_{n\to\infty}\left(1+\frac2n\right)^n&=\lim_{n\to\infty}\left(1+\frac2n\right)^{(n/2)2}\\\\ &=\lim_{n\to\infty}\left(\left(1+\frac2n\right)^{n/2}\right)^2\\\\ &=\left(\lim_{n\to\infty}\left(1+\frac2n\right)^{n/2}\right)^2\\\\ &=e^2\;. \end{align}$$
H: I just proved that $ℂ$ is not a field. What is the mistake in my reasoning? What is the mistake in my reasoning? Consider $(X^2+1)$ in $ℝ[X]$. Then $(X^2+1)⊂(X^2,1)$. Because if $f \in (X^2+1)$. Then $f=(X^2+1)g=X^2g+1⋅g$. So $f \in (X^2,1)$. Therefore $(X^2+1)$ not a maximal ideal. And therefore $ℝ[x]/(X^2+1)≅ℂ$ is not a field. AI: The mistake is that you didn't notice that the ideal $(X^2,1)$ contains $1$, and therefore is the entire ring $\mathbb{R}[X]$. Of course, the unit ideal is an ideal that properly contains all maximal ideals.
H: Discrete Math - Combinatorics - Trinomial Coefficients question Let $k,l,m,n \in Z \geq 0$ be such that $n=k+l+m$. The trinomial coefficient ${n \choose k,l,m}$ is given by the rules: for $k+l=n$, ${n\choose k,l,0} = {n \choose k,0,l} = {n \choose 0,k,l} = {n\choose k}$ ${n\choose k,l,m} = {n-1 \choose k-1,l,m} + {n-1 \choose k,l-1,m} + {n-1\choose k,l,m-1}$ The following questions use this definition. (a) What are all the trinomial coefficients for $n=1,2,3$? (b) Describe the "triangle" of trinomial coefficients (Hint: Think three dimensional Pascal's triangle). I don't understand the notation at all: n choose k,l,m. What does that even mean? And I don't get how to set up part (a) at all either. It's just very confusing. For (b) I can visualize a Pascal's "pyramid." One where the number below is a sum of the 3 above it. Something like that. But other than that, I'm not really sure what's going on. AI: The lines numbered (1) and (2) are the definition of $\binom{n}{k,\ell,m}$, so it means exactly what they say it means. It’s a recursive definition: it tells you how to compute trinomial coefficients with upper number $n$ if you already know how to compute them with upper number $n-1$. However, you’ve miscopied (2), unless there was a typo in your source: it should read $$\binom{n}{k,\ell,m}=\binom{n-1}{k-1,\ell,m}+\binom{n-1}{k,\ell-1,m}+\binom{n-1}{k,\ell,m-1}\;.$$ Here’s an example to illustrate how to use (1) and (2) to calculate a trinomial coefficient: $$\begin{align} \binom4{1,2,1}&=\binom3{0,2,1}+\binom3{1,1,1}+\binom3{1,2,0}&\text{using (2)}\\\\ &=\binom32+\binom3{1,1,1}+\binom31&\text{using (1)}\\\\ &=\binom32+\binom2{0,1,1}+\binom2{1,0,1}+\binom2{1,1,0}+\binom31&\text{using (2)}\\\\ &=\binom32+\binom21+\binom21+\binom21+\binom31&\text{using (1)}\\\\ &=3+2+2+2+3\\ &=12\;. \end{align}$$ The pyramid that you describe is exactly what’s wanted for (b). At the peak you have $\binom0{0,0,0}$, which is $1$. Below that you have a triangle of three $1$’s. The next layer down will have six entries forming a triangle, corresponding to $\binom1{1,0,0},\binom1{0,1,0}$, and $\binom1{0,0,1}$. The next layer will have ten entries in a triangle, corresponding to the ten possible trinomial coefficients $\binom2{k,\ell,m}$ with $0\le k,\ell,m$ and $k+\ell+m=2$. And so on.
H: Any Subset of a Set Containing No Accumulation Point is Closed Let $(X,\tau)$ be a topological space. Suppose that $B\subseteq X$ has no accumulation point. That is, for any $x\in X$ there exists some $U\subseteq X$ such that $x$ is in the interior of $U$ and $B\cap U\cap\{x\}^c$ is empty. Claim: If $C\subseteq B$, then $C$ is closed. Proof: If $C\subseteq B$, then, since $B$ does not have an accumulation point, $C$ does not have any, either. But any set that contains the set of its own accumulation points (which is the empty set in the case of $C$) must be closed. $\blacksquare$ Corollary If $(Y,\nu)$ is any topological space, then any function $f:B\to Y$ is continuous (in the relative topology). Proof: If $V$ is any closed set in $Y$, then $f^{-1}(V)\subseteq B$. Therefore, $f^{-1}(V)$ is closed in the topology $\tau$ and thus must be closed in the relative topology (generated by $B$) as well. $\blacksquare$ This set of results (especially the corollary) is so surprising to me that I thought I would seek some feedback. Thank you very much for sharing your thoughts. AI: It’s entirely correct. The hypothesis on $B$ amounts to saying that $B$ is a closed, discrete subset of $X$. Every subset of a closed, discrete set is also closed and discrete, and every function whose domain is a discrete space is continuous. A familiar example is the set $\Bbb Z$ in $\Bbb R$ with its usual topology: $\Bbb Z$ is a closed, discrete subset of $\Bbb R$, and therefore so is every subset of $\Bbb Z$. And any function $f:\Bbb Z\to Y$, where $Y$ is any space, is continuous, since every subset of $\Bbb Z$ is open.

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