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77307 | Existence of a sequence of continuous functions convergent pointwise to the indicator function of irrational numbers
Prove that there does not exist a sequence of continuous functions [imath] f_n :\left[ {0,1} \right] \to R [/imath] such that converges pointwise, to the function [imath]f(x)= \begin{cases} 0 & \text{if $x$ is rational},\\\\ 1 & \text{otherwise}. \end{cases}. [/imath] I have no idea How can I prove this. Prove that there no exist such sequence if the convergence is uniform, it's easy, because the limit would be continuous, but here I don't know How can I do. I suppose that some "nice" properties are "preserved" in the limit, in this kind of convergence, but I don't know any of them. | 1550796 | Exists continuous [imath]f_n: [0,1] \to \mathbb{R}[/imath] that converges pointwise, as [imath]n \to \infty[/imath], to [imath]\chi_\mathbb{Q}[/imath]?
Does there exist a sequence of continuous [imath]f_n: [0, 1] \to \mathbb{R}[/imath] that converges pointwise, as [imath]n \to \infty[/imath], to [imath]\chi_\mathbb{Q}[/imath], the characteristic function of the rationals in [imath][0, 1][/imath]? |
425959 | Right-hand side limit equals zero in whole interval
Suppose [imath]f[/imath] is continuous on [imath][a,b][/imath] and [imath]f_r'(x)=0[/imath] for all [imath]x\in[a,b)[/imath]. Prove that [imath]f[/imath] is constant on [imath][a,b][/imath]. The definition of [imath]f_r'(x)=0[/imath] is that [imath]\lim_{y\rightarrow x^+}\frac{f(y)-f(x)}{y-x} = 0[/imath] Now it's hard to apply mean-value theorem, since the limit is given only for one side for each point. | 418737 | Continuous right derivative implies differentiability
A book of mine says the following is true, and I am having some trouble proving it. (I've considered using the Lebesgue differentiation theorem and absolute continuity, as well as elementary analysis methods.) Let [imath]f: [0, \infty) \rightarrow \mathbb{R}[/imath] be continuous and have right derivatives at each point in the domain, with the right derivative function being continuous. Then [imath]f[/imath] is differentiable. |
426316 | Gambling expected value riddle
A friend of mine gave me this probability riddle i couldn't solve, Maybe you could help me. Say i go to a casino playing roulette. I always gamble that a black number would pop (probability is: [imath]\frac{18}{38}[/imath]) but i made up some rules: I will play some rounds until i'll win at least once. If i lose 11 times in a row, I quit. In the first game i'll play for [imath]1[/imath] coin, Second for [imath]2[/imath], And in the '[imath]i[/imath]' games i'll play for [imath]2^{i-1}[/imath] coins. If i win the casino gives me the same amount of coins i gambled on (And i get back the coins i placed on the roulette) What is the expected amount of games i would play? What's the probability i would gain coins? (At least [imath]1[/imath] coin) My tries were a complete failure, My main problem was that: I couldn't figure out how to do i 'stop' when either (1) or (2) is 'completed'. | 425777 | Roulette probability
In a roulette game, a player bet only on the Red color (we are given that the probability of Red is 18/38). The player will stop playing after at least one winning or losing 11 rounds in a row. On the first game he bets \[imath]1, in the 2nd game he bets \[/imath]2, and in the [imath]i[/imath]th game he bets [imath]2^{i-1}[/imath]. If he won, he will get the money that he bet on, else he lose the money he bet on. (meaning that if he won/lost a game he can start a new round and bet from \[imath]1 again) 1. How much he will earn if he wins the [/imath]i$th game? 2. What is the expected value of the number of rounds? |
412843 | variance inequality
Show that. for any discrete random variable X that takes on values in the range [0,1]. Var[X] [imath]\le[/imath] 1/4. I translate it into a inequality like this: [imath]x_1, x_2, x_3 \cdots ,x_n[/imath] where [imath]0 \le x_i \le 1[/imath], and [imath]p_1, p_2, p_3 \cdots ,p_n[/imath] where [imath]p_1+ p_2+ p_3 \cdots +p_n = 1[/imath], prove that [imath]\sum _1^nx_i^2p_i - (\sum _1^n x_ip_i)^2 \le {1\over 4}[/imath] , how to prove it? At first I tried cauchy inequatlity, but I fail :( | 810898 | Let Y be a random variable with [imath]0\le Y\le 1.[/imath]
Let Y be a random variable with [imath]0\le Y\le 1.[/imath]Show that [imath]var(Y)\le 1/4 [/imath] and that [imath]var(Y)= 1/4 [/imath] if and only if P(0)=1/2=P(1). |
426651 | Double Angle Identities
Given that [imath]\sin(2x)=\frac 5 {13}[/imath] and [imath]0^\circ < x < 45^\circ[/imath], I need to find [imath]\cos(x)[/imath] and [imath]\sin(x)[/imath]. If I work it out, I get [imath]\begin{gather} \frac 5 {13} = 2 \sin(x) \cos(x) \\ \frac 5 {26} = \sin(x) \cos(x) \end{gather}[/imath] I'm stuck at this point. I understand that [imath]\cos(x) = \sqrt{1-\sin^2(x)}.[/imath] The answers should be [imath]\frac {26}{\sqrt{26}}[/imath] and [imath]5\left(\frac {26}{\sqrt{26}} \right)[/imath] but I'm having trouble getting there. Thanks for the help! | 426516 | How to use double angle identities to find [imath]\sin x[/imath] and [imath]\cos x[/imath] from [imath]\sin 2x [/imath]?
If [imath]\sin 2x =\frac{5}{13}[/imath] and [imath]0^\circ < x < 45^\circ[/imath], find [imath]\sin x[/imath] and [imath]\cos x[/imath]. The answers should be [imath]\frac{\sqrt{26}}{26}[/imath] and [imath]\frac{5\sqrt{26}}{26}[/imath] Ideas The idea is to use double angle identities. One such identity is [imath]\sin 2x=2\sin x\cos x[/imath]. It's easy to use it to find [imath]\sin 2x[/imath] from known [imath]\sin x[/imath] and [imath]\cos x[/imath]. But here it's the other way around. |
426690 | Maximal ideals in rings of polynomials
Let [imath]k[/imath] be a field and [imath]D = k[X_1, . . . , X_n][/imath] the polynomial ring in [imath]n[/imath] variables over [imath]k[/imath]. Show that: a) Every maximal ideal of [imath]D[/imath] is generated by [imath]n[/imath] elements. b) If [imath]R[/imath] is ring and [imath]\mathfrak m\subset D=R[X_1,\dots,X_n][/imath] is maximal ideal such that [imath]\mathfrak m \cap R[/imath] is maximal and generated by [imath]s[/imath] elements, then [imath]\mathfrak m[/imath] is generated by [imath]s + n[/imath] elements. The days that I am trying to solve. Help me. | 422182 | Number of generators of the maximal ideals in polynomial rings over a field
Hi I'm trying to prove the following If [imath]K[/imath] is a field (not necessary algebraically closed) then every maximal ideal of [imath]K[x_{1},\dots,x_{n}][/imath] is generated by exactly [imath]n[/imath] elements. I know that if [imath]K[/imath] is algebraically closed by Hilbert basis the problem is done. But if [imath]K[/imath] is not algebraic closed, I know that the Krull dimension of [imath]K[x_{1},\dots,x_{n}][/imath] is [imath]n[/imath] and by the generalized Krull theorem we have that every ideal has height at most [imath]n[/imath], but I'm stucked in this point. Thank you for any help. |
87636 | On the equality case of the Hölder and Minkowski inequalites
I'm following the book Measure and Integral of Richard L. Wheeden and Antoni Zygmund. This is the problem 4 of chapter 8. Consider [imath]E\subseteq \mathbb{R}^n[/imath] a measurable set. In the following all the integrals are taken over [imath]E[/imath], [imath]1/p + 1/q=1[/imath], with [imath]1\lt p\lt \infty[/imath]. I'm trying to prove that [imath]\int \vert fg\vert =\Vert f \Vert_p\Vert g \Vert_q[/imath] if and only if [imath]\vert f \vert^p[/imath] is multiple of [imath]\vert g \vert^q[/imath] almost everywhere. To do this, I want to consider the following cases: if [imath]\Vert f \Vert_p=0[/imath] or [imath]\Vert g \Vert_q=0[/imath], we are done. Then suppose that [imath]\Vert f \Vert_p\ne 0[/imath] and [imath]\Vert g \Vert_q\ne 0[/imath]. If [imath]\Vert f \Vert_p=\infty[/imath] or [imath]\Vert g \Vert_q=\infty[/imath], we are done (I hope). If [imath]0\lt\Vert f \Vert_p\lt\infty[/imath] and [imath]0\lt\Vert g \Vert_q\lt\infty[/imath], proceed as follows. When we are proving the Hölder's inequality, we use that for [imath]a,b\geq 0[/imath] [imath]ab\leq \frac{a^p}{p}+\frac{b^q}{q},[/imath] where the equality holds if and only if [imath]b=a^{p/q}[/imath]. Explicitly [imath]\int\vert fg \vert\leq \Vert f \Vert_p \Vert g \Vert_q \int\left( \frac{\vert f \vert^p}{p\Vert f \Vert_p^p} + \frac{\vert g \vert^q}{q\Vert g \Vert_q^q}\right)=\Vert f \Vert_p \Vert g \Vert_q.[/imath] From here, we see that the equality in Hölder's inequalty holds iff [imath]\frac{\vert fg \vert}{\Vert f \Vert_p \Vert g \Vert_q}=\frac{\vert f \vert^p}{p\Vert f \Vert_p^p} + \frac{\vert g \vert^q}{q\Vert g \Vert_q^q}, \text{ a.e.}[/imath] iff [imath]\frac{\vert g \vert}{\Vert g \Vert_q}=\left( \frac{\vert f \vert}{\Vert f \Vert_p} \right)^{p/q},\text{ a.e.}[/imath] iff [imath]\vert g \vert^q\cdot \Vert f \Vert_p^p=\vert f \vert^p \cdot \Vert g \Vert_q^q,\text{ a.e.}[/imath] Q.E.D. But, assuming that [imath]\Vert f \Vert_p\ne 0[/imath] and [imath]\Vert g \Vert_q\ne 0[/imath], what about when [imath]\Vert f \Vert_p=\infty[/imath] or [imath]\Vert g \Vert_q=\infty[/imath]? How can I deal with it? In the case of Minkowski inequality, suppose that the equality holds and that [imath]g\not \equiv 0[/imath] (and then [imath]\left( \int \vert f+g \vert^p\right)\ne 0[/imath]). I need to prove that [imath]\Vert f \Vert_p[/imath] is multiple of [imath]\Vert g \Vert_q[/imath] almost everywhere. I can reduce to the "Hölder's equality case". I can get [imath]\vert f \vert^p=\left( \int \vert f+g \vert^p\right)^{-1}\Vert f \Vert_p^p\vert f+g \vert^p[/imath] [imath]\vert g \vert^p=\left( \int \vert f+g \vert^p\right)^{-1}\Vert g \Vert_p^p\vert f+g \vert^p[/imath] almost everywhere, but again, using the finiteness and nonzeroness of [imath]\Vert f \Vert_p[/imath] and [imath]\Vert g \Vert_p[/imath]. | 1298377 | How to prove that [imath]\|x+y\|_p<\|x\|_p+\|y\|_p[/imath]?
Let [imath]l_p=\{(x_n)\in\mathbb{R}^\mathbb{N}: \sum_{n=0}^\infty |x_n|^p<\infty\}[/imath] and consider the following norm in [imath]l_p[/imath]: [imath]\|x\|_p=\left(\sum_{n=0}^\infty|x_n|^p\right)^{1/p}[/imath] for [imath]x=(x_n)_{n\in\mathbb{N}}[/imath]. I'm studying this norm, and I have to prove that the inequality [imath]\|x+y\|_p<\|x\|_p+\|y\|_p[/imath] holds for [imath]x\neq \lambda y[/imath], [imath]\lambda>0[/imath]. My attempt: the restriction is needed because if [imath]y=\lambda x[/imath] then the equality [imath]\|x+y\|_p=\|(1+\lambda)x\|_p=|\|x\|_p+\lambda\|x\|_p=\|x\|_p+\|y\|_p[/imath] holds ever. Now I do the particular case [imath]y=-\lambda x[/imath], for [imath]0<\lambda\leq 1[/imath] (and [imath]x\neq 0[/imath]) then: [imath]\|x+y\|_p=\|x-\lambda x\|_p= \|(1-\lambda)x\|_p=(1-\lambda)\|x\|_p=\|x\|_p-\lambda\|x\|_p<\|x\|_p+\|y\|_p[/imath] (the ''[imath]<[/imath]'' is because [imath]\|x\|_p>0 [/imath]) Now if [imath]\lambda>1[/imath], I found that the inequality holds again with '''<'' just following a similar approach. But how can I proceed in the general case? Thanks in advance! |
426762 | Finding The Periodicity of [imath]f(x) = \sin(x/2)\cos(x/3)[/imath]
I want to find the periodicity of the following function: [imath]f(x) = \sin(x/2)\cos(x/3).[/imath] I have calculated the periodicity of the above functions which is [imath]\pi/6[/imath]. Is it correct? Can you please help me to solve this problem? | 127154 | What is the periodicity of the function [imath]\sin(ax) \cos(bx)[/imath] where [imath]a[/imath] and [imath]b[/imath] are rationals?
So, I have a general question first. What happens to the periodicity when we multiply two periodic trig functions with one another ? The next one is very specific, what is the period of the function [imath]g(x)=\sin{(ax)}\cos{(bx)}[/imath], where [imath]a[/imath] and [imath]b[/imath] are rational numbers ? I'd be interested in a proof of sorts. Cheers, Dave |
426962 | Ring and Subring with different Identities
Is there an example of a ring [imath]S[/imath] with identity [imath]1_S[/imath] containing a non-trivial subring [imath]R[/imath] which itself has an identity [imath]1_R[/imath], but [imath]1_R\neq 1_S[/imath] (or equivalently [imath]1_S\notin R[/imath]). I'd also like to know under what conditions the identities have to be equal. I know they must be equal if [imath]S[/imath] has no zero divisors, since for every [imath]r\in R[/imath] we have [imath](1_S-1_R)r=0[/imath] In other words: Is the category of unital rings with identity-preserving morphisms a full subcategory of the category of rings (where morphism are only required to respect addition and multiplication)? | 170953 | Nontrivial subring with unity different from the whole ring?
Is there an example of a ring [imath]R[/imath] with unity and a nontrivial subring [imath]J[/imath], such that [imath]1_J \ne 1_R[/imath]? |
427064 | If [imath]f\in\mathbb{F}_p[x][/imath] is irreducible and has a root in [imath]\mathbb{F}_{p^n}[/imath], then [imath]f[/imath] splits over [imath]\mathbb{F}_{p^n}[/imath]
Let [imath]f(X) \in \mathbb F_p[X][/imath] irreducible with [imath]p[/imath] prime and assume [imath]\exists \alpha \in \mathbb F_{p^n}: f(\alpha) = 0[/imath] where [imath]n \geq 1[/imath]. I then have to prove that [imath]f[/imath] splits over [imath]\mathbb F_{p^n}[/imath]. Some general thoughts are that [imath]\mathbb F_{p^n}[/imath] is the splittingfield of [imath]X^{p^n}-X \in \mathbb F_p[X][/imath]. I can look at [imath]\mathbb F_p(\alpha)[/imath] which must be a subfield of [imath]\mathbb F_{p^n}[/imath] since [imath]\alpha \in \mathbb F_{p^n}[/imath]. I further have that [imath]f[/imath] divides [imath]X^{p^n}-X[/imath] since [imath]f[/imath] is irreducible and [imath]\alpha[/imath] is a zero of both polynomials. I am quite confused. Help please :) Some more considerations: Assume [imath]f(\beta) = 0[/imath] Then with [imath]f| X^{p^n}-X[/imath] is get [imath]f(X)g(X) = X^{p^n}-X[/imath] so [imath]\beta^{p^n}-\beta = 0[/imath] s.t. [imath]\beta \in \mathbb F_{p^n}[/imath] and thus [imath]f[/imath] splits in [imath]\mathbb F_{p^n}[/imath] ? | 409681 | Is the splitting field equal to the quotient [imath]k[x]/(f(x))[/imath] for finite fields?
maybe that's an idiot question. Given a finite field [imath]k[/imath] and some irreducible polynomial [imath]f(x) \in k[x][/imath], then [imath]k_f \cong k[x]/(f(x))[/imath]? I know that it's true if [imath]k[/imath] is the prime field and I think that the statement is not true for general finite fields, however I could not find any counter example. Thanks in advance. |
427278 | A set theory problem - proof
Prove that [imath]X \cap (\bigcup_{i \in I} Y_i) = \bigcap_{i \in I} (Y_i \cap X) [/imath] Is the indices finite? How do I know it isn't [imath]1 \leq i < \infty[/imath]? Also, isn't the RHS just [imath]Y_1 \cap Y_2 \cap ....\cap X[/imath]? (assuming we have finite intersection). EDIT A quick counterexample from myself disproves this. [imath]Y_1 = \{1,2 \}[/imath] and [imath]Y_2 = \{ 2\}[/imath] and [imath]X = \{1\}[/imath] EDIT2 Yes, yes, yes; thank you all for your answers. | 66279 | Intersection distributing over an infinite union, union over an infinite intersection
I'm not really sure how to type this in notation...but I hope this makes sense. I help in proving that: [imath]F \cap \bigcup\limits_{i=1}^\infty E_i = \bigcup\limits_{i=1}^\infty (F\cap E_i)[/imath] And [imath]F \cup \bigcap\limits_{i=1}^\infty E_i = \bigcap\limits_{i=1}^\infty(F\cup E_i)[/imath] I really don't know where to begin with this one. |
126451 | Zeros of a differentiable function are all of finite order and isolated?
If you have a differentiable complex function [imath]f[/imath] in a domain [imath]D[/imath], which is not identically zero, why are the zeros necessarily all of finite order and isolated? | 389320 | Prove that the zeros of an analytic function are finite and isolated
Let us assume that the zeros of [imath]f = \{Z_1,\ldots,Z_n,a\}[/imath] are infinite and converge towards [imath]a[/imath]. The book which I am reading says that any neighborhood of [imath]a[/imath] will contain infinite zeros. Since [imath]f[/imath] is continuous, [imath]f[/imath] must be identically zero in this neighborhood. .....(A) Then as the neighborhood grows, it will be identically [imath]0[/imath] in that neighborhood too until it engulfs the whole of this region. Hence, the function becomes identically zero in the whole region. I don't quite understand (A). I am confused over (A) with this analogy. Consider [imath]f(z) = z-1[/imath]. This has a zero at [imath]z=1[/imath]. Since [imath]f[/imath] is continuous in the [imath]z[/imath] plane, this means that as per (A) points in the neighborhood of [imath]z =1[/imath] should also have mapped value [imath]= 0[/imath]. I am confused; where could I be making a mistake? Thanks |
427676 | Proving a sequence limit. Is this correct?
Let [imath]\{X_n\}[/imath] be a sequence of non-negative real numbers satisfying [imath]\lim_{n\to\infty}X_n=0[/imath] and [imath]X_n\neq 0[/imath] for all [imath]n\in\Bbb N[/imath]. Prove that [imath]\lim_{n\to\infty}X_n\sin(1/X_n)=0.[/imath] This what I do: Let [imath]\epsilon > 0[/imath] be given. Since [imath]\lim X_n[/imath] exists and is [imath]0[/imath], there is a positive integer [imath]N[/imath] with the property that whenever [imath]n \geq N[/imath] one has [imath]|X_n| < \epsilon[/imath]. For any [imath]n \ge N[/imath] we then also have [imath]|X_n \sin(1/X_n)| = |X_n|\cdot|\sin(1/X_n)| \leq |X_n|\cdot 1 = |X_n| < \epsilon.[/imath] [The first inequality in the above line holds because [imath]|\sin(t)| \le 1[/imath] holds for all real [imath]t[/imath], and the second inequality holds because [imath]n \ge N[/imath].] Since [imath]\epsilon[/imath] was an arbitrary positive number, this shows from the definition of the limit that [imath]\lim X_n \sin(1/X_n)[/imath] exists and is [imath]0[/imath]. [Slight generalization with very similar proof: If [imath]\lim X_n = 0[/imath], and [imath]B_n[/imath] is any bounded sequence, then [imath]\lim X_n B_n[/imath] exists and is [imath]0[/imath].] | 427598 | Limits of functions of sequences
1) Let [imath]\{a_n\}[/imath] be a sequence of non-negative real numbers with [imath]\lim_{n\rightarrow\infty}a_n= a>0[/imath]. Prove that [imath]\lim_{n\rightarrow\infty}\sqrt{a_n} = \sqrt{a}[/imath] 2) Let [imath]\{X_n\}[/imath] be a sequence of non-negative real numbers satisfying [imath]\lim_{n\rightarrow\infty}X_n=0[/imath] and [imath]X_n\neq0[/imath] for all [imath]n\in N[/imath]. Prove that [imath]\lim_{n\rightarrow\infty}X_n \sin\left(\frac{1}{X_n}\right)=0[/imath] |
423963 | Question about the Euclidean ring definition
I recently came across the following definition for a euclidean ring: There exists a function [imath]g:R\to\Bbb N_0[/imath] with the following properties: 1.) [imath]\forall x,y \in R[/imath] with [imath] y \neq 0[/imath] there exist [imath]q,r \in R[/imath] with [imath]x = qy + r[/imath] where [imath]r=0[/imath] or [imath]g(r) < g(y)[/imath]. 2.) [imath]\forall x,y\in R\setminus\{0\}[/imath] we have [imath]g(xy)\ge g(x)[/imath] What I don't understand, is the point of the second requirement. I have seen definitions where this is not a requirement. What does 2.) achieve? Thanks very much | 142969 | Problem about the definition of Euclidean domain
In the definition of domain, we first define a degree function [imath]\vartheta: R^\times \rightarrow \mathbb{N}[/imath] with such two constraints: (1) [imath]\vartheta(f)\leq \vartheta(fg)[/imath] for all [imath]f,g\in R^\times[/imath]. (2) for all [imath]f,g\in R[/imath] with [imath]f\in R^\times[/imath], there exist [imath]q,r\in R[/imath] with [imath]g=qf+r[/imath] and either [imath]r=0[/imath] or [imath]\vartheta(r)<\vartheta(f)[/imath]. I wonder why we need the first constraints? I think with only the second constraint, it is enough to prove the theorem: every Euclidean ring is a PID. Can anyone give me a example where the first constraint is used? |
427929 | why can't we divide by zero ?!
in arabic sites which is interested in maths , i find many topics like ,here is a proof that 0=2 . and we answer that the proof is wrong as we can't divide by zero . but i really wonder , why can't we divide by zero ? i think the reason that mathematicians refused dividing by zero is that make us into a contradictions like [imath]1=2=3=...[/imath] and things like this , but those are facts about the physical world , why should mathematicians obey the outside world ? i also have read a news in BBC about a new theorem which find special cases where we can divide by zero , but not details was mentioned i think , have any one had any idea about this ? | 1791907 | Why is [imath] (\frac{1}{0} )[/imath] not defined?
Why is [imath] (\frac{a}{0} ), a \in R [/imath] not defined? if :[imath]a=0[/imath] what? |
424854 | [imath]-1 = 0[/imath] by integration by parts of [imath]\tan(x)[/imath]
I had a calculus final yesterday, and in a question we had to find a primitive of [imath]\tan(x)[/imath] in order to solve a differential equation. A friend of mine forgot that such a primitive could easily be found, tried to integrate [imath]\tan(x)[/imath] by parts... and then arrived to the result [imath]0 = -1[/imath]. The kind of thing you're pretty satisfied to "prove", except during an important exam. :-° So afterwards I tried to do the same : [imath]\begin{align*} \int \tan(x)dx &= \int \sin(x) \times \frac{1}{\cos(x)}dx \\[0.1in] &= -\frac{\cos(x)}{\cos(x)} - \int - \frac{\cos(x) \times \sin(x)}{\cos(x)^2}dx \\[0.1in] &= -1 + \int \tan(x)dx \end{align*}[/imath] And therefore we get : [imath] \int \tan(x)dx = -1 + \int \tan(x)dx \implies 0 = -1[/imath] What? The reasoning sounds about right to me. Could someone explain where something went wrong? Thanks, Christophe. | 806375 | [imath]\int\tan x\ dx[/imath] by integration by parts
Can anyone help me figure out what is wrong in the following step: [imath]\begin{align}\int \tan x\ dx &= \int (\sec x \sin x)\ dx\\ &= -\sec x \cos x + \int \sec x\tan x\cos x\ dx\\ &= -1 + \int \tan x dx\end{align}[/imath] So I got [imath]\displaystyle\int \tan x\ dx = -1 + \int \tan x\ dx[/imath] that is [imath]0 = -1[/imath] ? Any help will be really appreciated! Now I understand. So an indefinite integral can be defined only upto a constant. Thanks a lot! |
428681 | If [imath]L[/imath] is a chain, prove it is finite.
I need ideas on the following problem. Suppose [imath]L[/imath] is a poset and every subset [imath]S[/imath] of [imath]L[/imath] has a top and bottom element. Prove [imath]L[/imath] is a finite chain. All I need to do is prove that [imath]L[/imath] is finite (I have already proved [imath]L[/imath] is a chain). Any ideas or suggestions on solving this problem would be great! Thanks. | 412698 | A set is a finite chain if every subset has a top and bottom element
I am presently attempting Exercise 2 in Kaplansky, Set Theory and Metric Spaces Exercise 2: Let [imath]L[/imath] be a partially ordered set in which every subset has a top and bottom element. Prove that [imath]L[/imath] is a finite chain. Proof: Denote a subset of [imath]L[/imath] by [imath]S[/imath]. If [imath]S[/imath] has a top and bottom element, then [imath]\sup S[/imath] and [imath]\inf S[/imath] exist and are elements of [imath]S[/imath]. Denote them [imath]a[/imath] and [imath]A[/imath] respectively. Since [imath]a,A \in S[/imath] then [imath]S[/imath] is finite. This means that all the elements in [imath]S[/imath] can be ordered from smallest to largest, thus we have: [imath]S= \{ a, ..., A \}[/imath]. Based upon this ordering, given any two elements, [imath]b,c \in S[/imath] one may determine that [imath]b \le c[/imath] or [imath]c \le b[/imath]. Since all the subsets of L are a chain, this implies that L must also be a chain. My Question: I am uncertain on whether the step in bold is valid. I believe it is based upon the transitivity condition in the definition for a partially ordered set, though I would appreciate some feedback on the matter. |
429035 | Find the distance between two lines in [imath] \Bbb R^3 [/imath]
There are two lines in [imath] \Bbb R^3 [/imath] given in parametric form: [imath] l_1: \left\{ \begin{aligned} x &= x_1 +a_1t\\ y &= y_1 +b_1t \\ z &= z_1 +c_1t \\ \end{aligned} \right. [/imath] [imath] l_2: \left\{ \begin{aligned} x &= x_2 +a_2s \\ y &= y_2 +b_2s \\ z &= z_2 +c_2s \\ \end{aligned} \right. [/imath] What's the simplest method (or formula) for finding (the shortest) distance beteen them? example I'm doing: [imath] l_1: \left\{ \begin{aligned} x &= 0 \\ y &= -1 - 2t \\ z &= -2t \\ \end{aligned} \right. [/imath] [imath] l_2: \left\{ \begin{aligned} x &= 3s \\ y &= 1 - s \\ z &= 2 + 4s \\ \end{aligned} \right. [/imath] | 210848 | Finding the shortest distance between two lines
I know how to find the distance between a point and a line, not between two lines. Find the shortest distance between the lines [imath](-1,1,4) + t(1,1,-1)[/imath] and [imath](5,3,-3) + s(-2,0,1)[/imath] Any help would be appreciated. |
340465 | Where can I learn about solving Big-Oh problems that are written in algebra?
Where can I learn about solving Big-Oh problems that are written in algebra? Such as this [imath]\sum_{i=1}^{n} (3i + 2n) = O(n^2)[/imath] | 340518 | [imath]\sum_{i=1}^{n} (3i + 2n)[/imath]
I want to verify what would be the simplified solved version of this summation. [imath]\sum_{i=1}^{n} (3i + 2n)[/imath] Would it be this? [imath] \frac32n^2 + \frac32n + 2n^2 [/imath] |
428327 | Generators of the cyclic group [imath]{\mathbb{Z}_p^*}[/imath]
Given two prime numbers [imath]{p, q > 2}[/imath], where [imath]{p=2q+1}[/imath], I have to show that the cyclic group [imath]{G = \mathbb{Z}_p^*}[/imath] has [imath]{p-1}[/imath] generators. I know that [imath]{|G| = p-1 = 2q}[/imath] and that [imath]{a \in G}[/imath] is a generator iff [imath]{a^2 \neq 1~\text{mod}~p}[/imath] and [imath]{a^q \neq 1~\text{mod}~p}[/imath]. So there have to be [imath]{2q}[/imath] solutions for these two equations, but I have no idea how to show that. Can you give me a hint? | 145578 | What are the generators for [imath]\mathbb{Z}_p^*[/imath] with p a safe prime?
lets consider [imath]\mathbb{Z}_p^*[/imath] with [imath]p = 2 \cdot q + 1[/imath] a safe prime ([imath]p[/imath] and [imath]q[/imath] have to be prime). Then [imath]\varphi\left(p\right) = 2 \cdot q[/imath] is the order of [imath]\mathbb{Z}_p^*[/imath], and [imath]\varphi\left(\varphi\left(p\right)\right) = q-1[/imath] the number of generators in [imath]\mathbb{Z}_p^*[/imath]. Also there are exactly [imath]\frac{p-1}{2} = q[/imath] quadratic residue and [imath]q[/imath] non quadratic residue in [imath]\mathbb{Z}_p^*[/imath]. Now my question: Is every non quadratic residue (except [imath]-1[/imath] if it is a nqr) a generator of [imath]\mathbb{Z}_p^*[/imath]? (due to the fact, that we have [imath]q-1[/imath] generators and quadratic residues cannot be generators) An affirmation or compelling arguments, why I am wrong, are very appreciated, thanks! |
429893 | How to integrate [imath]\int_{0}^{2\pi}\log{|a-e^{i\theta}|}d\theta[/imath]?
I need to calculate the integral [imath]\frac{1}{2\pi}\int_{0}^{2\pi}\log{|a-e^{i\theta}|}d\theta[/imath] When [imath]|a|> 1[/imath], I can show it equals [imath]\log|a|[/imath]. But I failed to handle the situation when [imath]|a|\leqslant 1[/imath]. I know it should be [imath]0[/imath]. I tried to use [imath]\log(a-z)[/imath], but when [imath]|a|\leqslant 1[/imath], such [imath]\log[/imath] can not be defined. So anyone can help? Updated: the second situation should be [imath]|a|<1[/imath]. | 387402 | An integral[imath]\frac{1}{2\pi}\int_0^{2\pi}\log|\exp(i\theta)-a|\text{d}\theta=0[/imath] which I can calculate but can't understand it.
I have encountered this integral:[imath]\frac{1}{2\pi}\int_0^{2\pi}\log|\exp(i\theta)-a|\text{d}\theta=0[/imath] where [imath]|a|<1[/imath] in proving Jensen's formula. because I am stupid, I don't know why it is equal [imath]0[/imath], I can't work it out. After a moment,I find a smart way in Stein's book. After the change of variable [imath]\theta \mapsto -\theta[/imath],the above integral becomes:[imath]\frac{1}{2\pi}\int_0^{2\pi}\log|1-a\exp(i\theta)|\text{d}\theta[/imath] and use Cauchy's integral formulas to [imath]F(z)=1-az[/imath] in [imath]\Omega \supset \overline{\Bbb{D}}=\{z\in\Bbb{C}:|z|\leq 1\}[/imath]:[imath]\log F(0)=\frac{1}{2\pi i}\int_{\partial \Bbb{D} }\frac{\log F(\zeta)}{\zeta}\text{d}\zeta[/imath] we get [imath]0=\log|F(0)|=\frac{1}{2\pi}\int_0^{2\pi}\log(1-a\exp(i\theta))\text{d}\theta[/imath] But I still feel uncomfortable. I think the reason maybe : I can't calculate it directly I don't know why it equal 0 (or I can't see it apparently) Please help me, thanks very much. |
95069 | How would one be able to prove mathematically that [imath]1+1 = 2[/imath]?
Is it possible to prove that [imath]1+1 = 2[/imath]? Or rather, how would one prove this algebraically or mathematically? | 1077300 | Is this a valid proof for [imath]1+1=2[/imath]?
I am extremely new to proofs, and quite bad at them. In studying and practicing the different types of proofs, I developed this very rough proof that [imath]1+1=2[/imath], one of the simplest mathematical truths I could think of, but due to my ignorance I cannot tell if it is valid. Is the following a valid proof, and if not, how can it be improved upon? Proof: Let [imath]a=1[/imath]. It is true that [imath]a \in \mathbb N \setminus \{0\}[/imath] and [imath]\not\exists x \in \mathbb N\setminus \{0\}[/imath] such that [imath]x < a[/imath]. Therefore, [imath]a[/imath] is the least element of [imath]\mathbb N \setminus \{0\}[/imath]. Let [imath]m[/imath] be the least element of [imath]\mathbb N \setminus \{0,1\}[/imath]. [imath]2[/imath] is the least element of [imath]\mathbb N \setminus \{0,1\}[/imath], therefore [imath]m=2[/imath]. Assume that [imath]1+1 \neq 2[/imath]. There are two possible cases: Case 1: [imath]2<1+1[/imath]. If this is true, then because a set of natural numbers only contains positive whole numbers, [imath]2=1[/imath]. Because [imath]2[/imath] is an element of [imath]\mathbb N \setminus \{0,1\}[/imath], [imath]2 \neq 1[/imath]. Therefore this case leads to a contradiction. Case 2: [imath]2>1+1[/imath]. If this is true, then there must exist a natural number [imath]z < 2[/imath] such that [imath]z=1+1[/imath]. Because two is the least element of [imath]\mathbb N \setminus \{0,1\}[/imath], [imath]2=z[/imath]. However, this contradicts the statement that [imath]z<2[/imath], and therefore this case leads to a contradiction. Because assuming that [imath]1+1 \neq 2[/imath] leads to contradiction, it must be true that [imath]1+1=2[/imath]. |
256275 | Is [imath]\{(0,0)\}\cup\{(x,\sin{1\over x}):x\in\mathbb{R},x>0\}[/imath] path connected?
Is [imath]\{(0,0)\cup\{(x,\sin{1\over x}):x\in\mathbb{R},x>0\}[/imath] path connected? I think it is path connected if we neglect the point[imath](0,0)[/imath] it is as we can define a continuous function easily from [imath][0,1][/imath] but if we included the point [imath](0,0)[/imath], then any continuous functions seems would disconnected at [imath](0,0)[/imath] as the [imath]\sin{1\over x}[/imath] vibrate very quick between the value [imath]0,1[/imath]for [imath]x[/imath] close to[imath]0[/imath] so for any [imath]\delta[/imath]-ball at the origin, then choose [imath]\epsilon={1\over2}[/imath], then there must exist some [imath]x\in B_{\delta}((0,0))[/imath] where [imath]\sin{1\over x}>\epsilon[/imath]. Is it correct? if not, is there any way can show it more clearly? Or how to present it in a better way? | 35054 | Topologist's sine curve is not path-connected
Is there a (preferably elementary) proof that the graph of the (discontinuous) function [imath]y[/imath] defined on [imath][0,1)[/imath] by [imath] y(x) =\begin{cases} \sin\left(\dfrac{1}{x}\right) & \mbox{if $0\lt x \lt 1$,}\\\ 0 & \mbox{if $x=0$,}\end{cases}[/imath] is not path connected? |
430257 | Proving [imath]\left\| \frac{\vec{v}}{\|\vec{v}\|}\right\| =1[/imath], [imath]\vec{v}\ne \vec{0}[/imath]
I've been trying to prove that [imath]\left\Vert\dfrac{\vec{v}}{\Vert\vec{v}\Vert}\right\Vert=1, \quad \vec{v}\ne \vec{0}[/imath]. This is my attempt: \begin{align} \vec{v}&\in \mathbb{R}^n, \quad \vec{v}=\begin{pmatrix}v_1 \\ \vdots \\ v_n \end{pmatrix} \\ \Vert\vec{v}\Vert&=\sqrt{\sum_{i=1}^nv_i^2} \\ \frac{\vec{v}}{\sqrt{\sum_{i=1}^nv_i^2}}&=\begin{pmatrix}\frac{v_1}{\sqrt{\sum_{i=1}^nv_i^2}} \\ \vdots \\ \frac{v_n}{\sqrt{\sum_{i=1}^nv_i^2}} \end{pmatrix} =\vec{u} \\ \\ \Vert\vec{u}\Vert^2&=\sum_{k=1}^n\left(\frac{v_k^2}{\sum_{i=1}^nv_i^2}\right) \end{align} I think it is the last step that is either overcomplicated or wrong. | 286913 | Question Concerning Vectors
I am given the information that [imath] \vec{u} = \langle 1, 1/2 \rangle[/imath] and [imath]\vec{v} = \langle 2,3 \rangle[/imath]. There are a few pieces I am asked to find, and these are the one I am having trouble with: [imath]\left\Vert \large \frac{\vec{u}}{|| \vec{u}||}\right\Vert[/imath] The magnitude of vector u is [imath] \sqrt{5}/2[/imath], of which I correctly calculated; the unit vector is then [imath] \langle \large \frac{2 \sqrt{5}}{5}, \frac{ \sqrt{5}}{5} \rangle[/imath]. However, when I go to find the magnitude of the unit vector, I get [imath] \sqrt \frac{3}{5}[/imath], which is clearly not one. What did I do wrong? |
430052 | Rees algebra of a monomial ideal
Let [imath]R=K[x_1,\ldots,x_n][/imath] be a polynomial ring over a field [imath]K[/imath] and [imath]I=(f_1,\ldots,f_q)[/imath] a monomial ideal of [imath]R[/imath]. If [imath]f_i[/imath] is homogeneous of degree [imath]d\geq 1[/imath] for all [imath]i[/imath], then prove that [imath] R[It]/\mathfrak m R[It]\simeq K[f_1t,\ldots, f_q t]\simeq K[f_1,\ldots,f_q] \text{ (as $K$-algebras).} [/imath] [imath]R[It][/imath] denotes the Rees algebra of [imath]I[/imath] and [imath]\mathfrak m=(x_1,\ldots,x_n)[/imath]. | 428689 | Rees algebra of a monomial ideal
User fbakhshi deleted the following question: Let [imath]R=K[x_1,\ldots,x_n][/imath] be a polynomial ring over a field [imath]K[/imath] and [imath]I=(f_1,\ldots,f_q)[/imath] a monomial ideal of [imath]R[/imath]. If [imath]f_i[/imath] is homogeneous of degree [imath]d\geq 1[/imath] for all [imath]i[/imath], then prove that [imath] R[It]/\mathfrak m R[It]\simeq K[f_1t,\ldots, f_q t]\simeq K[f_1,\ldots,f_q] \text{ (as $K$-algebras).} [/imath] [imath]R[It][/imath] denotes the Rees algebra of [imath]I[/imath] and [imath]\mathfrak m=(x_1,\ldots,x_n)[/imath]. |
375874 | show that [imath]f(z)[/imath] is a polynomial in [imath]z.[/imath]
Let [imath]f(z); z = x + iy,[/imath] be an analytic function and [imath]u[/imath] be its real part. If [imath]u[/imath] is a polynomial in the variables [imath]x[/imath] and [imath]y,[/imath] then show that [imath]f(z)[/imath] is a polynomial in [imath]z.[/imath] I need to confirm my attempt: [imath]u,v[/imath] must satifies the C-R equation. So, [imath]u_x=v_y\implies v=\int u_x dy,[/imath] a polynomial in [imath]x,y.[/imath] Consequently [imath]f=u+iv[/imath] is a polynomial in [imath]x+iy.[/imath] | 289020 | to show [imath]f[/imath] is a polynomial in [imath]z[/imath]
given that [imath]f[/imath] is an analytic function with real part [imath]u[/imath] is a polynomial in the variable of [imath]x,y[/imath], [imath]z=x+iy[/imath], we need to show [imath]f[/imath] is a polynomial in [imath]z[/imath], I am kind of puzzled to see the problem, first I would like to say:any hint please |
422926 | faithful representation related to the center
Let [imath]H[/imath] and [imath]K[/imath] be two finite groups, [imath]G = H \times K[/imath]. [imath]\phi[/imath] is an irreducible representation of [imath]H[/imath], and [imath]\psi[/imath] is an irreducible representation of [imath]K[/imath] (both representations are finite-dimensional, over [imath]\mathbb C[/imath]). Then [imath]\phi \sharp \psi[/imath] is faithful if and only if [imath](|Z(H)|,|Z(K)|) =1[/imath]. Relative Definitions: Tensor product of representations. Let [imath]G_i[/imath], [imath]i=1,2[/imath] be two finite groups, [imath](\rho_i, V_i)[/imath] be a representation of [imath]G_i[/imath], [imath]i=1,2[/imath]. Then the representation [imath](\rho_1 \otimes \rho_2, V_1 \otimes V_2)[/imath] is defined to be [imath](\rho_1 \otimes \rho_2)(g) = \rho_1(g) \otimes \rho_2(g)[/imath] for any [imath]g \in G[/imath]. Let [imath]G = G_1 \times G_2[/imath]. [imath]\rho_i[/imath] is a representation of [imath]G_i[/imath], [imath]i=1,2[/imath]. Let [imath]p_i[/imath] be the projection of [imath]G[/imath] to [imath]G_i[/imath], [imath]i=1,2[/imath]. Let [imath]\rho_i'=\rho_i \circ p_i[/imath], [imath]i=1,2[/imath]. Then [imath]\rho_1'[/imath] and [imath]\rho_2'[/imath] are representations of [imath]G[/imath]. Define [imath]\rho_1 \sharp \rho_2 : = \rho_1' \otimes \rho_2'[/imath]. MY PROBLEM: If I set [imath]H[/imath] to be a cyclic group of order [imath]2[/imath], and [imath]K[/imath] to be a cyclic group of order [imath]3[/imath]. [imath]\phi[/imath] and [imath]\psi[/imath] are both [imath]1[/imath]-dimensional representations sending everything to [imath]1[/imath]. Then [imath]\phi \sharp \psi[/imath] still sends everything to [imath]1[/imath], so it is not faithful. Where is wrong? How can I give the proof? | 430839 | [imath]\phi \times \theta[/imath] is faithful iff [imath](|Z(H)| ,|Z(K)|)=1[/imath] for faithful characters [imath]\phi \in Irr(H)[/imath] and [imath]\theta \in Irr(K)[/imath] .
Let [imath]G = H \times K[/imath]. Let [imath]\phi \in \operatorname{Irr}(H)[/imath] and [imath]\theta \in \operatorname{Irr}(K)[/imath] be faithful. Show that [imath]\phi \times \theta[/imath] is faithful iff [imath](|Z(H)| ,|Z(K)|)=1[/imath]. Problem 4.3 of Isaac's character theory. |
431243 | Proof Help: In a group [imath]G[/imath], there exists a [imath]g[/imath] such that [imath]g^2 = e[/imath]
I'm working through an Abstract Algebra book to teach myself, and came across the problem: Prove: If [imath]G[/imath] is a finite group of even order, then there exists a [imath]g\in G[/imath] such that [imath]g^2 = e[/imath] and [imath]g \ne e[/imath]. (In this book, [imath]e[/imath] is used as the identity element. I don't know if that's standard or not...) I have a proof outline, but don't really know how to write it in a formal way. My idea is as follows: First, note that this problem is equivalent to saying that "there exists a [imath]g\in G[/imath] such that [imath]g = g^{-1}[/imath]. Also note that the identity element is its own inverse. Since the identity element is its own inverse, we have an odd number of elements remaining in the group that need their inverse "assigned." Assign all but one of the remaining elements an inverse so that none satisfy [imath]g = g^{-1}[/imath]. You have one remaining element left; the rest of the elements already have inverses. As inverses are unique, this element must be its own inverse. My question is twofold: Am I even close to being on the right track as far as a proof outline goes? If so, what can I do to make this proof rigorous, and not just an outline? | 188519 | Prove that any finite group [imath]G[/imath] of even order contains an element of order [imath]2.[/imath]
Prove that any finite group [imath]G[/imath] of even order contains an element of order [imath]2.[/imath][ Let [imath]t(G)[/imath] be the set [imath]\{g\in G|g\neq g^{-1}\}[/imath]. Show that [imath]t(G)[/imath] has an even number of elements and every nonidentity element of [imath]G-t(G)[/imath] has order [imath]2[/imath].] My proof: Observe that for any [imath]g\in t(G)[/imath] there exists [imath]g^{-1}(\neq g)\in t(G)[/imath] since every element in a group has an inverse. Thus if there are [imath]n[/imath] elements that belong to [imath]t(G)[/imath] we must have another [imath]n[/imath] elements that belong to [imath]G[/imath]. Hence, [imath]t(G)[/imath] has an even number of elements. Now any [imath]g\in G-t(G)[/imath] must also have inverse [imath]g^{-1}(=g)\in G-t(G)[/imath] since all [imath]g\neq g^{-1}[/imath] belong to [imath]t(G)[/imath], so all the elements except the identity in [imath]G-t(G)[/imath] have [imath]g=g^{-1}[/imath] thus [imath]g^2=id[/imath] i.e., every element of [imath]G-t(G)[/imath] has order [imath]2.[/imath] Therefore, we can conclude that there is atleast one element of order [imath]2.[/imath] in [imath]G[/imath]. Can anyone check my proof? Thank you. |
431297 | Points at which function is continuous is countable intersection of open sets
Let [imath]f[/imath] be a real-valued function defined on [imath]\mathbb{R}[/imath]. Show that the set of points at which [imath]f[/imath] is continuous is a countable intersection of open sets. Not sure where to start on this one... what would be the open sets? | 138072 | How to show that the set of points of continuity is a [imath]G_{\delta}[/imath]
I am trying to solve this exercise from Royden's 3rd edition. The question is as follows: Let [imath]f[/imath] be a real-valued function defined for all real numbers. Show that the set of points at which [imath]f[/imath] is continuous is a [imath]G_{\delta}[/imath]. Let [imath]A_n = \{y : \text{there is a }~\delta_y \gt 0 : |f(s)-f(t)|\lt 1/n ~ \text{whenever}~ s,t \in (y-\delta, y+\delta)\}\;.[/imath] Then by the definition of open sets, [imath]A_n[/imath] is open. To complete the proof, I need help in showing that [imath]f[/imath] is continuous at say [imath]x[/imath] if and only if [imath]x\in \cap A_n[/imath]. If [imath]f[/imath] is continuous at [imath]x[/imath], the there is a [imath]\delta \gt 0[/imath] such that [imath]|f(x) - f(a)| \lt 1/n[/imath] whenever, [imath]x\in (a-\delta, a+\delta)[/imath]. so [imath]x \in A_n[/imath] son it must be in [imath]\cap A_n[/imath]. Thanks. |
431912 | An unknown limit with nth root: [imath]\lim\limits_{n\to\infty}n(x^{1/n}-1)[/imath]
How can I find such a limit: [imath]\lim_{n\to\infty}n(x^{1/n}-1)[/imath] I tried using a kind of binomial formulas. But nothing helped so far. Thank you! | 124667 | How to find [imath]\lim\limits_{n\to\infty} n·(\sqrt[n]{a}-1)[/imath]?
How to compute the following limit? [imath]\lim\limits_{n\to\infty} n·(\sqrt[n]{a}-1)[/imath] We know that it's got something to do with [imath]\ln[/imath] or [imath]\exp[/imath]. We know that [imath]\lim\limits_{n\to\infty} \sqrt[n]{a} = 1[/imath] but it seems to not to be true that therefore [imath]\lim\limits_{n\to\infty} n·(\sqrt[n]{a}-1) = \lim\limits_{n\to\infty} n·(1-1) = 0[/imath]. What we know is [imath]\lim\limits_{n\to\infty} (1-\frac{1}{n})^n = \lim\limits_{n\to\infty} (1 + \frac{1}{n})^n \lim\limits_{n\to\infty} (1-\frac{1}{n})^{n+1} = e[/imath] and [imath]\lim\limits_{n\to\infty} (1+\frac{x}{n})^n = \exp(x)[/imath] |
216657 | Convergence Problem.
Let [imath](a_k)[/imath] be a sequence of real numbers and let [imath]b_k=\frac{a_1+a_2+\dots a_k}{k}[/imath] for each [imath]k\in \mathbb{N}[/imath]. Prove that if [imath](a_k)[/imath] converges to [imath]\alpha\in \mathbb{R}[/imath], then the sequence [imath](b_k)[/imath] also converges to [imath]\alpha[/imath]. My answer: Let [imath]\epsilon > 0[/imath]. Since ([imath]a_k[/imath]) converges to [imath]\alpha[/imath], then there exists a [imath]N_{\epsilon1}\in\mathbb{N}[/imath] such that [imath]|a_k - \alpha | < \frac{\epsilon}{2}[/imath] for all [imath]k \geq N_{\epsilon1}[/imath]. Notice that [imath]\frac{a_1+...+a_n - n\alpha}{k}[/imath] converges to [imath]0[/imath] as [imath]k[/imath] approaches infinity. Thus, there exists [imath]N_{\epsilon2}\in\mathbb{N}[/imath] such that [imath]|\frac{a_1+...+a_n - n\alpha}{k}|<\frac{\epsilon}{2}[/imath] for all [imath]k>N_{\epsilon2}[/imath]. Now, let [imath]N= [/imath]max[imath] (N_{\epsilon1},N_{\epsilon2})[/imath]. Then for any [imath]k>N[/imath] we get, [imath]\begin{array}{rcll} |b_k-\alpha| &=& |b_k - \alpha*1|&\\ &=& |\frac{a_1+...+a_k}{k} - \frac{k\alpha}{k}|&\\ &=& |\frac{a_1+...+a_n}{k} + \frac{[a(n+1)+...+a_k]}{k} - \frac{(k-n)\alpha}{k} - \frac{n\alpha}{k}|& \text{(notice that $N_{\epsilon1}\leq N<k$)}\\ &=& |\frac{a_1+...+a_n - n\alpha}{k} + \frac{[a(n+1)+...+a_k - (k-n)\alpha]}{k}|&\\ &=& |\frac{a_1+...+a_n - n\alpha}{k} + \frac{[(a(n+1) -\alpha) + ... + (a_k - \alpha)]}{k}|&\\ &=& |\frac{a_1+...+a_n - n\alpha}{k} + \frac{(a(n+1) -\alpha)}{k} + ... + \frac{(a_k - \alpha)}{k}|&\\ &\leq& |\frac{a_1+...+a_n - n\alpha}{k}| + |\frac{(a(n+1) - \alpha)}{k}| + ... + |\frac{(a_k - \alpha)}{k}|& \text{(by the triangle inequality)}\\ &=&|\frac{a_1+...+a_n - n\alpha}{k}| + \frac{|(n+1) - \alpha|}{k} + ... + \frac{|a_k - \alpha|}{k}&\\ &<& \frac{\epsilon}{2} + [\frac{\epsilon}{2}/k + ... + \frac{\epsilon}{2}/k]& \\ &=& \frac{\epsilon}{2} + \frac{\epsilon}{2}*\frac{(k-n)}{k}&\\ &<& \frac{\epsilon}{2} + \frac{\epsilon}{2}& \text{(since $k-n<k$)}\\ &=& \epsilon& \end{array}[/imath] And since we found such an [imath]N[/imath] for arbitrary [imath]\alpha[/imath], then [imath](b_k)[/imath] converges to [imath]\alpha[/imath] by definition. I'm not sure i kept all my variables straightened out. How does this look and if you have a better idea let me know! | 883722 | Prove that {[imath]b_n[/imath]} is convergent with [imath] b_n \to L[/imath]
Let [imath]\lbrace a_n\rbrace[/imath] be a convergent sequence with [imath]a_n \to L[/imath] Define [imath] b_n = \frac{ a_1 + a_2 + ... a_n}{n} \forall n \in \mathbb Z_+ [/imath] Prove that [imath]\lbrace b_n\rbrace[/imath] is convergent with [imath] b_n \to L[/imath] I tried to go through [imath]\displaystyle \lim_{n \to \infty} f \{a_n \} = f \lim_ {n \to \infty} \{ a_n \} [/imath] but I can't figure out [imath]f[/imath]. |
432427 | Finding the inverse of a matrix involving reciprocals of positive integers
The result of Example 16 suggests that perhaps the matrix [imath]A=\begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} \ \ \cdots \ & \frac{1}{n} \\ \frac{1}{2}& \frac{1}{3} & \cdots & \frac{1}{n+1}\\ \frac{1}{3} & . & \cdots & .\\. & . & \cdots & .\\\frac{1}{n} & \frac{1}{n+1} & \cdots & \frac{1}{2n-1} \end{bmatrix}[/imath] is invertible and [imath]A^{-1}[/imath] has integer entries. Can you prove that? | 302581 | The inverse of the matrix [imath]\{1/(i+j-1)\}[/imath]
Let [imath]n[/imath] be a positive integer. Show that the matrix [imath]\begin{pmatrix} 1 & 1/2 & 1/3 & \cdots & 1/n \\ 1/2 & 1/3 & 1/4 & \cdots & 1/(n+1) \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1/n & 1/(n+1) & 1/(n+2) & \cdots & 1/(2n-1) \end{pmatrix}[/imath] is invertible and all the entries of its inverse are integers. This is an exercise in Hoffman and Kunze's linear algebra book. Any hints will be appreciated! |
432777 | Is [imath]2^{\aleph_0}=c[/imath]?
When I search about the cardinality of real number set in Wiki (http://en.wikipedia.org/wiki/Cardinality_of_the_continuum) I found: By the Cantor–Bernstein–Schroeder theorem we conclude that [imath]c=|P(\mathbb{N})|=2^{\aleph_0}[/imath] And the Cantor–Bernstein–Schroeder state that if [imath]A\preceq B[/imath] and [imath]B\preceq A[/imath], then [imath]A \sim B[/imath]. How can I use it to prove that [imath]c=|P(\mathbb{N})|=2^{\aleph_0}[/imath]. How to create a bijection function mapping [imath]P(\mathbb{N})[/imath] onto [imath]\mathbb{R}[/imath]? | 270671 | Easiest way to prove that [imath]2^{\aleph_0} = c[/imath]
[imath]\aleph_0[/imath] is the cardinality of the set of natural numbers, [imath]\aleph_0 = |N|[/imath]. [imath]c[/imath] is the cardinality of the continuum, i.e. the set of real numbers [imath]c = |R|[/imath]. I know that [imath]|P(A)| = 2^{|A|}[/imath]. This means that the cardinality of the power set of a set is 2 raised to the power of the cardinality of that set. This basically means that to prove [imath]2^{\aleph_0} = c[/imath], I need to prove [imath]c = |P(N)|[/imath]. I've seen the Cantor diagonal argument, although I have no idea on how to use it to prove this. Could anyone suggest ideas of the simplest way to prove that [imath]2^{\aleph_0} = c[/imath]? |
432847 | [imath]\sum_{r=0}^{k-1}4^r[/imath]
[imath]\sum_{r=0}^{k-1}4^r[/imath] Hi, I was wondering whether someone could explain to me how to work out the above series. I know it ends up equalling [imath]\frac{4^k-1}3[/imath] but I don't know why or how I get to it. Thank you :D | 432793 | Summing a series with a changing power
[imath]\sum_{r=0}^{k-1}4^r[/imath] Hi, I was wondering whether anyone could explain how to work this out. I know the end result is [imath]\frac{4^k-1}{3}[/imath], but I don't know why or how to get there. Thank you :D |
265908 | Prove that [imath]s[/imath] is finite and find and an [imath]n[/imath] so large that [imath]S_n[/imath] approximates [imath]s[/imath] to 3 decimal places.
Let [imath]S_n[/imath] represent its partial sum, and let [imath]s[/imath] represent its value. Prove that [imath]s[/imath] is finite and find and an [imath]n[/imath] so large that [imath]S_n[/imath] approximates s to 3 decimal places. [imath]\sum_{k=1}^{\infty}\left(\frac{k}{k+1}\right)^{k^2}[/imath] Solution we use root test i think. But how to calculate this? Thanks. | 263724 | Prove that s is finite and find an n so large that [imath]S_n[/imath] approximate s to three decimal places.
[imath]S=\sum_{k=1}^\infty\left(\frac{k}{k+1}\right)^{k^2};\hspace{10pt}S_n=\sum_{k=1}^n\left(\frac{k}{k+1}\right)^{k^2}[/imath] Let [imath]S_n[/imath] represent its partial sums and let [imath]S[/imath] represent its value. Prove that [imath]S[/imath] is finite and find an n so large that [imath]S_n[/imath] approximate [imath]S[/imath] to three decimal places. Solution: first of all, I think that we Will use L'hopital rule and then use root test while starting to solve this. But how? |
432786 | Hatcher Problem 1.3.15
Would just like a sanity check. I don't see the necessity of the locally path connected condition on [imath]A[/imath]. The proof that [imath]\tilde{A}[/imath] is a covering space seems straightforward. We use the path-connectedness of [imath]A[/imath] obviously to talk about a single [imath]\pi_1[/imath]. For the other part, take [imath]i[/imath] as the inclusion map of [imath]A[/imath] in [imath]X[/imath]. If we have [imath][f]\in p_*(\pi_1(\tilde{A}))[/imath], we have a homotopy to the constant map already since [imath]\tilde{X}[/imath] is simply connected, so [imath][f] \in \ker(i_*)[/imath]. For any [imath][f] \in \ker(i_*)[/imath], by Proposition 1.31 the image of [imath]f[/imath] in [imath]X[/imath] lifts to some loop [imath]\tilde{f}[/imath] in [imath]\tilde{X}[/imath] with basepoint in [imath]\tilde{A}[/imath]. Here the path connectedness of [imath]\tilde{A}[/imath] is useful. Then [imath][p(\tilde{f})] = [f][/imath], so [imath][f] \in p_*(\pi_1(\tilde{A}))[/imath]. And that should be it? | 76252 | Another Question in Hatcher
First of all, I apologize for asking yet another question about the hypotheses of a problem in Hatcher, but the statement of one of his problems has stumped me again. The problem is 1.3.15. It reads as follows: Let [imath]p:\widetilde{X}\rightarrow X[/imath] be a simply-connected covering space of [imath]X[/imath] and let [imath]A\subseteq X[/imath] be a path-connected, locally path-connected subspace, with [imath]\widetilde{A}[/imath] a path-component of [imath]p^{-1}(A)[/imath]. Show that [imath]p:\widetilde{A}\rightarrow A[/imath] is the covering space corresponding to the kernel of the map [imath]\pi _1(A)\rightarrow \pi _1(X)[/imath]. My question is: why do we need to assume that [imath]A[/imath] is path-connected and locally path-connected? Here is a sketch of the "proof" I have that does not make use of these hypotheses: If we have a loop in [imath]\widetilde{A}[/imath], this loop is nulhomotopic in [imath]\widetilde{X}[/imath], and hence will be nulhomotpic when mapped down to [imath]X[/imath] via [imath]p[/imath]. On the other hand, if you go the other way and first map the loop in [imath]\widetilde{A}[/imath] down to [imath]A[/imath] via [imath]p[/imath], and then include this loop into [imath]X[/imath], by commutativity of the corresponding diagram I can't draw here, you get the same loop as the first way, which is a nulhomtopic loop. This shows that [imath]\pi _1[p]\left( \pi _1\left[ \widetilde{A}\right] \right)[/imath] is contained in the kernel of the inclusion. Convserly, if we have a loop in [imath]A[/imath] that is nulhomotpic in [imath]X[/imath], we can lift this to a loop in [imath]\widetilde{X}[/imath], and hence we can choose such a lift that is contained in [imath]A'[/imath]. Pushing this lift down via [imath]p[/imath], shows that the containment is actually equality. Any idea where this proof goes wrong? Thanks in advance! |
433057 | If [imath]E_1, E_2[/imath] are connected and [imath]A_1\subseteq E_1[/imath] and [imath]A_2\subseteq E_2[/imath] then [imath](E_1\times E_2)-(A_1\times A_2)[/imath] is connected
Let [imath]E_1[/imath], [imath]E_2[/imath] connected metric spaces. Let [imath]A_1\subseteq E_1[/imath] and [imath]A_2\subseteq E_2[/imath] proper subsets. Show that the complement of [imath]A_1\times A_2[/imath] [imath](E_1\times E_2)-(A_1\times A_2)[/imath] is connected. I have tried to perform the test using the result that says if [imath]A, B[/imath] are connected and [imath]A\cap B\neq \varnothing[/imath] then [imath]A\cup B[/imath] is connected. Note that [imath](E_1\times E_2)-(A_1\times A_2) = \left[E_1\times(E_2-A_2)\right]\cup\left[(E_1-A_1)\times E_2\right][/imath] So it suffices to prove that the sets [imath]E_1\times(E_2-A_2)[/imath] and [imath](E_1-A_1)\times E_2[/imath] are related and non-empty | 214317 | Prove that [imath](X\times Y)\setminus (A\times B)[/imath] is connected
I'm reading topology of Munkres and I have a problem that stuck me for a while. I'm so greatful if anyone can help me with this. Let [imath]A[/imath] be a proper subset of [imath]X[/imath], and let [imath]B[/imath] is a proper subset of [imath]Y[/imath]. If [imath]X[/imath] and [imath]Y[/imath] are connected, show that [imath](X\times Y) \setminus (A\times B)[/imath] is connected. Thanks so much for your consideration ^^ |
237817 | Derivative of [imath]x^2[/imath]
This seems too easy, but here's the question: [imath]x^2[/imath] is [imath]x + x + ...+ x[/imath] (with [imath]x[/imath] terms). Its derivative is [imath]1 + 1 + ... + 1[/imath] (also [imath]x[/imath] terms). So the derivative of [imath]x^2[/imath] seems to be [imath]x[/imath]. And another expression: we know that if [imath]y = nx[/imath], then [imath]y' = n[/imath], so that if [imath]y = x * x[/imath] then [imath]y' = x[/imath]. But we know by formula that if [imath]y = x^2[/imath], then [imath]y' = 2x[/imath] So, how to prove [imath]y' = x[/imath] is wrong ? Thanks | 855526 | Why [imath]2x[/imath]? Can't it be [imath]x[/imath]?
So today in my school our neighbor class monitors were complaining to that few of our students were yelling and making noise. Actually the case was that we were having very aggressive debate over a question. The question is quite simple but there is lot of confusion regarding it's solution because according to different students there are two very different approach toward this question. The question is, Question: Find the derivative of [imath]f(x)=x+x+x+x+x+...\quad x \mathrm\quad {times}[/imath] So here are two approach that my classmates are pretending to do. First one (that I think is the correct approach) is that we can simplify the above function as [imath]x[/imath] times [imath]x[/imath] ( i.e. [imath]f(x)=x^2[/imath]**)** , so that equation becomes, [imath]f(x)=x^2[/imath] and simply we can write [imath]f'(x)=2x[/imath] But my classmates are pretending to say that there is another way ( that I think is wrong ) to do this. In this method without simplifying the [imath]f(x)[/imath] we can differentiate individual terms in the sum. i.e. [imath]f'(x)=\frac{d(x)}{dx}+\frac{d(x)}{dx}+\frac{d(x)}{dx}+\frac{d(x)}{dx}+...\quad x \quad\mathrm{times}[/imath] [imath]\implies f'(x)=1+1+1+1+1...\quad x \quad \mathrm{times}[/imath] [imath]\implies f'(x)=x[/imath] So in this way we are having two different answers (i.e [imath]2x[/imath] and [imath]x[/imath]). However I know that [imath]2x[/imath] is the correct answer but honestly speaking I am really tiered to explain them how. However I tried to tell them that in second method (that I think is wrong method) we unintentionally ignore that [imath]x[/imath] times term and didn't involve in the differentiation while we do take care of that in the first method (that I think is correct). Some are saying that [imath]x[/imath] is right while some are pretending to [imath]2x[/imath] to be right while some are saying that that both are wrong. I need a perfect answer that can satisfy my classmates. Thanks! |
276593 | Showing there is only one isomorphism between well ordered sets using transfinite induction
I need to show specifically using transfinite induction that given two well-ordered sets [imath]\left(A,<_{1}\right)[/imath] and [imath]\left(B,<_{2}\right)[/imath] there is only one isomorphism between them. To do this I want to show by induction that there is only one way to define the mapping and that is: [imath]\forall\, x\in A: \varphi\left(x\right)=\min\left(B\backslash\varphi\left(\left\{ a\in A\,:\, a<_{1}x\right\} \right)\right)[/imath] Showing the base of the induction is no problem and the induction hypothesis would be that [imath]\forall\, y<_{1}x\,:\,\varphi\left(y\right)=\min\left(B\backslash\varphi\left(\left\{ a\in A\,:\, a<_{1}y\right\} \right)\right)[/imath] My problem is that I can't seem to manage to use the hypothesis in order to prove the step. I've found a couple of ways to show this definition of the mapping is indeed necessary but none of those used the induction hypothesis. My plan was to mark [imath]c_{0}=\min\left(B\backslash\varphi\left(\left\{ a\in A\,:\, a<_{1}x\right\} \right)\right)[/imath] Using the fact [imath]\varphi[/imath] is surjective take [imath]y\in A[/imath] such that [imath]\varphi\left(y\right)=c_{0}[/imath] and show that it can't be that [imath]y<_{1}x[/imath] or [imath]x<_{1}y[/imath] and thus necessarily [imath]y=x[/imath]. If someone could give me a way to show either one of these are false using the induction hypothesis it would solve my problem. Thanks a lot! | 2618450 | Proving isomorphism between [imath]X[/imath] and [imath]Y[/imath]
let [imath](X,\leq)[/imath] and [imath](Y,\preceq)[/imath] be well-order sets and they are isomorphic. Prove that there is one isomorphism between [imath]X[/imath] and [imath]Y[/imath]. Notice that [imath]X[/imath] and [imath]Y[/imath] are sets, so by isomorphism I mean isomorphism between sets. |
434036 | [imath]\sum \limits_{n=1}^{\infty}{a_n^2}[/imath] converges [imath]\implies \sum \limits_{n=1}^{\infty}{\frac{a_n}{n}}[/imath]
I need some help to solve the following problem: *Show that if [imath]\sum \limits_{n=1}^{\infty}{a_n^2}[/imath] converges, then [imath]\sum \limits_{n=1}^{\infty}{\dfrac{a_n}{n}}[/imath] converges. I tried to solve it by using the Ratio Test, but got anything. Regards and thanks a lot. | 112579 | If [imath]\sum_{n=1}^{\infty} a_n^{2}[/imath] converges, then so does [imath]\sum_{n=1}^{\infty} \frac {a_n}{n}[/imath]
Let [imath]$a_1,a_2,a_3,\ldots$[/imath] be reals. Prove that if [imath]\sum_{n=1}^{\infty} a_n^{2}[/imath] converges, then so does [imath]\sum_{n=1}^{\infty} \frac {a_n}{n}[/imath]. For this I have shown the case for when [imath] a_n^{2} \le\frac {|a_n|}{n}[/imath] [imath]\Rightarrow[/imath] [imath] |a_n|\le\frac {1}{n}[/imath] [imath]\Rightarrow[/imath] [imath]\frac {|a_n|}{n} \le \frac{1}{n^{2}}[/imath] and we know that [imath]\sum_{n=1}^{\infty} \frac {1}{n^{2}}[/imath] converges and hence [imath]\sum_{n=1}^{\infty}\frac {a_n}{n}[/imath] converges by the comparison test. Now considering [imath] a_n^{2} \ge\frac {|a_n|}{n}[/imath] [imath]\Rightarrow[/imath] [imath]\frac {|a_n|}{n} \le a_n^{2}[/imath] [imath]\rightarrow[/imath] combining the two cases for any n we have: [imath]\frac {|a_n|}{n}\le\frac{1}{n^{2}}+a_n^{2}[/imath] Hence using the comparion test again we know that [imath]\sum_{n=1}^{\infty} a_n^{2}[/imath] converges and [imath]\sum_{n=1}^{\infty} \frac {1}{n^{2}}[/imath] converges hence the sum converges so we can conclude that [imath]\sum_{n=1}^{\infty} \frac {a_n}{n}[/imath] is absoluetly convergent [imath]\Rightarrow[/imath] convergent. Not to sure if this is correct, any help would be much appreciated, many thanks. |
434041 | The tangent space [imath]T_{(x,x)}(\triangle)[/imath] is the diagonal of [imath]T_x(X) \times T_x(X).[/imath] - Is this proof legit?
If [imath]\triangle[/imath] is the diagonal of [imath]X \times X[/imath], show that its tangent space [imath]T_{(x,x)}(\triangle)[/imath] is the diagonal of [imath]T_x(X) \times T_x(X).[/imath] Is the following proof legit? [imath]T_{(x,x)} \Delta \subseteq \Delta \subseteq T_x X \times T_x X[/imath] All tangent vectors at the point [imath](x,x)[/imath] in [imath]\Delta[/imath] are described by velocity vectors of curves passing through [imath](x,x)[/imath]. Suppose [imath]c(t)[/imath] is a curve through [imath](x,x)[/imath] such that [imath]c(0) = (x,x)[/imath], then the velocity vector is given by [imath]c'(0)[/imath]. Any [imath]c(t)[/imath] that lies in the diagonal looks like [imath]c(t) = (\gamma(t), \gamma(t))[/imath], where [imath]\gamma(t)[/imath] is a curve in [imath]X[/imath] through the point [imath]x[/imath] i.e., [imath]\gamma(0) = x[/imath] and [imath]\gamma'(0) = v \in T_x X[/imath]. Then [imath]c'(t) = (\gamma'(t), \gamma'(t))[/imath] and [imath]c'(0) = (v,v) \in \Delta \subseteq T_x X \times T_x X[/imath]. [imath]T_x X \times T_x X \subseteq \Delta \subseteq T_{(x,x)} \Delta[/imath] All tangent vectors at the point [imath]x[/imath] in [imath]X[/imath] are described by velocity vectors of curves passing through [imath]x[/imath]. Suppose [imath]r(t)[/imath] is a curve through [imath]x[/imath] such that [imath]r(0) = x[/imath], then the velocity vector is given by [imath]r'(0)[/imath]. Any element that lies in [imath]T_x \times T_x[/imath] looks like [imath](r^\prime(t), r^\prime(t))[/imath], which is element in [imath]T_{x,x}(\Delta)[/imath]. Thanks. This question intends to fill in the answer for the question GP 1.2.10(b) The tangent space [imath]T_{(x,x)}(\triangle)[/imath] is the diagonal of [imath]T_x(X) \times T_x(X).[/imath] And I do realize this is a very popular question, here's another identical problem again with partial solution: Show that the tangent space of the diagonal is the diagonal of the product of tangent space. | 429609 | GP 1.2.10(b) The tangent space [imath]T_{(x,x)}(\triangle)[/imath] is the diagonal of [imath]T_x(X) \times T_x(X).[/imath]
If [imath]\triangle[/imath] is the diagonal of [imath]X \times X[/imath], show that its tangent space [imath]T_{(x,x)}(\triangle)[/imath] is the diagonal of [imath]T_x(X) \times T_x(X).[/imath] I don't have the slightest idea on how to do this. By definition, the tangent space of [imath]X[/imath] at [imath]x[/imath] is the image of the map [imath]d\phi_0: \mathbb{R}^k \rightarrow \mathbb{R}^N[/imath], where [imath]x + d\phi_0(v)[/imath] is the best linear approximation to [imath]\phi: V \rightarrow X[/imath] at 0. [imath]\phi: V \rightarrow X[/imath] is a local parametrization around [imath]x[/imath], [imath]X[/imath] sits in [imath]\mathbb{R}^N[/imath] and [imath]V[/imath] is an open set in [imath]\mathbb{R}^k[/imath], and [imath]\phi(0) = x[/imath]. Thanks. |
434365 | Statements regarding ordinal numbers
Let [imath]m[/imath] and [imath]n[/imath] be infinite ordinal numbers Which of the following is true a) [imath]m<n \Rightarrow |m|^{|m|}<n^{|n|}[/imath] b) [imath]m+n[/imath]= Max{[imath]m,n[/imath]} c) [imath]m=n \Rightarrow |m|=|n|[/imath] d)[imath]|m|=|n| \Rightarrow m=n [/imath] e) Max{[imath]m,n[/imath]} <[imath]|m|+|n|[/imath] I could only understand that d) is false due to the cardinality of the two different ordinals can be the same and c is true! Please help on the rest of the statements | 433580 | Cardinal numbers
Suppose [imath]m, n[/imath] are infinite ordinal numbers. [imath]a) m=n → |m|=|n|[/imath] [imath]b)|m|=|n| →m=n[/imath] [imath]c)m<n→ |m|<|n|[/imath] [imath]d)|\max{(m,n)}|< |m|+|n|[/imath] [imath]e)|m|<|n| →|m|^{|n|}<|n|^m[/imath] Which of the above statements are true? (a) looks true but I do not know the way to work it out. Please help |
434480 | True/ False: If [imath]f '(c) < 0[/imath], then [imath]f[/imath] is concave down at [imath]x = c[/imath]?
Determine whether the following statement is true or false. Whether true or false, explain why. If false, also correct the statement. If [imath]f'(c) < 0[/imath], then [imath]f[/imath] is concave down at [imath]x = c[/imath]. I know the question is False because if [imath]f'(c) < 0[/imath], then the graph is always decreasing. How can I correct the question to be true? | 429377 | True or false: if [imath]f'(c)<0[/imath], then [imath]f[/imath] is concave down at [imath]x=c[/imath]?
How can I determine the following statement is true or false? If [imath]f'(c)<0[/imath], then [imath]f[/imath] is concave down at [imath]x=c[/imath] ? |
434597 | Set of finite Borel measures is Banach
Let [imath]\Omega \subset \mathbb{R}^n[/imath] and [imath]M(\Omega)[/imath] be the set of finite Borel measures on [imath]\Omega[/imath]. With the norm [imath]\lVert \mu \rVert = |\mu|(\Omega)[/imath], this space is Banach. Can someone show me how it's proven that it's Banach? In particular completeness. Also I'm not sure what [imath]|\mu|[/imath] means.. why absolute values? | 178921 | Space of Complex Measures is Banach (proof?)
How can we prove that the space of Complex Measures is Complete? with the norm of Total Variation. I have stuck on the last part of the proof where I have to prove that the limit function of a Cauchy sequence of measures has the properties of Complex measures. We are using the norm of total variation :[imath]\lVert \mu\rVert = \lvert \mu \rvert(X)[/imath] |
435031 | Prove that [imath]R[/imath] has unity
Let [imath]R[/imath] be a finite ring such that [imath]x^2=x[/imath] for all [imath]x[/imath] in [imath]R[/imath]. Prove that [imath]R[/imath] has unity. I was able to show that it was commutative. Proof: [imath]x^2=x[/imath] [imath]x^2-x = 0[/imath] [imath]x (x-1) =0 [/imath] thus [imath]x = 0[/imath] or [imath]x = 1[/imath]. Since [imath]x\cdot 1 = x[/imath]. 1 is the unity of [imath]R[/imath]. Thus [imath]R[/imath] has unity. I feel like there is something wrong with my solution. | 305337 | Do these two observations suffice to show that a finite boolean ring must be of the form [imath]\mathbb{Z}_2\times\cdots\times\mathbb{Z}_2[/imath]?
Question: If [imath]R[/imath] is a finite boolean ring, then show [imath]R \cong \mathbb Z_2 \times \mathbb Z_2 \times \cdots\times \mathbb Z_2[/imath]. I know that [imath]\mathrm{char}(R) =2[/imath] [imath]R[/imath] has [imath]2^k[/imath] elements for some [imath]k[/imath] (this is because, if [imath]p\neq 2[/imath] is a prime with [imath]p\mid|R|[/imath], then there exists a non-zero [imath]x \in R[/imath] s.t. [imath]px=0[/imath] and [imath]p [/imath] is odd [imath]\implies x=0[/imath], contradiction). Does this suffice to answer my question? |
435176 | How prove this [imath]|a-b||c|\le|a-c||b|+|b-c||a|[/imath] if [imath]a,b,c[/imath] are vectors?
Let [imath]a,b,c[/imath] are vector numbers,show that [imath]|a-b||c|\le|a-c||b|+|b-c||a|[/imath] my idea: [imath]a_{i},b_{i},c_{i}\in \mathbb{R}[/imath] let [imath]a=(a_{1},a_{2},\cdots,a_{n})[/imath] [imath]b=(b_{1},b_{2},\cdots,b_{n})[/imath] [imath]c=(c_{1},c_{2},\cdots,c_{n})[/imath] [imath]\Longleftrightarrow \sqrt{\sum_{i=1}^{n}(a_{i}-b_{i})^2}\sqrt{\sum_{i=1}^{n}c^2_{i}}\le\sqrt{\sum_{i=1}^{n}(a_{i}-c_{i})^2}\sqrt{\sum_{i=1}^{n}b^2_{i}}+\sqrt{\sum_{i=1}^{n}(b_{i}-c_{i})^2}\sqrt{\sum_{i=1}^{n}a^2_{i}}[/imath] I think follow is use Cauchy-Schwarz or Minkowski inequality. But I can't, I feel this inequality is interesting. | 426649 | For any three vectors [imath]x,y,z\in\mathbb{R}^d[/imath], we have [imath] \|y-z\|\cdot\|x\|\leq\|x-y\|\cdot\|z\|+\|z-x\|\cdot\|y\|[/imath]
Does anyone know a proof of the following problem? Problem: Show that for any three vectors [imath]{\bf x}, {\bf y}, {\bf z}\in \mathbb{R}^d[/imath] the following holds, [imath] \|{\bf y} - {\bf z}\|\cdot \|{\bf x}\| \leq \|{\bf x} - {\bf y}\|\cdot \|{\bf z}\| + \|{\bf z} - {\bf x}\|\cdot\|{\bf y}\|.[/imath] All of the norms are Euclidean 2-norm. |
435419 | Proving [imath]\sum_{k=1}^nk^3 = \left(\sum_{k=1}^n k\right)^2[/imath] using complete induction
I tried to prove the following statement using complete induction but I couldn't manage to solve it because I got a complex notation eventually. The statement is the following: [imath]\sum_{k=1}^nk^3 = \left(\sum_{k=1}^n k\right)^2[/imath] I hope someone can help me out! | 460774 | Prove that for all positive integers [imath]n........[/imath]
Prove that for all positive integers [imath]n, 1^3+2^3+\ldots+n^3=(1+2+\ldots+n)^2[/imath] |
435539 | Derive the centroid of an area from a limiting procedure
I wondered if the centroid of an area could be derived as a limit from the centroid of the solid of revolution built by the same area revolving around an axis of revolution. e.g.: the volume centroid of a spherical wedge is [imath]C_{V} = \frac{3\pi}{16}r\frac{\sin[\theta]}{\theta}[/imath], where [imath]r[/imath] is the radius of the sphere and [imath]\theta[/imath] is the dihedral angle. The centroid of a semicircular lamina is [imath]C_{A} = \frac{4r}{3\pi}[/imath] Now I tought it was possible a limit procedure like [imath]\lim\limits_{\theta\to 0} C_{V}[/imath]=[imath]C_{A}[/imath], but that's false and I would like to know why (I guess that has something to do with the passage from three-dimension to two-dimension) Thanks | 435403 | Derive the centroid of an area from a limiting procedure
I wondered if the centroid of an area could be derived as a limit from the centroid of the solid of revolution built by the same area revolving around an axis of revolution. e.g.: the volume centroid of a spherical wedge is [imath]C_{V} = \frac{3\pi}{16}r\frac{\sin[\theta]}{\theta}[/imath], where [imath]r[/imath] is the radius of the sphere and [imath]\theta[/imath] is the dihedral angle. The centroid of a semicircular lamina is [imath]C_{A} = \frac{4r}{3\pi}[/imath] Now I tought it was possible a limit procedure like [imath]\lim\limits_{\theta\to 0} C_{V}[/imath]=[imath]C_{A}[/imath], but that's false and I would like to know why (I guess that has something to do with the passage from three-dimension to two-dimension) Thanks |
392 | Intuitive understanding of the derivatives of [imath]\sin x[/imath] and [imath]\cos x[/imath]
One of the first things ever taught in a differential calculus class: The derivative of [imath]\sin x[/imath] is [imath]\cos x[/imath]. The derivative of [imath]\cos x[/imath] is [imath]-\sin x[/imath]. This leads to a rather neat (and convenient?) chain of derivatives: sin(x) cos(x) -sin(x) -cos(x) sin(x) ... An analysis of the shape of their graphs confirms some points; for example, when [imath]\sin x[/imath] is at a maximum, [imath]\cos x[/imath] is zero and moving downwards; when [imath]\cos x[/imath] is at a maximum, [imath]\sin x[/imath] is zero and moving upwards. But these "matching points" only work for multiples of [imath]\pi/4[/imath]. Let us move back towards the original definition(s) of sine and cosine: At the most basic level, [imath]\sin x[/imath] is defined as -- for a right triangle with internal angle [imath]x[/imath] -- the length of the side opposite of the angle divided by the hypotenuse of the triangle. To generalize this to the domain of all real numbers, [imath]\sin x[/imath] was then defined as the Y-coordinate of a point on the unit circle that is an angle [imath]x[/imath] from the positive X-axis. The definition of [imath]\cos x[/imath] was then made the same way, but with adj/hyp and the X-coordinate, as we all know. Is there anything about this basic definition that allows someone to look at these definitions, alone, and think, "Hey, the derivative of the sine function with respect to angle is the cosine function!" That is, from the unit circle definition alone. Or, even more amazingly, the right triangle definition alone. Ignoring graphical analysis of their plot. In essence, I am asking, essentially, "Intuitively why is the derivative of the sine the cosine?" | 800821 | Why is [imath]\cos(x)[/imath] the derivative of [imath]\sin(x)[/imath]?
The derivative of [imath]\sin(x)[/imath] is [imath]\cos(x)[/imath], and the derivative of [imath]\cos(x)[/imath] is [imath]-\sin(x)[/imath]. Is there a simple proof of this, preferably using pictures? |
435762 | [imath]\pi^4+\pi^5 < e^6 [/imath]
Any idea about this inequality: [imath]\large \pi^4+\pi^5<e^6[/imath] Any hints would be appreciated. | 198508 | How can one prove that [imath]\pi^4 + \pi^5 < e^6[/imath]?
A proof of the inequality using properties of [imath]\pi[/imath] and [imath]e[/imath], for example, is what I'm looking for. Not calculator approximations showing the inequality holds. |
413071 | Order of special linear group [imath]SL_2 (F)[/imath]
In ex. 1.2.3 on p.21 of Diamond's A first course in modular forms, I was trying to show that [imath]|SL_2(\mathbb{Z}/N \mathbb{Z})|=N^3 \prod_{p|N}(1-1/p^2).[/imath] First of all, I knew that If [imath]F[/imath] is a field with [imath]q[/imath] elements, then [imath]|GL_2(F)|=(q^2-1)(q^2-q)[/imath]. If we consider the surjective homomorphism [imath]\det : GL_2(F) \rightarrow F^*[/imath] with kernel [imath]SL_2(F)[/imath], then we have [imath]|SL_2(F)|=(q^2-1)(q^2-q)/(q-1)=q(q^2-1)=q^3-q[/imath]. Therefore if we replace [imath]F[/imath] by [imath]\mathbb{Z}/N \mathbb{Z}[/imath], we get [imath]|SL_2(\mathbb{Z}/N \mathbb{Z})|= N^3-N[/imath], but thats' not agree with the book, so what's wrong with me...? In fact, in the part (a) of the problem, it asks us show [imath]|SL_2(\mathbb{Z}/p^e \mathbb{Z})|=p^{3e}(1-1/p^2)[/imath] for [imath]p[/imath] prime (that's already doesn't agree with my conclusion...), and uses chinese remainder theorem to conclude. | 341033 | How to calculate [imath]|\operatorname {SL}_2(\mathbb Z/N\mathbb Z)|[/imath]?
the answer should be [imath]|\operatorname {SL}_2(\mathbb Z/N\mathbb Z)|=N^{3}\prod_{p|N}(1-{1 /p^2})[/imath] But first how to prove [imath]|\operatorname {SL}_2(\mathbb Z/p^e\mathbb Z)|=p^{3e}(1-{1 /p^2})[/imath] |
436089 | Show that the limit exists and find it's value?
The Fibonacci series defined recursively by [imath]x(1) = 1, x(2) = 2[/imath] and [imath]x(n+1) = x(n) + x(n-1)[/imath] Find[imath]\lim_{n\rightarrow\infty}\frac{x(n+1)}{x(n)}[/imath] | 132305 | How to prove that [imath]\lim \limits_{n\rightarrow \infty} \frac{F_{n+1}}{F_n}=\frac{\sqrt{5}+1}{2}[/imath]
How would one prove that [imath]\lim_{n\rightarrow \infty} \frac{F_{n+1}}{F_n}=\frac{\sqrt{5}+1}{2}=\varphi[/imath] where [imath]F_n[/imath] is the nth Fibonacci number and [imath]\varphi[/imath] is the Golden Ratio? |
436545 | Cluster point of a sequence [imath]\{x_n\}[/imath] is the limit of some subsequence - Axiom of Choice?
In a metric space, a cluster point of a sequence [imath]\{x_n\}[/imath] is the limit of some subsequence. The only proof that I know works like this: Construct a sequence [imath]\delta _k \to 0[/imath]. For each [imath]\delta _k[/imath] find a point [imath]x_{n_k}[/imath] in the sequence which is within [imath]\delta_k[/imath] of the cluster point. Make sure that [imath]n_{k} \lt n_{k+1}[/imath], so that you really have a subsequence, and you can do this because there are only finitely many terms before index [imath]n_{k}[/imath]. Question: Is this using Countable Choice? If so is it because we're "picking" a sequence element for each [imath]\delta_k[/imath]? If we are using Countable Choice, then is there a way to prove this without using CC? I apologize if this is a duplicate question - I know there have been previous posts where it's explained why Sequential Continuity Implies [imath]\delta - \epsilon[/imath] Continuity requires Countable Choice. The argument used in that proof seems similar to this one. Thank you very much. | 183404 | (ZF)subsequence convergent to a limit point of a sequence
Arthur's answer; (ZF) Prove 'the set of all subsequential limits of a sequence in a metric space is closed. Let [imath]\{p_n\}[/imath] be a sequence in a metric space [imath]X[/imath]. Let [imath]B=\{p_n|n\in\mathbb{N}\}[/imath] and [imath]p[/imath] be a limit point of [imath]B[/imath]. It first seemed obvious that there exists a subsequence convergent to [imath]p[/imath], but i realized that I can construct a 'infinite subset' of [imath]\{p_n\}[/imath] and form a new sequence convergent to [imath]p[/imath], but can't construct a 'subsequence'. Help me how to construct a subsequence. Additional Question; What is the precise definition of convergent? I think the definition should mention ordering of [imath]\mathbb{N}[/imath], but every definition i saw doesn't mention this. For example, let [imath]p_n = 1/n[/imath]. Say [imath]G[/imath] is the usual ordering of [imath]\mathbb{N}[/imath] (i.e. Well-ordered by [imath]\in[/imath]) If i follow this ordering, [imath]\{p_n\}[/imath] is convergent to 0. But if i follow [imath]G^{-1}[/imath], it's convergent to 1. Or is it defined by the point where is the most, unofficially speaking, 'dense'? |
437036 | Sturm-Liouville Questions (2)
In thinking about Sturm-Liouville theory a bit I see I have no actual idea what is going on. The first issue I have is that my book began with the statement that given [imath]L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)[/imath] the problem [imath]L[y] \ = \ f[/imath] can be re-cast in the form [imath]L[y] \ = \ \lambda y[/imath]. Now it could be a typo on their part but I see no justification for the way you can just do that! More importantly though is the motivation for Sturm-Liouville theory in the first place. The story as I know it is as follows: Given a linear second order ode [imath]F(x,y,y',y'') = L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)[/imath] it is an exact equation if it is derivable from a differential equation of one order lower, i.e. [imath]F(x,y,y',y'') = \frac{d}{dx}g(x,y,y').[/imath] The equation is exact iff [imath]a''(x) - b'(x) + c(x) = 0. [/imath] If [imath]F[/imath] is not exact it can be made exact on multiplication by a suitable integrating factor [imath]\alpha(x)[/imath]. This equation is exact iff [imath](\alpha(x)a(x))'' - (\alpha(x)b(x))' + \alpha(x)c(x) = 0 [/imath] If you expand this out you get the Adjoint operator [imath]L^*[\alpha(x)] \ = \ (\alpha(x)a(x))'' \ - \ (\alpha(x)b(x))' \ + \ \alpha(x)c(x) \ = 0 [/imath] If you expand [imath]L^*[/imath] you see that we can satisfy [imath]L \ = \ L^*[/imath] if [imath]a'(x) \ = \ b(x)[/imath] & [imath]a''(x) \ = \ b'(x)[/imath] which then turns [imath]L[y][/imath] into something of the form [imath]L[y] \ = \ \frac{d}{dx}[a(x)y'] \ + \ c(x)y \ = \ f(x).[/imath] Thus we seek an integratiing factor [imath]\alpha(x)[/imath] so that we can satisfy this & the condition this will hold is that [imath]\alpha(x) \ = \ \frac{1}{a(x)}e^{\int\frac{b(x)}{a(x)}dx}[/imath] Then we're dealing with: [imath]\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \alpha(x)f(x)[/imath] But again, by what my book said they magically re-cast this problem as [imath]\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \lambda \alpha(x) y(x)[/imath] Then calling [imath]\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ ( \alpha(x)c(x)y \ - \ \lambda \alpha(x) )y(x) \ = \ 0[/imath] a Sturm-Liouville problem. My question is, how can I make sense of everything I wrote above? How can I clean it up & interpret it, like at one stage I thought we were turning our 2nd order ode into something so that it reduces to the derivative of a first order ode so we can easily find first integrals then the next moment we're pulling out eigenvalues & finding full solutions - what's going on? I want to be able to look at [imath]a(x)y'' \ + \ b(x)y' \ + \ c(x)y \ = \ f(x)[/imath] & know how & why we're turning this into a Sturm-Liouville problem in a way that makes sense of exactness & integrating factors, thanks for reading! | 436516 | Sturm-Liouville Questions
In thinking about Sturm-Liouville theory a bit I see I have no actual idea what is going on. The first issue I have is that my book began with the statement that given [imath]L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)[/imath] the problem [imath]L[y] \ = \ f[/imath] can be re-cast in the form [imath]L[y] \ = \ \lambda y[/imath]. Now it could be a typo on their part but I see no justification for the way you can just do that! More importantly though is the motivation for Sturm-Liouville theory in the first place. The story as I know it is as follows: Given a linear second order ode [imath]F(x,y,y',y'') = L[y] = a(x)y'' + b(x)y' + c(x)y = f(x)[/imath] it is an exact equation if it is derivable from a differential equation of one order lower, i.e. [imath]F(x,y,y',y'') = \frac{d}{dx}g(x,y,y').[/imath] The equation is exact iff [imath]a''(x) - b'(x) + c(x) = 0. [/imath] If [imath]F[/imath] is not exact it can be made exact on multiplication by a suitable integrating factor [imath]\alpha(x)[/imath]. This equation is exact iff [imath](\alpha(x)a(x))'' - (\alpha(x)b(x))' + \alpha(x)c(x) = 0 [/imath] If you expand this out you get the Adjoint operator [imath]L^*[\alpha(x)] \ = \ (\alpha(x)a(x))'' \ - \ (\alpha(x)b(x))' \ + \ \alpha(x)c(x) \ = 0 [/imath] If you expand [imath]L^*[/imath] you see that we can satisfy [imath]L \ = \ L^*[/imath] if [imath]a'(x) \ = \ b(x)[/imath] & [imath]a''(x) \ = \ b'(x)[/imath] which then turns [imath]L[y][/imath] into something of the form [imath]L[y] \ = \ \frac{d}{dx}[a(x)y'] \ + \ c(x)y \ = \ f(x).[/imath] Thus we seek an integratiing factor [imath]\alpha(x)[/imath] so that we can satisfy this & the condition this will hold is that [imath]\alpha(x) \ = \ \frac{1}{a(x)}e^{\int\frac{b(x)}{a(x)}dx}[/imath] Then we're dealing with: [imath]\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \alpha(x)f(x)[/imath] But again, by what my book said they magically re-cast this problem as [imath]\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ \alpha(x)c(x)y \ = \ \lambda \alpha(x) y(x)[/imath] Then calling [imath]\frac{d}{dx}[\alpha(x)a(x)y'] \ + \ ( \alpha(x)c(x)y \ - \ \lambda \alpha(x) )y(x) \ = \ 0[/imath] a Sturm-Liouville problem. My question is, how can I make sense of everything I wrote above? How can I clean it up & interpret it, like at one stage I thought we were turning our 2nd order ode into something so that it reduces to the derivative of a first order ode so we can easily find first integrals then the next moment we're pulling out eigenvalues & finding full solutions - what's going on? I want to be able to look at [imath]a(x)y'' \ + \ b(x)y' \ + \ c(x)y \ = \ f(x)[/imath] & know how & why we're turning this into a Sturm-Liouville problem in a way that makes sense of exactness & integrating factors, thanks for reading! |
437205 | If [imath]X[/imath] is well-ordered set, how to prove that [imath]\mathcal{P}(X)[/imath] can be linearly ordered?
I'm having troubles solving the following exercise, proposed in T. Jech, 'Set theory' Exercises 5.4. Let [imath]X[/imath] is well-ordered set then [imath]\mathcal{P}(X)[/imath] can be linear ordered. [Let [imath]X<Y[/imath] if the least element of [imath]X\triangle Y[/imath] belongs to [imath]X[/imath].] I can prove that [imath]<[/imath] satisfies totality (i.e. [imath]X<Y[/imath] or [imath]X=Y[/imath] or [imath]Y<X[/imath]) and [imath]X\not< X[/imath], but I can't prove transitivity of [imath]<[/imath]. Thanks for any help. | 90078 | Totally ordering the power set of a well ordered set.
Let's say I take a set [imath]S[/imath], where [imath]S[/imath] can be well ordered. From what I understand, one can use that well ordering to totally order [imath]\mathscr{P}(S)[/imath]. How does a body actually use the well ordering of [imath]S[/imath] to construct a total ordering of [imath]\mathscr{P}(S)[/imath]? Perhaps construct is too strong a word. Is there a way to prove that a total ordering of [imath]\mathscr{P}(S)[/imath] just exists, no need to explicitly state what it is, since that might be very difficult. |
437281 | Determine all integers x,y that satisfy the equation [imath]x^3 = y^2 +2[/imath]
Determine all integers x,y that satisfy the equation [imath]x^3 = y^2 +2 [/imath] My attempt to the solution Taking mod 2 we get that x an y can't be even. How to proceed? By trial and error there is a solution [imath]y =\pm5[/imath] and [imath]x =3[/imath]. | 347425 | Integer solutions for [imath]x^3+2=y^2[/imath]?
I've heard a famous result that [imath]26[/imath] is the only integer, such that [imath]26-1=25[/imath] is a square number and [imath]26+1=27[/imath] is a cubic number.In other words, [imath](x,y)=(5,3)[/imath] is the only solution for [imath]x^2+2=y^3[/imath]. How if we make it like this [imath]x^3+2=y^2[/imath]? Are there any integral solutions? If so, finite or infinite many? I've checked first [imath]100[/imath] naturals and no solutions satisfy the equation. However, I have no ideas how to start the proof. |
437175 | Optimizing [imath]x^2+y^2[/imath] from two given equations?
What is the maximum value of [imath]x^2+y^2[/imath], where [imath](x,y)[/imath] are solutions to: [imath]2x^2+5xy+3y^2=2[/imath] and [imath]6x^2+8xy+4y^2=3[/imath] Note: Calculus is not allowed. I tried everything I could but whenever I got for example or [imath]x^2+y^2=f(y)[/imath] or [imath]f(x)[/imath] the function [imath]f[/imath] would always be a concave up parabola, so I could not find a maximum for either variable. And by the way I know that you can solve for [imath]x[/imath] and [imath]y[/imath] using the quadratic formula and get [imath]4[/imath] different solutions but I am looking for less messy way. I've asked this question before, but I didn't get the nice answer I wanted. Thanks. This question came from a math competition from the Math Honor Society, Mu Alpha Theta. | 407244 | Not so easy optimization of variables?
What is the maximum value of [imath]x^2+y^2[/imath], where [imath](x,y)[/imath] are solutions to [imath]2x^2+5xy+3y^2=2[/imath] and [imath]6x^2+8xy+4y^2=3[/imath]. (calculus is not allowed). I tried everything I could but whenever I got for example [imath]or[/imath] [imath]x^2+y^2=f(y)[/imath] or [imath]f(x)[/imath] the function [imath]f[/imath] would always be a concave up parabola, so I could not find a maximum for either variable. However, I also don't see how you could solve it if you leave both variables on one side. And by the way I know that you can solve for [imath]x[/imath] and [imath]y[/imath] using the quadratic formula and get [imath]4[/imath] different solutions but I am looking for a much more efficient way than that. This question came from a math competition from the Math Honor Society, Mu Alpha Theta. |
437767 | Why don't we define division by zero as an arbritrary constant such as [imath]j[/imath]?
Why don't we define [imath]\frac 10[/imath] as [imath]j[/imath] , [imath]\frac 20[/imath] as [imath]2j[/imath] , and so on? I know that by following the rules of math this eventually leads to [imath]1=2[/imath] , but we could make an exception and say that [imath]j[/imath] is the only number such that [imath]0*j \not= 0[/imath] , and put other restrictions necesary so that we don't get contradictions. We do this for [imath]i[/imath] , so why can't we do it here? For example, [imath]i^2[/imath] , is defined as [imath]-1[/imath] , but you could also say [imath]i^2=\sqrt {-1} *\sqrt {-1}=\sqrt {(-1)(-1)}=\sqrt{1}=1[/imath] , but we make an exception for this. | 259584 | Why don't we define "imaginary" numbers for every "impossibility"?
Before, the concept of imaginary numbers, the number [imath]i = \sqrt{-1}[/imath] was shown to have no solution among the numbers that we had. So we declared [imath]i[/imath] to be a new type of number. How come we don't do the same for other "impossible" equations, such as [imath]x = x + 1[/imath], or [imath]x = 1/0[/imath]? Edit: OK, a lot of people have said that a number [imath]x[/imath] such that [imath]x = x + 1[/imath] would break the rule that [imath]0 \neq 1[/imath]. However, let's look at the extension from whole numbers to include negative numbers (yes, I said that I wasn't going to include this) by defining [imath]-1[/imath] to be the number such that [imath]-1 + 1 = 0[/imath]. Note that this breaks the "rule" that "if [imath]x \leq y[/imath], then [imath]ax \leq ay[/imath]", which was true for all [imath]a, x, y[/imath] before the introduction of negative numbers. So I'm not convinced that "That would break some obvious truth about all numbers" is necessarily an argument against this sort of thing. |
438137 | Is a subset of [imath]\mathbb{R}^{n}[/imath] that is homeomorphic to [imath]\mathbb{R}^{n}[/imath] necessarily open?
Let [imath]A[/imath] be a subset of [imath]\mathbb{R}^{n}[/imath], such that [imath]A[/imath] is homeomorphic to [imath]\mathbb{R}^{n}[/imath]. Is [imath]A[/imath] open in [imath]\mathbb{R}^{n}[/imath]? | 208188 | Subset of [imath]\mathbb{R}^n[/imath] homeomorphic to [imath]\mathbb{R}^n[/imath]
Let [imath]X[/imath] be a subset of [imath]\mathbb{R^n}[/imath]. The topology on [imath]X[/imath] is induced by the topology of [imath]\mathbb{R^n}[/imath]. If there is an homeomorphism from [imath]X[/imath] onto [imath]\mathbb{R}^n[/imath], is it true that [imath]X[/imath] is open in [imath]\mathbb{R}^n[/imath] ? I think yes, but how can it be shown? (of course, [imath]X[/imath] is open in [imath]X[/imath]). If [imath]x \in X[/imath], [imath]H_{n-1}(\mathbb{R}^n-\{f(x)\})= \mathbb{Z}[/imath] so [imath]H_{n-1}(X-\{x\})= \mathbb{Z}[/imath], but I don't know how to continue. Thanks in advance. |
31725 | Intuition behind Matrix Multiplication
If I multiply two numbers, say [imath]3[/imath] and [imath]5[/imath], I know it means add [imath]3[/imath] to itself [imath]5[/imath] times or add [imath]5[/imath] to itself [imath]3[/imath] times. But If I multiply two matrices, what does it mean ? I mean I can't think it in terms of repetitive addition. What is the intuitive way of thinking about multiplication of matrices? | 1550010 | Why is the matrix multiplication defined as it is?
Matrix multiplication is defined as: Let [imath]A[/imath] be a [imath]n \times m[/imath] matrix and [imath]B[/imath] a [imath]m\times p[/imath] matrix, the product [imath]AB[/imath] is defined as a matrix of size [imath]n\times p[/imath] such that [imath](AB)_i,_j = \sum\limits_{k=1}^mA_i,_kB_k,_j[/imath]. For what good reason did mathematicians define it like this? |
438694 | Definition of opposite category
I don't quite understand the definition of opposite category. Wikipedia defines [imath]\mathcal{C}^\text{op}[/imath] to be the category formed by 'reversing' morphisms. So for instance, in Grps, given a groups [imath]G,H[/imath] and a homomorphism [imath]f: G \to H[/imath], what does reversing [imath]f[/imath] mean? Do we reverse [imath]f[/imath] into a homomorphism [imath]H \to G[/imath]? | 44768 | Maps in Opposite Categories
Given some category [imath]{\mathcal C}[/imath], then the opposite category will consist of the same objects with the morphisms "turned around." Given [imath]f:A\rightarrow B[/imath], for [imath]A,B[/imath] objects of [imath]{\mathcal C}[/imath], then in general do we have a canonical [imath]f^{op}:B\rightarrow A[/imath] which is induced by [imath]f[/imath]? Slight Motivation: This is something that's just been bothering me recently, and I feel like I am thinking about this concept incorrectly. To make this question slightly less vague, if we had, say, the category of rings and we had the mapping [imath]f:{\mathbb Z}\rightarrow {\mathbb Z}/2{\mathbb Z}[/imath] by [imath]f[/imath] sends even numbers to 0 and odd numbers to 1, then what is the corresponding map in the opposite category, [imath]f^{op}:{\mathbb Z}/2{\mathbb Z}\rightarrow {\mathbb Z}[/imath] induced by [imath]f[/imath]? Also, I always see this [imath]f^{op}[/imath] that I've mentioned just written as [imath]f[/imath]; is this shorthand, or is there something deeper there? |
438802 | Why does [imath]\int\limits_0^1 {\dfrac{{x - 1}}{{\ln x}}} \;\text{d}x=\ln2[/imath]?
I have found that [imath]\int\limits_0^1 {\dfrac{{x - 1}}{{\ln x}}} \;\text{d}x=\ln2[/imath] but I can't prove it. Any hint? Thank you in advance | 100495 | Showing that [imath] \int_{0}^{1} \frac{x-1}{\ln(x)} \mathrm dx=\ln2 [/imath]
I would like to show that [imath] \int_{0}^{1} \frac{x-1}{\ln(x)} \mathrm dx=\ln2 [/imath] What annoys me is that [imath] x-1 [/imath] is the numerator so the geometric power series is useless. Any idea? |
438832 | Prove that [imath]\gcd(a, b) = 1 ⇒ \gcd(a^2, b^2) = 1[/imath]
I have a question from a sample exam I find difficults to solve: Prove that if [imath]\gcd(a, b) = 1 ⇒ \gcd(a^2, b^2) = 1[/imath] . I don't have any idea how to start. I'd like to get helped. thanks! | 75676 | How can I prove that [imath]\gcd(a,b)=1\implies \gcd(a^2,b^2)=1[/imath] without using prime decomposition?
How can I prove that if [imath]\gcd(a,b)=1[/imath], then [imath]\gcd(a^2,b^2)=1[/imath], without using prime decomposition? I should only use definition of gcd, division algorithm, Euclidean algorithm and corollaries to those. Thanks for your help! |
438913 | Determine all units of the domain [imath]\mathbb{Z}[\sqrt{2}][/imath]
I would like to determine all units of the domain [imath]\mathbb{Z}[\sqrt{2}][/imath]. Recall that if [imath]D[/imath] is an integral domain, then a unit of [imath]D[/imath] is an associate of the multiplicative identity [imath]1[/imath], i.e. [imath]u \in D[/imath] is a unit iff [imath]u[/imath] divides [imath]1[/imath] and [imath]1[/imath] also divides [imath]u[/imath]. Equivalently, [imath]u \in D[/imath] is a unit iff it is a divisor of [imath]1[/imath], i.e. iff [imath]\exists x \in D[/imath] such that [imath]ux=1[/imath]. Thus, units are also called invertible elements. The set [imath]\mathbb{Z}[\sqrt{2}] := \{a+b \sqrt{2}: a,b \in \mathbb{Z} \}[/imath], under the usual addition and multiplication operations, forms a domain. How do I find all its units? Suppose [imath]a+b \sqrt{2}[/imath] is a unit. Then [imath](a+b \sqrt{2})(x+y \sqrt{2})=1[/imath] for some [imath]x,y \in \mathbb{Z}[/imath]. Thus, [imath]ax+2by=1[/imath] and [imath]ay+bx=0[/imath]. Solving for [imath]x,y[/imath], we obtain that [imath]x=\frac{a}{a^2 - 2b^2}[/imath] and [imath]y=\frac{-b}{a^2 - 2b^2}[/imath] must be integers. How do I obtain the set of values of [imath]a,b[/imath] for which [imath]x[/imath] and [imath]y[/imath] are integral? Since [imath](\sqrt{2}-1)(\sqrt{2}+1)= 2-1^2=1[/imath], the two factors [imath]\pm 1 + 1 \sqrt{2}[/imath] are units. Can we systematically find all the units of [imath]\mathbb{Z}[\sqrt{2}][/imath]? This answer given here to a related question shows only that if [imath]a+b \sqrt{2}[/imath] is a unit and is > 1, and if [imath]a,b>0[/imath], then [imath]a+b \sqrt{2}[/imath] is of the form [imath](1+\sqrt{2})^n[/imath]. The case where [imath]a,b[/imath] are of opposite signs, and the case where the unit is between 0 and 1 (such units exist, for eg, [imath]3-2 \sqrt{2}[/imath]) are open. | 280878 | The units of [imath]\mathbb Z[\sqrt{2}][/imath]
How can I show that the units [imath]u[/imath] of [imath]R=\mathbb Z[\sqrt{2}][/imath] with [imath]u>1[/imath] are [imath](1+ \sqrt{2})^{n}[/imath] ? I have proved that the right ones are units because their module is one, and it is said to me to do it by induction on [imath]b[/imath] and multiplication by [imath]-1+\sqrt{2}[/imath]. I have already shown that the units of this ring has norm [imath]1[/imath] and all the numbers with norm [imath]1[/imath] are units, this may help. |
68726 | A closed subset of an algebraic group which contains [imath]e[/imath] and is closed under taking products is a subgroup of [imath]G[/imath]
A closed subset of an algebraic group which contains [imath]e[/imath] and is closed under taking products is a subgroup of [imath]G[/imath]. Denote this set as [imath]X[/imath]. If the condition of [imath]X[/imath] being closed is dropped, this statement does not hold. The set of nonzero integers in [imath]\mathbb{G}_m[/imath] over [imath]\mathbb{C}[/imath] is a counterexample. It suffice to prove that for any [imath]x \in X[/imath], [imath]x^{-1}X = X[/imath]. As [imath]X[/imath] is closed under taking products, it is clear that [imath]X \subseteq x^{-1}X[/imath]. In order to prove the inverse inclusion, the closedness of [imath]X[/imath] (under Zariski topology) must be used. Let [imath]\phi: G \rightarrow G, y \mapsto x^{-1}y[/imath] is a homoemorphism of [imath]G[/imath] as an algebraic variety. So [imath]x^{-1}X = \phi(X)[/imath] is a closed subset of [imath]G[/imath] containing [imath]X[/imath]. But Why are they equal? Thanks very much. | 1070787 | Closed subset of a affine linear group
Let [imath]G\subseteq GL_n(\mathbb{C})[/imath] a Zariski-closed linear subgroup and [imath]X\subseteq G[/imath] closed with [imath]X*X\subseteq X[/imath] and [imath]e \in X[/imath]. Then [imath]X[/imath] is a subroup. I am not sure how to start here. I know that [imath]X^{-1}[/imath] is also closed and that [imath]e\in X\cap X^{-1}[/imath]. Is that going to help me? Thank you. |
439083 | Number theory: if [imath](a,b)=1[/imath], show that [imath](ac,b)=(c,b)[/imath]
Mostre que, se [imath](a, b) = 1[/imath], então [imath](a · c, b) = (c, b)[/imath] Como posso fazer isso usando [imath](a,b) = 1\implies[/imath] Existem [imath]m,n[/imath] naturais tais que [imath]am - bn = 1[/imath] Tentative translation: Show that if [imath](a,b)=1[/imath], then [imath](ac,b)=(c,b)[/imath]. How can this be done using [imath](a,b)=1\implies[/imath] there exist natural numbers [imath]m,n[/imath] such that [imath]am-bn=1[/imath]? | 20889 | Prove that if [imath]\gcd( a, b ) = 1[/imath] then [imath]\gcd( ac, b ) = \gcd( c, b ) [/imath]
I know it might be too easy for you guys here. I'm practicing some problems in the textbook, but this one really drove me crazy. From [imath]\gcd( a, b ) = 1[/imath], I have [imath]ax + by = 1[/imath], where should I go from here? The extra [imath]ac[/imath] is so annoying. Any hint? Thanks, Chan |
439318 | How determinate the torsion subgroup and the normal maximal [imath]\pi[/imath]-subgroup of [imath]\mathbb{R}/\mathbb{Z}[/imath]?
Let [imath]\mathbb{R}[/imath] and [imath]\mathbb{Z}[/imath] the real numbers and the integers, respectively. Consider [imath]G =\mathbb{R}/\mathbb{Z}[/imath]. Find [imath]O_{\pi}(G)[/imath] - ie, the (unique) normal maximal [imath]\pi[/imath]-subgroup of [imath]G[/imath]; [imath]T(G)[/imath] - ie, the torsion subgroup of [imath]G[/imath]. | 439142 | Maximal normal [imath]\pi[/imath]-subgroups and torsion subgroups
Let be [imath]\mathbb{R}[/imath] the real numbers and [imath]\mathbb{Z}[/imath] the integers. Let [imath]G = \mathbb{R}/\mathbb{Z}[/imath]. Determine [imath]O_\pi(G)[/imath] - the normal maximal [imath]\pi[/imath]-group of [imath]G[/imath], [imath]T(G)[/imath] - torsion subgroup. |
439349 | Prove [imath]2^n > n^3[/imath] for all [imath]n \ge10[/imath]
I am stuck with the this question: Prove by induction that [imath]2^n > n^3[/imath], for all [imath]n \ge 10[/imath] I got this far: Base: For [imath]P(10)[/imath]: [imath] 2^n > n^3 \\ 2^{10} > 10^3 \\ 1024 > 1000 [/imath] so, [imath]P(10)[/imath] is true Inductive steps: Need to show [imath]P(k+1)[/imath] is true, assuming [imath]P(k)[/imath] is true. For [imath]P(k+1)[/imath]: [imath] 2^{k+1} > (k+1)^3 \\ 2 \times 2^k > k^3 + 3k^2 + 3k + 1 \\ 2^k + 2^k > k^3 + 3k^2 + 3k + 1 [/imath] We assume [imath]2^k > k^3[/imath], so we only need to show [imath]2^k > 3k^2 + 3k + 1[/imath] ? But I am stuck here. Any idea/hints? Thanks a lot for the help. | 438260 | Proving by induction: [imath]2^n > n^3 [/imath] for any natural number [imath]n > 9[/imath]
I need to prove that [imath] 2^n > n^3\quad \forall n\in \mathbb N, \;n>9.[/imath] Now that is actually very easy if we prove it for real numbers using calculus. But I need a proof that uses mathematical induction. I tried the problem for a long time, but got stuck at one step - I have to prove that: [imath] k^3 > 3k^2 + 3k + 1 [/imath] Hints??? |
439316 | [imath]f[/imath] measurable, [imath]f=g[/imath] almost everywhere, complete measure space
Let [imath]X[/imath] be a nonempty set, [imath]\mathcal{X}[/imath] a [imath]\sigma[/imath]-algebra of subsets of [imath]X[/imath], and [imath]\mu[/imath] a measure on [imath]\mathcal{X}[/imath] (i.e., [imath]\mu:X\to[0,+\infty][/imath], [imath]\mu(\phi)=0[/imath], and [imath]\mu[/imath] is countably additive. Proposition: If [imath]f:X\to R[/imath] is a measurable function, [imath]f=g[/imath] almost everywhere, then [imath]g[/imath] is a measurable function. Can this proposition be proved in the the measure space is not complete ([imath]\mathcal{X}[/imath] contains all subsets of measure zero)? If not, can someone provide a counterexample? Thanks. | 344791 | Counterexample for a non-measurable function?
I am struggling to solve an exercise in my measure theory book and any help for solving it would be appreciated: Let [imath](\Omega,\mathcal{A},\mu)[/imath] be a measure space and let [imath]f:\Omega \to \mathbb{R}[/imath] be measurable. Find a function [imath]g:\Omega \to \mathbb{R}[/imath] which is equal to [imath]f[/imath] almost everywhere but is not measurable. |
438689 | Find integers [imath]a[/imath] and [imath]b[/imath] such that [imath]a^5b+3[/imath] and [imath]ab^5+3[/imath] are both perfect cubes of integers?
Are there integers [imath]a[/imath] and [imath]b[/imath] such that [imath]a^5b+3[/imath] and [imath]ab^5+3[/imath] are both perfect cubes of integers? [imath]a,b[/imath] are distinct integers. P.S.: I think trying to find some special cases would not be helpful. | 377800 | (USAJMO)Find the integer solutions:[imath]ab^5+3=x^3,a^5b+3=y^3[/imath]
Find the integer solutions: [imath]a·b^5+3=x^3,a^5·b+3=y^3[/imath] This is the first problem of today's USAJMO (has finished),I only find a trival result that [imath]x\equiv y \pmod6[/imath] and [imath]abxy≠0 \pmod 3[/imath]. Thanks in advance! |
439025 | Propriedades do MDC (Properties of Greatest Common Divisor)
Estou com uma grande lista de exercícios de PROPRIEDADES DO MDC (MÁXIMO DIVISOR COMUM), e não estou conseguindo entender quais os passos que tenho que seguir nas demonstrações, e gostaria muito de aprender este conteúdo, alguém me ajuda em uma questão para ver se eu consigo entender as outras? Questão: Mostre que, se (a, b) = 1, a|c e b|c, então a · b|c Added Translation from Portuguese I have a large list of exercises PROPERTIES MDC (greatest common divisor), and I am not able to understand what steps you have to follow in the statements, and would love to learn this content, someone help me on a question to see if I I can understand the other? question: Show that if [imath](a, b) = 1, a | c[/imath] and [imath]b | c[/imath], then [imath]a · b | c[/imath] | 62054 | Let [imath]a\mid c[/imath] and [imath]b\mid c[/imath] such that [imath]\gcd(a,b)=1[/imath], Show that [imath]ab\mid c[/imath]
Let [imath]a\mid c[/imath] and [imath]b\mid c[/imath] such that greatest common divisor (gcd) [imath]\gcd(a,b)=1[/imath], Show that [imath]ab\mid c[/imath]. |
439612 | Does [imath]\sum_{n=1}^{\infty} \frac{1}{n+1}[/imath] converge?
Sorry for the oversimplified question, but does the series [imath]\sum_{n=1}^{\infty} \frac{1}{n+1}[/imath] converge? The ratio test of it gives the result of "1". Thanks a lot. | 180903 | My proof that a harmonic series diverges..
Suppose [imath]\sum_{n=1}^\infty \frac{1}{n} = S[/imath] where [imath]S[/imath] is finite. Then [imath]S =\sum_{n=1}^\infty \frac{1}{n}= \sum_{n=1}^\infty \frac{1}{2n-1} + \frac{1}{2n} > \sum_{n=1}^\infty \frac{1}{2n} + \frac{1}{2n} = S[/imath] which is a contradiction. Is this valid? |
439589 | proving a sum of binomial coefficients
How can i prove that [imath]\displaystyle\sum_{k=0}^{n}{2n\choose 2k}=2^{2n-1}[/imath] I tried using induction and pascal's identity but it didn't help me. | 126613 | Sum with binomial coefficients: [imath]\sum_{k=0}^{n}{2n\choose 2k}[/imath]
I'm repeating material for test and I came across the example that I can not do. How to calculate this sum: [imath]\displaystyle\sum_{k=0}^{n}{2n\choose 2k}[/imath]? |
397010 | Calculating the Riemann sum [imath]\lim_{n \to \infty} { \sum_{k=1}^{n} { (\frac{nk-1}{n^3})\sin(\frac{k}{n}) } }[/imath]
We need to calculate this: [imath]\lim_{n \to \infty} { \sum_{k=1}^{n} { (\frac{nk-1}{n^3})\sin(\frac{k}{n}) } }[/imath] So I know this is Riemann sum. This is what I started doing: [imath]\sum_{k=1}^{n} { (\frac{nk-1}{n^3})\sin(\frac{k}{n})} = \sum_{k=1}^{n} { (\frac{k}{n^2 } - \frac{1}{n^3})\sin(\frac{k}{n})}[/imath] But then I noticed that we can not get [imath]\frac{1}{n}[/imath] outside. How do we exactly know what is the partition? | 439889 | Calculating this Riemann sum limit
Calculate the limit [imath]\lim_{n\to \infty} {\sum_{k=1}^{n} {\left(\frac{nk-1}{n^3}\right) \sin\frac{k}{n}}}[/imath] How exactly do we calculate this limit of the Riemann sum? I am never able to find what is the partition. I know that our [imath]f(x)[/imath] is [imath]\sin(x)[/imath]. |
440166 | Zero divisor polynomial
Let [imath]f(x)\in R[x][/imath] be a zero divisor. How to prove that there is an element [imath]0\neq a\in R[/imath] such that [imath]af(x) = 0[/imath]? If [imath]R[/imath] has no nilpotent elements, it is easy. What about the general case? Can anyone help me? Thanks. | 83121 | Zero divisor in [imath]R[x][/imath]
Let [imath]R[/imath] be commutative ring with no (nonzero) nilpotents. If [imath]f(x) = a_0+a_1x+\cdots+a_nx^n[/imath] is a zero divisor in [imath]R[x][/imath], how do I show there's an element [imath]b \ne 0[/imath] in [imath]R[/imath] such that [imath]ba_0=ba_1=\cdots=ba_n=0[/imath]? |
297949 | Sequences in [imath]C(X)[/imath]
How would I go about proving this? Let [imath]X[/imath] be a compact metric space. Let [imath](f_n)[/imath] be a sequence in [imath]C(X)[/imath]. If [imath](f_n)[/imath] is uniformly convergent, show that [imath](f_n)[/imath] is both uniformly bounded and equicontinuous. | 329247 | Sequences in [imath]C(X)[/imath] converging uniformly implies uniform boundedness and equicontinuity
Let [imath]X[/imath] be a metric space and let [imath]\{ f_n \}_{n \geq 1}[/imath] be a sequence in [imath]C(X)[/imath]. Show that if [imath]\{ f_n \}_{n \geq 1}[/imath] converges uniformly on [imath]X[/imath] then [imath]\{ f_n \}_{n \geq 1}[/imath] is both uniformly bounded and equicontinuous. Could I get a hint? |
441208 | Definition of reflexive spaces
I'm currently studying this for an exam and I was wondering why the definition states: "Let E be a Banach space, then it's called reflexive if the canonical injection [imath]J:E\rightarrow E^{**}[/imath] is surjective." Why must [imath]E[/imath] be a Banach space? My guess is that if [imath]E[/imath] isn't a Banach space then there's some way to see that [imath]J[/imath] can't be surjective, is it correct? How? Thank you in advance! | 398970 | Show reflexive normed vector space is a Banach space
[imath]X[/imath] is a normed vector space. Assume [imath]X[/imath] is reflexive, then [imath]X[/imath] must be a Banach space. I guess we only need to show any Cauchy sequence is convergent in [imath]X[/imath]. |
440454 | Сauchy integral formula
I would like to understand statement of Cauchy integral theorem,which says that Cauchy's integral theorem implies that the line integral of every holomorphic function along a loop vanishes: or where γ is a rectifiable path in a simply connected open subset U of the complex plane C whose start point is equal to its end point, and [imath]f : U → C[/imath] is a holomorphic function. so first as i know rectifiable means finite,also because we have such complex plane whose start point is the same as end point,then does it means that it is similar to such situation when we have [imath]\int f(x)dx[/imath] from [imath]a[/imath] to [imath]b[/imath] is equal to [imath]F(b)-F(a)[/imath] and because end and start point is same ,then this equals to zero? or there is two different path and they compensate each other?please help me to understand why this formula held? | 4054 | Intuitive explanation of Cauchy's Integral Formula in Complex Analysis
There is a theorem that states that if [imath]f[/imath] is analytic in a domain [imath]D[/imath], and the closed disc {[imath] z:|z-\alpha|\leq r[/imath]} contained in [imath]D[/imath], and [imath]C[/imath] denotes the disc's boundary followed in the positive direction, then for every [imath]z[/imath] in the disc we can write: [imath]f(z)=\frac{1}{2\pi i}\int\frac{f(\zeta)}{\zeta-z}d\zeta[/imath] My question is: What is the intuitive explanation of this formula? (For example, but not necessary, geometrically.) (Just to clarify - I know the proof of this theorem, I'm just trying to understand where does this exact formula come from.) |
441481 | Why does [imath]\sum_{n = 0}^\infty \frac{n}{2^n}[/imath] converge to 2?
Apparently, [imath] \sum_{n = 0}^\infty \frac{n}{2^n} [/imath] converges to 2. I'm trying to figure out why. I've tried viewing it as a geometric series, but it's not quite a geometric series since the numerator increases by 1 every term. | 458836 | Show that [imath]\sum_{n = 1}^{+\infty} \frac{n}{2^n} = 2[/imath]
Show that [imath]\sum_{n = 1}^{+\infty} \frac{n}{2^n} = 2[/imath]. I have no idea to solve this problem. Anyone could help me? |
441689 | evaluate the line integral
Evaluate the line integral [imath]\int_{C}^{ }xe^{y}ds[/imath], where [imath]C[/imath] is the line segment from [imath](-1,2)[/imath] to [imath](1,1)[/imath]. I know that I need to first get a parametrized equation. I got [imath]r(t) = [(2t - 1), (-t + 1)][/imath]. I am not sure how to proceed. | 437188 | Evaluate the line segment intergal
Evaluate the line integral [imath]\int_C xe^{y}\, {\rm d}s,[/imath] where [imath]C[/imath] is the line segment from [imath](-1,2)[/imath] to [imath](1,1)[/imath]. I do not get this part of calculus at all please show me how this is solved and if you give me hints is ok too. Please help me. Thanks |
441823 | [imath]a+b+c=n[/imath] find number of ways
Please tell me how to find the total number of intergral solutions of [imath]a+b+c=n[/imath] I already know that total number of solutions will be [imath](n+3−1)c(3−1)[/imath]. but if value of [imath]a[/imath] and [imath]b[/imath] and [imath]c[/imath] is given then what will be the answer?? | 150285 | What will be total number of solutions of [imath]a+b+c = n[/imath]?
Please tell me how to find the total number of intergral solutions of [imath] a+b+c=n [/imath] I already know that total number of solutions will be [imath](n+3-1)c(3-1)[/imath]. But what will be the case when a varies from let [imath]2[/imath] to [imath]6[/imath] , [imath]b[/imath] varies from [imath]2[/imath] to [imath]6[/imath], [imath]c[/imath] varies from [imath]2[/imath] to [imath]6[/imath].There can be other conditions also. Please give the generalized result. It will be really helpful. |
442008 | Prove that there exists walks that each edge is in [imath]G[/imath]
For some [imath]k \in\mathbb{N}[/imath], let [imath]G[/imath] be a connected graph with [imath]2k[/imath] odd-degree vertices, and any number of even-degree vertices. Prove that there exists [imath]k[/imath] walks such that each edge in [imath]G[/imath] is used in exactly one walk exactly once, assuming that the main theorem about Eulerian circuits is true for graphs with multiple edges. | 103536 | Prove a graph Containing [imath]2k[/imath] odd vertices contains [imath]k[/imath] distinct trails
I'm reading the book Graphs and Their Uses which contains the following theorem and proof: THEOREM 2.3. A connected graph with 2k odd vertices contains a family of k distinct trails which, together, traverse all edges of the graph exactly once. PROOF. Let the odd vertices in the graph be denoted by in some order. [imath]a_1,a_2,\dots,a_k[/imath] and [imath]b_1,b_2,\dots,b_k[/imath] When we add the [imath]k[/imath] edges [imath]a_lb_l, a_2b_2 ,\dots, a_kb_k[/imath] to the graph, all vertices become even and there is an Eulerian trail T. When these edges are dropped out again, T falls into k separate trails covering the edges in the original graph. However this doesn't seem to make sense since in the graph whose vertices have degrees 3, 1, 1, 1 there is no way to add 2 edges in such a way that the degree of all odd vertices becomes even. What am I missing here? |
435412 | induction proof: [imath]\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}[/imath]
I encountered the following induction proof on a practice exam for calculus: [imath]\sum_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}[/imath] I have to prove this statement with induction. Can anyone please help me with this proof? | 509046 | Basic Mathematical Induction
I'm not quite sure how to approach this question. I need to prove that for [imath]n\ge1[/imath] [imath]1^2+2^2+3^3+\dots+n^2=\frac16n(n+1)(2n+1)[/imath] Do I just plug [imath]1[/imath] and see if [imath]\frac16(1)((1)+1)(2(1)+1) = 1^2\text{ ?}[/imath] |
442312 | Integrate [imath]\int e^{-x} \cos x \,\mathrm{d}x[/imath]
I know that integration by parts leads to an infinite loops of sin and cos so what do I do? I can't do [imath]u[/imath] substitution because I can't get rid of all the variables. [imath]\int e^{-x} \cos x \,\mathrm{d}x[/imath] | 20952 | Integration by parts: [imath]\int e^{ax}\cos(bx)\,dx[/imath]
I need to evaluate the following function and then check my answer by taking the derivative: [imath]\int e^{ax}\cos(bx)\,dx[/imath] where [imath]a[/imath] is any real number and [imath]b[/imath] is any positive real number. I know that you set [imath]u=\cos(bx)[/imath] and [imath]dv=e^{ax} dx[/imath], and the second time you need to integrate again you set [imath]u=\sin(bx)[/imath] and [imath]dv=e^{ax}dx[/imath] again. It eventually simplifies down to [imath]\int e^{ax}\cos(bx)dx = \frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left(\frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a}\int e^{ax}\cos(bx)\,dx\right).[/imath] Now I know to move the integral on the left side to the right side so that I can just divide by the constant to solve. Here is my problem: I know I need to solve the right side to be: [imath]\frac{e^{ax}\left(a\cos(bx) + b\sin(bx)\right)}{a^2+b^2} + C.[/imath] To divide by the constant, I multiplied everything on the right side by [imath]\frac{a^2}{b^2+1}.[/imath] but this leads me to get [imath]b^2 + 1[/imath] on the bottom instead of [imath]a^2 + b^2[/imath]. I will show what I am doing in detail: After setting the initial integral equal to: [imath]\frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left(\frac{1}{a}e^{ax}\sin(bx) - \frac{b}{a}\int e^{ax}\cos(bx)\,dx\right) + C[/imath] I simplify: [imath]\int e^{ax}\cos(bx)\,dx = \frac{a^2}{b^2+1}\left(\frac{1}{a}e^{ax}\cos(bx) + \frac{b}{a}\left(\frac{1}{a}e^{ax}\sin(bx)\right)\right) + C[/imath] If this is already wrong, can someone point be in the right direction? If I have not gone wrong yet, I can edit to show the rest of my work. |
442197 | Weakly compact implies bounded in norm
The weak topology on a normed vector space [imath]X[/imath] is the weakest topology making every bounded linear functionals [imath]x^*\in X^*[/imath] continuous. If a subset [imath]C[/imath] of [imath]X[/imath] is compact for the weak topology, then [imath]C[/imath] is bounded in norm. How does one prove this fact? | 241604 | Does weak compactness imply boundedness in a normed vector space (not necessarily complete)?
Does this argument work? Let [imath]X[/imath] be a normed vector space, with [imath]A[/imath] weakly compact in [imath]X[/imath]. The collection of sets of the form [imath]\{x \in X: |f(x) - f(a)| < 1 \},f \in X^*,a \in A[/imath] forms a cover of [imath]A[/imath] consisting of weakly open sets in [imath]X[/imath], and so should has a finite subcover. |
267910 | Sum [imath]\cos x + \cos 2x + \cdots + \cos (n-1)x.[/imath]
Find the sum of the series [imath]\cos x + \cos 2x + \cdots + \cos (n-1)x.[/imath] You must calculate the sum of this series only by multiplying through by [imath]2\sin\left(\frac{x}{2}\right)[/imath]. Now I've heard of finding the sum of a trig series by finding real and imaginary parts etc., but I have no idea how to do it this way. | 192065 | Evaluation of [imath] \sum_{k=0}^n \cos k\theta [/imath]
I just wanted to evaluate [imath] \sum_{k=0}^n \cos k\theta [/imath] and I know that it should give [imath] \cos\left(\frac{n\theta}{2}\right)\frac{\sin\left(\frac{(n+1)\theta}{2}\right)}{\sin(\theta / 2)} [/imath] I tried to start by writing the sum as [imath] 1 + \cos\theta + \cos 2\theta + \cdots + \cos n\theta [/imath] and expand each cosine by its series representation. But this soon looked not very helpful so I need some clue about how this partial sum is calculated more efficiently ... |
358830 | About [imath]\lim \left(1+\frac {x}{n}\right)^n[/imath]
I was wondering if it is possible to get a link to a rigorous proof that [imath]\displaystyle \lim_{n\to\infty} \left(1+\frac {x}{n}\right)^n=\exp x[/imath] | 882741 | Limit of [imath](1+ x/n)^n[/imath] when [imath]n[/imath] tends to infinity
Does anyone know the exact proof of this limit result? [imath]\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n = e^x[/imath] |
442822 | Looking for an algebraic proof of an unusual equality
I have stumbled on the equality [imath][\sum i]^2 = \sum i^3[/imath] which holds when both sums are over the range 1 to k. The equality is readily demonstrated inductively, but I wonder if anyone can provide an algebraic proof? | 95047 | Combinatorial interpretation of sum of squares, cubes
Consider the sum of the first [imath]n[/imath] integers: [imath]\sum_{i=1}^n\,i=\frac{n(n+1)}{2}=\binom{n+1}{2}[/imath] This has always made the following bit of combinatorial sense to me. Imagine the set [imath]\{*,1,2,\ldots,n\}[/imath]. We can choose two from this set, order them in decreasing order and thereby obtain a point in [imath]\mathbb{N}^2[/imath]. We interpret [imath](i,*)[/imath] as [imath](i,i)[/imath]. These points give a clear graphical representation of [imath]1+2+\cdots+n[/imath]: [imath] \begin{matrix} &&&\circ\\ &&\circ&\circ\\ &\circ&\circ&\circ\\ \circ&\circ&\circ&\circ\\ \end{matrix} [/imath] Similar identities are: [imath]\sum_{i=1}^n\,i^2=\frac{n(n+1)(2n+1)}{6}=\frac{2n(2n+2)(2n+1)}{24}=\frac{1}{4}\binom{2n+2}{3}[/imath] [imath]\sum_{i=1}^n\,i^3=\frac{n^2(n+1)^2}{4}=\binom{n+1}{2}^2[/imath] I am aware of geometric explanations of these identities, but not combinatorial ones similar to the above explanation for summing first powers that make direct use of the "choosing" interpretation of the binomial coefficient. Can anyone offer combinatorial proofs of these? |
442945 | What's the cardinality of [imath]\{ (i,j,k) : i+j+k=10, i,j,k \in \mathbb{N} \}[/imath] ? General case.
How many triples of natural numbers are there of sum equal to 10? What about the cardinality of [imath]\{ (n_1,...,n_K) : \ n_i \in \mathbb{N} \ \wedge \sum_{i=1}^{K} n_i=N \ \} \ ?[/imath] Is there an intuitive way of thinking about this? I'm looking for a simple argument how to count this. | 271090 | Number of solutions for [imath]x[1] + x[2] + \ldots + x[n] =k[/imath]
Omg this is driving me crazy seriously, it's a subproblem for a bigger problem, and i'm stuck on it. Anyways i need the number of ways to pick [imath]x[1][/imath] ammount of objects type [imath]1[/imath], [imath]x[2][/imath] ammount of objects type [imath]2[/imath], [imath]x[3][/imath] ammounts of objects type [imath]3[/imath] etc etc such that [imath]x[1] + x[2] + \ldots x[n] = k\;.[/imath] Order of the objects of course doesn't matter. I know there is a simple formula for this something like [imath]\binom{k + n}n[/imath] or something like that i just can't find it anywhere on google. |
442920 | Complex Analysis Proof2
Let [imath]f(z)[/imath] be an entire function satisfying [imath]|f(z)|\leq k|z|[/imath] for some positive constants [imath]k[/imath] and all [imath]z[/imath]. Show that [imath]f(z)=az^2[/imath] for some constant [imath]a[/imath]. | 327360 | Question on application of Liouville's theorem
Let [imath]f[/imath] be an entire function such that [imath]\mid f(z) \mid \leq A \mid z \mid[/imath] for all [imath]z[/imath], where [imath]A[/imath] is a fixed number. Show that [imath]f(z)=a_1z[/imath], where [imath]a_1[/imath] is a complex constant. |
443510 | Counter-example of an affirmation
Does anyone know any example that invalidates the following affirmation: If a morphism [imath]f:A\to A[/imath] induces the identity [imath]\hat f:\operatorname{Spec} \left( A \right) \to \operatorname{Spec} \left( A \right)[/imath] then [imath]f = \operatorname{id} _A [/imath]. | 443506 | Counterexample to "if [imath]f:A\to A[/imath] induces the identity [imath]\hat f:\operatorname{Spec}(A)\to\operatorname{Spec}(A)[/imath], then [imath]f=\operatorname{id}_A[/imath]"
Does anyone know any example that invalidates the following affirmation: If a morphism [imath]f:A\to A[/imath] induces the identity [imath]\hat f:\operatorname{Spec} \left( A \right) \to \operatorname{Spec} \left( A \right)[/imath] then [imath]f = \operatorname{id} _A [/imath]. |
443585 | Determinant of a Matrix Proof: [imath]\;\det(qA) = q^n(\det A)[/imath]
I am required to show that: [imath]\det(qA) = q^n(\det A)[/imath], where [imath]A[/imath] is a real [imath]n\times n[/imath] Matrix, and [imath]q[/imath] is a constant I believe that this claim is true after doing few examples. However, but I do not know how to start the proof. | 442240 | Prove [imath]\det(kA)=k^n\det A[/imath]
Let [imath]A[/imath] be an [imath]n \times n[/imath] matrix and [imath]k[/imath] be a scalar. Prove that [imath]\det(kA)=k^n\det A[/imath]. I really don't know where to start. Can someone give me a hint for this proof? |
443586 | Prove that [imath]x^2<\sin x \tan x[/imath] as [imath]x \to 0[/imath]
[imath]x^2<\sin x \tan x \quad as \; x \to 0[/imath] I made the substitution [imath]x \to \arctan x[/imath] . [imath]\arctan^2 x<x\sin (\arctan x)[/imath] [imath]\arctan x < \large \frac{x}{(x^2+1)^{\frac 14}}[/imath] There are two functions [imath]f(x)[/imath] and [imath]g(x)[/imath] . [imath]f(0)=g(0)[/imath] . If [imath]f'(x)>g'(x)[/imath] on the interval [imath](0, a)[/imath] , then that implies that [imath]f(x)>g(x)[/imath] on the interval [imath](0, a)[/imath] . Therefore if [imath]RHS'>LHS'[/imath] , then [imath]RHS>LHS[/imath] . [imath]LHS'=\large \large \large \frac {1}{x^2+1} <RHS'=\frac {x^2+2}{2(x^2+1)^{\frac 54}}[/imath] [imath]\large \large \large \frac {1}{x^2+1}<\frac {x^2+2}{2(x^2+1)^{\frac 54}}[/imath] [imath]1<\large \frac {x^2+2}{2(x^2+1)^{\frac 14}}[/imath] Using standard techniques (such as first derivative test) we can show that the [imath]RHS[/imath] has a minimum at [imath](0, 1)[/imath] so we have proved the inequality. I have two questions: [imath]1)[/imath] Is my proof correct? [imath]2)[/imath] Are there nicer ways of doing this inequality (without involving math higher than calc 1)? I have tried AM-GM and others but I think it is very hard to do it very elegantly because the [imath]RHS[/imath] is such a good approximation of the [imath]LHS[/imath] as [imath]x \to 0[/imath] . | 442831 | Calculating [imath]\lim_{x\to0} \left\lfloor\frac{x^2}{\sin x \tan x}\right\rfloor[/imath]
Find [imath]\lim_{x\to0} \left\lfloor\frac{x^2}{\sin x \tan x}\right\rfloor[/imath] where [imath]\lfloor\cdot\rfloor[/imath] is greatest integer function I am a high school teacher. One of my students came up to ask this limit. For [imath]\lfloor\frac{\sin x}{x}\rfloor[/imath], I have used [imath]\sin x > x[/imath] using increasing decreasing functions. I tried to prove [imath]x^2 > \sin x \tan x[/imath] using increasing /decreasing function but I am not getting it. |
443957 | Strange Method of Differentiating [imath]x^2[/imath]
A book by the name of Calculus Super Textbook teaches the following method for differentiating [imath]x^2[/imath] with respect to [imath]x[/imath]: Take the equation: [imath]y=x^2[/imath] The author reasons that increasing both [imath]x[/imath] and [imath]y[/imath] by a differential amount preserves equality, so: [imath]y+dy=(x+dx)^2[/imath] Expanding the right-hand side gives: [imath]y+dy=x^2+2xdx+dx^2[/imath] Subtract off the original [imath]y=x^2[/imath]: [imath]dy=2xdx+dx^2[/imath] The author reasons that an infinitesimal change in [imath]x[/imath] multiplied by an infinitesimal change in [imath]x[/imath] is so small to be effectively zero, so any powers of [imath]dx[/imath] higher than [imath]1[/imath] are dropped: [imath]dy=2xdx[/imath] Then, dividing both sides by [imath]dx[/imath]: [imath]{dy \over dx}=2x[/imath] This is nothing like the methods I've been taught for finding a derivative (which I understand to be the limit of the difference quotient), but it at least appears to work for polynomial functions. Is this a valid method that I'm not aware of? If so, how would you use this to differentiate more complex functions like sine and cosine? | 441525 | Differentiating [imath]y=x^{2}[/imath]
I am reading in a book about differentiating, but I am confused with one of the steps he takes. We start with: [imath] \begin{align} y &= x^{2} \\ y + \mathrm{d}y &= (x + \mathrm{d}x)^2 \\ y + \mathrm{d}y &= x^2 + x\mathrm{d}x + x\mathrm{d}x + (\mathrm{d}x^2) \end{align} [/imath] Now the author simplifies this to: [imath]y + dy = x^2 + 2x\mathrm{d}x + (\mathrm{d}x^2)[/imath] I dislike how the middle term is simplified to [imath]2x\mathrm{d}x[/imath] instead of [imath]2(x\mathrm{d}x)[/imath], as I feel like it is more intuitive on what is going. As in, [imath]2[/imath] of the term [imath]x\mathrm{d}x[/imath], instead of [imath]2x\mathrm{d}x[/imath]. But I fear writing it as [imath]2(x\mathrm{d}x)[/imath] may result in an incorrect distributive property. Next, he omits the [imath](dx^2)[/imath]: [imath]y + \mathrm{d}y = x^2 + 2x \mathrm{d}x[/imath] Subtract the original [imath]y = x^2[/imath]: [imath]\mathrm{d}y = 2x \mathrm{d}x[/imath] Now here is where I get confused: [imath]\frac{\mathrm{d}y}{\mathrm{d}x} = 2x[/imath] How can he just divide both sides by [imath]\mathrm{d}x[/imath]!? If the original term was [imath]2[/imath] of [imath]x\mathrm{d}x[/imath], wouldn't it have to be written out as [imath]2x * 2\mathrm{d}x[/imath], and thus divide both sides by [imath]2\mathrm{d}x[/imath] instead? I think the root of my confusion is how to properly simplify: [imath]x\mathrm{d}x + x\mathrm{d}x[/imath] I trust that he is right, but I am looking for an explanation of why his simplification can work, and why [imath]2(x\mathrm{d}x)[/imath] would be incorrect! Thank you! |
423577 | How many irreducible factors of grade [imath]6[/imath] there is in [imath]\mathbb{F}_{2}\left[ x\right][/imath]?
How many irreducible factors of grade [imath]6[/imath] there is in the polynomial ring [imath]\mathbb{F}_{2}\left[ x\right][/imath]? I have solved this by using the fact that every irreducible polynomial of grad [imath]i[/imath] is a factor of the polynomial [imath]t_{i}=x^{p^{i}}-x[/imath], which reduces the problem to find all irreducible factors of grad [imath]6[/imath] in [imath]g_{6}=x^{2^{6}}-x[/imath], and to do this i will use algorithm which is used to factor any squarefree polynomial [imath]f[/imath] to a product [imath]f=g_{1}g_{2}...g_{k}[/imath] where every irreducible factor of [imath]g_{i}[/imath] is of grad [imath]i[/imath]. [imath]f_{0}=f[/imath] [imath]g_{1}=Gcd(f_{0},x^{2^{1}}-x)=x^{2}+x[/imath] [imath]f1:=f_{0}/g_{1}=x^{62}+x^{61}+ ...etc[/imath] [imath]g_{2}=Gcd(f_{1},x^{2^{2}}-x)= x^{2}+x+1[/imath] [imath]f_{2}=f_{1}/g_{2}=x^{60}+x^{57}+x^{54}+x^{51}+x^{48}+... etc[/imath] [imath]g_{3}=Gcd(f_{2},x^{2^{3}}-x)=x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+1[/imath] [imath]f_{3}=f_{2}/g_{3}=x^{54}+x^{54}+x^{51}+ ...etc[/imath] [imath]g_{4}:=Gcd(f_{3},x^{2^{4}}-x)=1[/imath] [imath]f_{4}=f_{3}[/imath] [imath]g_{5}:=Gcd(f_{4},x^{2^{5}}-x)=1[/imath] [imath]f_{5}=f_{4}[/imath] [imath]g_{6}:=Gcd(f_{5},x^{2^{6}}-x)=f_{5}[/imath] so [imath]g_{1}[/imath] is product of [imath]2[/imath] irreducible polynomials of degree [imath]1[/imath] and [imath]g_{2}[/imath] is irreducible of degree [imath]2[/imath] and [imath]g_{3}[/imath] is product of two irreducible polynomials of degree [imath]3[/imath] and also [imath]g_{6}[/imath] is product of [imath]9[/imath] irreducible polynomials of degree [imath]6[/imath] My question is: what is the relation between the number and irreducible factors of grad 6 and the subfields of [imath]\mathbb{F}_{2^{6}}[/imath], i mean there is a hint to solve the question says that every irreducible factor gives a subfield and i am wondering how can this help me solving the question? Many thanks for any responses | 152880 | How many irreducible polynomials of degree [imath]n[/imath] exist over [imath]\mathbb{F}_p[/imath]?
I know that for every [imath]n\in\mathbb{N}[/imath], [imath]n\ge 1[/imath], there exists [imath]p(x)\in\mathbb{F}_p[x][/imath] s.t. [imath]\deg p(x)=n[/imath] and [imath]p(x)[/imath] is irreducible over [imath]\mathbb{F}_p[/imath]. I am interested in counting how many such [imath]p(x)[/imath] there exist (that is, given [imath]n\in\mathbb{N}[/imath], [imath]n\ge 1[/imath], how many irreducible polynomials of degree [imath]n[/imath] exist over [imath]\mathbb{F}_p[/imath]). I don't have a counting strategy and I don't expect a closed formula, but maybe we can find something like "there exist [imath]X[/imath] irreducible polynomials of degree [imath]n[/imath] where [imath]X[/imath] is the number of...". What are your thoughts ? |
444026 | If [imath]f[/imath] is compact is [imath]f[/imath] continuous?
If [imath]f[/imath] is a compact function (image of every compact set is compact) is [imath]f[/imath] continuous? Attempt: I can't find a counterexample. I can't prove it. I only know how to prove the converse. | 226917 | Must [imath]f[/imath] be continuous?
Let [imath]f:\mathbb{R^n}\to\mathbb{R^m}[/imath] be a function such that the image of any closed bounded set is closed and bounded. Must [imath]f[/imath] be continuous? |
444106 | invertibility of a matrix with gaussian distances
Suppose we have have n points [imath]x_1, x_2, ..., x_n[/imath] and we create a matrix [imath]A[/imath] such that [imath]a_{(i,j)}=e^{−(||x_i−x_j||_2)/\epsilon}[/imath]. Is there a proof that A is invertible? | 248976 | How to prove that exponential kernel is positive definite?
The exponential kernel is defined by: [imath]k(x,z) = e^{-\alpha\|x-z\|}[/imath] where [imath]\alpha>0[/imath], [imath]x,z\in \Bbb{R}^d[/imath], [imath]\|x\|[/imath] is the 2-norm. The kernel matrix is defined by [imath]K_{ij} = k(x_i,x_j)[/imath], [imath]i,j\in[1\ldots n][/imath]. How to prove that [imath]K[/imath] is a positive (semi-positive) definite matrix? |
444147 | Discrete topology on infinite sets
I want to prove the following: Let [imath]X[/imath] be an infinite set and [imath]\tau[/imath] a topology on [imath]X[/imath]. If every infinite subset of [imath]X[/imath] is in [imath]\tau[/imath], then [imath]\tau[/imath] is the discrete topology on [imath]X[/imath]. Proof. Let [imath]x\in X[/imath]. There exist two infinite subsets [imath]A[/imath] and [imath]B[/imath] of [imath]X[/imath] such that [imath]\{x\}=A\cap B[/imath]. So every singleton is open in [imath]X[/imath]. It follows that [imath]\tau[/imath] is the discrete topology on [imath]X[/imath]. Edit. Justifying the existence of [imath]A[/imath] and [imath]B[/imath]. Assume that [imath]C[/imath] is a countable subset of [imath]X[/imath]. Let [imath]A[/imath] be the set of odd-numbered points in [imath]C[/imath] and [imath]B=(C-A)\cup\{x\}[/imath] for some [imath]x[/imath] in [imath]A[/imath] (i.e. [imath]B[/imath] contains the even-numbered points together with [imath]x[/imath]). [imath]A\cap B=\{x\}[/imath]. This holds for every [imath]x[/imath] in [imath]B[/imath] as well (by symmetry) so such [imath]A[/imath] and [imath]B[/imath] always exist. | 439399 | Topological proof on discrete topology where [imath]X[/imath] is infinite
How must I prove this problem? Let [imath]X[/imath] be an infinite set and let [imath]T[/imath] a topology in [imath]X[/imath] in which all infinite subsets of [imath]X[/imath] are open. Prove: [imath]T[/imath] is a discrete topology in [imath]X[/imath]. |
444134 | Let [imath]A[/imath] be a complex [imath]3 \times 3[/imath] matrix with [imath]A^3 = -I[/imath].
Let [imath]A[/imath] be a complex [imath]3 \times 3[/imath] matrix with [imath]A^3 = -I[/imath]. Which of the following statements are correct? [imath]A[/imath] has three distinct eigenvalues. [imath]A[/imath] is diagonalizable over [imath]\mathbb{C}[/imath]. [imath]A[/imath] ls triangularizable over [imath]\mathbb{C}[/imath]. [imath]A[/imath] is non-singular. If we take [imath]A = -I[/imath], then [imath]A^3 = -I[/imath] and has eigenvalues as [imath]-1, -1, -1[/imath]. so option [imath]1[/imath] is incorrect. Since [imath]A[/imath] is not equal to [imath]0[/imath], [imath]A[/imath] is non-singular and also option [imath]2[/imath] and [imath]3[/imath] are correct. Hence option [imath]2, 3[/imath], and should be options. Is it correct? | 433806 | [imath]A[/imath] be a complex [imath]3\times 3[/imath] matrix such that [imath]A^3=-I[/imath]
Let [imath]A[/imath] be a complex [imath]3\times 3[/imath] matrix such that [imath]A^3=-I[/imath], then we need to find out which of the following statements are correct? [imath]A[/imath] has three distinct eigenvalues; [imath]A[/imath] is diagonalizable over [imath]\mathbb{C}[/imath]; [imath]A[/imath] is triangulizable over [imath]\mathbb{C}[/imath]; [imath]A[/imath] is non singular. Wll, from minimal polynomial approach I have deduced that all are correct as minpoly has to be [imath]x^3+1[/imath] am I right? |
444191 | Explicit isomorphism [imath]S_4/V_4[/imath] and [imath]S_3[/imath]
Let [imath]S_4[/imath] be a symmetric group on [imath]4[/imath] elements, [imath]V_4[/imath] - its subgroup, consisting of [imath]e,(12)(34),(13)(24)[/imath] and [imath](14)(23)[/imath] (Klein four-group). [imath]V_4[/imath] is normal and [imath]S_4/V_4[/imath] if consisting of [imath]24/4=6[/imath] elements. Hence [imath]S_4/V_4[/imath] is cyclic group [imath]C_6[/imath] or a symmetric group [imath]S_3[/imath] (really, there are only two groups consisting of [imath]6[/imath] elements). It is easy to see, that an order of each element of [imath]S_4[/imath] is [imath]1,2,3[/imath] or [imath]4[/imath]. So, [imath]S_4/V_4[/imath] is isomorphic to [imath]S_3[/imath]. My question: how to build the isomorphism explicitly? | 106777 | An epimorphism from [imath]S_{4}[/imath] to [imath]S_{3}[/imath] having the kernel isomorphic to Klein four-group
Exercise [imath]7[/imath], page 51 from Hungerford's book Algebra. Show that [imath]N=\{(1),(12)(34), (13)(24),(14)(23)\}[/imath] is a normal subgroup of [imath]S_{4}[/imath] contained in [imath]A_{4}[/imath] such that [imath]S_{4}/N\cong S_{3}[/imath] and [imath]A_{4}/N\cong \mathbb{Z}_{3}[/imath]. I solved the question after many calculations. I would like to know if is possible to define an epimorphism [imath]\varphi[/imath] from [imath]S_{4}[/imath] to [imath]S_{3}[/imath] such that [imath]N=\ker(\varphi)[/imath]. Thanks for your kindly help. |