qid
stringlengths 1
7
| Q
stringlengths 87
7.22k
| dup_qid
stringlengths 1
7
| Q_dup
stringlengths 97
10.5k
|
|---|---|---|---|
1086260 | Is [imath]\pi[/imath] a rational multiple of e?
Does [imath]\pi = re[/imath] for some rational [imath]r[/imath]? I assume the answer is no but cannot prove so. | 460445 | Is [imath]\large \frac {\pi}{e}[/imath] rational, irrational, or trandescendal?
Is there an argument for why [imath]\large \frac {\pi}{e}[/imath] is rational, irrational, or trandescendal? Can the quotient of any two transcendental numbers (which are not rational multiples of each other) be rational, or at least irrational? Thanks. |
1081320 | If [imath]f'(z_0)\neq 0[/imath] then [imath]f[/imath] is one to one on some open disk [imath]D_r(z_0)[/imath]
This is what I am trying to prove Let [imath]D\subset\mathbb{C}[/imath] be open and [imath]f[/imath] be analytic in [imath]D[/imath]. If there is [imath]z_0\in D[/imath] such that [imath]f'(z_0)\neq 0[/imath] then there exists [imath]D_r(z_0)\subset D[/imath] and [imath]f[/imath] is one to one in [imath]D_r(z_0)[/imath]. I am finding it difficult to prove this but on some disk [imath]D_r(z_0)\subset D[/imath] [imath]f'=0[/imath] due to continuity. But I just can't go ahead and get anything out of this. ANy hints will be appreciated. Thanks | 333850 | Holomorphic function [imath]f[/imath] such that [imath]f'(z_0) \neq 0[/imath]
Let [imath]f: \mathbb{C} \rightarrow \mathbb{C}[/imath] be an holomorphic function such that [imath]f'(z_0) \neq 0[/imath] for some [imath]z_0 \in \mathbb{C}[/imath].Prove that there is [imath]r>0[/imath] such that, if [imath]|z-z_0|<r[/imath] and [imath]z \neq z_0 [/imath], then [imath]f(z) \neq f(z_0)[/imath]. It seems very trivial, but I can't solve it... |
783345 | Polar equation for a Heptagon
What is polar equation of a Heptagon ? I need to move some Android views in the form of a heptagon, for I need to have polar equations for Heptagon like for [imath]x= r\sin(\theta)[/imath] and [imath]y=r\cos(\theta)[/imath]. Is it possible to have polar equation for a heptagon too ? | 41940 | Is there an equation to describe regular polygons?
For example, the square can be described with the equation [imath]|x| + |y| = 1[/imath]. So is there a general equation that can describe a regular polygon (in the 2D Cartesian plane?), given the number of sides required? Using the Wolfram Alpha site, this input gave an almost-square: PolarPlot(0.75 + ArcSin(Sin(2x+Pi/2))/(Sin(2x+Pi/2)*(Pi/4))) (x from 0 to 2Pi) This input gave an almost-octagon: PolarPlot(0.75 + ArcSin(Sin(4x+Pi/2))/(Sin(4x+Pi/2)*Pi^2)) (x from 0 to 2Pi) The idea is that as the number of sides in a regular polygon goes to infinity, the regular polygon approaches a circle. Since a circle can be described by an equation, can a regular polygon be described by one too? For our purposes, this is a regular convex polygon (triangle, square, pentagon, hexagon and so on). It can be assumed that the centre of the regular polygon is at the origin [imath](0,0)[/imath], and the radius is [imath]1[/imath] unit. If there's no such equation, can the non-existence be proven? If there are equations, but only for certain polygons (for example, only for [imath]n < 7[/imath] or something), can those equations be provided? |
1086004 | Continuity and Convergent nets.
The original question before it was marked as duplicated question: I was wondering if one construct a non-continuous function [imath]f:X\to Y[/imath] between two topological spaces [imath]X[/imath] and [imath]Y[/imath], such that [imath]f[/imath] sends every convergent net into convergent net... I want to see an example... Thanks in advance. Edition: Can Someone construct a non-continuous function [imath]f:X\to Y[/imath] between two topological spaces [imath]X[/imath] and [imath]Y[/imath], such that [imath]f[/imath] sends every convergent net into convergent net... I want to see an example... Thanks in advance. My own Answer: Consider a two point set [imath]X:=\{x,y\}[/imath] with a topology [imath]\mathcal{T}=\{\emptyset,\{x\},\{x,y\}\}[/imath]. Define [imath]f:X\to X[/imath] by [imath]x\mapsto y[/imath] and [imath]y\mapsto x[/imath]. Then, since [imath]f^{-1}(\{x\})=\{y\}[/imath] where [imath]\{x\}\in\mathcal{T}[/imath], [imath]f[/imath] is not continuous. On the other hand, pick any convergent net [imath]\{x_\lambda\}_{\lambda\in\Lambda}[/imath] in [imath]X[/imath]. We claim that [imath]f(x_\lambda)[/imath] converges to [imath]y[/imath]. Note that there is only one neighborhood of [imath]y[/imath], [imath]\{x,y\}[/imath], since [imath]\{y\}[/imath] is not open. Then, it follows that [imath]\forall\mu\in\Lambda[/imath], [imath]\mu\prec\lambda\Rightarrow f(x_\lambda)\in\{x,y\}=X[/imath]. where [imath]\prec[/imath] is the preorder on the given directed set [imath]\Lambda[/imath]. | 360419 | [imath]f[/imath] brings convergent nets to convergent nets, is it continuous?
Let [imath]f:(X,\mathcal T)\to (Y,\mathcal S)[/imath] be a function between topological spaces. Let for any convergent net [imath](x_\alpha)[/imath] in [imath]X[/imath], [imath](f(x_\alpha ))[/imath] be convergent in [imath]Y[/imath]. Is [imath]f[/imath] continuous? (It seems to be true in completely regular spaces). |
1086123 | Polynomial with a root modulo every prime but not in [imath]\mathbb{Q}[/imath].
I recently came across the following fact from this list of counterexamples: There are no polynomials of degree [imath]< 5[/imath] that have a root modulo every prime but no root in [imath]\mathbb{Q}[/imath]. Furthermore, one such example is given: [imath](x^2+31)(x^3+x+1)[/imath] but I have not been able to prove that this does has that property above. How can such polynomials be generated and can we identify a family of them? | 608919 | Is it true that if [imath]f(x)[/imath] has a linear factor over [imath]\mathbb{F}_p[/imath] for every prime [imath]p[/imath], then [imath]f(x)[/imath] is reducible over [imath]\mathbb{Q}[/imath]?
We know that [imath]f(x)=x^4+1[/imath] is a polynomial irreducible over [imath]\mathbb{Q}[/imath] but reducible over [imath]\mathbb{F}_p[/imath] for every prime [imath]p[/imath]. My question is: Is it true that if [imath]f(x)[/imath] has a linear factor over [imath]\mathbb{F}_p[/imath] for every prime [imath]p[/imath], then [imath]f(x)[/imath] has a linear factor over [imath]\mathbb{Q}[/imath]? Edit: Thanks for @Jyrki Lahtonen's answer, I want to do some modifications: Is it true that if [imath]f(x)[/imath] has a linear factor over [imath]\mathbb{F}_p[/imath] for every prime [imath]p[/imath], then [imath]f(x)[/imath] is reducible over [imath]\mathbb{Q}[/imath]? Thanks in advance! |
1086891 | Functions such that [imath]f(f(n))=n+2015[/imath]
Is there a function [imath]f:\mathbb N \to \mathbb N[/imath] such that [imath]\forall n \in \mathbb N, f(f(n))=n+2015[/imath] ? Here's what I've done: Assuming such a function exists, [imath]f(f(f(n)))=f(\color{red}{f(f(n))})=f(n+2015)=\color{red}{f(f(}f(n)\color{red}{))}=f(n)+2015[/imath]. Hence [imath]\forall n, f(n+2015)=f(n)+2015[/imath]. A simple induction yields [imath]\forall n,\forall k, f(n+2015k)=f(n)+2015k[/imath]. This means that [imath]f(n) \operatorname{mod}2015[/imath] only depends on [imath]n \operatorname{mod}2015[/imath]. What then ? I can't reach a contradiction from here. | 325504 | IMO 1987 - function such that [imath]f(f(n))=n+1987[/imath]
Show that there is no function [imath]f: \mathbb{N} \to \mathbb{N}[/imath] such that [imath]f(f(n))=n+1987, \ \forall n \in \mathbb{N}[/imath]. This is problem 4 from IMO 1987 - see, for example, AoPS. |
1086908 | Prove that if [imath]A,B[/imath] are finite Abelian groups and for every [imath]n[/imath] they have the same amount of elements of order [imath]n[/imath], then [imath]A \simeq B[/imath]
Prove that if [imath]A,B[/imath] are finite Abelian groups and for every [imath]n[/imath] they have the same amount of elements of order [imath]n[/imath], then [imath]A \simeq B[/imath]. I know I have to use primary decomposition, but am not sure how exactly. Any guidance would be appreciated! | 1084570 | Finite Abelian groups with the same number of elements for all orders are isomorphic
Let [imath]A[/imath] and [imath]B[/imath] be finite abelian groups. Suppose that for every natural number [imath]m[/imath], the number of elements of order [imath]m[/imath] in [imath]A[/imath] is equal to the number of elements of order [imath]m[/imath] in [imath]B[/imath]. Prove that [imath]A[/imath] and [imath]B[/imath] are isomorphic. Idea Given that these groups are finite, I think you have to use the primary decomposition theorem somehow. |
1086864 | Question regarding a graded ring
Let [imath]S[/imath] be a finitely generated graded ring over [imath]A[/imath]. So degree [imath]0[/imath] elements of [imath]S[/imath] is [imath]A[/imath] and we assume that there is no negative degree elements here. Thus, we have that [imath]S[/imath] is generated over [imath]A[/imath] by a finite number of homogeneous elements of positive degree, say [imath]\{ t_1, ..., t_m \}[/imath]. Does it then follow that [imath](S_{t_i})_0[/imath], which is the degree [imath]0[/imath] elements of the ring [imath]S_{t_i}[/imath], is also finitely generated over [imath]A[/imath]? Thanks! By [imath]S_{t_i}[/imath] I mean the localization of [imath]S[/imath] at [imath]t_i[/imath]. | 233625 | If [imath]S[/imath] is a finitely generated graded algebra over [imath]S_0[/imath], [imath]S_{(f)}[/imath] is finitely generated algebra over [imath]S_0[/imath]?
Let [imath]S = \bigoplus_{n\ge 0} S_n[/imath] be a graded commutative ring. Let [imath]f[/imath] be a homogeneous element of [imath]S[/imath] of degree [imath]> 0[/imath]. Let [imath]S_{(f)}[/imath] be the degree [imath]0[/imath] part of the graded ring [imath]S_f[/imath], where [imath]S_f[/imath] is the localization with respect to the multiplicative set [imath]\{1, f, f^2,\dots\}[/imath]. Suppose [imath]S[/imath] is finitely generated algebra over [imath]S_0[/imath]. Then [imath]S_{(f)}[/imath] is a finitely generated algebra over [imath]S_0[/imath]. |
1086474 | Inclusions regarding the limsup and liminf of sets: [imath] \liminf E_n \subset \limsup E_n [/imath]
Let [imath]\{ E_n \}_{n \in \mathbb{N} }[/imath] be a sequence of sets in some ambient set [imath]\Omega [/imath]. I want to show that [imath] \liminf E_n \subset \limsup E_n [/imath] My attempt: IF [imath]x \in \liminf E_n = \bigcup_{k=1}^{\infty} \bigcap_{n \geq k} E_n [/imath], then there is some [imath]k_0 \in \mathbb{N}[/imath] so that [imath]x \in \bigcap_{n \geq k_0} E_n [/imath]. How can I show that [imath]x \in \bigcap_{k =1}^{\infty} \bigcup_{n \geq k} E_n = \limsup E_n [/imath] ?? | 472519 | Is this proof of [imath]\liminf E_k \subset \limsup E_k [/imath] correct?
I am wondering if my proof is correct? Thank you for whoever willing to take a look at it for me. Proof [imath]\liminf E_k \subset \limsup E_k [/imath] If [imath]\{E_k\}_{k=1}^\infty[/imath] is a sequence of sets, we define \begin{align*} \limsup E_k & = \bigcap_{j=1}^\infty\left(\bigcup_{k=j}^\infty E_k\right)\\ &= \bigcap_{j=1}^\infty(E_j \cup E_{j+1} \cup \cdots)\\ &= (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots. \end{align*} Therefore, [imath]\limsup E_k[/imath] consists of those points in [imath]\mathbb{R}^n[/imath] which belong to infinitely many [imath]E_k[/imath]. \begin{align*} \liminf E_k & = \bigcup_{j=1}^\infty\left(\bigcap_{k=j}^\infty E_k\right)\\ &= \bigcup_{j=1}^\infty(E_j \cap E_{j+1} \cap \cdots)\\ &= (E_1 \cap E_2 \cap \cdots) \cup (E_2 \cap E_3 \cap \cdots) \cup \cdots. \end{align*} Therefore, [imath]\liminf E_k[/imath] consists of those points in [imath]\mathbb{R}^n[/imath] which belong to all [imath]E_k[/imath] for [imath]k \geq k_0[/imath]. Now consider [imath]x \in \liminf E_k[/imath], then [imath]x \in E_1 \cap E_2 \cap \cdots[/imath], or [imath]x \in E_2 \cap E_3 \cap \cdots[/imath] and so on, that is to say, [imath]x \in E_{k_0} \cup E_{k_0+1} \cup \dots[/imath] for some [imath]k_0[/imath]. More specifically, [imath]x \in E_k \forall k \geq k_0[/imath]. Therefore, [imath]x \in (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots[/imath], so we showed [imath]x \in \limsup E_k[/imath]. |
1087046 | proof [imath][f(b)-f(a)]g'(c) = [g(b)-g(a)]f'(c)[/imath]
Let [imath]f,g:[a,b] \rightarrow \mathbb{R}[/imath] continuous functions and differentiable in [imath](a,b)[/imath] , show that [imath]\exists c \in (a,b)[/imath] such [imath][f(b)-f(a)]g'(c) = [g(b)-g(a)]f'(c)[/imath] I tried using the mean value theorem, but I can not relate the two functions with the same c. any help is appreciated | 296176 | Generalized mean value theorem
I know and understand the mean value theorem. But at the moment I don't have the intuition to understand the generalized mean value theorem If [imath]f[/imath] and [imath]g[/imath] are continuous on the closed interval [imath][a,b][/imath] and differentiable on the open interval [imath](a,b)[/imath], then there exists a point [imath]c\in(a,b)[/imath] where[imath][f(b)-f(a)]g'(c)=[g(b)-g(a)]f'(c).[/imath]If [imath]g'[/imath] is never zero on [imath](a,b)[/imath], then the conclusion can be stated as[imath]\frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)}.[/imath] What is the intuition? And how can you prove this (generally) ? |
1087285 | What is the product of [imath]p_i-1 \over p_i[/imath]
I am trying to find the value of [imath]\prod_{i=0}^{\infty}{p_i-1 \over p_i}[/imath] = [imath]{\lim_{x \to \infty}} {\phi(p_x!) \over p_x!}[/imath] Where [imath]p_x![/imath] is the [imath]x[/imath]th primorial, and [imath]p_i[/imath] is the [imath]i[/imath]th prime number. I guess I can honestly say I have no idea where to start, other than just iteriating it manually (around [imath]0.25[/imath] maybe?) | 976415 | Convergence of infinite product of prime reciprocals?
Where pn is the nth prime number, does the infinite product [imath]\prod_{n=1}^{\infty}\left(1-\frac{1}{p_n}\right)[/imath] converge to a nonzero value? (Any help would be much appreciated!) |
1087612 | Show that [imath]\Phi_{p^n}(X)=\Phi_p(X^{p^{n-1}})[/imath] with [imath]p[/imath] prime.
How to show that [imath]\Phi_{p^n}(X)=\Phi_p(X^{p^{n-1}})[/imath] with [imath]p[/imath] prime where [imath]\Phi_n(X)[/imath] is the cyclotomic polynomial define by [imath]\Phi_n(X)=\prod_{\underset{\gcd(k,n)=1}{k=1}}^n(X-e^{\frac{2ik\pi}{n}}).[/imath] I have shown already that [imath]X^n-1=\prod_{d\mid n}\Phi_d(X)[/imath] and that [imath]\Phi_p(X)=\prod_{k=1}^p(X-e^{\frac{2ik\pi}{p}})[/imath] but I'm not able to conclude that [imath]\Phi_{p^n}(X)=\Phi_p(X^{p^{n-1}})[/imath] with [imath]p[/imath] prime. | 1019633 | For [imath]p[/imath] prime, show that [imath]\Phi_{p^n}(x) = 1 + x^{p^{n-1}} + x^{2p^{n-2}} + \dots +x^{(p-1)^{p^n-1}}[/imath]
Well I attempted to try this but I failed to solve it: So [imath]\Phi_{p^{n}}(x)= \frac{x^{p^{n}} - 1}{\Phi_{1}(x) \Phi_{p^{2}}(x) \dots \Phi_{p^{n-1}}(x)}[/imath] Now I'm just stuck here. I saw a result somewhere on the net that said [imath]\Phi_{p^n}(x) = \Phi_p(x^{p^{n-1}})[/imath]. I would just use that, but I don't see why that result is true or know how to show that's true. How do I approach this? |
1087789 | Well-formed formulas: Difference between [imath]\forall x(x \in A \implies x \in B)[/imath] and [imath](\forall x \in A) x \in B[/imath]?
Let [imath]A[/imath] and [imath]B[/imath] be sets. There seem to be two ways of writing [imath]A \subseteq B[/imath]: \begin{equation} \forall x(x \in A \implies x \in B) \end{equation} or \begin{equation} (\forall x \in A) x \in B \end{equation} I usually see the first one written. I am curious to know if the second version is in proper form, logically. Is it a well-formed formula (wff)? Is there any way to prove these two formulas are logically equivalent? By the way, I am quite interested in this area of question. For example, I commonly want to know when it is better to write \begin{equation} (\forall x \in A) x \in B \end{equation} or \begin{equation} \forall x \in A (x \in B) \end{equation} is there anything I could read that would teach me the difference, and which is more preferable? | 79190 | Different standards for writing down logical quantifiers in a formal way
What are standard ways to write mathematical expressions involving quantifiers in a (semi)formal way ? In different posts of mine concerning similar question I have encountered for a generic expression of the type "for all [imath]x\in I[/imath] and [imath]y\in J[/imath] holds [imath]P(x,y)[/imath]" the following writing conventions 1) [imath]\forall y\in J \ \ \forall x \in I: P(x,y)[/imath] (this was how I would usually write statements in a formal way; sometimes also as [imath]\forall y\in J: \ \ \forall x \in I: P(x,y)[/imath] ;don't know if it standard) 2) [imath](\forall y)(\forall x)(x \in I \land y\in J \Rightarrow P(x,y))[/imath] (in the accepted answer from this question) 3) [imath](\forall y: \ y \in J) (\forall x: \ x \in I) (P(x,y))[/imath] (form the same answer as above) 4) [imath]\forall y \ \forall x\ (x \in I \land y \in J \Rightarrow P(x,y))[/imath] (in the accepted answer fromthis question) 5) [imath](\forall y\in J )(\forall x \in I) [P(x,y)] [/imath] (in the accepted answer from this question). ( (I hope I "generalised" them correctly, because at some points, they where written only for one variable, for example 3) was written just as [imath](\forall x: \ x \in I) (P(x))[/imath] ) Could you tell me which ones are generally accepted an if there is a standard to how these should be written ? (I think unique readability should be a criterion and I'm not sure the way I'm used to writing mathematical expression satisfies that) |
1087833 | Find all continuous functions [imath]f:\mathbb R\to\mathbb R[/imath] such that [imath]f(f(x))=e^{x}[/imath]
Find all continuous functions [imath]f:\mathbb R\to\mathbb R[/imath] such that [imath]f(f(x))=e^{x}[/imath] I didn't solve this problem, but I proved that f is increasing and [imath] x<f(x)<e^{x} [/imath] please help | 65876 | thoughts about [imath]f(f(x))=e^x[/imath]
I was thinking, inspired by mathlinks, precisely from this post, if there exists a continuous real function [imath]f:\mathbb R\to\mathbb R[/imath] such that [imath]f(f(x))=e^x.[/imath] However I have not still been able to come up with an answer. I would like to share this problem with you. I'm not aware of its level, though i wouldn't classify it as homework, so I'm not giving it the homework tag. If anybody feels that this problem is in reality easy or looks like an homework, please feel free to add that tag. EDIT for those interested in the complex case, I've found this on MathOverflow. It's the first answer. Wow! link |
1088054 | Simplifying composition of trigonometric functions
Sometimes when Integrating I end up with taking: [imath]\sin \left(\tan ^{-1}\left(\frac{x}{2}\right)\right)[/imath], or other similar combinations however thats when I use the computer to extract: [imath]\frac{x}{2 \sqrt{\frac{x^2}{4}+1}}[/imath] However, I want to find out how this is done and if there are any tricks I could use to evaluate similar expressions. | 426399 | How to derive compositions of trigonometric and inverse trigonometric functions?
To prove: [imath]\begin{align} \sin({\arccos{x}})&=\sqrt{1-x^2}\\ \cos{\arcsin{x}}&=\sqrt{1-x^2}\\ \sin{\arctan{x}}&=\frac{x}{\sqrt{1+x^2}}\\ \cos{\arctan{x}}&=\frac{1}{\sqrt{1+x^2}}\\ \tan{\arcsin{x}}&=\frac{x}{\sqrt{1-x^2}}\\ \tan{\arccos{x}}&=\frac{\sqrt{1-x^2}}{x}\\ \cot{\arcsin{x}}&=\frac{\sqrt{1-x^2}}{x}\\ \cot{\arccos{x}}&=\frac{x}{\sqrt{1-x^2}} \end{align}[/imath] |
1088076 | Convergence of a particular summation
Does summation [imath]\sum_{n=1}^\infty\frac{1}{n(\log n)^{a}}[/imath] converge if [imath]a>1[/imath]? | 658797 | Show that the series [imath]\sum_{n=2}^{\infty} \frac{1}{n\, (\log n)^s}[/imath] converges for [imath]s >1[/imath] and diverges for [imath]s=1[/imath]
Show that the series [imath]\sum_{n=2}^{\infty} \frac{1}{n\, (\log n)^s}[/imath] converges for [imath]s >1[/imath] and diverges for [imath]s=1[/imath]. |
9911 | Convergence of the series [imath]\sum \limits_{n=2}^{\infty} \frac{1}{n\log^s n}[/imath]
We all know that [imath]\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^s}[/imath] converges for [imath]s>1[/imath] and diverges for [imath]s \leq 1[/imath] (Assume [imath]s \in \mathbb{R}[/imath]). I was curious to see till what extent I can push the denominator so that it will still diverges. So I took [imath]\displaystyle \sum_{n=2}^{\infty} \frac{1}{n\log n}[/imath] and found that it still diverges. (This can be checked by using the well known test that if we have a monotone decreasing sequence, then [imath]\displaystyle \sum_{n=2}^{\infty} a_n[/imath] converges iff [imath]\displaystyle \sum_{n=2}^{\infty} 2^na_{2^n}[/imath] converges). No surprises here. I expected it to diverge since [imath]\log n[/imath] grows slowly than any power of [imath]n[/imath]. However, when I take [imath]\displaystyle \sum_{n=2}^{\infty} \frac{1}{n(\log n)^s}[/imath], I find that it converges [imath]\forall s>1[/imath]. (By the same argument as previous). This doesn't make sense to me though. If this were to converge, then I should be able to find a [imath]s_1 > 1[/imath] such that [imath]\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^{s_1}}[/imath] is greater than [imath]\displaystyle \sum_{n=2}^{\infty} \frac{1}{n (\log n)^s}[/imath] Doesn't this mean that in some sense [imath]\log n[/imath] grows faster than a power of [imath]n[/imath]? (or) How should I make sense of (or) interpret this result? (I am assuming that my convergence and divergence conclusions are right). | 1238695 | For which values of the parameter [imath]p[/imath] the following series is convergent?
For which values of the parameter [imath]p[/imath] the following series is convergent? [imath] \sum_{n=2}^{\infty} \frac{1}{n \ln^p n } [/imath] First, for [imath]p=0[/imath] , the series doesn't converge. For [imath]p<0[/imath] , the series also diverges, since: [imath]\ln^p n < n[/imath] which implies: [imath] \frac{1}{n \ln^p n } \geq \frac{n^{-p}}{n} [/imath] and the series in the RHS diverges, Will you please help me with the case [imath]p>0[/imath] ? Thanks |
1088109 | [imath]\Bbb F_p[/imath] where [imath]p[/imath] is an odd prime has exactly half the non-zero elements as a square
I want to show that in [imath]\Bbb F_p[/imath] where [imath]p[/imath] is an odd prime, that half the non-zero elements are squares. Now I know that all fields [imath]\Bbb F_p[/imath] where [imath]p[/imath] is prime are isomorphic to [imath]\Bbb Z / \Bbb p Z[/imath], and I have tested a few examples and I can see that it holds for them, so I don't doubt it is true, but I am unsure how to prove it. Can I please have a hint, but not a full solution. If you give me a full solution I won't get to have the eureka moment and I won't remember it! | 198875 | Find the number of non-zero squares in the field [imath]Zp[/imath]
Find the number of non-zero elements in the field [imath]Z_p[/imath], where [imath]p[/imath] is an odd prime number, which are squares, i.e. of the form [imath]m^2[/imath]; [imath]m \in Z_p[/imath]; [imath]m \neq 0[/imath]. please help how can i solve this problem? the number of nonzero element is [imath]p-1[/imath] here |
1088206 | Borel Set Example.
In terms of this definition, I still struggle very much with how the Borel [imath]\sigma[/imath]-Algebra is directly related to a Borel Set. Perhaps an example demonstrating how they are directly related? I always see the term B([imath]\mathbb R[/imath]) thrown around, but I still don't see a clear connection between that and the Borel Sets! A Borel Set is an element of the Borel [imath]\sigma[/imath]-Algebra, if that's the case then what does the Borel [imath]\sigma[/imath]-Algebra look like? B([imath]\mathbb R[/imath])= ? Definition: The Borel [imath]\sigma[/imath]-algebra on [imath]\mathbb R[/imath] is the [imath]\sigma[/imath]-algebra B([imath]\mathbb R[/imath]) generated by the [imath]\pi[/imath]-system [imath]\mathcal J[/imath] of intervals [imath]\ (a, b][/imath], where [imath]\ a<b[/imath] in [imath]\mathbb R[/imath] (We also allow the possibility that [imath]\ a=-\infty\ or \ b=\infty[/imath]) Its elements are called Borel sets. For A [imath]\in[/imath] B([imath]\mathbb R[/imath]), the [imath]\sigma[/imath]-algebra [imath]B(A)= \{B \subseteq A: B \in B(\mathbb R)\}[/imath] of Borel subsets of A is termed the Borel [imath]\sigma[/imath]-algebra on A. | 1080473 | Borel [imath]\sigma[/imath]-Algebra definition.
Definition: The Borel [imath]\sigma[/imath]-algebra on [imath]\mathbb R[/imath] is the [imath]\sigma[/imath]-algebra B([imath]\mathbb R[/imath]) generated by the [imath]\pi[/imath]-system [imath]\mathcal J[/imath] of intervals [imath]\ (a, b][/imath], where [imath]\ a<b[/imath] in [imath]\mathbb R[/imath] (We also allow the possibility that [imath]\ a=-\infty\ or \ b=\infty[/imath]) Its elements are called Borel sets. For A [imath]\in[/imath] B([imath]\mathbb R[/imath]), the [imath]\sigma[/imath]-algebra [imath]B(A)= \{B \subseteq A: B \in B(\mathbb R)\}[/imath] of Borel subsets of A is termed the Borel [imath]\sigma[/imath]-algebra on A. I struggle with this part especially "generated by the [imath]\pi[/imath]-system [imath]\mathcal J[/imath] of intervals (a, b]" In addition could someone please provide an example of a Borel set, preferably some numerical interval :) Also is [imath]\mathbb R[/imath] the type of numbers that the [imath]\sigma[/imath]-algebra is acting on? |
607300 | Isomorphic finite abelian groups
Let [imath]G[/imath] and [imath]H[/imath] be finite abelian groups. Show that if for any natural number [imath]n[/imath] the groups [imath]G[/imath] and [imath]H[/imath] have the same number of elements of order [imath]n[/imath], then [imath]G[/imath] and [imath]H[/imath] are isomorphic. I know, that for an infinity group doesn't work : [imath] \Bbb Z_{27}[/imath] It seems to me that I can use finitely-generated abelian group It is possible that this simple fact, but I would ask to write a proof . | 72944 | Two finite abelian groups with the same number of elements of any order are isomorphic
Suppose [imath]G, H[/imath] are finite abelian groups with the same number of elements of any given order. Show they are isomorphic. Since finite abelian groups are isomorphic if and only if their Sylow subgroups are, we may restrict our attention to the case where [imath]G, H[/imath], are [imath]p[/imath]-groups, but I can't quite make it past there... |
1088240 | If [imath]f:\mathbb{C}\to\mathbb{C}[/imath] is Entire and [imath]\vert f \vert\le A+B\vert z \vert^{3/2}[/imath], then [imath]f[/imath] is Linear
I am given that [imath]\vert f \vert\le A+B\vert z \vert^{3/2}[/imath] and I would like to show that [imath]f[/imath] is a linear polynomial. A generalization of Liouville's Theorem says that if [imath]\vert f\vert\le C\vert z \vert^k[/imath] for some positive integer [imath]k[/imath], then [imath]f[/imath] is a polynomial of degree at most [imath]k[/imath]. Since there is a power of [imath]3/2[/imath], this fact doesn't immediately apply. So we can write [imath]\vert fz^{1/2}\vert\le D\vert z^2\vert[/imath] for [imath]\vert z \vert>R[/imath] for some sufficiently large [imath]R[/imath]. Hence [imath]fz^{1/2}[/imath] is a polynomial of degree at most [imath]2[/imath]. Say, [imath]fz^{1/2}=a+bz+cz^2[/imath] ([imath]c[/imath] may be [imath]0[/imath]). Then [imath]f=az^{-1/2}+bz^{1/2}+cz^{3/2}[/imath], which is not a linear polynomial for two reasons (it's not a polynomial in [imath]z[/imath] and it's not linear in [imath]z[/imath]). Where am I going wrong? | 227226 | Entire function bounded by polynomial of degree 3/2 must be linear.
The problem is as follows: Suppose [imath]f[/imath] entire satisfying [imath] |f(z)| \leq A + B |z|^{3/2} [/imath] for some fixed [imath]A,B > 0[/imath]. Prove that [imath]f[/imath] is a linear polynomial. I know I want to reduce it to a function where I can use a Cauchy bound, but I'm not sure how. |
1086409 | Problem regarding the speed of two points [imath]A[/imath] and [imath]B[/imath] moving with constant speed in the plane
Consider a Point A that moves linearly on the positive x-axis with the speed 1 m/s and another Point B at a distance L from A with position (L,0). With each forward motion of point A the Point B moves in an arc consistently maintaining the distance L from point A. As soon as Point A reaches x=L, the Point B is perpendicular to Point A and its position is (L,L). Both the points move at a constant speed. How then do you determine the speed of B? Note that in this problem I considered the Center of Mass of the objects as points. Is it possible to determine the equation of the curve that point B traces and its arc length? If yes, then how? | 1085722 | How to determine the equation and length of this curve consistently formed by the intersection of Circles
Consider a Point [imath]A[/imath] that moves linearly on the positive [imath]x[/imath]-axis with the velocity [imath]1[/imath] m/s and another Point [imath]B[/imath] at a distance [imath]L[/imath] from [imath]A[/imath] with position [imath](L,0)[/imath]. With each forward motion of point [imath]A[/imath] the Point [imath]B[/imath] moves in an arc upward (i.e. along positive [imath]y[/imath]-axis) consistently maintaining the distance [imath]L[/imath] from point [imath]A[/imath]. As soon as Point [imath]A[/imath] reaches [imath]x=L[/imath], the Point [imath]B[/imath] is perpendicular to Point [imath]A[/imath] and its position is [imath](L,L)[/imath]. How do you then determine the equation and length of the path that [imath]B[/imath] traces during the period [imath]A[/imath] traverses from [imath]x=0[/imath] to [imath]L[/imath]? |
1088420 | Equivalent conditions between two ideals and a nilpotent ideal in a ring.
Let [imath]R[/imath] be a ring. Prove that for any two ideals [imath]I[/imath] and [imath]J[/imath] of [imath]R[/imath], the following conditions are equivalent: [imath](1)[/imath] [imath]IJ = (0) \implies I\cap J = (0) [/imath]; [imath](2)[/imath] [imath](0)[/imath] is the only nilpotent ideal of [imath]R[/imath]. I am able to prove [imath]1 \implies 2[/imath]. Let us assume that there exist another ideal [imath]K \neq (0) [/imath] and [imath]K[/imath] is nil-potent i.e. [imath]K^n = (0)[/imath] for some n. Then [imath]K^{n-1}K= (0)[/imath] and using the statement [imath](1)[/imath] we have [imath]K^{n-1} \cap K =(0)[/imath] and since [imath]K[/imath] being an ideal [imath]K^{n-1} \subset K[/imath]. Thus [imath]K^{n-1} = (0)[/imath]. In the same manner after going [imath]n-1[/imath] steps we can show that [imath]K = (0)[/imath]. Thus [imath](0)[/imath] is only the nil-potent ideal of [imath]R[/imath]. But I am not able to proof [imath]2\implies 1 [/imath]. Please help!! Thank You. | 992104 | An equivalent condition with [imath]\{0\}[/imath] being the only nilpotent ideal
In a ring [imath]R[/imath] prove that [imath]\{0\}[/imath] is the only nilpotent ideal if and only if for every ideals [imath]A[/imath] and [imath]B[/imath] from [imath]R[/imath], [imath]AB=\{0\}[/imath] implies [imath]A\cap B=\{0\} [/imath]. |
1088423 | How do you prove the eigenvalues and eigenvectors of matrices are the same?
I wondered if anyone could help me with a couple of proofs. The question is: Let [imath]A[/imath] be a given [imath]n \times m[/imath] matrix. The collection of scalars [imath]\lambda_i[/imath] and associated [imath]n \times 1[/imath] vectors [imath]q_i[/imath] that solve the equation [imath]Aq=\lambda q[/imath] are known as eigenvalues and eigenvectors of [imath]A[/imath], respectively show that: i. Suppose [imath]n=m[/imath]. Then for any non-singular [imath]n \times n[/imath] square matrix [imath]G[/imath] the eigenvalues of [imath]G^{-1}AG[/imath] are the same as those of [imath]A[/imath]. ii. If [imath]A^{-1}[/imath] exists then it shares the same eigenvectors [imath]q_i[/imath] as [imath]A[/imath] with corresponding eigenvalues [imath]\lambda_i^{-1}[/imath] Thanks in advance! | 110629 | similar matrices have same eigenvalue... stuck
So [imath]Ax=\lambda x[/imath] where [imath]A[/imath] is the matrix, [imath]\lambda[/imath] and [imath]x[/imath] its eigenvalue and eigenvector resptectively. Now I have another matrix similar to [imath]A[/imath] i.e. [imath]B=TAT^{-1}[/imath] I'll let a vector [imath]y[/imath] which is an eigenvector of [imath]B[/imath], i.e. [imath]By=\mu y[/imath] where [imath]\mu[/imath] is its ([imath]B[/imath]'s) eigenvalue. I have to now prove that [imath]By=\mu x[/imath]. Some maths leads me to [imath]By = TAT^{-1}y[/imath] but here I'm stuck. This answers wants me to assume [imath]y=Tx[/imath] which will essentially solve it but why/how is [imath]y=Tx[/imath]? |
1088422 | Integral [imath]\int_0^\infty \frac{\sqrt[3]{x+1} - \sqrt[3]{x}}{\sqrt{x}} \, \mathrm dx[/imath]
I have the following integral to solve: [imath]\int_0^\infty \frac{\sqrt[3]{x+1} - \sqrt[3]{x}}{\sqrt{x}} \, \mathrm dx[/imath] I tried substitution [imath]u= x+1[/imath] and [imath]u=\sqrt[3]{x+1}[/imath] then partial integration but that didn't really help because it didn't become any simpler. Thank you. | 159223 | Computing [imath] \int_0^{\infty} \frac{ \sqrt [3] {x+1}-\sqrt [3] {x}}{\sqrt{x}} \mathrm dx[/imath]
I would like to show that [imath] \int_0^{\infty} \frac{ \sqrt [3] {x+1}-\sqrt [3] {x}}{\sqrt{x}} \mathrm dx = \frac{2\sqrt{\pi} \Gamma(\frac{1}{6})}{5 \Gamma(\frac{2}{3})}[/imath] thanks to the beta function which I am not used to handling... [imath]\frac{2\sqrt{\pi} \Gamma(\frac{1}{6})}{5 \Gamma(\frac{2}{3})}=\frac{2}{5}B(1/2,1/6)=\frac{2}{5} \int_0^{\infty} \frac{ \mathrm dt}{\sqrt{t}(1+t)^{2/3}}[/imath] ...? |
1088481 | [imath]\int_{0}^{\infty}\frac{\sqrt{x}}{1+x^2}\,dx[/imath]
I want to evaluate the integral [imath]\displaystyle \int_{0}^{\infty}\frac{\sqrt{x}}{1+x^2}\,dx[/imath] using complex analysis methods. I know that I have to use a keyhole contour, but I don't know which function to integrate on the contour. For example if I had to integrate [imath]\displaystyle \int_{0}^{\infty}\frac{\ln x}{1+x^2}\,dx[/imath] then I would have chosen to integrate over a keyhole contour the function [imath]\displaystyle f(z)=\frac{\ln^2 z}{1+z^2}\,dx[/imath]. If i had known that function then the rest is routine, since if we declare the given integral as [imath]I[/imath] and the contour integral as [imath]J[/imath] , then there is a simple relationship of [imath]I, \; J[/imath]. It holds that [imath]J=2\pi i I [/imath]. Hence we have the result. However, I did not understand well the keyhole contour. If someone could show me how the other parts disappear I would be glad. | 97653 | What contour should be used to evaluate [imath]\int_0^\infty \frac{\sqrt{t}}{1+t^2} dt[/imath]
Could anyone help me decide what contour to use to evaluate this integral? [imath]\int_0^\infty \frac{\sqrt{t}}{1+t^2} dt[/imath] So we have simple poles at [imath]i[/imath],[imath]-i[/imath]. Why does using a quarter of a circle in the upper right quadrant not work -is it a problem having [imath]i[/imath] on the contour of integration? My second idea was to let [imath]\sqrt{t}=e^{\frac{1}{2}\operatorname{log}t}[/imath] And defining the branch of the logarithm to be the negative imaginary axis. Then using an upper semi-circular contour. With a hole at 0, which will tend to the point 0 as the radius of the semi circle tends to infinity. Then we'll divide by two to get the integral from [imath]0[/imath] to [imath]\infty[/imath]. Does this work, as this does not seem to be an even function we're integrating. I apolygise for the poor explanation of my contours, it's quite difficult without pen and paper! EDIT: Ok, so using an upper semi circle, with a semicircular hole about [imath]0[/imath]. let us call the contour [imath]\gamma[/imath], we have: [imath]\int_\gamma \frac{e^{\frac{1}{2}\operatorname{log}t}}{1+t^2}dt =2\pi i\Big(\frac{e^{\frac{1}{2}\operatorname{log}t}}{t+i}\Big)\Big|_{t=i}=\frac{e^{\pi/4}}{2i}[/imath] But by letting on the positive real axis [imath]z=x+iy[/imath] implies [imath]log(z)=log(x)+0[/imath] and on the negative axis [imath]log(z)=log(x)+\pi[/imath] Therefore if we tend the large arc's radius to infinity and the small arc about the origin's radius to zero we find the integral along the arc tends to zero. So [imath]\int_\gamma \frac{e^{\frac{1}{2}\operatorname{log}t}}{1+t^2}dt=\int_0^\infty \frac{\sqrt{x}}{1+x^2} dx+\int_0^\infty \frac{\sqrt{x}e^{\pi/2}}{1+x^2} dx=(1+e^{\pi/2})\int_0^\infty \frac{\sqrt{x}}{1+x^2} dx[/imath] [imath]\Rightarrow \int_0^\infty \frac{\sqrt{x}}{1+x^2} =\frac{e^{\pi/4}}{2i(1+e^{\pi/2})}[/imath] This is wrong but can anyone point out what mistake I have made here? |
1054423 | Proving [imath]f\in K(\beta)[X][/imath] irreducible iff [imath]g\in K(\alpha)[X][/imath] irreducible
Let [imath]K\subset L[/imath] a field extension and [imath]\alpha,\beta\in L[/imath] with minimal polynomials [imath]f,g\in K[X][/imath]. How to show that [imath]f\in K(\beta)[X][/imath] is irreducible iff [imath]g\in K(\alpha)[X][/imath] is irreducible? | 900877 | Proving that [imath]f(x)[/imath] is irreducible over [imath]F(b)[/imath] if and only if [imath]g(x)[/imath] is irreducible over [imath]F(a)[/imath]
Let [imath]f(x)[/imath] and [imath]g(x)[/imath] be irreducible polynomials over a field [imath]F[/imath] and let [imath]a,b \in E[/imath] where [imath]E[/imath] is some extension of [imath]F[/imath]. If [imath]a[/imath] is a zero of [imath]f(x)[/imath] and [imath]b[/imath] is a zero of [imath]g(x)[/imath], show that [imath]f(x)[/imath] is irreducible over [imath]F(b)[/imath] if and only if [imath]g(x)[/imath] is irreducible over [imath]F(a)[/imath]. Attempt: Since [imath]f(x),g(x)[/imath] are irreducible over [imath]F \implies a,b \notin F[/imath]. [imath]f(x)[/imath] is irreducible over [imath]F(b) [/imath] and [imath]f(x)[/imath] is irreducible over [imath]F \implies a \neq b[/imath] (As, [imath]a[/imath] is the zero of [imath]f(x)[/imath]) Which means [imath]b \notin F(a)[/imath] either [imath]\implies g(x)[/imath] is irreducible over [imath]F(a)[/imath]. Similarly, the other half can be proved in a similar way. Is my solution attempt correct? Thank you for your help.. |
1088684 | how does one integrate [imath]\int \sin(2x)\cos(4x)dx[/imath]
how does one integrate [imath]\int \sin(2x)\cos(4x)dx[/imath]? im looking for tips and hints on the matter ... | 703805 | Evaluate [imath]\int \cos(3x) \sin(2x) \, dx[/imath].
Evaluate the indefinite integral \begin{align} \int \cos(3x) \sin(2x) \, dx \end{align} Please see my work attempt as my answer below. |
1088796 | Proof that difference of compact and closed sets is also closed
I am trying to prove that for any [imath]A[/imath] compact, [imath]B[/imath] closed sets [imath]\Rightarrow A-B = \{a-b | a\in A, b\in B\}[/imath] is also closed, where A and B are subsets of a topological vector space [imath]X[/imath]. | 515496 | Sum of closed and compact set in a TVS
I am trying to prove: [imath]A[/imath] compact, [imath]B[/imath] closed [imath]\Rightarrow A+B = \{a+b | a\in A, b\in B\}[/imath] closed (exercise in Rudin's Functional Analysis), where [imath]A[/imath] and [imath]B[/imath] are subsets of a topological vector space [imath]X[/imath]. In case [imath]X=\mathbb{R}[/imath] this is easy, using sequences. However, since I was told that using sequences in topology is "dangerous" (don't know why though), I am trying to prove this without using sequences (or nets, which I am not familiar with). Is this possible? My attempt was to show that if [imath]x\notin A+B[/imath], then [imath]x \notin \overline{A+B}[/imath]. In some way, assuming [imath]x\in\overline{A+B}[/imath] should then contradict [imath]A[/imath] being compact. I'm not sure how to fill in the details here though. Any suggestions on this, or am I thinking in the wrong direction here? |
1088182 | A group of order [imath]pqr[/imath] (primes [imath]p > q > r[/imath]) has a subgroup of order [imath]qr[/imath]
I've done most of the following problem, but I can't seem to get part (d). Let [imath]G[/imath] be a group of order [imath]pqr[/imath] for primes [imath]p > q > r[/imath]. By a counting argument one can see that there is either a normal Sylow [imath]p[/imath]-subgroup or a normal Sylow [imath]q[/imath]-subgroup (you may assume this). Prove: (a) [imath]G[/imath] has a normal subgroup [imath]H[/imath] of order [imath]pq[/imath]. (b) Every subgroup of order [imath]p[/imath] or [imath]q[/imath] is contained in [imath]H[/imath]. (c) [imath]G[/imath] has a normal subgroup of order [imath]p[/imath]. (d) [imath]G[/imath] has a subgroup of order [imath]qr[/imath]. Proof: Let me first state the following lemma: if [imath]N_0[/imath] is a subgroup of a finite group [imath]G_0[/imath], and [imath][G_0 : N_0][/imath] is equal to the smallest prime which divides [imath]|G_0|[/imath], then [imath]N_0[/imath] is normal in [imath]G_0[/imath]. The proof is not difficult. It can be established by a counting argument, or by looking at the natural homomorphism from [imath]G_0[/imath] to the symmetric group on the left cosets of [imath]N_0[/imath] in [imath]G_0[/imath]. For (a), let [imath]N[/imath] be a Sylow [imath]p[/imath]-subgroup, and [imath]K[/imath] a Sylow [imath]q[/imath]-subgroup. One of these subgroups is normal by assumption. So [imath]H := NK[/imath] is a subgroup of [imath]G[/imath] with order [imath]pq[/imath]. Its index in [imath]G[/imath] is [imath]r[/imath], so it is normal in [imath]G[/imath] by the lemma. (b) If [imath]P[/imath] is a subgroup of order [imath]p[/imath], it is a Sylow [imath]p[/imath]-subgroup, so it equals [imath]gNg^{-1}[/imath] for some [imath]g \in G[/imath]. So [imath]g^{-1}Pg = N \subseteq H[/imath], whence [imath]P \subseteq gHg^{-1} = H[/imath]. Similarly for a subgroup of order [imath]q[/imath]. (c) Let [imath]P[/imath] be a subgroup of order [imath]p[/imath]. Then it is a Sylow [imath]p[/imath]-subgroup and it is contained in [imath]H[/imath]. Since [imath]|H| = pq[/imath], and [imath][H : P] = q[/imath], automatically [imath]P[/imath] is normal in [imath]H[/imath]. If [imath]P_1[/imath] is another Sylow [imath]p[/imath]-subgroup of [imath]G[/imath], then it is also contained in and normal in [imath]H[/imath], whence [imath]P = P_1[/imath]. (d) Let [imath]Q[/imath] be a Sylow [imath]q[/imath]-subgroup, and [imath]R[/imath] a Sylow [imath]r[/imath]-subgroup. If [imath]Q[/imath] is normal in [imath]G[/imath], we are done, since [imath]QR[/imath] is a subgroup of order [imath]qr[/imath]. Otherwise, [imath]n_q[/imath] (the number of Sylow [imath]q[/imath] subgroups) is either [imath]r, p,[/imath] or [imath]pr[/imath]. We cannot have [imath]n_q = r[/imath], since then [imath]q[/imath] divides [imath]r-1 < q[/imath]. If [imath]n_q = p[/imath], then [imath]|N_G(Q)| = qr[/imath] and we are done. The last possibility is that [imath]n_q = pr[/imath]. I don't know what to do here. I've tried making a counting argument involving the possibilities for [imath]n_r[/imath], but I haven't been able to make anything work. | 442289 | Group of order [imath]pqr[/imath], [imath]p < q < r[/imath] primes
Let [imath]G[/imath] be a group such that [imath]|G|=pqr[/imath], [imath]p<q<r[/imath] and [imath]p,q,r[/imath] are primes. i need to prove that: There exists a subgroup [imath]H[/imath] such that [imath]H\unlhd G[/imath] and [imath]|H|=qr[/imath]. [imath]G[/imath] is solvable. [imath]r[/imath]-Sylow subgroup of [imath]G[/imath] is normal. I know the proof that [imath]G[/imath] isn’t a simple group, and by that I think I can prove the second part of the question, but I can’t manage to prove the first and the third parts of the question. |
1088734 | How to find integral of [imath]\sin(x)x^2[/imath]
It's possible the integral bellow. What way I must to use for solve it. [imath]\int \sin(x)x^2dx[/imath] | 486062 | Integration of [imath]x^2 \sin(x)[/imath] by parts
How would I integrate the following? [imath]\int_0^{\pi/2} x^2\sin(x)\,dx[/imath] I did [imath]u=x^2[/imath] and [imath]dv=\sin(x)[/imath] I got [imath]x^2-\cos(x)+2\int x\cos(x)\,dx.\quad[/imath] I then used [imath]u=x[/imath] and [imath]dv=\cos(x).[/imath] I got [imath]x^2-\cos(x)+2[x-\sin(x)-\int\sin(x)][/imath] then [imath]x^2-\cos(x)+-2 \sin(x)(x)-\cos(x)\Big|_0^{\pi/2} =\dfrac{\pi^2}{4}-0-2[/imath] |
450028 | Is there a reference for compact imbedding of Hölder space?
Suppose [imath]0<\alpha <\beta[/imath]. Then, the Hölder space [imath]C^\beta[/imath] is compactly imbedded to [imath]C^\alpha[/imath]. See the Wikipedia article Hölder condition. However, I could not find precise reference from some books on functional analysis. Can anybody indicate a precise reference for this theorem? If possible, I would like to know a reference on the similar result on parabolic Hölder space. | 1806349 | Showing any bounded sequence in Holder space [imath]C^{1/2}[/imath] has a convergent subsequence in Holder space [imath]C^{1/3}.[/imath]
Prove that any bounded sequence in [imath]C^{1/2}([0,1])[/imath] admits a convergent subsequence in [imath]C^{1/3}([0,1]),[/imath] where we say that [imath]f \in C^{\alpha}([0,1])[/imath] if [imath]f[/imath] is Holder continuous of order [imath]\alpha.[/imath] The norm defined on [imath]C^{\alpha}([0,1])[/imath] is [imath] ||f||_{C^\alpha} = \sup{\{|f(x)| : x \in [0,1]\}} + \sup{\{\frac{|f(x) - f(y)|}{|x-y|^\alpha} : x,y \in [0,1] \text{ and } x \neq y \}}. [/imath] I'm pretty new to these types of problems, I've tried to work things out and write down what I think everything means: Letting [imath]\{f_n\}_{n=1}^\infty[/imath] be a sequence in [imath]C^{1/2}([0,1]),[/imath] the sequence is bounded if for all [imath]n[/imath], [imath]||f_n||_{C^{1/2}} < M[/imath] for some constant [imath]M < \infty.[/imath] Letting [imath]g_n = f_{\sigma(n)}[/imath] (the function [imath]\sigma(n)[/imath] just picks out the subsequence) be the elements of the subsequence of [imath]\{f_n\}_{n=1}^\infty,[/imath] this subsequence [imath]\{g_n\}_{n=1}^\infty[/imath] is convergent in [imath]C^{1/3}[/imath] if for [imath]\epsilon > 0, \exists N \text{ such that for } n,m>N, ||g_n - g_m||_{C^{1/3}} < \epsilon.[/imath] In other words, its required that [imath] \sup{\{|g_n(x) - g_m(x)| : x \in [0,1]\}} + \sup{\{\frac{|g_n(x) - g_m(x) - g_n(y) + g_m(y)|}{|x-y|^\alpha}}\} < \epsilon [/imath] for all [imath]n,m > N.[/imath] Im having a hard time seeing how boundedness in one space permits a convergent subsequence in another space. Any hints/solutions or explanations of why this makes sense to begin with would be greatly appreciated. |
1084891 | Why rationalize the denominator?
In grade school we learn to rationalize denominators of fractions when possible. We are taught that [imath]\frac{\sqrt{2}}{2}[/imath] is simpler than [imath]\frac{1}{\sqrt{2}}[/imath]. An answer on this site says that "there is a bias against roots in the denominator of a fraction". But such fractions are well-defined and I'm failing to see anything wrong with [imath]\frac{1}{\sqrt{2}}[/imath] - in fact, IMO it is simpler than [imath]\frac{\sqrt{2}}{2}[/imath] because 1 is simpler than 2 (or similarly, because the former can trivially be rewritten without a fraction). So why does this bias against roots in the denominator exist and what is its justification? The only reason I can think of is that the bias is a relic of a time before the reals were understood well enough for mathematicians to be comfortable dividing by irrationals, but I have been unable to find a source to corroborate or contradict this guess. | 26080 | No radical in the denominator -- why?
Why do all school algebra texts define simplest form for expressions with radicals to not allow a radical in the denominator. For the classic example, [imath]1/\sqrt{3}[/imath] needs to be "simplified" to [imath]\sqrt{3}/3[/imath]. Is there a mathematical or other reason? And does the same apply to exponential notation -- are students expected to "simplify" [imath]3^{-1/2}[/imath] to [imath]3^{1/2}/3[/imath] ? |
1089419 | trace of a matrix in [imath]\mathbb{Z}_{227}[/imath]
Let [imath]A[/imath] be a [imath]227 \times 227[/imath] matrix with entries from [imath]\mathbb{Z}_{227}[/imath]. If A has distinct eigenvalues what is the trace of [imath]A[/imath]? I am guessing the answer is zero. Since the eigenvalues of [imath]A[/imath] need not be in [imath]\mathbb{Z}_{227}[/imath], I dont know how to prove this. Give me some hint to prove this. Thanks in advance. | 479077 | Find the trace of the matrix?
Let [imath]A[/imath] be a [imath]227\times227[/imath] matrix with entries in [imath]\mathbb{Z}_{227}[/imath], such that all the eigenvalues are distinct. What is the trace of [imath]A[/imath]? |
173843 | Trace of a [imath]227\times 227[/imath] matrix over [imath]\mathbb{Z}_{227}[/imath]
well, I know that the trace is the negative of coefficient of [imath]x^{226}[/imath] of the characteristic polynomial the matrix, but I dont know how the Char.Poly looks like in this case.please give me some hint. Do I have to work in the splitting field of the characteristic polynomial and add the eigen values to get the trace?but I dont know how. | 143678 | Find out trace of a given matrix [imath]A[/imath] with entries from [imath]\mathbb{Z}_{227}[/imath]
Let [imath]A[/imath] be a [imath]227\times 227[/imath] matrix with entries in [imath]\mathbb{Z}_{227}[/imath] such that all of its eigen values are distinct. What would be its trace? I think it is zero by adding all 227 elements but i am not sure. Edited: Here I have assumed that eigenvalues are in a base field. |
1089458 | How can I prove the convergence of a power-tower?
In here, I saw that [imath]x^{x^{x^{x^{x^{x^{x^{.{^{.^{.}}}}}}}}}}[/imath] exists as a real number (convergent) if and only if [imath]x\in[e^{-e}, e^\frac{1}{e}].[/imath] How can I prove this?? | 890319 | Convergence of tetration sequence.
This question arose from here. I am interested to find a nice proof about the convergence of [imath]{^n}a=\underbrace{a^{a^{\ .^{\ .^{\ .^a}}}}}_{n\ \text{times}}.[/imath] I find with google a necessary and sufficient condition to have the convergence is [imath]\frac{1}{e^e} \leq a \leq e^{1/e}[/imath] but the part for [imath]a\le1[/imath] need some ugly work. Does anyone have an elegant/slick proof ? |
1090080 | Show the map from the product of disjoint subgroups of G to G is an isomorphism
Question: Show that if [imath]G[/imath] is a group with two normal subgroups [imath]H[/imath] and [imath]K[/imath] such that [imath]G=HK[/imath] and [imath]H\cap K=\{e\}[/imath], then the map [imath](h,k)\mapsto hk[/imath] is an isomorphism of groups from [imath]H\times K[/imath] to [imath]G[/imath] Comments: I would like to know where I can find a proof to this statement as it seems like a standard result. To me the statement seems intuitively obvious since [imath]H[/imath] will look like [imath]\{e,h_1,...,h_n\}[/imath] and [imath]K=\{e,k_1,...,k_m\}[/imath] and [imath]G=\{e,h_1,...,h_n,k_1,h_1k_1,...,h_nk_1,...,k_m,h_1k_m,...,h_nk_m\}[/imath] Therefore [imath]K[/imath] will have [imath]m+1[/imath] elements, [imath]H[/imath] will have [imath]n+1[/imath] elements and [imath]G[/imath] will have [imath](m+1)(n+1)[/imath] elements and there would be a 1-1 and onto mapping -but how do I prove this? The homomorphism property comes from taking [imath]a=(h_1,k_1)[/imath] and [imath]b=(h_2,k_2)[/imath] Then [imath]\varphi(ab)=\varphi(h_1h_2,k_1k_2)=h_1h_2k_1k_2=h_1k_1h_2k_2=\varphi(h_1,k_1)\varphi(h_2,k_2)=\varphi(a)\varphi(b)[/imath] This is my understanding of this question, how should I prove the bijective part of this map, also if possible what is the name of this proof or where can I find one. | 1089114 | Group Theory Isomorphism
Show that if [imath]G[/imath] is a group with two normal subgroups [imath]H[/imath] and [imath]K[/imath] such that [imath]G=HK[/imath] and [imath]H\cap K=\{e\}[/imath] then the map [imath](h,k)\rightarrow hk[/imath] is an isomorphism of groups from [imath]H\times K[/imath] to [imath]G[/imath]. I am not sure whether or not I should be using the homomorphism theorem or not. (If [imath]j:G\rightarrow H[/imath] is a homomorphism then [imath]G/\text{Ker}(j)[/imath] is isomorphic to [imath]\text{Im}(j)[/imath]) Thanks |
1081477 | Measurability of a function based on the slices
If [imath]X: \Omega_1 \times \Omega_2 \to \mathbb{R}[/imath] is measurable then each of the slices [imath]X_{\omega_1} : \Omega_2 \to \mathbb{R}[/imath], [imath]X_{\omega_1} : \Omega_2 \to \mathbb{R}[/imath] defined by [imath]X_{\omega_1}(\omega_2) = X(\omega_1,\omega_2)[/imath] etc. are measurable. I guess that the converse need not be true. Is it true that if [imath]X_{\omega_1}[/imath] is [imath]\textit{continuous}[/imath] for all [imath]\omega_1[/imath] while [imath]X_{\omega_2}[/imath] is [imath]\textit{measurable}[/imath] for all [imath]\omega_2[/imath] (e.g. for [imath]\Omega_1 = \Omega_2 = \mathbb{R}^d[/imath]) then [imath]X[/imath] is measurable? More generally, if [imath]X : \Omega_1 \times \Omega_2 \cdots \Omega_n \to \mathbb{R}[/imath] and [imath]X_{\omega_1},X_{\omega_2},\cdots,X_{\omega_{n-1}}[/imath] are continuous and [imath]X_{\omega_n}[/imath] is measurable can we say that [imath]X[/imath] is measurable? Thanks, Phanindra | 661087 | A function which is continuous in one variable and measurable in other is jointly measurable
Please help me to prove the following; Let [imath] \ f:[0,1]^2\longrightarrow\mathbb{R}[/imath] be such that: (i) [imath]\ f(x,\cdot)[/imath] is measurable for each fixed [imath]x\in[0,1][/imath]; (ii) [imath]\ f(\cdot,y)[/imath] is continuous for each fixed [imath]y\in[0,1][/imath]. Show that [imath]\ f[/imath] is measurable. |
763179 | How to prove that this sequence is Cauchy
How can I prove that [imath]f_n(x)=\left|x-\frac12\right|^{(n+1)/n}[/imath] is a Cauchy sequence in [imath]C^1[0,1][/imath] with sup norm? It seems obvious when we look at graphs but I need this formal way.I need this for my proof that [imath]C^1[0,1][/imath] with sup norm is not a Banach space. | 764245 | [imath]C^1[0,1][/imath] is not complete with respect to sup norm
I think the right sequence is [imath]f_n(t)=|t-\frac{1}{2}|^{(n+1)/n}[/imath] but I can't manage to prove it's Cauchy... (when I look at graph of [imath]f_n[/imath]s it seems obvious, but I need it formal way. |
1090428 | Solution for recurrence [imath]T(n+1) = T(n) + \lfloor \sqrt{n+1}\rfloor [/imath]
ould someone please give me an idea as to how the solve the following. [imath]T(1) = 1[/imath] [imath]T(n+1) = T(n) + \lfloor\sqrt{n+1}\rfloor[/imath] I converted the recurrence to [imath]T(n) = T(n-1) + \lfloor\sqrt{n}\rfloor[/imath] and then tried to solve it using the method taught in the "linear recurrences solving from Mathematics for computer science course" at MIT and got the solution as [imath]T(n) = n^\left(3/2\right) + 1[/imath]. But the page I found this problem in asks for [imath]T(n^2)[/imath] and gives the answer as [imath]\frac n6\left(4n^2 - 3n + 5\right)[/imath], which I'm not able to derive. Can someone please help me solve this recurrence? Thanks | 1075820 | If [imath]T(n+1)=T(n)+\lfloor \sqrt{n+1}, \rfloor[/imath] [imath]\forall n\geq 1[/imath], what is [imath]T(m^2)[/imath]?
[imath]T(n+1)=T(n)+\lfloor \sqrt{n+1} \rfloor[/imath] [imath]\forall n\geq 1[/imath] [imath]T(1)=1[/imath] The value of [imath]T(m^2)[/imath] for m ≥ 1 is? Clearly you cannot apply master theorem because it is not of the form [imath]T(n)=aT(\frac{n}{b})+f(n)[/imath] So I tried Back Substitution: [imath]T(n)=T(n-1)+\sqrt{n}[/imath] [imath]T(n-1)=T(n-2)+\sqrt{n-1}[/imath] therefore, [imath]T(n)=T(n-2)+\sqrt{n-1}+\sqrt{n}[/imath] [imath]T(n)=T(n-3)+\sqrt{n-2}+\sqrt{n-1}+\sqrt{n}[/imath] . . [imath]T(n)=T(n-(n-1))+...T(n-k)+\sqrt{n-(k-1)}+...+\sqrt{n-2}+\sqrt{n-1}+\sqrt{n}[/imath] . . [imath]T(n)=T(1)+...T(n-k)+\sqrt{n-(k-1)}+...+\sqrt{n-2}+\sqrt{n-1}+\sqrt{n}[/imath] [imath]T(n)=T(1)+...+\sqrt{n-2}+\sqrt{n-1}+\sqrt{n}[/imath] I'm stuck up here, how to solve this ahead? |
1090510 | Cauchy Continuous function is continuous??
Suppose [imath]X[/imath] and [imath]Y[/imath] be metric spaces.Let [imath]f:X\to Y[/imath] be function which is Cauchy Continuous.Show that [imath]f[/imath] is continuous. Cauchy Continuity:Let [imath]X[/imath] and [imath]Y[/imath] be metric spaces, and let [imath]f[/imath] be a function from [imath]X[/imath] to [imath]Y[/imath]. Then [imath]f[/imath] is Cauchy-continuous if and only if, given any Cauchy sequence [imath]<x_n>[/imath] in X, the sequence [imath]<f(x_n)>[/imath] is a Cauchy sequence in Y. I am planning to use sequential definition of Continuity but I am unable to produce a neat proof without completeness.Give me some idea! | 344452 | Is a Cauchy sequence - preserving (continuous) function is (uniformly) continuous?
Let [imath](X,d)[/imath] and [imath](Y,\rho)[/imath] be metric spaces and [imath]f:X\to Y[/imath] be a function and suppose for any Cauchy sequence [imath](a_n)[/imath] in [imath]X[/imath], [imath](f(a_n))[/imath] is a Cauchy sequence in [imath]Y[/imath]. Is [imath]f[/imath] continuous? Let [imath]f[/imath] be continuous, is it uniformly continuous? |
991291 | Let V be a vector space. If every subspace of V is T-invariant, prove that there exist a scalar multiple c such that T=c1v
I wrote a proof for the above question, but I am not sure whether it is right or not since I assumed linear independence. Here's the proof: Let [imath]u[/imath],[imath]v[/imath] be linearly independent vectors in [imath]V[/imath]. [imath]span(u)[/imath], [imath]span(v)[/imath], [imath]span(u+v)[/imath] are all [imath]T-invariant[/imath]. [imath]T(v)[/imath] is an element in [imath]span(v) \Longrightarrow T(v) = av[/imath] [imath]T(u)[/imath] is an element in [imath]span(u) \Longrightarrow T(u) = bu[/imath] [imath]T(u+v)[/imath] is an element in [imath]span(u+v) \Longrightarrow T(u+v) = T(u)+T(v)= bu+av = c(u+v)[/imath] Hence, [imath]bu+av-c(u+v) = 0 \Longrightarrow (b-c)u + (a-c)v = 0[/imath] Since u and v are linearly independent, [imath](b-c)=(a-c)=0 \Longrightarrow b=a[/imath] Hence, [imath]T(w) = kw[/imath], for all [imath]w[/imath] in [imath]V[/imath], and [imath]k[/imath] any scalar. Is it right to assume linear independence? And is there any problem with my proof? | 146416 | Demonstration: If all vectors of [imath]V[/imath] are eigenvectors of [imath]T[/imath], then there is one [imath]\lambda[/imath] such that [imath]T(v) = \lambda v[/imath] for all [imath]v \in V[/imath].
Let [imath]T: V \rightarrow V[/imath] be a linear operator. I need to demonstrate that if all nonzero vectors of [imath]V[/imath] are eigenvectors of [imath]T[/imath], then there is one specific [imath]\lambda \in K[/imath] such that [imath]T(v) = \lambda v[/imath], for all [imath]v \in V[/imath]. I understand that, if all nonzero vectors of [imath]V[/imath] are eigenvectors of [imath]T[/imath], then [imath]T[/imath] must be a scaling transformation. It just stretch or shrinks vectors, but doesn't change their directions. So, the statement says that if it happens, then, there is a single [imath]\lambda[/imath] such that [imath]T(v) = \lambda v[/imath]. In other words, if there is such transformation, then it scales all vectors by the same scalar [imath]\lambda[/imath]. Applying the transformation to our standard basis vectors, we have: [imath] T(e_1) = \lambda_1 e_1 \\ T(e_2) = \lambda_2 e_2 \\ \vdots \\ T(e_n) = \lambda_n e_n [/imath] I understand I need to prove that [imath]\lambda_1 = \lambda_2 = \dots = \lambda_n[/imath], but I can't see how! EDIT [imath] v = c_1e_1 + c_2e_2 + \dots + c_ne_n \\ T(v) = \mu v = \lambda_1c_1e_1 + \lambda_2c_2e_2 + \dots + \lambda_nc_ne_n \\ [/imath] Since what's multiplying [imath]v[/imath] coordinates is [imath]\lambda_i[/imath], then all of them must be [imath]\mu[/imath]. I'm not sure how to 'mathematize' this. Is this idea correct? EDIT 2 Extending the left hand side of EDIT 1, we have: [imath] \mu v = \lambda_1c_1e_1 + \dots + \lambda_nc_ne_n \\ \mu(c_1e_1 + \dots + c_ne_n) = \lambda_1c_1e_1 + \dots + \lambda_nc_ne_n \\ \mu c_1e_1 + \dots + \mu c_ne_n = \lambda_1c_1e_1 + \dots + \lambda_nc_ne_n \\ [/imath] And since [imath]e_i[/imath] are linearly independent, [imath]\mu = \lambda_1 = \lambda_2 = \dots = \lambda_n[/imath]. Is this proof correct? |
284926 | Proving metric( the triangle inequality)
In complex plane [imath]\mathbb{C}[/imath], how to prove that [imath]d(z_1,z_2)=\frac{|z_2-z_1|}{\sqrt{1+|z_1|^2}\sqrt{1+|z_2|^2}}[/imath] is a metric? I got stuck in the triangle inequality, and have no idea of proving it, any hint or solutions? Thanks for replying! | 1090222 | How to show that the spherical metric satisfies the triangle inequality?
For [imath]x,y\in \mathbb R^n[/imath] define [imath]d(x,y)={\|x-y\| \over \sqrt{1+\|x\|^2} \sqrt{1+\|y\|^2}}[/imath] Here [imath]\|x\|[/imath] is the euclidean norm of a vector. How to prove that [imath]d[/imath] (the spherical metric) is indeed a metric? Progress so far: [imath]d(x,y)\ge 0[/imath] is obvious. [imath]d(x,y) =0 \iff \|x-y\|=0 \iff x=y[/imath], so the positivity holds. [imath]d(x,y) = d(y,x)[/imath] is clear from the formula, so symmetry holds. But I am having difficulties with the triangle inequality. Writing it out in coordinates leads to a complicated inequality with square roots all over the place in denominators. Is there a better way? |
1091072 | Given a matrix [imath]A[/imath] of rank [imath]n[/imath], show that [imath]\det(\operatorname{adj}(A))=\det(A)^{n-1}[/imath] and [imath]\operatorname{adj}(\operatorname{adj}(A))=(\det A)^{n-2}A[/imath]
[imath]\newcommand{\adj}{\operatorname{adj}}\newcommand{\rank}{\operatorname{rank}}[/imath]If [imath]\adj(A)[/imath] denotes the classical adjoint and we are given that, for an [imath]n \times n[/imath] matrix [imath]A[/imath] over [imath]\mathbb{R}[/imath], [imath]\rank(\adj(A)) = \begin{cases} n& \rank(A)=n\\1& \rank(A)=n-1\\0& \rank(A)<n-1\end{cases}[/imath] show that for [imath]n \geq 2[/imath] [imath]\det(\adj(A)) = \det(A)^{n-1};[/imath] [imath]\adj(\adj(A)) = \det(A)^{n-2}A.[/imath] The first is easy, actually. If [imath]\rank(A) = n[/imath], [imath]\adj(A)A = \det(A)I[/imath]; taking determinants of both sides easily yields the result; if [imath]\rank(A) < n[/imath], [imath]\det(A) = 0 = \det(\adj(A))[/imath]. | 649187 | Prove that if [imath]A[/imath] is regular then [imath]\operatorname{adj}(\operatorname{adj}(A)) = (\det A)^{n-2} A[/imath]
[imath]\newcommand{\adj}{\operatorname{adj}}[/imath]Let [imath]A\in \mathbb{M}_n[/imath] ([imath]n \geq\ 2[/imath]) be a regular matrix and [imath]\adj(A)[/imath] its adjoint. Prove that if A is regular then [imath]\adj(\adj(A)) = (\det A)^{n-2} A[/imath] (where [imath]adj(adj(A))[/imath] is the adjoint of [imath]\adj(A)[/imath]). For now, I know, or think I do, that [imath]A^{-1} = \adj(A)/\det(A)[/imath] from where we can see that [imath]\adj(A) = A^{-1} \det(A)[/imath] And that's all I know |
1091004 | what is the geometical interpretation of [imath]\vec a.\vec b[/imath]?
what is the geometical interpretation of [imath]\vec a.\vec b[/imath]?(dot product) I know the projection of [imath]\vec a [/imath] on [imath]\vec b[/imath] is [imath]\vec a.\hat b[/imath]. But what is a projection here? | 348717 | Dot Product Intuition
I'm searching to develop the intuition (rather than memorization) in relating the two forms of a dot product (by an angle theta between the vectors and by the components of the vector ). For example, suppose I have vector [imath]\mathbf{a} = (a_1,a_2)[/imath] and vector [imath]\mathbf{b}=(b_1,b_2)[/imath]. What's the physical or geometrical meaning that [imath]a_1b_1 + a_2b_2 = |\mathbf{a}||\mathbf{b}|\cos(\theta)\;?[/imath] Why is multiplying [imath]|\mathbf{b}|[/imath] times [imath]|\mathbf{a}|[/imath] in direction of [imath]\mathbf{b}[/imath] the same as multiplying the first and second components of [imath]\mathbf{a}[/imath] and [imath]\mathbf{b}[/imath] and summing ? I know this relationship comes out when we use the law of cosines to prove, but even then i cant get a intuition in this relationship. This image clarifies my doubt: Thanks |
999752 | Showing a sequence of integrals converges.
I'm having trouble with this problem - I don't even know how to begin. Thoughts? Solutions with explanation? Please help! Let [imath]f[/imath] be a bounded continuous function on [imath]\mathbb{R}[/imath]. Prove that [imath] \lim_{n \to \infty} \frac{n}{\pi} \int_{\mathbb{R}} \frac{f(t)}{1+n^2t^2}dt = f(0).[/imath] | 997709 | Integral of bounded continuous function on [imath]R[/imath]
Let [imath]f[/imath] be a bounded continuous function on [imath]R[/imath]. Prove that [imath]\lim_{n \to \infty} \frac{n}{\pi} \int_{ R} \frac {f(t)}{1+n^{2}t^{2}} dt=f(0)[/imath] I solved this question as follows, but I ran into a problem: Solution: Since [imath]f[/imath] is continuous at [imath]0[/imath], given [imath]\epsilon>0[/imath], there is [imath]\delta >0[/imath] such that [imath]|f(x)-f(0)| \leq \epsilon, \forall x \in [-\delta \hspace{2 mm} \delta][/imath]. Now consider [imath]f_{n}(x)=\frac{nf(x)}{1+n^{2}x^{2}}[/imath],[imath]n \in N[/imath], is a sequence of continuous function that converges to [imath]0[/imath] on [imath]R-[-\delta \hspace{2 mm} \delta][/imath] when [imath]n \to \infty[/imath]. Also, according to guestion, [imath]f[/imath] is bounded (i.e.[imath]|f_{n}(x)|<g(x)[/imath]). Therefore according to Dominated Convergence Theorem [imath]\lim_{n \to \infty} \frac{n}{\pi} \int_{ R-[-\delta \hspace{2 mm} \delta]} \frac {f(t)}{1+n^{2}t^{2}} dt=\lim_{n \to \infty} \frac{n}{\pi} \int_{ R-[-\delta \hspace{2 mm} \delta]} 0 \hspace{2 mm} dt=0[/imath] Now I have a problem to solve the second part. I want to prove [imath]\lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(t)}{1+n^{2}t^{2}} dt = \lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(0)}{1+n^{2}t^{2}} dt [/imath] which [imath]\lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(0)}{1+n^{2}t^{2}} dt =\pi f(0)[/imath] If I can prove that, the rest of question is easy, because I can say: [imath]\lim_{n \to \infty} n \int_{ R} \frac {f(t)}{1+n^{2}t^{2}} dt=\lim_{n \to \infty} n \int_{ R-[-\delta \hspace{2 mm} \delta]} \frac {f(t)}{1+n^{2}t^{2}} dt+\lim_{n \to \infty} n \int_{ R} \frac {f(t)}{1+n^{2}t^{2}} dt=\lim_{n \to \infty} n \int_{ [-\delta \hspace{2 mm} \delta]} \frac {f(t)}{1+n^{2}t^{2}} dt=\pi f(0)[/imath] Now how can I prove: [imath]\lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(t)}{1+n^{2}t^{2}} dt = \lim_{n \to \infty} \int_{ -\delta}^{\delta} \frac {nf(0)}{1+n^{2}t^{2}} dt [/imath] |
995069 | e as sum of an infinite series
I read that $e = \sum_{i=0}^\infty[imath] 1\over n!$. This isn't immediately obvious to me, and I can't find proof of this. Can somebody explain to me, how do I prove this from definition $e = \lim_{n\to \infty}[/imath] (1 + {1\over n})^n $, or point me to one? | 1171482 | Prove [imath] \lim_\limits{n \to\infty}{(1 + \frac{1}{1!} + \frac{1}{2!} + \cdot\cdot\cdot + \frac{1}{n!})} = e [/imath]
Use that [imath] \lim_\limits{n \to \infty}{\left(1 + \frac{1}{n}\right)^n} = e [/imath] to show that [imath] \lim_\limits{n \to \infty}{\left(1 + \frac{1}{1!} + \frac{1}{2!} + \cdot\cdot\cdot + \frac{1}{n!}\right)} = e [/imath] I can't figure this one out. Can I get some help in solving this? |
649343 | Proving uniqueness (basics of group theory)
If [imath](G,*)[/imath] is a group, prove that the identity and the inverse elements are unique. What I did for the first one is: Suppose [imath]\exists e,g\in G[/imath] such that [imath]\forall a\in G a*e=e*a=a[/imath] and also that [imath]\forall a\in G a*g=g*a=a[/imath], then [imath]g=g*e=e[/imath], hence [imath]g=e[/imath]. The first equallity happens because [imath]e[/imath] is an identity, and the second happens because [imath]g[/imath] is an identity. Is this right? I checked some notes of another class and their proof is much more longer and complicated (although I understand every step), what am I doing wrong? For the second one, can I use that if [imath]a[/imath] is the inverse of [imath]b[/imath], then [imath]b[/imath] is the inverse of [imath]a[/imath]? I feel that I shouldn't, but I'm not sure why. | 80596 | Prove that identity is unique in a group
I was given a group and I have to prove a) If [imath]f\cdot g = f[/imath] or [imath]g\cdot f = f[/imath] then [imath]g = 1[/imath]. Is it right to do it using just Identity axiom of group: [imath]f \cdot g = f = f \cdot1 \Longrightarrow g = 1 ?[/imath] Thanks. |
433546 | Is a semigroup [imath]G[/imath] with left identity and right inverses a group?
Hungerford's Algebra poses the question: Is it true that a semigroup [imath]G[/imath] that has a left identity element and in which every element has a right inverse is a group? Now, If both the identity and the inverse are of the same side, this is simple. For, instead of the above, say every element has a left inverse. For [imath]a \in G[/imath] denote this left inverse by [imath]a^{-1}[/imath]. Then [imath](aa^{-1})(aa^{-1}) = a(a^{-1}a)a^{-1} = aa^{-1}[/imath] and we can use the fact that [imath]cc = c \Longrightarrow c = 1[/imath] to get that inverses are in fact two-sided: [imath] aa^{-1} = 1[/imath] From which it follows that [imath]a = 1 \cdot a = (aa^{-1})a = a (a^{-1}a) = a \cdot 1[/imath] as desired. But in the scenario given we cannot use [imath]cc = c \Longrightarrow c = 1[/imath], and I can see no other way to prove this. At the same time, I cannot find a counter-example. Is there a simple resolution to this question? | 194083 | Any concrete example of ''right identity and left inverse do not imply a group''?
In the abstract algebra class, we have proved the fact that right identity and right inverse imply a group, while right identity and left inverse do not. My question: Are there any good examples of sets (with operations on) with right identity and left inverse, not being a group? To be specific, suppose [imath](X,\cdot)[/imath] is a set with a binary operation satisfies the following conditions: (i) [imath](a\cdot b)\cdot c=a\cdot (b\cdot c)[/imath] for any [imath]a,b,c\in X[/imath]; (ii) There exists [imath]e\in X[/imath] such that for every [imath]a\in X[/imath], [imath]a\cdot e=a[/imath]; (iii) For any [imath]a\in X[/imath], there exists [imath]b\in X[/imath] such that [imath]b\cdot a=e[/imath]. I want an example of [imath](X,\cdot)[/imath] which is not a group. |
1091416 | weakly convergence imply strong convergence when [imath] \|f_n\| \rightarrow \|f\| [/imath] in [imath]l^2([0,1])[/imath]?
I know in general weakly convergence do not imply strong convergence in [imath]L^p[/imath],but in [imath]L^2[0,1][/imath] space which if we have additional condition do this condition plus the weak convergence will give us strong convergence? The additional condition is [imath]f \in L^2[0,1][/imath] and [imath] \|f_n\| \rightarrow \|f\| [/imath] | 536101 | Weak convergence, together with convergence of norms, implies strong convergence in a Hilbert space.
Let [imath](x_n)[/imath] be a weakly convergent sequence in a Hilbert space [imath]H[/imath]. If [imath]\| x_n \| \to \| x \|[/imath], show that [imath]x_n[/imath] converges strongly to [imath]x[/imath]. Context This problem comes from a question in my exam paper; the original problem was incorrect. |
1084217 | [imath]f:\mathbb R \to \mathbb R[/imath] is a differentiable function such that [imath]f'(x)\le r<1 [/imath] , does [imath]f[/imath] necessarily have a fixed point ?
Let [imath]f:\mathbb R \to \mathbb R[/imath] be a differentiable function . If [imath]\exists r \in \mathbb R [/imath] such that [imath]|f'(x)|\le r<1 , \forall x \in \mathbb R[/imath] then using Lagrange's theorem one can show [imath]f[/imath] is a Lipscitz contraction and then use Banach contraction principle to conclude [imath]f[/imath] has a unique fixed-point. My question is what happens if [imath]f'(x)\le r<1 , \forall x \in \mathbb R[/imath] ? Then does [imath]f[/imath] even have a fixed point ? | 331151 | Number of Fixed Point(s) of a Differentiable Function
For a differentiable function [imath]f:\mathbb R\to\mathbb R[/imath] I need to choose the correct statement(s): [imath]f'(x)\le r<1~\forall~x\in\mathbb R\implies f[/imath] has at least one fixed point. [imath]f'(x)\le r<1~\forall~x\in\mathbb R\implies f[/imath] has a unique fixed point. My attempt: [imath]g:\mathbb R\to\mathbb R:x\mapsto f(x)-x[/imath] is differentiable on [imath]\mathbb R.[/imath] The search is for the number of roots of [imath]g.[/imath] Now [imath]f'(x)\le r<1~\forall~x\in\mathbb R\implies g'(x)\leq r-1<0~\forall~x\in\mathbb R\implies g[/imath] is decreasing everywhere on [imath]\mathbb R.[/imath] Can I say something from this observation[imath]?[/imath] |
491518 | Second order linear homogeneous ODE with constant coefficients and repeated roots. Why second solution needed?
In this case of a second order linear homogeneous ODE with constant coefficients and repeated roots: [imath]ay'' + by' + cy = 0 [/imath] (and [imath]r1 = r2[/imath]) why is the solution [imath]y_1(t) = e^{-b/2a}[/imath] not enough? in other words, why do we need a second equation? | 1081902 | Second order ODE - why the extra X for the solution?
Assuming I have the following homogeneous ODE equation: [imath]a\cdot y'' + b\cdot y' + c \cdot y = 0[/imath] Why for [imath](b^2 - 4\cdot a\cdot c=0) \quad [/imath],(meaning, when [imath]m_1=m_2[/imath]) then the solution is: [imath]y = C_1\cdot e^{m_{1}x} + C_2\cdot x \cdot e^{m_{2}x}[/imath] Why isn't it simply: [imath]y = C_1\cdot e^{m_{1}x} + C_2 \cdot e^{m_{2}x}[/imath] ? Also, why did they choose to multiply [imath]C_2[/imath] with [imath]x[/imath]? Why not having a totally different approach for the solution when [imath](m1=m2)[/imath] (e.g. diving the equation with [imath]x[/imath])? |
1091358 | A function [imath]f[/imath] such that [imath]f \in L_1[/imath] but [imath]f \notin L_p[/imath] for [imath]p>1[/imath]
I want find a function [imath]f: [0,1] \mapsto \mathbb{R}[/imath] such that [imath]f \in L_1[0,1][/imath] but [imath]f \notin L_p[0,1][/imath] for all [imath]p>1[/imath]. My attempts: First I thought in the family of functions [imath]\frac{1}{x^\alpha}[/imath] but this function belongs to [imath]L_q[/imath] iff [imath]\alpha \cdot q \leqslant 1[/imath] so I need find [imath]\alpha[/imath] such that: [imath]\alpha <1 [/imath] and [imath]\alpha \cdot q \geqslant 1[/imath] for all [imath]q>1[/imath] but this its impossible!! After other attempts using variations and combinations of [imath]1/x[/imath], [imath]ln x[/imath] and [imath]e^x[/imath] I researched in the mathstack and found this questions: Prove that for any [imath]1 < p < ∞[/imath] there exists a function [imath]f ∈ L_p(μ)[/imath] such that [imath]f \notin L_q(μ)[/imath] for any [imath]q > p.[/imath] The kingkongdonutguy's question is exactly what I was looking for, but I do not understand very well the Tomas' (and of Davide) hint... My interpretation: Choice two sequences [imath]\{a_n\}_{n \in \mathbb{N}}[/imath] and [imath]\{t_n\}_{n \in \mathbb{N}}[/imath] com [imath]a_n,t_n \to 0[/imath] now make a sequence os disjoint intervals [imath]\{I_n\}_{n \in \mathbb{N}}[/imath] such that, for each [imath]n[/imath],[imath]0 < m(I_n) < t_n[/imath] and [imath]\bigcup I_n = [0,1][/imath]. Define a function: [imath]f(x)= \sum\limits_{n=1}^{\infty} a_n \cdot \chi_{I_n}(x)[/imath] Make a simple calculation: [imath]\int\limits_{0}^{1} f(x)dx = \sum\limits_{n=1}^{\infty} \int_{I_n} a_n dx = \sum\limits_{n=1}^{\infty} a_n\cdot m(I_n) \leqslant \sum a_n \cdot t_n[/imath] So I need choice [imath]\{a_n\}[/imath] and [imath]\{t_n\}[/imath] such that [imath]\sum a_n \cdot t_n[/imath] converges but [imath]\sum a_n ^{p} \cdot t_n[/imath] not converges for [imath]p>1[/imath]. The problem: using limit comparison test we have [imath]\lim_{n \to \infty} \frac{a_n^p \cdot t_n}{a_n\cdot t_n} = \lim_{n \to \infty} a_n^{p-1}=0 [/imath](because [imath]p>1[/imath]) so don't is possible this choice ... Also found this question Is it possible for a function to be in [imath]L^p[/imath] for only one [imath]p[/imath]? but I could not adapt for a finite domain.. Someone can give me a (other) hint to construct this function?? | 1039064 | [imath]f \in L^1[/imath], but [imath]f \not\in L^p[/imath] for all [imath]p > 1[/imath]
"Find an [imath]f \in [0,1][/imath] such that [imath]f \in L^1[/imath] but [imath]f \not\in L^p[/imath] for any [imath]p > 1[/imath]." I've thought about doing something like [imath]f(x) = \frac{1}{x}[/imath] where [imath]|f|^p = \frac{1}{x^p}[/imath] doesn't converge when [imath]p > 1[/imath]. But this function isn't itself in [imath]L^1[/imath]. Could someone please give me a hint for how to solve this problem? I wish there were a situation where you had convergence on the closed half disc [imath][0,1][/imath] and divergence on [imath](1, \infty)[/imath], rather than my current predicament where I have convergence on the open half-disc [imath][0, 1)[/imath] and divergence on [imath][1, \infty)[/imath]. |
1091631 | Sum [imath]S=\sum _{k=1}^{\infty } \frac{(-1)^k H_k}{k^3}?[/imath]
We know that [imath]\sum _{k=1}^{\infty } \frac{H_k}{k^3} = \frac{\pi^4}{72}.[/imath] Is there a closed form for the sum [imath]S=\sum _{k=1}^{\infty } \frac{(-1)^k H_k}{k^3}?[/imath] Mathematica doesn't give anything resembling a closed form and I have no idea if one exists. | 457371 | Alternating harmonic sum [imath]\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k[/imath]
How to analytically prove [imath]\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k=-\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3) [/imath] As O.L answer where [imath]H_k = \sum_{n\geq 1}^{k}\frac{1}{n}.[/imath] Addition So far I developed the following [imath]\sum_{k\geq 1} \frac{H_k}{k^2} \, x^{k} = \text{Li}_3(x)-\, \text{Li}_3(1-x)+\, \log(1-x) \text{Li}_2(1-x) +\frac{1}{2}\log(x) \log^2(1-x)+\zeta(3)[/imath] where [imath]\text{Li}_3(x)[/imath] is the trilogarithm . For the derivation see http://www.mathhelpboards.com/f10/interesting-logarithm-integral-5301/ Update A frined on another site gave the following answer |
1091771 | Is [imath]U(\mathbb{Z}_{54})[/imath] a cyclic group?
Is the group [imath]U(\mathbb{Z}_{54}) = \{1,5,7,11,13,17,19,23,25,29,31,35,37,41,43,47,49,53\}[/imath] a cyclic group? If so, how do I show this? This group was found as a group of units of the monoid [imath]\mathbb{Z}_{54}[/imath]. | 1078867 | Show that the group is cyclic.
I'm trying to show that the group [imath]U(Z_{54})[/imath] is cyclic. To start, I found the divisors of 54 = {1, 2, 3, 6, 9, 18, 27, 54} Then I started to find the elements using the powers of a. Where [imath]U(Z_{54})= <a>[/imath] Therefore, [imath]<a^{54/1}>[/imath]={ e } [imath] <a^{54/2}>[/imath]= { e, [imath] a^{36}[/imath] } and such, where the last would be < [imath]a^{54/54}[/imath]> = { e } The things I would like to know is, how does this process show that it's cyclic? Is it because the first divisor (1) and the last divisor (54) is equal to e? Therefore creating a cycle? Also, what would the < a > to the powers finding the elements process be called? Thank you! |
1091752 | Why is the volume one third of that? I mean, where's the fault in my logic?
The volume of a cuboid is [imath]l \times b \times h[/imath]. That is, it is equal to base area times height. I think it means that the base is added up height times, it becomes volume (It makes sense if we think about it) And if we think about cylinder, the above mentioned logic still holds. The volume of cylinder is [imath]\pi r^2 \times h[/imath] . Which is same as saying the area of bottom circle (i.e. [imath]\pi r^2[/imath]) times the height [imath]h[/imath]. But when it comes to a cone , it doesn't makes sense. In above examples 2-d shapes have been collected one upon another, a specific number of times(h). In a cone its kinda same except that now a triangle isn't collected one upon another (that would make a prism), but collected in a circular way. We can imagine a cone to be group of right triangles with their axis being one of the sides (except hypotenuse, that would make a compound shape of two cones ). We can also imagine it this way, A right triangle is rotated with one of the sides being axis. So now the volume should be, by above mentioned logic, the area of repeated shape times number of shapes. And there are 2pi x r triangles because, well, it is rotated that much number of times. Visualize it in your mind, it'll become clear. Hope you get the idea. Now, Volume of cone should be area of the triangle times the circumference of the base => [imath]1/2 \times b \times h \times 2\pi r[/imath] Now the base of triangle is same as the radius of the base and the 2s cancel each other out we are left with [imath]\pi r^2 h[/imath] But the volume of cone if one third of that value. My question is Why is that? | 860071 | Why isn't the volume formula for a cone [imath]\pi r^2h[/imath]?
So I understand that the volume formula of a cone is: [imath]\frac{1}{3}\pi r^2h[/imath], and when I read about how to derive this formula, it makes sense to me. Funny thing is, I'm stuck on why it ISN'T [imath]\pi r^2h[/imath] when I think about deriving the volume formula in a different way. Here's what I mean. Now we're told that to figure out the volume formula for something like a cylinder or cube, we simply multiply the area of the base by the height, and this makes intuitive sense. In the case of a cone, what if we took a triangle with base r and height h, and rotated it around some axis, producing a cone. When I think of it this way, it seems reasonable (in my mind) to calculate the volume of the produced cone by taking the area of the triangle, and multiplying it by the circumference of the circle that is the cone's base. So if the area of our triangle is: [imath]\frac12rh[/imath] and the circumference of the generated cone's circle-base is: [imath]2\pi r[/imath], then the volume should be: [imath]\pi r^2h[/imath]. This derivation seems intuitively correct to me. Now I know it isn't correct, and I almost feel silly asking, but I just can't see WHY it's not correct. What's the flaw in the reasoning that the volume of a cone can be derived by multiplying the area of the triangle by the circumference of its circe-base? |
1091911 | [imath]\lvert G\rvert=24[/imath] not simple by a counting arguments
I want to prove that [imath]\lvert G\rvert=24[/imath], then it has a non-trivial normal subgroup. Here is my attempt: [imath]n_2[/imath]: number of 2-Sylow subgroup; [imath]n_3[/imath]: number of 3-Sylow subgroup. [imath]n_2\in\{1,3\},\:\:n_3\in\{1,4\}[/imath] by Sylow's Theorem. Suposse [imath]n_2=3[/imath], [imath]n_3=4[/imath]. [imath]n_3=4[/imath] means there are 4 Sylow subgroups of order 3, which means 8 elements of order 3. [imath]n_2=3[/imath] means there are 3 Sylow subgroups of order 8. Suppose [imath]P_1,P_2,P_3[/imath]. Using [imath]\lvert P_1P_2\rvert\leq 24[/imath], then [imath]2<\lvert P_1\cap P_2\rvert <8[/imath], so it is 4. Suppose [imath]P_1=\{1,a_1,\ldots,a_7\}[/imath]. [imath]\lvert P_1\cap P_2\rvert=4 \Longrightarrow P_2=\{1,a_1,a_2,a_3,b_1,b_2,b_3,b_4\}[/imath] for example. [imath]\lvert P_1\cap P_3\rvert=4[/imath] and [imath]\lvert P_3\cap P_2\rvert=4[/imath] [imath]\Longrightarrow P_3=\{1,a_7,a_6,a_5,b_1,b_2,b_3,c\}[/imath] for example. Then we have 12 different elements of even order. Adding we get 21 different elements. | 1091857 | [imath]|G|=24[/imath] prove that [imath]G[/imath] is not simple
Let [imath]G[/imath] be a group of order 24, and we shall assume there there exist a non-normal 2-sylow group in [imath]G[/imath]. i want to show that it's not simple. first i have showed that there are exactly three 2-sylow group's. and now i am trying to show that there is a non trivial homomorphism from [imath]G[/imath] to [imath]S_3[/imath]. i know that [imath]S_3\cong D_3[/imath] and indeed there are three 2-sylow groups in [imath]D_3[/imath] , the reflection's. and there is one [imath]3-[/imath]sylow group in [imath]D_3[/imath] so i wanted to creat a homomorphism that take the different p-sylow groups in [imath]G[/imath] and set them in right [imath]p-[/imath]sylow group's in [imath]D_3[/imath] but for that i need to show that the intersection of every two [imath]2-[/imath]sylow gouts in [imath]G[/imath] is trivial. and i can't manage to do that.... |
1091979 | Is [imath](g \circ f)^{-1} (x)[/imath] equal to [imath](f^{-1}\circ g^{-1})(x)[/imath] or not?
Is [imath](g \circ f)^{-1} (x)[/imath] equal to [imath](f^{-1}\circ g^{-1})(x)[/imath] or not? | 2349 | How to prove [imath](f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}[/imath]? (inverse of composition)
I'm doing exercise on discrete mathematics and I'm stuck with question: If [imath]f:Y\to Z[/imath] is an invertible function, and [imath]g:X\to Y[/imath] is an invertible function, then the inverse of the composition [imath](f \circ\ g)[/imath] is given by [imath](f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}[/imath]. I've no idea how to prove this, please help me by give me some reference or hint to its solution. |
1092160 | How to evaluate [imath]\int_0^1 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}} dx[/imath]?
How to evaluate [imath]\int_0^1 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}} dx[/imath] I am clueless, I doubt if this has a closed form either. | 805893 | A logarithmic integral [imath]\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx[/imath]
How to prove the following [imath]\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx=\frac{\pi^2}{2}[/imath] I thought of separating the two integrals and use the beta or hypergeometric functions but I thought these are not best ideas to approach the problem. Any other ideas ? |
1092253 | Why is [imath]\langle p,q \rangle = 0[/imath]?
First of all, sorry for opening a new question about it, but I'm curious to understand: John Hughes claims that [imath]\langle p,q \rangle =0[/imath] (in the end of his answer) Why is it true? Prove that there is a unique inner product on V | 1092180 | Prove that there is a unique inner product on [imath]V[/imath]
Let [imath]V[/imath], a vector space over [imath]\mathbb{F}[/imath] and [imath]W_1,W_2 \subseteq V[/imath] such that [imath]W_1 \oplus W_2 = V[/imath]. For [imath]i=1,2[/imath], let [imath]\langle , \rangle [/imath] on [imath]W_i[/imath]. Prove that there is a unique inner product on [imath]V[/imath] such that: [imath]W_2 = W^\perp_1[/imath] For [imath]i=1,2[/imath] and for all [imath]v,u\in W_i[/imath]: [imath]\langle v, u \rangle = \langle v, u \rangle _i[/imath] Uniqueness: Let [imath]v = w_1 + w_2[/imath] and [imath]u = w_1' + w_2'[/imath] [imath]\langle u, v \rangle_V = \langle w_1,w_1' \rangle + \langle w_1,w_2' \rangle + \langle w_2,w_1' \rangle + \langle w_2,w_2' \rangle = \langle w_1,w_1' \rangle + \langle w_2,w_2' \rangle [/imath] So the inner product is determined uniquely by [imath]\langle, \rangle_i [/imath] for [imath]i=1,2[/imath]. How do I show the existence of such inner product? |
1092349 | Prove that [imath]n^4−1 [/imath] is divisible by 5 when n is not divisible by 5.
Apparently the easiest method is to use proof by exhaustion, but I've no idea how. Any ideas/solutions? Prove that [imath]n^4−1[/imath] is divisible by 5 when n is not divisible by 5. | 871353 | Prove that if [imath]n[/imath] is not divisible by [imath]5[/imath], then [imath]n^4 \equiv 1 \pmod{5}[/imath]
Suppose [imath]n[/imath] is an integer which is not divisible by [imath]5[/imath]. Prove that [imath]n^4 \equiv 1 \pmod{5}[/imath]. |
1092155 | Visualization of relation between integration and derivative operations
If I have a function [imath]f(x)[/imath] and I find the derivative I will get [imath]f'(x)[/imath]. Furthermore, if I do the integration of the derivative [imath]f'(x)[/imath], as a result I will get again my original function [imath]f(x)[/imath]. What I do not understand is a relation between integration and derivative operations. For me, the derivative is an operation where we calculate the slope of a function, and integration is an area under a curve. But, if we take an example where [imath]f(x) = x^3[/imath]. The derivative is [imath]f'(x) = 3x^2[/imath]. If we do the integration of the [imath]3x^2[/imath] we will get our [imath]x^3[/imath] back again. But, [imath]3x^2[/imath] is some function, and if we do the integration on it, we should get its area under the [imath]3x^2[/imath]. Is it possible to visualize that relation between derivative and integral operation, because I do not see it from the graph below? | 116620 | Understanding the relationship between differentiation and integration
I am trying to understand the relationship between differentiation and integration. Differentiation has been introduced to me by this diagram: Which displays that the derivative of a point [imath]x[/imath] on a continuous function [imath]f(x)[/imath] is the gradient of a line which is a tangent to that particular point, as shown in the diagram. This can be written as [imath]\lim_{h \to 0} \left(\frac{f\left(x+h\right)-f\left(x\right)}{h}\right)=f'\left(x\right)[/imath] and this gives the gradient of the point [imath](x, f(x))[/imath]. Next examine this diagram: The area under the continuous function from [imath]f(x)[/imath] from [imath]a[/imath] to [imath]x[/imath] is [imath]A(x)[/imath], shaded pink. The area of [imath]A(a)[/imath] is [imath]0[/imath] and the area from [imath]a[/imath] to [imath]b[/imath] is [imath]A(b)[/imath]. Following this convention the area of [imath]x[/imath] to [imath]x+h[/imath] is [imath]A\left(x+h\right)-A\left(x\right)[/imath] This section has with [imath]h[/imath] and the area is close to that of a rectangle so it can be said that [imath]A\left(x+h\right)-A\left(x\right)\approx hf(x)[/imath] If you divide this through by [imath]h[/imath] you get an equation which shows that the derivative of [imath]f(x)[/imath] is [imath]A'(x)[/imath]. [imath]\lim_{h \to 0} \left(\frac{A\left(x+h\right)-A\left(x\right)}{h}\right)=f\left(x\right)[/imath] and hence [imath]A'(x)=f(x)[/imath] This is where I have a problem, earlier it was stated that the derivative is the gradient of a line which is a tangent to a point on a continuous function. Here however, we have an area, which encloses many points, which are not a line which is a tangent to a continuous function. So I do not see how you can find the derivative of an area because it is not a point that you can find the gradient of a line which is a tangent to it.? (Images taken from course texts provided by the Open University, Chapters C1 & C2 of the course MST121). |
101062 | Is the product of two Gaussian random variables also a Gaussian?
Say I have [imath]X \sim \mathcal N(a, b)[/imath] and [imath]Y\sim \mathcal N(c, d)[/imath]. Is [imath]XY[/imath] also normally distributed? Is the answer any different if we know that [imath]X[/imath] and [imath]Y[/imath] are independent? | 1908692 | [imath]X,Y[/imath] are independent random variables, normal distribution, [imath]XY[/imath] has normal distribution
[imath]X,Y[/imath] are random independent variables with normal distribution. Can we conclude that [imath]XY[/imath] has also normal distribution ? I know definition of normal distributio, however I have a problem with it. Can you explain it ? |
1093094 | Are commuting endomorphisms polynomials of a same (third) endomorphism?
Let [imath]E[/imath] be a finite dimensional [imath]k[/imath]-vector space ([imath]k[/imath] may be assumed algebraically closed if needed) and [imath]u,v \in \mathcal L(E)[/imath] be two endomorphisms that commute ([imath]u\circ v=v \circ u[/imath]). Does there exist a third endomorphism [imath]w[/imath] and two polynomials [imath]P,Q\in k[X][/imath] such that [imath]u=P(w)[/imath] and [imath]v=Q(w)[/imath]? I think the heart of the problem is to deal with the nilpotent subcase (where u and v are supposed to be both nilpotent). If [imath]u[/imath] (or [imath]v[/imath]) is cyclic (ie only one Jordan block) this is classical, but I am not sure how to deal with the general case... | 326293 | Can commuting matrices [imath]X,Y[/imath] always be written as polynomials of some matrix [imath]A[/imath]?
Consider square matrices over a field [imath]K[/imath]. I don't think additional assumptions about [imath]K[/imath] like algebraically closed or characteristic [imath]0[/imath] are pertinent, but feel free to make them for comfort. For any such matrix [imath]A[/imath], the set [imath]K[A][/imath] of polynomials in [imath]A[/imath] is a commutative subalgebra of [imath]M_n(K)[/imath]; the question is whether for any pair of commuting matrices [imath]X,Y[/imath] at least one such commutative subalgebra can be found that contains both [imath]X[/imath] and [imath]Y[/imath]. I was asking myself this in connection with frequently recurring requests to completely characterise commuting pairs of matrices, like this one. While providing a useful characterisation seems impossible, a positive anwer to the current question would at least provide some answer. Note that in many rather likely situations one can in fact take [imath]A[/imath] to be one of the matrices [imath]X,Y[/imath], for instance when one of the matrices has distinct eigenvalues, or more generally if its minimal polynomial has degree [imath]n[/imath] (so coincides with the characteristic polynomial). However this is not always possible, as can be easily seen for instance for diagonal matrices [imath]X=\operatorname{diag}(0,0,1)[/imath] and [imath]Y=\operatorname{diag}(0,1,1)[/imath]. However in that case both will be polynomials in [imath]A=\operatorname{diag}(x,y,z)[/imath] for any distinct values [imath]x,y,z[/imath] (then [imath]K[A][/imath] consists of all diagonal matrices); although in the example in this answer the matrices are not both diagonalisable, an appropriate [imath]A[/imath] can be found there as well. I thought for some time that any maximal commutative subalgebra of [imath]M_n(K)[/imath] was of the form [imath]K[A][/imath] (which would imply a positive answer) for some [imath]A[/imath] with minimal polynomial of degree[imath]~n[/imath], and that a positive answer to my question was in fact instrumental in proving this. However I was wrong on both counts: there exist (for [imath]n\geq 4[/imath]) commutative subalgebras of dimension[imath]{}>n[/imath] (whereas [imath]\dim_KK[A]\leq n[/imath] for all [imath]A\in M_n(K)[/imath]) as shown in this MathOverflow answer, and I was forced to correct an anwer I gave here in the light of this; however it seems (at least in the cases I looked at) that many (all?) pairs of matrices [imath]X,Y[/imath] in such a subalgebra still admit a matrix [imath]A[/imath] (which in general is not in the subalgebra) such that [imath]X,Y\in K[A][/imath]. This indicates that a positive answer to my question would not contradict the existence of such large commutative subalgebras: it would just mean that to obtain a maximal dimensional subalgebra containing [imath]X,Y[/imath] one should in general avoid throwing in an [imath]A[/imath] with [imath]X,Y\in K[A][/imath]. I do think these large subalgebras easily show that my question but for three commuting matrices has a negative answer. Finally I note that this other answer to the cited MO question mentions a result by Gerstenhaber that the dimension of the subalgebra generated two commuting matrices in [imath]M_n(K)[/imath] cannot exceed[imath]~n[/imath]. This unfortunately does not settle my question (if [imath]X,Y[/imath] would generate a subalgebra of dimension[imath]{}>n[/imath], it would have proved a negative answer); it just might be that the mentioned result is true because of the existence of [imath]A[/imath] (I don't have access to a proof right now, but given the formulation it seems unlikely that it was done this way). OK, I've tried to build up the suspense. Honesty demands that I say that I do know the answer to my question, since a colleague of mine provided a convincing one. I will however not give this answer right away, but post it once there has been some time for gathering answers here; who knows somebody will prove a different answer than the one I have (heaven forbid), or at least give the same answer with a different justification. |
1093057 | How to prove that [imath]n[/imath] is a prime if [imath]2^n-1[/imath] is a prime
I got this by watching a video on youtube and they said that this is always true but I am wondering why. I have tried to use Fermat's little theorem (FLT) but got nowhere bcs it says that if p is a prime then [imath]p|{a^p-a}[/imath]. | 186587 | Prove that if [imath]2^n - 1[/imath] is prime, then [imath]n[/imath] is prime for [imath]n[/imath] being a natural number
Prove that if [imath]2^n - 1[/imath] is prime, then [imath]n[/imath] is prime for [imath]n[/imath] being a natural number I've looked at https://math.stackexchange.com/a/19998 It is known that [imath]2^n-1[/imath] can only be prime if [imath]n[/imath] is prime. This is because if [imath]jk=n[/imath], [imath]2^n-1=\sum_{i=0}^{n-1}2^i=\sum_{i=0}^{j-1} 2^i \sum_{i=0}^{k-1} 2^{ij}[/imath] So they will only continue to alternate at twin primes. In particular, [imath]2^{6k+2}-1, 2^{6k+3}-1[/imath] and [imath]2^{6k+4}-1[/imath] will all be composite What I dont understand is how do I get: [imath]\sum_{i=0}^{n-1}2^i=\sum_{i=0}^{j-1} 2^i \sum_{i=0}^{k-1} 2^{ij}[/imath] If I am looking at the wrong answer, how can I proof the above? Then again, maybe I am indeed looking at the wrong thing? In particular, [imath]2^{6k+2}-1, 2^{6k+3}-1[/imath] and [imath]2^{6k+4}-1[/imath] will all be composite This doesnt say why [imath]2^{6k+2}-1[/imath] is composite? |
1093097 | Problem on Integral Domain
Why [imath]n\mathbb{Z} \times n\mathbb{Z}[/imath] is not an integral domain? | 233316 | Zero Divisors in Direct Product of Rings
Suppose that [imath]R = S × T[/imath] is a direct product of rings with [imath]S[/imath] and [imath]T[/imath] each having at least two elements. Prove that [imath]R[/imath] has zero divisors. |
1093085 | Find all polynomials [imath]P(x)[/imath] has [imath]n[/imath] rational roots.
Find all polynomials [imath]P(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0[/imath] such as: 1) [imath](a_0,a_1,a_2,...,a_n)[/imath] is a permutation of [imath](0,1,2,...,n)[/imath] 2) [imath]P(x)[/imath] has [imath]n[/imath] rational roots. | 596589 | Polynomials with rational zeros
Find all polynomials [imath]F(x)={a_n}{x^n}+\cdots+{a_1}x+a_0[/imath] satisfying [imath]a_n \neq0[/imath]; [imath](a_0, a_1, a_2, \ldots ,a_n)[/imath] is a permutation of [imath](0, 1, 2 ... n)[/imath]; all zeros of [imath]F(x)[/imath] are rational. |
1093431 | Prove: A group of order 315 with a normal 3-Sylow group is Abelian.
Prove: A group of order 315 with a normal 3-Sylow group is Abelian. I know that [imath]315 = 3^2\times5\times7[/imath] I've tried using Sylow's theorems on 5, 7 but haven't got anywhere meaningful. I would appreciate any help. | 324886 | On Groups of Order 315 with a unique sylow 3-subgroup .
in Dummit and Foote , an exercise asked me to prove that , if [imath]G[/imath] is a group of order [imath]315[/imath] , [imath]G[/imath] has a normal sylow [imath]3[/imath]-subgroup then , [imath]G[/imath] is abelian . this is exercise number [imath]27[/imath] , section [imath]5[/imath] , chapter [imath]4[/imath] my method on proving this is showing that [imath] |G| = |Z(G)|[/imath] as follows i showed that [imath]|Z(G)|[/imath] [imath]\in[/imath] [imath]{ 9 , 63 , 45 , 315 } [/imath] and showed that[imath] |Z(G)|[/imath] is not [imath]63 , 45[/imath] or [imath]9[/imath] so it must be [imath]315 [/imath] but i showed this under two condition , first condition: if [imath]X[/imath] is the set of all elements of [imath]G[/imath] of order [imath]5[/imath] , [imath]H[/imath] is the subgroup generated by [imath]X[/imath] then [imath]|H|[/imath]is not [imath]315[/imath] second condition : also , if [imath]Y[/imath] is the set of all elements of [imath]G[/imath] of order [imath]7[/imath] , [imath]N[/imath] is the subgroup generated by [imath]Y[/imath] then [imath]|N|[/imath] is not [imath]315[/imath] so , if those two condition are then my method of proving works . but i don't know how to prove those conditions . any help ? |
1093296 | Maximization question
I'm trying to find the maximum value of the function [imath]f(x,y)=(ax+by)^p+x^p[/imath] subject to the constraint [imath]x^p+y^p=1[/imath]. Here, [imath]a,b[/imath] and [imath]p[/imath] are constants with [imath]a,b>0[/imath] and [imath]p>1[/imath], and [imath]x,y>0[/imath]. I have found the maximum in the special case [imath]p=2[/imath] and tried to use Lagrange multipliers in the general case but couldn't success. Any help? | 1090810 | A maximization problem
I'm trying to find the maximum value of the function [imath]f(x,y)=(ax+by)^p+x^p[/imath] subject to the constraint [imath]x^p+y^p=1[/imath]. Here, [imath]a,b[/imath] and [imath]p[/imath] are constants with [imath]a,b>0[/imath] and [imath]p>1[/imath], and [imath]x,y>0[/imath]. I have found the maximum in the special case [imath]p=2[/imath] and tried to use Lagrange multipliers in the general case but couldn't success. Any help? |
1093543 | Find a holomorphic function [imath]f[/imath] on the unit disc, such that [imath]f(1/n) =2^{-n}[/imath] for [imath]n\geq2[/imath], [imath]n\in\mathbb{}[/imath]N.
Find a holomorphic function [imath]f[/imath] on the unit disc, such that [imath]f\left(\frac{1}{n}\right)=2^{-n}[/imath] for [imath]n\geq2[/imath], [imath]n\in\mathbb{N}[/imath]. Is this possible? | 686040 | Holomorphic function interpolating [imath]e^{-n}[/imath]
Consider the question: does there exist a holomorphic function [imath]f[/imath] on the unit disk in the complex plane such that [imath]f\left({1 \over n}\right) = e^{-n}[/imath] ? I came up with an answer but I'd like to know if there is another argument. Here's my answer: let [imath]f[/imath] be such a function. [imath]f[/imath] must vanish at [imath]0[/imath] by continuity. Any holomorphic function vanishes at [imath]0[/imath] satisfies the following estimate : [imath]\left| f(z) \right|\sim_{z \to 0} \alpha \,|z|^m[/imath], where [imath]\alpha>0[/imath] and [imath]m[/imath] is a positive integer (this is easily derived from the expression of [imath]f[/imath] as a power series). This contradicts the assumption [imath]f\left({1 \over n}\right) = e^{-n}[/imath]. By curiosity, I would like to know if there is another argument (maybe something about [imath]f(z) - e^{1 \over z}[/imath])? |
1093656 | Compute the limit [imath]\lim_{n\to\infty}{(\sqrt[n]{e}-\frac{2}{n})^n}[/imath]
How can I compute this limit of the sequence? [imath]\lim_{n\to\infty}{(\sqrt[n]{e}-\frac{2}{n})^n}[/imath] | 1088938 | Find this limit: [imath] \lim_{n \to \infty}{(e^{\frac{1}{n}} - \frac{2}{n})^n}[/imath]
Can anyone help me with this limit: [imath] \lim_{n \to \infty}{\left(e^{\frac{1}{n}} - \frac{2}{n}\right)^n}[/imath] I think that I should expand [imath]e^{\frac{1}{n}}[/imath] as the sequence [imath]1 + \frac{1}{n} + \frac{1}{n^{2}2!} + \frac{1}{n^{3}3!} + \dots [/imath]. But I can't see how it helps me. Edit: or maybe use logarithm? |
1093774 | Trouble doing simple polynomial interpolation
I need to do a polynomial interpolation of a set of [imath]N[/imath] experimental points; the functional form I have to use to interpolate is this: [imath] f(x) = a + bx^2 + cx^4[/imath] as you can see the coefficient that I need to find are just 3: [imath]a, b, c[/imath]; however the points I have are [imath]N>3[/imath] and so it looks like the determination of the coefficients is impossible because over-determined. Does anyone have an idea of what should be done in such case (supposing it is even possible)? | 1093771 | Trouble doing polynomial interpolation
I need to do a polynomial interpolation of a set [imath]N[/imath] of experimental points; the functional form I have to use to interpolate is this: [imath] f(x) = a + bx^2 + cx^4,[/imath] as you can see the coefficient that I need to find are just 3: [imath]a, b, c[/imath]; however the points I have are [imath]\#N>3[/imath] and so it looks like the determination of the coefficients is impossible because is over-determined. Does anyone have an idea of what should be done in such case (supposing it is even possible)? |
932775 | kernel of homomorphism [imath]\mathbb{C}[x,y] \to \mathbb{C}[t][/imath] but in general case
Let [imath]f:\mathbb{C}[x,y] \to \mathbb{C}[t][/imath] be a homomorphism that is identity on [imath]\mathbb{C}[/imath] and sends [imath]x\to x(t),y \to y(t)[/imath] and such that [imath]x(t),y(t)[/imath] aren't both constant. Prove [imath]\ker(f)[/imath] is a principal ideal. | 244452 | Prove that the kernel of a homomorphism is a principal ideal. (Artin, Exercise 9.13)
I have been having trouble with an exercise in my abstract algebra course. It is as follows: Let [imath]f: \mathbb{C}[x,y] \rightarrow \mathbb{C}[t][/imath] be a homomorphism that is the identity on [imath]\mathbb{C}[/imath] and sends [imath]x[/imath] to [imath]x(t)[/imath] and [imath]y[/imath] to [imath]y(t)[/imath] such that [imath]x(t)[/imath] and [imath]y(t)[/imath] are not both constant. Prove that the kernel of [imath]f[/imath] is a principal ideal. |
1092868 | How prove that [imath]\int_0^\pi {x\,f(\sin x)\,} dx = \frac{\pi }{2}\int_0^\pi {f(\sin x)} \,dx[/imath]
To prove that [imath]\int_0^\pi {x\,f(\sin x)\,} dx = \frac{\pi }{2}\int_0^\pi {f(\sin x)} \,dx[/imath] is true, first I started calculating the integral of the left indefinitely [imath] \int {x\,f(\sin x)\,\,dx} [/imath] using substitution: [imath] \sin x = t,\quad x = \arcsin t, \quad {dx = \frac{{dt}}{{\sqrt {1 - {t^2}} }}}[/imath] is obtained: [imath] \int {x\,f(\sin x)\,\,dx} = \int {\arcsin t \cdot f(t) \cdot \frac{{dt}}{{\sqrt {1 - {t^2}} }}}[/imath] [imath] \qquad\quad = \int {\frac{{\arcsin t\,dt}}{{\sqrt {1 - {t^2}} }} \cdot f(t)} [/imath] Then using integration by parts: [imath] \begin{array}{*{20}{c}} {u = f(t)},&{dv = \frac{{\arcsin t\,dt}}{{\sqrt {1 - {t^2}} }}} \\ {du = f'(t)\,dt},&{v = \frac{{{{(\arcsin t)}^2}}}{2}} \end{array} [/imath] then: \begin{align*} \int {x\,f(\sin x)\,\,dx} &= f(t) \cdot \frac{{{{(\arcsin t)}^2}}}{2} - \int {\frac{{{{(\arcsin t)}^2}}}{2}} \cdot f'(t)\,dt \\ &= f(\sin x) \cdot \frac{{{x^2}}}{2} - \int {\frac{{{x^2}}}{2} \cdot f'(\sin x)\,\cos x\,dx} \\ \end{align*} Now, evaluating from 0 to [imath]\pi[/imath] \begin{align} \int_0^\pi {x\,f(\sin x)} \,dx & = \left[ {f(t) \cdot \frac{{{x^2}}}{2}} \right]_0^\pi - \int_0^\pi {\frac{{{x^2}}}{2} \cdot f'(\sin x)\,\cos x\,dx} \\ \int_0^\pi {x\,f(\sin x)} \,dx & = f(0) \cdot \frac{{{\pi ^2}}}{2} - \int_0^\pi {\frac{{{x^2}}}{2} \cdot f'(\sin x)\,\cos x\,dx} \qquad ..[1] \\ \end{align} On the other hand, doing the same process with the integral on the right side I get: \begin{equation}\int_0^\pi {f(\sin x)} \,dx = f(0) \cdot \pi - \int_0^\pi {x \cdot f'(\sin x)\,\cos x\,dx} \qquad ..[2] \end{equation} And even here I do not have enough data to say that equality [imath]\int_0^\pi {x\,f(\sin x)\,} dx = \frac{\pi }{2}\int_0^\pi {f(\sin x)} \,dx[/imath] is true. Can anyone suggest me what to do with the equalities [1] and [2]? Thanks in advance. | 159381 | Evaluate [imath]\int_0^\pi xf(\sin x)dx[/imath]
Let [imath]f(\sin x)[/imath] be a given function of [imath]\sin x[/imath]. How would I show that [imath]\int_0^\pi xf(\sin x)dx=\frac{1}{2}\pi\int_0^\pi f(\sin x)dx[/imath]? |
1095164 | prove [imath]f[/imath] is a constant
Lets's say we have a differentiable function [imath]f:[a,b]\to \mathbb{R}[/imath] with [imath]f^\prime\equiv0[/imath] How do I show that [imath]f\equiv C[/imath] by using the mean value theorem? | 425845 | [imath]f[/imath] is constant if derivative equals zero
Suppose [imath]f'(x)=0[/imath] for all [imath]x\in (a,b)[/imath]. Prove that [imath]f[/imath] is constant on [imath](a,b)[/imath]. This seems painfully obvious, but I can't prove it rigorously. [imath]f'(x)=0[/imath] for all [imath]x\in (a,b)[/imath] means that for any [imath]c\in(a,b)[/imath], we have [imath]\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}=0[/imath] So for any [imath]c\in (a,b)[/imath], for any [imath]\epsilon[/imath] there exists [imath]\delta[/imath] such that if [imath]|x-c|<\delta[/imath] then [imath]\left|\frac{f(x)-f(c)}{x-c}\right|<\epsilon[/imath] which also means [imath]|f(x)-f(c)|<\epsilon\delta[/imath] How to continue from here? |
1014103 | Greatest common divisor / euclidean algorithm linear combination proof
Consider integers [imath]m[/imath] and [imath]n[/imath], not both 0. Show that gcd[imath](m,n)[/imath] is the smallest positive integer that can be written as [imath]am + bn[/imath] for integers [imath]a[/imath] and [imath]b[/imath]. I'm confused on what exactly to do--I'm having trouble wrapping my head around the question. I know that using the euclidean algorithm backwards will provide me with an [imath]a[/imath] and a [imath]b[/imath] such that [imath]am + bn = gcd(m,n)[/imath]. But I'm not sure how to prove that it is the smallest positive integer. | 980426 | The greatest common divisor is the smallest positive linear combination
How to prove the following theorems about gcd? Theorem 1: Let [imath]a[/imath] and [imath]b[/imath] be nonzero integers. Then the smallest positive linear combination of [imath]a[/imath] and [imath]b[/imath] is a common divisor of [imath]a[/imath] and [imath]b[/imath]. Theorem 2: Let [imath]a[/imath] and [imath]b[/imath] be nonzero integers. The gcd of [imath]a[/imath] and [imath]b[/imath] is the smallest positive linear combination of [imath]a[/imath] and [imath]b[/imath]. Progress For Theorem 1 I have assumed that [imath]d[/imath] is the smallest possible linear combination of [imath]a[/imath] and [imath]d[/imath]. Then [imath]a = dq + r[/imath]. Solved it and found a contradiction. Is my method correct? Don't know what to do for Theorem 2. |
718783 | Greatest Common Divisor property: If [imath]\gcd(a, b) = 1[/imath] and [imath]a | c[/imath] and [imath]b | c[/imath], then [imath]ab | c[/imath]
Here is what I am trying to prove: Let [imath]a,b,c,d \in ℤ_+[/imath] with gcd[imath](a,b)=1[/imath]. If [imath]a|c[/imath] and [imath]b|c[/imath], prove that [imath]ab|c[/imath]. Does the result hold if gcd [imath](a,b)\neq 1[/imath] ? I know that gcd [imath](a,b)=1[/imath] can be written [imath]ax+by=1[/imath] where [imath]x,y[/imath] are integers. And [imath]a|c[/imath] can be written [imath]c=al[/imath] where [imath]l[/imath] is an integer, b|c can be written [imath]c=bk[/imath] where [imath]k[/imath] is an integer, and [imath]ab|c[/imath] can be written [imath]c=ab(m)[/imath] where [imath]m[/imath] is an integer. What do I do? | 1017556 | Proof that if [imath]\gcd(a,b) = 1[/imath] and [imath]a\mid n[/imath] and [imath]b\mid n[/imath], [imath]ab \mid n[/imath]
I'm learning about properties of greatest common divisors, specifically when two numbers are relatively prime. The exercise I'm working through is : Suppose that [imath]\gcd(a,b) = 1[/imath] and that [imath]a\mid n[/imath] and [imath]b\mid n[/imath]. Prove that [imath]ab\mid n[/imath] Proof(so far): [imath]\gcd(a,b)=1[/imath] implies [imath]am+bk = 1[/imath] for some integers [imath]m[/imath] and [imath]k[/imath]. (Given hint) Multiplying the equation by [imath]n[/imath] yields: [imath]amn+bkn=n[/imath] I'm not sure what to do after applying the hint. Am I supposed to show that [imath]ab[/imath] divides both [imath]amn[/imath] and [imath]bkn[/imath]? Thanks in advance. |
1095250 | Replacing Variables in Integration
I have posted questions about this, but they werent clear, here is my actual misunderstanding. [imath]I = \int_{-\infty}^{\infty} e^{-x^2} dx[/imath] I dont understand, we say: [imath]I = \int_{-\infty}^{\infty} e^{-x^2} dx[/imath] Then we say: [imath]I = \int_{-\infty}^{\infty} e^{-t^2} dt[/imath] I want to see how this process works? We consider [imath]f(x) = e^{-x^2}[/imath] Right? Then we say that: [imath]f(t) = e^{-t^2}[/imath] Right? Ideally we are replacing x with t correct? Then we say: where R is the whole real axis. [imath]I^2 = \int_{R} \int_{R} e^{-(t^2 + x^2)} dtdx[/imath] But the problem is that now you consider t and x different axes, while before they lay on the same axis. How does this work? | 1091925 | How do "Dummy Variables" work?
I do not understand how dummy variables work in math. Suppose we have: [imath]I_1 = \int_{0}^{\infty} e^{-x^2} dx[/imath] How is this equivalent to: [imath]I_2 = \int_{0}^{\infty} e^{-y^2} dy[/imath] How does this dummy variable system work? Since [imath]y[/imath] is the dependent variable for [imath]I_1[/imath] How can [imath]y[/imath] itself be and independent variable for [imath]I_2[/imath] ?? Thanks! |
157876 | Normal subgroups of [imath]S_4[/imath]
Can anyone tell me how to find all normal subgroups of the symmetric group [imath]S_4[/imath]? In particular are [imath]H=\{e,(1 2)(3 4)\}[/imath] and [imath]K=\{e,(1 2)(3 4), (1 3)(2 4),(1 4)(2 3)\}[/imath] normal subgroups? | 1108156 | Symmetric group [imath]S_4[/imath]
Let [imath]G[/imath] be the Symmetric group [imath]S_4.[/imath] Give a representative of each conjugacy class of [imath]G.[/imath] Then calculate the size of each conjugacy class. I have no idea how to do this. |
1093475 | Deduce that if [imath]G[/imath] is a finite [imath]p[/imath]-group, the number of subgroups of [imath]G[/imath] that are not normal is divisible by [imath]p[/imath]
Given: Let [imath]G[/imath] be a group, and let [imath]\mathcal{S}[/imath] be the set of subgroups of [imath]G[/imath]. For [imath]g\in G[/imath] and [imath]H\in S[/imath], let [imath]g\cdot H=gHg^{-1}[/imath] Question: Deduce that if [imath]G[/imath] is a finite [imath]p[/imath]-group, for some prime [imath]p[/imath], the number of subgroups of [imath]G[/imath] that are not normal is divisible by [imath]p[/imath]. Comments: The subgroups of [imath]G[/imath] that are normal have the property that [imath]g\cdot H=gHg^{-1}=H[/imath] The question is equivalent to showing [imath]p[/imath] divides [imath]\left|G\right|-\left|\{H\in S\vert gHg^{-1}\neq H\}\right|[/imath] [imath]p[/imath]-group: [imath]\forall g\in G,|g|=p^k[/imath] for [imath]k\in\mathbb{N}^+[/imath] I don't know where to start with this problem, been thinking about it for a while to no avail, I feel that there are too many definitions for me to consider when finding a solution. My problems I have considered and not been able to answer are: what is the order of [imath]|G|[/imath]? I think it must be of order that is divisible by [imath]p[/imath], hence the question becomes show [imath]p[/imath] divides the order of [imath]\left|\{H\in S\vert gHg^{-1}\neq H\}\right|[/imath]. How do you compute the order of this set? It seems like quite a standard problem, so I would really appreciate it if I could be directed to more information about whatever problem it is. | 2748882 | Let [imath]|G| = p^n[/imath] where [imath]p[/imath] is prime. Show that the number of non-normal subgroups of [imath]G[/imath] is [imath]pm[/imath].
Let [imath]|G| = p^n[/imath] where [imath]p[/imath] is prime. Show that the number of non-normal subgroups of [imath]G[/imath] is [imath]pm[/imath]. I am just studying about group theory so please try to use some simple knowledge. |
783152 | Induction proof [imath]F(n)^2 = F(n-1)F(n+1)+(-1)^{n-1}[/imath] for n [imath]\ge[/imath] 2 where n is the Fibonacci sequence
Prove that [imath]F{_n}^2 = F_{n-1}F_{n+1}+(-1)^{n-1}[/imath] for n [imath]\ge[/imath] 2 where n is the Fibonacci sequence F(2)=1, F(3)=2, F(4)=3, F(5)=5, F(6)=8 and so on. Initial case n = 2: [imath]F(2)=1*2+-1=1[/imath] It is true. Let k = n [imath]\ge[/imath] 2 To show it is true for k+1 How to prove this? | 523925 | Induction proof on Fibonacci sequence: [imath]F(n-1) \cdot F(n+1) - F(n)^2 = (-1)^n[/imath]
I can't seem to solve this problem. It is: The Fibonacci numbers [imath]F(0), F(1), F(2),\dots [/imath] are defined as follows: \begin{align} F(0) &::= 0 \\ F(1) &::= 1 \\ F(n) &::= F(n-1) + F(n-2)\qquad(\forall n \ge 2)\end{align} Thus, the first Fibonacci numbers are [imath]0, 1, 1, 2, 3, 5, 8, 13,[/imath] and [imath]21[/imath]. Prove by induction that [imath]\forall n \ge1[/imath], [imath]F(n-1) \cdot F(n+1) - F(n)^2 = (-1)^n[/imath] I'm stuck, as I my induction hypothesis was the final equation, and I replaced n in it with n+1, which gave me: [imath]F(n) \cdot F(n+2) - F(n+1)^2 = (-1)^{n+1}[/imath] I then tried simplifying this using the first equation, which gave me: [imath][(F(n-1) + F(n-2)]\cdot F(n+2) - F(n+1)^2 = (-1)^{n+1}[/imath] I then tried replacing [imath]n[/imath] in the first equation with [imath]n+1[/imath], but that just gave me [imath]2F(n-1) + F(n-2)[/imath] I'm really not sure how to proceed, and I was hoping for some help. I'm new to induction and I'm hoping this is just an algebra problem and not a problem with the method, but any help would be greatly appreciated. |
711936 | Mathematical Induction proof that [imath]\sum\limits_{i=1}^n \frac{1}{i^2} < 2 - \frac1n[/imath]
I am to use mathematical induction to prove: [imath]\sum_{i=1}^n\frac{1}{i^2}<2 - \frac{1}{n}[/imath] my base case is n = 3: LHS: [imath]\frac{1}{1}+\frac{1}{4}+\frac{1}{9}= \frac{49}{36}[/imath] RHS: [imath]2-\frac{1}{3}=\frac{5}{3}[/imath] base case holds true. assume [imath]n = k[/imath] [imath]2-\frac{1}{k}[/imath] now show [imath]k+1[/imath] [imath]2-\frac{1}{k+1}[/imath] Inductive Step: [imath]2-\frac{1}{k}+ \frac{1}{{k+1}^2}[/imath] Now I am stuck I have no clue what math I should be doing from here. I thought about trying to FOIL the [imath]\frac{1}{{k+1}^2}[/imath] but that didn't really give me anything I could work with. Any tips? Or general pointers when doing this stuff? Thanks guys! | 33741 | Proof by Induction for inequality, [imath]\sum_{k=1}^nk^{-2}\lt2-(1/n)[/imath]
Let [imath]n[/imath] be a positive natural number, [imath]n\ge 2[/imath]. Then [imath]\displaystyle\sum_{k=1}^n \frac{1}{k^2} \lt 2 - \frac{1}{n}.[/imath] The basis step was easy but could someone give me a hint in the right direction as to how to do the induction step? I tried this: [imath]\displaystyle\sum_{k=1}^k \frac{1}{k^2} + \frac{1}{(k + 1)^2} \lt 2 - \frac{1}{k + 1}[/imath] But it's getting me nowhere or I am doing something wrong. I am no expert so a clear explanation would be appreciated. Thanks. |
1095590 | prove by inductive step
I have some problem to prove this statement by the Principle of mathematical Induction. [imath]\sum_{i=0}^{n} \binom{n}{i} = 2^n.[/imath] So I begin with the base step. For [imath]n=0[/imath], [imath]\sum_{i=0}^{0} \binom{0}{i} = 2^0 =1.[/imath] Now could you help me to show the inductive step? Thank you in advance! | 734900 | proof by induction: sum of binomial coefficients [imath]\sum_{k=0}^n (^n_k) = 2^n[/imath]
Question: Prove that the sum of the binomial coefficients for the nth power of [imath](x + y)[/imath] is [imath]2^n[/imath]. i.e. the sum of the numbers in the [imath](n + 1)^{st}[/imath] row of Pascal’s Triangle is [imath]2^n[/imath] i.e. prove [imath]\sum_{k=0}^n \binom nk = 2^n.[/imath] Hint: use induction and use Pascal's identity What I have so far: [imath]p(n)= \sum^{n}_{k=0}{n\choose k} = 2^n[/imath] [imath]P(0)=\sum^{0}_{k=0}{0\choose k} = 2^0[/imath] LHS: [imath]{0\choose k}=1[/imath] RHS [imath]2^0=1[/imath] now of for the inductive step I am getting all tangled up in my terms. It seems like a silly thing to get confused on but every inductive proof I've done I do [imath]P(n) =[/imath] something to [imath]P(1)[/imath] then my inductive step is [imath]P(k)[/imath] and [imath]P(k+1)[/imath] but I don't think can use that terminology because those variables are in the problem. Also the two (i.e) are throwing me off, am i even setting it up right to begin with or am I missing something. How does [imath](x+y)^n[/imath] even fit in to here, I know how it connects to pascals because I googled it but I don't understand how it fits into my inductive proof. I need help getting on track form here. PS my professor is a stickler for set up so making sure I write it properly is a big part of it. |
815963 | Proof by induction that [imath]1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}[/imath]
I am trying to understand how to do proof by induction for inequalities. The step that I don't fully understand is making an assumption that n=k+1. For equations it is simple. For example: Prove that 1+2+3+...+n = [imath] \frac {n(n+1)}{2} [/imath] is valid for [imath] i \ge 1 [/imath] 1) Base case for n=1. (...) 2) Assume that equation is true for n=k. (...) 3) Proof for n=k+1. 1+2+3+...+k+(k+1) = [imath] \frac {(k+1)[(k+1)+1]}{2} [/imath] So we add +1 to k on both sides. For inequalities: Prove that [imath] 4n \lt 2^n [/imath] is valid for [imath] n \ge 5 [/imath] 1) Base case. (...) 2) n=k assumption. (...) 3) n=k+1 proof. How I would do this(add 1 to k on both sides): [imath] 4(k+1) < 2^{k+1} [/imath] How it's done in tutorials: [imath] 4(k+1) < 2^k + 4 [/imath] [imath] 4k+4 < 2^k + 4 [/imath] Why they add extra 4 to RHS instead of increasing k by 1? | 1316472 | Prove by induction that [imath]\sum_{i=1}^{n} 2i=(n+1)n[/imath], for every positive integer n.
Can anyone explain the concept behind this? I just don't get how I should proceed with it? Like each step, why and how is it done? Prove by induction that [imath]\displaystyle\sum_{i=1}^{n} 2i=(n+1)n[/imath], for every positive integer [imath]n[/imath]. |
962587 | Proof by induction steps
Today in class, the instructor is trying to show that for [imath]n \ge 0[/imath], [imath]n < 2^n[/imath]. And this are the steps he took: First we assume the inductive hypothesis i.e. [imath]0 < 2^0[/imath], and this is true. Then to prove that for any arbitrary integer [imath]k \ge 0[/imath], [imath]k < 2^k[/imath]. The proof proceeds as follows: Assume [imath]k < 2^k[/imath] is true for [imath]k ge 0[/imath]. Show that [imath](k + 1) < 2 ^{k + 1}[/imath]. By induction hypothesis, [imath]k < 2^k[/imath] thus [imath]k + 1 < 2^k + 1 < 2^k + 2^k = 2^{k + 1}[/imath]. Thus by mathematical induction the result has been proven My question now is how was step 3 derived? Where does the extra [imath]2^k[/imath] manifest from? | 928873 | Prove by induction that [imath]n < 2^n[/imath] for all [imath]n \ge 1[/imath]
I'm trying to do homework problems and for the most part I've been getting the results. For this one though, I am having some trouble since its [imath]2^n[/imath] and I can't relate it properly: Prove using simple induction that [imath]n < 2^n[/imath] for all [imath]n \ge 1[/imath]. So obviously, the basis step holds as [imath]1 < 3[/imath]. Now, I assume that [imath]n = k[/imath] holds as well and have to prove for [imath]n=k+1[/imath]. This is the step I am having trouble with as I cannot relate the induction hypothesis with what I want to end up with. Can anybody show me a model solution for this one? I think my trouble comes because of the [imath]2^n [/imath] |
1096003 | Why do we have [imath]ab=ba[/imath]?
Why is it that [imath]ab=ba[/imath] for positive integers [imath]a,b[/imath]? Is there an intuitive explanation or we just need to accept it as a given fact? | 111603 | On the commutative property of multiplication (domain of integers, possibly reals)
[imath]ab = ba[/imath] This is, inherently, true. Some texts drop it like an axiom without any justification. But I'm a bit curious where it stems from or basically why/how it works. If anyone could enlighten me a bit further, I'd be most grateful. Peano axioms? From set theory? Help. What bugs me with this definition is it's real world "application": Let's say I have a power output of 5 W ( [imath]kg\cdot m^2 \cdot s^{-3}[/imath]) and I want 30% of that power output (to have some units and natural context). Naturally, 30% is 30 1/100 (% is the unit). For simplicity, let's express that as: [imath]30[/imath]%[imath] = 30/100 = 3/10 = 3d[/imath] where [imath]d = 1/10[/imath], a simpler unit (because %W would look... Wrong?). So 30% of power is then [imath]5 W \cdot 3d[/imath], and we can force it to "make sense" if we associate d with W or basically scale the unit of power by d (which is intuitively understood as being 10 times smaller then the output of 1 W) [imath]dW[/imath]. And we want to scale these 5 units of [imath]dW[/imath] three times. [imath]5 dW \cdot 3[/imath] But the commutative property says it is the same as scaling 3 units of [imath]dW[/imath] 5 times or: [imath]3 dW \cdot 5[/imath] And the result is the same. This is the bit that hurts my head, the fact it is the same. I try to interpret the same as, for example 50 J of work, it's either applying 50 N over 1 m of distance or 1 N over 50 m of distance. Down here is additional stuff I think I have (you don't have to read it if you know how to answer immediately). I don't claim it's correct, if anyone parses this information, feel free to correct me. [imath]m\sum\limits_{i=1}^{n}{1} = n\sum\limits_{i=1}^{m}{1}[/imath] This is how I've been trying to break it down, as it is repeated addition. It is repeated [imath]n[/imath] times and multiplied by [imath]m[/imath], and reverse on the right side. There is [imath]x = m - n[/imath], so when I multiply [imath]n[/imath] [imath]m[/imath] times, each repeated addition of [imath]n[/imath] lacks exactly [imath]x[/imath] to [imath]m[/imath]. [imath]mn = nm[/imath] [imath]x = m-n[/imath] [imath]m(m-x) = (m-x)m[/imath] [imath]m^2 - mx = m^2 - mx[/imath] [imath] true [/imath] And yes, sadly, I realise that I am using the very property I'm trying to prove ([imath]xm[/imath] shifted as [imath]mx[/imath] on the right-hand side.) And it even calls in the distributive property of multiplication over addition which I derive (for myself, informally) from the very nature of multiplication and the way we "process" numbers: For example, number [imath]55[/imath] times [imath]2[/imath] [imath]55 = 5 \cdot 10^1 + 5 \cdot 10^0 = 5 \cdot 10 + 5 \cdot 1[/imath] [imath](5 \cdot 10 + 5 \cdot 1) \cdot 2[/imath] This is the basis of our positional notation which appends digits of varying orders of magnitude or units together in a way it makes "sense". Each one is b times bigger than the one to the right. Basic stuff. It is why I expressed 10 and 1 explicitly, those are what I consider units in this case. Now, from the definition of multiplication, which is at its heart just repeated addition, it is truly the same if you add together [imath]2[/imath] [imath]50[/imath] times and then again [imath]5[/imath] times or "all at once" (figuratively, we are always doing the former mentally) [imath]2[/imath] [imath]55[/imath] times. That added together gives 110, which is true. To me, the distributive property (in case of integer multiplication) is a repercussion of the very definition of multiplication which is repeated addition (evading strict concepts). I am not certain whether this is a good way to look at it. |
1095750 | Is my induction on euler's formula sufficient?
[imath]n-m+l=2[/imath], where n=#vertices, m=#edges, l=#faces. I've been asked to demonstrate an intuitive, inductive proof (whatever the hell that means). Anyway so I've shown it's true for a tree, where necessarily [imath]m \leq n-1[/imath], and also for when [imath]m=n[/imath]. Am I done, or do I need to show it also for when [imath]m \geq n+1[/imath]? | 252015 | Inductive Proof of Euler's Formula [imath]v-e+r=2[/imath]
I'm just studying for finals here. My professor told me that there would be an inductive proof on the final, and I've never done one before. He told me a good sample problem was to prove Euler's formula [imath]v-e+r=2[/imath] inductively. I've submitted my proof below. I'm just looking for criticism / corrections! Is it a proper inductive proof? If not, could you show me one / make corrections? |
1096349 | Examples of functions [imath]f:\mathbb Z \to \mathbb Z:\;[/imath] surective, injective, bijective?
I'm having a problem finding functions [imath]f:Z \rightarrow Z[/imath], such that: [imath]f[/imath] is onto but not one-to-one, [imath]f[/imath] is one-to-one but not onto, [imath]f[/imath] is neither one-to-one or onto, [imath]f[/imath] is both one-to-one and onto (bijective). Please help me thank you | 111851 | Surjective, Injective, Bijective Functions from [imath]\mathbb Z[/imath] to itself
Can you give me examples for the following [imath]f:\mathbb Z \to \mathbb Z[/imath] (for integer functions) injective but not surjective surjective but not injective surjective and injective neither surjective nor injective |
1096661 | A question about the orders of the elements of a group
Let [imath]m[/imath] and [imath]n[/imath] be to positive integers strictly larger than [imath]1[/imath]. Is it possible to find a group [imath]G[/imath] in which there are two elements, say [imath]a[/imath] and [imath]b[/imath], such that the order of [imath]a[/imath] is [imath]m[/imath], the order of [imath]b[/imath] is [imath]n[/imath] and their product has infinite order? If [imath]m[/imath] and [imath]n[/imath] are not coprime I think I have solved using a restricted wreath product (which is then directly summed to some cyclic group); but in the general case I could not achieve anything. | 1083311 | For any integers [imath]m,n>1[/imath] , does there exist a group [imath]G[/imath] with elements [imath]a,b \in G[/imath] such that [imath]o(a)=m , o(b)=n[/imath] but [imath]ab[/imath] has infinite order ?
For any integers [imath]m,n[/imath] , both greater than [imath]1[/imath] , does there exist a group [imath]G[/imath] with elements [imath]a,b \in G[/imath] such that [imath]o(a)=m , o(b)=n[/imath] but [imath]ab[/imath] has infinite order ? |
1096562 | Prove that there exists a linear transformation [imath]T: \mathbb R^2 \rightarrow \mathbb R^3[/imath] such that [imath]T(1,1)=(1,0,2)[/imath] and [imath]T(2,3)=(1,-1,4)[/imath]
Prove that there exists a linear transformation [imath]T: \mathbb R^2 \rightarrow \mathbb R^3[/imath] such that [imath]T(1,1)=(1,0,2)[/imath] and [imath]T(2,3)=(1,-1,4)[/imath] I know how to prove that a map is linear if I'm given the general rule the map is defined by. But that's not given here and I don't know how to find it from the two particular values given. Please help! | 107660 | How to prove there exists a linear transformation?
I've been given the following problem as homework: Prove that there exists a linear transformation [imath]T: \mathbb{R}^2 \to \mathbb{R}^3[/imath] such that [imath]T(1,1) = (1,0,2)[/imath] and [imath]T(2,3) = (1,-1,4)[/imath]. Since it just says prove that one exists, I'm guessing I'm not supposed to actually identify the transformation. One thing I tried is showing that it holds under addition/multiplication in the sense of: 1) [imath]T(x + y) = T(x) + T(y)[/imath] 2) [imath]T(cx) = cT(x)[/imath] 3) [imath]T((1,1) + (2,3)) = T(1,1) + T(2,3)[/imath]? But that's not necessarily true. I don't know how I'd accomplish this given my limited knowledge. What's the approach to solving a problem like this? |
1096590 | Do an infinite set and its double have the same cardinality?
My question was inspired by this answer. Suppose [imath]A[/imath] is an infinite set. Does its double, [imath]A\times\{0,1\}[/imath], always have the same cardinality? In my head I quickly spotted a simple proof that the answer is yes if we assume Axion of choice / Zorn's lemma. I have stumbled upon a couple of related questions, but the question looks so simple that I would like it answered without assuming AC. | 393196 | The relationship of [imath]{\frak m+m=m}[/imath] to AC
Two simple questions: (Of course [imath]{\frak m}[/imath] denotes a cardinal in the weak sense in the claims below.) Can we prove in ZF that [imath]\aleph_0\le{\frak m\Rightarrow m+m=m}[/imath]? If not, what is the relationship of this statement for all cardinals [imath]{\frak m}[/imath] to AC? The weaker principle [imath]\aleph_0\le{\frak m\Rightarrow m}+1={\frak m}[/imath] can be derived in ZF, and it is a well-known result of Tarski that [imath]\forall{\frak m}\ge\aleph_0,{\frak m\times m=m}[/imath] implies the axiom of choice, but I am not sure about the intermediate result with [imath]{\frak m+m=m}[/imath]. Edit: A related question: is it consistent with ZF that there exists a cardinal [imath]{\frak m}\ge\aleph_0[/imath] such that [imath]{\frak m=n+p}[/imath] with [imath]{\frak n<m}[/imath] and [imath]{\frak p<m}[/imath]? It's hard to say whether [imath]{\frak m+m=m}[/imath] precludes this scenario. |
1088493 | let [imath]G[/imath] to be group such that [imath]O(G)=p^2[/imath] where [imath]p[/imath] is prime,prove that [imath]G[/imath] is cyclic or [imath]G[/imath] is Direct product of two cyclic subgrops of order [imath]n[/imath].
the only hint that i got is Sylow's first theorem, which implies that if [imath]p^n[/imath] is any prime power dividing [imath]O(G)[/imath], then [imath]G[/imath] has a subgroup of order [imath]p^n[/imath]. in our case [imath]p[/imath] devides [imath]p^2[/imath], then we can say that we have a subgroup of order [imath]p[/imath] and we know that every group of prime order is cyclic. how can i continue from here ? | 1075067 | Group of order [imath]p^2[/imath] is commutative with prime [imath]p[/imath]
Please help me on this one: Let [imath]p[/imath] be a prime number, show that each group of order [imath]p^2[/imath] is commutative. If you do not mind at all, could you please not give me the elegant explanation, but instead only the dummies' way? I know it must be against your craft skill, but I am self-studying and my knowledge is spotty at the very best. Thank you for your patience. PS. The text then hints using these lemmas, but if you have simpler explanation, do please give me that one instead. Thanks again. Lemma 1: If [imath]G[/imath] finite and [imath]G/Z(G)[/imath] cyclic, [imath]G[/imath] is commutative. Lemma 2: [W. Burnside] [imath]G[/imath] is called primary if there exists a prime number [imath]p[/imath] such that [imath]G[/imath] is a [imath]p[/imath]-group. If [imath]G \neq \{1\},[/imath] then [imath]Z(G) \neq \{1\}.[/imath] |
1096761 | The identity [imath]a^n-b^n=(a-b) (\sum_{i=0}^{n-1}a^ib^{n-1-i})[/imath]
How do I use finite induction to prove that [imath]a^n-b^n=(a-b) (\sum_{i=0}^{n-1}a^ib^{n-1-i}), \forall a,b\in \Bbb{R}\space \text{and} \space \forall n \in \Bbb{N}?[/imath] Ok, for [imath]n=2[/imath] it's fine. [imath]a^2-b^2=(a-b)(b+a)=(a-b)(\sum_{i=0}^1a^ib^{n-1-i})[/imath]. Right, then suppose the equality valid for n. But, how to handle with [imath]a^{n+1}-b^{n+1}[/imath]? How to use the induction hypothesis here? Need some help! If this answers has already been answered, please, give me the link... Sorry about that. | 430645 | Prove that [imath](a-b) \mid (a^n-b^n)[/imath]
I'm trying to prove by induction that for all [imath]a,b \in \mathbb{Z}[/imath] and [imath]n \in \mathbb{N}[/imath], that [imath](a-b) \mid (a^n-b^n)[/imath]. The base case was trivial, so I started by assuming that [imath](a-b) \mid (a^n-b^n)[/imath]. But I found that: \begin{align*} (a-b)(a^{n-1}+a^{n-2}b + a^{n-3}b^2+...+b^{n-1}) &= a^n-b^n. \end{align*} Doesn't this imply that [imath](a-b) \mid (a^n-b^n)[/imath] as [imath]a^{n-1}+a^{n-2}b+\dots+b^{n-1}[/imath] is clearly an integer? This obviously isn't a proof by induction, but is there anything wrong with taking this approach to prove this result, other than the fact that it isn't what is being asked? Aside, I'm having trouble getting started with the proof by induction. I've tried making the above equivalence useful during the induction step, but it doesn't seem to be helpful, so any hints would be greatly appreciated! |
1097376 | Proving that [imath]A^2 x=x[/imath] implies [imath]Ax = x[/imath] for a matrix of [imath]n\times n[/imath]
Let [imath]A[/imath] be an [imath]n\times n[/imath] matrix and [imath]x\in \mathbb{R}^n[/imath], both with positive real entries. Prove that if [imath]A^2 x=x[/imath] then [imath]Ax = x[/imath]. | 271922 | Prove that if [imath]A^2x=x[/imath] then [imath]Ax=x[/imath]
I feel this should be easy but I cant solve this problem: Prove that if [imath]A[/imath] is a [imath]n\times n[/imath] matrix and [imath]x[/imath] a vector in [imath]\mathbb R^n[/imath] both with real positive entries and [imath]A^2x=x[/imath] then [imath]Ax=x[/imath]. I looking at the terms of the sum that defines the product [imath]AAx[/imath] and comparing with the entries of [imath]x[/imath] but I get nowhere. Can you give me any hints? |
806480 | Subgroup of abelian group
Let [imath]G[/imath] be abelian group and [imath]H=\{ x^{2}/x\in G\}[/imath] [imath]H[/imath] is it a subgroup of [imath]G[/imath] ? H is subset of a group G is a subgroup if : (i)[imath]\quad e \in H[/imath] (e is The identity element of G) (ii) if [imath]x, y \in H[/imath], then [imath]x\ast y \in H;[/imath] (iii) if [imath]x \in H[/imath], then [imath]x^{-1} \in H.[/imath] indeed, For ii) let [imath]x,y \in H[/imath] then [imath]x \ast y=(xy) \in H[/imath] i'm stuck in that level so i can't prove (i) or (ii) true or flase If H is subgroup of G can i general that for [imath]H^{n}=\{ x^{n}/x\in G\}\quad \forall n \in \mathbb{N} [/imath] | 537587 | Prove that [imath]H[/imath] is a subgroup of an abelian group [imath]G[/imath]
Let [imath]H = \{x \in G: x = y^2[/imath] for some [imath]y\in G\}[/imath]; that is, let [imath]H[/imath] be the set of all the elements of [imath]G[/imath] which have a square root. Prove that [imath]H[/imath] is a subgroup of [imath]G[/imath], where [imath]G[/imath] is an abelian group. So I'm pretty lost in group theory. I know that I need to prove that for any element, [imath]a,b \in H[/imath], that [imath](a^{-1} * b) \in H[/imath], where * is the group operation on [imath]G[/imath]. I know that I also need to use the fact that [imath]G[/imath] is abelian. So somehow commutativity is involved. But I admit that I have no idea what I am doing. I guess my primary question is: What does [imath]x^{-1}[/imath] look like in [imath]H[/imath]? Since [imath]x[/imath] = [imath]y^2[/imath] then [imath]x^{-1} = (y^2)^{-1}[/imath], which is what? |
1097668 | Fibonacci, prove that [imath]F_{n}\cdot F_{n+2}-({F_{n+1}})^2=(-1)^n[/imath] with induction
I need to prove by induction that: [imath]F_{n}\cdot F_{n+2}-({F_{n+1}})^2=(-1)^n[/imath] I did the following: Check if the statement holds for [imath]n=1[/imath]: [imath]1\cdot 3-(2)^2=(-1)^1[/imath] Check if the statement holds for [imath]n=p+1[/imath]: [imath]F_{p+1}\cdot F_{p+3}-({F_{p+2}})^2=(-1)^{p+1}[/imath] I am having some troubles with this. I don't know how to continue. | 1031575 | Proving this [imath]F_{n+1} \cdot F_{n-1} - F^2_n = (-1)^n[/imath] by induction
Where [imath]n \in \mathbb{N}[/imath] and [imath] F_n = \begin{cases} 0 & \text{ if } n = 0 \\ 1 & \text{ if } n = 1 \\ F_{n-1} + F_{n-2} & \text{ if } n > 1 \end{cases} [/imath] This is basically describing the famous Fibonacci sequence. If we try the base case with 1, it works (for [imath]0[/imath] I am not sure...) When [imath]n = 1[/imath] [imath]2 \cdot0 - 1^2 = (-1)^1 = -1[/imath] For the inductive hypothesis, we assume this [imath]F_{n+1} \cdot F_{n-1} - F^2_n = (-1)^n[/imath] is true. Now, for the inductive step, we try to prove for [imath]n+1[/imath], so for [imath]F_{n+2} \cdot F_{n} - F^2_{n+1} = (-1)^{n+1}[/imath]. Since [imath]n[/imath] is always a natural number, and it will be always or even or odd, the [imath]-1[/imath] raised to [imath]n[/imath] will be always either [imath]-1[/imath] (when [imath]n[/imath] is odd) or [imath]1[/imath] (when [imath]n[/imath] is even). Thus, [imath]F_{n+1} \cdot F_{n-1} - F^2_n[/imath] = -([imath]F_{n+2} \cdot F_{n} - F^2_{n+1}[/imath]). Or simply: [imath] (-1)^n = -(-1)^{n+1} [/imath] [imath] 1 \cdot (-1)^n = -(-1)\cdot(-1)^n [/imath] [imath] 1 = -(-1) [/imath] Which is true. I don't know if this is sufficient... I can arrive at the second step and say that we know this [imath](-1)^n[/imath] is equals to [imath]F_{n+1} \cdot F_{n-1} - F^2_n[/imath], but I am not sure... |
1097805 | Does [imath]\sum_{n = 2}^{\infty} \frac{\sqrt{n + 1}}{n(n-1)}[/imath] converge or diverge?
This is a homework question so I'll show what I've tried. The integral test gives convergence, but I have to use [imath]tanh[/imath], which is not part of our high-school syllabus. I've tried doing the limit comparison test with [imath]\frac{1}{n}[/imath] and [imath]\frac{1}{n^2}[/imath] both of which give zero as the limit. D'Almbert's criterion is also inconclusive. | 129547 | Determine the convergence of [imath]\sum\limits_{n=2}^\infty \frac{\sqrt{n+1}}{n(n-1)}[/imath]
So I need to know the convergence of this sum. The integral test does not seem to work (WolframAlpha gives a confusing answer which implies it's not the correct method). I also tried the limit comparison test for various harmonic series, but they are all inconclusive. I'm also not getting very far with the ratio test. [imath] \sum_{n=2}^\infty \dfrac{\sqrt{n+1}}{n(n-1)} [/imath] |
1096741 | A rank-nullity theorem between [imath]\mathbb Z^n[/imath] and [imath]\mathbb Z^k[/imath]
I think this is correct: If [imath]\phi:\mathbb Z^{n}\to\mathbb Z^{k}[/imath] is a group homomorphism then [imath]n=\operatorname{rank}\operatorname{im}\phi+\operatorname{rank}\ker\phi[/imath]. Here is my attempt at a proof: [imath]\phi[/imath] may be naturally and uniquely extended to a linear map between [imath]\mathbb Q[/imath]-vector spaces [imath]\widetilde{\phi}:\mathbb Q^{n}\to\mathbb Q^{k}[/imath], where [imath]n=\dim\operatorname{im}\widetilde{\phi}+\dim\ker\widetilde{\phi}[/imath]. My questions: Is this proof correct? I am concerned with "rank" either not making sense for subgroups or not coinciding with "dimension". Does this appear in books (maybe with different formulation)? I have not found it in some standard textbooks I've looked in. | 999806 | Rank-nullity theorem for free [imath]\mathbb Z[/imath]-modules
From linear algebra we know that given vector spaces [imath]V[/imath], [imath]W[/imath] over a field [imath]k[/imath] and a linear map [imath]f\colon V\to W[/imath] we have [imath]\dim V = \dim \operatorname{im} f + \dim \ker f.[/imath] Is this still true when we consider free [imath]\mathbb Z[/imath]-modules (i.e. free abelian groups) instead of vector spaces? Given a homomorphism [imath]f\colon G\to H[/imath] between free [imath]\mathbb Z[/imath]-modules, do we have [imath]\operatorname{rk}(G) = \operatorname{rk}\operatorname{im} f + \operatorname{rk}\ker f?[/imath] To provide some context: This question comes up when computing homology groups of free chain complexes, where we need to check if some generating set of a kernel is a basis. Using the rank nullity theorem for free [imath]\mathbb Z[/imath]-modules is an appealing way to do this. |
1090446 | All [imath]f[/imath] such that [imath](\exists k)(f^\prime(x) = f(x+k))[/imath]
I was wondering if there is a general way to solve the functional equation [imath](\exists k)(f^\prime(x) = f(x+k))[/imath] I know that this is true for certain functions: [imath](e^{cx})^\prime = e^{c(x+\frac{\ln c}{c})}[/imath] [imath]\sin(x)^\prime=\sin(x+\tfrac{\pi}{2})[/imath] but I was wondering if these functions take a general form or are related to the eigenvalues of the second derivative, because both of these are. | 815687 | Are Exponential and Trigonometric Functions the Only Non-Trivial Solutions to [imath]F'(x)=F(x+a)[/imath]?
Are exponential & trigonometric functions the only non-trivial solutions to [imath]F'(x)=F(x+a)[/imath]? [imath]F(x)=0[/imath] would be the trivial solution. Then, for [imath]a=0[/imath] (or [imath]a=2\pi i[/imath]), we have [imath]F(x)=e^x[/imath], and for [imath]a=\dfrac\pi2[/imath] there are [imath]F(x)=\sin x[/imath] and [imath]F(x)=\cos x[/imath]. But the three are connected by Euler's formula [imath]e^{ix}=\cos x[/imath] [imath]+i\sin x[/imath]. Indeed, on a more general note, letting [imath]F(x)=e^{\lambda x}[/imath], we have [imath]\lambda=\dfrac{W(-a)}{-a}[/imath] where W is the Lambert W function. My question would be if these are the only ones, due to the special properties of the number e and the exponential function, or if there aren't by any chance more, which do not belong in the same family or category as these, i.e., which are not exponential or trigonometric in nature ? Thank you. |
1098610 | How can I prove [imath]\arctan x < x[/imath] for [imath]x > 0[/imath]
How can I prove [imath]\arctan\ x \lt x[/imath] for x [imath]> 0[/imath]? I have tried different stuff without success. | 1098253 | how can I prove that [imath]\frac{\arctan x}{x }< 1[/imath]?
I have got some trouble with proving that for [imath]x\neq 0[/imath]: [imath] \frac{\arctan x}{x }< 1 [/imath] I tried doing something like [imath]x = \tan t[/imath] and playing with this with no success. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.