qid
stringlengths 1
7
| Q
stringlengths 87
7.22k
| dup_qid
stringlengths 1
7
| Q_dup
stringlengths 97
10.5k
|
|---|---|---|---|
1827496
|
Unitary operator
Let [imath]A:V \to V[/imath] be a linear operator in inner product space [imath]V[/imath] and [imath]\forall x,y \in V \\ x\cdot y=0 \Rightarrow Ax \cdot Ay =0[/imath] Show there exists [imath]\lambda \in \mathbb F[/imath] such that [imath]\lambda A[/imath] is unitary operator. I haven't done anything so far, and I would appreciate any hint you may have for me. Thanks in advance
|
1790611
|
Linear transformation preserving orthogonality
If [imath]A[/imath] is such that if [imath]u[/imath] and [imath]v[/imath] are orthogonal than [imath]A(u)[/imath] and [imath]A(v)[/imath] are also orthogonal (that is, their scalar multiple is [imath]0[/imath]) for any [imath](u,v)[/imath] pair and [imath]A \in Hom (V,V)[/imath] where V is a complex vector space then prove that A is a scalar (real,non-negative) multiple of a unitary transformation(so [imath]A=\lambda D[/imath] where [imath]D^*D=I[/imath] and [imath]\lambda \ge0[/imath] and [imath]\lambda \in \mathbb R[/imath]).
|
1828015
|
Solving an exponential inequality problem
How do I prove the following inequality : [imath]\Bigg(\frac{2}{\alpha^2} \, \big( e^{\alpha x} - e^{\alpha y} \big) \, + \, e^{\alpha y} (y^2 - x^2) \; \Bigg) \geq 0 [/imath] given, [imath]x, y \geq 0[/imath] ? Can anyone provide me with hints about this problem ? Here, [imath]\alpha[/imath] is a strictly positive constant.
|
1806070
|
Solving an exponential inequality problem
How do I prove the following inequality : [imath]\Bigg(\frac{2}{\alpha^2} \, \big( e^{\alpha x} - e^{\alpha y} \big) \, + \, e^{\alpha y} (y^2 - x^2) \; \Bigg) > 0 [/imath] given, [imath]x, y > 0[/imath] ? Can anyone provide me with hints about this problem ? Thanks in advance. Here, [imath]\alpha[/imath] is a positive constant.
|
1828057
|
Prove that an infinite chain of proper containments of compact sets is non empty
I need to prove that if [imath]K_1\supset K_2 \supset K_3 \supset K_4 \supset \ldots[/imath] is a chain of proper containments and each [imath]K_{i}\subseteq \mathbb{R}^{n}[/imath] is compact, then [imath]\bigcap_{i=1}^{\infty} K_{i} \neq \emptyset[/imath]. I understand that [imath]K_i \neq \emptyset, \forall i[/imath] because of the proper subset condition and since the empty set has no proper subsets. I also understand that the compact condition is necessary in order to prevent sets that limit, at infinity, to an empty set but for no [imath]i[/imath] are empty themselves (example: [imath]K_i=(0, 1/i)[/imath]). However, I'm not sure how to complete.
|
1709401
|
Let [imath]\{K_i\}_{i=1}^{\infty}[/imath] a decreasing sequence of compact and non-empty sets on [imath]\mathbb{R}^n.[/imath] Then [imath]\cap_{i = 1}^{\infty} K_i \neq \emptyset.[/imath]
Let [imath]\{K_i\}_{i=1}^{\infty}[/imath] a decreasing sequence of compact and non-empty sets on [imath]\mathbb{R}^n.[/imath] Then [imath]\cap_{i = 1}^{\infty} K_i \neq \emptyset.[/imath] I heard about a proof that take [imath]x_i \in K_i.[/imath] Then I have constructed a sequence. It is bounded because for every [imath]i[/imath], [imath]x_i \in K_1[/imath] that is bounded. Then there is a subsequence that converges. That is where I stopped. What indeed means that a subsequence converges? And why does the limit of such subsequence is in every [imath]K_i,[/imath] showing that such intersection is not empty. Thank you!
|
1828122
|
slope of a tangent line through point
I was given this calculus problem and am very stuck on it and cant get an answer The slope of the tangent line to the graph of [imath]f(x)[/imath] at each [imath]x\neq 0[/imath] is given by [imath]e^{6x}+\frac{6}{x}[/imath] and knowing that the graph contains the point [imath](1,e/3)[/imath], find [imath]f(x)[/imath]. I know you need to integrate but I am stuck and cant get the right answer can anyone walk me through it?
|
1827518
|
slope of a line tangent through a point
I have a problem I am stuck on. its really confusing and I don't know where to start The slope of the tangent line to the graph of [imath]f(x)[/imath] at each [imath]x\ne 0[/imath] is given by [imath]e^{6x}+{6\over x}[/imath] and knowing that the graph contains the point [imath](1,{e\over 3})[/imath], find [imath]f(x)[/imath]. How do I start this?
|
1806641
|
Show that [imath](\mathbb{Z}\oplus\mathbb{Z})[x][/imath] is not isomorphic to [imath]\mathbb{Z}[x]\oplus\mathbb{Z}[x] [/imath] .
Problem says: Show that [imath](\mathbb{Z}\oplus\mathbb{Z})[x][/imath] is not isomorphic to [imath]\mathbb{Z}[x]\oplus\mathbb{Z}[x][/imath]. But consider the map [imath]\varphi:(\mathbb{Z}\oplus\mathbb{Z})[x]\rightarrow\mathbb{Z}[x]\oplus\mathbb{Z}[x][/imath] defined by [imath](a_{m},b_{m})x^{m}+\cdots+(a_{1},b_{1})x+(a_{0},b_{0})\mapsto(a_{m}x^{m},b_{m}x^{m})+\cdots+(a_{1}x,b_{1}x)+(a_{0},b_{0}) .[/imath] It seems a well-defined isomorphism between those rings. So, I think that problem is wrong. Is it okay?
|
1823355
|
[imath](\Bbb Z\oplus \Bbb Z)[X] \not\cong \Bbb Z[X]\oplus \Bbb Z[X]\,[/imath]?
Well i tried to prove it but failed. They are both commutative and have the same set of units. So i can't see a property of one ring which the other ring does not possess. Thank you for any help.
|
1827985
|
Show that the unit sphere is connected
I need to show that [imath]\{(x,y,z)\in\mathbb{R}^{3}:x^2+y^2+z^2 = 1\}[/imath] is connected. Intuitively I understand that it is path connected and, therefore, connected. However, I don't understand how I would define such a path mathematically. For example, if I wish to show that every point on the unit sphere connected to the north pole and, therefore, to every other point, how would I show this?
|
1801973
|
Is the unit sphere in [imath]\Bbb R^4[/imath] is path connected?
I am asked whether [imath]X=\{(x,y,z,w)|x^2 + y^2 + z^2 + w^2 = 1 \}\subset \mathbb{R}^4[/imath], is path connected or not. I just know that [imath]X[/imath] is a closed subset. How can I answer this question? Is there any hint? Thank you very much.
|
1828253
|
Why is [imath]I[x][/imath] not maximal [imath]\mathbb{Z}[x][/imath]?
We have that [imath]I=(2)[/imath] is maximal in [imath]\mathbb{Z}[/imath] because [imath](2)\subseteq (4)\subseteq \dots \subseteq (2^k)[/imath], right? Why is [imath]I[x][/imath] not maximal [imath]\mathbb{Z}[x][/imath] ?
|
143136
|
If [imath]I = \langle 2\rangle[/imath], why is [imath]I[x][/imath] not a maximal ideal of [imath]\mathbb Z[x][/imath], even though [imath]I[/imath] is a maximal ideal of [imath]\mathbb Z[/imath]?
Let [imath]I = \langle 2\rangle[/imath]. Prove [imath]I[x][/imath] is not a maximal ideal of [imath]\mathbb Z[x][/imath] even though [imath]I[/imath] is a maximal ideal of [imath]\mathbb Z[/imath]. My professor mentioned that I should try adding something to it to prove that [imath]I[x][/imath] is not the maximal ideal. I want to assume he meant to add [imath] \langle 2 \rangle[/imath] to [imath]I[x][/imath]?
|
1828398
|
Disprove the statement [imath]f(A \cap B) = f(A) \cap f(B)[/imath]
If someone could walk me through this I would greatly appreciate it. Disprove the following statement: If [imath]f : X \rightarrow Y[/imath] is a function and [imath]A[/imath], [imath]B[/imath] are subsets of [imath]X[/imath] then [imath]f(A \cap B) = f(A) \cap f(B)[/imath].
|
170725
|
Do we have always [imath]f(A \cap B) = f(A) \cap f(B)[/imath]?
Suppose [imath]A[/imath] and [imath]B[/imath] are subsets of a topological space and [imath]f[/imath] is any function from [imath]X[/imath] to another topological space [imath]Y[/imath]. Do we have always [imath]f(A \cap B) = f(A) \cap f(B)[/imath]? Thanks in advance
|
1829068
|
Proving that there exists [imath]a\in R[/imath] such that [imath]a \equiv a_k \pmod{I_k}[/imath]
Let [imath]I_1,...,I_m[/imath] be ideals of a ring [imath]R[/imath] such that [imath]I_j+\cap_{k\neq j}I_k=R[/imath] for every [imath]j\in\{1,...,m\}[/imath]. Then if [imath]a_1,...,a_m\in R[/imath] there exists [imath]a\in R[/imath] such that [imath]a \equiv a_k \pmod{I_k}[/imath] for every [imath]k\in\{1,...,m\}[/imath]. Fix [imath]k\in\{1,...,m\}[/imath]. To show [imath]a \equiv a_k \pmod{I_k}[/imath] (for some [imath]a[/imath]) is equivalenltly to show that [imath]a_k\in I_k+a[/imath]. Since [imath]a_k\in R=I_k+\cap_{t\neq j}I_t[/imath], then [imath]a_k=b_k+c_k[/imath] for some [imath]b_k\in I_k[/imath] and [imath]c_k\in\cap_{t\neq j}I_t[/imath]. However, clearly this [imath]c_k[/imath] is not the same for all of the [imath]a_k's[/imath]. Would anyone give me a hint? Thank you.
|
90274
|
Question on Comaximal Ideals
There's a small detail in a proof of the Chinese Remainder Theorem for modules I don't understand when it comes to showing the normally constructed homomorphism is a surjection. Suppose [imath]R[/imath] is a commutative unital ring, with [imath]I_1,\dots,I_k[/imath] pairwise comaximal ideals. How then for any [imath]j[/imath] can I find [imath]r\in R[/imath] such that [imath]r\equiv 1\pmod{I_j}[/imath] but [imath]r\equiv 0\pmod{I_i}[/imath] for [imath]i\neq j[/imath]? Thank you!
|
236320
|
Jacobi theta function
This is a question from Stein & Shakarchi's complex analysis book. Show that if [imath]\rho[/imath] is fixed with [imath]Im(\rho)>0[/imath], then the Jacobi theta function [imath]\theta(z|\rho)=\sum_{n=-\infty}^\infty e^{\pi in^2\rho}e^{2\pi inz}[/imath] is of order 2 as a function of z. And here's the hint from the book: [imath]-n^2t + 2n|z| \le -(n^2t)/2[/imath] when [imath]t \gt 0[/imath] and [imath]n \ge 4|z|/t[/imath] I've been teaching myself complex analysis with Stein's book but it's pretty difficult to do so. I can't quite figure out the steps to do a problem. Any help?
|
74780
|
On the growth of the Jacobi theta function
So, I ran into this exercise from Stein & Shakarchi. CA, Chapter 5: Show that if [imath]\tau[/imath] is fixed with positive imaginary part, then the Jacobi theta function [imath]\theta(z | r) = \sum_{n=-\infty}^{\infty}{ e^{\pi i n^{2} \tau} e^{2\pi i n z} }[/imath] is of (growth) order [imath]2[/imath] as a function of z They give this hint: [imath]-n^{2}t + 2n|z| \leq - n^{2}t /2[/imath] when [imath]t > 0[/imath] and [imath]n \geq 4 |z|/t[/imath], but I don't understand how to use it. Any help is to be well received //or any reference of course. Thanks!
|
1217752
|
How to solve this limit: [imath]\lim_{x\to\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}[/imath]
Compute the limit [imath]\lim\limits_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}[/imath] my attempt: I tried to multiply top and bottom by the conjugate [imath]\begin{align} \lim_{x\to+\infty}\sqrt{x+\sqrt{x}}-\sqrt{x-1}&=\lim_{x\to+\infty}\left(\sqrt{x+\sqrt{x}}-\sqrt{x-1}\right)\frac{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\ &=\lim_{x\to+\infty}\frac{\left(\sqrt{x+\sqrt{x}}\right)^2-\left(\sqrt{x-1}\right)^2}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\ &=\lim_{x\to+\infty}\frac{(x+\sqrt{x})-(x-1)}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}}\\ &=\lim_{x\to+\infty}\frac{1+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x-1}} \end{align}[/imath] But I don't know what I can do after this.
|
524288
|
Find the value of : [imath]\lim_{x\to\infty} \sqrt{x+\sqrt{x}}-\sqrt{x}[/imath]
I tried to multiply by the conjugate: [imath]\displaystyle\lim_{x\to\infty} \frac{\left(\sqrt{x+\sqrt{x}}-\sqrt{x}\right)\left(\sqrt{x+\sqrt{x}}+\sqrt{x}\right)}{\sqrt{x+\sqrt{x}}+\sqrt{x}}=\displaystyle\lim_{x\to\infty} \frac{x-x+\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x}}=\displaystyle\lim_{x\to\infty} \frac{\sqrt{x}}{\sqrt{x+\sqrt{x}}+\sqrt{x}}[/imath] I don't even know if my rewriting has helped at all. How would I go about doing this?
|
1374810
|
Prove that orthonormalsystem is an orthonormalbasis
We have an orthonormalsystem in [imath]L^2(0, 2\pi)[/imath]: [imath]\{e^{ikx} : k \in \mathbb{Z}\}[/imath]. Now I want to show that it's also an orthonormalbasis. I thought the easiest way to do that would be to show that for every [imath]v \in L^2(0, 2\pi)[/imath] with [imath](v,\phi_k) = 0, \phi_k(x) = e^{ikx}[/imath] for all [imath]k \in \mathbb{Z}[/imath], [imath]v(x) = 0[/imath] for all [imath]x[/imath]. But I'm stuck showing that if [imath]\int_0^{2\pi} v(x)e^{ikx} dx = 0[/imath] for all [imath]k \in \mathbb{Z}[/imath], then [imath]v=0[/imath]. How do I proceed? Is there an easier way?
|
316235
|
Proving that the Fourier Basis is complete for C(R/[imath]2*\pi[/imath] , C) with [imath]L^2[/imath] norm
Let [imath]H[/imath] be the inner product space = [imath]\{f: \mathbb{R} \to \mathbb{C} \mid f \text{ is continuous and has period }2 \pi\}[/imath] where the inner product is: [imath]\langle f,g \rangle = \int_{0}^{2\pi}f(t)\overline{g(t)} dt [/imath] How do I prove that for [imath]n \in \mathbb{Z}[/imath] and [imath]e_n(t)=\dfrac{1}{\sqrt{2\cdot\pi}} e^{i n t}[/imath], [imath](e_n)_{n \in N}[/imath] is a basis of [imath]H[/imath] (that it is orthonormal is easy).
|
9269
|
How to solve inhomogeneous quadratic forms in integers?
If I have a quadratic form like [imath]y^2 - x^2 - x = k[/imath] none of the techniques I know work because of the nasty [imath]x[/imath]. Note that homogenizing doesn't work because a solution of [imath]Y^2 - X^2 - X Z = k Z^{(2)}[/imath] does not lead to a solution of the original equation in integers, at least as far as I have been able to determine. How does one understand the solution set of these equation? How can I solve them?
|
673836
|
Enumerating integer solutions to quadratic equations
Consider a quadratic equation with integer coefficients in two variables. [imath]ax^2+bxy+cy^2+dx+ey+f=0[/imath] I would like to know how to find the number of integer solutions [imath](x,y)[/imath] to this equations. Is there some closed formula?
|
1830158
|
Is function invertible?
Reflection on the unit circle: Let [imath]E=\mathbb R ^{2} - \left\{0,0\right\} [/imath] be perforated plane and [imath]f: E \mapsto E[/imath] defined by [imath]f\left(x,y\right)=\left(\frac{x}{x^{2}+y^{2} } , \frac{y}{x^{2}+y^{2} } \right) [/imath] Show with Jacobian matrix that [imath]f[/imath] is in all points local invertible. Show that [imath]f[/imath] is also global invertible. Find [imath]f^{-1}[/imath] and explain mapping geometrically. What I did: I started with determinante of Jacobian matrix to show that function is local invertible, but I got the following result. Is that enough for showing that function is local invertible? How do I do rest? [imath]Df=\frac{-x^{4}-y^{4}-2x^{2}y^{2} }{\left(x^{2}+y^{2} \right) ^{2} } [/imath]
|
1831109
|
How to show is function surjective?
[imath]f\left(x,y\right)=\left(\frac{x}{x^{2}+y^{2} } , \frac{y}{x^{2}+y^{2} } \right) [/imath] [imath]R^{2}-(0,0)[/imath] Can someone help me with shwoing that this function is surjective?
|
1829727
|
What is the dual category of topological spaces?
What is the dual category of topological spaces [imath]Top[/imath]? I know that the order theoretic dual of a topological space is a closed set system rather than an open set system. However, this doesn't answer my question (I think) since it is just a transformation of the objects, and not of the morphisms. I thought that maybe it would be the same as [imath]Top[/imath] but have morphisms involving closed sets instead of open sets somehow -- however the definition of a continuous function as one for which preimages of closed sets are closed is equivalent to the definition for which preimages of open sets are open, so this doesn't seem to lead anywhere unless [imath]Top[/imath] is self-dual, which seems unlikely. EDIT: let's see if we can start with a simple example and generalize. Consider two copies of the real line, one with the discrete topology (call it [imath]A[/imath]), and one with the regular Euclidean/metric topology (call it [imath]B[/imath]). The space of continuous functions from [imath]A[/imath] to [imath]B[/imath] is simply the space of all real-valued functions on the real line, i.e. [imath]Hom(A,B)=\{\text{real valued functions on the real line}\}[/imath]. So now for [imath]Top^{opp}[/imath], the dual category to topological spaces, [imath]Hom^{opp}(B,A)=\{\text{relations whose "inverses" are real valued functions on the real line}\}?[/imath] (since the range of a function doesn't have to be its codomain). This is a simple/special case of two objects and their morphisms in [imath]Top^{opp}[/imath], but already the morphisms seem likely to be pretty useless in general since they aren't even functions in general. Still, now what confuses me is that it seems like this problem wouldn't be unique to defining the dual category of [imath]Top[/imath]; any category whose morphisms aren't all invertible would seem to have the same problem. Thus it seems more likely that I am misinterpreting the definitions. EDIT: I am looking for a more precise definition than given in the answer to this question: What is the opposite category of [imath]\operatorname{Top}[/imath]? However, it seems that looking at that question (which I couldn't find before asking this one) that there is no simple answer to what the dual category of topological spaces [imath]Top[/imath] is, which perhaps is unsurprising considering how pathological the simple example given above seems to be. Therefore I am going to vote to close this question as a duplicate.
|
1711330
|
What is the opposite category of [imath]\operatorname{Top}[/imath]?
My question is rather imprecise and open to modification. I am not entirely sure what I am looking for but the question seemed interesting enough to ask: The opposite category of rings is the category of affine schemes. This is usually thought of as the category of spaces. Can we run the construction backwards for categories usually thought of as containing spaces? For instance, does [imath]\operatorname{Top}^{\operatorname{op}}[/imath] have a nice description as some "algebraic" category? Note that it does not seem easy to describe the opposite category of all schemes. Therefore, the above question might be asking too much. Perhaps the following is a more tractable (or not) question: Can we find an "algebraic" category [imath]C[/imath] such that we can embed [imath]C^{\operatorname{op}}[/imath] in [imath]\operatorname{Top}[/imath] such that every topological space can be covered by objects in [imath]C^{\operatorname{op}}[/imath]? Perhaps one would like to replace this criterion of being covered by objects by a more robust notion in general. One can repeat the question for other categories of spaces like: Category of manifolds (perhaps closer to schemes than general topological spaces) Compactly generated spaces Simplicial Sets and so on. A perhaps interesting example is the category of finite sets, it's opposite category is the category of finite Boolean algebras.
|
1830424
|
Help with proving if [imath]s_n[/imath] converges to [imath]0[/imath] and [imath]x_n[/imath] is a bounded sequence, then [imath]\lim(x_ns_n)=0[/imath]
Prove: If [imath]s_n[/imath] converges to [imath]0[/imath] and [imath]x_n[/imath] is a bounded sequence, then [imath]\underset{n \to \infty}{\lim}(x_ns_n)=0[/imath] I'm have trouble getting started on this proof. Since I know [imath]s_n[/imath] converges to [imath]0[/imath] I feel like I should start using the epsilon definition with [imath]x_n[/imath] Since [imath]\underset{n \to \infty}{\lim} s_n=0[/imath] So [imath]\left|x_n-s_n\right| \lt \epsilon \implies (=) \quad \left|x_n\right| \lt \epsilon[/imath] But I'm not sure if the aforementioned steps are correct and where to go from here.
|
633805
|
Prove that [imath] \lim (s_n t_n) =0[/imath] given [imath]\vert t_n \vert \leq M [/imath] and [imath] \lim (s_n) = 0[/imath]
Let [imath] (t_n) [/imath] be a bounded sequence, i.e., there exists [imath] M [/imath] such that [imath] \vert t_n \vert \leq M [/imath] for all [imath] n [/imath], and let [imath] (s_n) [/imath] be a sequence such that [imath] \lim s_n = 0 [/imath]. Prove [imath] \lim (s_n t_n) =0[/imath]. My attempt: The given limit, [imath] \lim (s_n) = 0 [/imath], implies that for any [imath] \epsilon \in \textbf{R}^+ [/imath] there exists a [imath] K\in\textbf{R} [/imath] such that for any [imath] n > K [/imath], [imath]\vert s_n - 0 \vert < \epsilon [/imath]. Thus if [imath] \vert s_n \vert = 0 [/imath], then [imath] \vert s_n t_n - 0 \vert = 0 < \epsilon [/imath]. Instead suppose [imath] \vert s_n \vert \neq 0 [/imath]. Then since [imath] 0 \leq \vert t_n \vert \leq M < M + 1 [/imath], [imath] \vert s_n t_n - 0\vert \leq \vert s_nM - 0 \vert \leq \vert s_n - 0 \vert M < \vert s_n - 0 \vert (M+1) < \epsilon (M+1) [/imath]. Then since [imath] \epsilon(M+1) \in \textbf{R}^+ [/imath], it follows that there is a satisfying [imath] K [/imath]. As this exhausts the cases, [imath] \lim (s_n t_n) =0[/imath]. Is this proof correct? What are some other ways of proving this? Thanks!
|
1830494
|
Calculate the series [imath]\sum_{n=1}^{\infty} \frac{x^n}{(3n)!}[/imath]
I can't find the series [imath]\sum_{n=1}^{\infty} \frac{x^n}{(3n)!}[/imath] But I have no idea how to find. Thanks for any hints or solutions.
|
1795437
|
Series [imath]\frac{x^{3n}}{(3n)!} [/imath] find sum using differentiation
Find sum of the series [imath]\sum_{n=1}^{\infty}\frac{x^{3n}}{\left(3n\right)!}[/imath] using differentiation. So far I found that [imath]S(x)+1=S'''(x)[/imath] but it does not help. Then I tried different interesting ideas like [imath]S(x)+S'(x)+S''(x)=e^x-1\,.[/imath]Maybe if I get the third equation it will allow me to construct a kind of differential equation. Then, by solving it, obtain [imath]S(x)[/imath].
|
1830637
|
How to find the General expression of [imath]\sum_{k=0}^ {\lfloor n/3\rfloor} {n \choose 3k}[/imath]
Well as the title says I'm having problems trying to derive a general expression for this sum which involves cubic roots of unity [imath]\sum_{k=0}^ {\lfloor \frac n 3\rfloor} {n \choose 3k}[/imath] Need help guys!
|
1563905
|
Roots of unity filter, identity involving [imath]\sum_{k \ge 0} \binom{n}{3k}[/imath]
How do I see that[imath]\sum_{k \ge 0} \binom{n}{3k} = (1 + 1)^n + (\omega + 1)^n + (\omega^2 + 1)^n,[/imath]where [imath]\omega = \text{exp}\left({2\over3}\pi i\right)[/imath]? What is the underlying intuition behind this equality?
|
1830769
|
Prove or disprove: if [imath]\int_1^\infty f(x)\,dx[/imath] converges, then [imath]\lim\limits_{x\to\infty}f(x) =0[/imath].
I received the following question: If [imath]\int_1^\infty f(x)\,dx[/imath] converges, then [imath]\lim\limits_{x\to\infty}f(x) =0[/imath]. Image. I know that it is false, but i can't come up with a counter example, Thanks in advance
|
1683426
|
Is there a continuous function such that [imath]\int_0^{\infty} f(x)dx[/imath] converges, yet [imath]\lim_{x\rightarrow \infty}f(x) \ne 0[/imath]?
Is there a continuous function such that [imath]\int_0^{\infty} f(x)dx[/imath] converges, yet [imath]\lim_{x\rightarrow \infty}f(x) \ne 0[/imath]? I know there are such functions, but I just can't think of any example.
|
1831104
|
How we can solve like this integral [imath]\int \frac{dx}{x^n+1}[/imath] (Hypergeometric function)
In the wolfram given equation [imath]\displaystyle\int \dfrac{dx}{x^n+1}=x\ {}_2F_1\left(1,\dfrac{1}{n};1+\dfrac{1}{n};-x^n\right)[/imath], But what it's mean? How we can solve without this.
|
873856
|
Evaluation of Indefinite Integral resulting in Hypergeometric Function
I am attempting to derive the result: [imath] \int \left(1+x^n\right)^{-1/m}dx= x\,_2F_1\left(\frac 1m,\frac 1n;1+\frac 1n;-x^n\right)[/imath] First, I start off with the binomial expansion of the integrand to get: [imath]\int\sum_{k=0}^{\infty}\frac{(1/m)_k}{k!}\left(-x^n\right)^kdx [/imath] Then, I pull out a [imath]-1[/imath] and interchange summation and integration: [imath]\sum_k\int (-1)^k\frac{(1/m)_k}{k!}x^{nk}dx [/imath] [imath] \sum_k (-1)^k\frac{(1/m)_k}{k!}\frac{x^{nk+1}}{nk+1} [/imath] [imath] x\sum_k \frac{(1/m)_k}{k!}\frac{(-x^n)^k}{nk+1} [/imath] So, I'm just right there, but not sure how to express this series in the general form of a hypergeometric series with the two additional Pochhammer symbols. I even typed this last series into Mathematica, and it returned the Hyper. Function. What am I missing, or how do I transform this last series into the desired form?
|
1831058
|
How to solve: [imath]\int\limits_{-\pi \cdot 0.5}^{\pi \cdot 0.5} \sin(x) \int\limits_{0}^{\cos(x)} e^{\sin(t)} \, dt \, dx[/imath]
Let [imath] f(t) = e^{\sin(t)} [/imath] and [imath] F(x) = \int\limits_0^x f(t) \, dt [/imath] and then one has to find: [imath]\int\limits_{-0.5 \pi}^{0.5 \pi} \sin(x) F(\cos(x)) \, dx [/imath] Thus far I got: [imath] I :=\int_{-0.5 \pi}^{0.5 \pi} \sin(x) F(\cos(x)) \, dx = -\int\limits_{-1}^{1} F(u) \, du \tag{$*$} [/imath] And: \begin{align} & \int_{-0.5 \pi}^{0.5 \pi} \sin(x) F(\cos(x)) \,dx \\[10pt] = {} & -\cos(x) F(\cos(x)) - \int_{-0.5 \pi}^{0.5 \pi} \sin(x) F'(\cos(x)) \cos(x) \, dx = 0 + \int_{-1}^1 F'(u)u \, du \\[10pt] = {} & \left[F(u)u \vphantom{\frac 1 1} \right]_{-1}^1 - \int_{-1}^1 F(u) \, du \end{align} Remebering [imath](*)[/imath], it is indicated that: [imath]\left[ F(u)u \vphantom{\frac \int \int} \right]_{-1}^1 = 0 [/imath], however this does not seem right. Does anyone have an idea, how to approach this problem best?
|
1831006
|
[imath] 2.28319 = 0 [/imath] !? - How do I spot the mistake?
I am given [imath] f(t) = e^{\sin(t)} [/imath] and [imath] F(x) = \int\limits_{0}^{x} f(t) dt [/imath] and have to compute: [imath]\int\limits_{-0.5 \pi}^{0.5 \pi} \sin(x) F(\cos(x)) dx [/imath] How do I go about this problem? Thus far I got: [imath] I :=\int\limits_{-0.5 \pi}^{0.5 \pi} \sin(x) F(\cos(x)) dx = -\int\limits_{-1}^{1} F(u) du [/imath] [imath] \ \ \ [/imath](*) And: [imath] \int\limits_{-0.5 \pi}^{0.5 \pi} \sin(x) F(\cos(x)) = -cos(x) F(cos(x)) - \int\limits_{-0.5 \pi}^{0.5 \pi} \sin(x) F'(\cos(x)) \cos(x) dx = 0 + \int\limits_{-1}^{1} F'(u)u du = [F(u)u]_{-1}^{1} - \int\limits_{-1}^{1} F(u)du = 2.28319 + I [/imath] Remembering (*) I get: [imath] 2.28319 = 0[/imath], which cannot be. I am have run my calculations down from top to bottom and up again and seem not able to find the mistake. I would be very happy, if someone could point it point it out in the comments.
|
1831504
|
Numbers divisible by [imath]11[/imath]
A number is divisible by [imath]11[/imath], when the difference between the sum of the digits in the odd positions counting from the left (the first, third, ....) and the sum of the remaining digits is either 0 or divisible by 11. Why is that?
|
328562
|
Divisibility criteria for [imath]7,11,13,17,19[/imath]
A number is divisible by [imath]2[/imath] if it ends in [imath]0,2,4,6,8[/imath]. It is divisible by [imath]3[/imath] if sum of ciphers is divisible by [imath]3[/imath]. It is divisible by [imath]5[/imath] if it ends [imath]0[/imath] or [imath]5[/imath]. These are simple criteria for divisibility. I am interested in simple criteria for divisibility by [imath]7,11,13,17,19[/imath].
|
891122
|
Probability that elements in a noncommutative group commute
We consider [imath]G[/imath], a noncommutative group which contains [imath]n[/imath] elements. Show the probability that [imath]2[/imath] elements of [imath]G[/imath] commute is lower than [imath]\frac{5}{8}[/imath].
|
846217
|
Prove: if [imath]a,b\in G[/imath] commute with probability [imath]>5/8[/imath], then [imath]G[/imath] is abelian
Suppose that [imath]G[/imath] is a finite group. If [imath]P( ab=ba ) >5/8[/imath], prove [imath]G[/imath] is abelian.
|
1832260
|
How to show [imath]\sqrt{3} + \sqrt{2}[/imath] is algebraic?
How to show [imath]\sqrt{3} + \sqrt{2}[/imath] is algebraic? Is there a way I can do this without trial and error? Thanks.
|
50296
|
How to show that [imath]\sqrt{2}+\sqrt{3}[/imath] is algebraic?
In Abbot's Understanding Analysis I am asked to show that [imath]\sqrt{2}+\sqrt{3}[/imath] is an algebraic number. I have shown that those two are algebraic separately (that was simple), but I can't figure out what to do to show that their sum is algebraic, too. For example, I tried [imath](\sqrt{2}+\sqrt{3})^{2}=5+\sqrt{24}[/imath] and then I tried to think of a polynomial of form [imath]ax^2-bx^{0}=0[/imath] e.g. [imath]c(\sqrt{2}+\sqrt{3})^{2}-c(5+\sqrt{24})=0[/imath] that would work, but couldn't find the [imath]c[/imath] value that would make [imath]b=c(5+\sqrt{24})[/imath] integer, but couldn't find one. Maybe I can somehow conjecture that the sum of two algebraic numbers must be algebraic, too, but I was wondering if there's a way to find a polynomial to show this. Thanks!
|
1624739
|
In a [imath]\triangle ABC,[/imath] Evaluation of minimum value of [imath]\cot^2 A+\cot^2 B+\cot^2 C[/imath]
In a [imath]\triangle ABC,[/imath] Evaluation of minimum value of [imath]\cot^2 A+\cot^2 B+\cot^2 C[/imath], Given [imath]A+B+C = \pi[/imath] [imath]\bf{My\; Try::}[/imath] Using [imath]\bf{A.M\geq G.M}[/imath] [imath]\frac{\cot^2 A+\cot^2 B}{2}\geq \cot A\cdot \cot B\Rightarrow \cot^2 A+\cot^2B \geq 2\cot A\cdot \cot B[/imath] Similarly [imath]\cot^2 B+\cot^2 C\geq 2\cot B\cdot \cot C[/imath] and [imath]\cot^2 C+\cot^2 A\geq 2\cot C\cdot \cot A[/imath] So [imath]\cot^2 A+\cot^2 B+\cot^2 C\geq \cot A\cdot \cot B+\cot B\cdot \cot C+\cot C\cdot \cot A=1[/imath] Using [imath]A+B+C = \pi\Rightarrow A+B=\pi-C[/imath] So [imath]\cot\left(A+B\right) = \cot\left(\pi-C\right)\Rightarrow \frac{\cot B\cdot \cot A-1}{\cot B+\cot A} = -\cot C[/imath] So we get [imath]\cot A\cdot \cot B+\cot B\cdot \cot C+\cot C\cdot \cot A=1[/imath] My question is Instead of using [imath]\bf{A.M\geq G.M}[/imath] Inequality, Can we use Jensen inequality directly [imath]\frac{\cot^2A+\cot^2 B+\cot^2 C}{3}\geq \cot^2\left(\frac{A+B+C}{3}\right)[/imath] So we can Write it as [imath]\cot^2A+\cot^2 B+\cot^2 C \geq 1[/imath] If no, then what wrong with it, Thanks
|
1827361
|
Minimizing [imath]\cot^2 A +\cot^2 B + \cot^2 C[/imath] for [imath]A+B+C=\pi[/imath]
If [imath]A + B + C = \pi[/imath], then find the minimum value of [imath]\cot^2 A +\cot^2 B + \cot^2 C[/imath]. I don't know how to solve it. And can you please mention the used formulas first. What I can see is that if one of the angles [imath]A[/imath], [imath]B[/imath], [imath]C[/imath] is small, then the value [imath]\cot^2A[/imath] or [imath]\cot^2B[/imath] or [imath]\cot^2C[/imath] will be big. So I want to make angles big (more precisely, close to [imath]\pi/2[/imath], where cotangent is zero), but the condition [imath]A+B+C=\pi[/imath] prevents me from making all three of them very big . I can see that if [imath]A=B=C=\frac\pi3[/imath], then I get [imath]\cot A=\cot B=\cot C=\frac1{\sqrt3}[/imath] and [imath]\cot^2A+\cot^2B+\cot^2C=1[/imath]. But I do not know whether this is indeed minimum. (According to WolframAlpha this is the minimum. However, I would like to see some proof of this fact.)
|
1832066
|
How does one find any polynomials common roots
I was wondering if there was a way to find the common roots of 2 polynomials. For example let [imath]P_1(x) = x^5+x^3+2x^4-5x^2-7[/imath] and [imath]P_2(x) = 2x^7 +3x^3+4x^6+6x^2-14x^4-21[/imath] Is there an algorithm or a method to find [imath]P_1[/imath] and [imath]P_2[/imath]'s common roots?
|
86265
|
How to find the GCD of two polynomials
How do we find the GCD [imath]G[/imath] of two polynomials, [imath]P_1[/imath] and [imath]P_2[/imath] in a given ring (for example [imath]\mathbf{F}_5[x][/imath])? Then how do we find polynomials [imath]a,b\in \mathbf{F}_5[x][/imath] so that [imath]P_1a+ P_2b=G[/imath]? An example would be great.
|
648728
|
Which of the following sets are compact?
The set of all upper triangular matrices in [imath]\mathbb M(n,\mathbb R)[/imath] such that all their eigenvalues satisfy [imath]|\lambda| \leq 2[/imath]. The set of all real symmetric matrices in [imath]\mathbb M(n,\mathbb R)[/imath] such that all their eigenvalues satisfy [imath]|\lambda| \leq 2[/imath]. The set of all diagonalisable matrices in [imath]\mathbb M(n,\mathbb R)[/imath] such that all their eigenvalues satisfy [imath]|\lambda| \leq 2[/imath].
|
282150
|
Which of the following sets are compact in [imath]\mathbb{M}_n(\mathbb{R})[/imath][NBHM_PhD Screening Test-2013, Topology]
Which of the following sets are compact in [imath]\mathbb{M}_n(\mathbb{R})[/imath] (a) The set of all upper triangular matrices all of whose eigenvalues satisfy [imath]|\lambda|≤2 [/imath]. (b) The set of all real symmetric matrices all of whose eigenvalues satisfy [imath]|\lambda|≤2 [/imath]. (c) The set of all diagonalizable matrices all of whose eigenvalues satisfy [imath]|\lambda|≤2 [/imath]. (a) & (c) are not true. But I am not sure about (b).can anybody help me please.
|
1833099
|
Does there exist a non constant entire function [imath]f: \mathbb C \to \mathbb C[/imath] which is square integrable?
Does there exist a non constant entire function [imath]f: \mathbb C \to \mathbb C[/imath] which is square integrable i.e. [imath]\int_{\mathbb C} \vert f(z)\vert^2 dz< \infty[/imath] ? I think that answer to above question is no and this should be direct application(probably by power series expansion of [imath]f[/imath]) of some theorem but i am unable to solve this.Any ideas?
|
1822732
|
Prove that a square-integrable entire function is identically zero
Suppose [imath]f[/imath] is entire and [imath]\iint_\mathbb{C}|f(z)|^2dxdy < \infty[/imath]Prove that [imath]f\equiv 0.[/imath] So far I have: Suppose [imath]f[/imath] is bounded. Then [imath]f[/imath] is constant by virtue of Liouville and so the conclusion is obvious. Thus, assume [imath]f[/imath] non-bounded. Then [imath]\Big|\iint_\mathbb{C}f^2(z)dA\Big| \leq \iint_\mathbb{C}|f(z)|^2dxdy < \infty[/imath] We can parameterize \begin{align*} \iint_\mathbb{C}f^2(z)dA &= \int_0^{\infty}\int_0^{2\pi}f^2(re^{i\theta})\;d\theta\;rdr \\ &= \int_0^{\infty}\int_0^{2\pi}f^2(re^{i\theta})\;ire^{i\theta}[-i\frac{1}{r}e^{-i\theta}]d\theta\;rdr \\ &= \int_0^{\infty}\oint_{C_R}f^2(z)\;[-i\frac{1}{z}]dz\;rdr \\ &= \int_0^{\infty}2\pi r dr\frac{1}{2\pi i}\oint_{C_R}\frac{f^2(z)}{z}dz \\ \\ &= 2\pi f^2(0)\int_0^\infty rdr \end{align*} Whence f(0) = 0. And I have no idea where to go from there. The "entire" bit seems to suggest Liouville but I have already dealt with the bounded case. Please advise ...
|
1828669
|
Estimate for [imath]\sum_{q=1}^{M}\frac{\varphi(q)}{q^{2}}[/imath] Related to Bourgain Paper
Let [imath]N\gg 1[/imath] be a large parameter, which I ultimately want to let tend to infinity. I am reading an old paper of Bourgain, where he claims the lower bound (Equation 2.50, pg. 118) [imath]\sum_{q=1}^{N^{1/2}-1}\sum_{{1\leq a < q}\atop{(a,q)=1}}\frac{N^{3}}{q^{2}}\geq c(\log N)N^{3}, \tag{1}[/imath] where [imath](a,q)[/imath] denotes the GCD of [imath]a[/imath] and [imath]q[/imath] and [imath]c>0[/imath] is some absolute constant, the value of which I don't care and may change from line to line. Using the following lower bound for Euler's totient function [imath]\varphi(n)[/imath] [imath]\varphi(n)>\frac{n}{e^{\gamma}\log\log n + \frac{3}{\log\log n}} \tag{2}[/imath] and the PNT, I only know how to show that \begin{align*} \sum_{q=1}^{N^{1/2}-1}\sum_{{1\leq a < q}\atop{(a,q)=1}}\frac{N^{3}}{q^{2}}&\geq c\sum_{{1\leq q<N^{1/2}-1}\atop{q=\mathrm{prime}}}\frac{N^{3}}{q^{2}}\cdot q\\ &+c\sum_{{1\ll q<N^{1/2}-1}\atop{q\neq\mathrm{prime}}}\frac{N^{3}}{q^{2}}\cdot\frac{q}{e^{\gamma}\log\log q+\frac{3}{\log\log q}}\\ &\geq cN^{3}\log\log N \tag{3} \end{align*} If I knew the asymptotics of the series [imath]\sum_{q=1}^{N^{1/2}-1}\frac{\varphi(q)}{q^{2}}[/imath] or [imath]\sum_{q=1}^{N^{1/2}-1}\frac{1}{\sigma(q)}[/imath], where [imath]\sigma[/imath] is the divisor sum function, then my problem would be solved. However, I have been unable to find such information. John M. has very kindly provided me with the necessary asymptotics. To see how to get that formula, first recall that [imath]\varphi(n)=\sum_{d\mid n}\mu(d)\frac{n}{d}[/imath], where [imath]\mu[/imath] is the Mobius function, and [imath]\sum_{n=1}^{\infty}\frac{\mu(n)}{n^{2}}=\frac{6}{\pi^{2}}[/imath]. Then [imath]\sum_{n\leq x}\frac{\varphi(n)}{n^{2}}=\sum_{n\leq x}\frac{1}{n^{2}}\sum_{d\mid n}\mu(d)\frac{n}{d}=\sum_{{q,d}\atop{qd\leq x}}\frac{\mu(d)}{d^{2}q}=\sum_{d\leq x}\frac{\mu(d)}{d^{2}}\sum_{q\leq\frac{x}{d}}\frac{1}{q} \tag{4}[/imath] Using the formula [imath]\sum_{q\leq k}\frac{1}{k}=\log(k)+\gamma+\epsilon_{k},[/imath] where [imath]\epsilon_{k}\sim\frac{1}{k}[/imath], and [imath]|\log(y)-\log([y])|=O(1/y)[/imath], we see that \begin{align*} \mathrm{(4)}&=\sum_{d\leq x}\frac{\mu(d)}{d^{2}}[\log(x)-\log(d)+\gamma+O(\frac{1}{x/d})]\\ &=(\log(x)+\gamma)(\frac{6}{\pi^{2}}-\sum_{d>x}\frac{1}{d^{2}})-\sum_{d=1}^{\infty}\frac{\mu(d)\log(d)}{d^{2}}+\sum_{d>x}\frac{\mu(d)\log(d)}{d^{2}}+O(\frac{\log x}{x})\\ &=\frac{6}{\pi^{2}}(\log(x)+\gamma)-A+O(\frac{\log x}{x}) \end{align*}
|
326147
|
Bounds on a sum of gcd's
Does there exist a positive real number [imath]C[/imath] and a positive integer [imath]M[/imath] such that for all [imath]n > M[/imath] we have: [imath]\sum_{i=1}^n\sum_{j=1}^n\gcd (i, j)\ge Cn^2 \log n[/imath] This originally appeared as an Olympiad problem in the case of [imath]4n^2[/imath] on the RHS. This case was much simpler; simply let [imath]i[/imath] only range on primes values and one quickly derives a [imath]\omega \left ( n^2 \log \log \frac{n}{\log n} \right )[/imath] bound if I didn't mess up. The motivation for above bound is derived from the fact that [imath]\sum_{j=1}^n \gcd(i,j) \approx \frac{n}{i} \sum_{d|i} \varphi(i/d) \cdot d \approx n \cdot \tau(i) \cdot \frac{6}{\pi^2}[/imath] So we have: [imath]\sum_{i=1}^n\sum_{j=1}^n\gcd (i, j) \approx \frac{6n^2}{\pi^2} \sum_{i=1}^n \frac{1}{i} \approx \frac{6}{\pi^2} n^2 \log n[/imath] However, this is pretty unrigorous reasoning. Does anybody have any ideas as to how to rigorously prove this bound? Of course, showing that the sum in fact gets "close" to [imath]\frac{6}{\pi^2} n^2 \log n[/imath] as [imath]n \to \infty[/imath] would be even better.
|
1788466
|
Integrate [imath] \int \frac{x^2 + x}{(e^x + x +1)^2}dx [/imath]
[imath] \int \frac{x^2 + x}{(e^x + x +1)^2}dx [/imath] I cant think of any substitution to start this question.
|
1778382
|
Evaluating the integral [imath]\int \frac{x^2+x}{(e^x+x+1)^2}dx[/imath]
Evaluate [imath]\int \frac{x^2+x}{(e^x+x+1)^2}dx[/imath] I tried converting in the form of Quotient rule(seeing the square in the denominator), neither am I able to make the denominators' derivative in the numerator. Some hints would be great. Thanks.
|
1834273
|
[imath]\int_0^4\frac{\log x}{\sqrt{4x-x^2}} dx=0[/imath]
I am having trouble proving that it is equal to zero analytically. I have tried plotting and know that for [imath]0<x<1[/imath] the integrand is negative and positive otherwise. I have tried substitution [imath]u\to \sqrt{x}[/imath] but I cannot proceed further.
|
1689682
|
Prove that [imath]\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}}~dx=0[/imath] (without trigonometric substitution)
The integral is from P. Nahin's "Inside Interesting Integrals...", problem C2.1. His proposed solution includes trigonometric substitution and the use of log-sine integral. However, I think the problem should have an easier solution (without appealing to another complicated integral at least). I have the following trick in mind. Let's introduce the substitution [imath]x=4-z[/imath] [imath]I=\int_0^4 \frac{\ln x}{\sqrt{4x-x^2}}~dx=\int_0^4 \frac{\ln (4-z)}{\sqrt{4z-z^2}}~d(4-z)=\int_0^4 \frac{\ln (4-z)}{\sqrt{4z-z^2}}~dz[/imath] [imath]2I=\int_0^4 \frac{\ln (4x-x^2)}{\sqrt{4x-x^2}}~dx[/imath] [imath]I=\int_0^4 \frac{\ln \sqrt{4x-x^2}}{\sqrt{4x-x^2}}~dx[/imath] And here I'm stuck. I'm not sure if this can go somewhere. Maybe partial integration can help, but I don't know how to choose the functions. What do you think? Here is a question about this integral. Only one answer does not use trig substitution, it used gamma function instead. If there are no other ways, I'm prepared to give up on my question. But I would be grateful if it's left open at least for several days Edit After many attempts, I conclude that there is no trick to this integral. The reason is: the general form of this integral in not zero, but has the same symmetry properties, as the above case: [imath]I(a)=\int_0^a \frac{\ln x}{\sqrt{ax-x^2}}~dx=\int_0^a \frac{\ln (a-x)}{\sqrt{ax-x^2}}~dx=\int_0^a \frac{\ln \sqrt{ax-x^2}}{\sqrt{ax-x^2}}~dx \neq 0[/imath] [imath]I(4)=0[/imath] So we will get nothing from symmetry considerations alone. There are two possible ways to solve this - either trigonometric substitution or gamma function. Edit 2 I was wrong it seems, see the accepted answer.
|
1822811
|
Continuity of [imath]\frac{x^3y^2}{x^4+y^4}[/imath] at [imath](0,0)[/imath]?
Suppose a function [imath]f[/imath] is defined as follows: [imath]f(x,y)=\begin{cases} \frac{x^3y^2}{x^4+y^4}&\text{ when }(x,y)\neq(0,0),\\0 & \text{ when }(x,y)=(0,0).\end{cases}[/imath] Is this function continuous at [imath](0,0)[/imath]? How is this shown? I've tried considering limits for different [imath]y=g(x)[/imath] functions and I am unable to find a counterexample. But I do not see how to prove continuity in general.
|
511518
|
Multivariable Delta Epsilon Proof [imath]\lim_{(x,y)\to(0,0)}\frac{x^3y^2}{x^4+y^4}[/imath] --- looking for a hint
I have the limit [imath]\lim_{(x,y)\to(0,0)}\frac{x^3y^2}{x^4+y^4},[/imath] and would like to show with an [imath]\epsilon-\delta[/imath] proof that it is zero. I know with a situation like [imath]\left|\frac{x^4y}{x^4+y^4}\right|\leq y[/imath] or something similar, but I can't find a way to do the same thing here, as no single term in the numerator is of sufficient degree, although I think I could get this with a small hint.
|
1834939
|
What is the algorithm to factor something like [imath]2+\frac{1}{x}+x?[/imath]
I came across this in homework but I'm interested in the general example, say [imath]ax+bx^{-1}+c.[/imath]
|
47144
|
Factoring Quadratics
Is there a method to find which numbers to use when simplifying quadratics? For example [imath]x^2 + 5x + 6[/imath] is easy enough to find, but what if I have bigger numbers, or I have this quadratic expression: [imath]x^2 + 7x – 6[/imath], then it gets very hard to think of a valid simplification. EDIT: Here's an example: [imath](x^2 + 5x + 6)/(x + 2)[/imath]. When simplifying the first half we get [imath](x + 2)(x + 3)/(x + 2)[/imath], then we can cancel out [imath](x + 2)[/imath] and the answer would be [imath](x + 3)[/imath]. The question is: How to always find the right numbers which to put in? (in this example they where 2 and 3)
|
1834567
|
How many different chains in a Poset?
I found that problem and I could use some help. I have a partial order [imath](2^S,⊆)[/imath] and |S| = n. How many different chains are there in that poset? If I had the Hasse diagram or knew the elements of S it would be easy to find out. But now with knowing just that |S| = n I have absolutely no idea. Could anyone help and provide a methodology? Thanks
|
1551286
|
How many chains are there in a finite power set?
Let [imath]A[/imath] be a finite set with [imath]n[/imath] elements. How many chains are there in [imath]\mathcal P(A)[/imath] -- that is, how many different subsets of [imath]\mathcal P(A)[/imath] are totally ordered by inclusion? It's easy enough to count maximal chains; there are [imath]n![/imath] of them. But counting all chains gets me into horrible inclusion-exclusion situations even if I try to do it by hand for small [imath]n[/imath]. Of course this is also the number of chains in a Boolean algebra with [imath]2^n[/imath] elements, or the number of chains in a finite lattice with [imath]2^n[/imath] elements. By hand computation, the number of chains for [imath]n[/imath] running from [imath]0[/imath] to [imath]6[/imath] is [imath]2, 4, 12, 52, 300, 2164, 18732[/imath]. This sequence appears to be unknown in OEIS.
|
1834778
|
Continuous function on the unit sphere
Let S[imath]^2[/imath] := [imath]\lbrace[/imath] x [imath]\in[/imath] $\mathbb{R}[imath]^3$ : $\Vert x\Vert[/imath]_2$ [imath]\rbrace[/imath] [imath]\subset[/imath] ($\mathbb{R}[imath]^3$, $\Vert .\Vert[/imath]_2$) and T: S[imath]^2[/imath] [imath]\to[/imath] ([imath]\mathbb{R}[/imath], [imath]\vert x\vert[/imath] ) a continuous function. a) Is there a point x[imath]_0[/imath] [imath]\in[/imath] S[imath]^2[/imath] such that T(x[imath]_0[/imath]) = T(-x[imath]_0[/imath]) ? b) Is there a value T [imath]\in[/imath] (Tmax,Tmin), which exactly one x [imath]\in[/imath] S[imath]^2[/imath] takes, with Tmax = sup [imath]\lbrace[/imath] T(x) : x [imath]\in[/imath] S[imath]^2[/imath] [imath]\rbrace[/imath] and Tmin = inf [imath]\lbrace[/imath] T(x) : x [imath]\in[/imath] S[imath]^2[/imath] [imath]\rbrace[/imath] ? I think I have a example of a function in a) : Let T(x) = cos (x[imath]_1[/imath]) with x = (x[imath]_1[/imath], x[imath]_2[/imath], x[imath]_3[/imath])[imath]^T[/imath], so take for example x[imath]_0[/imath]=(1,0,0)[imath]^T[/imath], so is T(x[imath]_0[/imath])=T(-x[imath]_0[/imath]), however I don't know how to start b)
|
1833272
|
Continuous map on [imath]S^2[/imath]
Can you help me with this? Let [imath]S^2 := \{x\in \mathbb R^3:||x||_2 = 1\} \subset (\mathbb R^3, ||\cdot||_2)[/imath] and [imath]T:S^2 \to (\mathbb R, |\cdot|)[/imath] be a continuous map. a) Why does T assume its maximum ([imath]T_{max}[/imath]) and its minimum ([imath]T_{min}[/imath])? b) Is there a point [imath]x_0 \in S^2[/imath] with [imath]T(x_0) = T(-x_0)[/imath]? c) Is there a value [imath]T\in]T_{min},T_{max}[[/imath] that is assumed at only one [imath]x\in S^2[/imath]? a) is easy, it is compact since it's closed and bounded (Heine-Borel). For b) I know it can't be injective since it maps from a higher to a lower dimension, but I don't know if I can use that to show what I need. For c) I assume I can use non-injectivity again, but don't know either how. Any hints?
|
1592503
|
Show that [imath]E =\{(x,\alpha )\mid|f(x)|>\alpha\geq 0 \}[/imath] is measurable.
Show that [imath]E=\{(x,\alpha )\mid|f(x)|>\alpha\geq 0 \}[/imath] is measurable, where [imath]f[/imath] is integrable. Is my proof correct? Let [imath]A[/imath] measurable and [imath]f=1_A[/imath]. Then [imath]E=A\times [0,1][/imath] which is measurable. If [imath]f=\sum \limits _{i=1}^n a_i 1_{A_i}[/imath] is simple, then [imath]E=\bigcup \limits _{i=1}^n A_i\times [0,a_i][/imath] which is measurable. If [imath]f\geq 0[/imath], there is an increasing sequence [imath](f_n)[/imath] of simple function s.t. [imath]f_n\to f[/imath]. If I denote [imath]E^n=\{x\mid |f_n(x)|>\alpha \geq 0\}[/imath] then [imath]E^n\subset E^{n+1}[/imath] and [imath]E=\bigcup \limits _{n=1}^\infty E^n[/imath]. Since [imath]E[/imath] are measurable and the union is countable, then [imath]E[/imath] is measurable. If [imath]f[/imath] is measurable, then we can take a sequence [imath]|f_n|\to |f|[/imath] of simple function that is increasing and do as we did previously.
|
1496524
|
Show that [imath]E=\{(x,\alpha)\mid 0\leq \alpha<|f(x)|\}[/imath] is measurable if [imath]f[/imath] is measurable.
I have to show that [imath]E=\{(x,\alpha)\mid 0\leq \alpha<|f(x)|\}[/imath] is measurable if [imath]f[/imath] is measurable. My attempt : If [imath]f=\boldsymbol 1_F[/imath] where [imath]f[/imath] is measurable, then [imath]E=\big((\mathbb R\cap F)\times [0,1[\big)\cup \big((\mathbb R\cap F^c)\times \{0\}\big)[/imath] which is measurable since it's intersection, union and product of measurable set. If [imath]f[/imath] is simple (remark that [imath]f[/imath] simple [imath]\iff[/imath] [imath]|f|[/imath] simple) i.e. [imath]|f|=\sum_{i=1}^n a_i\boldsymbol 1_{F_i}[/imath] where [imath]a_i>0[/imath] and [imath]F_i[/imath] measurable, we obtain in the same way [imath]E=\bigcup_{i=1}^n\big((\mathbb R\cap F_i)\times [0,a_i[\big)\cup \bigcap_{i=1}^n \big((\mathbb R \cap F_i^c)\times \{0\}\big)[/imath] which is also measurable since it's finite union, intersection and product of measurable set. Finally, if [imath]|f|\geq 0[/imath] we have a sequence of simple function [imath]\{\varphi_k\}_{k\in\mathbb N}[/imath] s.t. [imath]\varphi_k(x)\nearrow |f(x)|\quad\text{ for all }x.[/imath] Let [imath]E_n=\{(x,\alpha)\mid 0\leq \alpha\leq \varphi_n(x)\}.[/imath] Therefore [imath]E_n\subset E_{n+1}[/imath] and thus [imath]E_n\nearrow E[/imath] i.e. [imath]E=\bigcup_{n=1}^\infty E_n,[/imath] and since [imath]E_n[/imath] is measurable, the set [imath]E[/imath] is measurable. Is it correct ?
|
1834719
|
Is there only one complex structure on complex plane [imath]\mathbb{C}[/imath]?
There is a trivial complex structure on [imath]\mathbb{C}[/imath]. Do we have other complex structures on complex plane [imath]\mathbb{C}[/imath]? If not, how to prove it?
|
1530473
|
Is there an elementary way to see that there is only one complex manifold structure on [imath]R^2[/imath]?
Is there an elementary way to see that there is only one complex manifold structure on [imath]\mathbb{R}^2[/imath]? (Up to biholomorphism, naturally.) Elementary in the sense of not appealing to the uniformization theorem.
|
718827
|
Show [imath]GCD(a_1, a_2, a_3, \ldots , a_n)[/imath] is the least positive integer that can be expressed in the form [imath]a_1x_1+a_2x_2+ \ldots +a_nx_n[/imath]
Given [imath]a_1, a_2, a_3, \ldots , a_n[/imath] not all zero, show [imath]\gcd(a_1, a_2, a_3, \ldots , a_n)[/imath] is the least positive integer that can be expressed in the form [imath]a_1x_1+a_2x_2+ \ldots +a_nx_n[/imath]. Also deduce [imath]a_1x_1+a_2x_2+ \ldots +a_nx_n = b[/imath] has solutions [imath]\iff \gcd(a_1, a_2, a_3, \ldots , a_n)|b[/imath]. I know that [imath]ax+by=n[/imath] has integer solutions [imath]\iff \gcd(a,b)|n[/imath]. I'll need to use induction to show [imath]a_1x_1+a_2x_2+ \ldots +a_nx_n = \gcd(a_1, a_2, a_3, \ldots , a_n)[/imath] has solutions, and then I will be able to say that as [imath]\gcd(a_1, a_2, a_3, \ldots , a_n) | a_1x_1+a_2x_2+ \ldots +a_nx_n [/imath] for any [imath]x_1,x_2, x_3 \ldots x_n[/imath] then [imath]c \leq \gcd(a_1, a_2, a_3, \ldots , a_n)[/imath] and [imath]c \geq \gcd(a_1, a_2, a_3, \ldots , a_n)[/imath], and thus [imath]c=\gcd(a_1, a_2, a_3, \ldots , a_n)[/imath].
|
2045409
|
Prove that given [imath]a,b \in \mathbb{Z} : \gcd(a,b)=1 \Rightarrow \exists s,t : sa+tb=1[/imath]
(A) let [imath]S=\{xa+yb: x, y \in \mathbb{Z}\}[/imath]. Explain why [imath]S[/imath] has a smallest positive element. Denote the smallest element by [imath]u[/imath] Attempted Solution: I tried plugging in values for a and b, since they can not be [imath]0[/imath] but the only common positive divisor is [imath]1[/imath]. I assumed [imath]1[/imath] would be the only option for these values. I'm not sure if this is even what the question is asking.
|
1835304
|
Smallest positive element of [imath] \{ax + by: x,y \in \mathbb{Z}\}[/imath] is [imath]\gcd(a,b)[/imath]
Let [imath]S = \{ax + by: x,y \in \mathbb{Z}\}[/imath] and let [imath]e > 0[/imath] be the smallest element in [imath]S[/imath]. Prove that [imath]e \mid a[/imath], and hence prove that [imath]e = \gcd(a,b)[/imath] I'm afraid I can't provide much of my initial working here because I'm at a loss of how to start. I suppose that to show [imath]e\mid a[/imath] I would have to show that [imath]eg = a[/imath] for some [imath]g \in \mathbb{Z}[/imath]. But I have no idea how I would do this.
|
2945010
|
Please help me prove this question
The Question: We can define a set of integers [imath]X_a$$_,$$_b[/imath] = {∀u, v ∈ [imath]\Bbb Z[/imath], au + bv}. For example, if a = 6 and b = 8 then X[imath]_6$$_,$$_8[/imath] includes numbers like 20 = 2[imath]*[/imath]6 + 1[imath]*[/imath]8 and 4 = −2[imath]*[/imath]6 + 2[imath]*[/imath]8. Let c be the smallest positive integer in X[imath]_a$$_,$$_b[/imath]. Prove that every number in X[imath]_a$$_,$$_b[/imath] is a multiple of c.
|
1590783
|
Evaluating [imath]\int_{0}^{\pi}\ln (1+\cos x)\, dx[/imath]
The problem is [imath]\int_{0}^{\pi}\ln (1+\cos x)\ dx[/imath] What I tried was using standard limit formulas like changing [imath]x[/imath] to [imath]\pi - x[/imath] and I also tried integration by parts on it to no avail. Please help. Also this is my first question so please tell if I am wrong somewhere.
|
887069
|
How to find [imath]\int_{0}^{\pi/2} \log ({1+\cos x}) dx[/imath] using real-variable methods?
How do you find the value of this integral, using real methods? [imath]I=\displaystyle\int_{0}^{\pi/2} \log ({1+\cos x}) dx[/imath] The answer is [imath]2C-\dfrac{\pi}{2}\log {2}[/imath] where [imath]C[/imath] is Catalan's constant.
|
1835175
|
Prove that [imath]\Bbb{R}[\cos(\theta),\sin(\theta)]\cong\Bbb{R}[x,y]/(1-x^2-y^2)[/imath]
More precisely, given the ring homomorphism [imath]\phi:\Bbb{R}[x,y]\to\Bbb{R}^\Bbb{R}[/imath], with [imath]\phi(f(x,y)):\Bbb{R}\to\Bbb{R},\,\,\phi(f(x,y))(\theta)=f(\cos(\theta),\sin(\theta))[/imath], where [imath]\Bbb{R}[x,y][/imath] is the ring of formal polynomials in two variables on [imath]\Bbb{R}[/imath], I wish to show that [imath]\mathrm{ker}(\phi)=(1-x^2-y^2)[/imath]. So far the best I can do is to show that [imath]\mathrm{ker}(\phi)\supseteq(1-x^2-y^2)[/imath], which is pretty trivial anyway. In particular, I'm interested in a proof of this statement from the perspective of algebraic geometry.
|
937517
|
Is Pythagoras the only relation to hold between [imath]\cos[/imath] and [imath]\sin[/imath]?
Pythagoras says that [imath]\cos^2 \theta + \mathrm{sin}^2\theta = 1[/imath] for all real [imath]\theta[/imath]. (Vague) Question. Is this the only relationship between the functions [imath]\cos[/imath] and [imath]\sin[/imath]? More precisely: Let [imath]\langle \cos,\sin\rangle[/imath] denote the intersection of all subalgebras of the [imath]\mathbb{R}[/imath]-algebra of all functions [imath]\mathbb{R} \rightarrow \mathbb{R}[/imath] containing [imath]\{\cos,\sin\}[/imath]. (By default, all my algebras are unital, associative and commutative.) Let [imath]A[/imath] denote the [imath]\mathbb{R}[/imath]-algebra presented by the generators [imath]\{c,s\}[/imath] and the relation [imath]c^2+s^2=1[/imath]. There is a unique [imath]\mathbb{R}[/imath]-algebra homomorphism [imath]\varphi : A \rightarrow \langle \cos,\sin\rangle[/imath] given as follows. [imath]\varphi(c) = \cos, \,\,\varphi(s) = \sin[/imath] We know that [imath]\varphi[/imath] is surjective. Question. Is [imath]\varphi[/imath] injective? So consider [imath]f \in A[/imath]. Then [imath]f = \sum_{i,j : \mathbb{N}}a_{ij}s^ic^i[/imath] for certain choices of [imath]a_{ij} : \mathbb{R}[/imath]. Now suppose [imath]\varphi(f)=0[/imath]. We want to show that [imath]f=0[/imath]. Ideas, anyone?
|
1214667
|
[imath]F[/imath] is a field iff [imath]F[x][/imath] is a Principal Ideal Domain
A commutative ring [imath]F[/imath] is a field iff [imath]F[x][/imath] is a Principal Ideal Domain. I have done the part that if [imath]F[/imath] is a field then [imath]F[x][/imath] is a PID using the division algorithm and contradicting the minimality of degree of a polynomial. But I am facing difficulty to do the other part.
|
2183430
|
show that if [imath]K[/imath] is a field then [imath]K[x][/imath] is principal
I want to show that if [imath]K[/imath] is a field then [imath]K[x][/imath] is principal Here is what I did: I took 2 polynomial p(x) and q(x). Since we're on a field we can do a euclidean division of polynomials: [imath]q(x) = p(x)a(x) + r(x)[/imath] with [imath]deg(r) < deg(p)[/imath] Now if [imath]r= 0[/imath] we obviously have [imath]p \in (q)[/imath] But I don't know how to show that it is principal if [imath]r \neq 0[/imath] Also I would like to show that the assertion is false if we replace "[imath]K[/imath] is a field" by "[imath]K[/imath] is a principal domain and an integral domain". I can't figure out how to do that
|
1833004
|
when the sum of some fractions equal to 1.
[imath]r[/imath] is a number such that [imath]r=p^a[/imath]. If the sum of some fractions equal to [imath]1[/imath] and one of the denominators is divisible by [imath]r[/imath] then there is another denominators that is exactly divisible [imath]r[/imath]. It seems to be really easy but I cannot prove it for example: [imath]\frac{7}{12}+\frac{4}{15}+\frac{3}{20}=1[/imath] You can see here we have two denominators that are divisible by [imath]4[/imath] or two denominators that are divisible by [imath]3[/imath].Any hints?
|
1832975
|
when the sum of some fractions be [imath]1[/imath]
prove if we want that the sum of some fractions be [imath]1[/imath] and the denominators of one of them is [imath]d[/imath] then another denominators should divisible by [imath]d[/imath] or [imath]d[/imath] should be divisible to another denominators. It seems to be easy I tried to prove it.I first tried some cases. [imath]1=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}[/imath] Here we can see that [imath]6[/imath] is divisible by [imath]3[/imath]. Also here [imath]6[/imath] is divisible by [imath]2[/imath] .But I want to prove one of the denominators but here two of them is possible.After trying a lot I cannot found any proofs.Any hints? update1: the numerator should be prime
|
1835615
|
Continuous positive function such that [imath]\int\limits_1^{\infty}f(x)dx[/imath] converges, while [imath]\int\limits_1^{\infty}f^2(x)dx[/imath] diverges
Does there exist a continuous positive function such that [imath]\int\limits_1^{\infty}f(x)dx[/imath] converges, while [imath]\int\limits_1^{\infty}f^2(x)dx[/imath] diverges? I have proved that if [imath]f[/imath] is decreasing monotonically, then this is false and [imath]\int\limits_1^{\infty}f^2(x)dx[/imath], but what happens when the demand for monotonic drops?
|
1823185
|
Find a continuous function [imath]f:[1,\infty)\to\Bbb R [/imath] such that [imath]f(x) >0 [/imath], [imath]\int_1^\infty f(x)\,dx [/imath] converges and [imath]\int_1^\infty f(x)^2\,dx[/imath] diverges
I'm trying to find an example of a continuous function [imath]f:[1,\infty) \to \Bbb R [/imath] such that [imath]f(x) >0 [/imath] [imath]\int_{1}^{\infty} f(x) \ dx \ \text{converges and } \int_{1} ^{\infty} f(x)^2 dx \ \text{diverges}.[/imath]
|
603276
|
Volume of "deformed torus"
I'm trying to find explicit form of volume of "deformed torus": Suppose we have a curve [imath]\gamma(t)[/imath] in [imath]\mathbb{R}^n[/imath], [imath]t\in[0,1][/imath]. The curve closed and smooth : [imath]\gamma(0)=\gamma(1)[/imath],[imath]\gamma'(0)=\gamma'(1)[/imath]. I'm defining "deformed torus" as [imath]T_r=\{x\in\mathbb{R}^n \mid |\gamma(t)-x|\leq r\}[/imath] for some [imath]t[/imath]. I guess that volume must be something like [imath]\approx A r^{n-1}[/imath]. So I understand that I need to integrate volume form. I always do it by pullback volume form in parameter space and integrate then. But here I have difficulties with entering any parametrization. Could you help me here?
|
604079
|
Volume of neighborhood of the curve
Let [imath]\gamma:[0,1] \to \mathbb R^n[/imath] be a smooth closed curve, such that [imath]\gamma(0)=\gamma(1), \gamma'(0)=\gamma'(1), |\gamma'(a)|=1[/imath] and let [imath]B_r=\{x| \exists t, |\gamma(t)-x|<r\}[/imath]. How can i show that the volume of [imath]B_r[/imath] is equal to [imath]br^{(n-1)}[/imath] for [imath]r <\varepsilon[/imath] for some [imath]\varepsilon>0[/imath]. And how can I find [imath]b[/imath]? It is intuitively obvious. But how can one prove it? I have no idea.
|
1834917
|
Mean value theorem sum of [imath]\frac{1}{f ' (x _k)}[/imath]
Let [imath]f[/imath] be a function which is derivable in [imath][0,1][/imath] with [imath]f(0)=0[/imath] and [imath]f(1)=1[/imath]. Show that for every [imath]n[/imath] in [imath]\mathbb{N}[/imath], there exists numbers [imath]x_1,x_2,.....,x_n[/imath] all in [imath][0,1][/imath] such as \begin{equation} \sum_{k=1}^n \frac{1}{f'(x_k)}=n \end{equation} I think the mean value theorem should be applied: \begin{equation} \exists\; x_1 \in (0,1): f'(x_1)=\frac{f(1)-f(0)}{1-0}=1 \end{equation} \begin{equation} \exists\; x_2 \in (0,x_1): f'(x_2)=\frac{f(x_1)-f(0)}{x_1-0}=\frac{f(x_1)}{x_1} \end{equation} etc. We have the sum \begin{equation} 1+ \frac{x_1}{f(x_1)} + \frac{x_2}{f(x_2)} +....+\frac{x_{n-1}}{f'(x_{n-1})} \end{equation} But than I get stuck. I was wondering if anyone could be so kind to help Thanks in advance for your time and assistence
|
1835506
|
sum of series using mean value theorem
Let [imath]f(x)[/imath] be a function which is differentiable on [imath][0,1][/imath] with [imath]f(0)=0[/imath] and [imath]f(1)=1[/imath]. Show that for every [imath]n\in \Bbb N[/imath] there exists numbers [imath]x_1,x_2,\ldots,x_n\in [0,1][/imath] such as [imath] \sum_{k = 1}^n \frac{1}{f' (x_k)} = n [/imath] I think the mean value theorem should be applied. So there exists [imath]x_1[/imath] in [imath][0,1][/imath] such that [imath]f ' (x_1) = \frac{f(1) - f(0)}{1-0} =1[/imath] and there exists [imath]x_2[/imath] in [imath][0,x_1][/imath] such that [imath]f ' (x_2) = \frac{f(x1) - f(0)}{x1-0} = \frac{f(x1)}{x1}[/imath], so on and so forth for [imath]x_3 ,x_4, \ldots,x_n[/imath] and we have the sum [imath]1+\frac{x_1}{f(x1)} + \frac{x_2}{f(x2)} +\cdots+\frac{x_{n-1}}{f(x_{n-1})}[/imath] and from here I have no idea what to do . I was wondering if anyone could be so kind to help ?
|
1835912
|
How to sketch the region on the complex plane?
I am going through a basic course on complex analysis. I have a problem in understanding the following. E [imath]\subset\mathbb{C}[/imath] is defined as [imath]E := \{z\in\mathbb{C}:\vert z+i \vert = 2\vert z\vert \}[/imath] I want to know if this set is connected, closed, bounded but I do not know how to sketch this set or visualize it.
|
1824008
|
Geometric interpretation of a complex set
These usually aren't too bad but I had difficulties thinking of what the set [imath]\{z\in\mathbb{C}:|z+i|=2|z|\}[/imath] looks like in the complex plane. I got as far as [imath]|z+i|=2|z|\Rightarrow \sqrt{(z+i)(\overline{z+i})}=2z\overline{z}\Rightarrow \sqrt{z\overline{z}+i\overline{z}-iz+1}=2z\overline{z}\\ \Rightarrow z\overline{z}-i(z+ z\overline{z})+1=4z\overline{z}\Rightarrow 3z\overline{z}=1-i2\Re(z)[/imath] And rearranging a bit I guess I can see that [imath]z\overline{z}+\frac{2i\Re(z)}{3}=1/3[/imath]. If there weren't that second term on the left, this would be clear, but I am not sure how to think about the dependence on the real part of [imath]z[/imath] here. Having graphed in wolfram, I know this is just a circle of radius 2/3 centered at [imath]i/3[/imath], but I would appreciate help with the thought process.
|
1831157
|
Show norm preserving property and determine Eigenvalues
Can someone of you give me a solution for this? Let [imath]N\in \mathbb N[/imath]. a) We define the map [imath]\mathfrak F:(\mathbb C^N, ||\cdot||_2)\to(\mathbb C^N, ||\cdot||_2)[/imath] by [imath](\mathfrak F(x))_k := \frac{1}{\sqrt N}\sum_{j=1}^Nx_je^{2\pi i\frac{(j-1)(k-1)}{N}} \quad \forall k\in \{1,...,N\}[/imath] Show that [imath]\mathfrak F[/imath] is norm-preserving, i.e. [imath]||(\mathfrak F(x))_k||_2 = ||x||_2 \quad \forall k\in \{1,...,N\}[/imath] b) For [imath]n,m \in \{1,2,...,N\}[/imath] we define the entries of [imath]M\in \mathbb C^{N\times N}[/imath] by [imath]M_{nm} := \begin{cases} \frac{N+1}{2}, \quad \quad n=m \\ \frac{1}{e^{2\pi i \frac{m-n}{N}}-1}, \,\,n \neq m \end{cases}[/imath] Show that [imath]1, 2, 3, ..., N[/imath] are the Eigenvalues of M.
|
1831506
|
Show map is norm-preserving and determine Eigenvalues
Can someone of you give me a solution for this? Let [imath]N\in \mathbb N[/imath]. a) We define the map [imath]\mathfrak F:(\mathbb C^N, ||\cdot||_2)\to(\mathbb C^N, ||\cdot||_2)[/imath] by [imath](\mathfrak F(x))_k := \frac{1}{\sqrt N}\sum_{j=1}^Nx_je^{2\pi i\frac{(j-1)(k-1)}{N}} \quad \forall k\in \{1,...,N\}[/imath] Show that [imath]\mathfrak F[/imath] is norm-preserving, i.e. [imath]||(\mathfrak F(x))_k||_2 = ||x||_2 \quad \forall k\in \{1,...,N\}[/imath] b) For [imath]n,m \in \{1,2,...,N\}[/imath] we define the entries of [imath]M\in \mathbb C^{N\times N}[/imath] by [imath]M_{nm} := \begin{cases} \frac{N+1}{2}, \quad \quad n=m \\ \frac{1}{e^{2\pi i \frac{m-n}{N}}-1}, \,\,n \neq m \end{cases}[/imath] Show that [imath]1, 2, 3, ..., N[/imath] are the Eigenvalues of M.
|
1818382
|
Problem regarding isometric isomorphisms
I need help regarding the following two exercises: a) Show that [imath](\mathbb R^2, d_2)[/imath] and [imath](\mathbb R,d_1)[/imath] [imath]d_2,d_1[/imath] being the respective euclidean metrics, are not isometric isomorphic, i.e. there is no bijective isometric map between those metric spaces. b) Let d be the restriction of the euclidean metric of [imath]\mathbb R[/imath] on [imath](-1,1) \times (-1,1)[/imath]. Find a metric [imath]d':\mathbb R \times \mathbb R\to \mathbb R[/imath] such that the spaces [imath]((-1,1),d)[/imath] and [imath](\mathbb R,d')[/imath] are isometric isomorphic. For a) I think I have solution but I am not really sure: Does it suffice to say that for every possible candidate [imath]f[/imath] for an isomorphism it can't be injective since it must be norm preserving, i.e. [imath]||f(x)||_2 = ||x||_2, \quad ||f(-x)||_2 = ||-x||_2 = ||x||_2[/imath] with [imath]||\cdot||_2[/imath] being the respetive euclidean norm? For b) I don't really know how to start.
|
1822399
|
Isometric isomorphism between [imath]R^2[/imath] and [imath]R[/imath]
Can someone help me solving the following problems? [imath](\mathbb R^2,d_2)[/imath] and [imath](\mathbb R, d_1)[/imath], [imath]d_2, d_1[/imath] being the respective euclidean norms, are not isometric isomorphic, i.e. there is no distance preserving isomorphism between them. Let [imath]d[/imath] be the restriction of the euclidean metric on [imath](-1,1) \times(-1,1).[/imath] Find a metric [imath]d':\mathbb R \times \mathbb R \to \mathbb R[/imath] such that the spaces [imath]((-1,1),d)[/imath] and [imath](\mathbb R,d')[/imath] are isometric isomorphic. For 1.) I thought it suffices to prove that if there was a isomorphism between those two, it wouldn't be distance preserving since it then also must be norm preserving which can't be injective since [imath]||f(x)||_2 = ||x||_2 = ||f(-x)||_2[/imath]. I don't know if that suffices or if it is even true. I would be glad if someone could prove me why they can't be isomorphic in the first place. For 2.) I don't know what to do.
|
1836363
|
Show that there exists at most one extension of [imath]f[/imath] whose co-domain is a Hausdorff space
I want to show the following Suppose [imath]A \subset X, f: A \to Y[/imath] is continuous, [imath]Y[/imath] is Hausdorff. Show that there is at most one continuous extension [imath]g: \overline A \to Y[/imath] I feel like I am very close but I cannot finish. By way of contradiction, suppose that there exists another extension [imath]h: \overline A \to Y[/imath], take [imath]x \in \overline A \backslash A[/imath], we claim that [imath]h(x) \neq g(x)[/imath] Since [imath]Y[/imath] is Hausdorff, there exists open sets [imath]U, V, U \cap V = \varnothing[/imath] such that [imath]g(x) \in U[/imath], and [imath]h(x) \in V[/imath] Then [imath]x \in g^{-1}(U) \cap h^{-1}(v) \subset \overline A[/imath] How do I proceed from here?
|
305472
|
Uniqueness of a continuous extension of a function into a Hausdorff space
Suppose that [imath]A\subset X[/imath] and suppose that [imath]f : A \to Y[/imath] is a continuous function with [imath]Y[/imath] Hausdorff. Show that there is at most one continuous function [imath]g : \bar{A} \to Y[/imath]. My try: Suppose there are two extension [imath]g[/imath] and [imath]h[/imath], then for some [imath]a\in \overline{A}-A[/imath], we have [imath]g(a)\neq h(a)[/imath]. As [imath]Y[/imath] is Hausdorff, there are two disjoint open set [imath]U_{g(a)}[/imath] and [imath]U_{h(a)}[/imath] such that [imath]g(a) \in U_{g(a)}[/imath] and [imath]h(a) \in U_{h(a)}[/imath]. Now I have feeling somehow from here I have to construct open set [imath]U[/imath] of [imath]a[/imath] which does not intersect [imath]A[/imath]. But I am not able to do this. Can someone please help me.
|
1836102
|
If [imath]\text{Ext}_R^1(A,I) = 0[/imath] for all [imath]A\in ob(_R\text{Mod})[/imath], then [imath]I\in ob(_R\text{Mod})[/imath] is injective.
Let [imath]\text{Ext}^1(A,I)=0[/imath] for all [imath]A\in ob(_R\text{Mod})[/imath], then [imath]I\in ob(_R\text{Mod})[/imath] is injective. I got stuck by this problem. Any ideas?
|
363643
|
[imath]\operatorname{Ext}[/imath] and injectives, respectively projectives
If [imath]\operatorname{Ext}^{ 1}_{R}(A,B) = 0 [/imath] for all [imath]R[/imath]-Mod [imath]A[/imath] then [imath]B[/imath] is injective ? If [imath]\operatorname{Ext}^{ 1}_{R}(A,B) = 0 [/imath] for all [imath]R[/imath]-Mod [imath]B[/imath] then [imath]A[/imath] is projective ?
|
278534
|
Is the class of algebraic extensions distinguished?
In paragraph V.1 of Algebra proposition 1.7 Lang claims that the class of algebraic extensions is distinguished. I know that if [imath]F/k[/imath] and [imath]E/F[/imath] are algebraic extensions than so is the [imath]E/k[/imath] - that is easy to prove. However, the second required property is that if [imath]E/k[/imath] is algebraic and [imath]F/k[/imath] is arbitrary, then (assuming the compositum is defined) [imath]EF/F[/imath] is algebraic. Lang more or less skips the proof of this, only saying that "an element remains algebraic under lifting, and hence does the extension." However, there are no finiteness conditions here and no element given in advance. Although [imath]EF[/imath] is generated over [imath]F[/imath] by a set of elements which are algebraic over [imath]F[/imath] (namely, all the elements of [imath]E[/imath]), there are no facts known (up to this point in the book at least) about the infinitely generated extensions. What am I missing? Thank you.
|
243512
|
Algebraic Property inherited by a "lift"?
Suppose [imath]K[/imath] is a field, and [imath]K\subset E[/imath] is an algebraic extension, and [imath]K\subset F[/imath] is any extension. Suppose that [imath]L[/imath] is some field which contains both [imath]F[/imath] and [imath]E[/imath] as subfields. I want to show that [imath]FE[/imath] (the smallest subfield of [imath]L[/imath] containing both [imath]F[/imath] and [imath]E[/imath]) is algebraic over [imath]F[/imath]. Lang mentions that the result is true because an element remains algebraic under lifting, but I can't prove the claim with this. I interpret his comment to mean that if [imath]x\in E[/imath] is algebraic over [imath]K[/imath], then [imath]x\in FE[/imath] is algebraic over [imath]F[/imath]. This is certainly true. But this doesn't seem enough if I start with any element of [imath]FE[/imath].
|
1836758
|
Solve [imath]\sqrt{1-2x+x^2+o(x^3)}[/imath] with [imath]x \to 0[/imath]
I need help to solve [imath]\sqrt{1-2x+x^2+o(x^3)}[/imath] with [imath]x \to 0[/imath], I do not understand when and why I should stop. Here my steps: I can use Taylor formula for [imath]\sqrt{1+t}[/imath] so: [imath]\sqrt{1+t} = 1+ \frac{t}{2}-\frac{t^2}{8}+\frac{t^3}{16}+o(t^3)[/imath] where: [imath]t = -2x+x^2+o(x^3)[/imath] I start with the first order: [imath]1+\frac{t}{2}+o(t) = 1+\frac{-2x+x^2+o(x^3)}{2}+o(-2x+x^2+o(x^3))[/imath] [imath]1+\frac{t}{2}+o(t) = 1+\frac{-2x+x^2+o(x^3)}{2}+o(x)[/imath] [imath]1+\frac{t}{2}+o(t) = 1-x+\frac{x^2}{2}+\frac{1}{2}o(x^3)+o(x)[/imath] [imath]1+\frac{t}{2}+o(t) = 1-x+\frac{x^2}{2}+o(x^3)+o(x)[/imath] [imath]1+\frac{t}{2}+o(t) = 1-x+\frac{x^2}{2}+o(x)[/imath] This do not solve the problem so I think I should continue wih the second order. Using the same steps above I get: [imath]1+\frac{t}{2}-\frac{x^2}{8}+o(t^2) = 1-x+o(x)[/imath] is it correct? At this point I don't know how to procede, I don't think that this solve the problem, should I continue to the 3rd order?
|
1832965
|
How to solve asymptotic expansion: [imath]\sqrt{1-2x+x^2+o(x^3)}[/imath]
Determinate the best asymptotic expansion for [imath]x \to 0[/imath] for: [imath]\sqrt{1-2x+x^2+o(x^3)}[/imath] How should I procede? In other exercise I never had the [imath]o(x^3)[/imath] in the equation but was the maximum order to consider.
|
1836876
|
[imath] \frac{a^3}{c} + \frac{b^3}{d} \ge 1[/imath] prove inequality
Suppose that [imath] a,b,c,d \in \mathbb{R}[/imath], [imath]a,b,c,d \gt 0[/imath] and [imath] c^2 +d^2=(a^2 +b^2)^3[/imath]. Prove that [imath] \frac{a^3}{c} + \frac{b^3}{d} \ge 1.[/imath] If I rewrite the inequation like [imath] \frac{a^3}{c} + \frac{b^3}{d} \ge \frac{c^2 +d^2}{(a^2 +b^2)^3}[/imath] and manage to simplfy it brings me nowhere. I try with Cauchy-Schwarz Inequality but still can not solve it. Plese help
|
1836753
|
prove inequation
a,b,c,d [imath]\in \mathbb{R}[/imath] [imath]a,b,c,d \gt 0[/imath] and [imath] c^2 +d^2=(a^2 +b^2)^3[/imath] prove that [imath] \frac{a^3}{c} + \frac{b^3}{d} \ge 1[/imath] If I rewrite the inequation like [imath] \frac{a^3}{c} + \frac{b^3}{d} \ge \frac{c^2 +d^2}{(a^2 +b^2)^3}[/imath] and manage to simplfy it brings me nowhere. I try with Cauchy-Schwarz Inequality but still can not solve it. I would very much appreciate it if anyone could help me
|
1836917
|
Orthogonal Matrix 4
Let [imath]M_{32}[/imath] be vector space with inner product of [imath]AB[/imath] given by [imath]\text{tr}(B^TA)[/imath]. The question is to find a non-zero matrix B orthogonal to [imath]A=\left[\begin{matrix} 1 & 2 \\ 3 & 4 \\ 5 & 6 \end{matrix}\right][/imath] Any hints as to how I can start this?
|
1836867
|
Orthogonal matrices 5
The question is to find in the space [imath]\mathrm{Mat}_{3\times 2}(\mathbb{\mathbb{R}})[/imath] a non-zero matrix that is orthogonal to [imath] A= \begin{pmatrix} 1 & 2\\ 3 & 4\\ 5 & 6\\ \end{pmatrix}.[/imath] The inner product is [imath]\langle A, B \rangle = tr(A B^T) = tr(B^T A)[/imath]. Any hints or clues on how to start this?
|
1836958
|
Prove equinumerosity between [imath]2^\mathbb{N}[/imath] and R total orderings
T=[imath]\left\{R\vert R \text{ is a total order over } \mathbb{N}\right\}[/imath] Prove that T and [imath]2^\mathbb{N}[/imath] are equinumerous.
|
340747
|
How many total order relations on a set [imath]A[/imath]?
Let's define a set [imath]T_A[/imath] which is the set of all total order relations on [imath]A[/imath]. This set is a subset of the set of all [imath]2[/imath]-adic relations on [imath]A[/imath]: [imath]T_A \subset \mathcal P(A^2) [/imath] 1-Which is the cardinality of [imath]T_A[/imath]? 2-Which is the standard notation for the set [imath]T_A[/imath]?
|
1147312
|
Last digit of [imath]3^{459}[/imath].
I am supposed to find the last digit of the number [imath]3^{459}[/imath]. Wolfram|Alpha gives me [imath]9969099171305981944912884263593843734515811805621702621829350243852275145577745\\3002132202129141323227530694911974823395497057366360402382950449104721755086093\\572099218479513977932448616356300654729978057481366551670706\color{red}{\mathbf{7}}[/imath] Surely there's some sort of numerical trick to doing this. I thought maybe modular arithmetic was involved? Any ideas on how to approach this problem?
|
757894
|
Calculate the last digit of [imath]3^{347}[/imath]
I think i know how to solve it but is that the best way? Is there a better way (using number theory). What i do is: knowing that 1st power last digit: 3 2nd power last digit: 9 3rd power last digit: 7 4rh power last digit: 1 5th power last digit: 3 [imath]3^{347} = 3^{5\cdot69+2} = (3^5)^{69} \cdot3^2 = 3\cdot3^2=3^3=27 [/imath] so the result is [imath]7[/imath].
|
1832325
|
[imath]S[/imath] is continuous with Weak * topology from [imath]Y^*[/imath] to [imath]X^*[/imath] if an only if [imath]S=T^*[/imath] for some [imath]B(X,Y)[/imath]
How to prove that prove that [imath]S[/imath] is weak[imath]^*[/imath]-continuous from [imath]Y^*[/imath] to [imath]X^*[/imath] if an only if [imath]S=T^*[/imath] for some [imath]T\in B(X,Y)[/imath] Thanks for any hints. To show that [imath]T[/imath] is continuous is straight forward but inverse is hard.
|
1670753
|
Given [imath]S \in B(Y^{*}, X^{*})[/imath], does there exist [imath]T\in B(X,Y)[/imath] such that [imath]S=T^{*}[/imath]?
Let [imath]X, Y[/imath] be Banach spaces, [imath]S \in B(Y^{*}, X^{*})[/imath]. Does such operator [imath]T \in B(X, Y)[/imath] exist so that [imath]T^{*}=S[/imath]? I suppose that the answer should be - no. Are there any hints that might help in constructing a counterexample? Any help would be much appreciated.
|
1837016
|
Describe the prime elements of the ring [imath]\mathbb Z[\sqrt{-2}][/imath]
I have a ring [imath]\mathbb Z[\sqrt{-2}][/imath] and I need to describe all the prime numbers of that ring. How I can do that? Thank you
|
101537
|
Irreducible elements in [imath]\mathbb{Z}[\sqrt{-2}][/imath] and is it a Euclidean domain?
First of all I am new to this topic, algebraic number theory, so I only know a decent (not great) amount of abstract algebra. The question I have is that, given the imaginary quadratic field [imath]\mathbb{Z}[\sqrt{-2}][/imath], I want to find; (1) all irreducible elements of it, (2) show that it is a Euclidean domain, and (3) show that for an odd prime number [imath]p,\; \exists \;x,y\; \in \mathbb{Z}[/imath] s.t. [imath]p = x^2+2y^2[/imath] iff [imath]p=1,3(\textrm{mod}\; 8)[/imath]. I have been reading and have books but there are some things I am not getting. (a) My attempt at finding the units (I read that there are only [imath]{}^{\pm}1[/imath] for this integral domain (ID)); A unit is an element with an inverse, so for an element [imath]p_1 \in \mathbb{Z}[\sqrt{-2}][/imath], there is another element [imath]p_1'[/imath] s.t. [imath]p_1\,p_1' = p_1'\,p_1 = 1[/imath] (it is integral domain, not just domain). Let [imath]p_1 := a+b\sqrt{-2}[/imath] and [imath]p_1':=x+y\sqrt{-2}[/imath] and so [imath]p_1\,p_1' = 1[/imath] becomes [imath](a+b\sqrt{-2})(x+y\sqrt{-2}) = 1 = 1+0\sqrt{-2}[/imath] and into the two equations, [imath]1=ax-2by[/imath] and [imath]0=ay+bx[/imath]. Solving these leads to [imath]x=\frac{a}{a^2+2b^2}[/imath], [imath]y=\frac{-b}{a^2+2b^2}[/imath] and [imath]p_1'=\frac{a}{a^2+2b^2} + \left( \frac{-b}{a^2+2b^2} \right)\sqrt{-2}[/imath], which can only belong to [imath]\mathbb{Z}[/imath] if [imath]b=0,\;a={}^{\pm}1[/imath]. Is there a better way of determing the units of an ID? I read that the units, [imath]\epsilon[/imath], of a quadratic ID of the general form [imath]R[\sqrt{d}][/imath], where [imath]d[/imath] was square-free, were determined by [imath]Norm(\epsilon) = {}^{\pm}1[/imath] Is this general ? Is this for any [imath]d[/imath] that is square-free (though I see little difference between [imath]d=d[/imath] and [imath]d=z*d[/imath], as [imath]z\in \mathbb{Z}[/imath]) ? (1) I know, procedurally, how to do this for a given element, but I do not know of a better way to do it in general. Here is my attempt: I read that: An element [imath]p[/imath] of an ID is irreducible in R if it satisfies: (i) [imath]p \neq 0[/imath] and [imath]p[/imath] is not a unit, (ii) if [imath]p=ab[/imath] in R, then [imath]a[/imath] or [imath]b[/imath] is a unit in R. (maybe because I know the units in this ID I can say that all other non-zero elements are irreducible ? ) So if [imath]p = ab[/imath], with [imath]a = m+n\sqrt{-2}[/imath], then using the as-of-yet-unproved-homomorphism-norm-map, [imath]N(ab) = N(a)\,N(b)[/imath], [imath]N(p=x+y\sqrt{-2}) = x^2+2y^2 = N(a)\,N(b) = (m^2+2n^2)\,N(b)[/imath]. Now if I had a specific element, to determine if it was irreducible, I could then determine what values of [imath]N(a)[/imath] and [imath]N(b)[/imath] were valid so that their product equaled [imath]N(p) = N(x+y\sqrt{-2})[/imath], which this latter term would be an integer ([imath]N: \mathbb{Z}[\sqrt{-2}] \mapsto \mathbb{Z}[/imath]). The latter two questions I haven't got far with either but am wanting to get this initial question(s) understood first. Thanks all for your time reading my rather lengthy question!
|
1836323
|
In the answer to the question attached below, I don't quite see how step-3 is derived from step-2, Can anyone explain
Calculating the expected values of the min/max of 2 random variables Consider two fair [imath]k[/imath]-sided dice with the numbers 1 through [imath]k[/imath] on their faces, obtaining values [imath]X_1[/imath] and [imath]X_2[/imath]. What is [imath]\mathbb{E}[\max(X_1, X_2)][/imath] and what is [imath]\mathbb{E}[\min(X_1, X_2)][/imath]. And the answer to this question as posted in the link above is HINT: [imath]\min(x_1,x_2) + \max(x_1,x_2) = x_1 + x_2[/imath], so it is enough to evaluate either expected value. Also [imath]\max(x_1,x_2) - \min(x_1,x_2) = \vert x_2-x_1\vert[/imath]. Therefore, finding [imath]\mathbb{E}(\vert X_2-X1 \vert)[/imath] allows to determine expectations needed: [imath] \begin{eqnarray} \mathbb{E}\left( \vert X_2- X_1\vert \right) &=& \sum_{n_1=1}^k \sum_{n_2=1}^k \mathbb{P}(X_1=n_1) \mathbb{P}(X_2=n_2) \vert n_2 - n_1 \vert \\ &=& 2 \sum_{n_2=1}^k \sum_{n_1=1}^{n_2-1} \mathbb{P}(X_1=n_1) \mathbb{P}(X_2=n_2) (n_2 - n_1 ) \\ &=& 2 \sum_{n_2=1}^k \sum_{n_1=1}^{n_2-1} \mathbb{P}(X_1=n_1) \mathbb{P}(X_2=n_2-n_1) n_1 \end{eqnarray} [/imath] I don't understand how step three in the above answer is derived
|
82945
|
Calculating the expected values of the min/max of 2 random variables
Consider two fair [imath]k[/imath]-sided dice with the numbers 1 through [imath]k[/imath] on their faces, obtaining values [imath]X_1[/imath] and [imath]X_2[/imath]. What is [imath]\mathbb{E}[\max(X_1, X_2)][/imath] and what is [imath]\mathbb{E}[\min(X_1, X_2)][/imath].
|
1837256
|
Is f(x,y)=[imath]\frac{x^{2}y}{x^{2}+y^{4}} [/imath]with f(0,0)=0 continuous in (0,0)
I believe that the function: f(x,y)=[imath]\frac{x^{2}y}{x^{2}+y^{4}}[/imath] is continuous on the point (0,0) but i can't prove it. I know you have to choose something like [imath]x=cy^{2}[/imath](with c a constant) to prove this but it doesn't seem to work. Can anyone help me solve this?
|
168531
|
Multivariable Limits
Can someone help me calculate the following limits? 1) [imath] \displaystyle\lim _ {x \to 0 , y \to 0 } \frac{\sin(xy)}{\sqrt{x^2+y^2}} [/imath] (it should equal zero, but I can't figure out how to compute it ) . 2) [imath]\displaystyle\lim_ {(x,y)\to (0,\frac{\pi}{2} )} (1-\cos(x+y) ) ^{\tan(x+y)} [/imath] (it should equal [imath]1/e[/imath]). 3) [imath] \displaystyle\lim_{(x,y) \to (0,0) } \frac{x^2 y }{x^2 + y^4 } [/imath] (which should equal zero). 4) [imath]\displaystyle \lim_{(x,y) \to (0,1) } (1+3x^2 y )^ \frac{1}{x^2 (1+y) } [/imath] (which should equal [imath]e^{3/2}[/imath] ). Any help would be great ! Thanks in advance !
|
1837466
|
Is there a name for matrices that are symmetric along the cross diagonal?
Something like [imath] A= \begin{bmatrix} a & b & c\\ b & d & e\\ c & e & f \end{bmatrix} [/imath] would be a symmetric matrix because the values are reflected along the diagonal, and [imath]A=A^\intercal[/imath] Is there a name for a matrix that's symmetric along the cross diagonal? Something like [imath] B= \begin{bmatrix} c & b & a\\ e & d & b\\ f & e & c \end{bmatrix} [/imath]
|
378696
|
Name of a special matrix
I have a matrix which is kind of symmetrical with the other diagonal, i.e., something like [imath]A = \left[ \begin{array}{c c c c} a & b & c & d \\ e & f & g & c \\ h & i & f & b \\ j & h & e & a \end{array} \right][/imath] Does this matrix have a special name in literature? What are it's properties? And a matrix that is symmetrical by both diagonals [imath]A = \left[ \begin{array}{c c c c} a & b & c & d \\ b & e & f & c \\ c & f & e & b \\ d & c & b & a \end{array} \right][/imath] What's the name of it? Any interesting properties?
|
1837154
|
a problem concerning continuous functions of bounded variation
Here is a problem: Suppose [imath]f,g: [a,b]\rightarrow \mathbb{R}[/imath] are both continuous and of bounded variation. Show that the set [imath]\{(f(t),g(t))\in\mathbb{R}^2: t\in [a,b]\}[/imath] CANNOT cover the entire unit square [imath][0,1]\times [0,1][/imath]. Thanks for all helps!
|
1831586
|
Can [imath]\{(f(t),g(t)) \mid t\in [a,b]\}[/imath] cover the entire square [imath][0,1] \times [0,1][/imath] ?
Suppose [imath]f,g : [a, b] \to R[/imath] are both continuous and of bounded variation. Then can set [imath]\{(f(t),g(t)) \mid t\in [a,b]\}[/imath] cover the entire square [imath][0,1] \times [0,1][/imath] ? And what if we remove the condition both [imath]f,g[/imath] are of bounded variation ?
|
1837407
|
Let [imath] \omega = dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n} \in \mathbb{R}^{2n}[/imath]. Find [imath]\omega^n[/imath] (in respect to [imath]\wedge[/imath])
Let [imath] \omega = dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n} \in \mathbb{R}^{2n}[/imath]. Find [imath]\omega^{n}[/imath] (in respect to [imath]\wedge[/imath]) When I say "[imath]\omega^{n}[/imath] (in respect to [imath]\wedge[/imath])" I mean the exterior product of [imath]n[/imath] [imath]\omega[/imath]'s. I tried writing [imath](dx_1 \wedge dx_2) (v_1,v_2) = \det \begin{pmatrix} d{x_1}(v_1) & d{x_1}(v_2)\\ d{x_2}(v_1) & d{x_2}(v_2) \end{pmatrix} = d{x_1}(v_1)d{x_2}(v_2) - d{x_1}(v_2)d{x_2}(v_1) [/imath] [imath] (dx_3 \wedge dx_4) (v_1,v_2) = \det \begin{pmatrix} d{x_3}(v_1) & d{x_3}(v_2)\\ d{x_4}(v_1) & d{x_4}(v_2) \end{pmatrix} = d{x_3}(v_1)d{x_4}(v_2) - d{x_3}(v_2)d{x_4}(v_1) [/imath] but that wouldn't lead me nowhere. I know the property [imath] \omega \wedge (\phi_1 + \phi_2) = \omega \wedge \phi_1 + \omega \wedge \phi_2 [/imath] Then: [imath] \begin{align} \omega \wedge \omega = {} & (dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n}) \\ & {} \wedge (dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n}) \\[8pt] = {} & (dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n}) \wedge (dx_1 \wedge dx_2) \\ & {} + (dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n}) \wedge (dx_3 \wedge dx_4) + \cdots \\ & {} + (dx_1 \wedge dx_2 + dx_3 \wedge dx_4 + \cdots + dx_{2n-1} \wedge dx_{2n}) \wedge (dx_{2n-1} \wedge dx_{2n}) \end{align} [/imath] but I'm not sure that's what I'm supposed to do... Thanks.
|
372776
|
Computing [imath]n[/imath]-th external power of standard simplectic form
I need some help: Define a 2-form on [imath]R^n[/imath] by [imath]\omega=dx_1\wedge dx_2+dx_3\wedge dx_4+...+dx_{2n-1}\wedge dx_{2n}[/imath]. How to compute [imath]\omega^n:=\omega\wedge\omega\wedge\ldots\wedge\omega[/imath]?
|
1837641
|
Prove that [imath]\sin \theta=\frac{3 \sin \alpha+\sin^3 \alpha}{1+3\sin^2 \alpha}[/imath] using given condition
If [imath]\tan(\frac{\pi}{4}+\frac{\theta}{2})=\tan^3(\frac{\pi}{4}+\frac{\alpha}{2})[/imath], then prove that [imath]\sin \theta=\frac{3 \sin \alpha+\sin^3 \alpha}{1+3\sin^2 \alpha}[/imath] I tried using the fact that [imath]\frac{\cos A}{1-\sin A}=\tan(\frac{\pi}{4}+\frac{A}{2})[/imath] but now not able to eliminate [imath]\cos \alpha[/imath] and [imath]\cos \theta[/imath]. How should I proceed?
|
1720471
|
If [imath]\tan(\frac{\pi}{4}+\frac{y}{2})=\tan^3(\frac{\pi}{4}+\frac{x}{2})[/imath],then prove that [imath]\sin y=\frac{\sin x(3+\sin^2 x)}{(1+3\sin^2 x)}[/imath]
If [imath]\tan(\frac{\pi}{4}+\frac{y}{2})=\tan^3(\frac{\pi}{4}+\frac{x}{2})[/imath],then prove that [imath]\sin y=\frac{\sin x(3+\sin^2 x)}{(1+3\sin^2 x)}[/imath] [imath]\tan(\frac{\pi}{4}+\frac{y}{2})=\tan^3(\frac{\pi}{4}+\frac{x}{2})[/imath] [imath]\frac{1+\tan\frac{y}{2}}{1-\tan\frac{y}{2}}=(\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}})^3[/imath] Applying componendo and dividendo rule on both sides, [imath]\tan\frac{y}{2}=\frac{\tan\frac{x}{2}(3+\tan^2\frac{x}{2})}{(1+3\tan^2\frac{x}{2})}[/imath] I am stuck here.
|
1837857
|
Existence of analytic function from A to B .
Does there exists a non constant analytic map [imath]f:A\to B[/imath] . Where [imath]A=\{z\in \mathbb C~:~ |z|\neq 0\}[/imath] and [imath]B=\{z \in \mathbb C ~:~ |z|>1\}[/imath]. I am unable to construct one
|
1833688
|
Non constant analytic function from [imath]\{z\in\mathbb{C}:z\neq 0\}[/imath] to [imath]\{z\in\mathbb{C}:|z|>1\}.[/imath]
Does there is non constant analytic function from [imath]\{z\in\mathbb{C}:z\neq 0\}[/imath] to [imath]\{z\in\mathbb{C}:|z|>1\}?[/imath] According to me there is no such non constant analytic function because if there is any such function say [imath]f,[/imath] then [imath]f[/imath] can have either a pole or essential singularity at [imath]z=0[/imath]. In the case of pole Picard's theorem of meromoprphic function will work and in the case of essential singularity we know that image of any neighbourhood of essential singularity is dense in [imath]\mathbb{C}[/imath], so in both of the cases we get a contradiction. So no such non constant analytic function. Am i right? Please suggest me. Thanks.
|
1837794
|
How to solve this [imath]80^\circ[/imath]-[imath]80^\circ[/imath]-[imath]20^\circ[/imath] triangle ([imath]60^\circ+20^\circ[/imath] and [imath]70^\circ+10^\circ[/imath] variant)?
A friend of mine asked me for help with a math problem and I struggled with this for over an hour. I told him sorry, and I felt bad. It's been bugging me now for hours. I don't even so much care for the answer of A, I just want to know how it should be done.
|
1738400
|
Solve for [imath]x[/imath] in the [imath]80^\circ[/imath]-[imath]80^\circ[/imath]-[imath]20^\circ[/imath] triangle
I've been solving this for days but I still I couldn't solve this. Can I know how to solve [imath]x[/imath] ?
|
1837226
|
Calculating Categorical Quotients
I am trying to practice calculating categorical quotients and I ran into this example. I am unable to get the answer and was wondering if someone can help? Let [imath]G = Z/3Z =[/imath] [imath]\{1, \omega, \omega^2\}[/imath], where [imath]\omega[/imath] is a cube root of unity. Let [imath]G[/imath] act on [imath]^{2}[/imath] via [imath]\omega(x, y) = (\omega x, \omega y)[/imath]. Find the categorical quotient of [imath]^{2}[/imath] by G.
|
1837853
|
Categorical Quotient and group actions
I am trying to practice calculating categorical quotients and I ran into this example. I am unable to get the answer and was wondering if someone can help? Let [imath]G = Z/3Z =[/imath] [imath]\{1, \omega, \omega^2\}[/imath], where [imath]\omega[/imath] is a cube root of unity. Let [imath]G[/imath] act on [imath]^{2}[/imath] via [imath]\omega(x, y) = (\omega x, \omega y)[/imath]. Find the categorical quotient of [imath]^{2}[/imath] by G.
|
1838110
|
Does specific function exist?
Check if exists function [imath]f(x,y):R^2->R[/imath] such that f(x,y) has directional derivatives in point (0,0) in each direction and (0,0) is point of discontinuity.
|
280233
|
example of discontinuous function having direction derivative
Is there a function (non piece-wise unlike below) which is discontinuous but has directional derivative at particular point? I have a manual that says the function has directional derivative at [imath](0,0)[/imath] but is not continuous at [imath](0,0)[/imath]. [imath]f(x,y) = \begin{cases} \frac{xy^2}{x^2+y^4} & \text{ if } x \neq 0\\ 0 & \text{ if } x= 0 \end{cases}[/imath] Can anyone give me few examples which is not defined piece wise as above?
|
1611659
|
Inequality involving binomial coefficients and [imath]e[/imath]
I found this inequality in a textbook: [imath]\binom{n}{k} < \left(\frac{e\cdot n}{k}\right)^k \, .[/imath] I wonder how to prove it and why [imath]e[/imath] counts here?
|
132625
|
Inequality [imath]{n \choose k} \leq \left(\frac{en}{ k}\right)^k[/imath]
This is from page 3 of http://www.math.ucsd.edu/~phorn/math261/9_26_notes.pdf (Wayback Machine). Copying the relevant segment: Stirling’s approximation tells us [imath]\sqrt{2\pi n} (n/e)^n \leq n! \leq e^{1/12n} \sqrt{2\pi n} (n/e)^n[/imath]. In particular we can use this to say that [imath] {n \choose k} \leq \left(\frac{en}{ k}\right)^k[/imath] I tried the tactic of combining bounds from [imath]n![/imath], [imath]k![/imath] and [imath](n-k)![/imath] and it didn't work. How does this bound follow from stirling's approximation?
|
63553
|
Difference between [imath]:=[/imath] and [imath]=[/imath]
I am sorry if this is quite elementary question. But I always think, that why we use [imath]:=[/imath] at some places, instead of [imath]=[/imath]. Is there any fundamental difference between these two? Before reading Terry Tao's blog (4 months ago), I had never seen a symbol like this (:=).
|
2248833
|
What is the distinction between [imath]=[/imath] and [imath]:=[/imath] when writing math?
I am persistently confused by the usage of [imath]=[/imath] and [imath]:=[/imath] when reading papers and writing in math. Quick google search on "When to use defined as (symbol)", returned no useful result. The way I am currently using [imath]:=[/imath] is basically assign the name of a variable/quantity to another. For example, let [imath]f(x):= \sin(x)[/imath], then clearly [imath]f'(x) = \cos(x)[/imath]. But if I write it this way, then wouldn't it also be acceptable to write: [imath]f(x):= \sin(x) := \dfrac{\exp(ix) - \exp(-ix)}{2i}[/imath] After all, isn't [imath]\sin[/imath] defined as the difference of complex exponential over [imath]2i[/imath]? I have seldom seen anyone write this way (certainly not in any introductory calculus textbooks), therefore I am reluctant to follow this convention. As another example, On Wikipedia, it writes: But it doesn't follow this convention when defining functions, for instance: I have also seen very varied usage of [imath]:=[/imath] when introducing sets. Some authors prefer to say something like: let [imath]C:= \{x\in \mathbb{R}^2|x_1^2+x_2^2 = 1\}[/imath], other prefer just to say [imath]C= \{x\in \mathbb{R}^2|x_1^2+x_2^2 = 1\}[/imath] Can anyone elaborate on under which context I should use [imath]:=[/imath] over [imath]=[/imath]? What is a way to keep consistency?
|
1838849
|
Devise a test for divisibility of an integer by 11, in terms of properties of its digits
Using the fact that [imath]10 \equiv-1 \pmod{11}[/imath], devise a test for divisibility of an integer by [imath]11[/imath], in terms of properties of its digits. Approach: Let the number with its digits [imath]a_0\cdots a_n[/imath] be represented as [imath]f(x)=a_0x^n+\cdots+a_n[/imath]. By exhaustion [imath]f(-1)=0[/imath] if the number is divisible by [imath]11[/imath], so [imath]f(10)[/imath] is divisible by [imath]11[/imath]. Do I have to prove my conjecture? or it's trivial.
|
1789063
|
A positive integer (in decimal notation) is divisible by 11 [imath] \iff [/imath] ...
(I am aware there are similar questions on the forum) What is the Question? A positive integer (in decimal notation) is divisible by [imath]11[/imath] if and only if the difference of the sum of the digits in even-numbered positions and the sum of digits in odd-numbered positions is divisible by [imath]11[/imath]. For example consider the integer 7096276. The sum of the even positioned digits is [imath]0+7+6=13.[/imath] The sum of the odd positioned digits is [imath]7+9+2+6=24.[/imath] The difference is [imath]24-13=11[/imath], which is divisible by 11. Hence 7096276 is divisible by 11. (a) Check that the numbers 77, 121, 10857 are divisible using this fact, and that 24 and 256 are not divisible by 11. (b) Show that divisibility statement is true for three-digit integers [imath]abc[/imath]. Hint: [imath]100 = 99+1[/imath]. What I've Done? I've done some research and have found some good explanations of divisibility proofs. Whether that be 3,9 or even 11. But...the question lets me take it as fact so I don't need to prove the odd/even divisible by 11 thing. I need some help on the modular arithmetic on these. For example... Is 77 divisible by 11? [imath]7+7 = 14 \equiv ...[/imath] I don't know what to do next. Thanks very much, and I need help on both (a) and (b).
|
1802175
|
placing balls inside ball
Is it possible to put pairwise disjoint open 3d-balls with radii [imath]\frac{1}{2},\frac{1}{3},\frac{1}{4},\dots[/imath] inside a unit ball? not an original question, I found it somewhere in the internet once, but without any answer.
|
1491754
|
Can all circles of radius [imath]1/n[/imath] be packed in a unit disk, excluding the circle of radius [imath]1/1[/imath]?
This problem occurred to me when I came across a similar problem where the radii were taken over only the primes. That question was unanswered, but it seems to me infinitely many circles of radius [imath]1/2, 1/3, 1/4...[/imath] can fit into a unit disk. The area of all those circles would be [imath]\pi \sum_2^\infty 1/n^2 = \pi^3/6 -\pi[/imath], which is less than the area of the unit disk [imath]\pi[/imath]. But can the circles actually be packed with no overlaps?
|
1838916
|
Solving [imath]2x^4+x^3-11x^2+x+2 = 0[/imath]
I am having no idea how I can solve this problem. I need help! Here's the problem [imath]2x^4+x^3-11x^2+x+2 = 0[/imath] I am learning Quadratic Expressions and this is what I need to solve, and I can't understand how :C
|
499367
|
condition for a cubic polynomial to have a real root
Let [imath]a,b \in R[/imath] and assume that [imath]x=1[/imath] is a root of the polynomial [imath]p(x)= x^4+ax^3+bx^2+ax+1[/imath]. Find the ranges of values of [imath]a[/imath] for which [imath]p[/imath] has a complex root which is not real. Here first I factored out [imath]x-1[/imath] which left me with a cubic polynomial and then i thought of using the discriminant of a cubic polynomial, [imath]\Delta= 18abcd -4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2[/imath] [imath]< 0[/imath] to get the condition on a. But I don't know how this discriminant is derived. And I want to know if there is another method where we don't need to use the discriminant.
|
1827805
|
Proving continuity of a operator [imath]T\colon E \to E'[/imath]
Let be [imath]E[/imath] a Banach space over real numbers and [imath]T\colon E \to E'[/imath] linear such [imath]T(x)(x)\geq 0[/imath] for all [imath]x\in E[/imath], prove T is continuous. If [imath]x_n\to x[/imath] and [imath]T(x_n)\to \phi\in E'[/imath] then [imath]T(x_n)(y)\to\phi(y)[/imath] how to use [imath]T(x)(x)\geq 0[/imath] for all [imath]x\in E[/imath]?
|
216858
|
Positive operator is bounded
For a real Banach space [imath]X[/imath] let [imath]A:X\rightarrow X^*[/imath] be a positive operator in the sense that [imath](Ax)(x)\geq 0[/imath] for all [imath]x\in X[/imath]. Show that [imath]A[/imath] is bounded. I don't know how to do that, maybe it's an application of the closed graph theorem?
|
1839906
|
Inequality [imath]ab\le \frac{a^p}{p}+\frac{b^q}{q}[/imath]
If [imath]\frac {1}{p}+\frac {1}{q}=1[/imath] and [imath]a,b \ge 0[/imath] , then prove [imath]ab\le \frac{a^p}{p}+\frac{b^q}{q}[/imath] . I can't find a simple and short way to prove this. Any hint would work. Thanks in advance!
|
676803
|
Prove that [imath]xy \leq\frac{x^p}{p} + \frac{y^q}{q}[/imath]
OK guys I have this problem: For [imath]x,y,p,q>0[/imath] and [imath] \frac {1} {p} + \frac {1}{q}=1 [/imath] prove that [imath] xy \leq\frac{x^p}{p} + \frac{y^q}{q}[/imath] It says I should use Jensen's inequality, but I can't figure out how to apply it in this case. Any ideas about the solution?
|
1840078
|
[imath]R[/imath] integral domain, [imath]P[/imath] projective and injective module [imath]\implies P=0[/imath] or [imath]R[/imath] is field of fractions
I'm having difficulties proving the following: Let [imath]R[/imath] be an integral domain and [imath]P[/imath] a projective and injective [imath]R[/imath]-module. Show that [imath]P=0[/imath] or [imath]R=Q(R)[/imath], where [imath]Q(R)[/imath] denotes the field of fractions of [imath]R[/imath]. (I haven't found a similar question here on MSE, but maybe I'm searching for the wrong terms). This is how far I have come so far: I could prove that if [imath]M[/imath] is a free, divisible [imath]R[/imath]-module, then [imath]M=0[/imath] or [imath]R=Q(R)[/imath] I could prove that an injective [imath]R[/imath]-module is divisible. (my definition of 'divisible is': for each [imath]r \in R[/imath], where [imath]r[/imath] is not a zero divisor, the map [imath]M \to M, m \mapsto rm[/imath] is surjective) Then I wanted to use that if [imath]P[/imath] is projective, there exists a free module [imath]F[/imath] and a module [imath]Q[/imath], so that [imath]F = P \oplus Q[/imath], and then wanted to show that [imath]F[/imath] is also divisible. But I don't think that needs to be. Any help/hints/links in the right direction are much appreciated! Thanks in advance!
|
585495
|
A problem about an [imath]R[/imath]-module that is both injective and projective.
Let [imath]R[/imath] be a domain that is not a field, and let [imath]M[/imath] be an [imath]R[/imath]-module that is both injective and projective. Prove that [imath]M= \left \{ 0 \right \}[/imath]. This is exercise 7.52 of Rotman's Advanced Modern Algebra. Using theorems before exercises, because [imath]M[/imath] is injective and [imath]R[/imath] is a domain, I conclude that [imath]\forall m\in M ,\forall r\in R\ (r\neq 0) ,\exists {m}'\in M \Rightarrow m=r{m}'[/imath] and also because [imath]M[/imath] is projective there is a surjective [imath]\psi[/imath] from free [imath]R[/imath]-module [imath]F[/imath] with basis [imath]\left \{ e_{i} \right \}_{i\in I}[/imath] to [imath]M[/imath] and thus we can conclude that for every [imath]m\in M[/imath] we have [imath]m=\sum r_{i}\Psi (e_{i})[/imath] now I don't know how should I use these together. The idea of what is happening or a suggestion or a hint will be great.
|
1840816
|
When a prime ideal in polynomial ring over integers is principal
While dealing with a question about a prime ideal [imath]I\subset\mathbb{Z}[x][/imath] (with [imath]0[/imath] as the only constant polynomial in [imath]I[/imath]) I was asked to show that there exists [imath]f(x)\in\mathbb{Z}[x][/imath] such that [imath]I=\left<f(x)\right>[/imath] and got stuck. How do I know that some ideal is generated by only one element? What have i tried: i) Looked at [imath]I[/imath] and tried to adapt the proof that integers is a principal ideal domain and and so as [imath]I[/imath] is an ideal of itself I'd be done, but no success. ii) Ignored one operation and tried to see I as a subgroup, but that was a rather unsuccessful strategy. I appreciate any hints and book references where i can find proper explanations.
|
574535
|
Prime ideal in a polynomial ring over an integrally closed domain
Let [imath]R[/imath] be an integrally closed domain. Let [imath]P[/imath] be a prime ideal of [imath]R[x][/imath] with the property that [imath]P\cap R=\left\{ 0\right\} [/imath] and [imath]P[/imath] contains a monic polynomial. Prove that [imath]P[/imath] is principal. Help me a hint to prove. Thank for any insight.
|
1840505
|
Map from 2-sphere into [imath](\mathbb R^3, |\cdot|)[/imath]
Can you help me with this? Let [imath]S^2 := \{x\in \mathbb R^3:||x||_2 = 1\} \subset (\mathbb R^3, ||\cdot||_2)[/imath] and [imath]T:S^2 \to (\mathbb R, |\cdot|)[/imath] be a continuous map. Since [imath]S^2[/imath] is compact, [imath]T[/imath] attains its maximum [imath]T_{max}[/imath] and its minimum [imath]T_{min}[/imath]. Is there a value [imath]T\in]T_{min},T_{max}[[/imath] that is assumed at only one [imath]x_0\in S^2[/imath]? I think I have to use the fact that the image of a connected space under a continuous function is connected again, but I don't know how. Can you help?
|
1840027
|
[imath]S := \{x \in \Bbb R^3: ||x||_2 = 1 \}[/imath] and [imath]T: S^2 \to \Bbb R[/imath] is a continuous function. Is [imath]T[/imath] injective?
[imath]S := \{x \in \Bbb R^3: ||x||_2 = 1 \} \subset (\Bbb R^3, || \cdot ||_2 )[/imath] and [imath]T:S^2 \to (\Bbb R, |\cdot |)[/imath] is a continuous function. I've already shown that [imath]T_{\mathrm{max}} := \mathrm{sup}\{ T(x) : x \in S^2\} \: \: \mathrm{ and }\: \: T_{\mathrm{min}}:= \mathrm{inf}\{ T(x) : x \in S^2\}[/imath] are reached by [imath]T[/imath] with the argument that [imath]S^2[/imath] is closed and bounded and therefore after the Heine-Borel-Theorem compact, and therefore since T is continuous it assumes its max and min in [imath]S^2[/imath]. Now i have answer the following: 1) Is there a point [imath]x_0 \in S^2[/imath] with [imath]T(x_0)=T(-x_0)[/imath] ? 2) Is there a value [imath]T \in(T_{\mathrm{min}}, T_{\mathrm{max}})[/imath] which is only assumed at exactly one point of [imath]S^2[/imath] ? I wanted to show that [imath]T[/imath] is injective to get that 1) is false and 2) is true but i didn't get anywhere. Any ideas? Thanks in advance!
|
1841341
|
Freshman's Dream Quotient Rule
I have been researching the Freshman's Dream Quotient Rule for awhile now and have been unable to find an example for which the rule works. I want to know if anyone knows of 2 function which it works for. Rule is: [imath]\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)}{g'(x)}[/imath]
|
10927
|
When do the Freshman's dream product and quotient rules for differentiation hold?
This is motivated by looking at the calculus exams of some of my undergraduate students. A recurring mistake is assuming that the derivative of the product of functions is a product of derivatives and the derivative of the quotient of two functions is the quotient of their derivatives. This might be an ill formed question, but do their exist a pair of functions such that these rules hold. Explicitly, do there exist [imath]f(x),g(x)[/imath] satisfying all of: \begin{align*} f&\neq 0;\\ \frac{d(fg)}{dx}&=\frac{df}{dx}\cdot \frac{dg}{dx}\\ \text{and }\frac{d(f/g)}{dx}&=\frac{df}{dx}/ \frac{dg}{dx}\text{ ?} \end{align*} EDIT: Just to clarify my question, looking at the responses so far. I am looking for functions satisfying all three conditions above simultaneously.
|
1841365
|
What is the least possible value of [imath]n[/imath] such that [imath]n^2+n+17[/imath] is composite?
The question is- Find the least [imath]n[/imath] for which [imath]n^2+n+17[/imath] is composite. I tried to factorize it and show that it has a factor greater than [imath]1[/imath].But I could not factorize it and I also found that it has no real zeroes (roots).Now,how do I solve the problem? Thanks for any help!!
|
819706
|
Are numbers of the form [imath]n^2+n+17[/imath] always prime
Someone claimed that a number, multiplied by the number after it plus 17 is always prime, and showed several cases. I'm not a complete amateur in Number Theory, and I know that [imath]17*18+17=17*19[/imath], so it does not work for [imath]n\equiv0(\mod17)[/imath] but does it always work for other [imath]n[/imath] values? If not, can someone give me a counter example that I can show that person so they can correct their statement?
|
1841777
|
How to show convergence and evaluate [imath]\sum\limits_{n=0}^{\infty} \frac{n}{(n+1)!}[/imath]
I try to evaluate series [imath]{n\over(n+1)!}[/imath] obviously the term [imath]s_n = {n\over (n+1)!} = {1\over (n-1)! (n+1)}[/imath] converges, which has a limit = [imath]0[/imath] when [imath]n\to \infty[/imath] .Then I was stuck and I don't know how to sum this series and whether it converges.
|
1098034
|
Calculating [imath]\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}[/imath]
I want to calculate the sum:[imath]\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}=[/imath] [imath]\sum_{n=1}^\infty\frac{n}{(n+1)!}=[/imath] [imath]\sum_{n=1}^\infty\frac{n+1-1}{(n+1)!}=[/imath] [imath]\sum_{n=1}^\infty\frac{n+1}{(n+1)!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}=[/imath] [imath]\sum_{n=1}^\infty\frac{1}{n!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}.[/imath] I know that [imath]\sum_{n=0}^\infty\frac{1}{n!}=e[/imath] so [imath]\sum_{n=1}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{n!}-1=e-1[/imath] But, what can I do for [imath]\sum_{n=1}^\infty\frac{1}{(n+1)!}[/imath] ? Am I allowed to start a sum for [imath]n=-1[/imath] ? How can I bring to a something similar to [imath]\sum_{n=0}^\infty\frac{1}{n!}[/imath]?
|
23057
|
Under what condition we can interchange order of a limit and a summation?
Suppose f(m,n) is a double sequence in [imath]\mathbb R[/imath]. Under what condition do we have [imath]\lim\limits_{n\to\infty}\sum\limits_{m=1}^\infty f(m,n)=\sum\limits_{m=1}^\infty \lim\limits_{n\to\infty} f(m,n)[/imath]? Thanks!
|
1891015
|
Conditions for passing limit inside a series
Consider a succession [imath]\{f_{n,k}\}_{n,k}[/imath] s.t. for any fixed [imath]k[/imath], [imath]\lim_{n\rightarrow\infty}f_{n,k}=g_k[/imath]. Which are the conditions s.t. \begin{equation} \lim_{n\rightarrow\infty}\sum_{k}f_{n,k}=\sum_{k}g_k \; \;? \end{equation}
|
1841378
|
Integration of [imath]f(x)[/imath] [imath]:=[/imath] [imath]\sqrt{x^2 + 1} \over x[/imath]
Given, [imath]f(x)[/imath] [imath]:=[/imath] [imath]\sqrt{x^2 + 1} \over x[/imath], with [imath]x > 0,[/imath] I wondered which kind of integration would be the most clever one. Following the advice of Gilbert Strang, I would try to substitute [imath]x = \sin u,[/imath] but in the end, this would lead me to the integration of [imath]\csc u,[/imath] and that's not an easy way to do it, I guess. Another approach would be to substitute [imath]u = \sqrt{x^2 + 1}.[/imath] This substitution works much more straight-forward, but it includes a lot of smaller calculations that could cost me time in an exam. So, do you know another approach that works with a clever trick and goes much faster?
|
591073
|
How to integrate [imath]\int\frac{\sqrt{1+x^2}}{x}\,\mathrm dx[/imath]
I want to know how to integrate [imath]\int\frac{\sqrt{1+x^2}}{x}\,\mathrm dx[/imath] Could anyone solve it? Thanks
|
1841950
|
Implication in local noetherian domain of Krull's dimension 1.
Let [imath](A,m)[/imath] be a local noetherian domain with Krull dimension [imath]1[/imath]. Let [imath]k[/imath] be the field [imath]A/m[/imath]. I'm trying to prove that if [imath]m/m^2[/imath] has dimension [imath]1[/imath] as a [imath]k[/imath]-vector space, then every ideal [imath]I[/imath] of [imath]A[/imath] is a power of [imath]m[/imath]. I started the argument with; If [imath]I=A[/imath] then trivially [imath]I=m^0[/imath] and we're done. Otherwise it must be [imath]I\subset m[/imath] since every element of [imath]A\setminus m[/imath] is invertible; but I can't think of a way to use the hypothesis from this point. Could someone sketch a way to prove this? Thank in advance.
|
1818783
|
[imath]A[/imath] local Noetherian ring with principal maximal ideal implies PIR?
Suppose that [imath]A[/imath] is a local Noetherian ring with principal maximal ideal. Can we prove that every ideal of [imath]A[/imath] is principal? I tried to exploit the Noetherian property on the set of non-principal ideals obtaining that (if the statement is false) there must exist a prime non-principal ideal but I can't conclude nothing.
|
1842196
|
singular homology of a simplicial complex
On Page 108 of the book Algebraic Topology, Allen Hatcher, the singular homology of a topological space [imath]X[/imath] is defined to be the homology of the chain complex by setting the [imath]n[/imath]-chains [imath]C_n(X)[/imath] as the free abelian group with basis [imath]\text{Map}(\Delta^n,X)[/imath], where [imath]\Delta^n[/imath] is the standard [imath]n[/imath]-simplex, and the boundary map [imath]\partial_n[/imath] as [imath] \partial _n (\sigma)=\sum_i (-1)^i\sigma\mid_{[v_0,\cdots,\hat v_i,\cdots,v_n]}, [/imath] where [imath]\sigma[/imath] is a map from [imath]\Delta^n[/imath] to [imath]X[/imath] and [imath]\sigma\mid [/imath] is the restriction of [imath]\sigma[/imath]. Let [imath]K[/imath] be a simplicial complex and let [imath]|K|[/imath] be the geometric realization of [imath]K[/imath]. Suppose the vertices ([imath]0[/imath]-simplices) of [imath]K[/imath] are [imath]v_0,v_1,\cdots,v_k[/imath], [imath]k[/imath] finite. Let [imath] K_{n}= \text{Simplicial Map}(\Delta^n,K). [/imath] Then [imath] K_n\cong\{f\in \text{Map}(\mathbb{Z}_{n+1},\mathbb{Z}_{k+1})\mid \text{ there exists a simplex } \sigma\in K \text{such that Im}f \text{ is the set of vertices of } \sigma\}. [/imath] Here we use [imath]\mathbb{Z}_{n+1}[/imath] to represent the set of all vertices of [imath]\Delta^n[/imath] and use [imath]\mathbb{Z}_{k+1}[/imath] to represent the set of all vertices of [imath]K[/imath]. Let [imath]C_n(K)[/imath] be the free abelian group with basis [imath]K_n[/imath]. Analogous to the boundary map in singular homology, there is a boundary map [imath] \partial_n: C_n(K)\longrightarrow C_{n-1}(K). [/imath] Question: Is the singular homology of [imath]|K|[/imath] same or different with the homology of the chain complex [imath] \cdots C_n(K)\overset{\partial_n}{\longrightarrow} C_{n-1}(K)\overset{\partial_{n-1}}{\longrightarrow}\cdots\overset{\partial_1}{\longrightarrow} C_0(K) ? [/imath]
|
1241500
|
Homology of a simplicial set
Let [imath]X[/imath] be a simplicial set. Define the complex [imath](C^X_\bullet,D)[/imath] by [imath]C^X_n=\bigoplus_{X_n} \mathbb{Z}[/imath] and [imath]D_n=\sum_{i=0}^n (-1)^i d_i:C_n \to C_{n-1}[/imath] where the [imath]d_i[/imath]'s are the face maps. I wonder if there is a relation between the homology of [imath](C^X_\bullet,D)[/imath] and the singular homology of [imath]|X|[/imath], the geometric realization of [imath]X[/imath]. I know that there is a weak equivalence [imath]X \to S|X|[/imath] and that the singular homology of [imath]|X|[/imath] is exactly the homology of [imath](C^{S|X|},D)[/imath] as defined above. But I cannot conclude yet that there is a homotopy equivalence between the complexes [imath](C^X,D)[/imath] and [imath](C^{S|X|},D)[/imath], do I ?
|
1842348
|
Is [imath]\sqrt{2}^{\sqrt{2}}[/imath] rational?
In §2.2 of her essay on mathematical morality, Eugenia Cheng includes the following example: Why is it possible for an irrational to the power of an irrational to be rational? Here is a nice little proof that it is possible: Consider [imath]\sqrt{2}^{\sqrt{2}}[/imath]. If it is rational, we are done. If it is irrational, consider [imath] \left(\sqrt{2}^{\sqrt{2}}\right)^\sqrt{2} = \sqrt{2}^2=2.[/imath] However, as Cheng notes, this doesn't tell us whether [imath]\sqrt{2}^{\sqrt{2}}[/imath] itself is rational or not. So which is it?
|
955975
|
Can [imath]\sqrt{a}^\sqrt{b}[/imath] be rational if [imath]\sqrt{a}[/imath] and [imath]\sqrt{b}[/imath] are irrational?
Let [imath]a[/imath] and [imath]b[/imath] be rational numbers, such that [imath]\sqrt{a}[/imath] and [imath]\sqrt{b}[/imath] are irrational. Can [imath]\sqrt{a}^\sqrt{b}[/imath] be rational? I found examples, where the irrational power of an irrational number is rational, but in those examples at least one of those numbers (base and exponent) has not been a square root of a rational.
|
1842028
|
Relationship between circumscribed sphere radius and edge length of a dodecahedron?
Hello and I'm a secondary student doing a math exploration, but I'm currently stuck with this problem... Can anyone kind enough to show me the derivation of the relationship between the circumscribed sphere radius and the edge length in a dodecahedron? The formula I looked up is this... [imath]r = a \; \frac{\sqrt{3}}{2} \cdot \frac{1 + \sqrt{5}}{2}[/imath] where [imath]r[/imath] is the circumscribed sphere's radius and [imath]a[/imath] is the edge length. (Taken from this reference: http://www.fxsolver.com/browse/formulas/Regular+Dodecahedron+(+circumscribed+sphere+radius) I notice that the [imath](1 + \sqrt{5})∕2[/imath] is the golden ratio, but I do not understand the rest of the formula and how to give reasons for using the golden ratio. Can anyone kind enough to explain the reasoning for using golden ratio, and possibly the derivation for the whole formula? Thank you for your effort to help.
|
374692
|
How to find the maximum diagonal length inside a dodecahedron?
I am trying to find the maximum length of a diagonal inside a dodecahedron with a side length of [imath]2.319914107\times10^{89}[/imath] meters. I am not sure if any other information than that is needed, if it is I probably have it and will give it, but if you could help that would be great. I am really just trying to find the distance between opposite vertices.
|
1843095
|
Simplify [imath]\frac{1}{1\times2\times3} + \frac{1}{2\times3\times4} +\frac{1}{3\times4\times5}+\cdots+\frac{1}{n(n+1)(n+2)}[/imath]
This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with sequencing and series, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes: Simplify [imath]\frac{1}{1\times2\times3} + \frac{1}{2\times3\times4} +\frac{1}{3\times4\times5}+\cdots+\frac{1}{n(n+1)(n+2)}[/imath]
|
1837431
|
Prove that [imath]\sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}[/imath]
Prove that [imath]\displaystyle \sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}[/imath]. I tried using the partial fraction decomposition [imath]a_j = \frac{1}{2j} - \frac{1}{j+1} + \frac{1}{2(j+2)}[/imath], but I don't see how that helps.
|
1835727
|
Finding [imath]A[/imath] and [imath]B[/imath] using [imath] A \times B = \text{lcm}\,(A,B)\times\text{hcf}\,(A,B)[/imath]
The highest common factor and lowest common multiple of two numbers [imath]A[/imath] and [imath]B[/imath] are [imath]12[/imath] and [imath]168[/imath] respectively. Find the possible values of [imath]A[/imath] and [imath]B[/imath] with the exception of [imath]12[/imath] and [imath]168[/imath]. I know this question involves the concept of [imath] A \times B = \text{lcm}\,(A,B)\times\text{hcf}\,(A,B)[/imath] But I'm not sure how to apply it. Can someone guide me? Thanks!
|
1614491
|
Solve for [imath]A,B[/imath]: [imath]\mathrm{LCM}(A,B)=168[/imath], [imath]\mathrm{HCF(A,B)}=12[/imath]
The highest common factor and the lowest common multiple of two numbers [imath]A[/imath] and [imath]B[/imath] are [imath]12[/imath] and [imath]168[/imath] respectively. Find the possible values of [imath]A[/imath] and [imath]B[/imath] with the exception of [imath]12[/imath] and [imath]168[/imath]. Help me with this equation please. I kind of get it but I don't understand the concept behind it... I do not want to memorize.
|
1836841
|
find [imath]u,v\in \mathbb Z[/imath] such that [imath]231u+45v=1[/imath].
I have to find [imath]u,v\in \mathbb Z[/imath] such that [imath]231u+45v=1[/imath]. By Euclide algorithm, \begin{align*} 231&=5\cdot 45+6\\ 45&=6\cdot 7+3\\ 7&=3\cdot 2+1 \end{align*} The first equation gives [imath]6=231-5\cdot 45.[/imath] We put this in the second equation, and we get [imath]45=(231-5\cdot 45)\cdot 7+3\implies 3=36\cdot 45-7\cdot 231.[/imath] Now I replace [imath]3[/imath] in the last equation, and obtain [imath]7=2\cdot (36\cdot 45-7\cdot 231)+1\implies 7-72\cdot 45+14\cdot 231=1,[/imath] but the 7 disturbs me, did I do my algorithm wrong ?
|
148415
|
Given two coprime integers, find multiples of them that differ by 1
I'll describe the problem with an example. Find integers [imath]n[/imath] and [imath]m[/imath] such that [imath]n\cdot13 + m\cdot101 = 1[/imath]. This is the same as solving the equation [imath]n\cdot13 \equiv 1 \pmod {101}[/imath] I'm revising for a maths exam in a few days and it seems like a lot of the questions and examples rely on an ability to solve this type of problem. I can't seem to find a method in my notes, the solutions just get plucked out of thin air! What is a neat way of doing it on pen and paper with a calculator?
|
1843779
|
Number of distinct integer-valued vector solution for [imath]x_1 + x_2 + ... + x_r = n[/imath]
The Number Of Integer Solutions Of Equations [imath]x_1 + x_2 + ... + x_r = n[/imath] An approach is to find the number of distinct non-negative integer-valued vectors [imath](x_1,x_2,...,x_r)[/imath] such that [imath]x_1 + x_2 + ... + x_r = n[/imath] There are [imath]n+r-1\choose r-1[/imath] distinct non-negative integer-valued vectors [imath](x_1, x_2,...,x_r)[/imath] satisfying the equation [imath] x_1 + x_2 + ... + x_r = n [/imath] So there are [imath]n+r-1\choose r-1[/imath] solutions. My question is how to count number of solutions for above equation considering all vectors whose permutation is same. Now consider for [imath]n = 2[/imath] and [imath]r = 2[/imath] solutions are {(0,2), (2,0), (1,1)} as we can see (0,1) is a permutation of (1,0), so count them only once. Therefore total number of solutions for [imath]n = 2[/imath] and [imath]r = 2[/imath] is 2 instead of 3.
|
217597
|
Number of ways to write n as a sum of k nonnegative integers
How many ways can I write a positive integer [imath]n[/imath] as a sum of [imath]k[/imath] nonnegative integers up to commutativity? For example, I can write [imath]4[/imath] as [imath]0+0+4[/imath], [imath]0+1+3[/imath], [imath]0+2+2[/imath], and [imath]1+1+2[/imath]. I know how to find the number of noncommutative ways to form the sum: Imagine a line of [imath]n+k-1[/imath] positions, where each position can contain either a cat or a divider. If you have [imath]n[/imath] (nameless) cats and [imath]k-1[/imath] dividers, you can split the cats in to [imath]k[/imath] groups by choosing positions for the dividers: [imath]\binom{n+k-1}{k-1}[/imath]. The size of each group of cats corresponds to one of the nonnegative integers in the sum.
|
1843007
|
Map of degree two from [imath]S^2[/imath] to the torus [imath]T^2[/imath].
Prove that there is no map of degree two from [imath]S^2[/imath] to the torus [imath]T^2[/imath]. I'm struggling with this problem. I've tried lifting the map to the covering space but I'm not sure what to do from there. I keep getting results that I know are wrong. Most examples I can find of problems like this involve maps from [imath]T^2[/imath] to [imath]S^2[/imath], and I think those call for different techniques than what is needed here.
|
1836029
|
Why every map [imath]f : S^n \to T^n (n>1)[/imath] has topological degree zero?
Why every map [imath]f : S^n \to T^n (n>1)[/imath] has topological degree zero? I don't know anything about covering spaces, and has been told to me that this assertion comes from this theory! I do appreciate any help.
|
1844076
|
In the context of linear algebra, is it possible for a vector space or a subspace to have a finite number of elements?
A vector space must satisfy closure under addition and multiplication. Sorry if this is obvious but does that mean that, assuming the normal rules of arithmetic and excluding the trivial examples like [imath]V=\{0\}[/imath] it is impossible for a vector space to have a finite number of elements? (since for every two elements [imath]u,v \in V[/imath] there must be another number [imath]u+v \in V[/imath])
|
1343523
|
Are there nontrivial vector spaces with finitely many elements?
I have only seen infinite vector spaces and the one finite vector space i.e the trivial vector space [imath]\{0\}[/imath]. Is there any other finite vector space?
|
1843410
|
To show an entire function is zero
[imath]f[/imath] is entire, satisfying [imath]|f(z)|\le \frac1{|\text{Re}(z)|}.[/imath] Show that [imath]f[/imath] is identically zero. It suffices to show boundedness. Since [imath]f[/imath] is bounded on the real line, then if [imath]f[/imath] is unbounded on the complex plane, the infinity must be an essential singularity, which means, for an arbitrarily chosen point [imath]z_0\in\Bbb C[/imath], we can pick a sequence [imath]z_n[/imath] whose moduluses tend to infinity with [imath]f(z_n)\to z_0[/imath], and it follows that [imath]|z_0|\le \frac1{|\text{Re}(z_n)|}[/imath]. This is the best I can do. I'm not sure if I'm going in the correct direction. It however seems unlikely for me to deal with the problem on the [imath]y[/imath]-strip in this way.
|
987676
|
Suppose [imath]f[/imath] is entire and [imath]|f(z)| \leq 1/|Re z|^2[/imath] for all [imath]z[/imath]. Show that [imath]f [/imath] is identically [imath]0[/imath].
This is a problem from my complex analysis textbook. The hint is to consider [imath]g(z)=(z-iR)^2(z+iR)^2 f(z)[/imath] and to show that [imath]|g(z)| \leq 8R^2[/imath]. This is what i have tried: Consider [imath]Re z \geq 0[/imath], then [imath]|(z-iR)^2 f(z)|\leq \left(\frac{|z-iR|}{Re z}\right)^2\leq 2[/imath] But then I don't know what to do next. Any help I'd appreciate :)
|
1843973
|
Choosing integration strategy
I need to integrate this function: [imath]\int \frac {x^3}{\sqrt{4-x^2}} \;dx[/imath] I don't know how i should integrate, should i divide because the numerator has an higher grade? There is any strategy i should know for choosing the best way?
|
1826494
|
Integrating [imath]\int\frac{x^3}{\sqrt{9-x^2}}dx[/imath] via trig substitution
What I have done so far: Substituting [imath]x=3\sin(t)\Rightarrow dx=3\cos(t)dt[/imath] converting our integral to [imath]I=\int\frac{x^3}{\sqrt{9-x^2}}dx=\int \frac{27\sin^3(t) dt}{3\sqrt{\cos^2(t)}}3\cos(t)dt\\ \Rightarrow \frac{I}{27}=\int \sin^3(t)dt=-\frac{\sin^2(t)\cos(t)}{3}+\frac{2}{3}\int\sin(t)dt\\ \Rightarrow I=18\cos(t)-9\sin^2(t)\cos(t)+c[/imath] Which is where I'm stuck. I can substitute back in for the [imath]\sin(t)[/imath] terms but don't know hot to deal with the [imath]\cos(t)[/imath] terms and back out information about [imath]x[/imath]. edited: because I miswrote the problem, I am sorry.
|
1844803
|
Is [imath]p_1p_2p_3\cdots p_i > p_{i+1}^2[/imath]?
Is it always true that [imath]p_1p_2p_3\cdots p_i > p_{i+1}^2[/imath] where [imath]p_i[/imath] are the prime numbers listed in increasing order and [imath]i>3[/imath]? I was wondering if this is true because it would seem to depend on the prime gap between [imath]p_{i}[/imath] and [imath]p_{i+1}[/imath]. Testing a few examples seems like it is likely to be true.
|
1367485
|
The square of n+1-th prime is less than the product of the first n primes.
I wanted to prove the following question in an elementary way not using Bertrand postulate or analytic estimates like [imath]x/\log x[/imath]. The question is [imath] p_{n+1}^2<p_1p_2\cdots p_n,\qquad(n\geq4) [/imath] I made the following argument. Does anyone have some opinion or simpler ideas to complete. We consider two cases: Case 1: [imath]N=p_1p_2\cdots p_n-1[/imath] is composite: then there will be a prime factor [imath]q\leq\sqrt{N}[/imath], and of course [imath]q[/imath] is not any of the [imath]p_i[/imath]'s. therefore [imath]p_{n+1}\leq q\leq\sqrt{N}[/imath]. So [imath]p_{n+1}^2<N+1[/imath]. Case 2: [imath]N=p_1p_2\cdots p_n-1[/imath] is prime. Then???
|
1845498
|
Let [imath](f_1,\ldots,f_r)\subset k[x_1,\ldots,x_n],(g_1,\ldots,g_s)\subset k[y_1,\ldots,y_m][/imath] be radical ideals. Then their sum is radical.
Let [imath]\mathfrak{a}:=(f_1,\ldots,f_r)\subset k[x_1,\ldots,x_n],\mathfrak{b}:=(g_1,\ldots,g_s)\subset k[y_1,\ldots,y_m][/imath] be radical ideals. Then I wish to prove that [imath]\mathfrak{c}:=(f_1,\ldots,f_r,g_1,\ldots,g_s)\subset k[x_1,\ldots,x_n,y_1,\ldots,y_m][/imath] is a radical ideal. If we set [imath]x=(x_i)_i,\ y=(y_i)_i[/imath] for simplicity, then otherwise stated, I wish to prove that [imath]\sqrt{k[x,y]\mathfrak{a}+[x,y]\mathfrak{b}}=k[x,y]\mathfrak{a}+[x,y]\mathfrak{b}=\mathfrak{c}[/imath] Trivially we have [imath]\supseteq[/imath]. Suppose [imath]f^t\in k[x,y]\mathfrak{a}+[x,y]\mathfrak{b}[/imath] for some [imath]t\in\mathbb{N}[/imath], so let [imath]f^t=f'(x,y)f''(x)+g'(x,y)g''(y),\ f',g'\in k[x,y],\ f''\in\mathfrak{a},\ g''\in\mathfrak{b}[/imath]. However, I can't figure out the next step from here. Edit: [imath]k[/imath] is algebraically closed.
|
1662307
|
Ideal of product of affine varieties
Let [imath]X\in\mathbb{A}^n[/imath] and [imath]Y\in\mathbb{A}^m[/imath] be affine varieties (irreducible algebraic sets). Then, if we denote by [imath]i_X[/imath] and [imath]i_Y[/imath] their ideals in their respective affine spaces, we define their product in [imath]\mathbb{A}^{n+m}[/imath] as the set [imath] X\times Y = \{(x,y):x\in X\text{ and } y\in Y\} = \{(x,y):x\in Z(i_X)\text{ and }y\in Z(i_Y)\}. [/imath] Using the second characterization it's easy to show that in fact [imath]X\times Y[/imath] is closed in [imath]\mathbb{A}^{n+m}[/imath] and particularly that [imath]X\times Y = Z(I_X+I_Y)[/imath], where [imath]I_X[/imath] and [imath]I_Y[/imath] are the vanishing ideals of [imath]X\times\mathbb{A}^m[/imath] and [imath]\mathbb{A}^n\times Y[/imath] (these ideals clearly are in bijective correspondence with [imath]i_X[/imath] and [imath]i_Y[/imath], respectively). Then, by the Nullstellensatz, we get [imath]I(X\times Y) = \sqrt{I_X+I_Y}[/imath]. The problem is that the references I've seen claim that in fact [imath]I(X\times Y) = I_X+I_Y[/imath] (Gathmann's notes, example 2.3.9 in page 25, and Georges' answer in this math.se post, specially the last display). This would only be true if [imath]I_X+I_Y[/imath] is radical, which I haven't been able to show for this case and know isn't true in general (see this math.se post). Are my references wrong, or what am I missing?
|
698460
|
How could I show that if [imath]a[/imath] is an integer, then [imath]a^3 \equiv 1, 0, 6 \mod 7[/imath]?
I've literally tried everything including proofs by cases. If I were to try a an odd integer, then I would get [imath]8k^3 + 4k^2 + 8k^2 + 4k + 1[/imath], which obviously couldn't convert to a [imath]\mod 7[/imath], and I'm honestly not sure what I can do! Does anyone here have any tips? Note, this is a soft question.
|
1845171
|
The cube of any number not a multiple of [imath]7[/imath], will equal one more or one less than a multiple of [imath]7[/imath]
Yeah so I'm kind of stuck on this problem, and I have two questions. Is there a way to define a number mathematically so that it cannot be a multiple of [imath]7[/imath]? I know [imath]7k+1,\ 7k+2,\ 7k+3,\ \cdots[/imath], but that will take ages to prove for each case. Is this a proof? [imath](7k + a)^3 \equiv (0\cdot k + a)^3 \equiv a^3 \bmod 7[/imath] Thank you.
|
1845688
|
Naive understanding of choice axiom
It's written in many resources that if we consider only finite sets, that choice axiom can be skipped and be proven from other ZF axioms. I remember I've read the following explanation. If [imath]A[/imath] is a finite set, there is exists some natural [imath]n, |A|=n[/imath], therefore exists some bijection [imath]f:\{1,\cdots,n\}\to A[/imath]. And we can choose an element from [imath]A[/imath] by choosing [imath]f(1)[/imath] But there is a problem with this explanation. Actually there are [imath]n![/imath] of such bijections. So how can we choose one [imath]f[/imath] of them without this axiom of choice?! UPD: actually I can simplify my question heavily. Let's consider one-point set [imath]A, |A|=1[/imath]. How can I get an element from this set without using axiom of choice? We know that [imath]A=\{x\}[/imath] for some [imath]x[/imath]. But we don't know how to obtain it. What allows us to just say Let's pick [imath]x[/imath] from [imath]A[/imath]?
|
132717
|
Do We Need the Axiom of Choice for Finite Sets?
So I am relatively familiar with the Axiom of Choice and a few of its equivalent forms (Zorn's Lemma, Surjective implies right invertible, etc.) but I have never actually taken a set theory course. I know there are times when an explicit way of choosing some elements may not be provided, and instead we call upon the Axiom of Choice to do it for us. But one thing has always bothered me. I have heard that for a finite set we do not need the Axiom of Choice to choose an element. I find this a little confusing. How is choosing an element from a finite set any different than choosing from an infinite set? I have heard before "since the set it finite, there are finitely many orders, so order the set and choose the first element in the order." But how does one choose what order to use? This involves picking an element from a finite set, i.e. it presumes the conclusion. I'm guessing there is a good reason why everyone keeps insisting that the Axiom of Choice in not needed to choose from finite sets, so could someone please explain it to me? Edit: There seems to be confusion on what exactly I am asking. I know that we need the Axiom of Choice, sometimes, to make infinitely many choices, from finite or infinite sets. This is not what is confusing me. Allow me to be more precise: Suppose [imath]S[/imath] is a nonempty finite set. I want to pick one element from [imath]S[/imath]. Does choosing one element from [imath]S[/imath] require the Axiom of Choice? If not, how can the choice be made without the Axiom of Choice? Saying [imath]S[/imath] is not empty, therefore there is an [imath]x \in S[/imath], does not suffice. How do I choose such an [imath]x[/imath]? There is one, but how do I describe it without any information about the set, other than the fact that it nonempty and finite? Another clarification: I am not asking about using "Let ... be given" as a proof technique.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.