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1278170 | Prove that if an even function [imath]f(x)[/imath] is [imath]C^2[/imath], [imath]f(|x|)[/imath] is also [imath]C^2[/imath]
Let [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] be a [imath]C^2(\mathbb{R})[/imath] function which is also even (ie, [imath]f(-x) = f(x)[/imath]). Prove that the function [imath]F: \mathbb{R^n} \rightarrow \mathbb{R}[/imath] defined by [imath]F(x) = f(|x|)[/imath] (where [imath]|x|[/imath] denotes the Euclidean norm of the vector [imath]x[/imath]) is also [imath]C^2(\mathbb{R^n})[/imath]. It feels so simple, but I'm just not getting it, so any help would be appreciated. Thanks! | 1277242 | Proving function is [imath]C^k[/imath]
This question is from an exercise in Way of Analysis (section 10.2.4 problem 20). If [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] is [imath]C^k[/imath] and [imath]f[/imath] is even, then show [imath]F: \mathbb{R}^n \rightarrow \mathbb{R}[/imath] given by [imath]F(x) = f(|x|)[/imath] is also [imath]C^k[/imath]. [imath]C^k[/imath] just means the function has [imath]k[/imath] continuous derivatives, and [imath]|x|[/imath] is referring to the regular Euclidean norm. I am not sure how to use the fact [imath]f[/imath] is even in the proof and having trouble for a general [imath]k[/imath]. Any hints, even on a particular case like [imath]k=2[/imath], are welcome. My workings: As a comment suggests [imath]f(x) = F(x)[/imath] for [imath]n = 1[/imath]. For [imath]n = 2[/imath], [imath]F(x) = f(\sqrt {x_1^2 + x_2^2}).[/imath] I'm stuck on where to proceed from here. Is there some form of induction here that I am not seeing? |
1278268 | Prove that orthogonal n x n matrices form a [imath]C^1[/imath] surface of dimension n(n-1)/2 in [imath]\mathbb{R^n}^2[/imath]
Consider the function [imath]F:Mat_n → Sym_n[/imath] defined by the formula [imath]F(A) = A∗A[/imath]. [imath]Mat_n[/imath] denotes the vector space of n × n matrices with real entries, while [imath]Sym_n[/imath] denotes the vector space of symmetric [imath]n × n[/imath] matrices with real entries. We will also identify [imath]Mat_n[/imath] with [imath]\mathbb{R^n}^2[/imath], and [imath]Sym_n[/imath] with [imath]\mathbb{R^\frac{n(n+1)}{2}}[/imath] I have a general layout of the proof, but no idea how to actually implement it. First, I should show that for every [imath]A ∈ Mat_n[/imath], the differential map [imath]DF(A)[/imath] is given by [imath][DF(A)](B)=A∗B+B∗A[/imath]. Next, I should prove that for every invertible matrix [imath]A∈Mat_n[/imath], one has [imath]dim(ker(DF(A)))=\frac{n(n − 1)}{2}[/imath]. From all this, I should conclude that the orthogonal [imath]n×n[/imath] matrices form a [imath]C^1[/imath] surface of dimension [imath]\frac{n(n − 1)}{2}[/imath] in [imath]\mathbb{R^n}^2[/imath]. | 552293 | Use implicit function theorem to show [imath]O(n)[/imath] is a manifold
In class today our teacher mentioned that one can use the implicit function theorem to show that [imath]O(n) \subseteq \mathbb{R}^{n^2}[/imath] is a submanifold...that is, map [imath]A \mapsto A^* A[/imath], and set it equal to the identity matrix. There should be [imath]n(n+1)/2[/imath] equations in [imath]n^2[/imath] variables ([imath]A^*A[/imath] is always symmetric, so we only need [imath]n(n+1)/2[/imath] equations). When I asked how to show that the jacobian matrix of this transformation had full row rank (the condition of the implicit function theorem), my teacher said "take the derivative near the identity, and you see that the derivative is just a matrix plus its transpose ." I wasn't sure what he meant. Can someone clarify? |
1257259 | For groups [imath]A,B,C[/imath], if [imath]A\times B[/imath] and [imath]A\times C[/imath] are isomorphic do we have [imath]B[/imath] isomorphic to [imath]C[/imath]?
The motivation for this question is in my answer to this post : Compatibility of direct product and quotient in group theory (1) If I take [imath]A[/imath], [imath]B[/imath] and [imath]C[/imath] three groups such that [imath]A\times B[/imath] is isomorphic to [imath]A\times C[/imath] then do we have [imath]B[/imath] is isomorphic to [imath]C[/imath] ? In the post a counter example is given when [imath]A[/imath] is an infinitely generated group. However I cannot get a counter-example when [imath]A[/imath], [imath]B[/imath] and [imath]C[/imath] are assumed to be finite groups. Note that there is no reason a priori if [imath]\phi[/imath] is an isomorphism from [imath]A\times B[/imath] to [imath]A\times C[/imath] that [imath]\phi(A\times\{1_B\})=A\times\{1_C\}[/imath]. In the case of finite groups it is true because of the Krull-Schmidt theorem : http://en.wikipedia.org/wiki/Krull%E2%80%93Schmidt_theorem. A slightly more general question is the following : (2) If [imath]B[/imath] and [imath]C[/imath] are two finitely generated groups, can I always find a finitely generated group [imath]A[/imath] such that [imath]A\times B[/imath] is isomorphic to [imath]A\times C[/imath]? If [imath]A[/imath] is not assumed to verify any property but to be a group we can take : [imath]A=B^{\mathbb{N}}\times C^{\mathbb{N}} [/imath] Furthermore replacing :[imath] \prod_{\mathbb{N}}[/imath] by [imath]\bigoplus_{\mathbb{N}}[/imath] (or more likely the restricted product for groups but I am not sure of a well known notation, it is the subgroup of the product with all but a finite number components being trivial) will give you the same result for countably infinite groups. [imath]\underline{\text{Edit}}[/imath] : To be clear about what I am asking. (1) is kown to be true for finite groups whereas (2) is known to be true when we are considering groups without restriction (even just for countably infinite groups). Let us pretend we are working on the smaller class of groups : the finitely generated groups. Is there any counter example to [imath](1)[/imath]? Is there any counter example to [imath](2)[/imath]? [imath]\underline{\text{Edit 2}}[/imath]: Here is a partial answer, from the classification of finitely generated abelian groups [imath](1)[/imath] is always true if all groups are assumed to be abelian finitely generated groups. Now assume that [imath]A\times B=G=A\times C[/imath]. Then you have (derived groups) : [imath]D(G)=D(A)\times D(B)=D(A)\times D(C) [/imath] So the abelianizations verify : [imath]A^{ab}\times B^{ab}=A^{ab}\times C^{ab}[/imath] Because we are now in the realm of finitely generated abelian groups we get : [imath]B^{ab}=C^{ab} [/imath] So [imath](2)[/imath] cannot be true in general for instance you can take [imath]B:=\mathbb{Z}[/imath] and [imath]C=\mathbb{F}_5[/imath], those two groups have different abelianizations. The new question can be found here : About a relation of non-discernability between (classes of) finitely generated groups. | 882968 | [imath]A\oplus C \cong B \oplus C[/imath]. Is [imath]A \cong B[/imath] when [imath]C[/imath] is finite, A and B infinite.
So my question is simply that for groups [imath]A, B, C,[/imath] if C is finite, A and B infinite and [imath]A\oplus C \cong B \oplus C[/imath], is [imath]A \cong B[/imath]? My gut tells me this must be the case, and logically I can find no reason they shouldn't be, but I can not seem to derive a formal proof of this for the life of me. Note that I am well aware that this is not always true for infinite C. |
1278401 | What is [imath]13^{498}[/imath] (mod [imath]997[/imath])?
I have to determine [imath]13^{498} \pmod{997}[/imath] I know that it can only be [imath]1[/imath] or [imath]-1[/imath]. But I don't quite know which. How can I decide? | 195634 | How do you calculate the modulo of a high-raised number?
I need some help with this problem: [imath]439^{233} \mod 713[/imath] I can't calculate [imath]439^{223}[/imath] since it's a very big number, there must be a way to do this. Thanks. |
742858 | Descartes' Rule of Signs
I have read several places that Descartes' Rule of Signs was familiar to both Descartes and Newton, and that both considered it too "obvious" to merit a proof. I know how to prove it, but I would like to know how they intuitively sensed that it was true. Newton apparently used it in one of his books, so he must have had a good reason to believe it was true if he never bothered to attempt a proof. Just for clarification, I am referring to the theorem that the number of positive roots of the polynomial [imath]p(x)=a_nx^n+\cdots +a_1x+a_0[/imath] is equal to or less than by an even number the number of sign changes in [imath]p[/imath] as written in the order above (descending powers of [imath]x[/imath]). Thanks! | 745583 | Intuition behind Descartes' Rule of Signs
I have read several places that Descartes' Rule of Signs was familiar to both Descartes and Newton, and that both considered it too "obvious" to merit a proof. I know how to prove it, but I would like to know how they intuitively sensed that it was true. Newton apparently used it in one of his books, so he must have had a good reason to believe it was true if he never bothered to attempt a proof. Just for clarification, I am referring to the theorem that the number of positive roots of the polynomial [imath]p(x)=a_nx^n+⋯+a_1x+a_0[/imath] is equal to or less than by an even number the number of sign changes in p as written in the order above (descending powers of x). |
1245128 | A regular surface with non zero mean curvature is orientable
How can I prove that any regular surface with non zero mean curvature is orientable? UPDATE: The surface is embedded in [imath]\mathbb{R}^3[/imath]. | 389596 | Surface with non-zero mean curvature means orientable
Let [imath]M[/imath] be a surface in [imath]\Bbb R^3[/imath] with non-zero mean curvature for every point. How could I show that this implies that [imath]M[/imath] is orientable? By our definition, orientable means that an unitary, normal vector can be defined continuously for every point in the surface. This is a homework question, so just hints would be appreciated :) |
1279570 | For [imath]u,v[/imath] nonzero, show [imath]u\cdot v = 0[/imath] implies [imath]u[/imath] and [imath]v[/imath] are linearly independent.
I know that if [imath]u\cdot v = 0[/imath] then by definition, [imath]u_1v_1 + u_2v_2 + \cdots + u_nv_n = 0[/imath] I also know that if [imath]u[/imath] and [imath]v[/imath] are linearly independent and a matrix [imath]A[/imath] has [imath]u[/imath] and [imath]v[/imath] as its successive column vectors then the equation [imath]Ax = 0[/imath] will have only the trivial solution. However I can't think of anything else that might help me answer this question. | 953512 | If the dot product between two vectors is [imath]0[/imath], are the two linearly independent?
If we have vectors [imath]V[/imath] and [imath]W[/imath] in [imath]\mathbb{R^n}[/imath] and their dot product is [imath]0[/imath], are the two vectors linearly independent? I can expand [imath]V_1 \cdot V_2 = 0 \Rightarrow v_1w_1+...+v_nw_n = 0[/imath], but I don't understand how this relates to linear independence. |
1278528 | Showing that language [imath]L^{'}[/imath] is regular given [imath]L[/imath] is regular
Say [imath]L \subseteq \{a,b\}^*[/imath] is a regular language with words whose length is divisible by 3. Each word [imath]w \in L[/imath] has the form [imath]w=xyz[/imath] with [imath]|x|=|y|=|z|[/imath], where [imath]y[/imath] is then called the middle third of [imath]w[/imath]. I want to show that the language [imath]L^{'}=\{y \in \{a,b\}^*:y \text{ is the middle third of a word } w \in L\}[/imath] is regular. I could think of a description of language [imath]L[/imath] with the regular expression [imath]((a+b)(a+b)(a+b))^*[/imath]. Would this be correct? Unfortunately, I have no idea how to start to show that [imath]L^{'}[/imath] is regular. Could you please give me some hints to get started? I'd like to know how to solve this problem using an automaton, e.g. a NFA. Thank you very much in advance. | 1229040 | show that language [imath]L'[/imath] is regular (given [imath]L[/imath] regular)
Let [imath]L[/imath] be a regular language. Show that [imath]L'=\{x \mid\exists_{y,z} xyz\in L \text{ and }|x|=|y|=|z|\}[/imath] is also regular. Firstly I show my idea. When you accept it I will try to formalize it. Every automat can have exactly one accept state. So let automat for language [imath]L[/imath] has exactly one accept state. And now we start in two place - in [imath]q_0[/imath] and [imath]a_{accept}[/imath]. From [imath]a_{accept}[/imath] we guess symbol. For one symbol we do two steps. From [imath]q_0[/imath] we go according to symbol - one step. State accepting is when two "starts" meet in one state. What about idea ? |
1279745 | What is the adjoint of an inverse matrix?
What is the adjoint of an inverse matrix? Is [imath](T^{-1})^{*} = (T^{*})^{-1}[/imath]? | 1042065 | Adjoint of a matrix and inverse of a matrix
As everyone know that we can use a matrix [imath]A[/imath] to represent an operator [imath]T[/imath]. The adjoint of a matrix [imath]A[/imath] is denoted as [imath]A^*[/imath], which takes complex conjugate of [imath]A[/imath] and then transpose. My problem is, what is the relationship between [imath]A^*[/imath] and [imath]A^{-1}[/imath]? I mean as we know [imath]Ax = b[/imath], [imath]x = A^{-1}b[/imath]. (If not emphasize on the dimension of [imath]x[/imath] and [imath]b[/imath]. |
1151875 | prove that [imath]f_n = 37111111...111[/imath] is never prime
Let [imath]f_n = 37111111...111[/imath] with n 1's. Prove that [imath]f_n[/imath] will never be prime for [imath]n\ge1.[/imath] I tried to look [imath]f_n[/imath] in mod(p), assuming [imath]f_n[/imath] is prime, for the sake of contradiction. I also tried to apply Wilson and Fermat's small theorem. I'm sure there must be a simple factorization which I'm overseeing. | 1256191 | Prove that [imath]371\cdots 1[/imath] is not prime
Prove that [imath]371\cdots 1[/imath] is not prime. I tried mathematical induction in order to prove this, but I am stuck. My partial answer: To be proved is that [imath]37\underbrace{111\cdots 1}_{n\text{ ones}}[/imath] is never prime for [imath]n\geq 1[/imath]. Let [imath]P(n)[/imath] be the statement that [imath]37\underbrace{111\cdots 1}_{n\text{ enen}}[/imath] is not prime. For [imath]n=1[/imath], we can write [imath]371=7\cdot 53[/imath], and therefore [imath]P(1)[/imath] is true. Let [imath]P(k)[/imath] be true for [imath]k>1[/imath]. Then we now have to prove that [imath]P(k+1)[/imath] is true. I found that [imath]37\underbrace{111\cdots 1}_{k\text{ ones}}[/imath] can be written as [imath]37\cdot 10^k+10^{k-1}+10^{k-2}+\cdots +10^0[/imath] and that [imath]37\underbrace{1111\cdots 1}_{k+1\text{ ones}}[/imath] can be written as [imath]37\cdot 10^{k+1}+10^{k}+10^{k-1}+\cdots+10^0[/imath]. I think I'm pretty close to the answer know, but I don't know how to proceed. |
526063 | Prove that [imath]x^3 \equiv x \bmod 6[/imath] for all integers [imath]x[/imath]
Prove that [imath]x^3 \equiv x \bmod 6[/imath] for all integers [imath]x[/imath] I think I got it, but is this proof correct? We can write any integer x in the form: [imath]x = 6k, x = 6k + 1, x = 6k + 2, x = 6k + 3, x = 6k + 4[/imath], and [imath]x = 6k + 5[/imath]. If [imath]x = 6k[/imath], then [imath]x^3 = 216k^3[/imath]. Then [imath]x^3 - x = 216k^3 - 6k = 6(36k^3 - k)[/imath]. Thus, [imath]6 | (x^3 - x)[/imath]. Thus, [imath]x^3 \equiv x \pmod 6[/imath]. If [imath]x = 6k + 1[/imath]. Then [imath]x^3 - x = (216k^3 + 108k^2 + 18k + 1) - (6k + 1) = 6(36k^3 + 18k^2 + 2k)[/imath] and so [imath]6 | (x^3 - x)[/imath] and [imath]x^3 \equiv x \pmod 6[/imath]. If [imath]x = 6k + 2[/imath], then [imath]x^3 - x = (216k^3 + 216k^2 + 72k + 8) - (6k + 2) = 6(36k^3 + 36k^2 + 11k + 1)[/imath] and so [imath]6 | (x^3 - x)[/imath] and [imath]x^3 \equiv x \pmod 6[/imath] If [imath]x = 6k + 3[/imath], then [imath]x^3 - x = (216k^3 + 324k^2 + 162k + 27) - (6k + 3) = 6(36k^3 + 54k^2 + 26 + 4)[/imath] and so [imath]6 | (x^3 - x)[/imath] and [imath]x^3 \equiv x \pmod 6 [/imath] If [imath]x = 6k + 4[/imath], then [imath]x^3 - x = (216k^3 + 432k^2 + 288k + 64) - (6k + 4) = 6(36k^3 + 72k + 47k + 10)[/imath] and so [imath]6 | (x^3 - x)[/imath] and [imath]x^3 \equiv x \pmod 6[/imath]. If [imath]x = 6k + 5[/imath], then [imath]x^3 - x = (216k^3 + 540k^2 + 450k + 125) - (6k + 5) = 6(36k^3 + 90k + 74k + 20)[/imath] and so [imath]6 | (x^3 - x)[/imath] and [imath]x^3 \equiv x \pmod 6[/imath]. In all cases, we have shown that [imath]x^3 \equiv x \pmod 6[/imath]. QED. | 1280400 | Prove that [imath]n^3=n \text{ mod }6[/imath] for every integer [imath]n[/imath].
Prove that for every integer [imath]n[/imath] , [imath]n^3=n \text{ mod }6[/imath] I was having no clue how to do this, then I thought of case-by-case analysis and obviously it worked. The problem is that there were six case and together they are long. Is there any shorter method of proving this. Kindly provide some hints. |
1281321 | Show that [imath] \sum_{n\in \mathbb {S}} \frac{1}{n} [/imath] is convergent
Let [imath]\mathbb {S} =\left \{ 1,2,3,...,9,11,12,...,19,21,...99,111,112,113... \right \} [/imath] i.e, the positive integers set which contain zero digit is omitted. Now show that [imath] \sum_{n\in \mathbb {S}} \frac{1}{n} [/imath] is convergent . I really don't have no idea about how to prove this | 913566 | Sum over all non-evil numbers
I'm working on the following contest math problem: Define an evil number to be any positive integer that contains the digit [imath]9[/imath]. Show that [imath] \sum_{x} \frac{1}{x} < 80 [/imath] where the sum is over all non-evil positive integers [imath]x[/imath]. I'm very confused on where to begin. Initially, I tried to consider this sum as part of the sum of [imath]1/x[/imath] over all positive integers, namely by noting that [imath] \sum_{n=1}^{\infty} \frac{1}{n} = \sum_{x} \frac{1}{x} + \sum_{y} \frac{1}{y} [/imath] where the first summation on the right side is the same as in the problem statement and the second is the sum over all evil numbers. However, I can't seem to find a way to use this since the sum of the left diverges. Could anyone lend a helping hand? |
1281149 | Explain this proof by induction?
[imath]P(n)[/imath] is the statement [imath]n! < n^n[/imath], where [imath]n[/imath] is an integer greater than [imath]1[/imath]. I found a solution online here (https://people.cs.umass.edu/~barring/cs2... But I don't understand how they got from one step to the next. One user gave me the explanation: So we start with: [imath](k+1)! = (k+1) \cdot k![/imath] Now, since [imath]P(k)[/imath] is true, [imath]k! < k^k[/imath], therefore: [imath](k+1) \cdot k! < (k+1) \cdot k^k [/imath] Obviously, [imath]k < (k+1)[/imath], therefore: [imath](k+1) \cdot k^k < (k+1) \cdot (k+1)^k[/imath] And finally: [imath](k+1) \cdot (k+1)^k = (k+1)^1 \cdot (k+1)^k = (k+1)^{k+1} [/imath]. What I don't understand is how they got from the 2nd step to the third step. Please explain it. | 1260233 | Inductive Proof that [imath]k!, for k\geq 2.[/imath]
Call [imath]P(k): k!<k^k[/imath], for [imath]k\geq 2[/imath] Test it out with 2, and it's true ([imath]2<4[/imath]). Assume that [imath]P(k)[/imath] is true for some [imath]k\geq2[/imath]. Then show that [imath]P(k+1)[/imath] is true. [imath]P(k+1): (k+1)!<(k+1)^{k+1}[/imath] Ministep: [imath](k+1)!=k!(k+1)[/imath] Ministep: [imath](k+1)^{k+1}=(k+1)^{k}(k+1)[/imath] [imath]k!(k+1)<(k+1)^{k}(k+1)[/imath] Can I just pull out that (k+1) and call it a day? That will make [imath]k!<(k+1)^{k}[/imath]. We're already assuming that [imath]k!<k^k[/imath], for some integer [imath]k\geq 2[/imath], and I don't think I need to prove that [imath]k^{k}<(k+1)^{k}[/imath]. I mean, I guess I could go prove that as well, but let's assume for the moment that I don't have to. Since we're assuming the original [imath]P(k)[/imath] is true, I should be able to write [imath]k!<k^{k}<(k+1)^{k}[/imath] right? That would prove the [imath]k+1[/imath] case. Is this the correct way to go about things? It seems too easy, but makes sense. Thanks for confirmation and/or pointing out mistakes! |
1281169 | Sign of fractional exponent
What is the sign of [imath]-1^{\frac{2}{3}}[/imath]? I thought it was positive 1 because it involves squaring, but that doesn't seem to be the case. Why? | 215634 | What's [imath](-1)^{2/3}\; [/imath]?
I know that [imath]\left ( -1 \right )^{2/3}=\left ( \left ( -1 \right )^{2} \right )^{1/3}=1[/imath] But Matlab computes this as [imath]- 0.5 + 0.8660254038i[/imath] a complex number.Why? |
1277084 | A graph with exactly one vertex of degree one contains a cycle?
A graph [imath]G[/imath] with exactly one vertex of degree one contains a cycle. Is there any counter example? I could not find any graph for counterexample. Please help. | 1277197 | Need a counter example for cycle in a graph
Could anyone give a counter example for that theorem : A graph G has exactly one vertex of degree [imath]1[/imath], then it contains a cycle. I am so confused. I wonder that may I give a counter example which considers trees. |
1281624 | Check the identity [imath]A\cap (B - C)=(A\cap B) - (A\cap C)[/imath]
[imath]A\cap (B - C)=(A\cap B) - (A\cap C)[/imath] I am trying to prove this by set algebra from left to right. I can do it from right to left and also using mutual inclusions, but get stuck in [imath]\rightarrow[/imath]. This is what I got till now. [imath]A\cap(B-C)=A\cap(B\cap C^c)[/imath] By definition [imath]= (A\cap B)\cap C^c[/imath] by associativity [imath]=(A\cap B)-C[/imath] by definition Don't know how to get it.... | 9719 | Proving : [imath]A \cap (B-C) = (A \cap B) - (A \cap C)[/imath]
I have proved this using Venn diagram but when I am trying to prove this using the rule that "If [imath] A \subset B \text{ and } B \subset A [/imath] then [imath] A = B [/imath]", I am having some problems with my understanding of the same,here is how I did so far: Let [imath]x \in A \cap (B-C) \Rightarrow x \in A \text{ and } x \in (B-C) \Rightarrow x \in A \text{ and } (x \in B \text{ and } x \notin C) [/imath] How to proceed next? Since if I am do something like this: [imath] x \in A \text{ and } x \in B \text{ and } x \in A \text{ and } x \notin C [/imath], it's not giving the correct results, what exactly I am missing here? |
1281591 | The lower bound of the number of cubic residue mod n.
For arbitrary positive integer [imath]n[/imath] , Denote [imath]a\sim_n b \iff a^3\equiv b^3 \mod n[/imath], and [imath]P(n):=\mathrm{Card}\{\mathbb{Z}/\sim_n\}[/imath], How to calculate the value [imath]\displaystyle\inf_{n\in\mathbb{Z}^+}\frac{P(n)}{n}[/imath]? | 1249750 | Number of [imath]q[/imath]-th residues modulo [imath]n[/imath]
Let [imath]q[/imath] be a prime and [imath]n\ge 2[/imath] an integer. Moreover, define [imath]f_q(n)[/imath] as the number of [imath]q[/imath]-th residues modulo [imath]n[/imath]. Is it true that if [imath]K[/imath] is a positive constant then there exist infinitely many [imath]n[/imath] such that [imath] f_q(n)\le Kn? [/imath] I would like to add that the result holds if [imath]q=2[/imath], indeed it is known that [imath] \limsup_{n \to \infty}\frac{f_q(n)}{n}=\frac{1}{2} \,\,\text{ and }\,\,\liminf_{n \to \infty}\frac{f_q(n)}{n}=0. [/imath] |
1282722 | Finite or Infinite n in [imath]φ(n)[/imath]
Are there finitely or infinitely many integers n for which [imath]φ(n) = 1000[/imath]? I think that there are finite, but I don't know how to prove it. | 912137 | Is there an integer [imath]N>0[/imath] such that [imath]\varphi(n) = N[/imath] has infinitely many solutions?
Let [imath]\varphi: \mathbb{N} \to \mathbb{N}[/imath] be the totient function. Is there an integer [imath]N > 0[/imath] such that there are infinitely many integers [imath]n > 0[/imath] such that [imath]\varphi(n) = N?[/imath] |
1282913 | [imath]\leq[/imath] V.S. [imath]\leqslant[/imath]
Is there a substantial difference between [imath]\leq[/imath] and [imath]\leqslant[/imath]? My textbook uses both, but I could not tell why the authors selected one or the other. I asked my teacher, and she said that there was no difference between the two, but if there is no difference, why were both used? When I am writing solutions and proofs, is there any time when it would be considered better practice to use one than the other? | 247719 | Meaning of [imath]\geqslant[/imath], [imath]\leqslant[/imath], [imath]\eqslantgtr[/imath], [imath]\eqslantless[/imath]
What do slanted inequality signs mean? Specifically, these are [imath]\geqslant[/imath], [imath]\leqslant[/imath]; and the variation: [imath]\eqslantgtr[/imath], [imath]\eqslantless[/imath]. Is there any place I can look this up? I've searched Wikipedia and the web and can't find anything about them. The last two were found when looking up the first two. |
1283380 | Does "[imath](\exists f:A\twoheadrightarrow B)\implies(\exists f:B\hookrightarrow A)[/imath]" implies the axiom of choice?
Let [imath]P[/imath] denotes the property that if there exists a surjection from set [imath]A[/imath] to set [imath]B[/imath], then there exists an injection from [imath]B[/imath] to [imath]A[/imath]. It's apparent that [imath]P[/imath] can be proved in ZFC. My question is that if [imath]P[/imath] and AC are equivalent in the ZF theory? | 490995 | Is the following equivalent to the axiom of choice?
For sets [imath]A,B[/imath], let [imath]|A|\leq^*|B|[/imath] say that there exists an onto map [imath]f:B\rightarrow A[/imath] or [imath]A=\emptyset[/imath]. My question is, is [imath]\forall A,B(|A|\leq^*|B|\longrightarrow |A|\leq|B|)[/imath] equivalent to the axiom of choice? Thanks |
1282506 | Proving UNIT INTERSECTION NP-complete
I am working on some review problems right now and am extremely stuck on how to solve problem - any help would be so appreciated. We are told to consider the following combinatorial problem: Unit Intersection: Let X = {1, 2,...,n}. Given a family of subsets [imath]S_1,...,S_m[/imath] of X, determine whether there is a subset T of X such that for all i,|T ∩ [imath]S_i[/imath]| = 1 ? I am then trying to prove that Unit Intersection is NP-complete using a reduction from Exactly-One-3SAT. In the Exactly-One-3SAT problem, we are given a 3CNF formula, and need to decide whether there is an assignment to the variables such that every clause contains exactly one true literal. In a 3CNF formula, every clause has at most three literals. A clause in a 3CNF formula may contain repeated literals. I've been working for hours trying to find a way to solve this problem, but am so stuck right now. Thank you in advance for any help. | 1280874 | Proving that Unit Intersection is NP-complete
I am extremely stuck on how to go about this problem and any help would be so appreciated. We are told to consider the following combinatorial problem: Unit Intersection: Let X = {1, 2,...,n}. Given a family of subsets [imath]S_1,...,S_m[/imath] of X, determine whether there is a subset T of X such that for all i,|T ∩ [imath]S_i[/imath]| = 1 ? I am then trying to prove that Unit Intersection is NP-complete using a reduction from Exactly-One-3SAT. In the Exactly-One-3SAT problem, we are given a 3CNF formula, and need to decide whether there is an assignment to the variables such that every clause contains exactly one true literal. In a 3CNF formula, every clause has at most three literals. A clause in a 3CNF formula may contain repeated literals. I've been working for hours trying to find a way to even start this problem, but am so stuck right now. Thank you in advance for any help. |
1284265 | Problem with solving a conditional-probability question
I have a problem with a statistics question. The question is as follows: I want to go to location B, but I am lost. To find this location I will ask random people where it is. There are two kind of people. 1/3 of the people will always lie where the location is, and 2/3 of the people have the probability ¾ that they are telling the truth, also they have a really bad memory, the moment that they told you the answer they will forget what they have said, so the answers will be independent. I’ve drawn a tree diagram to show probabilities: T=true F=false I need to answer the following questions: You ask a random stranger where location b is. He says: "South". What is the chance this is the correct answer? You ask for the second time to the same random stranger the same question. He gives the same answer. Show that the chance that this answer is correct equals to 1/2. You ask the same question for the 3rd time to the same strangers. He gives the same answers. What is the chance that “South” is correct? The 4th time you will ask the same random stranger if location B is in the North or South. He gives the same answer. Show that the chance that this answer is correct equals to 27/70 Lets say he didn't said the 4th time "South", but he said "North". Show that the chance that this answer is correct equals to 9/10 I got the following answers: 1): [imath] {2\over3}*{3\over4}={1\over2} [/imath] 2) [imath] {({2\over3}*{3\over4} + {2\over3}*{3\over4})\over2}={1\over2} [/imath] 3) [imath] {({2\over3}*{3\over4} + {2\over3}*{3\over4} + {2\over3}*{3\over4})\over3}={1\over2} [/imath] 4) I do not know 5) I do not know I hope that someone can tell me the solution of the questions. | 1282008 | Probability of a repeatedly right answer
Suppose you forgot where the airport is on an island, north or south. You will ask people where to go. Two third of the people on the island are tourists, they give the right answer with a chance of [imath]\frac{3}{4}[/imath]. The local inhabitants [imath]\frac{1}{3}[/imath] of the people always give the wrong answer. When you repeatedly ask where the airport is to the same person, the answers are independent. Now I am investigating how the probability that an answer is right depends on the frequency of the question. For example, the first time I will ask a random passer-by where the airport is, the probability that his answer (i.e. "south") is correct is: \begin{align} P(R)= \frac{2}{3} \cdot \frac{3}{4} + \frac{1}{3} \cdot 0 = \frac{1}{2}. \end{align} How does this change when I again ask the same question to this same person? What is the probability that the answer (i.e. "south") is again correct? Well, since the events are independent. I will say that again the answer (i.e. "south") is correct will be: \begin{align} P(R)= \frac{2}{3} \cdot \frac{3}{4} + \frac{1}{3} \cdot 0 = \frac{1}{2}. \end{align} Is this supposition true? Furthermore, what can I say about the fourth time that I again ask this same person the same question, what is the probability that this answer (i.e. "south") is the correct answer? |
752726 | Power Set, Bijection Function, Equivalence Relation
Let [imath]S[/imath] be a set and [imath]P(S)[/imath] the power set of [imath]S[/imath]. For sets [imath]A,B⊆P(S)[/imath], we say that [imath]A \sim B[/imath] if there exists a bijective function [imath]f: A \rightarrow B[/imath]. Show that [imath]\sim [/imath] is an equivalence relation. | 773465 | Equivalence relations and power sets.
Let [imath]\mathcal{A}[/imath] be the class of all sets and define the relation [imath]R[/imath] on [imath]\mathcal{A}[/imath] as: [imath]A\space R\space B[/imath] iff there is a bijective function [imath]f:A \to B[/imath]. Prove that [imath]R[/imath] is an equivalence relation on [imath]\mathcal{A}[/imath]. I know I need to prove bijection by proving injection and surjection, but I don't even know where to start. Any help would be greatly appreciated. |
1281847 | How to prove [imath]f(x_1,\ldots,x_n) = \sum x_i\ln x_i - (\sum x_i )\ln(\sum x_i)[/imath] is convex on R++
How can i prove [imath]f(x_1,\ldots,x_n) = \sum x_i\ln x_i - (\sum x_i )\ln(\sum x_i)[/imath] is convex on R++ | 1281612 | convexity proof of a function including ln and sums
[imath]f(x_1,\dots,x_n)=\sum\limits_{i=1}^nx_i\ln x_i-\left(\sum\limits_{i=1}^nx_i\right)\ln\left(\sum\limits_{i=1}^nx_i\right)\rightarrow R_{++}^n[/imath] How can I prove this is convex on [imath]R_{++}^n[/imath]? I tried using the Hessian and couldn't prove it. There is a solution using the gradient and Jensen but very long and complicated. |
208191 | Cesaro summable implies that [imath]c_{n}/n[/imath] goes to [imath]0[/imath]
Theorem. If [imath]\sum_{n=1}^{\infty}c_{n}[/imath] is Cesaro summable, then [imath]c_{n}/n[/imath] tends to [imath]0[/imath]. How to prove it? | 1384226 | Cesaro sum of a series
[imath]\sum_{n=0}^{\infty}a_n[/imath] diverges in the regular term but is Cesaro summable Prove [imath]a_n/n\to 0[/imath] when [imath]n\to \infty[/imath] We used the definition of the Cesaro sum and obtained: [imath]\lim_{N\to \infty}\frac{1}{N+1}\sum_{n=0}^{N}S_n[/imath]=[imath]\lim_{N\to \infty}\frac{1}{N+1}\left ( \sum_{n=0}^{N-1}S_n+\sum_{k=0}^{N}a_k\right )[/imath]=[imath]\sum_{n=0}^{N-1}\frac{S_n}{N+1}[/imath][imath]+\sum_{k=0}^{N}\frac{a_k}{N+1}[/imath]=[imath]L[/imath] but from here we're kind of stuck *([imath]S_n[/imath] is the partial sums of [imath]a_n[/imath]) |
1285086 | Show 3 is a primitive root modulo [imath]7^n[/imath] for [imath]n\in \Bbb{N}[/imath].
I tried induction but got stuck. For [imath]n=1[/imath] it is true. Suppose it holds for [imath]n[/imath], i.e, the order of 3 is [imath]\phi(7^n)[/imath]. Now I should prove it for [imath]\phi(7^{n+1})=6\cdot 7^n[/imath]. [imath]7^n\mid3^{6\cdot 7^n}-1[/imath]. How do I derive that [imath]7^{n+1}\mid 3^{6\cdot 7^{n+1}} - 1[/imath]. I would appreciate your help. | 754187 | Prove that 3 is a primitive root of [imath]7^k[/imath] for all [imath]k \ge 1[/imath]
so I am trying to find out how to prove that 3 is a primitive root of [imath]7^k[/imath] for all [imath]k \ge 1[/imath]. I am trying to prove this via induction. Thanks. |
1285568 | the number of [imath]n\times n[/imath] matrices of 0's and 1's such that every row and column has three 1's
In the Example 1.1.3 of Stanley's book Enumerative Combinatorics Vol 1 (2nd edition), an explicit formula for the number [imath]f(n)[/imath] of [imath]n\times n[/imath] matrices of [imath]0[/imath]'s and [imath]1[/imath]'s such that every row and column has three [imath]1[/imath]'s is presented: [imath]f(n)=6^{-n}n!^2\sum\frac{(-1)^\beta(\beta+3\gamma)!\,2^\alpha\,3^\beta}{\alpha!\,\beta!\,\gamma!^2\,6^\gamma}\;,[/imath] where the sum ranges over all [imath](n+2)(n+1)2[/imath] solutions to [imath]\alpha+\beta+\gamma=n[/imath] in nonnegative integers. But, I would like to know how we can prove it? It seems that inclusion-exclusion principle is used. | 328277 | Number of certain (0,1)-matrices, Stanley's Enumerative Combinatorics
Stanley's Enumerative Combinatorics (http://www-math.mit.edu/~rstan/ec/ec1.pdf) contains next fact: 1.1.3 Example. Let f(n) be the number of n × n matrices M of [imath]0[/imath]’s and [imath]1[/imath]’s such that every row and column of M has three 1’s. For example, f(0) = 1, f(1) = f(2) = 0, f(3) = 1. The most explicit formula known at present for f(n) is [imath]f(n)=6^{-n}{(n!)}^2\sum\frac{(-1)^{\beta}(\beta+3\gamma)!2^\alpha 3^{\beta}}{\alpha!\beta!\gamma!^26^{\gamma}}[/imath] where the sum ranges over all (n + 2)(n + 1)/2 solutions to α + β + γ = n in nonnegative integers. I need proof of this fact. (i.e. reference to book or articles that contains proof this fact). |
1285541 | How to go from this equation in terms of cosine and sine to this one in terms of only one.
[imath]x=A\cos(\omega t)+B\sin(\omega t )\equiv\mu \cos(\omega t+\phi)[/imath] I'm thinking it must have something to do with the double angle forumla. Any help? | 729974 | Linear combinations of sine and cosine
If you take a linear combination of the cosine and sine function, then the result is again a sinusoid, but with a new amplitude and phase shift. [imath]a \cos(\theta) + b \sin(\theta) = A \cos(\theta + \theta_0)[/imath] This is one of those facts which I've always felt should be more obvious to me than it is. I can't help feeling that I might be missing out on some good "conceptual reason" for this to be true. I did a bit of thinking about it today and I came up with the following explanation which I'm not particularly satisfied with. I feel like there should be some perspective on this which renders the fact obvious. Multiplying the polar equation [imath]r = 2a \cos(\theta) + 2b \sin(\theta)[/imath] by [imath]r[/imath] yields [imath]x^2+y^2 = 2a x + 2by[/imath]. Rearranging terms and adding [imath]a^2 + b^2[/imath] to both sides gives [imath]x^2 - 2ax + a^2 +y^2 -2by + b^2 = a^2 + b^2[/imath] which is identical to [imath](x-a)^2 + (y-b)^2 = a^2 + b^2[/imath] which we recognize as the equation for the circle with centre at [imath](a,b)[/imath] and passing through the origin. This hints that [imath]r = 2a \cos(\theta) + 2b \sin(\theta)[/imath] should define a parametrization of this circle and, indeed, this holds. This circle can be rotated about the origin so as to have its centre on the [imath]x[/imath]-axis. Such a circle will have the simpler equation [imath]r = D \cos(\theta) [/imath] where [imath]D=2(a^2+b^2)[/imath] is the diameter of the circle. In polar coordinates, performing this rotation amounts to introducing a phase [imath]\theta_0[/imath]. In fact [imath]\theta_0[/imath] is an argument for the point [imath](a,b)[/imath]. Note I'm not really looking for rigorous proofs of this fact, more like interpretations, really. |
1286198 | Lemma 3.3-7 and Theorem 3.6-2 in Kreyszig's "Introductory Functional Analysis With Applications": What if completeness is lost?
Let [imath]X[/imath] be an inner product space, and let [imath]M[/imath] be a non-empty subset of [imath]X[/imath]. Then we have the following: (a) If the space of [imath]M[/imath] is dense in [imath]X[/imath], then [imath]M^\perp = \{0 \}[/imath], that is, [imath]x \in X[/imath], [imath]x \perp M[/imath] imply [imath]x =0[/imath]. (b) If [imath]X[/imath] is complete and [imath]M^\perp = \{0 \}[/imath], then the span of [imath]M[/imath] is dense in [imath]X[/imath]. In (b), what if the space [imath]X[/imath] is not complete? In such a case, can we have a set [imath]M[/imath] such that [imath]M^\perp = \{0\}[/imath], but the span of [imath]M[/imath] is not dense in [imath]X[/imath]? | 1302571 | Theorem 3.6-2 in Erwine Kreyszig's "Introductory Functional Analysis with Applications:" Does the converse hold if the space is not complete?
First, a definition: Let [imath]X[/imath] be a normed space. A subset [imath]M (\neq \emptyset) \subset X[/imath] is said to be total in [imath]X[/imath] if the span of [imath]M[/imath] is dense in [imath]X[/imath]. Now theorem 3.6-2 in Kreyszig states the following: (a) If a subset [imath]M \neq \emptyset[/imath] of an inner product space [imath]X[/imath] is dense in [imath]X[/imath], then [imath]M^\perp = \{ 0 \}[/imath]. (b) If [imath]X[/imath] is a Hilbert space and if [imath]M[/imath] ( [imath]\neq \emptyset[/imath] ) [imath] \subset X[/imath] such that [imath]M^\perp = \{0\}[/imath], then [imath]M[/imath] is total in [imath]X[/imath]. So far, so good! Now my question is, what if [imath]M^\perp = \{0\}[/imath], but [imath]X[/imath] is not a complete inner product space? Can we find a non-trivial example of such a set [imath]M[/imath] which is not total in [imath]X[/imath]? |
278907 | Continous periodic function is uniformly continuous?
[imath]f:\mathbb{R}\rightarrow\mathbb{R}[/imath] be a continuous function such that [imath]f(x+1)=f(x)\forall x[/imath] we need to show [imath]f[/imath] is uniformly continuous function. please give me some hint: | 775045 | How do I show that all continuous periodic functions are bounded and uniform continuous?
A function [imath]f:\mathbb{R}\to \mathbb{R}[/imath] is periodic if there exits [imath]p>0[/imath] such that [imath]f(x+P)=f(x)[/imath] for all [imath]x\in \mathbb{R}[/imath]. Show that every continuous periodic function is bounded and uniformly continuous. For boundedness, I first tried to show that since the a periodic function is continuous, it is continuous for the closed interval [imath][x_0,x_0+P][/imath]. I know that there is a theorem saying that if it is continuous on a closed interval, then it is bounded. However, I'm not allowed to state that theorem directly. Should I just aim for a contradiction by supposing f is not bounded on the interval stated above? |
1286791 | Diophantine equations in [imath]\Bbb Z[/imath]
[imath]x + 2y + 3z = 4[/imath] [imath]w = x + 2\times y[/imath], then the equation becomes [imath]w + 3z = 4[/imath]. [imath]\gcd(1, 3) = 1 | 4[/imath], so this two variable equation is solvable. [imath]w = -2, y = 2[/imath] i can't seem to pass this point | 126610 | Problem on Linear Diophantine Equation over 3 variables
How to solve [imath]ax+by+cz=d[/imath] over integers where [imath]a,b,c,d[/imath] are integers? |
1286071 | What is identity arrow in the category Set?
Given is category Set Given two objects from this category, [imath]A[/imath], and [imath]B[/imath], which are sets without any other structure, there is an arrow [imath]f: A \to B[/imath], from [imath]A[/imath] to [imath]B[/imath], which is any total function from [imath]A[/imath] to [imath]B[/imath]. For given set [imath]A[/imath], there is ( by axiom of CT ) identity arrow [imath]1_A[/imath] . Then it looks like identity arrow may be a permutation... Buy identity arrow for a given object has to be unique. And there is many permutations for given set [imath]A[/imath]. So do we have here "equivalence class" of identity arrow? | 21951 | The identity morphism in [imath]\mathbf{Set}[/imath] is the identity function
I've been trying to wrap my head around the basic concepts of category theory, and I thought I would attempt to illustrate what I understand with the category of sets, probably the easiest example. Particularly, I've been trying to prove that [imath]id_A[/imath] (the identity morphism on [imath]A[/imath], for all [imath]A \in Obj(\mathbf{Set})[/imath]) is [imath]1_A \colon A \rightarrow A, x \mapsto x[/imath]. This is a very intuitive and reasonable statement, and it's trivial to prove that [imath]1_A[/imath] is indeed an identity morphism on [imath]A[/imath], and I suppose uniqueness of [imath]id_A[/imath] can be demonstrated analogously to uniqueness of the identity element in a monoid (considering the subcategory which has [imath]A[/imath] as its only object, and endofunctions on [imath]A[/imath] as its only morphisms). In this manner, it is not hard to prove that the proposition in the title is true, but this demonstration requires to make an assumption or guess as to what could [imath]id_A[/imath] be. Specifically, the scheme of the proof is: assume [imath]id_A[/imath] = [imath]1_A[/imath], see that it works with the definition of an identity morphism, show that the identity morphism is unique, and, in conclusion, [imath]id_A[/imath] can only be [imath]1_A[/imath]. What I'm looking for, nonetheless, is a somehow more direct proof, that doesn't assume [imath]id_A[/imath] = [imath]1_A[/imath] at the start. I want to place myself in a state of little or no knowledge about sets and functions, and under this assumption, why would I assume [imath]id_A[/imath] = [imath]1_A[/imath] at first? Why not try with [imath]id_\mathbb{Z}[/imath] = [imath]f \colon \mathbb{Z} \to \mathbb{Z}, x \mapsto x^2 + 1[/imath], for example? It wouldn't work, but I don't have any reason to think that [imath]1_\mathbb{Z}[/imath] is a better guess for [imath]id_\mathbb{Z}[/imath]. I suppose that the proof for which I'm asking would work for categories of sets with additional structure, and probably for posets as well, although I'm not clear as to what modifications it would require to work. Thanks. |
1287299 | Are members of a diagonal positive definite matrix positive?
If square matrix [imath]M[/imath] is diagonal and positive definite, does it mean that all [imath]m_{ii}[/imath] (diagonal entries) of this matrix are necessarily positive? | 180173 | Prove: symmetric positive definite matrix
I'm studying for my exam of linear algebra.. I want to prove the following corollary: If [imath]A[/imath] is a symmetric positive definite matrix then each entry [imath]a_{ii}> 0[/imath], ie all the elements of the diagonal of the matrix are positive. My teacher gave a suggestion to consider the unit vector "[imath]e_i[/imath]", but I see that is using it. [imath]a_{ii} >0[/imath] for each [imath]i = 1, 2, \ldots, n[/imath]. For any [imath]i[/imath], define [imath]x = (x_j)[/imath] by [imath]x_i =1[/imath] and by [imath]x_j =0[/imath], if [imath]j\neq i[/imath], since [imath]x \neq 0[/imath], then: [imath]0< x^TAx = a_{ii}[/imath] But my teacher says my proof is ambiguous. How I can use the unit vector [imath]e_1[/imath] for the demonstration? |
1288595 | X a locally compact Haussdorf space, each singleton of which is the intersection of countably many open sets. Show that X is first countable.
Let [imath]X[/imath] be a locally compact Haussdorf space, each singleton of which is the intersection of countably many open sets. Show that X is first countable. | 240472 | [imath]G_\delta[/imath] singletons in compact Hausdorff and first countability
Given a compact Hausdorff space [imath]X[/imath], if [imath]a \in X[/imath] is a [imath]G_\delta[/imath] singleton (i.e., [imath]\{ a \}[/imath] is a [imath]G_\delta[/imath] set), then there is a countable local base at [imath]a[/imath]. The point [imath]a[/imath] can be written as countable intersection of open sets let's call [imath]U_{n}[/imath]. How can I use the compactness? What is the next step for proof? Could you give me hint please? |
1284490 | Minimizing a functional with a free boundary condition
Find the extremals of the functional [imath]\text{J}(y)= y^2(1) + \int_0^1 y'^2(x)dx , \ \ y(0)=1.[/imath] Answer: [imath]y(x)=1-\frac{1}{2}x[/imath] My solution: [imath] F (x,y,y')=y'^2(x)[/imath] After solving the Euler Lagrange equation we get [imath]\frac{\mathrm{d}}{\mathrm{d}x}(2y')=0[/imath] Which implies that [imath]y=\frac{a}{2}x+b[/imath] , using initial value condition we get [imath]b=1[/imath]. Could you please help me find value of [imath]a[/imath]? | 1179633 | Difficult Problem on Calculus of variation
Consider the functional [imath]J(y)=y^2(1)+\int_0^1 y'^2(x) \ dx[/imath] with [imath]y(0)=1[/imath] where [imath]y\in C^2[0,1][/imath]. If [imath]y[/imath] extremizes [imath]J[/imath] then find [imath]y[/imath]. Any hint will be appreciated. |
1288871 | series and dyadics
I'm in Calc II and am unfamiliar with what dyadics is but my teacher said that it's possible to find the sum of [imath]\frac{\cos n}{n^2}[/imath] by using dyadics. Would you mind laying out a step by step? Thanks. | 552640 | Series [imath]\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}[/imath]
Is it true that for [imath]x\in[0,2\pi][/imath] we have [imath]\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}=\frac{x^2}{4}-\frac{\pi x}{2}+\frac{\pi^2}{6}[/imath] How can I prove it? For other intervals what is the value of above series if is convergent? |
1289322 | What is zero? Irrational or rational or it have both the properties?
We say, A number is rational if it can be represented as [imath]\frac{p}{q}[/imath] with [imath]p,q \in \mathbb Z[/imath] and [imath]q\neq 0[/imath]. Any number which doesn't fulfill the above conditions is irrational. What about zero? It can be represented as a ratio of two integers as well as ratio of itself and an irrational number such that zero is not dividend in any case. People say that [imath]0[/imath] is rational because it is an integer. Which I find to be a lame reason. May be any strong reason is there. Can any one tell me please? | 522933 | Is zero irrational?
I think of the number zero as a whole number. It can certainly be a ratio = [imath]\frac{0}{x}, x \neq 0.[/imath] Therefore it is rational. But any ratio equaling zero involves zero, or is irrational, e.g.[imath]\frac{x}{\infty}, x \neq 0[/imath] is not a ratio of integers. Can a rational number that is rational only when the number itself is involved still be rational? I realize that it "should" be rational but it doesn't seem to fit into the same category with other rational numbers. (I'm glad mse has a soft question category for questions like this...). |
1289229 | Area of a curve [imath]\frac{x^2}{4}+\frac{(y-3)^2}{9}=1[/imath]
The equation of a curve is [imath]\frac{x^2}{4}+\frac{(y-3)^2}{9}=1[/imath] How to calculate the area of the region bounded by the curve? | 1289298 | Area of an ellipse. (Calculus)
This is the question extracted from a exam in my country. The equation of the curve is [imath]\frac{x^2}{4}+\frac{(y-3)^2}{9}=1[/imath] How to calculate the area of the region bounded by the curve? (9 Marks) How to solve this by calculus? It's 9 marks question. We didn't learn that the area of an ellipse is [imath]\pi ab[/imath]. |
1289031 | Evaluate [imath] \lim_{n\to\infty} \frac{(n!)^2}{(2n)!} [/imath]
I'm completely stuck evaluating [imath] \lim_{n\to\infty} \frac{(n!)^2}{(2n)!} [/imath] how would I go about solving this? | 1006352 | Calculate [imath]\lim_{n\rightarrow +\infty}\binom{2n} n[/imath]
Calculate [imath]\lim_{n\rightarrow +\infty}\binom{2n} n[/imath] without use Stirling's Formula. Any suggestions please? |
256100 | How can I find the points at which two circles intersect?
Given the radius and [imath]x,y[/imath] coordinates of the center point of two circles how can I calculate their points of intersection if they have any? | 1666720 | Find the coordinates of the points where the two circles intersect.
The two circles are: 1) [imath](x-2)^2 + (y+1)^2 = 25[/imath] 2) [imath](y-2)^2 + (x+1)^2 = 25[/imath] |
317953 | Pointwise convergence, bounded variation, and lim inf's
Suppose that [imath]\{f_n\}_{n = 1}^\infty[/imath] is a sequence of functions in [imath]BV[0, 1][/imath] that converges pointwise to a function [imath]f[/imath] on [imath][a, b][/imath]. Show that [imath]V_a^b f \leq \liminf_{n \to \infty} V_a^b f_n[/imath]. I know that [imath]\forall \varepsilon > 0[/imath], [imath]\exists N > 0[/imath] such that [imath]V_a^b f_n > \liminf_{n \to \infty} V_a^b f_n - \varepsilon[/imath]. I was hoping to get a hint. | 126625 | Pointwise convergent and total variation
I'm preparing for a test for real analysis and I came across this problem in Royden's book: Let [imath]\{f_n\}[/imath] be a sequence of real valued functions on [imath][a,b][/imath] that converges pointwisely on [imath][a,b][/imath] to the real valued function [imath]f[/imath]. Show that [imath]TV(f) \leq \liminf ~TV(f_n)?[/imath] This looks quite similar in form to Fatou's Lemma to me, but can't find any way to establish TV with integration, can anybody please help? (TV is short for total variation) |
1289456 | Show that [imath]\psi_a:G\rightarrow G',a\in G,\psi_a(g)=aga^{-1}[/imath] is homomorphism one to one and onto
Let [imath]G[/imath] be a group and [imath]\psi_a:G\rightarrow G,a\in G,\psi_a(g)=aga^{-1}[/imath], I need to show that [imath]\psi_a[/imath] is homomorphism one to one and onto It's not the same question like "Is the conjugation map always an isomorphism?" because here I need to show why [imath]\psi_a[/imath] is one to one and onto Attempt: [imath]\psi_a(g_1g_2)=ag_1g_2a^{-1}[/imath] [imath]\psi_a(g_1)\psi_a(g_2)=ag_1a^{-1}ag_2a^{-1}=ag_1g_2a^{-1}[/imath] [imath] \Rightarrow \psi_a [/imath] is hom' | 250264 | Is the conjugation map always an isomorphism?
Given a group [imath]G[/imath] with [imath]a\in G[/imath] fixed, define [imath]\phi: G \to G[/imath] by [imath]\phi(x) = axa^{-1}, x \in G[/imath], I am wondering when [imath]\phi[/imath] is an isomorphism. I think that it is always an isomorphism because it is one to one, and the order of the domain ([imath]|G|[/imath]) and the order of the range are equal. Am I wrong in thinking this? |
1289675 | Is this series computable?
I would like to compute the value of this series: \begin{equation*} \sum_{n = 0}^{+ \infty} n . e^{- \alpha n} \end{equation*} Where [imath]\alpha[/imath] is a constant. | 525021 | How to compute infinite series [imath]\sum_{n=0}^{\infty} ne^{-n}[/imath]
I'm trying to compute the infinite series [imath]\sum_{n=0}^{\infty} ne^{-n}[/imath]. I know the answer is [imath]e/(e-1)^2[/imath], but I don't understand how to find this result. Thanks for the help! |
1289878 | Each eigenvalue of [imath]A[/imath] is equal to [imath]\pm 1[/imath]. Why is [imath]A[/imath] similar to [imath]A^{-1}[/imath]?
[imath]A[/imath] is a non-singular matrix ([imath]n \times n[/imath]) and each eigenvalue of [imath]A[/imath] is equal to [imath]\pm 1[/imath]. Why is [imath]A[/imath] similar to [imath]A^{-1}[/imath]? (by Jordan form) | 1284461 | If each eigenvalueof [imath]A[/imath] is either [imath]+1[/imath] or [imath]-1[/imath] [imath] \Rightarrow[/imath] [imath]A[/imath] is similar to [imath]{A^{ - 1}}[/imath]
Let [imath]A \in {M_n}[/imath] is nonsingular and each eigenvalue of [imath]A[/imath] is either [imath]+1[/imath] or [imath]-1[/imath].Why [imath]A[/imath] is similar to [imath]{A^{ - 1}}[/imath]? |
1290071 | An inequality about the CDF of a random variable
I have to prove that for a random variable [imath]X\sim N(0,1)[/imath], \begin{equation*} \mathbb{P}(X>t) \leq \frac{1}{2}\,e^{-\frac{t^2}2}. \end{equation*} I tried using the fact that for [imath]h(t)=e^{\alpha t}[/imath], \begin{equation*} \mathbb{P}(|X|>t)\leq \frac{E(h(X))}{h(t)} \end{equation*} but didn't manage to advance that much. | 988822 | Normal distribution tail probability inequality
I am trying to show that [imath]P(X>t)\leq \frac{1}{2}e^\frac{-t^2}{2}[/imath] for [imath]t>0[/imath] where [imath]X[/imath] is a standard normal random variable. Perhaps this is simple. I have been starting with [imath] \int_{t}^{\infty} \frac{1}{\sqrt{2\pi}} e^\frac{-x^2}{2}dx \leq \int_{t}^{\infty}\frac{x}{t}\frac{1}{\sqrt{2\pi}}e^\frac{-x^2}{2}dx = \frac{1}{t\sqrt{2\pi}}e^\frac{-t^2}{2}[/imath] but this is not what I want...I am looking for a much stronger inequality. Any help in the direction of getting the [imath]1/2[/imath]? |
1289714 | Going into dual space for a vector product
I am a bid confused regarding the notation for tensor products when going into dual-space If [imath]\left| \Psi \rangle \right. = \left| A \rangle \right. \left| B \rangle \right.[/imath] is [imath] \left. \langle \Psi \right| = \left. \langle A \right| \left. \langle B \right|[/imath] or [imath]\left. \langle B \right| \left. \langle A \right|[/imath]? My guess is the last choice, since operators should act on [imath]\left| 0 \rangle \right. = \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) [/imath] [imath]\left| 1 \rangle \right. = \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) [/imath] so [imath]\left| 01 \rangle \right. = \left| 0 \rangle \right. \left| 1 \rangle \right. = \left( \begin{matrix} 0 \\ 1 \\ 0 \\ 0 \end{matrix} \right) [/imath] But then, for me, it should be [imath]\left( \begin{matrix} 0 & 1 & 0 & 0 \end{matrix} \right) = \left \langle 10 \right|[/imath] But this is not correct if i calculate the tensor product, since [imath]\left \langle 1 \right| \left \langle 0 \right| = \left( \begin{matrix} 0 & 0 & 1 & 0 \end{matrix} \right)[/imath] For instance, shoulden't it be [imath]\left| 01 \rangle \right. \left \langle 10 \right| = \left( \begin{matrix} 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{matrix} \right) [/imath] ? I hope I have made my confusion clear | 1289803 | Tensor product in dual-space
I am a bid confused regarding the notation for tensor products when going into dual-space If [imath]\left| \Psi \rangle \right. = \left| A \rangle \right. \left| B \rangle \right.[/imath] is [imath] \left. \langle \Psi \right| = \left. \langle A \right| \left. \langle B \right|[/imath] or [imath]\left. \langle B \right| \left. \langle A \right|[/imath]? My guess is the last choice, since operators should act on [imath]\left| 0 \rangle \right. = \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) [/imath] [imath]\left| 1 \rangle \right. = \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) [/imath] so [imath]\left| 01 \rangle \right. = \left| 0 \rangle \right. \left| 1 \rangle \right. = \left( \begin{matrix} 0 \\ 1 \\ 0 \\ 0 \end{matrix} \right) [/imath] But then, for me, it should be [imath]\left( \begin{matrix} 0 & 1 & 0 & 0 \end{matrix} \right) = \left \langle 10 \right|[/imath] But this is not correct if i calculate the tensor product, since [imath]\left \langle 1 \right| \left \langle 0 \right| = \left( \begin{matrix} 0 & 0 & 1 & 0 \end{matrix} \right)[/imath] For instance, shoulden't it be [imath]\left| 01 \rangle \right. \left \langle 10 \right| = \left( \begin{matrix} 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \end{matrix} \right) [/imath] ? I hope I have made my confusion clear |
583707 | Intuitive explanation for formula of maximum length of a pipe moving around a corner?
For one of my homework problems, we had to try and find the maximum possible length [imath]L[/imath] of a pipe (indicated in red) such that it can be moved around a corner with corridor lengths [imath]A[/imath] and [imath]B[/imath] (assuming everything is 2d, not 3d): My professor walked us through how to derive a formula for the maximum possible length of the pipe, ultimately arriving at the equation [imath]L = (A^{2/3} + B^{2/3})^{3/2}[/imath]. The issue I have is understanding intuitively why this formula works, and exactly what it's doing. I understand the steps taken to get to this point, but there's an odd symmetry to the end result -- for example, is the fact that [imath]\frac{2}{3}[/imath] and its inverse are the only constants used just a coincidence, or indicative of some deeper relationship? I also don't quite understand how the formula relates, geometrically, to the diagram. If I hadn't traced the steps myself, I would have never guessed that the formula was in any way related to the original problem. If possible, can somebody give an intuitive explanation as to why this formula works, and how to interpret it geometrically? Here's how he found the formula, if it's useful: The formula is found by finding the maximum possible length of the pipe by expressing the length in terms of the angle [imath]\theta[/imath] formed between the pipe and the wall, and by taking the derivative to find when [imath]\frac{dL}{d\theta} = 0[/imath], which is the minimum of [imath]\frac{dL}{d\theta}[/imath] and is therefore when [imath]L[/imath] is the smallest: [imath] L = \min_{0 \leq \theta \leq \frac{\pi}{2}} \frac{A}{\cos{\theta}} + \frac{B}{\sin{\theta}} \\ 0 = \frac{dL}{d\theta} = \frac{A\sin{\theta}}{\cos^2{\theta}} - \frac{B\cos{\theta}}{\sin^2{\theta}} \\ 0 = \frac{A\sin^3{\theta} - B\cos^3{\theta}}{\sin^2{\theta}\cos^2{\theta}} \\ 0 = A\sin^3{\theta} - B\cos^3{\theta} \\ \frac{B}{A} = \tan^3{\theta} \\ \theta = \arctan{\left( \frac{B}{A} \right)^{\frac{1}{3}}} \\ [/imath] At this point, we can substitute [imath]\theta[/imath] back into the original equation for [imath]L[/imath] by interpreting [imath]A^{1/3}[/imath] and [imath]B^{1/3}[/imath] as sides of a triangle with angle [imath]\theta[/imath] and hypotenuse [imath]\sqrt{A^{2/3} + B^{2/3} }[/imath]: [imath] \cos{\theta} = \frac{A^{1/3}}{ \sqrt{A^{2/3} + B^{2/3} }} \\ \sin{\theta} = \frac{B^{1/3}}{ \sqrt{A^{2/3} + B^{2/3} }} \\ \therefore L = A^{2/3} \sqrt{A^{2/3} + B^{2/3} } + B^{2/3} \sqrt{A^{2/3} + B^{2/3} } \\ L = (A^{2/3} + B^{2/3}) \sqrt{A^{2/3} + B^{2/3} } \\ L = (A^{2/3} + B^{2/3})^{3/2} \\ [/imath] The equation for the formula for the maximum length of the pipe is therefore [imath]L = (A^{2/3} + B^{2/3})^{3/2}[/imath]. | 1367531 | Longest pipe that fits around a corner.
While studying, I came upon the problem "Two corridors of widths [imath]a[/imath] and [imath]b[/imath] intersect at right angle. What is the length of the longest pipe that can be carried across the two corridors, touching the corner of the wall where the corridors meet?" The explanation is not detailed, but the answer is shown as [imath](a^{2/3} + b^{2/3})^{3/2}[/imath]. I have attempted the problem using trigonometry and derivatives, but got stuck early in. Could someone help with this problem? |
1290771 | What are the possible eigenvalues of a linear transformation [imath]T[/imath] satifying [imath]T = T^2[/imath]
Let [imath]T[/imath] be a linear transformation [imath]T[/imath] such that [imath]T\colon V \to V[/imath]. Also, let [imath]T = T^2[/imath]. What are the possible eigenvalues of [imath]T[/imath]? I am not sure if the answer is only [imath]1[/imath], or [imath]0[/imath] and [imath]1[/imath]. It holds that [imath]T = T^2[/imath], thus [imath]T(T(x)) = T(x)[/imath]. Let's call [imath]T(x) = v[/imath], so [imath]T(v) = v[/imath]. which means that [imath]\lambda=1[/imath]. But I am not sure about this, while I have seen a solution that says that [imath]0[/imath] is possible as well. Thanks in advance ! | 1157589 | Find the eigenvalues of a projection operator
A projection operator [imath]P[/imath] is defined as [imath]P^2[/imath]=[imath]P[/imath]. Use this definition to find the eigenvalues of this operator. In this question is it necessary to define what the projection operator is? And won't the eigenvalue just be zero? |
609082 | If [imath]a,b,c>0[/imath] and [imath]a+b+c=1[/imath], prove that [imath]\sqrt{\frac{ab}{c+ab}}+\sqrt{\frac{bc}{a+bc}}+\sqrt{\frac{ac}{b+ac}} \le \frac{3}{2}[/imath].
I can't see any ways I could use the fact that [imath]a+b+c=1[/imath]. I've tried solving the inequality using various AM-GM inequalities, but I just can't make it. I need some help. Thanks. | 2647707 | Proving an inequality using relationships between means
Given three positive numbers a, b, c with the constraint: [imath]a+b+c=1[/imath] prove the inequality [imath]\sqrt{{ab}\over{c+ab}}+\sqrt{{ac}\over{b+ac}}+\sqrt{{bc}\over{a+bc}}\le{3\over2}[/imath] |
1290853 | Why is this matrix invertible
I was wondering if there is a way to see why [imath](1+A)[/imath] invertible, if [imath]A[/imath] is a skew symmetric matrix. and I know that all eigenvalues of [imath]A[/imath] have zero real part and [imath]A[/imath] is unitarily diagonalisable. | 52593 | Problem in skew-symmetric matrix
Let [imath]A[/imath] be a real skew-symmetric matrix. Prove that [imath]I+A[/imath] is non-singular, where [imath]I[/imath] is the identity matrix. |
1291154 | Denote [imath]y_n=\int_0^1{\frac{f^{n+1}(x)}{g^n(x)}}dx[/imath] for all integer [imath]n \geq 0[/imath]. Prove that [imath](y_n)_{n \geq 1}[/imath] is an increasing and divergent sequence.
Let [imath]f,g:[0,1] \rightarrow (0,\infty)[/imath] be two distinct, continuous functions such that [imath]\int_0^1 f(x)dx=\int_0^1 g(x)dx[/imath] Denote [imath]y_n=\int_0^1{\frac{f^{n+1}(x)}{g^n(x)}}dx[/imath] for all integer [imath]n \geq 0[/imath]. Prove that [imath](y_n)_{n \geq 1}[/imath] is an increasing and divergent sequence. Here is the source of the problem. My attempt: Note that [imath]y_n -y_{n-1}=\int_0^1{\frac{f^{n+1}(x)}{g^n(x)}}dx-{\frac{f^{n}(x)}{g^{n-1}(x)}}dx = \int_0^1{\frac{f^{n+1}(x)-f^n(x)g(x)}{g^n(x)}}dx = \int_0^1{\frac{f^n(x)}{g^n(x)}(f(x)-g(x))}dx \geq A\int_0^1(f(x)-g(x)dx=0[/imath] where [imath]A>0[/imath]. Hence, [imath](y_n)_{n \geq 1}[/imath] is increasing. However, I have trouble to show the sequence diverges. Can anyone give some hint? | 515584 | Let [imath]f,g[/imath] be two distinct functions from [imath][0,1][/imath] to [imath](0, +\infty)[/imath] such that [imath]\int_{0}^{1} g = \int_{0}^{1} f [/imath].
Let [imath]f,g[/imath] be two continuous, distinct functions from [imath][0,1][/imath] to [imath](0, +\infty)[/imath] such that [imath]\int_{0}^{1} g = \int_{0}^{1} f [/imath]. Given [imath]n\in \mathbb{N},[/imath] let [imath]y_n = \int_{0}^{1} \frac{f^{(n+1)}}{g^{(n)}} [/imath] How do I show that [imath](y_n)[/imath] is increasing and divergent? Appreciate all advice. Thank you. |
1290763 | Find all complex matrices [imath]A[/imath] such that [imath]n\operatorname{Tr}(AB) = \operatorname{Tr}(A)\operatorname{Tr}(B)[/imath] for all [imath]B[/imath].
Consider a bilinear form [imath]f(A,B) = n\operatorname{Tr}(AB) - \operatorname{Tr}(A)\operatorname{Tr}(B)[/imath] defined on [imath]M_n(\mathbb{C})[/imath]. I need to find the set [imath]U^\perp[/imath] of all matrices [imath]A[/imath] such that [imath]f(A,B) = 0[/imath] for every [imath]B \in M_n(\mathbb{C})[/imath], or, more specifically, find [imath]\dim(U^\perp)[/imath]. Other than the zero matrix I can't find any other general matrix. So [imath]\dim (U^\perp) = 0[/imath] I'd assume? | 1290116 | Finding the dimension of the orthogonal complement
Let [imath]U=M_{n}(\mathbb{C})[/imath], and we define a bilinear form [imath]\xi (A,B)=n\cdot tr(AB)-tr(A)tr(B)[/imath]. How do I find [imath]dim(U_{\perp })[/imath]? I know that [imath]U_{\perp }=\{A \in M_{n}(\mathbb{C})|\ \forall B\in M_{n}(\mathbb{C}), n\cdot tr(AB)-tr(A)tr(B)=0 \}[/imath]. I have no idea how to continue. I'd appreciate any help. Thanks. |
1291647 | Find a Four-element Abelian Subgroup of [imath]S_5[/imath]
Prof. Charles Pinter's "A Book of Abstract Algebra" provides this exercise: Ch 7 (Groups of Permutations) Part B #3 - Find a four-element abelian sub-group of [imath]S_5[/imath]. Write its table. Please provide guidance on how to answer this question. I'm guessing that I need to come up with a permutation table: [imath]\begin{matrix} 1& 2& 3& 4& 5&\\ ?& ?& ?& ?& ?&\\ \end{matrix} [/imath] Then, I need to compose [imath]f[/imath] (where f defines the initial table) 5x in order to get [imath]S_5[/imath]. However, I'm unsure if that's correct, and how I'd proceed once I found [imath]S_5[/imath]. | 532626 | 4 Element abelian subgroup of S5.
I have a homework question from my intro to group theory class. Question: Find a 4 element abelian subgroup of [imath]S_5[/imath]. Write it's table. This is where I've gotten so far, but I don't even know if I'm on the right mental track. We know [imath]S_5 = \{1,2,3,4,5\}[/imath]. Abelian means the subgroup of [imath]S_5[/imath] is commutative. I am having trouble figuring out how to start this. Do I pick any 4 elements of [imath]S_5[/imath]? Or are there particular elements that must be chosen? Do I perform permutations on the group first? Or do I chose my 4 elements, and then perform the permutations? Thank you! |
1291548 | Stuck on a Limit Question in Multivariable Calculus
I have just started learning about limits in my multivariable class and I came to a problem: Let [imath]h(x,y)=\frac{x^5y}{2x^{10}+y^2}.[/imath] How would I prove that [imath] \lim_{(x,y) \to (0,0)} h(x,y) \text{ Does Not Exist} [/imath] When I used polar coordinates ([imath]x=r\cos(\theta),y=r\sin(\theta)[/imath]), I keep on getting that the limit is 0. I don't know how to do this. A friend said to look at the squeeze theorem, but I don't know how that applies here. I asked this question earlier and I didn't receive answers that were up to my level of learning. I mentioned this on there but the answer I got was wrong. | 1289995 | Prove that the [imath]\lim_{(x,y) \to (0,0)}h(x,y)[/imath] Does not Exist using Polar Coordinates
Let [imath]h(x,y)=\frac{x^5y}{2x^{10}+y^2}[/imath]. How would I prove that the [imath]\lim_{(x,y) \to (0,0)}h(x,y)[/imath] Does Not Exist? I think that we might use polar coordinates, but I am not sure. |
643122 | Splitting field containing [imath]n[/imath]th root
Let [imath]K[/imath] be a splitting field of a polynomial over [imath]\mathbb{Q}[/imath]. Suppose [imath]K[/imath] contains an [imath]n[/imath]th root of some number [imath]a[/imath]. Then how can we show that [imath]K[/imath] contains all the [imath]n[/imath]th roots of unity? I don't really see any relation between [imath]\sqrt[n]{a}[/imath] and the [imath]n[/imath]th roots of unity. Knowing that the former is in [imath]K[/imath] doesn't seem to imply anything about the latter. | 643237 | Splitting field of [imath]x^n-a[/imath] contains all [imath]n[/imath] roots of unity
This statement is suggested as a correction to this question: If [imath]K[/imath] is the splitting field of the polynomial [imath]P(x)=x^n-a[/imath] over [imath]\mathbb{Q}[/imath], prove that [imath]K[/imath] contains all the [imath]n[/imath]th roots of unity. How to prove it? |
1291775 | A question about polynomials
How can we prove that the following expression is a polynomial? [imath] \frac{1-x^{2^{n-1}}}{1-x} [/imath] I've asked this question just for learning the ways different from using [imath]1-x^n=(1-x)(1+x+x^2+···+x^{n-1})[/imath]. | 383379 | Prove that [imath]x-1[/imath] divides [imath]x^n-1[/imath]
In algebra & polynomials, how do we prove that [imath]x-1 \mid x^n -1?[/imath] |
1292405 | Non-constant entire function-bounded or not?
Show that if [imath]f[/imath] is a non-constant entire function,it cannot satisfy the condition: [imath]f(z)=f(z+1)=f(z+i)[/imath] My line of argument so far is based on Liouville's theorem that states that every bounded entire function must be a constant. So I try-to no avail-to show that if [imath]f[/imath] satisfies the given condition, it must be bounded. I haven't made much progress with this, so any hints or solutions are welcome. | 1244416 | Proving that a doubly-periodic entire function [imath]f[/imath] is constant.
Let [imath]f: \Bbb C \to \Bbb C[/imath] be an entire (analytic on the whole plane) function such that exists [imath]\omega_1,\omega_2 \in \mathbb{S}^1[/imath], linearly independent over [imath]\Bbb R[/imath] such that: [imath]f(z+\omega_1)=f(z)=f(z+\omega_2), \quad \forall\,z\in \Bbb C.[/imath]Prove that [imath]f[/imath] is constant. The intuition seems clear to me, we have the three vertices of a triangle given by [imath]0[/imath], [imath]\omega_1[/imath] and [imath]\omega_2[/imath]. All points in the plane are one of the vertices of that triangle under a suitable parallel translation. The constant value will be [imath]f(0)[/imath], fine. Throwing values for [imath]z[/imath] there, I have found that [imath]f(n\omega_1) = f(\omega_1) = f(0)=f(\omega_2) = f(n\omega_2), \quad \forall\, n \in \Bbb Z.[/imath] I don't know how to improve the above for, say, rationals (at least). Some another ideas would be: Checking that [imath]f' \equiv 0[/imath]. I don't have a clue of how to do that. Write [imath]w = a\omega_1+b\omega_2[/imath], with [imath]a,b \in \Bbb R[/imath], do stuff and conclude that [imath]f(w) = f(0)[/imath]. This approach doesn't seem good, because I only have a weak result with integers above. Finding that [imath]f[/imath] coincides with [imath]f(0)[/imath] on a set with an accumulation point. This seems also bad: the set on with [imath]f[/imath] coincides with [imath]f(0)[/imath] by which I found above is discrete. Nothing works and this is getting annoying... And I don't see how analyticity comes in there. I'll be very thankful if someone can give me an idea. (On a side note.. I know that this title is not informative at all. Feel free to edit if you come up with something better.) |
1292555 | Is the Sorgenfrey Line second countable?
The Sorgenfrey topology on [imath]\mathbb{R}[/imath] is the topology whose basic open sets are of the form [imath][a,b)[/imath] where [imath]a < b \in \mathbb{R}[/imath]. Does it have a countable base? (I suspect not.) Certainly it is separable and first countable. The reals with the usual topology is second countable because open intervals with rational endpoints form a countable base. However this approach does not work for the Sorgenfrey topology, since open sets of the form [imath][i,b)[/imath] where [imath]i[/imath] is irrational, cannot be written as the union of basic open sets [imath][p,q)[/imath] with rational endpoints. | 1135993 | [imath]\mathbb{R}[/imath] with the lower limit topology is not second-countable
I am trying to prove that [imath]\mathbb{R}[/imath] with the lower limit topology is not second-countable. To do this, I'm trying to form an uncountable union [imath]A[/imath] of disjoint, half-open intervals of the form [imath][a, b)[/imath], [imath]a < b[/imath]. Is this possible? I think this would imply the [imath]A[/imath] is open but no countable union of basis elements could coincide with [imath]A[/imath] therefore making the real numbers with the lower limit topology not second-countable. I think there must exist something like [imath]A[/imath] described above but I am having trouble visualizing it and coming up with a formula to represent it. Maybe there is some other way to show it is not second-countable. |
1293030 | Which of the following sets have the cardinality the same as [imath]R[/imath]
Which of the following sets has the same cardinality as that of [imath]\mathbb{R}[/imath]? [imath]W[/imath]=The set of constant functions on [imath]\mathbb{R}[/imath] [imath]X[/imath]=The set of polynomial functions on [imath]\mathbb{R}[/imath] [imath]Y[/imath]=The set of continuous functions on [imath]\mathbb{R}[/imath] [imath]Z[/imath]=The set of all functions on [imath]\mathbb{R}[/imath] Clearly [imath]W[/imath] has same cardinality as that of [imath]\mathbb{R}[/imath] (by using the inclusion map).I can give injective maps from [imath]\mathbb{R}[/imath] to [imath]X[/imath] (send [imath]r[/imath] to the constant polynomial [imath]r[/imath]).The other way around : i use the map [imath]X[/imath] to [imath]\mathbb{R}[/imath]. [imath]p(x)=a_0 + a_1x +a_2x^2 + a_3 x^3 +.....+a_n x^n[/imath] is sent to [imath]a_00a_10a_20...a_n[/imath]. By schroeder bernstein theorem i conclude that they have same cardinality.The others i can't conclude. | 940266 | Cardinality of sets regarding
Consider the following sets of functions on [imath]\mathbb{R}[/imath]. [imath]W=[/imath]The set of all constant functions on [imath]\mathbb{R}[/imath] [imath]X=[/imath]The set of polynomial functions on [imath]\mathbb{R}[/imath] [imath]Y=[/imath] The set of continuous functions on [imath]\mathbb{R}[/imath] [imath]Z=[/imath] The set of all functions on [imath]\mathbb{R}[/imath] Which of the sets has the same cardinality as that of [imath] \mathbb{R} [/imath]? Only [imath] W [/imath] Only [imath] W [/imath] and [imath]X[/imath] Only [imath] W, X[/imath] and [imath]Y[/imath] All of [imath] W,X,Y [/imath] and [imath]Z[/imath] I suppose [imath]W[/imath] is the correct answer as [imath]X[/imath] is equivalent to a subset of [imath]\mathbb R\times \mathbb R[/imath] and [imath]X\subset Y\subset Z[/imath]. Am I correct? |
1292152 | Prove or Disprove that if [imath]f_x(x_0,y_0)[/imath] and [imath]f_y(x_0,y_0)[/imath] both exist, then f is continuous at [imath](x_0,y_0)[/imath]
I have to either prove or disprove the fact that if [imath]f_x(x_0,y_0)[/imath] and [imath]f_y(x_0,y_0)[/imath] both exist, then f is continuous at [imath](x_0,y_0)[/imath]. What I thought: I thought that the best way to approach this is to use a function that does what we want to prove or disprove. So my attempt to finding a function is: [imath]f(x,y)=\left\{\begin{array}{cl} \frac{xy}{x^2+y^2} & \text{if } (x,y) \neq (0,0)\\ 0 & \text{ } \text{otherwise}\end{array}\right.[/imath] Is this correct to use? What do I do next? | 87688 | Does existence of partial derivatives implies continuity at a point [imath](x_0,y_0)[/imath]?
If [imath]F: \mathbb R^2 \to \mathbb R[/imath] and [imath]F_x[/imath] (partial derivative of [imath]F[/imath] wrt [imath]x[/imath]) and [imath]F_y[/imath] exist at [imath](x_0,y_0)[/imath] then the function is continuous at that point. Is this true? If not what could be a counter-example? |
1293099 | If [imath]A|B[/imath] and [imath]B|A[/imath] then prove [imath]A=\pm B[/imath]
If [imath]A|B[/imath] and [imath]B|A[/imath] then prove [imath]A=\pm B[/imath] So far I have [imath]A|B \iff AX=B[/imath] and [imath]B|A \iff BY=A[/imath] with [imath]X,Y \in \mathbb{Z}[/imath] Not sure how to finish, any help. | 940718 | Prove or disprove - If a divides b and b divides a does a=b
Prove or disprove: If a, b belong to the set of positive integers, and if a divides b and b divides a, then a=b. Does this hold if if a,b are not necessarily positive? Why or Why not? Here is what I have: If a and b are integers, we say a divides b if there exists an integer k so that b=ak. Thus: [imath]b=ak[/imath] [imath]a=bj[/imath] [imath]a=akj[/imath] [imath]kj=1[/imath] Since a divides b and b divides a, we know that k,j must both be integer, which means k and j must both equal 1 or -1. This means: [imath]a=b[/imath] This proves that a=b when we assume that a,b are both positive integers. For the second part of the question, I am asked if I can still say a=b if the assumption about a, b is relaxed so that a,b belong to the integers (not the positive integers.) I think The answer is no, I cannot say that a=b for every case when we allow a,b to also be negative integers. Here is my reasoning (I use the same idea that a=bj and b=ak here) [imath]+/-a=akj[/imath] [imath]+/-1=kj[/imath] So [imath] |a|=|b|[/imath] but not [imath]a=b[/imath] I'm very new to proofs so any suggestions or thoughts are greatly appreciated. |
219759 | Show that both mixed partial derivatives exist at the origin but are not equal
[imath]f(x,y) = \begin{cases} \displaystyle \frac{xy(x^2-y^2)}{x^2+y^2} & \text{if } (x,y) \neq (0,0), \\ 0 & \text{if } (x,y) = (0,0). \end{cases}[/imath] I tried finding both mixed partial derivatives but they ended up being the same for that function. I must not be taking into account something dealing with the fact that it is piece-wise. I still need to show the mixed partial derivatives exist. How can I do all of this? | 2881185 | Partial derivative of [imath]xy\frac{x^2-y^2}{x^2+y^2}[/imath]
I am asked to show, that [imath]f(x,y)=\begin{cases} xy\frac{x^2-y^2}{x^2+y^2}\space\text{for}\, (x,y)\neq (0,0)\\ 0\space\text{for}\, (x,y)=(0,0)\end{cases}[/imath] is everywhere two times partial differentiable, but it is still [imath]D_1D_2f(0,0)\neq D_2D_1 f(0,0)[/imath] But this does not make much sense in my opinion. Since [imath]f(0,0)=0[/imath] hence it should be equal? I calculated [imath]\frac{\partial^2 f(x,y)}{\partial x\partial y}[/imath] and [imath]\frac{\partial^2 f(x,y)}{\partial y\partial x}[/imath] and I got in both cases the same result (for [imath](x,y)\neq (0,0)[/imath]) which is: [imath]\frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}[/imath] Wolframalpha says, that this is correct. So is there a mistake in the task? Or do I not understand it? Also, when I want to show, that [imath]f[/imath] is everywhere two times partial differentiable. Is it enough to calculate [imath]\frac{\partial^2 f}{\partial^2 x}[/imath] and [imath]\frac{\partial^2 f}{\partial^2 y}[/imath] and not needed to go by the definition, since we know, that this function is differentiable as a function in [imath]\mathbb{R}[/imath]? What do you think? Thanks in advance. |
1292599 | Probability generating function of binomial distribution
In a population of [imath]2n[/imath] individuals there are [imath]n[/imath] infected individuals and [imath]n[/imath] uninfected. Suppose that [imath]X[/imath] of the n uninfected become infected, where [imath]X \sim \mathcal B(n, p)[/imath], and, then, given [imath]X = j[/imath], [imath]Y[/imath] of the [imath]n + j[/imath] infected recover, where [imath]Y\mid X=j \sim \mathcal B(n + j, q)[/imath], allowing [imath]Z = n + j − Y[/imath] infected. It is assumed the infection and recovery probabilities satisfy [imath]p, q ∈ (0; 1)[/imath]. First find the probability generating function of [imath]X[/imath] and the conditional probability generating function of [imath]Y[/imath] given [imath]X = j[/imath]. Then, use conditional expectation to find the probability generating function of [imath]Z[/imath]. Now, assume that [imath]Z[/imath] has the same distribution as the sum of two independent binomial random variables and thus comment on the effect of the infection/recovery process on the initial number of infected [imath]n[/imath]. I know that [imath]G_X(z)=(1-p+pz)^n[/imath] where [imath]z[/imath] is an element of [imath]\Bbb R[/imath] is the generating function for Binomial distribution but I'm unsure how to use this to start the question?? | 1287865 | Finding the PGF of [imath]Z[/imath] using conditional expectation.
I am working on this problem, where I am required to write down the pgf of [imath]X[/imath], as well as the pgf of [imath]Y[/imath] given [imath]X=j[/imath], and then using conditional expectation find the pgf of [imath]Z[/imath]. So far, I have \begin{align*} G_{X}(z)&=(pz+(1-p))^{n}\\ G_{Y|X=j}(z)&=(qz+(1-q))^{n+j} \end{align*} I'm new to generating functions and am having difficulty finding the pgf of [imath]Z[/imath]. I thought I had a breakthrough when I wrote down \begin{align*} G_{Z}(z)&=\mathbb{E}\left(z^{Z}\right)\\ &=\mathbb{E}\left(z^{n+j-Y}\right)\\ &=\frac{z^{n+j}}{G_{Y}(z)} \end{align*} But I'm quite sure this isn't correct as I am then required to deduce that [imath]Z[/imath] has the same distribution as the sum of two independent Binomial random variables. I don't know how to apply conditional expectation here. I'm sure this is a very simple problem, and I just want to understand it. Thanks for your patience and time. |
1287594 | Roots of [imath]X^2-z[/imath] in [imath]2[/imath]-adics
Using Hensel's lemma is it true that [imath]X^2-z[/imath], for [imath]z\in \mathbf Z[/imath] any integer have no roots in 2-adics ? Hensel's lemma only shows, if a root of a polynomial can be lifted to a root in [imath]\mathbf Z_p[/imath], so it doesn't say anything about the nonsolvability but if I look at the conditions: If there is an [imath]a[/imath] s.t. [imath]f(a)\equiv 0\mod p,\quad f'(a)\not\equiv 0\mod p[/imath] If [imath]p=2[/imath], then the derivative is always zero for a polynomial of the form [imath]X^2-z[/imath], Is there still a root ? | 473595 | Characterization of integers which has a [imath]2[/imath]-adic square root
Does anyone know an "elementary" proof of the following theorem? Let [imath]k \neq 0[/imath] be a rational integer. Then [imath]k[/imath] admits a square root in [imath]\mathbb{Z}_2[/imath] if [imath]k = 4^a (8b+1)[/imath] for some [imath]a \in \mathbb{N}[/imath], [imath]b \in \mathbb{Z}[/imath]. About [imath]p[/imath]-adic numbers I don't know anything more sophisticated than Hensel lemma. Thank you! |
1293757 | Brouwer fixed-point theorem on non-convex sets
I read about the Brouwer fixed-point theorem in Wikipedia, and got confused about whether it holds when the domain of the function is non convex. On one hand, convexity is explicitly mentioned as a pre-condition. On the other hand, they say that it holds also for domains that are homeomorphic to a disc, and there may be such domains which are not convex (e.g. a concave quadrangle). So, what is the largest family of sets of [imath]\mathbb{R}^n[/imath] on which the BFPT holds? | 323841 | Why is convexity a requirement for Brouwer fixed points? Shouldn't "no holes" be good enough?
Brouwer's fixed point theorem: Every continuous function [imath]f[/imath] from a convex compact subset [imath]K[/imath] of a Euclidean space to [imath]K[/imath] itself has a fixed point. I am wondering why the word "convex" is in there. It seems to me that it is necessary and sufficient for [imath]K[/imath] to have no holes, which is a weaker condition than convexity. Necessary: if [imath]K[/imath] has a hole, then the continuous mapping that simply rotates points around this hole has no fixed point. Sufficient: A unit disk in any number of dimensions has a Brouwer fixed point. Any compact, hole-less set [imath]K[/imath] is homeomorphic to the unit disk in some number of dimensions. If we map [imath]K[/imath] to the unit disk [imath]D[/imath] with a homeomorphism [imath]h[/imath], then consider the function from [imath]D[/imath] to [imath]D[/imath] given by [imath]h \circ f \circ h^{-1}[/imath]. This is a continuous function from the unit disk to itself (which is convex and compact), so it has a Brouwer fixed point [imath]x[/imath]. Then [imath]h(f(h^{-1}(x))) = x[/imath], so [imath]f(h^{-1}(x)) = h^{-1}(x)[/imath], so [imath]h^{-1}(x)[/imath] is a Brouwer fixed point of [imath]K[/imath]. What's wrong with this argument? |
1293722 | Is entire function a polynomial?
Let [imath]f:\mathbb{C}\to \mathbb{C}[/imath] be an entire function, and suppose that for every [imath]z\in \mathbb{C}[/imath] there exists [imath]n_z\in \mathbb{N}[/imath] such that [imath]f^{(n_z)}(z)=0[/imath]. Is [imath]f[/imath] necessarily a polynomial? | 995105 | Entire function with vanishing derivatives?
Let [imath]f:\mathbb{C}\rightarrow\mathbb{C}[/imath] be an entire function. And assume that at each point, one of it's derivatives vanishes. What can you say about [imath]f[/imath]? A hint suggests that [imath]f[/imath] must be a polynomial. |
1294268 | Finding value (Trigo Series)
Find the value of [imath]\cos ^2\theta+\cos^2 (\theta+1^{\circ})+\cos^2(\theta+2^{\circ})+......+\cos^2(\theta+179^{\circ})[/imath] Can anyone teach me where to start with? I've no idea. | 1294870 | Is there another way to solve this Trigo in series?
Find the value of [imath]\cos ^2\theta+\cos^2 (\theta+1^{\circ})+\cos^2(\theta+2^{\circ})+...... +\cos^2(\theta+179^{\circ})[/imath] Attempt, [imath]\cos x=-\cos(180^\circ-x),\sin x=\cos(90^\circ-x),\cos x=\sin(90^\circ-x),\sin x=\sin(180^\circ-x)[/imath] [imath]\cos^2\theta+\cos^2(\theta+1^\circ)+\cos^2(\theta+ 2^\circ)+ \dots+\cos^2(\theta+179^\circ)=\cos^2\theta+\sum_{ n=1}^{179}\cos^2(\theta+n^\circ)[/imath] [imath]\cos^2(\theta+1^\circ)=\cos^2\theta\cos^21^\circ-2\cos\theta\cos1^\circ\sin\theta\sin1^\circ+\sin^2 \theta\sin^21^\circ[/imath] [imath]\cos^2(\theta+179^\circ)=\cos^2\theta\cos^2179^ \circ-2\cos\theta\cos179^\circ\sin\theta\sin179^\circ+ \sin^2\theta \sin^2179^\circ=\cos^2\theta\cos^2179^\circ+2\cos \theta\cos1^\circ \sin\theta\sin1^\circ+\sin^2 \theta\sin^2179^\circ[/imath] [imath]\cos^2\theta+\sum_{n=1}^{179}\cos^2(\theta+n^\circ )=\cos^2\theta+\sum_{n=1}^{179}(\cos^2\theta\cos^2 n^\circ+\sin^2\theta \sin^2n^\circ)[/imath] [imath]=\cos^2\theta+\sin^2\theta+2\sum_{n=1}^{89}(\cos^2 \theta\cos^2n^\circ+\sin^2\theta\sin^2n^\circ)[/imath] [imath]=1+2\sum_{n=1}^{89}\cos^2n^\circ=1+2\left(44+ \dfrac12 \right)=90[/imath] |
1295633 | How to prove [imath]\lim_{x \to \infty}\frac{\log(x)}{x}=0[/imath]?
I will soon make a math exame where one can't use L'Hôpital's rule, integrals concept neither the formal limit definition. The most I can use is the derivative definition and the algebraic ways to solve limits. My thought was: Let [imath]\displaystyle \lim_{x \to \infty} \frac{\log(x)}{x}=y[/imath], so [imath]\displaystyle \lim_{x \to \infty}\log(x^{\frac{1}{x}})=y[/imath]. Then, [imath]\displaystyle \lim_{x \to \infty}e^{\log(x^{\frac{1}{x}})}=e^y \Leftrightarrow[/imath] [imath]\displaystyle \lim_{x \to \infty}x^{\frac{1}{x}}=e^y [/imath] I ended with an indetermination. How can I proof the [imath]\displaystyle \lim_{x \to \infty}\frac{\log(x)}{x}=0[/imath] with the restritions and as formal it can get? Thanks. | 127750 | Is this limit correct: $\lim_{x \to+\infty} \frac{\log_{2}(x-1)}{x} = 0$?
Find [imath]\space\ \begin{align*} \lim_ {x \to+\infty} \left [ \frac{\log_{2}(x-1)}{x}\right] \end{align*}[/imath]. After some minutes around this limit I did it this way: [imath]\log_{2}(x-1)=y \Leftrightarrow 2^y=x-1[/imath] So,[imath]\space x=2^y+1[/imath]. When [imath]x \to +\infty[/imath],[imath]\space y \to +\infty[/imath] also. By substitution: [imath]\begin{align*} \lim_ {y \to+\infty} \left [ \frac{\log_{2}(2^y+1-1)}{2^y+1}\right]=\lim_ {y \to+\infty} \left [ \frac{\log_{2}(2^y)}{2^y+1}\right]=\end{align*}[/imath] [imath]\begin{align*}\lim_ {y \to+\infty} \left [ \frac{y}{2^y+1}\right]=\lim_ {y \to+\infty} \left [ \frac{1}{\frac{2^y+1}{y}} \right]=\lim_ {y \to+\infty} \left [ \frac{1}{\frac{2^y}{y}+\frac{1}{y}}\right]= \frac{1}{+\infty+0}=0 \end{align*}[/imath] Is this correct?Are there any other easy way to find this limit?Thanks |
1295643 | Is the group [imath](\Bbb Z,+)[/imath] isomorphic to the the group [imath](\Bbb Q\setminus\{0\},\cdot)[/imath]?
Is the group [imath](\Bbb Z,+)[/imath] isomorphic to the the group [imath](\Bbb Q\setminus\{0\},\cdot)[/imath]? | 1109241 | Prove that [imath]\mathbb{Q}^{\times}[/imath] not isomorphic to [imath]\mathbb{Z}^{n}[/imath]
[imath]\mathbb{Q}^{\times}[/imath] is the group of rational number without [imath]0[/imath] under multiplication, and [imath]\mathbb{Z}^{n}[/imath] is the free abelian group of rank [imath]n[/imath]. Show that [imath]\mathbb{Q}^{\times}[/imath] not isomorphic to [imath]\mathbb{Z}^{n}[/imath]. I tried to assume that there is an isomorphism and to get a contradiction.. didn't succeed |
1295662 | [imath]\int _{ 0 }^{ 1 }{ \frac { { x }^{ t }-1 }{ \ln { x } } dx } [/imath]
How do I solve the following integral: [imath]\int _{ 0 }^{ 1 }{ \frac { { x }^{ t }-1}{ \ln { x } } dx } [/imath] | 1282315 | Computing [imath]\int^1_0 \frac{p^2-1}{\text{ln } p}dp[/imath]
Does anyone know the exact value of this integral? [imath]\int^1_0 \frac{p^2-1}{\ln p}dp[/imath] I know that the approximate value is [imath]1.09861...[/imath]. But I can't seem to get a figure on the exact value, or an antiderivative. |
1295945 | Prove: If [imath]p \in \mathbb{N}[/imath], [imath]p[/imath] is prime and [imath]p\mid ab[/imath] then [imath]p\mid a[/imath] or [imath]p\mid b[/imath].
Theorem 1. If [imath]p \in \mathbb{N}[/imath], [imath]p[/imath] is prime and [imath]p\mid ab[/imath] then [imath]p\mid a[/imath] or [imath]p\mid b[/imath]. I am stuck on this proof here is what I have done so far: Proof of Thm 1. [imath]p\mid ab \implies px=ab~(x\in \mathbb{Z})[/imath] ... I'm unsure how to proceed. | 1216213 | If a prime [imath]p\mid ab[/imath], then [imath]p\mid a[/imath] or [imath]p\mid b[/imath]
If a prime number [imath]p[/imath] is a divisor of a product [imath]ab[/imath], [imath]p[/imath] has to be a divisor of [imath]b[/imath] or [imath]a[/imath]. How can I demonstrate this theorem? I demonstrated this theorem on one way using Bezout's theorem in an answer below but are there other proofs?? |
1295986 | Is the Cartesian product of two uncountable sets uncountable?
Is Cartesian product of two uncountable sets uncountable? Suppose we have a set of real numbers [imath]R[/imath], Can't it be shown that [imath]R[/imath] is uncountable by Cantor's diagonalization method, so it follows that the Cartesian product of two sets of real numbers in uncountable? | 1294585 | Is the set of all pairs of real numbers uncountable?
My hypothesis is that [imath]\mathbb{R \times R}[/imath], the set of all pairs [imath](r_1, r_2)[/imath], of real numbers is uncountable. I understand that the set of all pairs of natural numbers is countable. But could someone explain why the set of all pairs of real numbers uncountable? I am having trouble proving it using diagolization |
1296054 | Prove [imath] \lim\limits_{n\to\infty}\int_0^1 f(x)g(nx)\,dx=\int_0^1 f(x)\,dx\int_0^1 g(x) \, dx [/imath]
Let [imath]f[/imath] and [imath]g[/imath] be a real valued continuous functions on [imath]\mathbb{R}[/imath] such that [imath]f(x+1)=f(x)[/imath] and [imath]g(x+1)=g(x)[/imath] for all [imath]x\in \mathbb{R}[/imath]. Prove that [imath] \lim_{n\to\infty}\int_0^1 f(x)g(nx)\,dx=\int_0^1 f(x)\,dx\int_0^1 g(x)\,dx. [/imath] | 406998 | Putnam 1967 problem, integration
I have some problems understanding this solution of this problem: Given [imath]f,g : \mathbb{R} \rightarrow \mathbb{R}[/imath] - continuous functions with period [imath]1[/imath], prove that [imath]\lim _{n \rightarrow \infty} \int_0^1 f(x)g(nx) \, dx = \int_0^1 f(x) \, dx \cdot \int_0^1g(x) \, dx[/imath] It says there that if we split this integral into [imath]n[/imath] parts to get [imath]\int_0^1 f(x)g(nx) \, dx = \sum_{r=0}^{n-1} \int_{\frac{r}{n}}^{\frac{r+1}{n}} f(x)g(nx) \, dx[/imath] we will have [imath]\int_0^1 f(x)g(nx) \, dx = \sum_{r=0}^{n-1} f\left(\frac{r}{n}\right) \int_{\frac{r}{n}}^{\frac{r+1}{n}}g(nx) \, dx[/imath] because for large [imath]n[/imath], [imath]f[/imath] is roughly constant over the range. Could you tell me why [imath]f[/imath] is roughly constant over the range for [imath]n[/imath] large enough? I would really appreciate all your insight. |
778063 | Prove that the center of group G is a subgroup of G.
By definition, the center of G is the set: [imath]Z(G)[/imath] = {[imath]g\in G|g^{-1}xg=x[/imath] for all [imath]x\in G[/imath]} We need to show that: The identity element exists It is closed under the operation For every element [imath]g[/imath], there exists in [imath]G[/imath] the inverse element [imath]g^-1[/imath] My attempt: Since it is defined that: [imath]g^{-1}xg=x[/imath] Thus implying that [imath]g^{-1}\in G[/imath]. For the identity element [imath]e[/imath] and any element [imath]x\in G[/imath], since the identity element is its own inverse, it then implies: [imath]e^{-1}xe=x[/imath] Hence [imath]e\in G[/imath]. Now by multiplying by [imath]g[/imath] on both sides, we get: [imath]g(g^{-1}xg)=gx[/imath] [imath](gg^{-1})xg=gx[/imath] [imath](e)xg=gx[/imath] [imath]xg=gx[/imath] Thus, it is closed under the operation, hence implying that [imath]Z(G)[/imath] is a subgroup of [imath]G[/imath]. Your help and feedback would be really appreciated. | 777946 | Prove that $Z(G)$ which is the center of $G$ is a subgroup of $G$
Question: Let [imath]$G$[/imath] be a group. Prove that [imath]$Z(G)$[/imath] is a subgroup of [imath]$G$[/imath]. If I want to show that [imath]$Z(G)$[/imath] is a subgroup of [imath]$G$[/imath] that means I have to show that it is closed under group operation? Here is my attempt. Let [imath]a,b[/imath] be elements in [imath]Z(G)[/imath] and [imath]x[/imath] be an element in [imath]G[/imath]. Then [imath]ax=xa[/imath] which is under group multiplication commutative and under inverse [imath](a^{-1}) x=x(a^{-1})[/imath]. And hence [imath](a^{-1})bx= (a^{-1})xb=x (a^{-1})b[/imath] which is under group operation so [imath](a^{-1}),b[/imath] are elements in [imath]Z(G)[/imath] thus a subgroup of [imath]G[/imath]. Really appreciate if anyone can help me by directing me if my attempt is not good. Thanks in advance. |
1296999 | Why is [imath] {n\choose k} \equiv 0 \pmod n[/imath] if [imath]n[/imath] is prime?
For all [imath]n>k[/imath], why is: [imath] {n\choose k} \equiv 0 \pmod n[/imath] if [imath]n[/imath] is prime? Any hints anyone? I am really puzzled. | 1291115 | [imath]\binom{p}{i}[/imath] divisible by [imath]p[/imath], with [imath]p[/imath] prime
Let [imath]p[/imath] be a prime. How do you show that the binomial coefficients [imath]\binom{p}{i}[/imath] are divisible by [imath]p[/imath] for [imath]1\leq i\leq p-1[/imath]? And how does this result in the congruency [imath](x+y)^p\equiv x^p+y^p\pmod p[/imath] for [imath]x,y\in\mathbb{Z}[/imath]? |
1296901 | Brauer Algorithm
I'm working on a problem and I'm looking at an equation: [imath]lg(n) + (1 + o(1))(lg(n))/(lg(lg(n))[/imath] All I want to know is, what does o(1) mean? Thanks! | 29100 | small o(1) notation
It's probably a vey silly question, but I'm confused. Does o(1) simply mean [imath]\lim_{n \to \infty} \frac{f(n)}{\epsilon}=0[/imath] for some [imath]n>N[/imath]? |
1297219 | prove that if [imath]N\lhd G[/imath], [imath] M\lhd G[/imath], [imath]M\bigcap N=\{e\}[/imath] so: [imath]mn=nm , \forall n \in N,\forall m\in M[/imath]
prove that if [imath]N\lhd G[/imath], [imath] M\lhd G[/imath], [imath]M\bigcap N=\{e\}[/imath] so: [imath]mn=nm , \forall n \in N,\forall m\in M[/imath] | 678991 | Proving [imath]nk = kn[/imath]
Let [imath]K[/imath] and [imath]N[/imath] be normal subgroups of [imath]G[/imath] such that [imath]K \cap N = \langle e \rangle[/imath]. Prove that [imath]nk = kn[/imath] for all [imath]n \in N[/imath] and for all [imath]k \in K[/imath]. Since [imath]K[/imath] and [imath]N[/imath] are normal, we have [imath]nK = Kn[/imath] and [imath]kN = Nk[/imath]. However, on their own they don't imply [imath]nk = kn[/imath]. How can I prove this? |
1297247 | Number of ways to connect sets of k vertices in a perfect n-gon
This is a copy of my post at Mathexchange.com, as my question is still not fully answered and I really wanna find a solution to this. Feel free to refer to there for useful comments and partial solutions: Number of ways to connect sets of [imath]k[/imath] dots in a perfect [imath]n[/imath]-gon Let Q(n,k) be the number of ways in which we can connect sets of k vertices (dots), in a given perfect n-gon, such that no two lines intersect at the interior of the n-gon and no vertice remains isolated. Intersection of the lines outisde the n-gon is acceptable. Obviously, k|n, and n can't be prime because otherwise there will be dots left unconnected. The n-gon itself is an acceptable solution to a connection of n vertices, and in the case of k>2, these aren't lines, but a set of connected lines, a sort of a network formed by connected planar graphs with straight edges with k vertices, which are required to be vertices of the n-gon itself. There must always be S=n/k sets of lines. For k=2, there are exactly n/k lines, and for k>2, there are exactly n/k, not lines but sets of such connected planar graphs. Take for example Q(6,2). We have a perfect hexagon. By brute-forcing with pencil and paper, I found that there are 5 ways to connect sets of 2 vertices (dots) such that no two lines intersect inside the hexagon. Hence, Q(6,2) = 5. The following image depicts the case of Q(6,2): Now let's move one step further: Let U(n,k) be the number of unique ways to connect sets of k dots/vertices in a perfect n-gon, such that no lines intersect inside the n-gon, and rotational symmetry is neglected, i.e, every possible arrangement is unique and can't be formed by rotating another arrangement in any way. U(6,2)=2. Note that U(6,2)=2 because the arrangements of the first line in the image are not unique, and can be formed by rotating one another. The same happens for the second line of arrangements. Hence U(6,2)=2. I'm pretty sure there's a pure combinatorial approach to this problem, perhaps involving Polya's Enumeration Theorem (PET). Is there an elegant solution to these functions? Can they even be solved for k>2? Any light shed on any of the functions will be very much appreciable, as I haven't been successful in deriving a formula for any of them. Both algorithms and formulas will be great! P.S: I can program in java and mathematica Thanks a lot in advance. EDIT - Temporary Solutions + Relevant questions AND progress Q(n,2) = C(n/2) where C(n) = 1/(n+1) * Binomial (2n , n) And C(n) denotes the n'th Catalan number. Now let us denote W(n) = U(n,2). Can you find a formula for W(n)? Perhaps a connection between Q(n,2) and W(n)? | 1294224 | Number of ways to connect sets of [imath]k[/imath] dots in a perfect [imath]n[/imath]-gon
Let [imath]Q(n,k)[/imath] be the number of ways in which we can connect sets of [imath]k[/imath] vertices in a given perfect [imath]n[/imath]-gon such that no two lines intersect at the interior of the [imath]n[/imath]-gon and no vertex remains isolated. Intersection of the lines outside the [imath]n[/imath]-gon is acceptable. Obviously, [imath]k|n[/imath], and [imath]n[/imath] can't be prime because otherwise there will be dots left unconnected. The [imath]n[/imath]-gon itself is an acceptable solution to a connection of [imath]n[/imath] vertices, and in the case of [imath]k>2[/imath], these aren't lines, but a set of connected lines, a sort of a network formed by connected planar graphs with straight edges with [imath]k[/imath] vertices, which are required to be vertices of the [imath]n[/imath]-gon itself. There must always be [imath]S =\frac nk[/imath] sets of lines. For [imath]k=2[/imath], there are exactly [imath]\frac nk[/imath] lines, and for [imath]k>2[/imath], there are exactly [imath]\frac nk[/imath], not lines but sets of such connected planar graphs. Take for example [imath]Q(6,2)[/imath]. We have a perfect hexagon. By brute-forcing with pencil and paper, I found that there are 5 ways to connect sets of 2 vertices (dots) such that no two lines intersect inside the hexagon. Hence, [imath]Q(6,2) = 5[/imath]. The following image depicts the case of [imath]Q(6,2)[/imath]: I know for sure that an elegant solution exists for [imath]k=2[/imath], but I can't figure it out. For generality I ask about any amount of [imath]k[/imath] dots. It's also extremely important to note we're not dividing the [imath]n[/imath]-gon into [imath]k[/imath]-gons, but connecting paths between [imath]k[/imath] nodes/vertices/dots. Now let's move one step further: Let [imath]U(n,k)[/imath] be the number of unique ways to connect sets of [imath]k[/imath] dots in a perfect [imath]n[/imath]-gon, such that no two lines intersect, and rotational symmetry is neglected, i.e, every possible arrangement is unique and can't be formed by rotating or flipping another arrangement in any way. [imath]U(6,2)=2[/imath]. Note that [imath]U(6,2)=2[/imath] because the arrangements of the first line in the image are not unique, and can be formed by rotating one another. The same happens for the second line of arrangements. This results in only 1 unique arrangement for each line. Hence [imath]U(6,2)=2[/imath]. I'm pretty clueless about both functions [imath]U[/imath] and [imath]Q[/imath], and I couldn't derive an algorithm or formula to any of them. Hence I'm posting this here. I'm pretty sure there's a pure combinatorial approach to this problem, perhaps involving Polya's Enumeration Theorem (PET). Is there an elegant solution to these functions? Can they even be solved for [imath]k>2[/imath]? Any light shed on any of the functions will be very much appreciable, as I haven't been successful in deriving a formula for any of them. EDIT - Temporary Solutions + Relevant questions AND progress [imath]Q(n,2) = C_{n \over 2}\quad\text{where}\quad C_n = \frac{1}{n+1} {2n\choose n}[/imath] And [imath]C_n[/imath] denotes the [imath]n[/imath]'th Catalan number. Now let us denote [imath]W(n) = U(n,2)[/imath]. Can you find a formula for [imath]W(n)[/imath]? Perhaps a connection between [imath]Q(n,2)[/imath] and [imath]W(n)[/imath]? Another EDIT: (2) [imath]Q(k,k) = k\cdot 2^{k-3}[/imath] EDIT 3 It seems like in general, [imath]Q(n,3)[/imath] is given here: https://oeis.org/search?q=3%2C27%2C324&sort=&language=english&go=Search Perhaps generalization to higher powers will result in the general [imath]Q(n,k)[/imath] function.. EDIT 4 Here is a small table of values for [imath]Q(n,k)[/imath], which seems to be correct, provided by fabian's algorithm: (Table is in the form [imath]Q(x\cdot k,k)[/imath]) [imath] \begin{array}{c|rrrrr} {_k\,\backslash\, ^n} & k & 2k & 3k & 4k & 5k \\ \hline 2 & 1 & 2 & 5 & 14 & 42\\ 3 & 3 & 27 & 324 & 4455 & 66339\\ 4 & 8 & 256 & 11264 & 573440 & 31752192\\ 5 & 20 & 2000 & 280000 & 45600000 & 8096000000\\ 6 & 48 & 13824 & 5640192 & 2686058496 & 1396580548608\\ \end{array} [/imath] EDIT 5 - [imath]Q(n,k)[/imath] SOLVED As beautifully found and explained by @CuddlyCuttlefish in his answer, the formula for [imath]Q(n,k)[/imath] is as follows: [imath]Q(n,k) = \frac{(n)_{\frac{n}{k}-1}}{\left(\frac{n}{k}\right)!}\cdot (k\cdot 2^{k-3})^{\frac{n}{k}} \quad\text{where}\quad (n)_j = n(n-1)...(n-(j-1))[/imath] And [imath](n)_j[/imath] is the falling factorial, defined as above. Now moving to [imath]U(n,k)[/imath] Now it only remains to find a formula or an algorithm for [imath]U(n,k)[/imath]. Personally I think it has to do with Polya's Enumeration Theorem and Burnside's lemma, combined with the cycle-index of the symmetric group, [imath]Z(S_n)[/imath]. I've touched upon something related to that, and thus I think it's related. I'm not 100% sure but my instincts tell me it's related. EDIT 6 In order to make it clear, I'm hereby adding a picture to describe the case of [imath]U(6,3)[/imath] and by that to clarify better what [imath]U(n,k)[/imath] means. [imath]U(6,3)=4[/imath], as shown in the picture above (@Marko Riedel has pointed out an additional arrangement which I had previously missed). There are four unique arrangements to connect sets of 3 vertices such that no vertice remains isolated, no lines intersect at the interior of the hexagon, and each arrangement is unique and can't be formed by rotating or flipping any other arrangement. There are two unique path-types, one that is introduced by connecting 3 adjacent vertices ([imath]p_1[/imath]), and another that connects 2 vertices with a little "jump", and the third vertice is then adjacent ([imath]p_2[/imath]). Hope it makes things a bit clearer.. EDIT 7 As provided by @Marko Riedel's algorithm, [imath]U(n,3)[/imath] sequence starts as follows: [imath]1, 4, 22, 201, 2244, 29096, 404064, 5915838,\ldots[/imath] Computation was very rough, and calculation times were long. That's about as efficient as it gets, as of now. Producing more values just consumes either too much memory, time or both. Refer to Marko Riedel's answers for more sequences and further explanations. Also if anybody can verify the above it would be great. |
1297263 | Find [imath]f[/imath] so that [imath]\int_{1}^{\infty}f(x)dx[/imath] exists, but [imath]\displaystyle \lim_{x\to\infty}f(x)[/imath] does not exist?
I need help finding an example of a function such that [imath]\int_{1}^{\infty}f(x)dx[/imath] converges, but [imath]\displaystyle \lim_{x\to\infty}f(x)[/imath] does not exist. I was trying to find examples of functions containing some trigonometric function and/or some exponential function, but all the functions I've managed to think of whose improper integral converges also have a well defined limit as [imath]x[/imath] goes to infinity. I'd appreciate any ideas and hints you might have! | 663230 | [imath]\int_{0}^{\infty} f(x) \,dx[/imath] exists. Then [imath]\lim_{x\rightarrow \infty} f(x) [/imath] must exist and is [imath]0[/imath]. A rigorous proof?
Let [imath]f: \mathbb R \rightarrow \mathbb R [/imath] be a continuous function such that [imath]\int_{0}^{\infty} \,f(x) dx[/imath] exists. Then Prove that incase (i) [imath]f[/imath] is a non negative function, then [imath]\lim_{x\rightarrow \infty} f(x) [/imath] must exist and is [imath]0[/imath]. (ii) [imath]f[/imath] is a positive differentiable function , [imath]\lim_{x\rightarrow \infty} f'(x) [/imath] must exist and is [imath]0[/imath] [imath]Attempt[/imath]: For the first part, i don't have a rigorous proof except for the fact that the given condition can be visualised geometrically. Since, the definite integral is actually calculating the area beneath the non negative function, the only way the given limit can exist when limit of f(x) itself tends to 0 at infinity. Please give me a direction so that i can make this proof rigorous enough. For the second part, i took an example. We know that ( leaving out the finite integration parts from [imath]0[/imath] to [imath]1[/imath] ..) [imath]\int_{1}^{\infty} e^{-x^2} dx \leq \int_{1}^{\infty} e^{-x} dx[/imath] and the latter converges. But the derivative of [imath]e^{-x^2} = (-2x)e^{-x^2}[/imath] whose integration does not exist when x [imath]\in~[1,\infty)[/imath] as it's a monotonic function after a finite [imath]x[/imath]. Any help in providing rigor to the above proof will be very helpful Thanks |
1297261 | Let [imath]{A}[/imath] be a [imath]2 \times 2[/imath] matrix. For every two-dimensional vector [imath]{v}[/imath], there...
Let [imath]{A}[/imath] be a [imath]2 \times 2[/imath] matrix. For every two-dimensional vector [imath]{v}[/imath], there exists a two-dimensional vector [imath]{w}[/imath] such that Aw = v. Show that [imath]{A}[/imath] is invertible. I'm not sure how to do this. | 1296214 | Showing a [imath]2\times2[/imath] matrix is invertible
Let [imath]{A}[/imath] be a [imath]2 \times 2[/imath] matrix. For every two-dimensional vector [imath]{v}[/imath], there exists a two-dimensional vector [imath]{w}[/imath] such that [imath]{A} {w} = {v}.[/imath] Show that [imath]{A}[/imath] is invertible. I have no idea on how to start this, but I have a hint: Which two vectors [imath]{v}[/imath] are we most interested in if we are trying to find the matrix [imath]{B}[/imath] such that [imath]{A}{B}= {I}[/imath]? Any help is appreciated! |
1297326 | Prove that A is invertible if [imath]A^2 - 4A -7I = 0[/imath].
The [imath]2 \times 2[/imath] matrix [imath]A[/imath] satisfies [imath]A^2 - 4A -7I = 0,[/imath] where [imath]I[/imath] is the identity matrix. Prove that [imath]A[/imath] is invertible. I'm not sure how to do this. Help would be appreciated. | 1297100 | Proving that matrix in equation is invertible
The [imath]2 \times 2[/imath] matrix [imath]{A}[/imath] satisfies [imath]{A}^2 - 4 {A} - 7 {I} = {0}[/imath] where [imath]{I}[/imath] is the [imath]2 \times 2[/imath] identity matrix. Prove that [imath]{A}[/imath] is invertible. I have tried to solve it like a quadratic, but that doesn't work. Any help is appreciated! |
1297553 | How to compute [imath]\lim_{n\rightarrow\infty}e^{-n}\left(1+n+\frac{n^2}{2!}\cdots+\frac{n^n}{n!}\right)[/imath]
There is a probabilistic method to solve it. But I am not familiar with probability. I am trying to compute it by analytic method, such as using L Hospital's rule or Stolz formula, but they are not working. | 1543967 | Find lim[imath]_{n \to \infty} \sum _{ k =0}^ n \frac{e^{-n}n^k}{k!}[/imath]
We need to find out the limit of, lim[imath]_{n \to \infty} \sum _{ k =0}^ n \frac{e^{-n}n^k}{k!}[/imath] One can see that [imath]\frac{e^{-n}n^k}{k!}[/imath] is the cdf of Poisson distribution with parameter [imath]n[/imath]. Please give some hints on how to find out the limit. |
287138 | Moment generating functions/ Characteristic functions of [imath]X,Y[/imath] factor implies [imath]X,Y[/imath] independent.
This is solely a reference request. I have heard a few versions of the following theorem: If the joint moment generating function [imath]\mathbb{E}[e^{uX+vY}] = \mathbb{E}[e^{uX}]\mathbb{E}[e^{vY}][/imath] whenever the expectations are finite, then [imath]X,Y[/imath] are independent. And there is a similar version for characteristic functions. Could anyone provide me a serious reference which proves one or both of these theorems? | 1615059 | Does [imath]\varphi_{X+Y}(t)=\varphi_X(t)\varphi_Y(t)[/imath] imply independence of [imath]X[/imath] and [imath]Y[/imath]?
It shouldn't, but I am blanking on a counterexample. ETA: Note that the [imath]t[/imath] is shared on both sides - which differentiates this from this question. Similarly [imath]F_{X,Y}(x,y)=F_X(x)F_Y(y)[/imath] implies independence, but [imath]F_{X,Y}(t,t)=F_X(t)F_Y(t)[/imath] doesn't. |
1161528 | Nontrivial chain rule diagrams, how to write chain rule for them and is there implications or constraints in f that arises in the process?
(This is a short version of the question intended to post, because the original is just TOO LONG) From this link and various others (including 2 links deleted by the system as part of the clean up process, unfortunately...) we learnt how to manipulate partial derivatives rigorously. We have learnt in undergraduate on how to write chain rules of multivariable functions using chain rule diagrams such as this one (variables that are kept constant are shown explicitly because the function is written in full) [imath]\frac{\partial f(A(B,C),D(B,C)) )}{\partial B}=\frac{\partial f(A(B,C),D(B,C))}{\partial A(B,C)}\frac{\partial A(B,C)}{\partial B}+\frac{\partial f(A(B,C),D(B,C))}{\partial D(B,C)}\frac{\partial D(B,C)}{\partial B}[/imath] Or using another notation according to the link [imath]\left.\frac{\partial f}{\partial B}\right|_C=\left.\frac{\partial f}{\partial A}\right|_D\left.\frac{\partial A}{\partial B}\right|_C+\left.\frac{\partial f}{\partial D}\right|_A\left.\frac{\partial D}{\partial B}\right|_C[/imath] Things looks fine. However when you have a function that look something like this Attempt to write a chain rule gives a bizarre result for at least one of the variables [imath]\left.\frac{\partial f}{\partial A}\right|_D[/imath] [imath]\color{red}{\left.\frac{\partial f}{\partial D}\right|_{A?}}=\left.\frac{\partial f}{\partial A}\right|_{D?}\frac{dA}{dD}+\color{red}{\left.\frac{\partial f}{\partial D}\right|_{A?}} \hspace{12 mm} [*][/imath] If we consider the red terms identical, then we get [imath]0=\left.\frac{\partial f}{\partial A}\right|_{D?}\frac{dA}{dD}[/imath] Why the bizarre result? How to interpret or explain it (both geometrically and mathematically?) | 180216 | Chain rule for multivariable functions confusion
Suppose [imath]f=f(x,y(x))[/imath]. Then applying the chain rule we get [imath]\frac{\partial f}{\partial x}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}[/imath]. From this it seems that it always holds that [imath]\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}=0[/imath]. Where's the mistake? |
1297981 | Show that If [imath]R[X][/imath] is Euclidean domain then [imath]R[/imath] is a field
Let [imath]R[/imath] is an integral domain . Show that If [imath]R[X][/imath] is Euclidean domain then [imath]R[/imath] is a field . I'll be waiting for your help. Thank you very much in advance! | 940443 | [imath]R[/imath] is a commutative integral ring, [imath]R[X][/imath] is a principal ideal domain imply [imath]R[/imath] is a field
I've just read a proof of the statement: Let [imath]R[/imath] be a commutative integral ring. If [imath]R[X][/imath] is a principal ideal domain, then [imath]R[/imath] is a field. In one part of the proof there is a step which I don't understand. I'll copy the proof: Let [imath]y\in R[/imath] be non-zero. Then using the principal ideal property, for some [imath]f\in R[X][/imath] we have [imath]\langle y,X\rangle = \langle f \rangle \subseteq R[X][/imath]. Therefore, for some [imath]p,q∈R[X], y=fp[/imath] and [imath]X=fq[/imath]. By Properties of Degree we conclude that [imath]f=a[/imath] and [imath]q=b+cX[/imath] for some [imath]a,b,c\in R[/imath]. Substituting into the equation [imath]X=fq[/imath] we obtain: [imath]X=ab+acX[/imath] which implies that [imath]ac=1[/imath], i.e. [imath]a\in R^\times[/imath], the group of units of [imath]R[/imath]. Therefore, [imath]\langle f\rangle =\langle 1\rangle=R[X][/imath]. Therefore, there exist [imath]r,s\in R[X][/imath] such that: [imath]ry+sX=1[/imath] and if [imath]d[/imath] is the constant term of [imath]r[/imath], then we have [imath]yd=1[/imath]. Therefore [imath]y\in R^\times[/imath]. Our choice of [imath]y[/imath] was arbitrary, so this shows that [imath]R^{\times}⊇R \setminus \{0\}[/imath], which says precisely that [imath]R[/imath] is a field. I don't understand this: "[imath]X=ab+acX[/imath] which implies that [imath]ac=1[/imath], i.e. [imath]a\in R^\times[/imath], the group of units of [imath]R[/imath]. Therefore, [imath]\langle f\rangle = \langle 1\rangle =R[X][/imath]." Why does the fact that [imath]a=f[/imath] is a unit implies that [imath]\langle f\rangle = \langle 1\rangle[/imath]? I would appreciate if someone could explain this to me. Thanks in advance. |
1298307 | Why this solution of the birthday problem is wrong?
If we have [imath]n[/imath] people there are [imath]n(n-1)/2[/imath] possible pairs that we can find. The probability that any two people have the same birthday is [imath]1/365[/imath]. So for [imath]n[/imath] people the probability of finding at least one pair with the same birthday is [imath]\frac{1}{365} \cdot \frac{n(n-1)}{2}[/imath]. Setting this to [imath]50\%[/imath] we get that [imath]n=20[/imath] and not [imath]23[/imath] as it should be. What is wrong with the solution? | 122399 | What's wrong with my solution for the birthday problem?
So what I decided to do is to start a small case with only 3 people. There are three possible combinations that I could pair up the people, using the symbols A, B, and C to represent the persons involved: A&B, A&C, B&C. The probability of each pair having the same birthday is [imath]1\over365[/imath] (the first guy gets any birthday, and the second guy only gets one birthday to choose from). What my intuition directed me to do next is to find the probability that any of the three events are true, or the union of the three events, which involved adding up the probabilities. To put this generally, the equation for the probability SHOULD be: [imath]{n \choose 2} \div 365[/imath] However, clearly this is not the case since [imath]n \choose 2[/imath] gets over 365 when [imath]n = 28[/imath]. And with 28 people, there is obviously still a chance that they all have different birthdays (Person 1 has Jan. 1, Person 2 has Jan. 2. ... Person 28 has Jan. 28). Could anyone tell me what's wrong with my intuition? I don't want to know what's the right solution, I just want to know what's wrong with my solution .. it makes sense to my intuition, even though it's incorrect. |
1298196 | Help solve [imath]{{z}^{3}}=\overline{z}[/imath] ([imath]z\in \mathbb{C}[/imath])
Me and my friend try to solve [imath]{{z}^{3}}=\overline{z}[/imath] where [imath]z \in \mathbb{C}[/imath]. My way to solve it was: [imath]\operatorname{cin}(\theta )=\cos(\theta)+\sin(\theta)i[/imath] \begin{align} & z=r \operatorname{cin}(\theta ),\overline{z}=r\operatorname{cin}(-\theta +2\pi k) \\ & {{z}^{3}}={{r}^{3}}\operatorname{cin}(3\theta ) \\ & {{z}^{3}}=\overline{z} \\ & {{r}^{3}}\operatorname{cin}(3\theta )=r\operatorname{cin}(-\theta +2\pi k) \\ & {{r}^{3}}=r\Leftrightarrow {{r}^{3}}-r=0\Leftrightarrow r({{r}^{2}}-1)=0\Leftrightarrow r=\pm 1,r=0\Leftrightarrow r=1 \\ & 3\theta =-\theta +2\pi k\Leftrightarrow 4\theta =2\pi k\Leftrightarrow \theta =\frac{\pi k}{2} \\ \end{align} So my solution for k=0,1,2 : [imath]\begin{align} & {{z}_{1}}=\cos (0)+\sin (0)i=1+0i \\ & {{z}_{2}}=\cos (\frac{\pi }{2})+\sin (\frac{\pi }{2})i=0+1i \\ & {{z}_{3}}=\cos (\pi )+\sin (\pi )i=-1+0i \\ \end{align}[/imath] My friend on the other hand solve it like this: [imath]\begin{align} & {{z}^{3}}=\bar{z} \\ & {{z}^{3}}\cdot z=\bar{z}\cdot z \\ & {{z}^{4}}=|z{{|}^{2}} \\ \end{align}[/imath] He got the same solutions expect he got one more solution than me [imath]{{z}_{4}}=\cos (\pi )+\sin (\pi )i=0-i \\[/imath] What is the right solution and why? | 1283252 | Solving complex numbers equation [imath]z^3 = \overline{z} [/imath]
We have the following equation: [imath]z^3 = \overline{z} [/imath] I set z to be [imath]z = a + ib[/imath] and since I know that [imath] \overline{z} = a - ib[/imath]. I was trying to solve it by opening the left side of the equation. [imath] z^3 = (a+ib)^3 \Rightarrow [/imath] [imath] [a^2+b^2+i(ab + ba)](a+ib) \Rightarrow [/imath] [imath] a^3 - b^2a - 2b^2a + i (2a^2b + b^2a - b^3) [/imath] but this is where I got so far and I'm not sure how continue and if my solution so far is even the right way to solve it. |
1298635 | Which definition of a neighborhood is more standard?
I came across the following two definitions of a neighborhood in a topological space [imath]X[/imath]. Definition: A set [imath]N\subset X[/imath] is a neighborhood of [imath]x\in X[/imath] if [imath]N[/imath] contains a open set in [imath]X[/imath] which contains [imath]x[/imath] On the other hand, some books (like Munkres' book Topology) define a neighborhood as: [imath]N[/imath] is a neighborhood of [imath]x[/imath] if it is an open set containing [imath]x[/imath]. Which definition is more standard, or widely used? | 279904 | What is your definition for neighborhood in topology?
As you know, Munkres-Topology and Rudin-Analysis are really widely using textbooks for undergraduates. They all define a 'neighborhood of [imath]x[/imath]' as an open set containing [imath]x[/imath], so i have followed this definition for 6 months. However, surprisingly, Wikipedia defines a 'neighborhood of [imath]x[/imath]' as a set containing an open set containing [imath]x[/imath]. This really makes me annoyed, since this means that whenever I find a definition referring to a neighborhood on Wikipedia, I have to check whether that definition is equivalent to my definition of a neighborhood. Which one is widely used? |
46195 | Series rearrangement and Riemann's theorem
I think understand that when I have a conditionally convergent series, it consists of series of positive and negative values which are divergent and thus one can find such permutation of indices [imath]\phi : \mathbb{N}\rightarrow\mathbb{N}[/imath] so the rearranged series sums up to an arbitrary value, diverges or oscillates. This is how I understand what Riemann's rearrangement theorem says, but how do I use it practically? For example, when I have the conditionally convergent series: [imath]\sum_{n=1}^{\infty}\frac{(-1)^n}{\sqrt{n}}[/imath] and I want to rearrange it to be divergent or to sum up to certain value M? How do I define such bijections [imath]\phi[/imath] ? I appreciate all help. | 1886500 | Show that for all [imath]x\in\mathbb R[/imath] there is a permutation [imath]\rho : \mathbb N \rightarrow \mathbb N[/imath] such that [imath]\sum^{\infty}_{n=1}x_{\rho(n)}=x.[/imath]
Assume that [imath]\sum x_n[/imath] is convergent but not absolutely convergent. Show that for all [imath]x\in\mathbb R[/imath] there is a permutation [imath]\rho : \mathbb N \rightarrow \mathbb N[/imath] such that [imath]\sum^{\infty}_{n=1}x_{\rho(n)}=x.[/imath] I don't really know where to start on this one. I'm also having difficulty wrapping my head around the fact that simply changing the order of the terms within the sum can change the sum itself. Of course I accept it as true but I can't quite understand why it would be the case and so an explanation of that would also be appreciated. Thank you |
1298683 | How to find the greatest prime number that is smaller than [imath]x[/imath]?
I want to find the greatest prime number that is smaller than [imath]x[/imath], where [imath] x \in N[/imath]. I wonder that is there any formula or algorithm to find a prime ? | 35300 | What is the most efficient algorithm to find the closest prime less than a given number [imath]n[/imath]
Problem Given a number n, [imath]2 \leq n \leq 2^{63}[/imath]. [imath]n[/imath] could be prime itself. Find the a prime [imath]p[/imath] that is closet to [imath]n[/imath]. Using the fact that for all prime [imath]p[/imath], [imath]p > 2[/imath], [imath]p[/imath] is odd and [imath]p[/imath] is of the form [imath]6k + 1[/imath] or [imath]6k + 5[/imath], I wrote a for-loop checking from [imath]n - 1[/imath] to 2 to check if that number is prime. So instead of checking for all numbers I need to check for every odd of the two form [imath]p \equiv 1 \pmod{k}[/imath], and [imath]p \equiv 5 \pmod{k}[/imath]. However, I wonder if there is a faster algorithm to solve this problem? i.e. some constraints that can restrict the range of numbers need to be checked? Any idea would be greatly appreciated. |
1298839 | Let S be the set of stars in our galaxy. There is same number subsets as functions f : S --> {a,b}?
Question: Let [imath]S[/imath] be the set of stars in our galaxy. There is exactly the same number of subsets of stars in our galaxy as there are functions [imath]f:S \to \{a,b\}[/imath] . My solution is; True.However, I'm confused with how to prove it. I know it can be proven by using the formula of the number of elements in a subset and for the formula of the number of elements in a function and proving that both values are equal. Can someone provide me with some feedback, please. | 1298778 | Proving question on sets
I am unable to understand this question. I have to say whether its true of false and then prove it but I can't proceed with the question unless I understand it. Let [imath]S[/imath] be the set of stars in our galaxy. There is exactly the same number of subsets of stars in our galaxy as there are functions [imath]f:S\rightarrow\{\alpha,\beta\}[/imath] |
1298931 | Proof that if [imath]a^3 \mid b^2[/imath] then [imath]a\mid b[/imath].
I am trying to prove that if [imath]a^3 \mid b^2[/imath] then [imath]a\mid b[/imath], where [imath]a,b \in \mathbb{Z}[/imath]. Let [imath]PDC(x)[/imath] be the set of all primes in the prime decomposition of [imath]x[/imath]. So far, I am using the fundamental theorem of arithmetic to try to see what I can do with it. Proving contrapositively, I have [imath]a\nmid \ b \implies \exists p [/imath] a prime, [imath]\exists k \geq 1[/imath] such that [imath]p^k \in PDC(a)[/imath] and [imath]\exists r \in \mathbb{Z}[/imath] such that [imath]p^r[/imath] with [imath]r < k[/imath]. So, [imath]p^{3k} \in PDC(a^3)[/imath] and [imath]p^{2r} \in PDC(b)[/imath] where [imath]2r < 3k[/imath] which implies that [imath]a^3\nmid \ b^2[/imath]. Is this valid? Thank you in advance. | 1236862 | Prove or disprove : [imath]a^3\mid b^2 \Rightarrow a\mid b[/imath]
I think it's true, because I can't see counterexamples. Here's a proof that I am not sure of: Let [imath]p_1,p_2,\ldots, p_n[/imath] be the prime factors of [imath]a[/imath] or [imath]b[/imath] \begin{eqnarray} a&=& p_1^{\alpha_1}\cdots p_n^{\alpha_n} \\ b&=& p_1^{\beta_1}\cdots p_n^{\beta_n} \\ a^3\mid b^2 &&\Rightarrow \frac{b^2}{a^3} \in \mathbb{Z} \\ &&\Rightarrow 2 \beta_i - 3\alpha_i \geq 0 \\ &&\Rightarrow 2 \beta_i \geq 3\alpha_i \text{ and } \beta_i \geq 0 \\ &&\Rightarrow 3 \beta_i \geq 3\alpha_i \\ &&\Rightarrow \beta_i \geq \alpha_i \\ &&\Rightarrow \beta_i - \alpha_i \geq 0 \\ &&\Rightarrow \frac{b}{a} \in \mathbb{Z}\\ &&\Rightarrow a\mid b \end{eqnarray} Is this proof correct ? |
1281226 | Given positive integers a and b such that [imath]a\mid b^2, \:b^2\mid a^3, \:a^3\mid b^4,\: b^4\mid a^5[/imath]..., prove [imath]a=b.[/imath]
Given positive integers a and b such that [imath]a\mid b^2, \:b^2\mid a^3, \:a^3\mid b^4,\: b^4\mid a^5[/imath]..., prove [imath]a=b.[/imath] I was able to show that a and b have the same primes in their factorizations, but I'm not sure how to show that the exponents on these primes are equivalent for all primes in their factorization. I tried contradiction but got stuck. | 1143562 | Let [imath]a,b[/imath] be positive integers such that [imath]a\mid b^2 , b^2\mid a^3 , a^3\mid b^4 \ldots[/imath] so on , then [imath]a=b[/imath]?
Let [imath]a,b[/imath] be positive integers such that [imath]a\mid b^2 , b^2\mid a^3 , a^3\mid b^4 \ldots[/imath] that is [imath]a^{2n-1}\mid b^{2n} ; b^{2n}\mid a^{2n+1} , \forall n \in \mathbb Z^+[/imath] , then is it true that [imath]a=b[/imath] ? |
1299160 | Integrating [imath]\operatorname{Log}(z+2)[/imath] along the unit circle
For the function [imath]f(z) = \operatorname{Log}(z + 2)[/imath], where we choose the principal branch of logarithm (namely, [imath]−\pi < \operatorname{Arg}(z) < \pi[/imath]), and the contour [imath]C := \{z \in \mathbb{C}: |z| = 1\}[/imath], oriented in counterclockwise direction, show that [imath]\int_C f(z) \, \mathrm{d}z =0 .[/imath] Can anyone give me a hint so that I can approach this question? | 1294667 | Apply the Cauchy-Goursat theorem to show that [imath]\int_C \operatorname{Log}(z+2)\, dz=0[/imath] on a unit circle.
Cauchy-Goursat theorem. If a function [imath]f[/imath] is analytic at all points interior to and on a simple closed contour [imath]C[/imath], then [imath]\int_C f(z) \,dz=0.[/imath] This is a problem from Churchill's Complex Variables. Problem: Apply the Cauchy-Goursat theorem to show that [imath]\int_C f(z) \,dz=0[/imath] when the contour [imath]C[/imath] is the unit circle [imath]|z|=1[/imath], in either direction, and when [imath]f(z)=\operatorname{Log}(z+2)[/imath], where [imath]\operatorname{Log}[/imath] is the principal branch. I don't understand how I can apply the theorem in this case, since the branch of [imath]f(z)[/imath] is not defined on the ray [imath]\theta=\pi[/imath]; hence does not satisfy the condition of analytic at all points interior to and on a simple closed contour [imath]C[/imath], of the theorem. How can I make sense of this problem? I would greatly appreciate any help. |
580165 | Proving composition of functions
I am trying to prove the following theorems: Let A, B, and C be nonempty sets and let [imath]f : A \rightarrow B[/imath] and [imath]g : B \rightarrow C[/imath]. If [imath]g \circ f : A \rightarrow C[/imath] is an injection, then [imath]f : A \rightarrow B[/imath] is an injection. If [imath]g \circ f : A \rightarrow C[/imath] is a surjection, then [imath]g : B \rightarrow C[/imath] is an surjection. Any suggestions? Thanks! | 229065 | If [imath]f \circ g[/imath] is onto then [imath]f[/imath] is onto and if [imath]f \circ g[/imath] is one-to-one then [imath]g[/imath] is one-to-one
I am trying to make a picture in my head so I can understand and remember the rules. So if [imath]f \circ g[/imath] is onto, it is onto because the function [imath]f[/imath] maps every element from a set [imath]B[/imath] to a set [imath]C[/imath] (thus [imath]f[/imath] is onto) and if [imath]f \circ g[/imath] is one-to-one then every element from set [imath]A[/imath] is mapping an element of set [imath]B[/imath] (and thus is one-to-one). If both [imath]f[/imath] and [imath]g[/imath] is onto then [imath]f \circ g[/imath] is onto and if both [imath]f[/imath] and [imath]g[/imath] is one-to-one then [imath]f \circ g[/imath] is one-to-one and if both [imath]f[/imath] and [imath]g[/imath] are bijective then [imath]f \circ g[/imath] is bijective? If [imath]f \circ g[/imath] is bijective, we can't say anything, but that [imath]f[/imath] is onto and that [imath]g[/imath] is one-to-one? If [imath]f[/imath] is onto and [imath]g[/imath] is one-to-one, nothing can be said? If [imath]g[/imath] is one-to-one and [imath]g[/imath] is onto, nothing can be said? |
1299266 | How many zeros are there in the number [imath]50![/imath]?
How many zeros are there in the number [imath]50![/imath]? My attempt: The zeros in every number come from the 10s that make up the number. The 10s are, in turn, made up of 2s and 5s. So: [imath]\frac{50}{5*2} = 5[/imath] zeros? | 1087629 | Number of zeroes at end of factorial
Question: How many zeroes will there be at the end of [imath](127)![/imath] Approach: Considering the fact that when two numbers ending in [imath]x[/imath] and [imath]y[/imath] zeroes are multiplied, the resulting number contains [imath]x+y[/imath] zeroes: The numbers to be multiplied that contain zeroes: [imath]120,110,100,90,80.....10[/imath] That comes out to be a total of 13 zeroes. However, that doesn't seem to be the correct answer. Rethinking it, it's obvious that I'm missing out on several cases. For example: [imath]25*22 = 550[/imath] This is just one of the cases which will add a zero. How would I account for all these cases? |
1299134 | If [imath]a_1,a_2,a_3[/imath] are roots [imath]x^3+7x^2-8x+3,[/imath] find the polynomial with roots [imath]a_1^2,a_2^2,a_3^2[/imath]
If [imath]a_1,a_2,a_3[/imath] are the roots of the cubic [imath]x^3+7x^2-8x +3,[/imath] find the cubic polynomial whose roots are: [imath]a_1^2,a_2^2,a_3^2[/imath] and the polynomial whose roots are [imath]\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}.[/imath] Not really sure where to go. Hints appreciated. | 708567 | Elementary Symmetric Polynomials, Roots of cubic polynomials
I'm given [imath]a_1, a_2, a_3[/imath] as roots of the equation [imath]x^3 + 7x^2 - 8x + 3[/imath] and need to find the cubic polynomials with roots [imath]a_1^2, a_2^2, a_3^3[/imath] and [imath]\frac{1}{a_1}, \frac{1}{a_2}, \frac{1}{a_3}[/imath]. It's in a chapter about Galois Theory and is preceded by a problem about elementary symmetric polynomials. I'm guessing I need to use elementary symmetric polynomials to solve for the two desired functions, but I really don't know where to start. Hints would be much appreciated. |
133599 | Conjugacy Classes of [imath]A_n[/imath]
Am I right that conjugacy classes of group [imath]A_n[/imath] can be obtained from conjugacy classes of [imath]S_n[/imath] (which are in correspondence with Young diagrams). If class [imath]C(h)=\{\sigma h {\sigma}^{-1}|\sigma\in S_n\}[/imath] contains independent cycles of only odd length and length of all cycles are different then [imath]C(h)[/imath] in [imath]A_n[/imath] split to two classes [imath]C_1(h)=\{\sigma h {\sigma}^{-1}|\sigma\in A_n\}[/imath] [imath]C_2(h)=\{\sigma \tau h {\tau}^{-1} {\sigma}^{-1}|\sigma\in A_n\}.[/imath] I only interested in the answer. Thanks a lot! | 404656 | Splitting of conjugacy class in alternating group
Browsing the web I came across this: The conjugacy class of an element [imath]g\in A_{n}[/imath]: splits if the cycle decomposition of [imath]g\in A_{n}[/imath] comprises cycles of distinct odd length. Note that the fixed points are here treated as cycles of length [imath]1[/imath], so it cannot have more than one fixed point; and does not split if the cycle decomposition of [imath]g[/imath] contains an even cycle or contains two cycles of the same length. Anybody with a proof? |
1299962 | I.N.Herstein Topics in algebra problem no 2.5.18
If [imath]H[/imath] is a subgroup of [imath]G[/imath].Let [imath]N= \cap\,\, xHx^{-1} \;\;\;\forall x\in G[/imath]. Prove that [imath]N[/imath] is a subgroup such that [imath]aNa^{-1}=N[/imath]. I've proved the subgroup part but couldn't show the second part. | 562108 | Prove that [imath]N[/imath] is normal
Let [imath]H[/imath] be a subgroup of [imath]G[/imath]. Consider the set [imath]N=\cap_{x\in G}xHx^{-1}[/imath]. Prove that [imath]N[/imath] is normal subgroup of [imath]G[/imath]. Using the fact that any (finite or infinite) intersection of subgroups is a subgroup I am able to prove [imath]N[/imath] is a subgroup of [imath]G[/imath] and even [imath]H[/imath]. |
1299390 | Trying to prove [imath]2( \sqrt{n+1}-\sqrt n )< \frac{1}{\sqrt n}<2( \sqrt{n}-\sqrt {n-1})[/imath] and use this to prove...
I am trying to prove this [imath]2( \sqrt{n+1}-\sqrt n )< \frac{1}{\sqrt n}<2( \sqrt{n}-\sqrt {n-1})[/imath] if [imath]n \ge 1[/imath] and using this to prove [imath]2\sqrt{m}-2<\sum^m_{n=1} \frac{1}{\sqrt n}<2( 2\sqrt{m}-1)[/imath] if [imath]m\ge 2[/imath] and in particular i want to show that [imath]m=10^6[/imath] the last inequality is between 1998 and 1999 i am using prove by induction ,first ,for the first statement but i having trouble establish for n=1 and i am confuse how to prove if n is true by hypothesis then n+1.for the second part i didn't get to that part Edit reading the answers and comments bellow now i find myself trying to prove the second part that is prove [imath]2\sqrt{m}-2<\sum^m_{n=1} \frac{1}{\sqrt n}<2( 2\sqrt{m}-1)[/imath]. My attempt [imath]2( \sqrt{n+1}-\sqrt n )< \frac{1}{\sqrt n}<2( \sqrt{n}-\sqrt {n-1})\Leftrightarrow \sum^m_{n=1}2( \sqrt{n+1}-\sqrt n )< \sum^m_{n=1}\frac{1}{\sqrt n}<\sum^m_{n=1}2( \sqrt{n}-\sqrt {n-1})\Leftrightarrow [/imath] i assume this to be true [imath]2\sum^m_{n=1}( \sqrt{n+1}-\sqrt n )< \sum^m_{n=1}\frac{1}{\sqrt n}<2\sum^m_{n=1}( \sqrt{n}-\sqrt {n-1})\Leftrightarrow [/imath] now using telescoping that is [imath]\sum^m_{n=1} (a_n - a_{n-1})=(a_m - a_0)[/imath] i can not make sense of it | 1211973 | Estimating partial sums [imath]\sum_{n = 1}^m \frac{1}{\sqrt{n}}[/imath]
Apostol's Calculus, exercise number I 4.7 13. Prove that if [imath]n \geq 1[/imath], then [imath] 2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1}) [/imath] and use this to prove that if [imath]m \geq 2[/imath], then [imath] 2 \sqrt{m} - 2 < \sum_{n = 1}^m \frac{1}{\sqrt{n}} < 2\sqrt{m} - 1 [/imath] In particular, when [imath]m = 10^6[/imath] the sum lies between [imath]1998[/imath] and [imath]1999[/imath]. After establishing the trivial case, for [imath]n=1[/imath], I can't think of a way to perform the inductive step: assuming [imath]2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1})[/imath] and using this, deducing the same for [imath]n+1[/imath] is true as well: [imath]2(\sqrt{n+2} - \sqrt{n+1}) < \frac{1}{\sqrt{n+1}} < 2(\sqrt{n+1} - \sqrt{n})[/imath] Any help is very appreciated, thanks |
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