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H: How do I find the base angles without a vertex angle in a isosceles triangle? How can I find the base angles in a isosceles triangle if the vertex angle is missing? Normally, I would go: 2x + vertex angle = 180, but now even the vertex is missing, the only thing I have is a line in the middle, with 90 degree angle. Maybe the 90 degree should be used to find one of the base angles and one of the vertex angles, any help is appreciated, thanks a lot, I am still learning this, any tips are very useful. AI: All you need to do is find one base angle, since in an isosceles triangle, both those angles are opposite equal sides, and are therefore equal. Edit: Now that I see your image, You have enough information to find everything. You can use the Pythagorean Theorem to obtain the altitude (the unknown side in the middle.) The vertex, B, is split into two equal angles, but you can easily find the vertex angle by first finding one of the base angles. You can find the base angle B by using the fact that $$\cos B = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}\implies B = \cos^{-1}\left(\dfrac{\text{Adjacent}}{\text{Hypotenuse}}\right) = \cos^{-1}\left(\frac{10.9}{25.6}\right) = 64.8$$ Each base angle, let's call them each of measure $x$, plus the vertex angle add up to $180$. So if you need to solve for the vertex angle $\theta$, we have that $\theta = 180 - 2x$. Also note, to find an angle, given none of the angles are known, you can to use the lengths of the sides of the triangle to solve for the base angles and/or vertex, and can use the Law of Cosines to do that.
H: Unary minus on squared number According to algebra as I know it, $-2^2 = 4$, but most calculators expand this to $-2 * 2 = -4$, which yields a different answer. This is because of the order of precedence. In traditional math, I've seen that the unary operator is applied before exponents. This is not the case for calculators. I have two questions: 1) What caused this difference and why does it persist? 2) Is one method "more accurate" than the other and why? AI: The point at issue, here, is whether $-2^2$ indicates the square of the opposite of $2$ (which is $4$), or whether it indicates the opposite of the square of $2$ (which is $-4$). As a rule, we indicate the former by $(-2)^2,$ and the latter by $-2^2.$ This is because of the order of operations. $-2^2=-4$ because exponentiation occurs before multiplication (by $-1$, in this case), while $(-2)^2=4$ because the multiplication inside the parentheses takes place before the exponentiation. Why would we do this, though? It isn't just convention. If we consider the general case, with an arbitrary real $x,$ and forget all rules of precedence except parentheses, then there are two possible meanings for $-x^2,$ namely: $(-x)^2$ $-(x^2)$ In the first case, this is simply $x^2,$ so not only was there no real need to write the parentheses, there wasn't a need to write the $-$ sign! However, we'd like to have a way to talk about a perfect square's opposite, which the first interpretation quite simply won't allow. That's why we take the second interpretation of $-x^2,$ and use parentheses when we need to square a number's opposite. Ultimately, the order of operations is done for precisely such reasons: to let us have flexibility enough to say what we mean without needing a whole bunch of parentheses.
H: Integer division I think I found a mistake in the princeton review "Cracking the GRE" 2014 edition on page 408. The problem is as follows: If $\frac{13!}{2^x}$ is an integer, which of the following represents all possible values of $x$? a.) $0\le x\le10$ b.) $0\lt x\lt9$ c.) $0\le x\lt10$ d.) $1\le x\le10$ e.) $1\lt x\lt10$ Now princeton review claims the correct answer is d, but $2^0$ is $1$ in my book and $\frac{13!}{1}$ is indeed an integer, thus I would say answer a is correct. Is this really such a glaring error in the book or am I missing something (obvious)? AI: What range of $x$ is permitted? $-1$ works, too. I would accept zero. Typos happen.
H: conditional probability of dependent events I posted another question earlier but i can't modify it so please bare with me. I have a problem where i need to find it's conditional probability: probability of Pc(B)=(B\|C) where the event of B={no two people are born in the same month} and event C={exactly three people were born in the summer of june, july august} and there are 9 people involved. These events are dependents so Pc(B)=P(B\|C)=P(B intersection C)/P(C) I got $P(B)=\frac 1{12} \frac 1{11} \frac 1{10} \frac 19 \frac18 \frac17 \frac 16 \frac 15 \frac 14 =\frac 1{79833600}$ $P(C)=\binom 9 3 \left(\frac 3{12}\right)^3\left(\frac 9{12}\right)^6=\frac{15309}{65536}$ Pc(B)=P(B intersection C)/P(C) where P(B intersection C) = (P(B)*P(C\|B)). The thing is how do i find P(C\|B)? This is still dependent so i feel like i'm right back to where i started. I saw examples talking about aces where a person draws an ace and doesn't put it back(this is event A) so we have 4/54 and then drawing another ace(this is event B) would be 3/53 so P(B\|A) is 3/53... this is simple but in my case i can find the way to do it. AI: Compute $\mathbb{P}(B\cap C)$ by counting. There are $12^9$ ways that each person could be assigned a birthday month. Presumably, we are assuming these events are equiprobable. Now how many of these events fall in $B\cap C$? Well, first choose the person born in June (there are 9 possible choices), then the person born in July (8 possibilities), then the one in August (7 possibilities). There are now 6 people left and 9 months with no one yet assigned. Choose the month for the first of these remaining 6 (9 possibilities), then the next (8 possibilities), etc. That last of these remaining 6 will have 4 months to choose from. So multiplying all the possibilities in all the decisions, we end up with $9\cdot 8\cdot 7 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4$. Clear enough?
H: When is $f_{xy}(x,y)\neq f_{yx}(x,y)?$ When is $f_{xy}(x,y)\neq f_{yx}(x,y)?$, where $f_{xy}$ and $f_{yx}$ denote the mixed (second) partial derivatives of a multivariable function $z=f(x,y)$. AI: It's worth studying the following results: Schwartz Theorem or Young Theorem or Clairault Theorem. And see examples where $f_{xy}\neq f_{yx}$ to see what happens (or not happens!).
H: Dual of this primal optimization problem? How would one find the dual of the following problem? $min_x 1/2 \|\|y-x\|\|_2^2 + \lambda\|\|x\|\|_1 $ Can someone please explain to me how to do this since there are no specific constraints? AI: You could reformulate your problem as \begin{align} \operatorname{minimize}_{x,u} &\quad \frac12 \\|u\\|_2^2 + \lambda \\|x\\|_1 \\ \text{subject to} &\quad u = y - x. \end{align} The Lagrangian is \begin{align} L(x,u,z) &= \frac12 \\|u\\|_2^2 + \lambda \\|x\\|_1 + \langle z,y-x - u \rangle \\ &= \frac12 \\|u\\|_2^2 - \langle z,u \rangle + \lambda \\|x\\|_1 - \langle z,x \rangle + \langle z,y\rangle. \end{align} The dual function is \begin{align} g(z) &= \begin{cases} -\frac12 \\| z \\|_2^2 + \langle z,y \rangle \quad \text{if } \\|z\\|_{\infty} \leq \lambda \\ -\infty \quad \text{otherwise}. \end{cases} \end{align} The dual problem is \begin{align} \operatorname{maximize}_z & \quad -\frac12 \\| z \\|_2^2 + \langle z,y \rangle \\ \text{subject to} &\quad \\|z\\|_{\infty} \leq \lambda. \end{align}
H: From a Hilbert space to another Hilber space, is a norm preserving invertible transformation also unitary? Given two Hilber spaces - $H_1, H_2$ and a transformation $T:H_1 \to H_2$ that is norm preserving and invertable, does this imply that $T$ is also unitary transformation, namely that it preserves the inner product? [or equivalently: is it true that - $a,b, \in H_1 \langle a,b\rangle_{H_1} = \langle T(a),T(b)\rangle_{H_2}$] Any comments are welcome! AI: Yes if $T$ is linear, because the norm determines the inner product via the polarization identity $$ \langle x,y\rangle = \frac14 \sum_{n=0}^3 i^n \\|x+i^ny\\|^2 $$ No if $T$ is allowed to be nonlinear: a counterexample is given by $(r,\theta)\mapsto (r,\theta +\frac12 \sin\theta)$ in polar coordinates in the plane.
H: Prove uniform converging for $\{f_n(x)\} = x^n$, The definition says that: Suppose $D \subseteq R$. $\{f_n(x)\}$ uniform converging to $f(x)$ in $D$ if any $\epsilon > 0$ , $\exists N(\epsilon)$, such that for any $n > N(\epsilon)$ and any $x \in D$: $\|f_n(x) - f(x)\| < \epsilon$ I need to prove uniform convergence for $\{f_n(x)\} = x^n$, where $0 \leq x \leq \frac{1}{2}.$ Can you please solve this in detail? I couldn't find examples on youtube so I'm here wish you can help me. thanks! AI: Here's a rough sketch for how you might determine $N(\epsilon)$. Note this is definitely not in the form of a rigorous proof and there are several unjustified statements left for you to think about. We can show that $f_n(x) \to 0$ as $n \to \infty$ for all $x$ in $D = \lbrack0, 1/2\rbrack$. Given $\epsilon > 0$, you need to determine an $N = N(\epsilon)$ such that $n \geq N$ implies for all $x \in D$ that $\|f_n(x) - f(x)\| = x^n < \epsilon$. Since $\|f_n(x) - f(x)\| = x^n$ is maximized in $D$ at $x = 1/2$ (why?), you need $N$ large enough that $(1/2)^N < \epsilon$ (which will in turn make $(1/2)^n < \epsilon$ for $n \geq N$). So $$ N > -\log_2 \epsilon $$ must hold. Hence you can choose $N = \max\{\lceil -\log_2 \epsilon \rceil, 1\}$.
H: The empty function and constants On Wikipedia and also after searching this forum there is stated that for each $A$ there exists a unique function $f : \emptyset \to A$ called the empty function for $A$. In the theory of Algebraic Data Types (ADT's) and in functional programming languages (like Haskell) where one paradigma is "everything is a function" it is common to interpret constants as $0$-ary functions. A nullary function is just a function of the form $f : \emptyset \to A$, and so for every $A$ there must be such a function under this interpretation (so $\|A\|$ such functions) and this contradicts the uniques of one empty function for every $A$. Also on Wikipedia I read. The n-ary cartesian power of a set X is isomorphic to the space of functions from an n-element set to X. As a special case, the 0-ary cartesian power of X may be taken to be a singleton set, corresponding to the empty function with codomain X. So $X^0 = \{ x \}$, but otherwise if I interpret $Y^X$ as the set of functions from $X$ to $Y$ and interpret $X^n$ as $X^{\{1,\ldots,n\}}$ then $X^0 = X^{\emptyset} = \{ f : \emptyset \to X \}$? Could someone please clarify, I am confused... EDIT: Okay guess I got it now, but still confused. For every $A$ there is a unique $f : \emptyset \to A$, but that is not equal to $f : A^0 \to A$, with $A^0 = \{ f : \emptyset \to A \}$ there are exactly $\|A\|$ functions $f : A^0 \to A$. But then I guess the following one statement about the nullary product on Wikipedia is wrong, that if the index set $I$ is empty, that $$ \prod_{\emptyset} = \{ f_{\emptyset} : \emptyset \to \emptyset \}. $$ because there the set over which the product is built is lost? AI: A function $f$ from set $X$ to set $A$ is a triple $\left(X,G,A\right)$ where $G\subset X\times A$ and such that $\forall x\in X\exists!a\in A\;\left(x,a\right)\in G$. If here $X=\emptyset$ then automatically $G=\emptyset$ and the mentioned condition is satisfied vacuously. So $\left(\emptyset,\emptyset,A\right)$is a function from $\emptyset$ to $A$ and is automatically unique. Your definition $A^{0}=A^{\emptyset}=\left\{ f\mid f:\emptyset\rightarrow A\right\} $ is correct. It contains for every $A$ (not every element of $A$) exactly one element. Elements of $A$ can be identified with functions $f:\rightarrow A$ where $$ denotes a fixed singleton (not the empty set). Taking for instance $*=\left\{ 0\right\} $ we have the function $\left(\left\{ 0\right\} ,\left\{ \left(0,a\right)\right\} ,A\right)$ corresponding with element $a\in A$. EDIT We can define $\sqcap_{i\in I}X_{i}$ as the set of functions $f:I\rightarrow\cup_{i\in I}X_{i}$ with $\forall i\in I\; f\left(i\right)\in X_{i}$, but doing so it must be kept in mind that here we are not defining the product, but a product of the family of $\left(X_{i}\right)_{i\in I}$ and this with projections $p_{i}$ defined by $f\mapsto f\left(i\right)$. If we loose that out of sight than ambiguity can arise. For instance if $\forall i\in I\; X_{i}=X$ then the product can be abreviated as $X^{I}=\mathbf{Set}\left(I,X\right)$ . If $I\neq\emptyset$ then this set equals the set $\sqcap_{i\in I}X_{i}$ defined as above. However in special case $I=\emptyset\wedge X\neq\emptyset$ it does not, because $\cup_{i\in I}X_{i}=\emptyset\neq X$. As magma states well in his comment: products are defined up to isomorphism. In category $\mathbf{Set}$ any singleton can serve as product of an empty family. The singletons are in fact the terminal objects there.
H: How to show that a given set is a vector space? I am having some issues with this problem in my Linear Algebra textbook. The goal is to either show that the given set, $W$, is a vector space, or to find a specific example to the contrary: \begin{Bmatrix} \begin{bmatrix} a\\ b\\ c\\ d \end{bmatrix} : \begin{matrix} 3a + b = c\\ a + b + 2c = 2d \end{matrix} \end{Bmatrix} I understand the basic properties of Vector Spaces - such as having to contain the zero vector, being closed under addition, and being closed under scalar multiplication. I have no problem proving when these sets are not vector spaces, for example if they do not contain the zero vector. This set appears to contain the zero vector (if you plug in $0$ for $a, b, c, d$ the equations are consistent). But I'm not quite sure how to prove that this set is a vector space, or how to prove that it is closed under addition and scalar multiplication. Thanks for your help. AI: If you have a vector satisfying the two constraints then multiplying by $k$ you get $$3a+b=c \implies 3(ka)+(kb)=(kc)$$ $$ a + b + 2c = 2d \implies (ka) + (kb) + 2(kc) = 2(kd)$$ so you have closure under scalar multiplication. Similarly for addition $$3a_1+b_1=c_1 \text{ and }3a_2+b_2=c_2 \\ \implies 3(a_1+a_2)+(b_1+b_2)=(c_1+c_2)$$ $$a_1 + b_1 + 2c_1 = 2d_1 \text{ and }a_2 + b_2 + 2c_2 = 2d_2 \\ \implies (a_1+a_2) + (b_1+b_2) + 2(c_1+c_2) = 2(d_1+d_2)$$
H: What could be the lead to prove $\|\|X\|\|_2 \leq \|\|X\|\|_F \leq \sqrt{rank(X)}\|\|X\|\|_2$? In the above statement, $\|\|X\|\|_2$ = $L_2$ norm of X and $\|\|X\|\|_F$ = $Frobenius$ norm of X. It appears to me that the $L2$ norm of X and $Frobenius$ norm of X are the same. How should i proceed to prove the above statement? AI: Consider the singular values $\sigma_1,\ldots,\sigma_r$ of $X$, where $r=\mathrm{rank}(X)$, such that $\sigma_1\geq\cdots\geq\sigma_r>0$ and the expressions $$ \\|X\\|_2=\sigma_1,\quad \\|X\\|_F=\sqrt{\sigma_1^2+\cdots+\sigma_r^2}. $$
H: Spectrum of a ring is irreducible if and only if nilradical is prime (Atiyah-Macdonald, Exercise 1.19) Can anyone help me with this exercise, please? A topological space $X$ is said to be irreducible if $X\neq\emptyset$ and if every pair of non-empty open sets in $X$ intersect, or equivalently, if every non-empty open set is dense in $X$. Show that $\text{Spec}(A)$ is irreducible if and only if the nilradical of $A$ is a prime ideal. Notation: $A$ is a commutative ring with $1$ (not necessarily $1\ne0$) $\eta= \text{nilradical of $A$ }= \bigcap\limits_{\mathscr{p}\text{ prime}}\mathscr{p}=\{a\in A:\text{$a$ is nilpotent}\}$ $\text{Spec}(A)=\{p\subset A:\text{$p$ prime}\}$, and the topology is such that $V(E)=\{p\subset A\text{ prime}:E\subset A\}$ is a basis for closed sets, for all subset $E\subset A$ (we can show that the complementar of these sets form a basis for open sets) If the nilradical $\eta=\mathscr{p}$ is prime, then every non-empty closed set $V(E)$ satisfy: "$p\in V(E)\implies V(E)=\text{Spec}(A)$" (since every prime contains $\eta=p$), hence, every non-empty open set contains $p$, so $\text{Spec}(A)$ is irreducible. The conversely is the problem... A previous exercise showed that there exists minimal prime ideals in every ring $A$. I assumed that $\eta$ is not a prime ideal, hence there exists at least two distinct minimal prime ideals. So, let $p$ be a minimal prime ideal and $E=\bigcap\{q\subset A:\text{$q$ is prime minimal, $q\ne p$}\}$. If there are a finite number of minimal prime ideals (for example, if $A$ is Noetherian), then the complementar of $V(E)$ is contained in $V(p)$ (since if a finite intersection of prime ideals is contained in any ideal $I$, then at least one of these prime ideal is contained in $I$), hence, $\text{Spec}(A)$ is not irreducible. But this argument seems not to work for general rings... Any help will be appreciated! Thanks! AI: I think the open sets definition of irreducibility is easier to work with. You should show these useful facts about $Spec(A)$: the sets $D(f) = \{\mathfrak{p} \in Spec(A) : f \notin \mathfrak{p}\}$ form a basis of the topology of the spectrum, and $D(f) \cap D(g) = D(fg)$. Then we suppose that $f \notin Nil(A)$ and $g \notin Nil(A)$. This means that $D(f)$ and $D(g)$ are then nonempty open sets, and so if $Spec(A)$ is irreducible, their intersection $D(fg)$ is nonempty.
H: Can the inverse of a function be the same as the original function? I was wondering if the inverse of a function can be the same function. For example when I try to invert $g(x) = 2 - x$ The inverse seems to be the same function. Am I doing something wrong here? AI: You're correct. A function that's its own inverse is called an involution. Edit: Oh let's have some fun. :) What are some other functions that are easy to check are involutions? I've cherry picked some of my favorites in what follows, both from memory and also the references I provide below. First, note that there's an easy test to determine whether or not $f$ is an involution. Namely, since $f^{-1} = f$, you just need to double check that $f(f(x)) = x$ for all $x$ in the domain of $f$. This can be used to verify all three of the following examples are actually involutions. Your function $g(x)$ generalizes to a whole class of involutions! Namely, $$ f(x) = a-x $$ is an involution for any real number $a$. In particular, $f(x) = 0 - x = -x$ is an involution (as is $f(x) = x$, of course). As someone already pointed out, $f(x) = 1/x$ (defined for all real $x \neq 0$) is also an involution. More generally, for any real $a$ and $b$ the function $$ f(x) = a + \frac{b}{x-a} = \frac{ax + (b-a^2)}{x-a} $$ satisfies $$ f(f(x)) = a + \frac{b}{a + \frac{b}{x-a} - a} = a + (x-a) = x $$ for all real $x \neq a$, and as such is also an involution on this domain. Here's a less obvious (but cool) example. Consider the function $f(x) = (a - x^3)^{1/3}$. You can check this directly that $f(f(x)) = (a - ((a-x^3)^{1/3})^3)^{1/3} = x$. This is an example of a large class of involutions generated by a special type of symmetric function $F(x,y)$ (as explained here). Fun Fact: The only continuous, odd ($f(-x) = -f(x)$ for all $x$) involutions with domain $(-\infty,\infty)$ are $f(x) = \pm x$. (A short proof of this fact is given here.) There are many, many, more of these functions, and they occur naturally/are useful tools in many branches of mathematics.
H: Computing a formula for $\partial^2 f/\partial{v}\partial{w}$ Let $f(x,y,z)$ be of class $C^2$. Putting $x = u + v - w$, $y = 2u - 3v$, $z = v + 2w$ makes f into a function of u, v, and w. Compute a formula for $$\frac{\partial^2f}{\partial{v}\partial{w}}$$ in terms of the partial derivatives of f with respect to x, y, and z. Here is my go at it: $$ = \frac{\partial}{\partial{v}}( \frac{\partial{f}}{\partial{w}}) \\ = \frac{\partial}{\partial{v}} (\frac{\partial{f}}{\partial{x}}\frac{\partial{x}}{\partial{w}} + \frac{\partial{f}}{\partial{z}} \frac{\partial{z}}{\partial{w}}) \\ = \frac{\partial}{\partial{v}} (-\frac{\partial{f}}{\partial{x}} + 2\frac{\partial{f}}{\partial{z}}) \\ = -(\frac{\partial}{\partial{v}}\frac{\partial{f}}{\partial{x}}) + 2(\frac{\partial}{\partial{v}}\frac{\partial{f}}{\partial{z}}) \\ = -(\frac{\partial^2{f}}{\partial{x^2}}1) + 2(\frac{\partial^2{f}}{\partial{z^2}}1) \\ = 2\frac{\partial^2{f}}{\partial{z^2}} - \frac{\partial^2{f}}{\partial{x^2}} $$ I have two questions. 1) Is the statement I wrote true? Are all the operations I did value? 2) Did I actually answer the original question? AI: Since $\dfrac{\partial f}{\partial w}$ is in terms of $x$, $y$, and $z$, you'll need to be careful with how you rewrite $\dfrac{\partial}{\partial v}$. Since we know that $$\begin{aligned}\frac{\partial f}{\partial w} &= \frac{\partial x}{\partial w}\frac{\partial f}{\partial x} + \frac{\partial y}{\partial w}\frac{\partial f}{\partial y} + \frac{\partial z}{\partial w}\frac{\partial f}{\partial z} \\ &= \left(\frac{\partial x}{\partial w}\frac{\partial}{\partial x}+\frac{\partial y}{\partial w}\frac{\partial}{\partial y} + \frac{\partial z}{\partial w}\frac{\partial}{\partial z}\right)f\end{aligned}$$ it follows that $$\frac{\partial}{\partial w} = \frac{\partial x}{\partial w}\frac{\partial}{\partial x} + \frac{\partial y}{\partial w}\frac{\partial}{\partial y} + \frac{\partial z}{\partial w}\frac{\partial}{\partial z}$$ Hence, we similarly have that $$\frac{\partial}{\partial v} = \frac{\partial x}{\partial v}\frac{\partial}{\partial x} + \frac{\partial y}{\partial v}\frac{\partial}{\partial y} + \frac{\partial z}{\partial v}\frac{\partial}{\partial z}$$ and thus $$\frac{\partial^2f}{\partial v\partial w} = \frac{\partial x}{\partial v}\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial w}\right) + \frac{\partial y}{\partial v}\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial w}\right) + \frac{\partial z}{\partial v}\frac{\partial}{\partial z}\left(\frac{\partial f}{\partial w}\right)$$
H: Convergence of "sliced" power series Let $\phi(t)=\sum_{k=1}^\infty a_k t^k$, $x=t^m \in \mathbb{C}$ for some fixed $m\in \mathbb{N}$ be a convergent power series. I guess that $a_0=0$. For $r=0,\ldots,m-1$ and $k=mq+r$, why are the power series $$\tilde\phi_r(x) = \sum_{q=0}^\infty a_{mq+r}x^q$$ convergent? The issue is that I cannot deduce from the convergence of the first series (and a probably associated convergence radius $r$) anything that could provide the convergence of the second one. AI: The Cauchy-Hadamard formula for the radius of convergence does the trick. Suppose $$\frac{1}{R_0} = \limsup_{k\to\infty} \lvert a_k\rvert^{1/k} < \infty$$ (so the original series converges somewhere). Then $$\begin{align} \limsup_{q\to\infty} \lvert a_{qm+r}\rvert^{1/q} &= \limsup_{q\to\infty} \left(\lvert a_{qm+r}\rvert^{1/(qm+r)}\right)^{m+r/q}\\ &\leqslant \limsup_{q\to\infty}\left(\lvert a_{qm+r}\rvert^{1/(qm+r)}\right)^m \cdot \limsup_{q\to\infty} \left(\lvert a_{qm+r}\rvert^{1/(qm+r)}\right)^{r/q}\\ &\leqslant \frac{1}{R_0^m}\cdot 1, \end{align}$$ thus the radius of convergence of the slice is at least the $m$-th power of the radius of convergence of the full series.
H: What is the easiest way to solve the following linear system What is the easiest way to solve the following linear system for $a,b,c,d,e$ in terms of $h$? I want to do this quickly. Are matrices the way to go? $a+b+c+d+e=0, (-a+c+2d+3e)h=1, a+c+4d+9e=0, -a+c+8d+27e=0, a+c+16d+81e=0$ Thanks! AI: Yeah: Gaussian elimination I reckon.
H: Establishing an Upper Bound for a Curious Function Suppose I have a sequence of positive real numbers $a_1, \, a_2, \, \dots \,a_n$ such that the following is satisfied: $\sum \limits_{i=1}^{n}a_i = 1$ I am trying to find the smallest value of $L$ such that the following is also true, for any arbitrary sequence: $L \geq - \log_n \left( \sum \limits_{i=1}^{n}a_i^{\, 2} \right)$ My question: Is $L$ a constant, or is it best represented as a function of $n$? AI: By the Cauchy-Schwarz inequality, the condition $\sum a_i = 1$ implies $$1 = \sum_{i=1}^n 1\cdot a_i \leqslant \sqrt{\sum_{i=1}^n 1^2} \cdot \sqrt{\sum_{i=1}^n a_i^2},$$ so $$\sum_{i=1}^n a_i^2 \geqslant \frac1n.$$ The bound is sharp, $a_i = \frac1n$ attains it. Therefore $$-\log_n \sum_{i=1}^n a_i^2 \leqslant - \log_n \frac1n = 1.$$
H: find infimum of $(n+1)^2 / 2^n$ (using Bernoulli's inequality) What is the infimum for: $$(n+1)^2 / 2^n$$ I've tried to simplify $2^n$ with Bernoulli's inequality $((1+1)^n \ge (1+n))$ but it didn't work out.. AI: Hint: $$\frac{(n+2)^2}{2^{n+1}}\cdot\frac{2^n}{(n+1)^2} \to \frac12 < 1.$$
H: Residue of a complex function at some pole. How can one visualize residue of a complex valued function at some given pole? I know how to find it. but I want to know its significance and its geometric nature. Why do we study it? thank you. AI: If you take a complex function $f(z)$, we call $\bar{f(z)}$ the polya vector field of $f(z)$. If $f$ is analytic, the polya vector field is divergenceless and irrotational, meaning that it can be thought of as fluid flow or an electric field. In this interpretation, a residue is just a point charge, with the magnitude of the residue measuring the magnitude of the charge. Similarly for fluid flow, a residue measures a source or sink at a given point. See this answer of mine for more information.
H: When does $f(a),f(f(a)),f(f(f(a)))...$ produce better and better approximations to $x=f(x)$? I tried to approximate the solution to $x=f(x)$ for some given $f$, by guessing $x=a$, then I observed that $x=f(a)$ was an even better approximation, and $x=f(f(a))$ and so on was even better, so why does this method work and for which f, is it sufficient that f is continuous? AI: This method certainly works if the function is Lipschitz continuous with Lipschitz constant $L<1$. In general, it is not true. For example let $f(x)=x+1$, then for any starting value $x_0$, the sequence $x_n=f(x_{n-1})$ ($n\geq1$) diverges.
H: How can I prove that this function is uniformly continuous? How to show that $f(x)=\frac{x}{1+\|x\|}$ is uniformly continuous? Thank you. Also, how do I become good at writing these proofs? AI: You can show that it is differentiable, and has bounded derivative. Then use Lagrange's mean value theorem.
H: Subgroup $\{(1),(12)\}$ in $S_3$ is not kernel of any homomorphism How do I prove the following statement Subgroup $\{(1),(12)\}$ in $S_3$ is not kernel of any homomorphism. AI: $(1\,2\,3)^{-1}\circ (1\,2)\circ (1\,2\,3)=(1\,3)$ is not in this subgroup, hence it is not normal.
H: Pointwise lower bound for holomorphic functions on the unit disk I am looking at Problem 8C in August 2012 complex analysis qualifying exam from University of Wisconsin-Madison (direct link to the file). Let $f$ be a holomorphic function on the unit disc $\mathbb{D}$. Fix $z_0\in\mathbb{D}$. Suppose that $f(0) = \frac{1}{2}$, $f$ does not vanish on $\mathbb{D}$ and $\|f(z)\|\leqslant 1$. Show that $\|f(z_0)\| > c$ for some positive constant $c$ independent of $f$. Here $\mathbb{D}$ denotes the open unit disk centred at the origin, as usual. The part "independent" of $f$ is the tricky part for me. So if I understand the problem correctly, suppose $z_0=i/2\in\mathbb{D}$. Then there is no way to find a sequence of holomorphic functions $f_{n}$ such that $f_{n}(0)=\frac{1}{2}$, $0<\|f(z)\|\leqslant 1$ for all $z\in\mathbb{D}$, and that $\|f_{n}(i/2)\|\to 0$ as $n\to\infty$. I tried using Schwarz Lemma, but to no avail. I would appreciate any hints. :) AI: Hints: Consider $\log f$, normalised to $\log f(0) = -\log 2 \in \mathbb{R}$. Compose with a Möbius transformation. Schwarz lemma. Transform back to get a bound on $\lvert \log f(z_0)\rvert$. Assuming the OP has meanwhile completed the proof, an elaboration won't harm: Since $\mathbb{D}$ is simply connected and $f$ has no zero, holomorphic logarithms of $f$ exist. Let $g$ be such that $g(0) = \log\frac12$ is real. Since $\lvert f(z)\rvert < 1$ for $z\in\mathbb{D}$, we have $\operatorname{Re} g(z) < 0$ for $z\in\mathbb{D}$. The Möbius transformation $$T(w) = \frac{w + \log 2}{w-\log 2}$$ maps the left half plane biholomorphically to the unit disk, and $T(\log\frac12) = 0$, so $h = T\circ g$ is a holomorphic function $\mathbb{D} \to\mathbb{D}$ with $h(0) = 0$. By the Schwarz lemma, we have $\lvert h(z_0)\rvert \leqslant \lvert z_0\rvert$. Then $$g(z_0) = T^{-1}(h(z_0)) = \log 2\cdot \frac{h(z_0)+1}{h(z_0)-1}$$ can be bounded $$\lvert g(z_0)\rvert \leqslant \log 2 \cdot \frac{1+\lvert z_0\rvert}{1-\lvert z_0\rvert}$$ and $$\left\lvert f(z_0)\right\rvert = \left\lvert e^{g(z_0)}\right\rvert = e^{\operatorname{Re} g(z_0)} \geqslant e^{-\lvert g(z_0)\rvert} \geqslant \exp \left(-\log 2 \cdot \frac{1+\lvert z_0\rvert}{1-\rvert z_0\rvert}\right).$$ The bound is sharp, it is attained for $f = \exp \circ T^{-1} \circ \rho$, where $\rho$ is the rotation mapping $z_0$ to $\lvert z_0\rvert$.
H: Limit of sequences. Could someone tell me how one could work out the limit of the following: 1) Ax= (1.001)^x) / (x^1000) as x->∞ 2.) Bx=( 1^100 +2^100+......n^100) ^1/n as n->∞ . I know the final answers but I can't really show how do you get there step by step. Any help? AI: Hint: $$1\le\left(1^{100}+2^{100}+\ldots +n^{100}\right)^{1/n}\le(n^{101})^{1/n}=\left(\sqrt[n]n\right)^{101}\ldots\text{and squeeze...}$$ For the first one, apply l'Hospital 1000 times: $$\lim_{x\to\infty}\frac{1.001^x}{x^{1000}}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{1.001^x\log 1000}{1000x^{999}}\stackrel{\text{l'H}}=\lim_{x\to\infty}\frac{1.001^x\log^21000}{1000\cdot999x^{998}}\stackrel{\text{l'H}}=\ldots$$
H: Sampling with replacement events vs. probability of coverage I have a deck of $N$ cards, when $k \leq N$ of the cards bear a mark. I sample from the deck uniformly and with replacement until I find a marked card. I then erase the mark, and place the card back in the deck. The expectation for the number of sampling events necessary to erase all of the marks is not too difficult to derive using the negative binomial distribution, but what is the probability that all $k$ marks are erased as a function of the number of sampling events? AI: If I understand you correctly, you're now talking about the situation where you take a sample of given size $M$ with replacement, erasing a mark whenever it is found, and you want the probability $P(M)$ that all $k$ originally-marked cards are sampled. The probability that a given set $S$ with $\|S\|$ members escapes the sampling is $(N - \|S\|)/N)^M$. By inclusion-exclusion the probability that at least one escapes is $$ \sum_S (-1)^{\|S\|-1} \left(\dfrac{N - \|S\|}{N}\right)^M $$ where the sum is over all nonempty subsets of $\{1,\ldots,k\}$ There are $k \choose s$ such subsets with $\|S\| = s$, $s = 1 \ldots k$, so $$P(M) = 1 - \sum_{s=1}^k (-1)^{s-1} {k \choose s} \left(\dfrac{N-s}{N}\right)^M$$
H: polynomial ring over finite field Can someone provide a proof for the following? I wrote a one page proof. I must be doing something wrong. There has to be a quicker way to prove this. Thank you ahead of time. I will highly rate for your help. AI: I'm curious what approach you took in your proof, as this would help in saying what you're doing wrong—though I've often seen students write very long proofs which are basically correct and elegant, but have unnecessary steps, or prove the same thing several times in different forms, et cetera. Anyway, in principle this should be easy. First, let's give this map a name: $\psi : \mathbb{F}_p [x] \to F$ is defined by $\psi(x) = a$. Surjectivity is easy: each nonzero element of $F$ is some power $a^k$, and $\psi(x^k) = a^k$. $\psi(0) = 0$, so $\psi$ is surjective. By the theorem that's usually called something like The First Isomorphism Theorem, we have $F \cong \mathbb{F}_p [x] / I$ for some ideal $I$, namely $I = \ker{\psi}$. Since $\mathbb{F}_p [x]$ is a principal ideal domain (which probably requires the Euclidean Algorithm to prove, if you don't have it already), $I = (g(x))$ for some $g$. Let $m = \deg{g}$. $I$ must be a prime ideal, since its quotient is an integral domain. This forces $g(x)$ to be irreducible; otherwise $g(x) = g'(x)g''(x) \in I$, and some polynomial of degree $<m$ would lie in $I$, contradiction. Figuring out the degree of $g$ is immediate with some basic field theory, but we can also get it from first principles. An element of $\mathbb{F}_p [x] / (g(x))$ is determined uniquely by its remainder on division by $g(x)$. So the size of this ring is exactly the number of polynomials of degree $<m$, which is $p^m$ (each of $m$ coefficients is arbitrarily chosen from $\mathbb{F}_p$). Since $\| F\| = p^n$, we have $p^n=p^m$, so $n=m$. Well, this is getting close to page maybe—there are a lot of tiny things to check. Verbosity can be a good thing. :) The final part is to show that $g$ divides $x^q-x$. The point is that $\psi(x^q-x) = a^q-a=0$. But earlier, we agreed that $I=(g(x))$ was the kernel, so $x^q-x \in (g(x))$, which is what we need. Clear?
H: Is $\{w \in \Sigma ^* : \|w\|_2 mod 4 = 2\}$ a regular language? Given alphabet $\Sigma = \{1,2,3\}$, is $\{w \in \Sigma^* : \|w\|_2 \bmod 4 = 2\}$ a regular language? I tried so hard on finding a regular expression but couldn't... AI: /([13]2[13]2[13]2[13]2)[13]2[13]2[13]/
H: Residue of $\frac{1}{(e^z-e)^3}$ at $z = 1$ I'm trying to calculate the residue of $\dfrac{1}{(e^z-e)^3}$ at $z = 1$. The answer is $\dfrac{1}{e^3}$, but having trouble seeing how one would arrive at that. Any hints? AI: If $z=1+u$ and $u\to0$, then $f(z)=(\mathrm e^z-\mathrm e)^{-3}=\mathrm e^{-3}(\mathrm e^u-1)^{-3}$, that is, $f(z)=\mathrm e^{-3}u^{-3}g(u)$ with $g(u)=(1+\frac12u+\frac16u^2+o(u^2))^{-3}$. Since $(1+v)^{-3}=1-3v+6v^2+o(v^2)$ when $v\to0$, this implies that $$ g(u)=1-\tfrac32u-\tfrac12u^2+6\left(\tfrac12u\right)^2+o(u^2), $$ that is, $$f(z)=\mathrm e^{-3}(u^{-3}-\tfrac32u^{-2}+u^{-1}+o(u^{-1})). $$ In particular the residue of the function $f$ at $z=1$ is the coefficient of $u^{-1}$ in its Laurent expansion, that is, $\mathrm e^{-3}\cdot1$.
H: Find the coordinates of intersection of a line and a circle There is a circle with a radius of $25$ ft and origin at $(0, 0)$ and a line segment from (0, -31) to (-37, 8). Find the intersections of the line and circle. I am asking for somebody to analyze what I am doing wrong in calculating the answer, given the question above in its exact format. I almost never get an answer correct, and I would like an explanation as to why this appears to be the case. I begin with finding the formula of the line I used the slope formula, $\Delta y \over{\Delta x}$ to get ${-31 - 8 \over{0 - (-37)}} = {-39 \over{37}}$, which becomes $1.054$, so that $y = 1.054x + b$ $(0, -31)$ is the y-intercept when $x = 0$, so $y = 1.054x - 31$. Formula of circle The format I learned is $(x - h)^2 + (y - k)^2 = r^2$. Substitute known info: $(x - 0)^2 + (y - 0)^2 = 25$ Simplify: $x^2 + y^2 = 25$ Substitution I took the circle's equation $x^2 + y^2 = 25$ and plugged in the value of y: $x^2 + (0.621x - 31)^2 = 25$. I expanded and rounded it into $x^2 + 1.111x^2 + -65.348x + 961 = 25$. It simplifies to $2.111x^2 - 65.348x + 936$. Quadratic formula The template: $x = {-b \pm \sqrt{b^2 - 4ac }\over{2a}}$ is used by plugging in values: ${65.348 \pm \sqrt{4270.361 - 42.111936 }\over{2*2.111}} = {65.348 \pm \sqrt{4270.361 - 7903.584}\over{4.222}} = {65.348 \pm \sqrt{-3633.223}\over{4.222}}$. Either I do not know what to do next or (as I suspect) this is the unfortunate answer to the problem. The problem continues with questions about how long somebody is in the circle if they follow the line at a certain rate, which leads me to believe that there are real roots, but I certainly cannot find them. Edit: slope and template errors corrected, answer is still incorrect. AI: Check your slope again: It is $-\frac{39}{37}\approx-1.05$. Check your circle, it is $$x^2+y^2=625,$$ that is, $r^2=25^2=625$. I think you can get it from there, although I recommend you use fractions throughout until the end. For my final gift, here is a plot. If you hover over the intersection points, it will tell you what they are so you can verify your work.
H: Using Wolfram Alpha to solve a system of linear equations How do I input the below system of equations in Wolfram Alpha in order to solve for the unknowns? I'm wondering if there's some kind of code that can be written in order to make wolfram alpha understand what I'm talking about. $$\left.\begin{matrix} a+b+c+d+e = 0\\ (-a+c+2d+3e) h = 1\\ (a+c+4d+9e)\frac{h^2}{2}= 0\\ (-a+c+8d+27e)\frac{h^3}{6}=0\\ (a+c+16d+81e)\frac{h^4}{24}=0\\ \end{matrix}\right\}$$ I've tried many things already, like solve{a+b+c+d+f=0, (-a+c+2d+3f)h=1, (a+c+4d+9f)h^2/2=0, (-a+c+8d+27f)h^3/6=0, (a+c+16d+81f)h^4/24=0,[a,b,c,d,e]} Thanks for any help. AI: You have 6 variables but only 5 equations, so there should be a free variable. Mathematica did this just fine for me. What I entered was Solve[ a + b + c + d + e == 0 && (-a + c + 2 d + 3 e) h == 1 && (a + c + 4 d + 9 e) h^2/2 == 0 && (-a + c + 8 d + 27 e) h^3/6 == 0 && (a + c + 16 d + 81 e) h^4/24 == 0, {a, b, c, d, e, h}] which gave me the solution $$a =\frac{-1}{4h},b=\frac{-5}{6h},c=\frac{3}{2h}d=\frac{-1}{2h},e=\frac{1}{12h}$$ with a simple warning. I believe it was just the syntax of how you entered it. Mathematica is a useful tool but is very picky. Their help function always has useful examples and a good place when you have trouble is to start by seeing how they demonstrate approaching things.
H: Where have you seen this topological ring used? Let $R$ be a ring and define the topology on $R$ to be that where the open sets are unions of ideals. Then this forms a topological ring - let me know if you need proof. Have you seen it used anywhere or what can you do with it now that you know it's a topological ring? Claim 1. The unions of ideals forms a topology on $R. \ $ Proof. $R$ itself and $\varnothing$ are both unions of ideals, $\varnothing$ considered as the empty union. Let $I = \bigcup I_i$ and $J = \bigcup J_i$, be unions of ideals. Then $I \cap J =\bigcup_{i,j} I_i \cap J_j$. Prove that as an exercise. Since an arbitrary union of an arbitrary union is an arbitrary union, we're done. Claim 2. The multiplication map on $R\times 2$, $(x,y) \mapsto xy$ is continuous. Note that since ideals form a basis for the topology, for this particular problem (inverse function image), it suffices to consider the inverse image of a single ideal and prove that it's a union of ideals in $R\times R$. As the inverse image of any open set would then be a union of ideals in $R\times 2$, or open. But are open sets in $R\times R$ ideals? It turns out that they are... Claim 3. The product topology of $R\times r$ induced by the unions of ideals topology of $R$, coincides with the unions of ideals topology of the ring $R\times R$. As an exercise prove that the ideals of $R\times R$ are of the form $I \times J$ for ideals $I,J$ in $R$. Proof. We can come up with an identity for $(\bigcup I_i )\times (\bigcup J_i)$, namely that it equals $\bigcup (I_i \times J_j)$. So by definition of product topology a basis for the product topology is the set of all $U\times V$ where $U,V$ are both unions of ideals of $R$. Then an open set in $R\times R$ is the same thing as a union of ideals of $R\times R$ as we've shown that ideals are of that form. Proof of Claim 2. AI: The problem with this is that $\{0\}$ is an ideal. So if all ideals are open, then all singletons are open and in fact $R$ has the discrete topology. If you want an interesting ring, you have to use a smaller set of open ideals. Edit: Ah, I misinterpreted your question. What you have written down is not even a ring topology, unless $R=0$. Indeed, in a topological ring, opens sets can be translated using addition. But a translate of the zero ideal in a nonzero ring is not an ideal. (Thanks to Ricky for pointing it out.)
H: Which method is the most practical for this volume problem? Find the volume of the solid generated by revolving the region bounded by $$ y = \frac{3}{\sqrt{x^2 + 9}}, x = 4, y-axis $$ Rotated about the y-axis. After drawing the graph, I wanted to simply take the integral in regards to y and do the disk method but then I looked at $\frac{3}{\sqrt{x^2 + 9}}$ and decided that there might be an easier way than that. Am I able to simply use the shell method here instead? Such that: $$ 2 \pi \int_0^4 4 * \frac{3}{\sqrt{x^2 + 9}}$$ Pulling out the numbers up front: $$ 24 \pi \int_0^4 \frac{1}{\sqrt{x^2 + 9}} $$ AI: The shell method should be $$ V = 2\pi \int_0^4 xf(x) \ \mathrm{d}x $$
H: Multiple integral over General Domains This is a rather general question but I am having trouble conceptualizing the integral. Say that we for the function $f(x,y)$ we integrate $y$ over $v(x)$ and $u(x)$ and $x$ over an interval of constants $\left[a,b \right]$. Then what exactly does the integral measure? Since we are integrating over a function is it the area of the function that lies between those points? Thanks in advance. AI: We measure the volume under the surface of $f(x, y)$. See this picture (courtesy of Wikipedia): The rectangular region at the bottom is the domain of integration. The surface at the top is the graph of the function $f(x, y) = x^2 - y^2$. The volume is the double integral.
H: What is the order of this pole at $\pi/2$? Consider: $$f(z)=\frac{\cos(z)}{(z-\pi/2)^4}$$ at $z = \pi/2$. What is the order of the pole? How can I see this? I keep getting $2$, but my textbook says that the answer is $3$. AI: $$ \frac{\cos z}{(z-\pi/2)^4}=\frac{-(z-\pi/2)+\frac1{3!}(z-\pi/2)^3+O((z-\pi/2)^5)}{(z-\pi/2)^4}=\\-\frac1{(z-\pi/2)^3}+\frac1{3!}\,\frac1{z-\pi/2}+O(z-\pi/2) $$
H: When is the formula for the infinite geometric series valid When is the formula $$S_{\infty} = \dfrac{a}{1-r}$$ valid? Does \|$r\| <1$? AI: Since I found this very useful when learning geometric series, I'll show you this to enhance your understanding of the formula: Given a finite geometric sequence, let's say exempli gratia that it has 5 terms. So the terms are $a, ar, ar^2, ar^3, ar^4$. We want to find the sum of this sequence: $x = a + ar + ar^2 + ar^3 + ar^4$ Multiply through by r: $xr = ar + ar^2 + ar^3 + ar^4 + ar^5$ Subtract (2) from (1) to get $x(1 - r) = a - ar^5 \rightarrow x = \frac {a(1 - r^5)}{1-r}$ So that's why the formula is why it is. Thus, when you have an infinite geometric series, it only holds when $\lim_{n \rightarrow \infty} r^n = 0$, which occurs when $\|r\| < 1$
H: Probability of finding empty seats Supposed we are using a round table with 10 seats, there are 7 people who pick their seats randomly. you and your friend are late and you two want to have 2 seats next to each other. Another person is also late and he will arrive before you. One friend call, he assures that there are 2 seats next to each other at that time. What is your probability of having 2 seats next to each other? No one calls, and you are not assured there are 2 seats next to each other at that time (you don't know). What is your probability of having 2 seats next to each other? My approach for 1 is to calculate the number of times that there are 3 seats, then divided by the total number of time that the seats are taken by 7 people. How do we solve /2/ then? Thank you AI: I'm assuming the friend who calls is not the same as the friend who is going to arrive late, while the "other person" is not your friend, will take a seat randomly just like the first $7$, and comes after your friend called. I'm also assuming that when the friend says there are two seats next to each other, that doesn't exclude the possibility that all three vacant seats are adjacent. If all three are adjacent, the eighth person prevents you from getting two together only by taking the middle seat of the three (thus with probability $1/3$), while if there are just two together and a third separate the eighth person has probability $2/3$ of preventing you from getting two together. Now there are just $10$ ways to have three adjacent vacant seats around a $10$-seat table, but $10 \times 6 = 60$ ways to have two adjacent and one separate (i.e. given the positions of the two adjacent vacant seats, there are $6$ possibilities for the third vacant seat). Since all ways of having three vacant seats are equally likely, the probability in (1) is $$ \dfrac{10}{70} \times \dfrac{2}{3} + \dfrac{60}{70} \times \dfrac{1}{3} = \dfrac{8}{21}$$ In (2), after 8 people have chosen seats randomly there are just $10$ cases where the remaining two seats are adjacent, out of a total of ${10}\choose 8$.
H: Convergence of an infinite series of conjugated complex numbers Take the closed form of this Hadamard/Weierstrass product: $$\displaystyle \frac{\sinh(\pi s)}{\pi s} := \prod_{n=1}^\infty \left(1- \frac{s}{0 + n i} \right) \left(1- \frac{s}{{0 - n i}} \right)$$ Easy to see that the complex zeros occur in conjugated pairs at $ni$ and $-ni$ and, similar to the prime-counting function, these can be plugged into an infinite series: $$\displaystyle F(x) := \sum_{n=1}^\infty \left(\frac{x^{ni}}{n i} + \frac{x^{-ni}}{{- n i}} \right)$$ Note that this series only converges for $x>0, x \in \mathbb{R}$. It is easy to see that $F(x)=-F(\frac{1}{x})$. I then put $F(x)$ into the following 'counting' function: $$\displaystyle N(x) := x + F(x)$$ and I like to conjecture that: $\displaystyle\frac{N(x)}{x} \sim 1$, but how could I prove this? AI: Writing $x = e^\xi$, we see that $$f(\xi) = F(e^\xi) = \sum_{n=1}^\infty \left(\frac{e^{in\xi}-e^{-in\xi}}{ni} \right) = 2\sum_{n=1}^\infty \frac{\sin (n\xi)}{n}$$ is periodic, with period $2\pi$. It is the Fourier series of a bounded, piecewise continuous function (the $2\pi$-periodic extension of $\pi-\xi$ on $[0,2\pi)$; thanks to Marko Riedel for remembering that), so the sum is bounded too. Hence $F$ is bounded, and $\dfrac{N(x)}{x} \sim 1$.
H: How to show $\Bbb R$ is Archimedean? Suppose $X$ is a real number such that $X > 0$. We want to show there exists and $n \in \mathbb{N}$ such that $X \geq \frac{1}{n} $. MY attempt: If $X < \frac{1}{n} \; \; \; \forall n $ then $X \leq 0 $ by passing to the limit. Contradiction. Can someone show me a way to show this from scratch? By actually finding an $n$ that does this job? thanks AI: The Archimedianity of $\Bbb R$ can be seen as a corollary of the unboundedness of $\Bbb N$ in $\Bbb R$ with the usual order. That is, pick $x\in\Bbb R_{>0}$. Then $x^{-1}>0$ is not an upper bound for $\Bbb N$, so there is $n\in\Bbb N$ such that $x^{-1}<n$. Inverting, gives $x>n^{-1}$.
H: If this warped coin is flipped 5 times, what is the probability that more heads than tails occur? A warped coin has probability of 0.5 of landing Heads, probability of 0.4 of landing Tails, and probability 0.1 of landing on its Edge. It is flipped 5 times. What is the probability that more Heads occur than Tails? AI: Hint: probability of 3-5 heads, plus probability of 2 heads and 0-1 tails, plus probability of 1 head and 0 tails.
H: finding determinant as an function in given matrix Calculate the determinant of the following matrix as an explicit function of $x$. (It is a polynomial in $x$. You are asked to find all the coefficients.) \begin{bmatrix}1 & x & x^{2} & x^{3} & x^{4}\\ x^{5} & x^{6} & x^{7} & x^{8} & x^{9}\\ 0 & 0 & 0 & x^{10} & x^{11}\\ 0 & 0 & 0 & x^{12} & x^{13}\\ 0 & 0 & 0 & x^{14} & x^{15} \end{bmatrix} Can someone help me with this question? AI: Hint: $$\det\begin{pmatrix}A & 0\\ C & D\end{pmatrix} = \det\begin{pmatrix}A & B\\ 0 & D\end{pmatrix} = \det(A) \det(D)$$ Where $A,B,C,D$ are block matrices.
H: Can check my arithmetic/work on this arc length problem? Find the length of the curve $y^2 = x^3$ from (0,0) to $(\frac{1}{4}, \frac{1}{8})$. I know that the arc length formula for rectangular problems is: $$ \int_a^b \sqrt{1 + (\frac{dy}{dx})^2} $$ So, to get a, I subtract $\frac{1}{4}$ from 0. To get b, I subtract $\frac{1}{8}$ from 0. So now I have my bounds. Now I need take the derivative of my function. $$ y = x^\frac{3}{2} $$ Derivative: $$ y = \frac{3}{2} x^\frac{1}{2} $$ Square it: $$ y = \frac{9}{4}x $$ Plug it into my original equation: $$ \int_\frac{1}{4}^\frac{1}{8} \sqrt{1 + \frac{9}{4}} x $$ This leaves me with $$ \int_\frac{1}{4}^\frac{1}{8} 1 + \frac{3}{2}x $$ Swap a and b around, add negative sign and set up the equation: $$ - [ 1 + \frac{3}{8} ] - [1 + \frac{3}{16} ] $$ Finals result: $$ -\frac{11}{8} - \frac{19}{16} == -\frac{30}{16} == -\frac{15}{8} $$ AI: Not quite. First, the arc length formula is $\displaystyle s = \int_a^b \sqrt{1+(f'(x))^2} dx$, where $a$ is the $x$ value you start with, $b$ is the $x$ value you end with. In this case, $y^2 = x^3$. This gives $y = x^{\frac{3}{2}}$, and $y' = \frac{3}{2} x^{\frac{1}{2}}$. Square that, and you get $(y')^2 = \frac{9}{4} x$. The starting point is $a = 0$, and the ending point is $b = \frac{1}{4}$. Using these, we get $\displaystyle s = \int_0^{\frac{1}{4}} \sqrt{1+\left(\frac{9}{4}x\right)} dx$, which, according to Wolfram Alpha turns out to be $\displaystyle \frac{61}{216}$.
H: How to deduce that $\gcd(a, b) = \gcd(a, a+b)$ given that the common divisors of $a$ and $b$ are the same as the common divisors of $a$ and $a+b$? This homework question asks to first prove that the common divisors of $a$ and $b$ are the same as the common divisors of $a$ and $a+b$, which I have done, and then deduce from this result that $\gcd(a, b) = \gcd(a, a+b)$. Let $c\|a$ and $c\|(a+b)$, therefore $c\|b$. I have tried expanding $\gcd(a,a+b)$ in terms of multiples of $c$ and considering the case when $c = \gcd(a,a+b)$, but it gets me nowhere. Am I overlooking something obvious? AI: If the two sets of common divisors are the same, then the greatest common divisors are the same.
H: who has the higher probability to buy pizza Supposed that A, B, C are given a bunch of homework (N). They decided to share the homework and work individually on each part. A will take 20%, B take 30% and C take 50% of the homework (N). Because A is a good student so he will make 1% error (doing it wrong) in the homework that he is assigned, B is lazy so he will make 5% and C will make 2%. Supposed that the grader is lazy too. He picks a random homework n in the total N homework without looking at it and marks it as wrong. What is the possibility that n (a random picked homework) is truly wrong? Supposed that the random picked homework n is truly wrong. What is the possibility that A, B, C did that? Sorry I paraphrase the question (2), it originally said that "Determine who has a higher possibility to buy pizza for other if he is the one who did that wrong homework." I cannot find a way to solve the number 1. Please advise and help. Thank you. AI: With probability $0.20$, the problem was done by A. Given that it was done by A, the probability it was wrong is $0.01$. So the probability the thing was done by A and wrong is $(0.2)(0.1)$. The other two ways the result can be wrong have probabilities $(0.3)(0.05)$ and $(0.5)(0.02)$. So the probability a randomly chosen question is wrong is $(0.2)(0.01)+(0.3)(0.05)+(0.5)(0.02)$. Remark: You did not ask about the second question. Let $W$ be the event the question is done incorrectly. Let $A$ be the event it was done by A. We want (among other things) $\Pr(A\|W)$. For this the conditional probability definition $\Pr(X\|Y)=\frac{\Pr(X\cap Y)}{\Pr(Y)}$ will be useful.
H: Basis of a subspace of all continuous real-valued functions This is a homework problem I have and I have the right answer, but I don't know how to do it in proof form . . . $W$ is the subspace of all continuous real functions spanned by $\{\cos^2(t),\ \sin^2(t),\ \cos(2t)\}$. Find a basis for $W$. I know the given set is linearly dependent, and I know an answer is the right side of the identity of $\cos^2(t) = \sin^2(t) + \cos(2t)$, but I don't know how to show this as a proof. Believe I found the answer . . . Create the subset of $W$, $\{\sin^2(t), \cos(2t)\}$, and show that it is linearly independent and that $v_1 +v_2 = v_3$. If both of these conditions hold, which they do for this example, then the subset is a basis for the original set. AI: Yes, so basically all we have to prove is that the functions $\cos(2t)$ and $\sin^2(t)$ are linearly independent. Neither of these functions are the zero element (constant $0$ function), so if these were dependent, we must have $\cos(2t)=\lambda\cdot\sin^2(t)$ for a $\lambda$, the equality holds between functions, i.e. holds for all $t$. Now just plug in some values for $t$ to get a contradiction.
H: Use the definition of continuity to prove a function is defined at every nonnegative real number Use the definition of continuity to prove that the function f defined by f(x) = (x)^(½) is continuous at every nonnegative real number. Our definition of continuity: Let I be an interval, let f:I-R, and let c(element of)I. The function f is continuous at c if for each (epsilon)>0 there exists (delta)>0 such that \|f(x)-f(c)\| < (epsilon) for all x (element of) I that satisfy \|x-c\|<(delta). The function f is continuous on I if f is continuous at each point of I. I'm not sure how to the nonnegative real number comes into play here in terms of proving the continuity. AI: Continuity at $c=0$ is continuity from the right, so one wants to show given $\varepsilon$ there is $\delta$ for which $0 \le x < \delta$ implies $0 \le \sqrt{x} < \varepsilon$. The natural choice here is $\delta=\varepsilon^2.$ Otherwise $c>0$ and we initially restrict to $\delta<c/2$ (and throw that in at the end via the usual "min" definition.) From this assumption we have $c/2<x<3c/2$ and so $$\sqrt{c/2}<\sqrt{x}<\sqrt{3c/2}.$$ Now we consider that $$\|\sqrt{x}-\sqrt{c}\|=\frac{\|x-c\|}{\sqrt{x}+\sqrt{c}}. \tag{1}$$ Given our bounds for $\sqrt{x}$ the right side of $(1)$ is at most $\|x-c\|/(\sqrt{c/2}+\sqrt{c}),$ so that if we finally define $\delta$ as the minimum of the two numbers $c/2$ and $\varepsilon \cdot (\sqrt{c/2}+\sqrt{c})$ we'll have $\|x-c\|<\delta$ implies $\|\sqrt{x}-\sqrt{c}\|<\varepsilon.$
H: $\int_A f dm \leq 0 $ for all $A$ lebesgue measurable implies $f \leq 0 $ a.e $$ \textbf{Problem} $$ $\int_A f dm \leq 0 $ for all $A$ lebesgue measurable set implies $f \leq 0 $ a.e $$ \textbf{Solution (Attempt)} $$ We want to show $X = \{ x : f > 0 \} $ is a null set, that is $m(X) = 0 $. Consider, $X_n = \{ x : f \geq \frac{1}{n} \} $. We show $$ X = \bigcup_{n=1}^{\infty} X_n $$ If $x \in X$, then $f > 0 $. By archimidean, can find $n$ such that $f \geq \frac{1}{n}$. therefore, $x \in X_n $ for some $n$. $\therefore$ by definition $x \in \bigcup_n X_n $. For the reverse direction, if $x \in \bigcup_n X_n $, then $x \in X_n$ for some $n$. Hence, $f \geq \frac{1}{n} > 0 \implies f > 0 \implies x \in X $ Now, we show $m(X_n) = 0$. but by hypothesis, $$ \int\limits_{X_n} \frac{1}{n} \leq \int\limits_{X_n} f \leq 0 \implies \frac{1}{n} m(X_n) \leq \int\limits_{X_n} f \leq 0 $$ This can only happen if $m(X_n) = 0$ Which implies by additivity of the measure that $m(X) = 0$. Therefore, $f \leq 0 $ a.e. Is this correct? AI: Put $E = [f > 0]$. This is measurable. Since $f >0$ on $E$, $$\int_Ef(x)\,dx \ge 0.$$ but by your hypothesis, this integral cannot be positive, so $$\int_E f(x)\,dx = 0$$ Necessarily, $f = 0$ a.e. on $E$, so $\|E\| = 0$. Since $[f>0]$ has measure zero, $f \le 0$ a.e.
H: Order of modular group Prove $\|(\mathbb{Z} / p^e \mathbb{Z} )^{\times}\| = p^e - p^{e-1}$ I know it has something to do with the fact that we have $p^e$ elements and we're substracting $p^{e-1}$ multiples of $p$, but I'd like to know how to formally prove this. AI: Let $a \in \mathbb{Z}$. First show that in $\mathbb{Z}/n\mathbb{Z}$, the residue class $\bar{a} \in \mathbb{Z}/n\mathbb{Z}$ is invertible if and only if $\gcd(a,n) = 1$. Edit: Here is the argument. Suppose $\gcd(a,n) = 1$. Then there exists $x, y \in \mathbb{Z}$ such that $ax + ny = 1$. Now look at this equation in $\mathbb{Z}/n\mathbb{Z}$. There it says that $\bar{a}\bar{x} + \bar{0} = \bar{1}$. By definition, this says that $\bar{a}$ is invertible. For the converse, suppose that there exists $\bar{x} \in \mathbb{Z}/n\mathbb{Z}$ such that $\bar{a}\bar{x} = \bar{1}$. Let $x \in \mathbb{Z}$ be any integer whose residue class is $\bar{x}$. Then the equation $\bar{a} \cdot \bar{x} = \bar{1}$ says that $ax = 1 + ny$ for some $y \in \mathbb{Z}$. Of course, this says that $ax -ny = 1$ which in turn implies that $\gcd(a,n) = 1$. Now note that $\gcd(p^e, a) = 1$ if and only if $p \nmid a$. Your claim now follows.
H: About counting number of n-tuples Let n-tuples be $(x_1,x_2,x_3,...x_n)$ and $0\le x_i<q$ ($x_i$ is integers) for $i=1,2,3,...,n$. First part of the question was about the number of n-tuples. I got this part right, (number of n-tuples)$=q^n$ But for the second part, it is asking the number of n-tuples considering the order of the n-tuples does not matter anymore. For example, (0,0,1,5) and (1,0,0,5) are considered as the same thing. So, I am thinking as picking $n$ numbers in range of $0\le x_i<q$ because the order does not matter. So, the answer that I got is $\frac {q!} {n!(q-n)!}$. Is this right?? AI: This can be done using the Polya Enumeration Theorem. Here the group is the symmetric group $S_n$ on $n$ elements and we seek to evaluate the substituted cycle index $$Z(S_n)(Y_1+Y_2+\cdots+Y_q)_{Y_1=Y_2=\cdots=Y_q=1}.$$ The recurrence for the cycle index of $S_n$ is $$Z(S_n) = \frac{1}{n} \sum_{l=1}^n a_l Z(S_{n-l}).$$ Let the substituted cycle index be $Q_n.$ The recurrence then becomes $$Q_n = \frac{1}{n} \sum_{l=1}^n (Y_1^l+Y_2^l+\cdots+Y_q^l) Q_{n-l} = q \frac{1}{n} \sum_{l=1}^n Q_{n-l}.$$ The convention is that $Z(S_0)=1$ and hence also $Q_0 = 1.$ Introduce the generating function $$Q(z) = \sum_{n\ge 0} Q_n z^n.$$ Rewrite the recurrence as $$n Q_n = q \sum_{l=1}^n Q_{n-l}$$ and multiply by $z^{n-1}$ and sum over $n\ge 1:$ $$\sum_{n\ge 1} n Q_n z^{n-1} = Q'(z) = q \sum_{n\ge 1} z^{n-1} \sum_{l=1}^n Q_{n-l} = q \sum_{n\ge 1} z^{n-1} [z^{n-1}] \frac{1}{1-z} Q(z). $$ This yields $$Q'(z) = q \frac{1}{1-z} Q(z).$$ Solving the DE produces $$ Q(z) = C \frac{1}{(z-1)^q}.$$ But we must have $Q_0 = 1$ for all $q$ so $C = (-1)^q$ and therefore $$ Q(z) = \frac{1}{(1-z)^q}.$$ It follows that $$Q_n = [z^n] Q(z) = [z^n] \frac{1}{(1-z)^q} = {n+q-1\choose q-1}.$$ We recognize the stars-and-bars method from this link.
H: Vector in a linearly dependent set is a linear combination of other vectors in that set? I have a true/false HW problem: If $S$ = {$v_{1}$, $v_{2}$, . . . $v_{n}$} is linearly dependent then any $v_{k}$, inf ≥ k ≥ 1, is a combination of the other vectors in $S$. The setup I did is: $v_{k}$ = -($a_1/a_k$)$v_1$ - ($a_2/a_k$)$v_2$ - . . . - ($a_{k-1}/a_k$)$v_{k-1}$ - ($a_{k+1}/a_k$)$v_{k+1}$ - . . . - ($a_{n}/a_k$)$v_{n}$ Since at least one of the $a$ terms must not be zero because $S$ is linearly dependent $v_{k}$ seems to me to be possibly the linear combination of the other terms in $S$. However, I know that linear dependence requires only one of the vectors to be a combination of the preceding vectors in its vector space, so that leads me to believe that while $v_k$ could be a linear combination of the other vectors in $S$ it doesn't need to be. I don't have a good idea on where to go from here. AI: No. Let $S:=\{0,v\}$ where $v\ne 0$. It is linearly dependent, but $v\ne \lambda\cdot 0$.
H: Taylor polynomial about the origin Find the 3rd degree Taylor polynomial about the origin of $$f(x,y)=\sin (x)\ln(1+y)$$ So I used this formula to calculate it $$p=f(0,0)+(f_x(0,0)x+f_y(0,0)y)+(\frac{1}{2}f_{xx}(0,0)x^2+f_{xy}(0,0)xy+\frac{1}{2}f_{yy}(0,0)y^2)+(\frac{1}{6}f_{xxx}(0,0)x^3+\frac{1}{2}f_{xxy}(0,0)x^2y+\frac{1}{2}f_{xyy}(0,0)xy^2+\frac{1}{6}f_{yyy}(0,0)y^3)$$ I get $x(\ln(1)+y-\frac{\ln(1)x^2}{6}-\frac{y^2}{2})$ But as you can see, this is a very tedious task (especially if I have to do this on my midterm). There exists a Taylor series for $\sin(x)$ and $\ln(1+y)$. If I only keep the terms with degree $\le 3$, I have $$\sin(x)\ln(1+y)=(x-\frac{x^3}{3!})(y-\frac{y^2}{2}+\frac{y^3}{3}) \\=xy-\frac{xy^2}{2}$$ (I multiply the two and remove terms with degree > 3 from the answer) The two polynomials are different. Is the second method even a valid way to determine Taylor polynomial? AI: The answers are the same. $\ln(1) = 0$. And yes, your technique is correct.
H: Help needed with first-order logic representation I'm very new to first-order logic. I've been working on some tasks below, and would appreciate if somone could check if I have understood and solved the questions correctly Task: Assume that $B$, $F$ and $K$ are relational symbols so that $Bx$ interpreted as "$x$ is a biologist $Fx$ interpreted as "$x$ is a philosopher" $Kxy$ interpreted as "$x$ knows $y$" Assume that $a, b$ and $c$ are constant symbols which represents Aristoteles, Bolzano and Copernicus. Find first-order logic formulas for following sentences: Aristotle is both a biologist and a philosopher All biologists are philosophers No philosophers are biologists Aristoteles knows a philosopher Bolzano knows all philosophers Copernicus knows only biologists Here are my answers: There exists one $x$ so that Aristoteles is a $Gx$ and a $Fx$, written as $\exists$$x(Bx(a)$ $\land$ $Fx(a)$ For all $x$ if $x$ are $Bx$ so $x$ are $Fx$, written as $\forall$$x$$(Bx(x$) $\rightarrow$ $Fx(x)$ It's not that there is an $x$ such that $x$ are $Fx$ and $x$ are $Bx$, written as $\lnot$$\exists$$x(Fx(x) \land Bx(x)$ There exists one $x$ such that Aristoteles knows a $Fx$, written as $\exists$$x(Kxy(a, Fx)$ Not sure Not sure I would appreciate any help and please feel free to correct me if I've done something wrong. Thank you. AI: For $(1)$, we do not need a quantifier: $(1) \quad B(a) \land F(a)$ For $(4)$, we need $\exists x(F(x) \land K(ax))$. $(5) \quad \forall x(F(x) \rightarrow K(bx))$ $(6) \quad \forall x(\lnot B(x)\rightarrow \lnot K(cx))\equiv \forall x(K(cx) \rightarrow B(x))$. "For all x, if x is not a biologist, then Copernicus does not know x" $\equiv$ "For all x, if Copernicus knows x, then x is a biologist." Also, be careful with parentheses on $(2), (3)$: you're missing closing parentheses. (Also on $(4)$, but I've included it above.)
H: Subgroups containing kernel of group morphism to an abelian group are normal. Let $\varphi:G\rightarrow H$ be a group homomorphism from group $G$ to group $H$. Show that, if $H$ is abelian, all subgroups of $G$ that contain $\mathrm{ker} (\varphi)$ are normal in $G$. AI: Hint: $G / \ker \varphi$ is abelian. Hence $G' \le \ker \varphi$ where $G'$ is the commutator subgroup. $G / \ker \varphi$ is isomorphic to a subgroup of $H$ by the first isomorphism theorem, hence abelian. By the familiar property of $G'$, we have $G' \le \ker \varphi$. It follows that any subgroup $K$ that contains $\ker \varphi$ also contains $G'$. Hence $K$ is normal by another familiar property of $G'$. If you're not familiar with the property in the last statement, here is how to show it: Since $G/G'$ is abelian and $K/G'$ is a subgroup of $G/G'$, we have $K/G' \trianglelefteq G/G'$. We conclude that $K$ is normal by the fourth isomorphism theorem.
H: $\lim_{(x,y)\to(0,0)} \frac{xy}{\sqrt{x^2+y^2}}$ Let me be honest, when some teacher ask me to calculate such limit, I always try to found two paths with different results to show that such limit don't exist. But, I try four paths all fall at zero... and wolframalpha (internet boss (after math.stackexchange.com)) tells me this limit don't exit. Stewart's (book boss) calculus book tells me this limit is zero. But I don't know how to show this... I hate limit calculus, because it's full of tricks... if you're not aware enough the day of your exam, you lost quickly some points... Could you light my path? thx to everybody! AI: Convert into polar coordinates: so $x=r\cos\theta\space$ and $y=r\sin\theta.$ $\displaystyle\lim_{(x,y)\to(0,0)} \dfrac{xy}{\sqrt{x^2+y^2}}\\=\displaystyle\lim_{r\to 0}\dfrac{r\cos\theta\cdot r\sin\theta}{\sqrt{r^2\cos\theta+r^2\sin\theta}}\\=\displaystyle\lim_{r\to 0}\dfrac{r^2\sin\theta\cos\theta}{\sqrt{r^2(\sin^2\theta+\cos^2\theta)}} \text{Recall the Pythagorean Identity,}\sin^2\theta+\cos^2\theta=1.\\=\displaystyle\lim_{r\to 0}\dfrac{r^2\sin\theta\cos\theta}{\sqrt{r^2}}\\=\displaystyle\lim_{r\to 0}\space r\sin\theta\cos\theta\\=0$
H: Is the intersection of a sequence of nested subspaces nonempty? Say $X$ is a topological space (compact). if $\{ A_n \} $ is a collection of nonempty closed subsets of $X$ such that $A_{n+1} \subseteq A_n $ for all $n$, then does it follow that $ \bigcap_n A_n $ is non-empty?? My try: Since $X$ is compact topological space and $A_n$'s are closed in $X$, then they must be compact. In particular, if $\mathcal{O}_{n}$ is open cover for each $A_n$, then can take open finite subcover $\{ O_{n,1}, ... , O_{n,k} \} $ for each of the $A_n's$. Since $A_{n+1} \subseteq A_n $ then this means that an open cover for $A_{n}$ is also an open cover for $A_{n+1}$ for all $n$. Im kinda of stuck here. Maybe I am taking the wrong route. Can someone help me? thanks AI: This can be solved by using the finite intersection property of compact spaces, that a space $X$ is compact iff for all collections $\mathscr{C}$ of closed sets in $X$ having the finite intersection property, you have $\bigcap_{C \in \mathscr{C}} C \neq \varnothing $. And your ascending chain collection has the finite intersection property.
H: $(0 \leq f' \leq f$ on $\mathbb R$ and $f(a)=0)$ $\implies f=0$? Given that $0 \leq f' \leq f$ on $\mathbb R$ and $f(a)=0\in\mathbb R$ for some $a\in\mathbb R$, how do I prove that $f$ is identically zero? Using the mean value theorem naively didn't really get me anywhere. AI: Since $f'\ge 0$, then $$ f(x)=f(a)+\int_a^x f'\ge 0 \mbox{ for any } x>a $$ For the other side, $$ f'-f\le0 $$ Then $$ 0\ge e^{-t}[\,f'-f\,]=(e^{-t}f)' $$ Therefore $$ e^{-x}f(x)=e^{-a}f(a)+\int_a^x (e^{-t}f(t))'\le 0 \mbox{ for any } x>a $$ So, for $x>a$, $0\le f(x)\le 0$. For $x<a$ same idea.
H: Show that there is no invertible continuous function $f:[0,1) \to (0,1)$ This is a problem I came across recently and have been trying to figure out, but I'm not having a lot of luck! By considering the restriction of $f$ to $(0,1)$ or otherwise, show that there is no invertible continuous function $f:[0,1) \to (0,1)$. Hint: there are bijections which map $[0,1)$ onto $(0,1)$. Can you construct one? AI: The main idea is that such a function has to be always increasing or always decreasing. Assume that there exist three points $a<b<c$ such that $f(a) \leq f(b)$ and $f(b) \geq f(c)$. If either $f(a)=f(b)$ or $f(b)=f(c)$ we can conclude that the function is not injective hence cannot be invertible. Now we can assume $f(a) < f(b)$ and $f(b) > f(c)$ Then $\exists p \in [f(a),f(b)]\cap[f(c),f(b)]$. Then by IVP $\exists m_{1},m_{2}$ such that $a<m_{1}<b$ and $b<m_{2}<c$ with $f(m_{1})=f(m_{2})=p$. Thus $f$ is not injective hence not invertible. This proves that f has to be always increasing or always decreasing. Assume $f$ is always increasing and $f(0)=t\in (0,1)$ what maps to the rest of the values in $(0,t)$? This shows that f cannot be bijective (because it is not surjective) hence not invertible.
H: What can be the value of $m$ in following equation During calculations I got this step $$(e^m/((m+1)^{m+1}) )^{3n/4} = 1/2^n$$ I want the value of m here?? AI: Well, you can eliminate $n$ by raising both sides to $4/(3n)$: $$\frac{e^m}{(m+1)^{m+1}} = \frac{1}{2^{4/3}}.$$ I don't think there's much hope of a closed-form solution, but Wolfram Alpha can easily find $m$ numerically. Note that more, (likely complex, depending on $n$) solutions also exist -- multiply the right-hand side by $4/(3n)$th roots of unity.
H: Using the Squeeze Theorem in Sequences My textbook has an example that says "Show that the sequence {${c_n}$} $= (-1)^n \frac{1}{n!} $ " converges, and find its limit. It tells me that I must "find two convergent sequences that can be related to the given sequence" which the textbook states that the two possibilities are $a_n = \frac{-1}{2^n}$ and $b_n = \frac{1}{2^n}$. My question is how did they find $a_n$ and $b_n$? Is there a way to find them algebraically since my text doesn't show or explain the process to do so? AI: The idea of the squeeze theorem is that you find two sequences, in your example $a_n$ and $b_n$, with whom you can bound the sequence $c_n$ (you are interested in), i.e., so that you can get for all $n>t$ ($t$ is some finite threshold) $$a_n\le c_n \le b_n.$$ If this holds, and you know that both $a_n$ and $b_n$ converge to the same limit $x$, then $c_n$ must converge to the same limit $x$ (after all, it is "sandwiched" from above and below by those two sequences; it has no choice but to converge, too). This is helpful if the limit of $c_n$ is hard to figure out, but it is easy to find sequences $a_n$ and $b_n$ that bound ("sandwich") $c_n$ and for which the limit can be found easily. This is also why the squeeze theorem is sometimes informally called "sandwich theorem". In your example, $$c_n=(-1)^n\frac{1}{n!}<b_n=\frac{1}{2^n} \text{ for $n>3$},$$ because $n!=n\cdot (n-1)\cdot (n-2)\cdot \ldots > 2^n=2\cdot2\cdot\ldots\cdot 2$ and $\max\{(-1)^n\}=1$. Similarly, $a_n<c_n$ if $c_n$ is negative. Now $a_n$ and $b_n$ are rather simple sequences, and their limit is $x=0$. The only difference is that $a_n<0$ converges from below and $b_n>0$ converges from above. By the squeeze theorem, we therefore know that $c_n\to 0$ as $n\to\infty$ as well. How do you find those bounding sequences? This often happens on a case-by-case basis. But as a general rule, if your sequence $c_n$ consists of several terms, then try to bound each of them. Your $c_n$ is a product, so the first bound is $\max\{(-1)^n\}=1$ and the second is $\text{something}\ge \frac{1}{n!}$. In your textbook, $\text{something}=\frac{1}{2^n}\ge \frac{1}{n!}$, which holds for $n>3$ as shown above. Since both factors are positive, the product of two factors (of $b_n$), each of whom are weakly greater than the factors of another product ($c_n$), is also weakly larger. Hence $b_n\ge c_n$. Again, the important thing is that the sequence you find is simpler, so you just look at your sequence and think about how to simplify it, ideally so that is remains a close approximation for $n$ large.
H: Show that $C[a,b]$ is a complete space under the metric $d(f,g)=\sup_{t\in [a,b]}\|f(t)-g(t)\|$. $C[a,b]$ is a normed vector space of all continuous complex valued functions on $[a,b]$, with supremum norm $$\\|f\\|_\infty=\sup_{t\in [a,b]}\|f(t)\|.$$ The metric induced by the norm is $$d(f,g)=\\|f-g\\|_\infty = \sup_{t\in [a,b]}\|f(t)-g(t)\|.$$ Show that $C[a,b]$ is a complete metric space under this metric induced by the supremum norm (i.e. show $C[a,b]$ is a Banach space). AI: If $\langle f_n\rangle$ is Cauchy then for any $\varepsilon >0$ there exists $N$ such that $n,m\geqslant N$ gives $$\sup_{x\in[a,b]}\|f_n(x)-f_m(x)\|<\varepsilon$$ But then the sequence of complex numbers $\langle f_n(x)\rangle$ is Cauchy for each $x\in [a,b]$, thus it converges. This means we can define $f:[a,b]\to \Bbb C$ by $$f(x)=\lim\limits_{n\to\infty}f_n(x)$$ It remains to show that $f_n\to f$ uniformly, and that $f$ is continuous. Can you do this?
H: $A$ is dense iff there is not open subset nonempty in $X \setminus A$ $A$ is dense iff there is not open subset nonempty in $X \setminus A$. Let $X$ be topological space. my try: If $A$ is dense then $X = Cl(A) $. we can assume there is an open $O$ such that $O \subseteq X \setminus A $. My question is : Can we assume that if $X = cl(A) \; \;then\; \; X = A $?? Im kind of stuck here. For the other direction, if there is not open subset nonempty in $X \setminus A$. we must show $X = Cl(A)$. In other wrods, we want to show that every $x \in X$ is a limit point. so pick $x \in X$ Take a neighborhood $N$ of $x$. We want to show $N \cap A \neq \varnothing$ if $N \cap A = \varnothing $, then $N \subseteq X \setminus A $. So we have an open set inside $X \setminus A$ which contradicts hypothesis. so $A$ must be dense. Is this correct? AI: For the first part, no, $X = \mathrm{cl} (A)$ does not imply $A = X$. For an example, consider the rationals $\mathbb{Q}$ as a subset of the real line. For this direction it might be easier to prove the contrapositive: if there is a nonempty open $U \subseteq X \setminus A$, then $A$ is not dense. (Hint: Can any $x \in U$ belong to $\mathrm{cl}(A)$? Prove it!) The main line of your proof of the opposite direction seems fine. (Just be sure to mention that you have a nonempty open set inside $X \setminus A$.)
H: Non-trivial open dense subset of $\mathbb{R}$. I recently found the following exercise real analysis: Let $A\subseteq\mathbb{R}$ be open and dense. Show that $$\mathbb{R}=\{x+y:x,y\in A\}$$ I think it is not too hard to prove. But do we have a non-trivial example of such a set? So my question is: Can we find an example of a subset $A\subset\mathbb{R}$ that is open and dense, but $A\neq \mathbb{R}$? I don't know a lot of examples of dense subset of $\mathbb{R}$. The rational and irrational numbers are dense, but clearly not open. AI: $A = \mathbb{R} \setminus \{0\}$ works for this purpose, and isn't equal to $\mathbb{R}$. But that's still fairly trivial. For a less trivial example, fix an enumeration $\{r_n\}_{n = 0}^{\infty}$ of rational numbers and a positive number $\epsilon$. Define open intervals $$\mathcal{O}_n = \left(r_n - \frac{\epsilon}{2^{n + 2}}, r_n + \frac{\epsilon}{2^{n + 2}}\right)$$ and define $\mathcal{O} = \bigcup_n \mathcal{O}_n$. Then $\mathcal{O}$ is an open, dense subset of $\mathbb{R}$ with Lebesgue measure at most $\epsilon$. In fact, we could (by dilating one of our intervals) make the measure of $\mathcal{O}$ equal to any given positive number $\epsilon$.
H: Convergence of $\sum_{n=1}^\infty 1/(n - c)^2$ I'm curious as to whether or not the following series converges, $$\sum_{n=1}^\infty \frac{1}{(n - c)^2},$$ where $c$ is some positive constant, $c \notin \mathbb{Z}_{>0}$. Initially my intuition lead me to believe that it did converge since, for large enough $n$, one could argue that it looks a lot like the series $\sum \frac{1}{n^2}$ , but so far i've been unable to throw together a formal proof. Any ideas or a proof of convergence/divergence would be much appreciated. AI: Assuming that $c$ is a non-integer constant, we have the following argument: There is some $n_0$ so that $n_0 - c>0$. With that, we may note that $$ \sum_{n=1}^\infty \frac{1}{(n-c)^2} = \sum_{n=1}^{n_0} \frac{1}{(n-c)^2} + \sum_{k=1}^\infty \frac{1}{(k+(n_0-c))^2} $$ Now, consider that second sum. We note that $$ \frac{1}{(k+(n_0-c))^2} < \frac{1}{k^2} $$ Since the sum $\sum_{k=1}^\infty \frac 1{k^2}$ converges, we may state by the comparison test that $$ \sum_{k=1}^\infty \frac{1}{(k+(n_0-c))^2} $$ converges, which means that our original sum must converge.
H: $\|G\| = pqr$ with $p$, $q$ and $r$ distinct primes. Show G is not simple. $\|G\| = pqr$ with $p$, $q$ and $r$ distinct primes. Show G is not simple. I know this might have been asked and answered before. I just wanted someone to tell me if my argument is OK: Let $\|G\| = pqr$, and assume $p < q < r$. We have at least one Sylow-$q$ subgroup (call it $Q$) and at least another Sylow-$r$ subgroup (call it $R$). Consider the subgroup $K = $ $<Q, R>$. Now $K$ is not the whole group because there is no element of order $p$ in there. Further, $\|K\|$ must equal $qr$ because nothing smaller is possible by Lagrange's theorem. Therefore [G : K] = p, which is the smallest prime dividing G. Hence by a standard theorem (for instance, see corollary 4.5, p. 44, Isaacs' Algebra), K must be normal. Hence $G$ is not simple. Thanks! WH AI: There is nothing stopping $K$ from being equal to $G$ - ie. $K$ might have elements of order $p$. However, I cannot think of a counterexample of the top of my head. Here is an alternate solution though. You can do this by counting elements of orders $p, q$ and $r$ : Let $n_s$ denote the number of $s$-Sylow subgroups, for $s\in \{p,q,r\}$. Since any two distinct $s-$Sylow subgroups intersect trivially (why?), it follows that The number of elements in $G$ of order $s$ is $n_s(s-1)$ Now assume $n_s \neq 1$ for all $s\in \{p,q,r\}$. Then $n_r \equiv 1\pmod{r}, n_r\mid pq \Rightarrow n_r = pq$ $n_q \equiv 1\pmod{q}, n_q \mid pr \Rightarrow n_q \in\{ r,pr\} \Rightarrow n_q \geq r$ $n_p \equiv 1\pmod{q}, n_p\mid qr \Rightarrow n_p \geq q$ Hence, $$ \|G\| \geq n_r(r-1) + n_q(q-1) + n_p(p-1) + 1 $$ $$ \Rightarrow pqr \geq pq(r-1) + r(q-1) + q(p-1) + 1 $$ Solving this gives you a contradiction, and so $G$ cannot be simple (in fact, one of these Sylow subgroups must be normal)
H: spanning over vectors, but this span has only 1 vector I have seen a problem that asks the following: Is the following a subspace: $$H = \text{span}\left\{\left[\begin{array}{c}2\\0\\3\\4\end{array}\right]\right\}$$ I know the answer is yes it is a subspace, because a SPAN is a subspace by a corollary and I have even proved this. BUT, I am having difficulty understanding how this one vector spans anything. Is it supposed to span $\Bbb R^4$, I doubt it. So if anybody has any ideas about how this span can span of some vector space, will really appreciate it. [Also another way to phrase this problem is : Why is every span of a subset of a vector space a subspace?] Palu AI: Take $R^3$ and cut your span to $span[2, 0, 3]$. Then we have the vector $2, 0, 3$, which spans multiples of $2, 0, 3$ so we have $n(2, 0, 3)$. Now for $R^4$, do the same. As Betty mentioned it is one-dimensional. If we had two lines, each with different one-dimensional span, we would end up getting a plane.
H: Jordan Normal Form and eigenvalue 0 I understand the processes of putting a matrix into Jordan normal form and forming the transformation matrix associated to "diagonalizing" the matrix. So here's my question: Why is it that when you have an eigenvalue x=0 with algebraic multiplicity greater than 1, that you don't put a 1 in the superdiagonal of the JNF matrix but when the eigenvalue is non-zero and satisfies the same properties, we put a 1 in the superdiagonal of the Jordan normal form? My professor posted solutions to an assignment involving finding a matrix exponential, but the JNF of a matrix had eigenvalue x=0 with algebraic multiplicity of 3,yet had no entries of 1 along the superdiagonal. In advance, I would like to thank you for your help. AI: We have a single eigenvalue of $\lambda_1 = 1$ and a triple eigenvalue of $\lambda_{2,3,4} = 0$. For $\lambda=0$, we need to find three linearly independent eigenvectors and can just use the null space of $A$ for this. We have: $$NS(A) = NS \left(\begin{bmatrix} 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0\end{bmatrix}\right)$$ This produces $v_{2,3,4} = (0,0,0,1), (0,0,1,0), (0,1,0,0)$ as three linearly independent eigenvectors, thus this matrix is diagonalizable and we can write the Jordan block using the eigenvalues down the main diagonal as: $$J = \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1\end{bmatrix}$$
H: The size of $\sigma$-field generated by finite sets If we have a sequence of sets $F_i$ $i=1,2,...,k$ and $F_i \subset F_{i+1}$. what is the size of the $\sigma$-field, $\sigma(F_i, i=1,2,..,k)$? I guess the size is $2^{2^{k}-1}$ but I can't prove it. How can I prove this result? AI: Assuming inclusions are proper, I think we would actually have exactly $2^k$ sets in our $\sigma$-algebra (the empty set being included). Since $F_i\subset F_{i+1}$, we could just as well take $F_1$ and $F_{i+1}\backslash F_i$ for $i=1,\ldots,k-1$ to be our generators. This gives us $k$ disjoint sets, and the $\sigma$-algebra generated by disjoint sets consists simply of unions of said sets. There are $2^k$ choices for subsets of these $k$ generators.
H: Critical Values of a Function I need to find the critical values of $h(t) = t^{3/4} - 2t^{1/4}.$ So I began by finding the derivative of the function and simplifying: \begin{align} h'(t) &= (3/4)t^{-1/4} - (2/4)t^{-3/4} \\ &= \frac{3}{4t^{1/4}} - \frac{2}{4t^{3/4}}\\ &= \frac{3t^{3/4} - 2t^{1/4}}{4t} \end{align} To find the first critical value, I set the denominator equal to 0 and got the first value of 0, but I'm stuck on the last critical value. The back of the book says the critical values are 0 and 4/9. How can I go from here to find the 4/9 critical value? Thanks for your time :) AI: Consider $$3t^{3\over 4}-2t^{1\over 4}=0.$$ Factoring out $t^{1\over 4}$ we obtain $$t^{1\over 4}(3t^{1\over 2}-2)=0.$$ We know that $t=0$ is a soltuion. So $$3t^{1\over 2}=2.$$ Squaring both sides we get $$9t=4.$$ Thus $$t={4\over 9}.$$
H: Differentiability of Linear Maps I am wondering whether all linear mappings have first-order partial derivatives (or stronger properties such as being continuously differentiable at all orders). Formally, suppose $A$ is an $m \times n$ matrix and define the mapping $F: \mathbb{R}^{n} \to \mathbb{R}^{m}$ by $$F(x) = Ax \text{ for every } x \in \mathbb{R}^{n}$$ It is a well known result that the above identity implies that $F$ is linear. Given this, do we immediately know that $F$ has first order partial derivatives? My thought process so far is that because $F$ must be linear in each of its components, and because a linear function is differentiable, then $F$ has first order partial derivatives. Am I on the right track? AI: You are correct. If you write your function F in terms of its components it sends $(x_1,...,x_n)$ to the vector with first component $A_{11}x_1+A_{12}x_2 +...+A_{1n}x_n$, second component $A_{21}x_1+A_{22}x_2 +...+A_{2n}x_n$, etc., where $A_{ij}$ is the entry of your matrix $A$ in the i-th row and j-th column. The derivative of a vector valued function is defined componentwise, and each of the components is differentiable.
H: Chance of marrying a girl My girlfriend's father has a magic - fair - coin, he agrees to let me marry his daughter if I play his game: I have to toss the coin couple times until I see the head comes up. Then if the number of times I have tossed is divisible by three, I cannot marry his daughter, otherwise I can marry his daughter. What is the possibility that I can marry my girlfriend? Suppose $0 < \alpha < 1$; can you design a game like this, such that the probability of winning is $\alpha$? Your game still requires the player to toss the coin but you are allow to change when is pass/no pass case. ($\alpha\in\mathbb{Q})$ This is a really good and interesting question but I cannot solve it. Can you please help? AI: The length of the longest initial run of tails is a random variable that can take on values $0,1,2,\ldots$ with probability $1/2, 1/4, 1/8,\ldots$. The probability of winning is generally $$ P_{\text{win}}(\mathbf{a}) = \sum_{i=0}^{\infty}\frac{a_i}{2^i}, $$ where $a_i=1$ if a run of $i$ tails followed by a head is a win and $a_i=0$ if it's a loss. You can see that this is just the number whose binary representation is $$ 0.a_0a_1a_2\ldots$$ For instance, the original question (total flips divisible by three is a loss) has winning probability $$ 0.110110\overline{110} =\frac{1}{2}+\frac{1}{4}+\frac{1}{16}+\frac{1}{32}+\ldots=\frac{3/4}{1-1/8}=\frac{6}{7}. $$ In general, if you want to winning probability to be any $0<\alpha<1$, declare that a run of $i$ tails followed by a head is a win if and only if the $(i+1)$-th binary digit of $\alpha$ is a $1$.
H: Limit point compactness implies sequential compactness I am trying to go through the proof of: Suppose $ X $ is metrizable space. If $X$ is limit point compact, then $X$ is sequentially compact. Proof: Let $(x_n)$ where $n\in \mathbb{N_0}$ be a sequence of points in $X$. We need to find a convergent subsequence. Case: $A=\{x_n: n\in \mathbb{N_0}\}$ is infinite. By limit point compactness there is a limit point $x$ of $\{x_n, n\in \mathbb{N}\}$. We can find a subsequence that converges to $x$. I don't understand/ have intuition when they say "By limit point compactness there is a limit point $x$ of $\{x_n, n\in \mathbb{N}\}$". I am well aware of the definition of limit points and limit point compactness. But in this case what if we choose a neighborhood of say $x_1$ which will contain only $x_2$. Now if we intersect that with $A$ how can we get $x$? Any help? AI: There’s no reason to choose a nbhd of $x_1$. When we say that $x$ is a limit point of $A$, we’re saying that every open nbhd of $x$ contains a point of $A$ different from $x$, so $x$ is the only point whose nbhds are of interest. It’s entirely possible that for each $n\in\Bbb N$ the point $x_n$ has an open nbhd that contains no other point of $A$ and does not contain the limit point $x$. Suppose, for example, that the space is $\Bbb R$, and $A=\left\{\frac1n:n\in\Bbb Z^+\right\}$; the only limit point of $A$ is $0$. If $n\ge 2$, $\left(\frac1{n+1},\frac1{n-1}\right)$ is an open interval around $\frac1n$ whose intersection with $A$ is just $\left\{\frac1n\right\}$, and $\left(\frac12,2\right)$ is an open interval around $1$ whose intersection with $A$ is just $\{1\}$, so each point of $A$ has an open nbhd that contains no other point of $A$. Morover, none of these nbhds contains the limit point $0$. The thing that makes $0$ a limit point of $A$ is that every open nbhd of $0$ contains points of $A$ different from $0$. Now go back to the general situation. If $A$ is infinite, limit point compactness says that it must have a limit point $x$. This means that for each $\epsilon>0$, $B(x,\epsilon)\cap(A\setminus\{x\})\ne\varnothing$: each $\epsilon$ ball centred at $x$ contains a point of $A$ other than $x$. (Of course it’s possible that $x\notin A$, as in the example above, but we have to cover the possibility that $x$ is in $A$ as well.) In particular, for each $k\in\Bbb Z^+$ there some $x_{n_k}\in B\left(x,\frac1k\right)\cap(A\setminus\{x\})$. Then $\langle x_{n_k}:k\in\Bbb Z^+\rangle$ is a sequence in $A$ converging to $x$. The only problem is that it might not be a subsequence of the original sequence $\langle x_k:k\in\Bbb N\rangle$, because the indices $n_k$ might not be strictly increasing. To complete the proof, you must show that the sequence $\langle n_k:k\in\Bbb Z^+\rangle$ of indices has a strictly increasing subsequence.
H: Please help on this matrix transformation problem Let A be the matrix below and define a transformation $T: \mathbb{R}^3 \to \mathbb{R}^3$ by $T(U) = AU.$ For each of the vectors $B$ below, find a vector $U$ such that $T$ maps $U$ to $B$, if possible. Otherwise state that there is no such $U$. $$ \begin{pmatrix} 1 & -3& 2 \\ 2& -4& 4 \\ 3& -8& 6\\ \end{pmatrix} =A $$ a)$$ \begin{pmatrix} 4\\ 6\\ 11\\ \end{pmatrix} =B $$ b)$$ \begin{pmatrix} -3\\ -2\\ -7\\ \end{pmatrix} =B $$ AI: Hint: write $U=(u_1,u_2,u_3)$ and then work out $AU$ in terms of these components. You now have a system of linear equations to solve in the form $AU=B$.
H: Show the sequence ${n^m + 1}\over {n^{m+1} + 1}$ with $m \in \mathbb{R}$ is decreasing Consider the sequence $\{a_n\}_1^\infty$ such that $ a_n = $ ${n^m + 1}\over {n^{m+1} + 1}$ and $m \in \mathbb{R}$ EDIT: This is incorrect for $m < -1$, Then add the condition $m \geq -1$ I want to show this sequence is monotonically decreasing. This is not homework, I've seen many specific examples of this however and graphed many cases, and I'd like to show this for the general case. I've tried considering the continuous case and taking the derivative, using the quotient rule I get a square in the denominator, and since I'm only interested in the sign of the derivative I can throw away the denominator in the quotient rule, I then have: $mx^{m-1}(x^{m+1} + 1) - (m+1)x^m(x^m + 1)$ Plugging in $x = 1$ gives a negative result for the derivative showing the continuous function is decreasing at that point, so I get a feeling my "conjecture" is correct. Not sure where I can go from here, and if there's a simpler way without going to the continuous case. Expanding the brackets didn't help much either AI: $$\frac{n^m + 1}{n^{m+1} + 1} \geq \frac{(n+1)^m + 1}{(n+1)^{m+1} + 1} \Leftrightarrow \\ (n^m + 1)((n+1)^{m+1} + 1) \geq ((n+1)^m + 1)(n^{m+1} + 1)\Leftrightarrow \\ n^m (n+1)^{m+1} + n^m +(n+1)^{m+1} + 1\geq (n+1)^m n^{m+1} + (n+1)^m + n^{m+1} + 1 \Leftrightarrow \\ n^m (n+1)^{m} +n(n+1)^{m} \geq n^m (n-1) \Leftrightarrow \\ (n+1)^{m} +n(1+\frac{1}{n})^{m} \geq (n-1)$$ As $m \geq -1$ and the exponentials $(n+1)^x$ and $(1+\frac{1}{n})^{x}$ are increasing, we have $$(n+1)^{m} +n(1+\frac{1}{n})^{m} \geq (n+1)^{-1} +n(1+\frac{1}{n})^{-1} = \frac{1}{n+1}+\frac{n^2}{n+1}> \frac{n^2-1}{n+1}=n-1 $$
H: $f : X \to Y $ continuous and surjective. $A $ dense in $X$ $\implies$ $f(A)$ dense in $Y$ $$ \textbf{PROBLEM} $$ $f : X \to Y $ continuous and surjective. $A $ dense in $X$ $\implies$ $f(A)$ dense in $Y$ $$ \textbf{SOLUTION(ATTEMPT)} $$ Suppose $A$ is dense in $X$. Then we must have by definition that $Cl(A) = X $. Since $f$ enjoys surjectivity, we must have $f(X) = Y $. This implies $f(Cl(A)) = Y $. We must show $Cl(f(A)) = Y$. So my claim is that $f( Cl(A)) = Cl( f(A) )$ if $f$ is continous. But I am stuck how can I prove this claim. Any ideas? AI: What you’re trying to prove isn’t true; as Pedro says in his answer, continuity of $f$ guarantees only that $\operatorname{cl}f[A]\supseteq f[\operatorname{cl}A]$ for sets $A\subseteq X$. The proof can be carried out that way, but I find other approaches easier. First show that if $X$ is any space, a set $D\subseteq X$ is dense in $X$ if and only if $U\cap D\ne\varnothing$ whenever $U$ is a non-empty open set in $X$. (Added: Reading back, I see from this question that you’ve already encountered this fact.) Then let $U$ be a non-empty open set in $Y$; you want to show that $U\cap f[A]\ne\varnothing$. What can you say about $f^{-1}[U]$?
H: legendre's formula in number theory Prove that $(k!)^{(k-1)!}$ divides $(k!)!$ (k is a non-zero positive integer). I Know Legendre's formula for counting p's power in $n!$ for any prime p,but here we have $(k!)!$ and can't establish a useful formula for that to help prove this problem. I would be grateful for your helps. AI: If $\ n=(k-1)!\ $ then$$\frac{(k!)!}{(k!)^{(k-1)!}}=\frac{(nk)!}{(k!)^n}=\binom{nk}{k,k,k,\dots,k}=\binom kk\binom{2k}k\binom{3k}k\cdots\binom{nk}k$$which is of course an integer.
H: Why is Wolfram giving me a different answer for standard deviation? I have the following set of data: Raw Scores x-x̄ (x-x̄)² ----------------------------- 7 -6 36 8 -5 25 10 -3 9 14 1 1 26 13 169 ----------------------------- 65 0 240 ----------------------------- From the raw scores we know that : $$ n=5 $$ and ∴ the mean, $$ x̄ = \frac{\sum}{n} = \frac{65}{5} = 13 $$ I then proceeded to complete the table above filling in $$ x-x̄ $$ & $$ (x-x̄ )² $$ I am now required to calculate the standard deviation & have the following eqation. $$ \sigma = \sqrt{\frac{\sum(x_i-\overline{x})^2}{n}} = \sqrt{\frac{240}{5}} = 4 \sqrt{15} $$ However when I submit my data to wolfram and query standard deviatioin it returns the answer$$2\sqrt{15}$$ Maybe this is out of the scope of my learning but being curious I opened the step by step answer and I see that wolfram calculates the standard deviation by $$ \sigma = \sqrt{\frac{\sum(x_i-\overline{x})^2}{n-1}} $$ My textbook accessible here and another textbook my teacher printed out have the eqation , I find it hard to believe that My teacher and 2 other textbooks can be wrong however Im pretty sure the developers at wolfram couldn't have got it wrong either. What am I missing? AI: The difference is that you are calculating the true standard deviation, and Wolfram\|Alpha is calculating the sample standard deviation. Here's the difference, if you use the formula $$ \sigma = \sqrt{\frac{\sum(x_i-\overline{x})^2}{n}} $$ then you are answering the question "how much on average does my data differ from its mean?" If you use the formula $$ \sigma_S = \sqrt{\frac{\sum(x_i-\overline{x})^2}{n-1}} $$ you are answering the question "if my data is a representative sample of a population, what is my best estimate of how much the population differs from its mean on average?"
H: A function can be extended to a uniformly continuous function Let $S$ be a subset of the metric space $E$ with the property that each point of $S^c$ is a cluster point of $S.$ Let $E'$ be a complete metric space and $f: S\to E'$ a uniformly continuous function. Prove that $f$ can be extended to a continuous function from $E$ into $E'$ in one and only one way, and that this extended function is also uniformly continuous. Let $s \in S^c$ then since $s$ is a cluster point of $S$ then there is a sequence $s_n \in S$ that converges to $s$ thus the sequence $s_n$ is a Cauchy sequence. Since $f$ is uniformly continuous and $E'$ is complete we have that $f(x_n)$ is also a Cauchy sequence which converges in $E'$ to some $x.$ How can I now porve that $f$ can be extended, in one and only one way, to a continuous function from $E$? AI: Continue along the same lines : Define a function $g:E\to E'$ as follows : a) If $s \in S$, write $g(s) = f(s)$ b) If $s\in S^c$, choose a sequence $s_n \to s$, with $s_n \in S$, and then you know that $f(s_n)$ converges in $E'$, so define $$ g(s) = \lim f(s_n) $$ Now you need to show that this definition of $g(s)$ is independent of choice of $s_n$. This will be an extension of $f$, and it will be unique since $S$ is dense in $E$
H: we need to find the points where $\|f(z)\|$ has maximum and minimum value $f(z)=(z+1)^2$ and $R$ be the triangle with vertices $(0,0),(0,1),(2,0)$, we need to find the points where $\|f(z)\|$ has maximum and minimum value so here $z=2$ is the point of maximum and $z=0$ is minimmum (intuitively), as $f$ is analytic so it must attain maxima minima on boundary, but my question is how can I show $f$ has no extrimum value on the other points of the boundary of the triangle? Thank you for help AI: On the lower edge of the triangle, we have $z = x$, and $$\|f(z)\| = (x + 1)^2$$ This is a strictly increasing function of $x$, and so its minimum is at $0$ and maximum at $1$. On the vertical edge, we have $z = iy$ and $$\|f(z)\| = \|(iy + 1)^2\| = 1 + y^2$$ Again, this is strictly increasing in $y$ (take a derivative), and so maximized at $y = 1$, but the maximum value is only $2$. The third edge is defined by the line $y = 1 - \frac{x}{2}$, so \begin{align} \|f(z)\|^2 &= \|1 + z\|^2 \\ &=\left\| 1 + x + i \left(1 - \frac{x}{2}\right)\right\|^2 \\ &= (1 + x)^2 + \left(1 - \frac{x}{2}\right)^2 \\ &= 1 + 2x + x^2 + 1 - x + \frac{x^2}{4} \\ &= 2 + x + \frac{x^2}{4} \end{align} This is strictly increasing in $x$ for $0 \le x \le 2$, and the maximum is obtained at the right endpoint, giving a maximum of $5$, as expected.
H: Show a subring contains certain elements. Show that the set of all real numbers of the form $a_0 + a_1\pi + a_2\pi^2 +\cdots+ a_n\pi^n$ with $n≥0$ and $a_i ∈ \mathbb{Z}$ is a subring of $R$ that contains $\mathbb{Z}$ and $\pi$. Proof. We show that the set of all real numbers of the form $a_0 + a_1\pi + a_2\pi^2 +\cdots+ a_n\pi^n$ (call it the set $A$) contains $\mathbb{Z}$, $\pi$, and the set is closed under subtraction and multiplication of its elements. Take $a_1 = 1$ and $(a_n)_{n \in \{0\} \cup \mathbb{N} \setminus \{1\}}$. Then $\pi$ is in $A$. Consider the set $D = \{\pi^n : n \in \mathbb{N}\}$. It becomes evident this can be re-written as $aD = Da : a \in A$. $A$ clearly has a bijection with $\mathbb{Z}$. Another way to show this is to construct a function $g$ such that it sends $\pi^n$ to $1$. Thus we will make $g: D \twoheadrightarrow 1$. In this fashion I want to show I can use the function $g$ on $C$ to get it to correspond to $c_0 + c_1 + c_2 + \dots + c_n$. In this sense we can correspond to each $c_i$ a natural number. Since the naturals share the same cardinality with the integers, we are done. Comment: I am annoyed at how I had to use set theory to arrive at my answer. Perhaps there is a better way, algebraically? To show that $A$ is closed under subtraction, take any real number $a_0 + a_1\pi + a_2\pi^2 +\cdots+ a_n\pi^n$ in $A$ and subtract $b_0 + b_1\pi + b_2\pi^2 +\cdots+ b_n\pi^n$ (also in $A$) from it to get $$(a_0 - b_0) + (a_1 - b_1)\pi + (a_2 - b_2)\pi^2 +\cdots+ (a_n - b_n)\pi^n$$ Clearly subtraction of integers yields an integer, and since $\mathbb{Z}$ is contained in $A$, $A$ is closed under subtraction. For multiplication, I have no idea. It gets quite messy in my head. Questions. First off, what is it called when a set contains a ring? Is it called an ideal? Or a coset? For example, when I said "We show that the set of all real numbers of the form $a_0 + a_1\pi + a_2\pi2 +\cdots+ a_n\pi^n$ (call it the set $A$) contains $Z$". I know $a_i \in Z$ and I have seen examples like $bZ = Zb : b \in B$ but I don't know what that's called. Secondly, what is $\mathbb{Z}$ called? A ring? What is $\mathbb{Z} \mod 2$ called? An ideal in the ring? Just making sure. Lastly, how was my proof written - subjectively, what did you like and dislike about the proof? Is it correct in its entirety? AI: You seem to be confused about various definitions : The set $$ A := \{a_0 + a_1\pi + a_2\pi^2 + \ldots + a_n\pi^n : a_i \in \mathbb{Z}\} $$ forms a ring because, as you say, it forms a group under addition, and for any $$ a = \sum_{i=0}^n a_i\pi^i, \text{ and } b:= \sum_{j=0}^m b_j\pi^j $$ you have $$ ab = \sum_{k=0}^{n+m} c_k\pi^k, \text{ where } c_k = \sum_{i=0}^k a_ib_{k-i} $$ This is called a "Cauchy product" of two polynomials. $\mathbb{Z}$ is a ring, $\mathbb{Z}/2\mathbb{Z}$ is a quotient ring (not an ideal).
H: Which of the following subsets of $R[x]$ are subrings of $R[x]$? Which of the following subsets of $R[x]$ are subrings of $R[x]$? Prove or disprove. All polynomials with constant term $0_R$. The elements of this set have the form $r_nx^n + r_{n-1}x^{(n-1)} + \dots + r_{1}x$. This is closed, since $R[x]$ does not contain zero divisors. All polynomials of degree $2$. This is the subset $F$ with the elements of form $a_2n^2 + a_1n + a_0$. It is clear it is closed under subtraction. It is not closed under multiplication, for multiplying the first two terms of any elements of this set together yields $(a_2n^2)(b_2n^2) = a_2b_2n^4$, a polynomial not in the set $F$. All polynomials of degree $\leq k$ where $k$ is a fixed positive integer. Consider the set $G$ of polynomials with $k = 2$. Then there are two polynomials with degree $2$ that yield, under multiplication, an element not in $G$. All polynomials in which the odd powers of $x$ have zero coefficients. (Consider the even powers as well.) Call the set of polynomials in which the odd powers of $x$ have zero coefficients $H$. Then $H$ has elements of the form $h_0 + h_2x^2 + h_4x^4 + \dots + h_{2n}x^{2n}$. It is clearly closed under subtraction. Since an even number added to an even number yields an even number, the set $H$ is a subring of $R[x]$. Call the set of polynomials in which the even powers of $x$ have zero coefficients $J$. Then $J$ has elements of the form $h_1x + h_3x^3 + \dots + h_{2n-1}x^{2n-1}$. It is clearly closed under subtraction. Since an odd number added to an odd number yields an even number, the set $J$ is NOT a subring of $R[x]$. AI: Answer updated to match changes to the question. The set of polynomials with constant term $0_R$ is a subring of $R[x]$, but you’ve not actually proved this. First, $R[x]$ does have zero divisors if $R$ does. Secondly, it’s not clear why zero divisors are relevant in the first place. The expression $a_2n^2+a_1n+a_0$ is not a polynomial of degree $2$ in $R[x]$; if $n\in R$, it’s simply a constant, and if $n\notin R$, it’s not in $R[x]$ at all. You mean $a_2x^2+a_1x+a_0$. Note that $R$ may have zero divisors, so it’s possible that $(a_2x^2)(b_2x^2)=0_R$, in which case $$(a_2x^2+a_1x+a_0)(b_2x^2+b_1x+b_0)$$ might possibly be a polynomial of degree $2$, depending on whether $a_2b_1+a_1b_2=0_R$ or not. It’s true that the polynomials of degree $2$ aren’t a subring of $R[x]$, but not for the reason that you give. HINT: Contrary to your assertion, this set is not closed under subtraction. Unlike the set of polynomials of degree $2$, the set of polynomials of degree at most $k$ is closed under subtraction. It’s not closed under multiplication, but your argument doesn’t establish this, for the reasons noted above. Assuming that your ring is unital, however, you could consider the product of $x^k$ with itself. Your argument to show that $H$ is closed under multiplication is incomplete: the fact that the sum of two even numbers is even is the key ingredient, but in order to make this step completely convincing, you need to show exactly how the exponents in a product of polynomials are related to the exponents in the factors. Finally, $J$ is not closed under subtraction: what is $x-x$? As you say, $J$ is not closed under multiplication, but it would be better to give an explicit counterexample.
H: What line are inverse functions on the complex plane reflected over? On the real plane (xy plane) inverse functions are reflections of their original functions over y=x. Is there such line for complex functions and their inverses? AI: Since we have more dimensions to deal with when working with functions of a complex variable, you cannot expect there to be a reflection in a line, but if you consider the 4-dimensional space needed to draw a "graph" of a function $w=f(z)$, then you should find that you get something similar by doing a reflection in the plane $w=z$. The terminology can be a bit confusing: for a graph of $w=f(z)$, you need 2 real dimensions or one complex dimension for domain ($z$) and 2 real dimensions, or one complex dimension for the range ($w$). It is because of the 4-dimension nature of things here that we tend not to attempt to draw graphs of complex functions of a complex variable, but rather work with things like mappings between the $z$ plane and the $w$ plane, as they are easier to visualise.
H: "Problem of points (POP)" explanation Quite recently, there was a question related to "Problem of Points" at MSE. I did some literature survey on POP in the internet and found explanations using an example of coin toss. I have been trying to get an explanation in terms of a rolling a die. Two players keep rolling a fair six sided die and player A wins a point if 2 appears and player B wins a point if 6 appears. The player who wins 3 points first wins the $100 wager. Suddenly the game gets interrupted and the score is 2-1. How do you divide the stakes. The solution somebody posted was a split of $75 to $25 and the rational was that they play two more games and the future states of winning the rounds of additional two games was (2,2),(2,6),(6,2) and (6,6) and that there is a 3:1 ratio split. But I wonder if we should count this way. I am not challenging the answer but seek explanation under the following rational A)Probability of a 2 = 1/6, B)Probability of a 6 = 1/6 and C)Probability no one wins = 2/3 Unlike coin toss where there are only two outcomes head or tail, rolling die has three outcomes 2 appears, 6 appears and other numbers appear. Could someone give me clarity of how this would change the way the stakes are divided or should we only be considering winning? AI: Let $p$ be the probability that Player $A$ wins if they continue playing from the point at which the game was interrupted. If the first roll is a $2$, Player $A$ wins immediately; if it’s a $6$, the score is tied at $2$-$2$; and if it’s anything else, the situation is unchanged. In the first case Player $A$ wins with probability $1$; in the second case the game is even, so each player wins with probability $\frac12$; and in the third case Player $A$ wins with probability $p$. Thus, $$p=\frac16\cdot1+\frac16\cdot\frac12+\frac23\cdot p=\frac23p+\frac14\;,$$ so $\dfrac13p=\dfrac14$, and $p=\dfrac34$. A $3:1$ split of the stakes in favor of Player $A$ is therefore fair.
H: $f_n \rightarrow 0$ in $L^{1}$ implies $\exists N$ such that $\lim_{ k\rightarrow 0 }\frac{1}{2k}\int_{-1/k}^{1/k} f_N(x) dx = 0$? Suppose $f_n \geq 0$ and $f_n \in L^1(\mathbb{R})$ for $n=1,2,\ldots.$ If $f_n \rightarrow 0$ in $L^{1}(\mathbb{R})$, must there be an $N$ such that $$ \lim_{ k\rightarrow 0 }\frac{1}{2k}\int_{-1/k}^{1/k} f_N(x) dx = 0? $$ AI: No, consider the functions $f_n(x)=1/n \cdot \chi(x)$. Where $\chi(x)$ is (e.g.) the characteristic function of some open and bounded interval containing zero. Then zero is a Lebesgue point for each $f_n$ and hence the limit you ask for is always $1/n$.
H: Determine truth value If $ x \in \bigcup\{A:A \in \mathscr{A}\} $, then $x \in A$ for some $A \in \mathscr{A}$. $\mathscr{A}$ is a nonempty collection of sets and I have to determine the truth value of the above statement. I know about truth tables. $p\text{ and }q$ is true if both $p$ and $q$ are true, stuff like that. I just don't know how to start to show the truth value of the statement. How do you make a table for this? AI: You don’t make a truth table: this is just a fancy way of asking you to determine whether the statement is true or false. In this case it’s true by definition: that’s the definition of the union of a collection of sets. That union can be written in several ways: $$\bigcup\{A:A\in\mathscr{A}\}=\bigcup_{A\in\mathscr{A}}A=\bigcup\mathscr{A}\;,$$ for instance.
H: We need to find the limit of $\sum_{n=0}^{\infty} (n+1)z^n$ We need to find the limit of $\sum_{n=0}^{\infty} (n+1)z^n$ for what values of $z$ does the series converges? for convergence we need $\limsup\|(n+1)z^n\|^{1\over n}<1$ i.e $\|z\|<{1\over (n+1)^{1\over n}}$ Thank you for help. AI: HINT: Your series is the derivative of $\sum_{n\ge 0}z^n$; do you know a closed form for that series? Your last statement isn’t quite correct: it should say that $$\|x\|<\limsup_n\frac1{(n+1)^{1/n}}\;.$$
H: How do mathematics define a point? I have a serious doubt. How do mathematicians define a 'point' in a space or a plot? If we have a clear explanation for a 'point' , I think my doubt on infinitesimals and infinity will be clarified. AI: Your serious doubt has been shared by many great thinkers throughout history! This is a heavily philosophical issue, and there are smart people who would disagree with much of what I'm about to write. In a nutshell, a geometric point is an abstraction of the idea of a location in space. However, once abstracted, the geometric point becomes a non-physical entity. It exists and is described only by its mathematical properties. In other words a mathematical point has no "essential nature" in the sense that we can say this is what a point is, rather we say these are the properties of points. For example, "between any two distinct points there is a line." This is a very deep philosophical issue, and depending on your level of sophistication and experience, you may have trouble grasping it. That's the traditional, geometric, axiomatic notion of point. Another view, more and more popular these days, is to take the real numbers as your "primitive" notion, and then define a point in terms of its coordinates. In other words, for plane geometry, you might see a definition such as "a point is an ordered pair of real numbers". These ideas are like, for example the physical locations on a sheet of paper when we draw coordinate axes. But that is best thought of as a physical approximation to an ideal, abstract mathematical structure. Or, inversely, the mathematical structure is the abstracted, idealized version of the physical system. It's worth noting that "point" has other, more abstract meanings in higher branches of mathematics, but I don't think that's relevant to your question. Edit: In reply to @Henry's comment, simply quoting Euclid's definition "a point is that which has no part": much can be said about why Euclid wrote this and how it was an improvement on his predecessors. Regardless, that ancient "definition" is sorely lacking, as has been noted time and again in commentaries. The most resounding objection, though not the most damning, is this: ought we really to to define something so fundamental to an entire discipline, solely in terms of what it is not?
H: Pairwise independence implies intependence of unions Is the following statement true? Let $\mathcal{A}$, $\mathcal{B}_1$, $\mathcal{B}_2$ be $\sigma$-algebras such that $\mathcal{A}$ is independent from $\mathcal{B_1}$ and $\mathcal{B}_2$. Then $\mathcal{A}$ is independent form $\sigma\{\mathcal{B}_1\cup\mathcal{B}_2\}$. Of course for any $B\in\mathcal{B}_1\cup\mathcal{B}_2$ and $A\in\mathcal{A}$ we have $P(A\cap B)=P(A)P(B)$ since $B\in\mathcal{B}_1$ or $B\in\mathcal{B}_2$. But form there I can't conclude the statement above since $\mathcal{B}_1\cup\mathcal{B}_2$ is not $\cap$-stable. I could consider sets of the form $B_1\cap B_2$ for $B_i\in\mathcal{B}_i$ but for those I can't apply the independence condition immediately. AI: If true this would imply that three random variables pairwise independent are always independent. One (should) know(s) this fails. Counterexample: $\mathcal A=\sigma(Y)$ with $Y=X_1X_2$, and $\mathcal B_i=\sigma(X_i)$ where $X_1$ and $X_2$ are independent symmetric Bernoulli random variables. Then $\mathcal A$ is independent of each $\mathcal B_i$ (can you check this?) but $\mathcal A\subset\sigma(\mathcal B_1\cup\mathcal B_2)$ (can you check this?) and $\mathcal A$ is not trivial hence $\mathcal A$ is not independent of $\sigma(\mathcal B_1\cup\mathcal B_2)$ (can you check this?).
H: Connectedness of the the punctured plane and the right open half-plane Show that any set obtained by removing a single point from $\mathbb{R}^2$ is still connected, where $\mathbb{R}$ is the real numbers. Then show that $\Bbb H = \{(x,y) : x>0\}$ is connected. By considering the function $$f(x, y)/x,$$ or otherwise, show that there are precisely two continuous functions $f : \Bbb H \to \Bbb R$ such that $$f(x, y)^2 = x^2$$ for all $(x, y) \in \Bbb H$. This is a problem I saw yesterday and it's quite interesting, but I'm not having much luck with solving it! Can anyone help out with a proof? AI: HINT: Let $p\in\Bbb R^2$, and let $X=\Bbb R^2\setminus\{p\}$. If $x$ and $y$ are distinct points of $X$ either the line segment $\overline{xy}$ lies entirely in $X$, or $p$ is on that line segment. Show that in the latter case there is a path from $x$ to $y$ consisting of two line segments that both lie entirely in $X$. Finally, show that if $X$ were not connected, there would be two points of $X$ with no path in $X$ connecting them.
H: Finding modulus and argument of $\,z³ - 4\sqrt3 + 4i = 0$. I think I am messing up somewhere as the principle argument should be a nice number from the standard triangles such as $\frac{\pi}{4}$, $\frac{\pi}{3}$ or $\frac{\pi}{6}$ or something close. (That's what we have mainly been working with) I have made $z^3 = w$ So $w =4\sqrt3 - 4i$ $\|w\| = \sqrt{4\sqrt3^2+(-4i)^2}$ =$\sqrt{48-16}$ =$2\sqrt8$ $\theta$ (which I make the angle from the $x$ axis to the $\|w\|$ line in the complex plane) = $\tan^{-1}\frac{4\sqrt3}{-4}$ In this case it is less than $\frac{\pi}{2}$ so is the same as the principle argument or $\operatorname{Arg} z$ (not sure if my thinking is correct on this, it's just how I see it) So I have a modulus of $2\sqrt8$, and a principle argument of $\tan^{-1}\frac{4\sqrt3}{-4}$which doesn't look right. Normally this would be a simple fraction with $\pi$ in it. So $w = \|w\|e^{i(\operatorname{arg}z+2k\pi)}$ = $2\sqrt8e^{i(\tan^{-1}\frac{4\sqrt3}{-4}+2k\pi)}$ Since $z^3 = w$ $z = 2\sqrt8^\frac13e^{i\frac{(\tan^{-1}\frac{4\sqrt3}{-4}+2k\pi)}3}$ where $k = 0, 1, 2$ Can someone please let me know where I have gone wrong. Sorry about the small font, first time formatting. Cheers AI: Your mistake is in calculating the absolute value: $$\|w\|=\sqrt{(4\sqrt3)^2+(-4)^2}=\sqrt{48+16}=\sqrt{64}=8.$$ The general rule is $$ \|a+bi\|=\sqrt{a^2+b^2}. $$ Think of the absolute value as distance from the origin. The complex number $a+bi$ corresponds to the point $(a,b)$. The distance of the point $(a,b)$ from the origin is gotten from Pythagoras theorem.
H: $f : [a, ∞) → R$ is a continuous function. If $\lim_{x→∞} f (x) = L$, prove that $f$ is uniformly continuous on $[a, ∞)$. Suppose that $f : [a, ∞) → R$ is a continuous function. If $\lim\limits_{x→∞} f (x) = L$, prove that $f$ is uniformly continuous on $[a, ∞)$. My attempt at the proof: Well since I have to use both facts, I think my proof needs to be divided into two parts: 1- I have to prove that $f$ is uniformly continuous on $(N,∞)$ where $N>a$ (by using the limit definition somehow) 2- Using compactness, I can easily show that $f$ is uniformly continuous on $[a,N]$ I am having trouble with $(1)$ because I can't find an $\delta>0$ that works for all $\epsilon>0$ Just need a hint in the right direction, thank you!! AI: Suppose $\lim_{x\to \infty} f(x) = L$, then for any $\epsilon >0$, there is $N \in \mathbb{R}$ such that $$ x \geq N \Rightarrow \|f(x) - L\| < \epsilon/3 $$ Hence, for any $x,y > N$ $$ \|f(x) - f(y)\| < \epsilon/3 $$ Now, $f$ is uniformly continuous on $[a,N]$, so there is a $\delta > 0$ such that $$ \|x-y\|<\delta, \text{ and } x,y \in [a,N]\Rightarrow \|f(x) - f(y)\| <\epsilon/3 $$ Hence, for any $x,y \in [a,\infty)$, if $x, y \in [a,N]$ or $x,y \geq N$, then $$ \|x-y\|<\delta \Rightarrow \|f(x) - f(y)\| < \epsilon \qquad\text{(1)} $$ Furthermore, if $x < N < y$, then $\|x-y\|<\delta$ implies that $$ \|x-N\| < \delta, \text{ and } \|f(y) - L\| < \epsilon/3, \|f(N) - L\|<\epsilon/3 $$ hence $$ \|f(x)-f(y)\| \leq \|f(x) - f(N)\| + \|f(N) - L\| + \|L - f(y)\| < \epsilon $$ Hence, (1) holds for all $x,y\in [a,\infty)$
H: On the five-point set $X=\lbrace a,b,c,d,e \rbrace$, construct two topologies, one that is Hausdorff and one that is not Hausdorff On the five-point set $X=\lbrace a,b,c,d,e \rbrace$, construct two topologies, one that is Hausdorff (other than the discrete topology) and one that is not Hausdorff (other than the trivial topology). I manage to construct a non-Hausdorff topology, which is $\lbrace \emptyset,X, \lbrace a \rbrace \rbrace$. But I can't find a Hausdorff topology after I try many posibilities. Can anyone guide me? AI: It can’t be done: the only Hausdorff topology on a finite set is the discrete topology. To see this, note that a Hausdorff topology is necessarily $T_1$, so if $X$ is the finite Hausdorff space, $\{x\}$ is closed for each $x\in X$. But then $$X\setminus\{x\}=\bigcup_{y\in X\setminus\{x\}}\{y\}$$ is the union of finitely many closed sets, so it’s a closed set itself, and $\{x\}$, its complement, is an open set. Thus, the topology is discrete.
H: Prove that $m=(x+y^2, y+x^2+2xy^2+y^4)$ is a maximal ideal of $\mathbb{C}[x,y]$. Prove that $m=(x+y^2, y+x^2+2xy^2+y^4)$ is a maximal ideal of $\mathbb{C}[x,y]$. I can show that the ideal $(x,y)$ of $\mathbb{C}[x,y]$ contains $m$ and $(x,y)$ is a maximal ideal. Therefore to show that $m$ is itself a maximal ideal I only need to show that $m=(x,y)$. I tried to prove it by looking for two polynomials $f,g$ such that $x=f(x,y)(x+y^2)+g(x,y)(y+x^2+2xy^2+y^4)$ but wasn't able to find them. And I'm not sure whether this is the best way of proving this. I would appreciate any help. Thank you. AI: As Tbrendle said in the comments, you can show that $\mathbb C[x,y]/m$ is a field. This is equivalent to $m$ being a maximal ideal of $\mathbb C[x,y]$. So consider the quotient $R:= \mathbb C[x,y]/m$. We want to compute some residue classes. Note that in this ring we have $x=-y^2$. This implies that $$0= y+y^4-2y^4+y^4=y.$$ So by these identifications, we have $$ R \simeq \mathbb C[y]/(y) \simeq \mathbb C.$$ This is a field, and we're done.
H: A closed set in $\mathbb A^2_k\times\mathbb P^1_k$ Let $k$ be an algebraically closed field and consider the Zariski topology on $\mathbb A^2_k$ and on $\mathbb P^1_k$. If $$X:=\left\{((x_0,x_1),(y_0:y_1))\in\mathbb A^2_k\times \mathbb P^1_k\,\bigg\| x_0y_1=x_1y_0\,\right\}\subseteq \mathbb A^2_k\times \mathbb P^1_k$$ I don't understand why $X$ is closed in $\mathbb A^2_k\times \mathbb P^1_k$. The structure of variety on $\mathbb A^2_k\times \mathbb P^1_k$ is obtained by patching of the affine varieties $\{\mathbb A^2\times U_i\}_{i=0,1}$ where $U_1=\{(x_0:1)\,:x_0\in k\}$ and $U_0=\{(1:x_1)\,:x_1\in k\}$. Now is there some kind of "characterization" for closed sets of $\mathbb A^2_k\times \mathbb P^1_k$? AI: Yes, there is a characterization of closed subsets of $\mathbb A^2_k\times \mathbb P^1_k$. Namely, they are the common zeros of a family $(f_i(x_0,x_1;y_o,y_1))_{i\in I}$ of polynomials $f_i(x_0,x_1;y_o,y_1)$, required to be homogeneous in the variables $y_0,y_1$ associated to $\mathbb P^1_k$. The degree of homogeneity of these polynomials may vary with $i\in I$. Your example is a little confusing (by no fault of yours!) because the polynomial $x_0y_1-x_1y_0$ happens to be homogeneous (of degree $1$) also in the variables $x_0,x_1$ of $\mathbb A^2_k$, which is absolutely not required in general. This means that your subvariety $X\subset \mathbb A^2_k\times \mathbb P^1_k$ descends to a subvariety $\mathbb P(X)\subset \mathbb P^1_k \times \mathbb P^1_k$. A geometric interpretation You have the Plücker embedding $$p:\mathbb P^1_k \times \mathbb P^1_k\hookrightarrow \mathbb P^3_k: ((x_0:x_1),(y_0:y_1))\mapsto (z_0:z_1:z_2:z_3)=(x_0y_0:x_0y_1:x_1y_0:x_1y_1)$$The image $p(\mathbb P(X))\subset \mathbb P^3_k$ of $\mathbb P(X)$ under that embedding is then the intersection of the quadric $z_0z_3-z_1z_2=0$ with the plane $z_1-z_2=0$. In other words, a good old projective conic!
H: 31,331,3331, 33331,333331,3333331,33333331 are prime 31,331,3331, 33331,333331,3333331,33333331 are prime. This law can continue it? Will there emerge a composite number? Without using a computer how to judge. AI: 333333331 is not prime; it is divisible by 17. This does not require a computer. Euler did calculations like this all the time. What's more, in your sequence 31, 331, 3331, 33331, …, every 15th number is divisible by 31. Proof: An noted in lab bhattacharjee's answer, the sequence has the form $$ a_n = \frac{10^{n+1}-7}{3} $$ Now, 15 is the multiplicative order of $10 \pmod{31}$, so $$ a_{15k+1} = \frac{10^{15k+2}-7}{3} \equiv \frac{10^2-7}{3} \equiv 0 \pmod{31}. $$ It has been proven that for all sequences that look like $ab$, $abb$, $abbb$, $abbbb$, … or $ab$, $aab$, $aaab$, $aaaab$, … where the $a$ and $b$ are digits, that periodically the numbers in the sequence are divisible by the first number $ab$. As an easy exercise, show that in the sequence 11, 111, 1111, 11111, …, that every second term is divisible by 11.
H: Calculate double integral of ... I was doing a homework problem but now I'm stuck. The problem says: Calculate $\iint_{S} \frac{dx dy}{\sqrt{2a - x}}$ where S is a circle of radius $a$ which is tangent to to both coordinate axes and is in the first quadrant The cartesian equation for a circle of radius a, and center C(a,a) is: $$(x-a)^{2}+(y-a)^{2} = a^{2}$$ Sketching this equation for different values of $a$ I obtained the limits of integration for the double integral: $$4\int_{a}^{2a} \int_{a}^{a+\sqrt{2ay-y^{2}}} \frac{1}{\sqrt{2a-x}}dx dy$$ It is multiplied by 4 because that double integral only calculates a quarter of the desired area. Then evaluating the double integral I obtained this: $$4 \left ( 2a^{3/2}+2\int_{a}^{2a} \sqrt{a+\sqrt{2ay-y^{2}}} dy \right )$$ But I was stuck on this (I've tried to put the indefinite integral$\int \sqrt{a+\sqrt{2ay - y^{2}}} dy$ into Wolfram Alpha and I obtained this ), then I said "why I don't use polar coordinates?" and I used them, and I obtained this double integral: $$4\int_{0}^{\pi /2} \int_{0}^{a}\frac{r}{\sqrt{2a-rcos\theta}} dr d\theta$$ I evaluated it, and I obtained: $$4\left ( \int_{0}^{\pi/2} -\frac{2}{3}\left ( \sec ^{2} \theta \sqrt{2a-\arccos {\theta}} \left ( 4a + acos\theta \right ) \right ) d\theta \right )$$ But Wolfram says this for the indefinite integral so... What can I do? Am I wrong in something? Is there another way to do it? I would appreciate any help. AI: Change the order of integration. $$\begin{align} \iint_S \frac{dx\,dy}{\sqrt{2a-x}} &= \int_0^{2a} \frac{1}{\sqrt{2a-x}}\left(\int_{a-\sqrt{a^2-(x-a)^2}}^{a+\sqrt{a^2-(x-a)^2}}dy\right) dx\\ &= \int_0^{2a}\frac{2\sqrt{2ax-x^2}}{\sqrt{2a-x}}\,dx\\ &= \int_0^{2a} 2\sqrt{x}\,dx. \end{align}$$ You can take it from there, I'm sure.
H: collection $\mathcal{B}$ of subsets $V = \{ x + yk : k \in \mathbb{Z} \} $ for $x,y \in \mathbb{Z}$ form a basis for some topology of $\mathbb{Z}$ Problem: The collection $\mathcal{B}$ of subsets of the form $V = \{ x + yk : k > \in \mathbb{Z} \} $ for $x,y \in \mathbb{Z}$ is a basis for some topology of $\mathbb{Z}$ Solution attempt: Pick $n \in \mathbb{Z}$. and Put $N = \{ nk : k \in \mathbb{Z} \} $. Clearly $n \in N$ and hence $N$ is a neighborhood of $n$. Now, suppose $N_1= \{kn \}, N_2= \{ k'n\}$ are two neighborhoods of some $n \in \mathbb{Z}$. We must find a nhbd $N_3$ such that $n \in N_3 \subseteq N_1 \cap N_2 $. If we take $N_3 = \{ \max\{k,k'\} n \} $ then it would do the job. Also, I would like to know if this topology is compact. How can we conclude it is not compact? Is this topology Hausdorff? AI: I’m going to assume that the definition of $\mathscr{B}$ given in the question is slightly incorrect, and that $\mathscr{B}$ is actually the collection of all sets of the form $a+b\Bbb Z=\{a+bk:k\in\Bbb Z\}$ such that $a,b\in\Bbb Z$ and $b\ne 0$; without that last restriction $\{a\}\in\mathscr{B}$ for each $a\in\Bbb Z$, and $\mathscr{B}$ is trivially a base for the discrete topology on $\Bbb Z$, which of course is not compact. Your argument that $\mathscr{B}$ covers $\Bbb Z$ is correct but unnecessarily complicated, since $\Bbb Z\in\mathscr{B}$. The next bit, however, does not make sense: $\{kn\}$ is a set containing one integer, $kn$, and is not in $\mathscr{B}$. If you really meant $\{kn:k\in\Bbb Z\}$, which is in $\mathscr{B}$, then $N_1$ and $N_2$ are the same set. Moreover, there are members of $\mathscr{B}$ containing $n$ that are not of this form. You need to start with completely arbitrary $N_1,N_2\in\mathscr{B}$ with $n\in N_1\cap N_2$. Let $N_1=a_1+b_1\Bbb Z$ and $N_2=a_2+b_2\Bbb Z$; then there are $k_1,k_2\in\Bbb Z$ such that $n=a_1+b_1k_1=a_2+b_2k_2$. Now you need to find $a,b\in\Bbb Z$ with $b\ne 0$ such that $n\in a+b\Bbb Z\subseteq\big((a_1+b_1\Bbb Z)\cap(a_2+b_2\Bbb Z)\big)$. This actually takes a little work. Perhaps the most straightforward approach is to begin by showing that if $n\in r+s\Bbb Z$, then $r+s\Bbb Z=n+s\Bbb Z$. Thus, if $n\in N_1\cap N_2$, then $$(a_1+b_1\Bbb Z)\cap(a_2+b_2\Bbb Z)=(n+b_1\Bbb Z)\cap(n+b_2\Bbb Z)\;,$$ and you need only show that $(n+b_1\Bbb Z)\cap(n+b_2\Bbb Z)=n+m\Bbb Z$, where $m=\mbox{lcm}(b_1,b_2)$. $\Bbb Z$ with this topology is not compact: the set of primes is an infinite, closed, discrete subset. (If $p$ is prime, find an open nbhd of $p$ that contains only multiples of $p$; if $n$ is composite, find an open nbhd of $n$ that contains only composite numbers.) Added: To show that the space is Hausdorff, let $m,n\in\Bbb Z$ with $m\ne n$. Let $p$ be any positive integer that does not divide $n-m$, e.g., a prime larger than $\|n-m\|$; then $m+p\Bbb Z$ and $n+p\Bbb Z$ are disjoint open nbhds of $m$ and $n$, respectively. They are clearly open nbhds of $m$ and $n$. If $k\in(m+p\Bbb Z)\cap(n+p\Bbb Z)$, then there are integers $r$ and $s$ such that $k=m+pr=n+ps$. But then $n-m=pr-ps=p(r-s)$, where $r-s$ is an integer, so $p$ divides $n-m$, contradicting the choice of $p$. Let $P$ be the set of positive primes. If $p\in P$, then $p\Bbb Z=0+p\Bbb Z$ is an open nbhd of $p$ that does not contain any other prime: if $q$ is a prime different from $p$, then $p$ is not a divisor of $q$, so $q\notin p\Bbb Z$. Thus, $(p\Bbb Z)\cap P=\{p\}$, each $p\in P$ is an isolated point of $P$, and $P$ is therefore discrete. Now suppose that $n\in\Bbb Z\setminus P$. If $n$ is composite, then every multiple of $n$ is either composite or $0$, so $n\Bbb Z$ is an open nbhd of $n$ disjoint from $P$. If $n=0$, then $4\Bbb Z$ is an open nbhd of $n$ disjoint from $P$, since every element of $4\Bbb Z$ is $0$ or a multiple of $4$ and therefore not prime. The only remaining possibility is that $n=-p$ for some $p\in P$. In that case $-p+3p\Bbb Z$ is an open nbhd of $n$ disjoint from $P$: every element of $-p+3p\Bbb Z$ is a multiple of $p$, so $p$ is the only prime that could possibly belong to $-p+3p\Bbb Z$, and it doesn’t, since $\frac23$ isn’t an integer.
H: diagonalizability and finding a basis for $\mathbb R^3$ My question is about matrices: let $T$ be a linear operator on the vector space $\mathbb R^3$ which is represented in the standard basis by the matrix $$ \begin{bmatrix} -9 & 4 & 4 \\ -8 & 3 & 4 \\ -16 & 8 & 7 \\ \end{bmatrix} $$ I need to prove that $T$ is diagonalizable and I need to find a basis for $\mathbb R^3$ each vector of which is a characteristic for $T$. AI: You want to find the eigen-values of $T$, so you solve $$ \begin{vmatrix} -9-\lambda & 4 & 4 \\ -8 & 3-\lambda & 4 \\ -16 & 8 & 7-\lambda \end{vmatrix} $$ When you simplify this, you get $$ (\lambda+1)^2(\lambda - 3) $$ and so the eigen-values of $T$ are $\{-1,3\}$ To find eigen-vectors associated to the eigen-value $-1$, you try to find a vector $(x_1,x_2,x_3)$ such that $$ \begin{pmatrix} -8 & 4 & 4 \\ -8 & 4 & 4 \\ -16 & 8 & 8 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0 $$ If you think about it long enough, you will see that $$ (1,1,1) \text{ and } (1,2,0) $$ are both linearly independent solutions, and hence eigen-vectors. Now do the same for the eigen-value $3$, and you will get a third non-zero vector $(x_1,x_2,x_3)$. Now check that the set $\{(1,1,1),(1,2,0), (x_1,x_2,x_3)\}$ forms a basis for $\mathbb{R}^3$ and write down the matrix of $T$ w.r.t. this basis.
H: Prove that there are no fixed points in a non-autonomous system Consider the non-autonomous dynamical system $$ \dot x = f(x,t) $$ with $x \in \mathbb{R}^n$. This may be converted to an autonomous system of dimension $n+1$ with $t = x_{n+1}$ and $\dot x_{n+1} = 1$. Question: how can one prove that the new system has no fixed points? AI: Hint: A fixed point $y_0 \in \mathbb R^{n+1}$ of the new system $\dot y = F(y)$ with $F(y) = \binom{f((y_1, \ldots, y_n), y_{n+1})}{1}$ has $F(y_0) = 0$.
H: Set theory: proving of set identities Suppose we have four sets A, B, C and X. How can one prove the following identity: $(A \cap B \cap C \cap \neg X) \cup ( \neg A \cap C) \cup ( \neg B \cap C) \cup (C \cap X)=C$ I tried to apply here any of basic set identities like Distributive Law or Associative Law but it led me to nothing. Any help would be appreciated, then. AI: Prahlad Vaidyanathan’s suggestion in the comments is probably the easiest way to go, but an algebraic proof is certainly possible. First note that $$(\neg A\cap C)\cup(\neg B\cap C)=(\neg A\cup\neg B)\cap C$$ by one of the distributive laws. Then $C\cap X=X\cap C$, and $$\big((\neg A\cup\neg B)\cap C\big)\cup(X\cap C)=\big(\neg A\cup\neg B\cup X\big)\cap C$$ by one of the distributive laws. Thus, the original lefthand side is equal to $$(A\cap B\cap C\cap\neg X)\cup\big((\neg A\cup\neg B\cup X)\cap C\big)\;.\tag{1}$$ Now use one of the De Morgan laws and the fact that $\neg(\neg Y)=Y$ to see that $$\neg A\cup\neg B\cup X=\neg(A\cap B\cap\neg X)\;,$$ so that $(1)$ is equal to $$\Big((A\cap B\cap\neg X)\cap C\Big)\cup\Big(\neg(A\cap B\cap\neg X)\cap C\Big)\;,$$ and use a distributive law to pull out the $C$ and get $$\Big((A\cap B\cap\neg X)\cup\neg(A\cap B\cap\neg X)\Big)\cap C\;.\tag{2}$$ Finally, use the fact that $Y\cup\neg Y$ is the universal set, which I’ll call $U$, so that $(2)$ is $$U\cap C=C\;.$$
H: Amount of points on an elliptic curve over $F_q$ Assume I have these two elliptic curves: \begin{align} E:Y^2&=X^3+b_2X^2+b_4X+b_6\\ E':Y^2&=X^3+gb_2X^2+g^2b_4X+g^3b_6, \end{align} over $\mathbb{F}_q$, where $g$ is not a square in $\mathbb{F}_q$, and $\mathbb{F}_q$ does not have characteristic $2$. I know that $\#E(\mathbb{F}_q)=q+1-t$ and am asked to prove that $\#E'(\mathbb{F}_q)=q+1+t$. I am however not really sure how to do this. I know that by definition $\#E(\mathbb{F}_q)=q+1-\tau$ and $\#E(\mathbb{F}_q)=q+1-\pi-\pi'$, where $\pi$ and $\pi'$ are the zeroes of $T^2-\tau T+q$. Any ideas on how I could approach this problem? AI: I think that the following trick is wanted. Consider the quantities $$ f(X)=X^3+b_2X^2+b_4X+b_6 $$ and $$ h(X')=X'^3+gb_2X'^2+g^2b_4X'+g^3b_6. $$ We see that $g^3f(X)=h(gX)$. Because $g^3$ is a non-square, if we fix the value $X=x\in F_q$ then one and only one of the following alternatives will occur: We have $h(gx)=f(x)=0$. $h(gx)$ is a non-zero square, and $f(x)$ is a non-zero non-square. $h(gx)$ is a non-zero non-square, and $f(x)$ is a non-zero square. In all cases the equations $$ y^2=h(gx)\qquad\text{and}\qquad y^2=f(x) $$ have exactly two solutions $y\in F_q$ between them. In respective cases 1) one solution $y=0$ to each, 2) two solutions to former, none to the latter, 3) none to the former, two to the latter. [Edit] Adding more details. Let $q_i,i=1,2,3$ be the number of those elements $x\in \Bbb{F}_q$ such that we are in case $i$. Taking into account the point at infinity we see that the numbers of points on the two curves are $$\begin{aligned} E(\Bbb{F}_q)&=q_1+2q_3+1,\\ E'(\Bbb{F}_q)&=q_1+2q_2+1. \end{aligned}$$ This is because if $x$ is in case 1, then there is one point of the form $(x,0)\in E$, and one point $(gx,0)\in E'$. If $x$ is in case 2, then there are two points of the form $(gx,y)\in E'$ but no points of the form $(x,y)\in E$. And if $x$ is in case 3, then the reverse holds. The claim follows from this as each $x$ falls into exactly one of the three cases, so $q_1+q_2+q_3=q$. [/Edit]
H: Ordinary generating function and 1/(1 - x) I do believe it's dummy question, but I would be grateful if one explains me why following generating function is valid. I'm novice in the topic and intuitively I can't understand why it's true. It's well known OGF with 1, 1, 1, .... is generating function for $\frac{1}{(1-x)} = 1 + x^2 + x^3 + ... + x^n = \sum_{i=0}^\infty x^i$ it's easily proved by $A = 1 + x + x^2... = 1 + x(1 + x + x^2...) = 1 + x*A$ The question is how to validate it using the real numbers. Suppose we have $x = 10$ then $\frac{1}{1 - 10} = \frac{1}{-9} = 1 + 10 + 100 + 1000... $ which doesn't seem to be true. Could you please point me to the error in my conclusions? AI: The series $\sum_{k\ge 0}x^k$ converges if and only if $\|x\|<1$, so the only real numbers that you can meaningfully substitute into the equation $$\frac1{1-x}=\sum_{k\ge 0}x^k$$ are those in the interval $(-1,1)$. But generating functions are formal power series; as such they are to be thought of as algebraic objects with an associated ‘arithmetic’, not as real- or complex-valued functions. As Herbert S. Wilf says in his book generatingfunctionology, ‘A generating function is a clothesline on which we hang up a sequence of numbers for display’. In most cases of interest, however, the power series actually does converge on some domain, and on that domain we can also treat it as an analytic object, the function that it represents on that domain. Outside that domain the formal operations on generating functions still make sense, but the series no longer represents the function. This turns out not to be a problem. Chapter $2$ of generatingfunctionology starts with a brief introduction to formal power series; you can freely download the whole book here. Section $2.4$ then gives an outline of the analytic theory.

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