Datasets:

approach0

/

MSE-duplicate-questions

like 0

Follow

approach0 1

3050539

[imath]\int_{0}^{2\pi}\frac{d\theta}{a+b\sin\theta}[/imath] where [imath]a,b>0[/imath] [imath]\int_{0}^{2\pi}\dfrac{d\theta}{a+b\sin\theta}[/imath] where [imath]a,b>0[/imath] Supposedly a very simple residue theorem problem but I'm stuck with the pole(s). Let [imath]z=e^{i\theta}[/imath] and [imath]d\theta=\frac{dz}{iz}[/imath]. Then, [imath]\begin{align*} \int_{\mathbb{D}}\frac{dz}{\left[a+b\left(\frac{z-z^{-1}}{2i}\right)\right]iz}&=\int_{\mathbb{D}}\frac{2idz}{(2ai+bz-bz^{-1})iz}\\ &=\int_{\mathbb{D}}\frac{2dz}{bz^2+2aiz-b} \end{align*}[/imath] By quadratic formula, we can deduce that we have pole at [imath]z=\frac{-ai\pm\sqrt{b^2-a^2}}{b}[/imath] But I'm having trouble figuring out which, of the two is in fact in the unit circle. All we have is [imath]a>0[/imath] and [imath]b>0[/imath] and I feel like it could go so many different ways depending on the actual values that they take. Can anyone help me to proceed?

2660875

Evaluating [imath]\int_0^{2\pi} \frac {cos(\theta)}{5-3cos(\theta)}[/imath] using residue theorem I am to evaluate the following integral using residue theorem: [imath]\int_0^{2\pi} \frac {cos(\theta)}{5-3cos(\theta)}[/imath] I know that I need to perform substitution with the following: [imath]d\theta = \frac 1{iz}[/imath] [imath]cos{\theta} = \frac 12(z+\frac1z)[/imath] Which yields: [imath]\frac 1i=\int_{|z|=1} \frac{z+\frac1z}{-3z^2+10z-3}[/imath] Finding the roots for the denominator yields two roots at 1/3 and 3. Only 1/3 falls within our unit circle so we ignore the root at 3. This is where I am a little stuck. I was taught to use the shortcut here to calculate the residue where I would leave the numerator [imath]p(z)[/imath] as is and take the derivative of the denominator [imath]q'(z)[/imath] then plug in the value of the singularity [imath]z=\frac13[/imath] but this approach doesn't give me the correct answer which I know is [imath]\frac\pi6[/imath] Am I making a mistake in setting up the problem? Thanks!

3051167

Proving Geometric Hahn-Banach Implies Analytic Hahn-Banach I've come across a proof online of the H.B theorem using the goemetric version. There is a step in the proof which im not sure why is true and it is as follows: Let [imath]X[/imath] be a linear space and [imath]M \subset X \times \Bbb R[/imath] be a maximal subspace. Then [imath]M = G(F)[/imath] for [imath]F :X \to \Bbb R[/imath] linear. Someone can show me why this is true? Maybe I misunderstood the proof, its the last paragraph of Matrin's answer here -https://mathoverflow.net/questions/134508/direct-proof-of-the-separation-theorem-of-hahn-banach. Thanks a lot for helping!

3051629

Geometric Hahn Banach implies Analytic Hahn Banach. I want to prove that the geometric Hahn Banach theorem implies the analytic one. Edit: To avoid confusion I will state the vesion of H.B theorems im familiar with: Analytic H.B: Let [imath]X[/imath] be a linear space(over [imath]\Bbb R[/imath]) and [imath]Y\subset X[/imath] a subspace. Let [imath]p:X \to \Bbb R[/imath] be a sub-additive function. Suppose [imath]f:Y\to \Bbb R[/imath] is a linear map s.t [imath]f(y) \le p(y)[/imath] for all [imath]y\in Y[/imath] then there exists an extention [imath]g:X\to \Bbb R[/imath] of [imath]f[/imath] s.t. [imath]g(x)\le p(x) [/imath] for all [imath]x\in X[/imath]. Geometric H.B: Let [imath]X[/imath] be a linear space. [imath]K \subset X[/imath] convex s.t. each point in [imath]K[/imath] is an internal point. Let [imath]D[/imath] be a plane disjoint from [imath]K[/imath] then there exists hyperplane that contains [imath]D[/imath] and disjoint from [imath]K[/imath]. In my problem [imath]X[/imath] is NOT a normed space so there are no open sets. Given [imath]X[/imath] a linear space and [imath]Y\subset X[/imath] a subspace [imath]p:X\to \Bbb R[/imath] sub-additive, and [imath]f:Y\to \Bbb R[/imath] linear s.t [imath]f(y)\le p(y)[/imath] for [imath]y\in Y[/imath] we need to extend [imath]f[/imath] to [imath]g:X\to \Bbb R [/imath] and [imath]g(x)\le p(x)[/imath] fo r all [imath]x\in X[/imath] So, we look at [imath]X \times \Bbb R [/imath]and define [imath]K = \{(x,t) : t>p(x)\}[/imath]. The fact that [imath]K[/imath] is convex is easy. How can I show directly that every point in [imath]K[/imath] is internal? (can't say that [imath]K[/imath] is open). Now after showing that, we can look at [imath]Graph(f)[/imath] and observe that [imath]Graph(f)\cap K = \emptyset[/imath]. So by the geometric H.B theorem we have a hyperplane (which is a maximal subspace in this case because [imath](0,0)\in Graph(f)[/imath] ) M containing [imath]Graph(f) [/imath] and disjoint from [imath]K[/imath]. Now my problem is to show that each [imath]x\in X[/imath] has a unique [imath]t\in \Bbb R[/imath] s.t. [imath](x,t) \in M[/imath]. (I need it in order to extend [imath]f[/imath]). I know that if we take [imath]v_0\notin M[/imath] then each [imath]v \in X \times \Bbb R[/imath] has a unique representation as [imath]v = \alpha v_0 + m[/imath] for [imath]m \in M[/imath] , this is because [imath]M[/imath] is a maximal subspace. not sure if that helps. Thanks for helping!

1805675

prove [imath]a \equiv b \pmod{p_1}[/imath] and [imath]a \equiv b \pmod{p_2} \Rightarrow a \equiv b \pmod{p_1\times p_2}[/imath] I am following a proof of an RSA algorithm and the proof states the following: [imath]p_1[/imath] and [imath]p_2[/imath] are distinct primes, [imath]a \equiv b \pmod{p_1}[/imath] and [imath]a \equiv b \pmod{p_2} \Rightarrow a \equiv b \pmod{p_1\times p_2}[/imath]. Let's call that FACT X. can someone give me a proof for this. i have no formal mathematical training so a detailed proof will be much appreciated.

935147

If [imath]x \equiv a \pmod {p_1}[/imath] and [imath]x\equiv a \pmod{p_2}[/imath], then is it true that [imath]x\equiv a \pmod{p_1p_2} ?[/imath] [imath]p_1, p_2[/imath] are distinct prime numbers I have just observed this pattern when solving this problem. Is there a simple way to prove/disprove it ?

2640631

If x is an odd number, how to prove [imath]x^2=8y+1[/imath]? I tried to let [imath]x=2k+1[/imath] but it ended up proving [imath]k^2+k=2y[/imath] for some [imath]k[/imath] and [imath]y[/imath]. What is the correct solution?

146417

If [imath]n[/imath] is an odd natural number, then [imath]8[/imath] divides [imath]n^{2}-1[/imath] I am trying to show that if [imath]n[/imath] is an odd natural number, then [imath]8[/imath] divides [imath]n^{2}-1.[/imath] I was able to prove that because I know that if [imath]n[/imath] is an odd natural number, then [imath]n^{2}[/imath] can be written as [imath]8k+1[/imath] for some [imath]k\in \mathbb{N}.[/imath] I would like to show this question by using the Euclidian division. Then I wrote [imath]n^{2}-1=8k+r[/imath], where [imath]0\leq r < 8.[/imath] When [imath]r=2[/imath] we get [imath]n^{2}-1=8k+2.[/imath] Since [imath]n[/imath] is odd, then [imath]n^{2}-1[/imath] is even and I got stuck. Is there way to fix that?

3055991

Problems understanding: [imath]\Sigma \vdash \theta \text{ iff }\Sigma \vdash \forall x \theta[/imath] I'm trying to make sense of a theorem I came across recently: [imath]\Sigma \vdash \theta \text{ iff }\Sigma \vdash \forall x \theta[/imath] Say [imath]\theta[/imath] is [imath](x=1)[/imath], then we have [imath]\Sigma \vdash (x=1)\text{ iff }\Sigma \vdash \forall x (x=1)[/imath]. Well, [imath]\Sigma \vdash \theta \Rightarrow \Sigma \models \theta[/imath], which means for every structure [imath](\Bbb A)[/imath] with universe [imath](A)[/imath] and every assignment function [imath](s)[/imath]: [imath](\forall \Bbb A)(\forall s)(\Bbb A \models \Sigma[s] \Rightarrow \Bbb A \models \theta[s])[/imath] and [imath](\forall \Bbb A)(\forall s)(\forall a \in A)(\Bbb A \models \theta[s(x|a)])[/imath] where [imath]s(x|a)(v) = \{s(v) \text{ if v is not x}, a \text{ if v is x}\}[/imath]. But then doesn't this mean that with ANY structure and ANY universe associated with it and EVERY assignment function that every [imath]x[/imath] is [imath]1[/imath]?

3056916

If [imath]\Sigma \vdash \varphi[/imath] then [imath]\Sigma \vdash \forall x \varphi[/imath]. Why when [imath](\varphi \implies \forall x \varphi)[/imath] is not true? In the book of first order logic I'm reading they say that if [imath](\varphi_1,...,\varphi_n)[/imath] is a proof from a set of formulae [imath]\Sigma[/imath], then so is [imath](\varphi_1,...,\varphi_n,\forall x \varphi_j)[/imath] for all [imath]1\leq j \leq n[/imath]. They assume that as an axiom but it doesn't make sense to me. Why if [imath]\varphi[/imath] is true then so is [imath]\forall x \varphi[/imath]? For example if [imath]\varphi[/imath] is [imath]x=1[/imath] that wouldn't mean that [imath]\forall x : x=1[/imath] , or would it? Maybe what I don't understand is: why [imath](\Sigma \vdash \varphi \implies \Sigma \vdash \forall x \varphi)[/imath] is true but [imath](\varphi \implies \forall x \varphi)[/imath] is not true? The book is this: https://www.springer.com/cda/content/document/cda_downloaddocument/9781447121756-c2.pdf?SGWID=0-0-45-1192238-p174141200

3056983

[imath]|G|=8[/imath], then [imath]G[/imath] has proper nontrivial subgroups of order [imath]4[/imath] Let [imath]G[/imath] be a group and [imath]|G|=8[/imath], then [imath]G[/imath] has proper nontrivial subgroups of order [imath]4[/imath]. I can't see why. Does it follow by Sylow Theorem?

1261707

Prove the existence of order 4 subgroups of order 8 groups I am participating in an Introductory course in groups and I have the following question: Let [imath]G[/imath] be a finite group of order [imath]8[/imath]. Prove that [imath]G[/imath] has a subgroup of order [imath]4[/imath] and a subgroup of order [imath]2[/imath]. i know how to prove that [imath]G[/imath] has a subgroup of order [imath]2[/imath] but i can't show the existence of a subgroup of order 4. I am only a begginer in groups so please take this into account.

3055456

Show that [imath]\|.\|_{1/2}[/imath] is not a norm. Show that [imath]\|.\|_{1/2}[/imath] is not a norm. would anybody guide me that how can i prove or disprove? thanks

1918657

[imath]l^p[/imath] not norm, [imath]p<1[/imath] please I tried to find counterexamples to see that [imath]l^p[/imath] is not norm with [imath]p<1[/imath] in the triangle inequality but I have problems with convergence when I choose some successions. Thanks.

3057739

Let [imath]G=G_1\times G_2[/imath], prove [imath]G'=G_1'\times G_2'[/imath] I'm reading Hans Kurzweil's "The Theory of Finite Groups", in 1.6.2 (b) it's claimed that let [imath]G=G_1\times G_2[/imath], one has [imath]G'=G_1'\times G_2'[/imath] because: [imath]\begin{align} G'&=[G_1G_2,G_1G_2]\\ &=\prod_{i,j}[G_i,G_j]\\ &=G_1'\times G_2'. \end{align}[/imath] Here [imath]G'[/imath] is the commutator subgroup defined as [imath]G':= \langle [x,y] | x,y \in G\rangle,[/imath] and [imath][x,y]:=x^{-1}y^{-1}xy.[/imath] However, I'm a bit confused here about the proof. First, what does it mean by [imath]\prod_{i,j}[G_i,G_j][/imath], is it [imath][G_1,G_1][G_1,G_2][G_2,G_1][G_2,G_2][/imath]? Second, why [imath][G_1G_2,G_1G_2]=\prod_{i,j}[G_i,G_j][/imath]? With [imath]G'=\langle [a_1b_1,a_2b_2] | a_1,a_2\in G, b_1, b_2 \in G_2 \rangle,[/imath] I can't move all [imath]G_1[/imath] part to the left side without changing the element [imath]a_1,a_2,b_1,b_2[/imath], because in general [imath]a_1b_1\ne b_1a_1[/imath].

2081328

Commutator Subgroup of Direct Product equals the Direct Product of the Commutator Subgroups I need to prove that [imath](G_{1}\times G_{2})^{\prime} = G_{1}^{\prime} \times G_{2}^{\prime}[/imath], where [imath]G^{\prime}[/imath] denotes the commutator subgroup of [imath]G[/imath] - i.e., [imath]G^{\prime}=[G,G][/imath], the subgroup generated by all commutators of elements of [imath]G[/imath]. Recall that the commutator of two elements [imath]x[/imath] and [imath]y[/imath] in a group [imath]G[/imath], denoted [imath][x,y][/imath] is defined to equal [imath]x^{-1}y^{-1}xy[/imath]. I have posted my proof as an answer below. Could somebody please take a look at it and let me know if it's okay? If not, please let me know what I need to do in order to fix it. Thank you. :) Edit: I have been informed that my answer given below actually shows only equality of the set of commutators, and that [imath]G^{\prime}[/imath] consists of products of commutators. So, this question is no longer a proof check, but I am asking specifically how to fix what I have written below in order to make it actually answer the thing I set out to prove. I am having a bit of trouble understanding some of the hints given me thus far, and am going to need more detail in any answers given in order for them to be helpful.