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2146007 | Stable and unstable points
Looking for some help understanding the following concept. I know when you are trying to determine the stability of fixed points for the system, [imath]x'=\sin x[/imath] You proceed with the following steps, [imath]x'=0[/imath] [imath]\sin x=0[/imath] [imath]x^{*}=k\pi[/imath] deriving [imath]x'[/imath] to determine stability would give, [imath]\cos x[/imath] plugging in the fixed point, [imath]\cos (k\pi)[/imath] Therefore, when [imath]k\pi[/imath] is even it is equal to [imath]1[/imath], so it would be unstable. On the other hand when [imath]k\pi[/imath] is odd it is equal to [imath]-1[/imath] so it would be stable. However what if you were given [imath]x'=\cos x[/imath]. The derivative would be [imath]-\sin x[/imath] and the fixed points would still be [imath]k\pi[/imath] but when [imath]k\pi[/imath] is odd and when it is even, is it stable or unstable? | 2132398 | Phase Potentials
Looking to get a explanation of the following solution. [imath]x'=sinx[/imath] I understand that you must integrate [imath]sinx[/imath] and then take the negative of it making it [imath]v(x)=cos(x) + c = cos(x)[/imath] with [imath]c=0[/imath]. However I am having problems understanding the equilibrium points. The equilibrium points of this system are, [imath]x^*=kz[/imath] [imath]x*=kz[/imath] k is even, this is unstable [imath]x^*=kz[/imath] k is odd, are stable Why is it unstable when even and stable when even? |
2146839 | Proving that not all roots of a cubic polynomial are real
Consider a cubic polynomial [imath]p(x)=ax^3+bx^2+cx+d[/imath][imath]a,b,c,d[/imath] are integers such that [imath]ad[/imath] is odd and [imath]bc[/imath] is even.The question is to prove that not all roots of [imath]p(x)[/imath] can be real. I tried differentiating p(x) to get [imath]p'(x)=3ax^2+2bx+c[/imath]Now for the cubic equation to have nonreal roots [imath]4b^2-12ac<0[/imath].But I donot see how I can apply the conditions as given in the problem.Any hints to proceed shall be highly appreciated.Thanks. | 2069284 | If [imath]ad[/imath] and [imath]bc[/imath] are odd and even, respectively, then prove that [imath]ax^3+bx^2+cx+d[/imath] has an irrational root.
[imath]\displaystyle ax^3+bx^2+cx+d[/imath] is a polynomial with integer coefficients. It is given that [imath]ad,\,bc[/imath] are odd and even respectively. Then prove that not all roots of the polynomial are rational. It is easy to see that none of the roots are integer. But how to tackle the rational case? Any help is appreciated. |
2145939 | Proof that an automorphic injective hol. mapping on a bounded domain given one point being the identity implies the whole mapping is too
Let [imath]\Omega[/imath] be a bounded domain and let [imath]\phi[/imath] be an injective holomorphic mapping of [imath]\Omega[/imath] to itself. The proposition I would like to prove is: If [imath] \exists a \in \Omega[/imath] s.t. [imath]\phi(a) = a[/imath] and [imath]\phi'(a) = 1[/imath], [imath]\phi[/imath] must be the identity. A given hint was: Write the power series for [imath]\phi[/imath] centred at [imath]a[/imath] as: [imath]\phi(z)=a+(z-a)+ \text{ higher order terms}[/imath] and consider [imath]\phi, \phi \circ \phi, \phi \circ \phi \circ \phi, \dots[/imath]. Estimate the coefficient of the first nonzero term in the power series after the linear term, assuming that it exists, and show that it must in fact be zero. Unfortunately however, I'm not quite certain how to proceed with this hint. | 906489 | [imath]f: \Omega \rightarrow \Omega[/imath] holomorphic, [imath]f(0) = 0[/imath], [imath]f'(0) = 1[/imath] implies [imath]f(z) = z[/imath]
Let [imath]\Omega[/imath] be a bounded connected open subset of [imath]\mathbb{C}[/imath] containing [imath]0[/imath]. Let [imath]f: \Omega \rightarrow \Omega[/imath] be holomorphic and [imath]f(0) = 0[/imath], [imath]f'(0) = 1[/imath]. The problem I am working on is to show that [imath]f(z) = z[/imath]. If [imath]\Omega = \mathbb{D}[/imath], then this follows from the Schwarz Lemma. I also know of a solution (posted here) which involves looking at the power series coefficients of [imath]f^n := f\circ f \circ f \circ \cdots \circ f[/imath] ([imath]n[/imath] times) and using the Cauchy estimates, but is there a different way of doing this problem that doesn't involve taking the [imath]k[/imath]th derivative of [imath]f^n[/imath]? |
2146686 | Showing the irreducibility of a polynomial
I am asked to show that [imath]f(X)=X^5 - aX -1[/imath] is irreducible over [imath]\mathbf{Z[X]}[/imath] unless [imath]a=0,2,-1[/imath]. ([imath]\mathbf{Z}[/imath] being the ring of integers.) I have been able to do this, but my solution is a lengthy one. I used the fact that if [imath]f[/imath] is not irreducible, then we have [imath]f(X)=g(X)h(X)[/imath], where [imath]g[/imath] and [imath]h[/imath] are polynomials over [imath]Z[/imath]. We can ignore the case where either one is constant, because the g.c.d. of the coefficients is 1. Hence,there are two possibilities - one where one of the polynomials is linear, and the second where one of them has degree 2 and the other has degree 3. For the first case, f must have a rational root, and it is easy to show that this is only possible if [imath]a[/imath] is either [imath]0[/imath] or [imath]2[/imath]. In the latter case, we can assume [imath]g(X)= AX^3 + BX^2 + CX + D[/imath] and [imath]H(X)= A'X^2 + B'X +C'[/imath], after which a laborious calculation involving comparing coefficients shows that that only possibility is [imath]a=-1[/imath]. However, I feel that there is a much more simple and obvious solution that I am missing. Since the coefficients of [imath]f[/imath] have g.c.d 1, f is irreducible over [imath]Z[/imath] iff it is irreducible over [imath]Q[/imath]. I was wondering if there was any way in which we could perhaps use that to find a shorter solution. Thanks in advance! | 193201 | Showing [imath]x^{5}-ax-1\in\mathbb{Z}[x][/imath] is irreducible
This is an exercise for the book Abstract Algebra by Dummit and Foote (pg. 519): show [imath]x^{5}-ax-1\in\mathbb{Z}[x][/imath] is irreducible for [imath]a\neq-1,0,2[/imath] I need help with this exercise, I don't have an idea on how to prove there are no quadratic factors for [imath]a\neq-1,0,2[/imath] (for [imath]a=-1[/imath] there is a quadratic factor). |
2147193 | Prove that :[imath]\tan 70 - \tan 20 = 2\tan 40 + 4\tan 10[/imath]
Prove that :[imath]\tan 70 - \tan 20 = 2\tan 40 + 4\tan 10[/imath] My Attempt, [imath]70-20=40+10[/imath] [imath]\tan (70-20)=\tan (40+10)[/imath] [imath]\dfrac {\tan 70 - \tan 20}{1+\tan 70. \tan 20 }=\dfrac {\tan 40 + \tan 10 }{1-\tan 40. \tan 10 }[/imath] How should I move on? Please help Thanks | 1861288 | On the proof [imath]\tan 70°-\tan 20° -2 \tan 40°=4\tan 10°[/imath]
I am currently studying in class 10 and I am unable to do this problem. [imath]\tan 70 ° -\tan 20° -2 \tan 40° =4\tan 10°[/imath] Can anybody please help me. Thanks! |
2147564 | [imath]\int_{-2}^2 \sin\left(x^5\right)\, e^{(x^8)\sin(x^4)}\,dx\;[/imath]: integration by parts or substitution?
[imath]\large\int_{-2}^2 \sin\left(x^5\right)\, e^{(x^8)\sin(x^4)}\,dx[/imath] I think I should solve it by parts but I'm really confused. It seems as a hard question. | 2142105 | [imath]\int_{-2}^{2} \sin(x^5)e^{(x^8\sin(x^4))} dx[/imath]
Okay I like have no idea how to do this. I tried integration by parts then everything came back uglier. I then put this in wolfram it says "No found in standard mathematical functions" Is there a easy way to do this? Maybe like FTOC II? |
2147768 | Proving [imath]\int_{0}^{a}f(x)dx+\int_{0}^{b}f^{-1}(y)dy\ge ab[/imath]
I want to prove the following inequality Let [imath]a[/imath] and [imath]b[/imath] be positive real numbers and [imath]f:\mathbb{R}\rightarrow\mathbb{R}[/imath] a continuous function. Then, [imath]\int_{0}^{a}f(x)dx+\int_{0}^{b}f^{-1}(y)dy\ge ab[/imath] I know that [imath](a-b)^2 \ge 0[/imath] and so I easily get [imath]2ab\le a^2 + b^2 [/imath] and thus [imath]ab\le a^2 + b^2 [/imath]. I feel like this should guide me to the answer, but I don't know how. | 1132153 | Proof of Young's inequality
The following problem is from Spivak's Calculus. Suppose that [imath]f[/imath] is a continuous increasing function with [imath]f(0)=0[/imath]. Prove that for [imath]a,b \gt 0[/imath] we have Young's inequality [imath] ab \le \int_0^af(x)dx+\int_0^bf^{-1}(x)dx[/imath], and that equality holds if and only if [imath]b=f(a)[/imath]. It is enough to consider the case [imath]f(a) \gt b[/imath], and show that the strict inequality occurs in this case. I've tried proving this using the theorem [imath] \int_a^bf^{-1}=bf^{-1}(b)-af^{-1}(a)-\int_{f^{-1}(a)}^{f^{-1}(b)}f[/imath] but I got stuck along the way. How may I show this rigorously using the definition or properties of integrals? Any hint, suggestions or solutions would be appreciated. |
2147638 | An Introduction to Theory of Groups by Joseph J. Rotman - Exercise 1.5
If [imath]1 \leq r \leq n,\;[/imath] then there are [imath]\quad\dfrac{n(n-1)\cdots(n-r+1)}{r}\quad r[/imath]-cycles in [imath]S_n[/imath]. I think that I used induction to prove this, but I don't know how to prove this. Thanks in advance! | 965596 | If [imath]n>m[/imath], then the number of [imath]m[/imath]-cycles in [imath]S_n[/imath] is given by [imath]\frac{n(n-1)(n-2)\cdots(n-m+1)}{m}[/imath].
Show that if [imath]n>m[/imath], then the number of [imath]m[/imath]-cycles in [imath]S_n[/imath] is given by [imath]\frac{n(n-1)(n-2)\cdots(n-m+1)}{m}.[/imath] My doubt Suppose I wish to count the number of [imath]m-[/imath]cycles. Then I will get [imath]\binom n m[/imath] ways of choosing such cycles. And the cycles which have the same representation must be divided so I will get [imath]\frac{\binom n m}{m}[/imath] What am I doing wrong? Kindly help me. |
2147130 | Proof that the sum of the first [imath]n[/imath] cubes is equal to the square of the [imath]n^{th}[/imath] triangular number
Show that the difference between the squares of two consecutive triangular numbers is a cube. Hence or otherwise, show that the sum of the first [imath]n[/imath] cubes is equal to the square of the [imath]n^{th}[/imath] triangular number. I have managed to show that the difference between the squares of two consecutive triangular numbers is a cube by subbing [imath]n[/imath] and [imath]n+1[/imath] into the formula for a triangular number and working through to get a result of [imath](n+1)^3[/imath] which is indeed a cubed number. However, I don not know how to use this to show that the sum of the first [imath]n[/imath] cubes is equal to the square of the [imath]n^{th}[/imath] triangular number, any advice? | 1328798 | Proving “The sum of [imath]n[/imath] consecutive cubes is equal to the square of the sum of the first [imath]n[/imath] numbers.”
This site states: Example [imath]\boldsymbol 3[/imath]. The sum of consecutive cubes. Prove this remarkable fact of arithmetic: [imath]1^3 +2^3 +3^3 +\ldots +n^3 =(1 +2 +3 +\ldots +n)^2.[/imath] “The sum of [imath]n[/imath] consecutive cubes is equal to the square of the sum of the first [imath]n[/imath] numbers.” In other words, according to Example [imath]1[/imath]: [imath]1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}.[/imath] Should: [imath]1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}[/imath] not be: [imath]1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^3 +(n + 1)^3}{2^3}[/imath] as everything in the left-hand side is cubed? |
2148141 | Divisibilty and eulers thereom
if [imath]\gcd(a,b)=1[/imath] is it true that, [imath]a^{\phi(b)}+b^{\phi(a)}=1\mod ab[/imath] I think the answer is yes, but I am not exactly sure about my reasoning [imath]a^{\phi(b)}=1\mod b[/imath] and [imath]b^{\phi(a)}=1\mod a[/imath] by Euler's theorem. The question asks if [imath]ab| a^{\phi(b)}+b^{\phi(a)}-1[/imath] So what would a better way of seeing this be? | 894433 | Proof of an equality involving [imath]\phi(n)[/imath]
I would like to prove the following equality: [imath]k^{\phi(l)} + l^{\phi(k)} \equiv1\pmod{lk}[/imath]if [imath]\gcd(l,k)=1[/imath]. What methods can I use? Thank you for your help. |
2144691 | Proving that the sequence [imath]a_0=1, a_{n+1}=\cos(a_n)[/imath] converges
Consider the following recursively defined sequence: [imath]a_0 = 1, a_{n+1}=\cos\left(a_n\right)[/imath] Having a look with Wolfram Alpha, it's fairly clear that this sequence converges to something in the neighborhood of [imath]0.74[/imath] or so. However, I have no clue how to actually prove that this sequence is convergent. I've thought about proving that it is a Cauchy sequence - that is, for any given [imath]\varepsilon>0[/imath], there exists an [imath]N\in\mathbb{N}[/imath] such that [imath]|a_n-a_m|<\varepsilon[/imath] for any [imath]n,m>N[/imath]. However, I have no clue how to approach to repeated application of the cosine function, especially since you don't know how many iterations of the cosine function there are between [imath]n[/imath] and [imath]m[/imath]. Looking at the Wolfram Alpha plot of the first 30 terms in the sequence also gave me the idea to separate the whole thing into two subsequences [imath]a_{2n}[/imath] and [imath]a_{2n+1}[/imath] and then prove that they are increasing and decreasing while being bounded, but again I had no idea how to proceed to due to the repeated iteration of the cosine function. I'd also be interested in a closed form of the limit if there is one, but I'd guess that none exists. | 1701935 | Let [imath]a_n=\cos(a_{n-1}), L=[a_1,a_2,...,a_n,...].[/imath] Is there an [imath]a_0[/imath] such that [imath]L[/imath] is dense in[imath][-1,1]?[/imath]
I've been experimenting with recursive sequences lately and I've come up with this problem: Let [imath]a_n= \cos(a_{n-1})[/imath] with [imath]a_0 \in \Bbb{R}[/imath] and [imath]L=[a_1,a_2,...,a_n,...].[/imath] Does there exist an [imath]a_0[/imath] such that [imath]L[/imath] is dense in [imath][-1,1]?[/imath] I know of [imath]3[/imath] ways of examining whether a set is dense: [imath]i)[/imath]The definition, that is, whether its closure is the set on which it is dense, in our case this means if: [imath]\bar L=[-1,1][/imath]. ii)[imath](\forall x \in [-1,1])(\forall \epsilon>0)(\exists b \in L):|x-b|<\epsilon[/imath] [imath]iii)[/imath] [imath](\forall x \in [-1,1])(\exists b_n \subseteq L):b_n\rightarrow x[/imath] So far I haven't been able to use these to answer the question. I tried plugging in different values of [imath]a_0[/imath] and see where that leads but I have not found any corresponding promising "pattern" for [imath]a_n[/imath]. Any ideas on how to approach this? |
2148585 | Solve the exponential
If [imath] A = \begin{bmatrix} 3 & 4 \\ 4 & -3 \end{bmatrix}[/imath] Can someone find [imath]\mathbf e^A[/imath] ? Edit 1:- I got this question in my fucntional analysis paper ! How is it even related to functional analysis ? Can somebody explain ? | 1044148 | For a 2x2 matrix A satisfying [imath]A^k=I[/imath], compute [imath]e^A[/imath]
For a 2x2 matrix A satisfying [imath]A^k=I[/imath], compute [imath]e^A[/imath] Oh, the exponential of a matrix is: [imath]e^A=\sum_{i=0}^\infty\frac{1}{i!}A^i[/imath] I thought I'd solved the [imath]e^A[/imath] form but I actually did something really silly, now I'm a little stuck. If [imath]A^k=I[/imath] then [imath]A^{k+1}=A[/imath] and we have a cycle forming, so we will get: [imath]\sum^{k-1}_{i=0}\sum^{\infty}_{j=1}\frac{1}{(kj+i)!}A^i[/imath] (Or something like this form, I don't have paper to hand and just spotted that now) Is this what the question wants? It doesn't use the 2x2 property. I believe the answer lies in finding an expression for the inner summation, I can do this for two terms (think of [imath]e+e^{-1}[/imath] all the odd power leave) |
2148893 | Why [imath]\gcd(0,0)=0[/imath] when [imath]0/0[/imath] is undefined?
Is this only a standard we have to follow? But Doesn't such acceptance manifests into a wrong solution eventually for some problem? If no, is there any proof? | 495119 | What is [imath]\gcd(0,0)[/imath]?
What is the greatest common divisor of [imath]0[/imath] and [imath]0[/imath]? On the one hand, Wolfram Alpha says that it is [imath]0[/imath]; on the other hand, it also claims that [imath]100[/imath] divides [imath]0[/imath], so [imath]100[/imath] should be a greater common divisor of [imath]0[/imath] and [imath]0[/imath] than [imath]0[/imath]. |
2147851 | What are covariant derivatives, and what is the intuition behind/for covariant derivatives?
Anyone reading this post might be interested in my related question here. This is not a duplicate of my related question, which was about affine connections, not covariant derivatives. Here is the definition of covariant derivative, as appears in Milnor's book Morse Theory. DEFINITION. An affine connection at a point [imath]p \in \text{M}[/imath] is a function which assigns to each tangent vector [imath]\text{X}_p \in \text{TM}_p[/imath] and to each vector field [imath]\text{Y}[/imath] a new tangent vector[imath]\text{X}_p \vdash \text{Y} \in \text{TM}_p[/imath]called the covariant derivative of [imath]\text{Y}[/imath] in the direction [imath]\text{X}_p[/imath]. (Note that our [imath]\text{X} \vdash \text{Y}[/imath] coincides with Nomizu's [imath]\nabla_\text{X} \text{Y}[/imath]. The notation is intended to suggest that the differential operator [imath]\text{X}[/imath] acts on the vector field [imath]\text{Y}[/imath].) This is required to be bilinear as a function of [imath]\text{X}_p[/imath] and [imath]\text{Y}[/imath]. Furthermore, if[imath]f: \text{M} \to \mathbb{R}[/imath]is a real valued function, and if [imath]f\text{Y}[/imath] denotes the vector field[imath](f\text{Y})_q = f(q)\text{Y}_q[/imath]then [imath]\vdash[/imath] is required to satisfy the identity[imath]\text{X}_p \vdash (f\text{Y}) = (\text{X}_p f)\text{Y}_p + f(p) \text{X}_p \vdash \text{Y}.[/imath] (As usual, [imath]\text{X}_p[/imath] denotes the directional derivative of [imath]f[/imath] in the direction of [imath]\text{X}_p[/imath].) I have two questions. This definition of covariant derivative is quite terse here (indeed, it seems like affine connections and not covariant derivatives are being made out to be the most important thing here)—I'm just seeing text on a page and not really understanding what is going on here. Is it possible somebody could help me parse through/explain what is really being said here with regards to covariant derivative? Could somebody supply their intuitions behind/for covariant derivatives? Thanks. | 2145255 | What is the affine connection, and what is the intuition behind/for affine connection?
Here is the definition of affine connection, as appears in Milnor's book Morse Theory. DEFINITION. An affine connection at a point [imath]p \in \text{M}[/imath] is a function which assigns to each tangent vector [imath]\text{X}_p \in \text{TM}_p[/imath] and to each vector field [imath]\text{Y}[/imath] a new tangent vector[imath]\text{X}_p \vdash \text{Y} \in \text{TM}_p[/imath]called the covariant derivative of [imath]\text{Y}[/imath] in the direction [imath]\text{X}_p[/imath]. (Note that our [imath]\text{X} \vdash \text{Y}[/imath] coincides with Nomizu's [imath]\nabla_\text{X} \text{Y}[/imath]. The notation is intended to suggest that the differential operator [imath]\text{X}[/imath] acts on the vector field [imath]\text{Y}[/imath].) This is required to be bilinear as a function of [imath]\text{X}_p[/imath] and [imath]\text{Y}[/imath]. Furthermore, if[imath]f: \text{M} \to \mathbb{R}[/imath]is a real valued function, and if [imath]f\text{Y}[/imath] denotes the vector field[imath](f\text{Y})_q = f(q)\text{Y}_q[/imath]then [imath]\vdash[/imath] is required to satisfy the identity[imath]\text{X}_p \vdash (f\text{Y}) = (\text{X}_p f)\text{Y}_p + f(p) \text{X}_p \vdash \text{Y}.[/imath] (As usual, [imath]\text{X}_p[/imath] denotes the directional derivative of [imath]f[/imath] in the direction of [imath]\text{X}_p[/imath].) I have two questions. This definition of affine connection is quite terse here—I'm just seeing text on a page and not really understanding what is going on here. Is it possible somebody could help me parse through/explain what is really being said here with regards to affine connection? Could somebody supply their intuitions behind/for affine connections? Thanks. |
2147802 | Compute the Jacobi fields for the geodesic variations
This is one of my exercise on my notes. I am still stuck and hope someone can help. We have [imath]p \in M[/imath] and [imath]V,W\in ˇT_p M[/imath]. Compute the Jacobi fields of the following geodesic variations at [imath]s=0[/imath]. [imath]γ_s(t) = exp_{η(s)}(tV_{η(s)})[/imath], where [imath]η(s) = exp_p (sW)[/imath] and [imath]V_{η(s)} ∈ T_{η(s)}M[/imath] is the parallel transported version of V along η. | 786204 | How to differentiate exponential map with parameter dependent basepoint
Let [imath](M,g)[/imath] be a Riemannian manifold, [imath]\gamma:I\rightarrow M[/imath] a geodesic. Let [imath]c:(-\varepsilon,\varepsilon)[/imath] is defined to be another geodesic with [imath]c(0)= \gamma(0),\ \ \ c^\prime(0)=X.[/imath] Let [imath]V(s)[/imath],[imath]W(s)[/imath] be parallel vector fields along [imath]c[/imath]. The derivative that I want to compute is [imath]\frac{\partial}{\partial s}\exp_{c(s)} \big(t\cdot(V(s)+sW(s))\big)[/imath] at [imath]s=0[/imath]. I looked at examples, but there is no such thing as the [imath]c(t)[/imath] there so I can not figure it out. Thank you very much, I hope it is not a too simple calculation question for this forum. |
2148928 | Inverse of [imath]2x + \cos{x}[/imath]
Let [imath]f(x) = 2x + \cos{x}[/imath]. Calculate [imath]f^{-1}(1)[/imath]; Supposing that [imath]f^{-1}[/imath] is differentiable, show that [imath](f^{-1})^{\prime}(1) = \dfrac{1}{2}[/imath]. Thanks for the help :] | 1049784 | Find the derivative of the inverse of this real function [imath]f(x) = 2x + \cos(x)[/imath]
I don't know how to attack this problem. The last I've tried is using a differential equation, but I don't know how to solve it. Let [imath]y[/imath] be [imath]f^{-1}(x)[/imath]. Knowing that [imath]x=f(y)= 2y + \cos(y)[/imath] and derivating I obtained the following non-linear first order differential equation: [imath]y' \cdot (2-\sin(y))=1[/imath] I would thank you if you can help me. Edit: I haven't said, but it is trivial to check that the function is injective, so it has an inverse, because [imath]\forall x \in \mathbb{R}[/imath] [imath]f'(x) \neq0[/imath] |
2132148 | If [imath]cos(A-B)+cos(B-C)+cos(C-A)=\frac {-3}{2}[/imath], prove that [imath]cosA+cosB+cosC=sinA+sinB+sinC=0[/imath]
If [imath]cos(A-B)+cos(B-C)+cos(C-A)=\frac {-3}{2}[/imath] prove that [imath]cosA+cosB+cosC=sinA+sinB+sinC=0[/imath] My Attempt: [imath]cos(A-B)+cos(B-C)+cos(C-A)=\frac {-3}{2}[/imath] [imath]cosA \cdot cosB+sinA \cdot sinB+cosB \cdot cosC+sinB \cdot sinC+cosC \cdot cosA+sinC \cdot sinA=\frac {-3}{2}[/imath]. How should I go further? Please help me with a simple method. | 1382047 | Trigonometry problem
Okay..this one simple problem but I am really stuck and have no idea how to start.. [imath]\cos(a-b)+\cos(b-c)+\cos(c-a)=-\frac32[/imath] we need to prove [imath]\cos(a)+\cos(b)+\cos(c)=\sin(a)+\sin(b)+\sin(c)=0 [/imath] |
2149754 | Show that no set of nine consecutive integers can be partitioned into two sets with equal products
This problem is meant to be proved using contradiction and I've admittedly been stuck on it for some time. I'd really appreciate a point in the right direction, because it honestly feels like I've been getting nowhere. This is all I've done: Assume [imath]x[/imath] denotes the set of nine consecutive integers, and [imath]x_1[/imath] and [imath]x_2[/imath] denote the two partitions. We would have: [imath]\prod_{a\ \in \ x_1} a= \prod_{a\ \in \ x_2} a[/imath] [imath]\implies[/imath] [imath]\prod_{a\ \in \ x_1} a[/imath] and [imath] \prod_{a\ \in \ x_2} a[/imath] are both even numbers, since both sets [imath]x_1[/imath] and [imath]x_2[/imath] would require to contain strictly odd numbers to have a product that is odd. [imath]\implies[/imath] [imath]x_1[/imath] has at least two elements, one odd and one even. [imath]\implies[/imath] [imath]x_2[/imath] has at most seven elements. And that's where I hit a dead end. I thought perhaps I could represent the set [imath]x[/imath] as [imath]x := \{ (a_0, \ (a_0 +1), \ (a_0 +2), \ \dots , \ (a_0 +8) \}[/imath] where [imath]a_0[/imath] is any integer, then multiply things out and show in each case one side is less than the other, but then I noticed that'd be an extremely tedious route and that some of the more ugly expansions deserve some sort of proof by induction themselves to show whether or not they actually hold. Please don't supply a proof that relies too heavily on a Number Theory-approach, as I don't have a background in that area therefore shouldn't rely on Theorems there to prove this. | 1479417 | Prove that a set of nine consecutive integers cannot be partitioned into two subsets with same product
Problem: Prove that a set of nine consecutive integers cannot be partitioned into two subsets with same product My attempt: This problem would be solved if it could be proven that 9 consecutive numbers cannot exist such that all the numbers are of the form [imath]2^a 3^b 5^c 7^d[/imath] where [imath]a,b,c,d[/imath] are integers. A suitable hint would be appreciated, I don not want a complete solution. Edit: If there existed a prime factor greater than [imath]7[/imath] of any number present in the set, then its multiple cannot exist b/c [imath]22-11>9[/imath] Edit 2: As mentioned by Erick, my reasoning is not valid for numbers [imath]1-9[/imath] and [imath]2-9[/imath], let this be an exception. |
2150295 | counter example of a free module whose rank is less than the number of linearly independent elements
There is a statement. Let [imath]R[/imath] be an integral domain and let [imath]M[/imath] be a free [imath]R[/imath]-module of rank [imath]n<\infty[/imath]. Then any [imath]n+1[/imath] elements of [imath]M[/imath] are [imath]R[/imath]-linearly dependent. (This is extracted from the book 'Abstract Algebra' written by David S. Dummit and Richard M. Foote.) Let's make the condition weaker as we assume [imath]R[/imath] is commutative ring with unity. Then, what is an example that the number of linearly independent element of [imath]M[/imath] exceeds the rank? | 1205340 | What would be an example of free module such that cardinality of linearly independent set is greater than the rank?
Let [imath]R[/imath] be a commutative ring and [imath]M[/imath] be a free [imath]R[/imath]-module. Since [imath]R[/imath] is commutative, [imath]R[/imath] has the IBN property, hence the rank of [imath]M[/imath] is uniquely well-defined. So set [imath]n:=\mathrm{rank}(M)[/imath]. Let [imath]A[/imath] be an [imath]R[/imath]-linearly independent subset of [imath]M[/imath]. What is an example of [imath]A[/imath] such that [imath]|A|>n[/imath]? If [imath]R[/imath] is a division ring, it must be [imath]|A|≦n[/imath], but if [imath]R[/imath] is commutative, I think it is possible that [imath]|A|>n[/imath]. |
2148654 | Possible values of [imath]|z_1+z_2+z_3| [/imath]?
Let [imath]z_1,z_2,z_3[/imath] be three complex number such that : [imath]| z_1| = | z_2| = | z_3| = 1[/imath] and [imath]z^3_1 + z^3_2 + z^3_3 +z_1z_2z_3=0[/imath] Then [imath]|z_1+z_2+z_3|[/imath] can take which of the following values ? (A) [imath]1[/imath] (B) [imath]2[/imath] (C) [imath]3[/imath] (D) [imath]4[/imath] Using the triangle inequality I get [imath]|z_1+z_2+z_3| \leq |z_1|+ |z_2| + |z_3| \leq 3[/imath]. Using the inequality the answers should be (A), (B) and (C). However, my book says that the answer is only (A) and (B). What is wrong in my method? | 1884048 | If [imath]\frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1.[/imath]Then [imath]|z_{1}+z_{2}+z_{3}|[/imath]
If [imath]z_{1},z_{2},z_{3}[/imath] are three complex number such that [imath]|z_{1}| = |z_{2}| = |z_{3}| = 1[/imath] and [imath]\displaystyle \frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1.[/imath]Then possible values of [imath]|z_{1}+z_{2}+z_{3}|[/imath] [imath]\bf{My\; Try::}[/imath] Given [imath]\displaystyle \frac{z^2_{1}}{z_{2}z_{3}}+\frac{z^2_{2}}{z_{3}z_{1}}+\frac{z^2_{3}}{z_{1}z_{2}} = -1\Rightarrow z^3_{1}+z_{2}^3+z_{3}^3=-z_{1}z_{2}z_{3}[/imath] Now Using [imath]a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)[/imath] So [imath]\left(z_{1}+z_{2}+z_{3}\right)\left[z^2_{1}+z^2_{2}+z^2_{3}-z_{1}z_{2}-z_{2}z_{3}-z_{3}z_{1}\right] = -4z_{1}z_{2}z_{3}[/imath] So [imath](z_{1}+z_{2}+z_{3})\cdot ((z_{1}+z_{2}+z_{3})^2-3(z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1})) = -4z_{1}z_{2}z_{3}.......(1)[/imath] Now Let [imath]k=z_{1}+z_{2}+z_{3}.[/imath] Taking Conjugate, we get [imath]\displaystyle k = \frac{1}{z_{1}}+\frac{1}{z_{2}}+\frac{1}{z_{3}} = \frac{z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}}{z_{1}z_{2}z_{3}}[/imath] So we get [imath]z_{1}z_{2}+z_{2}z_{3}+z_{3}z_{1}=kz_{1}z_{2}z_{3}[/imath] Put into [imath](1)\;,[/imath] We get [imath](z_{1}+z_{2}+z_{3})\cdot \left[(z_{1}+z_{2}+z_{3})^2-3kz_{1}z_{2}z_{3}\right] = -4z_{1}z_{2}z_{3}[/imath] Now How can I solve after that, Help Required, Thanks |
2149577 | Proving [imath]3^n>n^2[/imath] by induction
Prove that [imath]3^n>n^2[/imath] for any positive integer n>2 Let [imath]n=1[/imath] [imath]3^1>1^2[/imath] Let [imath]n=2[/imath] [imath]3^2>2^2[/imath] Assume P holds for n=k [imath]3^k>k^2[/imath] Let [imath]n=k+1[/imath] [imath] \begin{align} 3^{k+1}>(k+1)^2 \\3\times3^k>(k+1)^2 \end{align}[/imath] From here I can not find where to go to finish the proof. The main part of the question is the proof, however; I would like to also know if using [imath]n=k+1[/imath] is always the way to go? I have only done a few proofs by induction and so far to my understanding is that the whole point is to prove the function, series or statement for all positive integers. Is there a special way to go about these types of problems when only given an inequality or a single statement? In comparison to being give a sequence and told what that sequence as a function is, i.e [imath]1+2+3+...+n=\frac{n(n+1)}{2}[/imath] as an example as an easy sequence to prove with [imath]n=k+1[/imath] | 1236667 | How to Prove with Mathematical Induction [imath]3^n > n^2[/imath]
How do I prove that [imath]3^n > n^2[/imath] with mathematical induction? I thought I had the correct answer but my teacher says its wrong. I let [imath]n=1[/imath] for the initial case and it works. I then assumed [imath]n=k[/imath] works and went onto the [imath]n=k+1[/imath] case, but after that it goes a little awry. Thanks in advance! |
2149678 | Why is there a separate symbol for partial derivatives?
The concept of a partial derivative is very simple: for a multivariate function [imath]f[/imath], the partial derivative of [imath]f[/imath] with respect to a single variable [imath]x[/imath] is computed by treating the other variables as constants and differentiating [imath]f[/imath] with respect to [imath]x[/imath]. As a student of Calculus I, I do not fully understand the need for a [imath]\partial y / \partial x[/imath]. As far as I know, the special “partial” [imath]\partial[/imath] does not change the process of the computation. In addition, I have become confused as the calculus of my physics course increases in difficulty. Take a three-dimensional position vector [imath]\vec r = \vec x + \vec y + \vec z[/imath] and an electric field vector [imath]\vec E[/imath] that varies with [imath]\vec r[/imath]. If [imath]V[/imath] denotes electric potential, then [imath]\Delta V = -\int \vec E\cdot d\vec r[/imath]. On a review sheet that my teacher created, he expanded this, saying [imath]\begin{align} E_x &= -\partial V / \partial x \\ E_y &= -\partial V / \partial y \\ E_z &= -\partial V / \partial z \end{align}[/imath] Given the standard setup of a partial derivative, I see no issue with this. However, our standard formula chart reads that [imath]E_x = -\frac{dV}{dx}[/imath] and this genuinely confuses me since our calculations of field potential are almost always expanded to multiple dimensions. What is the need for a [imath]\partial f / \partial x[/imath] notation, and why are sources (at least in physics, the only application of partial derivatives I encounter during the course of the school day) inconsistent? | 320228 | The notation for partial derivatives
Today, in my lesson, I was introduced to partial derivatives. One of the things that confuses me is the notation. I hope that I am wrong and hope the community can contribute to my learning. In single-variable calculus, we know that, given a function [imath]y =f(x)[/imath], the derivative of [imath]y[/imath] is denoted as [imath]\frac {dy}{dx}[/imath]. I understand this as the relative change in [imath]y[/imath], [imath]\delta y[/imath] given a small change in [imath]x[/imath], [imath]\delta x[/imath]. However, in today's lesson on partial derivative, my professor constantly used this notation. Given a function [imath]z = f(x,y)[/imath], the first derivative with respected to [imath]x[/imath] is written as [imath] \frac{\partial z}{\partial x} [/imath] So, for example [imath] z = 5x+3y\\ \frac{\partial z}{\partial x} = 5 [/imath] Why can't I just write it as [imath] z = 5x+3y\\ \frac{d z}{d x} = 5 [/imath] Is it some convention or am I not understanding something in the notation? |
2148170 | limit [imath] \lim_{x\to\infty} \left( (x+2017)^{1+\frac{1}{x}} \: -\: x^{1+\frac{1}{x+2017}} \right) [/imath]
Find the following limit: [imath] \lim_{x\to\infty} \left( (x+2017)^{1+\frac{1}{x}} \: -\: x^{1+\frac{1}{x+2017}} \right) [/imath] I tried to exchange the infinity to zero by [imath]x:=\frac{1}{t}[/imath] and then use [imath]\lim_{t\to 0^+}t^t=1[/imath], but it doesn't lead to anything, I can't avoid having [imath]\infty-\infty[/imath]... The answer is 2017, as graph showed. | 1905240 | Limit of the difference between two exponential functions
Find the limit [imath]\lim_{x \to \infty}\left((x+3)^{1+1/x}-x^{1+1/(x+3)}\right).[/imath] I did nothing by now. |
2150412 | Minimize [imath]\int\limits _{0}^{1} f^2(x) dx[/imath] if [imath]\int\limits_{0}^{1} f(x) dx = 1[/imath], [imath]\int\limits_{0}^{1} xf(x)dx = 1[/imath]
Any hints? Using integration by parts didn't produce any result. Interestingly, I cannot find any function [imath]f(x)[/imath] which such a property. | 46590 | What's the minimum of [imath]\int_0^1 f(x)^2 \: dx[/imath], subject to [imath]\int_0^1 f(x) \: dx = 0, \int_0^1 x f(x) \: dx = 1[/imath]?
The question is as in the title: what's the minimum of [imath]\int_0^1 f(x)^2 \: dx[/imath], subject to [imath]\int_0^1 f(x) \: dx = 0, \int_0^1 x f(x) \: dx = 1[/imath]? (Assume suitable smoothness conditions.) A problem in the textbook for the course I am TEACHING (not taking) reduces to minimizing [imath]w_1^2 + w_2^2 + w_3^2[/imath] subject to [imath]w_1 + w_2 + w_3 = 0, w_1 + 2w_2 + 3w_3 = 1[/imath]. Of course there's nothing special about the number [imath]3[/imath] here, and so one can ask for the minimum of [imath]\sum_{i=1}^n w_i^2[/imath] subject to [imath]\sum_{i=1}^n w_i = 0, \sum_{i=1}^n iw_i = 1[/imath]. At least when [imath]n = 3, 4, 5[/imath], we get [imath]w_i = c_n(i-(n+1)/2)[/imath], for some constant [imath]c_n[/imath] which depends on [imath]n[/imath]. So for fixed [imath]n[/imath], [imath]w_i[/imath] is a linear function of [imath]i[/imath]. (This is a bit of an annoying computation, so I won't reproduce it here.) So it seems like there should be a continuous analogue of this. If we have [imath] \int_0^1 f(x) \: dx = 0, \int_0^1 x f(x) \: dx = 1 [/imath] and [imath]f(x)[/imath] is linear, then we get [imath]f(x) = 12(x-1/2)[/imath], and [imath]\int_0^1 f(x)^2 \: dx = 12[/imath]. Is this the function satisfying these integral conditions with smallest [imath]\int_0^1 f(x)^2 \: dx[/imath]? That is, is it the case that [imath] \int_0^1 f(x)^2 \: dx \ge 12 [/imath] for every [imath]f(x)[/imath] satisfying the two conditions above and whatever smoothness conditions are necessary? I've tagged this calculus-of-variations because that's what it looks like to me. But I don't know the calculus of variations, which is why I can't just solve the problem myself. |
2150695 | Show that [imath]L^p[/imath] norm is equivalent to [imath]L^q[/imath] norm iff [imath]p=q[/imath].
I want to show that [imath]\| \cdot \|_p[/imath] is equivalent to [imath]\| \cdot \|_q[/imath] iff [imath]p=q[/imath] on [imath]\mathbb{R}[/imath]. I think the best way to do this is via a contradiction. Supposing that [imath]\| \cdot \|_p \leq \alpha \| \cdot \|_q \leq \beta \| \cdot \|_p[/imath], we have [imath]\left( \int_{\mathbb{R}} \left| f \right|^p d\mu \right)^{1/p} \leq \alpha \left( \int_{\mathbb{R}} \left| f \right|^q d\mu \right)^{1/q} \leq \beta \left( \int_{\mathbb{R}} \left| f \right|^p d\mu \right)^{1/p}.[/imath] I'm then thinking of applying Holder's inequality, but haven't been successful . | 454058 | Are all the norms of [imath]L^p[/imath] space equivalent?
Are all the norms of [imath]L^p[/imath] space equivalent? That is, for any [imath]p,~q \in R^+[/imath], there exist two positive number [imath]C_1,~C_2[/imath] such that [imath] C_1\|u\|_{L^q} \le \|u\|_{L^p} \le C_2\|u\|_{L^q}. [/imath] |
2150438 | [imath]\frac{x}{1+x^4} + \frac{y}{1+y^4} + \frac{z}{1+z^4} \leq \frac{3}{2}[/imath]
I've came across this problem which seems to be a classic mean inequality problem, but I can't get it right. Let [imath]x[/imath], [imath]y[/imath] and [imath]z[/imath] be pozitive real numbers with [imath]xyz=1[/imath]. Prove that [imath]\frac{x}{1+x^4} + \frac{y}{1+y^4} + \frac{z}{1+z^4} \leq \frac{3}{2}[/imath] My first try was [imath]\sqrt{1 * x^2} \leq \sqrt{\frac{1+x^4}{2}}[/imath] [imath]2x^2 \leq 1+x^4[/imath] [imath]\frac{x}{1+x^4} \leq \frac{x}{2x^2} = \frac{1}{2x}[/imath] then, using geometric mean and arithmetic mean, we obtain the inequality inside out. A hint would be really apreciated. | 2081728 | Given that [imath]xyz=1[/imath], prove that [imath]\frac{x}{1+x^4}+\frac{y}{1+y^4}+\frac{z}{1+z^4}\le \frac{3}{2}[/imath]
Given that [imath]xyz=1[/imath], prove that [imath]\frac{x}{1+x^4}+\frac{y}{1+y^4}+\frac{z}{1+z^4}\le\frac{3}{2}.[/imath] I proved this with Muirhead's inequality, but: is there a better/more elegant way? |
2110631 | If for each [imath]x[/imath] there is [imath]n_x[/imath] such that [imath]f^{(n_x)}(x)=0,[/imath] then [imath]f[/imath] is a polynomial.
The following result due to Sunyer y Balaguer and Corominas is somewhat of a classic: Let [imath]f: \mathbb{R} \to \mathbb{R}[/imath] be of class [imath]C^{\infty}[/imath]. For every [imath]x \in \mathbb{R}[/imath], there exists [imath]n_x \in \mathbb{N^*}[/imath] such that [imath]f^{(n_x)}(x)=0[/imath]. Then [imath]f[/imath] is a polynomial. Here [imath]\mathbb{N^*}[/imath] denotes the set of non-negative integers. This was an exercise on my test, where you are given the steps of the proof and you must fill in each step. Proving each step individually is not too hard, but there were too many steps and so I lost sight of the big picture. -The first step is usually to consider the set [imath]S[/imath] of real [imath]x[/imath] such that [imath]f[/imath] coincides with a polynomial in some neighborhood of [imath]x[/imath]. (What's the intuition for considering this set?) -Then you prove that [imath]S[/imath] is open and the complement of [imath]S[/imath] has no isolated points, which is easy enough. -Then [imath]S[/imath] is an at-most-coutable union of disjoint open intervals, and we can show that on each of these intervals, [imath]f[/imath] coincides with a polynomial. -After that, we suppose that [imath]S[/imath] is not equal to [imath]\mathbb{R}[/imath] (or we'd be done), and we deduce a contradiction. It is this part of the proof, which uses Baire's category theorem where I got lost in the details. Can please someone provide the grand lines of the proof and the intuition? | 1372354 | Prove: [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] st for every [imath]x \in \mathbb{R}[/imath] there exists [imath]n[/imath] st [imath]f^{(n)}(x) = 0[/imath], f is a polynomial.
If [imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] is a smooth function such that for every [imath]x \in \mathbb{R}[/imath] there exists [imath]n[/imath] such that [imath]f^{(n)}(x) = 0[/imath], then f is a polynomial. I'm kind of lost on this one. I know that I have to use Baire's category theorem somewhere here, but I'm not sure exactly how. |
2151346 | Better explicit notation for [imath]\sqrt {(-2) ^2} = \sqrt {2^2}[/imath] seemingly implying [imath]-2=2[/imath] type questions
EDIT : This about explicit notation and clarity, usage of [imath]\sqrt{(-x)^2}= x[/imath] is for demonstrative purposes not for the problem itself. [imath]\sqrt {(-2) ^2} = \sqrt {2^2}[/imath] causing confusion [imath]-2=2[/imath] which is nonsense. What seems to be needed is somehow distinguishing between [imath]-2=2[/imath] and explicitly stating [imath]2,-2[/imath] are in the set [imath]S=[/imath]same equivalence class of numbers [imath]k[/imath] s.t. [imath]k^2[/imath] having the same value. Another observation is the order of operation is important, [imath]\sqrt {(-2)^2} = \sqrt {2^2}[/imath] can be evaluated as either [imath]\sqrt {4} = \sqrt {4} =2[/imath] or [imath] {-2} = {2}[/imath] , how ever implicitly it known that the square function performs first and only after that the square root function can operate, how could this implicit knowledge be worked into explicit notation? Are there any notations that can be used to make order of operation explicitly visible or instead of ending up with [imath]2=-2[/imath], somehow end up with [imath]2 \equiv-2[/imath] with respect to their squares having same values? | 487509 | General misconception about [imath]\sqrt x[/imath]
I noticed a large portion of general public (who knows what square root is) has a different concept regarding the surd of a positive number, [imath]\sqrt\cdot[/imath], or the principal square root function. It seems to me a lot of people would say, for example, [imath]\sqrt 4 = \pm 2[/imath], instead of [imath]\sqrt 4 = 2[/imath]. People even would correct a statement of the latter form to one with a [imath]\pm[/imath] sign. Some also claim that, since [imath]2^2 = 4[/imath] and [imath](-2)^2 = 4[/imath], [imath]\sqrt 4 = \pm 2[/imath]. Some people continue to quote other "evidences" like the [imath]y=x^2[/imath] graph. While most people understand there are two square roots for a positive number, some seem to have confused this with the surd notation. From an educational viewpoint, what might be lacking when teaching students about surd forms? Is a lack of understanding to functions a reason for this misconception? Now I have noticed another recent question that hinted that poster was confused. Following @AndréNicolas's comment below, might these confusion really come from two different communities using the same symbol? |
2150773 | Is the product [imath](p^2+1)/(p^2-1)[/imath] for all [imath]p[/imath] rational? For all [imath]n[/imath]?
Is this product: \begin{align*} \prod_{\text{[imath]p[/imath] prime}}\frac{p^2+1}{p^2-1} \end{align*} rational? What is the value of it? What about for the natural numbers [imath]p>1[/imath]? | 1709940 | What is the infinite product of (primes^2+1)/(primes^2-1)?
I have shown that the infinite product [imath]\prod_{p \in \mathcal{P}}\frac{p^2+1}{p^2-1}[/imath] is equal to [imath]\frac{5}{2}[/imath] (pretty remarkable!). I have checked this numerically with Wolfram Alpha for up to [imath]500000[/imath] primes and it seems true. I was wondering if this result is recorded anywhere? Also if true, does this mean that there aren't infinitely many primes of form [imath]p^2+1[/imath]? |
2134127 | Spheres and almost complex structures
It is known that [imath]S^6[/imath] admits an almost complex structure which is not integrable, recently Michael Atiyah posted a proof of non-existent of a complex structure on this sphere. I am curious now about the other spheres i.e what about [imath]S^{4}[/imath] and [imath]S^{2n}[/imath] for [imath]n\geq4[/imath]? Is there any good introduction reference where one at the undergraduate level can read about that? | 1285519 | Does every even-dimensional sphere admit an almost complex structure?
We know that there is an almost complex structure on [imath]S^6[/imath] which is not integrable. Is it always possible to find almost complex structures on [imath]S^{2n}[/imath]? In particular does [imath]S^4[/imath] admit one? |
2151387 | Can anyone tell me how they expanded [imath]x^n-a^n[/imath]?
I am not getting how they factorised the [imath]x^n-a^n[/imath] term? | 1812545 | How is [imath]\lim_{x \to a}\left(\frac{x^n - a^n}{x - a}\right) = n\times a^{n-1}[/imath]?
In my book this is termed as a theorem and the proof given is as follows :- [imath]\begin{align} \lim_{x \to a}\left(\frac{x^n - a^n}{x - a}\right) &=\lim_{x \to a}\left(\frac{(x - a)*(x^{x-1} + x^{n-2}*a + x^{n-3}*a^2 + x^{n-4}*a^3 + \cdots + x^1*a^{n-2} + a^{n-1})}{x - a}\right) \\ &=(a^{n-1} + a*a^{n-2} + \cdots + a^{n-1}) \\ &=(a^{n-1} + a^{n-1} + \cdots + a^{n-1}) \\ &=(n*a^{n-1}). \end{align}[/imath] Everything made sense to me except [imath]x^n - a^n = (x - a)*(x^{x-1} + x^{n-2}*a + x^{n-3}*a^2 + x^{n-4}*a^3 + \cdots + x^1*a^{n-2} + a^{n-1})[/imath] Somebody please enlighten me on this topic. |
2152024 | Proving that [imath]T^2-5T+6I = 0[/imath].
The question is: Suppose that T is a self-adjoint operator on a finie-dimensional inner product space and that 2 and 3 are the only eigenvalues of T. Prove that [imath]T^2-5T+6I = 0[/imath]. To prove this question, can I take the inner product of [imath](T^2-5T+6I)v[/imath] and [imath]Tv[/imath], where [imath]Tv = 2v[/imath] and [imath]v \neq 0[/imath], and then show that the inner product of those two vectors is equal to [imath]0[/imath]? Since [imath]v \neq 0[/imath], would that imply that [imath]T^2-5T+6I = 0[/imath]? Or am I coming at this question completely wrong? Thanks. | 2046008 | Let [imath]4[/imath] and [imath]5[/imath] be the only eigenvalues of [imath]T[/imath]. Show [imath]T^2-9T + 20I = 0[/imath] , T is self adjoint.
Let [imath]T[/imath] be a self adjoint operator on a finite dimensional inner product space. And let [imath]4[/imath] and [imath]5[/imath] be the only eigenvalues of [imath]T[/imath]. Show [imath]T^2-9T + 20I = 0[/imath] . proof:Let [imath]T[/imath] be a self adjoint. And let [imath]4 , 5[/imath] be the only eigenvalues of [imath]T[/imath]. Then [imath]Tv = 4v, Tv = 5v[/imath] for some [imath]v \neq 0[/imath]. Then [imath]Tv = 4v, Tv = 5v[/imath] if and only if [imath](T-4I)v = 0, (T-5I)v=0[/imath]. Then [imath] (T-4I) (T-5I)v=0[/imath] implies [imath](T^2 -5T -4T + 20I^2)v = 0[/imath], implies [imath](T^2 -9T + 20I)v = 0[/imath], so we have [imath]T^2 -9T + 20I = 0 [/imath]. Can someone please check this? if it's not correct, could I please have some help on it? Thank you |
2152310 | Holomorphic functions inside an Annulus
Given an Annulus with [imath]A(0,r,R)[/imath] show by considering Cauchy's Theorem for primitives that there is no holomorphic function with [imath]f'(z)=1/z[/imath] I am struggling to picture this since but it seems like there are issues because [imath]f(z)=log(z)[/imath] isn't well defined in the same range as [imath]1/z[/imath]. Is it worth considering [imath]f(z)[/imath] as anything other than just a function? e.g setting [imath]f(z)=log(z)[/imath] and proving it this way? | 2151918 | Primitive of holomorphic Function [imath]\frac{1}{z}[/imath] on an Annulus.
Given an Annulus with [imath]A(0,r,R)[/imath] show by considering Cauchy's Theorem for primitives that there is no holomorphic function with [imath]f'(z)=\dfrac{1}{z}[/imath]. I am struggling to picture this since but it seems like there are issues because [imath]f(z)=\log z[/imath] isn't well defined in the same range as [imath]\dfrac{1}{z}[/imath]. Am I looking to show that [imath]\int_Ag(z)dz = 0[/imath] where [imath]g(z) = \dfrac{1}{z}[/imath]? |
2152694 | Trace(AB) = Trace(BA) for rectangular matrices
Given A is [imath]m\times n[/imath] and B is a [imath]n \times m[/imath] matrix, can we say trace(AB) = trace(BA)? It worked for a few examples with [imath]m = 2[/imath] and [imath]n = 3[/imath]. | 252272 | Is trace invariant under cyclic permutation with rectangular matrices?
I'm working on trace of matrices. Trace is defined for square matrix and there are some useful rule to deal with calculus (i.e. [imath]tr(AB) = tr(BA)[/imath], with [imath]A[/imath] and [imath]B[/imath] square, and more in general trace is invariant under cyclic permutation). I was wondering if the formula [imath]tr(AB) = tr(BA)[/imath] holds even if [imath]A[/imath] and [imath]B[/imath] are rectangular, namely [imath]A[/imath] is n-by-m and [imath]B[/imath] is m-by-n. I figured out that if one completes the involved matrices to be square by adding "zeros" where it is needed, then the formula still works...but I want to be sure of this thing!!! :D |
2149613 | How to find the maximum possible value of [imath]\big| (z_1-z_2)^2+(1-z_1)(1-z_2) \big|[/imath] given that [imath]|z_1|=|z_2|=1[/imath]?
How to find the maximum possible value of [imath]\big| (z_1-z_2)^2+(1-z_1)(1-z_2) \big|[/imath] given that [imath]|z_1|=|z_2|=1[/imath] ? I cannot think of any easy method to do it. Any suggestion? P.S: Avoid using multivariable calculus. | 2149243 | How to find the maximum and minimum value of [imath]\left|(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2\right|[/imath] (where [imath]|z_1|=|z_2|=|z_3|=1[/imath])?
How to find the maximum and minimum value of [imath]\left|(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2\right|[/imath] (where [imath]|z_1|=|z_2|=|z_3|=1[/imath] are complex numbers.) ? My try: [imath]\begin{align}\left|(z_1-z_2)^2 + (z_2-z_3)^2 + (z_3-z_1)^2\right| &\leq |z_1-z_2|^2 + |z_2-z_3|^2 + |z_3-z_1|^2 \\ &\leq (|z_1|+|z_2|)^2 + (|z_2|+|z_3|)^2 + (|z_3|+|z_1|)^2 \\ &\leq 2^2+2^2+2^2 \leq 12\end{align}[/imath] However the answer given is [imath]8[/imath]. Where am I going wrong and how to do it correctly? |
2152830 | Inequality of 6 Positive Integers
Let [imath]a,b,c,d,e,f[/imath] be positive integers such that [imath]\frac{a}{b} < \frac{c}{d} <\frac{e}{f}.[/imath] Suppose [imath]af - be = -1[/imath]. Show that [imath]d\ge b + f[/imath]. Tried everything. I introduced constants [imath]K_{1},K_{2}[/imath] etc between [imath]\frac{a}{b} , \frac{c}{d},\frac{e}{f}[/imath] so that I have an equality, which can somehow be related to the equality given to us. I tried to use every special identity/inequality which I know, but to no avail. The problem always seems to be eliminating [imath]c[/imath] or [imath]d[/imath] from the inequality. I have been at it for 2 hours. This problem featured in a Regional Olympiad, using which after many stages the team for the IMO is selected. | 897178 | How prove that [imath]q \geq b+d[/imath] for [imath]ad-bc = 1[/imath] and [imath]\frac{a}{b} > \frac{p}{q} > \frac{c}{d}[/imath]?
Let [imath]a,b,c,d,p[/imath], and [imath]q[/imath] be natural numbers such that [imath]ad-bc = 1[/imath] and [imath]\frac{a}{b} > \frac{p}{q} > \frac{c}{d}[/imath]. How prove that [imath]q \geq b+d[/imath]? |
2152398 | [imath]\lim_{|P|\to0} \sum_{i=1}^n f(\xi_i)g(\eta_i)(t_{i}-t_{i-1})=\int_a^b f(x)g(x)dx[/imath]
Let [imath]f,g:[a,b]\to\Bbb{R}[/imath] integrable functions. Let's dot each partition [imath]P[/imath] in two different ways, choosing in all the interval [imath][t_{i-1},t_i][/imath] a point [imath]\xi_i[/imath] and a dot [imath]\eta_i[/imath]. Show that: [imath]\lim_{|P|\to0} \sum_{i=1}^n f(\xi_i)g(\eta_i)(t_{i}-t_{i-1})=\int_a^b f(x)g(x)dx[/imath] I need something as accurate as possible, because i really need to understand this question and i'm stuck at the very beginning. | 1944480 | Riemann and Darboux Integral of a product of two functions
I'm studying the Darboux definition of integrability, which I completely explained here. There's an exercise that asks me to prove that the Darboux Integrability is equivalent to Riemann Integrability, but this Riemann integral is defined as the following: It first defines a 'pointed' partition (I don't know how to say it in ensligh) by the following: a 'pointed' partition [imath][a,b][/imath] is a pair [imath]P^*=(E,ξ)[/imath], where [imath]P=\{t_0, t_1, \cdots, t_n\}[/imath] is a partition of [imath][a,b][/imath] and [imath]ξ = (ξ_1, \cdots, ξ_n)[/imath] is a list of [imath]n[/imath] chosen numbers such that [imath]t_{i-1}\le ξ_i\le t_i[/imath] for each [imath]i=1,\cdots ,n[/imath]. Now, the Riemann Integral is defined as: [imath]\sum(f,P^*) = \sum_{i=1}^n f(ξ_i)(t_i-t_{i-1})[/imath] (I didn't understand the notation for the left hand side of the equation, by the way) Finally, I'm asked to prove the following: given [imath]f,g:[a,b]\to \mathbb{R}[/imath] integrable functions, for the entire partition [imath]P=\{t_0, \cdots, t_n\}[/imath] of [imath][a,b][/imath] let [imath]P^* = (P,ξ)[/imath] and [imath]P^{\#} = (P, η)[/imath] be pointed partitions of [imath]P[/imath], then: [imath]\lim_{|P|\to 0}\sum f(ξ_i)g(η_i)(t_i-t_{i-1}) = \int_a^b f(x)g(x) \ dx[/imath] I guess here I need to prove that the Riemann Integral of the product of two functions if the darboux integral of [imath]f(x)g(x)[/imath], but it seems too obvious, I just need to verify that [imath]f,g[/imath] are integrable, then their product is too, isn't it? I'm pretty sure this should be a hard question. Is there another interpretation that I'm missing? |
2153395 | Real solutions of [imath]x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}[/imath]
Find all real solutions of the equation: [imath]x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}[/imath] My approach: One idea was square [imath]3[/imath] times untill get a equation and try factoring it. Another was try to get a system, calling, for example, [imath]y=\sqrt{2+x}[/imath], but I didn't have luck with any of them. Any idea? | 334720 | finding the real values of [imath]x[/imath] such that : [imath]x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}[/imath]
How to find the real values of [imath]x[/imath] such that : [imath]x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}[/imath] |
2152758 | Given [imath]f[/imath] is an even function of interval [imath](-a, a)[/imath], .... [imath]f'[/imath] is an odd function on [imath](-a,a)[/imath]
Given [imath]f[/imath] is an even function of interval [imath](-a, a)[/imath], [imath]a>0[/imath] and [imath]0<c < a[/imath]. Prove that if [imath]L'f(c)[/imath] exists, then [imath]Rf'(-c)[/imath] exists and [imath]Lf'(c)=-Rf'(-c)[/imath]. Deduce that if [imath]f[/imath] is differentiable on [imath](-a,a)[/imath], then [imath]f'[/imath] is an odd function on [imath](-a,a)[/imath]. [imath]f[/imath] is even then [imath]f(-x)=f(x)[/imath], then [imath]f'(-x)=-f'(x)[/imath] i.e [imath]f'[/imath] is odd function. But what exactly the question want? Please help with apropriate answer. | 194062 | Derivative of an even function is odd and vice versa
This is the question: "Show that the derivative of an even function is odd and that the derivative of an odd function is even. (Write the equation that says f is even, and differentiate both sides, using the chain rule.)" I already read numerous solutions online. This was the official solution but I didn't quite understand it (particularly, I'm not convinced why exactly [imath]dz/dx=-1[/imath]; even though [imath]z=-x[/imath]). Thanks in advance =] |
2153742 | Why is Re [imath]z[/imath] nowhere differentiable?
I have to use the following definition to prove that the function Re [imath]z[/imath] is nowhere differentiable: Let [imath]f[/imath] be a complex-valued function defined in a neighborhood of [imath]z_0[/imath]. Then the derivative of [imath]f[/imath] at [imath]z_0[/imath] is given by [imath]\frac{df}{dz}(z_0)=f'(z_0)=\lim_{\Delta z \to 0} \frac{f(z_0+ \Delta z)-f(z_0)}{\Delta z},[/imath] provided this limit exists. (Such an [imath]f[/imath] is said to be differentiable at [imath]z_0[/imath]) First I need to understand what Re [imath]z[/imath] means? This is just the real part of [imath]z[/imath], where [imath]z[/imath] is [imath]x+yi[/imath], so the real part is just [imath]x[/imath], but I don't understand how the the real part alone is considered a function? | 222901 | Where is [imath]f(z)=\Re (z)[/imath] differentiable?
Let [imath]f:\mathbb{C}\to\mathbb{C}[/imath] be given by [imath]z\mapsto\Re\left(z\right)[/imath] . Where is [imath]f[/imath] differentiable? [imath]f[/imath] is definitely differentiable on [imath]\mathbb{R}[/imath] because on these values [imath]f\left(z\right)=z[/imath] . As for [imath]\mathbb{C}\backslash\mathbb{R}[/imath] , I am not sure. I am guessing it's not, but I can't prove it. The way I'm trying to show it is by constructing a sequence such that [imath]z_{n}\to z[/imath] but [imath]\frac{\Re\left(z_{n}\right)-\Re\left(z\right)}{z_{n}-z}[/imath] does not converge. Any help in showing this would be appreciated. |
2153782 | If [imath]f[/imath] is twice derivable at [imath]a[/imath], compute the following limit.
If [imath]f[/imath] is twice derivable at [imath]a[/imath], compute [imath]\lim_{h\to 0} \frac{f(a+h)+f(a-h)-2f(a)}{h^2}[/imath] I tried to rearrange the terms in this way [imath]\lim_{h\to0} \frac{f(a+h)-f(a)}{h^2} + \lim_{h\to 0}\frac{f(a-h)-f(a)}{h^2}[/imath] but it didn't get me anywhere. I also realised I can't use l'Hôpital rule or Taylor's theorems, because the hypothesis is weaker. | 1809060 | Proof of the second symmetric derivative
Prove that if [imath]f''(a)[/imath] exists, then [imath]$$f''(a)=\lim_{h\to0}\frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}.$$[/imath] I really have no idea on this one. Am I supposed to apply the mean value theorem? |
2154246 | Validating the inequality.
[imath]x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x)<x-\frac{x^2}{2}+\frac{x^3}{3}[/imath], [imath]x>0[/imath] Here I can only see that the right side of second inequality i.e. [imath]x-\frac{x^2}{2}+\frac{x^3}{3}[/imath] comes in the expansion of [imath]\log(1+x)[/imath]. We have done the Lagrange's mean value theorem and intermediate value theorem, do these have anything to do with the inequality. Kindly provide some hint. | 1546172 | Taylor's Remainder [imath]x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x) [/imath]
Prove that [imath]\displaystyle x-\frac{x^2}{2}+\frac{x^3}{3(1+x)}<\log(1+x) <x-\frac{x^2}{2}+\frac{x^3}{3}[/imath] My attempt: I can prove this by taking one side at a time and assuming [imath]\displaystyle f(x) = \log (1+x)-x+\frac{x^2}{2}-\frac{x^3}{3}[/imath] and then proving that it is a decreasing function and then the same for the other side. But I am looking for a better solution using Lagrange's Mean Value Theorem or using Taylor's remainder. |
2153399 | Combinations of bijective derangement functions
Let [imath]S = \{1,2,3,4,5\}[/imath] How many bijective functions [imath]f:S \to S[/imath] do we have such that [imath]f(x) \neq x[/imath] for all [imath]x\in S[/imath]? I was trying to solve by taking the total number of such functions and subtract all the cases where [imath]f(x) = x[/imath]. [imath]5![/imath] is all cases [imath]4![/imath] is where one [imath]f(x)=x[/imath] and rest are not [imath]3![/imath] is where two [imath]f(x)=x[/imath] and rest are not [imath]2![/imath] is where three [imath]f(x)=x[/imath] and rest are not [imath]1![/imath] is where all five [imath]f(x)=x[/imath] so [imath]5!-(4!-3!-2!-1)[/imath]. But then I realized that, for example, [imath]4![/imath] contains also the case where all [imath]f(x)=x[/imath] which mean I am subtracting too much. | 749262 | Rearranging people so that no one is in the same spot
I am not sure how to approach this problem: [imath]n[/imath] people are seated in numbered chairs [imath]1[/imath] to [imath]n[/imath]. Let [imath]N[/imath] be the number of ways the people can be reseated so that no one is in the same chair as before. Show that [imath]N=n! \sum_{k=0}^n \frac {(-1)^k}{k!}[/imath]. I feel like the way to do this is to come up with a way to count all the seating arrangements, yet I am not sure how to take into account the seat where the person already sat into account when counting them. Because it would seem that there is [imath]n-1[/imath] possible places to sit at, yet placing the first person does not eliminate one possibility for everyone... Any help is appreciated. Thank you in advance. |
2154707 | Does existence of continous bijections imply homeomorphism?
Let [imath]X[/imath], [imath]Y[/imath] be topological spaces and assume there exists a continous bijection [imath]X\to Y[/imath] and there exists a continous bijection [imath]Y\to X[/imath]. Are [imath]X[/imath] and [imath]Y[/imath] homeomorphic? | 1125293 | Two non-homeomorphic spaces with continuous bijective functions in both directions
I was asked the following question: if two topological spaces [imath]X, Y[/imath] are such that there exist a function [imath]f:X\rightarrow Y[/imath] continuos and bijective and a function [imath]g:Y\rightarrow X[/imath] continuous and bijective, then are the two spaces homeomorphic? I would say that the answer is No, but I didn't find any counterexample so far. |
2154605 | value of trigonometric product
Finding [imath]\displaystyle \tan \left(\frac{\pi}{14}\right)\cdot \tan \left(\frac{3\pi}{14}\right)\cdot \tan \left(\frac{5\pi}{14}\right) [/imath] assume [imath]\displaystyle \frac{\pi}{14} = \theta[/imath] then [imath]\tan (14\theta) = 0[/imath] wan,t be able to go further, some help me | 1055379 | Evaluating a product of tangents
(1) Evaluation of [imath]\displaystyle \tan \left(\frac{\pi}{7}\right)\cdot \tan \left(\frac{2\pi}{7}\right)\cdot \tan \left(\frac{3\pi}{7}\right) = [/imath] (2) Evaluation of [imath]\displaystyle \tan\left(\frac{\pi}{11}\right)\cdot \tan\left(\frac{2\pi}{11}\right)\cdot \tan\left(\frac{3\pi}{11}\right)\cdot \tan\left(\frac{4\pi}{11}\right)\cdot \tan\left(\frac{5\pi}{11}\right) = [/imath] [imath]\bf{My\; Try::(1):}[/imath] Let [imath]\displaystyle \frac{\pi}{7} = \theta\Rightarrow \pi = 7\theta\Rightarrow (\pi-3\theta) = 4\theta\Rightarrow \tan(\pi-3\theta) = \tan(4\theta)[/imath] So [imath]\displaystyle \tan (3\theta) = -\tan (4\theta)\Rightarrow \frac{3\tan \theta - \tan^3\theta}{1-3\tan^2 \theta} = -\left(\frac{2\tan 2\theta}{1-\tan^2 2\theta}\right) = \frac{4\tan^3 \theta-4\tan \theta}{1+\tan^4 \theta-6\tan^2 \theta}[/imath] So [imath]\displaystyle \left(\frac{3-\tan^2 \theta}{1-3\tan^2 \theta}\right) = \left(\frac{4\tan^2 \theta-4}{1+\tan^4 \theta-6\tan^2 \theta}\right)\Rightarrow \tan^6\theta-21\tan^4 \theta+21\tan^2 \theta-7=0[/imath] Now Let [imath]\displaystyle \tan \theta = y\;,[/imath] then eqn. convert into [imath]\displaystyle y^6-21y^4+21y^2-7=0[/imath] and equation has a roots [imath]\displaystyle y = \tan \left(\frac{\pi}{7}\right)\;,\tan \left(\frac{2\pi}{7}\right)\;,\tan \left(\frac{3\pi}{7}\right)\;,\tan \left(\frac{4\pi}{7}\right)\;,\tan \left(\frac{5\pi}{7}\right)\;,\tan \left(\frac{6\pi}{7}\right)\;[/imath]. So Product of Roots is [imath]\displaystyle \tan \left(\frac{\pi}{7}\right)\cdot \tan \left(\frac{2\pi}{7}\right)\cdot \tan \left(\frac{3\pi}{7}\right)\cdot \tan \left(\frac{4\pi}{7}\right)\cdot \tan \left(\frac{5\pi}{7}\right)\cdot \tan \left(\frac{6\pi}{7}\right) = -7[/imath] Now [imath]\displaystyle \tan\left(\frac{4\pi}{7}\right) = \tan\left(\pi-\frac{3\pi}{7}\right)=-\tan\left(\frac{3\pi}{7}\right)[/imath] Similarly [imath]\displaystyle \tan\left(\frac{5\pi}{7}\right) = -\tan\left(\frac{2\pi}{7}\right)[/imath] and [imath]\displaystyle \tan\left(\frac{6\pi}{7}\right) = -\tan\left(\frac{\pi}{7}\right)[/imath] So [imath]\displaystyle \tan\left(\frac{\pi}{7}\right)\cdot \tan\left(\frac{2\pi}{7}\right)\cdot \tan\left(\frac{3\pi}{7}\right)=\sqrt{7}[/imath] I did not understand how can i calculate [imath](2)[/imath] one using the same method:, bcz it is very lengthy can we solve the [imath](1)[/imath] and [imath](2)[/imath] question any other method. If yes then plz explain me, Thanks :: Thanks |
2153300 | If [imath]f \circ g[/imath] is injective and [imath]g[/imath] is surjective - is [imath]f[/imath] injective?
Let [imath]f: X \rightarrow Y[/imath] and [imath]g: Y \rightarrow X[/imath] If [imath]f \circ g[/imath] is injective and [imath]g[/imath] is surjective - is [imath]f[/imath] injective? I am trying to learn the way of prooving things. So I'll give it a try and please correct at every wrong step. I assume that the given statement ist true. So first I give the condition that [imath]g[/imath] is surjective: [imath]\forall y \in Y : \exists x \in X:g(y)=x[/imath] So two different [imath]y[/imath]-values can point to the same [imath]x[/imath]-value: Let [imath]y_1, y_2 \in Y, y_1 \neq y_2 : \exists x \in X : g (y_1)=x [/imath] and [imath]g (y_2)=x[/imath] Next I want to connect the fact that [imath]f \circ g[/imath] is injective with the surjective condition of [imath]g[/imath]: Let [imath]z_1, z_2 \in X \quad[/imath] Because [imath]z_1, z_2 \in X[/imath] it is true that [imath]z_1, z_2 \in g(y)[/imath] thus: [imath]f \circ g(y_1) = f \circ g (y_2)[/imath] [imath]f(z_1) = f(z_2)[/imath] [imath]z_1 = z_2[/imath] So [imath]f[/imath] is injective. | 845284 | If [imath]g \circ f[/imath] is injective, so is [imath]g[/imath]
If [imath]g \circ f[/imath] is injective, so is [imath]g[/imath] I don't think this is true. I think that [imath]f[/imath] has to be surjective. So I am going to try to prove that: If [imath]g \circ f[/imath] is injective, and [imath]f[/imath] is surjective, then [imath]g[/imath] is injective. First off, [imath]g \circ f[/imath] means that [imath](g \circ f)(a) = (g \circ f)(b)[/imath] then [imath]a=b[/imath]. And [imath]f[/imath] being surjective means that [imath]\forall b \in B, \exists a\in A[/imath] such that [imath]f(a)=b[/imath] Proof Suppose that [imath]g[/imath] was not injective. [imath]g(f(a))= g(f'(a))[/imath] and [imath]f(a)\ne f'(a)[/imath] But [imath]g(f(a)) = g(f(b)) \to a=b[/imath] [imath]g(f(a))= g(f(b)) \to f(a)=f(b) \to a=b[/imath] Hence contradiction, since [imath]g[/imath] is not injective. So [imath]g[/imath] is injective. Is this an acceptable proof, I think my logic is iffy around the second last paragraph. |
2154904 | For matrices [imath]A, B, C,\,[/imath] If [imath]AC=BC[/imath], can we say that [imath]A=B[/imath]?
Suppose we have [imath]n\times n[/imath] matrices [imath]A, B[/imath], and there is no information about singularity of [imath]A, B[/imath]. Also [imath]C[/imath] is a [imath]n\times n[/imath] projection matrix, i.e. singular matrix and it is not the zero metrix. If [imath]AC=BC[/imath], can we say that [imath]A=B[/imath]? All comments would be appreciated. | 481504 | Cancelling matrices
I'm solving a exercise, and the following question has arisen: Let [imath]A,B,C[/imath] be non-zero matrices, if [imath]AB=CB[/imath] then is [imath]A=C?[/imath] I already have a partial solution when [imath]A,B,C[/imath] are square matrices and [imath]B[/imath] is invertible: in this case [imath]A=CBB^{-1}=CI_n=C[/imath]. This result is valid in general case? (I mean, when the matrices are arbitrary) Thanks! |
2155275 | Why is the rank of this matrix [imath]2[/imath] (solved, thanks)?
[imath]A= \begin{pmatrix} 1 &2 &3 \\ 0 &6 &4 \\ 0 &3 &2 \end{pmatrix}[/imath] "Rank is the highest possible number of linearly independent column / line vectors." (I hope I have translated correctly?) Now using that, I see that line 2 and 3 are multiples, thus they are linearly dependent. But line 1 isn't. Rank is 1. That was about line vectors. Looking at column vectors, there are no linearly dependent vectors, since no multiples. Rank is 3. But what is the rank now? Shouldn't the number of linearly dependent / independent in both line / column vectors be equal? So what is the rank of this matrix now? I'm very confused. Or is the definition wrong? | 2155095 | How can I quickly know the rank of this / any other matrix?
I have looked this up on several sites but they confused me because some of the given information was wrong / unclear / contradicting whatever. I hope you can tell me all / most important ways to calculate the rank of a matrix. As example, I take the matrix [imath]A = \begin{pmatrix} 1 & 2 & 3\\ 0 & 5 & 4\\ 0 & 10& 2 \end{pmatrix}[/imath] Now several sites included that info so it must be true: If, we are looking at this example, there is no line with zeroes only, the rank of this matrix will be [imath]3[/imath]. (?) Here is the problem. It will cost time to form this matrix to see if there will be lines with zeroes only. For this I can use Gaussian Elimination. I have tested it with that Gauss and I couldn't get a line with zeroes only, so I conclude that this matrix [imath]rank(A)=3[/imath]. This however seems very inefficient way, I hope you can tell me better ways? |
2154850 | Prove that [imath]a^2 + b^2 + c^2 \lt 2(ab + bc + ca)[/imath]
Given that [imath]a, b[/imath] and [imath]c[/imath] are the sides of a triangle. How to prove that [imath]a^2 + b^2 + c^2 \lt 2(ab + bc + ca)[/imath]? Maybe any hint? Am I going to wrong direction? [imath]2(ab + bc + ca)-a^2 + b^2 + c^2>0[/imath] [imath]2ab + 2bc + 2ca-a^2 + b^2 + c^2>0[/imath] [imath]2b(a+c) + 2ca-a^2 + b^2 + c^2>0[/imath] ...? | 1804220 | If [imath]x, y, z[/imath] are the side lengths of a triangle, prove that [imath]x^2 + y^2 + z^2 < 2(xy + yz + xz)[/imath]
Question: If [imath]x, y, z[/imath] are the side lengths of a triangle, prove that [imath]x^2 + y^2 + z^2 < 2(xy + yz + xz)[/imath] My solution: Consider [imath]x^2 + y^2 + z^2 < 2(xy + yz + xz)[/imath] Notice that [imath]x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)[/imath] Hence [imath](x+y+z)^2-2(xy+yz+xz) < 2(xy + yz + xz)[/imath] [imath] (x+y+z)^2 - 4(xy+yz+xz) < 0 [/imath] As [imath]x,y,z[/imath] are all greater than zero as a side length of a triangle can not be negative. And because [imath](x+y+z)^2 >0[/imath] for all real [imath]x,y,z[/imath] therefore the whole expression is less than zero. [imath]Q.E.D[/imath] Am I correct? Or could I be more "rigorous" I am a highschool student and getting into proof so any tips would be appreciated as well :) |
2148744 | Constructing a contradiction to prove limit of a recurrence relation
Let [imath] \Delta P = (b-P_n)/2 \\ P_{n+1} = P_n + \Delta P\\ P_n : \mathbb{Q} \to \mathbb{R} [/imath] We shall attempt to prove that [imath] \lim_{n \to \infty} P_n = b [/imath] I'll try in three parts [imath]P_n[/imath] is monotonic [imath]P_n[/imath] is bounded by [imath]b[/imath] Something along the lines of [imath]\sup P_n = b[/imath] First, prove [imath]P_n[/imath] is bounded by [imath]b[/imath]. Assume the contrary by taking [imath]P_n[/imath] to be the supremum of [imath]P_n < b[/imath] and therefore [imath]P_{n+1} > b[/imath]. Now that implies [imath]\Delta P > b-P_n[/imath] which violates the definition of the recurrence. Therefore, [imath]P_n[/imath] is bounded from above by [imath]b[/imath] Second, let us prove that [imath]P_n[/imath] is monotonic with respect to [imath]n[/imath]. Assuming [imath]b > P_0[/imath], [imath]P_1 > P_0[/imath]. Now choose an arbitrary [imath]P_{n}[/imath], where [imath]b > P_n[/imath] from the previous result. If [imath]P_{n+1} = P_n + \Delta P[/imath], [imath]\Delta P = \frac{b-P_n}{2} > 0[/imath] and therefore [imath]P_n[/imath] is increasing for [imath]P_n < b[/imath] Now let us prove that [imath]\lim_{n \to \infty} P_n = b[/imath]. That means that for all [imath]M, \epsilon > 0[/imath] [imath] n > M \implies |P_n - L| < \epsilon [/imath] To start, we know [imath]L \le b[/imath] by our first result. Now all we must prove is that [imath]L < b[/imath] is impossible, which is what I am having trouble with. What I am having trouble with Well if for all [imath]n>M[/imath], [imath]P_n < L[/imath] then for all [imath]P_n[/imath], [imath] \Delta P = \frac{b-P_n}{2}< L-P_n < \epsilon [/imath] I'm having trouble constructing a contradiction and was hoping for some help doing so. I specifically want to use a delta epsilon proof and not any other form of proof. | 2146663 | Verify proof that sequence converges
I'm taking Calculus II next semester so I don't have all the tools from it to back me up but something came up in a discussion this evening that asked me to prove that the following sequence converges. As I'm new to formal math I'm hoping someone can verify my proof, especially step [imath]6[/imath]. I must prove that [imath]\lim_{n\to\infty}P_n=b[/imath] where [imath]P_{n+1} = P_n + \frac{b-P_n}{2}[/imath] [imath]P_0 = a[/imath] and [imath]a,b \in \mathbb{R}[/imath] such that [imath]a>b[/imath] My proof goes as follows: We must prove that [imath]\lim_{n\to\infty}P_n=b[/imath] That is, [imath]\forall\epsilon>0[/imath] there exists a [imath]\delta[/imath] such that there exists an [imath]n[/imath] satisfying [imath]n>\delta \iff b-P_n<\epsilon[/imath] First, let us prove that [imath]P_n[/imath] is bounded by [imath]b[/imath] by contradiction. Let us choose an [imath]i[/imath] such that [imath]P_i < b[/imath] and [imath]P_{i+1}>b[/imath]. Then we have [imath]P_{i+1} = P_i + \Delta P[/imath] where [imath]\Delta P > b-P_i[/imath]. but this is impossible because by the definition of the recurrence [imath]\Delta P = \frac{b-P_i}{2} < b-P_i[/imath]. Therefore, [imath]P_i[/imath] is bounded from above by [imath]b[/imath]. Additionally, it is trivial to show that [imath]P_{i+1} > P_i[/imath]. With this in mind, choose an arbitrary [imath]\epsilon[/imath]. We need to show that there exists an [imath]n[/imath] such that [imath]P_n > b-\epsilon[/imath]. Assume the contrary, that [imath]P_n[/imath] is bounded by [imath]\epsilon[/imath]. Using a similar argument to what which was used before, assume [imath]P_n[/imath] is bounded by [imath]\epsilon[/imath] and then there exists a [imath]P_i[/imath] such that [imath]P_{i+1}\not > P_i[/imath]. But this results in a contradiction with (4) Therefore, [imath]P_n[/imath] is not bounded by [imath]\epsilon[/imath] and we have proved what we want to prove. |
2154991 | set theory proof (cartesian products/symmetric differences)
I have to prove this theorem - Let [imath]A[/imath], [imath]B[/imath], and [imath]C[/imath] be sets. Then, [imath]A \times (B \Delta C) = (A \times B) \Delta (A \times C)[/imath]. I was wondering if this was the right way to start the proof? Let [imath]m[/imath] be an arbitrary element of [imath]A \times (B \Delta C)[/imath]. Then, [imath]m=(x,y)[/imath] s.t. [imath]x \in A[/imath] and [imath]y \in (B \Delta C)[/imath] by the definition of cartesian products. Then, [imath](m \in A) \land (y \in B) \land (y \not\in C)[/imath] by the definition of set difference. So, [imath]((x,y) \in A \times B) \land ((x,y) \not\in A \times C)[/imath] by the definition of cartesian product. Thus, [imath](x,y) \in (A \times B) \Delta (A \times C)[/imath]. So basically the LHS = RHS because (x,y) is an element of both sides? | 2145141 | elementary set theory (cartesian product and symmetric difference proof)
I've figured out the following statement is true but I was wondering how you actually go about proving something like this? [imath]A \times (B \triangle C) = (A \times B) \triangle (A \times C)[/imath] |
2155630 | Show that there exist a polynomial [imath]q_n(x) \in \mathbb Q[x] [/imath] satisfying [imath]p_n(x)=(x-1)^2q_n(x)[/imath].
I have this problem, Let [imath]p_n(x)=nx^{n+1}-(n+1)x^n+1 \in \mathbb Q[x][/imath] for any positive integer [imath]n[/imath]. Show that there exist a polynomial [imath]q_n(x) \in \mathbb Q[x] [/imath] satisfying [imath]p_n(x)=(x-1)^2q_n(x)[/imath]. initially I thought that I could use the division algorithm, such that [imath]p_n(x)=(x-1)^2q_n(x)+v(x)[/imath] but is there a reason that we could say that [imath]v(x)=0[/imath]? | 2079719 | Prove that the polynomial [imath]f_n(x)=nx^{n+1}-(n+1)x^n+1[/imath] is divisible by [imath](x-1)^2[/imath]
Prove that the polynomial [imath]f_n(x)=nx^{n+1}-(n+1)x^n+1[/imath] is divisible by [imath](x-1)^2[/imath] where [imath]n\in\Bbb{N_1}[/imath]. My Attempt I've tried proving it by induction but can't find a way to prove it for [imath]n+1[/imath] after assuming it holds for [imath]n[/imath]; I've also checked it graphically, it seems that for all [imath]n[/imath] [imath]f_n[/imath] intersects the axes at [imath](0,1)[/imath] and [imath](1,0)[/imath]; I've also tried finding a factor [imath]k[/imath] such that [imath](x-1)^2k=f_n(x)[/imath] for [imath]n=3, n=4[/imath] on WA but with no luck. |
2155690 | Prove that [imath]n<2^n[/imath] for any natural number [imath]n[/imath].
How do I prove that [imath]n<2^n[/imath] for any natural number [imath]n[/imath], assuming basic facts about the algebra of exponents? | 1893092 | Show that [imath]2^n>n[/imath] for [imath]n\in \mathbb{N}[/imath]
Show that [imath]2^n>n[/imath] for [imath]n\in \mathbb{N}[/imath] I can solve this by mathematical induction. Is there any other method to solve. |
1369444 | Find all [imath]2 \times 2[/imath] matrices [imath]A[/imath] such that [imath]AB = BA[/imath] for every [imath]2 \times 2[/imath] matrix [imath]B[/imath]
Find all possible [imath]2 \times 2[/imath] matrices A that for any [imath]2 \times 2[/imath] matrix B, AB = BA. Hint: AB = BA must hold for all B. Try matrices B that have lots of zero entries. I'm clueless as to how to solve this problem. How should I start it? I tried plugging in values for B that "have lots of zero entries" but didn't seem to see anything that could help. | 1107111 | All matrices which commute with all [imath]2\times 2[/imath] matrices
I would like to find all matrices which commute with all [imath]2\times2[/imath] matrices. I started solving problem in this way: 1) I have this matrix [imath]A[/imath] with real numbers: [imath]A=\left[\begin{array}{cc}a &b\\c &d\end{array}\right][/imath] 2) Matrix which commute with matrix [imath]A[/imath] is matrix [imath]B[/imath]: [imath]B = \left[\begin{array}{cc}e& f\\ g &h\end{array}\right][/imath] 3) When i solve the equation [imath]AB=BA[/imath], I get this: [imath]\left[\begin{array}{cc}ae+bg&af+bh\\ce+dg&cf+dh\end{array}\right]=\left[\begin{array}{cc}ea+cf&eb+df\\ga+ch&gb+dh\end{array}\right][/imath] How I can get the general look of wanted matrix? |
2153809 | Is this a new operation?
I was thinking "What is before addition", and came up with this. This analogy describes it: Addition is to multiplication as [operation] is to addition. On wikipedia it says it is just "1+b". I came up with an alterantive; the symbol that I'll use for it will be [imath]@[/imath]. [imath]b@b=b+2[/imath] [imath]b@b@b=b+3[/imath] [imath]b@b@b@b=b+4[/imath]... For example: [imath]3@3=5[/imath] If you try to compute it you run into problems. For example: [imath]3@2=(0@0@0)@(0@0)[/imath] but it also equals [imath](-1@-1@-1@-1)@(-1@-1@-1)[/imath] [imath](0@0@0)@(0@0)=(-1@-1@-1@-1)@(-1@-1@-1)[/imath] [imath]0@0@0@0@0=-1@-1@-1@-1@-1@-1@-1[/imath] [imath]0+5=-1+7[/imath] [imath]5\neq6[/imath] So, you can't take away or add parentheses. 1)Is this new? 2)How do you compute it? | 2154623 | what operation repeated [imath]n[/imath] times results in the addition operator?
I had a difficult time in phrasing my question. But I was wondering if there is an operation that, when repeated n times, results in the addition operator. Same way as repeating addition n times results in the multiplication operator, and repeating multiplication n times results in the exponentiation operator etc. So [imath]n[/imath] times addition of a number [imath]x[/imath] results in [imath]x\times n[/imath]. And [imath]n[/imath] times multiplication of a number [imath]x[/imath] results in [imath]x^n[/imath] Then my question is [imath]n[/imath] times ...what... results into the number [imath]x+n[/imath]. Let's call this operator: [imath]@[/imath]. For example, the following would then hold: [imath]a\times a=a^2[/imath] [imath]a+a=a\times 2[/imath] [imath]a@a=a+2[/imath] My question is, does it make any sense thinking of such an operator, is there anything known about it, can it be followed through even further like [imath]a\sim a = a@2[/imath]? |
2156410 | Prove that these definitions of [imath]e[/imath] are equivalent
Im sorry if this is stupid or obvious, but why [imath] e=\lim_{n\rightarrow \infty}\left(1+\frac{1}{n}\right)^n[/imath] AND [imath]e[/imath] is the unique positive number for which [imath] \lim_{h\rightarrow 0}\frac{e^h-1}{h}=1[/imath]? I mean, how do we know that these two definitions are equivalent? Again, maybe it's easy, but I'm just beginning calculus and our teacher just dropped those two definitions... | 2149106 | Particular definition of e
Show that, [imath]e=\lim_{x\to \infty} \left(1+\frac{1}{x}\right)^x [/imath] Is the same number that satisfies, [imath]\lim_{h\to0} \frac{e^{h}-1}{h} = 1 \tag{*}[/imath] You don't have to do that if it's too cumbersome, but (*) is used to find the derivatives of real exponential functions. I know that definitions are not to be proven but I'm looking for some derivation or intution for the claim that (*) is a legitimate definition for e, and furthermore, a proof that limit in (*) exists. |
2156173 | How to prove [imath]n[/imath] is even in this trigonometric equation?
I have this trigonometric equation- [imath]\bigg(\frac {\cos a+\cos b}{\sin a-\sin b}\bigg)^n+\bigg(\frac {\sin a+\sin b}{\cos a-\cos b}\bigg)^n=2\cot^n \frac {a-b}{2}[/imath]. How do I prove [imath]n[/imath] is even in this? | 2148166 | Prove that [imath]n[/imath] is even.
Prove that [imath]n[/imath] is even [imath]\left(\dfrac {\cos A + \cos B}{\sin A-\sin B}\right)^n + \left(\dfrac {\sin A+\sin B}{\cos A- \cos B}\right)^n = 2\cot^n \left(\dfrac {A-B}{2}\right)[/imath] Please help. I didn't get any idea regarding this. |
2156639 | Can the prime ideals of [imath]A_a[/imath] be identified with those prime ideals of [imath]A[/imath] which do not contain [imath]a[/imath]?
Let [imath]A[/imath] be a ring, and let [imath]a \in A[/imath]. Can the prime ideals of [imath]A_a[/imath] be identified with those prime ideals of [imath]A[/imath] which do not contain [imath]a[/imath]? | 765864 | Localisation and prime ideals
If [imath]A[/imath] is a ring and [imath]S=\{1,f,f^2,f^3,...\}[/imath] a multiplicative set of [imath]A[/imath], prove that [imath]\mathrm{Spec}(A_f)=\mathfrak{V}((f))^c[/imath]. Notation: [imath]A_f=S^{-1}A[/imath] and [imath]\mathfrak{V}((f))=\{P \in \mathrm{Spec}(A): P \supset (f)\}[/imath] My attempt: On one hand, if [imath]P \in \mathrm{Spec}(A)[/imath] and [imath]P \cap S[/imath] is empty then we identify [imath]\mathrm{Spec}(A_f)=\mathrm{Spec}(S^{-1}A)=\{P \in \mathrm{Spec}(A): P\cap S= \emptyset\}[/imath]. On the other hand, [imath]\mathfrak{V}((f))^c=\{P \in \mathrm{Spec}(A): P \nsupseteq (f)\}[/imath]. So the exercise is equivalent to prove: [imath]\{P \in \mathrm{Spec}(A): P\cap S =\emptyset\}=\{P \in \mathrm{Spec}(A): P \nsupseteq (f)\},[/imath] but I can't continue so I'd appreciate if somebody could help me. Thanks in advance. |
2156837 | Does there exist a function which is holomorphic on the open unit disc but goes to infinity on the boundary?
Does there exist a function [imath]f(z)[/imath] holomorphic on the disc [imath]|z| < 1[/imath] but such that [imath]\lim_{|z| \rightarrow 1} |f(z)| = \infty[/imath]? I think the answer is no because [imath]1/f(z)[/imath] would go to [imath]0[/imath] as [imath]|z| \to 1[/imath]. | 1468895 | Does there exist a holomorphic function with the following property?
Does there exist a holomorphic function [imath]f[/imath] defined over [imath]D = \{ z : |z| < 1 \}[/imath] such that [imath]|f| \rightarrow \infty[/imath] when [imath]|z| \rightarrow 1[/imath]? My approach: If such an [imath]f[/imath] exists, then for a given [imath]\varepsilon > 0[/imath], there exists an [imath]M > 0[/imath] such that [imath]|f(z)|>M[/imath] if [imath]1-\varepsilon < |z| < 1[/imath] Since [imath]f[/imath] has no zeros in that set, I can consider [imath]g = 1/f[/imath], I would like to apply Maximum Modulus Principle and get a contradiction, but actually I am not sure I can. |
2156406 | let [imath] a,b \in (0,1)[/imath] ,[imath]a+b>1[/imath] then : prove that : [imath]2^a+3^b<3a+4b[/imath]
let [imath] a,b \in (0,1)[/imath] ,[imath]a+b>1[/imath] then : prove that : [imath]2^a+3^b<3a+4b[/imath] my try : [imath]f(x):=e^x\\ e^{(2^a+4^b)}<e^{ (3a+4b)}\\ e^{2a}e^{3^b}<e^{3a}e^{4b}[/imath] now ? | 158628 | Prove that: [imath]2^a+3^b<3a+4b[/imath]
Let be [imath]a, b[/imath] in [imath](0,1)[/imath] such that [imath]a+b>1[/imath]. I need to prove that: [imath]2^a+3^b<3a+4b[/imath] I'm looking for an elementary proof that doesn't resort to the calculus tools. |
2156206 | Product in finite abelian groups
Product in finite abelian groups Let [imath]a\in A[/imath] where [imath]A[/imath] is a finite abelian group. How do I prove that [imath]\Pi_{x\in A}ax=\Pi_{x\in A}x[/imath]? Thoughts: If every element is not its own inverse, we see that [imath]\Pi_{x\in A}x=e[/imath]. If there are elements that are their own inverse, then it is the product of these elements. How do I get the identity above? | 2155374 | Product of Finite Abelian Group
I need to proof for a finite abelian group [imath]G[/imath] that for all [imath]x\in G[/imath] we have [imath]\prod_{g\in G}xg=\prod_{g\in G}g[/imath]. I figured that using the commutative property [imath]\prod_{g\in G}xg=x^n\prod_{g\in G}g[/imath]. Which would leave us to proof [imath]x^n=e[/imath], where [imath]e[/imath] is the identity element. Hopefully I'm just missing something obvious here, because I feel like smashing my head against the table now. |
2157016 | Extension ideal
Let [imath]R[/imath] be ring and let [imath]I[/imath] be a nonzero proper ideal of [imath]R[/imath]. If [imath]B[/imath] is a field and [imath]f:R\rightarrow B[/imath] is a ring homomorphism, and [imath]f(x)\in{I^e}[/imath] (where [imath]{I^e}[/imath] is extension ideal of [imath]I[/imath]), then [imath]x\in{rad(I)},[/imath] where [imath]rad(I)[/imath] is the radical of [imath]I[/imath]. It seems to me it is easy question but I just proved with assuming that [imath]f[/imath] is surjective. Any insight is much appreciated. | 2133840 | question about extension ideal
Suppose [imath]R[/imath] is a commutative ring with identity and let [imath]I[/imath] be a nonzero proper ideal of [imath]R[/imath]. Prove that if [imath]f: R \to k[/imath] is a ring map ([imath]k[/imath] is field) and [imath]f(x)\in\left<f(I)\right>[/imath], then [imath]x\in\mathrm{rad}(I)[/imath]. I believe this is not correct unless [imath]f[/imath] is surjective. My attempted. I assumed that [imath]f[/imath] is surjective. [imath]f(x)\in\left<f(I)\right>[/imath], and we will get [imath]x\in\mathrm{rad}(I)[/imath] since [imath]I\subseteq\mathrm{rad}(I)[/imath]. My question is this question can be true without assumption that [imath]f[/imath] must be surjective, if not, could give a counterexample. By the way, ring map is homomorphism. Any help will appreciated . |
2157844 | How to solve [imath]x^2 +\left(\frac{x}{x-1}\right)^2 =8[/imath]?
I tried to solve this question but it turns into a 4th degree equation and I could only get one solution for this equation,i.e., 2. It is to be evaluated for solutions. Thanks. | 2020139 | Solve the equation [imath]x^2+\frac{9x^2}{(x+3)^2}=27[/imath]
Problem Statement:- Solve the equation [imath]x^2+\dfrac{9x^2}{(x+3)^2}=27[/imath] I have tried to turn it into a quadratic equation so as to be saved from solving a quartic equation, but have not been able to come up with anything of much value. These are the things that I have tried to turn the given equation into a quadratic equation. [imath]x^2+\dfrac{9x^2}{(x+3)^2}=27\implies 1+\dfrac{1}{\left(\dfrac{x}{3}+1\right)^2}=3\left(\dfrac{3}{x}\right)^2[/imath] [imath]\text{OR}[/imath] [imath]x^2+\dfrac{9x^2}{(x+3)^2}=27\implies 1+\dfrac{\left(\dfrac{3}{x}\right)^2}{\left(1+\dfrac{3}{x}\right)^2}=3\left(\dfrac{3}{x}\right)^2[/imath] |
1982462 | How to prove this Tree/Set claim?
If [imath]T[/imath] is a tree and [imath]T_1, T_2, \ldots , T_k[/imath] are all subtrees of [imath]T[/imath] such that the vertices of [imath]T_i[/imath] are [imath]V_i[/imath]. Suppose [imath]V_i\cap V_j[/imath] is non-empty for all [imath]i[/imath] and [imath]j[/imath]. Show that there is a vertex that exists in all [imath]T_i[/imath]. I'm honestly not even sure what the question is asking. This is just a completion grade, but I honestly do want to know how I should go about proving this. What method to use? | 857698 | For all [imath]1 \leq i < j \leq k[/imath], the subtrees [imath]T_i[/imath] and [imath]T_j[/imath] have a vertex in common. Show that [imath]T[/imath] has a vertex which is in all of the [imath]T_i[/imath].
Can someone please verify my proof or offer suggestions for improvement? I am aware that there is a similar question elsewhere, but I want help with my proof in particular. Let [imath]T_1, \ldots, T_k[/imath] be subtrees of a tree [imath]T[/imath] such that for all [imath]1 \leq i < j \leq k[/imath], the trees [imath]T_i[/imath] and [imath]T_j[/imath] have a vertex in common. Show that [imath]T[/imath] has a vertex which is in all of the [imath]T_i[/imath]. Suppose, for the sake of contradiction, that for every vertex [imath]v \in V(T)[/imath], there is a subtree [imath]T'[/imath] in the list [imath]T_1, T_2, \ldots, T_k[/imath] such that [imath]v \notin V(T')[/imath]. Pick [imath]v_1 \in T_1[/imath]. There is a subtree [imath]T^{(1)}[/imath] such that [imath]v_1 \notin T_1[/imath]. [imath]T_1[/imath] and [imath]T^{(1)}[/imath] have a vertex [imath]y_1[/imath] in common. Now, there exists a tree [imath]T^{(2)}[/imath] such that [imath]y_1 \notin T^{(2)}[/imath]. But now, [imath]T^{(2)}[/imath] and [imath]T^{(1)}[/imath] have a vertex [imath]y_2[/imath] in common. Also, [imath]T_1[/imath] and [imath]T^{(2)}[/imath] have a vertex [imath]y_3[/imath] in common. There exists a path [imath]p_1[/imath] in [imath]T_1[/imath] from [imath]y_3[/imath] to [imath]y_1[/imath]. Also, there is a path [imath]p_2[/imath] in [imath]T^{(1)}[/imath] from [imath]y_1[/imath] to [imath]y_2[/imath] and a path [imath]p_3[/imath] in [imath]T^{(2)}[/imath] from [imath]y_2[/imath] to [imath]y_3[/imath]. Note that the path formed by the union of [imath]p_1[/imath] and [imath]p_2[/imath] passes through [imath]y_1[/imath], while [imath]p_3[/imath] does not, since [imath]y_1 \notin T^{(2)}[/imath]. Therefore, the union of [imath]p_1, p_2[/imath], and [imath]p_3[/imath] forms a cycle, contradicting the fact that [imath]T[/imath] is a tree. Therefore, it must be the case that there is a vertex [imath]x[/imath] in all of the [imath]T_i[/imath]. |
2158899 | GCD question in PID's
Let [imath]S[/imath] be a ring with subring [imath]R[/imath], both PIDs. Let [imath]a, b \in R[/imath] with gcd [imath]r \in R[/imath]. If [imath]s \in S[/imath] is the gcd of [imath]a[/imath] and [imath]b[/imath] when considered in the larger ring [imath]S[/imath], prove that [imath]r = su[/imath] for some unit [imath]u \in S[/imath]. I am trying to exploit the fact that the ideal [imath](a, b) = (s)[/imath] and any other generator of the ideal [imath](a, b)[/imath] is a unit multiple of [imath]s[/imath]. It is clear from the definition of gcd that [imath]r | s[/imath] which implies that [imath](s) \subset (r)[/imath]. These ideals are generated in [imath]S[/imath], I have given some thought as to what I get from considering them as ideals of [imath]R[/imath]. Still I can not see how to proceed. Any suggestions are appreciated. Thanks in advance. | 713290 | GCD in a PID persists in extension domains
Let [imath]R[/imath] be subring of integral domain [imath]S.[/imath] Suppose [imath]R[/imath] is [imath]\text{PID}.[/imath] Let [imath]a\in R[/imath] be a greatest common divisor of [imath]r_1,r_2[/imath] in [imath]R[/imath]. ([imath]r_1,r_2 \in R[/imath], not both zero). Could anyone advise me on how to prove [imath]a[/imath] is greatest common divisor of [imath]r_1,r_2[/imath] in [imath]S?[/imath] Hints will suffice, thank you. |
2158508 | Where does the [imath]\sqrt{N}[/imath] come from in standard error of the mean formula??
Where does the [imath]\sqrt{N}[/imath] come from in standard error of the mean formula?? When calculating z scores for multiple samples and want to describe the standard deviation of those sample means I know the formula is z = [imath]\frac{(x - \mu)}{\frac{\sigma}{\sqrt{N}}}[/imath] where N is our sample size. Intuitively why does dividing by [imath]\sqrt{N}[/imath] make sense?? | 1570915 | Why the standard deviation of the sample mean is calculated as [imath]\frac{\sigma}{\sqrt{n}}[/imath]?
According to Wikipedia, the standard deviation of a sample mean is calculated as follows [imath]\frac{\sigma}{\sqrt{n}}[/imath] Why is that? Why do we need to divide the standard deviation of the population by the square root of [imath]n[/imath] (which should I think be the size of the sample)? Why should that make sense? |
2157931 | How to check if a point is inside a square (2D Plane)?
There is a point [imath](x,y)[/imath], and a rectangle [imath]a(x_1,y_1),b(x_2,y_2),c(x_3,y_3),d(x_4,y_4)[/imath], how can one check if the point inside the rectangle? [imath]M[/imath] of coordinates [imath](x,y)[/imath] is inside the rectangle iff [imath](0<\textbf{AM}\cdot \textbf{AB}<\textbf{AB}\cdot \textbf{AB}) \land (0<\textbf{AM}\cdot \textbf{AD}<\textbf{AD}\cdot \textbf{AD})[/imath] (scalar product of vectors) I found this formula but I don't understand how I have to calculate it. I haven't done much math in a long time. I don't understand why this is considered a duplicate when I'm asking how to solve the equation provided by the answer of that question. | 190111 | How to check if a point is inside a rectangle?
There is a point [imath](x,y)[/imath], and a rectangle [imath]a(x_1,y_1),b(x_2,y_2),c(x_3,y_3),d(x_4,y_4)[/imath], how can one check if the point inside the rectangle? |
2159428 | on the series [imath]\sum {\sin(nz)\over n}[/imath]
Could anyone help me to show that [imath]\sum {\sin(nz)\over n}[/imath] diverges if [imath]Im(z)\ne 0[/imath]? Thank you | 2153390 | Show that[imath]\displaystyle\sum_{n=1}^\infty\frac{\sin( nz)}{n}[/imath] diverges if [imath]\mathfrak{Im} \,(z)\neq0[/imath]
We have to show that[imath]\displaystyle\sum_{n=1}^\infty\frac{sin( nz)}{n}[/imath] diverges if [imath]Im\,(z)\neq0[/imath] My try:[imath]\sin (nz) = \mathfrak{Im}(e^{nzi})[/imath] [imath]\sum_{n=1}^{\infty} \frac{\sin nz}{n} = \mathfrak{Im} \sum_{n=1}^{\infty} \frac{(e^{zi})^n}{n} = - \mathfrak{Im} \ln\left(1-e^{zi}\right). [/imath] Then how to proceed? |
2159398 | Find elements [imath]f,g\in\mathrm{End}(V)[/imath] such that [imath]f\circ g=id_V[/imath] and [imath]f,g\not\in \mathrm{End}(V)^{\times}[/imath]
Let [imath]K[/imath] be a field. Given [imath]V[/imath] a [imath]K[/imath] vector-space containing all sequences [imath]a=(a_n)_{n\in \mathbb N}[/imath] with values from [imath]K[/imath], such that [imath]a_n\neq 0[/imath] for only finitely many [imath]n\in \mathbb N[/imath]. I want to to find [imath]f,g\in\mathrm{End}(V)[/imath] such that [imath]f\circ g=id_V[/imath] and [imath]f,g\not\in\mathrm{End}(V)^{\times}.[/imath] | 352992 | Does [imath]ab=1[/imath] imply [imath]ba=1[/imath] in a ring?
Let [imath]R[/imath] be a ring with unity [imath]1[/imath] and let [imath]a,b\in R[/imath] be not zero divisors. Is there any counterexample for: [imath]ab=1\quad\Rightarrow \quad ba=1[/imath] |
2159974 | Prove that for every [imath]n[/imath] there is a multiple of [imath]5^n[/imath] which does not contain a zero.
How to prove that for every [imath]n[/imath] there is a multiple of [imath]5^n[/imath] which does not contain a zero in its decimal representation? | 2067752 | Does every number not ending with zero have a multiple without zero digits at all?
Does every number not ending with zero have a multiple without zero digits at all? We consider numbers in their usual base 10 representation. we don't consider numbers ending with [imath]0[/imath], because all multiples of such numbers end with [imath]0[/imath]. It can be proven that all powers of [imath]2[/imath] have such a multiple: Indeed, for every number of the form [imath]2^n[/imath], we can find a multiple [imath]l[/imath] of it whose last [imath]n[/imath] digits are nonzero (This can be done by repeatedly multiplying a number by [imath]10^k+1[/imath], as explained here), and then forget all other digits; these last [imath]n[/imath] digits constitute a number that is still divisible by [imath]2^n[/imath], since it is obtained from [imath]l[/imath] by subtracting a multiple of [imath]10^n[/imath] which is in particular a multiple of [imath]2^n[/imath]. The same argument shows that every power of [imath]5[/imath] has a multiple without zeros, but this does not seem to generalize to other numbers. My intuition is actually that the claim is not true, since the density of numbers without any zero digits approaches [imath]0[/imath] for big numbers. This is because the probability of a random number with at most [imath]n[/imath] digits to contain only nonzero digits is [imath](\frac{9}{10})^n[/imath]. Thus big numbers "nearly always" contain a zero. This hints on the existence of a counterexample, but does not prove its existence. |
2159661 | For [imath]a, b, c[/imath] is the length of three sides of a triangle. Prove that [imath]\left|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right|<\frac{1}{8}[/imath]
For [imath]a, b, c[/imath] is the length of three sides of a triangle. Prove that [imath]\left|\frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a}\right|<\frac{1}{8}[/imath] | 1621173 | Prove that [imath]\left | \frac{a-b}{a+b}+\frac{b-c}{b+c}+\frac{c-a}{c+a} \right | < \frac{1}{8}.[/imath]
Let [imath]a,b,[/imath] and [imath]c[/imath] be the lengths of the sides of a triangle. Prove that [imath]\left | \dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a} \right | < \dfrac{1}{8}.[/imath] The best idea I had was to expand the fractions to get something nicer. So we get [imath]\dfrac{a-b}{a+b}+\dfrac{b-c}{b+c}+\dfrac{c-a}{c+a} = \dfrac{a^2 b-a^2 c-a b^2+a c^2+b^2 c-b c^2}{(a+b) (a+c) (b+c)}.[/imath] Then I would try to relate this to side lengths of a triangle to prove the desired inequality but that is the step where I am unsure. |
2160285 | Compact manifold without boundary is boundary of another manifold
Is it true, that a compact (topological?) manifold without boundary is a boundary of another compact manifold? Intuitively it looks like for most manifolds there must be some relation similar to [imath]\mathbb{S}^n=\partial D^{n+1}[/imath]. It looks like a very trivial question, though I can not find any reference. Thanks. | 1385708 | Examples of manifolds that are not boundaries
What are some examples of manifolds that do not have boundaries and are not boundaries of higher dimensional manifolds? Is any [imath]n[/imath]-dimensional closed manifold a boundary of some [imath](n+1)[/imath]-dimensional manifold? |
2159976 | Prove by induction that [imath]1+3+3^2+3^3+...+3^n=\frac{1}{2}(3^{n+1}-1)[/imath]
I have [imath]\frac{1}{2}(3^{k+1}-1)+3^{k+1}=\frac{1}{2}3^{k+2}-1[/imath] Not sure where to go from here. First post on Mathematics, please correct me if I did anything wrong. | 1096214 | Simplifying a geometric sum
For some choice of [imath]\lambda \in \mathbb{R}[/imath] and [imath]n[/imath], what would be a general formula for [imath] \sum_{k=1}^{n} \lambda^k [/imath] as a function of [imath]\lambda[/imath] and [imath]n[/imath]? |
2145366 | If [imath]\left|z^3 + {1 \over z^3}\right| \le 2[/imath] then [imath]\left|z + {1 \over z}\right| \le 2[/imath]
[imath]\displaystyle \left|z^3 + {1 \over z^3}\right| \le 2[/imath] prove that [imath]\displaystyle \left|z + {1 \over z}\right| \le 2[/imath] [imath]\left|z^3 + {1 \over z^3}\right| = \left(z^3 + {1 \over z^3}\right)\left(\overline z^3 + {1 \over \overline{z}^3}\right) = \left(z + {1\over z}\right)\left(z^2 - 1 + {1\over z^2} \right)\left(\overline z + {1\over \overline z}\right)\left(\overline z^2 - 1 + {1\over\overline z^2} \right)[/imath] [imath]=\left|z + {1\over z}\right|^2\left|z^2 - 1 + {1\over z^2} \right|^2 \le 2[/imath] [imath]\therefore \left|z + {1\over z}\right| \le \sqrt{2}[/imath] where [imath]\displaystyle \left|z^2 - 1 + {1\over z^2} \right| \ge 1[/imath]. But I am not able to prove [imath]\displaystyle \left|z^2 - 1 + {1\over z^2} \right| \ge 1[/imath], need some help on this. | 1799712 | Let [imath]z \in C^*[/imath] such that [imath]|z^3+\frac{1}{z^3}|\leq 2[/imath] Prove that [imath]|z+\frac{1}{z}|\leq 2[/imath]
Problem : Let [imath]z \in C^*[/imath] such that [imath]|z^3+\frac{1}{z^3}|\leq 2[/imath] Prove that [imath]|z+\frac{1}{z}|\leq 2[/imath] My approach : Since : [imath](a^3+b^3)=(a+b)^3-3ab(a+b)[/imath] [imath]\Rightarrow z^3+\frac{1}{z^3}=(z+\frac{1}{z})^3-3(z+\frac{1}{z})[/imath] Now I don't know further about this problem whether or not this will help here , please guide to solve this problem will be of great help , thanks. |
2160570 | Prove that [imath]\phi=\frac{13}{8} + \sum \limits_{n=0}^{\infty}\frac{(-1)^{(n+1)}(2n+1)!}{n!\,(n+2)!\,4^{(2n+3)}}[/imath].
Prove that [imath]\phi=\dfrac{13}{8} + \sum \limits_{n=0}^{\infty}\dfrac{(-1)^{(n+1)}(2n+1)!}{n!\,(n+2)!\,4^{(2n+3)}}[/imath],where [imath]\phi[/imath] is the golden ratio ([imath]\approx1.6180[/imath]). Evaluating the first few few partial sums makes it obvious that the expression approaches [imath]\approx 1.61[/imath] as the limit of the infinite sum is [imath]\approx -0.007[/imath]. But is there any algebraic method to prove the same ? Stirling's approximation seems to work a little but not much as the limit of the sum has to calculated and not the limit of the fraction. If there is no other known way than Stirling's, how should I properly implement it ? I have tried my best, but there does not seem any way out. As far as I think, only numerical method will help here. | 1470099 | A golden ratio series from a comic book
The eighth installment of the Filipino comic series Kikomachine Komix features a peculiar series for the golden ratio in its cover: That is, [imath]\phi=\frac{13}{8}+\sum_{n=0}^\infty \frac{(-1)^{n+1}(2n+1)!}{(n+2)!n!4^{2n+3}}[/imath] How might this be proven? |
2160022 | Limit of nth root of n!
I am asked to determine if a series converges or not: [imath]\displaystyle\sum\limits_{n=1}^{\infty} (2^n)n!/(n^n)[/imath] So I'm using the nth root test and came up with [imath]\lim_{n \to {\infty}}(2/n)*(\sqrt[n]{n!})[/imath] I know that the limit of [imath]2/n[/imath] goes to [imath]0[/imath] when [imath]n[/imath] goes to infinity but what about the [imath](\sqrt[n]{n!})[/imath]? | 1931236 | Convergent or divergent: [imath]\sum_{k=1}^{\infty}\frac{2^{k}\cdot k!}{k^{k}}[/imath]
Convergent, absolutely convergent or divergent: [imath]\sum_{k=1}^{\infty}\frac{2^{k}\cdot k!}{k^{k}}[/imath] I have used ratio test because we got a fraction here and I think I did alright till the end: [imath]\lim_{k\rightarrow \infty}\frac{\frac{2^{k+1}\cdot (k+1)!}{(k+1)^{k+1}}}{\frac{2^{k}\cdot k!}{k^{k}}} = \lim_{k\rightarrow \infty}\frac{2^{k+1}\cdot (k+1)! \cdot k^{k}}{(k+1)^{k+1}\cdot 2^{k}\cdot k!} = \lim_{k\rightarrow \infty}\frac{2^{k}\cdot 2^1\cdot k! \cdot (k+1)\cdot k^{k}}{(k+1)^{k}\cdot (k+1)\cdot 2^{k}\cdot k!} = \lim_{k\rightarrow \infty}\frac{2k^{k}}{(k+1)^{k}}[/imath] [imath]=2\lim_{k\rightarrow \infty} \left( \frac{k}{k+1}\right)^{k}[/imath] Now I don't know (without a calculator) to what this would converge / diverge to... In the exam we are not allowed to use a calculator... So what to do? The denominator will be greater than the enumerator by 1, so dividing each other we got something [imath]<1[/imath]. We take exponent [imath]k[/imath] which is [imath]\geq 1[/imath], so we will end up with [imath]<1[/imath] again. Multiply this with 2 we get something [imath]< 1[/imath] but [imath]> 0[/imath] and thus the series is convergent...? I hope I have described it well? Would you give me full points to this task? :D Edit: I haven't described it well in the end. See the accepted answer and its comments! Thanks a lot to everyone - from every question I ask here, I always learn new things :-) |
2162237 | Suppose [imath]A[/imath] is a 2 x 2 matrix, prove that if [imath]A^{3} = 0[/imath], then [imath]A^{2} = 0[/imath]
Suppose [imath]A[/imath] is a 2 x 2 matrix, prove that if [imath]A^{3} = 0[/imath], then [imath]A^{2} = 0[/imath] I was given this equation as a hint: [imath]A^{2} - tr(A)A + det(A)I = 0[/imath] where [imath]tr(A)[/imath] is the sum of the diagonal entries of [imath]A[/imath]. My attempt: [imath]A^{2} - tr(A)A = -det(A)I[/imath] [imath]A(A - tr(A)) = -det(A)I[/imath] [imath]A[-\frac{1}{det(A)}A(A - tr(A))] = I[/imath] [imath]A[/imath] is invertible, thus [imath]det(A)[/imath] is not equal to [imath]0[/imath]. Multiplying [imath]A^{2}[/imath] to both sides of equation 2: [imath]A^{3}(A - tr(A)) = A^{2}(-det(A)I)[/imath] Since [imath]A^{3} = 0[/imath] and since [imath]-det(A)I[/imath] is not equal to [imath]0[/imath], then [imath]A^{2}[/imath] must be equal. QED. Is this proof correct? EDIT: There seems to be a similar question, but my question is more specific in that I have to use the equation given in the hint. | 525213 | A 2x2 matrix [imath]M[/imath] exists. Suppose [imath]M^3=0[/imath] show that (I want proof) [imath]M^2=0[/imath]
I'm sure I've done this before in abstract algebra. Regardless it's escaped me now. I have proved that for [imath]T:U\rightarrow V[/imath], with [imath]dim(U)=m[/imath] and [imath]dim(V)=n[/imath] that [imath]rank(T)\le m[/imath] which is obvious, but I have proved it none the less. I want to.... this is where I get stuck. I'm not quite sure how to say it. Suppose we have a [imath]T[/imath] for example that takes [imath]U\subset\mathbb{R}^3\rightarrow V\subset\mathbb{R}^3[/imath], we could still have a [imath]T[/imath] that maps a plane to something (a plane, line or point) in this case rather than [imath]rank(T)\le 3[/imath] I can state [imath]rank(T)\le 2[/imath]. (Correct my notation here, I don't like writing subset, I mean number of dimensions of the space!) To show this I can say [imath]F:P\subset\mathbb{R}^2\rightarrow U\subset\mathbb{R}^3[/imath]. Then let [imath]G:P\rightarrow V[/imath] and that [imath]G=TF[/imath], now [imath]rank(G)\le dim(P) = 2[/imath] Now, if [imath]U,P,Q,R[/imath] are spaces of dimension [imath]\le 2[/imath] and P,Q,R are subspaces of U [imath]A:U\rightarrow P[/imath] [imath]B=A:P\rightarrow Q[/imath] [imath]C=A:Q\rightarrow R[/imath] We can now say [imath]A^2=BA[/imath] then [imath]A^3=CBA[/imath] We know that for [imath]A[/imath] rank(A)[imath]\le 2[/imath], this thus provides an upper-bound for rank([imath]A^3[/imath]) I want to show now that if I keep applying a linear transformation that the rank is a monotonically decreasing ([imath]\le[/imath]) sequence. I am unsure on a proof that the rank of a transformation cannot be greater than the domain's dimensions. Although this is trivial. (NOTE: I have proven that if a set of n vectors, R, span a vector space V, and you take a set W of m linearly independent vectors in V that m[imath]\le[/imath]n) If rank(A)=2 then rank([imath]A^2[/imath])=2 and so rank([imath]A^3[/imath])=2 if rank(A)=1 then rank([imath]A^2[/imath])=0 and rank([imath]A^3[/imath])=0 if rank(A)=0 then rank([imath]A^2[/imath])=0 and rank([imath]A^3[/imath])=0 But again, I can't prove this, or at least I am unsure of how to write it. something to clarify: For a map [imath]T:U\rightarrow V[/imath] how do we distinguish (using notation) whether or not we actually use all dim(U)? For example we might have a T that maps a plane in R^3 to a line in R^3, this can be expressed as a composition of maps, one that takes a 2 dimensional vector space (coordinates of the plane in 3 space) to a 3 dimensional point in T's domain, which T then maps to a line. We have rank - the dimensions of the image - to make this distinction on the map's target (the rank of T is 1 if it maps something to a line, but the dimensions of the target is 3) How do we write this? What do I say to describe this? How do I distinguish between the number of vectors in a basis and the number of dimensions the space that basis is in has? Sorry for the length of this! I tried to talk though my problem in the hope of seeing it myself, no such luck, thanks. |
1849697 | whats the proof for [imath]\lim_{x → 0} [(a_1^x + a_2^x + .....+ a_n^x)/n]^{1/x} = (a_1.a_2....a_n)^{1/n}[/imath]
This equation is directly given in my book and I am don't know anything about its proof.I tried L'Hospital rule by differentiating the both numerator as well as denominator(division rule), but the result is still coming in indeterminate forms.I am a beginner , and haven't practiced limits that much. This formula is really confusing me. | 2333868 | How do I evaluate [imath]\lim_{x\to 0}\left( \frac{1^x+2^x+3^x+\cdots+ n^x}{n}\right)^{a/x} [/imath]?
[imath]\lim_{x\to 0}\left( \frac{1^x+2^x+3^x+\cdots+ n^x} n \right)^{a/x} [/imath] |
2162640 | how to show that X is a random variable
for example you toss a coin three times then let [imath]X[/imath] be the number of times you get heads. so since the coin is tossed three times it means the sample space is [imath] \Omega = \{HHH,HHT,HTH,HTT,TTT,TTH,THT,THH\}[/imath] with each sample point from the sample space above we can associate a number for X (either 3 or 2 or 1 or 0). this means X is a function from [imath]\Omega[/imath] to the real numbers so It follows that X is a random variable. but how can we prove that formally, I mean is there a more formal proof of it. | 694548 | Formal definition of a random variable
I'm not new to the concept of random variable and I know the measure theory. Anyway, I started reading the book "Stochastic Differential Equation" by B. Oksendal, and I'm having some problem in understanding. Given a probability space [imath](\Omega, \mathcal{F}, P)[/imath], [imath]X : \Omega \rightarrow \mathbb{R}^n[/imath] is a random variable if [imath]X[/imath] is a [imath]\mathcal{F}[/imath]-measurable function, or equivalently, if [imath] X^{-1}(U) = \{\omega \in \Omega; X(w) \in U\} \in \mathcal{F} [/imath] for all the open set [imath]U \subset \mathbb{R}^n[/imath]. My main problem is that I always thought that [imath]X \in \Omega[/imath] while the definition on the book assert that [imath]X : \Omega \rightarrow \mathbb{R}^n[/imath] and hence [imath]X \in \mathbb{R}^n[/imath]. Addition I'll try to better explain my problem. Suppose to have a dice. Then, [imath]\Omega = \{1, 2, 3, 4, 5, 6\}[/imath]. One can be concerned to evaluate the probability that the dice result is even. In this case, I write: [imath]P(X \in \{2, 4, 6\}) = ...[/imath] In general, the probability of an event [imath]U \subset \Omega[/imath] ([imath]U \in \mathcal{F}[/imath]) is normally written as follows: [imath]P(X \in U)[/imath] This notation suggest me that [imath]X \in \Omega[/imath]. Again, I don't understand what means that [imath]X : \Omega \rightarrow \mathbb{R}^n[/imath]. |
2162772 | Correspondence Theorem
Let [imath]G[/imath] be a group, [imath]N \triangleleft G[/imath] and [imath]\bar{G} = G/N[/imath]. Prove that for every two elements [imath]\bar{a},\bar{b} \in \bar{G}[/imath] the following is true: [imath]\bar{a}\bar{b} = \bar{b}\bar{a} \Leftrightarrow a^{-1}b^{-1}ab \in N[/imath] So this looks a lot like the Correspondence Theorem, so I have an understanding as to how these are connected, but I'm at a loss when it comes to the proof of this. I am supposed to prove this without having that G/N is Abelian. | 374607 | Prove [imath]aba^{-1}b^{-1}\in{N}[/imath] for all [imath]a,b[/imath]
I am trying to prove the above with the conditions that N is normal to G and [imath]G/N[/imath] is abelian. So then for [imath]Na,Nb\in{G/N}, NaNb=NbNa[/imath]. But [imath]NaNb=Nab=Nba[/imath], so then we know [imath]ab=ba[/imath], so G is abelian. So if [imath]ab\in{G}, N=abN(ab)^{-1}=abN(ba)^{-1}=baNa^{-1}b^{-1}....[/imath] and i'm getting lost... |
2164322 | Evaluate [imath]\lim_{n \to \infty} \frac{1}{n}\int_{0}^{n}\frac{x\ln(1+\frac{x}{n})}{1+x}dx[/imath]
I'd like to find [imath]\lim_{n \to \infty} \frac{1}{n}\int_{0}^{n}\frac{x\ln(1+\frac{x}{n})}{1+x}dx[/imath] I am wondering if it is true that it converges to [imath]\ln2[/imath]? thanks in advance | 2159872 | fine the limit : [imath]\lim_{ n \to \infty }\frac{1}{n}\int_{0}^{n}{ \frac{x\ln(1+\frac{x}{n})}{1+x}}=?[/imath]
fine the limit : [imath]\lim_{ n \to \infty }\frac{1}{n}\int_{0}^{n}{ \frac{x\ln(1+\frac{x}{n})}{1+x}}=?[/imath] My Try: in the http://www.integral-calculator.com [imath]I=\int_{}^{}{ \frac{x\ln(1+\frac{x}{n})}{1+x}}=\left(x+n\right)\ln\left(\left|x+n\right|\right)+\left(\ln\left(n\right)-\ln\left(n-1\right)\right)\ln\left(\left|x+1\right|\right)+\operatorname{Li}_2\left(-\dfrac{x+1}{n-1}\right)+\left(-\ln\left(n\right)-1\right)+c[/imath] now? |
2164198 | Solve the system of equations
I found system of equations on internet, I want to practice math and to solve it. All [imath]x[/imath],[imath]y[/imath] and [imath]z[/imath] are integers. [imath]\begin{cases} x+y+z=3 \\ x^3+y^3+z^3=3 \end{cases} [/imath] Should I start by watching all combinations that sum up to 3, please give me some hint to start. | 1312183 | Find all Integral solutions to [imath]x+y+z=3[/imath], [imath]x^3+y^3+z^3=3[/imath].
Suppose that [imath]x^3+y^3+z^3=3[/imath] and [imath]x+y+z=3[/imath]. What are all integral solutions of this equation? I can only find [imath]x=y=z=1[/imath]. |
2163113 | How do you get [imath]{X_i}+{X_j}{\sim}Binom(m,{p_i}+{p_j}) [/imath]
How do you get[imath] {X_i}+{X_j}{\sim}Binom(m,{p_i}+{p_j}) \\ from \\f(x_i,x_j)={m!\over{x_i!x_j!(m-x_i-x_j)!}}(p_i)^{x_i}(p_j)^{x_j}(1-p_i-p_j)^{m-x_i-x_j}\large? [/imath] [imath] My \; best \; try:[/imath] The only thing I thought of was manipulating the formula to [imath]f(x_i,x_j)=({m \choose x_i+x_j}(p_i+p_j)^{x_i+x_j}(1-(p_i+p_j)^{m-(x_i+x_j)})({ {x_i+x_j}\choose{x_i}}({p_i\over p_i+p_j})^{x_i}(1-{p_i\over p_i+p_j})^{(x_i+x_j)-x_i})=f(x_i+x_j)f(x_i|x_{k\neq i,j}) \\ with \\ f(x_i+x_j)\sim Binom(m,x_i+x_j) \; and \; f(x_i|x_{k\neq i,j}) \sim Binom(x_i+x_j,({p_i \over p_i+p_j})) \\[/imath] Though I feel like this is wrong and I need something more to show the distribution for [imath]X_i+X_j[/imath] | 571392 | Finding a distribution from the sum of two random variables in a joint trinomial distribution
Let X and Y have a trinomial distribution with join probability function: [imath] f(x,y) = \frac{n!}{x!y!(n-x-y)!}p^x q^y (1-p-q)^{n-x-y}; x=0,1,...n, y=0,1,...n[/imath] Let [imath]T=X+Y[/imath] I can't seem to figure out where to go exactly. Here's where I am at: [imath]\sum_{x=0}^{n} \frac{n!}{x!y!(n-x-y)!}p^x q^y (1-p-q)^{n-x-y}\mbox{ substitute in Y=T-X}[/imath] [imath]\sum_{x=0}^{n} \frac{n!}{x!(t-x)!(n-t)!}p^x q^{t-x} (1-p-q)^{n-t}[/imath] I think I could I figure out the algebra.. if I could just see what to do with the summation. I know I have to involve a t somewhere, but I'm not sure. Maybe somebody could just show me how to set up my summation correctly with the substitution? |
2165118 | Can these two limits be used to directly check the twice differentiability of a function?
Can the equality of these two be used as a second-order differentiability check?: [imath]\lim_{h\downarrow 0}\frac{f(x+2h)-2f(x+h)+f(x)}{h^2}[/imath] And, [imath]\lim_{h\downarrow 0}\frac{f(x-2h)-2f(x-h)+f(x)}{h^2}[/imath] I think that just like when the limits [imath]\lim\limits_{h\downarrow 0}\frac{f(x+h)-f(x)}{h}[/imath] and [imath]\lim\limits_{h\downarrow 0}\frac{f(x-h)-f(x)}{-h}[/imath] are equal, then it means that the function is differentiable, the equality of these two limits should mean that the function is twice differentiable. Please note that [imath]h[/imath] itself is assumed to be positive in the limits. | 855391 | Can the limit [imath]\lim_{h \to 0}\frac{f(x + h) - 2f(x) + f(x - h)}{h^2}[/imath] exist if [imath]f'(x)[/imath] does not exist at [imath]x[/imath]?
The second derivative of [imath]f[/imath] can be written as [imath]f''(x) = \lim_{h \to 0}\frac{f(x + h) - 2f(x) + f(x - h)}{h^2}[/imath] while it can also be written as (in fact, I believe this is the definition of [imath]f''(x)[/imath]): [imath]f''(x) = \lim_{h \to 0}\frac{f'(x + h) - f'(x)}{h}[/imath] From the second expression, it seems clear that if [imath]f'(x)[/imath] does not exist, then [imath]f''(x)[/imath] cannot exist as well. However, the first expression does not involve [imath]f'(x)[/imath], so this got me wondering: If [imath]f'(x)[/imath] does not exist at [imath]x[/imath] , can the limit [imath]\lim_{h \to 0}\frac{f(x + h) - 2f(x) + f(x - h)}{h^2}[/imath] at [imath]x[/imath] exist? How would one prove/disprove this? If this were to be true, what would be some examples? |
2164940 | Proving that function is bounded when its continous and its limits at infinity are bounded.
Let f be a function from real numbers to real numbers and let f(x) be a continous function, such that: [imath]\lim_{x\rightarrow\infty}f(x)[/imath] [imath]\lim_{x\rightarrow-\infty}f(x)[/imath] are bounded. How do I prove that f(x) is bounded? | 1118478 | Prove that a continuous real function with finite limits is bounded
[imath]f: \mathbb{R} \rightarrow \mathbb{R}[/imath] is a continuous function. Assume that [imath]\lim_{x \rightarrow \pm \infty} f(x)[/imath] exist and are finite. Prove that [imath]f[/imath] is bounded. So to show that [imath]f[/imath] is bounded, I must show that [imath]\exists M \in \mathbb{R}[/imath] such that [imath]|f(x)| \leq M[/imath] for all [imath]x \in Dom(f)[/imath]. Assuming that [imath]f(x)[/imath]'s limit exist, then I know that [imath]\forall \epsilon > 0[/imath], [imath]\exists N[/imath] such that for [imath]n > N[/imath], [imath]|f(n) - L| < \epsilon[/imath]. But I am having trouble with figuring out how to connect the two ideas together. |
2165015 | Prove If [imath]S \subset R[/imath] is a nonempty set, bounded from above, then for every [imath]\epsilon >0[/imath] there exists [imath] x\in S[/imath]
Prove If [imath]S \subset R[/imath] is a nonempty set, bounded from above, then for every [imath]\epsilon >0[/imath] there exists [imath] x\in S[/imath] such that [imath](\sup S) - \epsilon < x \le \sup S[/imath] How can I prove for this.. ? I always use this proposition, but I dont know how to prove. | 2158894 | Let [imath]A[/imath] be a nonempty subset of [imath]\mathbb{R}[/imath]. Prove...
I'm suppose to prove the following... Let [imath]A[/imath] be a nonempty subset of [imath]\mathbb{R}[/imath]. If [imath]\alpha = \sup A [/imath] is finite show that for each [imath] \epsilon > 0 [/imath] there is an [imath]a \in A [/imath] such that [imath]\alpha - \epsilon < a\leq \alpha [/imath]. This is what I have so far... Proof: We claim that [imath]\alpha = \sup A [/imath] Then by definition of a supremum [imath]\alpha [/imath] is the least upper bound of set A Thus [imath]\forall a \in A[/imath], [imath]a \leq \alpha[/imath] Suppose [imath]\epsilon > 0[/imath] Then [imath]\alpha - \epsilon < \alpha [/imath]. Thus [imath]\alpha - \epsilon < a\leq \alpha[/imath]. I feel like I'm missing some step between my last and second to last step so what am I forgetting? Also is the rest of my proof right? |
2164773 | General equation of a conic.
I would like to know the derivation of the general equation of a conic: [imath]Ax^{2} + Bxy + Cy^{2} + Dx + E y + F = 0[/imath] I have searched over the internet, but I did not find any resource which doesn't make use of trigonometry. | 2096193 | Confusion about the conic equation
Let me start with some basic definitions: Definition 1. A conic section is the curve resulting from the intersection of a plane and a cone. Definition 2. A conic section is the set of all points in a plane with the same eccentricity with respect to a particular focus and directrix. Definition 3. A conic section is the set of points [imath](x,y)[/imath] satisfying the implicit formula [imath]Ax^2+Bxy+Cy^2+Dx+Ey+F=0[/imath] All these definitions are familiar to me and they are taken from a paper of Mzuri S. Handlin, Conic Sections Beyond [imath]\mathbb{R}^2[/imath]. Now, in by university textbook I have the following theorem, Theorem. Any conic has an equation of the form [imath]Ax^2+Bxy+Cy^2+Dx+Ey+F=0[/imath] where [imath]A,B,C,D,E[/imath] and [imath]F[/imath] are real numbers and [imath]A,B[/imath] and [imath]C[/imath] are not all zero. Conversely, the set of all points in [imath]\mathbb{R}^2[/imath] whose coordinates [imath](x,y)[/imath] satisfy an equation of the form that above, is a conic. Proof to this theorem is omitted, and reference are not included. I've been searching for a proof, but failed to find one. It seems that the above theorem make motivate us to define definition 3, right? If one can provide me a reference for the proof of the theorem or sketch guidelines for possible proof? Thank you. |
2163599 | Comparing a sequence of fractions with a fraction
Let's assume that every a belongs to real numbers set and s is a sum of all the numbers from [imath]a_1, a_2...a_n[/imath]. How would you prove this? [imath]\frac{a_1}{s - a_1}[/imath] + [imath]\frac{a_2}{s - a_2}[/imath] ...+[imath]\frac{a_n}{s - a_n}\ge \frac{n}{n - 1}[/imath] | 125010 | Prove [imath]\sum_{i=1}^{i=n}\frac{a_i}{S-a_i}\geqslant \frac{n}{n-1}[/imath] and [imath]\sum_{i=1}^{i=n}\frac{S-a_i}{a_i}\geqslant n(n-1)[/imath].
Let [imath]a_1,a_2,\ldots ,a_n[/imath], be positive real numbers and S = [imath]a_1+a_2+a_3+\cdots+a_n[/imath]. Use the Cauchy-Schwarz inequality to prove that [imath]\sum_{i=1}^{i=n}\frac{a_i}{S-a_i}\geqslant \frac{n}{n-1}[/imath] and [imath]\sum_{i=1}^{i=n}\frac{S-a_i}{a_i}\geqslant n(n-1)[/imath] Hint: Apply the inequality to: [imath]\sum \frac{S}{a_i}[/imath] and [imath]\sum \frac{a_i}{S}[/imath] for one case, and [imath]\sum a_i(S-a_i)[/imath] and [imath]\sum \frac{a_i}{S-a_i}[/imath] for the other case. |
2148081 | Proof that all 2 dimensional vectors can be written as...
(a) Show that any two-dimensional vector can be expressed in the form [imath]s \begin{pmatrix} 3 \\ -1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 7 \end{pmatrix},[/imath]where [imath]s[/imath] and [imath]t[/imath] are real numbers. (b) Let u and v be non-zero vectors. Show that any two-dimensional vector can be expressed in the form [imath]s u + t v,[/imath]where [imath]s[/imath] and [imath]t[/imath] are real numbers, if and only if of the vectors [imath]u[/imath] and [imath]v[/imath], one vector is not a scalar multiple of the other vector. I know that we have to prove that [imath]3s+2t=a[/imath] [imath]-s+7t=b[/imath] for any integers a and b. But I don't know how to prove it. As for part(b), I do not have a starting point. | 1427377 | Two-dimensional Vectors Problem
(a) Show that any two-dimensional vector can be expressed in the form [imath]s \begin{pmatrix} 3 \\ -1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 7 \end{pmatrix},[/imath] where [imath]s[/imath] and [imath]t[/imath] are real numbers. (b) Let [imath]\vec{u}[/imath] and [imath]\vec{v}[/imath] be non-zero vectors. Show that any two-dimensional vector can be expressed in the form [imath]s \vec{u} + t \vec{v},[/imath] where [imath]s[/imath] and [imath]t[/imath] are real numbers, if and only if of the vectors [imath]\vec{u}[/imath] and [imath]\vec{v}[/imath], one vector is not a scalar multiple of the other vector. I've already gotten part (a) but (b) is a challenge. I have solved part a hough: Let's say that [imath] \begin{pmatrix} 3 \\ -1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 7 \end{pmatrix}[/imath] is equal to [imath]\binom{a}{b}[/imath]. \begin{align*} \begin{pmatrix} 3 \\ -1 \end{pmatrix} + t \begin{pmatrix} 2 \\ 7 \end{pmatrix} &= \binom{a}{b}\\ \binom{3s}{-s}+\binom{2t}{7t}&=\binom{a}{b}\\ \binom{3s+2t}{-s+7t}&=\binom{a}{b}\\ \Rightarrow 3s+2t&=a\\ -s+7t&=b \end{align*} |
2163939 | Simplification of a special iterated sum
Is it possible to simplify this iterated sum [imath]f(x)=\sum_{i_{u-1}=u}^{x}\sum_{i_{u-2}=u-1}^{i_{u-1}} \cdots \sum_{i_2=3}^{i_3} \sum_{i_1=2}^{i_2} \sum_{k=1}^{i_1} 1[/imath] by using a binomial coefficient? Is this iterated sum still equivalent of [imath]\frac{x^u}{u!}[/imath] ? This question is related to Iterated sums and asymptotics In this question, the initial indices are different. Thank you | 2149230 | Iterated sums and asymptotics
Let [imath]x[/imath] be a positive integer. I have the following iterated sums: [imath]f(x)=\sum_{i_{u-1}=1}^{x}\sum_{i_{u-2}=1}^{i_{u-1}} \cdots \sum_{i_2=1}^{i_3} \sum_{i_1=1}^{i_2}i_1[/imath] What is the asymptotic behavior of the total? Thank you |
2166367 | Solving trigonometric derivative problems
I have a function [imath]F(θ) = \sin^{−1} \sqrt{\sin(11θ)}[/imath] I derived the following answer using basic trigonometric and quotient rules. [imath]\dfrac{11\csc \left(\left(11\theta \right)^{\left(\frac{1}{2}\right)}\right)\cos \left(\left(11\theta \right)\right)}{2\sqrt{1-\sin \left(11\theta \right)}}[/imath] My answer however is wrong. I am having a very bad time with getting the solution for it. Can anyone outline how to go about getting the correct solution to a problem such as this? | 2166251 | Solving Trigonometric Derivatives
I have a function [imath]F(θ) = \sin^{−1} \sqrt{\sin(11θ)}[/imath] I derived the following answer using basic trigonometric and quotient rules. [imath]\dfrac{11\csc \left(\left(11\theta \right)^{\left(\frac{1}{2}\right)}\right)\cos \left(\left(11\theta \right)\right)}{2\sqrt{1-\sin \left(11\theta \right)}}[/imath] My answer however is wrong. Can anyone outline how to go about getting the correct solution to a problem such as this? |
2166381 | Mathematical Grammar, Converges to or Converges at, or both?
I lost a point on my test because I put after all my math work, "converges at [imath]0[/imath]" not "converges to [imath]0[/imath]". Should I argue with my teacher about this or is there a reason why I am wrong? | 76403 | Is "converges at" idiomatic English in some regions?
Some students write, e.g., "[imath]\sum(1/n^2)[/imath] converges at [imath]\pi^2/6[/imath]", where I would write "converges to". Are there regions of the English-speaking world where it is standard to say "converges at"? Or should this be considered a mistake made by non-native speakers of English? |
2166649 | Show the automorphism [imath]\phi={}^t(A^{-1})[/imath] is not inner for [imath]|F|>3[/imath]
I have a question that asks to show the automorphism [imath]\phi= {}^t(A^{-1})[/imath] is not inner for [imath]|F|>3[/imath] (The trace of the inverse of [imath]A[/imath]). For [imath]\phi[/imath] is be inner, it would have to be the case that there exists some [imath]B\in GL_n(F)[/imath] such that \begin{equation}\phi_B(A)=BAB^{-1}= {}^t(A^{-1})\quad\quad \forall A\in GL_n(F)\end{equation} The key to this problem lies in the fact that our field [imath]F[/imath] has more than 3 elements. I tried to use the above equation to show a contradiction, but I ended up with pages upon pages of calculations without anything becoming obvious to me about where the contradiction was, so then I thought perhaps I'm approaching this the wrong way. Can anyone offer a hint? | 1979382 | Transpose inverse automorphism is not inner
Consider the transpose inverse automorphism on [imath]GL_n(\mathbb F)[/imath] where [imath]n\geq2[/imath] and [imath]|\mathbb F|>2[/imath]. (i.e. [imath]\mathbb F[/imath] is a field, possibly infinite, with three or more elements). I want to show this automorphism is not inner. I was told to consider [imath]\det(BAB^{-1}) = \det(A)[/imath] and [imath]\det(\,^TA^{-1})=\det(A)^{-1}[/imath] and derive a contradiction. However, in some fields (where the result holds) I fail to see the contradiction. What about [imath]\mathbb F_3[/imath]? An element of [imath]GL_2(\mathbb F_3)[/imath] either has determinant [imath]1[/imath] or [imath]2[/imath], both of which are their own multiplicative inverses in [imath]\mathbb F_3[/imath]. I'm not sure if my knowledge of fields and determinants is flawed or the hint is. |
2166966 | Showing [imath]f(x)=ax+b[/imath] is a bijection for [imath]a\ne 0[/imath]
Let [imath]a,b\in \mathbb R[/imath] with [imath]a\neq 0[/imath]. We shall denote by [imath]f_{a,b}[/imath] the map from [imath]\mathbb R[/imath] to [imath]\mathbb R[/imath] such that [imath]f(x)=ax+b[/imath] is a bijection. How can I show that [imath]f_{a,b}[/imath] is a bijection? I understand for something to be bijective it must be injective and surjective, so do I need to show it has those properties instead? | 1230226 | How to prove that [imath]G(x)=ax+b[/imath] is a one-to-one correspondence where [imath]a\neq0[/imath] and [imath]a,b\in\mathbb{R}[/imath].
[imath]G(x) = ax + b[/imath] where [imath]a[/imath] is not equal to [imath]0[/imath] and [imath]b[/imath] are real numbers. Prove [imath]G[/imath] is a one-to-one correspondence. I understand that for every [imath]a[/imath] there is a corresponding [imath]b[/imath]-value that does not repeat; however, I do not understand how I can prove this. |
2167235 | Induction - request for help in proving lemma
Could anyone help with proving the following lemma, please? Let: [imath]n\in \mathbb{N}[/imath], [imath]Z_{n}^{*}:=\{k\in\mathbb{N}: k\in\{1,\dots,n\} \wedge \space GCD(k,n)=1\}[/imath]. Then: [imath]\forall n\in \mathbb{N} \space \forall p \in \mathbb{P}: |Z_{p^{n}}^{*}|=p^{n}-p^{n-1}[/imath] I tried to prove this by induction with respect to [imath]n[/imath], but I stuck at general case. I know how induction works, but I can't see how to do main point... | 2144324 | New, elegant proofs for [imath]\varphi(p^{k})=p^{k}-p^{k-1}[/imath]
Are there any short, elegant proofs known for the identity [imath]\varphi(p^{k})=p^{k}-p^{k-1}[/imath] ? (Here [imath]\varphi[/imath] is Euler's totient function and [imath]p[/imath] is a prime.) The standard combinatorial proof goes like this: In the set [imath]\left\{ 1,2\ldots,p^{k}\right\} [/imath] there in total [imath]p^{k}[/imath] number. Split this set into [imath]p[/imath] subsets [imath]\left\{ 1,\ldots,p\right\} [/imath], [imath]\left\{ p+1,\ldots,2p\right\} \ldots[/imath] Then in each of these sets there is only one number -- namely the one of the form [imath]m\cdot p[/imath] for some suitable [imath]m[/imath], that divides [imath]p^{k}[/imath]. There are in total [imath]\frac{p^{k}}{p}=p^{k-1}[/imath] such sets, so in total [imath]p^{k-1}[/imath]-many number from [imath]\left\{ 1,2\ldots,p^{k}\right\} [/imath] divide [imath]p^{k}[/imath]. Thus [imath](p^{k}-p^{k-1})[/imath]-many numbers are coprime to [imath]p^k[/imath], which proves the identity. [imath]\square[/imath] (A different proof that is often encountered assumes that we know that [imath]\varphi(n)=n\prod_{p\mid n}(1-\frac{1}{p})[/imath], from which our identity follows immediately. But this is actually a longer proof, since proving the auxiliary identity is longer.) Surprisingly, I would have imagined that there are tons of wildly different proofs of such a basic fact out there, but a preliminary internet seach as well as book skimming returned only (minor variations of) these two proofs. EDIT The present proofs are more or less reformulations (very polished with details hidden as good as possible - but still reformulations) of my first proofs. What I'm looking for are more radically different approaches (if these exist). |
2165386 | Show that [imath]R[/imath] is a ring
Let [imath]R[/imath] a set with [imath]+[/imath] and [imath]\cdot [/imath] s.t. 1) [imath](R,+)[/imath] is a group (not necessarily comutative) 2) [imath]\cdot [/imath] is associative, and distributive for [imath]+[/imath], i.e. [imath]a\cdot (b\cdot c)=(a\cdot b)\cdot c[/imath], [imath]a\cdot (b+c)=a\cdot b+a\cdot c[/imath] and [imath](a+b)\cdot c=a\cdot c+b\cdot c[/imath] 3) there is [imath]1\in R[/imath] s.t. [imath]1\cdot x=x\cdot 1=x[/imath]. Show that [imath]R[/imath] is a ring. I really don't know how to did it. I think that I have to show that [imath](R,+)[/imath] is commutative, but I didn't success. | 346375 | Commutative property of ring addition
I have a simple question answer to which would help me more deeply understand the concept of (non)commutative structures. Let's take for example (our teacher's definition of) a ring: Let [imath]R\neq \emptyset[/imath] be a set, let [imath]\oplus:A\times A \to A[/imath] and [imath]\bullet :A\times A \to A[/imath] be binary operations. Moreover, let [imath](R, \oplus)[/imath] be a commutative group, [imath](R, \bullet)[/imath] be a monoid and following property holds for all [imath]a, b, c\in R[/imath]: [imath]a\bullet(b\oplus c) = (a\bullet b)\oplus(a \bullet c)[/imath] [imath](b\oplus c)\bullet a = (b\bullet a)\oplus(c \bullet a)[/imath] Then ordered triple [imath]\mathbf R = (R, \oplus, \bullet \mathbf)[/imath] is called a (unitary) ring. Moreover, we call ring [imath]\mathbf R[/imath] commutative iff [imath](R, \bullet)[/imath] is a commutative monoid. Commutativity of a ring is always a matter of its multiplicative operation because the additive operation is always assumed to be commutative. Could anyone explain me the bold part? Why do we even in non-commutative rings (and fields etc.) assume the addition to be always commutative? Is there some serious reason? Would it make any trouble? Or studying of structures with non-commutative addition just doesn't give us anything new so we can take addition as commutative simply because of our comfort? |
2161380 | Subspace of dual space is annihilator?
Let [imath]V[/imath] be a finite dimensional vector space. I'm trying to see if any subspace [imath]A[/imath] of [imath]V^*[/imath] must be an annihilator of some subspace in [imath]U[/imath] of [imath]V[/imath]. I.e. I'm trying to see if [imath]A=U^0[/imath] for some subspace [imath]U\subseteq V[/imath]. So first I let [imath]A[/imath] be a subspace of [imath]V^*[/imath]. Then I denote the set of all [imath]v\in V[/imath] such that [imath]\phi(v)=0[/imath] for every [imath]\phi\in A[/imath] by [imath]U[/imath]. Lemma: [imath]U[/imath] is a subspace. Proof: Clearly [imath]0\in U[/imath]. Let [imath]u,v \in U[/imath] and [imath]\lambda \in \Bbb F[/imath]. Then for every [imath]\phi \in A[/imath] [imath]\phi(u+\lambda v) = \phi(u) + \lambda\phi(v) = 0 \implies u+\lambda v\in U[/imath] So [imath]U[/imath] is indeed a subspace of [imath]V[/imath]. [imath]\square[/imath] Because [imath]V[/imath] is finite dimensional, so is [imath]V^*[/imath] and thus [imath]A[/imath]. Let [imath]\{\phi_1, \dots, \phi_k\}[/imath] be a basis for [imath]A[/imath]. Let [imath]u\in U[/imath]. Then [imath]\phi_i(u)=0[/imath] by definition. Thus [imath]\phi_i\in U^0[/imath] for all [imath]i=1,\dots k[/imath]. So [imath]A\subseteq U^0[/imath]. So the only part that I still need is to show the reverse inclusion or that [imath]\dim A = \dim U^0[/imath]. But I can't figure out how to do it. Here are my problems with the duplicate question (or more specifically its answers): user218931 uses the double dual which is introduced in a later exercise. Rafael Deiga uses [imath]\dim(\operatorname{null}\phi_1 \cap \cdots \cap \operatorname{null}\phi_m) = \dim V-m[/imath] which is proven in a later exercise. Aweygan uses some functional analysis theorem called the Hahn-Banach theorem. So all of the answers use tools that I don't currently have in my mathematical toolkit. And yes, I realize that there's nothing theoretically wrong with using a result from later in the problem set as long as the proof of that result doesn't rely on this one, but I feel like Axler wouldn't have written out the exercises so that you needed to do that. So to me it feels sorta like cheating to do them out of order. Does anyone see a way to finish this problem without using the above three results? | 1879498 | Every subspace of the dual of a finite-dimensional vector space is an annihilator
Exercise 26 page 115 of Linear Algebra Done Right by Sheldon Axler is the following: Suppose [imath]V[/imath] is finite-dimensional and [imath]\Gamma[/imath] is a subspace of [imath]V'[/imath]. Show that [imath]\Gamma=\{v\in V:\varphi(v)=0\text{ for every }\varphi\in\Gamma\}^0[/imath] where [imath]V'[/imath] is the dual space of [imath]V[/imath] and, for any [imath]S\subset V[/imath], [imath]S^0[/imath] is the annihilator of [imath]S[/imath]. Attempt: Let [imath]S=\{v\in V:\varphi(v)=0\text{ for every }\varphi\in\Gamma\}[/imath]. Clearly, [imath]\Gamma\subset S^0[/imath].I tried to show that [imath]S^0\subset\Gamma[/imath] using some bases of [imath]V[/imath] and [imath]V'[/imath], but I failed. I also tried to show that [imath]\dim{\Gamma}=\dim{V}-\dim{S}=\dim{S^°}[/imath]. If [imath]s_1,\dots,s_n[/imath] is a basis of [imath]S[/imath] and [imath]s_1',\dots,s_n'[/imath] its dual, and if [imath]\psi_1,\dots,\psi_p[/imath] is a basis of [imath]\Gamma[/imath], it's easy to see that [imath]s_1',\dots,s_n',\psi_1,\dots,\psi_p[/imath] is a linearly independent list of [imath]V'[/imath]. Also, if you extend [imath]s_1,\dots,s_n[/imath] to a basis [imath]s_1,\dots,s_n,v_1,\dots,v_m[/imath] of [imath]V[/imath], then its dual [imath]s_1',\dots,s_n',v_1',\dots,v_m'[/imath] is a basis of [imath]V'[/imath]; in fact [imath]S^0=\text{span}\{v_1',\dots,v_m'\}[/imath]. Two remarkable facts:[imath]\forall i\in[1,m],\,\exists j\in[1,p],\,\psi_j(v_i)\neq0[/imath] because [imath]v_j\notin S[/imath], and [imath]\forall i\in[1,p],\,\exists j\in[1,m],\,\psi_i(v_j)\neq 0[/imath] because [imath]\psi_i\neq 0[/imath]; in other words the matrix of the inclusion map from [imath]\Gamma[/imath] to [imath]V'[/imath] with respect to the basis of [imath]\Gamma[/imath] and the dual base of the chosen basis of [imath]V[/imath] has no [imath]0[/imath] row nor [imath]0[/imath] column. But this doesn't seem to provide a way to prove that any linear comination of the [imath](v_i')_{1\le i\le m}[/imath] is a linear combination of the elements of the basis of [imath]\Gamma[/imath]. I believe that [imath]v_1,\dots, v_m[/imath] should be chosen more carefuly but I fail to. Could you please help me? Thank you in advance! |
554809 | Proving that the Intersection of any Collection of Compact Sets is Compact
I was trying to do this problem this way: Let [imath]\mathcal{B}=\{B_i\}[/imath] be a collection of compact sets. By Heine-Borel, each of the [imath]B_i[/imath]'s are closed and bounded. We already know that the intersection of a collection of closed sets is once again closed, so [imath]\cap \mathcal{B}[/imath] is closed. If I can prove [imath]\cap\mathcal{B}[/imath] is bounded as well, I can use Heine-Borel to say it is compact. So I was not sure how to show it was bounded, but a friend showed me his way of doing it. He did it the following way: Let [imath]b,a\in\mathbb{R}[/imath] be an upper and lower bound for [imath]B_1[/imath], respectively. Since [imath]\cap\mathcal{B}\subset B_1, a[/imath] is a lower bound for [imath]\cap \mathcal{B}[/imath] and [imath]b[/imath] is an upper bound for [imath]\cap\mathcal{B}[/imath]. Therefore, [imath]\cap\mathcal{B}[/imath] is bounded and then by Heine-Borel, it is compact. My question is on the paragraph right above. I don't understand why what he did means that [imath]a[/imath] and [imath]b[/imath] are lower and upper bounds for [imath]\cap\mathcal{B}.[/imath] Can anyone explain why this is? Thanks. | 3012490 | If [imath](X,d)[/imath] is a metric space, prove that intersection of any collection of compact sets in [imath](X,d)[/imath] is compact.
If [imath](X,d)[/imath] be a compact metric space then, arbitrary intersection of compact subsets is compact. Is it true if [imath](X,d)[/imath] is a metric space, but not compact? There is similar type of questions already have been asked before, but those are from real-analysis where any closed and bounded set is compact. But in metric space closed and bounded set may not be compact. Hence this question is different from previously asked questions. |
2168340 | Prove: [imath]f(x)=0, \forall x\in [0,\frac{1}{2}][/imath] if [imath]|f'(x)|\le |f(x)[/imath] and [imath]f(0)=0[/imath]
Let [imath]f:[0,\frac{1}{2}] \rightarrow\mathbb{R}[/imath] be a differentiable function such that [imath]|f^\ {'}(x)|\leq |f(x)|[/imath] and [imath]f(0)=0.[/imath] Prove: [imath]f(x)=0, \forall x\in [0,\frac{1}{2}][/imath]. So I tried to work by the definition of the derivative at [imath]0[/imath] and somehow squeeze it, but got stuck. Any help appreciated. | 473342 | Differentiable and Continuous functions on [0,1] with 'weird' conditions.
I've been stuck on this one for a while. Comes from an analysis qual question. Let f be a function that is continuous on [imath]\left[0,1\right][/imath] and differentiable on [imath](0,1)[/imath]. Show that if [imath]f(0)=0[/imath] and [imath]|f'(x)| \leq |f(x)|[/imath] for all [imath]x \in (0,1)[/imath], then [imath]f(x)=0[/imath] for all [imath]x \in \left[0,1\right][/imath]. What I've tried doing so far is see if there was anything I could do with MVT. I didn't really see anything to do with definitions either..to which I have a feeling I'll be playing around with them. Drawing a picture was a little difficult with these conditions as well Any hints/suggestions? |
1117732 | Order of [imath]\frac{2}{3}+\mathbb{Z}[/imath] in [imath]\mathbb{Q}/\mathbb{Z}[/imath]
Let [imath]\mathbb{Q}/\mathbb{Z}[/imath] be the quotient group of the additive group of rational numbers. Find the order of the element [imath]\frac{2}{3}+\mathbb{Z}[/imath] in [imath]\mathbb{Q}/\mathbb{Z}[/imath]. I tried it by using facts that any quotient [imath]G/H[/imath] of [imath]G[/imath] has induced operation from [imath]G[/imath]. So I can do [imath]\frac{2}{3}+\mathbb{Z} + \frac{2}{3}+\mathbb{Z} = \frac{4}{3}+\mathbb{Z} + \frac{2}{3}+\mathbb{Z} = 2+\mathbb{Z} =\mathbb{Z},[/imath] and [imath]\mathbb{Z}[/imath] being identity we get order three. But this way, to manually start computing elements, if correct is very unreliable in case of more difficult problem. So is there a generalized approach for quotient groups? | 197421 | how to find the order of an element in a quotient group
Consider the quotient group [imath]\mathbb{Q}/\mathbb{Z}[/imath] of the additive group of rational numbers. Then how to find the order of the element [imath]2/3 + \mathbb{Z} [/imath] in [imath]\mathbb{Q}/\mathbb{Z}[/imath]. |
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