qid
stringlengths 1
7
| Q
stringlengths 87
7.22k
| dup_qid
stringlengths 1
7
| Q_dup
stringlengths 97
10.5k
|
|---|---|---|---|
2054397
|
Can the product of [imath]k[/imath] conseuctive integers be a perfect square?
Let [imath]k\ge 2[/imath] be an integer. Can a product of [imath]k[/imath] consecutive integers be a perfect square?
|
33338
|
Product of Consecutive Integers is Not a Power
Is it true that the product of [imath]n>1[/imath] consecutive integers is never a [imath]k[/imath]-th power of another integer for any [imath]k \geq 2[/imath]? I can see this is true in certain cases. For instance if the product ends on a prime, But how would one prove this in general? Thanks for any help or suggestions.
|
2053768
|
Eigenvalue and eigenvector [imath]Av=λv[/imath] , [imath]A^{-1}v=\frac{1}{\lambda}v[/imath]
Suppose that [imath]A[/imath] is invertible, and [imath]v[/imath] is an eigenvector of [imath]A[/imath], corresponding to some eigenvalue [imath]λ \in \Bbb R[/imath]. Show that [imath]v[/imath] is an eigenvector of [imath]A^{−1}[/imath] , corresponding to the eigenvalue [imath]µ = 1/λ[/imath]
|
1852035
|
Is this how eigenvalues of some matrix [imath]A[/imath] are related to the inverse of [imath]A[/imath]?
Let [imath]A[/imath] be an invertible [imath]n\times n[/imath] matrix. If [imath]Av = \lambda v \qquad (1)[/imath] for some [imath]v[/imath] and [imath]\lambda[/imath] then [imath]\lambda[/imath] is an eigenvalue of [imath]A[/imath] and [imath]v[/imath] a corresponding eigenvector. Equation [imath](1)[/imath] may be written as [imath]Av = \lambda v = \lambda Iv \quad\Leftrightarrow\quad Av - \lambda I v = 0 \quad\Leftrightarrow\quad (A - \lambda I)v = 0[/imath] where [imath]I[/imath] is the identity matrix. The last equation can be used to find correct [imath]\lambda[/imath]'s via the fundamental theorem of algebra (first take determinant). On the other hand, it is also possible to write [imath]Av = \lambda v \quad\Leftrightarrow\quad A^{-1}Av = A^{-1}\lambda v \quad\Leftrightarrow\quad Iv - A^{-1}\lambda v = 0 \quad\Leftrightarrow\quad (I - A^{-1}\lambda)v = 0[/imath] since it was assumed [imath]A[/imath] is invertible. Now, since [imath](A - \lambda I)v = 0[/imath] and [imath](I - A^{-1}\lambda)v = 0[/imath] it must be that [imath]A - \lambda I = I - A^{-1}\lambda[/imath] which connects the matrix [imath]A[/imath] to its inverse via the eigenvalues of [imath]A[/imath]. I did some calculations and it seems this works for some invertible matrices [imath]A[/imath] but not all. Am I doing a mistake somewhere?
|
2055063
|
Is it possible to compute [imath]x^{1389}+(1/x)^{1389}[/imath] from some premises?
Suppose that we have [imath] x+(1/x)=1[/imath], can we compute the expression [imath]x^{1389}+(1/x)^{1389}[/imath] from that? and how?
|
1465320
|
[imath]x^{2000} + \frac{1}{x^{2000}}[/imath] in terms of [imath]x + \frac 1x[/imath].
If [imath]x + \frac{1}{x} = 1[/imath], then what is [imath] x^{2000} + \frac{1}{x^{2000}} = ?[/imath]
|
2055065
|
Show that [imath]\mathbb{T}^1 \cong \mathbb{C}^*/\mathbb{R}_{>0}[/imath]
I'm trying to use the First Isomorphism Theorem to show that [imath]\mathbb{T}^1 \cong \mathbb{C}^*/\mathbb{R}_{>0}[/imath] by constructing a surjective group homomorphism from the nonzero complex numbers to the circle group whose kernel is the set of positive reals. I haven't yet taken a course in complex variables, so I did some digging around and could only find a homomorphism from [imath]\mathbb{R}^*[/imath] to [imath]\mathbb{T}^1[/imath] defined by [imath]f(\theta) = e^{i\theta}[/imath]. I don't really know what that means, so I'm struggling to construct any homomorphism [imath]\phi : \mathbb{C}^* \to \mathbb{T}^1[/imath], let alone one whose kernel is [imath]\mathbb{R}_{>0}[/imath]. EDIT: Does [imath]f(z) = \frac{z}{|z|}[/imath] work? EDIT2: Probably not. I think the image of this is actually just [imath]\{\pm 1\}[/imath]. I really have no idea how to work with complex numbers at all.
|
2050423
|
Prove Quotient Group Isomorphism
Let [imath]G = (\mathbb{C} - \{0\}, \cdot)[/imath], a subgroup [imath]U = \{x+yi \mid x^2 + y^2 = 1\}[/imath]. Use the Fundamental Theorem to show that [imath]G/(\mathbb{R}_{> 0},\cdot)[/imath] is isomorphic to [imath]U[/imath] where [imath]\mathbb{R}_{> 0}[/imath] denotes the positive reals. I am struggling to find the correct mapping from [imath]G[/imath] to [imath]U[/imath] that results in the the kernel being the nonnegative reals under multiplication. I'm hoping to get help with this mapping then I can show it is a homomorphism and onto and can use the Fundamental Theorem.
|
2055094
|
Arbitrarily big prime gap
There is a well-known proof of the fact that the prime gaps can be arbitrarily large. Namely, for any natural number [imath]n[/imath], the consecutive integers [imath](n+1)!+2,...,(n+1)!+(n+1)[/imath] are never prime. But clearly, this proof does not guarantee that [imath](n+1)!+1[/imath] and [imath](n+1)!+(n+2)[/imath] are prime. How can we show that the prime gap can be exactly [imath]n[/imath] for any even [imath]n[/imath]?
|
1634110
|
Are there prime gaps of every size?
Is it true that for every even natural number [imath]k[/imath] there exists some [imath]n \in \mathbb{N}[/imath] such that [imath]g_n = p_{n+1} - p_n = k[/imath]? I don't know how to approach the problem at all, and in fact I don't even know enough about prime gaps to even form a conjecture as to the answer. I feel like the answer is "yes", but only because that would be "nicer" than having some even integers never appear in sequence of prime gaps. I hope it's not an unsolved problem! Edit: My question is distinct from Polignac's Conjecture, since I ask if there is at least one prime gap, instead of infinitely many prime gaps, for every size.
|
2054809
|
Fibonacci sequence induction question, [imath]F_1+F_2+\ldots+F_{n-1}=F_{n+1}-1[/imath]
The Fibonacci sequence is defined as the sequence where [imath]F_1 = 1,F_2= 1[/imath] and [imath]F_i=F_{i-1}+ F_{i-2}[/imath]. Use induction to prove the that for [imath]n\ge 2[/imath], [imath]F_1+F_2+ \ldots+F_{n-1}=F_{n+1}-1[/imath]
|
2053325
|
How do I add summation formula's like this: [imath]F(n+2) = 1 + \sum\limits_{i=0}^{n} F(i)[/imath]?
I have a Fibonacci problem ... basically I have to show that: [imath]F(n+2) = 1 + \sum_{i=0}^{n} F(i)[/imath] Using strong induction, I must now show that: [imath]F(n+2) = F(n+1) + F(n)[/imath] [imath]F(n+2) = (1 + \sum_{i=0}^{n-1} F(i)) + (1 + \sum_{i=0}^{n-2} F(i))[/imath] [imath]F(n+2) = 2 + \sum_{i=0}^{n-1} F(i) + \sum_{i=0}^{n-2} F(i)[/imath] How do I add these two summations to show that they equal the first equation (at the top)? Please provide the steps with explanation as to what you are doing. Thanks
|
2050601
|
How to prove the diagonal subgroup is a normal subgroup implies that group is abelian?
Let [imath]G[/imath] be a group and consider the subgroup [imath]H=\{(g,g)\mid g \in G\}[/imath] of [imath]G\times G[/imath]. Show that if [imath]H[/imath] is a normal subgroup of [imath]G\times G[/imath], then [imath]G[/imath] is abelian. (Hint: consider [imath](a,e) \in G\times G[/imath] and [imath](h,h) \in H[/imath].) I'm not sure how to prove this since [imath](a,e)(h,h)=(h,h)(a,e)[/imath] isn't implied even if its a normal subgroup. Is there another way to attack this with the hint given?
|
1999629
|
Prove that [imath]G[/imath] is an abelian group if [imath]\{(g, g):g\in G\}[/imath] is a normal subgroup.
Let [imath]G[/imath] be a group and let [imath]D=\{(g, g):g\in G\}[/imath]. If [imath]D[/imath] is a normal subgroup of [imath]G\times G[/imath], prove that [imath]G[/imath] is an abelian group. My attempt: [imath]D[/imath] is a normal subgroup of [imath]G\times G[/imath]. [imath]\implies(a, b)D=D(a, b)\ \forall(a, b)\in G\times G[/imath] So for a given [imath](a, b)\in G[/imath] and [imath](g, g)\in D[/imath], [imath]\exists(g', g')\in D[/imath] such that [imath](a, b)(g, g)=(g', g')(a, b)[/imath] [imath](ag, bg)=(g'a, g'b)[/imath] [imath]ag=g'a[/imath] and [imath]bg=g'b[/imath] I feel like I've used all of the information given but don't know how to conclude that [imath]G[/imath] is abelian. Any suggestions?
|
2055068
|
Residue calculus
How to solve the following integrand: [imath]\int^\infty_{-\infty}\frac{\sin x}{x} dx\;?[/imath] I know how to evaluate it by using the path of a rectangle, but it is too difficult and would make some mistakes. I am wondering if there is a simpler way to solve questions like this. Thanks！
|
219755
|
Showing [imath]\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}[/imath] using complex integration
Recently I had to use the fact that the Dirichlet integral evaluates as [imath]\int_{0}^{\infty} \frac{\sin{x}}{x} \ dx = \frac{\pi}{2}[/imath] a couple of times. There already is a question that specifically ask for methods to show this result [imath]\textbf{not}[/imath] using complex integration. In this question I am interested in seeing the derivation via contour integration. ( I am aware of the wikipedia entry, but am looking for more detail )
|
2055037
|
What happens when you don't require rings to be abelian groups?
My question is as stated in the title: What do we lose when we drop the condition that a ring [imath]R[/imath] be an abelian group (i.e. it may be the case that [imath]a+b \neq b+a[/imath] for some [imath]a,b \in R[/imath]). If we allow this condition then we may consider, for example, any group [imath]G[/imath] to be a (noncommutative) ring with multiplication being conjugation in [imath]G[/imath] and addition being multiplication in [imath]G[/imath]. In such a "ring" [imath]G[/imath], we always have [imath]0=1[/imath], so I suspect things might get strange. Has anyone thought at all about such objects? Are they entirely uninteresting?
|
262206
|
What happens if we remove the requirement that [imath]\langle R, + \rangle[/imath] is abelian from the definition of a ring?
Ever since I learned the definition of a ring, I've wondered why the additive group is required to be abelian. What happens if we allow [imath]\langle R, + \rangle[/imath] to be nonabelian as well as [imath]\langle R, \cdot \rangle[/imath]? Is this impossible? If not, what is the this type of algebraic structure called? Here is the definition of a ring that I am using: [imath]\langle R , + \rangle[/imath] is an abelian group with identity [imath]0[/imath]. [imath]\langle R , \cdot \rangle[/imath] is associative. [imath]a\cdot (b + c) = a\cdot b + a \cdot c[/imath]. [imath](a + b)\cdot c = a\cdot c + b \cdot c[/imath].
|
2055058
|
Prove gcd is one of the following {1, 3}
Prove that if [imath]\gcd(a, b) = 1[/imath] then [imath]\gcd(2a + b, a + 2b) \in \{1, 3\}[/imath] I can get up to: [imath]\gcd(2a + b, a + 2b) | 3(a + b)[/imath] But I am unable to move forwards. I would like a hint? BY Bezout's, there exists integers [imath]x, y[/imath] such that [imath]ax + by = 1[/imath]. [imath]d = \gcd(2a + b, a + 2b) \implies (2a + b)u + (a + 2b)t = d[/imath] for integers existing [imath]u, t[/imath]. But this isn't helping much. What can I do?
|
1976143
|
If [imath]a[/imath] and [imath]b[/imath] are relatively prime, prove that [imath]a + 2b[/imath] and [imath]2a+b[/imath] are also relatively prime or have a gcd of [imath]3[/imath].
If [imath]a[/imath] and [imath]b[/imath] are relatively prime, prove that [imath]a + 2b[/imath] and [imath]2a+b[/imath] are also relatively prime or have a [imath]\gcd[/imath] of [imath]3[/imath]. I'm very new to number theory so please don't assume I am familiar with some of the terminology.
|
2055355
|
Find [imath]P(n+1)[/imath] where [imath]P(x)[/imath] is a [imath]n[/imath] degree polynomial.
A polynomial [imath]P(x)[/imath] of [imath]n[/imath] degree satisfies [imath]P(k)=2^k[/imath] for [imath]k = 0,1,2,3......,n[/imath]. Find the value of [imath]P(n+1)[/imath]. How can I proceed in solving such problems.
|
1244188
|
If for a polynomial [imath]P(k) = 2^k[/imath] for [imath]k = 0, 1, . . . , n[/imath], what is [imath]P(n+1)[/imath]?
For a polynomial [imath]P(x)[/imath] of degree [imath]n[/imath], [imath]P(k) = 2^k[/imath] for [imath]k = 0, 1, 2, . . . , n[/imath]. Find [imath]P(n+1)[/imath]. If [imath]n=1[/imath], [imath]P(x)=x+1[/imath] and [imath]P(2)=3[/imath]. If [imath]n=2[/imath], [imath]P(x)=0.5x^2+0.5x+1[/imath] and [imath]P(3)=7[/imath]. How to approach further cases? I am stuck.
|
2055480
|
Conditional convergence of a power series on the [imath]z \in \mathbb{C}[/imath] for which [imath]|z|=\rho[/imath]
Consider the following complex power series [imath]\sum_{n \geq 1} \frac{z^n}{n} \,\,\,\,\,\,\, z \in \mathbb{C}[/imath] It surely converges conditionally for [imath]z=-1[/imath] (for alternating series test) and for [imath]z=1[/imath] it diverges (it is the harmonic series). My question is: how can one show that the power series converges conditionally for any [imath]z \in \mathbb{C}[/imath] such that [imath]|z|=1[/imath] (except for [imath]z=1[/imath])?
|
193542
|
Behavior of [imath]\sum_{n=1}^\infty n^{-1}z^n[/imath] on the circle of convergence
Consider the following complex power series :[imath]\sum_{n=1}^\infty\frac{z^n}{n}[/imath] The radius of convergence of this series is [imath]1[/imath] and the series is divergent for [imath]z=1[/imath]. I want to know what are the values of [imath]z\in C:=\lbrace z\in\mathbb{C}: |z|=1\rbrace[/imath], the circle of convergence, for which the given series converges.
|
2055983
|
Proving that [imath]\mathbb{Z}_2\times \mathbb{Z}_2[/imath] and [imath]\mathbb{Z}_4[/imath] are not isomorphic
How can I show that [imath]\mathbb{Z}_2[/imath][imath]\times[/imath] [imath]\mathbb{Z}_2[/imath] and [imath]\mathbb{Z}_4[/imath] are not isomorphic? I don't even know how to start Thanks for any help
|
1626908
|
Prove [imath]\mathbb{Z}_2 \times \mathbb{Z}_2[/imath] is not isomorphic to [imath]\mathbb{Z}_4[/imath]
In trying to understand why there are two sets of groups of order 4. I know that there exists the Vierergruppe of [imath]\mathbb{Z}_2 \times \mathbb{Z}_2[/imath] and the group [imath]\mathbb{Z}_4[/imath] but I do not understand why they are not isomorphic
|
1281804
|
A Boundary Value Problem [imath]y^{''} + xy = 0, x \in [a, b][/imath]
The following problem is an exam problem, which I could not do. But I attempted it as far as I can do. Let [imath]y[/imath] be a nontrivial solution of the boundary value problem [imath]y^{''} + xy = 0, x \in [a, b], y(a) = y(b) = 0.[/imath] Then [imath]b > 0[/imath] [imath]y[/imath] is monotonic in [imath](a, 0)[/imath] if [imath]a < 0 < b.[/imath] [imath]y^{'}(a) = 0.[/imath] [imath]y[/imath] has infinitely many zeros in [imath][a, b].[/imath] It is a Sturm Liouville Problem. If [imath]y[/imath] is a nontrivial solution to the problem, then [imath]y[/imath] is an eigen function corresponding to the eigen value 0. Then we know that the eigen values of a Sturm Liouville Problem can be arranged in an strictly increasing sequence [imath]\lambda_1 < \lambda_2 < ... < \lambda_n < ... [/imath] which tends to infinity. Also we know that corresponding to each eigen value [imath]\lambda_n [/imath], the eigen function [imath]y_n(x)[/imath] has exactly [imath]n - 1[/imath] roots in [imath](a, b).[/imath] Then the option 4 is False. But I do not know what the [imath]n[/imath] is ? I do not have any idea about other options. I could not even solve the [imath]y^{''} + xy = 0[/imath]. Can it be done by Power series method?
|
819550
|
Properties of [imath]y[/imath] if [imath]\frac{d^2y}{dx^2}+xy=0[/imath], [imath]x \in [a,b],[/imath] and [imath]y(a)=y(b)=0.[/imath]
Let [imath]y[/imath] be a nontrivial solution of the boundary value problem [imath]\frac{d^2y}{dx^2}+xy=0,\ x \in [a,b][/imath] [imath]y(a)=y(b)=0[/imath] Then: [imath]b>0[/imath] [imath]y[/imath] is monotone in [imath](a,0)[/imath] if [imath]a<0<b[/imath]. [imath]y'(a)=0[/imath]. [imath]y[/imath] has infinitely many zeros in [imath][a,b][/imath] I have no idea how to solve it and which options is/ are correct. please help me.
|
2055097
|
General Form for a Parametric Parabola
This question is an inverse of this other question on the Parametric Form for a General Parabola. What is the general form, ie. [imath](Ax+Cy)^2+Dx+Ey+F=0[/imath] , for the parabola given in parametric form as follows: [imath]\big(at^2+bt+c,\;\; pt^2+qt+r\big)[/imath] In other words, find [imath]A,C,D,E,F[/imath] in terms of [imath]a,b,c,p,q,r[/imath]. I've posted a solution, but would like to see other approaches to this problem. NOTE (Added April 2018) See Latest Addendum to solution which provides Cartesian equation in neat matrix determinant form.
|
530310
|
Parabola in parametric form
Show that the following system of parametric equations describes a line or a parabola: [imath]\begin{cases} x=a_1t^2+b_1t+c_1 \\ y=a_2t^2+b_2t+c_2 \end{cases}, t\in\mathbb{R}.[/imath]
|
2055955
|
rank(AB) = rank(A) if B is invertible
If [imath]B[/imath] is invertible, show that rank([imath]AB[/imath]) = rank([imath]A[/imath]). I've seen this question asked elsewhere but all had answers I didn't understand. I know how to solve the following problem If [imath]A[/imath] is invertible, then rank([imath]AB[/imath]) = rank([imath]B[/imath]) Because if [imath]Bx=0[/imath], then [imath]ABx = A0 = 0[/imath], and when [imath]ABx=0[/imath] then [imath]Bx=0[/imath] because [imath]A[/imath] is invertible, so null([imath]AB[/imath])=null([imath]A[/imath]), and by the rank-nullity theorem, rank([imath]A[/imath]) = rank([imath]AB[/imath]). However when [imath]B[/imath] is invertible, as in the problem we have to tackle, I don't know how to use that fact. [imath]ABx = 0[/imath], but [imath]B[/imath] is in the middle so we can't simply get rid of it to get a meaningful expression. Does someone know how to tackle this?
|
1200565
|
Prove $\operatorname{rank}(AB) = \operatorname{rank}(B)$
I am given the info that A is an invertible m by m matrix and B is a m by n matrix. By solving for the null space [imath]ABx=0[/imath], I get [imath]A^{-1}ABx=A^{-1}0[/imath], therefore [imath]Bx=0[/imath]. I concluded that the null spaces of A and AB must be equivalent, and therefore the nullity(AB) = nullity(B) = some number p. Therefore rank(AB) = n-p = rank(B) (because both AB and B are m by n matrices). Is this proof valid, specifically the null space step?
|
2055576
|
How to factor this expression?
Factor: [imath]x^{5}+a x^{3}+b x^{2}+cx+d[/imath] While: [imath]d^{2}+c b^{2}=abd[/imath] Please help me with this question. I don't have any clue. EDIT: I saw this but it was not explained what do we do after substitution. I still need help.
|
1847005
|
factorize [imath]x^5+ax^3+bx^2+cx+d[/imath] if [imath]d^2+cb^2=abd[/imath]
I want to factorize [imath]x^5+ax^3+bx^2+cx+d[/imath] if [imath]d^2+cb^2=abd[/imath] but I don't know how to use the second equality.I tried a lot but I cannot know how to use it for example it is [imath]d^2[/imath] but we have [imath]d[/imath] and if we calculate [imath]d[/imath] using that equality it would be harder.Any hints?
|
2056483
|
How to solve given Recurrence Equation [imath]a_{n} = 6n^2 + 2n + a_{n - 1}[/imath]?
How to solve given Recurrence Equation [imath]a_n = 6n^2 + 2n + a_{n - 1}[/imath] ? Given [imath]a_{1} = 8[/imath] Find the value of [imath]a_{99}[/imath] ? Is there any procedure or approach for such questions ?
|
1668618
|
Solving recurrence relation?
Consider the recurrence relation [imath]a_1=8,a_n=6n^2+2n+a_{n−1}[/imath]. Let [imath]a_{99}=K\times10^4[/imath] .The value of [imath]K[/imath] is______ . My attempt : [imath]a_n=6n^2+2n+a_{n−1}[/imath] [imath]=6n^2+2n+6(n−1)^2+2(n−1)+a_{n−2}[/imath] [imath]=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+a_1[/imath] [imath]=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+6.1^2+2.1[/imath] [imath]=6(n^2+(n−1)^2+...+2^2+1^2)+2(n+(n−1)+...+2+1)[/imath] [imath]=6×n(n+1)(2n+1)/6+2×n(n+1)/2[/imath] [imath]=n(n+1)(2n+1+1)[/imath] [imath]=2n^3+2n^2+2n^2+2n[/imath] [imath]=2n(n^2+n+n+1)[/imath] [imath]=2n(n^2+2n+1)[/imath] [imath]a_n=2n(n+1)^2[/imath] for [imath]n=99, a_{99}=2×99×(99+1)^2=198×10^4[/imath] I'm looking for short trick or alternative way, can you explain please?
|
2055161
|
If [imath]u\equiv 1[/imath] mod [imath]m[/imath] and [imath]n[/imath] any natural number [imath]\neq[/imath] char[imath](K)[/imath], then [imath]u[/imath] is a [imath]n[/imath]-th power in [imath]K[/imath].
let [imath]k[/imath] be a complete, non-Archimedean field, [imath]R[/imath] the corresponding valuation ring, [imath]m[/imath] its maximal ideal. If [imath]u\equiv 1[/imath] mod [imath]m[/imath] and [imath]n[/imath] any natural number [imath]\neq[/imath] char[imath](K)[/imath], then [imath]u[/imath] is a [imath]n[/imath]-th power in [imath]K[/imath]. I am stuck at the condition [imath]u\equiv 1[/imath] mod [imath]m[/imath]. What does it mean?
|
1080201
|
Elements with infinite roots in [imath]p[/imath]-adic integers
Let [imath]\mathbb{Q}_p[/imath] the [imath]p[/imath]-adic completion of [imath]\mathbb{Q}[/imath] and [imath]S=\{x\in\mathbb{Q}_p:1+x\mbox{ has $n^{th}$ roots in }\mathbb{Q}_p\mbox{ for infinitely many }n\in\mathbb{N}\}[/imath] I have to show that [imath]p\mathbb{Z}_p\subseteq S\subseteq \mathbb{Z}_p[/imath] where [imath]\mathbb{Z}_p=\{x\in\mathbb{Q}_p:|x|_p\le 1\}[/imath]. Any hint on how to start? I'm totally stuck at the moment, trying to find out the structure of those [imath]n^{th}[/imath] roots but going nowhere.
|
2056174
|
How to prove the rank of the conjugate transpose equals the original matrix?
Given [imath]A\in\Bbb C^{n \times n}[/imath],I want to prove [imath]\text{rank} A=\text{rank} A^H[/imath] where [imath]A^H[/imath] is the conjugate transpose matrix of [imath]A[/imath]. We only need to prove: [imath]\text{rank} A=\text{rank} \bar A[/imath] where [imath]\bar A[/imath] is the conjugate matrix of [imath]A[/imath]. I have tried but was not able to solve the problem. I found a similar question here rank of complex conjugate transpose matrix property proof. I know rank-nullity theorem, but I have not learn any theorem about maps that preserve the dimension of nullity. How do we prove [imath]A \to \bar A[/imath] map preserves dimension of nullity?
|
490434
|
rank of complex conjugate transpose matrix property proof
I have a question about complex conjugate matrix. Prove that for any rectangular matrix [imath]A[/imath], rank [imath]A[/imath]=rank [imath]A^*[/imath] where [imath]A^*[/imath] is complex conjugate transpose of A.
|
2057296
|
Prove that for all [imath]n \ge 1[/imath], [imath]1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots + \frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\cdots+\frac{1}{2n}[/imath]
Looked for a post regarding this specific proof by induction, and can't seem to find one... please see below for the proposition and my attempt. It seems I'm getting stuck when it comes to manipulating the fractions during the inductive step, but I'm now concerned that my approach is somehow flawed. Prove that for all [imath]n \ge 1[/imath], [imath]1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots + \frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+\cdots + \frac {1}{2n}[/imath]. [imath]Proof.[/imath] We will prove this via mathematical induction. Base Case. For [imath]n=1[/imath] we see that [imath]\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{2(1)-1}-\frac{1}{2(1)}=1-\frac{1}{2}=\frac{1}{2}=\frac{1}{2(1)}=\frac{1}{2n}[/imath]. Inductive Step. Assume our proposition holds for [imath]n=k\ge 1[/imath]. Now observe that [imath]\begin{align*}1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots + \frac{1}{2k-1}-\frac{1}{2k}+\frac{1}{2(k+1)-1}-\frac{1}{2(k+1)}&=\\ \frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+\cdots + \frac {1}{2k} + \frac{1}{2(k+1)-1}-\frac{1}{2(k+1)} \end{align*}[/imath] Stuck here ^^^ Wanting to get to here: [imath]\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+\cdots + \frac {1}{2(k+1)}.[/imath]
|
30975
|
Proof by induction: [imath]\sum\limits_{i=1}^{n} \frac{1}{n+i} = \sum\limits_{i=1}^{n} \left(\frac{1}{2i-1} - \frac{1}{2i}\right)[/imath]
How can the following be proved by induction? [imath]\sum\limits_{i=1}^{n} \frac{1}{n+i} = \sum\limits_{i=1}^{n} \left(\frac{1}{2i-1} - \frac{1}{2i}\right)[/imath] I am out of ideas after practicing for a while: [imath]\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\dots+\left(\frac{1}{2n-1}-\frac{1}{2n}\right) [/imath] Does this involve telescoping series?
|
775107
|
Prove that limit [imath] a^x \text{is } a^c[/imath]
Prove that [imath]\lim\limits_{x\to c} a^x = a^c[/imath] strictly by the epsilon-delta definition. I know it is quite easy to prove this using logarithm but that is assuming that we already know the continuity of logarithm function and I think that requires a separate set of proof. I mean the reason is that I want to use this to show that logarithm is continuous instead of the other way round. Help appreciated thanks.
|
762614
|
How to prove continuity of [imath]e^x[/imath].
I simply want a proof that [imath]e^x[/imath] is continuous. I have never really been able to find something satisfying these points: [imath]e[/imath] is defined to be the limit [imath]\lim_{n\to\infty}\left(1+{1\over n}\right)^n[/imath]. This is not essential, and another reasonable definition may be used if that significantly simplifies things. EDIT: In fact, it does not really matter what the base is taken to be. Any number, or a general [imath]a^x[/imath], would be fine. For positive integers, we define [imath]a^n:=a\times a\times\cdots\times a[/imath] where there are [imath]n[/imath] multiplications. We then let [imath]a^0=1[/imath] and [imath]a^{-n}=1/a^n[/imath]. For roots, [imath]a^{1/n}[/imath] is the (principal) number [imath]x[/imath] such that [imath]x^n=a[/imath]. For general rationals, [imath]a^{b/c}=(a^{1/c})^b[/imath] is the [imath]c[/imath]'th root brought to the [imath]b[/imath]'th power. And for irrational numbers [imath]x[/imath], [imath]a^x=\lim{a^n}[/imath] where [imath]n[/imath] is a sequence of rational numbers which converge to [imath]x[/imath]. In other words, I want a proof of the exponential's continuity derived from the "arithmetic" definition itself. The reason I ask is because as a student I never had an intuition for why a number brought to two close exponents would have to yield similar answers. For example, why would [imath]2^{1/2}\approx 2^{47/99}[/imath] I thought? After all, one has a [imath]99[/imath]'th root to a [imath]47[/imath]'th power, and the other is just a square root. I now know that I was looking for a proof of continuity... and I'm still looking. Please do not use anything advanced - the kid in me still groans when I see people 'define' [imath]e^x[/imath] as a power series, because I know students everywhere (like I was) are getting confused. Also, I want the proof to be convincing, as in I can sleep well at night, but it need not be headache-inducingly thorough. For example, I didn't derive some of the laws of exponents. I didn't explain what I mean by an irrational number, or prove that the limit must exist and be unique, etc. These you should take as granted - what I want is a straightforward proof of continuity based on the "actual" (!) definition that could be explained to an advanced calculus student frustrated about this question and ready to push themselves to learn - in other words, to my past self.
|
2032726
|
Multiplicative inverse of [imath]4x+3[/imath] in the ring [imath] \mathbb Z_{11}[x]/(x^2+1)[/imath]
How to find the multiplicative inverse of [imath]4x+3[/imath] in the ring [imath] \mathbb Z_{11}[x]/(x^2+1)[/imath]? I understand that any element belonging to the ring [imath] \mathbb Z_{11}[x]/(x^2+1)[/imath] is of the form [imath]ax+b[/imath] where [imath]a,b \in \mathbb Z_{11}[/imath]. How to proceed?
|
1624660
|
Compute the (multiplicative) inverse of [imath]4x+3[/imath] in the field [imath]\frac {\Bbb F_{11}[x]}{\langle x^2+1 \rangle}[/imath]?
So I am finding a polynomial [imath]px+q[/imath] ([imath]p,q \in \Bbb F_{11}[/imath]) which is multiplicative inverse of [imath]4x+3[/imath] in [imath]\frac {\Bbb F_{11}[x]}{\langle x^2+1 \rangle}[/imath]. i.e. [imath][(4x+3)+\langle x^2+1 \rangle][(px+q)+\langle x^2+1 \rangle]=1+\langle x^2+1 \rangle[/imath] [imath]\Rightarrow[/imath] [imath](4x+3)(px+q)+\langle x^2+1 \rangle=1+\langle x^2+1 \rangle[/imath] [imath]\Rightarrow[/imath] [imath]4px^2+(4q+3p)x+3q+\langle x^2+1 \rangle=1+\langle x^2+1 \rangle[/imath]. We see that the remainder,when [imath](4x+3)(px+q)[/imath] is divided by [imath]x^2+1[/imath], is [imath]1[/imath]. So by Division algorithm, [imath] \require{enclose} \begin{array}{r} 4p \\[-3pt] x^2+1 \enclose{longdiv}{4px^2+(4q+3p)x+3q} \\[-3pt] \underline{4px^2+4p} \\[-3pt] (4q+3p)x+(3q-4p) \\[-3pt] \end{array} [/imath] So I equate [imath](4q+3p)x+(3q-4p)=1[/imath] and solve the simultaneous linear equations [imath]4q+3p=0, 3q-4p=1[/imath]. I get [imath]p=6,q=1[/imath] Hence [imath]6x+1[/imath] is the required inverse. I am pretty sure that the answer is correct but is the method to achieve it right?
|
2057208
|
Show that there is a continuous function [imath]g:\Bbb R\to \Bbb R[/imath] with [imath]g(x)=f(x)[/imath] for every [imath]x\in C[/imath].
Let [imath]C[/imath] be closed set in [imath]\Bbb R[/imath] and let [imath]f:C\to \Bbb R[/imath] be continuous. Show that there is a continuous function [imath]g:\Bbb R\to \Bbb R[/imath] with [imath]g(x)=f(x)[/imath] for every [imath]x\in C[/imath]. I am unable to view how to construct [imath]g[/imath] from [imath]f[/imath]. Please give some hints.
|
219512
|
Continuous extension of a real function
Related; Open set in [imath]\mathbb{R}[/imath] is a union of at most countable collection of disjoint segments This is the theorem i need to prove; "Let [imath]E(\subset \mathbb{R})[/imath] be closed subset and [imath]f:E\rightarrow \mathbb{R}[/imath] be a contiuous function. Then there exists a continuous function [imath]g:\mathbb{R} \rightarrow \mathbb{R}[/imath] such that [imath]g(x)=f(x), \forall x\in E[/imath]." I have tried hours to prove this, but couldn't. I found some solutions, but ridiculously all are wrong. Every solution states that "If [imath]x\in E[/imath] and [imath]x[/imath] is not an interior point of [imath]E[/imath], then [imath]x[/imath] is an endpoint of a segment of at most countable collection of disjoint segments.". However, this is indeed false! (Check Arthur's argument in the link above) Wrong solution Q4.5; http://www.math.ust.hk/~majhu/Math203/Rudin/Homework15.pdf Just like the argument in this solution, i can see that [imath]g[/imath] is continuous on [imath]E^c[/imath] and [imath]Int(E)[/imath]. But how do i show that [imath]g[/imath] is continuous on [imath]E[/imath]?
|
2057766
|
If [imath]\lvert f(x)-f(y) \rvert \le \lvert x-y \rvert ^2[/imath] for all [imath]x[/imath] and [imath]y[/imath] in [imath]\mathbb{R}[/imath] then [imath]f[/imath] is a constant function
Assume that [imath]\lvert f(x)-f(y) \rvert \le \lvert x-y \rvert ^2[/imath] for all [imath]x,y\in \mathbb{R}[/imath]. Prove that [imath]f[/imath] is the constant function. My first thought was to use continuity of [imath]f[/imath], so that we get that [imath]\forall x\in\mathbb{R},\forall \epsilon>0, \exists \delta>0, \forall y \in \mathbb{R}, \lvert x-y \rvert<\delta \implies \lvert f(x)-f(y) \rvert < \epsilon[/imath]. So choosing [imath]y[/imath] so that the above property holds, we get that [imath]\lvert f(x)-f(y) \rvert \le \lvert x-y \rvert ^2 <\delta^2[/imath] Using the inequality I found (supposing it is correct), how would I continue to prove that [imath]f(x)=f(y)[/imath], and so their difference must be zero.
|
396205
|
To show that function is constant
Let [imath]f[/imath] be defined on [imath]\mathbb{R}[/imath] and suppose that |[imath]f(x)[/imath] - [imath]f(y)[/imath]| [imath]\leq[/imath] [imath](x-y)^2[/imath] [imath]x,y \in\mathbb{R}[/imath]. Here I have to show that [imath]f[/imath] is a constant function. I think I have to show that [imath]f'(x)[/imath] = 0 for all [imath]x[/imath]. But I don't know from where to start this. I tried taking it as (|[imath]f(x)[/imath] - [imath]f(y)[/imath]|/|[imath]x[/imath]-[imath]y[/imath]|) [imath]\leq[/imath] |[imath]x[/imath] - [imath]y[/imath]|. Am I right in doing so? Any hint or suggestion will be helpful. Thanks
|
2058269
|
Entire function and constant
Is it true that an entire function, whose [imath]Re(s)\gt0[/imath], then it has to be a constant? I think this is true and we could use Louville's theorem to prove this, but not sure.
|
561410
|
Entire function with positive real part is constant (no Picard)
A problem asks to show that an entire function on [imath]\mathbb{C}[/imath] with positive real part must be a constant. I spoke to a professor, and asked why not just use the Picard theorem. He said that we should try to aim the solution at the level of the problem, and that Picard was a little too high-powered for this problem. How would I solve it in the absence of Picard's theorem? A related question: suppose we have an isolated singularity, near which [imath]\Re (f)[/imath] (alternately, [imath]\Im (f)[/imath]) is bounded. How can we show the singularity is removable? Yet another related problem: why is a positive harmonic function on Rn a constant? Mean value property seems not to be the way...
|
2058189
|
continuity of [imath]\sqrt{x}[/imath] at [imath]x=0[/imath]
My book (Calculus by Howard Anton) says that [imath]\sqrt{x}[/imath] is continuous at [imath]x=0[/imath]. But [imath]\lim_{h\to 0} \sqrt{h}[/imath] is defined easily whereas but [imath]\lim_{x\to 0^{-}} \sqrt{-h}[/imath] cannot be defined. So if limits of both sides isn't equal then how can we say it's continous at [imath]x=0[/imath]?
|
1970235
|
Why is [imath]f(x) = \sqrt{x}[/imath] is continuous at [imath]x=0[/imath]? Does [imath]\lim_{x \to 0^{-}} \sqrt{x}[/imath] exist?
If I remember right, [imath]f(x)[/imath] is continuous at [imath]x=a[/imath] if [imath]\lim_{x \to a} f(x)[/imath] exists [imath]f(a)[/imath] exists [imath]f(a) = \lim_{x \to a} f(x)[/imath] So [imath]\lim_{x \to 0^{-}} \sqrt{x}[/imath] exists? Thus [imath]\lim_{x \to 0^{-}} \sin(\sqrt{x})[/imath] exists?
|
1617628
|
Polynomials and Arithmetic: If [imath]p(x_i) = 7[/imath] for four integers, then [imath]p(z)\ne14[/imath]
Consider the polynomial [imath]p(x) = a_0 + a_1x + a_2x^2 + · · · + a_nx^n[/imath] where [imath]a_0, a_1, . . . , a_n ∈ \Bbb Z[/imath]. Show that if [imath]p(x_i) = 7[/imath] for 4 distinct integers [imath]x_0, x_1, x_2, x_3[/imath], then [imath]p(z) \neq 14[/imath] for any [imath]z ∈ \Bbb Z[/imath]. How do I start this question?
|
757135
|
integral roots for [imath]f(x) = 41[/imath] if [imath]f(x) = 37[/imath] has 5 distinct integral roots.
Given a polynomial [imath]f(x)[/imath] with integral coefficients and [imath]f(x) = 37[/imath] has 5 distinct integral roots, find the number of integral roots of [imath]f(x) = 41[/imath]? My Approach: Say [imath]f(x) = (x-r_1)(x-r_2)(x-r_3)(x-r_4)(x-r_5)g(x) + 37[/imath], where [imath]r_i[/imath] are the distinct integers. Now for [imath]f(x) = 41[/imath] we have [imath](x-r_1)(x-r_2)(x-r_3)(x-r_4)(x-r_5)g(x) = 4[/imath], so the factors can be [imath]\pm 1, \pm2[/imath] or [imath]\pm 4[/imath]. Given [imath]r_i[/imath] are distinct at most two of them will give [imath]\pm1[/imath], then there can be both of [imath]\pm 2[/imath] or one of [imath]\pm 4[/imath]. This is where I get lost, since even if I use all of [imath]\pm1, \pm2[/imath], I will be still be left with one [imath]x-r_i[/imath] factor. What about that? Does it matter that 37 and 41 are primes, or is it just a coincidence? Thanks in advance.
|
2055031
|
Let [imath]w>0[/imath].compute the matrix [imath]e^{A}[/imath].
Let [imath]w>0[/imath].compute the matrix [imath]e^{A}[/imath],where [imath]A=\begin{bmatrix} 0 & w \\ -w& 0 \\ \end{bmatrix}[/imath]
|
2038351
|
[imath]e^A[/imath], where [imath]A= \begin{bmatrix} 0 & \omega \\ -\omega& 0 \end{bmatrix}[/imath]
[imath]\omega>0[/imath], Then Compute the matrix: [imath]e^A[/imath], where [imath]A= \begin{bmatrix} 0 & \omega \\ -\omega& 0 \end{bmatrix}[/imath] I have never seen any problem like this. So Please help me solving this. Any link for theoretical reading is welcome.
|
987488
|
Equation for a plane perpendicular to a line through two given points
The following type of question is quite popular with examiners at the institution where I study. Find an equation of the plane containing the point [imath](0, 1, 1)[/imath] and perpendicular to the line passing through the points [imath](2, 1, 0)[/imath] and [imath](1, -1, 0)[/imath] I start by calculating the the parametric equation of the line passing through [imath](2, 1, 0)[/imath] and [imath](1, -1, 0)[/imath], but after that I am lost. I realize that I have to find the equation for the second line, and that the first plane will be perpendicular to it if the dot product of the vectors equal zero, but I cannot seem to put the pieces together. Can someone please guide me in the correct direction for solving this type of problem? Much appreciated.
|
554773
|
Equation of a plane containing a point and perpendicular to a line
Find an equation of the plane containing the point [imath](1, 1, -1)[/imath] and perpendicular to the line through the points [imath](2, 0, 1)[/imath] and [imath](-1, 1, 0)[/imath] This is what I have: I first find the vector between the points: [imath]\vec{n}^{\ } = (2,0,1) - (-1, 1, 0) = (3, -1, 1)[/imath] Next I find the formula for the plane that is perpendicular to the vector by taking the dot product of the vector and the given point + another point of the plane [imath](3, -1, 1) \cdot (1 + x, 1 + y, z -1 )\\ 3 + 3x -1 - y -1 -z\\ 3x - y - z +1 = 0[/imath] I am not at all sure if I followed the correct steps
|
2058906
|
Prove [imath]\frac{\ln x}{x-1} \leq \frac{1}{\sqrt{x}}[/imath] using derivatives
[imath]\frac{\ln x}{x-1} \leq \frac{1}{\sqrt{x}} \qquad x > 0,\ x \neq 1[/imath] From here we have that [imath]0 \leq x-1-\sqrt{x}\ln x[/imath] So I was trying to prove that [imath]f(x) = x-1-\sqrt{x}\ln x\ \geq\ 0,\ \forall \ x > 0,\ x \neq 1[/imath]. By finding the first derivative we have [imath]f'(x) = 1 - \frac{\ln \sqrt{x}}{\sqrt{x}} - \frac{1}{\sqrt{x}}[/imath] and the second [imath]f''(x) = \frac{\ln x}{4x^{3/2}}[/imath] By analyzing both derivatives we see that in the interval [imath](0,1),\ f' > 0[/imath]. It reaches its minimum at [imath]x = 1[/imath] (not in the domain of the function) and, since [imath]f'' > 0[/imath] for [imath]x > 1[/imath], we have that [imath]f' > 0[/imath] for any value of [imath]x[/imath] in the domain, which in turn means that f is an strictly increasing function. But, how can I show that [imath]f \geq 0[/imath] in the first place? I haven't been able to do that. I graphed [imath]f(x)[/imath] and get that [imath]f[/imath] is indeed negative for values of [imath]x[/imath] near zero, so I don't know what to do.
|
489017
|
Prove [imath]\frac{1}{\sqrt{x}}\geq \frac{\ln x}{x-1}[/imath]
I am trying to show that, for all [imath]x>0:[/imath] [imath]\frac{1}{\sqrt{x}}\geq \frac{\ln x}{x-1}[/imath] This inequality is closer than I expected. I have tried exponentiating, power series, and have achieved nothing. I would really appreciate some help. Below is a graph of the two functions for small [imath]x:[/imath]
|
2058829
|
The function [imath]\sqrt{z-a}\cdot \sqrt{z-a}[/imath] has a branch in [imath]C\setminus{[a,b]}[/imath] for all [imath]a,b\in \Bbb{C}[/imath] using [imath]z-a\over z-b[/imath]
Show that the function [imath]f(z)=\sqrt{z-a}\cdot \sqrt{z-a}[/imath] has a branch in [imath]C\setminus{[a,b]}[/imath] for all [imath]a,b\in \Bbb{C}[/imath]. Hint: [imath]\phi={z-a\over z-b}[/imath] maps [imath][a,b][/imath] to [imath]\Bbb{C}\setminus(-\infty,0][/imath]. I tried to use the hint and also found [imath]\phi^{-1}[/imath] so as to get close to a holomorphic composition to fit [imath]f(z)[/imath], but I am stuck here, and I can't quite see how I can form it. What should be the approahc here or the explanation for what makes a function that defines a continuous subset of [imath]f(z)[/imath] a branch...I could use some help here. This question is referred to another question that is identical in result to this one. However, these are not identical in the attempt or in the solution, as they do not rely on the hint I am trying to utilize and connect to the given case, and in any case I do not quite understand the reference's comments and solutions, whose authors would not necessarily be as responsive to me as they might be to the referrence's OP.
|
2057679
|
Prove that there exist a branch
The question is- Prove that there is a branch for the function [imath]\sqrt{z-a}\sqrt{z-b}[/imath] where [imath]a,b[/imath] are complex numbers, in [imath]C\backslash [a,b][/imath]. Generally,I have no idea how to prove that there is a branch for a function... Any idea?
|
2058364
|
Complex Number Circumcenter
How would I go about finding the formula for the complex number representation (affix) of the circumcenter of [imath]\triangle abc[/imath] given the affixes [imath]a,b,c[/imath]?
|
481357
|
Finding center and radius of circumcircle
Find the center and radius of the circle which circumscribes the triangle with (complex) vertices [imath]a_1,a_2,a_3[/imath]. Express the result in symmetric form. I'm not sure where to start in this question. The circumcenter is the intersection of the perpendicular bisectors, but I don't know how to calculate the line equation of the perpendicular bisectors yet. As for the radius, after we find the circumcenter [imath]c[/imath] we can calculate it using [imath]|c-a_1|[/imath] (or [imath]|c-a_2|[/imath] or [imath]|c-a_3|[/imath]; all three should be equal.) So how can I calculate the circumcenter?
|
2044325
|
Is it true that [imath]\gcd(ta,tb)=t\gcd(a,b)[/imath] in a domain?
Is it true that in a domain [imath]R[/imath], [imath]\gcd(ta,tb)=t\gcd(a,b)[/imath]? I know how to prove it in [imath]\mathbb{Z}[/imath], but what about if it's an arbitrary domain [imath]R[/imath]?
|
1932541
|
For [imath]D[/imath] a GCD domain , let [imath]a,b,x \in D \setminus \{0\}[/imath] , then is it true that [imath]\gcd (ax,bx)=x \cdot \gcd (a,b)[/imath]?
Let [imath]D[/imath] be a GCD domain, and let [imath]a,b,x \in D \setminus \{0\}[/imath]. Then is it true that [imath]\gcd (ax,bx)=x \cdot \gcd (a,b)[/imath] ? Let [imath]c=\gcd (a,b)[/imath] and [imath]d=\gcd(ax,bx)[/imath], then as [imath]cx|ax[/imath] and [imath]cx|bx[/imath] so [imath]cx|d[/imath]. We would be done if we could show [imath]d|cx[/imath], but I am unable to show that. Please help me to solve this problem.
|
2060034
|
Find all prime solution
Find all prime [imath]p,q[/imath] such that [imath]p^2+7pq+q^2[/imath] is a perfect square. I got a family of solutions as [imath]p=q[/imath].
|
478802
|
Prime Numbers And Perfect Squares
Find all primes [imath]p[/imath] and [imath]q[/imath] such that [imath]p^2 + 7pq + q^2[/imath] is a perfect square. One obvious solution is [imath]p = q[/imath] and under such a situation all primes [imath]p[/imath] and [imath]q[/imath] will satisfy. Further if [imath]p \neq q[/imath] then we can assume without the loss of generality that [imath]p > q[/imath]. Assuming this and that there exists at least one such perfect square I have tried to show some contradiction modulo [imath]4[/imath] as any odd perfect square leaves a remainder of [imath]1[/imath] when divided by [imath]4[/imath], but it is not working. However I firmly believe that [imath]p = q[/imath] is the only solution, but I have failed to prove this.
|
2059998
|
Find the value of expression
I was preparing for Olympiad then I got this question and I don't understand how to solve. [imath]a,b,c[/imath] are real no such that [imath]a-7b+8c=4[/imath]and [imath]8a+4b-c=7[/imath] find[imath]a^2-b^2+c^2[/imath]
|
1805350
|
Find the value of [imath]a^2-b^2+c^2[/imath]
Let [imath]a[/imath], [imath]b[/imath], and [imath]c[/imath] be real numbers such that [imath]a − 7b + 8c = 4[/imath] and [imath]8a + 4b − c = 7[/imath]. What is the value of [imath]a^2-b^2+c^2[/imath] ?
|
2057813
|
Riemann zeta function on the line Re(s) = 1
How can I prove that for [imath]s \in \mathbb{C}[/imath], with real part of [imath]s[/imath] being equal to 1, \begin{equation} \sum_{n=1}^{\infty}\frac{1}{n^{s}} \end{equation} diverges? Thanks a lot!
|
1909006
|
Does [imath]\sum_{k=1}^\infty\frac1{k^n}[/imath] converge for [imath]\Re(n)=1,\Im(n)\ne0[/imath]?
Does [imath]\sum_{k=1}^\infty\frac1{k^n}[/imath] converge for [imath]\Re(n)=1,\Im(n)\ne0[/imath]? The ratio test is inconclusive. It passes the term test for [imath]\Re(n)=1[/imath], but this is not sufficient to prove convergence. Since we are dealing with so many complex numbers, I do not know of any convergence tests for this. I know that if it is convergent, it is conditionally convergent.
|
2060490
|
Terrence Tao Real Analysis exercise: Prove that all Cauchy sequences [imath](a_n)[/imath] are bounded
I've been struggling for one and a half hours trying to think of a proof for this Lemma, I haven't made any progress at all. It's in Terrence Tao's lectures notes for his Real Analysis course (week 2, lemma 9). "Prove that all Cauchy sequences [imath](a_n)[/imath] are bounded". If any could give me a hint that would be really helpful, thanks.In particular, something I am having an issue with is that most proofs I have seen involve the Max function which hasn't been defined by Tao yet.This makes me think it's not the easiest way of doing it.
|
1999633
|
lemma: Cauchy sequences are bounded.
Lemma: Cauchy sequences are bounded. Proof: Given Cauchy sequence [imath](s_n)[/imath], for [imath]\varepsilon=1[/imath] we obtain [imath]N\in\Bbb N[/imath] such that [imath]m, n > N[/imath] implies [imath]|s_n − s_m| < 1[/imath]. (2) implies [imath]|s_n − s_{N+1}| < 1[/imath] for all [imath]n\geq N[/imath], (from this step (2) to the next step (3) how does one get there? is it through the reverse triangle inequality ? do we simply say that [imath]||s_n|-|s_{N+1}||\leq|s_n-s_{N+1}|<1 [/imath] then [imath]||s_n|-|s_{N+1}|| < 1[/imath] for all [imath]n\geq N[/imath] then add [imath]|s_{N+1}|[/imath] to both sides of the inequality? is it that simple?) (3) implies [imath]|s_n| < |s_{N+1}| + 1[/imath] for all [imath]n>N[/imath]. If [imath]M = \max\{|s_{N+1}| + 1, |s_1|, |s_2|,\ldots, |s_N |\}[/imath], then [imath]|s_n| \leq M[/imath] for all [imath]n \in\Bbb N[/imath] hence the sequence [imath](s_n)[/imath] is bounded. QDE
|
2060904
|
Prove or disprove that [imath]f[/imath] is identity function given that [imath]f[/imath] is continuous and [imath]f(f(f(x)))=x[/imath]
Let [imath]f [/imath] be continuous on [imath]\mathbb{R}[/imath] satisfying [imath]f(f(f(x)))=x[/imath] for all [imath]x\in \mathbb{R}[/imath]. Prove/disprove that [imath]f[/imath] is an identity function. So far I have figured out that any point [imath]x[/imath] such that [imath]f(x)\neq x [/imath] is not isolated(if it exists). i.e every neighbourhood of such a point contains other points where function is not identity. Cant proceed any further!
|
718327
|
Show that [imath]f(x) = x[/imath] if [imath]f(f(f(x))) = x[/imath].
If [imath]f: \mathbb{R} \to \mathbb{R}[/imath] and [imath]f[/imath] is strictly increasing, show that [imath]f(x) = x[/imath] if [imath]f(f(f(x))) = x[/imath]. So this compulsorily ESTABLISHES that [imath]f(x) = x[/imath] only, and no other solution. So, merely substituting [imath]f(x) = x[/imath] and hence showing the given equality holds will not earn any credit. I was proceeding via inverses, but then I got confused with the notation. And for that, the problem got tricky.
|
1017881
|
Weak Mathematical Induction for Modulo Arithmetic [imath]8\mid 3^{2n}-1[/imath]
Using Weak Mathematical Induction, I have to show that, for all integers [imath]n \geq 1[/imath], [imath]8|3^{2n} -1[/imath] I really don't know how to go about solving this problem. Currently I only have the base case and the Inductive Hypothesis: Base Case: For [imath]n = 1[/imath] [imath]8|3^{2(1)}-1 = 8|8 = 0[/imath] Inductive Hypothesis: Assume true for [imath]n = k[/imath] [imath]8|3^{2k}-1[/imath] Inductive Step: I want to show that the statement is true for [imath]n = k+1[/imath] so [imath]8|3^{2(k+1)}-1[/imath] This is where I am currently stuck. Any help would be really helpful. Thanks
|
1014307
|
Mathematical Induction divisibility [imath]8\mid 3^{2n}-1[/imath]
So I'm trying to use mathematical induction to show that for all integers [imath]n \ge 1[/imath] , [imath] 8|(3^{2n} - 1)[/imath] (is divisible by 8) I have my base case: [P(1)], [imath]3^2 - 1 = 9 - 1 = 8[/imath], since [imath]8|8[/imath], the base case proves true Assume [P(k)], [imath] 8 | (3^{2k} - 1)[/imath]. I know that I need to show [P(k+1)], [imath] 8| (3^{2(k+1)}-1) [/imath], but I'm not sure how to prove this. I've only been using induction for summation, so how could I prove divisibility?
|
2061323
|
Trigonometry question with a given value
Let [imath](1+\sin\theta)(1+\cos\theta) = \frac 54[/imath] Then what will be the value of [imath](1-\sin\theta)(1-\cos\theta)[/imath]. I tried Squaring and Expanding the terms and Tried to replace the value with the expanded value from first equation but it got complicated further more.
|
1061928
|
If [imath]\left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\;,[/imath] Then [imath]\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right)[/imath]
If [imath]\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\;,[/imath] Then [imath]\left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) = [/imath] [imath]\bf{My\; Try::}[/imath] Given [imath]\displaystyle \left(1+\sin \phi\right)\cdot \left(1+\cos \phi\right) = \frac{5}{4}\Rightarrow 1+\sin \phi\cdot \cos \phi+\sin \phi+\cos \phi = \frac{5}{4}.[/imath] So [imath]\displaystyle \sin \phi+\cos \phi+\sin \phi \cdot \cos \phi = \frac{1}{4}\Rightarrow \left(\sin \phi+\cos \phi\right) = \frac{1}{4}-\sin \phi \cdot \cos \phi.[/imath] Now [imath]\displaystyle \left(1-\sin \phi\right)\cdot \left(1-\cos \phi\right) = 1-\left(\sin \phi+\cos \phi\right)+\sin \phi \cdot \cos \phi =\frac{3}{4}+\sin 2\phi[/imath] Now How can I calculate [imath]\sin 2\phi.[/imath] Help me, Thanks
|
2061487
|
Why is [imath]\cos(\arctan(3x)) = \frac{1}{\sqrt{1 + 9x^2}}?[/imath]
Why is [imath]\cos(\arctan(3x)) = \frac{1}{\sqrt{1 + 9x^2}}[/imath]? I've tried messing around with identities but still I'm not sure what's going on here. Thanks.
|
1894265
|
How do I show this [imath]\cos^2(\arctan(x))=\frac{1}{1+x^2}[/imath]
How do I show this? [imath]\cos^2(\arctan(x))=\frac{1}{1+x^2}[/imath] I have absolutely no idea. Thank you.
|
2060839
|
What is the permutation of word "MISSISSIPPI"?
What is the total no. of permutations of the letters of the word MISSISSIPPI in which no four "I"s come together? My try-: [imath]7!/4!\times 2! \times 8!/4![/imath] But not getting the right answer. Please help
|
1418717
|
Permutations on word [imath]MISSISSIPPI[/imath].
In how many ways can the letters of the word [imath]MISSISSIPPI[/imath] be rearranged ? I am confused on whether it is [imath]\dfrac{11!}{4!4!2!}[/imath] or [imath]\dfrac{11!}{4!4!2!}-1[/imath] since it is given rearranged and not arranged.
|
2060002
|
Matrix as a sum of two commuting matrices
The problem I'm attempting is to prove that, given an n x n matrix [imath]A[/imath], we can find matrices [imath]B[/imath] & [imath]C[/imath] such that [imath]BC=CB[/imath], [imath]A=B+C[/imath], [imath]B[/imath] is diagonalizable, and [imath]C^n = 0[/imath]. I first tried simply breaking the matrix into upper & lower triangular matrices, but one is not guaranteed to be diagonalizable and the two don't necessarily commute. I'm considering having [imath]B[/imath] simply be [imath]\lambda I[/imath], but guaranteeing that [imath]C^n=0[/imath] seems tricky. Any suggestions?
|
1157625
|
Every [imath]n\times n[/imath] matrix is the sum of a diagonalizable matrix and a nilpotent matrix.
I would like to prove that every [imath]n\times n[/imath] matrix is the sum of a diagonalizable matrix and a nilpotent matrix. How is this possible? I'm not sure where to begin really- I know that a nilpotent matrix is one of which some power is the zero matrix. I also know that a matrix A can be written as [imath]AP=PJ[/imath] with [imath]P[/imath] invertible and [imath]J[/imath] of Jordan form. I have proven that any strictly upper triangular matrix is nilpotent, so [imath]J[/imath] can be written as [imath]D+N [/imath], with D diagonal and [imath]N[/imath] nilpotent, but how can I change this for A? Thank you!
|
2061653
|
[imath]a \in \mathbb{N}[/imath] and [imath]p \in \mathbb{P}[/imath] with [imath]gcd(a, p-1)=1[/imath], then for every integer [imath]b[/imath], then [imath]x^a \equiv b[/imath] mod [imath]p[/imath] has a solution
I'd like to show the following: Let [imath]a \in \mathbb{N}[/imath] and [imath]p \in \mathbb{P}[/imath] with [imath]gcd(a, p-1)=1[/imath], then for every integer [imath]b[/imath], the congruence relation [imath]x^a \equiv b[/imath] mod [imath]p[/imath] has a solution. If [imath]p \mid b[/imath], then this trivially holds. For the other case, I've started with showing that there exists a [imath]c[/imath] such that [imath]a*c \equiv 1[/imath] (mod [imath]p-1[/imath]) and I think that [imath]x = b^c[/imath] should satisfy the relation, but I've not yet been able to show this. Can someone help me? Thanks!
|
2058505
|
Proof of Unique Solution for Modular Exponentiation In Cryptography
The question is about the encryption equation in asymmetric encryption. For [imath]c\equiv m^e\bmod n[/imath], prove that for every unique [imath]m[/imath] there is a unique [imath]c[/imath]. Here [imath]n=pq[/imath] where [imath]p[/imath] and [imath]q[/imath] are primes larger than [imath]m[/imath] and the exponent [imath]e[/imath], is [imath]\geq3[/imath] and coprime with [imath]p-1[/imath] and [imath]q-1[/imath]. Thanks!
|
2061325
|
Random placement of rooks on a chessboard
[imath]8[/imath] rooks are placed randomly on an [imath]8\times 8[/imath] chess board. What is the probability of having exactly one rook each row and each column? I guess there is no meaningful order here?
|
679529
|
What is the probability that when you place 8 towers on a chess-board, none of them can beat the other.
What is the probability that when you place 8 towers on a chess-board, none of them can beat the other. Attempt: [imath]{64 \choose 8}^{-1} \approx1[/imath] in [imath]4\ 400\ 000\ 000[/imath] Correct answer: [imath]{64 \choose 8}^{-1} \cdot 8! \approx 1[/imath] in [imath]9\ 000\ 000[/imath]. I disagree with the [imath]8![/imath]. If there's combinations (binomial coefficient) in the denominator, why would there be permutations i.e. the order counts, in the numerator?
|
2061055
|
Find [imath]\lim_{n \to \infty}\left(1+\frac{1}{n^2}\right)\left(1+\frac{2}{n^2}\right)\cdots\left(1+\frac{n}{n^2}\right)[/imath]
Problem: find [imath]\lim_{n \to \infty}\left(1+\frac{1}{n^2}\right)\left( 1+\frac{2}{n^2}\right)\cdots\left(1+\frac{n}{n^2}\right)[/imath] My thought: Taking the log and we get [imath]\sum_{k=1}^{n}\ln(1+\frac{k}{n^{2}})[/imath]. There is an inequality in another problem, which says: [imath]\frac{k}{n+k}\lt \ln\left(1+\frac{k}{n}\right)\lt\frac{k}{n}, \forall k \in N^{+} [/imath]. So I think maybe I can use it here. Plug in [imath]n^2[/imath] and sum over k, we get [imath]\sum_{k=1}^{n}\frac{k}{n^2+k}\lt \sum_{k=1}^{n}\ln\left(1+\frac{k}{n^{2}}\right) \lt\sum_{k=1}^{n}\frac{k}{n^2}=\frac{1}{n^2}\frac{n(n+1)}{2}=\frac{n+1}{2n}[/imath] Anyway, the limit is [imath]\sqrt{e}[/imath](So if we take the limit of both sides of the above inequality, it should be[imath]\frac{1}{2}[/imath] and the right hand side is just right), but I cannot go further with the left side. Any hint would be appreciated!
|
389155
|
How to calculate [imath]\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)[/imath]?
How can we compute the following limit: [imath]\lim_{n\to\infty}(1+1/n^2)(1+2/n^2)\cdots(1+n/n^2)[/imath] Mathematica gives the answer [imath]\sqrt{e}[/imath]. However, I do not know to do it.
|
2061971
|
[imath]x^2+3x+24[/imath] is a perfect square then find the value of [imath]x[/imath]
Find all integer values of [imath]x[/imath] where [imath]x^2+3x+24[/imath] is a perfect square. By guessing I found one solution that [imath]x=5[/imath].I found the problem in the exercise of a book in the chapter of polynomials. So please help me.
|
1128670
|
Find all integers [imath]x[/imath] such that [imath]x^2+3x+24[/imath] is a perfect square.
Find all integers [imath]x[/imath] such that [imath]x^2+3x+24[/imath] is a perfect square. My attempt: [imath]x^2+3x+24=k^2[/imath] [imath]3(x+8)=(k+x)(k-x)[/imath] Now, do I find solution treating cases? But that doesn't seem very easy. Please help.
|
1939203
|
Integral domain( not merely [imath]\mathbb{Z}[/imath] ). Prove that [imath]\gcd(ca, cb) = c\gcd(a, b)[/imath]
In an integral domain [imath]\gcd[/imath] is only defined up to associate relation. It's defined as follows: [imath]d = \gcd(a,b) \Leftrightarrow d|a, d|b[/imath] and [imath]c|a, c|b[/imath] implies [imath]c|d[/imath]. Or we can say that [imath](d)[/imath] is minimal among principal ideals containing [imath](a,b)[/imath]. All the same. But how to prove [imath]\gcd(ca,cb) = c\gcd(a,b)[/imath]? I found similar questions, but they were about [imath]\mathbb{Z}[/imath] and not an arbitrary integral domain, I don't think it counts as a duplicate.
|
1725148
|
In an Integral Domain is it true that [imath]\gcd(ac,ab) = a\gcd(c,b)[/imath]?
In my algebra class I was given as homework assignment to prove that: Given an integral domain [imath]A[/imath] and [imath]a,b,c,d,e \in A[/imath]. Then if [imath]d = \gcd(b,c)[/imath] and [imath]e = \gcd(ac, ab)[/imath] then [imath]e = ad[/imath]. It is easy to see that [imath]ad \mid ab[/imath] and [imath]ad \mid ac[/imath], so this implies that [imath]\exists q \in A; e = q(ad)[/imath]. Now I'm having problems showing that [imath]q = 1_A[/imath]. Working through some equalities, I proved that [imath]\forall n\in \mathbb{N}, q^n \mid d[/imath] so this gives me the light suspicion that I might be going the right way, but I'm pretty positive that this does not imply [imath]q = 1_A[/imath]. Does anyone want to give me a hint? Thanks in advance :)
|
2062086
|
pls notice that r is a rational number:) how to solve [imath]\lim_{x\rightarrow 1}\frac{x^r -1}{x-1}[/imath] without using L Hospital?
[imath]\lim_{x\rightarrow 1}\frac{x^r -1}{x-1}[/imath] r is Rational_number, with L Hospital rule is this easy to solve, but how could I solve it in another way?
|
1789815
|
How to solve [imath]\lim_{x \rightarrow 0} {\frac{(1+x)^a-1}{x}}[/imath]?
How to solve this limit without using L'Hospital rule? [imath]\lim_{x \rightarrow 0}{\frac{(1+x)^a-1}{x}}[/imath]
|
2062085
|
[imath]I[X,Y] :=- \sum_{x,y} p(x,y)\ln{p(x,y)}[/imath], Show that [imath]I[X, Y] \le I[X]+I[Y][/imath]
Let [imath]x,y[/imath] be stochastic variables and [imath]p[/imath] be a probability measure. [imath]I[X,Y] :=- \sum_{x,y} p(x,y)\ln{p(x,y)}[/imath] with [imath]p(x,y) := P(X=x, Y=y)[/imath] I have to show that [imath]I[X, Y] \le I[X]+I[Y][/imath] I mustn't assume that [imath]X[/imath] and [imath]Y[/imath] are independent. My problem is how to write [imath]p(x,y)[/imath] into [imath]p(x)[/imath] and [imath]p(y)[/imath]. Any tipps or ideas on how to separate the p(x,y) and show the inequality? Thanks in advance! Conditional Entropy is less than entropy is similar, but it doesn't fully answer my question (their definition of [imath]I[/imath] is different: over there: [imath]I[X,Y] = \sum_{x,y} p(x,y) \ln \frac{{p(x,y)}}{p(x)p(y)}[/imath]
|
70136
|
Conditional Entropy is less than entropy
Let [imath]p_{ij}=P(X=i,Y=j)[/imath] be the joint distribution, [imath]P(X=i)=p_i=\sum_j p_{ij}, P(Y=j)=q_j=\sum_i p_{ij}[/imath] be the marginal distributions, and [imath]p_{i|j}=\frac{p_{ij}}{q_j}[/imath] be the conditional distribution. Then the conditional entropy is defined as [imath]E[h(X|Y)]=-\sum_j \sum_i p_{i|j} \ln p_{i|j} q_j [/imath] Show [imath]E[h(X|Y)]\leq h(X)[/imath] where [imath]h(X)[/imath] is the entropy of [imath]X[/imath]. I'm at a lost as to even know where to begin.
|
2062172
|
An example of a domain whose complement has a different boundary
Can someone give an example of a domain [imath]G[/imath] which has a boundary different from the boundary of its complement [imath]G^c[/imath]?
|
514167
|
Show that the boundary of a set equals the boundary of its complement
[imath]\newcommand{\bdy}{\operatorname{bdy}}[/imath] I'm trying to show that [imath]\bdy(A) = \bdy(A^c)[/imath]. I know that [imath]\bdy(A) = \operatorname{closure} A \setminus \operatorname{int}(A) = (\operatorname{int}(A^c))^c \setminus \operatorname{int}(A)[/imath], but I don't know where to go from there. Any help or hints would be very much appreciated.
|
664594
|
Why [imath]\{\mathbf{0}\}[/imath] has dimension zero?
According to C.H. Edwards' Advanced Calculus of Several Variables: The dimension of the subspace [imath]V[/imath] is defined to be the minimal number of vectors required to generate [imath]V[/imath] (pp. 4). Then why does [imath]\{\mathbf{0}\}[/imath] have dimension zero instead of one? Shouldn't it be true that only the empty set has dimension zero?
|
1669647
|
If the kernel of a matrix is the 0 vector, why is the basis of the kernel non existent?
If the kernel of a matrix is the [imath]0[/imath] vector, why is the basis of the kernel non existent? If I have the matrix [imath]\begin{bmatrix}1 & 2\\3 & 4\end{bmatrix}[/imath] The kernel is the [imath]0[/imath] vector Why is the basis of the kernel non-existent? Shouldn't the basis of the kernel be [imath]0[/imath]?
|
2062456
|
Polynomials with integer coefficients such that [imath]p(a)=b,p(b)=c,p(c)=a[/imath]
While studying polynomials a question struck in my mind that To have a polynomial [imath]p(x)[/imath] with integer coefficients and [imath]a,b,c[/imath] are three distinct integers, then is it possible to have [imath]p(a)=b[/imath][imath]p(b)=c[/imath][imath]p(c)=a[/imath]
|
387721
|
Polynomial [imath]P(a)=b,P(b)=c,P(c)=a[/imath]
Let [imath]a,b,c[/imath] be [imath]3[/imath] distinct integers, and let [imath]P[/imath] be a polynomial with integer coefficients.Show that in this case the conditions [imath]P(a)=b,P(b)=c,P(c)=a[/imath] cannot be satisfied simultaneously. Any hint would be appreciated.
|
2015767
|
Solve a congruence relation
Solve the congruence relation [imath]1978^{20}\equiv x \pmod{125}[/imath] We have [imath]125=5^3[/imath] [imath]1978^{2}\equiv -1 \pmod 5\Rightarrow 1978^{20}\equiv 1978^{2\cdot10+0}\equiv 1\pmod 5[/imath] The remainder [imath]x[/imath] should be [imath]26[/imath]. If the remainder [imath]\bmod 5[/imath] is [imath]1[/imath], how to evaluate remainder [imath]\bmod 25[/imath] and combine them to give [imath]x=26[/imath]?
|
2022772
|
Solve a congruence [imath]1978^{20}\equiv x\pmod{125}[/imath]
I have checked the solution ([imath]x=26[/imath]). Solving modulo [imath]5[/imath] gives [imath]1978^{20}\equiv 1978^{2\cdot 10}\equiv 1\pmod{5}[/imath] Solving modulo [imath]25[/imath] also gives [imath]1978^{20}\equiv 1\pmod{5}[/imath] How to evaluate the remainder [imath]x[/imath]?
|
2062908
|
[imath](S \circ T)^t = S^t \circ T^t[/imath]
I am trying to understand an elementary idea in the transpose of a linear operator, namely that if [imath]S \in L_K(U,V)[/imath] and [imath]T \in L_K(V,W)[/imath] for vector spaces [imath]U,V,W[/imath] over a field [imath]K[/imath] that [imath](S \circ T)^t = T^t \circ S^t[/imath]. So it seems I should first take a linear form say [imath]g:W^* \rightarrow U^*[/imath] then [imath](S \circ T)^t (g) = g(S \circ T)[/imath]. Here is where I am stuck. Any hints/tips appreciated.
|
382580
|
[imath](U\circ T)^{*} = T^{*}\circ U^{*}[/imath]
Let [imath]T : V \longrightarrow W[/imath] and [imath]U : W \longrightarrow Z[/imath] be linear maps. How do I prove that [imath](U\circ T)^{*} = T^{*}\circ U^{*}[/imath]? I'm used to seeing [imath]V^{*}[/imath] not [imath](U\circ T)^{*}[/imath]. Any help is appreciated. [imath]^{*}[/imath] denotes the dual map (transpose).
|
2062468
|
The set of functions [imath]P[/imath] for which [imath]P(x)^2=P(x^2)[/imath] for all [imath]x\in \mathbb{R}[/imath]
We showed that all of polynomials satisfy the equation [imath]P(x)^2 = P(x^2)[/imath] are [imath]P(x) = x^n, n \in \mathbb{N}[/imath]. If [imath]P(x)[/imath] is only a continuous function, can we describe the set of solutions of the above equation, i.e., [imath]P(x)^2 =P(x^2)[/imath]?
|
1957418
|
Find all continuous functions from positive reals to positive reals such that [imath]f(x)^2=f(x^2)[/imath]
Find all continuous functions [imath]f:\mathbb R^{+}\to\mathbb R^{+}[/imath] such that [imath]f(x)^2=f(x^2)[/imath] for all positive reals [imath]x[/imath]. If [imath]f[/imath] is a constant function, then [imath]f(x)=1[/imath]. If [imath]f[/imath] is non-constant, I'm suspecting the only solutions are of the form [imath]f(x)=x^k[/imath], where [imath]k[/imath] is a constant, but I have no idea how to prove this. Thanks in advance.
|
2060743
|
What is the proof of divisibility by [imath]13[/imath]?
While solving the question of [imath]26^{th}[/imath] digit given by @The Lone Wolf a question arrived in my mind; the question was Everyone knows the divisibility rule of [imath]13[/imath]. Test for divisibility by [imath]13[/imath]: Add four times the last digit to the remaining leading truncated number. If the result is divisible by 13, then so was the first number. For Example : [imath]50661[/imath] [imath]5066+4=5070[/imath] [imath]507+0=507[/imath] [imath]50+28=78[/imath] and [imath]78[/imath] is [imath]6\times13[/imath], so [imath]50661[/imath] is divisible by [imath]13.[/imath] Please help me!!!
|
1961065
|
Rules of thumb for divisibility
Given a number presented in decimal form [imath]n=\sum_{k=0} d_k10^k[/imath] there are some nice thumb rules to decide whether it is divisible with a prime number [imath]p\in\{7,11,13,17,19,23,29\}[/imath]. For each of this numbers there is a number [imath]a_p[/imath] such that [imath]p|n\iff p|(a_p\cdot d_0+\sum_{k=0} d_{k+1}10^k)[/imath]. [imath]a_7=-2,\;a_{11}=-1,\;a_{13}=4,\;a_{17}=-5,\;a_{19}=2,\;a_{23}=7,\;a_{29}=3[/imath] E.g. [imath]\;19\cdot 17=323[/imath] and [imath]32+2\cdot 3=38=2\cdot 19[/imath] and [imath]32-5\cdot 3=17[/imath]. Is there a number [imath]a_{31}[/imath] that works the same way? Or [imath]a_p,\;p\ge37[/imath]?
|
2062636
|
Proving [imath]\lim \limits_{n\to +\infty } \left(1+\frac{x}{n}\right)^n=\text{e}^x[/imath]
Here: Link between [imath]\lim \limits_{n \to \infty} (1+{1/n})^n[/imath] and [imath]\lim \limits_{n \to \infty} (1+{x/n})^n[/imath] is a proof of the formula for [imath]e^x[/imath]. The thing causing my doubt is that we cannot assume that [imath]u[/imath] is an integer. The most common definition of limit of sequence concerns integers only. How to fix the original proof?
|
420018
|
Proving [imath]\lim \limits_{n\to +\infty } \left(1+\frac{x}{n}\right)^n=\text{e}^x[/imath].
I knew that [imath]e^x=\lim \limits_{n\to+\infty }{\left(1+\frac{x}{n}\right)^n}[/imath]. But I've never seen its proof. So I tried to prove it using [imath]\exp(\ln x)=\ln(\exp(x))=x[/imath]. Here is what I've tried so far : [imath] \left(1+\frac{x}{n}\right) ^n=e^{n\ln(1+\frac{x}{n})}[/imath] [imath]\text{I'll now study just } {n\ln\left(1+\frac{x}{n}\right)}.[/imath][imath] \text{If this function has the line }y=x \text{ as oblique asymptote, then the equality is proven.}[/imath] [imath] n\ln\left(1+\frac{x}{n}\right) = n\ln\left(\frac{n+x}{n}\right)[/imath] [imath]=n[\ln(n+x)-ln(n)][/imath] [imath]=n\left[\int_1^{n}\frac{dt}{t}+\int_{n}^{x}\frac{dt}{t}-\int_1^{n}\frac{dt}{t}\right][/imath] [imath]=n[\ln(x)-\ln(n)][/imath] But I just don't know how to show that this expression has an oblique asymptote [imath]y=x[/imath]. I've thought that if there is an oblique asymptote as [imath]n[/imath] goes to infinity, than for a huge [imath]n[/imath], we have : [imath]\ln\left(1+\frac{x}{n}\right)\approx \frac{x}{n}\approx0[/imath] Which looks correct but we could have any other function [imath]f(x)[/imath], [imath]\ln\left(1+\frac{x}{n}\right)\approx\frac{f(x)}{n}\approx 0[/imath]. Which doesn't prove the oblique asymptote because [imath]x[/imath] is constant. So how can prove [imath]e^x=\lim \limits_{n\to +\infty } \left(1+\frac{x}{n}\right)^n[/imath]? And where did I go wrong?
|
2062861
|
completeness in banach-steinhauss theorem
I would like to see an example (with some explanation) which shows that in Banach-Steinhaus theorem space [imath]X[/imath] must be really a complete space.
|
99987
|
What facts about the weak topology fail in spaces that aren't Banach?
I'm learning about the weak and weak* topologies on a normed vector space following the book of Brezis. He limits his discussion to case where [imath]E[/imath] is a Banach space, and my question is most simply stated as, "Why?". I can't find an example of a theorem where the completeness of [imath]E[/imath] is a necessary hypothesis. Most of the basic results on weak and weak* topologies proceed by applications of Hahn-Banach, which holds for a much wider class of spaces than just Banach spaces. Are there any examples of reasonably elementary (i.e. of relevance to a first-year graduate student who does not anticipate having heavy contact with functional analysis in the future) facts about weak or weak* topologies that are true for Banach spaces but not all normed vector spaces? EDIT: I should add that, as Yemon Choi points out, dual spaces of normed vector spaces are complete, so the weak-star topology will never be defined on a space that isn't Banach. With regards to the weak-star topology, then, my question should refer to aspects of the weak-star topology on a space [imath]E^*[/imath] where the original [imath]E[/imath] is not Banach.
|
2063804
|
[imath]f(x + y) = f(x) + f(y)[/imath]. Show that [imath]f[/imath] is continuous.
Let [imath]f:\mathbb{R} \rightarrow \mathbb{R}[/imath] and [imath] f(x + y) = f(x) + f(y)[/imath]. How can I show that [imath]f[/imath] is continuous, when [imath]f[/imath] is continuous at [imath]f(0)[/imath]?
|
983493
|
Show a function for which [imath]f(x + y) = f(x) + f(y) [/imath] is continuous at zero if and only if it is continuous on [imath]\mathbb R[/imath]
Suppose that [imath]f: \mathbb R \to\mathbb R[/imath] satisfies [imath]f(x + y) = f(x) + f(y)[/imath] for each real [imath]x,y[/imath]. Prove [imath]f[/imath] is continuous at [imath]0[/imath] if and only if [imath]f[/imath] is continuous on [imath]\mathbb R[/imath]. Proof: suppose [imath]f[/imath] is continuous at zero. Then let [imath]R[/imath] be an open interval containing zero. Then [imath]f[/imath] is continuous at zero if and only if [imath]f(x) \to f(0)[/imath] as [imath]x \to 0[/imath]. Then [imath]|f(x) - f(0)| < \epsilon[/imath]. Can anyone please help me? I don't really know how to continue. Thank you.
|
2060060
|
Find quadruple solutions
How to solve equation like this [imath](b+c+d)^{2010}=3\times a[/imath] [imath](a+c+d)^{2010}=3\times b[/imath] [imath](a+b+d)^{2010}=3\times c[/imath] [imath](a+b+c)^{2010}=3\times d[/imath].
|
2053864
|
Find all the possible values of [imath](a,b,c,d)[/imath].
Find all quadruples of real numbers [imath](a,b,c,d)[/imath] satisfying the system of equations [imath](b+c+d)^{2010}=3a[/imath] [imath](a+c+d)^{2010}=3b[/imath] [imath](a+b+d)^{2010}=3c[/imath] [imath](a+b+c)^{2010}=3d[/imath] I tried to find the solutions using hit and trial and by using some logic also I find two solutions which are [imath](0,0,0,0)[/imath] and [imath](\frac{1}{3},\frac{1}{3},\frac{1}{3},\frac{1}{3})[/imath] I think these are the only solutions that exist.
|
2064007
|
Uniformly equivalent metrics
Could I ask for help in the following homework question? Given any metric space [imath](M, d)[/imath], show that the metric [imath]\rho = d /(1 + d)[/imath] is always uniformly equivalent to [imath]d[/imath] but that there are cases in which the inequality [imath]d < C \rho[/imath] may fail to hold. Does equivalent imply uniformly equivalent?
|
1950208
|
Let [imath]\rho(x,y)= \frac{d(x,y)}{1+d(x,y)}[/imath], show that [imath]\rho[/imath] is a metric equivalent to [imath]d[/imath]
Let [imath](E,d)[/imath] be a metric space and [imath]\rho(x,y)= \frac{d(x,y)}{1+d(x,y)}[/imath]. Show that [imath](E,\rho)[/imath] is a metric and [imath]\rho[/imath] and [imath]d[/imath] are equivalent. I have proved the part about the metric space. However, I cant remember how to prove the equivalence. Also I'm stuck with a follow up which says: Show that there are no constants [imath]c,C >0[/imath] such that for all [imath]x,y \in E:[/imath] [imath]cd(x,y) \leq \rho(x,y) \leq Cd(x,y)[/imath]
|
672174
|
Continuity of an inverse function.
Theorem. Let [imath]f\colon I \to J[/imath], where [imath]I[/imath] is an interval and [imath]J[/imath] is the image [imath]f(I)[/imath], be a function such that: [imath]f[/imath] is strictly increasing on [imath]I[/imath]. [imath]f[/imath] is continuous on [imath]I[/imath]. Then [imath]J[/imath] is an interval, and [imath]f[/imath] has an inverse function [imath]f^{-1}\colon J\to I[/imath] such that: [imath]f^{-1}[/imath] is strictly increasing on [imath]J[/imath]. [imath]f^{-1}[/imath] is continuous on [imath]J[/imath]. I need to prove only that [imath]f^{-1}[/imath] is continuous on [imath]J[/imath]. Proof. Let [imath]y\in J[/imath], and (for simplicity) assume that [imath]y[/imath] is not an end-point of [imath]J[/imath]. Then [imath]y=f(x)[/imath], for some [imath]x\in I[/imath], and we want to prove that [imath]y_n\to y\implies f^{-1}(y_n)\to f^{-1}(y)=x.[/imath] Thus we want to deduce that for each [imath]\varepsilon>0[/imath], there is some number [imath]X[/imath] such that [imath]x-\varepsilon<f^{-1}(y_n)<x+\varepsilon\qquad \forall n>X\qquad(1).[/imath] Since [imath]f[/imath] is strictly increasing, we know that [imath]f(x-\varepsilon)<f(x)<f(x+\varepsilon).[/imath] also, since [imath]y_n\to y=f(x)[/imath], there is some number [imath]X[/imath] such that [imath]f(x-\varepsilon)<y_n<f(x+\varepsilon), \qquad\forall n>X.[/imath] Then, applying the strictly increasing function [imath]f^{-1}[/imath] to these inequalities, we obtain (1). Edit I don't understand why, since [imath]y_n\to y=f(x)[/imath], there is some number [imath]X[/imath] such that [imath]f(x-\varepsilon)<y_n<f(x+\varepsilon), \qquad\forall n>X.[/imath] I would have written the last line as follows: [imath]f(x)-\varepsilon<y_n<f(x)+\varepsilon, \qquad\forall n>X.[/imath] Applying the [imath]f^{-1}[/imath] to my last line, which I think it's correct, I obtain [imath]f^{-1}[f(x)-\varepsilon]<f^{-1}[y_n]<f^{-1}[f(x)+\varepsilon], \qquad\forall n>X.[/imath] which is not (1). What am I misunderstanding? Why the author of this proof wrote that last line and not my version? Thank you.
|
2558902
|
does inverse of continuous function is also continuous?
Suppose that [imath]f:R\rightarrow R [/imath] is continuous and does have an inverse function. How can I prove that [imath]f^{-1}[/imath] is also continuous?
|
92556
|
Operator norm. Alternative definition
Let [imath]T\colon X\to Y[/imath] be a linear operator with norm [imath]\|T\|=\sup_{\|x\|=1}\|Tx\|.[/imath] Prove that [imath]\|T\|=\sup_{\|x\|\leq 1}\|Tx\|.[/imath]
|
1559379
|
Equivalent definition operator norm
Let [imath]T: X \to Y[/imath] be a bounded linear map between normed spaces. The operator norm is defined by [imath]\sup_{\|x\| = 1} \|T(x)\|[/imath] Is this equivalent to [imath]\sup_{x \in B(0, 1)} \|T(x)\|[/imath] where [imath]B(0, 1)[/imath] is the open unit ball?
|
2064700
|
Algebra and remainder with division
What is the largest perfect square that divides [imath]2014^3-2013^3+2012^3-2011^3+\cdots+2^3-1^3[/imath]
|
1978304
|
What is the largest perfect square that divides [imath]2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3[/imath]
I've tried this but didn't get the answer : Let [imath]S=2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3[/imath] Using [imath]n^3-(n-1)^3 = 3n^2-3n+1[/imath], \begin{align} S &= 3(2014^2)-3(2014)+1+3(2012^2)-3(2012)+1+\ldots+3(2^2)-3(2)+1 \\&= 3\left ( 2014(2013)+2012(2011)+2010(2009)+ \ldots+2(1) \right ) + 1(1007) \\&= 3\left ( \sum_{n=1}^{1007}2n(2n-1) \right )+1007\\ =& \left ( \sum_{n=1}^{1007}4n^2-\sum_{n=1}^{1007}2n \right )+1007 \\=&\frac{12(1007)(1008)(2015)}{6}-\frac{2(1007)(1008)(3)}{2}+1007 \end{align} This is divisible by [imath]1007[/imath] but not by [imath]1007^2[/imath] which is the correct answer. Where have I gone wrong ?
|
2064622
|
Suppose that every subsequence of [imath]X=\{x_n\}[/imath] has a subsequence that converges to [imath]0[/imath]. Show that [imath]\lim X= 0[/imath]
Suppose that every subsequence of [imath]X=\{x_n\}[/imath] has a subsequence that converges to [imath]0[/imath]. Show that [imath]\lim X= 0[/imath]. I want to solve this question with the following hints: Prove [imath]\varlimsup x_n[/imath], [imath]\varliminf x_n[/imath] are both finite. Assume [imath]\varlimsup x_n=l[/imath] and [imath]\varliminf x_n=m[/imath]. Show that [imath]l=m=0[/imath] I have seen it here Suppose every subsequence of X converges to 0. Show that lim(X)=0 But I want to solve as per the hint given above. Please help me. A complete solution will help me a lot.
|
1695273
|
Suppose every subsequence of X converges to 0. Show that lim(X)=0
Question: Suppose that every subsequences of [imath]X = (x_n)[/imath] has a subsequence that converges to [imath]0[/imath]. Show that [imath]\lim(X) = 0[/imath] My attempt: Suppose that [imath]\lim(X) =L \neq 0[/imath] Let [imath]\epsilon > 0[/imath], and let [imath]K(\epsilon)[/imath] be s.t. [imath]n \ge K(\epsilon)[/imath] then [imath]|x_n-L| < \epsilon[/imath] let [imath]n_k \ge k[/imath], and then [imath]k \ge K(\epsilon)[/imath] and [imath]n_k \ge k \ge K(\epsilon)[/imath] s.t. [imath]|x_{n_{k}}-L| < \epsilon[/imath], which is a contradiction. therefore [imath]\lim(x) =0[/imath]
|
2065016
|
Why if [imath]a =b[/imath] then [imath]a = 0[/imath] is not a correct statement
Bogus Claim: If [imath]a[/imath] and [imath]b[/imath] are two equal real numbers, then [imath]a = 0[/imath] [imath]a = b[/imath] [imath]a^2 = ab[/imath] [imath]a^2 - b^2 = ab - b^2[/imath] [imath](a-b)(a+b) = (a-b)b[/imath] [imath]a + b = b[/imath] [imath]a = 0[/imath] I found this in my proof handouts, and correct me if I'm wrong,but is it wrong because after line 4 we divide both sides by [imath](a - b)[/imath] which would be [imath]0[/imath] if [imath]a = b[/imath] ?
|
2026631
|
Why can a = b imply 2 = 1?
[imath] a = b [/imath] [imath] a^2 = ab [/imath] [imath] a^2 - b^2 = ab - b^2 [/imath] [imath] (a+b)(a - b) = b(a - b) [/imath] [imath] a + b = b [/imath] [imath] 2b = b [/imath] [imath] 2 = 1 [/imath] Does the self-reference to the original formula make this path of argument invalid? I am confused as to what's going on here. Would this be a better continuation from [imath]a + b = b[/imath]: [imath] a + b = b \implies a = 0 \implies b = 0[/imath]
|
2065027
|
Prove: if [imath]d=\text{gcd}(m,n)[/imath] so [imath]\text{gcd}\left(\frac{m}{d},\frac{n}{d}\right)=1[/imath]
[imath]\newcommand{\gcd}{\text{gcd}}[/imath] Prove: if [imath]d=\gcd(m,n)[/imath] so [imath]\gcd\left(\frac{m}{d},\frac{n}{d}\right)=1[/imath] Intuitively it is obvious, but I am having a hard time to formalize the proof, what I have came to this: [imath]d=\gcd(m,n)[/imath] so [imath]d|m[/imath] and [imath]d|n[/imath] therefore [imath]m=dx[/imath] and [imath]n=dy[/imath] now if [imath]\gcd\left(\frac{m}{d},\frac{n}{d}\right)\neq 1[/imath] that mean that [imath]m[/imath] and [imath]n[/imath] have a common factor, after the division in [imath]d[/imath] which is the greatest common divisor is contradiction.
|
752928
|
Proving [imath]\gcd \left(\frac{a}{\gcd (a,b)},\frac{b}{\gcd (a,b)}\right)=1[/imath]
How would you go about proving that [imath]\gcd \left(\frac{a}{\gcd (a,b)},\frac{b}{\gcd (a,b)}\right)=1[/imath] for any two integers [imath]a[/imath] and [imath]b[/imath]? Intuitively it is true because when you divide [imath]a[/imath] and [imath]b[/imath] by [imath]\gcd(a,b)[/imath] you cancel out any common factors between them resulting in them becoming coprime. However, how would you prove this rigorously and mathematically?
|
1950495
|
Normed space where all absolutely convergent series converge is Banach
Let [imath]A[/imath] be a normed space where every absolutely convergent series converges in [imath]A[/imath]. How do I prove that [imath]A[/imath] is Banach? Let [imath]\sum_{n=1}^\infty x_n[/imath] be an absolutely convergent series in [imath]A[/imath], then we know that [imath]\sum_{n=1}^\infty ||x_n||[/imath] is convergent in [imath]\mathbb{R}[/imath] and that [imath]\sum_{n=1}^\infty x_n[/imath] is convergent in [imath]A[/imath]. Now using this, we want to show that all Cauchy sequences [imath]\{x_n\}[/imath] in [imath]A[/imath] converge. How would I do this?
|
1692697
|
If every absolutely convergent series is convergent then [imath]X[/imath] is Banach
Show that A Normed Linear Space [imath]X[/imath] is a Banach Space iff every absolutely convergent series is convergent. My try: Let [imath]X[/imath] is a Banach Space .Let [imath]\sum x_n[/imath] be an absolutely convergent series .Consider [imath]s_n=\sum_{i=1}^nx_i[/imath]. Now [imath]\sum \|x_n\|<\infty \implies \exists N[/imath] such that [imath]\sum_{i=N}^ \infty \|x_i\|<\epsilon[/imath] for any [imath]\epsilon>0[/imath] Then [imath]\|s_n-s_m\|\le \sum _{i=m+1}^n \|x_i\|<\epsilon \forall n,m>N[/imath] So [imath]s_n[/imath] is Cauchy in [imath]X[/imath] and hence converges [imath]s_n\to s[/imath] (say). Thus [imath]\sum x_i[/imath] converges. Conversely, let [imath]x_n[/imath] be a Cauchy Sequence in [imath]X[/imath]. Here I can't proceed how to use the given fact. Any help will be great.
|
2064719
|
An arctan exponential integral
Prove the following for [imath]\Re(z) >0[/imath] [imath] \log(\Gamma(z)) = 2\int_{0}^{\infty} \tan^{-1} \left(\dfrac{t}{z}\right)\dfrac{\mathrm{d}t}{e^{2\pi t} - 1} + \dfrac{\log(2z)}{2} + \left( z - \dfrac{1}{2} \right)\log (z) -z [/imath] I found this identity on Wolfram Functions here. I can't see where to start solving this problem. I'm looking for elementary solution, I haven't learned about Complex Analysis or Abel Plana formula yet. Please help. Thanks.
|
1604840
|
A [imath]\log \Gamma [/imath] identity: Where does it come from?
[imath]\log \Gamma (n)=n\log n -n +\frac{1}{2} \log \frac{2\pi}{n}+\int_0^\infty \frac{2\arctan (\frac{x}{n})}{e^{2\pi x}-1} \,\mathrm{d}x[/imath] Is an identity that is derived from using Sterling's approximation. I can't quite figure out how it was used, and was wondering for a proof.
|
2056157
|
Show that if [imath]\sum_{n=1}^{\infty}a_n^2[/imath] converges, then [imath]\sum_{n=1}^{\infty}{a_n \over n}[/imath] converges absolutely
I am trying to show that if the series [imath]\sum_\limits{n=1}^{\infty}a_n^2[/imath] converges, then the series [imath]\sum_\limits{n=1}^{\infty}{a_n \over n}[/imath] converges absolutely. How might I approach this?
|
2783429
|
If [imath]\sum\limits_{n=1}^∞u_n^2[/imath] is convergent, then [imath]\sum\limits_{n=1}^∞\frac{u_n}n[/imath] is absolutely convergent
If [imath]\{u_n\}[/imath] is a sequence of real numbers and the series [imath]\displaystyle\sum_{n=1}^{\infty}u_n^2[/imath] is convergent, prove that the series [imath]\displaystyle\sum_{n=1}^{\infty}\frac{u_n}{n}[/imath] is absolutely convergent. I have proved the series [imath]\displaystyle\sum_{n=1}^{\infty}\frac{u_n}{n}[/imath] is convergent, but how to prove it is absolutely convergent?
|
2065710
|
Determine the Largest Number in the Infinite Sequence
Determine the largest number in the infinite sequence [imath]1,\sqrt{2},\sqrt[3]{3},\sqrt[4]{4},....\sqrt[n]{n}[/imath]. I found [imath]3^{\frac{1}{3}}[/imath] is largest. I am confused how to start the question. Please help me
|
2051312
|
Greatest number in the sequence
I have seen this problem in a book . But I don't know what should be the solution . Question is There is a sequence defined by [imath]\sqrt[1]{1},\sqrt[2]{2},\sqrt[3]{3},\sqrt[4]{4},\cdots,\sqrt[n-1]{n-1},\sqrt[n]{n}[/imath] We have to find the largest number in this sequence. What I did is as usual what I do when comparing two irrational numbers I compared [imath]\sqrt[1]{1} [/imath] & [imath]\sqrt[2]{2}[/imath] and found that [imath]\sqrt[2]{2}[/imath] is greater. Then I compared [imath]\sqrt[3]{3} [/imath] &[imath]\sqrt[2]{2}[/imath] and found [imath]\sqrt[3]{3}[/imath] is greater. Then I compared [imath]\sqrt[3]{3} [/imath]& [imath]\sqrt[4]{4}[/imath] and found [imath]\sqrt[3]{3}[/imath] is greater. And then I compared [imath]\sqrt[5]{5}[/imath]&[imath]\sqrt[4]{4}[/imath] and found [imath]\sqrt[4]{4}[/imath] is greater. And finally I compared [imath]\sqrt[5]{5}[/imath]&[imath]\sqrt[6]{6}[/imath] and found [imath]\sqrt[5]{5}[/imath] is greater. I cannot guarantee will it work up to infinity or not and found [imath]\sqrt[3]{3}[/imath] is greatest term. And now I have to prove my result. More generally I have to prove that [imath]\sqrt[n]{n}\gt \sqrt[n+1]{n+1}[/imath] for all [imath]n\in\mathbb N[/imath] and [imath]n\gt 2[/imath] I applied induction to prove it but didn't got the solution. Please help me in proving this . Or please tell me an alternative way to tell which number is largest in the sequence.
|
2065836
|
Question on units of [imath]\mathbb{Z}[i\sqrt{5}][/imath]
Find the units of [imath]\mathbb{Z}[i\sqrt{5}] := \{a+bi\sqrt{5}:a,b \in\mathbb{Z}\} \subseteq \mathbb{C}[/imath]. By considering the modulus, we get that if [imath]x + yi\sqrt{5} \in \mathbb{Z}[i\sqrt{5}][/imath] is invertible, we must have that [imath]\frac{1}{a^2 + 5b^2} \in \mathbb{Z}[/imath] But this would mean [imath]a^2 + 5b^2 = 1[/imath] which is only possible if [imath]a = \pm 1[/imath] and [imath]b = 0[/imath]. Am I doing something wrong?
|
612520
|
Describe the units in [imath]\mathbb{Z}[\sqrt{-d}][/imath]
I need help describing all the units in the ring [imath]\mathbb{Z}[\sqrt{-d}]=\{a+b\sqrt{-d}|a,b\in\mathbb{Z}\}[/imath] where [imath]d>0[/imath] and is square-free. Thanks.
|
2042296
|
If [imath]q(A) = 0[/imath], then [imath]q[/imath] is divisible by the minimal polynomial of [imath]A[/imath]
If [imath]A[/imath] is a linear transformation with minimal polynomial [imath]p(x)[/imath], and if [imath]q(x)[/imath] is a polynomial such that [imath]q(A)=0[/imath], then show that [imath]q[/imath] is divisible by [imath]p[/imath].
|
1188497
|
Uniqueness of minimal polynomial: [imath]f(x)[/imath] divides [imath]g(x)[/imath]
so I am trying to show that [imath]f(x)[/imath] divides [imath]g(x)[/imath] for all polynomials [imath]g(x)[/imath] satisfying that [imath]g(A)=0[/imath] where [imath]f(x)[/imath] is the minimal polynomial of a square matrix [imath]A[/imath]. I know from my professor that [imath]f(x)[/imath] is the minimal polynomial of [imath]A[/imath] so [imath]f(A)=0[/imath] so then [imath]g(A)=0[/imath]. Therefore, [imath]f(x)[/imath] divides [imath]g(x)[/imath], i.e. [imath]g(x)=f(x)h(x)[/imath] where [imath]h(x)[/imath] is some other polynomial. My prof told us to solve this question we need to let [imath]b(x)=gcd((f(x),g(x))[/imath] and by using Bezout's Identity, we need to show that [imath]b(A)=0[/imath]. I am just having a little trouble trying to prove this. I know that this identity says: For two polynomials [imath]f_1(x)[/imath] and [imath]f_2(x)[/imath], there exist polynomials [imath]g_1(x)[/imath] and [imath]g_2(x)[/imath] such that [imath]g_1(x)f_1(x) + g_2(x)f_2(x) = > gcd(f_1(x), f_2(x))[/imath]. And the uniqueness of minimal polynomial theorem is: For a square matrix [imath]A[/imath], if [imath]f_1(A) = f_2(A) = 0[/imath] for two polynomials [imath]f_1(x)[/imath] and [imath]f_2(x)[/imath], then [imath]g(A) = 0[/imath] for [imath]g(x) = gcd(f_1(x), > f_2(x))[/imath]. So the minimal polynomial of [imath]A[/imath] is unique up to a scalar. So since [imath]f(x)[/imath] is already the minimal polynomial, I know that the [imath]deg(b(x))=deg(f(x))[/imath]. I know how to prove the Bezout identity using the Euclidean algorithm, I'm just not exactly sure how I would show all of this using the notation he gave us of [imath]b(x)=gcd((f(x),g(x))[/imath]. Any help at all would be appreciated.
|
1995581
|
Vector bundle and principal bundle
First, why the vector bundle, structured from principal bundle, has structure group [imath]G[/imath] ? Second, how to prove the principal [imath]G[/imath]-bundle, structured from vector bundle, is principal bundle according the Definition 2.3.3 ? Picture below is from the 65 of Jost's Riemannian Geometry and Geometric Analysis. In fact ,I think the definition 2.3.3 is not a good definition.
|
1994121
|
Vector bundle and principal bundle
Why fiber of [imath](P\times V)/G\rightarrow P/G[/imath] isomorphic to [imath]V[/imath] ? I think the fiber should be [imath]V/G[/imath], but it is not isomorphic to [imath]V[/imath] Picture below is from the 66 page of Jost's Riemannian Geometry and Geometric Analysis
|
2065800
|
Irreducible polynomial of degree [imath]5[/imath] in [imath]F_2 [x] [/imath] has distinct roots in any algebraic closure of[imath] F2[/imath]
Is it true that any irreducible polynomial of degree [imath]5[/imath] in [imath]F_2[x][/imath] has distinct roots in any algebraic closure of [imath]F_2[/imath]? [imath]F_2[/imath] : field of characteristic [imath]2[/imath].
|
396232
|
Irreducible polynomials have distinct roots?
I know that irreducible polynomials over fields of zero characteristic have distinct roots in its splitting field. Theorem 7.3 page 27 seems to show that irreducible polynomials over [imath]\Bbb F_p[/imath] have distinct roots in its splitting field (and all the roots are powers of one root). Is the proof correct? I have never seen this result anywhere else. The proof is very convincing to me. Does the result hold for [imath]\Bbb F_q[/imath] where q is a power of prime? I don't think it holds because I've heard there are irreducible polynomials with repeated roots? Please help.
|
2065577
|
Expectation of random variable.
Somewhere I found the definition of expectation of random variable as [imath]E[X]=\int_{0}^{\infty}(1-F(x))\,dx-\int_{-\infty}^0F(x)\,dx[/imath] But the definition I know is [imath]E[X]=\int_{-\infty}^{\infty}xf(x)\,dx[/imath] My Question is why [imath]\int_{-\infty}^{\infty}xf(x)\,dx=\int_{0}^{\infty}(1-F(x))\,dx-\int_{-\infty}^0F(x)\,dx[/imath]
|
1939758
|
Expectation formula of an integrable random variable
Given [imath]X[/imath] is integrable. Prove that [imath]E(X) = \int_{0}^{\infty} [1-F(x)]dx - \int_{-\infty}^{0}F(x)dx[/imath]. My 2ND solution Since [imath]X[/imath] is integrable, [imath]E(X) = E(X^{+}) - E(X^{-}) = \int_{0}^{\infty} xdF_{X^{+}}(x) - \int_{0}^{\infty} xdF_{X^{-}}(x) = \int_{0}^{\infty} [F_{X^{-}}(x) - F_{X^{+}}(x)]dx[/imath]. Now, since [imath]F_{X^{-}}(x) = P(-min(X,0)\leq x) = 1 - P(\min(X,0)<\ -x) = 1 - P(X<-x)[/imath] (since for [imath]x\in (0,\infty), P(0<-x) = 0[/imath]) [imath]= 1 - F_{X}(-x)[/imath], [imath]E[X] = \int_{0}^{\infty} [1 - F_{X}(-x) - F_{X^{+}}(x)] dx = \int_{0}^{\infty} 1 - F_{X}(x) - \int_{0}^{\infty} F_{X}(-x)dx[/imath] (the last equality is because for [imath]x\in (0,\infty), F_{X^{+}}(X) = F_{X}(x)[/imath]). Finally, by using the change of variable [imath]u=-x[/imath] for the 2nd integral, with [imath]u[/imath] as a dummy variable, we easily get the RHS that we need (Q.E.D) My question: Could someone help review the solution above to see if it's correct this time? Would really appreciate any input.
|
2065958
|
How is the infinite product 2?
I found this problem and according to Wolfram Alpha, the answer is 2. [imath]\prod_{n=0}^{\infty} \left (1+\frac{1}{2^{2^n}}\right)[/imath] Please do the favor of explaining me how the product is 2.
|
1922298
|
Closed form of infinite product [imath] \prod\limits_{k=0}^\infty \left(1+\frac{1}{2^{2^k}}\right)[/imath]
What will be the value of the following Infinite Product : [imath]\displaystyle \prod_{k=0}^\infty \left(1+\dfrac{1}{2^{2^k}}\right)[/imath] It would be nice if anyone could spare the time and boil down to the absolute basics and tell how they reached the solution.
|
2066097
|
Show that [imath]e^n>\frac{(n+1)^n}{n!}[/imath]
My brother give me a question that was To show that [imath]e^n>\frac{(n+1)^n}{n!}[/imath] where [imath]n\in \mathbb Z^+[/imath] Please help me!!!
|
2051203
|
Show that [imath]e^n>\frac{(n+1)^n}{n!}[/imath] without using induction.
I have got an inequality problem which is as follow: Show that [imath]e^n>\frac{(n+1)^n}{n!}[/imath] I can do it by induction but I have been told to prove it without induction. My Work: [imath]e^n=1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+........[/imath] [imath]e^n>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+........+\frac{n^n}{n!}[/imath] [imath]e^n>\frac{n^n}{n!}+\frac{n^{n-1}}{(n-1)!}.......+\frac{n^2}{2!}+n+1[/imath] From here I can't go further. I shall be thankful if you guys can provide me a complete solution/proof of this inequality. A hint will also work. Thanks in advance.
|
2065136
|
Function in [imath]C^{\infty} (\mathbb{R})[/imath] with geometric sequence
Let [imath]f\in C^{\infty}(\mathbb{R})[/imath]: 1) [imath]\exists[/imath] [imath]L>0[/imath] : [imath]\forall[/imath] [imath]x\in \mathbb{R}[/imath] and [imath]\forall[/imath] [imath]n\ge 1[/imath] [imath]|f^{(n)}(x)| \le L,[/imath] 2) [imath]f\left(\frac{1}{n}\right)=0, \quad \mbox{$\forall$} n\ge 1.[/imath] With these conditions, prove that: [imath]f(x)\equiv 0, \quad \mbox {in}\quad \mathbb{R}.[/imath]
|
2066972
|
How to do this...
Let [imath]f\in C^{\infty}(\mathbb{R})[/imath] verifying, [imath]a)[/imath] exists [imath]L>0[/imath] : [imath]\forall x\in \mathbb{R}[/imath] and [imath]\forall n\ge 1[/imath] [imath]|f^{(n)}(x)| \le L,[/imath] [imath]b)[/imath] [imath]f\left(\frac{1}{n}\right)=0 \quad \forall n\ge 1.[/imath] Prove that, [imath]f(x)\equiv 0, \quad \mbox {on} \quad \mathbb{R}.[/imath]
|
2064833
|
The support of a finitely generated module and its naturality
Reference: Atiyah and Macdonald, Introduction to Commutative algebra, page 46 Let [imath]A[/imath] be a commutative ring with 1 and [imath]M[/imath] a [imath]A[/imath]-module. Then we define [imath]Supp(M)[/imath] to be the set of prime ideals [imath]\mathfrak p[/imath] of [imath]A[/imath] such that [imath]M_\mathfrak p \neq 0[/imath]. It is known that if [imath]M[/imath] is finitely generated then [imath]Supp(M)=V(Ann(M))[/imath] where [imath]Ann(M)=\{f\in A |~ f M=0 \}[/imath] and [imath]V(E)=\{\mathfrak p \in Spec(A)|~ \mathfrak p \supset E\}[/imath] for a subset [imath]E[/imath] of [imath]A[/imath]. (1) What is the relationship between [imath]Supp(M)[/imath] and [imath] V(Ann(M))[/imath] if we drop the finitely generated condition, and do we have good examples to illustrate this? (2) Another question is about the "naturality" of the support. Given a ring homomorphism [imath]f:A\to B[/imath] and a finitely generated [imath]A[/imath]-module [imath]M[/imath], then we have [imath] Supp(B\otimes_A M) = f^{*~-1}(Supp(M)) [/imath] where [imath]f^*: Spec B \to Spec A[/imath] is induced by [imath]f[/imath]. I can only prove [imath] Supp(B\otimes_A M) \supset f^{*~-1}(Supp(M)) [/imath] by using (1) and observing that [imath]f(Ann(M))\subset Ann(B\otimes_A M)[/imath].
|
4260
|
Support of a module with extended scalars
I have a question which should be pretty basic commutative algebra, but I can't find a reference and I'm stuck on proving a result myself, so here it goes: Let [imath]\varphi: S \to R[/imath] be a morphism of commutative rings and let [imath]M[/imath] be a finitely generated [imath]S[/imath]-module. Then, we can "extend the scalars of [imath]M[/imath]" by considering [imath]M \otimes_S R[/imath], which is naturally an [imath]R[/imath]-module. I would like to get a description of [imath]\text{Supp}(M \otimes_S R)[/imath] (as an [imath]R[/imath]-module) involving [imath]\text{Supp}(M)[/imath] and [imath]\varphi[/imath]. Maybe this helps: by Eisenbud's Commutative algebra with a view toward algebraic geometry, Corollary 2.7, the problem can probably be solved by finding a description of [imath]\text{Ann}(M \otimes_S R)[/imath] and it is not difficult to see that [imath]\varphi(\text{Ann}(M))R \subset \text{Ann}(M \otimes_S R)[/imath], but I fail to prove the other inclusion.
|
2066277
|
Is there an alternative form/identity for [imath]\prod_{k=1}^{m} \cos(2^kx)[/imath]?
I am working with a family of integrals that involve a product of cosines of the form [imath]\prod_{k=1}^{m} \cos(2^kx)[/imath]. Is there a formula or identity for simplifying [imath]\prod_{k=1}^{m} \cos(2^kx)[/imath] to a polynomial number of terms in [imath]x[/imath]? Or [imath]\cos x[/imath], [imath]\sin x[/imath]?
|
8439
|
Proving: [imath]\cos A \cdot \cos 2A \cdot \cos 2^{2}A \cdot \cos 2^{3}A ... \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin A } [/imath]
[imath]\cos A \cdot \cos 2A \cdot \cos 2^{2}A \cdot \cos 2^{3}A ... \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin A } [/imath] I am very much inquisitive to see how this trigonometrical identity can be proved. PS:I am not much of interested about an inductive proof.
|
2066516
|
If [imath]|G|< \infty[/imath] and [imath]N\lhd G[/imath] such that [imath]G/N[/imath] is nilpotent, then there exists a nilpotent subgroup [imath]H\leq G[/imath] such that [imath]G=HN[/imath].
Let [imath]G[/imath] be a finite group and [imath]N[/imath] be a normal subgroup [imath]N\lhd G[/imath] such that the factor group [imath]G/N[/imath] is nilpotent. Then there exists a nilpotent subgroup [imath]H[/imath] of [imath]G[/imath] such that [imath]G=HN[/imath]. proof: I consider an arbitrary nilpotent subgroup [imath]H \leq G[/imath]. Then, there exists a number [imath]t[/imath] such that [imath]H^{(t)}=\{e\}[/imath]. Since [imath]G/N[/imath] is nilpotent, there exists a number [imath]k[/imath] such that [imath](G/N)^{(k)}=\{e\}[/imath] implying that [imath]G^k/N = \{e\}[/imath]. Also, since [imath]N\lhd G[/imath] and [imath]H\leq G[/imath], [imath]HN[/imath] is a subgroup of [imath]G[/imath]. Then, we can write that [imath]|HN|\leq |G|[/imath]. If I can show the inequality [imath]|G|<|HN|[/imath], then the proof will be done. How can I show this equality [imath]|G|<|HN|[/imath]?
|
2035468
|
An exercise about nilpotent quotient groups
Let [imath]G[/imath] be a finite group and [imath]N[/imath] be a normal subgroup of [imath]G[/imath] such that [imath]G/N[/imath] is nilpotent. a) Prove that there exists a nilpotent subgroup [imath]K[/imath] of [imath]G[/imath] such that [imath]G=NK.[/imath] b) Suppose that [imath]N[/imath] is abelian and [imath]Z\left(G\right) =\lbrace e \rbrace.[/imath] Show that if [imath]K[/imath] is nilpotent and [imath]G=NK[/imath] then [imath]K=N_G\left(K\right)[/imath] and [imath]K \cap N =\lbrace e \rbrace.[/imath] Could anyone help me to give a solution for the problem! thank you!
|
2066334
|
The sum of the fourth powers of the first [imath]n[/imath] positive integers
I am studying mathematical induction and most of the times I have to prove something. Like, for example: [imath]1 + 4 + 9 + ...+ n^2 = \frac{n(n+1)(2n+1)}{6}[/imath] This time I found a question that ask me to find a formula for [imath]1 + 16 + 81 + .... + n^4[/imath] How can I do this with induction? And is there really a formula for this sum?
|
2035188
|
Methods to compute [imath]\sum_{k=1}^nk^p[/imath] without Faulhaber's formula
As far as every question I've seen concerning "what is [imath]\sum_{k=1}^nk^p[/imath]" is always answered with "Faulhaber's formula" and that is just about the only answer. In an attempt to make more interesting answers, I ask that this question concern the problem of "Methods to compute [imath]\sum_{k=1}^nk^p[/imath] without Faulhaber's formula for fixed [imath]p\in\mathbb N[/imath]". I've even checked this post of common questions without finding what I want. Rule #1: Any method to compute the sum in question for arbitrary [imath]p[/imath] is good, either recursively or in some manner that is not in itself a closed form solution. Even algorithms will suffice. Rule #2: I don't want answers confined to "only some values of [imath]p[/imath]". (A good challenge I have on the side is a generalized geometric proof, as that I have not yet seen) Exception: If your answer does not generalize to arbitrary [imath]p[/imath], but it still generalizes to an infinite amount of special [imath]p[/imath]'s, that is acceptable. Preferably, the method is to be easily applied, unique, and interesting. To start us off, I have given my answer below and I hope you all enjoy.
|
20948
|
Fibonacci identity: [imath]f_{n-1}f_{n+1} - f_{n}^2 = (-1)^n[/imath]
Consider this Fibonacci equation: [imath]f_{n+1}^2 - f_nf_{n+2}[/imath] The problem asked to write a program with given n, output the the result of this equation. I could use the formula [imath]f_n = \frac{(1+\sqrt{5})^n - ( 1 - \sqrt{5} )^n}{2^n\sqrt{5}}[/imath] However, from mathworld, I found this formula Cassini's identity [imath]f_{n-1}f_{n+1} - f_{n}^2 = (-1)^n[/imath] So, I decided to play around with the equation above, and I have: [imath] \text{Let } x = n + 1 [/imath] [imath] \text{then the equation above becomes } f_x^2 - f_{x-1}f_{x+1} [/imath] [imath] \Rightarrow -( f_{x-1}f_{x+1} - f_x^2 ) = -1(-1)^x = (-1)^{x+1} = (-1)^{n+1+1} = (-1)^{n+2}[/imath] So this equation either is 1 or -1. Am I in the right track? Thanks, Chan
|
2238986
|
Deducing an answer on whether n is odd or even
How can I deduce the following: \begin{align*} F_{n}{^2} - F_{n+1}* F_{n-1}= +1 \text{ or} -1 \end{align*} Just from knowing if [imath]n[/imath] is even or odd? I have worked out by hand that if [imath]n[/imath] is odd, the result is +1 and if [imath]n[/imath] is even the result is -1, but I need to know how to work it out without just doing it. I was given this beforehand and it may or may not be related: \begin{align*} F_{n+1}{^2} - F_{n+2}* F_{n}= F_{n+1}*F_{n-1}-F_{n}{^2} \end{align*}
|
2066434
|
amount of non-abelian groups of a certain order
I know that there exists a unique group of order [imath]n[/imath] if and only if [imath](n, \phi(n))=1[/imath]. I also know how to obtain the number of abelian groups of a certain order There are exactly 5 groups of order 8, two of them are non-abelian and there are exactly 5 groups of order 12, three of them are non-abelian. I noticed that [imath]\phi(8)=4=(8,4)[/imath], [imath]\phi(12)=4=(12,4)[/imath] and that [imath]2=\frac{8}{4}[/imath],[imath]3= \frac{12}{4}[/imath]. So, my question is are there exactly [imath]\frac{n}{\phi(n)}[/imath] or [imath]\frac{n}{(\phi(n),n)}[/imath] non abelian groups of order [imath]n[/imath]? Or is there any way to know how many non-abelian groups of a certian order there are?
|
650772
|
How many groups exist with order [imath]n[/imath] (two isomorphic groups are treated as the same group)
I have tried to solve it for the case when [imath]n=1,2,3,4[/imath] which i found that there is only one group for [imath]n=1[/imath], one group for [imath]n=2,[/imath] one group for[imath] n=3 [/imath] and two groups for [imath] n=4[/imath], then my question come, if i were given a number [imath]n[/imath] which is larger, how do i know how many different groups are there? (I treat two isomorphic groups as the same group in this question.)
|
2056733
|
If [imath]f[/imath] is analytic in [imath]D_r(z_0)-\{z_0\}[/imath] and [imath]\Re f(z)>0[/imath] for all [imath]z\in D_r(z_0)-\{z_0\}[/imath], then [imath]z_0[/imath] is a removable singularity of [imath]f[/imath].
Show that if [imath]f[/imath] is analytic in a punctured disc [imath]D_r(z_0)-\{z_0\}[/imath] and we have [imath]\Re f(z)>0[/imath] for all [imath]z\in D_r(z_0)-\{z_0\}[/imath], then [imath]z_0[/imath] is a removable singularity of [imath]f[/imath]. I showed that [imath]z_0[/imath] cannot be an essential singularity of [imath]f[/imath]. But I cannot figure out a way to show that [imath]f[/imath] doesn't have a pole at [imath]z_0[/imath]. Can someone give me a hint?
|
1041034
|
If [imath]f[/imath] is ananlytic in [imath]D_r(z_0)[/imath]\ {[imath]z_0[/imath]} and [imath]Ref(z)>0[/imath] for all [imath]z\in D_r(z_0)[/imath]\ {[imath]z_0[/imath]} then [imath]z_0[/imath] is a removable singularity
If [imath]f[/imath] is ananlytic in [imath]D_r(z_0)[/imath]\ {[imath]z_0[/imath]} and [imath]Ref(z)>0[/imath] for all [imath]z\in D_r(z_0)[/imath]\ {[imath]z_0[/imath]} then [imath]z_0[/imath] is a removable singularity. I want to prove this statement but I just cannot seem to find a way. Clearly [imath]f(z)\neq0[/imath] in [imath]D_r(z_0)[/imath]\ {[imath]z_0[/imath]}. I just feel like this can be proved with this fact. Can someone give me a hint? Assistance is appreciated. Thanks a lot
|
2066963
|
Find : [imath] \int^\infty_0 \frac{1}{(1+x^{2015})(1+x^2)}dx [/imath]
Find: [imath] \int^\infty_0 \frac{1}{(1+x^{2015})(1+x^2)}dx [/imath] My initial thoughts were to use a trigonometric substitution of [imath]x=\tan(\theta)[/imath] which implies that [imath] dx= \sec^2(\theta) d\theta[/imath] this transforms our integral into [imath] \int^\infty_0 \frac{1}{(1+x^{2015})(1+x^2)}dx = \int^\frac{\pi}{2}_0 \frac{1}{1+(\tan(\theta))^{2015}} d\theta [/imath] But I'm having trouble evaluating it after. I also tried a substitution of [imath]y=\frac{1}{x}[/imath] but that did not get me anywhere either. Thanks for any help in advance :)
|
1885291
|
Computing [imath]\int_0^\infty\frac{1}{(1+x^{2015})(1+x^2)}[/imath]
How can I compute [imath]\int_0^\infty\frac{1}{(1+x^{2015})(1+x^2)}\quad?[/imath] My attempt: Looking at the limits of the integration I see that we should induce some [imath]\tan^{-1}(x)[/imath] so if we put [imath]\infty[/imath] we would get something like [imath]\frac{\pi}{2}[/imath] . But I am not sure how to proceed . Partial fractions don't yield good integral.
|
2055450
|
Graph Theory (Connectivity)
Suppose G is simple with degree sequence [imath]d_1 \leq d_2\leq ....\leq d_n[/imath], and for [imath]k \leq n-d_n-1[/imath], [imath]d_k \geq k[/imath]. Show G is connected. I posted this question before, and got an answer, but I couldn't understand it fully. (Especially about why [imath]d_1 \geq n-d_n-1[/imath].) Here's a link to that answer. Graph theory (Graph Connectivity) Anybody, help me, please.
|
2052344
|
Graph theory (Graph Connectivity)
I'm studying graph theory and got stuck in one of the problems. It's about graph connectivity with specific degree sequence condition. Please help me! Number 3 in original problem image My Try: Vertex of degree [imath]d_n[/imath] and [imath]d_n[/imath] other vertices which is connected to it by [imath]d_n[/imath] edges form a connected graph of [imath]d_n+1[/imath] vertices. Then a vertice [imath]d_{n-d_n-_1}[/imath] should be also in a component that contains vertice of [imath]d_n[/imath] degree.(Since there are at most [imath]n-d_n-1[/imath] vertices.) I don't know how to progress from here... Q: Suppose [imath]G[/imath] is simple with degree sequence [imath]d_1\le d_2\le \cdots d_n[/imath], and for [imath]k\le n-d_n-1, d_k\ge k[/imath]. Show [imath]G[/imath] is connected.
|
2066873
|
Problem in solving a question concerning real analysis.
The question is : Does there exist any function [imath]f : \mathbb R \longrightarrow \mathbb R[/imath] such that [imath]f(1) = 1[/imath], [imath]f(-1) = -1[/imath] and [imath]|f(x) - f(y)| \leq |x - y|^{\frac {3} {2}}[/imath]? It is clear that [imath]f[/imath] is continuous over [imath]\mathbb R[/imath] by the given condition and hence it attains all the values between [imath]-1[/imath] and [imath]1[/imath] in [imath](-1,1)[/imath].Now how can I proceed?Please help me. Thank you in advance.
|
1223377
|
[imath]f[/imath] defined [imath]|f(x) - f(y)| \leq |x - y|^{1+ \alpha}[/imath] Prove that [imath]f[/imath] is a constant.
Let [imath]f[/imath] be defined for all real [imath]x[/imath] and suppose that [imath]|f(x) - f(y)| \leq |x - y|^{1+ \alpha}[/imath] for all real [imath]x[/imath] and [imath]y[/imath], where [imath] \alpha > 0[/imath]. Prove that [imath]f[/imath] is constant. Proof: I shall show that the derivative of [imath]f[/imath] is zero. We have that [imath]0 \leq | {f(y) - f(x)} | \leq |x-y|^{1+ \alpha}[/imath]. Dividing through by [imath]|x-y|[/imath] we have [imath]0 \leq |\frac{f(y) - f(x)}{y - x}| \leq |y - x|^{\alpha}[/imath] and letting [imath]y \rightarrow x[/imath], we have shown that [imath]f'(x) = \lim_{x \to y}|\frac{f(y) - f(x)}{y - x}| = 0[/imath]. Therefore [imath]f[/imath] is a constant.
|
2067032
|
Number of nondecreasing functions from [imath]\{1,2,\dots,n\}[/imath] to [imath]\{1,2,\dots,n\}[/imath]
How many functions [imath]f: \{1,2,\dots,n\} \to \{1,2,\dots,n\}[/imath] are there such that for every [imath]i \leq j[/imath] we have [imath]f(i) \leq f(j)[/imath]? The only way I could think of was removing all the "bad" cases from [imath]n^n[/imath], but that wouldn't be very effective. What better way could I use?
|
1396896
|
Number of non-decreasing functions?
Let [imath]A = \{1,2,3,\dots,10\}[/imath] and [imath]B = \{1,2,3,\dots,20\}[/imath]. Find the number of non-decreasing functions from [imath]A[/imath] to [imath]B[/imath]. What I tried: Number of non-decreasing functions = (Total functions) - (Number of decreasing functions) Total functions are [imath]20^{10}[/imath]. And I think there are [imath]{20 \choose 10}[/imath] decreasing functions. Since you choose any [imath]10[/imath] codomain numbers and there's just one way for them to be arranged so that the resultant is a decreasing function. However my answer doesn't match. Where am I going wrong? How can I directly compute the non-decreasing functions like without subtracting from total?
|
2056650
|
Can you help to prove this property?
Let [imath]T : [0, 1]^2 → [0, 1][/imath]. Consider the following properties: [imath]T_1 : T(x, 1) = x[/imath] [imath]T_2 : T(x, y) = T (y, x) [/imath] [imath]T_3 : T(x, T (y, z)) = T(T (x, y), z) [/imath] [imath]T_4 [/imath]: If [imath]x ≤ u , y ≤ v \Rightarrow T(x, y) ≤ T(u, v) [/imath] Function [imath]T: [0, 1]^2 → [0, 1][/imath] that satisfies [imath]T_1 −T_4[/imath] is T-norm. consider [imath]T_L(x,y) = max(x+y-1,0) [/imath] we can show that [imath]T_L(x,y)[/imath] is a T norm. Conditions 1, 2 and 4 are obviously. Can you help me to prove condition 3?
|
2055122
|
Can you prove this property?
Let [imath]T : [0, 1]^2 → [0, 1][/imath]. Consider the following properties: [imath]T_1 : T(x, 1) = x[/imath] [imath]T_2 : T(x, y) = T (y, x) [/imath] [imath]T_3 : T(x, T (y, z)) = T(T (x, y), z) [/imath] [imath]T_4 [/imath]: If [imath]x ≤ u , y ≤ v \Rightarrow T(x, y) ≤ T(u, v) [/imath] Function [imath]T: [0, 1]^2 → [0, 1][/imath] that satisfies [imath]T_1 −T_4[/imath] is T-norm . For [imath]T(x) = \begin{cases}min(x,y),&max(x,y)=1,\\0,&o.w.\end{cases} [/imath] conditions 1, 2 and 4 obviously. Can you help me prove condition 3.
|
601809
|
Every two consecutive integers are coprime.
I start with knowing that two numbers are coprime if: [imath]n*k + m*j = 1[/imath] So, setting [imath]k = a[/imath] and j = [imath]a+1[/imath] I can solve as follows: [imath]n*a + m*a + a[/imath] Then, [imath]a(n+m) + m = 1[/imath] Where can I go from here?
|
296747
|
are two consecutive numbers relatively prime?
I have a question. I have been given this proof: "For any [imath]n[/imath] in the integers where [imath]n>2[/imath], show there are at least [imath]2[/imath] elements in [imath]U(n)[/imath] that satisfy [imath]x^2=1[/imath]." I have gone through and actually proved this, (that the numbers are [imath]1[/imath] and [imath]n-1[/imath]) but i didn't' know how to prove that [imath]n-1[/imath] is in fact in the set [imath]U(n)[/imath]. Is it because two consecutive numbers are always relatively prime?
|
2067240
|
prove that it's a constant function
Let [imath]~f~[/imath] be a function from [imath]\mathbb{R}[/imath] to [imath]\mathbb{R}[/imath] such that : [imath]~~\forall x \in \mathbb{R}~~ [/imath] , [imath]~~f(x) = f(2x) [/imath] and [imath]f[/imath] is continuous at [imath]~0[/imath] . prove that [imath]f[/imath] is a constant function .
|
1097122
|
proof that function is constant
I'm annoyed by quite a simple problem in calculus (I apologize in advance if I'm not using adequate terms in English, I don't take the course in English nor am I a native speaker): Let [imath]f:\mathbb R\longrightarrow \mathbb R[/imath] be a function which is continuous at [imath]0[/imath] and satisfies [imath]f(x)=f(2x)\,\forall x\in\mathbb R[/imath]. My question is how do I formally prove that for every [imath]x<0[/imath] [imath]f(x)=c_1[/imath], and for every [imath]x>0[/imath] [imath]f(x)=c_2[/imath], where [imath]c_1,c_2\in \mathbb R[/imath]? After proving those two statements (or at least one of them WLOG) it would be quite easy to prove that [imath]f(0)=c_1=c_2[/imath] by the definition of continuity of [imath]f(0)[/imath] and thus [imath]f(x)[/imath] is constant. Those two statements are just so obvious to see that I can't think of any formal way of proving so without "cheating". Thanks in advance!
|
2067257
|
Is there a bipartite graph with degrees of vertices [imath](3,3,3,3,3,3,3,3,3,5,6,6,6,6)[/imath]
I have no idea how to check this? Could someone give a hint?
|
1600289
|
Prove that there exists bipartite graph with this degree sequence: [imath](3,3,3,3,3,5,6,6,6,6,6,6,6,6)[/imath]
How do I prove that there exists (or does not exist) bipartite graph with this degree sequence: [imath](3,3,3,3,3,5,6,6,6,6,6,6,6,6)[/imath] ? The sum of the degree sequence is even and simplfied it looks like this: [imath](1,1,1,1,1,1)[/imath] => it's a degree sequence of a graph. The simplified degree sequence might suggest that it could be bipartite graph, but I have no idea how to prove it.
|
2067331
|
About Derivative.
What is derivative of [imath]\sin\sqrt x[/imath] using the definition of the derivative ? The definition of derivative is:- [imath]\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}[/imath]
|
1629163
|
How to find derivative of [imath]\sin\sqrt{x}[/imath] using difference quotient?
The definition of derivative of a function [imath]f(x)[/imath] is [imath]\lim_{h\to0} \frac{f(x+h)-f(x)}{h}[/imath] Using this definition, the derivative of [imath]\sin\sqrt{x}[/imath] will be: [imath]\lim_{h\to0} \frac{\sin\sqrt{x+h}-\sin\sqrt{x}}{h}[/imath] [imath]\lim_{h\to 0} \frac{\cos\left(\frac{\sqrt{x+h}+\sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x+h}-\sqrt{x}}{2}\right)}{h}[/imath] Now i got stuck. How to find the limit or simplify this expression? I get intuition that we have to use [imath]\lim_{x\to0}\frac{\sin x}{x} = 1[/imath] but that too is leading no where. I am unable to remove h from denominator. NOTE I know the derivative of [imath]\sin\sqrt{x}[/imath] is [imath]\frac{\cos\sqrt{x}}{2\sqrt{x}}[/imath] using chain rule, but this exercise was given to us for practice using division quotient.
|
2067440
|
Does having a countable subbasis also satisfy the second axiom of countability?
My guess is "yes". Unfortunately, the course I've taken this semester on "point-set topology" did not cover the concept of subbasis, however I have a few words on it in mind, subbasis unlike basis uses "intersection" instead of "union" of its elements to build the topology. I need this proposition (If it's true, otherwise can you give me a hint on proving the following ) to prove this; [imath]\mathbb{Z}[/imath] with the finite-complement topology is second-countable. My insight into proving this proposition is this: [imath]S=\{\{x\}^c| x \textrm{ is a point in the space}\}[/imath] and [imath]|S|=\aleph_0[/imath], thus since any open set can be created by such subsets, then the space is second-countable.
|
1823470
|
If [imath](X, \mathcal{T})[/imath] has a countable subbasis, then it has a countable basis
Given [imath](X, \mathcal{T})[/imath] a topological space. Let [imath]\mathcal{S}[/imath] be a subbasis on [imath](X, \mathcal{T})[/imath] Claim: If [imath]\mathcal{S}[/imath] is countable, then [imath]\mathcal{T}[/imath] has a countable basis [imath]\mathcal{B}[/imath] I am not sure how to go about approaching this quetion but here's my attempt: I want to show that there exists a surjection [imath]g[/imath] from [imath]\mathcal{S}[/imath] to [imath]\mathcal{B}[/imath], and there is an injection [imath]f[/imath] from [imath]\mathcal{B}[/imath] and [imath]\mathcal{S}[/imath] thus [imath]|\mathcal{S}| = \aleph_0 = |\mathcal{B}|[/imath] Define [imath]g[/imath] as [imath]g(S_1, S_2, \ldots, S_n) = S_1 \cap S_2 \cap \ldots \cap S_n = B[/imath] where [imath]S_1, \ldots, S_n \in \mathcal{S}[/imath], and [imath]B \in \mathcal{B}[/imath] But how does the countability of [imath]\mathcal{S}[/imath] come in? I'm really lost Per Henno's suggestion re-attempt: Let [imath]\mathcal{S}[/imath] be a countable subbase of [imath](X, \mathcal{T})[/imath]. Then [imath]\mathcal{S}[/imath] can be listed as [imath]\{S_1, S_2, \ldots \}, S_i \in S, i \in \mathbb{N}[/imath] By definition, each basis element is the finite intersection of subbasic elements written as [imath]\bigcap\limits_{i \in F_n} S_i[/imath], where [imath]F_n[/imath] is a finite set in [imath]\mathbb{N}[/imath]. Since there exists countably many finite sets in [imath]\mathbb{N}[/imath], we can list all the finite sets as [imath]\{F_1, F_2, \ldots\}[/imath] Then correspondingly we can list all the basis elements as:[imath]\{\bigcap\limits_{i \in F_1} S_i, \bigcap\limits_{i \in F_2} S_i, \ldots\}[/imath] which is a countable set. Hence [imath]\mathcal{B}[/imath] is countable.
|
2067757
|
How to find all integer pairs [imath](x,y)[/imath] that satisfy the equation [imath]21x+47y=1[/imath] without using modular arithmetic?
Find all integer pairs [imath](x,y)[/imath] that satisfy the following equation: [imath] 21x+47y=1. [/imath] I know the solution with modular arithmetic: [imath]\begin{align} 0x + 5y &= 1 \pmod{21} \\ 5y &= -20 \pmod{21}\\ y = -4 \pmod{21} &\Longleftrightarrow y = -4 + 21c \\ 21x + 47(-4 + 21c) &= 1\\ 21x &= 189 - 21 * 47c \\ x &= 9 - 47c. \\ \end{align}[/imath] Therefore [imath](x, y) = (9 - 47c, -4 + 21c)[/imath] Is there a way to do this without modular arithmetic? Thanks in advance!
|
2067728
|
Prove that the equation [imath]18x+42y=22[/imath] has no integer solution?
Can you help me to prove? Should I prove it by contradiction and suppose that [imath]x[/imath], y are solutions ?
|
2067849
|
Why does [imath]\lim_{n\to\infty}(1+\frac{1}{n})^n = e[/imath] instead of [imath]1[/imath]?
On Wikipedia, it says that [imath]\lim_{n\to\infty}(1+\frac{1}{n})^n = e[/imath] : It [e] is approximately equal to 2.71828,[1] and is the limit of (1 + 1/n)n as n approaches infinity, ... (Source) When I evaluate [imath](1+\frac{1}{n})^n[/imath] for [imath]n = 10^8[/imath], I get approximately [imath]2.718281798347[/imath] which indeed is pretty close to [imath]e[/imath]. But when I try to "solve" the limit using the laws of limits, I get [imath]\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = \left(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)\right)^n[/imath] because of the power law [imath] = \left(\lim_{n\to\infty}(1)+\lim_{n\to\infty}\left(\frac{1}{n}\right)\right)^n[/imath] because of the addition law [imath]=\left(1+0\right)^n = 1[/imath] but that would mean that [imath]e=1[/imath], which is obviously not true. What am I missing / doing wrong? Thanks in advance.
|
2156677
|
Strange limit : [imath]e=1[/imath]
Why the following is not correct? : [imath]\lim_{n \to \infty} (1+\frac1n)^n = \lim_{n \to \infty} (1+\frac1n) \lim_{n \to \infty} (1+\frac1n) \dots \lim_{n \to \infty} (1+\frac1n) = 1 \times 1 \times \dots 1 = 1.[/imath] I guess that the problem (as usual) lies in infinity, but if so how? I checked all the theorems in Fitzpatrick's book of Calculus, nothing prevents to do the limits infinitely many times (esp. for a convergent one)!
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.