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1197967 | If [imath]f:\omega_1 \to \mathbb{R}[/imath] is continuous, then there is [imath]a \in \omega_1[/imath] such that if [imath]x\geq a[/imath], then [imath]f(x) = f(a)[/imath].
Show that if [imath]f:\omega_1 \to \mathbb{R}[/imath] is continuous, then there is [imath]a \in \omega_1[/imath] such that if [imath]x\geq a[/imath], then [imath]f(x) = f(a)[/imath]. I'm stuck on this question for a while, although it seems to me that the proof will be related to the fact that [imath]f[/imath] is bounded (I proved this as folows: [imath]f[/imath] is continuous and [imath]w_1[/imath] has the least element of the set of its limit point, say [imath]a[/imath], then [imath]f(a)[/imath] should be the upper bound of [imath]f:\omega_1 \rightarrow \mathbb{R}[/imath]. Is this correct?). Can anyone please help on this problem? | 428344 | show that every continuous real-valued function defined on [imath]S_{\mathbb{\Omega}}[/imath] is eventually constant
show that every continuous real-valued function defined on [imath]S_{\mathbb{\Omega}}[/imath] is eventually constant.Where [imath]S_{\mathbb{\Omega}}[/imath] denote the first uncountable ordinal. There is a hint that for each [imath]\epsilon[/imath],there is an element [imath]\alpha[/imath] of [imath]S_{\mathbb{\Omega}}[/imath] such that [imath]|f(\alpha)-f(\beta)|<\epsilon[/imath] for all [imath]\beta>\alpha[/imath]. However, I couldn't figure out how to prove this statement. |
1197888 | Invariant Subspace of Two Operators
Let [imath]S[/imath], [imath]T[/imath] be linear operators on a finite-dimensional vector space [imath]V[/imath] over [imath]\mathbb{C}[/imath]. Suppose [imath]S^2 = T^2 = I.[/imath] Show that there exists either a [imath]1[/imath]-dimensional or [imath]2[/imath]-dimensional subspace of [imath]V[/imath] which is invariant under [imath]S[/imath] and [imath]T[/imath]. Ok so, since [imath]S^2 = T^2 = I[/imath], either [imath]1[/imath] or [imath]-1[/imath] are Eigenvalues of S and T. i,e The Minimal Polynomial [imath]M_T[/imath] and [imath]M_S[/imath] satisfy [imath]M_T \; | \; (x+1)(x-1)[/imath] [imath]M_S \; | \; (x+1)(x-1)[/imath] Edit : Found the same question elsewhere. For those of you who are interested in the answers. Invariant Subspace of Two Linear Involutions | 31363 | Invariant Subspace of Two Linear Involutions
I'd love some help with this practice qualifier problem: If [imath]A[/imath] and [imath]B[/imath] are two linear operators on a finite dimensional complex vector space [imath]V[/imath] such that [imath]A^2=B^2=I[/imath] then show that [imath]V[/imath] has a one or two dimensional subspace invariant under [imath]A[/imath] and [imath]B[/imath]. Thanks! |
1148682 | How are first digits of [imath]\pi[/imath] found?
Since Pi or [imath]\pi[/imath] is an irrational number, its digits do not repeat. And there is no way to actually find out the digits of [imath]\pi[/imath] ([imath]\frac{22}{7}[/imath] is just a rough estimate but it's not accurate). I am talking about accurate digits by either multiplication or division or any other operation on numbers. Then how are the first digits of [imath]\pi[/imath] found - 3.1415926535897932384626433832795028841971693993... In fact, more than 100,000 digits of [imath]\pi[/imath] are found (sources - 100,000 digits of [imath]\pi[/imath]) How is that possible? If these digits of [imath]\pi[/imath] are found, then it must be possible to compute [imath]\pi[/imath] with some operations. (I am aware of breaking of circle into infinite pieces method but that doesn't give accurate results.) How are these digits of [imath]\pi[/imath] found accurately? Can it be possible for a square root of some number to be equal to [imath]\pi[/imath]? | 1723504 | Ways to determine [imath]\pi[/imath]
I have read that it is possible to determine the value of a single digit, say the 874th of [imath]\pi[/imath]. I know that it is a trascentental number, how is that possible? How many ways are there to determine the value of [imath]\pi[/imath]? What is the simplest? |
1198353 | Are functors that are right-cancellable full, or do they have other characterizations?
In a former question it has become clear to me that a functor is left-cancellable if and only if it is injective on morphisms. This provides a nice characterization of monomorphisms in category [imath]{Cat}[/imath]. I would like to know whether there is also a nice characterization for epimorphisms in that category. I allready found out that functors that are surjective on morphisms are right-cancellable, but also that the opposite of this is not true. Are there any 'weaker' conditions then? Are these functors necessarily full for instance? Uptil now I couldn't manage to find that out, so my first question is: Are functors that are right-cancellable necessarily full? More generally my second question is: What characteristics can be mentioned of a functor that is right-cancellable? | 693970 | Monomorphisms and Epimorphisms in the category of small categories
Denote by [imath]\mathbf{Cat}[/imath] the category of small categories, i.e. categories whose class of arrows is a set. I am working with classes but I suppose it works the same if one is working in a given universe following the Grothendieck approach. My question is whether monomorphisms and epimorphisms in [imath]\mathbf{Cat}[/imath] can easily be classified. An arrow between two small categories is just a functor. In particular I was wondering if the subcategories of a given small category correspond to the category-theoretic subobjects in [imath]\mathbf{Cat}[/imath]. Given a fixed object [imath]X[/imath] in a category I call an isomorphism class of monomorphisms with codomain [imath]X[/imath] a [imath]\textit{subobject}[/imath] of [imath]X[/imath]. |
1023039 | If [imath]|f(x)-f(y)|\geq \frac12|x-y|[/imath], must [imath]f[/imath] be bijective?
Let [imath]f:\mathbb R\rightarrow \mathbb R[/imath] be a continuous function such that [imath]|f(x)-f(y)|\geq \frac12|x-y|[/imath] for all[imath]x,y\in \mathbb R[/imath]. Then is [imath]f[/imath] one-one and onto? Let [imath]f(x)=f(y)[/imath] i.e. [imath]0=|f(x)-f(y)|\geq (1/2)|x-y|[/imath] i.e [imath]x=y[/imath] Hence [imath]f[/imath] is injective. But I am unable to conclude whether [imath]f[/imath] is onto.Any help | 197285 | [imath]|f(x)-f(y)|\geq k|x-y|[/imath].Then [imath]f[/imath] is bijective and its inverse is continuous.
My exercise says: Let [imath]f:\mathbb{R} \rightarrow \mathbb{R}[/imath] a continuous function e suppose that exists [imath]k[/imath] such that: [imath]|f(x)-f(y)|\geq k|x-y|[/imath] Then [imath]f[/imath] is bijective and its inverse is continuous. Well, there's a Theorem , Invariance of domain, that says "If [imath]U[/imath] is an open subset of [imath]\mathbb {R^n}[/imath] and [imath]f : U \rightarrow\mathbb{R}[/imath] is an injective continuous map, then [imath]V = f(U)[/imath] is open and [imath]f[/imath] is a homeomorphism between [imath]U[/imath] and [imath]V[/imath]". But I'm not knowing how to proceed...need a clue...Thanks for attention!!! |
890050 | What can be the possible rank of adjoint of matrix of order n?
Let [imath] A [/imath] be matrix of order [imath] n [/imath]. What may the possible ranks of [imath]\mathop{\rm adj} (A) [/imath]? I think the possible answers are [imath]0[/imath], [imath]1[/imath], and [imath]n[/imath]. | 892039 | Is adjoint of singular matrix singular? What would be its rank?
Let [imath]A[/imath] be a square and singular matrix of order [imath]n[/imath]. Is [imath]$\operatorname{adj}(A)$[/imath] necessarily singular? What would be the rank of [imath]$\operatorname{adj}(A)$[/imath]? |
1198654 | With out the axiom of choice is the following true: If a topology on [imath]X[/imath] contains every infinite subset of [imath]X[/imath], then this is the discrete topology.
Is there a way to prove the following without having to use the axiom of choice or its countable version? Let [imath]X[/imath] be an infinite set and [imath]T[/imath] be a topology on [imath]X[/imath]. If [imath]T[/imath] contains every infinite subset of [imath]X[/imath], then [imath]T[/imath] is the discrete topology. | 233350 | Is the Discrete Topology on [imath]X[/imath] the Only One Containing All Infinite Subsets of [imath]X[/imath]?
Prove or find counterexamples. Let [imath]X[/imath] be an infinite set and [imath]T[/imath] be a topology on [imath]X[/imath]. If [imath]T[/imath] contains every infinite subset of [imath]X[/imath], then [imath]T[/imath] is the discrete topology. |
621160 | Number of divisors of a product
Let [imath]a[/imath] and [imath]b[/imath] be the number of divisors of two positive integers , is it possible to explicitly express the number of divisors of their product only in terms of [imath]a[/imath] and [imath]b[/imath]? If not , how can it be calculated efficiently without actually calculating the product? | 1198639 | Is there any formula for number of divisors of [imath]a \times b[/imath]?
Let [imath]a[/imath] and [imath]b[/imath] be two numbers, Number of divisors of [imath]a[/imath] is [imath]n_1[/imath]; Number of divisors of [imath]b[/imath] is [imath]n_2[/imath]; How to find the number of divisors [imath]N[/imath] of product [imath]a \times b[/imath] by using known number of divisors of factors [imath]n_1[/imath] and [imath]n_2[/imath]? If [imath]a = 20[/imath] and [imath]b = 21[/imath], then [imath]n_1 = 6[/imath] (number of divisors of [imath]a[/imath]), [imath]n_2 = 4[/imath] (number of divisors of [imath]b[/imath]. Then [imath]a \times b = 420[/imath] and [imath]N = 32[/imath] (number of divisors of [imath]a \times b[/imath]). Is there a formula to find [imath]N[/imath] in terms of [imath]n_1[/imath] and [imath]n_2[/imath]? |
1192212 | Help me Evaluate this Integral
I have this problem [imath]\int \frac {-5x + 11}{{(x+1)(x^2+1)}} \text{d}x[/imath] And i'm not sure how to deal with this. I've tried substitution and getting nowhere. I've peeked at the answer and there a trigonometric part with arctan. Do i use Partial fraction expansion here? Thanks | 20963 | Integration by partial fractions; how and why does it work?
Could someone take me through the steps of decomposing [imath]\frac{2x^2+11x}{x^2+11x+30}[/imath] into partial fractions? More generally, how does one use partial fractions to compute integrals [imath]\int\frac{P(x)}{Q(x)}\,dx[/imath] of rational functions ([imath]P(x)[/imath] and [imath]Q(x)[/imath] are polynomials) ? This question is asked in an effort to cut down on duplicates. See Coping with *abstract* duplicate questions. Also see List of Generalizations of Common Questions. |
1199663 | express tan(x) as a power series using maclauran's theorem.
the theorem states that if f(x) can be expanded as a power series for a given range of values of x then: [imath]f(x)=f(0)+xf'(0)+\frac{x^2}{2!}f''(0)+\frac{x^3}{3!}f'''(0)+\cdots[/imath] ([imath]'[/imath] means derivative) if [imath]f(x)=\tan(x)[/imath], what is the power series? | 770197 | Taylor Series of [imath]\tan x[/imath]
I found a nice general formula for the Taylor series of [imath]\tan x[/imath]: [imath]\tan x = \sum_{n\,=\,1}^\infty \frac {(-1)^{n-1}2^{2n} (2^{2n}-1) B_{2n}} {(2n)!} x^{2n - 1} [/imath] where [imath]B_n[/imath] are the Bernoulli numbers and [imath]|x| < \dfrac {\pi} 2[/imath]. I've tried Googling for a proof but didn't find anything. Hints would be appreciated too. I am using the typical definition of the Bernoulli numbers: [imath]\frac x {e^x - 1} = \sum_{n\,=\,0}^\infty \frac {B_n x^n} {n!}[/imath] |
1199734 | If there is an [imath]\epsilon_x >0[/imath] such that [imath]cl(B(x,ϵ_x))[/imath] is compact for each [imath]x \in X[/imath], then [imath]X[/imath] is complete.
If there is an [imath]\epsilon_x >0[/imath] such that [imath]cl(B(x,ϵ_x))[/imath] is compact for each [imath]x \in X[/imath], then [imath]X[/imath] is complete. We first take a Cauchy sequence [imath](x_n)[/imath] in [imath]X[/imath] and since it is bounded we get an [imath]x[/imath] and an [imath]\epsilon >0[/imath] such that except for finitely many terms the sequence will belong to [imath]cl(B(x,ϵ))[/imath] but can we say from here that [imath]cl(B(x,ϵ))[/imath] is compact? I am stuck here. How will the proof be? | 1066258 | A metric space is complete if for some [imath]\epsilon \gt 0[/imath], every [imath]\epsilon[/imath]-ball in [imath]X[/imath] has compact closure.
This is a problem from Munkres' Topology. Let [imath]X[/imath] be a metric space. (a) Suppose that for some [imath]\epsilon \gt 0[/imath], every [imath]\epsilon[/imath]-ball in [imath]X[/imath] has compact closure. Show that [imath]X[/imath] is complete. (b) Suppose that for each [imath]x \in X[/imath], there is an [imath]\epsilon \gt 0[/imath] such that the ball [imath]B(x, \epsilon)[/imath] has compact closure. Show by means of an example that [imath]X[/imath] need not be complete. I'm currently stuck on (a), I'm trying to show that given a Cauchy sequence ([imath]x_n[/imath]), some tail of the sequence belongs entirely on a ball [imath]B(x,\epsilon)[/imath] and thus since a compact set is sequentially compact in metric space, the sequence has a convergent subsequence and so the Caucy sequence converges to the same limit. However, I'm having trouble constructing this process. Also, for (b), I cannot think of any such example. Any solution, hint, or suggestions would be appreciated. |
1199587 | Supremum vs. Maximum in the definition of the Lp norm
The [imath]L_p[/imath] norm [imath]||A||_p[/imath] is defined as [imath]\sup_{x \neq 0} \frac{||Ax||_p}{||x||_p} = \max_{||x||_p = 1} ||Ax||_p \tag{1}[/imath] I'm not quite getting why the LHS uses [imath]\sup[/imath] but the RHS uses [imath]\max[/imath]. I know how both are defined but somehow I don't get it. The maximum requires the solution to be contained in the candidate set while the supremum does not. Thus, why does (1) hold? | 671311 | How can we replace [imath]\sup[/imath] with [imath]\max[/imath] in definition of subordinate norm for finite-dimensional vector space?
I have been given the definition of a subordinate (operator or matrix) norm: [imath]\lvert\lvert A \rvert\rvert=\sup_{x\in V}\frac{\lvert\lvert Ax \rvert\rvert}{\lvert\lvert x \rvert\rvert},[/imath] where [imath]V[/imath] is a vector space, and [imath]A[/imath] is a linear operator. I have been told that in the case of a finite-dimensional vector space, we can replace [imath]\sup[/imath] with [imath]\max[/imath], giving [imath]\lvert\lvert A \rvert\rvert=\max_{x\in V}\frac{\lvert\lvert Ax \rvert\rvert}{\lvert\lvert x \rvert\rvert},[/imath] and I am wondering why this is true. It seems reasonable to me, but how can I be convinced that it's true? |
1199078 | Proof of the definition of cardinal exponentiation
I really cannot seem to get my head around the definition of cardinal exponentiation with regards to finite sets: [imath]|X|^{|Y|}=|X^Y|[/imath] How would one even begin to prove this? Isn't [imath]X^Y[/imath] the set of all functions with the domain [imath]Y[/imath] and the codomain [imath]X[/imath]? | 89211 | How to Understand the Definition of Cardinal Exponentiation
I'm having trouble understanding the definition of cardinal exponentiation. Let's start with the definitions / claims I've been given: For any finite sets [imath]A,B[/imath], such that if [imath]|A|=a[/imath] and [imath]|B|=b\neq 0[/imath] then the number of functions from [imath]A[/imath] to [imath]B[/imath] is [imath]b^a[/imath]. ( I'm guessing this is only used when both sets are finite ). For any sets [imath]A,B[/imath], [imath]B^A[/imath] is the set of functions from [imath]A[/imath] to [imath]B[/imath]. For every set [imath]A[/imath], [imath]|P(A)|=|\left \{ 0,1 \right \}^A|[/imath] Now for the definition and why I don't understand it: Let [imath]a,b[/imath] be cardinal numbers. Let [imath]A,B[/imath] be sets such that [imath]|A|=a, |B|=b[/imath]. Then [imath]|B^A|=b^a[/imath]. How analagous to normal exponentiation is this and where does in differ when we introduce cardinals like [imath]\aleph_0[/imath] or [imath]C[/imath]? Is [imath]2^2=4[/imath] and is [imath]\frak c^2=c\cdot c=c\times c[/imath]? |
1199716 | Proving some property of a Formal Logic Language
I am stuck at this problem: Let [imath]\Sigma = \{\lnot,\lor,\land,\rightarrow,\leftrightarrow,(,),P_1,...,P_n\}[/imath] be an alphabet. Now let's define the set of logical expressions [imath]\mathscr{L} \subseteq \Sigma^\ast[/imath] recursively as follows: Rule #1: For each [imath]i\in\{1,2,...,n\}[/imath], [imath]P_i\in \mathscr{L}[/imath] Rule #2: For each [imath]\phi \in \mathscr{L}[/imath], [imath]\lnot \phi \in \mathscr{L}[/imath] Rule #3: For each [imath]\phi,\psi \in \mathscr{L}[/imath] and [imath]@\in\{\lor,\land,\rightarrow,\leftrightarrow\}[/imath], [imath](\phi @ \psi)\in\mathscr{L}[/imath] No strings other than those derived from Rules #1, #2 and #3 are in [imath]\mathscr{L}[/imath]. Prove that for all [imath]\phi,\phi' \in \mathscr{L}[/imath], [imath]\psi,\psi'\in\Sigma^\ast[/imath] and [imath]@\in\{\lor,\land,\rightarrow,\leftrightarrow\}[/imath], If [imath](\phi @\psi)=(\phi' @\psi')[/imath] Then it must be the case that [imath]\phi = \phi'[/imath] and [imath]\psi = \psi'[/imath]. I tried to prove it using structural induction and ordinary induction, but I wasn't able to prove it. What we need to show is that the @ in the strings [imath](\phi @\psi)[/imath] and [imath](\phi' @\psi')[/imath] lies at the same index position, and then we will get immediately that [imath]\phi = \phi'[/imath] and [imath]\psi = \psi'[/imath] by definition of string equality. Thanks for any help. | 1199447 | Proving a property of a Logic Formal Language
I am stuck at this problem: Let [imath]\Sigma = \{\lnot,\lor,\land,\rightarrow,\leftrightarrow,(,),P_1,...,P_n\}[/imath] be an alphabet. Now let's define the set of logical expressions [imath]\mathscr{L} \subseteq \Sigma^\ast[/imath] recursively as follows: Rule #1: For each [imath]i\in\{1,2,...,n\}[/imath], [imath]P_i\in \mathscr{L}[/imath] Rule #2: For each [imath]\phi \in \mathscr{L}[/imath], [imath]\lnot \phi \in \mathscr{L}[/imath] Rule #3: For each [imath]\phi,\psi \in \mathscr{L}[/imath] and [imath]@\in\{\lor,\land,\rightarrow,\leftrightarrow\}[/imath], [imath](\phi @ \psi)\in\mathscr{L}[/imath] No strings other than those derived from Rules #1, #2 and #3 are in [imath]\mathscr{L}[/imath]. Prove that for all [imath]\phi \in \mathscr{L}[/imath] and [imath]\psi\in\Sigma^\ast[/imath], If [imath](\phi\land\psi)\in\mathscr{L}[/imath] Then it must be the case that [imath]\psi\in\mathscr{L}[/imath]. I tried to prove it using structural induction, contradiction and several other ways but I wasn't able to prove it. Thanks for any help. |
1200622 | Group of all [imath]2\times2[/imath] matrices where [imath]a[/imath], [imath]b[/imath], [imath]c[/imath], and [imath]d[/imath] are integers modulo [imath]p[/imath], Herstein Q[imath]26[/imath] Page [imath]37[/imath]
Let [imath]G[/imath] be group of all square matrices of order [imath]2[/imath] [imath]\begin{bmatrix} a & b\\ c & d \end{bmatrix}[/imath] such that [imath]a[/imath], [imath]b[/imath], [imath]c[/imath], and [imath]d[/imath] are integers modulo a prime number [imath]p[/imath], such that [imath]ad-bc\ne0[/imath]. [imath]G[/imath] forms a group relative to matrix multiplication. What is order of [imath]G[/imath]? I have tried for smaller values of [imath]p[/imath] by explicitly writing matrices. But how do I do this for general [imath]p[/imath]. Thanks | 901654 | Order of group [imath]GL_{2}\left( \mathbb{F}_{p}\right) [/imath]
I'm having a hard time counting. I need to count the number of elements for the multiplicative group of invertible [imath]2\times 2[/imath] matrices [imath]GL_{2}\left( \mathbb{F}_{p}\right) [/imath] with elements from the field [imath]\mathbb{F}_{p}[/imath] where [imath]p[/imath] is any prime number. David S. Dummit and Richar M. Foote's Abstract Algebra (3rd edition) has a solution on page 413 but that approach is by counting the possible number of basis of vector spaces but I wanted to see if there was a direct way. The book says that [imath]\left\vert GL_{n}\left( \mathbb{F}_{p}\right) \right\vert =\left( p^{n}-1\right) \left( p^{n}-p\right) ...\left( p^{n}-p^{n-1}\right) [/imath] so my case reduces to [imath]p^{4}-p^{3}-p^{2}+p[/imath]. For any matrix [imath]\left( \begin{array}{cc} a & b \\ c & d% \end{array}% \right) [/imath] with [imath]a,b,c,d\in \mathbb{F}_{p}[/imath], I understand that [imath]p^{4}[/imath] is the total number of 2x2 matrices and that we need to subtract the possible combinations for [imath]ad=bc[/imath] (non-invertible elements) but I can't account for the rest. In particular, that plus sign is absolutely baffling. Any hint would be appreciated. I'd then take this up for the general [imath]n[/imath]x[imath]n[/imath] case but first I need to dispose the 2x2 case. P.S. I tried looking for a pre-existing answer and a related answer is present here but it's by the same approach as the book and I'd like to have an approach without resorting to linear independence. This answer, too, was a case for [imath]p=3[/imath] but I couldn't see the pattern. |
1201059 | Models of the empty theory T, and proof that T [imath]\kappa[/imath] categorical for every cardinality.
Bombarding stack exchange with model questions today I am tackled with the following problem: Note this is the same question as posted by B0bg0blin's here, i just need a bit more clarity. In the language [imath]L[/imath] with no relation, function, or constant symbols, let [imath]T[/imath] be the empty theory, that is, the empty set of [imath]L[/imath]-sentences. What is a model of [imath]T[/imath]? Show that [imath]T[/imath] is [imath]\kappa[/imath]-categorical for every cardinal [imath]\kappa[/imath]. Problem 1: What is a model of [imath]T[/imath]? Are all sets are a model of [imath]T[/imath]? Problem 2: I know a theory [imath]T[/imath] is categorical in cardinality [imath]\kappa[/imath] (ie [imath]\kappa[/imath] categorical) if and only if there is a model of [imath]T[/imath] of cardinality [imath]\kappa[/imath] and any 2 models of cardinality [imath]\kappa[/imath] are isomorphic. This is my standard problem with model theory, I get all the definitions but I just dont know how to piece things together to prove what I am trying to prove. So any help would be appreciated. | 1192679 | Models of the empty theory
so throughout my reading of model theory the idea of the "empty" theory has been put down as trivial, however I am curious as to why. Let us look at the following. Suppose We have [imath]L_=[/imath], the language with equality but no other relation/function/constant symbol. Let [imath]T[/imath] be the empty theory (so it contains no [imath]L_=[/imath] sentences). Is it true that any two models of [imath]T[/imath] of the same cardinality [imath]\kappa[/imath] are isomorphic? That is, show that [imath]T[/imath] is [imath]\kappa[/imath]-categorical for all cardinals [imath]\kappa[/imath]. My intuition wants me to say that the models of [imath]T[/imath] of cardinality [imath]\kappa[/imath] are just the underlying sets of a model, given the lack of relations/formulas. For countable models I can see how this would lead to an isomorphism [imath]\phi : A \to B[/imath] by setting [imath]\phi(a_i)=b_i[/imath] for all [imath]i \in \mathbb{N}[/imath], [imath]a_i \in A[/imath], [imath]b_j \in B[/imath]. For uncountable sets I have having difficulty I am having slightly more issues though, due to the inability to index the elements, and due to my lack of set theory knowledge (attempting to look purely from a model theory perspective). Any help would be much appreciated. |
481502 | [imath]az+b\overline{z}+c=0[/imath] represents a line
When does [imath]az+b\overline{z}+c=0[/imath] represent a line? All of [imath]a,b,c[/imath] are complex numbers. I know that a line in the complex plane is usually represented by [imath]z=a+bt[/imath], where the parameter [imath]t[/imath] runs through all real values. So suppose we're given the equation [imath]az+b\overline{z}+c=0[/imath]. If [imath]z[/imath] represents a line [imath]d+et[/imath], we must have [imath]a(d+et)+b(\overline{d}+\overline{e}t)+c=0[/imath]. Since the equation holds for all real values [imath]t[/imath], it must be that [imath]ae+b\overline{e}=0[/imath], and as a result [imath]ad+b\overline{d}+c=0[/imath]. How can we get to a condition involving only [imath]a,b,c[/imath] from here? | 374659 | When does [imath]az + b\bar{z} + c = 0[/imath] represent a line?
[imath]a,b,c[/imath] and [imath]z[/imath] are all complex numbers. My idea was to show that it passes through the point [imath]\infty[/imath] in the extended complex plane, but I'm not quite sure how to execute that. Update: It says in the text that a straight line can be represented by a parametric equation [imath]z = a+bt[/imath], where [imath]a[/imath] and [imath]b[/imath] are complex numbers, and [imath]b\neq 0[/imath], [imath]t\in\mathbb{R}[/imath] |
1200266 | Quick question about relation between Nullspace and Eigenspace
I have a question about a note given in a linear algebra textbook. It is just given as a remark, with no proof or explanation so I want to make sure I understand it correctly. First, it gives the definition of the eigenspace of A to be, The set, [imath]E_{\lambda}=\{{v \in \mathbb R^n : Av= \lambda v}\}[/imath] and it then notes, from this we can see that because [imath]E_{0}=N(A)[/imath], this [imath]\rightarrow[/imath] a matrix is singular if and only if [imath]\lambda=0[/imath] is an eigenvalue. But I am still a little confused. If I am recalling correctly, a transformation is singular if there exists a [imath]v \neq 0[/imath] for which [imath]T(v)=0[/imath] I guess in this case it is in regard to if a matrix is invertible or not? I think I may be getting confused between the two differences ( if there is one) and the differences between talking about just A and a transformation in general. Anyone have any advice /explanations for this? Thank you! | 1105353 | If the [imath]\det(A)=0[/imath] why does the matrix [imath]A[/imath] have an eigenvector?
If the [imath]\det(A)=0[/imath] why does the matrix [imath]A[/imath] have an eigenvector? Explain why there is a basis [imath]B[/imath] in [imath]R^n[/imath] so that the matrix [imath][A]_B[/imath] has the zero vector as its first column I know that if the [imath]\det(A)=0[/imath] of a matrix is zero then it is singular and thus not invertible. That in turn means that it doesn't have one solution. Why does that mean that it has an eigenvector? As for the second I have no idea. |
1201869 | Linear Algebra Question Vector Question
A 3 × 3 matrix [imath]A[/imath] is such that for [imath]X = [2, 3, 4]^\top[/imath], [imath]AX = 0[/imath]. Show that [imath]\det(A) = 0[/imath]. | 1200496 | Show that [imath]det(A)=0[/imath] for a Matrix [imath]A[/imath] Such That for [imath]X=[2,3,4]^{T}[/imath], [imath]AX=0[/imath]
Show that [imath]det(A)=0[/imath] for a [imath]3\times3[/imath] matrix [imath]A[/imath] such that for [imath]X=[2,3,4]^{T}[/imath], [imath]AX=0[/imath] It seems most intuitive to me that [imath]A[/imath] would just be a matrix composed of zeros, but is 1) this what the question is really asking (as that seems like a bit of an easy way out) and 2) a way to prove this for [imath]A[/imath] without it being composed entirely of zeros? For point 2 I'm unsure of what can be done other than just making arbitrary matrices that have a [imath]det=0[/imath], which doesn't seem like a very rigorous way to solve the problem. |
1201899 | Rational Function Residue Formula Proof
I came across a theorem used to calculate residues of rational functions that states that if f and g are analytic functions at [imath]z_k[/imath] and [imath]g'(z_k)[/imath] isn't 0, then the residue of [imath]f(z)/g(z)[/imath] at [imath]z_k[/imath] is [imath]f(z_k)/g'(z_k)[/imath]. I want to prove this using Laurent series (taking out the coefficient of [imath](z-z_k)^-1[/imath]) but I can't get around the algebra of actually dividing the two Laurent series. Can someone guide me through this? | 1191889 | Residue of two functions
Let be [imath]f,g[/imath] functions analytic in [imath]z_0[/imath], with [imath]z_0[/imath] a zero of order one of [imath]g[/imath] and [imath]f(z_{0})\neq 0[/imath]. Show that [imath] \operatorname{Res}\Bigl(\frac{f}{g},z_0\Bigr)=\frac{f(z_{0})}{g'(z_{0})} [/imath] My attemp... If [imath]z_{0}[/imath] is a zero of order one of [imath]g[/imath], then [imath]g(z)=(z-z_{0})h(z)[/imath], with [imath]h(z)[/imath] analytic and [imath]h(z_{0})\neq 0[/imath], then by analyticity of [imath]g[/imath] and [imath]h[/imath]: [imath]g'(z)=h(z)+(z-z_{0})h'(z)[/imath]. So, [imath] \frac{f(z_{0})}{g'(z_{0})}=\frac{f(z_{0})}{h(z_{0})+0}=\lim_{z\to z_{0}}{(z-z_{0})\frac{f(z)}{g(z)}} [/imath] But in the ultimate point I'm not sure |
823161 | Recurrence Relation Challenges
as i ask question and answered by some Clever people at this topic: Recurrence Relation Solving Problem i try to learn new thing with new question very similar to get familiar with recurrence relation and order complexity: [imath]T(n)=n + \sum\limits_{k=1}^n [T(n-k)+T(k)] [/imath] [imath]T(1) = 1[/imath]. We want to calculate order of T? i think the order is O(3^n)? every one could add any detail or step by step solving? | 811486 | Recurrence Relation Solving Problem
Can anyone help me in solving this complex recurrence in detail? [imath]T(n)=n + \sum\limits_{k-1}^n [T(n-k)+T(k)] [/imath] [imath]T(1) = 1[/imath]. We want to calculate order of T. I'm confused by using recursion tree and some maths induction. |
1201873 | How do I solve this quadratics problem?
The problem: A jet flew from Tokyo to Bangkok, a distance of 4800km. On the return trip, the speed was decreased by 200 km/h. If the difference in the times of the flights was 2 hours, what was the jet's speed from Bangkok to Tokyo? From the problem, we can see that the variables are: Distance: T to B = 4800km B to T = 4800km Speed: T to B = [imath]x[/imath]km/h B to T = [imath]x-200[/imath]km/h Time: T to B = [imath]x[/imath]h B to T = [imath]x-2[/imath]h The problem definitely needs to be solved using the distance formula, [imath]d=s*t[/imath]. Rearranging it for time, we get [imath]t=d/s[/imath] What do I do from here? I know from solving similar previous problems that I have to factor it and get the x-intercepts (one of them will be the jet's speed) but how to I solve up to that point when there are two unknowns? Thank you! | 1202050 | Find the speed of a jet given the time of travel back and forth
The problem: A jet flew from Tokyo to Bangkok, a distance of 4800km. On the return trip, the speed was decreased by 200 km/h. If the difference in the times of the flights was 2 hours, what was the jet's speed from Bangkok to Tokyo? From the problem, we can see that the variables are: Distance: T to B = 4800km B to T = 4800km Speed: T to B = [imath]x[/imath]km/h B to T = [imath]x-200[/imath]km/h Time: T to B = [imath]x[/imath]h B to T = [imath]x-2[/imath]h The problem definitely needs to be solved using the distance formula, [imath]d=s*t[/imath]. Rearranging it for time, we get [imath]t=d/s[/imath] What do I do from here? I know from solving similar previous problems that I have to factor it and get the x-intercepts (one of them will be the jet's speed) but how to I solve up to that point when there are two unknowns? I would appreciate if you provide the full solution. Thank you! |
1161049 | Evaluating [imath]\int_{0}^{\infty}\frac{e^{-x^2}-e^{-x}}{x}dx [/imath]
I am working on this improper integral, [imath]\int_{0}^{\infty}\frac{e^{-x^2}-e^{-x}}{x}dx[/imath] First, I separate the integral into two pieces, I have [imath]\int_{0}^{\infty}\frac{e^{-x^2}}{x}dx -\int_{0}^{\infty}\frac{e^{-x}}{x}dx=I_1+I_2[/imath] I know that [imath]I_2[/imath] can use double integral, [imath] I_2=\int_0^\infty \int _1^\infty e^{-tx} dt dx = \int _1^\infty \frac{1}{t}dt [/imath] [imath] \int_0^\infty (\frac{e^{-tx}}{-x}) \Bigg|_{1}^\infty dx = \int_1^\infty \frac{1}{t}dt =\ln t \Bigg|_{1}^{\infty}=\infty[/imath] But I don't know how to do [imath]I_1[/imath], can someone give me a hit or suggestion? Thanks! | 1120895 | Unusual integral
I got a clock as a gift recently. It has a very novel face in that the hour positions are given by a complex formula. For the most part, I have been able to verify the calculations presented as accurate, but the two o'clock identity has me stumped. [imath] \frac{\gamma}{\displaystyle{\int_0^\infty\frac{e^{-x^2}-e^{-x}}{x}\,dx}} [/imath] I know that [imath]\gamma[/imath] is the Euler-Mascheroni constant. And WolframAlpha tells me that [imath]\int_0^\infty\frac{e^{-x^2}-e^{-x}}{x}\,dx=\frac{\gamma}{2}[/imath] which makes sense because [imath] \frac{\gamma}{\displaystyle{\int_0^\infty\frac{e^{-x^2}-e^{-x}}{x}\,dx}}=\frac{\gamma}{\displaystyle{\frac{\gamma}{2}}}=2 [/imath] It is not clear how I would show [imath]\int_0^\infty\frac{e^{-x^2}-e^{-x}}{x}\,dx=\frac{\gamma}{2}[/imath]. Could anyone shed some light on this or point me to a source (book, article, etc...) where I can read up on this. The usual internet (listed above) resources have not been helpful to me. |
1159722 | Normal subgroup question/example
Give an example(s) of a group [imath]G[/imath] and [imath]H\leq G[/imath] (i.e., [imath]H[/imath] is a subgroup of [imath]G[/imath]) where [imath]H[/imath] is not normal in [imath]G[/imath]. What about [imath]S_3[/imath] and [imath]\langle(1\;2)\rangle=\{(1),(1\;2)\}[/imath]? [Since [imath](1\;2\;3)(1\;2)(1\;3\;2)=(2\;3)\not\in\langle(1\;2)\rangle[/imath]] Does this work? Any other ideas? | 532191 | Example of a Subgroup That Is Not Normal
Can you kindly provide an example of a subgroup that is not normal? I have been told many times that, for coset multiplication to be defined, the subgroup must be normal. I have seen the proof and examples of quotient group multiplications. Now, I am trying to find out where the process will break down if the subgroup is not normal. Suppose that [imath]x \in a_1H = a_2H, y \in b_1H = b_2H[/imath], where [imath]H[/imath] is not normal. So [imath]x = a_1h_1 = a_2h_2[/imath] for some [imath]h_1, h_2 \in H[/imath], and [imath]y = b_1h_1^* = b_2h_2^*[/imath] for some [imath]h_1, h_2 \in H[/imath]. [imath]xy = (a_1b_1)(h_1h_1^*) = (a_2b_2)(h_2h_2^*)[/imath]. However, both [imath]h_1h_1^*, h_2h_2^*[/imath] are still elements of [imath]H[/imath]. So I don't see how [imath]a_1b_1H \neq a_2b_2H[/imath]. |
1193064 | Assume that [imath] 1a_1+2a_2+\cdots+na_n=1[/imath], where the [imath]a_j[/imath] are real numbers.
Assume that [imath] 1a_1+2a_2+\cdots+na_n=1, [/imath] where the [imath]a_j[/imath] are real numbers. As a function of [imath]n[/imath], what is the minimum value of [imath]1a_1^2+2a_2^2+\cdots+na_n^2?[/imath] | 2287034 | Assume that [imath]1a_1+2a_2+\cdots+na_n=1[/imath], where the [imath]a_j[/imath] are real numbers...
Assume that [imath] 1a_1+2a_2+\cdots+na_n=1, [/imath] where the [imath]a_j[/imath] are real numbers. As a function of [imath]n[/imath], what is the minimum value of [imath]1a_1^2+2a_2^2+\cdots+na_n^2[/imath] There is a similar question on Math StackExchange, but nobody gave a solid and conclusive answer, so I felt the need to repeat it here. I also think this question should be solved using Cauchy-Schwarz, Mean Inequality Chain, or Trivial Inequality, but I would appreciate any answer. Thanks! |
1138797 | How to prove if [imath]u\in W^{1,p}[/imath], then [imath]|u|\in W^{1,p}[/imath]?
How to prove if [imath]u\in W^{1,p}[/imath], then [imath]|u|\in W^{1,p}[/imath]? Since [imath]|u|\in L_p[/imath], I only need to show weak derivative of [imath]|u|[/imath] exists and [imath]D|u| \in L_p[/imath]. Can anyone give me some hint? Thanks! | 1639274 | Help proving a map between Sobolev spaces is continuous
We are trying to show that the map [imath]f\mapsto|f|[/imath] is a continuous (nonlinear) map from [imath]W^{1,p}(\Omega)\to W^{1,p}(\Omega)[/imath] for any bounded/open region [imath]\Omega[/imath] and for [imath]p\in[1,\infty)[/imath]. We have tried using the regular definition of continuity using the standard norm on Sobolev spaces for both [imath]f[/imath] and [imath]|f|[/imath]: [imath]\displaystyle\|u\|^p=\sum_{\alpha:|\alpha|\leq1}\|\partial^\alpha u\|^p_{L^p(\Omega)}[/imath] where [imath]\alpha[/imath] is a multi-index and [imath]|\alpha|[/imath] denotes the sum of its components, but we unable to make any progress. Is there another approach we should try, or some trick that we seem to be forgetting? Thank you for your time. |
1202625 | real-anylsis: topology, continuous maps, and metric spaces
I am so lost trying to figure out this question: I'm not even sure what it is asking, let alone where to start. I basically only have definitions to use; but I can't seem to fit them all together to make this work: Let [imath]\\X[/imath] be an infinite set, [imath]\\a\in X[/imath], [imath]\Sigma[/imath] be a set of all finite subsets of [imath]\\X[/imath] that do not contain [imath]\\a[/imath]. Prove that [imath]\Sigma[/imath] is the set of all closed sets for some topological structure on [imath]\\X[/imath] and that any two continuous maps of [imath]\\X[/imath] with this topological structure to the same metric space [imath]\\Y[/imath] coincide with each other on [imath]\\a[/imath] if they coincide with each other on every [imath]b \ne a[/imath]. Keep in mind that, even though this is an analysis course, I can only use the topological definitions and theorems from the notes, which include the basics of topological spaces, metric spaces, continuous functions, compact spaces, and connected spaces. So while there might be some elegant way of doing this with a theorem, I have to resort to using the bare-bones definitions. | 1198141 | showing a collection of sets contain all closed sets
Let [imath]X[/imath] be infinite set and [imath]a \in X [/imath]. Let [imath]\mathscr{F} = \{ A \subset X: |A| \; \; finite \; \;, a \notin A \}[/imath] Then [imath]\mathscr{F}[/imath] is the set all of all closed sets for some topology [imath]\mathcal{T} [/imath] on [imath]X[/imath]. Also, any two continuous maps [imath]f,g: (X, \mathscr{T}) \to Y[/imath] ( [imath]Y[/imath] is metric space) have the property that if [imath]f(b) = g(b)[/imath] for all [imath]b \neq a [/imath], then [imath]f(a) = g(a)[/imath]. In my notes my professor posted this theorem without proof, so I was wondering what is best way to proof this theorem. |
1203676 | Is an empty set reflexive? Symmetric? Transitive?
Suppose [imath]A\neq\emptyset[/imath] Since, [imath]\emptyset\subseteq A * A[/imath] the set [imath]R=\emptyset[/imath] is a relation on A. Is [imath]R[/imath] reflexive? symmetric? transitive? I remember hearing something can be "vacuously" true. So the empty set would be reflexive, symmetric and transitive because it doesn't meet the definition? So there is no [imath](x,x)[/imath] that can exist in [imath]R[/imath] therefore vacuously reflexive. There is no [imath](x,y)[/imath] that can exist in [imath]R[/imath] therefore vacuously symmetric. There is no [imath](x,y)[/imath] that can exist in [imath]R[/imath] therefore vacuously transitive. Is my reasoning correct here? | 1081333 | Prove that the empty relation is Transitive, Symmetric but not Reflexive
Question: Let [imath]R[/imath] be a relation on a set [imath]A[/imath]. Prove that if [imath]A[/imath] is non-empty, the empty relation is not reflexive on [imath]A[/imath]. the empty relation is symmetric and transitive for every set [imath]A[/imath]. My Solution: For a relation to be reflexive: For all elements in A, they should be related to themselves. ([imath]x[/imath] [imath]R[/imath] [imath]x[/imath]). Now in this case there are no elements in the Relation and as A is non-empty no element is related to itself hence the empty relation is not reflexive. Now for a set to be symmetric and transitive: As these are conditional statements if the antecedent is false the statements would be true. And as the relation is empty in both cases the antecedent is false hence the empty relation is symmetric and transitive. (I just want to know if my solutions are correct. - I've given brief solutions without using any notation here.) |
1203868 | Problem 8 (a) in Exercises after Sec. 18 in Munkres' Topology, 2nd ed.: How to show this set is closed?
Let [imath]X[/imath] be an arbitrary topological space, let [imath]Y[/imath] be an ordered set in the order topology, and let the maps [imath]f, g \colon X \to Y[/imath] be continuous. Then how to show that the set [imath]S[/imath] given by [imath]S \colon= \{ \ x \in X \ \colon f(x) \leq g(x) \ \}[/imath] is closed in [imath]X[/imath]? | 287305 | Given a pair of continuous functions from a topological space to an ordered set, how to prove that this set is closed?
Given that [imath]X[/imath] is an arbitrary topological space, [imath]Y[/imath] is a totally ordered set in the order topology, and [imath]f[/imath], [imath]g \colon X \to Y[/imath] are continuous functions, how to show that the subset [imath]A[/imath] of [imath]X[/imath] given by [imath] A \colon= \left\{ \ x \in X \ \colon \ f(x) \leq g(x) \ \right\} [/imath] is closed in [imath]X[/imath]? Edit based on the answer by Hagen von Eitzen: Let [imath]U \colon= \{ u \times v \in Y \times Y \ \colon \ u < v \ \}[/imath]. We show that [imath]U[/imath] is open in [imath]Y \times Y[/imath]. Let [imath]u \times v \in U[/imath]. Then [imath]u < v[/imath]. Case I. If there is some [imath]y \in Y[/imath] such that [imath]u < y < v[/imath], then [imath]u \times v \in (-\infty, y) \times (y, +\infty) \subset U[/imath]. Case 2. If [imath](u,v)[/imath] is empty, then [imath]u \in (-\infty, v)[/imath] and [imath]v \in (u, +\infty)[/imath], and so [imath]u \times v \in (-\infty, v) \times (u, +\infty)[/imath]. Moreover, if [imath]a \times b \in (-\infty, v) \times (u, +\infty)[/imath], then we must have [imath]a < v[/imath] and [imath]b> u[/imath]. So [imath]a \leq u < v \leq b[/imath], which implies that [imath]a < b[/imath] and so [imath]a \times b \in U[/imath]. Thus, [imath]u \times v \in (-\infty, v) \times (u, +\infty) \subset U[/imath]. So [imath]U[/imath] is open in [imath]Y \times Y[/imath]. Similarly, we can show that the set [imath]V \colon= \left\{ \ u \times v \in Y \times Y \ \colon \ u > v \ \right\}[/imath] is open in [imath]Y \times Y[/imath]. Thus, it follows that the set [imath]A \colon= \{ \ u \times v \in Y \times Y \ \colon \ u \leq v \ \}[/imath] is closed in [imath]Y \times Y[/imath]. Now since the maps [imath]f \colon X \to Y[/imath] and [imath]g \colon X \to Y[/imath] are continuous, so is the map [imath]f \times g \colon X \to Y \times Y[/imath] defined as [imath](f \times g)(x) \colon= f(x) \times g(x) \ \mbox{ for all } \ x \in X.[/imath] Thus, the inverse image under [imath]f \times g[/imath] of the set [imath]A[/imath] is closed in [imath]X[/imath]. But [imath] \begin{align} (f \times g)^{-1} (A) &= \left\{ \ x \in X \ \colon \ (f \times g)(x) \in A \ \right\} \\ &= \left\{ \ x \in X \ \colon \ f(x) \times g(x) \in A \ \right\} \\ &= \left\{ \ x \in X \ \colon \ f(x) \leq g(x) \ \right\}. \end{align} [/imath] |
1203783 | If [imath]A[/imath] is any matrix then [imath]A^*A[/imath] and [imath]AA^*[/imath] are Hermitian with non-negative eigenvalues
How can I show that if [imath]A[/imath] is any matrix then [imath]A^*A[/imath] and [imath]AA^*[/imath] are Hermitian with non-negative eigenvalues? I am stuck and any help will be greatly appreciated. | 99630 | Is the matrix [imath]A^*A[/imath] and [imath]AA^*[/imath] hermitian?
As the title says. Is the matrices [imath]A^*A[/imath] and [imath]AA^*[/imath] hermitian (symmetric if [imath]A[/imath] is real)? |
1203271 | [imath]A\subseteq B\to A\setminus B = \emptyset[/imath]. Is this true?
If [imath]A\subseteq B[/imath], is then [imath]A\setminus B = \emptyset[/imath]? If we take a some [imath]x\in A[/imath], then also [imath]x\in B[/imath]. Right? So, if we take all the elements of [imath]B[/imath] away from [imath]A[/imath], we have nothing left? | 1170190 | [imath]A ⊂ B[/imath] if and only if [imath]A − B = ∅[/imath]
I need to prove that [imath]A ⊂ B[/imath] if and only if [imath]A − B = ∅[/imath]. I have the following "proof": [imath] A \subset B \iff A - B = \emptyset[/imath] proof for [imath]\implies:[/imath] [imath]\forall x \in A, x \in B[/imath] Therefore, [imath]A - B = \emptyset[/imath] proof for [imath]\impliedby[/imath]: If [imath]A - B = \emptyset[/imath] then [imath]\forall x \in B, x \in A[/imath] since [imath]\forall x \in B, x \in A[/imath], [imath] A \subset B [/imath] However the whole thing seems to be incredibly "fragile" and relies on circular logic (see how I just switched the sets in the for all statements) Is this a valid proof? Is there a better way to write it? |
1203910 | Suppose [imath]R[/imath] is commutative, and let [imath]M[/imath] be a free [imath]R[/imath]-module of finite rank. Prove that if [imath]A[/imath] and [imath]B[/imath] are bases of [imath]M[/imath], then [imath]|A| = |B|[/imath].
Suppose [imath]R[/imath] is commutative, and let [imath]M[/imath] be a free [imath]R[/imath]-module of finite rank. Prove that if [imath]A[/imath] and [imath]B[/imath] are bases of [imath]M[/imath], then [imath]|A| = |B|[/imath]. I can't really figure out how to prove this theorem. | 1075709 | Example of a ring without Invariant Basis Property
Let [imath]A[/imath] be a ring and consider the free modules [imath]A^{\oplus n}[/imath], [imath]A^{\oplus k}[/imath], with [imath]n,k\in \mathbb{N}[/imath]. Can [imath]A^{\oplus n}[/imath] be isomorphic to [imath]A^{\oplus k}[/imath] if [imath]k\neq n[/imath]? Thanks in advance for the help. |
1204285 | Suppose that [imath]A[/imath] is a connected subset of a space [imath]X[/imath] and that [imath]A\subseteq B \subseteq \bar A[/imath]. Prove [imath]B[/imath] is connected.
I think I can prove the closure of [imath]A[/imath], that is [imath]\bar A[/imath], is connected, as there are many other threads on this site. I am then just not sure how to make the jump to show formally that B is connected. | 284330 | Prove that if [imath]X[/imath] and it's closure [imath]\overline X[/imath] are connected and if [imath]X\subset Y \subset \overline X[/imath], show that Y is also connected.
Can anyone give me a proof of the statement that if [imath]X[/imath] and its closure [imath]\overline X[/imath] are connected and if [imath]X\subset Y \subset \overline X[/imath], show that Y is also connected? Thank you. |
1204332 | Can we find the integral [imath]\int_{0}^{2\pi} \cos(\cos x)\,dx[/imath]?
Can we find the integral [imath]\int_0^{2\pi} \cos(\cos x)\,dx\;?[/imath] | 117536 | Evaluate $\int \cos(\cos x)~dx$
Evaluate [imath]$\int \cos(\cos x)~dx$[/imath] I tried to use chain rule but failed. Can anyone help me please? |
1204454 | Prove that [imath]{f_n}[/imath] converges uniformly to f
Let [imath]f:\mathbb{R}→\mathbb{R}[/imath] be uniformly continuous; let [imath]\{y_n\}⊂\mathbb{R}[/imath] be such that [imath]\lim y_n=0[/imath]; and define the sequence [imath]f_n:\mathbb{R}→\mathbb{R}[/imath] by [imath]f_n(x)=f(x+y_n)[/imath]. Prove that [imath]{f_n}[/imath] converges uniformly to [imath]f[/imath]. Please explain and walk me through proving this statement. | 350582 | Prove that [imath]f_n[/imath] converges uniformly on [imath]\mathbb{R}[/imath]
Suppose that [imath]f[/imath] is uniformly continuous on [imath]\mathbb{R}[/imath]. If [imath]\displaystyle \lim_{n\to\infty} y_n =0[/imath] and [imath]f_n(x) := f(x + y_n)[/imath] for [imath]x \in \mathbb{R}[/imath], prove that [imath]f_n[/imath] converges uniformly on [imath]\mathbb{R}[/imath]. |
1204128 | Related Rate Question: Two sides of a triangle are 6 m and 8 m in length and the angle between them is increasing at a rate of 0.06 rad/s
Two sides of a triangle are 6 m and 8 m in length and the angle between them is increasing at a rate of [imath]0.06[/imath] rad/s. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is [imath]\large\frac {\pi}{3}[/imath] rad. | 534359 | Related rates: Find dA/dt of triangle, given d(theta)/dt -- Can't come to textbook answer
The question: ABC is a triangle in which the lines [imath]\overline {AB} = 20cm[/imath], [imath]\overline {AC} = 32cm[/imath] and [imath]\angle BAC = \theta[/imath]. If [imath]\theta[/imath] is increasing at the rate of 2° per minute, determine the rate at which the triangle's area is changing when [imath]\theta = 120°[/imath]. Here's my attempt: Let [imath]t[/imath] be time in minutes. We're given [imath]\frac{d\theta}{dt} = 2°[/imath], Area of triangle with two sides and included angle: [imath]A = \frac12\overline {AB}[/imath] [imath]\overline {AC}[/imath] [imath]sin\theta[/imath], i.e. [imath]A = \frac12(20)(32)sin\theta[/imath] [imath]= 320sin\theta[/imath] [imath]\therefore \frac{dA}{dt} = \frac{dA}{d\theta}\frac{d\theta}{dt} = 320cos\theta\frac{d\theta}{dt}[/imath] i.e. at [imath]\theta = 120°[/imath]: [imath]\frac{dA}{dt} = 320cos(120) * (2)[/imath] [imath]= -320 cm^2/min[/imath] The textbook answer: Decreasing at [imath]5.59cm^2/min[/imath] The question is exactly as above, and I've double checked the units. Can anyone please point out where I've gone wrong, or if I haven't? Thank you! |
1205253 | Prove tensor multiplication is bilinear
I am reading these online notes(http://people.virginia.edu/~mve2x/7752_Spring2010/lecture5.pdf), and on the bottom of page 2 the following exercise occurs. Define tensor multiplication of two [imath]R[/imath]-algebras [imath]A[/imath] and [imath]B[/imath] as: [imath](a_1 \otimes b_1) \cdot (a_2 \otimes b_2) = a_1 a_2 \otimes b_1 b_2[/imath]. Prove this is bilinear. The first property of a function being bilinear is [imath](u+w) \cdot v = (u \cdot v) + (w \cdot v)[/imath], so I tested the following: [imath](a_1 \otimes b_1 + a_3 \otimes b_3) \cdot (a_2 \otimes b_2) = ???[/imath] Tensor addition is not defined when one of the elements differs, so I am unsure of where to go beyond this. | 979655 | Show structure of a commutative ring in a tensor product
I need some help with this: Let [imath]R[/imath] be a commutative ring and [imath]S[/imath] and [imath]T[/imath] be commutative [imath]R[/imath]-algebras. Show that [imath] S \otimes T [/imath] has the structure of a commutative ring with multiplication: [imath] (s \otimes t)(s' \otimes t'):=(ss' \otimes tt') [/imath] How do I show here the ring axioms? |
1204406 | If [imath]X[/imath] is a metric space such that any metric space [imath]Y[/imath] , which is a homeomorphic image of [imath]X[/imath] , is complete , then is [imath]X[/imath] compact?
Let [imath]X[/imath] be a compact metric space , then it is easy to show that every homeomorphic image metric space of [imath]X[/imath] is complete . Is the reverse true ? That is if [imath]X[/imath] is a metric space such that any metric space [imath]Y[/imath] , which is a homeomorphic image of [imath]X[/imath] , is complete , then is [imath]X[/imath] compact ? I know that [imath](X,d)[/imath] is homeomorphic with the bounded metric space [imath](X , \dfrac{d}{1+d})[/imath] and also [imath](X , \min \{1,d\})[/imath] , but I cannot get a totally bounded homeomorphic image . Please help . | 587855 | Metrizable topological space [imath]X[/imath] with every admissible metric complete then [imath]X[/imath] is compact
How to prove: If [imath]X[/imath] is a metrizable topological space and every admissible metric on [imath]X[/imath] is complete then [imath]X[/imath] is compact. I was trying with an idea of contradiction and thereby to construct an incomplete metric on [imath]X[/imath]. But not able to construct. |
1200998 | For what [imath]n[/imath] does [imath][\log_21]+[\log_22]+[\log_23]+\dotsb+[\log_2n] = 1538[/imath]?
I just can't solve this problem in spite of doing a whole book on logs and inequalities Where [imath][\dotsc][/imath] denotes the greatest integer function, what is the value of the natural number [imath]n[/imath] satisfying the equation [imath] [\log_21]+[\log_22]+[\log_23]+\dotsb+[\log_2n] = 1538 [/imath] A. 255 B. 256 C. 254 D. 313 The answer is A. Please give an easy solution as I am a high school student. And I do fully understand the meaning of the question but I can't solve it. Since stack exchange forces me to provide an attempt here it goes- I wrote the equation as [imath]log_2n!=1538[/imath] after this I feel n! should be [imath]2^{1538} [/imath] and now i am confused... I hope you got your attempt for the people who put the question on hold. nice you pointed out chris.. yeah the question is same but the person who posted it has accepted the answer to it not me!....the answer maybe sufficient to understand for him.. but it may not be for me... PS-whats the point of giving a fake attempt and not speaking the truth?!! | 703895 | Find [imath]n[/imath] satisfying the equation [imath][\log_21]+[\log_22]+[\log_23]+\dots[\log_2n]=1538 [/imath]
If [imath][\cdot][/imath] denotes greatest integer function, then what is the value of natural number [imath]n[/imath] satisfying the equation [imath][\log_21]+[\log_22]+[\log_23]+\dots[\log_2n]=1538 ?[/imath] My try: Note that [imath]0+1\times2+2\times2^2+3\times2^3+4\times2^4+5\times2^5+6\times2^6+7\times2^7=1538[/imath] But the answer is [imath]n=\sum_{i=0}^72^i=255.[/imath] How is it derived? |
1205435 | How to start proof of triangular inequality?
[imath]\left| {\left| a \right| - \left| b \right|} \right| \le \left| {a \pm b} \right| \le \left| a \right| + \left| b \right| [/imath] | 1197201 | Prove triangle inequality using the properties of absolute value
So I was given the task of proving the following variant of the triangle inequality using only the properties of the absolute value: [imath]\vert\lvert x\rvert -\lvert y \rvert \rvert \leq \lvert x+y\rvert[/imath] Now my initial instinct was to divide it to four cases based on the possible values of x and y, and show that in each case I get a truth expression, but it seems a bit "brute force" to me and I'm not sure this is what my instructor meant. Since I'm not allowed to use the actual triangle inequality and must adhere to absolute value properties, is there any other solution? (Of course he'll accept this answer if it's correct, but I want to learn what it is I'm missing). |
1205828 | Angles inequality in acute triangle
Let [imath]\alpha[/imath], [imath]\beta[/imath], [imath]\gamma[/imath] be angles of acute triangle. How to prove that [imath](\tan(\frac{\alpha}{2}))^2 + (\tan(\frac{\beta}{2}))^2 + (\tan(\frac{\gamma}{2}))^2 \ge 1[/imath]? Does left side of equation have bigger than 1 bound? | 1203137 | Prove that if [imath]\alpha, \beta, \gamma[/imath] are angles in triangle, then [imath](tan(\frac{\alpha}{2}))^2+(tan(\frac{\beta}{2}))^2+(tan(\frac{\gamma}{2}))^2\geq1[/imath]
We have an acute triangle with angles [imath]\alpha, \beta, \gamma[/imath] and we need to tell whether the following inequality is true for all such triangles: [imath](\tan(\frac{\alpha}{2}))^2+(\tan(\frac{\beta}{2}))^2+(\tan(\frac{\gamma}{2}))^2\geq1[/imath] We are also asked if [imath]1%[/imath] is the greatest number, for which it is true. I'm having huge problems when trying to solve such optimization calculus problems. I think we must somehow find triangle for which we can prove, that the upper sum is minimum, than evaluate it and prove that it is equal to [imath]1[/imath]. As with most optimization problems, it turns out that if [imath]\alpha=\beta=\gamma=\frac{\pi}{3}[/imath], then the sum evaluates to [imath]1[/imath]. But how to prove that this is infimum of this sum? |
1205862 | Game Theory question about a financial pyramid scheme.
I have to solve 6 Game Theory problems and feel almost hopeless. Would appreciate any guidance. A company Zest is actively promoting its services. Everyone who invests in Zest will receive their money back as soon as the next client makes his investment. Besides, 90% of the investment made by the [imath]i+1[/imath] client will be forwarded to the [imath]i[/imath] client. All the residents of the city decide by turn whether they'll invest and if yes, how much. Every resident knows about all the decisions made by other people. If the Company can't fulfill its promises it'll default and it'll be closed and those who didn't receive their money before the default won't ever receive their money back. The company has no own money at all. There are 12 million residents of different wealth in the area. Questions: How many residents will invest their money in subgame perfect Nash Equilibrium? Does the answer depend on the sequence the investors will come? The answer for the 2nd question (I think) is No, because even if all the wealthier investors had already invested, a resident still would invest if the following investor invests more than 0, because all his money will be returned +90% of the next investor's input. How should I solve the 1st question? I probably should start from the end. The 12 000 000th resident won't invest because he's the last one, the 11 999 999th won't invest because he knows that the last resident won't invest but... what do I do next? | 1205934 | Game Theory question about a financial pyramid scheme
Salut, fellow game theorists. I have to solve 6 Game Theory problems and fell almost hopeless. Would appreciate any guidance with this one. A company Zest is actively promoting its services. Everyone who invests in Zest will receive their money back as soon as the next client makes his investment. Besides, 90% of the investment made by the [imath]i+1[/imath] client will be forwarded to the [imath]i[/imath] client. All the residents of the city decide by turn whether they'll invest and if yes, how much. Every resident knows about all the decisions made by other people. If the Company can't fulfill its promises it'll default and it'll be closed and those who didn't receive their money before the default won't ever receive their money back. The company has no own money at all. There are 12 million residents of different wealth in the area. Questions: How many residents will invest their money in subgame perfect Nash Equilibrium? Does the answer depend on the sequence the investors will come? The answer for the 2nd question(I think) is No, because even if all the wealthier investors had already invested, a resident still would invest if the following investor invests more than 0, because all his money will be returned +90% of the next investor's input. How should I solve the 1st question? I probably should start from the end. The 12 000 000th resident won't invest because he's the last one, the 11 999 999th won't invest because he knows that the last resident won't invest but... what do I do next? |
1204775 | Quaternion ^ Quaternion
I was looking at Quaternions at Wikipedia - I was trying to find the value of [imath]i^j[/imath] etc... Wikipedia lists [imath]q^\alpha[/imath] where [imath]\alpha[/imath] is real, but I can't find the value of [imath]i^j[/imath]. Any clues? The answer given here doesn't give an explicit solution. | 703593 | How to raise a number to a quaternion power
Now, I know that it's (relatively) easy to calculate, say, [imath]r^{a+bi}[/imath] (using the fact that, for [imath]z_1, z_2\in \mathbb{C}, {z_1}^{z_2}=e^{z_2\ln(z_1)}[/imath] and [imath]\ln(z_1[/imath]) can just be found using: [imath]\ln(a+bi)=\ln[\sqrt{a^2+b^2}]+i \cdot \arctan(\frac{b}{a})[/imath] ). Anyway, how would I go about calculating, say, [imath]i^j, k^i[/imath] etc., or, more generally, [imath](a_1+b_1i+c_1j+d_1k)^{a_2+b_2i+c_2j+d_2k}[/imath] (I know that exponentiating a complex number (to another non-real complex number) produces a non-unique result, so I assume the same would apply further up the hypercomplex ladder; if that's the case, I'm only concerned with the 'principal' value)? I obviously don't want a general formula or anything like that; just some intuition and a method by which I could calculate such a thing. And, finally (because I really like to push my luck), can this method for quaternions be extended to higher number systems (i.e. [imath]\mathbb{O, S},[/imath] etc.) to give an analogous result? Thanks |
1206192 | How to compute linear recurrence of a sum of binomial-multiplied linear recurrences
I have [imath]g(n) = \sum_{k=1}^{n} \binom{n}{k}f(k)[/imath] where [imath]f(k)[/imath] is a large linear recurrence. [imath]g(n)[/imath] is also a linear recurrence as well. Normally, when computing the value of a linear recurrence, I use matrix exponentiation. But to compute [imath]g(n)[/imath] this means computing [imath]f(k)[/imath] for every single [imath]k[/imath]. Is there a way to take the matrix associated with [imath]f(k)[/imath] and somehow use it to compute [imath]g(n)[/imath]? | 1205998 | Simplying linear recurrence sum with binomials
Is there a way to simplify [imath]\sum_{k=1}^{n} \binom{n}{k}f(k)[/imath] Where [imath]f(k)[/imath] is a large linear recurrence? |
1206423 | Showing that [imath]x^n+x+3[/imath] is irreducible
How do you show that [imath]x^n+x+3[/imath] is irreducible for all [imath]n \ge 2[/imath]? | 1198973 | Show that [imath]x^n + x + 3[/imath] is irreducible for all [imath]n \geq 2.[/imath]
So first of all, I used the following from my lecture notes: If [imath]f \in \mathbb{Z}[x][/imath] is primitive (gcd of all the coefficients is 1) - [imath]f[/imath] irreducible in [imath]\mathbb{Z}[x] \Leftrightarrow[/imath] f irreducible in [imath]\mathbb{Q}[x][/imath]. So now I only have to show that [imath]f_n = x^n + x + 3[/imath] is irreducible in [imath]\mathbb{Z}[x][/imath] for all [imath]n \geq 2. [/imath] Is there a pattern I should be appealing to in some way? |
1206496 | Stuck on this integration [imath]\int_0 ^\infty \frac{1}{1+x^2} \cos(kx) dx =\frac{\pi}{2}e^{-k}[/imath]
I'm not sure how to show this [imath]\int_0 ^\infty \frac{1}{1+x^2} \cos(kx) \ \mathrm dx =\frac{\pi}{2}e^{-k}[/imath] I tried by parts but I'm not getting anywhere, I'd really appreciate the help | 658397 | Compute [imath]\int_0^\infty\frac{\cos(xt)}{1+t^2}dt[/imath]
Let [imath]x \in \mathbb R[/imath] Find a closed form of [imath]\int_0^\infty\dfrac{\cos(xt)}{1+t^2}dt[/imath] . Let me give some context: this is an exercise from an improper integrals course for undergraduates. My teacher was not able to give a valid solution without resorting to Fourier series. He actually wanted to prove first that the function is a solution of the ODE [imath] y-y'' =0[/imath] without succeeding in proving it. Anyway the result sould be [imath]\dfrac{\pi}{2}e^{-x}[/imath] Thanks for any help on this subject. |
1203839 | Assume [imath]\arctan[/imath] is differentiable. Prove that [imath]\arctan(x) =1-x[/imath] has a solution in [imath](0,1)[/imath].
Assume [imath]\arctan[/imath] is differentiable. Prove that [imath]\arctan(x) =1-x[/imath] has a solution in [imath](0,1)[/imath]. The hint given is to use the intermediate value theorem (IVT). I do not understand how to show there is a solution using the intermediate value theorem. Could I get help to better understand how to apply IVT in this case? | 1204240 | Solutions of [imath]\arctan x = 1 - x[/imath]. Proofs?
In this question, we examine the equation [imath]\arctan x = 1 - x[/imath]. You may assume without proof that [imath]\arctan x[/imath] is continuous on [imath]R[/imath]. a) Prove that there is a solution to the equation in the interval [imath](0,1)[/imath]. [Hint: IVT] b) Prove that there can't be a second solution in R. [Hint: Use a proof by contradiction using Rolle's Theorem] |
1206871 | Can the sum of reciprocals of a set without density converge?
It is known that if a set of natural numbers has positive asymptotic density then the sum of the reciprocals of those elements diverge. Let [imath]\{a_n\}[/imath] be an increasing sequence of natural numbers where the density of the set [imath]\{a_n | n\in \mathbb N\}[/imath] does not exist. Can the series [imath]\sum_{i=0}^n \frac{1}{a_i}[/imath] converge? All the examples that I have of sets where the density does not exist do not converge. | 5932 | If [imath]A\subseteq\mathbb N[/imath] and [imath]\sum\limits_{a\in A}\frac1a[/imath] converges then [imath]A[/imath] has natural density [imath]0[/imath]
In this answer to a question about a series, a theorem was stated: If [imath]A= \{a_i \}[/imath] is a set such that [imath]\sum_{i = 1}^{\infty} \frac{1}{a_i}[/imath] converges, then [imath]d(A) = 0[/imath], where [imath]d(A)[/imath] is the natural density of the set. My background in number theory is basically zero and all my attempts to prove this have been utterly unseccessful; would anyone please help me, or at least provide me with a hint to prove it? |
1207056 | Discontinuous function proof using [imath]\epsilon - \delta[/imath]
[imath]f:\Bbb R \to \Bbb R[/imath] Showing [imath]f(x)=\left\{\begin{array}{cc} 3x,&x\in\Bbb Q\\-3x,&x\in \Bbb I\end{array}\right.[/imath] I want to show that [imath]f(x)[/imath] is discontinuous for all [imath]x\ne0[/imath] using [imath]\epsilon - \delta[/imath] So I want to show: [imath]\exists \epsilon \gt 0 , \forall \delta \gt 0| |x-l|\lt \delta, |f(x)-f(l)|\gt \epsilon[/imath] Now I believe I should pin this to four cases: 1)[imath]x,l\in \Bbb Q[/imath],[imath]\quad[/imath]2)[imath]x\in \Bbb Q,l\in\Bbb I[/imath],[imath]\quad[/imath]3)[imath]l\in \Bbb Q,x\in \Bbb I[/imath],[imath]\quad[/imath]4)[imath]x,l\in \Bbb I[/imath] So case [imath]1)[/imath]: [imath]|x-l|\lt \delta, |2x-2l|\gt \epsilon[/imath] [imath]|x-l|\lt\delta,2|x-l|\gt\epsilon[/imath] [imath]\implies 3\delta \gt |xl| \gt \epsilon[/imath] Now I have no idea what to do for case 1 and I haven't attempted other cases yet since I assume a similar problem will occur. I put my delta after my epsilon, so I can't choose an epsilon relying on my delta, hence I am confused. | 123712 | [imath]\epsilon{\rm -}\delta[/imath] proof of the discontinuity of Dirichlet's function
How can I prove that the function [imath] f(x) = \left\{\begin{array}{l l} x &\text{if }x \in \mathbb{Q} \\ -x & \text{if } x \notin \mathbb{Q} \end{array} \right. [/imath] is discontinuous for [imath]x \neq 0[/imath], using [imath]\epsilon[/imath]'s and [imath]\delta[/imath]'s? I see that is truth. But I cannot prove using only [imath]\epsilon[/imath]'s and [imath]\delta[/imath]'s. |
1207358 | Theory automata. Proof.
Let [imath]A/B= \{ w : wx \in A \}[/imath] for some [imath]x \in B [/imath]. Show that if [imath]A[/imath] is regular and [imath]B[/imath] is any language, then [imath]A/B[/imath] is regular. Please hint me with doing it using Myhill-Nerod's theorem. I observed that: [imath]\bigcup A/x = A/B [/imath] and that: [imath]A/x \subset K [/imath] where [imath]K[/imath] is equivalence class on [imath]A[/imath]. The class I mean is described here: ( from Wikipedia) Given a language [imath]L[/imath], and a pair of strings [imath]x[/imath] and [imath]y[/imath], define a distinguishing extension to be a string z such that exactly one of the two strings [imath]xz[/imath] and [imath]yz[/imath] belongs to [imath]L[/imath]. Define a relation [imath]R_L[/imath] on strings by the rule that [imath]x R_L y[/imath] if there is no distinguishing extension for [imath]x[/imath] and [imath]y[/imath]. It is easy to show that [imath]R_L[/imath] is an equivalence relation on strings, and thus it divides the set of all strings into equivalence classes. | 1184235 | Prove that [imath]A/B[/imath] is regular when [imath]A[/imath] is regular and [imath]B[/imath] is regular or not regular.
Prove that [imath]A/B[/imath] is regular when [imath]A[/imath] is regular and [imath]B[/imath] is regular or not regular. [imath]A/B=\{w: wx\in A\ \text { for some }\ x\in B\}[/imath] Please give me a clue. |
1208549 | every symmetric matrix has at least one real eigenvalue
let [imath]A \in {M_{n \times m}}[/imath] why every real symmetric matrix has at least one real eigenvalue? . | 1124875 | Eigenvalues of symmetric matrices are real without (!) complex numbers
Is there any proof of the fact that the eigenvalues of symmetric matrices (i.e. [imath]A\in\mathbb{R}^{n\times n}[/imath] with [imath]A^t=A[/imath]) are real without the use of the concept of complex numbers? |
1208675 | Existence of a nonsingular submatrix
Why is the following true? Given a matrix [imath]A \in \mathbb R^{n\times m}[/imath] with rank [imath]r \le \min(n,m)[/imath], there are row and column sets [imath] \{ i_1, \ldots, i_r \}, \quad \{ j_1, \ldots, j_r \}, [/imath] such that the [imath]r\times r[/imath] submatrix [imath] (\hat A_{s,t})_{s,t=1}^r = A_{i_s,j_t} [/imath] is nonsingular. Of course, with the column rank of [imath]A[/imath] being [imath]r[/imath], we can find [imath]r[/imath] linearly independent columns of [imath]A[/imath]. But how can we know that they stay linearly independent after restricting them to the [imath]r[/imath] similarly chosen rows of [imath]A[/imath]? | 368408 | An [imath]r\times r[/imath] submatrix of independent rows and independent columns is invertible (Michael Artin's book).
Let [imath]A[/imath] be an [imath]m \times n[/imath] matrix of rank [imath]r[/imath], let [imath]I[/imath] be a set of row indices such that the corresponding rows of [imath]A[/imath] are independent and let [imath]J[/imath] be a set of [imath]r[/imath] column indices such that the corresponding columns of [imath]A[/imath] are independent. Let [imath]M[/imath] denote the [imath]r \times r[/imath] submatrix obtained by taking rows from [imath]I[/imath] and columns from [imath]J[/imath]. Then [imath]M[/imath] is invertible. So far I've got: Let [imath]B[/imath] be the [imath]m \times r[/imath] matrix obtain from taking the rows from [imath]J[/imath]. [imath]B[/imath] is row-reducible to having a pivot in each column or else [imath]Ax = 0[/imath] has more than one solution, a contradiction. Also, the dimension of the row space of [imath]B[/imath] equals [imath]r[/imath] as well. |
804556 | How to prove Ass[imath](R/Q)=\{P\}[/imath] if and only if [imath]Q[/imath] is [imath]P[/imath]-primary when [imath]R[/imath] is Noetherian?
Let [imath]R[/imath] be a Noetherian ring, [imath]P[/imath] be a prime ideal, and [imath]Q[/imath] an ideal of [imath]R[/imath]. How to prove that [imath] \text{Ass}(R/Q)=\{P\} [/imath] if and only if [imath]Q[/imath] is [imath]P[/imath]-primary? Update In fact, I have proved that if [imath]Q[/imath] is primary, then Ann[imath](R/Q)[/imath] is primary. Let [imath]P=\text{rad}(\text{Ann}(R/Q))[/imath], we have [imath]Q[/imath] is [imath]P[/imath]-primary and [imath]\text{Ass}'(R/Q))=\{P\}[/imath] (where [imath]\text{Ass}'[/imath] consists of all primes occur as the radical of some [imath]\text{Ann}(x)[/imath] ) | 723419 | Primary ideals in Noetherian rings
For an [imath]R[/imath]-module [imath]M[/imath] I have the following definition for a submodule [imath]N\subset M[/imath] to be [imath]\mathfrak{p}[/imath]-primary: this is the case when [imath]\text{Ass}(M/N) = \{\mathfrak{p}\}[/imath], that is, [imath]M/N[/imath] is coprimary or [imath](0)[/imath] is a primary submodule. Now for an ideal [imath]\mathfrak{q} \subset R[/imath] this definition gives that [imath]\mathfrak{q}[/imath] is [imath]\mathfrak{p}[/imath]-primary iff [imath]\text{Ass}(R/\mathfrak{q}) = \{\mathfrak{p}\}[/imath]. It can be shown that this is equivalent with [imath]\mathfrak{p}^n \subset \mathfrak{q}[/imath] for some [imath]n\in \mathbb{N}[/imath] and the condition \begin{align*} \forall r,s \in R: rs \in \mathfrak{q}, r\not\in \mathfrak{p}\Rightarrow s \in \mathfrak{q}. \end{align*} Now I suspect that if [imath]R[/imath] is Noetherian, and if [imath]\mathfrak{q}[/imath] is [imath]\mathfrak{p}[/imath]-primary, then we have [imath]\sqrt{\mathfrak{q}} = \mathfrak{p}[/imath]. I have shown this for PID's, but does this generaly hold? And if so, is every ideal [imath]\mathfrak{q}\subset R[/imath] then [imath]\sqrt{\mathfrak{q}}[/imath]-primary? Many thanks. |
1207482 | Why is the number of conjugacy classes modulo 16 equal to the order for a finite group of odd order?
Let [imath]G[/imath] be a group of odd order [imath]n[/imath] and suppose [imath]|Con(G)| = k[/imath] ( Con(G) is the set of conjugacy classes of G), prove that [imath]k \equiv n \pmod{ 16}.[/imath] How do I proceed on this? Thanks. | 38694 | Surprising but simple group theory result on conjugacy classes
I have read that for any group [imath]G[/imath] of order [imath]2m+1[/imath] (odd) with [imath]n[/imath] conjugacy classes, it is always the case that [imath]16[/imath] divides the value [imath](2m+1)-n = |G|-n[/imath]. This seems to me like an astonishing result: what on earth would [imath]16[/imath] have to do with every single odd group and its conjugacy classes? At any rate, I am wondering, assuming it is true, how would you go about proving it? I would like to show it in a reasonably simple way, but nothing whatsoever comes to mind, could anyone help? Many thanks - M. |
1209553 | Construction Of A Cantor-like set with certain property
Is there a way to construct a measurable set [imath]E \subset [0,1][/imath] with the property that for every interval [imath][a,b] \subset [0,1][/imath], both [imath][a,b]\cap E [/imath] and [imath][a,b][/imath]\ [imath]E[/imath] have positive Lebesgue measure? I think you have to construct a cantor set of positive measure and add in each of the intervals that are ommited another cantor like set with positive measure, and continue this procedure indefinitely but I cannot rigorously prove that its correct. Any ideas? Thanks in advance. | 961745 | Construct a Borel set on R such that it intersect every open interval with non-zero non-"full" measure
This is from problem [imath]8[/imath], Chapter II of Rudin's Real and Complex Analysis. The problem asks for a Borel set [imath]M[/imath] on [imath]R[/imath], such that for any interval [imath]I[/imath], [imath]M \cap I[/imath] has measure greater than [imath]0[/imath] and less than [imath]m(I)[/imath]. I was thinking of taking the Cantor approach: taking [imath]R[/imath] to be the union of [imath][a,b][/imath] with [imath]a[/imath] and [imath]b[/imath] rationals, and for each [imath][a,b][/imath] we construct Cantor sets inside it. During theconstruction of each Cantor set, in order to have positive measure on it, we need to take off smaller and smaller intervals from it, namely the proportion goes to [imath]0[/imath]. As a result, these Cantor sets are extremely "dense" on their ends. If for an interval [imath]I[/imath] it intersects with the Cantor set on [imath][a,b][/imath] while [imath]b-a>>m(I)[/imath], we shall expect the measure of intersection to be rather close to [imath]m(I)[/imath] and then we lose control on these cases. Is there any way to fix this or shall I consider other approaches? Thank you |
1210385 | Proof of [imath]\lim_{n\to\infty}\left(1+\frac1n\right)^n=e[/imath]
I know that this sequence converges because it is increasing and bounded (well, this is the usual way to prove it). In some books, the number [imath]e[/imath] is defined to be this limit. But in other books the number [imath]e[/imath] is defined this way: [imath]e=\sum_{n=0}^\infty\frac1{n!}[/imath] and the exponential function [imath]e^x=\sum_{n=0}^\infty\frac x{n!}[/imath] But my problem is to prove that both definitions are consistent. Assumig the second definition of [imath]e[/imath] and [imath]e^x[/imath], I have done this: I define [imath]f_n(x)=\left(1+\frac xn\right)^n[/imath]. Then I take any compact interval [imath][a,b][/imath] and I show that the numeric sequence [imath]\left\{\left(1+\frac xn\right)^n\right\}_{n=1}^\infty[/imath] is monotonic and bounded for every [imath]x\in[a,b][/imath]. So the sequence of functions [imath]f_n[/imath] converges -pointwisely at least- in that interval. Let be [imath]f[/imath] the limit. Now [imath]f'_n(x)=\frac1nn\left(1+\frac xn\right)^{n-1}[/imath] and we see that [imath]\lim f'_n=f[/imath]; therefore, [imath]f'=f[/imath]. Since [imath]f(0)=1[/imath], we deduce that [imath]f(x)=e^x[/imath]. This is not an exercise or anything like that, but I'd like to ensure if vevery step in my proof is correct. I am specially dubious with the derivative of the sequence. Do I need uniform convergence to do that? How could I show it in this case? | 860796 | How to show that $ \sum_{n = 0}^{\infty} \frac {1}{n!} = e$?
How to show that [imath]$\sum\limits_{n = 0}^{\infty} \frac {1}{n!} = e$[/imath] where [imath]$e = \lim\limits_{n\to\infty} \left({1 + \frac 1 n}\right)^n$[/imath]? I'm guessing this can be done using the Squeeze Theorem by applying the AM-GM inequality. But I can only get the lower bound. If [imath]S_n[/imath] is the [imath]n[/imath]th partial sum of our series then, [imath]$$ \left({1 + \frac 1 n}\right)^n = 1 + n\cdot\frac{1}{n} + \frac{n(n - 1)}{2}\cdot \frac{1}{n^2} + \cdots + \frac{n(n - 1)\ldots(n - (n -1))}{n!} \cdot \frac{1}{n^n} $$[/imath] [imath]$$ \le 1 + \frac{1}{2!} + \cdots + \frac{1}{n!} =S_n $$[/imath] How can I show that [imath]S_n \le [/imath] a subsequence of [imath]$\left({1 + \frac 1 n}\right)^n$[/imath] or any sequence that converges to [imath]e[/imath]? |
414350 | Showing a compact metric space has a countable dense subset
I know that this is separable, but that has not been covered by the text at this point. I believe my proof (below) to be correct, but not very rigorous. Assuming that it actually is correct, could someone please help me make it more formal? (If it's not correct, please help with the proof). Thanks. Proof: Let [imath]X[/imath] be a compact metric space. Let [imath]B_n = \{B(x,\frac1{n}) : x\in X\}[/imath]. be the collection of open balls centered at [imath]x[/imath]. Let [imath]\{x\}^{(n)}[/imath] be the set of centers for some particular [imath]n[/imath]. Then by compactness of [imath]X[/imath], there are also a finite number of centers for a particular [imath]n[/imath] (corresponding to each open ball). [This is where I have trouble properly saying what's on my mind:] Consider [imath]B(x,r_0)[/imath], where [imath]r_0 = \frac{1}{n_0}[/imath]. When we decrease the magnitude of [imath]r_0[/imath] to [imath]r_1[/imath], we must still be able to cover [imath]X[/imath]. Meaning some open ball(s) must be "formed" where [imath]B(x,r_0)[/imath] was once "covering". So by adding open balls, we also add to our collection of centers. This means at [imath]B(x,r_0)[/imath], there was originally a point [imath]y \in B(x,r_0)[/imath] s.t. [imath]B(y,r') \subset B(x,r_0)[/imath] for some [imath]r'[/imath]. That is, [imath]B(x,r_0)[/imath] contained another point of [imath]X[/imath]. Thus, by definition, [imath]X[/imath] contains a countable dense subset. | 1935110 | Compactness & countable dense subset
I have a question in real analysis! If [imath]K[/imath] is compact metric space, show that there is a countable dense subset [imath]E[/imath] of [imath]K[/imath] How can i prove it? Could you help me for the idea or proof? |
55165 | Eigenvalues of the rank one matrix [imath]uv^T[/imath]
Suppose [imath]A=uv^T[/imath] where [imath]u[/imath] and [imath]v[/imath] are non-zero column vectors in [imath]{\mathbb R}^n[/imath], [imath]n\geq 3[/imath]. [imath]\lambda=0[/imath] is an eigenvalue of [imath]A[/imath] since [imath]A[/imath] is not of full rank. [imath]\lambda=v^Tu[/imath] is also an eigenvalue of [imath]A[/imath] since [imath]Au = (uv^T)u=u(v^Tu)=(v^Tu)u.[/imath] Here is my question: Are there any other eigenvalues of [imath]A[/imath]? Added: Thanks to Didier's comment and anon's answer, [imath]A[/imath] can not have other eigenvalues than [imath]0[/imath] and [imath]v^Tu[/imath]. I would like to update the question: Can [imath]A[/imath] be diagonalizable? | 806762 | Problem about real square matrix with rank 1
Given [imath]A \in \mathbb{R}^{n \times n}[/imath] and [imath]\text{rank}(A) = 1[/imath]. By working only on real field, show that [imath]A[/imath] is diagonalizable if and only if [imath]\text{tr}(A) \neq 0[/imath]. Here, [imath]\text{tr}(A)[/imath] is the sum of all eigenvalues of [imath]A[/imath]. |
1211662 | Why are the elements of GF(q), whose characteristic is 2, all squares?
If [imath]a\in GF(2^n)[/imath], then there is the element in [imath]GF(2^n)[/imath] such that [imath]x^2 = a[/imath]. Why? | 742926 | [imath]F[/imath] is a finite field of size [imath]s[/imath]. Prove that if [imath]s=2^m[/imath] for some [imath]m>0[/imath], then all elements of F have a square root.
[imath]F[/imath] is a finite field of size [imath]s[/imath]. Prove that if [imath]s=2^m[/imath] for some [imath]m>0[/imath], then all elements of [imath]F[/imath] have a square root. (Hint: For some integer [imath]i[/imath], ([imath]1\leq i \leq s-1[/imath]), [imath]i[/imath] is odd if and only if [imath](s-1)+i[/imath] is even.) My attempt: If [imath]s=2^m[/imath], then [imath]s[/imath] is even. This means [imath]s-1[/imath] is odd. So the multiplicative group has an odd number of elements. Let [imath]x[/imath] be the primitive element of [imath]F[/imath] such that: [imath]F/\{0\}=\{x, x^2, x^3, ... , x^{s-1}\}[/imath]. I tried to arrive at a contradiction assuming there existed an element without a square root, but I couldn't think of anything. Also I'm not quite sure how to use the hint. |
1211626 | Find all [imath]n\in\mathbb{Z}^+[/imath] such that the sum of the digits of [imath]5^n[/imath] equals [imath]2^n[/imath]
Find all [imath]n\in\mathbb{Z}^+[/imath] such that the sum of the digits of [imath]5^n[/imath] equals [imath]2^n[/imath] Starting with a table of values, I found that [imath]n=3[/imath] works. Beyond this, it's hard to imagine any other number working. As [imath]n[/imath] gets larger, it seems to me that [imath]2^n[/imath] grows far more rapidly than the sum of the digits of [imath]5^n[/imath] would ever be able to "catch up" to. My reasoning: multiplying by each [imath]2[/imath] doubles [imath]2^n[/imath], but multiplying by each [imath]5[/imath] adds (I think) at most [imath]1[/imath] digit to [imath]5^n[/imath], which would add at most [imath]9[/imath] to the sum of the digits of [imath]5^n[/imath]. Another way of looking at it, suppose [imath]n=1000000[/imath]. Notice that [imath]5^{1000000}<10^{1000000}[/imath], which is [imath]1[/imath] with [imath]1000000[/imath] zeros. Thus [imath]5^{1000000}[/imath] has at most [imath]1000000[/imath] digits, and the sum of its digits would be maximized if those digits were all [imath]9[/imath]s. Then the sum of the digits of [imath]5^n[/imath] would be [imath]9\cdot1000000=9000000[/imath], which is far less than the number [imath]2^{1000000}[/imath]. But is there a way to prove this using theory? | 685386 | Find all [imath]n\in\mathbb N^+[/imath] such that the sum of the digits of [imath]5^n[/imath] is equal to [imath]2^n[/imath].
Find all [imath]n\in\mathbb N^+[/imath] such that the sum of the digits of [imath]5^n[/imath] is equal to [imath]2^n[/imath]. I've solved this, but I think the proof is a bit weird and I wonder if there's a better one. My proof: We can see that [imath]n\neq 1[/imath], [imath]n\neq 2[/imath], [imath]n=3[/imath], [imath]n\neq 4[/imath] and [imath]n\neq 5[/imath]. Let [imath]n>5[/imath]. Notice that [imath]5^6[/imath] has [imath]5[/imath] digits, while [imath]\frac{2^6}{9}=7\frac{1}{9}[/imath], so [imath]5^6[/imath] has to have at least [imath]8[/imath] digits in order for the sum of the digits to be able to at least reach [imath]2^6[/imath]. Everytime [imath]1[/imath] is added to the exponent of [imath]5[/imath], the amount of digits [imath]5^n[/imath] gets is at most [imath]1[/imath], while [imath]\frac{2^{n+1}}{9}-\frac{2^n}{9}=\frac{2^n}{9}>3[/imath], so the lowest amount of digits it has to have to reach [imath]2^n[/imath] becomes higher by at least [imath]4[/imath]. When [imath]n=6[/imath], the amount of digits [imath]5^n[/imath] has is already lower than the amount it has to have in order to reach [imath]2^n[/imath], and I've just proved that the amount it has to have in order to reach [imath]2^n[/imath] grows faster than the amount of digits [imath]5^n[/imath] has, therefore [imath]5^n[/imath] will never be able to reach [imath]2^n[/imath] when [imath]n>5[/imath]. Hence the only solution is [imath]n=3[/imath]. |
1211317 | Does [imath]^nC_{n+1}[/imath] exist?
is there any value assigned to [imath]^nC_{n+1}[/imath]? My teacher wrote it equal to [imath]0[/imath], but what will negative factorial mean? [imath]^nC_{n+1} = \frac{n!}{(n+1)!\cdot(n-n-1)!} = \frac{n!}{(n+1)!\cdot(-1)!}[/imath] Thanks | 709343 | How are lopsided binomials (eg [imath]\binom{n}{n+1})?[/imath] defined?
For instance is [imath]\binom{n}{n+1}=0[/imath] always or something else? |
182720 | Prove that [imath] \frac{n}{\phi(n)} = \sum\limits_{d \mid n} \frac{\mu^2(d)}{\phi(d)} [/imath]
I am trying to show that: \begin{equation} \frac{n}{\phi(n)} = \sum_{d \mid n} \frac{\mu^2(d)}{\phi(d)} \end{equation} where [imath]\phi(n)[/imath] is Euler's totient function and [imath]\mu[/imath] is the Möbius function. The identity clearly holds for [imath]n=1[/imath], so if we write [imath]n=p_1^{a_1} \cdots p_k^{a_k}[/imath] for the prime decomposition of [imath]n[/imath], the left-hand side becomes: \begin{eqnarray*} \frac{n}{\phi(n)} & = & \frac{p_1^{a_1} \cdots p_k^{a_k}}{\phi(p_1^{a_1}) \cdots \phi(p_k^{a_k})} \\ & = & \frac{p_1^{a_1} \cdots p_k^{a_k}}{(p_1^{a_1-1})(p_1-1) \cdots (p_k^{a_k})(p_k-1)} \\ & = & \frac{p_1 \cdots p_k}{(p_1-1)\cdots (p_k-1)} \end{eqnarray*} Whereas the right-hand side becomes: \begin{eqnarray*} \sum_{d \mid n} \frac{\mu^2(d)}{\phi(d)} & = & \sum_{\substack{d \mid n \\ p^2 \nmid d}} \frac{1}{\phi(d)} \\ & = & 1 + \frac{1}{\phi(p_1)} + \dots + \frac{1}{\phi(p_k)} + \frac{1}{\phi(p_1 p_2)} + \dots + \frac{1}{\phi(p_{k-1}p_k)} \\ & + & \dots + \frac{1}{\phi(p_1 \cdots p_k)} \\ & = & 1 + \frac{1}{p_1 - 1} + \dots + \frac{1}{p_k - 1} + \frac{1}{(p_1 - 1)(p_2 - 1)} + \dots + \frac{1}{(p_{k-1}-1)(p_k-1)} \\ & + & \dots + \frac{1}{(p_1 - 1) \cdots (p_k - 1)} \end{eqnarray*} I am unsure of how to proceed from this point however. Any help would be much appreciated. | 497732 | For a positive integer [imath]n[/imath], prove that [imath]\sum_{d|n}{\mu^2(d)/\phi(d)=n/\phi(n)}[/imath]
For a positive integer [imath]n[/imath], prove that [imath]\sum_{d|n}{\mu^2(d)/\phi(d)=n/\phi(n)}[/imath] I suspect something is missing here. I guess there is on extra condition on [imath]n[/imath], which is [imath]n[/imath] must ne square free. Because when I substitute [imath]n=20[/imath], the equality does not hold. Can anyone clarify for me ? |
1212450 | Combinational proof problem
I'm having trouble finding a combinatorial argument for [imath]\sum_{k=m}^n {k \choose m} = {n+1 \choose m+1}[/imath] The right side is just choosing m+1 things from a set of n+1 things, but I can't see any way to relate this to the left side, where you're choosing m from m things, m from m+1 things, m from m+2 things and so on... | 321022 | Give the combinatorial proof of the identity [imath]\sum\limits_{i=0}^{n} \binom{k-1+i}{k-1} = \binom{n+k}{k}[/imath]
Given the identity [imath]\sum_{i=0}^{n} \binom{k-1+i}{k-1} = \binom{n+k}{k}[/imath] Need to give a combinatorial proof a) in terms of subsets b) by interpreting the parts in terms of compositions of integers I should not use induction or any other ways... Please help. |
1212338 | Why if the antecedent P is false, and the consequence Q true, then the implication P [imath]\Rightarrow[/imath] Q is true?
I know that that's the definition but I wonder why logicians choose that thefinition to be true. It sounds strange to me and I cant make sense of it if someone tell me 'if the sky is red, then I'm Marco Polo'. | 439987 | Assumed True until proven False. The Curious Case of the Vacuous Truth
Given two statements, [imath]P[/imath] and [imath]Q[/imath], and the logical connective, [imath]\implies[/imath], the truth table for [imath]P \implies Q[/imath] is: [imath]\begin{array}{ c | c || c | } P & Q & P\Rightarrow Q \\ \hline \text T & \text T & \text T \\ \text T & \text F & \text F \\ \text F & \text T & \text T \\ \text F & \text F & \text T \end{array}[/imath] Lines one and two are quite clear. What is ambiguous is lines three and four. One explanation as to why [imath]P \implies Q[/imath] is true when [imath]P[/imath] is false, provided by Velleman goes: Let [imath]P(x)[/imath] be the statement [imath]x>2[/imath] and [imath]Q(x)[/imath], [imath]x^2 > 4[/imath]. When [imath]x=3[/imath], [imath]P[/imath] is true, and [imath]Q(x) = 9[/imath] thus [imath]Q[/imath] is true. If [imath]P(x) = 1[/imath] then [imath]Q(x) = 1[/imath] and they are both false. If [imath]P(x) = -3[/imath] then [imath]Q(x) = 9[/imath] and so the statement is true. This explanation was quite unsatisfactory to me. Looking at Enderton, we have: For example, we might translate the English sentence, ”If you're telling the truth then I'm a monkey's uncle,” by the formula ([imath]V \implies M[/imath]). We assign this formula the value [imath]T[/imath] whenever you are fibbing. In assigning the value [imath]T[/imath], we are certainly not assigning any causal connection between your veracity and any simian features of my nephews or nieces. The sentence in question is a conditional statement. It makes an assertion about my relatives provided a certain condition — that you are telling the truth — is met. Whenever that condition fails, the statement is vacuously true. Very roughly, we can think of a conditional formula ([imath]p\implies q[/imath]) as expressing a promise that if a certain condition is met (viz., that [imath]p[/imath] is true), then q is true. If the condition [imath]p[/imath] turns out not to be met, then the promise stands unbroken, regardless of [imath]q[/imath]. Though a significant improvement over the Velleman explanation, I still feel uncomfortable with it. Really, it seems we can conjure up as many silly counter-examples as we like, such as: If pigs can fly, then I can walk on water. Though following the above truth table, the implication is that my ability to walk on water is true. After considering it, it seems to me that [imath]\implies[/imath] is only meaningful when [imath]P[/imath] is true, then we can look at it in relation to [imath]Q[/imath]. However, if [imath]P[/imath] is false, then we actually know nothing about the relationship between [imath]P[/imath] and [imath]Q[/imath]. This would give a truth table of: [imath]\begin{array}{ c | c || c | } P & Q & P\Rightarrow Q \\ \hline \text T & \text T & \text T \\ \text T & \text F & \text F \\ \text F & \text T & \text ? \\ \text F & \text F & \text ? \end{array}[/imath] where the [imath]?[/imath] denotes that given [imath]P[/imath] we actually don't know anything about [imath] \implies [/imath]. Thus. one way to clear this up would be to assume that [imath]? = T[/imath]. Thus the vacuous truth is a "definition of convenience" in a sense. The above is my take on the matter. Could someone provide some clarification on the conditional logical connective? |
1212562 | Evaluate:[imath]\int_0^{\pi/2}\frac{x \sin x \cos x}{(a^2 \cos^2 x+b^2 \sin^2 x)^2}dx[/imath]
How to evaluate the integral of [imath]\int_0^{\pi/2}\frac{x\sin x\cos x}{(a^2 \cos^2 x+b^2 \sin^2 x)^2}dx[/imath] | 1200858 | Integral of [imath]\int \frac{x \sin{x} \cos{x}}{(a^2 \sin^2 x + b^2 \cos^2 x)^2} dx [/imath]
Please help me to find the integral of [imath]\int \frac{x \sin{x} \cos{x}}{(a^2 \sin^2 x + b^2 \cos^2 x)^2} dx [/imath] There is a problem in having x in the numerator. Please guide me . Thanks in advance. |
1212597 | How do I prove function convexity?
I have the following task: Prove that if [imath] f : I \rightarrow \mathbb{R} [/imath] is continuous ([imath] I [/imath] is a range) and [imath] \forall {x,y \in I} \qquad f\left(\frac{x+y}{2}\right) \leq \frac{f(x) + f(y)}{2} [/imath] then [imath]f[/imath] is a convex function. Can somebody give a hint what can I use to prove this? This is a homework assignment, so I'd like to try solve it myself. | 777066 | How to prove that [imath]f[/imath] is convex function if [imath]f(\frac{x+y}2)\leq \frac12f(x) + \frac12f(y)[/imath] and [imath]f[/imath] is continuous?
Let [imath]f:(a,b) \mapsto \mathbb{R}[/imath] be continuous function such that [imath]f(\frac{x+y}{2})\leq \frac{1}{2}f(x) + \frac{1}{2}f(y) \;\; \forall x,y \in (a,b)[/imath] Show that [imath]f[/imath] is convex function. Please give me hints to prove it. |
1207240 | (x,y) coordinates from gluing together a sequence of right triangles with arbitrary angles
I have been scratching my head all day over this question for one of my assignments. I haven't made any progress and I'm at the point of giving up. Here's what I need help with. Start by gluing together a sequence of right angled triangles as shown in the following diagram. The angles [imath]\theta_i[/imath], [imath]i = 1, 2, 3, \dots[/imath] can be freely chosen. This construction produces a sequence of points [imath](x,y)_n[/imath] in the plane. Show that \begin{align} x_{n+1} &= x_n - y_n \tan \theta_{n+1} \tag{1} \\ y_{n+1} &= y_n + x_n \tan \theta_{n+1} \tag{2} \end{align} I have no idea how to even begin to work this out. Any help will be greatly appreciated! | 1207134 | Recurrence relation for right-angled triangles stuck-together
Given the image: and that [imath]x_0 = 1, y_0=0[/imath] and [imath]\text{angles} \space θ_i , i = 1, 2, 3, · · ·[/imath] can be arbitrarily picked. How can I derive a recurrence relationship for [imath]x_{n+1}[/imath] and [imath]x_n[/imath]? I actually know what the relationship is, however, don't know how to derive it. Although it is obvious to see that [imath]x_0 = x_1[/imath] so for [imath]n = 0[/imath] We have [imath]x_{1} = x_{0}+0 [/imath] The recurrence relations are [imath]x_{n+1}=x_n−y_ntan(θ_{n+1})[/imath] and [imath]y_{n+1}=y_n+x_ntan(θ_{n+1})[/imath] But I can't get the derivation. I tried taking some arbitrary right angle triangle and constructing two vectors. [imath]\mathop r_{\sim} = \langle x_n,y_n \rangle [/imath] and a vector perpendicular to [imath]\mathop r_\sim[/imath], [imath]\mathop d_{\sim} = \langle a,b\rangle[/imath] such that [imath]\mathop r_{\sim} \cdot \mathop d_{\sim} = 0 [/imath] Then we can let construct a unit vector for [imath]\mathop d_\sim[/imath] and eventually construct a line through a point? |
1213186 | Is there another way to solve this integral? (A bit long for my method)
[imath]\int \frac{1}{\sqrt{e^x+e^{2x}+1}}dx[/imath] My attempts, Substitute [imath]u=e^x[/imath] and [imath]du=e^xdx[/imath] [imath]=\int \frac{1}{u\sqrt{u^2+u+1}}[/imath] [imath]=\int \frac{1}{u\sqrt{(u+\frac{1}{2})^2+\frac{3}{4}}}[/imath] Substitute [imath]s=u+\frac{1}{2}[/imath] and [imath]ds=du[/imath] [imath]=\int \frac{1}{(s-\frac{1}{2})\sqrt{s^2+\frac{3}{4}}}[/imath] Substitute [imath]s=\frac{1}{2}\sqrt{3}\tan (p)[/imath] and [imath]ds=\frac{1}{2}\sqrt{3}\sec ^2(p)dp[/imath]. Then [imath]\sqrt{s^2+\frac{3}{4}}=\sqrt{\frac{3\tan ^2(p)}{4}+\frac{3}{4}}=\frac{1}{2}\sqrt{3}\sec (p)[/imath] and [imath]p=\tan ^{-1}(\frac{2s}{\sqrt{3}})[/imath] [imath]=\frac{\sqrt{3}}{2}\int \frac{2\sec (p)}{\sqrt{3}(\frac{1}{2}\sqrt{3}\tan (p)-\frac{1}{2})}dp[/imath] [imath]=\int \frac{\sec (p)}{\frac{1}{2}\sqrt{3}\tan (p)-\frac{1}{2}}dp[/imath] Substitute [imath]w=\tan (\frac{p}{2})[/imath] and [imath]dw=\frac{1}{2}dp\sec ^2(\frac{p}{2})[/imath] [imath]=\int \frac{2}{(1-w^2)(-\frac{\sqrt{3}w}{w^2-1}-\frac{1}{2})}dw[/imath] [imath]=\int \frac{4}{w^2+2\sqrt{3}w-1}dw[/imath] [imath]=4\int \frac{1}{(w+\sqrt{3})^2-1}dw[/imath] Substitute [imath]v=w+\sqrt{3} and [/imath]dv=dw$ [imath]=4\int \frac{1}{v^2-4}dv[/imath] [imath]=4\int -\frac{1}{4(1-\frac{v^2}{4}})dv[/imath] [imath]=-\int \frac{1}{1-\frac{v^2}{4}}dv[/imath] Substitute $z_1=\frac{v}{2}[imath] and [/imath]dz_1=\frac{1}{2}dv$ [imath]=-2\int \frac{1}{1-z_1^2}dz_1[/imath] [imath]=-2\tanh^{-1}(z_1)+c[/imath] [imath]=-2\tanh^{-1}(\frac{v}{2})+c[/imath] [imath]=-2\tanh^{-1}(\frac{1}{2}(w+\sqrt{3}))+c[/imath] [imath]-2\tanh^{-1}(\frac{1}{2}(\tan (\frac{p}{2})+\sqrt{3}))+c[/imath] [imath]-2\tanh^{-1}(\frac{3(\sqrt{4s^2+3}+\sqrt{3}+2\sqrt{3}s}{2\sqrt{12s^2+9}+6})+c[/imath] [imath]=-2\tanh^{-1}(\frac{3\sqrt{u^2+u+1}+\sqrt{3}u+2\sqrt{3}}{2\sqrt{3}\sqrt{u^2+u+1}+3})+c[/imath] [imath]=-2\tanh^{-1}(\frac{3\sqrt{e^x+e^{2x}+1+\sqrt{3}e^x+2\sqrt{3}}}{2\sqrt{3}\sqrt{e^x+e^{2x}+1}+3})+c[/imath] How to simplify my answer to $x-\ln(2\sqrt{e^x+e^{2x}+1}+e^x+2)+c$? Is there another way to solve this integral as this method is a bit long. ? | 1212692 | how to integrate [imath]\frac{1}{\sqrt{e^{2x}+e^x+1}}[/imath]
I am having troubles with this integral: [imath]\int \frac{1}{\sqrt{e^{2x}+e^x+1}}dx.[/imath] Could anyone help me? |
1213341 | Prove that [imath]Z(S_n)[/imath] is trivial for [imath]n \geq 3[/imath]
Hi I am trying to solve this question I don't know where to begin but I have an idea so we have [imath]Z(G) = \{x \in G \space |\space xy = yx \space\forall y \in G\}[/imath] We must show that for all [imath]\sigma[/imath] [imath]\in S_n[/imath] such that [imath]\sigma \neq (1)[/imath] Then there exists y [imath]\in S_n[/imath] such that [imath]\sigma * y[/imath] [imath]\neq[/imath] [imath]y * \sigma[/imath]. However I don't know how to proceed further. I was thinking I could use conjugation somehow but I don't know what would I do with them. | 719915 | Find the center of the symmetry group [imath]S_n[/imath].
Find the center of the symmetry group [imath]S_n[/imath]. Attempt: By definition, the center is [imath]Z(S_n) = \{ a \in S_n : ag = ga \forall\ g \in S_n\}[/imath]. Then we know the identity [imath]e[/imath] is in [imath]S_n[/imath] since there is always the trivial permutation. Suppose [imath]a[/imath] is in [imath]S_n[/imath], but not equal to identity. Now we can imagine the permutation as bijective function that maps from [imath]\{1,2,\dotsc,n\}[/imath] to [imath]\{1,2,\dotsc,n\}[/imath]. So suppose [imath]p[/imath] is a permutation map. Then [imath]p[/imath] maps from a location [imath]i[/imath] to a location [imath]j[/imath]. Take [imath]p(i) = j[/imath] where [imath]i\neq j[/imath]. Let [imath]k[/imath] be in [imath]\{1,2,\dotsc,n\}[/imath], where [imath]k[/imath], [imath]i[/imath] and [imath]j[/imath] are all different elements. The cycle [imath]r = (jk)[/imath], then we will see if this commutes. [imath]rp(i) = rj[/imath] Can someone please help me, I am stuck? Thank you. |
1041487 | A question about compact sets: how to prove [imath]g[/imath] must be an isometry
Let [imath](X,p)[/imath] be a compact metric space. Suppose that [imath]g:X\rightarrow X[/imath] is a function such that for all [imath]x_1,x_2\in X[/imath] we have [imath]p(g(x_1),g(x_2))\geq p(x_1,x_2)[/imath]. Prove that, in fact, [imath]g[/imath] must be an isometry ([imath]p(g(x_1),g(x_2))=p(x_1,x_2)[/imath] for all [imath]x_1,x_2\in X[/imath]) and that g is bijective. I am stuck here, and I have tried many times, but still have no idea how to solve the question. Can someone tell me how to solve this question? | 1201072 | Let [imath](M,d)[/imath] be a compact metric space and [imath]f:M \to M[/imath] such that [imath]d(f(x),f(y)) \ge d(x,y) , \forall x,y \in M[/imath] , then [imath]f[/imath] is isometry?
Let [imath](M,d)[/imath] be a compact metric space and [imath]f:M \to M[/imath] such that [imath]d(f(x),f(y)) \ge d(x,y) , \forall x,y \in M[/imath] ; then how to prove that [imath]d(f(x),f(y))=d(x,y) , \forall x,y \in M[/imath] i.e. that [imath]f[/imath] is an isometry ? I wanted to prove by contradiction , that suppose [imath]\exists a,b \in M[/imath] such that [imath]d(f(a),f(b)) > d(a,b)[/imath] ; I wanted to use the sequences [imath]\{f^n(a)\}[/imath] and [imath]\{f^n(b)\}[/imath] , these have convergent subsequences [imath]\{f^{r_n}(a) \}[/imath] and [imath]\{f^{r_n}(b)\}[/imath] ( without loss of generality we have extracted a common subsequence ) converging to say [imath]x,y \in M [/imath] respectively . Then [imath]d (f^{r_n}(a) , f^{r_n}(b))[/imath] converges to [imath]d(x,y) >d(a,b)[/imath] ; but I don't know what to do next , please help . Thanks in advance |
1212923 | A parabola and circle
A circle touches the parabola [imath]y^2=4ax[/imath] at [imath]P[/imath]. It also passes through the focus [imath]S[/imath] of the parabola and intersects the axis of the parabola at [imath]Q[/imath]. If [imath]\angle SPQ[/imath] is [imath]90^\circ[/imath], what is the equation of circle ? Attempt:-assuming the equation of circle as [imath](x-h)^2 + (y-k)^2 = R^2[/imath] and using the condition that it passes through [imath]S\equiv(a\,,0)[/imath] and since [imath]\angle SPQ[/imath] is [imath]90^\circ[/imath], [imath]SQ[/imath] is the diameter. But still not getting the answer... | 962367 | A circle touches the parabola [imath]y^2=4ax[/imath] at P. It also passes through the focus S of the parabola and int......
Problem : A circle touches the parabola [imath]y^2=4ax[/imath] at P. It also passes through the focus S of the parabola and intersects its axis at Q. If angle SPQ is [imath]\frac{\pi}{2}[/imath] find the equation of the circle. Solution : Let the point P ([imath]at^2,2at)[/imath] If the circle touches the parabola [imath]y^2=4ax[/imath] at [imath] P (at^2,2at)[/imath] , they must have a common tangent at that point, and hence a common normal. The centre of the circle must lie on that normal. Let [imath](h,k)[/imath] be the coordinates of the centre and r be the radius of the circle. Then its equation can be written as [imath]x^2+y^2-2hx-2ky+c=0......(i)[/imath] The equation to the normal at [imath](at^2,2at)[/imath] is [imath]y=-tx +2at +at^3 ......(ii)[/imath] As the centre [imath](h,k)[/imath] lies on the normal (ii) , hence [imath]k =-th +2at +at^3[/imath]....(iii) Focus of the parabola is [imath](a,0)[/imath] As the circle passes through (a,0) and [imath](at^2,2at)[/imath] the distance of these points from the centre (h,k) must each be equal to the radius. so [imath]r^2=(h-a)^2+k^2= (h-at^2)^2 +(k-2at)^2[/imath] or [imath]-2ah +a^2=-2aht^2-4akt+a^2t^4+4a^2t^2[/imath] or [imath]4kt =h(1-t^2)+at^4 +4at^2-a[/imath].....(iv) Solving (iii) and (iv) , [imath]2h= a(3t^2+1) .....(v) [/imath] and [imath]2k =a(3t-t^3).....(vi)[/imath] Putting the values of h and k in (i) we get the equation of the circle [imath]x^2+y^2-a(3t^2+1)x -a(3t-t^3)y+c=0[/imath] .....(vii) As this circle passes through the focus (a,0) the coordinates will satisfy it : Therefore, [imath]a^2-a^2(3t^2+1)+c=0[/imath] [imath]\therefore c =3a^2t^2[/imath] Putting this value of c in (vii) the circle is [imath]x^2+y^2-ax(3t^2+1)-ay(3t-t^3)+3a^2t^2=0[/imath] But I am unable to get the equation of the circle which is [imath]x^2+y^2-10ax +9a^2=0[/imath] Please help . Thanks. |
1213363 | How do we prove such integral?
[imath]\int_0^\infty\frac 1{1+x^n}dx=\frac\pi n\csc\left(\frac\pi n\right)[/imath] If we want to prove the left side equal to the right side in this case, how do we start? How do we prove this definite integral? | 432371 | Prove [imath]\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^n}=\frac{\pi}{n \sin\frac{\pi}{n}}[/imath] using real analysis techniques only
I have found a proof using complex analysis techniques (contour integral, residue theorem, etc.) that shows [imath]\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^n}=\frac{\pi}{n \sin\frac{\pi}{n}}[/imath] for [imath]n\in \mathbb{N}^+\setminus\{1\}[/imath] I wonder if it is possible by using only real analysis to demonstrate this "innocent" result? Edit A more general result showing that [imath]\int\limits_{0}^{\infty} \frac{x^{a-1}}{1+x^{b}} \ \text{dx} = \frac{\pi}{b \sin(\pi{a}/b)}, \qquad 0 < a <b[/imath] can be found in another math.SE post |
1213800 | Prove if, [imath]2^n - 1[/imath] is prime, then [imath]n[/imath] is prime.
Prove, when [imath]n[/imath] is a positive integer, if [imath]2^n - 1[/imath] is prime, then [imath]n[/imath] is prime. I did read some sort of proving on the web, but I could not understand it... Any help? And if possible, could the explanations be very clear? Thank you guys :) So, I've found this from somewhere: Theorem. If 2n−1 is prime then n is prime. Proof. Suppose that 2n−1 is prime, and write n=st where s,t are positive integers. Since xs−1=(x−1)(xs−1+xs−2+⋯+1) , we can substitute x=2t to see that 2t−1 is a factor of 2n−1. Since 2n−1 is prime there are only two possibilities, 2t−1=1or2t−1=2n−1 . Therefore t=1 or t=n. We have shown that the only possible factorisations of n are n×1 and 1×n. Hence, n is prime. However, I have a few questions: 1) how do you know xs−1=(x−1)(xs−1+xs−2+⋯+1) 2) why is '2t−1' a factor of '2n−1' 3) how do you determine '2t−1=1or2t−1=2n−1' from 'Since 2n−1 is prime' 4) 'the only possible factorisations of n are n×1 and 1×n. ' how? 5) how did 'Hence, n is prime' come out as result? | 929707 | Are Mersenne prime exponents always odd?
I have been researching Mersenne primes so I can write a program that finds them. A Mersenne prime looks like [imath]2^n-1[/imath]. When calculating them, I have noticed that the [imath]n[/imath] value always appears to be odd. Is there confirmation or proof of this being true? |
709 | Derivation of the formula for the vertex of a parabola
I'm taking a course on Basic Conic Sections, and one of the ones we are discussing is of a parabola of the form [imath]y = a x^2 + b x + c[/imath] My teacher gave me the formula: [imath]x = -\frac{b}{2a}[/imath] as the [imath]x[/imath] coordinate of the vertex. I asked her why, and she told me not to tell her how to do her job. My smart friend mumbled something about it involving calculus, but I've always found him a rather odd fellow and I doubt I'd be able to understand a solution involving calculus, because I have no background in it. If you use something you know from calculus, explain it to someone who has no background in it. Because I sure don't. Is there a purely algebraic or geometrical yet elegant derivation for the [imath]x[/imath] coordinate of a parabola of the above form? | 2632010 | Vertex of a parabola without calculus?
I am curious if there is a way to get the vertex of a parabola without calculus? I know the calculus approach has you take the derivative set to [imath]0[/imath] (for [imath]0[/imath] slope) so if [imath]f(x) = ax^2 + bx + c[/imath] then [imath]f'(x) = 2x + b[/imath] so [imath]0 = 2x + b[/imath] implies [imath]x = -\frac{b}{2a}[/imath] with [imath]y[/imath] coordinate [imath]f(-\frac{b}{2a})[/imath]. But how would we get the vertex without calculus? |
985015 | A inverse Trigonometric multiple Integrals
How to calculate the closed form of the integral [imath]\int\limits_0^1 {\frac{{\int\limits_0^x {{{\left( {\arctan t} \right)}^2}dt} }}{{x\left( {1 + {x^2}} \right)}}} dx[/imath] | 1032483 | How to evaluate [imath]\int_0^1 (\arctan x)^2 \ln(\frac{1+x^2}{2x^2}) dx[/imath]
Evaluate [imath] \int_{0}^{1} \arctan^{2}\left(\, x\,\right) \ln\left(\, 1 + x^{2} \over 2x^{2}\,\right)\,{\rm d}x [/imath] I substituted [imath]x \equiv \tan\left(\,\theta\,\right)[/imath] and got [imath] -\int^{\pi/4}_{0}\theta^{2}\,{\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right) \over \cos^{2}\left(\,\theta\,\right)}\,{\rm d}\theta [/imath] After this, I thought of using the Taylor Expansion of [imath]\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right)[/imath] near zero but that didn't do any good. Please Help! |
1214399 | Is there any group with these conditions?
Is there any infinite group [imath]G[/imath], such that [imath]G^\prime[/imath] is Abelian, and any Abelian and normal subgroup of [imath]G[/imath] is finite? In other words: [imath]G^\prime[/imath] is Abelian [imath]\forall H \trianglelefteq G[/imath] : ([imath]H[/imath] is Abelian) [imath]\Rightarrow |H| < \infty[/imath] | 454953 | A group [imath]G[/imath] with [imath]G'[/imath] abelian and every abelian normal subgroup finite
Let [imath]G[/imath] be a group such that [imath]G'[/imath] abelian and any abelian normal subgroup of [imath]G[/imath] is finite. Show that [imath]G[/imath] is finite. |
746388 | Calculating [imath]1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? [/imath]
How to find infinite sum How to find infinite sum [imath]1+\dfrac13+\dfrac{1\cdot3}{3\cdot6}+\dfrac{1\cdot3\cdot5}{3\cdot6\cdot9}+\dfrac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? [/imath] I can see that 3 cancels out after 1/3, but what next? I can't go further. | 1699983 | find [imath]\frac{3}{6}+\frac{3\cdot5}{6\cdot9}+\frac{3\cdot5\cdot7}{6\cdot9\cdot12}+\cdots[/imath]
find [imath]\frac{3}{6}+\frac{3\cdot5}{6\cdot9}+\frac{3\cdot5\cdot7}{6\cdot9\cdot12}+\cdots[/imath] I had [imath](1-x)^{-\frac{p}{q}}[/imath] in mind. [imath]S=\frac{3}{6}+\frac{3\cdot5}{6\cdot9}+\frac{3\cdot5\cdot7}{6\cdot9\cdot12}+\cdots[/imath] [imath]S+1=1+\frac{3}{6}+\frac{3\cdot5}{6\cdot9}+\frac{3\cdot5\cdot7}{6\cdot9\cdot12}+\cdots[/imath] [imath]S+1=1+\frac{3}{2!\cdot3}+\frac{3\cdot(3+2)}{3!\cdot9}+\frac{3\cdot(3+2)\cdot(3+4)}{4!\cdot27}+\cdots[/imath] [imath]S+1=1+\frac{3}{2!}\left(\frac{\frac{2}{3}}{2}\right)+\frac{3\cdot(3+2)}{3!}\left(\frac{\frac{2}{3}}{2}\right)^2+\frac{3\cdot(3+2)\cdot(3+4)}{4!}\left(\frac{\frac{2}{3}}{2}\right)^3+\cdots[/imath] [imath]S+1=\left(1-\frac{2}{3}\right)^\frac{-3}{2}[/imath] I got [imath]S=3\sqrt{3}-1[/imath] But answer given is [imath]S=3\sqrt{3}-4[/imath] |
1207432 | Two fundamental questions about convexity of a function (number2)
The second question is as follows (the first one is here): Consider a function of two variables [imath]f(x,y)[/imath] on some compact domain. Let it be convex in [imath]x[/imath] and concave in [imath]y[/imath]. Is it true that [imath]f(x,y)[/imath] has a uniqe saddle point (stationary point)? Is [imath](\hat x,\hat y)=\arg \min_x\max_y f(x,y)[/imath], the argument of saddle point, unique? Is there any role of being strictly concave (or convex) in this case? My guess is that [imath]f(x,y)[/imath] owns a unique saddle point because of the convexity and concavity of the function. I also guess that if the function is strictly concave-convex, then, the argument of the saddle point is also unique. | 1214440 | Is there a unique saddle value for a convex/concave optimization?
Here is the question: Consider a function of two variables [imath]f(x,y)[/imath] on some compact domain. Let it be convex in [imath]x[/imath] and concave in [imath]y[/imath]. Is it true that [imath]f(x,y)[/imath] has a uniqe saddle point (stationary point)? Is [imath](\hat x,\hat y)=\arg \min_x\max_y f(x,y)[/imath], the argument of saddle point, unique? Is there any role of being strictly concave (or convex) in this case? My guess is that [imath]f(x,y)[/imath] owns a unique saddle point because of the convexity and concavity of the function. I also guess that if the function is strictly concave-convex, then, the argument of the saddle point is also unique. Any reference for a proof will also suffice. Thx. |
1214054 | Relation between Hermite polynomials and Brownian motion (on martingale property)
Let us define Hermite polynomials as [imath]H_n(x)=(-1)^n e^{x^2/2}\frac{d^n}{dx^n} e^{-x^2/2}.[/imath] One can prove that [imath]e^{\theta x-\frac{1}{2} \theta^2}=\sum_{n=0}^{\infty} \frac{1}{n!}H_n(x)\theta^n \quad (*).[/imath] Now denote [imath]H_n(t,x):=t^{n/2}H_n(x/\sqrt t)[/imath]. We want to prove that [imath]H_n(t,B(t))[/imath] is a martingale. Now, I've thought to use the identity [imath]H_{n+2}(t,x)=xH_{n+1}(t,x)-(n+1)tH_{n}(t,x)[/imath], so [imath]\mathbb E(H_{n+2}(t,B(t))|\mathcal F_s)=\mathbb E(B(t)H_{n+1}(t,B(t))|\mathcal F_s)-(n+1)t \mathbb E(H_{n}(t,B(t))|\mathcal F_s),[/imath] but that seems not to be the right way to solve the exercise. So I guess one shall use [imath](*)[/imath], but I don't know how to do it. Can someone help me? If it is possible, I'de like to prove it without using the fact that [imath]dH_{n+1}(t,B(t))= H_{n}(t,B(t))[/imath], because the exercise I proposed here is from a sheet that contains also the exercise [imath]e^{\theta x-\frac{1}{2} \theta^2}=\sum_{n=0}^{\infty} \frac{1}{n!}H_n(x)\theta^n[/imath], and I think that I should use that fact, but I'm not sure. | 981606 | Hermite Polynomials and Brownian motion
I am asked to prove the following : Let [imath]B_t[/imath] be a standard brownian motion. The [imath]n[/imath]th Hermite polynomial is [imath]\displaystyle H_n(t,x)=\frac{(−t)^n}{n!} e^{x^2/(2t)} \frac{d^n}{dx^n}e^{-x^2/(2t)}[/imath]. Show that the [imath]H_n[/imath] play the role that the monomials [imath]\dfrac{x_n}{n!}[/imath] play in ordinary calculus. In other words show : [imath]dH_{n+1}(t, B_t) = H_n(t, B_t)[/imath] Now my guess is that we can take the integral of both sides and use ito integration ? But I am not sure how to proceed from there. |
1214721 | Can [imath][0,1][/imath] be partitioned with the following property?
Let [imath]I:= [0,1][/imath]. Does there exist a partition [imath]\{I_1,I_2\}[/imath] of [imath]I[/imath] such that for all Borel subsets [imath]A[/imath] of [imath][0,1][/imath] we have: [imath] \mu (I_1 \cap A) = \frac{1}{2} \mu(A) = \mu (I_2 \cap A)[/imath] where [imath]\mu[/imath] is the Lebesgue measure? Many thanks for your help. | 1088944 | [imath]E[/imath] measurable set and [imath]m(E\cap I)\le \frac{1}{2}m(I)[/imath] for any open interval, prove [imath]m(E) =0[/imath]
Ran across this problem and need some help. Let [imath]E[/imath] be a measurable subset of the real numbers and suppose that for any open interval [imath]I[/imath] one has [imath]m(E\cap I)\le \frac{1}{2}m(I)[/imath], where [imath]m[/imath] is the Lebesgue measure. Prove that [imath]m(E)=0[/imath]. |
1214970 | Prove [imath]d = \gcd(a,b) \iff 1= \gcd (k_1, k_2)[/imath].
This is the assumption they give me: Let [imath]a, b[/imath] be integers and [imath]d[/imath] a positive integer. Let [imath]d|a[/imath] and [imath]d|b[/imath] so there there exists [imath]a=dk_1[/imath] and [imath]b=dk_2[/imath]. I can go the backwards direction but I'm going in circles when I go forward. Any hints? | 1213430 | $ d = \gcd(a,b)\Rightarrow\ \gcd(a/d,b/d) = 1$
Could someone please help me with this proof? Suppose that [imath]a, b \in N[/imath], and [imath]d = \gcd(a, b)[/imath]. Since [imath]d[/imath] divides [imath]a[/imath], we have [imath]a = de[/imath] for some integer [imath]e,[/imath] and similarly [imath]b = df[/imath] for some integer [imath]f[/imath]. Prove that [imath]\gcd(e, f) = 1[/imath]. I understand why it works. Since d is all the common factors of [imath]a[/imath] and [imath]b, e[/imath] and [imath]f[/imath] had no common factors, therefore the [imath]\gcd(e,f) = 1[/imath]. But how do I prove this? Thanks in advance. |
927293 | Understanding a lemma regarding subspace topology
I am learning topology and am struggling with some of the concepts of open sets regarding subspaces. The problem I am working on says, that Y is a subspace of X, and A is a subset of Y, then show that the topology A inherits as a subspace of Y is the same topology it inherits as a subspace from X. now what I am trying to prove is that A is a basis for the subspace topology of Y of X, and A is the basis for the subspace topology on X, then these two topologies are the same. Here is the lemma I am struggling to utilize , it says, "Let Y be a subspace of X, if U is open in Y and Y is open in X, then U is open in X." Since A is a subset of Y, does this imply that A is open in Y? (i think so). Y is open in X, so does this imply that A is open in X? (I think so) If A is open in Y, and A is open in X, does this mean that A is open in [imath]Y \bigcap U[/imath] for [imath]U \subset X[/imath] ? (This I am not sure of). Further, assuming my intuition is right, if A is open in [imath]Y \bigcap U[/imath] does this mean for [imath]x \in Y \bigcap U[/imath] that [imath]x \in A \bigcap U[/imath] since A is open in both Y and X? this yields my needed argument saying that [imath]A \bigcap U \subset Y \bigcap U[/imath] this can then be shown that [imath]\mathbb{B}_{A}[/imath] is a basis for both topologies. my question is that just because A is a subset of Y and A is open in X, and [imath]x \in Y \bigcap U[/imath] , I am not sure if this implies that [imath]x \in A \bigcap U[/imath]. Thank you again so very much and I will greatly appreciate any clarification. | 622212 | Subspace Topology of a subset
Show that if [imath]Y[/imath] is a subspace of [imath]X[/imath], and [imath]A[/imath] is a subset of [imath]Y[/imath], then the topology [imath]A[/imath] inherits as a subspace of [imath]Y[/imath] is the same as the topology it inherits as a subspace of [imath]X[/imath]. First some notational specifications: Let us define [imath]\tau[/imath] as the topology of [imath]X[/imath] [imath]\tau_{Y}=\{U \cap Y; U \in \tau \}[/imath]; [imath]\tau_{A}^Y=\{V \cap A; V \in \tau_{Y} \}[/imath] and [imath]\tau_{A}^X=\{U \cap A; U \in \tau \}[/imath] Let us show that [imath]\tau_{A}^Y=\tau_{A}^X[/imath] [imath]\subset[/imath]: Let [imath]V \cap A[/imath] be an element of [imath]\tau_{A}^Y[/imath] such that [imath]V \in \tau_{Y}[/imath] Then, there exists [imath]U\in \tau[/imath] such that [imath]V=U\cap Y[/imath] [imath]V\cap A=(U\cap Y)\cap A=U\cap (Y\cap A)= U\cap A[/imath] Hence: [imath]V\cap A \in \tau_{A}^X[/imath] and so: [imath]\tau_{A}^Y \subset \tau_{A}^X[/imath] [imath]\supset[/imath]: Let [imath]U \cap A \in \tau_{A}^X[/imath] [imath]U\in \tau \iff U \text{ open in } X[/imath] But [imath]U \cap Y[/imath] open in [imath]Y \iff U \cap Y \in \tau_{Y}[/imath] Hence [imath](U \cap Y) \cap A \in \tau_{A}^Y[/imath] since [imath]U \cap Y \in \tau_{Y}[/imath] Therefore [imath]U \cap A \in \tau_{A}^Y[/imath] Hence: [imath]\tau_{A}^X \subset \tau_{A}^Y[/imath] Therefore: [imath]\tau_{A}^Y=\tau_{A}^X[/imath] |
1215150 | Proof about prime numbers
Show that if [imath]n[/imath] is composite then there exists a prime [imath]p \leq n^\frac{1}{2}[/imath] such that [imath]p\mid n[/imath]. I would like to use contradiction to prove this claim but I'm not sure about how I should contradict this statement. | 431930 | Prove that if a number [imath]n > 1[/imath] is not prime, then it has a prime factor [imath]\le \sqrt{n}[/imath].
Prove that if a number [imath]n > 1[/imath] is not prime, then it has a prime factor [imath]\le \sqrt{n}[/imath]. My answer is that this is not always true, because you can pick a non-prime number that is greater than [imath]1[/imath] and its factor is equal to its square root. |
1215471 | Integral [imath]\int \frac{dx}{x^4+1}[/imath]
Find [imath]\int \frac{dx}{x^4+1}[/imath] I found a possible solution to this question here However, I was wondering if there is a "nicer" solution, that would more understandable to a person who isn't advanced in complex numbers. Thank you! | 333611 | Evaluating [imath]\int \frac{1}{{x^4+1}} dx[/imath]
I am trying to evaluate the integral [imath]\int \frac{1}{1+x^4} \mathrm dx.[/imath] The integrand [imath]\frac{1}{1+x^4}[/imath] is a rational function (quotient of two polynomials), so I could solve the integral if I can find the partial fraction of [imath]\frac{1}{1+x^4}[/imath]. But I failed to factorize [imath]1+x^4[/imath]. Any other methods are also wellcome. |
358999 | generalized inequalities defined by proper cones
The generalized inequality defined by a proper cone [imath]K[/imath] is that [imath]x \ge_{K} y[/imath] if [imath]x-y \in K[/imath] for [imath]x,y \in K[/imath]. Does this means that for any [imath]x \in K[/imath], we have [imath]x \ge_{K} 0[/imath] since [imath]x - 0 = x \in K[/imath] ? EDIT: My definitions of cones and proper cones follow the ones in "Convex Optimization" by S. Boyd, in which 0 is contained in cones. On page 53 of the book, there is a statement about generalized inequalities: [imath]x \le_{K} y[/imath] if and only if [imath]\lambda^T x \le \lambda^T y[/imath] for all [imath]\lambda \ge_{K^*} 0[/imath] where [imath]K^*[/imath] is the dual cone of the proper cone [imath]K[/imath], so [imath]K^*[/imath] is also a proper cone. The reason I ask this question is that if for any [imath]\lambda \in K^*[/imath], we have [imath]\lambda \ge_{K^*} 0[/imath], why not just using [imath]\lambda \in K^*[/imath] instead of [imath]\lambda \ge_{K^*} 0[/imath] in the above statement? | 669085 | What does "curly (curved) less than" sign [imath]\succcurlyeq[/imath] mean?
I am reading Boyd & Vandenberghe's Convex Optimization. The authors use curved greater than or equal to (\succcurlyeq) [imath]f(x^*) \succcurlyeq \alpha[/imath] and curved less than or equal to (\preccurlyeq) [imath]f(x^*) \preccurlyeq \alpha[/imath] Can someone explain what they mean? |
1215047 | Nonsigular curve of degree [imath]3[/imath] in [imath]\mathbb P^2[/imath] over a field of characteristic [imath]3[/imath]
I am trying to do problem [imath]1.5.5[/imath] from Algebraic Geometry by Robin Hartshorne. The problem states: For every degree [imath]d>0[/imath], and every [imath]p=0[/imath] or a prime number, give the equation of a nonsingular curve of degree [imath]d[/imath] in [imath]\mathbb{P^{2}}[/imath] over a field of characteristic [imath]p[/imath]. I have polynomials that work for every case except for the case where [imath]p=d=3[/imath] .I have tried several different polynomials of degree [imath]3,[/imath] but they always seems to have a singular point.Any help/ hint would be greatly appreciated. | 705657 | Nonsingular projective variety of degree [imath]d[/imath]
For each [imath]d>0[/imath] and [imath]p=0[/imath] or [imath]p[/imath] prime find a nonsingular curve in [imath]\mathbb{P}^{2}[/imath] of degree [imath]d[/imath]. I'm very close just stuck on one small case. If [imath]p\nmid d[/imath] then [imath]x^{d}+y^{d}+z^{d}[/imath] works. If [imath]p\mid d[/imath] then I have chosen the curve [imath]zx^{d-1}+xy^{d-1}+yz^{d-1}[/imath]. After some work using the Jacobian criterion for nonsingularity I arrive at [imath]3z=0[/imath]. As long as [imath]p\neq 3[/imath] this curve is nonsingular. But I haven't been able to deal with the [imath]p=3[/imath] case. Any ideas? Thanks in advance. |
652245 | Convergence Proof: [imath]\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x}[/imath]
I have to check whether the following expression converges; if yes I have to give the limit. [imath]\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x}[/imath] Now I did the following: [imath]\lim_{x\rightarrow\infty} \sqrt{4x+x^2}- \sqrt{x^2+x}[/imath] [imath]\lim_{x\rightarrow\infty} x\sqrt{\frac{4}{x}+1}- x\sqrt{1+\frac{1}{x}}[/imath] [imath]\lim_{x\rightarrow\infty} x \lim_{x\rightarrow\infty}(\sqrt{\frac{4}{x}+1}- \sqrt{1+\frac{1}{x}})[/imath] That confuses me. The left limit approaches [imath]\infty[/imath] while the right approaches [imath]0[/imath]. What is wrong here or what can I conclude from that? Thank you very muvh for your help in advance! FunkyPeanut | 962458 | [imath] \lim\limits_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})[/imath]
Find the limit: [imath] \lim_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})[/imath] I did the following: \begin{align} (\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x}) = \frac{(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})}{1} \cdot \frac{(\sqrt{x^2 + 2x} + \sqrt{x^2 - 7x})}{(\sqrt{x^2 + 2x} + \sqrt{x^2 - 7x})} \end{align} I know the final answer is [imath]\frac{9}{2}[/imath]. After multiplying by the conjugate, I see where the [imath]9[/imath] in the numerator comes from. I just can't remember how I solved the rest of the problem. |
1216626 | How do you find [imath]\sum_{j=0}^{p-1}\left(\frac{j(j+1)}p\right)[/imath],p = prime, where [imath]\left(\frac{j(j+1)}p\right)[/imath] is the Legendre symbol?
I'm having trouble finding a pattern by calculating each step of the sum from [imath]j=0[/imath] to [imath]j = p-1[/imath]. | 333704 | sum of the product of consecutive legendre symbols is -1
How do I prove the formula [imath]\newcommand{\jaco}[2]{\left(\frac{#1}{#2}\right)}\sum\limits_{a=1}^{p-2} \jaco{a(a+1)}p = -1[/imath] where a varies from 1 to p-2 and p is a prime I got as far as [imath]\jaco{p-a}p = \jaco{-1}p \jaco ap[/imath] so I can reduce the sum but that does not seem of much help. |
1143514 | [imath]n,a,b \mathbb \in \mathbb Z^+[/imath] , such that [imath]n\mid a^n-b^n[/imath] ; to show [imath]n\mid \frac {a^n-b^n}{a-b}[/imath]
Let [imath]n,a,b \in \mathbb Z^+[/imath] be such that [imath]n\mid a^n-b^n[/imath] , then how to prove that [imath]n\mid {\dfrac {a^n-b^n}{a-b}}[/imath] ? My try : [imath]d=\gcd(n,a-b),[/imath] so [imath]d \mid{\dfrac {a^n-b^n}{a-b}}.[/imath] Also [imath]\,n \mid(a-b)\left({\dfrac {a^n-b^n}{a-b}}\right),\;[/imath] so [imath]\;\dfrac nd\mid\dfrac {(a-b)}d\Big({\dfrac {a^n-b^n}{a-b}}\Big)[/imath]. Now, since [imath]\gcd\left(\dfrac nd ,\dfrac {(a-b)}d \right)=1,\;[/imath] so [imath]\dfrac nd\mid{\dfrac {a^n-b^n}{a-b}},\;[/imath] it follows that [imath]\;\operatorname{lcm}\left(d, \dfrac nd \right)\mid {\dfrac {a^n-b^n}{a-b}}.[/imath] But then I am stuck , Please help . Thanks in advance | 6310 | [imath]n \mid (a^{n}-b^{n}) \ \Longrightarrow[/imath] [imath]n \mid \frac{a^{n}-b^{n}}{a-b}[/imath]
How does one prove that if [imath]n \mid (a^{n}-b^{n}) \ \Longrightarrow[/imath] [imath] \displaystyle n \mid \frac{a^{n}-b^{n}}{a-b}[/imath] where [imath]a,b, n \in \mathbb{N}[/imath]. What i thought of is to consider [imath](a-b)^{n} \equiv a^{n} + (-1)^{n}b^{n} \ (\text{mod} \ n)[/imath] and if we suppose that [imath]n[/imath] is odd then we have, [imath](a-b)^{n} \equiv a^{n} -b^{n} \ (\text{mod} \ n)[/imath] and since [imath]n \mid (a^{n} - b^{n})[/imath] we have [imath](a-b)^{n} \equiv 0 \ (\text{mod} \ n) [/imath] I think i am far away from the conclusion of the problem, but this is what i could work on regarding the problem. |
1216763 | Density of [imath]H^1_0(\mathbb{R}^n)[/imath] in [imath]H^1(\mathbb{R}^n)[/imath]
[imath]H^1_0(\mathbb{R}^n)[/imath] is defined as the space of all functions in the first Sobolev space [imath]H^1(\mathbb{R}^n)[/imath] that are compactly supported. My question is: is [imath]H^1_0(\mathbb{R}^n)[/imath] dense in [imath]H^1(\mathbb{R}^n)[/imath]? Thanks! | 572418 | [imath]C_c^{\infty}(\mathbb R^n)[/imath] is dense in [imath]W^{k,p}(\mathbb R^n)[/imath]
As the title say, I want to prove that [imath]C_c^{\infty}(\mathbb R^n)[/imath] is dense in [imath]W^{k,p}(\mathbb R^n)[/imath] i.e. [imath]\displaystyle{ W^{k,p} (\mathbb R^n) = W_0^{k,p}(\mathbb R^n) \quad (\star)}[/imath]. In a book I found that in order to prove it, we need the following claim: Claim: Let [imath]\displaystyle{\zeta \in C^{\infty} ([0, \infty)) }[/imath] such that [imath]\zeta (t) =1 [/imath] for [imath] t \leq 1 [/imath] and [imath]\displaystyle{ \zeta (t) =0 }[/imath] for [imath] t \geq 2[/imath]. For [imath]R>0[/imath] and [imath]x \in \mathbb R^n[/imath] define [imath]\displaystyle{ \eta_R (x) = \zeta \left( \frac{|x|}{R} \right)}[/imath]. If [imath] \displaystyle{ u \in W^{k,p} ( \mathbb R^n) }[/imath] with [imath]1 \leq p < \infty [/imath] then, [imath]\displaystyle{ \eta_R u \to u }[/imath] in [imath]W^{k,p}(\mathbb R^n) [/imath] as [imath] R \to \infty[/imath]. If I prove the claim then the [imath](\star)[/imath] follows using mollifiers. But unfortuently I have stuck, with the proof of the claim. Any ideas? |
1216472 | Gamma Distribution and Probability
The lifetimes of batteries are independent exponential random variables, each having parameter λ. A flashlight needs 2 batteries to work. If one has a flashlight and a stockpile of n batteries, what is the distribution of time that the flashlight can operate? My friend said this was a gamma distribution with parameters (n/2, λ). But what does that even mean? I honestly have no idea where to begin. I know the distribution for an exponential random variable, [imath]λe^{-λx}[/imath], but other than that, not much. Help | 1181457 | Distribution of time that a flashlight can operate
The lifetimes of batteries are independent exponential random variables , each having parameter [imath]\lambda[/imath]. A flashlight needs two batteries to work. If one has a flashlight and a stockpile of n batteries, What is the distribution of time that the flashlight can operate? What I have so far: Let [imath]Y[/imath] be the lifetime of the flashlight; [imath]Y_1=min(X_1,...,X_n)[/imath] where [imath]X_i[/imath] is the lifetime of a battery ([imath]1\le i\le n[/imath]), and [imath]Y_2[/imath] the second smallest of the [imath]X_i[/imath] (so [imath]Y_1\le Y_2[/imath]) I wanted to compute: [imath]P[Y\le t]=P[Y_2\le k+m|Y_1\le m][/imath] where [imath]k+m=t[/imath] then we have that [imath]P[Y_2\le k+m|Y_1\le m]={P[Y_2\le k+m, Y_1\le m]\over P[Y_1\le m]}={P[Y_2\le k+m] P[Y_1\le m]\over P[Y_1\le m]}=P[Y_2\le k+m=t][/imath] (because of the independence of the random variables) So [imath]P[Y_2\le t]=P[min(X_1,...,X_{j-1},X_{j+1},...X_n)\le t][/imath] (assuming [imath]X_j=min(X_1,...,X_n)[/imath]) hence: [imath]P[min(X_1,...,X_{j-1},X_{j+1},...X_n)\le t]= 1-P[min(X_1,...,X_{j-1},X_{j+1},...X_n)\ge t]=1-P[X_1\ge t,...,X_{j-1}\ge t, X_{j+1}\ge t,... X_n\ge t]=1-e^{(n-1)\lambda t}[/imath] I would really appreciate if you can tell me if this is the correct approach :) |
1217093 | How we can prove that the logical result of a set is effectively enumerable?
How we can prove that the logical result of [imath]\{(p_i \vee [/imath]~ $p_{i+1}[imath]) [/imath]: i \in \mathbb{N} \}$ is effectively enumerable ? Update: as one user requests, I add my method. I use truth table for check the result but I'm sure there is a flaw and other method must be used. | 1204879 | Completness and Set of Result of One Set ?!?
Dear Everyone on this Wonderful Sites: I'm so glad to participate on this site and ask the first question that mentioned on the Contest some days ago. I ran into a question that wrote this set: [imath]\{(p_i \vee [/imath]~ $p_{i+1}[imath]) [/imath]: i \in \mathbb{N} \}$ and ask this question, Which of the following is False: -) this is Satisfiable --) this is Complete ---) this is Decidable ----) set of logical result of this set is effectively enumerable. Solution of answer sheet is (2). We try to solve it that without any detail is mentioned in the contest. one of our TA says, It is satisfiable because the TFTF... will satisfy it. (We couldent get it) and it is decidable because propositional logic is decidable. We read some useful website for completeness (--) and for (----), but we get stuck in it until yet. any useful hint or idea for this question will be highly appreciated. |
1217378 | How do I prove the function [imath]f(z) = \sin \bar{z}[/imath] is not differentiable on the disc [imath]D(0,1)[/imath]?
How do I prove the function [imath]f(z)=\sin \bar{z}[/imath] is not differentiable on the disc [imath]D(0,1)[/imath]? I originally used cauchy reimann equations to prove it is not differentiable everywhere, but I need to be able to show it's not differentiable specifically for the disc. Any help would be greatly appreciated. | 133226 | Showing [imath]\sin(\bar{z})[/imath] is not analytic at any point of [imath]\mathbb{C}[/imath]
Use the Cauchy-Riemann equations to show that the function [imath]g(z) = \sin (\bar z)[/imath] is not analytic at any point of [imath]\mathbb{C}[/imath]. Here's as far as I got - [imath]\sin \left(\frac{\bar z}{1}.\frac{z}{z}\right) =\sin \left(\frac{|z|^2}{z}\right) =\sin \left(\frac{x^2 + y^2}{x+iy}\right)[/imath] I can't see how to separate the real and imaginary parts so that I can apply the Cauchy-Riemann equations. |
1217867 | How can I prove like this exercise
Let we have [imath]X=R[/imath] the set of real numbers and [imath]B_R[/imath] is borel algebra and [imath]2^X=P(X)[/imath] How can I prove that [imath]B_R[/imath] is not equal to [imath]P(X)[/imath] | 1209126 | Borel sigma algebra not containing all subsets of [imath]\mathbb{R}[/imath]?
Consider the smallest sigma algebra [imath]\mathscr{B}[/imath] generated by all open subsets of [imath]\mathbb{R}[/imath]. One would expect that [imath]\mathscr{B}[/imath] contains all subsets of [imath]\mathbb{R}[/imath], but as it turns out, if we assume the axiom of choice to be true, there are some subsets of [imath]\mathbb{R}[/imath] which don't belong to [imath]\mathscr{B}[/imath]. I was hoping if someone could help me out with a proof of the above statement, i.e, [imath]\mathscr{B}[/imath] does not contain all subsets of [imath]\mathbb{R}[/imath]. |
1217644 | How to prove that [imath]\lfloor a\rfloor+\lfloor b\rfloor\leq \lfloor a+b\rfloor[/imath]
We have the floor function [imath]F(x)=[x][/imath] such that [imath]F(2.5)=[2.5]=2, F(-3.5)=[-3.5]=-4, F(2)=[2]=2[/imath]. How can I prove the following property of floor function: [imath][a]+[b] \le [a+b][/imath] | 963245 | [imath]\lfloor x\rfloor + \lfloor y\rfloor \leq \lfloor x+y\rfloor[/imath] for every pair of numbers of [imath]x[/imath] and [imath]y[/imath]
Give a convincing argument that [imath]\lfloor x\rfloor + \lfloor y\rfloor \leq \lfloor x+y\rfloor[/imath] for every pair of numbers [imath]x[/imath] and [imath]y[/imath]. Could someone please explain how to prove this? I attempted to say that the largest values that could be added to [imath]x[/imath] and [imath]y[/imath] is [imath]0.99[/imath] and that doing so still made [imath]\lfloor x\rfloor + \lfloor y\rfloor < \lfloor x+y\rfloor[/imath]. However, my answer was not accepted. |
1218410 | If [imath]S=\sum (\frac{n}{p})\zeta^n[/imath] then how to prove that [imath]S^2=(\frac{-1}{p})p[/imath]?
Here [imath]\zeta[/imath] is a primitive [imath]n[/imath]-th root of unity and ([imath]\frac{n}{p}[/imath]) denotes the Legendre symbol. Can someone please give a proof of this fact? I tried writing [imath]S^2[/imath] as the product of two sums [imath]S=\sum (\frac{m}{p})\zeta^m[/imath] and [imath]S=\sum (\frac{n}{p})\zeta^n[/imath] and use the fact that the Legendre symbol is multiplicative in its top argument, but I can't complete the argument and/or manipulations. As a follow-up question to this, is there any way in which I can use this identity to prove that any quadratic extension is contained in a cyclotomic extension? My teacher says so, but I don't see how this follows from the main identity. | 775320 | Quadratic Gauss Sum
The wikipedia page for Quadratic Gauss Sum say that the fact that [imath]g(a;p)^2 = \left(\frac{-1}{p}\right)p[/imath] is simple, yet I am having trouble coming up with a proof. Are there any references I could be pointed to? Thanks! |
1218041 | Exact values of [imath]x[/imath] for [imath]2\cos^2x=1+\sin x[/imath]
This question involves finding the exact values of [imath]x[/imath] such that [imath]0 \leqslant x \leqslant 2\pi[/imath]. So far I have subtracted everything to the left side of the equation and then used the pythagorean identities to change [imath]2\cos^2x\ \ \text{to}\ \ 2-2\sin^2x[/imath], leaving [imath]2-2\sin^2x-\sin x-1=0\,.[/imath] However, I need to further factor this, and I am not sure about what to do next. Could someone please help me? | 1218005 | Determine the exact value of equations involving more two trig variables
[imath]2\cos^2x=1+\sin x[/imath]. Determine the exact values of [imath]x[/imath] such that [imath]0 \leq x \leq 2\pi[/imath]. I am experiencing problems with factoring this question. First I started by getting everything on to the left side so: [imath]2\cos^2x-\sin x-1=0[/imath]. Then seeing as there are two trig variable here ([imath]\cos[/imath] and [imath]\sin[/imath]), I used the pythagorean identity and replaced [imath]2\cos^2x[/imath] with [imath]1-2\sin^2x[/imath] to make it into a one variable question: [imath]1-1-2\sin^2x-\sin x-1=0[/imath]. By collecting like terms it then becomes [imath]-\sin^2x-\sin x=0[/imath]. Now you could factor these with [imath]0[/imath] and [imath]-1[/imath], but then it creates uneven brackets so this musn't work. I don't think the quadratic formula is necessary, but I am wondering if this could be treated as a difference of squares question. If you replace the [imath]\sin x[/imath] and make the question into [imath]-x^2-x=0[/imath] it is a little easier to look at, however I am not sure if this is the right approach. If someone could please clarify my answer, that would be appreciated! |
1218430 | Simple proof that if [imath]a^2[/imath] is divisible by a prime [imath]b[/imath] then [imath]a[/imath] is divisible by [imath]b[/imath].
I know it's obvious that if [imath]a^2[/imath] is divisible by [imath]b[/imath], given that [imath]b[/imath] is prime ,then [imath]a[/imath] is divisible by [imath]b[/imath]. Is there any way to prove this without going into the fundamental theorem of arithmetic? I was told it can be proved starting from the fact that every number is the product of distinct primes, but is there a simpler way? | 711452 | Given that [imath]p[/imath] is a prime and [imath]p\mid a^n[/imath], prove that [imath]p^n\mid a^n[/imath].
Given that [imath]p[/imath] is a prime and [imath]p\mid a^n[/imath], prove that [imath]p^n\mid a^n[/imath]. I know that the fundamental theorem of arithmetic states that any positive integer can be represented as a product of primes but how do I apply this to the proof? |
417479 | If [imath]\gcd(a,b)= 1[/imath] and [imath]a[/imath] divides [imath]bc[/imath] then [imath]a[/imath] divides [imath]c\ [/imath] [Euclid's Lemma]
Well I thought this is obvious. since [imath]\gcd(a,b)=1[/imath], then we have that [imath]a[/imath] does not divide [imath]b[/imath] AND [imath]a[/imath] divides [imath]bc[/imath]. this implies that [imath]a[/imath] divides [imath]c[/imath]. done. but apparently this is wrong. help explain why this way is wrong please. the question tells you give me two relatively prime numbers [imath]a[/imath] and [imath]b[/imath] such that [imath]a[/imath] divides the product [imath]bc[/imath], prove that [imath]a[/imath] divides [imath]c[/imath]. how is this not obvious? explain to me how my "proof" is not correct. | 1538789 | Elementary proof of [imath]\gcd(a, b) = 1 \wedge a\ |\ b\ c \Rightarrow a \ | \ c [/imath]
How does on prove [imath]\gcd(a, b) = 1 \wedge a\ |\ b\ c \Rightarrow a \ | \ c [/imath] with as elementary steps as possible (i.e. not using the fundamental theorem of arithmetic (unique prime factorization))? EDIT: I saw that this theorem is called Gauss Theorem and is proved formally for integers [imath]\mathbb Z[/imath] in Coq, https://coq.inria.fr/library/Coq.ZArith.Znumtheory.html#Gauss EDIT: Clarification: I forgot to tell that I want to prove this for the natural numbers [imath]\mathbb N \geq 0[/imath]. Is Bezout's lemma applicable for the natural numbers, or is some other method needed? |
1218129 | Suppose that [imath]a[/imath] and [imath]b[/imath] belong to a field of order [imath]8[/imath] and that [imath]a^2 + ab + b^2 =0[/imath] then [imath]a=0[/imath] and [imath]b=0[/imath] .
Suppose that [imath]a[/imath] and [imath]b[/imath] belong to a field of order [imath]8[/imath] and [imath]a^2 + ab + b^2 =0[/imath]. Then [imath]a=0[/imath] and [imath]b=0[/imath]. Do the same when the field has order [imath]2^n[/imath] with [imath]n[/imath] odd? If one of the term is zero, i.e. let [imath]b=0[/imath] then [imath]a^2 =0 \implies a =0[/imath]. We also note that since field is of order [imath]8[/imath], [imath]a^7 = 1 = b^7[/imath]. But how to bring a contradiction if I consider any one to be not equal to [imath]0[/imath]? I want to use only basics. | 263733 | A property of finite field of order [imath]2^n[/imath]
Suppose [imath]a[/imath] and [imath]b[/imath] are elements of a finite field of order [imath]2^n[/imath] with [imath]n[/imath] odd and [imath]a^2+ab+b^2=0[/imath]. Is it necessary that both [imath]a[/imath] and [imath]b[/imath] must be zero ? I understand that the field has characteristic [imath]2[/imath] but don't know how to use the fact that [imath]n[/imath] is odd, please help. |
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