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1338965 | Show that $(r!)^2 ≡ (−1)^{r−1} \pmod p,\ r = (p-1)/2$
I need to prove that if p is an odd prime and [imath]r = (p-1)/2[/imath] then [imath](r!)^2 ≡ (−1)^{r−1} \pmod p[/imath] I think it has something to do with gauss's lemma https://en.wikipedia.org/wiki/Gauss%27s_lemma_(number_theory) but I tried a lot and couldn't find a way to break it . any help or hint will be appreciated. | 24095 | $(p\!-\!1\!-\!h)!\,h! \equiv (-1)^{h+1}\!\!\pmod{\! p}\,$ [Wilson Reflection Formula]
Suppose that [imath]p[/imath] is a prime. Suppose further that [imath]h[/imath] and [imath]k[/imath] are non-negative integers such that [imath]h + k = p − 1[/imath]. I want to prove that [imath]h!k! + (−1)^h \equiv 0 \pmod{p}[/imath] My first thought is that by Wilson's theorem, [imath](p-1)! \equiv -1 \pmod{p}[/imath], and [imath]h!k![/imath] divides [imath](p-1)![/imath] (definition of a binomial). Where would I go from here? |
1339306 | If [imath]S[/imath] and [imath]T[/imath] are closed vector subspaces then [imath]S+T[/imath] is closed
Let [imath]V[/imath] be a Banach normed space, [imath]S,T \subset V[/imath] be closed vector subspaces. Assume [imath]\operatorname{dim}(T)<\infty[/imath]. Show that [imath]S+T[/imath] is closed. So I encountered this problem trying to use induction in [imath]\operatorname{dim}(T)[/imath]. So first of all, for [imath]\operatorname{dim}(T)=1[/imath], take [imath]span(t_0)=T[/imath] Take [imath]x_n \subset S+T[/imath], [imath]x_n=s_n + \lambda_n t_0 \to x[/imath], with [imath]s_n \in S[/imath], [imath]\lambda_n \in \Bbb{R}[/imath] for every [imath]n\in \Bbb{N}[/imath], I want to see that [imath]x \in S+T[/imath] I proceeded by dividing in two cases, [imath]\lambda_n \to \infty[/imath], in this case [imath]\|x_n - x\|=\|s_n+\lambda_n t_0 - x\|<\epsilon[/imath] for [imath]n[/imath] sufficiently big. Dividing each term by [imath]\lambda_n[/imath] (wich we suppose different than [imath]0[/imath]) we obtain [imath]\|\frac{s_n}{\lambda_n}+t_0-\frac{x}{\lambda_n}\|=\|s'_n+t_0-\frac{x}{\lambda_n}\|<\frac{\epsilon}{\lambda_n}[/imath] Taking [imath]\lambda_n \to \infty[/imath] we have that [imath]s'_n \to t_0[/imath] and since [imath]s'_n \in S[/imath] for all [imath]n[/imath] then [imath]t_0 \in S[/imath] and therefore [imath]T\subset S[/imath] wich implies, [imath]S=S+T[/imath] closed. So, now I have to take care of the other case, where [imath]\lambda_n[/imath] is bounded. However, Im having trouble, since I can't prove that [imath]\lambda_n[/imath] actually converges. (I don't even know if this is true). However I'm not being able to prove in this case that [imath]S+T[/imath] is closed. And this is just the base step. So when trying to prove for [imath]n \Rightarrow n+1[/imath] I don't have much expectations.. Am I on the right track? Is there another, more elegant way to solve this? | 548840 | Let [imath]E[/imath] be a Banach space, prove that the sum of two closed subspaces is closed if one is finite dimensional
Let [imath]E[/imath] be a Banach space and let [imath]S[/imath] and [imath]T[/imath] be closed subspaces, with dim[imath]\space T<\infty[/imath]. Prove that [imath]S+T[/imath] is closed. To prove that [imath]S+T[/imath] is closed I have to show that for any limit point [imath]x[/imath] of [imath]S+T[/imath], [imath]x \in S+T[/imath]. So let [imath]x[/imath] be a limit point that subspace. Then, there is a sequence [imath]\{x_n\}_{n \in \mathbb N} \subset S+T[/imath] : [imath]x_n \rightarrow x[/imath]. [imath]x_n \in S+T \space \forall \space n \in \mathbb N[/imath]. This means that [imath]x_n=s_n+t_n[/imath] with [imath]s_n \in S[/imath] and [imath]t_n \in T[/imath] [imath]\forall \space n \in \mathbb N[/imath]. Well, I couldn't go further than this. [imath]\{s_n\}_{n \in \mathbb N}[/imath] and [imath]\{t_n\}_{n \in \mathbb N}[/imath] are sequences in [imath]S[/imath] and [imath]T[/imath] respectively. I would like to prove that both of them converge to some points [imath]s[/imath] and [imath]t[/imath], for if this is the case, the hypothesis would assure [imath]s \in S[/imath] and [imath]t \in T[/imath]. Then, I would have to prove that [imath]x=lim x_n=lim s_n+t_n=lim s_n+limt_n=s+t[/imath] and from all the previous steps I would conclude [imath]x \in S+T[/imath]. [imath]$[/imath][imath]Maybe I could start by proving that [/imath]\{t_n\}_{n \in \mathbb N}[imath] is convergent. But I don't see why this is true, I've tried to prove that [/imath]\{t_n\}_{n \in \mathbb N}$ is a Cauchy sequence (in a Banach space this would imply that the sequence converges) but I couldn't. |
1339359 | Uniform distribution on [imath]\{1,\dots,7\}[/imath] from rolling a die
This was a job interview question someone asked about on Reddit. (Link) How can you generate a uniform distribution on [imath]\{1,\dots,7\}[/imath] using a fair die? Presumably, you are to do this by combining repeated i.i.d draws from [imath]\{1,\dots,6\}[/imath] (and not some Rube-Goldberg-esque rig that involves using the die for something other than rolling it). | 901222 | Simulate repeated rolls of a 7-sided die with a 6-sided die
What is the most efficient way to simulate a 7-sided die with a 6-sided die? I've put some thought into it but I'm not sure I get somewhere specifically. To create a 7-sided die we can use a rejection technique. 3-bits give uniform 1-8 and we need uniform 1-7 which means that we have to reject 1/8 i.e. 12.5% rejection probability. To create [imath]n * 7[/imath]-sided die rolls we need [imath]\lceil log_2( 7^n ) \rceil[/imath] bits. This means that our rejection probability is [imath]p_r(n)=1-\frac{7^n}{2^{\lceil log_2( 7^n ) \rceil}}[/imath]. It turns out that the rejection probability varies wildly but for [imath]n=26[/imath] we get [imath]p_r(26) = 1 - \frac{7^{26}}{2^{\lceil log_2(7^{26}) \rceil}} = 1-\frac{7^{26}}{2^{73}} \approx 0.6\%[/imath] rejection probability which is quite good. This means that we can generate with good odds 26 7-die rolls out of 73 bits. Similarly, if we throw a fair die [imath]n[/imath] times we get number from [imath]0...(6^n-1)[/imath] which gives us [imath]\lfloor log_2(6^{n}) \rfloor[/imath] bits by rejecting everything which is above [imath]2^{\lfloor log_2(6^{n}) \rfloor}[/imath]. Consequently the rejection probability is [imath]p_r(n)=1-\frac{2^{\lfloor log_2( 6^{n} ) \rfloor}}{6^n}[/imath]. Again this varies wildly but for [imath]n = 53[/imath], we get [imath]p_r(53) = 1-\frac{2^{137}}{6^{53}} \approx 0.2\%[/imath] which is excellent. As a result, we can roll the 6-face die 53 times and get ~137 bits. This means that we get about [imath]\frac{137}{53} * \frac{26}{73} = 0.9207[/imath] 7-face die rolls out of 6-face die rolls which is close to the optimum [imath]\frac{log 7}{log6} = 0.9208[/imath]. Is there a way to get the optimum? Is there an way to find those [imath]n[/imath] numbers as above that minimize errors? Is there relevant theory I could have a look at? P.S. Relevant python expressions: min([ (i, round(1000*(1-( 7**i ) / (2**ceil(log(7**i,2)))) )/10) for i in xrange(1,100)], key=lambda x: x[1]) min([ (i, round(1000*(1- ((2**floor(log(6**i,2))) / ( 6**i )) ) )/10) for i in xrange(1,100)], key=lambda x: x[1]) P.S.2 Thanks to @Erick Wong for helping me get the question right with his great comments. Related question: Is there a way to simulate any [imath]n[/imath]-sided die using a fixed set of die types for all [imath]n[/imath]? |
1338801 | How to prove Raabe's Formula
For quite some time, I've been trying to prove Raabe's Formula, or in other words: [imath]\int_a^{a+1} \ln\bigg(\Gamma(t)\bigg)dt=\dfrac{1}{2}\ln(2\pi)+a\ln(a)-a[/imath] This is how I tried: [imath]I(s)=\int_a^{a+1}\ln\bigg(s\Gamma(t)\bigg)dt[/imath]Differentiating with respect to [imath]s,[/imath] [imath]I'(s)=\int_a^{a+1}\dfrac{\Gamma(t)}{s\Gamma(t)}dt=\int_a^{a+1} \dfrac{dt}{s}[/imath] However, at this point I stopped thinking I must have made a mistake because I was told that proving Raabe's Formula was really difficult, and this seemed too simple a method to prove Raabe's Formula. [imath]$[/imath]$ I would be grateful if somebody would be so kind as to tell me how to prove this result, as well as what went wrong with my method. Many, many thanks in advance! | 784529 | Integral [imath]\int_0^1 \log \left(\Gamma\left(x+\alpha\right)\right)\,{\rm d}x=\frac{\log\left( 2 \pi\right)}{2}+\alpha \log\left(\alpha\right) -\alpha[/imath]
Hi I am trying to prove[imath] I:=\int_0^1 \log\left(\,\Gamma\left(x+\alpha\right)\,\right)\,{\rm d}x =\frac{\log\left(2\pi\right)}{2}+\alpha \log\left(\alpha\right) -\alpha\,,\qquad \alpha \geq 0. [/imath] I am not sure whether to use an integral representation or to somehow use the Euler reflection formula [imath] \Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin \pi z} [/imath] since a previous post used that to solve this kind of integral. Other than this method, we can use the integral representation [imath] \Gamma(t)=\int_0^\infty x^{t-1} e^{-x}\, dx. [/imath] Also note [imath]\Gamma(n)=(n-1)![/imath]. |
1339258 | How to derive the formula for the sum of the first [imath]n[/imath] perfect squares?
How do you derive the formula for the sum of the series when [imath]S_n = \sum_{j=1}^n j^2[/imath]? The relationship [imath](n, S_n)[/imath] can be determined by a polynomial in [imath]n[/imath]. You are supposed to use finite differences to determine the degree and then use a calculator to derive the function. But how? | 1692308 | Derive a polynomials formula by setting [imath]n = 1[/imath], [imath]n = 2[/imath], [imath]n = 3[/imath] to determine the coefficients.
Derive a formula for the sum of squares [imath]1^2 + 2^2 + 3^2 + … + n^2[/imath]. Hint: assume the formula is a polynomial of degree 3, i.e. [imath]an^3 + bn^2 + cn + d[/imath], and use the cases of [imath]n=0, n=1, n=2[/imath], and [imath]n=3[/imath] to determine its coefficients. |
1339094 | Find the Nearest Point of $[imath]y[/imath]$ from a Probability Simplex.
I need to compute the nearest point of [imath]x[/imath] from a probability simplex. Formally, I want to ask if there is a close form for the solution to the following optimization problem: for [imath]y\in \mathbb{R}^k[/imath], [imath]\min\limits_{x} \|x-y\|^2_2 [/imath] [imath]s.t. \ \sum\limits_{i=1}^k x_i =1,\ x_i \geq 0 ,\ \forall i \in[k][/imath] If not, how can I derive an iterative algorithm to find the unique solution? Thanks! | 46129 | Projecting a Non Negative Vector onto the Simplex
Given an elementwise nonnegative vector [imath]y[/imath], I'd like to find the projection of [imath]y[/imath] onto the simplex [imath]S: \{ (x_1, \ldots, x_n) ~|~ \sum_{i=1}^n x_i=1, x_i \geq 0 \mbox{ for all } i \}[/imath]. Is there a closed form expression for this? If not, I need to write a computer program which will compute this projection; is there something simple I could do to compute this? Simplicity is more important to me than running time; I don't want to spend a long time coding this. I do realize this is a convex optimization problem and could be solved by using various optimization solvers, but that seems like overkill. |
1138950 | Mean value theorem for the second derivative, when the first derivative is zero at endpoints
Suppose [imath]f:[a,b]\to \mathbb R[/imath] has derivative up to order [imath]2[/imath], and [imath]f'(a)=f'(b)=0[/imath]. Prove there is a point [imath]c[/imath] at [imath](a,b)[/imath] such that [imath] |f''(c)|\geq 4\frac{|f(b)-f(a)|}{(b-a)^2}. [/imath] If it was factor 2, not 4, then I could use a Taylor expansion with Lagrange residue. | 2331402 | Spivak Calculus- Chapter 11, question 25.
I'm working through the Spivak Calculus book, and am having some difficulty with the following problem. Assume that [imath]f[/imath] is a second differentiable function, [imath]f(0) = 0[/imath], [imath]f(1) = 1[/imath], and [imath]f'(0) = f'(1) = 0[/imath]. Prove that [imath]\exists c\in [0,1][/imath] such that [imath]|f''(c)|\geq 4[/imath]. The hint that the text provides is prove that either [imath]f''(x) > 4[/imath] for some [imath]x\in [0, \frac{1}{2}][/imath] or [imath]f''(x) < -4[/imath] for some [imath]x\in [\frac{1}{2}, 1][/imath]. Using the Mean Value theorem, it's easy enough to prove that that [imath]\exists c_1\in (0,1)[/imath] and [imath]f'(c_1) = \frac{f(1) - f(0)}{1-0} = 1[/imath]. From here, you can apply MVT to [imath]f'[/imath] on [imath][0,c_1][/imath] and [imath][c_1, 1][/imath] giving [imath]\exists c_2\in (0,c_1)[/imath] where [imath]f''(c_2) = \frac{1}{c_1}[/imath] and [imath]\exists c_3\in (c_1, 1)[/imath] where [imath]f''(c_3) = \frac{-1}{1-c_1}[/imath]. From here, using the assumption that [imath]c_1 \leq \frac{1}{4}[/imath] yields [imath]f''(c_2) \geq 4[/imath] or [imath]c_1 \geq \frac{3}{4}[/imath] yields [imath]f''(c_3) \leq -4[/imath]. However, I am unsure of what to do if [imath]\frac{1}{4} < c_1 < \frac{3}{4}[/imath]. I have been told that Cauchy's Mean Value Theorem (sometimes called the Generalized Mean Value Theorem) is the way to go concerning this problem, but I'm unable to make it work. Any thoughts would be greatly appreciated. |
1339805 | How to prove this sequence tends to zero
Suppose [imath]a_n>0[/imath],[imath]\sum a_n[/imath] converges, [imath]\{na_n\}_{\Bbb N}[/imath] is monotonic, prove [imath]\lim na_n\ln n=0[/imath] My attempt so far has shown that [imath]\{na_n\}[/imath] is decresing: otherwise, [imath]na_n\ge a_1\implies a_n\ge\frac{a_1}{n}\implies \sum a_n=+\infty[/imath] which contradicts the convergence of [imath]\sum a_n[/imath]. My friend has got a bit further: [imath](n+1)a_{n+1}< na_n\implies \frac{a_{n+1}}{a_n}<\frac{n}{n+1}[/imath]. He then let [imath]b_n:=\frac{a_n}{\frac{1}{n\ln n}}[/imath] and therefore [imath]\frac{b_{n+1}}{b_n}=\frac{a_{n+1}}{a_n}\cdot\frac{(n+1)\ln(n+1)}{n\ln n}<\frac{\ln(n+1)}{\ln n}[/imath] However he can't prove the monotony of [imath]b_n[/imath] because the RHS of the inequality is too weak. In fact it only shows [imath]\limsup \frac{b_{n+1}}{b_n}\le 1[/imath], which is not effective here. And I can't come up with a stronger version, either. Can you help me? Any direct help or hint will be appreciated. Thanks! | 134604 | Prove [imath]\lim\limits_{n\to\infty}na_n\ln(n)=0[/imath]
Let [imath]a_n>0[/imath],[imath]\sum a_n[/imath] is convergent,[imath]na_n[/imath] is monotone, Prove: [imath]\lim\limits_{n\to\infty}na_n\ln(n)=0[/imath] I try to prove [imath]a_n=o(n\ln(n))[/imath], but it doesn't work. |
416581 | How [imath]\frac{1}{n}\sum_{i=1}^n X_i^2 - \bar X^2 = \frac{\sum_{i=1}^n (X_i - \bar X)^2}{n}[/imath]
How [imath]\frac{1}{n}\sum_{i=1}^n X_i^2 - \bar X^2 = \frac{\sum_{i=1}^n (X_i - \bar X)^2}{n}[/imath] i have tried to do that by the following procedure: [imath]\frac{1}{n}\sum_{i=1}^n X_i^2 - \bar X^2[/imath] =[imath]\frac{1}{n}(\sum_{i=1}^n X_i^2 - n\bar X^2)[/imath] =[imath]\frac{1}{n}(\sum_{i=1}^n X_i^2 - \sum_{i=1}^n\bar X^2)[/imath] =[imath]\frac{1}{n} \sum_{i=1}^n (X_i^2 - \bar X^2)[/imath] Then i have stumbled. | 2589248 | square of sum inside summation notation
I am looking to find out how the derivation went in this computation [imath] \frac{1}{n-1} \sum (x_i - \bar{x})^2 = \frac{1}{n-1} \left( \sum x_i^2 - n\bar{x}^2 \right) [/imath] The exercise belongs to sample distribution section but that's not what bothers me. As you can see there's a square notation inside a summation notation. Now I would take square of sum's here as in [imath](a-b)^2 = a^2 - 2ab + b^2[/imath] . Yet the result in the picture seems to ommit [imath]2ab[/imath] or probably say simplify. My question is how does he get to such a derivation. Is there any sort of special formula or some point that I am severely missing :/ ? |
1340414 | How do I integrate [imath]\frac{\sin x+\cos x}{\sin^4 x+\cos^4 x}[/imath]
How do I integrate [imath]\frac{\sin x+\cos x}{\sin^4 x+\cos^4 x}[/imath] ? Tried different ways including the tangent half-angle substitution (which seems to be disastrous). | 1331762 | How to evaluate [imath]\int \frac { \sin x+\cos x }{ \sin^4 x+\cos^4x}\, dx[/imath]?
How can one find [imath]\int \frac { \sin x+\cos x }{ \sin^4 x+\cos^4x}\, dx?[/imath] |
1340156 | Calculation of determinant using its properties
The task is to calculate the following determinant by using the properties of a determinant: [imath]\begin{vmatrix} n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2 \\ (n+2)^2 & (n+3)^2 & (n+4)^2 \\ \end{vmatrix}[/imath] Hint: The result is [imath]-8[/imath]. | 676586 | How to find the determinant of this matrix?
Today at my linear algebra exam, there was this question that I couldn't solve. There was a matrix [imath]A[/imath] [imath]A=\begin{bmatrix} n^{2} & (n+1)^{2} &(n+2)^{2} \\ (n+1)^{2} &(n+2)^{2} & (n+3)^{2}\\ (n+2)^{2} & (n+3)^{2} & (n+4)^{2} \end{bmatrix} [/imath] and we had to prove that [imath]\det(A)=-8[/imath]. Clearly, calculating the determinant, with the matrix as it is, wasn't the right way. The calculations went on and on. But I couldn't think of any other way to solve it. Is there any way to simplify [imath]A[/imath], so as to calculate the determinant? |
1267788 | Why do we invert then multiply when dividing fractions?
I have looked up various sites online. They only explain it in a very basic arithmetic way. Would someone explain to me why is [imath]\displaystyle\frac{x}{\frac{1}{2}}[/imath] is [imath]\displaystyle\frac{2x}{1}[/imath]? | 2029069 | Why do we 'invert and multiply' when dividing fractions?
I read in a textbook that the explanation to this rule lies in the fact that division is the reverse operation to multiplication. Unfortunately, the author did not elaborate on this point. Based on this, can someone help me to understand why [imath]\frac{a}{b}[/imath] / [imath]\frac{c}{d}[/imath] = [imath]\frac{a}{b}[/imath] * [imath]\frac{d}{c}[/imath]? |
1340617 | Basis for [imath]\mathbb{R}^{\infty}[/imath]
It follows from Zorn's Lemma that every vector space [imath]V[/imath] has a basis (this means, a subset [imath]B[/imath] of [imath]V[/imath] that generate any [imath]v \in V[/imath] by means a finite linear combination, and such that [imath]B[/imath] is LI) . But, seems quite impossible to me to find a basis for [imath]\mathbb{R}^{\infty}[/imath]; I have listened somewhere that is really impossible find an actual basis for this space, because of something that I don't remember. There exist a concrete basis? There exists some reason for we never shall find one? | 122571 | Is there a constructive way to exhibit a basis for [imath]\mathbb{R}^\mathbb{N}[/imath]?
Assuming the Axiom of Choice, every vector space has a basis, though it can be troublesome to show one explicitly. Is there any constructive way to exhibit a basis for [imath]\mathbb{R}^\mathbb{N}[/imath], the vector space of real sequences? |
1340192 | Very easy question about infinitesimals
how can I prove that: [imath] \lim_{x\to 0} \frac{e^{-1/x^2}}{x} = 0 ? [/imath] I suppose that the exponential "goes" to [imath]0[/imath] faster than linear, but I'm not sure. | 1337367 | Solution of limit [imath]\lim\limits_{x\to 0} \frac {e^{-1/x^2}}{x} [/imath]
Small question, I'm trying to solve this limit but I just can't wrap my head around this problem. [imath]\lim_{x\to 0} \frac {e^{-1/x^2}}{x} [/imath] L'Hopital just seems to make it messier. It's probably pretty simple - I'd like to hear what I'm missing. |
1340700 | A ring isomorphic to its square
Is there an example of a ring [imath]A[/imath] (with unity) which is isomorphic as unital rings to [imath]A\times A[/imath]? Any such ring can't have invariant basis number so in particular can't be commutative. | 895428 | Can [imath]R \times R[/imath] be isomorphic to [imath]R[/imath] as rings?
I know from this question that [imath]R \times R[/imath] can be isomorphic to [imath]R[/imath], as [imath]R[/imath]-modules. But can they ever be isomorphic as rings? |
1340125 | Find inverse [imath]f^{-1}[/imath] of a function [imath]f(x,y)=(x-y,x-10y)[/imath]
I know how to find inverse function if the given function is in the explicit form. Could someone show on this example how to find [imath]f^{-1}[/imath]? Thanks for replies. | 1339987 | Examine if function [imath]f:\mathbb{R^2}\rightarrow \mathbb{R^2}[/imath] which is defined as [imath]f(x,y)=(2x-y,x-4y)[/imath] is bijective. If bijective, find [imath]f^{-1}[/imath].
Function is bijective when it is injective and surjective. Function is injective if [imath](\forall x_1,x_2 \in A)f(x_1)=f(x_2)\Rightarrow x_1=x_2[/imath] and surjective if [imath](\forall y \in B)(\exists x \in A)y=f(x)[/imath] Is it possible to transform a function, so that it is given in explicit form? I don't know how to manipulate with function in this form ([imath]f(x,y)=(2x-y,x-4y)[/imath]). Thanks for replies. |
1340738 | Why is [imath]\sqrt{x^2}= |x|[/imath] rather than [imath]\pm x[/imath]?
Shouldn't the square root of a number have both a negative and positive root? According to Barron's, [imath]\displaystyle \sqrt{x^2} = |x|[/imath]. I don't understand how. | 258876 | Proving square root of a square is the same as absolute value
Lets say I have a function defined as [imath]f(x) = \sqrt {x^2}[/imath]. Common knowledge of square roots tells you to simplify to [imath]f(x) = x[/imath] (we'll call that [imath]g(x)[/imath]) which may be the same problem, but it isn't the same equation. For example, say I put [imath]-1[/imath] into them: [imath]\begin{align} f(x) &= \sqrt {x^2} \\ f(-1) &= \sqrt {(-1)^2} \\ f(-1) &= \sqrt {1} \\ f(-1) &= 1 \end{align}[/imath] [imath]\begin{align} g(x) &= x \\ g(-1) &= -1 \end{align}[/imath] thereby, we conclude that [imath]f(x)[/imath] and [imath]g(x)[/imath] do not produce the same results even though they are mathematically the same. This is also shown when we try to graph the functions: [imath]y = \sqrt {x^2}[/imath]: [imath]y = x[/imath]: [imath]y = \mid x \mid[/imath]: From this, we can see that given [imath]f(x) = \sqrt {x^2}[/imath], when simplified is not the same as [imath]f(x) = x[/imath]. So, is there any way to prove that [imath]y = \sqrt {x^2}[/imath] is not the same as [imath]y = x[/imath] for negative values, but is infact the same as [imath]y = \mid x \mid[/imath]? |
1339190 | Are there fewer positive integers than all integers?
In our 6th grade math class we got introduced to the concept of integers. With all the talk about positive and negative, it got me wondering. Is the amount of elements in [imath]\mathbb{Z^+}[/imath] less than the amount of elements in [imath]\mathbb{Z}[/imath]? Here is how I thought of it. If we have [imath]\mathbb{Z^+}[/imath] and add one element to the "back" of it ([imath]\mathbb{Z}^{\geq-1}[/imath]) there are certainly more elements in that new set, so there must be more elements in [imath]\mathbb{Z}[/imath] than in [imath]\mathbb{Z}^+[/imath] but on the other side if we try to express the amount of elements in them "numerically" (In a loose sense of the word) they both have [imath]\infty[/imath] elements. So what is the right answer? Is there even one? (Please note I am a bit of a layman when it comes to mathematics) Edit: About the possible duplicate, I am not looking for a bijection between the two sets, I am looking for if they even have the same number of elements and why a bijection shows that they do/don't | 873927 | How to show the integers have same cardinality as the natural numbers?
How would I show the following have a bijection. Which is one to one and onto ofcourse. [imath]\mathbb{Z}\rightarrow \mathbb{N}[/imath] I know I need to find a function. But I do not see the pattern that makes integer go to natural. I already did this problem from natural to integer but now I want to see it done this way. I have listed some numbers [imath],..,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,..[/imath] [imath]1,2,3,4,5,6,7,8,9,10,11,12,13,14,.[/imath] |
1339941 | Using a number system other than the decimal
Travelling in the Kingdom of Crystal Skull, Indiana Jones discovered a small box with notation [imath] 3 x^2 - 25 x + 66 = 0 \implies x_1 = 4,\; x_2 = 9, [/imath] which seems to be incorrect. However after some reflection he realised a number system other than the decimal was used to determine the roots of the quadratic equation. What is the base of that number system? | 460729 | What would have been our number system if humans had more than 10 fingers? Try to solve this puzzle.
Try to solve this puzzle: The first expedition to Mars found only the ruins of a civilization. From the artifacts and pictures, the explorers deduced that the creatures who produced this civilization were four-legged beings with a tentatcle that branched out at the end with a number of grasping "fingers". After much study, the explorers were able to translate Martian mathematics. They found the following equation: [imath]5x^2 - 50x + 125 = 0[/imath] with the indicated solutions [imath]x=5[/imath] and [imath]x=8[/imath]. The value [imath]x=5[/imath] seemed legitimate enough, but [imath]x=8[/imath] required some explanation. Then the explorers reflected on the way in which Earth's number system developed, and found evidence that the Martian system had a similar history. How many fingers would you say the Martians had? [imath](a)\;10[/imath] [imath](b)\;13[/imath] [imath](c)\;40[/imath] [imath](d)\;25[/imath] P.S. This is not a home work. It's a question asked in an interview. |
1341026 | construction field [imath]F_n[/imath]
I know that if [imath]F_n[/imath] is a finite field, then [imath]n[/imath] should be a prime power. I want some sources in order to learn how I can construct finite fields [imath]F_n[/imath] for such an [imath]n[/imath]. | 51738 | Constructing a finite field
I'm looking for constructive ways to obtain finite fields, for any given size [imath]q=p^n[/imath]. For example, I know it suffices to find an irreducible polynomial of degree [imath]n[/imath] over [imath]\mathbb{Z}_p[/imath] (and then obtain the field as its quotient ring), but how can such polynomial be efficiently found? Also, I know there are more ways to represent elements of finite fields - are they easier to use than the irreducible polynomial method? What is done in practice in computational mathematical libraries? |
1341311 | Finding value of an exponential equation
If [imath] x^2=3x-1[/imath] then find the value of [imath](x^6+1)/x^3[/imath] I used the quadratic but it became too complicated | 1341237 | Finding value of an expression
If [imath]x^2-3x-1=0[/imath] then find the value of [imath](x^6+1)/x^3[/imath] I tried to solve the quadratic but it became too complicated any way of doing this without a calculator |
1341985 | Prove that the following function is [imath]C^\infty[/imath]
Prove that the following function is [imath]C^\infty[/imath] (and in the point [imath]ξ=0[/imath]) : [imath]f:\Bbb R\to\Bbb C:\xi\mapsto {e^{i\cdot\xi\cdot λ}-1\over i\cdot\xi}-λ[/imath] for whichever [imath]λ>0[/imath] I am trying to find a proof for this elementary problem that has been asked in a mathematical competition in France in 2004.Maybe can we find a proof? | 1296668 | Prove that the function [imath]\xi\in R \mapsto {e^{i\cdot \xi\cdot λ}-1\over i\cdot \xi}-λ[/imath] is [imath]C^{\infty}[/imath]
Prove that the following function is [imath]C^\infty[/imath] in the point [imath]\xi=0[/imath]: [imath]f:\Bbb R\to\Bbb C:\xi\mapsto {e^{i\cdot\xi\cdot λ}-1\over i\cdot\xi}-λ[/imath] Any ideas how to prove this? I am trying to think some ideas but I cannot find any way to prove it. |
1341964 | Can we generalize Aleph numbers to non integer values?
I'm really new to those kind of arguments so don't call me mad but I was wondering if there is a way to define an infinite set which cardinality is an Aleph-number like: [imath]\aleph_{\frac 12}[/imath]. I have a sort of passion to generalization of math objects and I've been really astonished when I learned about fractals, whose dimensions are not an integer number so I was thinking if there is some way to build infinite sets (maybe with an unusual list of axioms) whose cardinality is not an Aleph-number with integer index and how those numbers and sets would work. Any help or reference is appreciated :) | 215977 | "Real" cardinality, say, [imath]\aleph_\pi[/imath]?
Is there any meaningful definition to afford for [imath]\aleph_r[/imath] (as in cardinality) where [imath]r\in\mathbb{R}^+[/imath]? [imath]r\in\mathbb{C}[/imath]? What about [imath]\aleph_{\aleph_0}[/imath]? Can we iterate this? [imath]\aleph_{\aleph_{\aleph_{\cdots}}}[/imath] I may be throwing in bunch of rather naive/basic questions, for I haven't learnt much about infinite cardinalities. If I am referring to bunch of stuff abundantly dealt in established areas, please kindly point out. |
1340347 | If [imath]a,b,c>0, a+b+c=3[/imath], minimize [imath]\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{3-c^3}{c}[/imath]
Let [imath]a,b,c[/imath] be positive real numbers such that [imath]a+b+c=3[/imath]. Find the minimum value of the expression [imath]A= \frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{3-c^3}{c}[/imath] I tried solving it, but I got nothing | 1341100 | Find the minimum value of [imath]A=\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}[/imath]
Let [imath]a, b[/imath] and [imath]c[/imath] three positive real numbers such that [imath]a+b+c=3[/imath]. Find the minimum value of [imath]A=\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}.[/imath] Here is my attempt. By symmetry we can assume that [imath]a\leq b\leq c[/imath]. The function [imath]f(x)=\frac{2-x^3}{x}[/imath] is decreasing and convex on [imath]]0,2^{1/3}][/imath]. So if [imath]c\leq 2^{1/3}[/imath] then [imath]3f(1)=3f(\frac{a+b+c}{3})\leq f(a)+f(b)+f(c)=A[/imath] with equality if [imath]a=b=c=1[/imath]. If [imath]c>2^{1/3}[/imath] I don't see how to proceed. |
1342373 | If [imath]R[/imath] is a commutative ring, the nilpotents form necessarily an ideal of [imath]R[/imath]?
This is an algebra question from an exam a few years ago: Let [imath]R[/imath] be a ring, and let [imath]N = \{a \in R: a^n = 0 \text{ for some } n \in \mathbb{N}, (n \text{ depends on } a) \}[/imath]. Prove or disprove: If [imath]R[/imath] is commutative, then [imath]N[/imath] is an ideal of [imath]R[/imath]? So, clearly [imath]N[/imath] is nonempty and it is easy to show that it is closed under multiplication. However, for showing that [imath]N[/imath] is closed under subtraction, I believe this is not possible for we do have that [imath]R[/imath] is commutative, however, we don't have that [imath]R[/imath] is of characteristic [imath]p[/imath] and so the freshman's dream is just that...a dream. So, I would conclude that [imath]N[/imath] is not an ideal. Am I right about this? I don't really understand the solution given in: If [imath]x,y[/imath] are nilpotent and commute, [imath]x+y [/imath] is nilpotent. and also isn't it not necessarily true that [imath]a^m + b^n = (a+b)^{m+n}[/imath]? | 953170 | If [imath]x,y[/imath] are nilpotent and commute, [imath]x+y [/imath] is nilpotent.
Let [imath]A[/imath] a ring. Supose that [imath]x,y \in A[/imath] are nilpotents elements and that [imath]xy=yx[/imath]. Prove that [imath]x+y[/imath] is nilpotent. |
1342385 | The limit of iterated square root with multiplication under the root, [imath]\sqrt{ a \sqrt{ a \sqrt{a \cdots}}}[/imath]
[imath] \sqrt{ a \sqrt{ a \sqrt{a \cdots}}}=\text{ ?} [/imath] options were given as [imath]0[/imath] [imath]-a[/imath] [imath]a[/imath] [imath]1[/imath] i did not know how to solve it or what it was related to. Could anyone please explain the concept and/or provide helpful references or links? | 589288 | [imath]\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}[/imath] approximation
Is there any trick to evaluate this or this is an approximation, I mean I am not allowed to use calculator. [imath]\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}[/imath] |
1343293 | Calculating [imath]\lim_{x\to\infty} (\sin\frac{1}{x}+\cos\frac{1}{x})^x[/imath] without l'Hopital
Calculate [imath]\lim_{x\to\infty} (\sin\frac{1}{x}+\cos\frac{1}{x})^x[/imath] without using l'Hopital's rule. I attempted pulling something out to get a limit that resembles [imath]e[/imath], this gave me: [imath]\sin^x\frac{1}{x}(1+\frac{1}{\tan\frac{1}{x}})^x=\sin^x\frac{1}{x}\bigg[(1+\frac{1}{\tan\frac{1}{x}})^{\tan\frac{1}{x}}\bigg]^{\frac{x}{\tan\frac{1}{x}}}[/imath] But then the first part goes to [imath]0[/imath] and the exponent goes to infinity, and I'm kinda stumped. | 1310890 | What is [imath]\lim_{x\to\infty}\left(\sin{\frac 1x}+\cos{\frac 1x}\right)^x[/imath]?
[imath]\lim_{x\to\infty}\left(\sin{\frac 1x}+\cos{\frac 1x}\right)^x[/imath] It is about the [imath](\to1)^{(\to\infty)}[/imath] situation. Can we find its limit using the formula [imath]\lim_{x\to\infty}(1+\frac 1x)^x=e[/imath]? If yes, then how? |
1343701 | process stochastics and branching process
Consider a discrete time branching process [imath]X_{n}[/imath] with [imath]X_{0}=1.[/imath] Establish the simple inequality [imath]P\{X_{n}>L\ \textrm{for some}\ 0\leq n\leq m\ |\ X_{m}=0 \}\leq [P\{X_{m}=0\}]^L[/imath] Note: This issue is the second edition of the book "The First Course in Stochastic Processes" Samuel Karlin author. Could anyone help me to solve it? The only idea I have is: [imath]P(X_{n}>L|X_{m}=0)= \dfrac{P(X_{n}>L,X_{m}=0)}{P(X_{m}=0)}=\dfrac{P(X_{m}=0|X_{n}>L)P(X_{n}>L)}{P(X_{m}=0)}=?[/imath] | 1343236 | Conditional probability branching process
Consider a discrete time branching process [imath]X_{n}[/imath] with [imath]X_{0}=1.[/imath] Establish the simple inequality [imath]P\{X_{n}>L\ \textrm{for some}\ 0\leq n\leq m\ |\ X_{m}=0 \}\leq [P\{X_{m}=0\}]^L[/imath] Note: This issue is the second edition of the book "The First Course in Stochastic Processes" Samuel Karlin author. Could anyone help me to solve it? I have no idea to answer it! I thank you! |
1343637 | Implies in a truth table, unclear.
In my textbook, we have the following truth table: [imath]P[/imath] true and [imath]Q[/imath] true means that "[imath]P \implies Q[/imath]" is true. [imath]P[/imath] true and [imath]Q[/imath] false means that "[imath]P \implies Q[/imath]" is false. [imath]P[/imath] false and [imath]Q[/imath] true means that "[imath]P \implies Q[/imath]" is true. [imath]P[/imath] false and [imath]Q[/imath] false means that "[imath]P \implies Q[/imath]" is true. I'm confused by what this means? I don't see why [imath]P[/imath] true and [imath]Q[/imath] false is the only combination that gets us that false and the rest of them get us true. Could anyone help clarify what's going on? Why is what is in the textbook correct? Thanks. | 243949 | Equivalence of [imath]a \rightarrow b[/imath] and [imath]\lnot a \vee b[/imath]
Is there a proof for the logical equivalence of [imath]a \rightarrow b[/imath] and [imath]\lnot a \vee b[/imath]? |
1343633 | Primitive roots of unity
Let [imath]R[/imath] be the set of primitive [imath]42^{\text{nd}}[/imath] roots of unity, and let [imath]S[/imath] be the set of primitive [imath]70^{\text{th}}[/imath] roots of unity. How many elements do [imath]R[/imath] and [imath]S[/imath] have in common? How would you work this out? Thanks | 647031 | Complex numbers and Roots of unity
I have no clue how to begin these problems. How do I start? I don't think I should pound em out...Thanks. Let P be the set of [imath]42^{\text{nd}}[/imath] roots of unity, and let Q be the set of [imath]70^{\text{th}} [/imath] roots of unity. How many elements do P and Q have in common? Let P be the set of [imath]42^{\text{nd}} [/imath]roots of unity, and let Q be the set of [imath]70^{\text{th}} [/imath]roots of unity. What is the smallest positive integer n for which all the elements in P and all the elements in Q are [imath]n^{\text{th}}[/imath] roots of unity? |
1343975 | Why does this hyperboloid change into a surface?
Given this equation [imath]x^2+y^2+z^2+2xy+2xz+2yz-x-y-z=6[/imath] and the corresponding quadric: If I rearrange the equation to [imath](x+y+z-3)(x+y+z+2)=0[/imath] (which is equivalent), I get: So, which is the right quadric to my equation? Why does it have two representations? | 1343257 | What geometric object is given by this equation?
What geometric object is given by this equation? [imath]x^2+y^2+z^2+2xy+2xz+2yz-x-y-z-6=0[/imath] Maple says it's a hyperboloid of one sheet, but is there a way to show it without going the long way by using the principal axis theorem? |
895580 | what is the fundamental group of a torus with [imath]k[/imath] points removed
I was about to use the Seifert-van Kampen to compute the fundamental group of a torus of genus 2. In the process, I need to know the fundamental group of a torus (of genus one) with a hole removed. is there an obvious deformation retraction ? (I don't see it ... ) In general : How can I find the fundamental group of a torus with [imath]k[/imath] holes removed ? Thank you in advance, | 22980 | Fundamental group of a torus with points removed
Question 5.33 from "Topology and its Applications" by Baesner is to compute the fundamental group of the torus ([imath]T^2[/imath]) with [imath]n[/imath] points removed. I can "see" in my mind that if we remove one point we get a bouquet of two circles. Less clear is what happens when we remove two (or more) points. Any hints? |
1344594 | Set intersection of finite,nested sets of real numbers
I'm currently trying to write up a solution to the following problem: If [imath] \displaystyle A_1 \supseteq A_2 \supseteq A_3 ... [/imath] where each [imath] \displaystyle A_j [/imath] is a non-empty, finite set of real numbers, prove (or disprove) the intersection [imath]\displaystyle \bigcap_{n \in \mathbb{N}} A_n [/imath] is non-empty and finite. My thoughts: this is obviously true, as for some point onwards we must, for some [imath] \displaystyle N \in \mathbb{N}[/imath], have the following equality for all [imath] \displaystyle k \ge N[/imath]: [imath]\displaystyle A_k = A_N[/imath]. If this did not hold, then the only way it could fail would be for us to have strict super-set relations (at some stage), which can't continue indefinitely, for we know [imath] \displaystyle A_1[/imath] is finite, and hence each set is finite. However, I'm having trouble writing this up as a proper mathematical rigorous direct proof; my thought for a proof is something like follows. Proof. We prove this by contrapostion. It follows that we should consider either (i) the intersection is infinite, or (ii) the intersection is empty. If (i), then as the intersection is a subset of each [imath]\displaystyle A_j[/imath] for every [imath] \displaystyle j \in \mathbb{N}[/imath], it follows that, in particular, [imath] \displaystyle A_1[/imath] must be infinte, because it has an infinite subset. We now suppose (ii); then for any element, [imath] \displaystyle x [/imath], we cannot have [imath] \displaystyle x \in \bigcap_{N \in \mathbb{N}} A_n[/imath]. In particular, if [imath] \displaystyle A_1 = \{a_1, ..., a_k\}[/imath], we can find a [imath] \displaystyle j_1 \in \mathbb{N}[/imath] such that [imath]\displaystyle a_1 \not\in A_{j_1}[/imath], and generally find a [imath] \displaystyle j_s \in \mathbb{N}[/imath], with [imath]\displaystyle j_s > j_t[/imath] if [imath] \displaystyle t < s[/imath] such that [imath]\displaystyle a_s \not\in A_{j_s}[/imath], where [imath] \displaystyle 1 \le s \le k[/imath]. In particular, for [imath] \displaystyle s = k[/imath], by the fact the [imath] \displaystyle j's[/imath] are increasing, it follows [imath] \displaystyle a_1, ..., a_k \not\in A_{j_k}[/imath], and from the original superset relation, this set is empty. [imath] \displaystyle \square[/imath] My concerns: I think this is a valid proof, but I'd much rather prove it directly by using the first observation that eventually the subsets of [imath] \displaystyle A_1[/imath] are all equal. However, I'm finding this hard to explicitly, and basically end up writing the above argument to show that eventually the sets are empty. How can one prove this directly? | 47908 | Seeking a direct proof about nested inclusions
Statement of the problem: Prove that if [imath]A_1 \supseteq A_2 \supseteq A_3 \cdots[/imath] are all finite, nonempty sets of real numbers, then the intersection [imath]\bigcap_{n=1}^{\infty} A_n[/imath] is also finite and nonempty. My solution: [imath]\bigcap_{n=1}^{\infty} A_n = \emptyset.[/imath] Then at least one [imath]A_n[/imath] is disjoint from the rest - contradiction, since each [imath]A_n \subseteq A_{n-1}.[/imath] Next assume that the intersection has infinite cardinality - contradiction, since [imath]A_n \supseteq \bigcap_{n=1}^{\infty} A_n,[/imath] but we have [imath]A_n[/imath] finite. This is a homework problem and this is what I will be turning in regardless of any answers I get. I would like to know if there is a direct proof of this claim. I had considered trying to show that [imath]\exists x \in A_n[/imath] for each choice of [imath]n,[/imath] which would show that the intersection is nonempty, but: a) I don't know how to do this, and b) I still would not be able to prove the intersection was finite except by contradiction. Thanks! edit: Moreover, we were told not to use induction on this problem. But if [imath]A_1[/imath] is finite, can't we reduce the collection of [imath]A_n's[/imath] to a subset of the power set of [imath]A_1[/imath] and then induct? |
578312 | If [imath]0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0[/imath] is exact and [imath]B\simeq A\oplus C[/imath] as a [imath]R[/imath]-module, does this sequence split?
Suppose [imath]0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0[/imath] is a short exact sequence that [imath]B\simeq A\oplus C[/imath] as a [imath]R[/imath]-module. Does this short exact sequence split? I think the answer is no, but I must have a counterexample. I could not find any. Can you help me? | 135444 | A nonsplit short exact sequence of abelian groups with [imath]B \cong A \oplus C[/imath]
A homework problem asked to find a short exact sequence of abelian groups [imath]$$0 \rightarrow A \longrightarrow B \longrightarrow C \rightarrow 0$$[/imath] such that [imath]B \cong A \oplus C[/imath] although the sequence does not split. My solution to this is the sequence [imath]$$0 \rightarrow \mathbb{Z} \overset{i}{\longrightarrow} \mathbb{Z} \oplus (\mathbb{Z}/2\mathbb{Z})^{\mathbb{N}} \overset{p}{\longrightarrow} (\mathbb{Z}/2\mathbb{Z})^{\mathbb{N}} \rightarrow 0$$[/imath] with [imath]i(x)=(2x,0,0,\dotsc)[/imath] and [imath]p(x,y_1,y_2,\dotsc)=(x+2\mathbb{Z},y_1,y_2,\dotsc)[/imath]. My new questions: Is there an example with finite/finitely generated abelian groups? If the answer to[imath](1)[/imath]is negative, will it help to pass to general [imath]R[/imath]-modules for some ring [imath]R[/imath]? |
1344517 | After switching a lamp on and off infinitely many times in one minute, is it on or off?
So we have a lamp. It's switched on. let's represent its state of being switched on with associating it with [imath]1[/imath] and being off with [imath]-1[/imath]. after half a minute passes, you turn it off, after another quarter of a minute passes you turn it on and so on. Now this process will take : [imath]1/2+1/4+1/16+...=1[/imath] minute and this is the states of the lamp put into a sequence: [imath](1,-1,1,-1,1,...)[/imath] After a minute of switching it on an off passes, will it be on or off? I'm well aware that the limit of this sequence does not exist, however, my intuition dictates that after a minute passes, it is either on or off. So does this lamp have a final state(on or off) which is impossible for us to know? or it does not have a finals state(it seems to me, it must have one!)? | 445817 | Thomson's Lamp and the possibility of supertasks
The Thomson's Lamp paradox: A mad scientist owns a desk lamp. It begins in the toggled on position. The scientist toggles the lamp off after one minute, then on after another half-minute. After a quarter-minute the lamp is toggled off, then the scientist waits an eigth-minute and turns the lamp on again. The scientist continues toggling the lamp, waiting one-half of the previously waited time between toggles. After a total sum of two minutes of toggling, what is the state of the lamp (on or off)? The Wiki article states that supertasks are impossible and the lamp is neither on nor off after the two minutes. This does not make sense to me, as this would mean that the lamp is in a superposition of two states. Proof (1 is on, 0 is off): [imath]S = \sum \limits_{i=0}^n {(-1)^i}[/imath] [imath]S=1-1+1-1+1...[/imath] [imath]S=1-(1-1+1-1+1...)[/imath] [imath]S=1-S[/imath] [imath]S=\frac1 2[/imath] How can a macroscopic object like a lamp exist in such a state? |
1259433 | Prove that a continuous function of compact support defined on [imath]R^n[/imath] is bounded.
I am working through a few sets of notes I found on the internet and I came across this exercise. How do I prove that a continuous function [imath]f[/imath] of compact support defined on [imath]R^n[/imath] is bounded? It seems believable that it is true for [imath]f[/imath] in [imath]R[/imath] because I can visualize it but how do I prove this properly for [imath]R^n[/imath]? Any ideas? | 1344706 | Are continuous functions with compact support bounded?
While studying measure theory I came across the following fact: [imath]\mathcal{K}(X) \subset C_b(X)[/imath] (meaning the continuous functions with compact support are a subset of the bounded continuous functions). This seems somehow odd to me; I've tried to prove it but did not succeed. Could someone help me out here? Thanks! |
982422 | Ladder against a wall.
Having a bit of a problem with a question. There is a 4m ladder leaving against a wall. There is a box in between The ladder and wall. The box is a cubic metre. I have found a quartic to find the length up the wall the ladder will reach. But its proving difficult for me to solve. [imath]x^4 +2x^3 -14x^2 +2x +1=0[/imath] | 1344991 | The position of a ladder leaning against a wall and touching a box under it
I was reading a newspaper and there was a little math riddle, I thought "how funny, that's gonna be easy, let's do it" and here am I... The problem goes as follow : in a barn, there is a 1 meter cubic box against a wall and a 4 meter ladder is leaning against the wall, touching the box at its corner. Here is a picture : So, the big triangle has a hypotenuse [imath]FE[/imath] of [imath]4[/imath], the square [imath]ABDC[/imath] has sides of length 1 and is basically "insquared" at the right angle, i.e. [imath]D\in \overline{FE}[/imath]. The question is "what is the length of the biggest cathetus", here [imath]AF[/imath]. So far, no problem. Now here are my solutions: By Thales' intercept theorem, [imath]\frac{FB}{FA}=\frac{BD}{AE}[/imath], by hypothesis, [imath]FB=FA-1[/imath] and [imath]BD=1[/imath]. Now by Pythagoras, [imath]FA^2+AE^2=FE^2[/imath]; by hypothesis, [imath]FE=4[/imath], so we end up with a system of equations, letting [imath]h=FA, d=AE[/imath]: [imath] \begin{align} &\frac{h-1}{h}=\frac{1}{d} \\ &h^2+d^2=4^2 \end{align} [/imath] Which solves (removing 3 non-relevant solutions) into [imath]d \cong 1.3622[/imath] and [imath]h \cong 3.76091[/imath]. Now, if I consider the "function" of the line : [imath]f(x)=\frac{-h}{d}x+h[/imath], I know that [imath]f(1)=1[/imath] and I end up with Pythagoras with the system :[imath] \begin{align} &\frac{-h}{d}+h=1 \\ &h^2+d^2=4^2 \end{align} [/imath] it solves again into the same, again removing 3 non-relevant solutions Okay, this means that using Pythagoras is no good since it ends up giving a quartic equation (4 answers of which 3 are "non-relevant"). Now if I consider the length of the arc [imath]f(x)[/imath] between [imath]0[/imath] and [imath]d[/imath] it has to be [imath]4[/imath] and again [imath]f(1)=1[/imath] I end up with the system: [imath] \begin{align} &\frac{-h}{d}+h=1 \\ &\int_0^d \sqrt{1+(f'(x))^2} dx =\int_0^d \sqrt{1+\left(\frac{-h}{d}\right)^2} dx = d \sqrt{1+\left(\frac{-h}{d}\right)^2} \end{align} [/imath] Which solves again into the same answers, but this time removing only 2 non-relevant solutions (i.e. it gives a cubic equation instead of a quartic). I tried also using the areas and the smaller trangles [imath]FAD[/imath] and [imath]AED[/imath] for example : [imath]\frac{h \cdot d}{2} = \frac{h\cdot 1}{2}+\frac{d\cdot 1}{2}[/imath] Yet I wasn't able to get to any "hand solvable" solution : if I were able to bring it down to some quadratic equation, that would be nice, since it is a common assumption, here, that everybody has seen the "general formula for solving quadratic equations" in school and so would be able to solve this, I may then see how it is seen as a funny riddle in the newspaper. My best trial, with "just" a cubic equation, is way too complicated for the normal readers of this newspaper, so it's bugging me. What am I missing? Some basic properties maybe? It's really bugging me, not being able to solve this without Wolfram. |
1345142 | Prove that [imath]2AB[/imath] is square
Let [imath]A= 1! \cdot 2! \cdot 3! \cdots 1002![/imath] [imath]B= 1004!\cdot 1005! \cdots 2006![/imath] Prove that [imath]2AB[/imath] is square. Help guys, I tried, I really did but I couldn't. | 197555 | [imath]2AB[/imath] is a perfect square and [imath]A+B[/imath] is not a perfect square
If :[imath]A=1!2!\cdots 1002![/imath], and [imath]B=1004 ! 1005!\cdots2006![/imath], how to prove that: a) [imath]2AB[/imath] is a perfect square b) [imath]A+B[/imath] is not a perfect square |
1345645 | Show that the equation has a natural solution
let [imath]n[/imath] be a natural number and [imath]r[/imath] , [imath]s[/imath] be rational such that [imath]n=s^2+r^2[/imath] show that there are natural numbers a,b such that [imath]n=a^2+b^2[/imath] | 582188 | Is an integer a sum of two rational squares iff it is a sum of two integer squares?
Let [imath]a\in \mathbb Z[/imath]. Is it true that [imath]a[/imath] is a sum of two squares of rational numbers if and only if it is a sum of two squares of integers? I came to face this problem while dealing with quaternion algebras. |
1345628 | Finding [imath]\lim\limits_{n\to \infty}({1\over n+1}+{1\over n+2}+...+{1\over n+n})[/imath] using integrals
Finding [imath]\lim\limits_{n\to \infty}\left({1\over n+1}+{1\over n+2}+\dots+{1\over n+n}\right)[/imath]. I tried many things but it would work out. I am now studying calculus 2 (In my country the first calculus course is split, and I saw a solution while studying calculus 1.) However, I understood that questions with sum are usually solved like that: For example, [imath] \lim_{n\to \infty}{1\over n^{k+1}}({1^k+2^k+ \ldots +n^k})= \ldots =\lim_{n\to \infty}{1\over n}{\sum_{j=1}^{n}\Bigl({j\over n}\Bigr)^k}=\int_{0}^{1}{x^k}\,dx [/imath] but now I can't bring it to that form. I also can't verify my answer because I can't find other answers for it. I would really appreciate your help. Please don't mark my question is a duplicate before making sure that either the origin involves integrals or that it cannot be solved with integrals. | 1343499 | Monotonicity and convergence of the sequence [imath]a_n=\sum_{k=1}^{n}\frac{1}{k+n}[/imath]
Let we have the following sequence [imath](a_n)[/imath] such that [imath]a_n=\sum_{k=1}^n\frac{1}{n+k}[/imath] How can I prove that [imath](a_n)[/imath] is increasing bounded sequence, then prove it is convergent and find its limit? |
73296 | Example of a Borel set that is neither [imath]F_\sigma[/imath] nor [imath]G_\delta[/imath]
I'm looking for subset [imath]A[/imath] of [imath]\mathbb R[/imath] such that [imath]A[/imath] is a Borel set but [imath]A[/imath] is neither [imath]F_\sigma[/imath] nor [imath]G_\delta[/imath]. | 1499577 | A subset of [imath]\mathbb{R}[/imath] that neither is an [imath]F_\sigma[/imath] set nor a [imath]G_\delta[/imath] set
Can anyone give me a simple example of a subset of [imath]\mathbb{R}[/imath] which is neither an [imath]F_\sigma[/imath] set nor a [imath]G_\delta[/imath]. |
1345896 | Power set of [imath]\{\emptyset,\{\emptyset\}\}[/imath]
For writing the power set of [imath]\{\emptyset,\{\emptyset\}\}[/imath], do I have to consider [imath]\emptyset[/imath] as null set or as a member of the given set? If I consider [imath]\emptyset[/imath] as a member, then the power set is [imath]\{\{\emptyset,\{\emptyset\}\}, \emptyset, \{\emptyset\}, \{\{\emptyset\}\}\}[/imath]. If my assumption is right, is there a better way to present the answer? Thanks in advance! | 67496 | Powerset of set with empty sets
I am working on a problem with the following set: [imath]S = \{\varnothing,\{\varnothing\}\}[/imath] My solution was: [imath]P(S) = \{\varnothing, \{\varnothing\},\{ \varnothing , \{ \varnothing \}\}\}[/imath], but the solution on the textbook shows: [imath]P(S) = \{ \varnothing , \{ \varnothing \}, \{\{\varnothing\}\}, \{ \varnothing , \{ \varnothing \}\}[/imath] Where does the [imath]\{\{\varnothing\}\}[/imath] come from? Thanks |
1346027 | Evaluate the improper integral [imath]\int_{0}^{\infty}{f(x)-f(2x)\over x}dx[/imath], where [imath]\lim_{x \to \infty} f(x) = L[/imath]
Find [imath]\int_{0}^{\infty}{f(x)-f(2x)\over x}\, \mathrm{d}x[/imath] if [imath]f\in C([0,\infty])[/imath] and [imath]\lim\limits_{x\to \infty}{f(x)=L}[/imath]. I tried denoting [imath]\displaystyle \int{f(x)\over x}dx=F(x)[/imath], but I don't know what to do with [imath]F(\infty)-F(0)[/imath]. I also tried somehow to use l'Hopital's rule but don't seem to come by a plausible justification for it, nor do I arrive at a clear expression. I would really appreciate your help in this. | 2579722 | Solution of [imath]\int_0^{\infty} \frac{\arctan(x)-\arctan(2x)}x\mathrm dx[/imath] : Why Wolfram Alpha shows a different answer than mine?
For, [imath]\int_0^{\infty} \frac{\arctan(x)-\arctan(2x)}x \mathrm dx[/imath] Let, [imath]I_1=\int_0^{\infty} \frac{arctan(x)}x\mathrm dx[/imath] and [imath]I_2=\int_0^{\infty} \frac{\arctan(2x)}x\mathrm dx[/imath] For, [imath]I_1[/imath] let [imath]x=\frac 1u [/imath] so [imath]I_1=\int_{\infty}^{0} \frac{\arctan(\frac1u)}{\frac1u}\left(-\frac1{u^2}\right)\mathrm du[/imath] [imath]=\int_0^{\infty} \frac{arccot(u)}{u}du[/imath] From which it follows, [imath]I_1=\int_0^{\infty} \frac{\frac\pi2-\arctan(x)}x\mathrm dx[/imath] So, [imath]2I_1=\int_0^{\infty} \frac{\frac \pi2}x\mathrm dx[/imath] In the same way it can be shown that, [imath]2I_2=\int_0^{\infty} \frac{\frac\pi2}x\mathrm dx[/imath] So, [imath]2I=2I_1-2I_2=0[/imath] But, Wolfram Alpha says, So, where is my error? |
590215 | If [imath]0\leq A\leq B[/imath] on Hilbert space and [imath]A^{-1}[/imath] exists, show that [imath]A^{-1}\geq B^{-1}[/imath]
Does anyone know how to show this? Let [imath]H[/imath] be a Hilbert space and [imath]A[/imath], [imath]B[/imath] bounded positive operators defined on [imath]H[/imath] such that [imath]A^{-1}: H \rightarrow H[/imath] exists and hence bounded and [imath]A \leq B[/imath]. Therefore, [imath]B^{-1}: H \rightarrow H[/imath] exists and bounded. Show that [imath]A^{-1} \geq B^{-1}[/imath]. I tried to use this theorem "If two bounded self-adjoint linear operators [imath]S[/imath] and [imath]T[/imath] on a Hilbert space [imath]H[/imath] are positive and commute, then their product [imath]ST[/imath] is positive." However, I don't know whether [imath]A[/imath] and [imath]B[/imath] commute with each other. Are there any other ways to deal with it? I am using book of Kreyszig. | 95862 | For positive invertible operators [imath]C\leq T[/imath] on a Hilbert space, does it follow that [imath]T^{-1}\leq C^{-1}[/imath]?
I need the following result. I think it's quite obvious but I don't know how to prove that: Let [imath]C, T : \mathcal{H} \rightarrow \mathcal{H}[/imath] be two positive, bounded, self-adjoint, invertible operators on a Hilbert space [imath]\mathcal{H}[/imath] such that [imath]C \leq T[/imath]. Then it follows also that [imath]T^{-1} \leq C^{-1}[/imath]. Or maybe this is even not true? If this is not true can somebody give me a counter-example or if it is true some strategy how to solve this? I would be very thankful. mika |
1346935 | What does [imath]A^{B}[/imath] mean?
Assume, that A and B are finite sets. What notion [imath]A^{B}[/imath] does mean? Have been looking for awhile now. | 901735 | Meaning of a set in the exponent
Let [imath] D = 2^\mathbb{N} [/imath], i.e., D is the set of all sets of natural numbers. What's the meaning of this definition? Intuitively, I would suggest that [imath] D = \{1,2,4,...\} [/imath] but the explanation "set of all sets" leads me to the guess that this is wrong. |
1346738 | Find the integral closure of an integral domain in its field of fractions
Let [imath]k[/imath] be a field and let [imath]R = k[x,y]/(x^2-y^2+y^3)[/imath]. Note that [imath]R[/imath] is an integral domain. Let [imath]F[/imath] be the field of fractions of [imath]R[/imath]. How to determine the integral closure of [imath]R[/imath] in [imath]F[/imath]? I have no idea how this integral closure looks like. But I find that [imath]F = k(\overline{x}/\overline{y})[/imath], because [imath]\overline{y} = -(\overline{x}/\overline{y})^2 + 1[/imath] and hence [imath]\overline{x} = -(\overline{x}/\overline{y})^3 + (\overline{x}/\overline{y})[/imath]. Also, a general method for find the integral closure of an integral domain in its field of fractions is strongly desirable. Many thanks. | 1057771 | Normalisation of [imath]k[x,y]/(y^2-x^2(x-1))[/imath]
I am trying to figure out the normalisation of [imath]k[x,y]/(y^2-x^2(x-1))[/imath], for an algebraically closed field [imath]k[/imath]. I can show that it is not normal and I have the information that the normalisation is [imath]k[t][/imath], where [imath]t=\frac{y}{x}[/imath]. I can't figure out how to prove that this is the normalisation i.e. that the fraction fields of [imath]k[x,y]/(y^2-x^2(x-1))[/imath] and [imath]k[t][/imath] are the same, that [imath]k[t][/imath] is integrally closed and that [imath]k[t][/imath] is an integral extension of [imath]k[x,y]/(y^2-x^2(x-1))[/imath]. Also, how does one get to the idea of [imath]t=\frac{y}{x}[/imath]? Algebraically if possible, since this is from a commutative algebra class, not algebraic geometry. |
1347241 | Help finding the limit of the sequence?
I was given the sequence [imath]\frac{1}{2}[/imath], [imath]\frac{1}{4}[/imath],[imath]\frac{1}{8}[/imath], [imath]\frac{1}{16},\ldots[/imath] . I created an equation to represent the sequence [imath]a_n=\dfrac{1}{2^{n+1}}[/imath]. Now how do I go about finding the limit? | 1347223 | Help finding the limit of this series [imath]\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots[/imath]
How can I go about finding the limit of [imath]\frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} + \cdots = \sum_{k = 1}^{\infty} \frac{1}{2^{k+1}}?[/imath] Could I use the absolute value theorem? I have a feeling it converges to [imath]0[/imath] but I am not sure. |
421987 | choosing [imath]5[/imath] non consecutive books from a shelve of [imath]12[/imath]
In how many ways can you pick five books from a shelve with twelve books, such that no two books you pick are consecutive? This is a problem that I have encountered in several different forms ("In how many ways can you paint five steps of a twelve step staircase, without any two consecutive steps being painted?", etc.) but the idea is the same. Here is how I solved it: Consider a shelve with [imath]13[/imath] books, and then calculate the number of ways you arrange [imath]5[/imath] pairs and [imath]3[/imath] single books on that shelve. The answer to that question is [imath]{8 \choose 5} = 56[/imath]. This question is equivalent to the original, because the pairs prevent you from choosing two consecutive books, and a shelve of [imath]13[/imath] books rather than [imath]12[/imath] is considered because otherwise, it would not be possible to choose the last book on the shelve. This approach has its downside: writing a formal proof of this is not easy, as you can see. I tried to explain this proof to a friend of mine in order to give him the intuition [imath]{8 \choose 5}[/imath] is the right answer (which is usually easier than a formal proof) but I already had a hard time doing that. My question is: Is there a more rigorous proof this is the right answer? Not only would that help me explain my answer, but it would probably allow me to use that technique to a broader range of problems. | 20132 | How many ways can [imath]r[/imath] nonconsecutive integers be chosen from the first [imath]n[/imath] integers?
During self-study, I ran across the question of how many ways six numbers can be chosen from the numbers 1 - 49 without replacement, stipulating that no two of the numbers be consecutive. I can obtain a simple lower bound by saying that, in the worst-case scenario, when you choose a particular number, there are now three numbers that cannot be chosen next. For example, if I first pick 17, then I can't choose 16, 17, or 18 for the remaining choices. This gives me the lower bound [imath]\frac{49*46*43*40*37*34}{6!} = 6,773,770 \frac{8}{9}[/imath] This is about 48% of [imath]{}_{49}C_6 = 13,983,816[/imath]. The real answer must be bigger (and an integer). I haven't found a way to calculate it, though. The original problem asked to show that the probability of having non-consecutive integers when you choose six is greater than 50%, so if the problem is complicated to count exactly, better approximations that show the answer is above 50% would also be appreciated. Of course, I can use a computer to do the counting, but I'm interested in learning what methods I'm missing. |
1347875 | Element invertible in integral extension of ring implies invertible in ring
Please excuse some minor hiccups in terminology, I am primarily reading this in Swedish so feel free to correct any. Let [imath]A\subseteq B[/imath] be an integral extension and [imath]\alpha\in A[/imath] an invertible element in [imath]B[/imath]. Show that [imath]\alpha^{-1}\in A[/imath]. This is the question I am currently struggling with. I know that [imath]\alpha\in B[/imath] from the question, which means that there exists a monic polynomial [imath]p\in A[x][/imath] such that [imath]p(\alpha)=0[/imath]. I have also considered that [imath]p(x)=x-z=0[/imath] with [imath]z\in A[/imath] would naturally always imply that [imath]x\in A[/imath], but other monic polynomials can also do that, so I feel it doesn't contribute anything of real substance. So I am primarily struck on how to approach this to reach the answer; any advice? | 732676 | If ring [imath]B[/imath] is integral over [imath]A[/imath], then an element of [imath]A[/imath] which is a unit in [imath]B[/imath] is also a unit in [imath]A[/imath].
Let [imath]A[/imath] be a subring of ring [imath]B[/imath], with [imath]B[/imath] integral over [imath]A[/imath]. If [imath]x\in A[/imath] is a unit in [imath]B[/imath], then it is a unit in [imath]A[/imath]. I know that [imath]f(t) = t - x[/imath] is in [imath]A[t][/imath] with [imath]f(x) = 0[/imath], and that there exists some monic polynomial [imath]g(t) = t^n + a_{n-1}t^{n-1}... + a_0[/imath] also in [imath]A[t][/imath] so that [imath]g(x^{-1}) = 0[/imath], and as I understand it, this proof comes down to showing that [imath]x^{-1}[/imath] must be an element of A. But I'm not sure what kind of manipulation of these polynomials will do that for me. I tried examining [imath]x^n g(t)[/imath] and saw that [imath]0 = x^ng(x^{-1}) = 1 + xa_{n-1} + ... + x^na_0[/imath], but I can't seem to get anything useful out of this. Any hints toward the right direction would be appreciated. |
831905 | Finding the inhomogeneous solution
[imath]x_{n+2} = x_{n+1} + 20x_n + n^2 + 5^n \text{ with } x_0 = 0 \text{ and } x_1 = 0[/imath] How would I find the inhomogeneous solution for this since the homogenous solution is 0 given initial conditions? | 830929 | Help solving recursive relations
[imath]x_{n+2} = x_{n+1} + 20x_n + n^2 + 5^n \text{ with } x_0 = 0 \text{ and } x_1 = 0[/imath] How would you solve this recursive relation? I have the homogenous solution, but am having issues with the in-homogenous one. |
1348178 | An exercise question in Linear Algebra Done Right by Axler
Prove or give a counterexample: if [imath]U_1[/imath], [imath]U_2[/imath], [imath]W[/imath] are subspaces of [imath]V[/imath] such that [imath]V[/imath] = [imath]U_1\oplus W[/imath] and [imath]V = U_2 \oplus W[/imath], then [imath]U_1 = U_2[/imath]. I'm a beginner in linear algebra and I'm frustrated at these types of questions, could you please offer an example solution? Thanks in advance. | 1064508 | Is it true that [imath] U \oplus W_1 = U \oplus W_2 \implies W_1 = W_2 [/imath]?
Is it true that if [imath] U \oplus W_1 = U \oplus W_2 [/imath], then [imath] W_1 = W_2 [/imath]? I think that if [imath] U \oplus W_1 = U \oplus W_2 [/imath], then u+w1=u+w2, so W1=W2. But did I make any mistakes? |
1348224 | Finding the convergence radius of [imath]\sum_{n=1}^\infty n! x^{n!}[/imath]
I need help finding the covergence radius of [imath]\sum_{n=1}^\infty n! x^{n!}[/imath] . The factorials make me think that I need to use derivatives/ integrals but I don't quite know how. I'd love any help, thanks. | 1339237 | Find the radius of convergence of [imath]\sum_{n=1}^{\infty}{n!x^{n!}}[/imath]
Find the radius of convergence of [imath]\sum_{n=1}^{\infty}{n!x^{n!}}[/imath]. Should I look at this series as: [imath]\sum_{n=1}^{\infty}({n!x^{(n-1)!})x^{n}}[/imath]? I am really confues here. In addition, any attempt to compute any kind of "lim sup" fails. I would appreciate your help in this. |
1278169 | number of subsets of even and odd
Let [imath]A[/imath] be a finite set. Prove or disprove: the number of subsets of [imath]A[/imath] whose size is even is equal to the number of subsets of [imath]A[/imath] whose size is odd. Example: [imath]A = {1,2}[/imath]. The subsets of [imath]A[/imath] are {},{1},{2}, and {1,2}. Since there are two subsets of odd size ({1} and {2}) and two subsets of even size ({} and {1,2}) the claim holds for this particular example. | 557494 | How to prove even subsets equal to odd subsets?
There is question that I don't know how to prove. we have set [imath]A=\{1,2,3,\ldots,n\},\; O=\{B\mid B⊆A,\text{ odd }B\},\; E=\{B\mid B⊆A,\text{ even }B\}[/imath] it ask to prove that subsets even equal to subsets odd by proving that [imath]f:O\to E[/imath] is an injective and surjective function |
1348310 | If [imath]x_{n+1}\leq x_n + 1/n^2[/imath] then [imath]x_n[/imath] converges
Let [imath]x_n[/imath] be a sequence of non-negative real numbers such that [imath]\forall n, x_{n+1}\leq x_n+ \frac{1}{n^2}[/imath] Prove that [imath]x_n[/imath] converges. The problem is trivial whenever [imath]x_{n}[/imath] is an increasing sequence. In the general case I managed to prove that [imath](x_n)[/imath] is bounded. It remains to prove that [imath](x_n)[/imath] has only one accumulation point, but that seems difficult. Any hint is welcome. | 185989 | Convergence of a sequence of non-negative real numbers [imath]x_n[/imath] given that [imath]x_{n+1} \leq x_n + 1/n^2[/imath].
Let [imath]x_n[/imath] be a sequence of the type described above. It is not monotonic in general, so boundedness won't help. So, it seems as if I should show it's Cauchy. A wrong way to do this would be as follows (I'm on a mobile device, so I can't type absolute values. Bear with me.) [imath]\left|x_{n+1} - x_n\right| \leq \frac{1}{n^2}.[/imath] So, we have [imath] \left|x_m -x_n\right| \leq \sum_{k=n}^{m} \frac{1}{k^2}[/imath] which is itself Cauchy, etc., etc. But, of course, I can't just use absolute values like that. One thing I have shown is that [imath]x_n[/imath] is bounded. Inductively, one may show [imath] \limsup_{n \to \infty} x_n \leq x_k + \sum_{k=n}^{\infty} \frac{1}{k^2},[/imath] although I'm not sure this helps or matters at all. Thanks in advance. Disclaimer I've noticed that asking a large number of questions in quick succession on this site is often frowned upon, especially when little or no effort has been given by the asker. However, I am preparing for a large test in a few days and will be sifting through dozens of problems. Therefore, I may post a couple a day. I will only do so when I have made some initial, meaningful progress. Thanks. |
1265979 | Analysis: Basic Sequence Proof
Prove that, if [imath]\left\{a_n\right\}_{n=1}^{\infty}[/imath] converges to A, then [imath]\left\{|a_n|\right\}_{n=1}^{\infty}[/imath] converges to |A|. Is the converse true? | 667084 | Proving that [imath]x_n\to L[/imath] implies [imath]|x_n|\to |L|[/imath], and what about the converse?
Problem 3. Show that for a sequence [imath](x_n)[/imath] the following are true: (i) [imath]\lim x_n=0[/imath] if and only if [imath]\lim |x_n|=0[/imath]. (ii) [imath]\lim x_n=L[/imath] implies [imath]\lim |x_n|=|L|[/imath]. Is the converse true? Prove or give a counterexample. (i) is already done, easy. I'm halfway done with (ii), I split it into three cases: [imath]L = 0[/imath], [imath]L > 0[/imath] and [imath]L < 0[/imath]. For [imath]L = 0[/imath] I just refer to part (i). For [imath]L > 0[/imath], we know that [imath]|x_n - L| < \epsilon[/imath] if lim [imath]|x_n|[/imath] = [imath]|L|[/imath], then we must have [imath]||x_n| -|L|| < \epsilon[/imath] but by the reverse triangle equality, [imath]||x_n| - |L|| < |x_n - L|[/imath] so clearly [imath]||x_n|-|L|| < \epsilon[/imath], thus lim [imath]x_n = L[/imath] [imath]\implies[/imath] lim [imath]|x_n| = |L|[/imath] for [imath]L > 0[/imath] for some reason I'm confused as to part three, [imath]L<0[/imath]. I'm having trouble seeing the difference between [imath]L>0[/imath] and [imath]L<0[/imath], and i'm starting to think that splitting it up like that isn't necessary at all. |
1347397 | If [imath]u \in L^2(0,T;L^2(\Omega))[/imath] is [imath]\int_{\Omega}\int_0^T |u(t,x)|^2[/imath] defined?
Let [imath]u \in L^2(0,T;L^2(\Omega))[/imath] on some domain [imath]\Omega[/imath]. We know that [imath]\int_0^T \int_{\Omega}|u(t,x)|^2[/imath] is defined, but is it equal to [imath]\int_{\Omega}\int_0^T |u(t,x)|^2?[/imath] Can I interchange the integrals? In order to apply Fubini--Tonelli's theorem, I need [imath]u[/imath] to be measurable on [imath][0,T]\times \Omega[/imath], I think, so I don't know if works! | 1133346 | Is [imath]L^2(0,\infty;L^2(\Omega)) = L^2((0,\infty)\times \Omega)[/imath]?
If [imath]\Omega[/imath] is a bounded [imath]C^1[/imath] domain, is [imath]L^2(0,\infty;L^2(\Omega)) = L^2((0,\infty)\times \Omega)[/imath]? Are they the same? I know this is true when instead of [imath](0,\infty)[/imath] we have a bounded interval. |
1348867 | L(V,W) is Banach, then W is Banach
Let [imath]V,W[/imath] normed vector spaces, [imath]V[/imath] not empty and with a finite dimension. Prove that [imath]L(V,W)[/imath] is Banach, then [imath]W[/imath] is also Banach. | 602041 | If space of bounded operators L(V,W) is Banach, V nonzero, then W is Banach (note direction of implication)
Let [imath]V,W[/imath] be normed vector spaces, and [imath]L(V,W)[/imath] be the space of bounded linear operators. Usually I would only see the statement "If [imath]W[/imath] is Banach, then [imath]L(V,W)[/imath] is Banach.". But Wikipedia writes that there is a converse: "If [imath]L(V,W)[/imath] is Banach, and if [imath]V[/imath] is non-trivial, then [imath]W[/imath] is Banach". This is pretty interesting since I never seen a converse before. I was wondering if anyone has a nice proof. I tried reversing the proof for the usual direction, but the inequalities can't be reversed. I also tried to start with a Cauchy sequence in [imath]W[/imath], and construct linear operators (using Hahn-Banach to control the operator norms), but alas I cannot say much about the distance between these operators (much less say that it is Cauchy) |
1349235 | every [imath]F[/imath]-algebraic homomorphism of [imath]K[/imath] is [imath]1-1[/imath] and onto?
Suppose that [imath]K|F[/imath] is a field extension of finite degree. We know that every [imath]F[/imath]-algebraic homomorphism of [imath]K[/imath] is [imath]1-1[/imath] and onto. Also we know that every finite field extension is algebraic extension. What can we say if [imath]K|F[/imath] is just algebraic extension? Can we say every [imath]F[/imath]-algebraic homomorphism of [imath]K[/imath] is [imath]1-1[/imath] and onto? | 152494 | Showing a homomorphism of a field algebraic over [imath]\mathbb{Q}[/imath] to itself is an isomorphism.
Suppose [imath]F[/imath] is algebraic over [imath]\mathbb{Q}[/imath] and [imath]\varphi : F\to F[/imath] is a homomorphism. Prove [imath]\varphi[/imath] is an isomorphism. Showing injectivity follows from the fact that the only ideals in a field are [imath](0)[/imath] and [imath]F[/imath]. But how do you show surjectivity? |
1348757 | Showing there is no triplet of positive integers [imath](a,b,c)[/imath] satisfying [imath]a^7+b^7=7^c[/imath]
Show that [imath]a^7+b^7=7^c[/imath] has no positive integer solutions [imath](a,b,c)[/imath]. I've posted a general and way too long approach as an answer. How may one prove the claim more briefly and specifically? | 689159 | Solving [imath]x^p + y^p = p^z[/imath] in positive integers [imath]x,y,z[/imath] and a prime [imath]p[/imath]
The question is from Zeitz's ''The Art and Craft of Problem Solving:" Find all positive integer solutions [imath]x,y,z,p[/imath], with [imath]p[/imath] a prime, of the equation [imath]x^p + y^p = p^z[/imath]. One thing I noticed is that [imath] \frac{x^p + y^p}{x + y} = \sum_{i=0}^{p-1}x^{(p-1)-i}(-y)^i \implies (x+y) |p^z \implies x + y = p^n, \text{ }n < z [/imath] It is also not hard to determine all solutions for [imath]p =2[/imath]. After this, however, I am at a loss. I don't know what restrictions I can impose to try to narrow down the solution set; for example, the class of solutions [imath] p=3, x = 3^n, y = 2\cdot3^n, z = 2 + 3n [/imath] for [imath]n \ge 0[/imath] show that [imath]x+y = p^n[/imath] cannot be sharpened. (Let me note that these are the only other solutions I have found besides the ones for [imath]p = 2[/imath].) Could anybody give me a push in the right direction on this problem? |
654923 | How do I start from a 10% discount and find the original price?
I have a database of prices that already have a 10% discount. For example a product could be [imath]100 after a 10% discount. Is there a reusable formula I can use to determine what the original price was of all the 10% discounted prices in the database?[/imath] Edit: Thank you for the fast responses. Is there any way to account for rounding errors? A real example is a product with a discounted price of \129.00 Using the X/.9 formula, I get \[imath]143.33 as the original price, which does not actually work out. To have had \[/imath]129.00 as the discount price, the original price would have needed to have been \$143.34. | 1378481 | If a given # is [imath]70[/imath]% of [imath]X[/imath]. How do you determine what [imath]X[/imath] is?
Given I have the number [imath]50,000[/imath] which is [imath]70[/imath]% of [imath]X[/imath]. How do I calculate what [imath]X[/imath] is without guessing. Thanks |
1256915 | Why is [imath]n^2+4[/imath] never divisible by [imath]3[/imath]?
Can somebody please explain why [imath]n^2+4[/imath] is never divisible by [imath]3[/imath]? I know there is an example with [imath]n^2+1[/imath], however a [imath]4[/imath] can be broken down to [imath]3+1[/imath], and factor out a three, which would be divisible by [imath]4[/imath]. | 1255318 | Proving that [imath]x^2 + 4[/imath] is not divisible by [imath]3[/imath]
I need to show the following: For any integer [imath]x[/imath], [imath]x^2 + 4[/imath] is not divisible by [imath]3[/imath]. I was trying proof by contraposition, but I do not believe that is the most efficient way to go about this. Can anybody point me in the right direction, please? |
910647 | Determinants of Matrices det(4A) equals?
Suppose A is a 4 x 4 matrix such that [imath]\det(A) = 1/64[/imath]. What will [imath]\det(4A^{-1})^T[/imath] be equal to? Here's my thinking, [imath]\det(A^T) = \det(A)[/imath] I has no effect on the determinant. And [imath]\det(A^{-1}) = 1/\det(A)[/imath] so [imath]\det(4A^{-1}) = \det(4\times64)[/imath] = can't be 256? I don't understand, Please help me! THANKS! | 909990 | Suppose [imath]A[/imath] is a 4x4 matrix such that [imath]\det(A)=\frac{1}{64}[/imath]
Suppose A is a 4x4 matrix such that [imath]\det(A)=\frac{1}{64}[/imath] then [imath]\det(4A^{-1})^T[/imath] I created a 2x2 matrix [imath]B[/imath] and transposed it both had the same determinant I then found [imath]\det(B)[/imath] and [imath]\det(B^{-1})[/imath] the results were inverses of each other I then tried found the [imath]\det(2B)[/imath] but I cannot see what is the relationship between the determinant of a matrix and the determinant of that matrix multiplied by a constant? |
1350585 | Fibonacci series generated with division [imath]1/999999999999999999999998999999999999999999999999[/imath]
I tested the suggestion of finding the Fibonacci series by division, which sounded very surprising to me. I therefore used a simple sympy script to test it and found that it works as advertised. #!/usr/bin/python2 # Prints the Fibonacci series up to 24 digits by means of division! import sympy # note the single '8' in here \\ divisor = 999999999999999999999998999999999999999999999999 # // # more than '2761' overflows! digits = 2761 chunkSize = 24 # calc 1/'divisor' with 'digits' accuracy result = sympy.N(sympy.Integer(1)/sympy.Integer(divisor),digits) # convert to string and strip '0.' resultStrippedStr = str(result)[2:] # http://stackoverflow.com/a/434328 def chunker(seq, size): return (seq[pos:pos + size] for pos in xrange(0, len(seq), size)) for item in chunker(resultStrippedStr, chunkSize): print item # check, starting from 2nd number m1 = 1 m2 = 0 # fails for last non overflowing value for item in list(chunker(resultStrippedStr, chunkSize))[2:]: if not m2 + m1 == int(item): print "" print "{0} + {1} \\neq {2}".format(m2, m1, int(item)) raise ValueError ("Fibonacci test failed ") m2 = m1 m1 = int(item) In the article, the last, non-overflowing number 781774079430987230203438 is not shown and also does not pass the Fibonacci test. Checking manually by adding the last digits 2 and 5 certainly does not equal 8. Why does this division reveal the Fibonacci series and which limit is reached before the obvious overflow (number greater than 24 digits). | 656183 | Why does [imath]\frac{1 }{ 99989999}[/imath] generate the Fibonacci sequence?
[imath]\frac{1}{99989999} = 1.00010002000300050008001300210034005500890144... \times 10^{-8}[/imath] (Link), which includes the Fibonacci sequence [imath](1\ 1\ 2\ 3\ 5\ 8\ 13\ 21\ 55\ 89\ 144\ldots )[/imath]. This is fascinating to me but I can't figure out how this works, and a Google search doesn't seem to reveal anything. |
1349830 | prove the continuity of [imath]T_\phi f=\int_0^1 f(x)\phi(x) \,dx\\[/imath]
Let [imath]\phi\in C[0,1][/imath] and let [imath]T_\phi~:C[0,1]\rightarrow\mathbb{R}[/imath], defined as [imath]T_\phi f=\int_0^1 f(x)\phi(x) \,dx\\[/imath]. How can i prove that it's a continuous operator? | 1349416 | Show that [imath]Kf(x,y)=\int_0^1k(x,y) f(y) \,dy\\[/imath] is linear and continuous
Let [imath]k:[0,1]\times[0,1]\rightarrow \mathbb{R}[/imath] continuous and [imath]K:C[0,1]\rightarrow C[0,1][/imath], given by [imath]Kf(x,y)=\int_0^1k(x,y) f(y) \,dy\\[/imath]. Prove that [imath]K[/imath] is continuous. I try to see continuity in [imath]0[/imath], but i can't find [imath]\delta[/imath]. |
1350544 | Why a change of a brownian motion does not depend on the past values of it?
[imath](B_t)_{t \in \mathbb R_0^+}[/imath] are random variables on [imath](\Omega,\mathcal A,P)[/imath]. [imath]\forall r \le s, t > s, B_t-B_s,B_r[/imath] are independent (i.e. [imath]\sigma(B_t-B_s)[/imath] and [imath]\sigma(B_r)[/imath] are independent, where with [imath]\sigma(X)[/imath] I denote the [imath]\sigma[/imath]-algebra produced by the variable [imath]X[/imath]). Can one show that [imath]B_t-B_s[/imath] and [imath]\sigma(\{B_i| r \le s\})=\sigma(\bigcup_{r \le s}\sigma(B_r))[/imath] are independent? Does one need to use the fact that any endless sequence of random variables from [imath](B_t)_{t \in \mathbb R_0^+}[/imath] is multivariate normally distributed? | 918943 | Simple question about the definition of Brownian motion
I have a question concerning the definiton of Brownian motion. Usually (e.g. on Wikipdia) one demands a brownian motion [imath]\lbrace B_t\rbrace_{t\in[0,\infty)}[/imath] to satisfy the following condition: [imath]\forall 0\leq t_0<t_1<...<t_m: B_{t_1}-B_{t_0},...,B_{t_m}-B_{t_{m-1}}[/imath] are independent random variables. But today I come across with formally another definition of Brownian motion! Instead of the above condition the book demands the following property: [imath]\forall t,h\geq 0[/imath], [imath]B_{t+h}-B_t[/imath] is independent of [imath]\lbrace B_u : 0\leq u\leq t \rbrace[/imath]. Now I'm wondering whether both properties are indeed equivalent or not?! Hopefully someone can help me. |
1350338 | how to prove that [imath]\lim\limits_{n\to\infty} \left( 1-\frac{1}{n} \right)^n = \frac{1}{e}[/imath]
I need to prove [imath]\lim\limits_{n\to\infty} \left( 1-\frac{1}{n} \right)^n = \frac{1}{e}[/imath] For now, I only know the e definition: [imath]\lim\limits_{n\to\infty} \left ( 1+\frac{1}{n} \right)^n = e [/imath], and I need to proof all the partial results in between to reach the wanted result. | 295584 | Proof for [imath]e^z = \lim \limits_{x \rightarrow \infty} \left( 1 + \frac{z}{x} \right)^x[/imath]
Q1: Could someone provide a proof for this equation (please focus on this question): [imath]e^z = \lim \limits_{x \rightarrow \infty} \left( 1 + \frac{z}{x} \right)^x[/imath] Q2: Is there any corelationn between above equation and the equation below (this would benefit me a lot because i at least know how to proove this one): [imath]e = \lim \limits_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right)^x[/imath] |
1349801 | finding the limit for a martingale
I have trouble finding the exact limit for a martingale: Let [imath]\{\xi_n\}_{n\in\mathbb{N}}[/imath] be a Markov chain with [imath]\xi_0[/imath] uniformly distributed in [imath][0,1][/imath] and [imath] P(\xi_{n+1}=\alpha\xi_n+(1-\alpha)|\xi_n)=\xi_n [/imath] [imath] P(\xi_{n+1}=\alpha\xi_n|\xi_n)=1-\xi_n, [/imath] for some [imath]\alpha\in(0,1)[/imath]. It's clear that [imath]\xi_n[/imath] is a martingale and [imath]\xi_n\leq1[/imath] for all n, so by Martingale Convergence theorem [imath]\xi_n[/imath] have a almost surely limit [imath]\xi[/imath], which I'm trying to find . In Durrett's book, this kind of question normally ends when we see the stationary point by observation (if I'm not mistaken), while here it seems like [imath]0[/imath] and [imath]1[/imath] are both stable. How to get the almost surely limit formally? Thanks for any help. Let me add some of my thoughts I just came up: For those events always go left, the probability is [imath]\prod(1-\xi_n)=\prod(1-\alpha^{n}\xi_0)[/imath], which is convergent to a nonzero constant since [imath]\xi_0\Sigma\alpha^{n}[/imath] converges. That is to say, there is a positive possibility that [imath]\xi_n[/imath] converges to [imath]0[/imath]. As [imath]\xi_n[/imath] converges to [imath]\xi[/imath] almost surely, it forces [imath]\xi[/imath] to be [imath]0[/imath]. Is my argument above correct? Please let me know if there are any problems in it. Thanks a lot. As being pointed out, it's similar to the other question while I want to find the limit [imath]\xi[/imath]. With the help of the other question, it can be seen that [imath]P(\xi=0)=P(\xi=1)=1/2.[/imath] I would guess [imath]\xi=1_{(1/2,1]}[/imath], but how to prove it formally? | 243554 | Probability that a sequence of random variables converges to 0 or 1
Fix [imath]\alpha , \beta > 0[/imath] with [imath]\alpha + \beta = 1[/imath] and consider the sequence of random variables defined by [imath]X_0 = \theta \in (0,1)[/imath] and [imath]P(X_n = \alpha + \beta X_{n-1} \mid X_{n-1} , ... , X_0) = X_n , \qquad P(X_n = \beta X_{n-1} \mid X_{n-1} , ... , X_0) =1- X_n.[/imath] I am trying to prove that [imath]P(\lim_{n\rightarrow \infty} X_n = 1) = \theta[/imath] and [imath]P(\lim_{n\rightarrow \infty} X_n = 0) = 1-\theta[/imath]. It seems that there should be a way to do this using the result that [imath]P(E \mid \mathcal F_n) \rightarrow P(E \mid \mathcal F)[/imath], whenever [imath]\mathcal F_n \uparrow \mathcal F[/imath] is an increasing sequence of [imath]\sigma[/imath] algebras. But, I am not seeing a solution. Any suggestions? By the way, this is an exercise on page 224 of Durrett's Probability: Theory and Examples. |
1351379 | A question on the representation of all integers in terms of the sum of other interger cubes
The question is from a book used for transition between high school mathematics and university mathematics, which states: Prove the following statement or give a counterexample [imath]\forall n \in \mathbb Z, \exists a,b,c,d,e,f,g,h \in \mathbb Z[/imath] such that [imath]n=a^3+b^3+c^3+d^3+e^3+f^3+g^3+h^3[/imath] After several trials I believe the statement is true but find it difficult to prove. Could somebody offer a solution without reference to congruence? I can see that if one choose [imath]a,b,c,d,e,f,g[/imath] equals from [imath]0 \:to\: 1[/imath] each then [imath]n \in {1,2,3,4,5,6,7,8} [/imath] can be represented. For the next eight numbers, by choosing [imath]a=2[/imath] and adjusting the remaining parameters from [imath]0\:to\:1[/imath], can also be achieved. Thus I suspect all numbers can be represented this way, but, having said that, I still feel hard to find a proof. | 207366 | Can any integer can be written as the sum of 8 integer cubes?
Can anyone offer an elementary proof why: [imath]\forall n \in {\mathbb Z} \space \exists a, b, c, d, e, f, g, h \in {\mathbb Z}[/imath]such that[imath]n=a^3+b^3+c^3+d^3+e^3+f^3+g^3+h^3{\rm ?}[/imath] In other words, why every integer is the sum of eight cubes integers. My first thought was that since the gaps between [imath]p^3[/imath] and [imath](p+1)^3[/imath] increase with [imath]p,[/imath] this would only hold for small integers. This is wrong; can anyone tell me why? N.B. This is different to Waring's problem because [imath]n,a,b,c...[/imath] dot not have to be natural numbers, so we can have [imath]27=3^3+(-1)^3+(-1)^3+(-1)^3+(-1)^3[/imath] |
801963 | Random walk around circle
For one of the exercises of my homework I need to answer the following question, but I am not sure how I should apply gamblers ruin theory to solve this problem (it is stated as a hint, not that I must use it). The problem is as follows: A mouse is in a room in a circular hallway with [imath]N+1[/imath] rooms. In the the other [imath]N[/imath] rooms there is a cheese. He has a chance of [imath]\frac{1}{2}[/imath] to go either way every step. If he enters a room with a cheese he eats the cheese. Prove that all cheeses have an equal chance of being eaten last, namely [imath]\frac{1}{N}[/imath]. I see that eating the [imath]i^{\textrm{th}}[/imath] cheese as last has two possibilities, first mouse walks to [imath]i+1[/imath] and then goes to [imath]i-1[/imath] without passing [imath]i[/imath], and the other way around. But I am at loss what to do next. | 116446 | Random walk on [imath]n[/imath]-cycle
For a graph [imath]G[/imath], let [imath]W[/imath] be the (random) vertex occupied at the first time the random walk has visited every vertex. That is, [imath]W[/imath] is the last new vertex to be visited by the random walk. Prove the following remarkable fact: For the random walk on an [imath]n[/imath]-cycle, [imath]W[/imath] is uniformly distributed over all vertices different from the starting vertex. |
666997 | How many different spanning trees of [imath]K_n \setminus e[/imath] are there?
I need to know how many different spanning trees of [imath]K_n \setminus e[/imath] are there. [imath]K_n \setminus e[/imath] is a graph created by removing one of the edges of a full graph [imath]K_n[/imath]. Well as we all know the number of spanning trees of a full graph is [imath]n^{n-2}[/imath]. But graphs fulfill the deletion-contraction rule, meaning that [imath]\tau[/imath] being the number of spanning trees fulfills the following equality. [imath]\tau(K_n)-\tau(K_n / e)=\tau(K_n \setminus e)[/imath], where [imath]K_n / e[/imath] is a graph made by joining two vertices at the end of edge [imath]e[/imath] and connecting it to all neigbours of edges that lie on [imath]e[/imath]. But [imath]K_n / e[/imath] is [imath]K_{n-1}[/imath], right? It has [imath]n-1[/imath] vertices and each an every one of them connects to each other. Therefore [imath]\tau(K_n \setminus e)=n^{n-2}-(n-1)^{n-3}[/imath], right? Or does deletion contraction rule not apply to simple graphs? | 3014542 | How many trees are there on vertex set [imath][n][/imath] that contain a given edge [imath]uv[/imath]?
How many trees are there on vertex set [imath][n][/imath] that contain a given edge [imath]uv[/imath]? If we glue the vertex [imath]u[/imath] and [imath]v[/imath] with an edge then there are [imath]n-1[/imath] vertices and using the Cayley's formula there are total [imath](n-1)^{(n-3)}[/imath] trees with vertex set [imath][n-1][/imath] and a given edge [imath]uv[/imath]. Is this correct? |
241741 | Simple algebra question - proving [imath]a^2+b^2 \geqslant 2ab[/imath]
How to prove that [imath]a^2+b^2 \geqslant 2ab[/imath] is true, where [imath]a[/imath] and [imath]b[/imath] are both real numbers? | 943994 | Show that for all real numbers [imath]a[/imath] and [imath]b[/imath], [imath]\,\, ab \le (1/2)(a^2+b^2)[/imath]
so as in the title, I have the following theorem to prove. Theorem Show that for all [imath]a[/imath], [imath]b\in \mathbb R[/imath], that the following inequality holds, [imath]\begin{equation} ab \leq \frac{1}{2}(a^2 + b^2) \end{equation}[/imath]. My attempt So if [imath]a=b=0[/imath], then the inequality is trivially [imath]0 \leq 0[/imath] and we are ok. If [imath]a > 0[/imath] and [imath]b < 0[/imath], then the product on the right is less than zero, while the one on the right is greater than [imath]0[/imath], so the inequality holds. The problem that I am having is that if the numbers are the same sign, we have that both products will be positive and now we have to the show the inequality is true in this case. I tried making the following re-arrangement as a starting point [imath]\begin{equation} \\ 2ab \leq (a^2 + b^2) \end{equation}[/imath] But I am not really sure where to go from here. I've been staring at it for a while and I just can't seem to get anything out of it. I noticed that it looks rather similar to the law of cosines for theta equal to 90 degrees and [imath]c = 0[/imath], though that would be wierd indeed. Anyway that hasn't been talked about in the book yet so I don't think it matters. We have been introduced to the binomial theorem, difference of powers and geometric sum theorems, but I couldn't find any enlightenment from these. I don't really want an answer if you can help it, but a hint in the right direction would be appreciated if you could offer one. Thanks. |
1351278 | On the space [imath]l_2[/imath] we define an operator [imath]T[/imath] by [imath]Tx=(x_1, {x_2\over2}, {x_3\over3}, . . . )[/imath]. Show that [imath]T[/imath] is bounded, and find its adjoint.
On the space [imath]l_2[/imath] we define an operator [imath]T[/imath] by [imath]Tx=(x_1, {x_2\over2}, {x_3\over3}, . . . )[/imath]. Show that [imath]T[/imath] is bounded I know that [imath]||T||\leq 1[/imath], but I don't know how to show this. Any solutions or hints are appreciated. | 502262 | [imath]T:\ell_2\to \ell_2, T(x_1,x_2,\dotsc)=(x_1,x_2/2,x_3/3,\dotsc)[/imath]
[imath]T:\ell_2\to \ell_2, T(x_1,x_2,\dotsc)=(x_1,x_2/2,x_3/3,\dotsc)[/imath] I need to know whether it is self adjoint and unitary operator given that [imath]x_i\in\mathbb C[/imath] I am not able to do it please tell me how to proceed. In my earlier post one said eigenvalues of a self adjoint operator must be real, here I tried like say [imath]\lambda[/imath] be an eigenvalue for [imath]T[/imath] then [imath]T(x)=\lambda x[/imath] but then [imath](x_1,x_2/2,x_3/3,\dotsc)=\lambda(x_1,x_2,x_3,\dotsc)[/imath], I am lost here, |
1352384 | If [imath]G[/imath] is abelian, it has a subgroup in every order of [imath]|G|'s[/imath] divisors?
Assume that [imath]G[/imath] is an abelian group, I read somewhere that it can be derived from Lagrange's theorem that it has a number of subgroups that is equal to the number of [imath]G[/imath]'s divisors. Why does it hold? I'm confused... | 910426 | every Abelian group is a converse lagrange theorem group
Let [imath]G[/imath] be a finite abelian group, then [imath]G[/imath] has a subgroup of order [imath]n[/imath] if and only if [imath]n\mid G[/imath]. Proof: by Lagrange if [imath]H\leq G[/imath] then [imath]|H|[/imath] divides [imath]|G|[/imath] so this proves one of the implications. We prove the other implication by strong induction on the order of [imath]G[/imath], for primes this is just Cauchy, so we can take these like base cases. Now let [imath]n[/imath] be a proper divisor of [imath]|G|[/imath], take [imath]p[/imath] a prime in the factorization of [imath]|G|[/imath] (instead of [imath]|G|[/imath] is should be [imath]n[/imath], I had made a mistake when typing the first time), then by cauchy there is an element [imath]x[/imath] of [imath]G[/imath] of order [imath]p[/imath], since [imath]G[/imath] is abelian [imath]\langle x\rangle[/imath] is normal in [imath]G[/imath] so we can look at the quotient group [imath]G/\langle x\rangle[/imath], this group has order [imath]\frac{|G|}{p}[/imath], now take a prime [imath]p_1[/imath] in the factorization of [imath]\frac{|G|}{n}[/imath]. By induction the quotient group has a subgroup [imath]E[/imath] of order [imath]G/pp_1[/imath] now look at the preimage of [imath]E[/imath] under the natural homomorphism from [imath]G[/imath] to [imath]G/\langle x\rangle[/imath], it is easy to see this is a subgroup of [imath]G[/imath], we call it [imath]I[/imath], consider the restriction of the natural projection to [imath]I[/imath], this is a surjective homomorphism from [imath]I[/imath] to [imath]E[/imath] with kernel [imath]\langle x\rangle[/imath] so by the first isomorphism theorem [imath]I/x\cong E[/imath] so [imath]\frac{|I|}{|\langle x\rangle|}=|E|\implies |I|=\frac{|G|}{pp_1}{p}=\frac{|G|}{p_1}[/imath] Notice [imath]\frac{|G|}{p_1}[/imath] is a multiple of [imath]n[/imath] since [imath]p_1[/imath] was in the fatorization of [imath]\frac{|G|}{n}[/imath]. So by induction [imath]I[/imath] has a subgroup of order [imath]n[/imath] which is also a subgroup of [imath]G[/imath]. |
1349037 | In how many ways can the committee be selected if the girls must include either Roberta or Priya but not both?
A committee of three boys and three girls is to be selected from a class of [imath]14[/imath] boys and [imath]17[/imath] girls. In how many ways can the committee be selected if the girls must include either Roberta or Priya but not both? I did [imath](14 C 3) \times (15 C 2) \times 2 = 76,440[/imath] However, the correct answer is [imath]65,520[/imath]. Any help is much appreciated, thanks in advance. | 1224768 | In how many ways can you select one of the two but not both?
For this question: A committee of three boys and three girls is to be selected from a class of 14 boys and 17 girls. In how many ways can the committe be selected if: a.) Ana has to be on the committee. Ana is a girl so: [imath](^{14}C_3)(^{16}C_2)=43680[/imath] b.) the girls must include either Roberta or Priya, but not both. I basically needed help in this question (Question B) Attempts I made so far: 1.) [imath](^{14}C_3)*2(^{16}C_2)-(^{15}C_1)=87345[/imath] 2.) [imath](^{14}C_3)*2[(^{16}C_2)-(^{15}C_1)]=76440[/imath] I need help in what I've been doing wrong in these workings for question B |
1351444 | Finding the factorial moment generating function
I need help finding [imath]G_x(t)[/imath] [imath]f(x)= pq^{x-1}[/imath] for x = 1, 2,... and 0 otherwise. I know [imath]G_x(t)= M_x(ln t)[/imath] I have started the following [imath]\sum_{x=1}^\infty e^{xlnt}f(x)[/imath] [imath]\sum_{x=1}^\infty e^{xlnt}pq^{x-1}[/imath] [imath]pq^{-1}\sum_{x=1}^\infty (e^{lnt}pq)^x[/imath] [imath]pq^{-1}\sum_{x=1}^\infty t^xpq^x[/imath] [imath]=pq^{-1}{tpq\over 1-tpq}[/imath] Am I on the right track? Have I simplified the sum correctly? | 38184 | Factorial Moment of the Geometric Distribution
I am trying to caclulate the Factorial Moment of the Geometric Distribution #2 with parameter [imath]p[/imath]. Therefore I set [imath]\Omega = \mathbb{N}_0[/imath] and have by using the pochhammer symbol and setting [imath]q=1-q[/imath] that [imath]E((k)_l)= \sum _{k=0}^{\infty } (k)_l p q^k = p^{-l} q \cdot l! \sum _{k=0}^{\infty } (\frac{(k+l-1)!}{(k-1)! \cdot l!}\cdot p^{l+1} q^{k-1}) [/imath] Now Mathematica tells me that [imath]\sum _{k=0}^{\infty } (\frac{(k+l-1)!}{(k-1)! \cdot l!}\cdot p^{l+1} q^{k-1})=1[/imath], but I cannot see why this identity is true. Also when using FactorialMoment[GeometricDistribution[p], l] Mathematica suggests that [imath]E((k)_l)=(\frac{q}{p})^l l![/imath]. Thank you in Advance for your help. |
94722 | Stirling's formula: proof?
Suppose we want to show that [imath] n! \sim \sqrt{2 \pi} n^{n+(1/2)}e^{-n}[/imath] Instead we could show that [imath]\lim_{n \to \infty} \frac{n!}{n^{n+(1/2)}e^{-n}} = C[/imath] where [imath]C[/imath] is a constant. Maybe [imath]C = \sqrt{2 \pi}[/imath]. What is a good way of doing this? Could we use L'Hopital's Rule? Or maybe take the log of both sides (e.g., compute the limit of the log of the quantity)? So for example do the following [imath]\lim_{n \to \infty} \log \left[\frac{n!}{n^{n+(1/2)}e^{-n}} \right] = \log C[/imath] | 1628202 | Limit of Stirling's approximation as n goes to infinity.
I would like to see some detailed solution for [imath]\frac{n!}{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n}[/imath] as [imath]n\to\infty[/imath]. I know that the answer is 1 but i am not sure why? Here is what is tried: I rewrote the stirling's formula like this. [imath]\frac{(e/n)\cdot (2e/n)\cdot (3e/n)\cdots (ne/n)}{\sqrt{2\pi n}}\to 0[/imath] as [imath]n\to \infty[/imath]. I am not sure where I went wrong. |
1352897 | How many solutions exists for this equation?
[imath]x_1 + x_2 + x_3 + x_4 = 28[/imath] I tried to solve it with generating functions. Is it correct to get to the form of [imath]{(1 + x + {x^2} + {x^3} + ....)^4}[/imath] and this equals to: [imath]{(1 - x)^{ - 4}}[/imath] and then solve it with binomial expansion ? Is this correct ? every question similar to "how many solutions are axsist to this equation?" can be solved this method? I know there're a lot of methods and manipulations for this kind of problem. Can you confirm it's the right path for this kind of problems? It's like distribute 28 things into 4 objects. This is why I used generating functions. | 1065770 | The number of nonnegative integer triples with sum equal to [imath]12[/imath]
How many triples [imath]x,y,z[/imath] satisfy [imath]x+y+z=12[/imath], where [imath]x,y,z\ge 0[/imath] are integers? Note that these are not ordered: [imath] (12,0,0)[/imath] and [imath](0,12,0)[/imath] are treated as same. Progress I have tried the formula [imath]{13\choose 2}[/imath] to select values of [imath]x[/imath] and [imath]y[/imath] from [imath]0[/imath] to [imath]12[/imath] as [imath]z[/imath] would keep changing according to [imath]x[/imath] and [imath]y[/imath]. (That is, [imath]z=12-x-y[/imath].) The problem is that that certain combinations of [imath]x[/imath] and [imath]y[/imath] would give value of [imath]x+y[/imath] greater than [imath]12[/imath]. I can't seem to get around this. |
1352539 | No of ways of selecting r objects from n distinct objects, allowing repeated selections
I'm self studying discrete math from a books which states the formula for No of ways of selecting r objects from n distinct objects, allowing repeated selections as [imath]C(n+r-1, r)[/imath]. I couldn't understand completely how they derived this formula. Could someone please explain this to me? Kindly also do give some examples to help me understand how to use this formula. The book is elements of discrete mathematics by KC liu and DP mohapatra | 208377 | Combination with repetitions.
The formula for computing a k-combination with repetitions from n elements is: [imath]\binom{n + k - 1}{k} = \binom{n + k - 1}{n - 1}[/imath] I would like if someone can give me a simple basic proof that a beginner can understand easily. |
1353057 | Show that a prime p divides [imath]a^p - a[/imath].
Let [imath]p[/imath] be a prime. I want to show that [imath]p | a^p - a[/imath]. I want to use induction. I showed that this is true for the case [imath]a = 1[/imath]. I'm having trouble with the bridge. | 1331815 | [imath]p[/imath] divides [imath]n^p-n[/imath]
Its very easy to prove [imath]p\mid n^p-n[/imath] for p=3,5,7, it fails for p=9 because [imath] (n+1)^9-(n+1)= n^9+9n^8+36n^7+84n^6+126n^5+126n^4+84 n^3+36n^2+8n [/imath] and [imath]84= 2²\times 3\times7[/imath]. Is it true for any [imath]p[/imath] prime? I have checked many primes, but have no idea for the general case. |
1352431 | Prove the Axiom of Replacement implies the Axiom of Specification.
The Axiom of Specification says that: For a set [imath] A[/imath] and objects [imath]x\in A[/imath],for any object [imath]y[/imath] there exists a set: [imath] y \in \{x: P(x) \wedge x \in A\} \implies y \in A \text{ and } P(y)[/imath] Where [imath]P(x)[/imath] is a statement pertaining to [imath]x[/imath]. The Axiom of Replacement says that: For a set [imath] A[/imath] and objects [imath]x\in A[/imath],for any object y, suppose we have a statement [imath]P(x,y)[/imath], pertaining to [imath]x[/imath] and [imath]y[/imath] such that for each [imath]x[/imath] there is at most one [imath]y[/imath] for which [imath]P(x,y)[/imath] is true, then there exists a set, such that for any object z, [imath] z\in \{y:P(x,y) \text{ s.t } x \in A \} \implies P(x,z) \text{ for } x \in A[/imath] I can see that the AOR is stronger than the AOS. If we let [imath]P(x,y)[/imath] be whether [imath]y[/imath] is in the same set as [imath]x[/imath] then [imath]P(x,y)[/imath] is a true of false statement and hence it will follow that if [imath]P(x,y)[/imath] is true then [imath] y \in A \text{ and } P(y)[/imath]. Does this hold any water? I have sketched diagrams, to try and motivate a solution but have failed. I am editing this question simply because I don't understand the answers to the other posts. ( I have tried...) So I am going to add an answer to mine. MY ATTEMPT: Let A be a set, then for any object [imath]y[/imath] let [imath]\gamma[/imath] be defined as follows: [imath]\gamma(x) = x \text{ if } \theta(x) \text{ for } x \in A[/imath] [imath]\gamma(x) = y \text{ if } !\theta(x) \text{ for } x \in A[/imath] Then by the Axiom of replacement There exists a set [imath]B[/imath] such that [imath]B = \{\gamma(x): x \in A, \gamma(x) = x\} = \{x:\theta(x), x \in A\}[/imath] The last set in the above statement implies the Axiom of separation. | 32483 | How do the separation axioms follow from the replacement axioms?
It has come to my attention that the pairing axiom and the separation axiom schema are rarely listed since they follows from the replacement axioms. I see how this works for the pairing axiom, since we can take some set, say [imath]A=\{\emptyset,\{\emptyset\}\}[/imath], which exists by the power set axiom and the existence of the emptyset, and define a nominating formula on [imath]A[/imath] by [imath] \phi(u,v)\colon (u=\emptyset\land v=a)\lor(u\neq\emptyset\land v=b) [/imath] to see that [imath]\{a,b\}[/imath] is a set for any sets [imath]a[/imath] and [imath]b[/imath]. However, I don't see it for the separation schema. Could someone enlighten me on how they follow? Thanks. |
1352268 | Under what conditions is [imath]|x+y|=|x|+|y|[/imath] true?
What instance that this equation would be true? [imath]|x+y|=|x|+|y|[/imath] Given that [imath]x[/imath], [imath]y[/imath] are elements of real numbers. | 226569 | Equality holds in triangle inequality iff both numbers are positive, both are negative or one is zero
How do we show that equality holds in the triangle inequality [imath]|a+b|=|a|+|b|[/imath] iff both numbers are positive, both are negative or one is zero? I already showed that equality holds when one of the three conditions happens. |
1353387 | Prove that [imath]\{a_n\}[/imath] is bounded
Let [imath]\{a_n\}[/imath] be a sequence an [imath]L[/imath] a real number such that [imath]\lim_{n\to\infty} a_n = L[/imath] Prove that [imath]\{a_n\}[/imath] is bounded This reminds me of the bounded monotone convergence theorem (BMCT) but in reverse. So I was thinking of proving by somehow reversing the proof of the BMCT. Am I approaching this correct? And are there alternative ways of doing this that may be simpler? | 213936 | Prove: Convergent sequences are bounded
I don't understand this one part in the proof for convergent sequences are bounded. Proof: Let [imath]s_n[/imath] be a convergent sequence, and let [imath]\lim s_n = s[/imath]. Then taking [imath]\epsilon = 1[/imath] we have: [imath]n > N \implies |s_n - s| < 1[/imath] From the triangle inequality we see that: [imath] n > N \implies|s_n| - |s| < 1 \iff |s_n| < |s| + 1[/imath]. Define [imath]M= \max\{|s|+1, |s_1|, |s_2|, ..., |s_N|\}[/imath]. Then we have [imath]|s_n| \leq M[/imath] for all [imath]n \in N[/imath]. I do not understand the defining [imath]M[/imath] part. Why not just take [imath]|s| + 1[/imath] as the bound, since for [imath]n > N \implies |s_n| < |s| + 1[/imath]? |
1336830 | In most geometry courses, we learn that there's no such thing as "SSA Congruence".
In most geometry courses, we learn that there's no such thing as "SSA Congruence". That is, if we have triangles [imath]ABC[/imath] and [imath]DEF[/imath] such that [imath]AB = DE[/imath], [imath]BC = EF[/imath], and [imath]\angle A = \angle D[/imath], then we cannot deduce that [imath]ABC[/imath] and [imath]DEF[/imath] are congruent. However, there are a few special cases in which SSA "works". That is, suppose [imath]ABC[/imath] is a triangle. Let [imath]AB = x[/imath], [imath]BC = y[/imath], and [imath]\angle A = \theta[/imath]. For some values of [imath]x[/imath], [imath]y[/imath], and [imath]\theta[/imath], we can uniquely determine the third side, [imath]AC[/imath]. (a) Use the Law of Cosines to derive a quadratic equation in [imath]AC[/imath]. (b) Use the quadratic polynomial you found in part (a) in order to find conditions on [imath]x, y,[/imath] and [imath]\theta[/imath] which guarantee that the side [imath]AC[/imath] is uniquely determined. | 1353031 | In most geometry courses, we learn that there's no such thing as "SSA Congruence".
In most geometry courses, we learn that there's no such thing as "SSA Congruence". That is, if we have triangles [imath]ABC[/imath] and [imath]DEF[/imath] such that [imath]AB = DE[/imath], [imath]BC = EF[/imath], and [imath]\angle A = \angle D[/imath], then we cannot deduce that [imath]ABC[/imath] and [imath]DEF[/imath] are congruent. However, there are a few special cases in which SSA "works". That is, suppose [imath]ABC[/imath] is a triangle. Let [imath]AB = x[/imath], [imath]BC = y[/imath], and [imath]\angle A = \theta[/imath]. For some values of [imath]x[/imath], [imath]y[/imath], and [imath]\theta[/imath], we can uniquely determine the third side, [imath]AC[/imath]. (a) Use the Law of Cosines to derive a quadratic equation in [imath]AC[/imath]. (b) Use the quadratic polynomial you found in part (a) in order to find conditions on [imath]x, y,[/imath] and [imath]\theta[/imath] which guarantee that the side [imath]AC[/imath] is uniquely determined. I have solved part (a) and gotten [imath]z^2-2xz\cos (\theta) + (x^2-y^2)=0[/imath]. I don't quite get part (b). It is obvious [imath]90^{\circ}[/imath] is a solution but do I have to prove that this is right? Or do I need to prove that this is the only solution? Or is there other answers? My work: All I know is that the discriminant would be [imath]4x^2z^2\cos^2 \theta-4(-x^2+y^2)[/imath] P.S. My answer for part(a) is correct, right? |
1353440 | Compare between [imath]\ln(2)[/imath], [imath]\ln(-2)[/imath]
[imath]x^2 =(-x)^2, \;\forall x \in \mathbb{R}^+[/imath] [imath]\begin{align*} \therefore \ln(x^2) &=\ln(-x)^2\\ 2\ln(x)&=2\ln(-x)\\ \ln(x)&=\ln(-x) \end{align*}[/imath] If the statement above is correct, then compare between: [imath]\ln(2)[/imath] and [imath]\ln(-2)[/imath]? My try is, I think the statement is wrong, because [imath]x[/imath] must be positive? any help? | 799200 | What did Johann Bernoulli wrong in his proof of [imath]\ln z=\ln (-z)[/imath]?
Some people say, Johann Bernoulli has proven [imath]\ln z=\ln (-z)[/imath] in the following way [imath]\ln ((-z)^2 )=\ln(z^2)\;\;\;\Rightarrow\;\;\;2\ln(-z)=2\ln z\;\;\;\Rightarrow\;\;\;\ln (-z)=\ln z[/imath] While the statement is not true, I'm unable to figure out what exactly went wrong: Since [imath](-z)^2=z^2[/imath] for all [imath]z\in\mathbb{C}[/imath], it holds [imath]\ln ((-z)^2 )=\ln(z^2)[/imath], too. By definition of [imath]a^b=e^{b\ln a}[/imath] for all [imath]a\in\mathbb{C}^{-},b\in\mathbb{C}[/imath] it follows [imath]\ln (e^{2\ln z})=\ln (e^{2\ln (-z)})[/imath] for all [imath]z\in\mathbb{C}\setminus\mathbb{R}[/imath] Further, for [imath]z=re^{i\varphi}[/imath] we've got [imath]\ln (e^{2\ln z})=\ln (r^2)+i\pi +i\text{arg}_0(e^{i(2\varphi -\pi)})=2\ln r+2i\varphi[/imath] and [imath]\ln z=\ln r+i\varphi[/imath] So, it seems like [imath]2 \ln z=\ln (z^2)[/imath] holds, too (at least for all [imath]z\in\mathbb{C}\setminus\mathbb{R}[/imath]). So, what is wrong here? PS: Let's define [imath]\mathbb{C}^{-}:=\left\{z\in\mathbb{C} : \text{Re }z>0\vee \text{Im}\ne 0\right\}[/imath] |
1282622 | The box has minimum surface area
Show that a rectangular prism (box) of given volume has minimum surface area if the box is a cube. Could you give me some hints what we are supposed to do?? [imath]$[/imath][imath] [/imath] EDIT: Having found that for z=\frac{V}{xy}[imath] the function [/imath]A_{\star}(x, y)=A(x, y, \frac{V}{xy})[imath] has its minimum at [/imath](\sqrt[3]{V}, \sqrt[3]{V})[imath], how do we conclude that the box is a cube?? [/imath] We have that x=y[imath]. Shouldn't we have [/imath]x=y=z$ to have a cube?? | 569308 | Minimum area of the parallelepiped surface
Among all the retangular parallelpipeds of volume [imath]V[/imath], find one whose total surface área is minimum Using the Lagrange Multipliers method, I've found that it is a cube with dimensions [imath] \sqrt[3]{V} [/imath]. But I don't know how to prove that it is, indeed, a cube with those dimensions, since I couldn't prove that the function [imath]S_A(x,y,z)=2xy + 2xz + 2yz[/imath] (surface total area) have a minimum. Can you help me with it? Thanks in advance |
1351966 | Let [imath]f:K\to K[/imath] with [imath]\|f(x)-f(y)\|\geq ||x-y||[/imath] for all [imath]x,y[/imath]. Show that equality holds and that [imath]f[/imath] is surjective.
[imath]K[/imath] is a compact subset of [imath]\Bbb R^n[/imath] and [imath]f:K\rightarrow K [/imath] satisfies : [imath]\|f(x)-f(y)\|\geq \|x-y\|[/imath] Show that [imath]f[/imath] is bijective, and that : [imath]\|f(x)-f(y)\| = \|x-y\| [/imath] It's easy to show that [imath]f[/imath] is injective. But I can't think of a way to prove surjectivity. | 495648 | Can this intuition give a proof that an isometry [imath]f:X \to X[/imath] is surjective for compact metric space [imath]X[/imath]?
A prelim problem asked to prove that if [imath]X[/imath] is a compact metric space, and [imath]f:X \to X[/imath] is an isometry (distance-preserving map) then [imath]f[/imath] is surjective. The official proof given used sequences/convergent subsequences and didn't appeal to my intuition. When I saw the problem, my immediate instinct was that an isometry should be "volume-preserving" as well, so the volume of [imath]f(X)[/imath] should be equal to the volume of [imath]X[/imath], which should mean surjectivity if [imath]X[/imath] is compact. The notion of "volume" I came up with was the minimum number of [imath]\epsilon[/imath]-balls needed to cover [imath]X[/imath] for given [imath]\epsilon > 0[/imath]. If [imath]f[/imath] were not surjective, then because [imath]f[/imath] is continuous this means there must be a point [imath]y \in X[/imath] and [imath]\delta > 0[/imath] so that the ball [imath]B_\delta(y)[/imath] is disjoint from [imath]f(X)[/imath]. I wanted to choose [imath]\epsilon[/imath] in terms of [imath]\delta[/imath] and use the fact that an isometry carries [imath]\epsilon[/imath]-balls to [imath]\epsilon[/imath]-balls, and show that given a minimum-size cover of [imath]X[/imath] with [imath]\epsilon[/imath] balls, that a cover of [imath]X[/imath] with [imath]\epsilon[/imath]-balls could be found with one fewer ball if [imath]f(X) \cap B_\delta(y) = \emptyset[/imath], giving a contradiction. Can someone see a way to make this intuition work? |
1353647 | Limit of ratio of consecutive terms equals lmit of [imath]n^{th}[/imath] roots?
I got the following solved exercise from an old tutorial sheet. Evaluate the limit [imath]\lim_{n \rightarrow \infty} \Bigg{(}\frac{(2n)!}{(n!)^2}\Bigg{)}^{\frac{1}{n}}[/imath] So here is now the solution goes let [imath]a_n = \frac{2n}{(n!)^2}[/imath]. Then [imath]\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}=4[/imath]. Therefore [imath]\lim_{n \rightarrow \infty} \Bigg{(}\frac{2n}{(n!)^2}\Bigg{)}^{\frac{1}{n}}=4[/imath]. I dont understand how the limit of ratio of consecutive terms equals lmit of [imath]n^{th}[/imath] roots? Is this true in general? | 700370 | Relationship Between Ratio Test and Power Series Radius of Convergence
Let [imath] \{a_k\} [/imath] be a sequence of positive real numbers. Why does it hold that [imath] \liminf \frac{a_{k+1}}{a_{k}} \leq \liminf (a_k)^{\frac{1}{k}}\leq\limsup (a_k)^{\frac{1}{k}} \leq \limsup \frac{a_{k +1}}{a_ {k}} [/imath] This question has arisen from me trying to understand why [imath] \limsup \left| \frac{a_{k+1}}{a_k}\right| = R \implies \frac{1}{\limsup |a_k|^{1/k}} = R [/imath] |
1353683 | Example of open cover of (0,1) which has no finite subcover
Give an example of an open cover of the segment [imath](0,1)[/imath] which has no finite subcover. Example: Taking [imath]G_n=(0,1-1/n)[/imath] for [imath]n>1[/imath]. It is obvious that [imath](0,1)\subset \cup_{n=2}^{\infty}G_n[/imath] but [imath]\{G_n\}[/imath] has no finite subcover of [imath](0,1)[/imath]. Does this example satisfy the conditions of problem? | 257514 | Cover of (0,1) with no finite subcover & Open sets of compact function spaces
I just got back from my exam and these questions' solutions eluded me, it would be great to use the rest of my evening figuring these out... Q1: Find an open covering of the set [imath](0,1) \subset \mathbb{R}[/imath], say [imath]G =\{U_\alpha\}_{\alpha \in A}[/imath], (where [imath]A[/imath] is some indexing set) such that [imath]G[/imath] has no finite subcover. Q2: Let [imath]f: [0,1] \to [0,\infty) [/imath] be a continuous function. Let there be some [imath]c\in [0,1][/imath] such that [imath]f(c)[/imath] is non-zero. Prove that there exists an [imath]\epsilon \gt 0[/imath] such that the set: [imath]X_1=\{\ x\in[0,1]\ | \ f(x)\gt\epsilon\ \}[/imath] is non-empty, and open. |
1353504 | Question on proving quotient space is homeomorphic to circle
I am new to quotient spaces and was given this in class. I really have no idea how to solve. I tried one approach that my teacher said to be incorrect so I'd really appreciate the help on this. I am given [imath]S^1= \{x \in \mathbb{R}^2 : \|x\|=1 \}[/imath] i.e., the unit circle in [imath]\mathbb{R}^2 [/imath], and I am asked to show the quotient space [imath] S^1 / {\sim}[/imath], where [imath]\sim[/imath] is the equivalence relation [imath]x\sim{-x}[/imath], is homeomorphic to [imath] S^1 [/imath]. My attempt was to look at the quotient space as the set of points of the unit circle with only one of [imath](1,0)[/imath] and [imath](0,1)[/imath], which I now realize is incorrect as I know quotient space means gluing as opposed to omitting points. I am stumped as I cannot really figure out how to show an explicit homeomorphism between the quotient space and the unit circle in [imath]\mathbb{R}^2[/imath], or if I am even on the right track (maybe I need to use a theorem or something without a constructive proof). I really would appreciate the help on this. Thanks all. | 433912 | The space obtained by identifying the antipodal points of a circle
Take the unit circle [imath]S^1[/imath] in [imath]\Bbb{R}^2[/imath] and partition it into subsets which contain exactly two points, the points being antipodal (at opposite ends of a diameter). [imath]P[/imath] is the resulting identification space. Now, what is homeomorphic to [imath]P[/imath]? My intuition says that it must be a semicircle. But couldn't construct a homeomorphism. |
1354570 | Show that [imath]f(x) =\frac{1}{1+x^2}[/imath] is Lipschitz continuous on [imath]\mathbb{R}[/imath]?
Show that [imath]f(x) = \frac{1}{1+x^2}[/imath] is Lipschitz continuous on [imath]\mathbb{R}[/imath]? I know [imath]0<f(x)<1[/imath] but [imath]x[/imath] and [imath]y[/imath] are any number in [imath]\mathbb{R}[/imath]. How do you find a constant [imath]c[/imath] such that [imath]|f(x) - f(y)| < c |x - y| \ ?[/imath] | 997602 | Show [imath]1/(1+ x^2)[/imath] is uniformly continuous on [imath]\Bbb R[/imath].
Prove [imath]\frac 1{1+ x^2}[/imath] is uniformly continuous on [imath]\Bbb R[/imath]. Attempt: By definition a function [imath]f: E →\Bbb R[/imath] is uniformly continuous iff for every [imath]ε > 0[/imath], there is a [imath]δ > 0[/imath] such that [imath]|x-a| < δ[/imath] and [imath]x,a[/imath] are elements of [imath]E[/imath] implies [imath]|f(x) - f(a)| < ε.[/imath] Then suppose [imath]x, a[/imath] are elements of [imath]\Bbb R. [/imath] Now \begin{align} |f(x) - f(a)| &= \left|\frac1{1 + x^2} - \frac1{1 + a^2}\right| \\&= \left| \frac{a^2 - x^2}{(1 + x^2)(1 + a^2)}\right| \\&= |x - a| \frac{|x + a|}{(1 + x^2)(1 + a^2)} \\&≤ |x - a| \frac{|x| + |a|}{(1 + x^2)(1 + a^2)} \\&= |x - a| \left[\frac{|x|}{(1 + x^2)(1 + a^2)} + \frac{|a|}{(1 + x^2)(1 + a^2)}\right] \end{align} I don't know how to simplify more. Can someone please help me finish? Thank very much. |
394363 | Homology of [imath]S^1 \times (S^1 \vee S^1)[/imath]
I'm trying to solve question 2.2.9(b) in Hatcher's Algebraic Topology. Question: Calculate the homology groups of [imath]X=S^1 \times (S^1 \vee S^1)[/imath]. My attempt: I try to use the Mayer-Vietoris sequence. Let [imath]A=S^1 \times (S^1 \vee \text{small bit}) =S^1\times S^1[/imath] and [imath]B=S^1 \times ( \text{small bit} \vee S^1)=S^1 \times S^1[/imath]. (Unsure how to express this, but hopefully this is clear enough.) Then [imath]A\cap B=S^1\times \{\text{point}\}=S^1[/imath] and [imath]A\cup B=X[/imath]. The Mayer-Vietoris sequence then gives us an exact sequence in reduced homology as follows: [imath]0 \to \mathbb{Z}^2 \to \tilde{H}_2(X) \to \mathbb{Z} \to \mathbb{Z}^2 \to \tilde{H}_1(X) \to 0.[/imath] Is it possible from this exact sequence to determine the homology groups, in some algebraic fashion? Or do I need to determine what the maps are? How do I determine the maps? | 1790129 | Homology groups of [imath]S^1 \times (S^1 \vee S^1)[/imath]
How to compute all homology groups of [imath]S^1 \times (S^1 \vee S^1)[/imath]? Thank you. I don't see how to define a simplicial structure. |
1355135 | holomorphic functions and boundedness at the boundary
Let [imath]f[/imath] be a holomorphic function on a bounded region [imath]U=\{z:|z|<1\}.[/imath] Show that there is a sequence [imath]\{z_n\}[/imath] in [imath]U[/imath] such that [imath]|z_n|\rightarrow 1[/imath] and [imath]\{f(z_n)\}[/imath] is bounded. How can I approach this problem, without any continuity condition at the boundary? | 620881 | Behavior of holomorphic functions on the boundary of the unit disk
[imath]\textbf{Problem.}[/imath] Suppose [imath]f[/imath] is holomorphic on the unit disk [imath]\mathbb{D}[/imath]. Show there are points [imath]a_n\in \mathbb{D}[/imath], [imath]a\in \partial \mathbb{D}[/imath], and [imath]b\in \mathbb{C}[/imath] such that [imath]a_n\to a[/imath] and [imath]f(a_n)\to b[/imath], as [imath]n\to \infty.[/imath] [imath]\textbf{Attempt at a Solution:}[/imath] First, if [imath]f[/imath] has an essential singularity at some point [imath]a\in \partial\mathbb{D}[/imath], I think the statement follows easily. To contradiction suppose that the statement is false. That is, suppose that [imath]f(a_n)\to \infty[/imath] for all sequences [imath]\{a_n\}[/imath] in [imath]\mathbb{D}[/imath] where [imath]a_n\to a[/imath], [imath]a\in \partial\mathbb{D}.[/imath] Then [imath]\frac{1}{f(z)}\to 0, \hspace{1cm} \mbox{as} \hspace{1cm}|z|\to 1.[/imath] [imath]\textbf{ Case 1}[/imath]: If [imath]f[/imath] is nowhere vanishing on [imath]\mathbb{D}[/imath], then [imath]1/f(z)[/imath] would be holomorphic in [imath]\mathbb{D}[/imath] and can be extended continuously on [imath]\bar{\mathbb{D}}[/imath]. Then, by the maximum principle this would imply that [imath]1/f(z)\equiv 0[/imath] on all of [imath]\mathbb{D}[/imath]. But this contradicts the fact that [imath]f[/imath] is holomorphic (non-constant) in [imath]\mathbb{D}.[/imath] [imath]\textbf{Case 2}[/imath]: Suppose [imath]f[/imath] has a finite number of zeros in [imath]\mathbb{D}[/imath]. That is, let [imath]\{b_1,...,b_n\}[/imath] be points of [imath]\mathbb{D}[/imath] such that [imath]f(b_i)=0.[/imath] Let [imath]r=\min \{|z-b_i| \forall z\in \partial\mathbb{D}\}[/imath], and consider the annulus [imath]A=\{z\in\mathbb{C}: r<r_1<|z|<1\}[/imath] Then, [imath]1/f(z)[/imath] is holomorphic on [imath]A[/imath], and can be continuously extended to [imath]\bar A[/imath]. (Note that, [imath]1/f(z)[/imath] is already holomorphic for [imath]|z|=r_1[/imath]. Let [imath]g(z)=\frac{1}{f(z)}\frac{1}{f(r_1/z)}.[/imath] Then [imath]g(z)=0[/imath] for all [imath]z\in \partial \bar{A}[/imath]. By the maximum principle it follows that [imath]g(z)\equiv 0[/imath] on [imath]A[/imath], and by the identity theorem it follows that [imath]1/f(z)\equiv 0[/imath] on [imath]A[/imath]. But this, again, contradicts the fact that [imath]f[/imath] is holomorphic on [imath]\mathbb{D}[/imath] and thus on [imath]A[/imath] as well. [imath]\textbf{Case 3:}[/imath] This is the case where i'm not so sure what i'm doing is correct. Suppose, now, that [imath]f[/imath] has an infinite number of zeros in [imath]\mathbb{D}[/imath]. That is, let [imath]E=\{b_i\in \mathbb{D}: g(b_i)=0\}[/imath]. Then, [imath]E[/imath] cannot have an accumulation point in [imath]\mathbb{D}.[/imath] My idea was to take neighborhoods, [imath]\mathbb{B}_{r_i}(b_i)[/imath], and consider [imath]g(z)=\frac{1}{f(z)}\prod_{w\in A_w}\frac{1}{f(z/w)},[/imath] where [imath]A_w=\{w\in \mathbb{D}: |w-b_i|=r_i\},[/imath] for all [imath]z\in \overline{\mathbb{D}\setminus\bigcup \mathbb{B}_{r_i}(b_i)}.[/imath] Then, I would argue similarly as in Case 2. There is probably a much shorter method. I'd appreciate any thoughts or suggestions!! |
1355382 | If [imath]n[/imath] is an integer then [imath]n^2[/imath] is the same as [imath]0[/imath] or [imath]1\pmod 4[/imath]?
I have been stuck on this problem for awhile. How would i go about solving it, an explanation would be helpful as well. Show that if [imath]n[/imath] is an integer then [imath]n^2 \equiv 0[/imath] or [imath]1 \pmod 4[/imath]? | 99716 | The square of an integer is congruent to 0 or 1 mod 4
This is a question from the free Harvard online abstract algebra lectures. I'm posting my solutions here to get some feedback on them. For a fuller explanation, see this post. This problem is from assignment 6. The notes from this lecture can be found here. a) Prove that the square [imath]a^2[/imath] of an integer [imath]a[/imath] is congruent to 0 or 1 modulo 4. b) What are the possible values of [imath]a^2[/imath] modulo 8? a) Let [imath]a[/imath] be an integer. Then [imath]a=4q+r, 0\leq r<4[/imath] with [imath]\bar{a}=\bar{r}[/imath]. Then we have [imath]a^2=a\cdot a=(4q+r)^2=16q^2+8qr+r^2=4(4q^2+2qr)+r^2, 0\leq r^2<4[/imath] with [imath]\bar{a^2}=\bar{r^2}[/imath]. So then the possible values for [imath]r[/imath] with [imath]r^2<4[/imath] are 0,1. Then [imath]\bar{a^2}=\bar{0}[/imath] or [imath]\bar{1}[/imath]. b) Let [imath]a[/imath] be an integer. Then [imath]a=8q+r, 0\leq r<8[/imath] with [imath]\bar{a}=\bar{r}[/imath]. Then we have [imath]a^2=a\cdot a=(8q+r)^2=64q^2+16qr+r^2=8(8q^2+2qr)+r^2, 0\leq r^2<8[/imath] with [imath]\bar{a^2}=\bar{r^2}[/imath]. So then the possible values for [imath]r[/imath] with [imath]r^2<8[/imath] are 0,1,and 2. Then [imath]\bar{a^2}=\bar{0}[/imath], [imath]\bar{1}[/imath] or [imath]\bar{4}[/imath]. Again, I welcome any critique of my reasoning and/or my style as well as alternative solutions to the problem. Thanks. |
1355632 | Apply Stone- Weierstrass Theorem
Suppose [imath]f: [0,1]\to \mathbb R[/imath] is continuous and [imath]\int_{0}^{1} f(x)e^{nx} \mathsf dx=0[/imath] for every [imath]n[/imath]. Prove that [imath]f(x)=0[/imath] for all [imath]x \in[0,1][/imath]. Since [imath]f[/imath] is continuous on [imath][0,1][/imath], by Stone-Weierstrass theorem, we can say that there exists a sequence of polynomials [imath]P_n[/imath] that converges uniformly to [imath]f[/imath] on [imath][0,1][/imath]. What the next step? | 1102388 | Proving that [imath]\int_0^1 f(x)e^{nx}\,{\rm d}x = 0[/imath] for all [imath]n\in\mathbb{N}_0[/imath] implies [imath]f(x) = 0[/imath]
I'm trying to show that if [imath]f[/imath] is a continuous function on [imath][0,1][/imath] and [imath]\int_0^{1} f(x)e^{nx}\,{\rm d}x = 0[/imath] for all [imath]n = 0, 1, 2, \dots[/imath], then [imath]f(x) = 0[/imath]. I'd like to use Weierstrass approximation theorem to find a sequence of polynomials [imath]p_m[/imath] that converge uniformly to [imath]f(x)[/imath]. Then we could say [imath]\lim\limits_{m\to \infty} \int p_m(x)e^{nx}\,{\rm d}x = \int f(x)e^{nx}\,{\rm d}x = 0[/imath], but I'm struggling to deduce that then all the [imath]p_m[/imath] are zero which would give the result. |
1355736 | Prove/Disprove existence of a set
I would like to know if my arguments are correct. Prove/Disprove existence of a set X [imath]\subset[/imath] P([imath]\mathbb{N}[/imath]) , [imath] |X|=\aleph[/imath] and for every [imath] A,B \in X [/imath] ,[imath] A \subset B [/imath] or [imath] B \subset A[/imath] I thought about define a bijection function [imath] f:X \to \mathbb{N}\bigcup [/imath] {[imath]\aleph_0[/imath]} A [imath]\mapsto[/imath] [imath]|A|[/imath] and [imath] \mathbb{N}\bigcup [/imath] {[imath]\aleph_0[/imath]} is countable [imath]\to[/imath] [imath]|X|\neq \aleph [/imath] thanks | 253192 | Chain of length [imath]2^{\aleph_0}[/imath] in [imath] (P(\mathbb{N}),\subseteq)[/imath]
How can I find a chain of length [imath]2^{\aleph_0}[/imath] in [imath] (P(\mathbb{N}), \subseteq )[/imath]. The only chain I have in mind is [imath]\{\{0 \},\{0,1 \},\{0,1,2 \},\{ 0,1,2,3\},...,\{\mathbb{N} \} \}[/imath] But the chain is of length [imath]\aleph_0[/imath], right? |
1355171 | trigonometric finite series equals to polynomial function
I am interest to prove the equation below : [imath] \sum_{k=1}^m \tan^2\left(\frac{k\pi}{2m+1}\right) = m(2m+1) [/imath] you can understand better the first member of the equation here: WolframAlpha (mark the whole url with your mouse because i dont know why the link isn't blue at all) sorry but i am not familiar writing equations here . i hope to understand. so what's the problem in formula i cannot eliminate the trigonometric functions to prove this series is a polynomial . any idea how to manipulate the formula ? | 173447 | Proving [imath]\sum\limits_{l=1}^n \sum\limits _{k=1}^{n-1}\tan \frac {lk\pi }{2n+1}\tan \frac {l(k+1)\pi }{2n+1}=0[/imath]
Prove that [imath]\sum _{l=1}^{n}\sum _{k=1}^{n-1}\tan \frac {lk\pi } {2n+1}\tan \frac {l( k+1) \pi } {2n+1}=0[/imath] It is very easy to prove this identity for each fixed [imath]n[/imath] . For example let [imath]n = 6[/imath]; writing out all terms in a [imath]5 \times 6[/imath] matrix, we obtain: [imath]\begin{matrix} \tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13} & \tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13} & \tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13} & \tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13} & \tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13} \\ \tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13} & \tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13} & \tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13} & \tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13} & \tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13} \\ \tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13} & \tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13} & \tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13} & \tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13} & \tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13} \\ \tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13} & \tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13} & \tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13} & \tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13} & \tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13} \\ \tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13} & \tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13} & \tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13} & \tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13} & \tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13} \\ \tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13} & \tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13} & \tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13} & \tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13} & \tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13} \end{matrix}[/imath] one can notice then, that the first column vanish the fourth one : [imath]\tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13}[/imath] [imath]\tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13}[/imath] [imath]\tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13}[/imath] [imath]\tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13}[/imath] [imath]\tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13}=-\tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13}[/imath] [imath]\tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13}[/imath] and the third column vanish the fifth one : [imath]\tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}[/imath] [imath]\tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}[/imath] [imath]\tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13}[/imath] [imath]\tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13}[/imath] [imath]\tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13}[/imath] [imath]\tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13}=-\tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13}[/imath] while the second column is self-vanishing: [imath]\tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13}[/imath] [imath]\tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13}[/imath] [imath]\tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13}[/imath] . So the equality occurs. But how to generalize the proof? |
1355435 | limit of quadartic variation
I am trying to understand why : [imath][\int_{0}^{t}a_s dB_s]=\int_{0}^{t}a_s^2 ds[/imath] [] is the 2-variation process, [imath]B[/imath] is brownian motion in the proof I have seen they used Riemman-sums to get an expression to the 2-variation of LHS as limit of: [imath] \sum_{j=1}^{n-1} \alpha_s(t_j)^2(B_j-B_{j-1})^2 [/imath] [imath]t_j[/imath] is parition points of [0,t] now why is this expression equals to Riemman sum of [imath]\int_{0}^{t}a_s^2 ds[/imath] thank you | 822284 | (Ito lemma proof): convergence of [imath]\sum_{i=0}^{n-1}f(W(t_{i}))(W(t_{i+1})-W(t_{i}))^{2}.[/imath]
The purpose of this question is to complete my personal exposition on the rigorous proof of Ito's lemma. I have consulted more than half a dozen mathematical finance texts and not a single one, for all its importance, rigorously proves this. The closest I've seen is Shreve's treatment in Stochastic Calculus for Finance Vol II Let [imath]\{W(t)\}_{t\geq0}[/imath] be a standard Brownian motion. Let [imath]f[/imath] be a smooth function, [imath]T>0[/imath], and [imath]\Pi=\{t_{0}=0,t_{1},\ldots,t_{n}=T\}[/imath] be a partition of [imath][0,T][/imath]. Consider the random variable [imath]X(\omega)=\lim_{||\Pi||\to0}\;\sum_{i=0}^{n-1}f(W(t_{i}))(W(t_{i+1})-W(t_{i}))^{2}.[/imath] If [imath]f\equiv1[/imath], then almost surely [imath]X\equiv T[/imath]. This is the standard result that Brownian motion accumulates quadratic variation at a unit rate. If we make the substitution [imath](W(t_{i+1})-W(t_{i}))^{2}\approx(t_{i+1}-t_{i})[/imath], then after [imath]||\Pi||\to0[/imath] and summing, the errors committed with this approximation go to [imath]0[/imath] and we recover almost surely [imath]X\equiv \int_{0}^{T}f(W(s))\;ds.[/imath] Is there a way to make this more precise? I would like to adapt the usual proof that [imath]\lim_{||\Pi||\to0}\;\sum_{i=0}^{n-1}(W(t_{i+1})-W(t_{i}))^{2}=T,[/imath] which proceeds by showing that the expectation of the sampled quadratic variation (i.e. the sum on any partition) is [imath]T[/imath] and that the variance tends to [imath]0[/imath] as the partition becomes finer. Indeed, [imath]\mathbb{E}[(W(t_{i+1})-W(t_{i}))^{2}]=\text{Var}[W(t_{i+1})-W(t_{i})]=t_{i+1}-t_{i}[/imath] and [imath]\text{Var}[(W(t_{i+1})-W(t_{i}))^{2}]=2(t_{i+1}-t_{i})^{2}[/imath] and the claim follows after summing, taking limits, and performing a simple estimate on the variance. Based on my efforts, it does not seem like such simple formulas hold for the expectation and variance when the quantity is multiplied by [imath]f(W(t_{i})).[/imath] A heuristic for the approximation above can be obtained by noting that (if [imath]\Pi[/imath] is a uniformly spaced partition) [imath](W(t_{i+1})-W(t_{i}))^{2}=\frac{T}{n}\left(\frac{W(t_{i+1})-W(t_{i})}{\sqrt{t_{i+1}-t_{i}}}\right)^{2}.[/imath] Thus if [imath]Y_{i}:=\left(\frac{W(t_{i+1})-W(t_{i})}{\sqrt{t_{i+1}-t_{i}}}\right)^{2},[/imath] then (being a collection of squared normal iid random variables with mean [imath]1[/imath]) the law of large numbers implies [imath]\sum_{i=0}^{n-1}(W(t_{i+1})-W(t_{i}))^{2}=T\sum_{i=0}^{n-1}\frac{Y_{i+1}}{n}\to T\;\text{as}\;n\to\infty.[/imath] Thus, while [imath](W(t_{i+1})-W(t_{i}))^{2}\neq t_{i+1}-t_{i}[/imath], the law of large numbers says that the error committed with this approximation cancels out in the limit. In other words, we may write [imath](W(t_{i+1})-W(t_{i}))^{2}=(t_{i+1}-t_{i})+\phi(t_{i})[/imath] where [imath]\sum_{i=0}^{n-1}\phi(t_{i}))\to0\;\text{as}\;n\to\infty.[/imath] What I'm trying to rigorously establish is that the above equations/limits all remain valid if we multiply them by [imath]f(W(t_{i}))[/imath]. In other words, is it correct to write [imath]f(W(t_{i}))(W(t_{i+1})-W(t_{i}))^{2}=f(W(t_{i}))(t_{i+1}-t_{i})+\phi(t_{i}))[/imath] where [imath]\sum_{i=0}^{n}f(W(t_{i}))\phi(t_{i})\to0\;\text{as}\;n\to\infty?[/imath] The obvious problem with the above heuristic is that the random variables [imath]Y_{i}':=f(W(t_{i}))Y_{i}:=f(W(t_{i}))\left(\frac{W(t_{i+1})-W(t_{i})}{\sqrt{t_{i+1}-t_{i}}}\right)^{2}[/imath] are no longer independent, so the law of large numbers cannot be used in the manner above. |
1356418 | Prove that [imath]5\cos A+3\cos\left(A+\frac{\pi}{3}\right)+3[/imath] lies between [imath]-4[/imath] and [imath]10[/imath]
After much of solving, I arrive at the expression [imath]\frac{1}{2}\left[13 \cos A-3\sqrt{3}\sin A +6\right].[/imath] I can't conclude anything from this. Please help. Hints are welcome. | 1323759 | How to find the maximum value for the given $x\cos{\lambda}+y\sin{\lambda}$?
How do I find the maximum value for [imath]$x\cos {λ}+y\sin {λ}$[/imath] for fixed [imath]x,y[/imath] and varying [imath]$λ$[/imath] ? I would appreciate if only hint is given. |
1356043 | Construction of a triangle with given angle bisectors
given three distinct lines [imath]g,h,l[/imath] meeting in one point [imath]P[/imath]. I want to construct a triangle with vertices on [imath]g,h,l[/imath] such that those lines [imath]g,h,l[/imath] become its angle bisectors. In general, if we consider a triangle [imath]\Delta(ABC)[/imath] with angles [imath]\alpha,\beta,\gamma[/imath] whose angle bisectors meet at a point [imath]P[/imath], then I recognized the following relations between the angles: [imath]\angle (APB)=\frac{1}{2}(\pi+\gamma), \angle (BPC)=\frac{1}{2}(\pi+\alpha), \angle (CPA)=\frac{1}{2}(\pi+\beta).[/imath] Now suppose we choose points [imath]C,C'\in g[/imath] on different sides of [imath]g[/imath] with respect to [imath]P[/imath] and let [imath]Q\in l,R\in h[/imath] such that [imath]C'[/imath] lies in the interior of the angle [imath]\angle (QPR)[/imath]. I started the construction as follows: Construct the angle [imath]\angle (QPR)-\pi/2[/imath] on one side of [imath]g[/imath] at the point C. Then we get a intersection point [imath]A[/imath] with the ray [imath]\overrightarrow{PQ}[/imath]. Construct the angle [imath]\angle (QPR)-\pi/2[/imath] on the other side of [imath]g[/imath] at the point C. Then we get a intersection point [imath]B[/imath] with the ray [imath]\overrightarrow{PR}[/imath]. Now connect the points [imath]A,B[/imath] But now I'm worried about the angles at [imath]A,B[/imath]. I do not know if the angles have the right size and the lines [imath]l,h[/imath] through [imath]A,B[/imath] are indeed angle bisectors? Maybe my construction is wrong. Do you know how to do it right? Best wishes | 1341896 | Construct the triangle with given angle bisectors
given three lines [imath]\ell_1,\ell_2, \ell_3 [/imath] which intersect in one point [imath]P[/imath]. How can one construct a triangle such that the given lines become its angle bisectors? So far I tried to find conditions on how the three given lines have to intersect such that the desired triangle exists. There are altogether six rays [imath]\omega_1^1, \omega_1^2; \omega_2^1,\omega_2^2;\omega_3^1,\omega_3^2[/imath] - any line [imath]\ell_j[/imath] determines the two rays [imath]\omega_j^1,\omega_j^2[/imath]- emanating from the common point [imath]P[/imath]. Any of the three vertices [imath]V_1,V_2,V_3[/imath] of the desired triangle has to lie on exactly one line, i.e. [imath]V_j\in \ell_j=\omega_j^1\cup \omega_j^2[/imath]. Hence we have either [imath]V_j\in \omega_j^1[/imath] or [imath]V_j\in \omega_j^2[/imath]. Therefore the triangle should exist if and only if it is possible to choose three rays [imath]\tilde{\omega}_1\in \lbrace \omega_1^1, \omega_1^2\rbrace [/imath], [imath]\tilde{\omega}_2\in \lbrace \omega_1^2, \omega_1^2\rbrace[/imath], [imath]\tilde{\omega}_3\in \lbrace \omega_1^3, \omega_1^3\rbrace[/imath] such that any three of the rays [imath]\tilde{\omega_1}, \tilde{\omega_2}, \tilde{\omega_3}[/imath] form an angle less than [imath]\pi[/imath]. But I do not know how to construct the triangle?! I started to draw a circle around [imath]P[/imath] with any radius. Now I can choose a Point [imath]V_1[/imath] on the ray [imath]\tilde{\omega_1}[/imath] outside the circle. And now I can draw the tangents through [imath]V_1[/imath] to the circle. I think those tangents will meet the other rays at some points [imath]V_2,V_3[/imath] (Why?). Now we can draw the line [imath]V_2,V_3[/imath]. But this line needs to be tangent to the circle and I'm not sure about that. Will this construction work, or is it done in a different way? Best regards |
1356474 | How many arrangements exist (a + b + c = 4)
For example, [imath]a + b + c = 4[/imath] Solving this using stars and bars You have [imath]4[/imath] stars and [imath]2[/imath] bars: [imath] x | x | xx[/imath] For example. Then what does [imath]\binom{6}{2}[/imath] mean? The number of arrangements of the two bars out of [imath]6[/imath] objects? If so, is this a general case? | 432541 | Solutions to [imath]a+b+c=12[/imath], [imath]a, b, c \in \mathbb{N}_0[/imath]
Let [imath]a, b, c \in \mathbb{N}_0[/imath]. If [imath]a+b+c=12[/imath], how many solutions [imath](a,b,c)[/imath] satisfy the equation? Is the answer: [imath]729[/imath] |
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