qid
stringlengths 1
7
| Q
stringlengths 87
7.22k
| dup_qid
stringlengths 1
7
| Q_dup
stringlengths 97
10.5k
|
|---|---|---|---|
2225690 | Calculate the coordinates of one of the vertices of a rectangular triangle with data on the length of one side
I am using a cartesian coordinate system, and I have the coordinates of two points (red dots in the figure). By Pythagoras I can calculated the distance between those two points. However I want to know the coordinates of the green dot, which is perpendicular the side [imath]b[/imath], given that I know the distance [imath]a= 80.[/imath] I cannot find a way to calculate the coordinates of the green point. Any suggestions will be most welcomed. | 145443 | Finding point coordinates of a perpendicular
Given that I know the point coordinates of point [imath]A[/imath] and point [imath]B[/imath] on segment [imath]AB[/imath] and the expected length of a perpendicular segment [imath]CA[/imath] (perpendicular over [imath]AB[/imath]), how do I calculate the point coordinates of the point [imath]C[/imath]? |
2226003 | irreducibility of multivariate integral polynomial
Let [imath]f[/imath] be a polynomial in the variables [imath]x_1, \ldots, x_n[/imath], where [imath]n > 1[/imath], and with coefficients in [imath]\mathbb{Z}[/imath]. Are there convenient assumptions on [imath]f[/imath] to make the following implication? Assume that [imath]f[/imath] is irreducible in the ring [imath]{\mathbb Z}[x_1, \ldots, x_n][/imath], then [imath]f[/imath] is irreducible in [imath]\mathbb{Q}[x_1, \ldots, x_n][/imath]. For instance, if [imath]n = 1[/imath], you would require that the leading coefficient of [imath]f[/imath] is equal to [imath]1[/imath] (Gauss lemma). Are you aware of a similar condition in case [imath]n > 1[/imath] ? Thank you. | 213505 | Multivariable Gauss's Lemma
Gauss's Lemma for polynomials claims that a non-constant polynomial in [imath]\mathbb{Z}[X][/imath] is irreducible in [imath]\mathbb{Z}[X][/imath] if and only if it is both irreducible in [imath]\mathbb{Q}[X][/imath] and primitive in [imath]\mathbb{Z}[X][/imath]. I wonder if this holds for multivariable case. Is it true that a non-constant polynomial in [imath]\mathbb{Z}[X_1,\dots,X_n][/imath] is irreducible in [imath]\mathbb{Z}[X_1,\dots,X_n][/imath] if and only if it is both irreducible in [imath]\mathbb{Q}[X_1,\dots,X_n][/imath] and primitive in [imath]\mathbb{Z}[X_1,\dots,X_n][/imath]? Thank you for your help. |
2226431 | Can six "[imath]1\times1\times1[/imath] pyramids" fit inside a [imath]1\times1\times1[/imath] cube?
Consider the pyramid [imath]\left \{ (x, y, z): x,y,z\geq 0, x+y+z\leq 1 \right \}[/imath]. This is a pyramid that has [imath]3[/imath] pairwise-orthogonal edges, all of length [imath]1[/imath] (just like a corner of a room): From now on I'll call it a "unit pyramid" (UP). One can easily check that the volume of a UP is exactly [imath]\frac{1}{6}[/imath]. So that raises the question: Can we take six UPs and fit them all perfectly inside a unit cube? When I thought about it, my initial answer was an obvious yes. So I tried to imagine such arranging, but I couldn't. I was quite surprised since I usually have a very good spatial vision. Then I tried drawing a solution and failed. Since I'm not tech-savvy enough to use a 3D graphics software, I decided to take this problem into the real world. So here's a unit pyramid: It's composed of 3 edges of length 1 (which I set to be [imath]5[/imath] cm) and 3 edges of length [imath]\sqrt2[/imath] ([imath]\approx 7[/imath] cm). Now, there's only one reasonable way to attach another one to it, if our goal is a unit cube: (since now we have our square base). Now I added all the edges necessary to form a cube: And as your 3D mind can see, now we have four unit pyramids: It's pretty hard to see the region that has stayed uncovered by our 4 UPs, so here it is, emphasized: And what do you know - that's a regular tetrahedron! A quick reality check: a regular tetrahedron with edges of length [imath]\sqrt 2[/imath] has a volume of [imath]\sqrt 2 (\sqrt 2)^3/12=4/12=1/3[/imath], which suits the fact that we entered 4 UPs inside. But now, I'm pretty certain that you can't fit another 2 of them inside. To be sure, I constructed them seperately (last picture, I promise!): On the left there's a UP, and on the right - our regular tetrahedron (would you believe that one is twice the volume of the other?) Now you'll just have to trust me that it won't fit. Well, that convinced me that the mission of fitting 6 unit pyramids inside a unit cube is impossible, but of course I wouldn't call it a mathematical proof. Does anyone have a more mathematically-convincing argument? I'd be curious to hear your ideas and thoughts about this. Thank you for your time reading the question! | 26052 | Equidecomposability of a Cube into 6 Trirectangular Tetrahedra
In grade school we learn that a square can be equidecomposed into two congruent right isosceles triangles. Does the following three dimensional generalization hold? Consider a trirectangular tetrahedron with vertices at [imath](0,0,0)[/imath], [imath](1,0,0)[/imath], [imath](0,1,0)[/imath] and [imath](0,0,1)[/imath] and volume [imath]\frac{1}{6}[/imath]. (It has three orthogonal right triangular faces and one equilateral triangular face opposite the (tri-right-angled) common vertex.) The inscribing cube of the tetrahedron has unit side length and unit volume. Is there a dissection of this cube into 6 tetrahedra of the above form? Provided that such a dissection exists and this is the way it is to be done, it's easy to see how 4 tetrahedra can be glued together to give the boundary of the cube, but I don't quite see how the other two should fit inside. |
2179834 | Weird large K symbol
Take a look at this symbol: [imath] \pi=3 + \underset{k=1}{\overset{\infty}{\large{\mathrm K}}} \frac{(2k-1)^2} 6 [/imath] Does it look familiar to you? If so please help me! | 2295917 | What does the operator [imath]\mathop{\large\mathrm K}_{i=1}^\infty[/imath] stand for?
There are many different operators, for example sums [imath]\sum_{i=1}^\infty[/imath] or products [imath]\prod_{i=1}^\infty[/imath] But what does this operator stand for? What does it mean? [imath]\mathop{\LARGE\mathrm K}_{i=1}^\infty[/imath] I am sorry for this might being a dumb question, but I tried googling things like "big K operator math" and didn't found anything. |
2226741 | How to determine the Galois Group of a polynomial with "ugly" roots?
This is related to my previous question. Suppose I have the polynomial [imath]p(x) = x^2 - 2 \in \Bbb Q[x][/imath]. This is irreducible over [imath]\Bbb Q[/imath] so we need to move up to the extension field [imath]K/\Bbb Q = \Bbb Q(\sqrt{2})[/imath]. Now [imath]p[/imath] splits over [imath]K[/imath] into [imath](x - \sqrt{2})(x+\sqrt{2})[/imath] and it can be shown that the Galois Group [imath]\operatorname{Gal}(K/\Bbb Q) \cong \Bbb Z/2\Bbb Z[/imath] since it contains only the identity automorphism and one involution that sends [imath]\sqrt 2[/imath] to its conjugate [imath]-\sqrt 2[/imath]. My question is this; what if I have a polynomial with ugly roots? Say, for instance, [imath]p(x) = x^5 - x^3 - 2x^2 - 2x - 1[/imath] This was shown in my previous question to be irreducible over [imath]\Bbb Q[/imath] so the splitting field of [imath]p[/imath] is simply [imath]\Bbb Q[/imath] adjoin the roots of [imath]p[/imath]. These roots are, however, [imath]\alpha \approx 1.73469\ \ \ \beta \approx 0.79645e^{0.49411\pi i}\ \ \ \gamma \approx 0.9533e^{5.6494\pi i}[/imath] and the conjugates [imath]\overline \beta, \overline \gamma[/imath] (do we get the conjugates for free when we adjoin complex roots or do we explicitly have to adjoin them?). Then the splitting field [imath]F[/imath] of [imath]p[/imath] is [imath]\Bbb Q(\alpha, \beta, \gamma)[/imath] and their conjugates (depending on the answer to the question just asked). At this point I'm not sure where to go; I've been trying to list automorphisms that fix [imath]\Bbb Q[/imath], obviously the identity is included, it seems also that the autmorphisms [imath]\varphi_1 : \begin{cases} \beta \to \overline \beta \\ \gamma \to \gamma \end{cases}\ \ \ \text{and}\ \ \ \varphi_2 : \begin{cases} \beta \to \beta\\ \gamma \to \overline \gamma \end{cases}[/imath] are in the galois group. Also, [imath]\varphi_2 \circ \varphi_1 = \psi[/imath] should be in the group and this sends both [imath]\beta[/imath] and [imath]\gamma[/imath] to their conjugates. Any composition of [imath]\psi[/imath] with either of the others I've listed gives the other. I'm stuck for finding more now though. Since conjugation is an involution automorphism these are self inverse. I'm not sure where to go from here! Is what I've done thus far even correct or have I totally misunderstood the concepts I'm trying to understand here? | 45893 | How to find the Galois group of a polynomial?
I've been learning about Galois theory recently on my own, and I've been trying to solve tests from my university. Even though I understand all the theorems, I seem to be having some trouble with the technical stuff. A specific example would be how to find the Galois group of a given polynomial. I know some tricks, and I manage to solve some of those questions, but some not. For example, one of the tests asks to find the Galois group of [imath]x^{4}-4x+2[/imath]. I can see that it is irreducible over [imath]\mathbb Q[/imath] (Eisenstein), but I have no clue as to how to find its Galois group over [imath]\mathbb Q[/imath]. Can someone tell me how to do this? General techniques concerning this sort of problems are also welcome :). Thanks! |
2226797 | Prove that if [imath]X\cap Y=\emptyset[/imath], then [imath]X\cup Y[/imath] has [imath]n+m[/imath] elements.
Let [imath]X[/imath] and [imath]Y[/imath] be sets with [imath]n[/imath] and [imath]m[/imath] elements, respectively. Prove that if [imath]X\cap Y=\emptyset[/imath], then [imath]X\cup Y[/imath] has [imath]n+m[/imath] elements. It's clear to me intuitively how this is true. I understand that if two sets have no elements in common, then there is no 'repetition' in the union and the number of elements of the union is the sum of the number of elements in each set. How could I formalize this? Is there a way to do it by induction? I am not sure. | 527811 | Prove that if [imath]A[/imath] is a set with [imath]m[/imath] elements, [imath]B[/imath] is a set with [imath]n[/imath] elements, and [imath]A \cap B = \emptyset[/imath], then [imath]A \cup B[/imath] has [imath]m+n[/imath] elements.
I'm working through a real analysis textbook, so I don't want the full answer. I'm only looking for a hint on this problem. The book starts the proof like this: let [imath]f[/imath] be a bijection of [imath]\mathbb{N}_m[/imath] onto [imath]A[/imath], and let [imath]g[/imath] be a bijection of [imath]\mathbb{N}_n[/imath] onto [imath]B[/imath]. Define [imath]h[/imath] on [imath]\mathbb{N}_{m+n}[/imath] as [imath]h(i):=f(i)[/imath] for [imath]i=1, ..., m[/imath], and [imath]h(i):=g(i)[/imath] for [imath]i=m+1,...,m+n[/imath]. Show that [imath]h[/imath] is a bijection from [imath]\mathbb{N}_{m+n}[/imath] onto [imath]A \cup B[/imath]. Ok, so with that, I started out proving that [imath]h[/imath] is an injection into [imath]A \cup B[/imath]. The usual method for this is to prove that if [imath]h(i)=h(j)[/imath], where [imath]i,j=1,...,m+n[/imath], then [imath]i=j[/imath]. I think I proved this by looking at four cases. [imath]1 \leq i,j \leq m[/imath]. In this case, [imath]h(i)=f(i)[/imath] and [imath]h(j)=f(j)[/imath], so because [imath]f[/imath] is a bijection, we know that [imath]f(i)=f(j)[/imath] implies that [imath]i=j[/imath]. [imath] 1 \leq i \leq m[/imath] and [imath]m+1 \leq j \leq m+n[/imath]. We know that [imath]i \neq j[/imath] (obviously), so we want to prove that in this case, [imath]h(i) \neq h(j)[/imath]. By definition, [imath]h(i)=f(i)[/imath] and [imath]h(j)=g(j-m)[/imath]. Because [imath]f[/imath] is a bijection into [imath]A[/imath] and [imath]g[/imath] is a bijection into [imath]B[/imath], and [imath]A \cap B = \emptyset[/imath], we know that [imath]f(i) \neq g(j-m)[/imath], so [imath]h(i) \neq h(j)[/imath]. This case is identical to case 2, except that [imath]m+1 \leq i \leq m+n[/imath] and [imath]1 \leq j \leq m[/imath], so the logic is reversed. The final case, where [imath]m+1 \leq i,j \leq m+n[/imath], is similar to case 1, because since [imath]h(i)=g(i)[/imath] and [imath]h(j)=g(j)[/imath], we can use the fact that [imath]g[/imath] is a bijection to prove that [imath]g(i)=g(j)[/imath] implies that [imath]i=j[/imath]. This is where I got stuck. I know I need to prove that [imath]h[/imath] is surjective, but am I on the right track for proving that it's injective? I think proving that [imath]h[/imath] is a surjection shouldn't be too hard, but I don't know if I'm approaching this part of the proof correctly. |
2226424 | If [imath]G_1, G_2[/imath] are two nonzero subgroups of [imath](\mathbb{Q},+)[/imath], prove that [imath]G_1\cap G_2\neq\{0\} [/imath].
Let [imath](\mathbb{Q},+)[/imath] be additive group of rational numbers. If [imath]G_1, G_2[/imath] are two nonzero subgroup of [imath](\mathbb{Q},+)[/imath], prove that [imath]G_1\cap G_2\neq\{0\}[/imath]. My approach : Take [imath]p/q\in G_1[/imath] and [imath]a/b \in G_2[/imath]. Then [imath]p/q+p/q+..=p \in G_1 [/imath] ([imath]q[/imath] times addition). Similarly [imath]a\in G_2[/imath]. Now we can always choose [imath]p[/imath] and [imath]a[/imath] as positive because in group their inverse are of opposite parity. Now set [imath]l=lcm(p,a)[/imath]. Since [imath]l=\frac{pa}{d}[/imath] where [imath]d=\gcd(p,a)[/imath] one finds [imath]l\in G_1[/imath]. Reason : Just [imath]p+...p=l[/imath] where (addition is done [imath]\frac{a}{d}[/imath] times). similarly [imath]l\in G_2[/imath]. So [imath]l\in G_1\cap G_2 [/imath]. Hence [imath]G_1\cap G_2\neq \{0\}[/imath]. Proved I would like to know if my solution is correct. This was supposed to be a tough problem. So it feels like i have committed some mistake in proof. Thank you. | 2013483 | Prove that any two nontrivial subgroups of [imath]\mathbb{Q}[/imath] have nontrivial intersection
I need to prove that any two nontrivial subgroups of [imath]\mathbb{Q}[/imath] have a nontrivial intersection as part of a larger proof that [imath]\mathbb{Q}[/imath] cannot be represented as a nontrivial direct product. (Yes, I realize that there are other ways to prove that [imath]\mathbb{Q}[/imath] cannot be represented as a nontrivial direct product, but I am not interested in those, and will not accept answers showing me how to do that). Based on the density of the rationals, I guess this idea kind of makes sense, but I am doing this proof for an abstract algebra class, and I don't think real analysis proofs would be accepted. Everywhere I've looked online has said this is "trivial" or "easy to see" without bothering to actually prove it for those of us mere mortals who don't see it as such. As for myself, I have literally no idea where to even get started proving this. Could somebody help me figure this out (without giving me cutesy examples that aren't really all that related with the expectation that I should be able to figure it out from looking at those - that's not really how my brain works)? Thank you. |
2225298 | Metric on a Quotient of the Riemann Sphere (Revised)
Let [imath]P[/imath] denote the quotient space obtained by the action of [imath]\mathbb{Z}\backslash2\mathbb{Z}[/imath] by the map [imath]z\mapsto\frac{1}{z}[/imath] on the riemann sphere [imath]\hat{\mathbb{C}}[/imath] (identified here with [imath]\mathbb{C}\cup\left\{\infty\right\}[/imath]). I identify [imath]P[/imath] with the set: [imath]Q=\left\{ z\in\mathbb{C}:\left|z\right|<1\right\} \cup\left\{ e^{it}:0\leq t\leq\pi\right\} [/imath] that is to say, I use elements of [imath]Q[/imath] as the representatives for the equivalence classes in [imath]P[/imath]. Note that this is a valid identification, seeing as every element of the orbit space of [imath]z\mapsto\frac{1}{z}[/imath] on [imath]\hat{\mathbb{C}}[/imath] is attested to by exactly one point in [imath]Q[/imath]. I need a formula for a function [imath]f:Q\times Q\rightarrow[0,\infty)[/imath] such that [imath]f[/imath] is a metric on [imath]P[/imath]. Just to be clear, I do not want an explanation of how to get such a formula; I want the formula. An analogy for you if I have yet to make myself clear: suppose I was asking for the area of a square with side length [imath]s[/imath]. The answers I have received so far for my question are akin to saying "multiply [imath]s[/imath] by itself" or "use the area formula for a square". The answer I am looking for is akin to saying "[imath]s^{2}[/imath]". I need the formula. And please, no expressions with differentials, nor matrices, or any of that. I need to know how to compute the metric by using the complex numbers in the indicated set that I have identified with [imath]P[/imath]. As an aside, I found something called the "Fubini-Study metric" on wikipedia. I feel as if this might be close to what I need, but unfortunately, the use of notations and conventions of differential geometry and multilinear algebra makes the wikipedia article virtually unintelligible to me. If someone could give me a formula for how to compute this metric using the complex numbers in the set [imath]Q[/imath], it would be much appreciated—assuming that this Fubini-Study metric is even close to what I am actually asking for. This is not for homework or anything. I am doing research (in analytic number theory), and I need to know a formula for such a metric so that I can go ahead and show whether or not a certain map is a contraction. I have been making stellar progress with my research, and so it is extremely frustrating to have my work ground to a halt for want of a simple formula. For your answers, assume that I know nothing of differential geometry, riemannian geometry, tensors, riemannian metrics, pullbacks, metric tensors, projective geometry, multilinear algebra, differential forms, tangent spaces, tangent bundles, covering maps, stereographic projections (and so on). This is (obviously) not true, but I say so because I want to be absolutely certain that a prospective responder to my question understands the kind of answer I am looking for. Dumb it down. If there is no such formula, then try to explain what I need to do as methodically and algorithmically as possible, and please, do not leave out any steps. Please... help. | 2222974 | Metric on a Quotient of the Riemann Sphere
Let [imath]P[/imath] denote the quotient space obtained by the action of [imath]\mathbb{Z}\backslash2\mathbb{Z}[/imath] by the antipodal map [imath]z\mapsto\frac{1}{z}[/imath] on the riemann sphere [imath]\hat{\mathbb{C}}[/imath] (identified here with [imath]\mathbb{C}\cup\left\{\infty\right\}[/imath]). I identify [imath]P[/imath] with the set: [imath]\left\{ z\in\mathbb{C}:\left|z\right|<1\right\} \cup\left\{ e^{it}:0\leq t\leq\pi\right\} [/imath] that is to say, I use elements of the above set as the representatives for the equivalence classes in [imath]P[/imath]. I am looking for formula for a function [imath]f:P\times P\rightarrow[0,\infty)[/imath] such that [imath]f[/imath] is a metric on [imath]P[/imath]. Specifically, I would like a formula for [imath]f[/imath] that I can evaluate by plugging in representative elements in the above set (or something like that, more or less). Thanks in advance. Edit: forgive me for sounding desperate, but I cannot make due with an explanation of how to obtain such a formula. I want the formula. An analogy for you if I have yet to make myself clear: suppose I was asking for the area of a square with side length [imath]s[/imath]. The answers I have received so far for my question are akin to saying "multiply [imath]s[/imath] by itself" or "use the area formula for a square". The answer I am looking for is akin to saying "[imath]s^{2}[/imath]". I need the formula. And please, no expressions with differentials, nor matrices, or any of that. I need to know how to compute the metric by using the complex numbers in the indicated set that I have identified with [imath]P[/imath]. |
2226857 | Order type of subsets of [imath]\mathbb{R}[/imath]
I need to show that there is no subsets of [imath]\mathbb{R}[/imath] of order type [imath]\omega_1[/imath]. This is a very tricky one, because I was first trying to prove that there are no subsets with cardinality [imath]\omega_1[/imath] by showing that every uncountable subset of [imath]\mathbb{R}[/imath] has cardinality of [imath]P(\omega)[/imath] and thus not [imath]\omega_1[/imath], but I that was not true as there are subsets with cardinality [imath]\omega_1[/imath]. The trick for me is to work with order type instead of cardinality. | 1754463 | Is there any well-ordered uncountable set of real numbers under the original ordering?
I know that the usual ordering of [imath]\mathbb R[/imath] is not a well-ordering but is there an uncountable [imath]S\subset \mathbb R[/imath] such that S is well-ordered by [imath]<_\mathbb R[/imath]? Intuitively I'd say there is no such set but intuitively I'd also say there is no well-ordered uncountable set at all, which is obviously wrong. I still struggle to grasp the idea of an uncountable, well-ordered set. |
2227136 | How would I show that [imath]2^n - 1[/imath] is a solution to the recurrence relation:
[imath]a_n[/imath] = 2[imath]a_{n-1}[/imath] + 1, with the initial conditions [imath]a_0[/imath] = 0, [imath]a_1[/imath] = 1 Apparently the solution is from the Tower of Hanoi problem, but having trouble coming with the this on my own. | 2225544 | How to solve [imath]a_n = 2a_{n-1} + 1, a_0 = 0, a_1 = 1[/imath]?
[imath]a_n = 2a_{n-1} + 1, a_0 = 0, a_1 = 1[/imath] So to get the closed form of this recurrence relation, I would usually try to get it into the form of [imath]a_n = ra_{n-1}[/imath] and then [imath]a_n = r^na_0[/imath]. But what am I supposed to do with the [imath]1[/imath]? Thanks! |
2221752 | Discrete math statement [imath]\rightarrow[/imath] plain English. [imath]\left(Longer\right)[/imath]
I found this one more challenging, I think the length of it confused me the most. Given Statement: [imath]\forall x \in \Bbb N \left[\left(\exists y \in \Bbb N \left(2 \le y \land y \lt x \land y \mid x \right)\right) \to \left(\exists z \in \Bbb N \left(2 \le z \land z \le \sqrt x \land z \mid x\right)\right)\right][/imath] So far I have written it as: For all [imath]x[/imath] in the natural numbers and there exists [imath]y[/imath] for all natural numbers where [imath]z[/imath] is smaller or equal to [imath]y[/imath] if and only if [imath]y[/imath] is smaller than [imath]x[/imath] if and only if [imath]y[/imath] is a divisible of [imath]x[/imath] then, [imath]\cdot \cdot \cdot[/imath] I only wrote it up to the [imath]\rightarrow[/imath], but not sure if it makes sense and furthermore if it's correct. Thus, can I continue like this for the next part. Any edits of my existing statement and any help in converting to plain English is welcomed. Thank you in advance. | 2219022 | [imath]\forall x\in\mathbb{N}[(\exists y\in\mathbb{N}(2\leq y\land y[/imath]
Please help me to prove this statement. I'm very confused, Thank you! I Have tried prove it using examples but it needs to be proved using mathematical method |
2223584 | Matrix reciprocal positive [imath]5\times 5[/imath], [imath]6\times 6[/imath] or bigger
Suppose we have [imath]n\times n[/imath] matrix [imath]A[/imath] having only positive elements and satisfying the property [imath]a_{ij}=\dfrac{1}{a_{ji}}[/imath] (a matrix satisfying this property is called a reciprocal matrix). If its largest eigenvalue [imath]λ_{max}[/imath] is equal to [imath]n[/imath], then the matrix [imath]A[/imath] satisfies the property (consistency property) [imath]a_{ij}a_{jk}=a_{ik}[/imath] where [imath]i,j,k=1,2,...,n[/imath]. I already found an example of [imath]5\times 5[/imath] reciprocal matrix to show this condition: \begin{bmatrix}1&1/2&1&1&1/4\\2&1&2&2&1/2\\1&1/2&1&1&1/4\\1&1/2&1&1&1/4\\4&2&4&4&1\end{bmatrix} It has simple characteristic equation [imath]λ^5-5λ^4=0[/imath], so the eigenvalues of matrix [imath]A[/imath] are [imath]λ=0[/imath] and [imath]λ=5[/imath] and it's shown that [imath]λ_{max}=n=5[/imath]. Is there anyone can help me to give another positive reciprocal matrix example that has simple characteristic equation and integer number as its eigenvalues ([imath]5\times 5[/imath] or [imath]6\times 6[/imath] or bigger size matrix)? | 2184282 | Matrix reciprocal positive to prove [imath]λ_{max}=n[/imath]
Suppose we have [imath]n\times n[/imath] matrix [imath]A[/imath] having only positive elements and satisfying the property [imath]a_{ij}=1/a_{ji}[/imath] (a matrix satisfying this property is called a reciprocal matrix). If its largest eigenvalue [imath]λ_{max}[/imath] is equal to [imath]n[/imath], then the matrix [imath]A[/imath] satisfies the property (consistency property) [imath]a_{ij}a_{jk}=a_{ik}[/imath] where [imath]i,j,k=1,2,...,n[/imath]. I already found an example of [imath]5\times 5[/imath] reciprocal matrix to show this condition: \begin{bmatrix}1&1/2&1&1&1/4\\2&1&2&2&1/2\\1&1/2&1&1&1/4\\1&1/2&1&1&1/4\\4&2&4&4&1\end{bmatrix} It has simple characteristic equation [imath]λ^5-5λ^4=0[/imath], so the eigenvalues of matrix [imath]A[/imath] are [imath]λ=0[/imath] and [imath]λ=5[/imath] and it's shown that [imath]λ_{max}=n=5[/imath]. Is there anyone can help me to give another positive reciprocal matrix example that has simple characteristic equation and integer number as its eigenvalues ([imath]4\times 4, 5\times 5[/imath] or [imath]6\times 6[/imath] matrix)? |
2226850 | Cardinality of a set of all subsets of [imath]P(\omega)[/imath] with cardinality [imath]\omega[/imath]
I am stuck on this one, do not know where to start. Any hint would be appreciated. What is the cardinality of [imath][P(\omega)]^{\omega}[/imath]? Is it equal to [imath]|P(w)|[/imath]? How do I show that? | 184746 | The cardinality of the set of countably infinite subsets of an infinite set
Let [imath]A[/imath] be a set with card([imath]A[/imath])=[imath]a[/imath]. What is the cardinal number of the set of countably infinite subsets of [imath]A[/imath]? I see that this problem is equivalent to finding the cardinal number of the set of injective functions from [imath]\mathbb{N}\rightarrow{A}[/imath]. I also know that the cardinal number of the set of bijections from [imath]A\rightarrow{A}[/imath] is [imath]a^{a}[/imath]. Hints and general heurisitcs would be greatly appreciated. |
2227483 | Find the number of real roots of the following equation
Question: The number of real roots of the equation [imath]2\cos\left( \frac{x^2+x}{6}\right)=2^x+2^{-x}[/imath] is (i) [imath]0[/imath] [imath]\qquad[/imath] (ii) [imath]1[/imath] [imath]\qquad[/imath](iii) [imath]2[/imath] [imath]\qquad[/imath] (iv) [imath]\infty.[/imath] I have tried in the following way: [imath]2\cos\left( \frac{x^2+x}{6}\right)=2^x+2^{-x}\implies \cos\left( \frac{x^2+x}{6}\right)=\frac{2^x+2^{-x}}{2}.[/imath] What can I do after this? | 1121453 | Number of real roots of [imath]2 \cos\left(\frac{x^2+x}{6}\right)=2^x+2^{-x}[/imath]
Find the number of real roots of [imath] \cos \,\left(\dfrac{x^2+x}{6}\right)= \dfrac{2^x+2^{-x}}{2}[/imath] 1) 0 2) 1 3) 2 4) None of these My guess is to approach it in graphical way. But equation seems little difficult. |
2227138 | Existence of a measurable cardinal [imath]\kappa[/imath] by the existance of [imath]\kappa[/imath]-complete ultrafilter
Let [imath]U[/imath] be a countably complete free ultrafilter over S. Prove that there is an uncountable cardinal [imath]\kappa, \kappa \le |S|[/imath] with a [imath]\kappa[/imath]-complete free ultrafilter over [imath]\kappa[/imath]. Is the following attempt correct: I say that [imath]\kappa[/imath] is the smallest cardinal such that the intersection of [imath]\kappa[/imath] many members of [imath]U[/imath] is not a member of [imath]U[/imath] and assume a sequence {[imath]A_{\alpha} : \alpha \in \kappa[/imath]} and [imath]B_{\beta}[/imath]= [imath]\bigcap[/imath] {[imath]A_{\alpha} : \alpha \lt \beta[/imath]}\ [imath]A_{\beta}[/imath]. Now I need to prove that there is an ultrafilter that contains [imath]B[/imath] and I am stuck there. | 134251 | Measurable cardinal existence condition
We know that a cardinal [imath]k > \omega[/imath] is measurable if there is a measure function [imath]\mu :2^k\mapsto \{0, 1\}[/imath] that satisfies the following 3 conditions: 1.[imath]<k[/imath] -additive: for every set I of indices with card(I) < k, and for every family of pairwise disjoint sets [imath]z_i[/imath], where [imath]i\in I[/imath], we have [imath]\displaystyle \mu(\bigcup_{i\in I} z_i)=\sum_{i\in I} \mu(z_i)[/imath]. 2.[imath]\mu(k)=1[/imath] 3.[imath]\mu (s)=0[/imath] for every singleton. Ok, now what if there exists a cardinal [imath]k>\omega[/imath], and a measure function [imath]\mu :2^k\mapsto \{0, 1\}[/imath] that satisfies conditions 2 and 3 above, but is only [imath] <\omega_1[/imath] -additive ? How can we show that this weaker condition still implies the existence of a measurable cardinal? Thanks a lot! |
2226730 | Given 3 line equations, find incenter(center of the inscribed circle) of the triangle formed
Let [imath]4x -y+2=0, x-4y-8=0, x+4y-8=0[/imath] make a triangle. Find the [imath](x,y)[/imath] coordinates of the triangle's incenter(center of the inscribed circle). I know that the intersection of the bisectors make the triangle's incenter. I have also tried to calculate each of the triangle's vertices, but I can't make it any further. | 1908347 | Can we find incentre of a triangle by using equation of lines?
My question is, in the manner which we find the orthocentre of a triangle merely by using it's side's line equations, can we also find the incentre? For example, consider, [imath]3x+4y-7=0[/imath] [imath]4x-3y+19=0[/imath] [imath]18x-6y+7=0[/imath] We can find the orthocentre of the triangle made by these lines using the fact that in a right angled triangle, the vertex containing the right angle is the orthocentre. So, for any triangle made by three straight lines, can we find the incentre as well? I will appreciate both short and long answers. Any thoughts are welcome. Thanks! EDIT :I will appreciate if someone gives me a method where I don't have to solve for the vertices. |
2221049 | Condition for a stochastic matrix to be a second order transition probability matrix of a DTMC.
Let [imath]A[/imath] be a [imath]2\times 2[/imath] stochastic matrix [imath]A=\left( \begin{array}{cc} 1-a & a \\ b & 1-b\\ \end{array}\right). [/imath] Then A is a second order transition probability matrix of a DTMC, i.e. there is a DTMC with probability matrix P such that [imath]P^{2}=A[/imath] iff [imath]a+b\leq 1[/imath]. Attempt. The desired property holds iff [imath]A=\left( \begin{array}{cc} 1-a & a \\ b & 1-b\\ \end{array}\right) = \left( \begin{array}{cc} 1-p & p \\ q & 1-q\\ \end{array}\right)^2 [/imath] for some [imath]p,~q\in [0,1][/imath], that is [imath]a=2p-pq-p^2,~b=2q-pq-q^2[/imath] for some [imath]p,~q\in [0,1][/imath]. I am not sure how this leads to the equivalent desired condition [imath]a+b\leq 1[/imath]. Thank you in advance. | 32000 | Characterization of two-step 2x2 stochastic matrices
Show that: A 2 x 2 stochastic matrix is two-step transition matrix of a Markov chain if and only if the sum of its principal diagonal terms is greater than or equal to [imath]1[/imath]. |
2228413 | Sum of nil left ideals
Let [imath]I,J[/imath] be left ideals in a ring [imath]A[/imath] such that [imath]\forall x\in I \ \ \forall y \in J[/imath] we have that [imath]x,y[/imath] are nilpotent. Is then true that every element of [imath]I+J[/imath] is nilpotent? Now, if the ideals are bilateral, expanding [imath](x+y)^{n_0}[/imath] and "playing" a little bit with the exponent [imath]n_0[/imath] we've proved that the answer is: yes. But I think a different approach is needed with the hypothesis that [imath]I,J[/imath] are left ideals. Any suggestion? | 1725829 | Sum of nil right ideals
Is an arbitrary sum of nil (nilpotent) right ideals a nil (nilpotent) right ideal? If [imath]I=\sum I_i[/imath] is a sum of nil ideals then each element [imath]x[/imath] of [imath]I[/imath] is a finite sum [imath]x=x_1+...+x_n[/imath] of elements [imath]x_k\in I_k[/imath] a power [imath]{x_k}^{p_k}[/imath]of each of which is zero. But, since each power of [imath]x[/imath] is a finite sum constituting of summands of the form [imath]∏{x_i}^{t_i} [/imath]( of course, in this product the [imath]x_i[/imath]'s may be repeated with different powers), I guess that one could choose the power of [imath]x[/imath] so "big" that it be annihilated. Is my guess true? How? Thanks! |
838309 | Evaluate [imath]\sum_{j=0}^{k} (-1)^{j} \binom{n}{j}[/imath]
I want to evaluate [imath]\sum_{j=0}^{k} (-1)^{j} \binom{n}{j}[/imath] obviously if [imath]k[/imath] goes up to [imath]n[/imath] then this is quite a common question. My question is how to deal with this question when the sum is truncated (stops at [imath]k \le n[/imath]? Do we use the answer for [imath]n=k[/imath] and the fact that there's some symmetry in binomial coefficients? (that is, [imath]\binom{n}{k} = \binom{n}{n-j}[/imath]?) any help hugely appreciated. Mark | 887960 | Truncated alternating binomial sum
It is easily checked that [imath]\displaystyle\sum_{i\ =\ 0}^{n}\left(\, -1\,\right)^{i} \binom{n}{i} = 0[/imath], for example by appealing to the binomial theorem. I'm trying to figure out what happens with the truncated sum [imath]\displaystyle\sum_{i\ =\ 0}^{D}\left(\, -1\,\right)^{i}\binom{n}{i}[/imath]. How far away from [imath]0[/imath] can this get, as a function of [imath]D[/imath] ?. I'm mostly interested in the case of when [imath]D \ll n[/imath], such as [imath]D \sim \,\sqrt{\,n\,}\,[/imath]. Thanks ! |
2229148 | Z[x] is not isomorphic to Q[x] as a rings
The problem (from D&F) is to prove that the ring [imath]\mathbb{Z}[x][/imath] is not isomorphic to the ring [imath]\mathbb{Q}[x][/imath].I can't come up to any idea, so I'm asking for a hint. | 382654 | Proving that [imath]\mathbb{Z}[x][/imath] and [imath]\mathbb{Q}[x][/imath] are not isomorphic.
I am trying to prove that [imath]\mathbb{Z}[x][/imath] and [imath]\mathbb{Q}[x][/imath] are not isomorphic. I was thinking of arguing the following: Suppose there exists an isomorphism [imath]\varphi: \mathbb{Z}[x]\rightarrow\mathbb{Q}[x][/imath]. Because isomorphisms are by definition surjective, there exist [imath]x, y \in\mathbb{Z}[x][/imath] such that [imath]\varphi(x) = c \in \mathbb{Q}[x][/imath] and [imath]\varphi(y) = d \in \mathbb{Q}[x][/imath] for any [imath]c, d\in\mathbb{Q}[x][/imath]. Because [imath]\varphi[/imath] is an isomorphism we must have [imath]\varphi(x+y) = \varphi(x) + \varphi(y)[/imath] for all [imath]x, y \in \mathbb{Z}[x][/imath]. Namely, because polynomial addition is defined componentwise, we must have that the constant term of [imath]\varphi(a + b) = c_{0} + d_{0}[/imath] (where [imath]c_{0}, d_{0}[/imath] are the constant terms of [imath]c[/imath] and [imath]d[/imath] respectively. I would then argue that because [imath]\mathbb{Z}[/imath] and [imath]\mathbb{Q}[/imath] are not isomorphic as additive groups, no such isomorphism [imath]\varphi[/imath] exists. Is this a valid proof? I've seen proofs that argue that because [imath]\varphi(1) = 1[/imath] for any homomorphism we have [imath]1 = \varphi(2(1/2)) = 2(\varphi(1/2))[/imath] so [imath]\varphi(1/2)[/imath] must be contained in [imath]\mathbb{Z}[x]^{\times}[/imath]. Then because [imath]\mathbb{Z}[x]^{\times} = \mathbb{Z}^{\times} = \{\pm1\}[/imath] we have [imath]2 \times \pm1 \neq1[/imath], a contradiction. Is this any different than arguing that [imath]\mathbb{Z}[x][/imath] and [imath]\mathbb{Q}[x][/imath] each have a different number of units? |
2229222 | Are these the only solutions to [imath]a!b!=c![/imath]?
Curious about this post on, [imath]6!7!=10!\tag1[/imath] a little Mathematica expt yielded, [imath]r!(r!-1)!=(r!)!\tag2[/imath] Q: Are the only non-trivial integer solutions to [imath]\color{blue}{a!b!=c!}[/imath] with [imath]a,b>1[/imath] given by [imath](1)[/imath] and [imath](2)[/imath]? | 150192 | Solutions of [imath]p!q! = r![/imath]
The title says it all, more or less. Obviously, there are infinitely many "trivial" integral solutions of the form [imath]p=n, q=(n!-1), r= n![/imath]. How many non-trivial solutions are there? I came across this about ten years ago; as far as I can tell, it hasn't appeared here before, so I thought that it might be of interest. I'm actually most interested in finding whether there was any progress made since Florian Luca's 2007 article. |
1452762 | Order of group element divides order of finite group
Proving this can be done as follows: consider a finite group G and elements [imath]g_i \in G[/imath] for some integer [imath]i[/imath]. Now consider [imath]\langle g_i \rangle = \{g_i^n: n\geq 0\}[/imath], a generator. It can be proved that [imath]\langle g_i \rangle \leq G[/imath] and that the order of [imath]g_i[/imath] is equal to the order of [imath]\langle g_i \rangle[/imath], so [imath]|\langle g_i \rangle| \leq |G|[/imath]. We can now use Lagrange's theorem which states that if [imath]H \leq G[/imath] then [imath]|H|[/imath] divides [imath]|G|[/imath] and we're done. But... Is there a simpler way to prove this fact? | 2559750 | Proof Help: Order of element divides order of group
I'm stuck in proving that the order of an element divides the order of the group. I've already proved Lagrange's theorem that for [imath]H<G, |H|||G|[/imath] And I know that for any [imath]a\epsilon G, |a|=|<a>|[/imath] The problem is that though I know this for a fact, I need to prove that the order of the element equals the order of the cyclic subgroup it generates but how do i prove this result to finally conclude that the order of a divides G using lagrange? Please help. |
2225788 | Let [imath]f\in L^1(\mathbb{R}^n)[/imath] and [imath]g\in L^p(\mathbb{R}^n)[/imath], then [imath]\lVert f*g\rVert_p\le\lVert f\rVert_1\lVert g\rVert_p[/imath].
Simply stated, I wish to show that convolution behaves nicely with respect to norm, i.e. that for [imath]f\in L^1(\mathbb{R}^n)[/imath] and [imath]g\in L^p(\mathbb{R}^n)[/imath], then [imath]\lVert f*g\rVert_p\le\lVert f\rVert_1\lVert g\rVert_p[/imath]. I can prove this for [imath]p=1[/imath], but when [imath]p>1[/imath], the power of [imath]p[/imath] seems to get in the way. So, how is this proved? If this isn't true, then I wish to prove the weaker statement that for a mollifier [imath]\phi\in C_0^\infty(\mathbb{R}^n)[/imath] such that [imath]\lVert\phi\rVert_1=1[/imath] and [imath]\phi\ge0[/imath], we have [imath]\lVert g*\phi\rVert_p\le\lVert g\rVert_p[/imath]. | 1733782 | Why is a convolution of an [imath]L^1[/imath] and an [imath]L^p[/imath] functions well-defined?
I was reading the wikipedia article about convolution and I found this one: If [imath]f \in L^1(\mathbb{R}^n)[/imath] and [imath]g\in L^p(\mathbb{R}^n)[/imath], then [imath]||f\ast g||_p \leq ||f||_1||g||_p[/imath]. But to [imath]f\ast g[/imath] be well-defined, [imath]y\mapsto f(x-y)g(y)[/imath] must be of [imath]L^1[/imath]. Right? But why? How do I prove this? |
1802233 | [imath]C_2 \times C_2[/imath] and [imath]C_4[/imath]
With the cyclic group [imath]C_n [/imath], why can't we say [imath]C_4[/imath] is isomorphic to [imath]C_2 \times C_2[/imath]? Is it because 2 and 2 are not coprime? Are there any other types i should watch out for? | 2684259 | How to solve the exercise: "Find out if the groups [imath](\mathbb{Z}_2\times\mathbb{Z}_2,+)[/imath] and [imath](\mathbb{Z}_4,+)[/imath] can or not form an isomorphism"?
I need to find out if the groups [imath](\mathbb{Z}_2\times\mathbb{Z}_2,+)[/imath] and [imath](\mathbb{Z}_4,+)[/imath] can or not form an isomorphism. I know that you need to find a bijective function that is also a morphism, but I do not have any idea how to show if there is or not any function. Can you help me? |
2228917 | Prove that [imath]f: [a,b]\to \mathbb R[/imath], [imath]|f(x)-f(y)| \leq 4|x-y|[/imath] is integrable, when [imath]f[/imath] reaches its maximum in any closed subinterval.
Let [imath]f:[a,b]\to \mathbb R[/imath] such that it reaches its maximum and minimum in any closed sub-interval of [imath][a,b][/imath]. In addition: [imath]|f(x)-f(y)| \leq 4|x-y|[/imath] holds for any [imath]x,y[/imath] in the interval. How would you prove that the function is integrable using the fact that Riemann's condition (i.e. [imath]U-L < \epsilon[/imath] for any epsilon)? | 1563771 | Prove Lipschitz function [imath]f[/imath] with constant [imath]K[/imath] is integrable on [imath][0, 1].[/imath]
We suppose that [imath]f : [0, 1] \rightarrow \mathbb{R}[/imath] is a Lipschitz function with constant [imath]K[/imath]. We want to show that [imath]f[/imath] is integrable on [imath][0, 1].[/imath] I've been trying to use the Darboux criterion of integrability by trying to show that, for [imath]\epsilon > 0[/imath], [imath]S^*(f,P) - S_*(f, P) < \epsilon,[/imath] where [imath]S^*(f, P)[/imath] and [imath]S_*(f, P)[/imath] are the Upper and Lower Riemann Sums, respectively. This is what I have so far: [imath]\sum_{i=1}^n M_i (x_i - x_{i-1})- \sum_{i=1}^n m_i (x_i - x_{i-1}) = \sum_{i=1}^n (M_i - m_i) (x_i - x_{i-1})[/imath] [imath]\le \sum_{i=1}^n K (x_i - x_{i-1}).[/imath] because [imath]|f(x) - f(u)| \le K|x-u| \le K |1|[/imath], since [imath]x\in [0, 1][/imath]. Then, [imath]\sum_{i=1}^n K (x_i - x_{i-1}) = K.[/imath] However, this is only a single upper bound, not an arbitrary [imath]\epsilon.[/imath] Can I get some pointers about how to proceed? Thanks! |
2229188 | What is the formula to solve a cubic equation?
I know that to solve [imath]ax^2+bx+c=0[/imath] you have to use the formula [imath]x= \frac{-b \pm \sqrt{(b^2-4ac)}}{2a}[/imath] What about more complex ones, like [imath]ax^3+bx^2+cx+d=0[/imath]? And what about representing it on a cordinate plane, [imath]2D[/imath] or [imath]3D[/imath]? | 2052616 | Is there anything like “cubic formula”?
Just like if we have any quadratic equation which has complex roots, then we are not able to factorize it easily. So we apply quadratic formula and get the roots. Similarly if we have a cubic equation which has two complex roots (which we know conjugate of each other) and one fractional root, then we are not able to find its first root by hit & trial. So my question is like quadratic formula, is there exist any thing like cubic formula which help in solving cubic equations? For example, I have an equation [imath]2x^3+9x^2+9x-7=0\tag{1}[/imath] and I have to find its solution which I am not able to find because it has no integral solution. Its solutions are [imath]\dfrac {1}{2}[/imath], [imath]\dfrac{-5\pm \sqrt{3}i}{2} [/imath], I know these solutions because this equation is generated by myself. So how can I solve equations like these? Also while typing this question, I thought about the derivation of quadratic formula, which is derived by completing the square method. So I tried to apply ‘completing the cube’ method on the general equation [imath]ax^3+bx^2+cx+d=0[/imath] but it didn't help. So please help me in finding a cubic formula or to solve the equations like given in example by an alternative method. |
843518 | Solving a cubic polynomial equation.
Overview I have tried finding a solution to this problem myself and I have flailed. Its just a challenge for me. could you please tell me how far am I in solving this question? My approach for quadratic equation : I am still a student, and when I first saw the quadratic formula I wanted to derive it myself and I did it. This was my logic: the equation [imath]x^2 + bx + c=0[/imath] is obviously supposed to contain 2 roots (real or imaginary). So any 2 points in an axes can be represented as: [imath]n+k[/imath] or [imath]n-k[/imath]. so (1) : putting [imath]x = n+k[/imath], (2): putting [imath]x=n-k[/imath] we get two equations; on subtracting [imath](1) - (2)[/imath] we get an equation in [imath]n[/imath] and [imath]k[/imath] as, [imath]4nk+2kb=0[/imath] which gives [imath]n=-b/2[/imath]. Now substituting this value of [imath]n[/imath] in [imath](1)[/imath] and [imath](2)[/imath] and adding them we get [imath]k= (+or-) \sqrt{b^2 -4c}/2 [/imath]. My approaches for trying to solve cubic equations: Let [imath]x^3+bx^2+cx+d = 0[/imath] be the equation. My first trials were focused on extracting the roots by various substations. Let me explain some of them. Originally, when I derived the quadratic equation, I wanted to literary "represent the solution" by [imath]n+k[/imath] and [imath]n-k[/imath]. most methods that I have thought of for cubic equation similar to that of the quadratic (like assigning 3 unambiguous variables to uniquely represent 3 roots- [imath]n[/imath] and [imath]k[/imath] represents unique values), by considering 3 unique values, but they end up being ambiguous. Then I was suddenly struck with the idea that I could define certain random operators to "transform" the equation in some way changing the co-coefficients of [imath]x^2, x, x^0[/imath]. I have defined the first transformation as: [imath]x^3+bx^2+cx+d=0 \to x^3+ b'x^2 + c'x+d' =0 [/imath] then [imath]b' = b[/imath], [imath]c' =b^2+c[/imath], [imath]d' = bc-d[/imath]. Applying this operation changes the modified equation's zeros as follows: [imath]\alpha'=(\alpha+\beta)/2, \beta' = (\beta+\gamma)/2, \gamma' = (\alpha+\gamma)/2 [/imath] if the initial zeros are [imath]\alpha, \beta, \gamma[/imath]. My latest assumption is that creating [imath]n[/imath] number of transformations and applying them can help us obtain the roots in some combination. The other operation I have defined is "increment roots by by [imath]\theta[/imath] operation". in this case, [imath]x^3+ bx^2+ cx+ d \to x^3+b'x^2+ c'x+d=0[/imath]. here, [imath]b' = b + 3\theta, c' = 3\theta^2+ 2\theta b+ c, d' = \theta^3 +b\theta^2 + c\theta + d[/imath]. This modification transforms the original roots [imath]\alpha, \beta, \gamma[/imath] to [imath]\alpha' = \alpha + \theta, \beta' = \beta + \theta, \gamma' = \gamma + \theta[/imath] Now for my questions: How far am I from solving the cubic polynomial problem myself without any external help? More over, I give up. How to find the roots of a cubic equation? Could my latest idea for solving cubic equations actually be applied (the transformation method)? Is it possible to find the roots with my transformation method? (my second transformation definition was successful in converting the coefficient b or c to zero by transforming the roots linearly) | 2631239 | How to derive cubic equation?
Using completing the square method we can derive a formula [imath]ax^2+bx+c=0[/imath] for the quadratic equation to [imath]x={-b\pm \sqrt{b^2-4ac}\over 2a}[/imath] I am sure I haven't sees it in any book talking about deriving the cubic equation [imath]ax^3+bx^2+cx+d=0[/imath] using the method of completing the square or the cube. So how did they derive the cubic formula to solve the cubic equation? |
1893836 | Can't understand the limit solution to [imath]\lim_{n\to\infty}\sqrt[n]{2^n-n^2}[/imath]
I trying to solve the question: [imath]\lim_{n\to\infty}\sqrt[n]{2^n-n^2}[/imath] I know that i can proof [imath]2^n>n^2[/imath] in induction and i calculated the limit as "[imath]∞^0[/imath]" that equals to [imath]1[/imath] but then i saw that the answer should be 2. How is it 2? | 1515063 | Calculating [imath]\lim_{n\to\infty}\sqrt[n]{2^n-n^2}[/imath]
I'm having quite hard time proving the following statement: [imath]\lim_{n\to\infty}\sqrt[n]{2^n-n^2}=2[/imath] I though about using the squeeze theorem. So I started to look for upper and lower expressions and quickly found an upper limit to the expression, [imath]\sqrt[n]{2^n-n^2}<\sqrt[n]{2^n}=2[/imath] However, I found it very difficult to find the lower limit. Help would be appreciated!. |
2229672 | Quadratic with integer roots and coefficients in arithmetic progression
Question : Suppose the quadratic polynomial [imath]p(x)=ax^2 + bx + c~[/imath] has positive coefficients [imath]a, b, c[/imath] in arithmetic progression in that order.If [imath]p(x)=0[/imath] has integer roots [imath]\alpha[/imath] and [imath]\beta[/imath], then what is the value of [imath]\alpha + \beta + \alpha \beta[/imath] ? My attempt: I am not aware of any inequality of some sort if it exists, but I could not get far by trying substitutions over these unknown variables. | 2068329 | A.P. terms in a Quadratic equation.
The terms [imath]a,b,c[/imath] of quadratic equation [imath]ax^{2}+bx+c=0[/imath] are in A.P. and positive. Let this equation have integral root [imath]\alpha,\ \beta[/imath]. Then find the value of [imath]\alpha+ \beta + \alpha \cdot \beta[/imath] ? please point where I'm wrong: Let common difference be [imath]d[/imath] [imath]\implies \alpha+ \beta + \alpha \cdot \beta=\dfrac{c-b}{a}=\dfrac{d}{a} \implies a|d \ \ \ \ \longrightarrow \ \ \ \ \ \because (b=a+d[/imath], [imath]c=a+2d)[/imath] Also , [imath]ax^{2}+(a+d)x+(a+2d)=0[/imath]. [imath]\implies[/imath] [imath]\alpha,\ \beta=\dfrac{-(a+d) \pm \sqrt{(a+d)^{2}-4\cdot a \cdot (a+2d)}}{2a}[/imath]. For this to be integer [imath]\sqrt{(a+d)^{2}-4\cdot a \cdot (a+2d)}[/imath] must be perfect square. [imath]\implies[/imath] [imath]{(a+d)^{2}-4\cdot a \cdot (a+2d)}=p^{2}[/imath] for some [imath]p[/imath]. [imath]\implies -3a^{2}+d^{2}-6ad=p^{2}[/imath] [imath]\implies -3a^{2}+a^{2}q^{2}-6a^{2}q=p^{2}[/imath] [imath]\because[/imath] [imath]a|d \implies aq=d[/imath] for some [imath]q[/imath]. [imath]\implies a^{2}(-3+q^{2}-6q)=p^{2}[/imath] [imath]\implies -3+q^{2}-6q\ [/imath] has to be perfect square. By trial [imath]q=7[/imath] But I need to get this without trial, please help. |
878752 | Is the cartesian product of two polytopes again a polytope?
Given two polytopes [imath]P_1, P_2[/imath] then define the graph on [imath]P_1 \times P_2[/imath] as follows: [imath][(a,b),(x,y)][/imath] is an edge in [imath]P_1 \times P_2[/imath] iff [imath]a=x[/imath] and [imath][b,y][/imath] is an edge in [imath]P_2[/imath] or [imath]b=y[/imath] and [imath][a,x][/imath] is an edge in [imath]P_1[/imath]. Is this new structure a polytope? | 2150214 | What is the Cartesian product of [imath]n[/imath] polygons?
How do we calculate the Cartesian product between [imath]n[/imath] polygons? Is it always a polytope? If so, can we say anything about its faces? For example, [imath]n[/imath] squares: [imath]\{S_1:[a_1, a_2] \times [b_1, b_2], \ldots, S_n:[a_k, a_{k+1}] \times [b_{k}, b_{k+1}]\}[/imath] so, [imath]S_1 \times \ldots \times S_n \in \mathcal{R}^n[/imath] forms a [imath]n-[/imath]dimensional box. |
2230199 | Derived distribution: PDF of [imath]\sin^{2}(x)[/imath]
I want to find the PDF of [imath]Y = \sin^2(X)[/imath], where [imath]X ∼ \text{Exponential}(\lambda)[/imath] is a random variable. I thought i had solved this correctly by simply finding the inverse [imath]h(Y) = X[/imath] and taking the derivative via the chain rule. However, I was told that the solution of this problem would involve an infinite sum, but my solution did not, so I assume it was wrong. I can see why there would be an infinite sum due to the adding up of all the sine intervals, but I could not seem to figure out how to find an expression for each of these intervals that I can then sum up. Any help would be greatly appreciated, thank you very much in advance! | 2230176 | PDF of Y=sin^2(X)
Suppose X ∼ Exponential(λ). Find the PDF of [imath]Y = sin^2(X)[/imath] Note: While the solution involves an infinite sum, it is an example of a geometric series and so can be evaluated. My answer so far: I was able to go as far as finding the bounds, and now I am stuck on the integral itself. As in the directions, we are told to use a geometric series, but I don't know how to proceed from here. The integral which will give me the CDF I also have the piece [imath]P(0 <= x <= arcsin(\sqrt t))[/imath] I know this does not included in the integral, but I don't know how to incorporate it into my CDF. [imath]\int_a^b \lambda\ e^-\lambda x\ \,dx,[/imath] [imath] a = k \pi\ - arcsin(\sqrt t)[/imath] [imath] b = k \pi\ + arcsin(\sqrt t)[/imath] |
2230478 | [imath]\phi: M_2(\mathbb{Z})\to \mathbb{Z}[/imath] is necessarily trivial
Let [imath]\varphi: M_2(\mathbb{Z})\to \mathbb{Z}[/imath] be a ring homomorphism. Then [imath]\varphi[/imath] is necessarily trivial. My reasoning: Suppose not. Then [imath]\varphi(I)=1\implies \forall A\in M_2(\mathbb{Z}),\exists A^{-1}\in M_2(\mathbb{Z})[/imath] such that [imath]\varphi(I)=\phi(A)\phi(A^{-1})=\phi(A)\phi(A)^{-1}=aa^{-1}[/imath]. However, only [imath]1[/imath] and [imath]-1[/imath] have multiplicative inverses in [imath]\mathbb{Z}[/imath]. Hence, if [imath]A\ne I[/imath], then [imath]\phi(A)^{-1}[/imath] may not exist. This implies that [imath]\phi[/imath] must be trivial (that is, map everything to [imath]1[/imath]). [For if [imath]\phi(A)=0[/imath] for all [imath]A\ne I[/imath], then [imath]\phi(A)^{-1}[/imath] admits division by [imath]0[/imath]]. Do you think this is correct? | 1334276 | Is there a ring homomorphism [imath]M_2(\mathbb Z)\to \mathbb Z[/imath]?
I have the following problem: Is it possible to construct ring homomorphism from [imath]M_2(\mathbb Z)\to \mathbb Z[/imath], or in other words, a homomorphism from ring of all [imath]2\times2[/imath] matrices over the integers into integers? I tried determinant, trace and mapping that maps matrix to it's element in the position (1,1) but non of that obviously works, which led me to believe there might not be such homomorphism. |
2229616 | Prove that either [imath]a=2l[/imath] and [imath]b=m[/imath] or [imath]b+m=al[/imath]
If by eliminating [imath]x[/imath] between the equations [imath]x^2+ax+b=0[/imath] and [imath]xy+l(x+y)+m=0[/imath] then a quadratic in [imath]y[/imath] is formed whose roots are the same as the original quadratic in [imath]x[/imath]. Then prove that either [imath]a=2l[/imath] and [imath]b=m[/imath] or [imath]b+m=al[/imath]. It seems a very lengthy process to eliminate [imath]x[/imath]. Does anyone know of any shorter method? | 1876276 | Method to eliminate [imath]x[/imath] between the equation [imath]x^2 + ax + b = 0[/imath] and [imath]xy+ l(x + y) + m = 0[/imath]
If by eliminating [imath]x[/imath] between the equation [imath]x^2 + ax + b = 0[/imath] and [imath]xy+ l(x + y) + m = 0[/imath], a quadratic in [imath]y[/imath] is formed whose roots are the same as those of the original quadratic in [imath]x[/imath]. Then prove either [imath]a = 2l[/imath] & [imath]b = m[/imath] or [imath]b + m = al[/imath]. Initially, I thought of eliminating [imath]x[/imath] but that method seems to be very lengthy. Is there any trick that can help to solve the problem faster? |
2230826 | If [imath]n[/imath] is not prime, can [imath]X^{n-1}+\dotsb+X+1[/imath] be irreducible?
If [imath]n[/imath] is not prime, can [imath]X^{n-1}+\dotsb+X+1[/imath] be irreducible over [imath]\mathbb Q[/imath] ? If [imath]n[/imath] is even, there is always [imath]-1[/imath] as solution, so it's necessarily reducible, but what happen iff [imath]n[/imath] is odd and not prime ? | 1460834 | Show that [imath]x^{n-1}+\cdots +x+1[/imath] is irreducible over [imath]\mathbb Z[/imath] if and only if [imath]n[/imath] is a prime.
I proved that if [imath]n[/imath] is a prime, then [imath]p(x)=x^{n-1}+\cdots+x+1[/imath] is irreducible over [imath]\mathbb Z[/imath]. But, I don't know how to prove that if [imath]p(x)[/imath] is irreducible over [imath]\mathbb Z[/imath], then [imath]n[/imath] is prime. Can you give me a hints? |
2232096 | To find an example for which [imath]closure(\cup A_\alpha)\ne\cup closure(A_\alpha)[/imath]
Give an example where closure of [imath](\cup A_\alpha)\ne \cup closure (A_\alpha)[/imath] provided [imath]A_\alpha[/imath] are subsets of a topological space. If I consider [imath]R[/imath] to be a topological space with standard topology and [imath]A_n=(n,n+1)[/imath] where [imath]n\in N[/imath] then will the equality hold? | 1851655 | Closure of Union contains Union of Closures
I'm teaching my self topology using a book I found. This is the second part of a 4 part question. First part is here. I'm trying to prove the following problem from a book I found: Let [imath]X[/imath] be a topological space and let [imath]\mathscr{A}[/imath] be a collection of subset of [imath]X[/imath]. Prove [imath]\overline{ \bigcup \limits_{A \in \mathscr{A}} A}\supseteq \bigcup \limits_{A \in \mathscr{A}} \overline{A}[/imath] *where the over line indicates closure. First I'm going to show an example the them being equal, and then show an example of a proper superset. I know the examples are not needed for the proof, but I find them insightful, and these posts are like a notebook that I can refer back to later. Then I will attempt to prove this. (I) Example of [imath]\overline{ \bigcup \limits_{A \in \mathscr{A}} A} = \bigcup \limits_{A \in \mathscr{A}} \overline{A}[/imath] I find that with "most" collections of subsets, the two sides come up equal. Let [imath]X[/imath] be [imath]\mathbb{R}[/imath] with the standard Euclidean metric. Also let, [imath]\mathscr{A}=\{(1,6),(3,10)\}[/imath]. So then [imath] \bigcup \limits_{A \in \mathscr{A}} A= (1,10)[/imath], and [imath]\overline{ \bigcup \limits_{A \in \mathscr{A}} A} =[1,10][/imath]. Next the right side would be: [imath]\bigcup \limits_{A \in \mathscr{A}} \overline{A}= [1,6] \cup [3,10]=[1,10][/imath]. So it is possible for the two sides to be equal. (II) Example of [imath]\overline{ \bigcup \limits_{A \in \mathscr{A}} A} \supset \bigcup \limits_{A \in \mathscr{A}} \overline{A}[/imath] The only way I know of involves a collection of subsets that "expands" in a way that converges - ie [imath]\bigcup_{n=1}^{\infty} \overline {B_{\frac{R \cdot n}{n+1}}}(x)=B_{R}(x)[/imath]. Let [imath]X[/imath] be [imath]\mathbb{R}[/imath] with the standard Euclidean metric. Also let, [imath]\mathscr{A}=(1,\frac{10n}{n+1}), n\in\mathbb N[/imath]. So then [imath] \bigcup \limits_{A \in \mathscr{A}} A= (1,10)[/imath], and [imath]\overline{ \bigcup \limits_{A \in \mathscr{A}} A} =[1,10][/imath]. Next the right side would be: [imath]\bigcup \limits_{A \in \mathscr{A}} \overline{A}= \lim \limits_{n \rightarrow \infty} [1, \frac{10n}{n+1}]= [1,10)[/imath] . And [imath][1,10] \supset [1,10)[/imath]. So it is possible for Closure of Unions to contain Union of Closures. Proof: [imath]\overline{ \bigcup \limits_{A \in \mathscr{A}} A}\supseteq \bigcup \limits_{A \in \mathscr{A}} \overline{A}[/imath] (II) illustrates that the infinite union of closed sets can make an open (or half open) set. So: * [imath]\bigcup \limits_{A \in \mathscr{A}} \overline{A}[/imath] can be (half)open. * [imath]\overline{ \bigcup \limits_{A \in \mathscr{A}} A}[/imath] must be closed. Since we're dealing with the same collection of subsets, then the difference will be at the boundary points. And thus, the closed set will contain the (half)open set. QED The part of the proof i'm not sure about is at the end where I say, "the difference will be at the boundary points." In (II) that is the case, but will that always be the case when dealing with the most general topological spaces? |
2222270 | entire function with constant modulus on circle -Schwarz reflection principle
Let [imath]f(z)[/imath] be an entire function whose modulus is constant on some circle. Show that [imath]f(z)=f(z_0) + c(z-z_0)^n[/imath] for some [imath]n \geqslant 0[/imath] and some constant [imath]c[/imath], where [imath]z_0[/imath] is the center of the circle. I only know a sketch of the proof, using Schwarz reflection Principle. I would seriously appreciate giving details to it. | 626488 | Find all entire functions [imath]f(z)[/imath] such that [imath]|f(z)|=1[/imath] for [imath]|z|=1[/imath]
Find all entire functions [imath]f(z)[/imath] such that [imath]|f(z)|=1[/imath] for [imath]|z|=1[/imath] Hint: First show that [imath]f(z)[/imath] is a polynomial. Clearly one can not use Cauchy Estimates to prove that [imath]f(z)[/imath] is a polynomial, the other way is to prove that [imath]f(z)[/imath] has a pole at infinity which I am not sure how to prove that, are there any other ways to prove that a given entire function is a polynomial ? Thank you ! |
2231031 | What do the eigenvectors of the Hessian matrix actually represent?
It is known in multivariate calculus that, at a critical point [imath]p_c = (x_{1c}, x_{2c}, ... , x_{nc})[/imath] of the function [imath]f(x_1, x_2, ... , x_n)[/imath], if the Hessian is positive definite we have a local minimum and if it is negative definite we have a local maximum. However, I'm curious about the eigenvectors of the Hessian. Do they represent anything useful? For instance, do they represent an ellipsoid inside of which [imath]f(p_c)[/imath] is the global maximum/minimum? | 1519353 | Interpretation of eigenvectors of Hessian in context of local min/max/saddle?
Say [imath]f \in C^2[/imath] so we can possibly use its Hessian [imath]H[/imath] to determine whether [imath]f[/imath] has a local max, min, or saddle at a critical point [imath]x_0[/imath]. Since [imath]H(x_0)[/imath] is real and symmetric, it is diagonalizable, say with eigenvector-eigenvalue pairs [imath](v_1,\lambda_1),\ldots,(v_n,\lambda_n)[/imath]. The second derivative test asserts that if all the [imath]\lambda_i[/imath] are strictly positive, then [imath]f[/imath] has a local min, if they are all strictly negative, then [imath]f[/imath] has a local max, and if there are at least one strictly positive and one strictly negative, then [imath]f[/imath] has a saddle point. Is there some geometric interpretation to what the [imath]v_i[/imath] are? Are the [imath]v_i[/imath] somehow directions in which the function restricted to that direction has concavity [imath]\lambda_i[/imath]? |
2232434 | Find the number of distinct arrangements of 5 unequal positive integers such that their sum is 20
Let the integers be [imath]n_1, n_2, n_3, n_4[/imath] and [imath]n_5[/imath]. It's given in the question that [imath]n_1 + n_2 + n_3 + n_4 + n_5 = 20[/imath]. I thought of taking [imath]n_2[/imath] as [imath]n_1 + a[/imath], [imath]n_3[/imath] as [imath]n_1 + a + b[/imath] and so on... where a, b,... are not equal to 0. So I got this expression: [imath]5n_1 + 4a + 3b + 2c + d = 20[/imath] After this, I'm not able to continue. How do I proceed? Thanks in advance. | 2179424 | Number of distinct arrangements {[imath]n_i[/imath]} [imath]n_1 such that \sum n_i=20[/imath]
Let [imath]n_1<n_2<n_3<n_4<n_5[/imath] be positive integers such that [imath]n_1+n_2+n_3+n_4+n_5=20[/imath].Then what is the number of such distinct arrangements [imath](n_1,n_2,n_3,n_4,n_5)[/imath]? My Approach : I assumed [imath]n_1=t_0+1[/imath] [imath]n_2=n_1+t_1+1[/imath] [imath]n_3=n_2+t_2+1[/imath] [imath]n_4=n_3+t_3+1[/imath] [imath]n_5=n_4+t_4+1[/imath] Where [imath]t_0,t_1,t_2,t_3,t_4 \ge 0[/imath] Now the sum becomes : [imath]5t_0+4t_1+3t_2+2t_3+t_4=5[/imath] After this, I put the values of [imath]t_i[/imath]s [imath]0,1,..[/imath] and so on, and therefore found [imath]7[/imath] Solutions. My question : Is there another way to solve this question, because as this question was asked in a competitive exam (JEE Advanced), This a very long solution. |
2232752 | How to prove the set of functions from the natural numbers to the natural numbers is equinumerous to the set of real numbers
I know how to prove that $^\omega[imath]\omega$ is uncountable, but how can we prove that it is, in fact, equinumerous to $\Bbb R$? Or at least that $^\omega[/imath]\omega$ is equinumerous to [imath]^\omega[/imath][imath]2[/imath]? | 2146599 | Encode each [imath]n_1,n_2,n_3,...∈N^N[/imath] by an infinite sequence of 0s and 1s with infinitely many 0s, and give a proof that [imath]N^N[/imath] is equinumerous with [imath]R[/imath].
Encode each [imath]n_1,n_2,n_3,...∈N^N[/imath] by an infinite sequence of 0s and 1s with infinitely many 0s, and hence give a proof that [imath]N^N[/imath] is equinumerous with [imath]R[/imath]. Background: Here the set [imath]N^N=N \times N \times N \times N \times...[/imath] is the set of finite sequences of positive integers. Each infinite sequence [imath]n_1,n_2,n_3,...[/imath] of positive integers gives an irrational number (with a infinite continued fraction) between 0 and 1, because each rational has a finite continued fraction. Conversely, each irrational number between 0 and 1 has a continued fraction of the above form, and hence gives an infinite sequence of positive integers. Thus, we immediately have a bijection between [imath]N^N[/imath] and the irrational numbers in (0,1). The latter set is equinumerous with (0,1), and hence with [imath]R[/imath]. So far I tried to encode each n in N^N as either ones and zeros, where comas are represented as zeros. For example <3,2,3,0,1> ---> 1110110111001 TIA |
2232117 | Is the answer 2 or 4?
This is my first post on this forum so I apologise for not knowing how to ask this question in the desired format however I will do my best to make it clear. First consider: [imath] y = \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{...}}}} [/imath] Now, we can rewrite this as simply: [imath] y = (\sqrt{2})^y [/imath] Then further rearrange this to: [imath] y^{1/y} = \sqrt{2} [/imath] Thus leaving y to equal both [imath]2[/imath] and [imath]4[/imath], which obviously can't be correct. However, which is correct and why? Thanks! | 1445547 | Evaluate the expression [imath]\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdots}}}[/imath]
Is there a way to check whether the expression below converges to a specific number. [imath] \sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdots}}} [/imath] or in other words does the sequence defined by [imath]x_0=\sqrt{2}[/imath] and [imath]x_n=\sqrt{2}^{x_{n-1}}[/imath] converges? Trying with a calculator to evaluate an eight-length [imath]\sqrt{2}[/imath] construct I got [imath]1.9656648865173187[/imath] but I couldn't yet confirm convergence. |
2232739 | collection of all non invertible elements in the ring is the maximal ideal
Let [imath]R[/imath] be a ring with unit and let [imath]M(R)[/imath] be a collection of all non invertible elements in [imath]R[/imath]. If [imath]M(R)[/imath] is a ideal in [imath]R[/imath], prove that this is the only and maximal ideal My thoughts: suppose there is a bigger ideal choose some invertible element in the bigger ideal so the bigger ideal is the whole ring because there are two elements such that [imath]u\cdot v=1[/imath]. | 1667100 | Showing [imath]R[/imath] is a local ring if and only if all elements of [imath]R[/imath] that are not units form an ideal
My question is how would I go about proving this? Prove that [imath]R[/imath] is a local ring if and only if all elements of R that are not units form an ideal. I understand that I need to prove both directions so: [imath](\Rightarrow)[/imath] Local ring means has a unique maximal ideal, so I want to show that this implies the elements are not units. [imath](\Leftarrow)[/imath] non unit elements of [imath]R[/imath] form an ideal, so if I show this is unique maximal ideal I can then conclude local ring? Any hints would be appreciated. |
2232832 | Very interesting method for multiplication of natural numbers
In some book I saw this example on how to multiply numbers [imath]83[/imath] and [imath]157[/imath]. [imath]83 \qquad 157[/imath] [imath]41 \qquad 314[/imath] [imath]20 \qquad 628[/imath] [imath]10 \qquad 1256[/imath] [imath]5 \qquad 2512[/imath] [imath]2 \qquad 5024[/imath] [imath]1 \qquad 10048[/imath] The procedure is that in the left column we divide numbers by two and write down the result under the number we divided (if the number is odd then we take integer which is closest to the result and less than the result) and we repeat the procedure until we arrive at the number one. In the right column we just take the next number to be twice as big as the previous one. Now we look at the two columns and discard those numbers in the right column if they correspond to the even number in the left column, so we discard [imath]628[/imath] and [imath]1256[/imath] and [imath]5024[/imath]. If we now add the remaining numbers in the right column we obtain [imath]157+314+2512+10048=13031[/imath] but [imath]13031=83 \cdot 157[/imath]. So my questions would be: 1) Does this method works for every two natural numbers [imath]m,n[/imath]? 2) Can someone explain why this method works? | 2099095 | Proof of Russian Peasant Multiplication
I'm studying a combinatorics text (Cameron) in preparation for taking discrete math this upcoming semester. I've taken a fair amount of math in college, through Calculus II, but this is my first 500+ level math course. In chapter 2 of the Cameron book, the author introduces an algorithm for multiplication called Russian Peasant Multiplication. I want to prove that this algorithm will always result in the product of two numbers. I've been chewing on the proof for a while now, and just can't seem to get any traction. Thanks in advance for any help. Description of Algorithm I'll try to explain the algorithm clearly. The goal of the algorithm is to find the product of two numbers [imath]m[/imath] and [imath]n[/imath]. The below table is filled out for [imath]m[/imath] and [imath]n[/imath]: [imath]\begin{array}{c} m&&n&\\ \hline \bigl\lfloor \frac{m}{2} \bigr\rfloor&& 2n \\ \bigl\lfloor \frac{m}{4} \bigr\rfloor&& 4n \\ \bigl\lfloor \frac{m}{8} \bigr\rfloor&& 8n \\ \bigl\lfloor \frac{m}{16} \bigr\rfloor&& 16n \\ \bigl\lfloor \frac{m}{32} \bigr\rfloor&& 32n \\ \end{array}[/imath] The sequence of numbers in both columns is the sequence [imath]a_n = 2a_{n-1}\text{ where }a_1=2[/imath]. The columns are extended down until the value in the left column is equal to the number one. After filling out the table, the product is found by adding the values of the right column where the left column is an odd number. For example, to multiply 18 and 37 with the algorithm: [imath]\begin{array}{c} m&&n&\\ \hline 18 && 37 \\ 9 && 74 \\ 4 && 148 \\ 2 && 296 \\ 1 && 592 \\ \end{array}[/imath] To find the product, cross out values of the left column where the value of the right column is even: [imath]\begin{array}{c} m&&n&\\ \hline \require{enclose} \enclose{horizontalstrike}{18} && \enclose{horizontalstrike}{37} \\ 9 && 74 \\ \enclose{horizontalstrike}{4} && \enclose{horizontalstrike}{148} \\ \enclose{horizontalstrike}{2} && \enclose{horizontalstrike}{296} \\ 1 && 592 \\ \end{array}[/imath] And add the remaining numbers in the right column to get the result: [imath]18\times37=74+592=666[/imath] Alternate Description of Algorithm As an aside, here is a shorter way to describe the algorithm, repeat the below steps until the left column == 1: • Halve the last number in the first column • Write the result below (discard the remainder) • Double the last number in the first column. • Write the result below. Now, for every even number in the first column, delete the adjacent number in the second column. Add the remaining numbers in the second column. What I've done so Far I haven't been able to get much of anywhere on the proof. One thing I did was create a bit of a different way of arriving at the same outcome: [imath]\begin{array}{c} m&&n&&\text{index}&&\text{remainder}&&\text{if remainder == 1, }n\times2^{Index}&\\ \hline 18 && 37 && 0 && 0 \\ 9 && 74 && 1 && 1 && 74\\ 4 && 148 && 2 && 0 \\ 2 && 296 && 3 && 0 \\ 1 && 592 && 4 && 1 && 592\\ \end{array}[/imath] Which brings to light the relationship between this algorithm and binary numbers (the fourth column in the above table is the product in binary, read bottom to top). I also wrote some Python functions to help me explore the problem, Github repo. |
2231831 | Can [imath]k\mathbb{Z}[/imath] and [imath]m\mathbb{Z}[/imath] ever be isomorphic as rings?
Can [imath]k\mathbb{Z}[/imath] and [imath]m\mathbb{Z}[/imath] , [imath]k, m \in\mathbb{Z}[/imath],ever be isomorphic as rings ? This question came upon me when I'm doing/proving that [imath]2\mathbb{Z}[/imath] and [imath]3\mathbb{Z}[/imath] are not isomorphic as rings, but the proof is rather ad-hoc. Hence this leads me to wonder does there exist a [imath]m, k,m\neq k[/imath] such that [imath]k\mathbb{Z}[/imath] and [imath]m\mathbb{Z}[/imath] is isomorphic rings? Thank you in advance. :) | 1835336 | Is the ring [imath]m\mathbb{Z}[/imath] isomorphic to the ring [imath]n\mathbb{Z}[/imath]?
I came over a question in ring theory which I am not being able to proceed upon: When is the ring [imath]m\mathbb{Z}[/imath] isomorphic to the ring [imath]n\mathbb{Z}[/imath], where [imath]m, n \in \mathbb{N}[/imath]? I know that to show isomorphism, I need to show that it is onto and the kernel consists of only [imath]\{0\}_{m\mathbb{Z}}[/imath] but I am not being able to write things down properly. Will someone help? What is the relation between [imath]m[/imath] and [imath]n[/imath]? |
2232668 | [imath]\text {let}\ p(z)=a_0+a_1z+\dotsb+a_{n-1}z^{n-1}+z^n,a_i\in\mathbb{C}.[/imath] If [imath]|p(z)|=1 \forall |z|=1[/imath] is it true that [imath]p(z)=z^n[/imath]?
[imath]\text {let}\ p(z)=a_0+a_1z+......+a_{n-1}z^{n-1}+z^n,a_i\in\mathbb{C}.[/imath] if [imath]|p(z)|=1 \forall |z|=1[/imath] is it true that [imath]p(z)=z^n[/imath] for n=1,this is true.but is it also for higher value of n ??? | 177177 | Complex polynomial and the unit circle
Given a polynomial [imath] P(z) = z^n + a_{n-1}z^{n-1} + \cdots + a_0 [/imath], such that [imath]\max_{|z|=1} |P(z)| = 1 [/imath] Prove: [imath] P(z) = z^n [/imath] Hint: Use cauchy derivative estimation [imath] |f^{(n)} (z_0)| \leq \frac{n!}{r^n} \max_{|z-z_0|\leq r} |f(z)| [/imath] and look at the function [imath] \frac{P(z)}{z^n} [/imath] It seems to be maximum principle related, but I can't see how to use it and I can't understand how to use the hint. |
2233310 | Property of [imath]T^2=I_V[/imath]
Let [imath]V[/imath] be a vector space over [imath]\mathbb{C}[/imath] and [imath]T:V\to V[/imath] a linear transformation such that [imath]T^2 = T\circ T = I_V[/imath]. Define: [imath]V_1=\{v\in V\mid T(v)=v\}[/imath] and [imath]V_2=\{v\in V\mid T(v)=-v\}[/imath] Prove that [imath]V=V_1\oplus V_2[/imath]. Disjoint union is easy to show. However, how can I show every [imath]v[/imath] = [imath]v_1 + (v-v_1)[/imath], [imath]v-v_1[/imath] is in [imath]V_2[/imath], where [imath]v_1\in V_1[/imath]? | 1742136 | Prove that [imath]V = V_1 \oplus V_2[/imath] in the following
Let [imath]V[/imath] be a vector space over [imath]\mathbb{C}[/imath] such that [imath]T^2=1_V[/imath]. Define [imath]V_1 = \left\{v\in V |\ T(v) = v \right\}[/imath], [imath]\ V_2 = \left\{v\in V |\ T(v) = -v\right\}[/imath] prove that [imath]V = V_1 \oplus V_2[/imath] For start, I've shown that [imath]V_1\ ,V2[/imath] are subspace of [imath]V[/imath]. However, I'm stuck trying to show that [imath]\forall \ v\in V,\ v = v_1 + v_2[/imath] where [imath]v_1 \in V1, v_2 \in V_2[/imath]. I know at some point I have to make use of [imath]T^2=1_V[/imath], but cant figure out how. Any help or insight is deeply appreciated. |
2233324 | Noob question: Why do trignometric functions repeat on a graph?
When a sine wave (or a cosine wave) is graphed, it repeats itself at a fixed period. However, the value [imath]\sin(\frac{\pi}{2})[/imath] is geometrically undefined because you can't have a triangle with two right angles. The same argument is true for all obtuse angles. So why do we still graph these and use these values as valid inputs to trignometric functions? | 794201 | Why does [imath]\sin(0)[/imath] exist?
I can't understand why should [imath]\sin(0)[/imath] exist, because if an angle is [imath]0^{\circ}[/imath], then the triangle doesn't exist i.e. there is no perpendicular or hypotenuse. However, if we take [imath]\lim_{x \to 0} \sin(x)[/imath], then I can understand [imath]\lim_{x \to 0} \sin(x) = 0[/imath] since perpendicular [imath]\approx[/imath] 0. So although [imath]\lim_{x \to 0} \sin(x)[/imath] is [imath]0[/imath], I can't understand how [imath]\sin(0)=0[/imath], and if [imath]\sin(0)[/imath] is not defined, then why is graph of [imath]\sin(x)[/imath] continuous ? |
2227837 | A continuous function on a compact set which takes any value finitely many times.
Let [imath]f : [0,1] \to \mathbb{R}[/imath] be continuous function, such that it takes any value finitely many times and [imath]f(0) \neq f(1)[/imath], then does there exist a point [imath]r_{0}[/imath] in the range such that the cardinality of [imath]f^{-1}(r_{0})[/imath] is odd? (Note that the domain is compact.) I feel that the answer is affirmative but not been able to come up with any concrete argument. I feel this because, as the function is continuous on [imath][0,1][/imath], it would attain its maximum and minimum. With out loss of generality, we can assume that the minimum is attained at [imath]0[/imath] and maximum is attained at [imath]1[/imath]. In between, [imath]0[/imath] and [imath]1[/imath] it might oscillate any number of times. But the number of times minimum is attained and maximum is attained must have different parity. | 1394264 | Construct a function that takes any value even number of times.
I'm looking for a continous function [imath]f: [0,1] \to \mathbb{R}[/imath] such that it takes any value even (thus finite) number of times. I suppose that it exists in the class of Lipschitz functions. All my approaches have led to some value which is taken infinite number of times. |
2233875 | Why [imath](-1)\cdot (-1)=1[/imath]
One of my friends teaches mathematics in primary school. He was asked a question that Why [imath](-1)\cdot (-1)=1[/imath] ... At higher level we can answer this question saying that it is a definition and we want different things such as associative and distributive law to hold but how we convince a primary level student. I don't want any proof as I know very well how to prove..I want an intuition for a primary level student. I know how to prove it. | 539351 | how to prove [imath](-1)\cdot(-1)=1[/imath] based only on the field axioms?
How do I prove [imath](-1)\cdot(-1)=1[/imath] (or in other words that [imath]-1[/imath] is the multiplicative inverse of itself), based only on the field axioms? Thanks |
2234646 | Prove there are no positive integers [imath]x[/imath] and [imath]y[/imath] such that [imath]x^3 + y^3 = 10^3[/imath].
I am familiar with Fermat's Last Theorem that there are no integers such that [imath]x^3+y^3=z^3[/imath], but I need a simpler proof that demonstrates that fact when we know that [imath]x[/imath] and [imath]y[/imath] are positive, and [imath]z=10^3[/imath]. Thank you! | 2230179 | Prove that there are no positive integers [imath]x[/imath] and [imath]y[/imath] such that [imath]x^3 + y^3 = 10^3[/imath].
Prove that there are no positive integers [imath]x[/imath] and [imath]y[/imath] such that [imath]x^3 + y^3 = 10^3[/imath]. This is a homework question, and I understand that its part of Fermat's Last Theorem, but when I looked that up to try to figure out the homework, I realized that it is way further than what we've learned in class so far. We have to prove this, but the only way I can think to do it is by exhaustion, which would be extremely lengthy. Any little shove in the right direction would be appreciated. Thanks! |
2234709 | Equivalence of ideals of rings
Let [imath]R[/imath] be a ring and let [imath]I[/imath] be an ideal in [imath]R[/imath]. Prove that the following statements are equivalent. (1) [imath]I = R[/imath] (2) There exists [imath]u \in R[/imath] such that [imath]u[/imath] is a unit and [imath]u[/imath] belongs to [imath]I[/imath] (3) [imath]1[/imath] belongs to [imath]I[/imath] I only can get that From (1) I know the ring is closed under subtraction and left-right multiplication. And I know I need to prove that equivalent shows that (1) implies (2). Then show (2) implies (3). Could someone help me to finish it? I'd appreciate it. | 692703 | Proper Ideal and Units Proof
Show that an ideal of I R is proper if and only if it does not contain 1 iff and only if it does not contain any units. (1 is the identity element) I'll need to show 3 parts: (1) [imath]\implies[/imath] (2): Let I be a proper ideal and assume [imath]1 \in I[/imath]. I want to show that [imath]1 \notin[/imath] I. (2) [imath]\implies[/imath] (3): Let I be a proper ideal that doesn't contain 1 and assume there are units in R. By the definition, units of R are elements with multiplicative inverses. So I need to show that there aren't multiplicative inverses. (3) [imath]\implies[/imath] (1): Assume there are no units in R and then show that I is a proper ideal. For this part I was thinking of doing something like this- Let [imath]u \in I[/imath] such that u is a multiplicative inverse. Let [imath]x \in R[/imath]. Then [imath]u^{-1}x \in R[/imath] [imath]x=u(u^{-1}x) \in I[/imath] so [imath]R \subseteq I[/imath] which implies that [imath]I = R[/imath] But this is a contradition so therefore I must be a proper ideal. I just need help completing the proof. |
2235112 | Set of sequences summing to [imath]0[/imath] is dense in [imath]\ell_2[/imath]
Let [imath]M_2 := \{ (x_i) \in \ell_2 : \sum_{i=1}^\infty x_i = 0\}[/imath]. I want to show that [imath]\operatorname{cl} M_2 = \ell_2[/imath] using the following definition of closure: [imath]\operatorname{cl} A := \{x\in \ell_2 : \forall \varepsilon > 0 : B_{\varepsilon}(x) \cap A \neq \emptyset\}[/imath] where [imath]B_{\varepsilon}(x)[/imath] is the open ball of radius [imath]\varepsilon[/imath] around [imath]x[/imath]. So given [imath]x\in \ell_2[/imath] and [imath]\varepsilon > 0[/imath] I need to find a [imath]y\in M_2[/imath] with [imath]\| x - y\|_2 <\varepsilon[/imath]. I don't have any clue what kind of theorem I could use to establish the existence of such a [imath]y[/imath]. How can I prove this? | 213867 | The subspace of [imath]\ell_2[/imath]
[imath]M = \{(x_n) \in \ell_2 :\sum_{n=1}^\infty x_n = 0\}[/imath] It is obvious that it is a linear set. But I don't know how to prove it is closed. I try to prove the complement is open, but it doesn't work. Can anyone help me? |
2234867 | Prove that [imath]2^{n}-1\nmid3^{n}-1[/imath] for all odd integers [imath]n\geq3[/imath]
Prove that [imath]2^{n}-1\nmid3^{n}-1[/imath] for all odd integers [imath]n\geq3[/imath]. After factoring the numbers for some small [imath]n[/imath]'s, I guess [imath]\gcd(2^n-1,3^n-1)=1[/imath] for all odd integers [imath]n\geq3[/imath], but I am unable to prove this. I tried infinite descent but to no avail. It seems to have something to do with Mersenne primes. Any hints will be appreciated. | 116978 | Divisibility of integers
Let [imath]n > 1[/imath] be an integer. Then [imath]2^n - 1\nmid 3^n - 1[/imath]. I don't know how to prove it. Can anybody help me, please? In general, for a fixed positive integer [imath]a > 1[/imath], has [imath]a^n - 1|(a +1)^n - 1[/imath] any integer solutions? |
2233823 | Integer solutions for Diophantine equation
Find all pairs [imath](p, n)[/imath] of positive integers where [imath]p[/imath] is a prime number and [imath]p^3 − p = n^7 − n^3[/imath]. Attempt at question: I have factorised both the sides of the equation [imath](p-1)(p)(p+1)=(n-1)(n+1)(n^3)(n^2+1)[/imath] and tried equating terms under different cases. I am not sure if I am procedding in the right way or in the shortest way possible? | 2010543 | Unique pair of positive integers [imath](p,n)[/imath] satisfying [imath]p^3-p=n^7-n^3[/imath] where [imath]p[/imath] is prime
Q. Find all pairs [imath](p,n)[/imath] of positive integers where [imath]p[/imath] is prime and [imath]p^3-p=n^7-n^3[/imath]. Rewriting the given equation as [imath]p(p+1)(p-1)=n^3(n^2+1)(n+1)(n-1)[/imath], we see that [imath]p[/imath] must divide one of the factors [imath]n,n+1,n-1,n^2+1[/imath] on the [imath]\text{r.h.s}[/imath]. Now, the [imath]\text{l.h.s}[/imath] is an increasing function of [imath]p[/imath] for [imath]p\ge1[/imath]. This implies that for any given [imath]n\ge1[/imath], there is exactly one real [imath]p[/imath] for which [imath]\text{l.h.s}=\text{r.h.s}[/imath]. For [imath]p=n^2[/imath], we get [imath]\text{l.h.s}=n^6-n^2<n^7-n^3=\text{r.h.s}.[/imath] This means that either [imath]p>n^2[/imath] or [imath]p<n^2[/imath] must hold. Assuming [imath]p>n^2[/imath], it follows that the prime [imath]p[/imath] cannot divide any of [imath]n,n+1,n-1[/imath]. So [imath]p[/imath] must divide [imath]n^2+1[/imath] and hence [imath]p=n^2+1\quad (\because p>n^2)[/imath]. Substituting the value of [imath]p[/imath] in the given equation we get, [imath]n^2+2=n^3-n\implies n^3-n^2-n=2[/imath]. As the factor [imath]n[/imath] on the [imath]\text{l.h.s}[/imath] must divide [imath]2[/imath], the above equation has a unique integer solution [imath]n=2[/imath]. Finally, we get [imath](5,2)[/imath] as the solution to the given equation. But how do I conclude this is the only solution possible? Also, why does'nt [imath]p<n^2[/imath] (the case which I ignored) hold? As a bonus question, I would like to ask for any alternative/elegant solution (possibly using congruence relations) to the problem. |
2235300 | On closed convex set containing the extreme points of the unit ball of [imath]C(X)[/imath]
Let [imath]X[/imath] be a compact Hausdorff space, [imath]C(X)[/imath] denote the set of all complex-valued continuous functions from [imath]X[/imath]. Show that the smallest closed convex set containing the extreme points of the unit ball of [imath]C(X)[/imath] is the unit ball. The problem is from Douglas's book Banach Algebra Techniques in Operator Theory, problem 1.8. For the case [imath]X=[0,1][/imath], I have shown that the set of all extreme functions are those that have modulus value [imath]1[/imath], but still have no clue how to proceed, even for the case [imath]X=[0,1][/imath]. Please help. | 237277 | Is [imath]{\rm conv}({\rm ext}((C(X))_1))[/imath] dense in [imath](C(X))_1[/imath]?
Let [imath]X[/imath] be compact Hausdorff, and [imath]C(X)[/imath] the space of continuous functions over [imath]X[/imath]. Denote the closed unit ball in [imath]C(X)[/imath] by [imath](C(X))_1[/imath], then it can be shown [imath]f[/imath] is an extreme point of [imath](C(X))_1[/imath] if and only if [imath]|f(x)|=1[/imath] for all [imath]x\in X[/imath]. In Douglas's book, we are asked to show that the convex hull of extreme points of [imath](C(X))_1[/imath] is dense in [imath](C(X))_1[/imath]. By using a theorem of Fejer we can show this is true when [imath]X=[0,1][/imath]. But I do not know how to do it for general [imath]X[/imath]. Since it is only the 7th problem in the first chapter, I am guessing the solution should be elementary. Thanks! |
2226259 | An exercise using uniform boundedness principle
Let [imath](V, \| \cdot \|)[/imath] be a normed space and suppose that [imath](x_n)_{n\in\mathbb{N}} \subset V[/imath] has the property [imath]\sum_{n=1}^{\infty} |\phi(x_n)|<\infty[/imath] for each [imath]\phi \in V^*[/imath]. Prove that [imath] \sup_{\|\phi \| \leq 1} \sum_{n=1}^{\infty} |\phi (x_n) | < \infty. [/imath] I know that I'm supposed to show what work I have done. But I couldn't get started on this exercise. So it would be helpful if it could be pointed out where to start. | 1753566 | Application of Banach-Steinhaus theorem
Let [imath](x_n)[/imath] be a sequence in a Banach space [imath]E[/imath] such that [imath]\sum_{j=1}^{\infty} |\varphi (x_j) |<\infty[/imath], [imath]\forall \varphi \in E'.[/imath] Then [imath]\sup \limits_{\|\varphi\| \leq 1} \sum_{j=1}^{\infty}|\varphi (x_j)| <\infty [/imath]. My attempt: For all [imath]n \in \mathbb{N}[/imath], define [imath]f_n: E' \to \mathbb{K}[/imath], [imath]f_n(\varphi) = \sum_{j=1}^{n} \varphi (x_j)[/imath]. ([imath]\mathbb{K} = \mathbb{R}[/imath] or [imath]\mathbb{C}[/imath]) Each [imath]f_n[/imath] is a continuous linear functional, since: [imath]|f_n(\varphi)| = \bigg| \sum_{j=1}^{n} \varphi (x_j) \bigg| \leq \sum_{j=1}^{n} |\varphi (x_j)| \leq \sum_{j=1}^{n} \|\varphi\|\|x_j\| = (n \max \{\|x_j\|\} ) \|\varphi\|[/imath] For each [imath]\varphi \in E'[/imath], [imath](|f_n(\varphi)|)[/imath] is bounded since [imath]|f_n(\varphi)| \leq \sum_{j=1}^{n} |\varphi (x_j)| \leq \sum_{j=1}^{\infty} |\varphi (x_j)| = M_\varphi \in \mathbb{R} [/imath] By Banach-Steinhaus theorem, there exists [imath]M>0[/imath] such that [imath]\sup \limits_{n \in \mathbb{N}} \|f_n\|<M[/imath]. For all [imath]n \in \mathbb{N}[/imath], we have: [imath]M> \|f_n\| = \sup \limits_{\|\varphi\| \leq 1} |f_n(\varphi)| = \sup \limits_{\|\varphi\| \leq 1} \bigg|\sum_{j=1}^{n}\varphi (x_j)\bigg| [/imath] Then [imath]\sup \limits_{\|\varphi\| \leq 1} \bigg|\sum_{j=1}^{\infty}\varphi (x_j)\bigg| \leq M [/imath] Unfortunately, this is not what we want to prove. How can I fix it? |
1026366 | Distribution of sine of uniform random variable on [imath][0, 2\pi][/imath]
Let [imath]X[/imath] be a continuous random variable having uniform distribution on [imath][0, 2\pi][/imath]. What distribution has the random variable [imath]Y=\sin X[/imath] ? I think, it is also uniform. Am I right? | 2979315 | PDF of [imath]\sin(2\pi X)[/imath] where [imath]X[/imath] is uniform random variable
Let [imath]X[/imath] be a continuous random variable with uniform distribution between [imath]0[/imath] and [imath]1[/imath]. Compute the distribution of [imath]Y = \sin(2\pi X)[/imath]. [imath]\sin(2\pi \cdot0)[/imath] and [imath]\sin(2\pi \cdot1) =0[/imath]. So, the inverse image of the function has multiple roots. How can I find the PDF of [imath]Y[/imath] then? |
2234744 | Value of [imath]5050\cdot\int^1_0(1-x^{50})^{100} dx/\int^1_0(1-x^{50})^{101} dx[/imath]
Please help me find the value of the following integral: [imath]\frac{(5050)\int^1_0(1-x^{50})^{100} dx}{\int^1_0(1-x^{50})^{101} dx}[/imath] I tried solving both numerator and denominator via by-parts but it isn't giving me a conclusive solution. Any other suggestions? | 1710265 | Find the ratio of [imath]\frac{\int_{0}^{1} \left(1-x^{50}\right)^{100} dx}{\int_{0}^{1} \left(1-x^{50}\right)^{101} dx}[/imath]
[imath]I_1=\int_{0}^{1} \left(1-x^{50}\right)^{100} dx[/imath] and [imath]I_2=\int_{0}^{1} \left(1-x^{50}\right)^{101} dx[/imath] Then find [imath]\frac{I_1}{I_2}[/imath] I tried by subtracting [imath]I_1[/imath] and [imath]I_2[/imath] [imath]I_1-I_2=\int_{0}^{1}\left(1-x^{50}\right)^{100}\left(1-(1-x^{50}\right))dx[/imath] so [imath]I_1-I_2=\int_{0}^{1} \left(1-x^{50}\right)^{100} x^{50} dx[/imath] Now using Integration by Parts we get [imath]I_1-I_2= \left(1-x^{50}\right)^{100} \times \frac{x^{51}}{51}\bigg|_{0}^{1} -\int_{0}^{1} 100 \left(1-x^{50}\right)^{99} \times -50 x^{49} \times \frac{x^{51}}{51} dx[/imath] So [imath]I_1-I_2=\frac{5050}{51} \times \int_{0}^{1}\left(1-x^{50}\right)^{99} x^{100} dx[/imath] Now [imath]x^{100}=\left(1-x^{50}\right)^2-(1-x^{50}-x^{50})[/imath] so [imath]\frac{51}{5050}(I_1-I_2)=\int_{0}^{1} \left(1-x^{50}\right)^{99} \times \left(\left(1-x^{50}\right)^2-(1-x^{50})+x^{50}\right)dx=I_2-I_1+\int_{0}^{1} \left(1-x^{50}\right)^{99} x^{50} dx[/imath] Need a hint to proceed further. |
2228633 | Prove that [imath](A\cap B)\cup(B\cap C)\cup(C\cap A)=(A\cup B)\cap(B\cup C)\cap(C\cup A)[/imath]
I have used distributive law and absorption law combined but no result. I have tried to complement the LHS to intersect with the RHS, and if we get the empty set then they are equal. But none of the attempts helped me prove this. | 1928849 | Proving [imath](A\cup B)\cap(B\cup C)\cap(C\cup A)=(A\cap B)\cup (A\cap C)\cup (B\cap C)[/imath]?
Proving [imath](A\cup B)\cap(B\cup C)\cap(C\cup A)=(A\cap B)\cup (A\cap C)\cup (B\cap C)[/imath]? I managed to prove one side by saying: Let E=LHS and F=RHS Let [imath] x \in E[/imath]. Then [imath]x \in (A\cup B) \lor \ x \in (B\cup C) \lor \ x \in (C\cup A).[/imath] If [imath]x\notin A[/imath], then [imath]x[/imath] must be in [imath]B[/imath] and [imath]C[/imath]. If [imath]x\notin B[/imath], then [imath]x[/imath] must be in [imath]A[/imath] and [imath]C[/imath]. If [imath]x \notin C[/imath], then [imath]x[/imath] is in [imath]B[/imath] and [imath]A[/imath]. Therefore [imath]x \in F[/imath] and [imath] E \subset F [/imath]. I'm having trouble proving the other direction. Can I prove it in a similar way saying that let [imath]x \in F[/imath], if [imath]x \notin A[/imath], then [imath]x[/imath] would be in [imath]B[/imath] and [imath]C[/imath]? That doesn't make sense to me. Another way I thought of solving it would be: Let [imath]x \in F. [/imath] Then if [imath]x \in A[/imath], then [imath]x \in B \cup C[/imath], If [imath]x \in B[/imath], then [imath]x \in A \cup C[/imath]. if [imath]x \in C[/imath], then [imath]x \in B \cup A[/imath] And then is this sufficient to show that [imath] x \in F, thus \ F \subset E \ and \ E=F [/imath]? |
787299 | Isomorphism between finite fields adjoining a root
Let [imath]p(x)=x^3+x^2+1[/imath] and [imath]q(x)=x^3+x+1[/imath] be polynomials over the field [imath]\mathbb{Z}_2[/imath]. Let [imath]\alpha[/imath] be a root of [imath]p(x)[/imath] and [imath]\beta[/imath] be a root of [imath]q(x)[/imath]. Now let [imath]K=\mathbb{Z}_2(\alpha)[/imath] and [imath]F=\mathbb{Z}_2(\beta)[/imath]. I am asked to find an explicit isomorphism [imath]\phi: K \rightarrow F[/imath]. I know that since [imath]p(\alpha)=0[/imath] and [imath]q(\beta)=0[/imath] then [imath]\alpha[/imath] and [imath]\beta[/imath] are algebraic over [imath]\mathbb{Z}_2[/imath]. I'm rather stuck after this point, and my other problems are somewhat similar to this one, so if there is a key fact that I'm missing then it might allow me to solve the others. | 1559599 | What is the isomorphism between the fields [imath](Z_2[x]^{<3},+_{x^3+x^2+1},\times_{x^3+x^2+1})[/imath] and [imath](Z_2[x]^{<3},+_{x^3+x+1},\times_{x^3+x+1})[/imath]?
They are both Galois fields of order 8. I'm not exactly sure what the question means - how does one determine/describe an isomorphism? |
2236234 | The function [imath]f: (0,1) \to \mathbb R [/imath] defined by [imath]f(x) := 1/x[/imath] is not uniformly continuous, but it is continuous.
The function [imath]f: (0,1) \to \mathbb R [/imath] defined by [imath]f(x) := 1/x[/imath] is not uniformly continuous, but it is continuous. Proof: given [imath]\epsilon > 0[/imath], then for [imath]\epsilon > \mid 1/x - 1/y \mid[/imath] to hold we must have [imath]\epsilon > \mid 1/x - 1/y \mid = \frac{\mid y-x \mid }{\mid xy \mid } = \frac{\mid y-x\mid }{xy}[/imath] [imath]\mid x -y \mid < xy\epsilon[/imath] Therefore to satisfy the definition of uniform continuity we would have to have [imath]\delta \leq xy\epsilon[/imath] for all [imath]x, y[/imath] in (0,1). but that would mean that [imath]\delta \leq 0[/imath] Therefore there is no single [imath]\delta >0[/imath] I don't understand [imath]\delta \leq 0[/imath] part... why it is less than or equal to 0 ? | 1243384 | prove that [imath]\frac{1}{x}[/imath] is not uniformly continuous on [imath](0,1)[/imath]
I would like to show that the function [imath]\frac{1}{x}[/imath] is not uniformly continuous on [imath](0,1)[/imath] using two approaches. First Approach: We have the fact that if a function [imath]f[/imath] is uniformly continuous on an open interval [imath](a,b)[/imath], then the function [imath]f[/imath] is bounded on [imath](a,b)[/imath]. By using its contrapositive, since [imath]\frac{1}{x}[/imath] is not bounded on [imath](0,1)[/imath], it is not uniformly continuous. Second Approach: Proof by definition. Note that the definition of non-uniform continuity is: There exists an [imath]\epsilon_0>0[/imath] such that for all [imath]\delta>0[/imath], there exists [imath]x,y \in (0,1)[/imath] such that [imath]|x-y|<\delta[/imath] but [imath]|\frac{1}{x} - \frac{1}{y}| \geq \epsilon_0[/imath]. Assume that [imath]\delta < 1[/imath]. Let [imath]x = y + \frac{\delta}{2} \in (0,1), y < \frac{\delta}{2} \in (0,1), \epsilon_0 = 1[/imath]. Then we have [imath]|\frac{1}{x} - \frac{1}{y}| = |\frac{1}{y + \frac{\delta}{2}} - \frac{1}{y}| = \frac{\frac{\delta}{2}}{|y(y + \frac{\delta}{2})|} > \frac{\frac{\delta}{2}}{(\frac{\delta}{2})(\delta)} = \frac{1}{\delta} > 1 = \epsilon_0 [/imath] If [imath]\delta \geq 1[/imath], then let [imath]\delta^{\prime} = \frac{1}{\delta}[/imath] so that [imath]\delta^{\prime} \leq 1[/imath]. The rest proceeds the same as above. Question: Are my two approaches correct? |
2237083 | Question about notation of subgroups of [imath]S_3[/imath]
sorry if this is a basic question, but I was reading about the permutation group [imath]S_3[/imath], and I read that it has three subgroups: [imath]\langle y\rangle[/imath], [imath]\langle xy\rangle[/imath], [imath]\langle x^2y\rangle[/imath] of order 2 and [imath]\langle x\rangle[/imath] of order 3. Can someone mention which subgroups these are, and how to determine their order? are any of these groups normal? | 1361388 | Finding a normal and not normal subgroup of [imath]S_3[/imath]
I'm being asked to find 2 subgroups of [imath]S_3[/imath], one of which is normal and one that isn't normal. I guess, to find the non normal subgroup is easier. I would do this by trial and error, but since the group is [imath]S_3[/imath] and easily visualizable, I guess that there should be a geometrical property that makes it easy to find a normal subgroup. Any of you guys know one? Or should I try this by brute force? |
2236859 | Possible trivial counterexample to exam question
So doing an exam question for set theory this was the question: Let X be a set. Prove there is no injection [imath]f : P(X) → X.[/imath] Would X being an empty set be a counter example because in this case the powerset is a set of a single element [imath]\{\}[/imath] hence it is mapped to a null element. However would this be the case that [imath]f[/imath] would not be a function, since there isn't [imath](a,b)\in f[/imath] hence [imath]f[/imath] isn't defined for the whole of the domain hence [imath]f[/imath] is not a function. So yes I understand that this isn't a proof for this problem but while thinking about the problem I was trying some examples and non-examples and couldn't work out why this failed until I actually put it into words. Sorry for a convoluted question, I don't really have the correct notation to explain myself. | 284812 | A "reverse" diagonal argument?
Cantor's diagonal argument can be used to show that a set [imath]S[/imath] is always smaller than its power set [imath]\wp(S)[/imath]. The proof works by showing that no function [imath]f : S \rightarrow \wp(S)[/imath] can be surjective by constructing the explicit set [imath]D = \{ x \in S | x \notin f(s) \}[/imath] from a function [imath]f[/imath] and showing that no element of [imath]S[/imath] maps to [imath]D[/imath]. This proof works because all bijections are surjections, so if no surjection from [imath]S[/imath] to [imath]\wp(S)[/imath] exists, then there cannot be a bijection between [imath]S[/imath] and [imath]\wp(S)[/imath]. My question is whether it is possible to run a "reverse diagonalization" that works by instead showing that there is no injection from [imath]\wp(S)[/imath] to [imath]S[/imath]. I am curious about this because I have never seen Cantor's theorem proved this way (or, more generally, any diagonal argument structured like this). Is it possible to take Cantor's diagonal argument and "reverse" it to show that there cannot be an injection from [imath]\wp(S)[/imath] to [imath]S[/imath], rather than showing that there can be no surjection from [imath]S[/imath] to [imath]\wp(S)[/imath]? Thanks! |
2237194 | (Revised) Let [imath]\phi : \Bbb Z \rightarrow \Bbb Z[/imath] be given by [imath]\phi(n) = 7n[/imath]. Find the image of [imath]\phi[/imath].
Quoting " Let [imath]\phi : Z \rightarrow Z[/imath] be given by [imath]\phi(n) = 7n[/imath]. [...] Find the image of [imath]\phi[/imath]." My understanding (revised): I claim that the only possible image-elements of [imath]\phi[/imath] are the multiples of 7 in z denoted 7Z. To prove this I have to show that Im[imath](\phi) \subset 7 \Bbb Z[/imath] and that [imath] 7 \Bbb Z \subset [/imath] Im[imath](\phi)[/imath]. Proving Im[imath](\phi) \subset 7 \Bbb Z[/imath] : [imath]\forall y \in [/imath] Im[imath](\phi)[/imath] such that [imath] y =\phi(x)[/imath] where [imath]x \in 7 \Bbb Z[/imath]. It follows that [imath]y \in 7 \Bbb Z[/imath]. Therefore, Im[imath](\phi) \subset 7 \Bbb Z[/imath] Proving [imath] 7 \Bbb Z \subset [/imath] Im[imath](\phi)[/imath]: [imath]\forall y \in 7\Bbb Z[/imath] such that [imath] y =7x[/imath] where [imath]x \in \phi(x)[/imath]. It follows that [imath]y \in \phi(x)[/imath]. Therefore, [imath] 7 \Bbb Z \subset [/imath] Im[imath](\phi)[/imath]. Any input to my understanding is much appreciated. | 2236143 | (Revised) Prove that [imath]\phi[/imath] is a group homomorphism and find the kernel.
Quoting " Let [imath]\phi : \Bbb Z \rightarrow \Bbb Z[/imath] be given by [imath]\phi(n) = 7n[/imath]. Prove that [imath]\phi[/imath] is a group homomorphism. Find the kernel, and the image of [imath]\phi[/imath]." My understanding: Part 1: Given [imath]n,m \in \Bbb Z[/imath], let's check that [imath]\phi(n+m) =\phi(n)+\phi(m)[/imath] [imath]\phi(n+m) = 7(n+m) = 7n + 7m = \phi(n)+\phi(m)[/imath] Therefore as the group operation in [imath]\phi[/imath] is preserved, [imath]\phi[/imath] is a group homomorphism. Part 2 (revised): [imath]\phi[/imath] is one-to-one as: [imath]\space \space \phi (n) = \phi(m) \Rightarrow 7n =7m \Rightarrow n=m [/imath]. We also know that [imath]\phi(e)=e[/imath] by the property of homomorphism. Therefore, no other object than [imath]0[/imath] in the domain can map to [imath]0[/imath] in the codomain. It follows that: [imath]\ker(\phi)=\{ x \in \Bbb Z : \phi(x)=e\}=\{e\}[/imath] Part 3 (revised): I claim that the only possible image-elements of [imath]\phi[/imath] are the multiples of [imath]7[/imath] in [imath]\mathbb Z[/imath] denoted [imath]7\mathbb Z[/imath]. To prove this I have to show that Im[imath](\phi) \subset 7 \Bbb Z[/imath] and that [imath] 7 \Bbb Z \subset [/imath] Im[imath](\phi)[/imath]. Proving Im[imath](\phi) \subset 7 \Bbb Z[/imath] : [imath]\forall y \in [/imath] Im[imath](\phi)[/imath] such that [imath] y =\phi(x)[/imath] where [imath]x \in 7 \Bbb Z[/imath]. It follows that [imath]y \in 7 \Bbb Z[/imath], therefore Im[imath](\phi) \subset 7 \Bbb Z[/imath] Proving [imath] 7 \Bbb Z \subset [/imath] Im[imath](\phi)[/imath]: [imath]\forall y \in 7\Bbb Z[/imath] such that [imath] y =7x[/imath] where [imath]x \in \phi(x)[/imath]. It follows that [imath]y \in \phi(x)[/imath], therefore [imath] 7 \Bbb Z \subset [/imath] Im[imath](\phi)[/imath]. Any input to my understanding is much appreciated. |
2235886 | How to determine optimal control law
I was given the state space equation, [imath]\dot{x} = -2x + u[/imath] and told to determine the optimal control law [imath]u=-kx[/imath] which minimizes the performance index [imath]J = \int_{0}^{\infty} x^{2}\,dt.[/imath] My approach was to find the state feedback [imath]k[/imath]. But since the value of [imath]R[/imath] (positive definite Hermitian) is not given. That means [imath]R=0[/imath]. How do I determine the optimal control for this system where [imath]R=0[/imath]? | 2236785 | How to determine the optimal control law?
Given the differential equation [imath]\dot x = -2x + u[/imath] determine the optimal control law [imath]u = - kx[/imath] that minimizes the performance index [imath]J = \int_0^{\infty} x^2 \, \mathrm d t[/imath] My approach was to find the state feedback [imath]k[/imath]. But since the value of [imath]R[/imath] (positive semidefinite Hermitian) is not given, that means [imath]R=0[/imath]. How do I determine the optimal control for this system where [imath]R=0[/imath]? |
1129157 | Prove that if rank(A + B) = rank(A) + rank(B), then col(A) ∩ col(B) = {0}
Let A,B be in [imath]M_{mxn}(\mathbb{R})[/imath] Prove that if rank(A + B) = rank(A) + rank(B), then col(A) ∩ col(B) = {0} I started with a proof by contradiction, since we know that rank(A + B) ≤ rank(A) + rank(B). I assume that if rank(A + B) < rank(A) + rank(B), then col(A) ∩ col(B) = {0}. I realize that we don't have to do the > case. However I am a little confused on how to show col from rank. | 2252161 | Let [imath]f,g: V \rightarrow W[/imath] be linear operators. Prove that [imath]r(f + g) \leq r(f) + f(g)[/imath]
Let [imath]f,g: V \rightarrow W[/imath] be linear operators. Prove that [imath]r(f + g) \leq r(f) + f(g)[/imath] Note: r = rank My idea was to use matrix representation and to prove that the rank of matrix C ([imath]C= F+G[/imath]) can't be bigger than the sum of [imath]r(F) + r(G)[/imath],but to use that idea I must prove that [imath]r(f) = r(g)[/imath] which I don't have an idea how to do. Other idea was to prove that [imath](f+g)(v) = f(v) + g(v)[/imath] but also with no luck. |
2237589 | given a finite group [imath]G[/imath], is it possible to determine the exact number of one-dimensional representations of [imath]G[/imath]
Let's say [imath]G[/imath] has order [imath]n[/imath]. If this is all the information we have to go by, can we say conclusively what the number of one-dimensional representations of [imath]GT[/imath] is? Or do we need more information? Like the type of group? In general, how would one go about finding this number? | 133122 | Why are there [imath]|G/G'|[/imath] 1-dimensional representations of [imath]G[/imath]?
Let [imath]G'[/imath] be the derived subgroup of a finite group [imath]G[/imath]. We have a correspondence [imath]\{\mathrm{reps \ of \ G/G'}\} \longleftrightarrow \{\mathrm{reps \ of \ G \ with \ kernel \ containing \ G' }\} [/imath] If we restrict to 1-dimensional reps, we get: [imath]\{\mathrm{1\ dimensional \ reps \ of \ G/G'}\} \longleftrightarrow \{\mathrm{1 \ dimensional \ reps \ of \ G \ with \ kernel \ containing \ G' }\} [/imath] Now my notes say that there are [imath]|G/G'|[/imath] 1-dimensional reps of [imath]G[/imath]. Since there are [imath]|G/G'|[/imath] 1-dimensional reps of [imath]G/G'[/imath], this must mean that all 1-dimensional reps of [imath]G[/imath] have kernel containing [imath]G'[/imath]. Why is this so? Thanks |
2236907 | Find the limit [imath]\lim_{n\to \infty}\{{n!}^{1/n}\}/n[/imath]
Find the limit [imath]\lim_{n\to \infty}\{{(n!)}^{1/n}\}/n[/imath] I took exp log but getting answer as 1 but it should be 1/e. Required a nice approach. | 1598508 | I need help to advance in the resolution of that limit: [imath] \lim_{n \to \infty}{\sqrt[n]{\frac{n!}{n^n}}} [/imath]
how I can continue this limit resolution? The limit is: [imath] \lim_{n \to \infty}{\sqrt[n]{\frac{n!}{n^n}}} [/imath] This is that I have done: I apply this test: [imath] \lim_{n \to \infty}{\sqrt[n]{a_n}} = \frac{a_{n+1}}{a_n} [/imath] Operating and simplifying I arrive to this point: [imath] \lim_{n \to \infty}{\frac{n^n}{(n+1)^n}} [/imath] I've done something wrong? Thanks! |
2238419 | How to integrate [imath]\int^{2\pi}_{0} \frac{dx}{\mathrm{sin}^4x +\mathrm{cos}^4x}[/imath]?
there's: [imath]\int^{2\pi}_{0} \frac{dx}{\mathrm{sin}^4x+ \mathrm{cos}^4x}[/imath] I was always confused if it is possible to solve this integral not using Tangent half-angle substitution? Because it takes much time for calculus only, is there another way? | 820830 | How to integrate [imath]\int \frac{1}{\sin^4x + \cos^4 x} \,dx[/imath]?
How to integrate [imath]\int \frac{1}{\sin^4x + \cos^4 x} \,dx[/imath] I tried the following approach: [imath]\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \int \frac{1}{\sin^4x + (1-\sin^2x)^2} \,dx = \int \frac{1}{\sin^4x + 1- 2\sin^2x + \sin^4x} \,dx \\ = \frac{1}{2}\int \frac{1}{\sin^4x - \sin^2x + \frac{1}{2}} \,dx = \frac{1}{2}\int \frac{1}{(\sin^2x - \frac{1}{2})^2 + \frac{1}{4}} \,dx[/imath] The substitution [imath]t = \tan\frac{x}{2}[/imath] yields 4th degree polynomials and a [imath]\sin[/imath] substitution would produce polynomials and expressions with square roots while Wolfram Alpha's solution doesn't look that complicated. Another approach: [imath]\sin^4x + \cos^4 x = (\sin^2 x + \cos^2x)(\sin^2 x + \cos^2 x) - 2\sin^2 x\cos^2 x = 1 - 2\sin^2 x\cos^2 x = (1-\sqrt2\sin x \cos x)(1+\sqrt2\sin x \cos x)[/imath] and then I tried substituting: [imath]t = \sin x \cos x[/imath] and got [imath]\int\frac{t\,dt}{2(1-2t^2)\sqrt{1-4t^2}}[/imath] Another way would maybe be to make two integrals: [imath]\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \int \frac{1}{(1-\sqrt2\sin x \cos x)(1+\sqrt2\sin x \cos x)} \,dx = \\ \frac{1}{2}\int \frac{1}{1-\sqrt2\sin x \cos x} \,dx + \frac{1}{2}\int\frac{1}{1+\sqrt2\sin x \cos x} \,dx[/imath] ... and again I tried [imath]t = \tan\frac{x}{2}[/imath] (4th degree polynomial) and [imath]t=\sqrt2 \sin x \cos x[/imath] and I get [imath]\frac{\sqrt 2}{2} \int \frac{\,dt}{(1-t)\sqrt{1-2t^2}}[/imath] for the first one. Any hints? |
2238513 | Let [imath]p[/imath] be an arbitrary prime number and [imath]G[/imath] be an abelian group, how is [imath]pG[/imath] defined?
Let [imath]p[/imath] be a fixed prime in [imath]\Bbb Z[/imath] and [imath]Ab[/imath] represent the category of abelian groups. Define a functor [imath]F : Ab \rightarrow Ab[/imath] by [imath]F(G) = G/pG[/imath] and [imath]F(f) = x + pG \rightarrow f(x) + pH[/imath] (where [imath]f:G \rightarrow H[/imath] is a homomorphism) What does [imath]pG[/imath] mean for an abelian group? I am familiar with the notation [imath]aG = \{ag : g \in G\}[/imath], but [imath]G[/imath] isn't specified to be an integer group which would be compatible with a prime [imath]p[/imath]. | 232526 | Is [imath]G/pG[/imath] is a [imath]p[/imath]-group?
Jack is trying to prove: Let [imath]G[/imath] be an abelian group, and [imath]n\in\Bbb Z[/imath]. Denote [imath]nG = \{ng \mid g\in G\}[/imath]. (1) Show that [imath]nG[/imath] is a subgroup in [imath]G[/imath]. (2) Show that if [imath]G[/imath] is a finitely generated abelian group, and [imath]p[/imath] is prime, then [imath]G/pG[/imath] is a [imath]p[/imath]-group (a group whose order is a power of [imath]p[/imath]). I think [imath]G/pG[/imath] is a [imath]p[/imath]-group because it is a direct sum of cyclic groups of order [imath]p[/imath]. But I cannot give a detailed proof. |
2238618 | Concept of a normal subgroup
Question: Give an example of a normal subgroup of [imath]A_4[/imath]. Is the subgroup [imath]H[/imath]={[imath](1),(25)(34),(35)(24),(45)(23)[/imath]} of [imath]A_5[/imath] a normal subgroup?I am having a hard time understanding this concept because the question itself is confusing me. Do I have to come up with a group showing [imath]A_4[/imath]? Is [imath]A_4[/imath]=[[imath](1),(12)(34),(13)(24),(123)[/imath]}? Need help | 121493 | How to show that Klein four-group is a normal subgroup of the alternating group [imath]A_4[/imath]
I want to show that the Klein four-group is a normal subgroup of the alternating group [imath]A_4[/imath]. I am using the information in this link, that shows explicitly [imath]A_4[/imath], and Klein four-group as a subgroup. I know that there is the direct way, by definition, but is there a way that does not require actually multiplying so many permutations ? |
2238838 | Fourier Transform of a Derivative
I'm trying to prove that: [imath]$$F\,\{f'(x)\} = -i\omega F(\omega) \qquad (1) $$[/imath] where [imath]\, F(\omega) = F\,\{f(x)\}[/imath] This is my procedure so far: [imath]F\,\{f'(x)\} = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f'(t)e^{i\omega t} dt[/imath] Integrating by parts I obtained: [imath]=\frac{1}{\sqrt{2\pi}} \big[ \space f(t) e^{i\omega t} \space \big|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} i\omega f(t)e^{i\omega t} dt \space \big][/imath] Now, in order to (1) to be true I need to get: [imath] f(t)e^{i\omega t} \space \big|_{-\infty}^{\infty} =0 \qquad (2)[/imath] I developed it and got the following: [imath] f(\infty)e^{i\omega \infty} - f(-\infty)e^{-i\omega \infty } = f(\infty)e^{i\omega \infty} - 0= f(t)e^{i\omega \infty} [/imath] I undarstand that the second term is [imath]0[/imath], since f(t) must have a value to accomplish Dirichlet conditions (and [imath]e^{-\infty} = 0[/imath]), but I don't see how [imath] f(\infty)e^{i\omega \infty}[/imath] is [imath]0[/imath]. | 430858 | Fourier Transform of Derivative
Consider a function [imath]f(t)[/imath] with Fourier Transform [imath]F(s)[/imath]. So [imath]F(s) = \int_{-\infty}^{\infty} e^{-2 \pi i s t} f(t) \ dt[/imath] What is the Fourier Transform of [imath]f'(t)[/imath]? Call it [imath]G(s)[/imath].So [imath]G(s) = \int_{-\infty}^{\infty} e^{-2 \pi i s t} f'(t) \ dt[/imath] Would we consider [imath]\frac{d}{ds} F(s)[/imath] and try and write [imath]G(s)[/imath] in terms of [imath]F(s)[/imath]? |
227145 | Function [imath]f:[0,1] \to [0,1][/imath] taking on each value in [imath][0,1][/imath] exactly twice
I want to find a function [imath]f:[0,1] \to [0,1][/imath] such that [imath]f[/imath] takes on each value in [imath][0,1][/imath] exactly twice. I think this means there are an infinite number of discontinuities. Can anyone help me figure this one out? Anyone have any pointers? | 2890637 | Function on [imath][0,1][/imath] which takes every value twice has infinitely many discontinuities.
I know that there is no continuous function on [imath][0,1][/imath] which takes each value exactly twice. I know how to prove this fact using Intermediate value property and attainment of maximum value on compact set. I am interested to show that for such function number of discontinuties are infinitely many? I had no clue how to proceed for this example. Any help will be appreciated. |
2238624 | Cartesian product proof question
I need to prove that if [imath]A,B,C,D[/imath] are sets, [imath](A \times B) \cap (C \times D)=(A \times D) \cap (C \times B)[/imath] I am trying to show that if for all [imath]x,y[/imath] in LHS, it is also an element of RHS. [imath](x,y) \in (A \times B) \cap (C \times D) \iff[/imath] [imath](x,y) \in (A \times B)[/imath] and [imath](x,y) \in (C \times D)[/imath] [imath]\iff x \in A, y \in B[/imath] and [imath]x \in C, y \in D[/imath] [imath]\iff x \in A, y \in D[/imath] and [imath]x \in C, y \in B[/imath] (Am I allowed to change the order of [imath]y[/imath] as above?) and if so, what law allows me to do that?. | 1643496 | Proof of Sets involving the Cartesian Product
The question goes: Let [imath]A,B,C,D[/imath] be sets. Prove that [imath]\big(A\times B\big)\bigcup \big(C\times D\big)=\big(A\bigcap C\big)\times \big(B\bigcap D\big)[/imath] I started with the definition of the cartesian product [imath]A \times B=\left\{\big(a,b\big): a\in A \wedge b \in B\right\}[/imath] The union of [imath]A\times B[/imath] and [imath]C \times D[/imath] would be a set of coordinates where [imath]\big(x,y\big): x\in A\bigcup C \wedge y\in B\bigcup D[/imath] I am stuck here because the union of the two cartesian products on the left doesn't seem to imply that [imath]x\in A\bigcap C[/imath] and [imath]y \in B\bigcap D[/imath]. Could anyone give me a hint or a possible step I might not be seeing? Please and thank you. |
2239250 | If [imath](a,n)=d>1[/imath] and [imath]k[/imath] is any positive integer, then [imath]a^{k} \not\equiv 1 \pmod n[/imath]
Show that if [imath](a,n)=d>1[/imath] and [imath]k[/imath] is any positive integer, then [imath]a^{k} \not\equiv 1 \pmod n[/imath]. I know that the order divisibility property states the following. If [imath](a,n)=1[/imath] and [imath]k[/imath] is the order of [imath]a \pmod n[/imath], then for any positive integer [imath]j[/imath], we know that [imath]a^{j} \equiv 1 \pmod n[/imath] if and only if [imath]k \mid j[/imath]. It seems like I could somehow use the "if an only if" part to prove that [imath]k[/imath] does not divide [imath]j[/imath], which would imply the conclusion. But how can I show it formally? | 2237389 | Show that if [imath](a,n)=d>1[/imath] and [imath]k[/imath] is any positive integer, then [imath]a^k\not\equiv 1\pmod{n}.[/imath]
So I know that if [imath](a,n)=1[/imath], then [imath]a^{\phi(n)}≡1\pmod{n}[/imath] by Euler's Theorem. I also know that [imath]d \mid a^k[/imath] and obviously [imath]d \mid n[/imath]. I know I probably need to use this in the proof, but I just don't know how to get started. |
2239972 | The only global section of the tautological line bundle on [imath]\mathbb{P}^n[/imath] is the zero section.
Let [imath]L(-1)[/imath] be the tautological line bundle on [imath]\mathbb{P}^n[/imath]. I would like to understand the proof of the following: Fact. The only global section on [imath]L(−1)[/imath] is the zero section. Proof. Let [imath]s : \mathbb{P}^n \to L(−1)[/imath] be a holomorphic section. For any [imath]l ∈ \mathbb{P}^n[/imath] we have [imath]s(l)=(l,z_l)[/imath] for some [imath]z_l \in \mathbb{C}^{n+1}[/imath] lying on the line [imath]l[/imath]. Thus, [imath]l \to z_l[/imath] is a holomorphic map [imath]\mathbb{P}^n \to \mathbb{C}^{n+1}[/imath]. By the maximum principle this map must be constant, so [imath]z_l \equiv w \in \mathbb{C}^{n+1}[/imath]. On the other hand [imath]s[/imath] is fiber preserving, so [imath]w \in l[/imath] for each line [imath]l[/imath] through the origin of [imath]\mathbb{C}^{n+1}[/imath]. Hence [imath]w = 0[/imath]. How does the maximum principle imply that the map is constant? Thanks a lot. EDIT: my question is slightly different from the linked answer as that result is for holomorphic functions (to [imath]\mathbb{C}[/imath]), whereas here I have a holomorphic map (to [imath]\mathbb{C}^{n+1}[/imath]). | 881742 | Holomorphic functions on a complex compact manifold are only constants
Is there a simple proof that every holomorphic function [imath]M\to\mathbb{C}[/imath] on a compact complex manifold [imath]M[/imath] is constant? |
2240349 | Identification of the polynomial given symmetric relations
Let [imath]p(x)[/imath] be an eighth degree polynomial. Given that: [imath]p(1) = 1[/imath], [imath]p(2) = 1/2[/imath], [imath]p(3) = 1/3[/imath], .... [imath]p(9) = 1/9[/imath] Find the value of [imath]p(10)[/imath]. I tried to approach this by taking [imath]h(x) = p(x) - 1/x[/imath] but then [imath]h(x)[/imath] won't be a polynomial. I tried to manipulate the equation but I failed to find anything helpful. I have found a relation between the coefficients and some values; 9 equations and 9 variables but it would be too cumbersome to solve. Some guidance would be appreciated. Thanks! (Sorry for the poor title; if anyone can improve it please do so) | 640372 | Identify [imath]P(2014)[/imath] if [imath]P(i)=1/i[/imath] for every positive integer [imath]1\le i\le2013[/imath]
Suppose [imath]P(x)[/imath] is a polynomial of degree [imath]2012[/imath] and [imath]P(x) = 1/x[/imath] when [imath]x[/imath] takes the integer values [imath]1\cdots2013[/imath] (inclusive). What is the value of [imath]P(2014)[/imath]? I get [imath]1/1007[/imath] but I'm not sure if it's right, and my method is very inelegant. |
2240133 | Number of words that can be formed using "DAUGHTER" so that vowels never come together.
How many words can be formed from the letters of the word "d a u g h t e r" so that the vowels never come together? There are [imath]3[/imath] vowels and [imath]5[/imath] consonants. I first arranged [imath]5[/imath] consonants in five places in [imath]5![/imath] ways. [imath]6[/imath] gaps are created. Out of these [imath]6[/imath] gaps, I selected [imath]3[/imath] gaps in [imath]{}_6C_3[/imath] ways and then made the vowels permute in those [imath]3[/imath] selected places in [imath]3![/imath] ways. This leads me to my answer [imath]5!\cdot {}_6C_3 \cdot 3! = 14400[/imath]. The answer given in my textbook is [imath]36000[/imath]. Which cases did I miss? What is wrong in my method? | 162764 | How many words can be formed from the letters of the word 'DAUGHTER' so that the vowels never come together?
How many words can be formed from the letters of the word 'DAUGHTER' so that the vowels never come together ? The answer is obviously [imath]8!-6!\cdot3![/imath]. My question is that if we ponder from a different perspective, that is taking [imath]5[/imath] consonants first and arranging them ([imath]5![/imath] ways of doing that) and then placing the [imath]3[/imath] vowels in the [imath]6[/imath] places created due to the arrangement of consonants ([imath]\frac{6!}{3!}[/imath] ways to do that), the answer should be [imath]5!\frac{6!}{3!}[/imath]. What is wrong with this? |
2240339 | How [imath]\left(1+\frac{1}{x}\right)^x [/imath]is differentiable
I try to use [imath]f(x)= \frac{1}{x} [/imath] and [imath]g(x) = \left(1+x\right)^{\frac{1}{x}} [/imath] . edit:for all [imath]x \in \mathbb{N}[/imath] Try to show [imath]g \circ f[/imath] differentiable. I can get [imath]f(x)[/imath] is differentiable but I can't get [imath]g(x)[/imath] differentiable. In other way I use [imath] f(x) = 1 + \frac{1}{x}[/imath] and [imath]g(x) = (x)^{\frac{1}{x-1}}[/imath]. Have same problem It won't work on [imath]g(x)[/imath]. What should I do choose new [imath]f(x),g(x)[/imath] or prove in any other way. | 1073662 | Differentiating [imath] \left( 1 - \frac {1}{x} \right)^x [/imath]
I have a calculus question. How does one differentiate [imath]\left(1-\frac{1}{x}\right)^x[/imath], for x>1? It should be positive right? |
1379045 | How to factor intricate polynomial [imath] ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c [/imath]
I would like to know how to factor the following polynomial. [imath] ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c [/imath] What is the method I should use to factor it? If anyone could help.. Thanks in advance. | 1158190 | How do you factorise [imath]x^3z - x^3y - y^3z + yz^3 + xy^3 - xz^3[/imath]?
I'm trying to factorise [imath] x^3z - x^3y - y^3z + yz^3 + xy^3 - xz^3 [/imath] into four linear factors. By plugging it into WolframAlpha I've learned that it's [imath]-(x-y)(x-z)(y-z)(x+y+z)[/imath] My question is: what are the steps involved in factorising the expression? Is there a method I don't know about that I'd have access to with my limited maths? Really appreciate any help! |
2238417 | Evaluate [imath]\lim_{x\to 0} \frac{a^x -1}{x}[/imath] without applying L'Hopital's Rule.
The questions is: Evaluate [imath]\lim_{x\to 0} \frac{a^x -1}{x}[/imath] without applying L'Hopital's Rule. Does this question fundamentally same as asking if the [imath]\lim_{x\to 0} \frac{a^x -1}{x}[/imath] exists? rather than straightway asking to find the limit. That means are questions (1) proving if the limit of a function exists and (2) asking what is the limit of that function, essentially same question? | 1491220 | Show [imath]\lim\limits_{h\to 0} \frac{(a^h-1)}{h}[/imath] exists without l'Hôpital or even referencing [imath]e[/imath] or natural log
Taking as our definition of exponentiation repeated multiplication (extended to real exponents by continuity), can we show that the limit [imath]\lim_{h\to 0}\dfrac{a^h-1}{h}[/imath] exists, without l'Hôpital, [imath]e[/imath], or even natural logarithm? Sure, l'Hôpital will work, but that's circular if we're developing calculus of transcendentals from first principles. There is a good answer to this question already by user Neal, but he uses the exponential function with base [imath]e[/imath] (it's been answered many times: see also here, here, and here). But using the special properties of [imath]e[/imath] strikes me as circular too; not literally logically circular, in the sense that we are invoking results we're trying to prove, since there are definitions of [imath]e^x[/imath] which make it trivial to verify the derivative. But perhaps pedagogically circular for a complete novice, the special properties of [imath]e[/imath] appear unmotivated because they cannot be justified without reference to the very derivative we are trying to compute (or else a detour through logarithms, but let's not). Can we find nice squeeze theorem bounds like Neal has, but for the function [imath]a^x[/imath] instead of [imath]e^x[/imath], with the additional handicap that we can't just write [imath]a^x=e^{x\log a}[/imath]? I thought to substitute a series expansion for [imath]\log a[/imath], but didn't come up with any bounds that were nicely polynomial in both [imath]x[/imath] and [imath]a[/imath]. I wonder whether the geometric proof of [imath]\lim (\sin x)/x[/imath] (see for example, robjohn's answer here) could be adapted. Obviously without a reference to natural logarithm, we cannot compute the value of the limit. But I just want to show it exists (via squeeze theorem or monotone convergence). Once we know this limit exists, we can show it behaves like a logarithm, whence there is a unique base for which the limit is 1, which we call [imath]e[/imath]. The rest of the development of calculus of exponentials and logs follows easily. This seems like the approach that would appear the most accessible yet motivated to a novice calculus student. An analogous limit to [imath]\lim\dfrac{a^h-1}{h}[/imath] for understanding to differentiating exponential functions, are the limits [imath]\lim\limits_{n\to\infty} (1+\frac{1}{n})^n[/imath] and [imath]\lim\limits_{n\to\infty} (1+\frac{1}{n})^{n+1}[/imath] for differentiating the logarithm, if you prefer to start with that as your primitive concept. Both limits are shown to exist using Bernoulli's inequality (see WimC's answer here for the first limit, and see David Mitra's answer here for the second limit). I tried without success to use Bernoulli's inequality to show my sequence was monotone. This limit can also be analyzed using the AM-GM inequality as seen in user94270's answer to this question. So that inequality may help here. I would also accept an explanation of why the limit cannot be computed without transcendental techniques, or an opinion why this is not a pedagogically sound approach to introducing the calculus of exponentials and logarithms. Edit: This question has a nice solution by Paramanand Singh to a closely related problem. |
2240607 | Proving an Olympiad type inequality [imath]\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{a+c}}+\sqrt{\frac{2c}{a+b}}\le\sqrt{3(\frac{a}b+\frac{b}c+\frac{c}a)}[/imath]
Let [imath]a,b,c>0[/imath] be real numbers. Prove that: [imath]\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{a+c}}+\sqrt{\frac{2c}{a+b}}\leq\sqrt{3\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)} [/imath] | 189140 | Proving inequality [imath]\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq \sqrt{3 \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}[/imath]
In the pdf which you can download here I found the following inequality which I can't solve it. Exercise 2.1.11 Let [imath]a,b,c \gt 0[/imath]. Prove that [imath]\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq \sqrt{3 \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}.[/imath] Thanks :) |
2241299 | Intuition explaining [imath]\frac{\partial^2 f}{\partial x \partial y}[/imath]
This is probably something I should already know, but I would like to know an intuitive way to think about the partial x, partial y derivative of [imath]f[/imath]. I understand that [imath]\frac{\partial ^2 f}{\partial x^2} [/imath] can be understood as the concavity in the x-direction. Is there a similar way of thinking about [imath]\frac{\partial^2 f}{\partial x \partial y}[/imath]? | 29561 | Geometric interpretation of [imath]\frac {\partial^2} {\partial x \partial y} f(x,y)[/imath]
Are there any geometric interpretation for the second partial derivative? i.e. [imath]f_{xy} = \frac {\partial^2 f} {\partial x \partial y}[/imath] In particular, I'm trying to understand the determinant from second partial derivative test for determining whether a critical point is a minima/maxima/saddle points: [imath]D(a, b) = f_{xx}(a,b) f_{yy}(a,b) - f_{xy}(a,b)^2[/imath] I have no trouble understanding [imath]f_{xx}(x,y)[/imath] and [imath]f_{yy}(x,y)[/imath] as the of measure of concavity/convexity of f in the direction of x and y axis. But what does [imath]f_{xy}(x,y)[/imath] means? |
2241505 | Calculating the following series: [imath]\sum_{k=0}^{\infty}(−1)^k2k+1[/imath]
Here are the series: [imath] \sum_{k=0}^{\infty}\frac{(−1)^k}{2k+1} [/imath] I don't know how to start because obviously can't compute all the values up to infinity. My only thought is to use d'alembert ratio? but doesn't this only tell me if the series converges? not what the actual sum of the series is? Thank you for your help. | 2232150 | calculate the sum of an infinite series
Here is the series: [imath] \sum_{k = 0}^{\infty} \frac{(-1)^k}{2k + 1}. [/imath] I don't know how to start at all. Thank you for your help. |
2241480 | Order of a factor group and whether it is cyclic
Question: Determine the order of [imath]\left ( \mathbb{Z} \bigoplus \mathbb{Z}\right )/\left \langle \left ( 2,2 \right ) \right \rangle[/imath].Is it cyclic? The elements in the factor group are of the form [imath]\left ( a,b \right )+\left \langle \left ( 2,2 \right ) \right \rangle[/imath]. The subgroup [imath]\left \langle \left ( 2,2 \right ) \right \rangle[/imath] would absorb any elements of the form [imath]n\left ( 2,2 \right )[/imath] by the property of cosets. Why wouldn't the element (1,2)H be in the factor group? At this point I would appreciate hints. My intuition tells me a bit of number theory would be in use here. Thanks in advance. | 539224 | What is the order of [imath](\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle[/imath] and is it cyclic?
Evidently, [imath](\mathbb{Z} \oplus \mathbb{Z})/ \langle (2,2) \rangle[/imath] has order [imath]4[/imath], but I think it's infinite. The four cosets are listed as [imath](0,0) + \langle (2,2) \rangle[/imath], [imath](0,1) + \langle (2,2) \rangle[/imath], [imath](1,0)+ \langle (2,2) \rangle[/imath] and [imath](1,1) + \langle (2,2) \rangle[/imath]. However, [imath](2,0)[/imath] doesn't appear to be in any of these cosets. Maybe the answer I'm being told is wrong. |
2241473 | How many triangles with same color are there using [imath]6[/imath] and [imath]n[/imath] points?
How many triangles with same color are there using [imath]6[/imath] points? How many triangles with same color are there using [imath]n[/imath] points? It seems to be Ramsey numbers but in that we never prove there are two we prove there is minimum [imath]1[/imath] such triangle.For the second problem the book gave the answer [imath]\binom{n}{3}-\frac{n*\lfloor{\frac{n-1}{2}}\rfloor*\lceil{\frac{n-1}{2}}\rceil}{2}[/imath] Which is colmpletly odd for me.Any hints? Edit:We use two colors.None of three points are on one line and by same colore I mean both are blue or both are red. | 2216943 | counting triangles in a graph or its complement
Given a simple graph [imath]G[/imath] [no loops or parallel edges] on six vertices, let [imath]G^c[/imath] denote its complement. It is known that either [imath]G[/imath] or [imath]G^c[/imath] must contain a triangle [imath]T[/imath] in it. [An example of a Ramsey number I think.] My question came up when I tried to find such a graph on six vertices which had only one [imath]T[/imath] in it or its complement. I believe if I did things right there are none of these. This I tried to check by drawing a copy of each graph [imath]G[/imath] on six vertices which has a [imath]T[/imath] in it, and then looking at the complements to see if any had no [imath]T[/imath]'s in them. I found none. One way to pose my question for a general number [imath]n[/imath] of vertices is then as follows: Consider the set of all graphs [imath]G[/imath] on [imath]n[/imath] vertices, and for each one construct [imath]G^c.[/imath] Count the number of [imath]T[/imath] in [imath]G[/imath] and add that to the number of [imath]T[/imath] in [imath]G^c.[/imath] Finally find the minimal such sum for the given [imath]n[/imath] and call that say [imath]M(n).[/imath] So my above six vertex conjecture is that [imath]M(6)=2.[/imath] By considering a cycle graph on five vertices we find [imath]M(5)=0.[/imath] So going further one could ask if anything is known about this minimum function [imath]M(n)[/imath] for larger [imath]n[/imath] [such as bounds, etc.] And does it have a name? I would also like to see a simpler way to handle the six vertex case, one not involving drawing all the graphs. |
2242152 | Let G be a non abelian group then its order can be...
Let [imath]G[/imath] be a non abelian group.Then , its order can be:(which will be correct options) [imath](1).~25[/imath] [imath](2).~55[/imath] [imath](3).~125[/imath] [imath](4).~35[/imath] I know ,what is abelian group,also I know order of a group.but I can't ans this question. Please help me. | 1126354 | order of non abelian group can't be what?
Let [imath]G[/imath] be a non abelian group; then its order can be: [imath]25[/imath] [imath]55[/imath] [imath]35[/imath] [imath]125[/imath] I think the order cannot be [imath]25[/imath] and [imath]35[/imath]. But from option [imath]55[/imath] and [imath]125[/imath] which one is not possible? Why not [imath]25[/imath] because every group of order [imath]p^2[/imath] is abelian. Here [imath]p[/imath] is [imath]5[/imath]. Why not [imath]35[/imath] because [imath]35=5*7[/imath] and [imath]5[/imath] does not divide [imath]7-1[/imath]. There there is only one group which is cyclic up to isomorphism of order [imath]35[/imath]. |
1853062 | Maximum variance
Consider a random variable [imath]X[/imath] with continuous probability density [imath]f(x)[/imath] and compact support, say [imath][a,b][/imath] with [imath]a<b[/imath]. Moreover, let [imath]f(x)[/imath] vanish at the boundary, i.e. [imath]f(a) = f(b) = 0[/imath]. Question: What is the maximum variance which such a random variable can take? EDIT: It should have been mentioned here, that this problem is already solved in literature for probability densities without any further restrictions and the solution is given by a disrete (non-continuous) probability with weight 1/2 at the boundary. See comments and the "duplicate-link". However, I'm asking for a continuous probability density instead, which is zero at the boundary. With these two additionnal conditions, the question is completely different. Maybe, a solution does not exist. Then, maybe there is a sequence of continuous probability densities [imath]f_n(x)[/imath], [imath]n=1,2,3...[/imath], whose limes has maximum variance. But I don't know how to construct such a sequence which gets peaked at the boundary. | 1455058 | What is the max variance distribution with fixed mean and support
Is there a upper bound on the variance for a distribution that is confined say, in [imath][0,1][/imath], and has fixed mean [imath]\mu \in [0, 1][/imath] |
2242786 | Group homomorphism is also ring homomorphism
Let [imath]A[/imath] and [imath]B[/imath] be two rings (not necessarily with unit, so rng i guess?) and [imath]f[/imath] a group homomorphism from [imath](A,+)[/imath] to [imath](B,+)[/imath]. We also have the property that [imath]f(xy)[/imath] is [imath]f(x)f(y)[/imath] or [imath]f(y)f(x)[/imath] for all x,y in [imath]A[/imath]. Prove that [imath]f[/imath] is a ring homomorphism if and only if there exists at least one pair [imath](a,a')[/imath] of distinct elements in [imath]A[/imath] such as [imath]f(aa')=f(a)f(a')[/imath]. Essentially, we have to prove that if there is one pair for which the "good order" holds, then it holds for all pairs. I have no idea how to approach this problem. | 398795 | "Almost" ring homomorphism
This is an exercise out of Herstein which seems pretty straightforward but is eluding me. Let [imath]R,R'[/imath] be rings and let [imath]\phi:R\to R'[/imath] be a mapping such that, for every [imath]x,y\in R[/imath]: [imath]{\rm 1.}\qquad\phi(x+y)=\phi(x)+\phi(y)[/imath] [imath]{\rm 2.}\qquad\phi(xy)=\phi(x)\phi(y)\qquad{\rm or}\qquad\phi(xy)=\phi(y)\phi(x)[/imath] I'm to show that one of the conditions holds uniformly. That is, for all [imath]x,y\in R[/imath], [imath]\phi(xy)=\phi(x)\phi(y)[/imath] or, for all [imath]x,y\in R[/imath], [imath]\phi(xy)=\phi(y)\phi(x)[/imath]. Of course, this doesn't exclude the possibility that both conditions hold uniformly (which happens when e.g. [imath]R'[/imath] is commutative). Herstein hints to fix [imath]a\in R[/imath] and to consider the sets [imath]W_a=\{x\in R\mid\phi(ax)=\phi(a)\phi(x)\},[/imath] [imath]V_a=\{x\in R\mid\phi(ax)=\phi(x)\phi(a)\}.[/imath] I have shown that one of [imath]W_a[/imath] or [imath]V_a[/imath] must be the entire ring [imath]R[/imath]. I think I'm missing something simple to complete the argument. Any hints? Edit I forgot to mention that in my copy of the text, the condition to be shown is stated as "[imath]\phi(xy)=\phi(x)\phi(y)[/imath] or [imath]\phi(x)=\phi(y)\phi(x)[/imath]". I am pretty confident that it's a typo, but maybe not? |
2242393 | Let [imath]\{x_n\}[/imath] be a sequence defined by [imath]x_1 = 3, x_{n+1} = \frac{1}{4-x_n}[/imath], for n [imath]\geq 1[/imath]. prove it converges
Let [imath]\{x_n\}[/imath] be a sequence defined by [imath]x_1 = 3, x_{n+1} = \frac{1}{4-x_n}[/imath], for n [imath]\geq 1[/imath]. prove it converges (a) Prove that [imath]\{x_n\}[/imath] is decreasing and bounded (below by [imath]0[/imath] and above by [imath]4[/imath]), i.e., prove that [imath]0 < x_{n+1} < x_n < 4[/imath], for all [imath]n \in \mathbb N[/imath]. (b) Does [imath]\{x_n\}[/imath] converge or diverge? Justify My attempt: (Could someone check) Claim: [imath]\{x_n\}[/imath] is bounded below 0 and bounded above by 4 (1), [imath]\{x_n\}[/imath] is strictly decreasing (2), for all [imath]n \in \mathbb N[/imath] Proof of (1) WTS: [imath]0 < x_n < 4, \forall n \in \mathbb N[/imath] Since the function is not continuous at x = 4 I will find its limit for a better bound [imath]x_n = \frac{1}{4-x_n} \leftrightarrow \left(x_{n} - \frac{4-\sqrt{12}}{2}\right)\left(x_n - \frac{4+\sqrt{12}}{2}\right)[/imath] Since the function is decreasing this will be the new bound [imath]0 < x_n < \frac{4-\sqrt{12}}{2}[/imath] I will prove this by induction: Base case (n=1) 0 < 3 < [imath]\frac{4-\sqrt{12}}{2} < 4[/imath] holds Inductive hypothesis: Let [imath]k \in \mathbb N[/imath] be arbitrary. Assume [imath]0 < a_k < \frac{4-\sqrt{12}}{2}[/imath] Inductive step: [imath]\forall k \in \mathbb N[/imath], [imath]0 < a_k < \frac{4-\sqrt{12}}{2} \to 0 < a_{k+1} < \frac{4-\sqrt{12}}{2}[/imath] [imath]a_{k+1} = \frac{1}{4-x_n} \text{ By definition}[/imath] [imath]< \frac{1}{4-\frac{4-\sqrt{12}}{2}} \text{ By I.H.}[/imath] [imath]= \frac{2}{4-\sqrt{12}}[/imath] [imath]= \frac{2(4+\sqrt{12})}{16-12}[/imath] [imath]= \frac{4+\sqrt{12}}{2} < 4[/imath] Now for bounded below [imath]a_{k+1} = \frac{1}{4-x_n} \text{ By def}[/imath] [imath]> \frac{1}{4-0} \text{ By I.H.}[/imath] [imath]> 0[/imath] Therefore by induction [imath]0 < x_n < \frac{4-\sqrt{12}}{2} < 4, \forall n \in \mathbb N[/imath] Proof of (2) WTS: [imath]x_{n} > x_{n+1}, \forall n \in \mathbb N[/imath] [imath]x_n > x_{n+1}[/imath] [imath]\leftrightarrow x_n > \frac{1}{4-x_n} \text{ By definition}[/imath] [imath]\leftrightarrow 0 > \left(x_{n} - \frac{4-\sqrt{12}}{2}\right)\left(x_n - \frac{4+\sqrt{12}}{2}\right)[/imath] By 1 we know that the first one is > 0 and the second < 0. Then (+)(-) = - < 0 Therefore it is decreasing. For (b) by the bounded monotone convergence theorem we know that this converges since it is bounded below and strictly decreasing. Is this right? I think I did the choosing the limit part wrong. If its decreasing I choose the smaller one as the upperbound right? | 2195654 | If [imath]x_1 = 3[/imath], [imath]x_{n+1} = \frac{1}{4-x_n}[/imath] for [imath]n \geq 1[/imath], prove the sequence is bounded below by [imath]0[/imath], above by [imath]4[/imath].
If [imath]x_1 = 3[/imath], [imath]x_{n+1} = \dfrac{1}{4-x_n}[/imath] for [imath]n \geq 1[/imath], prove the sequence is decreasing and bounded below by [imath]0[/imath], above by [imath]4[/imath]. Want to show: [imath]0 < x_{n+1} < x_n < 4[/imath] I decided to do the bounded part first. Base Case, [imath]n=1[/imath] [imath]0 < 3 < 4[/imath] [works] IH Suppose [imath]0 < x_k < 4[/imath], for some [imath]k\in\mathbb{N}[/imath] IS Show [imath]0 < x_{k+1} < 4[/imath] We know [imath]x_{k+1} = \dfrac{1}{4-x_k}[/imath] We also know that [imath]0 < x_k < 4[/imath], by IH. So we have: [imath]\dfrac{1}{4} < x_{k+1} < \infty[/imath] But this doesn't show the boundness. Where have I went wrong? |
2242888 | Determining [imath]\ {\displaystyle\lim_{n\to\infty}}\left(\frac{(n+1)^{n+1}}{n^n}-\frac{n^n}{(n-1)^{n-1}}\right)[/imath]
This limit was a trivia question. The trivia league is set up so you have all day to mull over questions and submit your answer, but it's not "trivia for mathematicians." The questions tend to be tough, but I wouldn't expect them to ask a really really hard Calculus problem... [imath]\lim_{n\to\infty}\frac{(n+1)^{n+1}}{n^n}-\frac{n^n}{(n-1)^{n-1}}[/imath] I can provide the correct answer but it's easy to guess by plugging in a large number for [imath]n[/imath]. No calculators allowed! | 2206370 | Find [imath]\lim_{n \to \infty} \left[\frac{(n+1)^{n + 1}}{n^n} - \frac{n^{n}}{(n-1)^{n-1}} \right][/imath] (a question asked at trivia)
My friend's trivia league had this math question: [imath]\lim_{n \to \infty} \left[\frac{(n+1)^{n + 1}}{n^n} - \frac{n^{n}}{(n-1)^{n-1}} \right][/imath] After computing a few values, one could guess the answer is [imath]e[/imath] = 2.718...But how can we prove that is the limit? Someone offered up a hand-wavy proof like this: \begin{align} \lim_{n \to \infty} \left[\frac{(n+1)^{n + 1}}{n^n} - \frac{n^{n}}{(n-1)^{n-1}} \right] & = \lim_{n \to \infty} \left[\frac{(n+1)(n+1)^{n}}{n^n} - \frac{n \cdot n^{n-1}}{(n-1)^{n-1}} \right] \\ &= \lim_{n \to \infty} \left[(n+1)\frac{(n+1)^{n}}{n^n} - n\frac{n^{n-1}}{(n-1)^{n-1}} \right] \\ &= \lim_{n \to \infty} \left[(n+1)\left(1 + \frac{1}{n} \right)^n - n\left(\frac{n - 1 + 1}{n-1} \right)^{{n-1}} \right] \\ &= \lim_{n \to \infty} \left[(n+1)\left(1 + \frac{1}{n} \right)^n - n\left(1 + \frac{1}{n-1} \right)^{n-1} \right] \\ &= \lim_{n \to \infty} \left[(n+1)e - n \cdot e \right] \\ &= \lim_{n \to \infty} e \\ &= e \end{align} The part about substituting [imath]e[/imath] is hand-wavy since technically this is an indeterminate form of [imath]\infty - \infty[/imath]. And using [imath]e[/imath] as as upper bound did not lead me to an easy proof either. Is there a way to rigorously prove the limit? I tried a few approaches: (a) sandwiching the limit--I could prove [imath]e[/imath] was a lower bound, but I could not find a suitable upper bound converging to [imath]e[/imath], (b) using L'Hopital's rule with no luck, (c) using the mean value theorem--but that also got complicated. So this is a pretty tough problem to ask at trivia! Is there a way to prove this limit formally? Sources Trivia question: http://learnedleague.com/question.php?72&16&4 Thread on trivia: http://learnedleague.com/viewtopic.php?f=10&t=7961&hilit=euler Hand-wavy proof: http://imgur.com/rIXghhw Idea for mean value theorem: http://www.pharout.com/trickylimitproblem.pdf |
2240746 | Derivative of [imath]\sin x^{\cos x}[/imath]
I found this problem while studying online. When I take the derivative of [imath]\sin x^{\cos x}[/imath] using the rule: [imath](a^x)'=a^x\ln a,[/imath] and using the chainrule I get: [imath]\sin x^{\cos x} \ln \sin x\cdot(-\sin x).[/imath] However in the page I found this problem on, the solution is listed as: [imath]\sin x^{\cos x}\left(\frac {\cos^2x} {\sin x}-\sin x\ln(\sin x)\right)[/imath] I don't believe my solution was wrong but I'd like to know how to get to the solution that's on the page. Thanks in advance :-) | 545203 | How would I differentiate [imath]\sin{x}^{\cos{x}}?[/imath]
How can I differentiate [imath]\displaystyle \sin{x}^{\cos{x}}[/imath]? I know the power rule will not work in this case, but logarithmic differentiation would. I'm not sure how to start the problem though and I'm not too comfortable with logarithmic differentiation. |
2243219 | Rank of matrix whose coefficients are [imath]a_{i,j}=\gcd(i,j)[/imath]
What are he possible methods for showing that the [imath]n[/imath] by [imath]n[/imath] matrix [imath]A[/imath] whose coefficients are given by [imath]a_{i,j}=\gcd(i,j)[/imath] is inversible? | 1000026 | Determinant value of a square matrix whose each entry is the g.c.d. of row and column position
Let [imath]A=(a_{ij})[/imath] be a [imath]n \times n[/imath] matrix with [imath]a_{ij}=\gcd(i,j) , \forall i,j=1,2, \cdots, n[/imath] , then how do we prove [imath]\det A=\prod_{i=1}^n \phi(i)[/imath] ? , where [imath]\phi[/imath] is the Euler's phi function |
2242105 | [imath]2 + 2 = 5[/imath] (Fake proof, or ?)
Basically, there is no error in the following steps(as it seems), but there is some error due to which 2 + 2 = 5. What is it? -20 = -20 16-36 = 25-45 16-36+(81/4) = 25-45+(81/4) (4^2)-(2*4*9/2)+((9/2)^2) = (5^2)-(2*5*(9/2))+((9/2)^2) (4-(9/2))^2 = (5-(9/2))^2 4-(9/2) = 5-(9/2) 4 = 5 2+2 = 5 | 2449459 | What's wrong with this fake proof that [imath]2+2=5[/imath]?
Can anybody please describe how and why it is even possible? If there is anything wrong what it is,? Prove: [imath]2+2=5[/imath] [imath]-20=-20[/imath] [imath]16-36=25-45[/imath] [imath]16-36+\frac{81}{4}=25-45+\frac{81}{4}[/imath] [imath]4^2-2\cdot4\cdot\frac92+\left(\frac92\right)^2=5^2-2\cdot5\cdot\frac92+\left(\frac92\right)^2[/imath] [imath]\left(4-\frac92\right)^2=\left(5-\frac92\right)^2[/imath] [imath]4=5-\frac92+\frac92[/imath] [imath]4=5[/imath] [imath]2+2=5[/imath] |
2111032 | Pointwise convergence and convergence of integrals implies [imath]L^1[/imath] convergence
From Donald Cohn's Measure Theory, section 2.4, exercise 10. Let [imath](X, A, \mu)[/imath] be a measure space, and let [imath]f[/imath] and [imath]f_1, f_2, \dots[/imath] be non-negative functions that belong to [imath]L^1(X, A, \mu, R)[/imath] and satisfy (i) [imath]\{f_n\}_n[/imath] converges to [imath]f[/imath] almost everywhere; (ii) [imath]\int fd\mu = \lim_n\int f_nd\mu[/imath]. Show that [imath]\lim_n\int |f_n - f|d\mu = 0[/imath]. I let [imath]f_n = 1/n[/imath] on [imath][n, n+1][/imath] and [imath]0[/imath] elsewhere, and let [imath]f=0[/imath], then all the convergence theorems in that section (dominated convergence, and monotone convergence) failed. | 2733636 | Convergence of integrals implies convergence in [imath]L^1[/imath]
I have this problem which is exactly the opposite of what I would find easy to prove! Consider a sequence of Lebesgue measurable non-negative functions [imath](f_n)[/imath] such that [imath]f_n\rightarrow f[/imath] pointwise for some [imath]f\in L^1[/imath], and also that [imath]\int_{\mathbb{R}}f_n\rightarrow \int_{\mathbb{R}}f[/imath]. Prove [imath]||f_n-f||_1 \rightarrow 0[/imath]. This is actually giving me a hard time because it goes against my intuition. How does pointwise convergence and integral convergence imply convergence in [imath]L^1[/imath]? Why? How? Any help would be appreciated... Thanks. |
2239203 | Why does [imath]\lim\limits_{x\to\infty}(x!)^{1/x}\neq 1?[/imath]
Why does [imath]\lim\limits_{x\to\infty}(x!)^{1/x}\neq 1?[/imath] As far as I know, anything to the power of [imath]0[/imath] is [imath]1[/imath]. We have a factorial raised to [imath]1/\infty=0[/imath], but the limit is not [imath]1[/imath]? I don't even know what the limit is. But it seems like infinity? Why is this? | 1967813 | Evaluate [imath]\lim_{x\to \infty}\ (x!)^{1/x}[/imath]
Here's the problem... [imath]\lim\limits_{x\to \infty}\ (x!)^{1/x}[/imath] I've deduced the answer to this is [imath]\infty[/imath], but haven't exactly shown that. I'm getting [imath]\infty^0[/imath], so did both tricks where you change it to [imath]\exp\left({\lim_{x\to\infty}} \frac{1/x}{1/\ln(x!)}\right) = \exp\left(\frac{0}{0}\right)[/imath] And [imath]\exp\left({\lim_{x\to\infty}} \frac{\ln(x!)}{1/(1/x)}\right) = \exp\left(\frac{\infty}{\infty}\right)[/imath] And for both of which, I need to do L'Hospital on this [imath]\ln(x!)[/imath] which I haven't the slightest idea how to do... Any suggestions that don't include Stirling's Approximation? My prof eluded to this being out of the question unless we were going to prove Stirling's first |
1013351 | Need help with [imath]\int_0^1 \frac{\ln(1+x^2)}{1+x} dx[/imath]
Can the definite integral [imath]\int_0^1\dfrac{\ln(1+x^2)}{1+x}dx[/imath] be evaluated using the technique of “differentiation under integral sign”. I don't want a complete solution, just the parameter would do. PS: An alternative approach (preferably simple) would also be welcome as long as it doesn't involve contour integration. | 406719 | integral of [imath]\int_{0}^{1}\frac{\ln(x^{2}+1)}{x+1}dx[/imath] using contour integration?
I have an interest in contour integration. I am not that good at it, but I enjoy learning what I can about it. Here is a version of a rational log integral rarely encountered. [imath]\displaystyle \int_{0}^{1}\frac{\ln(x^{2}+1)}{x+1}dx=\frac{3}{4}\ln^{2}(2)-\frac{{\pi}^{2}}{48}[/imath] I can do this using real methods (via double integral and substitution). I can post my workings if anyone would be interested. My question is, can this be evaluated using contour integration due to the limits being [imath][0,1][/imath] instead of [imath][0,\infty)[/imath]?. Contours may not be the most efficient way to go about it, but what is the course of action when the limits are 0 to 1 instead of 0 to infinity?. |
2243586 | Finding [imath]\lim_{n\rightarrow \infty}\frac{1+2^2+3^3+4^4+\cdots +n^n}{n^n}[/imath]
Finding [imath]\lim_{n\rightarrow \infty}\frac{1+2^2+3^3+4^4+\cdots +n^n}{n^n}[/imath] Attempt: [imath]\lim_{n\rightarrow \infty}\bigg[\frac{1}{n^n}+\frac{2^2}{n^n}+\frac{3^3}{n^n}+\cdots \cdots +\frac{n^n}{n^n}\bigg] = 1[/imath] because all terms are approaching to zero except last terms but answer is not [imath]1[/imath] , could some help me to solve it , thanks | 2221844 | Computing [imath] \lim_{n\rightarrow \infty}\left(\frac{1^n+2^n+\cdots \cdots +n^n}{n^n}\right)[/imath]
Computing [imath]\displaystyle \lim_{n\rightarrow \infty}\left(\frac{1^n+2^n+\cdots \cdots +n^n}{n^n}\right)[/imath] Attempt: [imath]\displaystyle \lim_{n\rightarrow \infty}\bigg[\left(\frac{1}{n}\right)^n+\left(\frac{2}{n}\right)^n+\cdots \cdots +\left(\frac{n-1}{n}\right)^n+\left(\frac{n}{n}\right)^n\bigg] = 1[/imath] Is my answer right? If not, then could someone help me? Thanks. |
2243379 | Number of real or purely imaginary solution
What are the number of real or purely imaginary solution of the equation , [imath]z^3 +iz -1=0[/imath] In this I tried to substitute [imath]z=x+iy[/imath] and solve it, but it is getting too long, and the answer is also not coming | 1871407 | The real or purely imaginary solutions of the equation, [imath]z^3+iz-1=0[/imath] are?
The number of real or purely imaginary solutions of the equation, [imath]z^3+iz-1=0[/imath] is? I substituted [imath]x+iy=z[/imath] and am getting two equations which seem impossible to solved.What would be the correct approach ? |
2243704 | Maximum value of |Z|
Question :What is the maximum value of [imath]\lvert z\rvert[/imath] when [imath]z[/imath] satisfies the condition [imath]\lvert z+\frac{2}{z}\rvert=2[/imath] My try: By removing modulus, and simplifying as [imath]-2<z+\frac{2}{z}<2[/imath], I am not getting the required answer. Someone please help. | 853992 | Max. and Min. value of [imath]|z|[/imath] in [imath]\left|z+\frac{2}{z}\right| = 2\,[/imath]
If [imath]z[/imath] is a complex no. such that [imath]\displaystyle \left|z+\frac{2}{z}\right| = 2\,[/imath] Then find max. and min. value of [imath]\left|z\right|[/imath]. [imath]\bf{My\; Try:}[/imath] Given [imath]\displaystyle \left|z+\frac{2}{z}\right| = 2\Rightarrow \left|z+\frac{2}{z}\right|^2 = 2^2=4[/imath]. So [imath]\displaystyle \left(z+\frac{2}{z}\right)\cdot \left(\bar{z}+\frac{2}{\bar{z}}\right) = 4\Rightarrow \left|z\right|^2+2\left(\frac{z}{\bar{z}}+\frac{\bar{z}}{z}\right)+\frac{1}{|z|^2} = 4[/imath]. Now how can I find the max. and min. values of [imath]|z|[/imath]? Help me please. Thanks |
2244394 | Suppose our ring [imath]R = k[t][/imath]. Let [imath]M[/imath] be the [imath]R[/imath]-module [imath]R[x]/(tx - t)[/imath]. Prove that [imath]M[/imath] isn't flat.
I proved this problem before by the following theorem. Over PID Flat iff torsion free. Then, from this fact, we know that M isn't flat as it has torsion element. In the problem we were given hint that to consider the following sequence [imath]0 \rightarrow (t) \rightarrow R[/imath]. I am guessing, we then proceed to show that [imath]M \otimes_R (t)[/imath] doesn't inject into [imath]M \otimes_R R[/imath]. I couldn't show that this doesn't inject. If someone could help that would be nice. | 2220392 | Example of non-flat modules
Let [imath]R = \mathbb{C}[t][/imath] be a ring of polynomials in variable [imath]t[/imath] with coefficients in the field of complex numbers [imath]\mathbb{C}[/imath] and let [imath]N = R[x]/(tx-t).[/imath] I claim that [imath]N[/imath] is not a flat [imath]R[/imath]- module. If we consider the exact sequence [imath]0 \rightarrow(t) \rightarrow R[/imath] such that the ideal [imath](t)[/imath] is viewed as an [imath]R[/imath]-module. We know that the above sequence exact iff the map [imath](t) \rightarrow R[/imath] is injective. Let us assume that N is flat then this means that [imath]N \otimes (t) \rightarrow N \otimes R[/imath] has to be injective. How do I go from here? Thanks. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.