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H: Eigenvalues of a particular matrix Given two vectors $\boldsymbol\alpha=\left(\alpha_1,...,\alpha_N\right)$ and $\boldsymbol\beta=\left(\beta_1,...,\beta_N\right)$, is there an easy way to compute the eigenvalues of a matrix $M_{k,q}$ whose entries are expressed as $$ M_{k,q}=\frac{\alpha_k}{\beta_q},\quad k=1,...,N,\quad q=1,...,N\quad ? $$ AI: Hint: $M$ has rank $1$, and $M {\bf \alpha}^T = \ldots$
H: Arithmetic problem Prove that $(a^2:a^2+16)=1 \lor 4 \lor 16$ I know that $(a:b)=(a:b+ka)$ with $k\in\Bbb Z$ So: $(a^2:a^2+16)=(a^2:a^2+16-a^2)=(a^2:16)=1\lor4\lor16$ This is enough to prove this? AI: Maybe you need to explain a little more why $(a^2:16)=1\lor4\lor16$. For example why $(a^2:16)=2$ is impossible? Hint: $2=2^1$.
H: What is the equation for a line tangent to a circle from a point outside the circle? I need to know the equation for a line tangent to a circle and through a point outside the circle. I have found a number of solutions which involve specific numbers for the circles equation and the point outside but I need a specific solution, i.e., I need an equation which gives me the $m$ and the $b$ in $f(x) = mx + b$ for this line. AI: HINT: Let $C(a,b)$ be the point outside the circle Equation of any line passing through $C$ will be $$\frac{y-b}{x-a}=m\implies mx -y+b-am=0$$ where $m$ is the gradient. Now, for tangency, the distance of the tangent from the center of the circle will be equal to the radius of the circle. This condition will give us two values of $m$ As $C$ is outside the circle, we shall have two distinct real values of $m,$ resulting in two real tangents Alternatively, you find the intersection of the given circle with the line by eliminating $x$ or $y$ to form a quadratic equation of the surviving variable . For tangency, both roots must be same i.e., the discriminant must be $0$
H: What is the order of $e^{\large \frac{4\pi i}{5}}$ in the circle group $U_{20}$? We talk about the Circle group. What is the order of $e^{\large \frac{4\pi i}{5}}$? The power is $\frac{4\pi i}{5}$ if it's not clear... Thank you! AI: If this question stems from your earlier question, where you were talking about $\omega^{8}_{20}$ in the circle group of order $20$, recall that the circle group is cyclic, and so this group is isomorphic to $\mathbb Z_{20}$, the additive group of integers modulo $20$. Now, we have that $\left(\omega^{8}\right)^5 = \omega^{40} = e^{4\pi i} = 1$, and $5$ is the least integer $k$ such that $(\omega^8)^k = 1$. Hence $\operatorname{ord}(\omega^8) = \operatorname{ord}(e^{4\pi i/5})= 5$.
H: Inferenecs from the given: x = nonhomogenous + homogenous solution solve Ax = b. (GStrang P161, Ex 3.4B.4) Given: All solutions to $\mathbf{Ax = b}$ have the form $\mathbf{x} = (1, 1, 0)^T + c(1, 0, 1)^T$. $\Large{\color{red}{1. [}}$ Then $A$ must have $n = 3$ columns. $\Large{\color{red}{]}}$ With $(1, 0, 1)$ in the $\ker A$, since $\mathbf{Ax = 0} \iff RREF(A)\mathbf{x = 0}$, column $1$ [of both $A$ and $RREF(A)]+$ column $3$ [of both $A$ and $RREF(A)$] $ = 0$. Column $2$ must NOT be a multiple of Column $1$, or [else] the nullspace would contain another special solution. $\Large{\color{red}{2. [}}$ So the rank of $A = 3 - 1 =2$. $\Large{\color{red}{]}}$ $\Large{\color{red}{3. [}}$ Necessarily, A has $m \geq 2$ rows. $\Large{\color{red}{]}}$ The right side $b$ = column $1$ + column $2$. Independence, basis, dimension, orthogonality, determinants succeed P161 in Strang's Intro to Linear Algebra, 4th ed, so will you please to exclude/pretermit these concepts. $1.$ $\mathbf{x}$ is a $3 \times 1$ vector. By virtue of the accord of sizes in matrix multiplication, $\mathbf{x}$ is a $3 \times 1$ vector $ \Longrightarrow \mathbf{A}$ has $3$ columns. Are there other less pedestrian ways to deduce this? $2.$ I register $(1, 1, 0)^T$ as $1$ solution to the nonhomogenous linear system. Then how to proceed? $3.$ Since $\mathbf{A \neq 0}$, I thought $m \geq 1$ row(s). How and why $m \geq 2$? AI: Here's a (non-rigorous) way to look at it. Try working backwards. We are given that: $$ \begin{cases} x_1 = 1+c \\ x_2 = 1 \\ x_3 = c \end{cases} \implies \begin{cases} x_1 = 1+x_3 \\ x_2 = 1 \\ x_3 \text{ is free} \end{cases} $$ Since $x_1,x_2$ are basic variables and $x_3$ is a free variable, the smallest possible row-reduced echelon form of the augmented matrix $[A\|b]$ would have the form: $$\left[\begin{array}{ccc\|c} 1 & 0 & -1 & 1 \\ 0 & 1 & 0 & 1 \\ \end{array}\right]$$ We know that $m \geq 2$ because the number of rows in the matrix must be greater than or equal to the number of pivots in the matrix (that is, the rank of the matrix). To make the inequality strictly greater, we can add any number of zero rows to the bottom of the augmented matrix.
H: Urn probability function Suppose I have an urn with an infinite number of balls which can be either red or white. I do not know what the proportion of each colour is, but I do know it's a fixed proportion. After drawing $N$ balls, I have observed $r$ red ones and $w$ white ones. I believe the probability that I will observe a red ball on the next draw is given by Laplace's Law of Succession, $\frac{r+1}{N+2}$. However, how sure should I be of that? That is, before I drew any balls, I believed any proportion other than $0$ or $1$ was the true one. After I drew those $N$ balls, what should be my estimated pdf over the possible values for the proportion of red balls in the urn? AI: Okay, so, my first idea was using Jaynes' $A_p$ distribution, in which case we can define $A$ = next draw will be red and $N_r$ = out of $N$ draws, $r$ were red. With an ignorant prior distribution $(A_p\|X) = 1$, I get that $$(A_p\|N_rX) = (A_p\|X)\frac{P(N_r\|A_p)}{P(N_r\|X)}$$ We know that $$P(N_r\|A_p) = \binom{N}{r}p^r(1-p)^{N-r}$$ And we can find $$P(N_r\|X) = \int^1_0(N_rA_p\|X)dp = \int^1_0P(N_r\|A_p)(A_p\|X)dp = \int^1_0\binom{N}{r}p^r(1-p)^{N-r}dp$$ from which $$P(N_r\|X) = \frac 1 {N+1}, 0 \leq r \leq N$$ And then $$(A_p\|N_r) = (N+1)\binom{N}{r}p^r(1-p)^{N-r}$$ And that looks like my distribution. Is that correct?
H: $p$-polynomial of $n$'th degree, $q(x)=p[x,x_1,x_2,...,x_k]$, prove that q has the same leading coefficient. So I have a polynomial $p$ of $n$'th degree and q given by $q(x)=p[x,x_1,x_2,...,x_k]$, meaning that for $x$ it gives back the leading coefficient in interpolation of $p$ on points $x,x_1,...,x_k$. Prove that q is of degree $n-k$, and the coefficient beside $x^{n-k}$ is the same, as the one beside $x^n$ in $p$. AI: For the original question, a counterexample. $f(x)=x^2$ interpolated at two points is linear, $Ax+B$, and $A$ is dependent on the two points, not only the coefficient of $x^2$ in $f$. The statement in the new question is correct, because if you allow the interpolation points to include $x$, and substitute $x$ into the result, you recover the original polynomial completely, not only its highest degree term, and $c(x) \prod (x -x_i)$ is the unique part of this expression with highest degree of $x$.
H: "Rising sun" function Let $f:[0,1] \to \mathbb R$ be bounded and $$ f_\odot :[0,1] \to \mathbb R: x \mapsto \sup \{f(y) : y \in [x,1] \} $$ This is well defined since $f$ is bouned. Claim: If $f$ is continuous then $f_\odot$, too. I begun as follows: Let $x_0 \in [0,1]$ and $\epsilon > 0$. First: Find $y_0 \in [x_0,1]$ s.t. $f_\odot(x_0) - \epsilon < f(y_0) \leq f_\odot(x_0)$. Then using continuity of $f$ I get a $\delta > 0$ s.t. $\forall a \in (y_0-\delta,y_0+\delta) : f(y_0) - \epsilon < f(a) < f(y_0) + \epsilon$ and thus also $\|f(a)-f_\odot(x_0)\|<2\epsilon$ for $a \in (y_0-\delta,y_0+\delta)$. Now I get suck. I think that this $\delta$ also works for $f_\odot$ somehow but can't figure it out. AI: Note that by continuity of $f$ we have $\sup=\max$ on compact intervals. Thus we may assume that in fact $f_\odot(x_0)=f(y_0)$. If $f(x_0)<f_\odot(x_0)$, then $y_0>x_0$ and we have $f(x)<f_\odot(x_0)$ in some $\delta$-neighbourhood of $x_0$. Without loss of generality, $\delta<y_0-x_0$ and hence $f_\odot(x)=f(y_0)$ for $x\in(x_0-\delta,x_0+\delta)$. If on the other hand $f(x_0)=f_\odot(x_0)$, then $f(x)\le f_\odot(x)\le f(x_0)$ for $x\ge x_0$ and if we pick $\delta>0$ such that $\|f(x)-f(x_0)\|<\epsilon$ for $\|x_0-x\|<\delta$, then $f(x_0)-\epsilon<f(x)\le f_\odot(x)\le f_\odot(x_0)$ for $x_0\le x<x_0+\delta$ and also $f(x_0)\le f_\odot(x)<f(x_0)+\epsilon = f_\odot(x_0)+\epsilon$ for $x_0-\delta<x\le x_0$.
H: How to evaluate the limit $\lim_{x\to0}\frac{\sqrt{x+1}-\sqrt{2x+1}}{\sqrt{3x+4}-\sqrt{2x+4}}$ First I tried direct substitution, which resulted in the indeterminate form. Then because of the square roots, I tried rationalizing (both numerator and denominator), but still get the indeterminate form. I can't use L'Hopital's rule because we haven't learned that yet (not sure if it would work anyway). Don't really know what else to try algebraically... Any hints? AI: $\lim_{x\to0}\frac{\sqrt{x+1}-\sqrt{2x+1}}{\sqrt{3x+4}-\sqrt{2x+4}}\cdot\frac{\sqrt{3x+4}+\sqrt{2x+4}}{\sqrt{3x+4}+\sqrt{2x+4}}\cdot\frac{\sqrt{x+1}+\sqrt{2x+1}}{\sqrt{x+1}+\sqrt{2x+1}}=\lim_{x\to0}\frac{-x(\sqrt{3x+4}+\sqrt{2x+4})}{x(\sqrt{x+1}+\sqrt{2x+1})}=-\frac{4}{2}=-2$.
H: What is the difference between logarithmic decay vs exponential decay? I am a little unclear on whether they are distinctly different or whether this is a 'square is a rectangle, but rectangle is not necessarily a square' type of relationship. AI: The "Square is a rectangle" relationship is an example where the square is a special case of a rectangle. "Exponential decay" gets its name because the functions used to model it are of the form $f(x)=Ae^{kx} +C$ where $A>0$ and $k<0$. (Other $k$'s above $0$ yield an increasing function, not a decaying one.) Similarly for "logarithmic decay," it gets its name since its modeled with functions of the form $g(x)=A\ln(x)+C$ where $A<0$. These two families of functions do not overlap, so neither is a special case of the other. The giveaway is that the functions with $\ln(x)$ aren't even defined on half the real line, whereas the exponential ones are defined everywhere.
H: Reducing a fraction, divisibility and indeterminate symbol Quick question about validity, just to make sure. When I have a fraction in a form: $$\frac{3a + 3b}{a+b}$$ and I extract the common factor 3 out to get: $$\frac{3(a+b)}{a+b} \;=\; 3\frac{a+b}{a+b}$$ is it valid now to reduce the fraction to 1 to end up with just 3? Could it change anything if $b = -a$ ? Because then I would have: $$3\frac{a-a}{a-a} \;=\; 3 \frac{0}{0}$$ and mathematicians say that $0/0$ is an indeterminate symbol. Does it make the whole product indeterminate too? And what if the common factor is not 3 but some unknown symbol, say $c$? What can we say about the divisibility of the numerator by $a+b$ in that particular situation? Is it safe to say that the numerator is divisible by $a+b$ even if $a+b$ can be 0 sometimes? Or should I first enforce a condition on $a+b$ that it is never 0? Edit: I know about the general case when I have $a+b$ in the denominator, that then it cannot be 0. But I wonder if it changes anything if $a+b$ is also a factor of the numerator. Because then, we have somewhat different situation: division by 0 is undefined, but $0/0$ is just undetermined, that is, it can be anything (that's what mathematicians usually say to me). AI: The original fraction $\dfrac{3a+3b}{a+b}$ is defined if and only if $a+b\ne 0$. When that’s the case, then you have $$\frac{3a+3b}{a+b}=\frac{3a+3b}{a+b}\cdot1=\frac{3a+3b}{a+b}\cdot\frac{\frac1{a+b}}{\frac1{a+b}}=\frac31=3\;.$$ What you can say, then, is that if $\dfrac{3a+3b}{a+b}$ is defined in the first place, it’s equal to $3$. If, for instance, you have a function $f$ of two variables defined by $$f(a,b)=\frac{3a+3b}{a+b}\;,$$ this function is equal to the constant function $g(a,b)=3$ wherever it is defined, but while $g$ is defined on the whole plane, $f$ is defined only where $a+b\ne 0$, i.e., where $a\ne -b$.
H: Calculating percentage to compensate for percent discount. Missing something very basic here and cannot pin point it. We need to charge a client \$100 for a product. Let's say our payment processor charges us 10% on every transaction. We make this transparent to the client and charge them accordingly: $\$x = \$100 / (1 - (10 / 100)) = $111.11111..$ Now, we want to offer to contribute to transaction fees such that they only have to pay half of what the payment processor charges. But clearly, $(\$100 + ((\$111.11111... - \$100) / 2))$ is not the answer. The question is, how do we calculate the selling price such that the transaction fee is split half way between us and the client? UPDATE: Based on the answers, here is the clarification that should have been part of the original question. $\$x=\$100$ is the pre-determined "selling price" to the client. We add whatever the payment processor charges us (10% in this case) and increase the selling price such that we receive $\$x=\$100$. Since we now want to "split" the processing fee with the client, we should lose from $\$100$ exactly how much extra the client is paying above $\$100$. AI: As I understand it, you will charge the customer $p$ and receive $0.9p$. You want to pay half the processing fee, so $0.95p=100$, $p=\frac {100}{0.95}\approx 105.263$ You receive about $94.737$, so you are paying $5.263$ as are they.
H: $\sum\frac{(k!)^2}{2k!}$ serie converge or diverge Does the following series diverge or converge. $\sum\frac{(k!)^2}{2k!}$ I decide to apply the powerful ratio test. I did $\frac{(k+1)!^2}{2(k+1)!}$ which then becomes $\frac{(k+1)!(k+1)!}{(2k+2)!}$*$\frac{2k!}{(k!)(k!)}$ I get the $k\rightarrow\infty$ $\frac{(k+1)(k+1)}{(2k+2)(2k+2)}$ =$\frac{1}{4}$ $\frac{1}{4}$<1 convergent? AI: Assuming you're working with the series: $$\sum\frac{(k!)^2}{(2k)!},$$ then you should be able to reduce the limit of the ratio to $$\lim_{k \to \infty} \frac{(k+1)(k+1)}{(2k+2)(2k+1)}$$ The limit is still $\frac 14$, and so, yes, by the ratio test, we have that the series converges.
H: almost everywhere convergence Let $E\subseteq\mathbb{R}^l$ be s.t. $E$ is Lebesgue measurable and $m(E)>0$. Let for all $k\in\mathbb{N}$, $f_k:E\rightarrow \mathbb{R}$ be measurable functions. If for all $\epsilon>0$ we can find $F\subseteq E$ s.t. $F$ is closed and $m(F)\leq \epsilon$ AND $f_k\rightarrow f$ uniformly on $E-F$, is it true that $f_k\rightarrow f$ uniformly a.e. on $E$? AI: No. Take $f_n(x) = x^n$ on $[0,1]$. Then $f_n \to 0$ on any closed set that does not contain $1$. However, $f_n$ does not converge uniformly to zero on the complement of any null set. To see this, note that $m \{ x \| f_n(x) > \frac{1}{2} \} = 1-\frac{1}{\sqrt[n]{2}} > 0$ for all $n$, so there are always 'lots' of points (from a measure zero perspective) that satisfy $f_n(x) > \frac{1}{2}$.
H: $\sum\frac{k^k}{3^{k^2}}$ series be convergent or divergent Would the following series be convergent or divergent. $\sum\frac{k^k}{3^{k^2}}$ I applied the root test to the following series and got. $(\frac{k^k}{3^{k^2}})^{1/k}$ I got $\lim$ $k\rightarrow\infty$ $\frac{k}{3^k}$ using Le Hospitals rule I got $\frac{1}{3^k}$ which is zero 0<1 convergent? AI: Correct limit and correct conclusion; you simply have a slight error in the derivative of the denominator, $(3^k)' = (\log 3)3^k$.
H: Inner Product vs. Integrals with Fourier Series, When to include 1/2pi? I am confused about when to include a prefactor of $\frac{1}{2\pi}$ when dealing with integrals of functions that are expressed as fourier series. This is what I understand (please correct me if I'm wrong). Assume I have a square integrable function, $f\left(x\right)$. I can express it as a fourier series according to: $$f\left(x\right)=\sum_{k=-\infty}^{\infty} c_k e^{ikx}$$ Where the coefficients, $c_k$, are obtained via the following inner-product: $$c_k=\frac{1}{2\pi}\int_{0}^{2\pi} e^{-ikx} f\left(x\right)\text{d}x$$ My understanding of the standard inner-product between functions is, that it is defined by: $$\left\langle h\left(x\right),g\left(x\right)\right\rangle \equiv \int_{\Omega} h^\left(x\right) g\left(x\right) \text{d}\Omega$$ Where the $^$ denotes complex conjugation, $\Omega$ is the domain, and $\text{d}\Omega$ is the invariant measure. So, using this definition, I would intuitively think that the coefficients for a fourier series ought to be computed from: $$c_k=\int_{0}^{2\pi} e^{-ikx} f\left(x\right)\text{d}x$$ i.e. without the prefactor of $\frac{1}{2\pi}$. Question 1: So, why is the prefactor of $\frac{1}{2\pi}$ included in the special case when the basis functions are complex exponentials, i.e. for Fourier series'? I know that the complex exponentials are not orthonormal, only orthogonal, but why should the normalization convention of the basis functions change the definition of the inner-product? I don't think it comes from the invariant measure, since for a circle the invariant measure should be $\text{d}x$, not $\frac{1}{2\pi}\text{d}x$ Now, say that I have an equation that is defined using an integral whose arguments include functions expressed as Fourier series. Specifically: $$\bar{S} = \int_{0}^{2\pi} f\left(x\right)S\left(x\right)\text{d}x$$ Where $\bar{S}$ represents the average value of $S\left(x\right)$ weighted by the probability density function $f\left(x\right)$. If I express $f\left(x\right)$ and $S\left(x\right)$ as Fourier series, and use the fact that $f\left(x\right)=f^\left(x\right)$, then the equation should be: $$\bar{S} = \sum_{k=-\infty}^{\infty} \sum_{m=-\infty}^{\infty} c_k^ s_m\int_{0}^{2\pi} e^{-ikx} e^{imx} \text{d}x=2\pi \sum_{k=-\infty}^{\infty} c_k^* s_k$$ Question 2: In this case the integral amounts to doing an inner-product but without the prefactor of $\frac{1}{2\pi}$. So, when doing integrals of functions that are expressed as Fourier series do you need to always throw in that prefactor or not? When should you? AI: It depends on the inner product and the basis used. Suppose you have an inner product $\langle \cdot, \cdot \rangle$ and an orthogonal (Schauder) basis $b_k$ for a Hilbert space $\mathbb{H}$. If you represent some $f \in \mathbb{H}$ in terms of the basis, you will have $f = \sum_k \hat{f_k} b_k$, where $\hat{f_k}$ are the Fourier coefficients with respect to the basis $b_k$. Then you have no choice over the $\hat{f_k}$, as they must satisfy $\langle b_k, f \rangle = \hat{f_k} \\|b_k\\|^2$, or $\hat{f_k} = \frac{\langle b_k, f \rangle}{\\|b_k\\|^2}$. To answer your first question. Presumably you are dealing with $L^2[0,2 \pi]$, with $\langle f, g \rangle = \int_0^{2 \pi} \overline{f}(t)g(t) dt$, and the basis $b_k(t) = e^{i k t}$. Since $\\|b_k\\|^2 =\langle b_k, b_k \rangle = 2 \pi$, the coefficients are given by $\hat{f_k} = \frac{\langle b_k, f \rangle}{\\|b_k\\|^2} = \frac{1}{2\pi} \int_0^{2 \pi} e^{-ikt} f(t) dt$, whence the $2 \pi$. For your second question, you have $\overline{S} = \int_{0}^{2\pi} f(x) S(x) dx$. In terms of the inner product from the previous question, this becomes $\overline{S} = \langle f, S \rangle$. If $\hat{f_k}, \hat{S_k}$ are the Fourier coefficients (with respect to the basis and inner product above), then an easy computation (Parseval's theorem) shows that $\langle f, S \rangle = \sum_k \\|b_k\\|^2 \overline{\hat{f_k}} \hat{S_k}$, and so we obtain the result above (remember $\\|b_k\\|^2 = 2 \pi$).
H: Power Series Expansion Asymptotics From my text: Given $\cos^n(x),$ set $x=\frac{\omega}{\sqrt{n}}$, then a local expansion yields: $\displaystyle\cos^n(x)=e^{n\log\cos(x)}=e^{(-\frac{\omega^2}{2}+O(n^{-1} \omega^4))}$ the approximation being valid as long as $\omega = O(n^{1/4})$ I do get the approximation as indicated, but I don't understand why it's valid only if $\omega = O(n^{1/4})$ AI: It might mean "as long as" more literally, not in the sense "only if", but in the sense of "if". The asymptotic term $n^{-1}\omega^4$ will be asymptotically smaller than the leading term $\omega^2$ for $\omega = o(n^{1/2})$.
H: Is $ n^2-14n+24 $ a prime number? How many are those positive integers n such that Is $ n^2-14n+24 $ is prime ? I have tried to solve this problem by putting different values of natural number . Is it a right way ? AI: $$n^2−14n+24=(n-2)(n-12)=p \text{ (p prime number)}$$ so $$(n-2)(n-12) = p1\text{ or }1p$$ so $$n-2=1\text{ and }n-12 =p$$ $n=3$ but it's not answer (it's not prime) $n-12=1$ and $n-2=p$ then $n=13$ and $$(n-2)(n-12) =11 *1$$ I think that is only answer
H: discrete mathematics , sequences, characteristic equation Today in my discrete mathematics class we started combinatorics and also solved some some recurrence relation sequences using the characteristic equation. So my question is can you guys point me to some exercises about recurrence relation sequences using the characteristic equation, because in my text book there are any, but there will be problems of this kind on my exam. Thanks in advance. AI: Search "Introductory Combinatorics 6th edition by Richard Brualdi" chapter $7.3$ and $7.4$, "Discrete Mathematics with Combinatorics and Graph Theory by Kenneth Rosen", and "Discrete Mathematics with Graph Theory by Edgar G. Goodaire and Michael M. Parmenter" chapter $5.2$, $5.3$, and $5.4$. All of these books have many problems of the type you seek.
H: $\sum\sin(\frac{k\pi}{4})$ absolute convergent, conditional convergent,divergent? Would the following be abs convergent, conditional convergent or divergent. $\sum\sin(\frac{k\pi}{4})$ I know sin(x) is between $-1<x<1$ y=sin(x) is oscillating would it be $(-1)^n$ AI: Sorry about the old answer, that was just garbage. The real answer is that it's not convergent. You can try grouping them in terms of $8$ like I had done, but the problem is that, because it ends at $\infty$, you don't have a point to end at. That means the answer could be the same as any of the following eight: $\displaystyle \sum_{k=1}^1 \sin \left(\frac{k \pi}{4}\right) = \frac{\sqrt{2}}{2}$. $\displaystyle \sum_{k=1}^2 \sin \left(\frac{k \pi}{4}\right) = \frac{\sqrt{2}}{2} + 1$. $\displaystyle \sum_{k=1}^3 \sin \left(\frac{k \pi}{4}\right) = \frac{\sqrt{2}}{2}$. $\displaystyle \sum_{k=1}^4 \sin \left(\frac{k \pi}{4}\right) = 0$. $\displaystyle \sum_{k=1}^5 \sin \left(\frac{k \pi}{4}\right) = -\frac{\sqrt{2}}{2}$. $\displaystyle \sum_{k=1}^6 \sin \left(\frac{k \pi}{4}\right) = -\frac{\sqrt{2}}{2} - 1$. $\displaystyle \sum_{k=1}^7 \sin \left(\frac{k \pi}{4}\right) = -\frac{\sqrt{2}}{2}$. $\displaystyle \sum_{k=1}^8 \sin \left(\frac{k \pi}{4}\right) = 0$. Because they're different, and because you have no sure way of determining which is the right answer, the answer is that the series does not converge. This is actually a classic problem of trying to find the sum of an alternating series. Let's take a similar problem, where you're trying to find the answer to $$1 - 1 + 1 - 1 + \cdots$$ (This is called Grandi's Series, btw.) There are multiple answers, depending on how you group them. You can group them as $(1-1) + (1-1) + \cdots = 0 + 0 + \cdots = 0$, as $1 + (-1 + 1) + (-1 + 1) + \cdots = 1 + 0 + 0 + \cdots = 1$, or even as $\displaystyle \frac{1}{2} + \left(\frac{1}{2} - \frac{1}{2}\right) + \left(-\frac{1}{2} + \frac{1}{2}\right) + \cdots = \frac{1}{2} + 0 + 0 + \cdots = \frac{1}{2}$. Since there's no "right" way of doing this, the answer is DNE. Hence, it does not converge.
H: Prove « If P(A) is a subset of P(B) => A is a subset of B » I need to prove «If P(A) is a subset of P(B) => A is a subset of B», generally, I understand the main way I should prove it, but the problem is in the formal, pedantic language I have to use to prove such statement. General proof is: 1. Let suppose «a» is any element of A 2. If «a» is in A, thus «a» is a subset of P(A)… The problem is in the formal and pedantic way to right it, can I just write, that element «a» is a subset of P(A) or should I define additional set, which will be a subset of P(A) and element «a» will be a part of this new set. Please, help me to write such prove in a formal way with the correct syntax. Thanks. AI: You need to distinguzuish better between elements and sets, that is between $\in $and $\subseteq$. So assume $P(A)\subseteq P(B)$. Let $a\in A$. Then $\{a\}\subseteq A$, hence $\{a\}\in P(A)$, hence $\{a\}\in P(B)$, hence $\{a\}\subseteq B$, hence $a\in B$.
H: Compute $ I(n) = \int^{+\infty}_{-\infty} \frac{e^{n x}}{1+e^x}\, dx$ I need compute this definite integral for values 0 < n < 1. I am not sure how to begin at all. I was able to compute the indefinite integral for the special case of n = 1/2, but I am unable to use the same strategies of substitution. $I(n) = \int^{+\infty}_{-\infty} \dfrac{e^{nx}}{1+e^x}\, dx$ AI: $$ \begin{align} \int_{-\infty}^\infty\frac{e^{nx}}{1+e^x}\mathrm{d}x &=\int_{-\infty}^\infty\frac{e^{(n-1)x}}{1+e^x}\mathrm{d}e^x\\ &=\int_0^\infty\frac{t^{n-1}}{1+t}\mathrm{d}t\tag{1}\\[5pt] &=\mathrm{B}(n,1-n)\tag{2}\\[10pt] &=\Gamma(n)\Gamma(1-n)\tag{3}\\[5pt] &=\frac{\pi}{\sin(\pi n)}\tag{4} \end{align} $$ $(1)$: $t=e^x$ $(2)$: Beta Function identity $(3)$: Beta Function identity using $\Gamma(n+(1-n))=\Gamma(1)=1$ $(4)$: Gamma function identity The identities in both $(2)$ and $(3)$ are justified in this answer. The identity in $(4)$ is justified in this answer. The special case of $n=\frac12$ thus yields a value of $\pi$. If that is all you are really interested in, we can simplify the preceding argument as follows: $$ \begin{align} \int_{-\infty}^\infty\frac{e^{x/2}}{1+e^x}\mathrm{d}x &=\int_{-\infty}^\infty\frac{e^{-x/2}}{1+e^x}\mathrm{d}e^x\\ &=\int_0^\infty\frac{1/t}{1+t^2}\mathrm{d}t^2\tag{5}\\ &=2\int_0^\infty\frac1{1+t^2}\mathrm{d}t\tag{6}\\[9pt] &=\pi\tag{7} \end{align} $$ $(5)$: $t^2=e^x$ $(7)$: substitute $t=\tan(\theta)$
H: $x^{311} \equiv 662 \mod{713}$ We have to find $x$ such that $x^{311} \equiv 662 \mod{713}$ The example given in my notes has a few typos and the professor is unavailable. AI: $$713=23 \cdot 31 $$ By Chinese Remainder Theorem, you need to solve $$x^{311} \equiv 662 \pmod{23} \\ x^{311} \equiv 662 \pmod{31}$$ or equivalently $$x^{311} \equiv 18 \pmod{23} \\ x^{311} \equiv 11 \pmod{31}$$ It follows that $x$ is invertible in both arithmetics, and then by Fermat Little Theorem we have $$x^3\equiv 18 \pmod{23} \Rightarrow x^{-1} \equiv x^{21} \equiv 18^7 \pmod{23}\Rightarrow x \equiv (18^7)^{-1} \pmod{23}$$ and $$x^{11} \equiv 11 \pmod{31} \Rightarrow x^{3} \equiv x^{33} \equiv 11^{3} \pmod{31}\,.$$ Now use the CRT to find $x \pmod{713}$. Second solution As 662 is invertible, it follows that $x$ is invertible and hence, by Euler Theorem we have $$x^{660} \equiv 1 \pmod {713} \,.$$ now, find $a,b$ so that $$a\cdot 311 +b \cdot 660=1$$ Then $$x= x^{1}=x^{a\cdot 311 +b \cdot 660}=(x^{311})^a\cdot(x^{660})^b=662^a \pmod{713} \,.$$
H: Probability: Random Variables Let's $T_1$ be a random variable with pdf: $$f(t) = \frac{6+2t}{7}$$ and $T_2 \sim Exp(\frac{1}{3})$ Knowing that $T_1$ and $T_2$ are independent calculate $$P(T_1 + T_2 > 1) $$ During my classes we produced the following result: $$\int_0^1 \int_{1-t}^{\infty} \frac{6+2t}{7} \frac{1}{3} e^{-\frac{1}{3}x} dxdt $$ Could somebody explain me how it was obtained (espacially the limits of integragtion)? Thanks in advance AI: I suspect that the pdf of $T_1$ is supposed to be $$f(t_1)=\begin{cases}\cfrac{6+2t_1}7 & \text{for }0\le t\le1\\0 & \text{otherwise},\end{cases}$$ since without that requirement, it is not a probability distribution, and the limits of integration later make no sense. Since $T_1,T_2$ are independent, then their joint pdf is simply the product of their individual pdfs, namely: $$g(t_1,t_2)=\begin{cases}\cfrac{6+2t_1}7\cdot\cfrac13e^{-\frac13t_2} & \text{for }0\le t\le1,t_2>0\\0 & \text{otherwise}.\end{cases}$$ Now, observing that $T_1+T_2>1$ if and only if $T_2>1-T_1,$ we have $$P(T_1+T_2>1) = \int_{-\infty}^\infty\int_{1-t_1}^\infty g(t_1,t_2)\,dt_2\,dt_1.$$ Since $g(t_1,t_2)=0$ for $t_1<0$ and for $t_1>1,$ then $$P(T_1+T_2>1) = \int_0^1\int_{1-t_1}^\infty g(t_1,t_2)\,dt_2\,dt_1.$$ Since $1-T_1\ge 0$ when $T_1\le 1,$ then $T_2>0$ when $T_2>1-T_1$ and $T\le 1,$ and so $$P(T_1+T_2>1) = \int_0^1\int_{1-t_1}^\infty \cfrac{6+2t_1}7\cdot\cfrac13e^{-\frac13t_2}\,dt_2\,dt_1.$$
H: Preservation of moments under convergence in distribution Let $X_n$ be a sequence of random variables with first moment uniformly bounded. If this sequence of random variables converges to $X$ in distribution, then we have the inequality $ \lim E (\|X_n\| ) \geq E (\|X\|). $ I am aware of examples where the inequality is strict; intuitively, the "mass" can escape to infinity as $n$ increases. But I heard that with the additional assumption $E(\|X_n\|^{1+\delta}) <C $ for some positive constants $\delta, C$, then we have $ \lim E (\|X_n\| ) = E (\|X\|). $ How can we prove this? AI: The inequality should read $\mathbb E\|X\|\leqslant \liminf_{n\to \infty}\mathbb E\|X_n\|$. The limit in the RHS may not exist (take $X,X_1,X,X_2,\dots,X_n,X,X_{n+1},\dots$ where $(X_j)$ is such that the inequality is strict). We can show the following: If there is $\delta$ such that $\sup_n\mathbb E\|X_n\|^{1+\delta}$ is finite, then $\{X_n,n\geqslant 1\}$ is uniformly integrable. If $\{X_n,n\geqslant 1\}$ is uniformly integrable and $X_n\to X$ in distribution, then $\mathbb E(X_n)\to\mathbb EX$. For the second fact, consider $f_R(x):=\|x\|\chi_{(-R,R)}(x)$. Fix $\varepsilon\gt 0$ and $R$ such that $\mathbb P\{X\in\{-R,R\}\}=0$ and $\mathbb E[\|X\|\chi_{\|X\|\gt R}]+\sup_n\mathbb E[\|X_n\|\chi_{\|X_n\|\gt R}]\lt \varepsilon$. Then $\mathbb E[f_R(X_n)]\to \mathbb E[f_R(X)]$. We also have $$\|\mathbb E\|X_n\|-\mathbb E\|X\|\|\leqslant \|\mathbb E[f_R(X_n)]-\mathbb E[f_R(X)]\|+2\varepsilon.$$
H: Radius and Interval of Convergence of $\sum (-1)^n \frac{ (x+2)^n }{n2^n}$ I have found the radius of convergence $R=2$ and the interval of convergence $I =[-2,2)$ for the following infinite series: $\sum_{n=1}^\infty (-1)^n \frac{ (x+2)^n }{n2^n}$ Approach: let $a_n = (-1)^n \frac{ (x+2)^n }{n2^n}$ Take the ratio test $\lim_{x\to -\infty} \left\| \frac{a_{n+1}}{a_n} \right\| = \left\| -\frac{1}{2}(x+2)\right\|$ So it converges for $-2 < (x+2) < 2, R = 2$ Test end points if $x = -2$ $\sum_{n=1}^\infty (-1)^n(0) =0 < \infty$ , converges if $x = 2$ $\sum_{n=1}^\infty (-1)^n \frac{4^n}{n2^n} = \sum_{n=1}^\infty (-1)^n \frac{2^{2n}}{n2^n} = \sum_{n=1}^\infty (-1)^n \frac{2^n}{n}$ let $b_n = \frac{2^n}{n}$ $b_{n+1} > b_n$ for every value of $n$ By the Alternating Series Test, $\sum_{n=1}^\infty (-1)^n b_n$ diverges Hence, $R = 2$ and $I = [-2, 2)$ Is my approach correct? Are there any flaws, or better ways to solve? AI: The radius of convergence is right. The interval of convergence started out OK, you wrote that for sure we have convergence if $-2\lt x+2\lt 2$. But this should become $-4\lt x\lt 0$. The endpoint testing was inevitably wrong, since the incorrect endpoints were being tested. We have convergence at $x=0$ (alternating series) and divergence at $x=-4$ (harmonic series).
H: Where is the tensor product of two unit vectors projection onto? I know that $\bar{e} \otimes \bar{e}$ is a projection onto $\bar{e}$. Then, I start to think where is then $\bar{e}_{i} \otimes \bar{e}_{j}$ projection onto. Where is the expression $\bar{e}_{i} \otimes \bar{e}_{j}$ projection onto? It is likely a third vector like $\bar{e}_{k}$ that is orthogonal to the two unit vectors, like in vector calculus for cross-product. But then again, this is not possible if $\bar{e}$ is in line with the vectors in $\bar{e} \otimes \bar{e}$. AI: If what you mean by "tensor product" is the outer product (i.e., for vectors $a$ and $b$, the product $a\otimes b$ is the matrix $ab^{T}$, with elements $a_i b_j$), then you can write the following in general: $$ (a\otimes b)\cdot c = \sum_{i, j}\left(a_ib_jc_j\right)e_i. $$ If $a$ and $b$ are unit vectors, then you have $$ (e_a\otimes e_b)\cdot c = \sum_{i, j}\left((e_a)_i (e_b)_jc_j\right)e_i=\sum_{i,j}\delta_{ai}\delta_{bj}c_je_i=(c \cdot e_b)e_a. $$ That is, the product projects $c$ onto the $b$-axis, then rotates the result onto the $a$-axis. In general, $\left(a\otimes b\right)$ isn't a projection, but a projection followed by a rotation and dilation.
H: Showing S is an equivalence relation in X when we know R is a reflexive and transitive relation in X. I have this question and can't quite grasp it..I'll write down what it says then go through what I've tried. Let $R$ be a reflexive and transitive relation in $X$. Let $S$ be a relation in $X$ such that $(x,y) \in S \iff (x,y) \in R \land (y,x) \in R$. We need to prove $S$ is an equivalence relation, i.e it is reflexive, symmetric, and transitive...I know the definition of all of these (Reflexive is $\forall x(x,x) \in R$ , symmetric is $(x,y) \in R \implies (y,x) \in R$ and transitive is if $(x,y) \in R \land (y,z) \in R \implies (x,z) \in R$ Okay...right now I'm trying to prove S is reflexive but I can't think of any ordered pair that would give me $(x,y) \in R \land (y, x) \in R$ they would have to be the same variable but R isn't antisymmetric so I'm not sure I can say that. I think I might need some pair of ordered pairs to make this work but once again not sure what.. I'm pretty new to this stuff so please be kind =x I don't want hand holding but I'm just not sure how to approach this problem. I know my definitions but I'm not sure how they'll apply..thanks for the time and help. AI: You’re outthinking yourself: antisymmetry has nothing to do with it. $R$ is reflexive, so if $x\in X$, then $\langle x,x\rangle\in R$. The reversed pair is identical — reversing $\langle x,x\rangle$ gives you $\langle x,x\rangle$ — so it’s in $R$ as well, and therefore by definition $\langle x,x\rangle\in S$. This shows that $S$ is reflexive. It’s clear from the definition that $S$ is symmetric: that’s built into it. Thus, all that’s left is to show that $S$ is transitive. Suppose, then, that $\langle x,y\rangle\in S$ and $\langle y,z\rangle\in S$; you need to show that $\langle x,z\rangle\in S$, which means that you need to show that $\langle x,z\rangle\in R$ and $\langle z,x\rangle\in R$. See if you can do that on your own, but feel free to leave a comment if you get stuck.
H: Contrapositive clarification Let's say I have this statement: ∀ real numbers x, if −x is not irrational, then x is not irrational. Which one of the following statements is equivalent to this? [because −(−x) = x], 1.∀ real numbers x, if x is rational, then −x is rational. 2.∀ real numbers x, if -x is rational, then x is rational. I thought it was the second one..but apparently it's the first one. If it's the second, can someone explain why? thank you. AI: It's because "Not irrational" means rational. And the "not" sign is a negation of the entire clause. So -(x is not irrational) $\implies$ x is not irrational $\equiv$ -(x is rational) $\implies$ (x is rational).
H: Values of $\frac{e^{-x}}{1+x}$ For $$f(x)={\frac{e^{-x}}{1+x}}$$ where $ x\not=-1$ decide which values $f(x)$ could take. Should I take the limit as $x$ → -1 in each direction? And then as x → ± $\infty$? If so, I'm not sure of how that would be done. AI: You want to find the range of the function $f$: $$ f(x) = \frac{e^{-x}}{1+x}. $$ Note that $e^{-x}$ is always positive. And that $$ \lim_{x\to -1^-} f(x) = -\infty $$ $$ \lim_{x\to -1^+} f(x) = \infty. $$ Note that $$ f'(x) = \frac{-e^{-x}(1+x) - e^{-x}}{(1+x)^2} = \frac{-2e^{-x} - xe^{-x}}{(1+x)^2} = \frac{-e^{-x}(2+ x)}{(1+x)^2}. $$ so $f$ is increasing on the interval $(-\infty, -2]$ and decreasing on $[-2, -1) $ and $(-1,\infty)$. So we have a local maximum at $x=-2$. Now also note that $\lim_{x \to \infty} f(x) = 0$ so $f$ attains all values in the interval $(0,\infty)$. Now $f$ is only negative when $x < -1$ and has a local maximum at $x=-2$, so the interval $(-\infty, f(-2)]$ is in the range. In all you get the range is $(-\infty, f(-2)] \cup (0,\infty) = \mathbb{R} \setminus (f(-2), 0]$.
H: Lagrange Multipliers where no solutions(s) satisfy the constraints $f(x) = x-y$ subject to constraint $x^2-y^2=2$ Using the method of Lagrange Multipliers, we get: $(1, -1) = \lambda(2x, -2y)$ which gives $x=y$ but this does not satisfy $x^2-y^2=2$ Is this because the Lagrange method is not applicable here? Or is it valid for the Lagrange method to not produce any solutions? AI: Lagrange multipliers work in the following situation. Suppose $g(x,y)=0$ defines a bounded level curve. Suppose also that $\nabla g \neq 0$, that is the gradient is non-zero, on this level set. Then the max an min of $f$ on $g=0$ occurs at a point that satisfies $\nabla f = \lambda \nabla g$ and $g=0$. What if you just assume that $g(x,y)=0$ and allow for $\nabla g=0$? Then the min and max of a $f$ will occur at a point where $\nabla g =0 $ or where $\nabla f = \lambda \nabla g$ in addition to $g=0$. However, your problem is that your constraint does not define a bounded set. Think about it, if you have an unbounded set and you are trying to maximize the distance to the origin, is this possible? No. This is the sort of situation you have.
H: What is the difference between `convergence radius` and `convergencee interval`? I have a power series $ \sum^\infty_0 = a_nx^n $ , and I have to find the convergence interval and convergence radius. The convergence radius is $\lim_{n \to \infty} \frac{1}{\sqrt[n]a_n} $, but what is the convergence interval? AI: The convergence interval is the interval upon which the power series converges. The radius of convergence (convergence radius) is the radius of this interval. So for example, the series $$\sum_{n = 0}^{\infty} x^n$$ converges iff $-1 < x < 1$, so the interval is $(-1, 1)$ and the radius is $1$.
H: Relation between blowing up at a point and at a variety maybe this is an idiot question, but I have to ask it. Usually, in the classical background, one defines the blow up $Bl_Y(X)$ at a variety $Y$ in as the closure of the graph of the function $f: X/Z(f_1, f_2, ...f_m) \longrightarrow \mathbb{P}^{m-1}$ where $f$ evaluate the polynomials. However, when one blow up at a point, the definition seens kinda different, it's the subvariety of $X \times \mathbb{P}^{n - 1}$ determined by $x_i t_j = x_j t_i$ where $t_i$ is the homogeneous coordinate. The definitions seems to not match each other. Thanks in advance. AI: The two definitions match in the following sense: Let $X\subset \mathbb{A}^n$ be an affine variety. Let $f_0,\cdots,f_r\in k[x_1,\cdots,x_n]$ where they don't vanish identically on $X.$ Then $U=X \setminus Z(f_0,\cdots,f_r)$ is a non-empty open subset of $X.$ Define the well-defined morphism $$f:U \to \mathbb{P}^r \; \text{given by}\; p \mapsto [f_0(p):\cdots,f_r(p)]$$ Now the closure of the graph of $f$ inside $X \times \mathbb{P}^r$ is the blow up of $X$ in $(f_0,\cdots,f_r).$ In particular, let $X=\mathbb{A}^2$ with coordinates $x,y$ and take $f_0=x,f_1=y.$ The blow up of $\mathbb{A}^2$ in $(x,y)$ or at zero is then the following. The morphism $$f:U \to \mathbb{P}^1 \; \text{given by} \; (x,y) \mapsto [x:y]$$ is well defined, where $U=\mathbb{A}^2 \setminus \{(0,0)\}.$ The graph of $f$ on $U$ is $$\{((x,y),[s:t])\|xt=sy \} \subset U \times \mathbb{P}^1$$ and its closure is simply $\{((x,y),[s:t])\|xt=sy\} \subset \mathbb{A}^2 \times \mathbb{P}^1.$ The reason for the existence of the relation $xt=sy$ is to make sure that $[s:t]$ is in the image of $f.$
H: If E is a Hilbert space and $T \in B(E)$ is compact, show $T(E)$ does not contain a closed infinite dimensional subspace It's the problem from "Essential Results of Functional Analysis," R.J. Zimmer, Chapter 3, problem 3.1. I try to prove this problem and I am confused with the condition "closed infinite dimensional." Consider a compact operator T, given by diagonal matrix $T_{ij} = \lambda_i\delta_{ij}$, with $\lambda_i \to 0$. Then for Hilbert space $E$, we have $T(E) = E$. That means $T(E)$ has a closed infinite dimensional subspace($E$ itself). Where did I go wrong? AI: Hint: For each closed subspace $F \subset E$, there is an orthogonal projection $P_F$ onto $F$. The composition $P_F \circ T$ is also compact. So if $T(E)$ contained a closed infinite-dimensional subspace $F$, we'd have a compact map onto an infinite-dimensional Hilbert space. Why can that not happen?
H: If $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous, 1-1, and on-to, then f maps Borel Sets to Borel Sets Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be continuous, on-to, and 1-1. Prove that if $A$ is a Borel set, then $f(A)$ is a Borel set. AI: Let's not worry about whether $f$ is a homeomorphism, it ends up not playing a role. First, $f(C)$ is Borel when $C$ is closed. This is clear if $C$ is compact, because continuous maps send compact sets to compact (and therefore closed) sets. If $C$ is unbounded, we cannot quite say that $f(C)$ is closed by the same argument, but note that $C=\bigcup_n(C\cap[-n,n])$, so $f(C)=\bigcup_n f(C\cap[-n,n])$ is a countable union of closed sets. It follows that $f(O)$ is Borel is $O$ is open, because $f(O)=f(\mathbb R)\setminus f(\mathbb R\setminus O)$ is a difference of Borel sets (since both $\mathbb R$ and $\mathbb R\setminus O$ are closed). Note that we do not even need to use the assumption that $f$ is onto, but we are definitely using that $f$ is injective. Let $\mathcal F$ be the collection of Borel sets $A$ such that $f(A)$ is also Borel. We have shown that $\mathcal F$ contains the open sets. Note now that $\mathcal F$ is closed under complements. This is because if $f(A)$ is Borel, then so is $f(\mathbb R\setminus A)=f(\mathbb R)\setminus f(A)$ (again, the equality holds because $f$ is injective). Finally, $\mathcal F$ is closed under countable unions, because if $f(A_n)$ is Borel for each $n$, then so is $f(\bigcup_n A_n)=\bigcup_n f(A_n)$. This means that $\mathcal F$ is the $\sigma$-algebra of all Borel sets of reals, that is, $f(A)$ is Borel for each Borel set $A$. Hmm... It turns out the question is a duplicate so it may end up being closed. Let me add some remarks just in case: If you want to argue that $f$ is a homeomorphism, which would certainly simplify things, note that since $f$ is injective, it must be monotone, and therefore (by continuity) it maps open intervals to open intervals. It follows that $f^{-1}$ is continuous. This simplifies the argument, because we get not just that $f(A)$ is Borel if $A$ is Borel, but in fact we see that $f(A)$ has the same Borel complexity as $A$: If $A$ is open, so is $f(A)$, if $B$ is $F_\sigma$, so is $f(B)$, etc. Trevor hints in a comment that the above is anyway not the most direct approach. As you can see in the answers to the other question (the one this duplicates), descriptive set theory provides a very elegant alternative approach, via a theorem of Suslin, that if both a set $A$ and its complement are continuous images of a Borel set, then in fact $A$ is Borel.
H: Use single variable calculus methods to find the area of the region Use single variable calculus methods to find the area of the region in the first quadrant bounded by the curves $y^2=3x$, $y^2=4x$, $x^2=3y$, $x^2=4y$ Could someone please show me how to go about this? AI: First, express all the functions as functions of either $x$ or of $y$. Since we are interested in the first quadrant only, we can take the positive roots where necessary. Since you're likely more comfortable using formulas expressed as functions of $x$, express each formula as $y = f(x)$. $y^2=3x\implies y = \sqrt{3x}$, $y^2=4x \implies y = \sqrt{4x} = 2 \sqrt x$, $x^2=3y\implies y = \frac 13 x^2$, $x^2=4y \implies y = \frac 14 x^2$. Next, graph the functions, together, in one graph! GRAPH ADDED: My guess is that you are being asked to find the area of the "parallelogram-like" figure in shown in the upper-left portion of the graph below, each side belonging to one and only one of the curves. Find the exact points of intersection to help determine bounds of integration (you will likely need to sum different integrals to find the total area, as the upper and lower curves will change). You'll need to find the $x$-values for each of the four points of intersection, so the area will be determined by the sum of four integrals with bounds from $x_0$ to $x_1$, $x_1$ to $x_2$, $x_2$ to $x_3$, and $x_3$ to $x_4$, respectively. You'll also need to determine which functions to use as your upper and lower curve on each interval over which you are integrating.
H: Will somebody look over my real-analysis/calculus solution of $\lim\limits_{x\to 0}\frac{1}{x} \int_0^{x} \sqrt{9+t^2}\mathrm{d}t$ Evaluate $\lim_{x\to 0}\frac{1}{x} \int_0^{x} \sqrt{9+t^2}\mathrm{d}t$. Immediately, taking, $$\begin{align} &\lim_{x\to 0}\frac{1}{x} \int_0^{x} \sqrt{9+t^2}\mathrm{d}t \\ =&\frac{\lim_{x\to 0} \int_0^{x} \sqrt{9+t^2}\mathrm{d}t}{ \lim_{x\to 0}\frac{1}{x}} \end{align}$$ which yields the indeterminate form $\frac{0}{0}$. Recall the Fundamental Theorem of Calculus (1): If $f$ is Riemann integrable on $[a,b]$ for $a\leq x\leq b$, put $F(x)=\int_0^{x}f(t)\mathrm{d}t$. Then $F$ is continuous on $[a,b]$: furthermore, if $f$ is continuous at a point $x_0$ of $[a,b]$, then $F$ is differentiable at $x_0$, and $F'(x)=f(x)$. From this point, using L'Hospital's Rule yields $$\begin{align} &\lim_{x\to 0} \frac{1}{\frac{d}{dx}x} \cdot \lim_{x\to 0}\frac{d}{dx}\int_0^{x}\sqrt{9+x^2}\mathrm{d}t\\ \rightarrow& 1 \cdot \lim_{x\to 0}F'(x)\\ \rightarrow& \lim_{x\to 0} \sqrt{9+x^2}\\ =& 3. \end{align}$$\ I feel like I have assumed something that I shouldn't have. Could somebody look this over for me? AI: Your answer is correct but you do not need to use L'Hospital's rule. Let $F(x)=\int_{0}^x \sqrt{t^2+9}dt$. Then you have: $$\lim_{x\rightarrow 0} \dfrac{F(x)-F(0)}{x-0}=F^\prime(0)$$ by definition of the derivative and the FTC gives you that $F^\prime(t)=\sqrt{t^2+9}$.
H: Is this ODE solvable? this equation popped up when I was trying to apply math: $$Af^2+B\left(\frac{df}{dx}\right)^2=C$$ Where $A,B,C$ are positive constants. Also, $~f(0)=D$, another positive constant. What are the solutions? Is this solvable? If not, are there any texts you recommend to read? BTW you might want to know background: I am solving for the maximum velocity profile $v(s)$ for a car traveling on a curved path $\gamma(s)$ under the constraint that the max passenger g force, call it $\sqrt{C}$, is not exceeded. This results in $$\sqrt{g_l^2+g_r^2}=\sqrt{C}~~\to~~\left(\frac{v^2}{r}\right)^2+\left(\frac{dv}{dt}\right)^2=C$$ Where $r$ is the radius of curvature at the point $\gamma(s)$. I then made the substitution $f=v^2$ to get to an equation analogous to the one above. Also, I took $r$ to be piecewise (which is a good approximation in my usage case), which is why $A$ is constant. AI: Maple (software) gives these 4 solutions: f(x) = sqrt(AC)/A, f(x) = -sqrt(AC)/A, f(x) = sqrt(A(tan(sqrt(BA)(_C1-x)/B)^2+1)C)tan(sqrt(BA)(_C1-x)/B)/(A(tan(sqrt(BA)(_C1-x)/B)^2+1)), f(x) = -sqrt(A(tan(sqrt(BA)(_C1-x)/B)^2+1)C)tan(sqrt(BA)(_C1-x)/B)/(A(tan(sqrt(BA)(_C1-x)/B)^2+1)) It is solving by straightforward quadrature (integration) without imposing any initial condition. _C1 is an arbitrary constant. If you impose the initial condition $f(0)=D$ then you get only two implicit solutions: $$x-B\cdot \arctan(\sqrt{B\cdot A}\cdot f(x)/\sqrt{-A\cdot B\cdot f(x)^2+B\cdot C})/\sqrt{B\cdot A}+B\cdot \arctan(\sqrt{B\cdot A}\cdot D/\sqrt{-A\cdot B\cdot D^2+B\cdot C})/\sqrt{B\cdot A} = 0$$ and $$x+B\cdot \arctan(\sqrt{B\cdot A}\cdot f(x)/\sqrt{-A\cdot B\cdot f(x)^2+B\cdot C})/\sqrt{B\cdot A}-B\cdot \arctan(\sqrt{B\cdot A}\cdot D/\sqrt{-A\cdot B\cdot D^2+B\cdot C})/\sqrt{B\cdot A} = 0$$
H: Does the Gauss' Trick Really Belong to Gauss? We all have heard the story of the young Guass, summing 1 to 100 by writing the sum backward below the original one. In this article, just two books are referred for the trick. I looked at both of them but the story was just mentioned briefly without any firsthand reference. Is this story true and reliable? Is there any good reference proving this? Thanks. AI: Brian Hayes, who does the "Computing Science" column for American Scientist, wrote about the Gauss legend in one of his columns in 2006. Brian was particularly interested in variations on the story. A pdf reprint of the column is available here, and additional material, which he's gathered at his blog, is available here and here. It's worth noting that there is no universal agreement as to exactly what problem Gauss is said to have solved, nor the precise means by which he solved it.
H: Closed ball is not compact Show that the closed ball in $C([0,1])$ of center $0$ and radius $1$ is not compact. I thought it will be compact since every closed and bounded set in $\mathbb{R}$ is compact? Why is it not compact and how can I prove it? AI: Let $f_n$ be zero except on $[\frac{1}{2}(\frac{1}{n+1}+\frac{1}{n}), \frac{1}{2}(\frac{1}{n}+\frac{1}{n-1})]$, where the graph is described by joining the points $(\frac{1}{2}(\frac{1}{n+1}+\frac{1}{n}),0), (\frac{1}{n},1), (\frac{1}{2}(\frac{1}{n}+\frac{1}{n-1}),0)$. Then $\operatorname{supp} f_n = [\frac{1}{2}(\frac{1}{n+1}+\frac{1}{n}), \frac{1}{2}(\frac{1}{n}+\frac{1}{n-1})]$, and $f_n(\frac{1}{n}) = 1$. Hence $\\|f_n-f_m\\| = \delta_{mn}$. The collection $\{ B(x, \frac{1}{2}) \}_x$ is an open cover of $C[0,1]$. If we take any finite sub-collection, then at most one of the $f_k$ can be contained in each one, so the finite sub-collection can contain only a finite number of $f_n$. It follows that $C[0,1]$ is not compact.
H: Notation in group theory? I have three questions on notation. What does the squiggly line mean in the following: $Aut(G) \cong Aut(G) \wr \mathbb{Z}_2$ What does $\rtimes$ mean in the following: $ \varphi :(Aut(G)\times Aut(G))\rtimes\mathbb{Z}_2\rightarrow Aut(G)$ What does $\succeq$ mean in the following: For some monotonically increasing function $f$, $f\succeq f^m, m>1$ AI: These are all in the paper you mentioned. In order: $Aut(G) \cong Aut(G) \wr \mathbb{Z}_2$, and in particular the $\wr$ part, is defined at the top of page 6 as the wreath product. This is a common notation, though - so that's ok. $\rtimes$ is the semidirect product. (To be fair, this is mentioned on page 6, but it's a bit unclear that their mentioning of this relates to this symbol in particular). This is also a common notation. $\succeq$ is often used to denote some sort of general ordering (as opposed to $\geq$), especially in partial orders. In this paper, it's defined after exercise 1.2 on page 2. For two functions $f, g: \mathbb{N} \to \mathbb{N}$, they say that $f \succeq g$ if $g(n) \leq Cg(\alpha n)$ for all $n$ and some particular $C,\alpha$. This is not at all a common notation.
H: Reducibility of Equivalent Polynomials This feels like a trivial question but somehow I couldn't come up with an immediate solution. Given polynomial $P$ in a multivariate polynomial ring over some base field $F$, if $P$ is irreducible over $F$, does it follow that any equivalent polynomial is also irreducible over $F$? If not, is there an explicit counterexample? By equivalent we mean taking the same value as $P$ on all possible variable assignments. Just realized this probably won't work for finite field (e.g. $(x^2 + 1)^2 \equiv 1$ in $F_3$), but what about characteristic 0 fields? AI: If two polynomials are equivalent, then their difference is identically zero. There is no such univariate polynomial over an infinite field, except the zero polynomial. By induction on the number of variables, I expect you can prove a similar result in general.
H: Combinatorics: complete set of solution to the congruence $$111x \equiv 112 \bmod 113$$ I've tried all the theorems. The only thing I found is that there is only one solution. Other than trying all 1-113 possible values that x takes, is there any efficient way to do it? AI: Use negative numbers. We have $111\equiv -2\pmod{113}$ and $112\equiv -1\pmod{113}$. Thus we want $-2x\equiv -1\pmod{113}$, that is, $2x\equiv 1\equiv 114\pmod{113}$. That gives $x\equiv 57\pmod{113}$.
H: How do you prove translation invariance of Fourier transform? Let $f$ be a rapidly decreasing function in the sense that it lies in the Schwartz space $\mathcal{S}(\Bbb{R})$. Then $\widehat{f(x+h)} = \hat{f}(\omega) e^{i 2 \pi h \omega}$, where $\hat{f}(\omega)$ is the Fourier transform of $f(x)$. How do I prove that? This might not depend on the rapidly decreasing part. AI: Let $g(x) = f(x+h)$. Then we calculate $$\hat{g}(x) = \int g(t)e^{-2 \pi i x t} \mathrm{d}t = \int f(t + h) e^{-2 \pi i x t} \mathrm{d}t $$ Then make the $u$-sub $s = t + h$ and you get $$ \hat{g}(x) = \int f(s) e^{-2\pi i(s - h)x} \mathrm{d}s = e^{2\pi i h x}\int f(s) e^{-2\pi i s x} \mathrm{d}s = e^{2 \pi i h x} \hat{f}(x) $$
H: Showing that $\|\phi(\mathcal{N})\| = \kappa$ s.t. $\mathcal{M} \equiv \mathcal{N}$ with $\|\mathcal{N}\| = \kappa$ Problem: Suppose $\mathcal{M}$ is an $L$-structure and $\phi \in L_n$ ($n > 0$) is such that $\phi(\mathcal{M})$ is infinite. Then show that for every cardinal $\kappa$ with $\kappa \ge \|L\|$ there is an $L$-structure $\mathcal{N} \equiv \mathcal{M}$ of power $\kappa$ and such that $\|\phi(\mathcal{N})\| = \kappa$. Attempt: Let $\kappa \ge \|L\|$. By Lowenheim-Skolem (upwards or downwards, depending on the size of $\|\mathcal{M}\|$), we know that there exists a model $\mathcal{N}$ s.t. $\|\mathcal{N}\| = \kappa$ and whereby $\mathcal{M} \equiv \mathcal{N}$. It remains to be shown that $\|\phi(\mathcal{N})\| = \kappa$. Now we can note that $\|\phi(\mathcal{N})\| \not < \omega$, since otherwise $\mathcal{N} \not \equiv \mathcal{M}$. This is because the statement "$\phi(\mathcal{M})$ is not finite" is first-order, and hence since it's true in $\mathcal{M}$ it must also be true in $\mathcal{N}$ in order for $\mathcal{M} \equiv \mathcal{N}$. Hence $\|\phi(\mathcal{N})\|$ is infinite in $\mathcal{N}$. But of course, $\|\phi(\mathcal{N})\|$ could still be greater or less than the size of $\kappa$. So suppose that $\|\phi(\mathcal{N})\| > \kappa$. This leads to a contradiction though, for the size of $\|\mathcal{N}\| = \kappa$ implies that any set of $n$-tuples from $N^n$ must also be at most size $\kappa$. Hence since $\phi(\mathcal{N}) \subseteq N^n$ we have that $\|\phi(\mathcal{N})\| \le \kappa$. Question: Finally, suppose that $\omega \le \|\phi(\mathcal{N})\| < \kappa$. Then how would I either (i) show that this is impossible, or (ii) show that we can adjust $\mathcal{N}$ in such a way to make it impossible? AI: Hint: I do not believe we can use Löwenheim-Skolem directly, we need to use the proof of Löwenheim-Skolem. We sketch the upward part. Add to the language $\kappa$ constant symbols $c_\alpha$, axioms that say they are different, and axioms that say that $\phi(c_{\alpha})$ holds for all $\alpha\lt \kappa$. In the usual way, we show this has a model $\mathcal{N}$ of cardinality $\kappa$. It is automatic that $\|\phi(\mathcal{N})\|=\kappa$.
H: Bound of power series coefficients of a growth-order-one entire function An entire function $f(z)$ satisfies $$\|f(z)\| \leq A_\varepsilon e^{2\pi(M+\varepsilon)\|z\|}$$ for every positive $\varepsilon$. I want to show that $$\limsup_{n \to \infty}\ [f^{(n)}(0)]^{1/n} \leq 2\pi M.$$ Alternatively, we can state the result as $$\limsup_{n \to \infty}\ (n! \|a_n\|)^{1/n} \leq 2\pi M,$$ where $f(z) = \sum_{n=0}^\infty a_n z^n$ is the power series expansion. I can prove this result with Paley-Wiener theorem when $f$ is of moderate decrease; I've no idea otherwise. Note that this is a pretty good bound, since equality holds when $f(z)=e^{2\pi Mz}$, just to give one example. P.S. Sorry for the bad question title, since I can't really come up with an illuminating summarization of this problem. Please feel free to edit if you think of something better. AI: We have $$\begin{align} \left\lvert f^{(n)}(0)\right\rvert &= \frac{n!}{2\pi} \left\lvert\int_{\lvert z\rvert = R} \frac{f(z)}{z^{n+1}}\,dz\right\rvert\\ &\leqslant \frac{n!}{2\pi} \int_0^{2\pi} \frac{\lvert f(Re^{i\varphi})\rvert}{R^n}\,d\varphi\\ &\leqslant \frac{n! A_\varepsilon e^{2\pi(M+\varepsilon)R}}{R^n} \end{align}$$ for all $R > 0$. The right hand side is minimised for $R = \dfrac{n}{2\pi(M+\varepsilon)}$, which gives the estimate $$\left\lvert f^{(n)}(0)\right\rvert \leqslant A_\varepsilon \frac{n!e^n\bigl(2\pi(M+\varepsilon)\bigr)^n}{n^n},$$ and taking $n$-th roots, $$\left\lvert f^{(n)}(0)\right\rvert^{1/n} \leqslant \sqrt[n]{A_\varepsilon}\sqrt[n]{n!}\frac{e}{n}2\pi(M+\varepsilon),$$ from which we deduce $$\limsup_{n\to\infty} \left\lvert f^{(n)}(0)\right\rvert^{1/n} \leqslant 2\pi(M+\varepsilon).$$ Since that holds for all $\varepsilon > 0$, we have indeed $$\limsup_{n\to\infty} \left\lvert f^{(n)}(0)\right\rvert^{1/n} \leqslant 2\pi M.$$
H: $E[X]=1.8$, where $X$ is the total number of successes of 3 trials. What is the largest/smallest $P\{X=3\}$ can be? Since each trial has the same probability of success, $p$, can you not uniquely solve for $p$? I.e: Let, $X_{i} = 1$ if the $i^{th}$ trial is a success ($0$ otherwise). Then, $X=\sum_{i=1}^{3}X_{i}$, and $E[X] = E[\sum_{i=1}^{3}X_{i}] = \sum_{i=1}^{3}E[X_{i}] = \sum_{i=1}^{3}p =3p =1.8$ So, $p=0.6$, and $P\{X=3\}=0.6^{3}$ I thought what I did was sound, but the textbook says the answer to (a) is $0.6$ and (b) is $0$. Their reasoning (for (a)) is as follows: However, how can the above be true if all three trials have the same probability of success? That is, how can $P\{X=1\}$ and $P\{X=2\}$ be zero, when $P\{X=3\}$ is nonzero? AI: Hint: Nothing in the problem said that the trials are independent.
H: Limits of integration for a joint PDF I have $f_{X,Y}(x,y) = \lambda^2e^{-\lambda y}$ for 0 < x < y. If I want to show that this is a joint PDF, I need to do a double integral and show that it is equal to 1. Do I set my integration limits up as 0,y and x,$\infty$ for x and y respectively? AI: There are two ways. One is: $$ \int_{0}^{+\infty} \int_{0}^{y} \lambda^2\exp(-\lambda\cdot y)\,dx\,dy$$ and the other is: $$ \int_{0}^{+\infty} \int_{x}^{+\infty} \lambda^2\exp(-\lambda\cdot y)\,dy\,dx.$$
H: Confusion in Burnside's proof that any $2$-generated group of exponent $3$ is finite? I'm reading a proof of Burnside's theorem that groups of exponent $3$ are finite, but have some problems. Let $G=\langle x,y:z^3=1\rangle$ be a group generated by $x$ an $y$ with exponent $3$. Let $a=yxy^{-1}$ and $b=y^{-1}xy$. Then $$ xab=xyxy^{-1}y^{-1}xy=xyxyxy=(xy)^3=1. $$ Similarly, $bax=1$, so $ab=ba$. Then $1=xab=abx=xba=bax$, so from $abx=xba$, we have $bx=xb$, and similarly $ax=xa$. Let $N=\langle x,a,b\rangle$, which is normal and abelian, hence finite. If $b=x$, then $x=y^{-1}xy$ implies $yx=xy$ so $G$ is abelian and we are done. If $b=x^2$, then $x=1$. And if $b=x^3$, $x=1$, and there is nothing to prove. So suppose $b$ is not a power of $x$. Then $x$ and $b$ are independent commuting elements in $N$ , and since $a=x^{-1}b^{-1}$, $N=\langle x,b\rangle$, so $\|N\|=3^2$. Then $G/N\simeq\langle y\rangle$... I have two issues: Why does $b=x^2$ imply $x=1$? This means $x^2=y^{-1}xy$, or $yx^2=xy$ but I have no reason to see $x=1$. Also, why is $G/N\simeq\langle y\rangle?$ I think it is clear $G/N$ is isomorphic to a subgroup of $\langle y\rangle$, but couldn't $G/N$ be trivial? AI: If you only want to prove G is finite, your confusion is not necessary. When you got N is finite normal subgroup of G, yo know that G/N is a group of order no greater than 3, and G is finite.
H: How to parametrize the curve of intersection of two surfaces in $\Bbb R^3$? I have to parametrize the curve of intersection of two surfaces. The surfaces are: $$y^2 = z \text{ and } x + y = 4$$ Could someone please show me how to do this step by step? Thanks. AI: Let $y=t$. Then $x=4-t$ and $z=t^2$ and so $\vec r(t)= (x(t), y(t), z(t))= (4-t, t, t^2)$, $t\in\mathbf R$.
H: Existence of certain set Problem: Let $0<a<1$. Let $\lambda$ be the Lebesgue measure on $\mathbb{R}$. Show that: (i) There exists a closed set $A\subseteq[0,1]$, which does not contain any non-empty open sets, such that $\lambda(A)=a$. (ii) There exists an open, dense set $B\subseteq[0,1]$ such that $\lambda(B)=a$. Ideas: (i): This means the interior of A has to be the empty set. I'm stuck here. (ii): If we have a set A like in (i) where $\lambda(A)=1-a$, then $B=[0,1]\setminus A$ is obviously open, and it is also dense, since the closure of $B$ would be $[0,1]$. Furthermore $\lambda(B)=\lambda([0,1]\setminus A)=\lambda([0,1])-\lambda(A)=1-(1-a)=a$. As always, I don't want a full solution, just hints to guide me. Thanks in advance! AI: Hint for (ii): Enumerate the rationals as $\{r_n\}$ and put little intervals of really quickly decreasing length around each rational. This gives density and openness. For (i), try modifying the construction of the Cantor set to remove less at each step. Search term for (i): Fat Cantor set
H: Let m be an odd integer. Since gcd(2,m) = 1,2 is invertible modulo m. What is the inverse of 2 modulo m? Let m be an odd integer. Since gcd(2,m) = 1,2 is invertible modulo m. What is the inverse of 2 modulo m? Justify your answer. I know that m being odd has a crucial part in this solution and what does the notation 1,2 is invertible modulo m mean? AI: If $m$ is odd, $m + 1$ is even, so we see that $$2 \cdot \frac{m + 1}{2} \equiv 1 \pmod{m}$$
H: Stabilizer subgroups - proof verification I have a problem that I would like help on. I'm preparing for an exam, and I have provided my work below. Problem statement: Let $G$ act on $X$, and suppose $x,y\in X$ are in the same orbit for this action. How are the stabilizer subgroups $G_x$ and $G_y$ related? My attempt: $G_x = G_y = \{g \in G: gx=x\} = \{g \in G: gy=y\} \forall x,y \in X, \forall g \in G$. Thanks. AI: We have $G_{gx}=g G_x g^{-1}$ for any $g\in G,x\in X$. The inclusion $G_{gx}\supseteq g G_x g^{-1}$ can be shown by applying some $ghg^{-1}$ to $gx$ for any $h\in G_x$ For the other direction show that $g^{-1}kg\cdot x=x$ for any $k\in G_{gx}$.
H: Differential Equations/IVP: $\frac{dy}{dt} = 4 - y^3$ and $y(-1)=2$ Transform the given initial value problem into an equivalent problem with the initial point at the origin. $$\cfrac {dy}{dt} = 4 - y^3 \\ y(-1)=2$$ I have no idea about how to solve it. Could you please help? AI: Your I.V.P. looks like $y'=f\circ y, y(t_0)=y_0$. This is called an autonomous differential equation. Read its properties. Let $\varphi$ be a solution to the I.V.P. defined on an interval $I$. Then $\forall t\in I\left(\varphi '(t)=f(\varphi(t))\right)$ and $\varphi(t_0)=y_0$. Let $J=\{t-t_0\colon t\in I\}$ and define $\psi \colon J\to \Bbb R, x\mapsto \varphi(x+t_0)$. Check that $\psi$ is a solution to the I.V.P..
H: Non-negative, real matrix $\Rightarrow$ non-negative, real eigenvalues? Does a matrix with all non-negative, real entries have all non-negative, real eigenvalues? Where might I find a proof of such? Ideas: Perhaps we can multiply a prospective eigenvector so its biggest entries are positive, and then show that it is a contradiction for it to have a negative eigenvalue? I am currently looking at the Perron-Frobenius theorem on Wikipedia, but it seems not to mention this issue. (I suspect my conjecture is not true.) AI: Hint: Not true. Think of a $2\times 2$ matrix with non-negative entries and negative determinant. It looks like $I_2$.
H: How do you generalize the Laplace transform to more variables? Just what the title says. How can I take the Laplace transform of $f(x,y,z)$ ? AI: For multivariable version of Laplacian transformation, assume $t=(t_1,t_2,t_3)$, $X=(x,y,z)$, then $\mathcal L\{f(x,y,z)\}=\int e^{t\cdot X}f(X)$.
H: Any two paths in $X = \mathbb{R}^n$ having same initial and end point are homotopic Suppose $X = \mathbb{R}^n$. Let $\gamma, \alpha : [0,1] \to X $ be to paths such that $\gamma(0) = \alpha(0) = x_0 , \; \; \gamma(1) = \alpha(1) = x_1$. We want to show $\gamma$ and $\alpha$ are homotopic. My try: Take $F(s,t) = f_t(s) = (1-t)\gamma(s) + t\alpha(s)$ Clearly, $F(0,t) = x_0$ and $F(1,t) = x_1$. Since $\alpha, \gamma, t, 1$ are continuous, then $F$ must be continuous function. Hence it is a homotopy. Therefore $\alpha$ and $\gamma$ are homotopic IS this enough to show they are homotopic? AI: Yes, it's perfect. Note that this proof would also work if $\mathbb{R}^n$ were replaced by any convex subset.
H: How to find a transformation matrix, given coordinates of two triangles in $R^2$ I am an undergraduate student, and today I was given two triangles, $T_1$ (green) and $T_2$ (blue) in $R^2$: I was then asked to find the transformation matrix transforming $T_1$ to $T_2$. What I understand from this is, that I need to find $F$ in the following matrix equation: $T_2 = F \cdot T_1$. where $T_1= \begin{bmatrix}2&6&8\\2&-2&6\end{bmatrix}$ $T_2= \begin{bmatrix}-2&-10&-14\\-2&-4&10\end{bmatrix}$ which are the coordinates of the corners of the triangles. The above matrix equation however, is inconsistent, so how can I find $F$? It has to be a linear mapping, not an affine one. Have tried a lot, any help greatly appreciated! AI: The vertices of $T_1$ are $(2,2)$, $(6,-2)$, and $(8,6)$. The vertices of $T_2$ are $(-2,-2)$, $(-10,-4)$, and $(-14,10)$. We want a transformation mapping $T_1$ to $T_2$. So the vertices of $T_1$ must map to the vertices of $T_2$. Let's factor out the 2 to save us some trouble. We seek a linear transformation $L:\mathbb{R}^2 \to \mathbb{R}^2$ with $$\{(1,1), (3,-1), (4,3)\} \mapsto \{(-1,-1), (-5,-2), (-7,5)\} $$ The map $L$ is determined by its action on the two linearly independent vectors $(1,1)$ and $(4,3)$. It must map them to two of the vectors in $\{(-1,-1), (-5,-2), (-7,5)\}$. Using this we can easily calculate a matrix. For example, the matrix mapping $(1,1) \mapsto (-1,-1)$ and $(4,3) \mapsto (-5,-2)$ is $$ \begin{pmatrix} -2 & 1 \\ 1 & -2 \end{pmatrix}. $$ This matrix also happens to map $(3,-1)$ to the remaining vector $(-7,5)$ and so we are done. We got lucky this time; if this hadn't mapped to the right vector we could have kept choosing different pairs of vectors until we found the correct map.
H: Is it possible that all subseries converge to irrationals? Does there exists a positive decreasing sequence $\{a_i\}$ with $\sum_{i\in\mathbb{N}} a_i$ convergent, such that $\forall I\subset\mathbb{N},\sum_{i\in I}a_i$ is an irrational number? Such an example would give rise to a closed perfect set containing no rationals. I can only do it for infinite $I$ (for example let $a_i=10^{-p_i}$, where $p_i$ is the $i$th prime.), but the set of infinite sums is not closed. AI: Let $$a_n=\frac{\sqrt{2}}{10^{n!}}.$$ The sum of any finite (non-zero!) number of the $a_i$ is irrational. The sum of an infinite number of the $a_i$ is transcendental, since $\sum_{i=1}^\infty \frac{1}{10^{n_i!}}$ is a Liouville number.
H: Show that if $R_n$ is prime then $n$ must be prime. this is an exam practice question: For each positive $n$ define $R_n = \frac{1}{9}(10^n-1) $ (so that in the usual base 10 notation, $R_n = 111,\ldots,1$ where there are n digits). Show that if $R_n$ is prime then $n$ must be prime. Here is what I have so far: Suppose that $R_n$ is prime. Let $R_n = p$ $$9 \cdot p=(10^n-1)$$ $$ \implies 10^n \equiv 1 \pmod p$$By Fermat's theorem, $$ 10^{p-1} \equiv 10^n \equiv 1 \pmod p$$ and then if $ord_p(10) = j $, it follows that $$ p-1 \equiv n \pmod j$$ But now I'm not too sure where to go from here. I've tried manipulating and rearranging things but I don't really know what the way forward is from here. Any help would be greatly appreciated!! AI: Suppose that $n=ab$, where $a\gt 1, b\gt 1$. Note that $10^a-1$ is a proper divisor of $10^{ab}-1$. For let $x=10^a$. then $10^{ab}-1=x^b-1=(x-1)(x^{b-1}+x^{b-2}+\cdots +x+1)$. If $a\gt 1$ and $b\gt 1$ then both $\frac{x-1}{9}$ and $x^{b-1}+x^{b-2}+\cdots +x+1$ are greater than $1$. For completeness, note that if $n=1$, than $\frac{10^n-1}{9}$ is not prime.
H: How to prove even subsets equal to odd subsets? There is question that I don't know how to prove. we have set $A=\{1,2,3,\ldots,n\},\; O=\{B\mid B⊆A,\text{ odd }B\},\; E=\{B\mid B⊆A,\text{ even }B\}$ it ask to prove that subsets even equal to subsets odd by proving that $f:O\to E$ is an injective and surjective function AI: Let $$ f(B) = \begin{cases} B\smallsetminus\{1\} & \text{if }1\in B, \\ B\cup\{1\} & \text{if }1\not\in B. \end{cases} $$ This works unless $n=0$.
H: Differential Equation: $y'=ty+1$ , $y(0)=0$ For $y'=ty+1$, $y(0)=0$ determine $w(n)$ for an arbitrary value of $n$. (Picard's iteration method). I found a solution, but not sure if it true or not. $$w(n) = \sum_{n = 1}^{\infty} \frac{t^{2n - 1}}{3 \cdot 5 \cdot 7 \cdots (2n - 1)}$$ Need some help. AI: First write out the associated integral equation: $$y(t)=\int_0^t(sy(s)+1)ds.$$ The first step in the iteration is $y_0(t)=y_0=0$, so $y_1(t)=\int_0^t(s\cdot 0+1)ds=t.$ Continuing in this manner, the first few terms are: $$y_0(t)=0$$ $$y_1(t)=t$$ $$y_2(t)=t+\frac{t^3}{3}$$ $$y_3(t)=t+\frac{t^3}{3}+\frac{t^5}{3\cdot5}$$ From here it's probably easy enough to guess the general form of the $n$th term that you already noticed, $$y_n(t)=\sum_{k=1}^{n}\frac{t^{2k-1}}{3\cdot5\cdot7\cdots(2k-1)}.$$ We can prove that this is correct by induction. $$y_{n+1}=\int_0^t(1+s\cdot y_n(s))ds$$ $$=\int_0^t(1+s\cdot \sum_{k=1}^{n}\frac{s^{2k-1}}{3\cdot5\cdot7\cdots(2k-1)})ds$$ $$=t+\int_0^t(\sum_{k=1}^{n}\frac{s^{2k}}{3\cdot5\cdot7\cdots(2k-1)})ds$$ $$=t+\sum_{k=1}^{n}\frac{t^{2k+1}}{3\cdot5\cdot7\cdots(2k-1)(2k+1)}$$ $$=t+\sum_{k=2}^{n+1}\frac{t^{2k-1}}{3\cdot5\cdot7\cdots(2k-1)}$$ $$=\sum_{k=1}^{n+1}\frac{t^{2k-1}}{3\cdot5\cdot7\cdots(2k-1)},$$ which is the correct form we were looking for.
H: Set of orthonormal vectors can be identified with set of matrices Let $V_k(\mathbb{R}^n)$ be the set of all orthonormal $k$-tuples of vectors $v_1,\ldots,v_k\in\mathbb{R}^n$. Let $M_{k,n}$ be the set of all $k\times n$ matrices. Let $W=\{A\in M_{k,n}\mid AA^t=I_k\}$, where $I_k$ is the identity $k\times k$ matrix. Prove that $V_k(\mathbb{R^n})$ can be identified with $W$. (I'm not sure what "can be identified with" means. Maybe it just means showing there is a bijection.) We just put in the vectors $v_1,\ldots,v_k$ as rows of $A$. Then these vectors are columns of $A_t$, so that $AA^t=I_k$ by definition of orthonormality. Conversely, given any $A$, we can let $v_1,\ldots,v_k$ be its rows. Is there anything more to say about this? AI: It is exactly as you say. One says "identified" because you are not defining a bijection at random (after all, both sets have the same cardinality as $\mathbb R$), but rather you are creating your bijection in terms of the properties of the objects you are using.
H: Show that $(0,1)$ is open in $\mathbb{R}$ and that $[0,1]$ is not open in $\mathbb{R}$ My Question: I want to show that the interval $(0,1)$ is open in $\mathbb{R}$ and that $[0,1]$ is not open in $\mathbb{R}$. I proved the latter, but I felt that it was clumsy and could be refined. If you could also show that $(0,1)$ is open that would be great; these are the simplest problems that I can think of, and I am concerned with the format of the proof (my intuition tells me that they are both true). I am working with the following definitions (I may have made an error while translating the definitions into symbolic logic, if so please correct me). Thanks! Definitions: A set $G$ in $\mathbb{R}^{p}$ is said to be open in $\mathbb{R}^{p}$ if for all $x$ in $G$ ,there exists an $r>0$ , such that for each $y$ in $\mathbb{R}^{p}$ satisfying the condition $\\|x-y\\|<r$ also belong to $G$. Or, even more formally we could write $$G\text{ is an }\textbf{open set} \iff G\subset\mathbb{R}^{p}\wedge\forall x\in G\exists r>0\in\mathbb{R}\left(\forall y\in\mathbb{R}^{p}\left(\\|x-y\\|<r\iff y\in G\right)\right).$$ Using the definition of an open ball $\mathcal{B}(x,r)=\{y\in\mathbb{R}^{p}\::\:\\|x-y\\|<r\}$ , where $x$ is the center of the ball and $r>0$ is the radius of the ball, then we can rephrase the definition of an open set by saying that $G$ is open if every point $x$ in $G$ is the center of some open ball contained entirely in $G.$ Again more formally we can write $$G \text{ is an }\textbf{open set}\iff G\subset\mathbb{R}^{p}\wedge\forall x\in G\exists r>0\in\mathbb{R}\left(\mathcal{B}(x,r)\subset G\right).$$ My work: Consider the set $F=\{x\in\mathbb{R}\::\:0\leqslant x\leqslant1\}\subset\mathbb{R}$ , is $F$ an open set? No! proof. Since, we know that $G$ is a subset of $\mathbb{R}^{p}$ , we need only consider the second conjunct in the bi-conditional above. The negation of the statement $\forall x\in G\exists r>0\in\mathbb{R}\left(\mathcal{B}(x,r)\subset G\right)$ is, $\exists x\in G\forall r>0\in\mathbb{R}\left(\mathcal{B}(x,r)\not\subset G\right)$ . Now let us consider the point $x=1$ in $F$ then we must consider all of the balls centered at $1$ with a radius $r>0$, i.e. ${\cal B}(1,r)=\{y\in\mathbb{R}\::\:\|1-y\|<r\}$. If we want to show that the negation is false then we need to find a ball such that for each $y\in{\cal B}(1,r)$, $y\in G$. However, by the Archimedes property we have that we know that for all $r>0$ that there will always be a number $y^=1/n$ such that $0<y^<r$ , and that $(1+y^)\in{\cal B}(1,r)$ while $(1+y^)\notin G$ . Therefore given the element $1$ in $F$ , ${\cal B}(1,r)\not\subset G$ for all $r>0$ ; hence, $F$ is not an open set. AI: Yes, you've got exactly the right idea: Any ball centered at $1$ must contain a number greater than $1$. For a more direct proof that's constructive (as opposed to just invoking the Archimdean property), consider: Suppose that $[0, 1]$ is open. Then there exists an $r > 0$ such that whenever $\|y - 1\| < r$, we have $y \in [0, 1]$. Choose $$y = 1 + \frac{r}{2}$$ This leads to a contradiction since $\|y - 1\| = \|r/2\| < r$ but $y \notin [0, 1]$. As a hint to show that $(0, 1)$ is open, let $x \in (0, 1)$ be given. Consider $$r = \frac{\min(x, 1 - x)}{2}$$ and show that that $\mathcal{B}(x, r) \subseteq (0, 1)$.
H: A question about reduciblility Why a polynomial $f(x)＝2x^2+4$ is reducible over $\Bbb C$? Isn't 2 a unit on $\Bbb C$? Hop someone can explain it clearly. Thanks. AI: Our polynomial factors for example as $$2x^2+4=(2x-2\sqrt{2}i)(x+\sqrt{2}i).$$ Remark: It is true that $2x^2+4=2(x^2+2)$ does not prove reducibility, since $2$ is a unit.
H: Computing derivative of function between matrices Let $M_{k,n}$ be the set of all $k\times n$ matrices, $S_k$ be the set of all symmetric $k\times k$ matrices, and $I_k$ the identity $k\times k$ matrix. Let $\phi:M_{k,n}\rightarrow S_k$ be the map $\phi(A)=AA^t$. Show that $D\phi(A)$ can be identified with the map $M_{k,n}\rightarrow S_k$ with $B\rightarrow BA^t+AB^t$. I don't really understand how to compute the map $D\phi(A)$. Usually when there is a map $f:\mathbb{R}^s\rightarrow\mathbb{R}^t$, I compute the map $Df(x)$ by computing the partial derivatives $\partial f_i/\partial x_j$ for $i=1,\ldots,t$ and $j=1,\ldots,s$. But here we have a map from $M_{k,n}$ to $S_k$. How can we show that $D\phi(A)\cdot B=BA^t+AB^t$? AI: The derivative at $A$ is a linear map $D\phi(A)$ such that $$ \frac{\\|\phi(A+H)-\phi(A)-D\phi(A)H\\|}{\\|H\\|}\to0\ \ \mbox{ as } H\to0. $$ (the spirit of this is that $\phi(A+H)-\phi(A)\sim D\phi(A)H$, where one thinks of $H$ as the variable). In our case, we have $$ \phi(A+H)-\phi(A)=(A+H)(A+H)^T-AA^T=AH^T+HA^T+HH^T. $$ So $D\phi(A)H=AH^T+HA^T$ as the term $HH^T$ satisfies $\\|HH^T\\|=\\|H\\|^2$.
H: Limit of Fuctions Let $f(x)= \left \{ \begin{array}{cc} x & x\in \mathbb{Q}\\ 0 & \,\,\,\,\,\,x\in \mathbb{R}\setminus\Bbb{Q} & \end{array} \right . $ Determine all $a \in \mathbb{R}$ for which $\lim_{x \rightarrow a} f(x)$ exists. I see that the answer is $a=0$, but I don't know how to prove it. AI: Consider $x \in \Bbb{R}$ given. Choose a sequence $\{r_n\} \subseteq \Bbb{Q}$ such that $r_n \to x$; that is, a sequence of rational numbers converging to $x$; such a sequence exists by density. Likewise, choose $\{t_n\}$ a sequence of irrationals converging to $x$. Now suppose that $f$ is continuous at $x$; by the sequential definition of continuity, we have $$\lim_{n \to \infty} f(r_n) = f(x) = \lim_{n \to \infty} f(t_n)$$ But $f(r_n) = r_n$, so $f(r_n) \to x$ as $n \to \infty$. Where does $f(t_n)$ converge to?
H: Show that $\frac{a+b}{2} \ge \sqrt{ab}$ for $0 \lt a \le b$ I have to prove that $$\frac{a+b}{2} \ge \sqrt{ab} \quad \text{for} \quad 0 \lt a \le b$$ The main issue I am having is determining when the proof is complete (mind you, this is my first time). So I did the following steps: $$\begin{align} \frac{a+b}{2} &\ge \sqrt{ab} \\ \left(\frac{a+b}{2}\right)^2 &\ge \left(\sqrt{ab}\right)^2 \\ \frac{a^2+2ab+b^2}{4} &\ge ab \\ a^2+2ab+b^2 &\ge 4ab \\ a^2-2ab+b^2 &\ge 0\\ (a-b)^2 &\ge 0 \\ \end{align}$$ Now this is where I stopped because if I square root each side, I will be left with $a-b \ge 0$ or in other words, $a \ge b$ which doesn't make a whole lot of sense to me. So ultimately the question is: how do I know when I'm done? and is what I did above correct? Thanks! AI: You're working backward, but all of your steps are actually equivalent, so that's okay. When you take the square root, though, you get $\|a-b\|\ge 0.$ This is true, so you're fine. I would start with a true statement like $\|a-b\|\ge 0,$ and proceed through these steps in reverse, to prove the desired inequality. Or, more simply, note that $(a-b)^2\ge 0$ for all real $a,b$, so we can get there even more quickly starting from that point.
H: sum of independent random variables where $N$ is a random variable I want to show $E[S_N]=E[N]E[X_j]$ where: $X_1,X_2,\ldots$ is a sequence of independent random variables, and $N$ is a random variable independent of the sequence. $S_n=\sum_{i=1}^n X_i$, $S_N=X_1+X_2+\ldots+X_N$, So far, I have $$E[S_N]=E\left[\sum_{n=0}^{\infty} S_n1_{\{N=n\}}\right]=\sum_{n=0}^{\infty}E[S_n1_{\{N=n\}}]$$ by monotone convergence theorem, $$=\sum_{n=0}^{\infty}E[S_n]P(N=n)$$ by the fact that $S_n$ and $1_{N=n}$ is independent since $N$ is independent from the sequence so $N$ is independent from their sum. $$=\sum_{n=0}^{\infty}E\left[\sum_{j=1}^n X_j\right] P(N=n)$$ $$=\sum_{n=0}^{\infty}\left[\sum_{j=1}^n E[X_j]\right] P(N=n)$$ since $X_j$ are independent. I'm not sure if I can switch the summations here since the one inside goes to $n$. AI: The proposition only makes sense if all the X_j have the same expectation. Then the sum over j is just a multiplication by n. This completes the proof.
H: modular arithmetic with exponents I'm looking at the solution manual of a book and it lists a solution for $$[19^3\mod {23}]^2 \pmod {31} \equiv [(-4)^3\mod {23}]^2 \pmod {31} \equiv [-64\mod {23}]^2 \pmod {31}\equiv 5^2 \pmod{31} = 25$$ How does it get from $[19^3\mod {23}]^2 \pmod {31}$ to $[(-4)^3\mod {23}]^2 \pmod {31}$? in other words where does the transformation from $19^3$ to $(-4)^3$ come from? AI: The important thing is that it is from $19^3 \pmod {23}$ to $(-4)^3 \pmod {23}$ This works because $19 \equiv -4 \pmod {23}$ You lost that in going to the final question.
H: smooth bijective functon's derivative $\ne 0$? $f\colon (\alpha,\beta)\to (a,b)$ is a smooth bijective functon, that is, the derivative $f^{(n)}(x)$ exists for all $n\ge 1$. 1) If the inverse function $f^{-1}\colon (a,b)\to (\alpha,\beta)$ is also smooth, does it follow that the derivative $f^\prime(x)\ne0$ for all $x\in (\alpha,\beta)$? 2) If the derivative $f^\prime(x)\ne0$ for all $x\in (\alpha,\beta)$, Can we conclude that the inverse function $f^{-1}$ is also smooth? AI: Yes and Yes. The first one follows from chain rule: $f^{-1}(f(x)) = x$ for all $x\in (a, b)$ implies $$\frac{df^{-1}}{dx}(f(x)) \cdot \frac{df}{dx}(x) = 1 \Rightarrow \frac{df}{dx}(x)\neq 0\ .$$ For the second one, it follows from the inverse function theorem. The two results are true in any dimensions.
H: Placing K knights in an nxn board such that no two attack each other This is a problem from spoj A and B are playing a very interesting variant of the ancient Indian game 'shatranj(also known as chess)' on a 'maidaan'(chessboard) n×n in size. They take turns to put game pieces called 'ghoda'(knight) so that no two 'ghodas'(knights) could threat each other. A 'ghoda' located in square (a,b) can threat squares (a+1,b+2),(a-1,b+2),(a+1,b-2),(a-1,b-2),(a+2,b-1),(a+2,b+1),(a-2,b-1),(a-2,b+1). The player who can't put a new 'ghoda' during his move loses. Find out which player wins considering that both players play optimally well and A starts. I cannot use brute force as n is very large, Is there an O(n) or O(1) algorithm for this problem? AI: Hint: consider a strategy when one player answers the moves of his opponent by central symmetry. For which $n$ does it work for the second player? When it doesn't work for the second player, can the first one use the strategy instead?
H: Finding the number of solutions to this equation in range 50 to 100 $x^{2}\; -\; \mbox{floor}\left( x \right)\cdot x\; =\; 83.26$ I was able to find it by graphing and counting the number of intersections within the range, but there has to be an easier way to solve this. AI: On Ross Millikan's suggestion, I'm posting this as an answer: Hint: $x^2 - \mbox{floor{x}}\cdot x = x(x-\mbox{floor{x}})$ Now think about all the possible values that $x-\mbox{floor{x}}$ can take.
H: $f:X\rightarrow Y$, $g:Z\rightarrow Y$ then $h:X\times Z\rightarrow Y$ continuous $\textbf{Lemma}$: If $f:X\rightarrow Y$ and $g:Z\rightarrow Y$ are continuous functions with $X,Y,Z\subseteq \mathbb{R}$ then $h:X\times Z\rightarrow Y$ via $h(x,z)=f(x)g(z)$ is continuous. Is there a way to show the above that uses a combination of results that makes it not too difficult to prove? I am trying to construct a homotopy and I have a continuous function $f:[0,1]\rightarrow Y$, where $Y$ is some topological space, and I am composing the function $$h:\left[0,1\right]\times \left[0,\frac{1}{3}\right]\rightarrow [0,1] \text{ by } h(s,t)=3s(\frac{1}{3}-t) \text{ with } f.$$ I think it would be nice to prove the above general result and then apply it to the function $h$ in my problem and then have the lemma for future problems. This is mostly for my own confidence to be able to show such a thing is true, I doubt I need to go into details like this for my written turned in work. I am hoping I am overlooking some obvious things that will make proving the lemma easy. I know that if $f,g:X\rightarrow Y$ are continuous where $X,Y\subseteq \mathbb{R}$ then $fg:X\rightarrow Y$ is continuous. AI: If you have established that projections are continuous, then you can consider the two compositions $$ X \times Z \to X \overset{f}{\to} Y $$ and $$ X \times Z \to Z \overset{g}{\to} Y $$ and use your result that the pointwise product of the two maps is also continuous.
H: Demonstrating that $f(x) = x^2 + 1$ is bijective and calculating $f \circ f^{-1}(x)$ I did this exercise. I am not sure if my surjective proof is right - is it good? In fact, I'm not even sure what exactly am I trying to prove (I know that surjective means that each element in the codomain should have a preimage in the domain, but I don't really see how to prove that - I was just mimicking some example I had seen before). Could you give more insight on how to prove a function surjective? The rest of the exercise is here as well, in case I did a related mistake. Let $f:[0,+\infty[ \rightarrow [1,+\infty[$ defined by $$f(x) = x^2 + 1$$ Demonstrate that it is bijective. Injective $$f(a) = f(b)$$ $$a^2+1 = b^2 + 1$$ $$a^2 = b^2$$ Since $a,b >= 0$, it is safe to affirm that $$a = b$$ Surjective $$b = f(a)$$ $$b = a^2 + 1$$ $$b - 1 = a^2$$ $$a = \sqrt{b - 1}$$ Calculate $f^{-1}$ and verify that $f \circ f^{-1}(x) = x$. $$f^{-1}(x) = \sqrt{x - 1}$$ And $$f \circ f^{-1}(x)$$ $$f(f^{-1}(x))$$ $$(\sqrt{x - 1})^2+1$$ $$x - 1 + 1$$ $$x$$ AI: Yes, you're exactly correct, for we have $$f(a) = a^2 + 1 = \Big(\sqrt{b - 1}\Big)^2 + 1 = b - 1 + 1 = b$$ as desired. This is the important thing to prove. The general strategy to show that a function is surjective is two-fold: Find a candidate for the preimage by working backwards (i.e. scratchwork) Show that the candidate actually works Now in a case such as this where all the steps in solving for $a$ in terms of $b$ are reversible, these steps are really equivalent.
H: Map to symmetric matrices is surjective. Let $M_{k,n}$ be the set of all $k\times n$ matrices, $S_k$ be the set of all symmetric $k\times k$ matrices, and $I_k$ the identity $k\times k$ matrix. Suppose $A\in M_{k,n}$ is such that $AA^t=I_k$. Let $f:M_{k,n}\rightarrow S_k$ be the map $f(B)=BA^t+AB^t$. Prove that $f$ is onto (surjective). (Note: all matrices have entries in $\mathbb{R}$.) Clearly the matrix $BA^t+AB^t$ is symmetric, since $$(BA^t+AB^t)^t=(BA^t)^t+(AB^t)^t=AB^t+BA^t.$$ We want to show that for every $C\in S_k$, there exists $B\in M_{k,n}$ such that $$C=BA^t+AB^t.$$ How can we do that? AI: Let $B = \frac{1}{2}CA \in M_{k, n}$.
H: In a triangle $\angle A = 2\angle B$ iff $a^2 = b(b+c)$ Prove that in a triangle $ABC$, $\angle A = \angle 2B$, if and only if: $$a^2 = b(b+c)$$ where $a, b, c$ are the sides opposite to $A, B, C$ respectively. I attacked the problem using the Law of Sines, and tried to prove that if $\angle A$ was indeed equal to $2\angle B$ then the above equation would hold true. Then we can prove the converse of this to complete the proof. From the Law of Sines, $$a = 2R\sin A = 2R\sin (2B) = 4R\sin B\cos B$$ $$b = 2R\sin B$$ $$c = 2R\sin C = 2R\sin(180 - 3B) = 2R\sin(3B) = 2R(\sin B\cos(2B) + \sin(2B)\cos B)$$ $$=2R(\sin B(1 - 2\sin^2 B) +2\sin B\cos^2 B) = 2R(\sin B -2\sin^3 B + 2\sin B(1 - \sin^2B))$$ $$=\boxed{2R(3\sin B - 4\sin^3 B)}$$ Now, $$\implies b(b+c) = 2R\sin B[2R\sin B + 2R(3\sin B - 4\sin^3 B)]$$ $$=4R^2\sin^2 B(1 + 3 - 4\sin^2 B)$$ $$=16R^2\sin^2 B\cos^2 B = a^2$$ Now, to prove the converse: $$c = 2R\sin C = 2R\sin (180 - (A + B)) = 2R\sin(A+B) = 2R\sin A\cos B + 2R\sin B\cos A$$ $$a^2 = b(b+c)$$ $$\implies 4R^2\sin^2 A = 2R\sin B(2R\sin B + 2R\sin A\cos B + 2R\sin B\cos) $$ $$ = 4R^2\sin B(\sin B + \sin A\cos B + \sin B\cos A)$$ I have no idea how to proceed from here. I tried replacing $\sin A$ with $\sqrt{1 - \cos^2 B}$, but that doesn't yield any useful results. AI: $$a^2-b^2=bc\implies \sin^2A-\sin^2B=\sin B\sin C\text{ as }R\ne0$$ Now, $\displaystyle\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)=\sin(\pi-C)\sin(A-B)=\sin C\sin(A-B)\ \ \ \ (1)$ $$\implies \sin B\sin C=\sin C\sin(A-B)$$ $$\implies \sin B=\sin(A-B)\text{ as }\sin C\ne0$$ $$\implies B=n\pi+(-1)^n(A-B)\text{ where }n\text{ is any integer} $$ If $n$ is even, $n=2m$(say) $\implies B=2m\pi+A-B\iff A=2B-2m\pi=2B$ as $0<A,B<\pi$ If $n$ is odd, $n=2m+1$(say) $\implies B=(2m+1)\pi-(A-B)\iff A=(2m+1)\pi$ which is impossible as $0<A<\pi$ Conversely, if $A=2B$ $\displaystyle\implies a^2-b^2=4R^2(\sin^2A-\sin^2B)=4R^2\sin C\sin(A-B)$ (using $(1)$) $\displaystyle=4R^2\sin C\sin(2B-B)=2R\sin B\cdot 2R\sin C=\cdots$
H: Properties of $A + B$ related to $A$ and $B$. (GS2010) Let $A$ and $B$ be subsets or $\mathbb{R}$. Define $A + B = \{ a + b : a \in A, b \in B\}$. Then which of the followings are true and which are false? Why? Please give a proof for truth and a counterexample or false. Will the situation be different if we consider $[0, \infty)$ instead of $\mathbb{R}$? $A + B$ is bounded when $A$ and $B$ are bounded. $A + B$ is open when $A$ and $B$ are open. $A + B$ is closed when $A$ and $B$ are closed. $A + B$ is connected when $A$ and $B$ are connected I have tried as follows. is true as $a < M_1$ and $b < M_2$ gives $a + b < M_1 + M_2$. For any element $a+b$ we shall get open nbds $A'$ and $B'$ s.t. $a \in A' \subset A$ and $b \in B' \subset B$. So we shall get a nbd $A' + B'$ of $a + b$ in $A + B$. So $A + B$ is open. I have doubt. If $A$ and $B$ are finite and closed so $A + B$. When at least one of $A$ or $B$ will be infinite $A + B$ will have a limit point. I am not sure if it will stay in $A + B$. I have no idea about it. Thank you for your help. If this question is discussed earlier please give the link. AI: Let me omit 1. It is true. In 2, you need only one of A or B to be open. For example, if A is open, then $A+B = \cup_{b \in B} \{A+b\}$ but each $A+b$ is an open set. For 3, see Closed sum of sets. For 4, a connected subset of the real line is necessarily an interval, and you can prove that addition of two intervals gives us an interval.
H: If $f$ is injective, demonstrate that $f \circ g = f \circ h \implies g = h$ I'm trying to do this exercise: With functions: $$f : A \rightarrow B$$ $$g : C \rightarrow A$$ $$h : C \rightarrow A$$ Demonstrate that if $f$ is injective, then $f \circ g = f \circ h \implies g = h$ So we have two premises: $f$ is injective $f\circ g = f\circ h$ $f$ being injective means that $(\forall a,b \in A)(f(a) = f(b) \implies a = b)$. Different elements in $A$ map to different elements in $B$. I am not sure what can I infer from the fact that $f\circ g = f\circ h$. The composition $f\circ h$ goes something like $C \rightarrow A \rightarrow B$. Since $f$ is injective and it was used to form the composition, can I say that $f\circ h$ is also injective? Well, even so I am not sure what would I do next. Although I'm presenting this exercise, the real problem is that I don't quite get how to use function properties (injective or surjective) to deduce what happens to such function when it acts in an operation (in this case $\circ$) with another function. Is the result still injective? It depends? Etc. It would be great to get more feedback on that. AI: If $f$ is injective then $$f(g(x))=f(h(x))\implies g(x)=h(x)$$ for all $x$.
H: How should I express one $\log$ in terms of others? Can someone please help me with this logarithmic question? I know it’s easy, but I need to refresh my memory on how to do it. If $X=\log2$ and $Y=\log3$, express $\log0.6$ in terms of $X$ and $Y$ (assume all logs have a base of $10$). AI: All you need to know is $\log(ab) = \log a + \log b$. Then, $\log 0.6 = \log(2\times 3\times \frac{1}{10}) = \log 2 + \log 3 + \log \frac{1}{10} = X+Y-1.$
H: reducing enemies - geometry puzzle You are given an irreflexive symmetric (but not necessarily transitive) "enemies" relation on a set of people. In other words, if person A is an enemy of a person B, then B is also an enemy of A. How can you divide up the people into two houses in such a way that every person has at least as many enemies in the other house as in their own house? AI: Divide the people up however you like. Then, if there is anyone with more enemies in their own house than the other house, make them switch houses. (If there are multiple candidates, pick any one you like.) Keep doing this until everyone has at least as many enemies in the other house as in their own. Because each switch (by construction) reduces the total number of pairs of enemies that share a house, the procedure must terminate. This is known as the "Happynet" problem. While it is simple to find a solution, as just described, no polynomial-time algorithm for finding one is known. (The algorithm I described can take exponential time in the worst case.)
H: $\mathcal{C}_1 \subseteq \mathcal{C}_2 \implies \sigma( \mathcal{C}_1) \subseteq \sigma( \mathcal{C}_2) $ $\mathcal{C}_1$, $\mathcal{C}_2$ are collections of subsets of $X$,then Im having hard time seeing why the following is true. Can someone explain them to me? $\mathcal{C}_1 \subseteq \mathcal{C}_2 \implies \sigma( \mathcal{C}_1) \subseteq \sigma( \mathcal{C}_2) $ $\sigma(\sigma( \mathcal{C} )) = \sigma( \mathcal{C} ) $ where $\sigma( \mathcal{F} ) $ is the sigma field generated by the collection $\mathcal{F}$ of subsets of $X$ thanks AI: It involves two steps: $$ \mathcal{C}_1\subseteq\mathcal{C}_2\;\;\Longrightarrow\;\;\mathcal{C_1}\subseteq\sigma(\mathcal{C_2})\;\;\Longrightarrow\;\;\sigma(\mathcal{C}_1)\subseteq\sigma(\mathcal{C}_2). $$ The first implication is obvious since $\mathcal{C}_2\subseteq\sigma(\mathcal{C}_2)$. The second implication holds by definition of $\sigma(\mathcal{C}_1)$, i.e. it is the smallest sigma-algebra that contains $\mathcal{C}_1$. Since $\sigma(\mathcal{C}_2)$ is sigma-algebra that also contains $\mathcal{C}_1$, then it necessarily has to be larger than $\sigma(\mathcal{C}_1)$ (or equal to). The second implication explained more mathematically: the definition of $\sigma(\mathcal{C}_1)$ is $$ \sigma(\mathcal{C}_1)=\bigcap_{\mathcal{E}\in\Sigma(\mathcal{C}_1)}\mathcal{E}, $$ where $\Sigma(\mathcal{C}_1)=\{\mathcal{E}\mid \mathcal{E}\text{ is a sigma-algebra and }\; \mathcal{C}_1\subseteq\mathcal{E}\}$. If $\mathcal{C}_1\subseteq\sigma(\mathcal{C}_2)$, then $\sigma(\mathcal{C}_2)\in\Sigma(\mathcal{C}_1)$ and thus $\sigma(\mathcal{C}_1)\subseteq\sigma(\mathcal{C}_2)$ obviously. This also explains why $\sigma(\sigma(\mathcal{C}))=\sigma(\mathcal{C})$, since $\sigma(\sigma(\mathcal{C}))$ is the smallest sigma-algebra containing $\sigma(\mathcal{C})$, and this is of course $\sigma(\mathcal{C})$ itself.
H: Question on series Suppose $ a_i $ be a sequence of positive real numbers such that $ \sum_{i=1}^{\infty}a_i < \infty $. Is it true that $ \sum_{i=1}^{\infty} \log(i) \cdot a_i < \infty $? Thanks AI: No, a counterexample is $a_n = \frac{1}{n \log^2 n}$.
H: The Limitations of Vieta's Formula I was attempting to find the roots of $f(x)=2x^3+10x^2+5x−12$ and since the the OP had already found one of the roots, I tried to recall a relation to help me find the other two easily. The first one that popped to mind were Vieta's formulas for Cubic Polynomials: $$x_1+x_2+x_3=-\frac{b}{a}$$$$x_1x_2+x_1x_3+x_2x_3=\frac{c}{a}$$$$x_1x_2x_3=-\frac{d}{a}$$ But when I tried applying them, I found that they don't work. I've never actually used the relations to find roots like this before and frankly I've never seen anyone do it. As stupid a question as this is, Why doesn't it work for the cubic I tried it on?And also, in what other circumstances doesn't it work? Please help me understand the limitations of these relations and try to be as simple as you can. EDIT:These relations do work in helping to solve for the rest of the roots and it was sheer idiocy on my part that made me suspect that they didn't. AI: The relations work just fine. We get $x_1+x_2+x_3=-5$. One of the roots, say $x_3$, is $-4$, so $x_1+x_2=-1$. Similarly, $x_1x_2x_3=6$, so $x_1x_2=-\frac{3}{2}$. Now $(x_1-x_2)^2=(x_1+x_2)^2-4x_1x_2=7$. So $x_1-x_2=\pm\sqrt{7}$. We know $x_1+x_2$, so we can easily find $x_1$ and $x_2$.
H: Evaluation of $\int\frac{1}{(x^3+1)^2}\mathrm dx \cdots$ $\displaystyle \int\frac{1}{(x^3+1)^2} \mathrm dx$ $\bf{My\; Try}::$ Using Integration by parts Let $\displaystyle I = \int\frac{1}{(x^3+1)}\cdot 1\; dx = \frac{1}{(x^3+1)}\cdot x + \int\frac{3x^2\cdot x}{(x^3+1)^2}dx$ $\displaystyle I = \frac{1}{(x^3+1)}\cdot x + 3\int\frac{(x^3+1)-1}{(x^3+1)^2}dx$ $\displaystyle I = \frac{1}{(x^3+1)}\cdot x+3\int\frac{1}{x^3+1}dx-3\int\frac{1}{(x^3+1)^2}dx$ $\displaystyle I = \frac{1}{(x^3+1)}\cdot x+3I-3\int\frac{1}{(x^3+1)^2}dx$ $\displaystyle \int\frac{1}{(x^3+1)^2}dx = \frac{x}{3(x^3+1)}+\frac{2}{3}\int\frac{1}{x^3+1}dx$ Now Help Required Thanks AI: Good Work so far. $$ \begin{equation} \begin{split} \int\frac{1}{x^3+1} \mathrm{d}x & = \int\frac{1}{3(x+1)} \mathrm{d}x + \int\frac{2-x}{3(x^2-x+1)} \mathrm{d}x \\ & = \frac{1}{3} \ln\left\|x+1\right\| - \frac{1}{6} \int \frac{2x-1 -3}{x^2-x+1}\mathrm{d}x \\ & = \frac{1}{3} \ln\left\|x+1\right\| - \frac{1}{6} \int\frac{2x-1}{x^2-x+1}\mathrm{d}x + \frac{1}{2} \int\frac{1}{x^2-x+1}\mathrm{d}x \\ & = \frac{1}{3} \ln\left\|x+1\right\| - \frac{1}{6} \ln\left\|x^2-x+1\right\| + \frac{1}{2} \int\frac{1}{\left(x-\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\mathrm{d}x \\ & = \frac{1}{3} \ln\left\|x+1\right\| - \frac{1}{6} \ln\left\|x^2-x+1\right\| + \frac{1}{\sqrt{3}} \tan^{-1}\left(\frac{2x-1}{\sqrt{3}}\right) \\ \end{split} \end{equation} $$
H: Product of randomly drawn numbers Here are two code line to run in R: prod(rnorm(100, mean=1, sd=0)) # (1) prod(rnorm(100, mean=1, sd=0.2)) # (2) $prod(..)$ returns the product of a sequence. The sequence it given by $rnorm(n, mean, sd)$. This function $rnorm(...)$ return n values randomly drawn from a normal distribution with $mean$ and $sd$. The only things that differ between the two lines in the $sd$. Line (1) is totally deterministic as $sd=0$ and therefore it always returns the value $1$. Line (2) is not determinstic and almost always (in 90% of the cases I'd say!) return a value that is lower than $1$. Why is it so? What is the expected value of line (2)? In what percentage of the cases does line(2) returns values greater than $1$ AI: If the sequence $(x_k)$ is i.i.d. and $P[x_1=0]=0$, when $n$ is large, $\left\|\prod\limits_{k=1}^nx_k\right\|=\exp(n\xi_\sigma+o(n))$ almost surely, where $\xi_\sigma=E[\log \|x_1\|]$. This follows from the law of large numbers applied to the i.i.d. sequence $(\log \|x_k\|)$, plus the identity $\left\|\prod\limits_{k=1}^nx_k\right\|=\exp\left(\sum\limits_{k=1}^n\log \|x_k\|\right)$. In your case (2), one assumes that each $x_k$ is a normal random variable $(1,\sigma^2)$ with $\sigma=0.2$, then $\xi_\sigma=E[\log\|1+\sigma z\|]$ where $z$ is standard normal. What happens is that $\xi_\sigma\lt0$ when $\sigma=0.2$ hence $\prod\limits_{k=1}^nx_k$ converges to zero almost surely. We checked numerically that $\xi_\sigma\lt0$ when $\sigma=0.2$. Note that the function logarithm is concave, hence $E[\log u]\lt\log E[u]$ for every nondegenerate positive random variable $u$ (this is Jensen inequality, with a strict inequality sign when the distribution of $u$ is not Dirac). But this only shows that $\xi_\sigma$ is "not too positive" when $\sigma$ is "not too large", not that $\xi_\sigma\lt0$ always. As a word of caution against some too simple arguments, note finally that, for some larger values of $\sigma$, indeed $\xi_\sigma\gt0$, the transition occuring between $\sigma=1.55$ and $\sigma=1.56$ (and surely the zero of the function $\sigma\mapsto E[\log\|1+\sigma z\|]$ has a name...).
H: where to start reading theory of logics? I am a student who is working Lie Theory. I want to start read theory of logics. I just need some reference and I have few questions regarding this, i) will studying theory of logics will improve my theorem proving ability in 'other branch' of mathematics? ii) is there any results in mathematics, (I am sure there are) (say in algebra particularly in group theory) proof of which is made easy through the approaches and techniques in theory of logics? iii) where to start reading logic theory and which are good text books in logic theory for the beginners. I have read Paul R.Halmos 's Naive Set Theory book completely. But I haven't done all the exercises. This is the amount of knowledge I have regarding theory of logics. I hope I am not confusing between theory of sets and theory of logics. Thanks for your valuable suggestions. AI: Yes, there are for example results in algebra that are proved by model-theoretic means (and model theory is usually taken to be a core part of mathematical logic). A classic early example is touched on here http://bit.ly/JKyuuZ And for lots of references to different areas where model theory has got applied more recently, see this MO post and the answers: http://bit.ly/4IXUHa The logical resources needed for such applications are pretty sophisticated, but if you want to make a start learning some logic I'm inclined to think that (if you are a good mathematics student) Peter Hinman's Fundamentals of Mathematical Logic is perhaps the go-to book if you want a single textbook to get you into the field. It is very lucid. True, it is a Very Big Book, but that's because it covers a lot at an accessible pace -- and the preface explains very well how to steer a course through the book to suit particular interests. For an extensive annotated reading Guide to logic more generally, looking at lot of textbooks at different levels, you can download the PDF http://bit.ly/18t5CUD (which doesn't yet given an account of Hinman's book -- a major shortcoming that will be corrected in the next version). The Guide has a section on model theory in particular.
H: If $G$ and $H$ are nonisomorphic group with same order then can we say that $\operatorname{Aut}(G)$ is not isomorphic to $\operatorname{Aut}(H)$? We know that nonisomorphic groups may have isomorphic automorphism groups. As an example, you can think klein four group and $S_3$ since their automorphism group is isomorphic to $S_3$. Now,I wonder If $G$ and $H$ are nonisomorphic group with same order then can we say that $\operatorname{Aut}(G)$ is not isomorphic to $\operatorname{Aut}(H)$ or can we find two nonisomorphic groups with same order and their automorphism groups are isomorphic? AI: No. You can check that Automorphism group of both Dihedral group($D_8$) and Direct product of $Z_2$ and $Z_4$ is Dihedral group($D_8$). So we have two non-isomorphic groups with order $8$ and their Automorphism groups are the same group. This is the smallest example of such groups. http://groupprops.subwiki.org/wiki/Endomorphism_structure_of_direct_product_of_Z4_and_Z2 http://www.weddslist.com/groups/misc/autd8.html
H: Linear subspace of Banach space containing unit ball Am I right that any linear subspace of Banach space which contain unit ball is whole space? AI: Yes, take any vector in the space and divide it by (a little more than) its and you'd end up with a vector within the unit ball. Hence, every vector in a Banach space is conlinear to a vector in the unit ball and so the unit ball spans the entire space.
H: Inverse Percentage. Sorry for asking this foolish question. Here is the data i have. I purchased the product as $5 and additional fee is 2%. So Here is the total dollor $total = 5 + (5*2/100) = 5.1 total dollor i have. But now i want to revert back to original price in this case i have the data as only total amount $5.1 and 2% fee. How can i revert back to $5 based on the above data. Thanks AI: It appears from your calculation that the additional fee is $2$ percent of the price, not $\$2$. That means that if the base price is $p$, the total cost $t$ is $t=p+0.02p=1.02p$. Thus, $p=\frac{t}{1.02}$. In your example, for instance, this leads to the calculation $$p=\frac{5.1}{1.02}=5\;,$$ as it should.
H: Stokes' theorem for an annulus From Wiki, I'm looking at this definition: "If the surface is not closed, it has an oriented curve as boundary. Stokes' theorem states that the flux of the curl of a vector field is the line integral of the vector field over this boundary. This path integral is also called circulation, especially in fluid dynamics. Thus the curl is the circulation density." I'm looking to take the flux of an area between two oriented concentric circles in the plane $z=0$. Would I be right in thinking I can apply Stokes' theorem in this case as it's an open surface? If that's right, as there's a 'hole' in the middle, would the flux be the circulation of the outer boundary minus the circulation of the inner boundary? Thanks! AI: Yes, that is right. The boundary of the annulus between the two concentric circles is the union of the two circles, and the natural orientation is such that the outer circle is positively oriented, and the inner circle negatively, so $$\int_{r < \lvert (x,y)\rvert < R} \operatorname{curl} F \,dS = \int_{\lvert (x,y)\rvert = R} F\cdot d\vec{s} - \int_{\lvert (x,y)\rvert = r} F\cdot d\vec{s}.$$
H: Integral of $dy/dx$ confusion Why is $\displaystyle \int \dfrac{dy}{dx} dx = y + c$, but for example $\displaystyle \int \dfrac{dy}{dx} y dx = \dfrac{1}{2} y^2 + c$ instead of $\dfrac{1}{2} y^3 + c$? AI: Using and abusing the mathematical notation as sometimes is done when dealing with differential equations, what you really have here is $$\int\frac{dy}{\color{red}{dx}}\color{red}{dx}=\int 1\cdot dy=y+C\;,\;\;\text{since}\;\;\frac{d}{dy}(y+C)=1$$ OTOH, $$\int\frac{dy}{\color{red}{dx}}y\,\color{red}{dx}=\int y\,dy=\frac{y^2}2+C\;,\;\;\text{since}\;\frac d{dy}\left(\frac12y^2+C\right)=y$$
H: Question on pointwise convergence of a function Let $f_n:\mathbb R^+\to\mathbb R$ for $n \in \mathbb N$ be given by $$f_n(x)=\begin{cases}n & x \in \left(0,\frac1n\right)\\ \frac1x & x \in \left[\frac1n,\infty\right) \end{cases}$$ I have to show that it would converge pointwise to $\frac1x$ for all $x \in \mathbb R$. Even though this seems to be an easy problem, but my concepts regarding this topic are a bit rusty, so it would be nice if someone could help me out ! I don't want the entire proof, just give me a headstart. AI: Let $x>0$ then there is a $n_0=n_0(x):\frac {1}{n_0}<x.$ If we take $n\geq n_0$ we have that $\frac {1}{n}<x$ and thus for every $n\geq n_0:f_n(x)=\frac {1}{x}\to \frac {1}{x}$ Thus $f_n(x)\to \frac {1}{x}$
H: Angle between $\vec a$ and $\vec b $. We got the same size vector $\vec a$ and $\vec b $. We know that the vector $\vec a +2\vec b$ and $5\vec a-4\vec b$ are perpendicular? $(\vec a +2\vec b) \perp (5 \vec a-4\vec b)$ What is angle between $\vec a$ and $\vec b $? AI: $$ (\vec{a} + 2 \vec{b})\cdot(5\vec a-4\vec b) = 0\\ 5 \\|\vec a\\|^2-4(\vec a\cdot\vec b)+10(\vec a\cdot\vec b)-8\\|\vec b\\|^2 = 0\\ 6(\vec a\cdot\vec b)=3\\|\vec a\\|^2 \\ \frac{(\vec a\cdot\vec b)}{\\|\vec a\\|\\|\vec b\\|} = \frac{1}{2} \\ \cos \theta = \frac{1}{2} \\ $$ So angle between the vectors is $\frac{\pi}{3}$. (Note that I have used the fact that they are of same size twice.)
H: Set Theory Symbols I get easily confused when it comes to the symbol based terminology in Set Theory. Could someone please elaborate on what the following expressions mean? It would really help me out. $S_1$ = knowledge of a subject matter $S_2$ = problem solving related to this subject matter $S_3$ = ability to adapt properly the already existing knowledge for use in analogous similar cases set $U = \{a, b, c, d, e\}$. set MAi = subset of U Denote by $a, b, c, d$, and $e$ the linguistic labels (fuzzy expressions) of very low, low, intermediate, high and very high success respectively of a student in each of the $S_i$s and set $U = \{a, b, c, d, e\}$. Question 1 What exactly is $S_i$s? I realize the $i$ is a subscript and the $S$ represents the 'array' of characteristics. But what's up with the trailing s? There is another statement that says: We define the membership function $m_{A_i}$ for each $x$ in $U$ as ... What exactly is $x$? Since the elements are not integers, would $x$ be the index number and therefore represent $a$ as $0$, $b$ as $1$ and so on? I am terribly sorry about asking these really basic questions. I have gaps in the fundamentals of my knowledge which are hampering me. I would be very grateful if someone could help me bridge them. Thanks. AI: ‘$S_i$s’ is just the plural of $S_i$: the $S_i$s are the characteristics $S_1,S_2$, and $S_3$. In the second question, $x$ ranges over the set $U$: its possible values are $a,b,c,d$, and $e$. The objects $A_1,A_2$, and $A_3$ aren’t ordinary sets; they’re so-called fuzzy sets. Instead of objects being definitely in or definitely not in them, objects have some degree of membership between $0$ and $1$. The idea is that for $i=1,2,3$ the fuzzy subset $A_i$ of $U$ represents the distribution of students who scored very low, low, intermediate, high, and very high on characteristic $S_i$. If over $80\%$ of the group scored at level $x$ (where $x$ is one of $a,b,c,d,e$), then $m_{A_i}(x)=1$; if more then $60\%$ but at most $80\%$ scored at level $x$, then $m_{A_i}(x)=0.75$, and so on, according to the definition of the function on the third page of the PDF. For example, if $20\%$ of the students scored very low on $S_1$, $30\%$ scored intermediate, and the remaining $50\%$ very high, then the membership function for the fuzzy set $A_1$ would be $$f(x)=\begin{cases} 0,&\text{if }x=a\\ 0,&\text{if }x=b\\ 0.25&\text{if }x=c\\ 0,&\text{if }x=d\\ 0.5,&\text{if }x=e\;, \end{cases}$$ since $20\%$ is no more than $\frac15$, $30\%$ is between $\frac15$ and $\frac25$, and $50\%$ is between $\frac25$ and $\frac35$.
H: Is it possible to generalize the Mean value theorem for integral not on compact set? I wonder it is possible to extend the mean value theorem not on compactness. In more detail, Let $f : A \rightarrow \mathbb{R}$ be continuous on $A \subset \mathbb{R}^n$. The mean value theorem for integral states that if $A$ is connected and compact, there exists $x_{0} \in A$ such that $$ \frac{1}{\nu(A)}\int_{A}f(x)dx = f(x_{0}) \;\;\; (\nu(A)\text{ is a volume of A}) $$ Now, the point I am curious is that if I can generalize the mean value theorem for integrals to the case $A$ is not compact, just bounded and connected. I think it is possible, if I can take the supremum and infimum of $f(x)$, which is treaky for me. Could anybody help to extend the idea? Thanks in advance. AI: You can generalise it without much problem. If $A \subset \mathbb{R}^n$ is a connected set, and $f\colon A \to \mathbb{R}$ is continuous, then $f(A)$ is a connected subset of $\mathbb{R}$, i.e. an interval, so $\left(\inf f(A),\, \sup f(A)\right)\subset f(A)$. If $A$ also has finite (positive) measure, then $\inf f(A) \leqslant f(x) \leqslant \sup f(A)$ for all $x\in A$ yields $$\nu(A)\inf f(A) \leqslant \int_A f(x)\,dx \leqslant \nu(A)\sup f(A),$$ i.e. $$\inf f(A) \leqslant \frac{1}{\nu(A)}\int_A f(x)\,dx \leqslant \sup f(A).$$ If the mean value lies strictly between the infimum and the supremum, we know it is an attained value. If the mean value is the supremum, say, and it is finite, then we know that we must have $f(x) = \sup f(A)$ almost everywhere on $A$, so the mean value is again an attained value. Similar for the infimum. The only possibility for the mean value to not be an attained value is when $f$ is not integrable over $A$, so: If $A\subset \mathbb{R}^n$ is connected and has finite positive measure $0 < \nu(A) < \infty$, and $f\colon A \to \mathbb{R}$ is continuous and integrable over $A$, then there is an $x_0 \in A$ with $$\frac{1}{\nu(A)}\int_A f(x)\,dx = f(x_0).$$
H: Ordinals: if $\alpha < \omega^{\beta}$ then $\alpha + \omega^{\beta} = \omega^{\beta}$ I'm trying to proof that if $\alpha < \omega^{\beta}$ then $\alpha + \omega^{\beta} = \omega^{\beta}$, where $\omega$ is the least infinite ordinal. I started with transfinite induction on $\beta$ by proving it for $\beta = 0$ and $\beta =1$. But now I'm having problems with the successor and limit cases. Any hints? Thanks. AI: Note that if $\beta=\gamma+1$ then $\alpha+\omega^{\gamma+1}=\alpha+\omega^\gamma\cdot\omega$. Now if $\alpha<\omega^\gamma$ we are done; otherwise we can write it as $\omega^\gamma\cdot n+\delta$, such that $\delta<\omega^\gamma$. And so $$\alpha+\omega^{\gamma+1}=\omega^\gamma\cdot n+\delta+\omega^\gamma\cdot\omega=\omega^\gamma\cdot n+(\delta+\omega^\gamma\cdot\omega)=\omega^\gamma\cdot n+\omega^\gamma\cdot\omega=\omega^\gamma(n+\omega)=\omega^{\gamma+1}.$$ The limit case can be solved using the fact that $\omega^\delta=\sup\{\omega^\gamma\mid\gamma<\delta\}$.
H: finding the lim of the following set I'm given: $$a_0,\ldots,a_k\ge0$$ for the following set: $$\lim_{x \to \infty} \sqrt[n]{a_kn^k+a_{k-1}n^{k-1}+\cdots+a_1n+a_0}$$ I need to find the limit of the set. Any help appreciated. AI: Let assume $a_k\ne 0$ so since $$n^i=_\infty o(n^k)\quad \forall i<k$$ then $$\sqrt[n]{a_k n^k+\cdots+a_1n+a_0}\sim_\infty \sqrt[n]{a_k n^k}=a_k^{1/n} n^{k/n}\to 1 $$
H: Proving that $R_1 = R_2$ (linearly ordered sets) Let $R_1$ and $R_2$ be linearly ordered sets in set $X$. Prove that if $R_1R_2$ is linearly ordered set, then $R_1 = R_2$ I understand the defnitions of $R_1R_2$ and totally ordered sets, but when I start to prove, it seems that I still lack some knowledge. Definitions don't seem to be enough. AI: $xR_{1}R_{2}z \iff xR_{2}y$ and $yR_{1}z$ for some $y$. Assume that $xR_{2}z$. It is to be shown that $xR_{1}z$ and since $R_{1}$ is linear we have $xR_{1}z$ or $zR_{1}x$. In the first case we are ready. If $zR_{1}x$ then we combine this with $zR_{2}z$ and find $zR_{1}R_{2}x$.Next to that we have $xR_{2}z$ and $zR_{1}z$ leading to $xR_{1}R_{2}z$. Then $x=z$ and consequently $xR_{1}z$. Proved is now that $R_{2}\subset R_{1}$ and likewise it can be shown that $R_{1}\subset R_{2}$.
H: Homework - How many non negative solutions? $a+b+c+d+e = 30$ we know that $10\leq e$ and that $4\leq d \leq 7$ How many non negative solutions does this equation have? AI: Let $a'=a$, $b\,'=b$, $c'=c$, $d\,'=d-4$, and $e'=e-10$; $30-(4+10)=16$, so you’re trying to count solutions in non-negative integers to $$a'+b\,'+c'+d\,'+e'=16\tag{1}$$ with the extra condition that $d\,'\le 3$. Without the extra conditions this is just a stars-and-bars problem; $(1)$ has $$\binom{16+5-1}{5-1}=\binom{20}4$$ unrestricted solutions in non-negative integers. (Both the formula and a decent derivation of it can be found at the link.) Of course some of these violate the upper bounds on the variables and have to be subtracted from this count. That’s a basic inclusion-exclusion problem; my answer to this question illustrates the method in a similar but slightly more difficult problem.

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