qid
stringlengths 1
7
| Q
stringlengths 87
7.22k
| dup_qid
stringlengths 1
7
| Q_dup
stringlengths 97
10.5k
|
|---|---|---|---|
2635764 | Derivative of [imath]|x|[/imath] using definition of the derivative.
Using the definition: [imath]f’(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}[/imath] i try to find the derivative of the absolute value function. My first two steps are, [imath]f’(x)=\lim_{h\to 0} \frac{|x+h|-|x|}{h}[/imath] rewritten as [imath]\lim_{h\to 0} \frac{\sqrt{(x+h)^2}-\sqrt{x^2}}{h}[/imath] It’s at this point that I’m not sure about my last step, and if it’s correct then I’m lost on how to proceed in order to get the h cancellation to evaluate the limit. I apologize if this is a duplicate. Thank you. | 83861 | Finding the Derivative of |x| using the Limit Definition
Please Help me derive the derivative of the absolute value of x using the following limit definition. [imath]\lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} [/imath] I have no idea as to how to get started.Please Help. Thank You |
2635187 | Unbiasedness and Consistency of the following Estimator
Let [imath]T(x)=\frac{\sum_{i=1}^{n}X_{i}+\frac{\sqrt{n}}{2}} {n+\sqrt{n}}[/imath] Where, [imath]X_{i}[/imath]'s are from [imath]B(1,\theta)[/imath] or Bernaulli. Then, comment on the Unbiasedness and Consistency of the given estimator. My approach I think the given estimator is biased and Consistent. The Estimator is biased which directly follows from structure of estimator. As far as consistency is concerned, the estimator can be written in the following form: [imath]T(x)=\frac{\overline{X}}{1+\frac{1}{\sqrt{n}}}+\frac{\sqrt{n}}{2(n+\sqrt{n})}=a_n\overline{X}+b_n[/imath] [imath]\overline{X}[/imath] is consistent for [imath]\theta[/imath] and [imath]a_n \rightarrow 1[/imath] and [imath]b_n \rightarrow 0[/imath]. Hence, the given estimator is consistent. But my manual, gives the answer that given estimator is biased and inconsistent. How can this be possible? I hope I used the theorem correctly. Thanks | 1091493 | Checking the consistency and Bias of [imath]\frac{\sum X_i +\sqrt{n}/2}{n+\sqrt{n}}[/imath]
Let [imath]X_1,\ldots,X_n[/imath] be i.i.d. [imath]B(1,\theta)[/imath] random variables, [imath]0<\theta<1[/imath]. Then, as an estimator [imath]\theta[/imath], check if [imath]T(X_1,\ldots,X_n)= \dfrac{\sum_{i=1}^n X_i +\sqrt{n}/2}{n+\sqrt{n}}[/imath] is consistent and/or unbiased. [imath]T=\frac{\frac{1}{n}\sum_{i=1}^n X_i +\frac{1}{2\sqrt{n}}}{1+\frac{1}{\sqrt{n}}}[/imath] [imath]T=\frac{\bar{X} +\frac{1}{2\sqrt{n}}}{1+\frac{1}{\sqrt{n}}}[/imath] [imath]E(T)=\frac{E(\bar{X} +\frac{1}{2\sqrt{n}})}{1+\frac{1}{\sqrt{n}}}=\dfrac{\theta +\frac{1}{2\sqrt{n}}}{1+\frac{1}{\sqrt{n}}}.[/imath] So, [imath]T[/imath] is biased. Consistency: [imath]\lim\limits_{n \to\infty}E(T)=\lim\limits_{n \to\infty}\dfrac{E(\theta +\frac{1}{2\sqrt{n}})}{1+\frac{1}{\sqrt{n}}}[/imath] So [imath]T[/imath] is consistent. But The given answer is "neither unbiased nor consistent" Where did I go wrong ? Please advise. |
2635953 | Is the gradiant a column or a row?
Suppose we have [imath]f:\mathbb{R^2}\rightarrow \mathbb{R}[/imath]. Vectors which [imath]f[/imath] act on are column vectors i.e a [imath]2 \times 1[/imath] matrix. Is the gradiant [imath]\nabla f[/imath] then a row vector? And why is this logical? | 54355 | The Gradient as a Row vs. Column Vector
Kaplan's Advanced Calculus defines the gradient of a function [imath]f:\mathbb{R^n} \rightarrow \mathbb{R}[/imath] as the [imath]1 \times n[/imath] row vector whose entries respectively contain the [imath]n[/imath] partial derivatives of [imath]f[/imath]. By this definition then, the gradient is just the Jacobian matrix of the transformation. We also know that using the Riesz representation theorem, assuming [imath]f[/imath] is differentiable at the point [imath]x[/imath], we can define the gradient as the unique vector [imath]\nabla f[/imath] such that [imath] df(x)(h) = \langle h, \nabla f(x) \rangle, \; h \in \mathbb{R}^n [/imath] Assuming we ignore the distinction between row vectors and column vectors, the former definition follows easily from the latter. But, row vectors and column vectors are not the same things. So, I have the following questions: Is the distinction here between row/column vectors important? If (1) is true, then how can we know from the second defintion that the vector in question is a row vector and not a column vector? |
2635805 | Prove continuity of a function defined as a Lebesgue integral
Let [imath]f:\mathbb{R} \mapsto [0,+\infty][/imath] , [imath]\lambda[/imath] the Lebesgue measure in [imath]\mathbb{R}[/imath] and [imath]f\in L^1(\lambda)[/imath]. Show that the function [imath]F(x)=\int \limits_{-\infty}^xf(t)dt[/imath] is continuous. I tried to use the [imath]\epsilon- \delta[/imath] definition of continuity but I am not sure that will work, or how it will work. Any ideas? | 1775965 | If [imath]f[/imath] is nonnegative and integrable then [imath]F(x) = \int_{-\infty}^x f[/imath] is continuous.
I'm learning about measure theory, specifically the Lebesgue integral of nonnegative functions, and need help with the following problem. Let [imath]f:\mathbb{R}\to[0,\infty)[/imath] be measurable and [imath]f\in L^1[/imath]. Show that [imath]F(x)=\int_{-\infty}^x f[/imath] is continuous. I know is isn't much but the only thing a could think of is that given [imath]x, y \in \mathbb{R}[/imath] with [imath]x < y[/imath] we note that [imath]F(x) \leq F(y)[/imath], i.e. [imath]F[/imath] is increasing. So maybe we can apply one of the convergence theorems of Lebesgue integration theory. I was also wondering if this problem can be solved using only Riemann integration theory. |
2635523 | The integral [imath]\int \sqrt{\ x \sqrt[3]{\ x \sqrt[4]{\ x \sqrt[5]{\ x \cdots}}}} dx[/imath]
I have no idea how to solve the second question from 2018 MIT integration bee, [imath]\int \sqrt{\ x \sqrt[3]{\ x \sqrt[4]{\ x \sqrt[5]{\ x \cdots}}}} dx = \frac{x^{e-1}}{e-1}.[/imath] I tried to to find a value for the roots and i tried converting it to a power, and I tried making a [imath]u[/imath] substitution but nothing worked out, while i think that the right approach to this question is to convert the roots to a power, i cant find the right approach to the question. | 2635600 | Computing [imath]\int \sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x\cdots}}}} \,\mathrm{d}x[/imath]
I've just read a post that has been put on hold. The question was about computing [imath] \int_{ }^{ }\sqrt{x\sqrt[3]{x\sqrt[4]{x\sqrt[5]{x \cdots}}}}\, \mathrm{d}x [/imath] My attempt was to find explicitely the integrand hence I introduced the sequence [imath] a_1=1\ \text{ and }\ a_{n+1}=\left(xa_n\right)^{1/(n+1)} [/imath] hence [imath] a_2=\sqrt{xa_1}=\sqrt{x}, \ a_3=\sqrt[3]{xa_2}=\sqrt[3]{x\sqrt{x}} \ \dots [/imath] But the power is reversed, how can I ffind the corresponding [imath](a_n)_{n\in \mathbb{N}}[/imath] ? |
2637015 | Find the limit for the sum of a series with [imath]a_k = \frac{1}{k\cdot 2^k}[/imath]
I have to find the following limit: [imath]\lim_{n \to \infty} \sum_{k=1}^n\frac{1}{k\cdot 2^k}[/imath] The exercise also provides me with the following function definition: [imath] f_n(x) = \frac{x^n}{1-x} [/imath] I tried to integrate this function by parts on the interval [imath][0, \frac12][/imath] and it led me something like this: [imath] I_n = \int_0^\frac12 \frac{x^n}{1-x} \, dx = \log2 - \frac{1}{2^n} + n\int_0^\frac12 x^{n-1} \log(x-1) \, dx [/imath]. This seems to look like a part of my sum terms but I don't know how to finish the integration and I cannot complete the limit. Could you help me please? | 1106587 | Compute the sum of the series [imath]\sum_{n=1}^{\infty} \frac{1}{n \cdot 2^n}[/imath]
What would be the sum of the following series? [imath]\sum_{n=1}^{\infty} \frac{1}{n \cdot 2^n}[/imath] Thanks |
2637199 | "Such that" symbol in logic
So I am taking a logic course and I supposed to write down a couple of math statements using only mathematical logic symbols (the most basic ones but I can use disjunction and there exists). There are statements that require "such that" symbol in my opinion but I am pretty sure that is implied and I am not sure how to proceed . For example, Any two real numbers can divide some one number. I don't have the symbol for divide so I would have to work around that. What I would do is write for all [imath]a,b[/imath], there exists [imath]x[/imath] "such that" there exists [imath]v_1,v_2[/imath] "such that" [imath]x=av_1[/imath] and [imath]x=bv_2[/imath]. Is this correct? Do I just ignore the such that and not write it when I use symbols? | 309506 | Symbol for “such that” (not in set)
If [imath]A[/imath] is a set, we can use the set notation [imath]A= \{ b \mid\text{property $p_1$ of $b$}\}[/imath] But say [imath]A[/imath] is an element like [imath]b[/imath], [imath]A = b \mid \text{property $p_1$ of $b$}[/imath] is this a usual notation? I am trying to say that [imath]A[/imath] is a [imath]b[/imath] that such that( [imath]\mid[/imath] ) it satisfies property [imath]p_1[/imath] of [imath]b[/imath], and assume that exactly one [imath]b[/imath] satisfies property [imath]p_1[/imath]. Otherwise, is there a more usual convention to express this? |
2637325 | On an example in Macdonald's Symmetric Functions and Hall Polynomials on Paritions and their Frobenius Notation
I am following Macdonald's Symmetric Functions and Hall Polynomials, and in his example-section. He stated the following without any proof, and I'm trying to understand how he reached the expression. Any assistance will be appreciated. A partition [imath]\lambda = (\lambda_1, \lambda_2, \cdots, \lambda_n)[/imath] can be written in the Frobenius notation [imath](\alpha_1,\cdots,\alpha_r \ \vert \ \beta_1,\cdots,\beta_r)[/imath], where [imath]\alpha_i = \lambda_i - i[/imath], [imath]\beta_i = \lambda'_i - 1, \ [/imath] [imath](1 \leq i \leq r)[/imath]. [imath]r[/imath] is the length of the intersection of the young diagram of [imath]\lambda[/imath] with the diagonal [imath]i=j[/imath]. Essentially, in the Young Diagram of [imath]\lambda[/imath], if we discard the diagonal, which contains [imath]r[/imath] squares, counting left to right to the right of the diagonal gives us [imath]\alpha[/imath], and counting downwards to the left of diagonal gives us [imath]\beta[/imath]. Macdonald states: [imath]\sum_{i=1}^{n}t^i\big(1-t^{-\lambda_i}\big)=\sum_{j=1}^{r}\big(t^{\beta_j + 1}-t^{-\alpha_j}\big)[/imath] | 2619907 | Permutation induced by a partition
Let [imath]\lambda[/imath] be a partition of length [imath]n[/imath] and suppose its largest diagonal block, the Durfee square of [imath]\lambda[/imath], has size [imath]r[/imath]. By this I mean that [imath]\lambda = (\lambda_1,\ldots,\lambda_n)[/imath] is a non-increasing sequence of numbers, which I depict by the following diagram \begin{align*} &\square \cdots \square \square \quad (\lambda_1 \text{ squares })\\ &\square \cdots \square \quad (\lambda_2 \text{ squares }) \\ &\quad\vdots \\ &\square \quad(\lambda_n \text{ squares }) \end{align*} and the largest [imath]i\times i[/imath] block one can fit to the topmost left is of size [imath]r\times r[/imath]. The conjugate partition [imath]\lambda'[/imath] is given by reflecting the drawing above along the diagonal. If [imath]\alpha_i[/imath] and [imath]\beta_i[/imath] denote the sequence of numbers of blocks to the right of the diagonal in the [imath]i[/imath]th row, and below the diagonal in the [imath]i[/imath]th column, we write [imath]\lambda = (\alpha_1,\ldots,\alpha_r\mid \beta_1,\ldots,\beta_r)[/imath]. For example, for the partition [imath](5,4,2,1,1)[/imath] has diagram [imath] \begin{align} &\blacksquare\square\square\square\square\\ &\square\blacksquare\square\square\\ &\square\square\\ &\square\\ &\square \end{align} [/imath] and its conjugate is [imath](5,3,2,2,1)[/imath]. Its diagonal has length [imath]2[/imath], and in Frobenius notation we have [imath]\lambda = (4,2\mid 4,1)[/imath]. How can one show that the numbers [imath]\lambda_1',\lambda_2'-1,\ldots,\lambda_r'-r+1,r+1-\lambda_{r+1},\ldots,n-\lambda_n[/imath] form a permutation of [imath]1,\ldots,n[/imath]? If [imath]\lambda = (\alpha\mid \beta)[/imath] in Frobenius notation, this is equivalent to the identity [imath]\sum_{i=1}^n t^i (1-t^{-\lambda_i}) = \sum_{j=1}^r (t^{\beta_j+1}-t^{-\alpha_j})[/imath] which is Example 4 in page 11 of MacDonald's Symmetric Functions and Hall Polynomials, which he states without proof, so presumably this is easy. Continuing with the example, we compute that the sequence for [imath]\lambda = (5,4,2,1,1)[/imath] is [imath]5,3-1,3-2,4-1,5-1=5,2,1,3,4[/imath] a permutation of [imath]1,2,3,4,5[/imath]. In [imath](1.7)[/imath] MacDonald proves that if we take [imath]m\geqslant \lambda_1[/imath] and [imath]n\geqslant\lambda_1'[/imath] then the numbers [imath]\lambda_i+n-i,1\leqslant i\leqslant n,\quad n-1+j-\lambda_j',1\leqslant j\leqslant m[/imath] are a permutation of [imath]0,\ldots,m+n-1[/imath] by labelling the vertical and horizontal edge-lines on the diagram of [imath]\lambda[/imath] fitted inside the diagram of [imath](m^n)[/imath], but I haven't been able to come up with a proof similar to this. |
2049852 | If [imath]p + q = 1[/imath] prove that for any natural [imath]n, m[/imath] following is true: [imath](1 - p^n)^m + (1 - q^m)^n \ge 1[/imath]
Let [imath]p, q \in \mathbb R[/imath] be positive reals for which [imath]p + q = 1[/imath]. How to prove that for any two natural numbers [imath]n, m[/imath] the following inequality is true? [imath](1 - p^n)^m + (1 - q^m)^n \ge 1[/imath] I don't have a big knowledge about solving inequalities, so I tried to use cauchy-schwarz inequality, binomial theorem and some other baisc techniques, but it lead me nowhere. I've been thinking about it for a long time and now I'm completly stuck. | 2639594 | Prove that for positive integers [imath]m, n[/imath] and two positive numbers [imath]p, q[/imath] satisfying [imath]p+q = 1[/imath] we have [imath](1 − p^n)^m + (1 − q^m)^n \ge 1[/imath].
Prove that for positive integers [imath]m, n[/imath] and two positive numbers [imath]p, q[/imath] satisfying [imath]p+q = 1[/imath] we have [imath](1 − p^n)^m + (1 − q^m)^n \ge 1.[/imath] Using binomial theorem, we have [imath]1-C^m_1*p^n+C^m_2*p^2n-\cdots [/imath] for [imath](1 − p^n)^m[/imath]. And same logic for the second term of the equation but they don't cancel each other out. Is there anything I am doing wrong? (this is a practice from a probability class) |
2639066 | Find all [imath]n[/imath] by [imath]n[/imath] matrices [imath]A[/imath] such that [imath]A^2 = A.[/imath]
It isn't too hard to show that there are infinitely many non-invertible matrices, and finitely many invertible ones. But how do we find all the possible matrices? Edit : The other question has not given all the matrices. | 2011127 | What is the square matrix [imath]A[/imath] equal to if [imath]A=A^2[/imath]
The question is: If A is a square matrix such that [imath]A^2=A[/imath] then [imath]A^n =A[/imath] for all natural numbers [imath]n[/imath] greater than one. What is [imath]A[/imath] if [imath]A \ne 0[/imath] and [imath]A \ne I[/imath]. I figured out an answer but I can't tell if that's the only answer. Let's say that [imath]a_{kk}[/imath] is a value in [imath]A[/imath]. Every value in [imath]A[/imath] is [imath]0[/imath] except for [imath]a_{kk}[/imath] and [imath]a_{00}[/imath], they're [imath]1[/imath]. I haven't gotten this answer mathematically. I've tried a few approaches but ended up with the identity matrix. |
2637877 | Question about a submodule
Let [imath]R=\prod_{n\in\mathbb{N}}\mathbb{Z}[/imath] and [imath]M[/imath] be [imath]R[/imath] regarded as [imath]R[/imath]-module is usual way. The submodule [imath]N=\bigoplus_{n\in\mathbb{N}}\mathbb{Z}[/imath] is it finitely generated? Thanks | 2640916 | Submodule and direct sum
Please help me to prove the following resut: Let [imath]R=\prod_{n\in\mathbb{N}}\mathbb{Z}[/imath] and [imath]M[/imath] be [imath]R[/imath] regarded as [imath]R[/imath]-module is usual way. Then the submodule [imath]N=\bigoplus_{n\in\mathbb{N}}\mathbb{Z}[/imath] is not finitely generated Hint: for every [imath]n\in\mathbb{N}[/imath], [imath]e_n=(0,…,0,1,0,…)\in N[/imath] (the [imath]1[/imath] is on the [imath]n-[/imath]th position). Thanks in advance |
2638849 | Deriving the power series for [imath]e[/imath] simply?
The definition of [imath]e[/imath] is: [imath]e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n[/imath] If we use the Binomial Theorem on the function itself: [imath]\left(1 + \frac{1}{n}\right)^n = \sum_{k=0}^{n} \binom{n}{k}\frac{1}{n^k} = \sum_{k=0}^{n} \frac{n!}{k!(n-k)! \cdot n^k}[/imath] At this point I am not sure where to go, how to both manipulate this expression and also have it be a valid input to the limit definition. Is there a simple approach that does not require Calc II knowledge, or is this one of those problems where there's really just one way to do it that everyone uses? | 2502963 | Infinite series for [imath]e[/imath]...
How do you prove that [imath]e=\sum_{n=0}^{\infty}\frac{1}{n!}[/imath]? Here I am assuming [imath]e:=\lim_{n\to\infty}(1+\frac{1}{n})^n[/imath]. Do you have any good PDF file or booklet available online on this? I do not like how my analysis text handles this... |
2639239 | Show that [imath]A\times B=B\times A[/imath] if and only if [imath]A=B.[/imath]
Could you tell me if this all right. And how is the best way to go backwards? it is by cases? Thanks. Sorry is not in latex, but I´m just learning. | 2238832 | Under what conditions on the sets [imath]A[/imath] and [imath]B[/imath] will it be true that [imath]A\times B= B\times A[/imath]
Under what conditions on the sets [imath]A[/imath] and [imath]B[/imath] will it be true that [imath]A\times B= B\times A[/imath]? I think it should be set [imath]A[/imath] [imath]=[/imath] set [imath]B[/imath] but don't know how to prove it or is there any other way for those two cross product to be true? |
2640009 | Find the limit as x approaches zero
[imath]\sin(x)^{200}\over x^{199}\sin(4x)[/imath] as x approaches zero. I applied three times l'Hopital rule but i do not know if it is right. Any help appreciate. [imath]200(199(198(\sin^{197}(x)\cos^{3}(x)-\sin^{198}(x)\sin(2x))-200\sin^{199}(x)\cos(x))\over -64x^{199}\cos(4x)-9552x^{198}\sin(4x)+472824x^{197}\cos(4x)+7762164x^{196}\sin(4x)[/imath] And if it is right what next?Again l'Hopital rule? | 2638190 | Find limit of [imath]\lim_{x\to 0}\frac{\sin^{200}(x)}{x^{199}\sin(4x)}[/imath], if it exists
I'm practising solving limits and the one I'm currently struggling with is the following: [imath]\ell =\lim_{x\to 0}\frac{\sin^{200}(x)}{x^{199}\sin(4x)}[/imath] What I've done: Since this is an obvious [imath]0/0[/imath] , I tried using de L'Hospital's Rule consecutively only to see both the numerator and the denominator grow so much in size that each couldn't fit in one row. [imath] \begin{align} l & =\lim_{x→0}{{\sin^{200}(x)}\over{x^{199}\sin(4x)}}\\ & = \lim_{x→0}{{200\sin^{199}(x)\cos(x)}\over{x^{198}\left(199\sin\left(4x\right)+4x\cos\left(4x\right)\right)}}\\ & = \lim_{x→0}{{39800\cos^2\left(x\right)\sin^{198}\left(x\right)-200\sin^{200}\left(x\right)}\over{x^{198}\left(800\cos\left(4x\right)-16x\sin\left(4x\right)\right)+198x^{197}\left(199\sin\left(4x\right)+4x\cos\left(4x\right)\right)}} \end{align} [/imath] Another solution I tried was through manipulation and the use of trigonometric identities and formulae but to no avail. I tried substituting: [imath]\color{red}{\sin(4x)}[/imath] with [imath]\color{blue}{4\sin(x)\cos(x) - 8\sin(3x)\cos(x)}[/imath] and then [imath]\color{red}{8\sin(3x)\cos(x)}[/imath] with [imath]\color{blue}{4\sin(4x)+4\sin(2x)}[/imath]. [imath] \begin{align} l & =\lim_{x→0}{{\sin^{200}(x)}\over{x^{199}\sin(4x)}}\\ & =\lim_{x→0}{{\sin^{200}(x)}\over{x^{199}(4\sin(x)\cos(x) - 8\sin(3x)\cos(x))}}\\ & =\lim_{x→0}{{\sin^{200}(x)}\over{x^{199}(4\sin(x)\cos(x) - 4\sin(4x)+4\sin(2x))}}\\ \end{align} [/imath] No matter what I try, the limit remains [imath]0/0[/imath]. Question: Does the above limit exist? If so, what I path should I follow to work out a solution? |
2640011 | [imath]f|_{A_i} [/imath]is continuous, [imath]A_i[/imath] closed [imath]\implies[/imath] [imath]f:X\to Y[/imath] is continous
Take topological spaces [imath]X[/imath] and [imath]Y[/imath], a function [imath]f:X \to Y[/imath] and a family [imath](A_i)_{i\in I}[/imath] os subsets of [imath]X[/imath] and [imath]X= \cup A_i[/imath]. Assume that [imath]f|_{A_i} [/imath]is continuous for each [imath]i \in I[/imath]. Show that if I is finite and each [imath]A_i[/imath] is closed in [imath]X[/imath], then [imath]f[/imath] is continous. [imath]f[/imath] is continous if preimage of every closed set in [imath]Y[/imath] is closed in [imath]X[/imath]. I know that a union of finitely many closed sets is closed but I am not sure how to mix it. For each [imath]i \in I[/imath] [imath]f_i : A_i \to Y[/imath] is continous, that means for each closed [imath]C \subset Y[/imath], [imath]f^{-1}(C)[/imath] is closed in [imath]A_i[/imath]. Can I write something like this [imath]f^{-1}(C_i)=A_i [/imath]? | 2494899 | Function is Continuous When Restricted to Closed Sets Which Cover the Space
Suppose that [imath]\{A_i\}[/imath] is a finite collection of closed sets in [imath]X[/imath] such that [imath]X = \bigcup A_i[/imath], and [imath]f : X \to Y[/imath] restricted to each [imath]A_i[/imath] is continuous. Then [imath]f[/imath] is continuous. Okay. The base case ([imath]n=2[/imath]) is just the pasting lemma. I am having a little trouble with the inductive case, specifically with what the hypothesis is and how to deal with continuity with respect to subspaces. I believe the inductive hypothesis is to assume that any function over a topological space that is the union of [imath]n-1[/imath] closed sets such that the function restricted to each of these [imath]n-1[/imath] closed set is continuous must be continuous over the whole space. Then I assume I have a topological space [imath]X = \bigcup_{i=1}^n A_i[/imath], where each [imath]A_i[/imath] is closed, and a function [imath]f : X \to Y[/imath] such that [imath]f|_{A_i}[/imath] is continuous for each [imath]i[/imath]. Thus [imath]X = \bigcup_{i=1}^{n-1} A_i \cup A_n[/imath], where [imath]C = \bigcup_{i=1}^{n-1} A_i[/imath] is closed since it is a finite union of closed sets. If I can show [imath]f|_C[/imath] is continuous, then I simply apply the pasting lemma to finish the problem. Give [imath]C[/imath] the subspace topology, and notice that each [imath]A_i[/imath] is closed in [imath]C[/imath]. Thus we have a topological space [imath]C[/imath] that is the union of [imath]n-1[/imath] closed. But do I still have continuity [imath]f[/imath] restricted to [imath]A_i[/imath] since I am no longer considering them subspaces of [imath]X[/imath] but of [imath]C[/imath]? When one says [imath]f|_{A_i}[/imath] is continuous, is one saying that [imath]f[/imath] is continuous when restricted to [imath]A_i[/imath] and [imath]A_i[/imath] is given the subspace topology? EDIT: I think I figured out the lemma I need. Claim: Let [imath]A[/imath] and [imath]C[/imath] be subspaces of [imath]X[/imath] such that [imath]A \subseteq C[/imath]. Topologize [imath]C[/imath] with subspace topology with respect to [imath]X[/imath]. Then the subspace topology on [imath]A[/imath] as a subspace of [imath]X[/imath] (call it [imath]\tau_X[/imath]) is the same as the subspace topology of [imath]A[/imath] as a subspace of [imath]C[/imath] (call it [imath]\tau_C[/imath]). Proof: Let [imath]U[/imath] be open in [imath]X[/imath]. Then [imath]A \cap U[/imath] is some open set in [imath]A[/imath] as a subspace of [imath]X[/imath]. Since [imath]C[/imath] is a subspace of [imath]X[/imath], [imath]C \cap U[/imath] is open in [imath]C[/imath] as a subspace of [imath]X[/imath]. Then [imath]A \cap (C \cap U) = (A \cap C) \cap U = A \cap U[/imath] is open in [imath]A[/imath] as a subspace of [imath]C[/imath], thereby proving [imath]\tau_X \subseteq \tau_C[/imath]. Now let [imath]O[/imath] be some open set in [imath]C[/imath] as a subspace of [imath]X[/imath]. Then [imath]A \cap O[/imath] is some open subset of [imath]A[/imath] as a subspace of [imath]C[/imath]. But [imath]O = C \cap U[/imath] for some open set in [imath]X[/imath]. Therefore [imath]A \cap O = A \cap (C \cap U)= A \cap U[/imath], which is also open in [imath]A[/imath] as a subspace of [imath]X[/imath]. Hence [imath]\tau_C \subseteq \tau_X[/imath] Kind of obvious. But it was hard to spot the need for this lemma when I was considering subspace of subspaces and continuous functions, etc. Am I right in thinking this is the lemma I need? |
2639359 | Find, [imath]\sum_{k=0}^{n}4^k \binom{n}{k}[/imath]
Reading through my textbook I came across the following problem, and I am looking for some help solving it. I believe I have done the workings correctly but I just wanted to make sure as it seemed a little to easy. Find, [imath]\sum_{k=0}^{n}4^k \binom{n}{k}[/imath] Consider the expansion [imath](1+x)^n[/imath] using binomial theorem. [imath](1+x)^n = \sum_{k=0}^{n} x^k \binom{n}{k}[/imath] Then substitute [imath]x = 4[/imath]. That should lead you to the answer [imath]5^n[/imath]. Therefore [imath]\sum^{n}_{k=0}4^k\binom{n}{k} = 5^n[/imath] | 2638566 | Find [imath]\sum_{k=0}^{n}4^k \binom{n}{k}[/imath]
Reading through my textbook I came across the following problem, and I am looking for some help solving it. In the back of the textbook they have shown the final answer but I'm not quite sure how to get there. If I could get some help that would be great, thanks! Find, [imath]\sum_{k=0}^{n}4^k \binom{n}{k}[/imath] The solution given is, [imath]C(7,1)C(4,1)+C(7,2)C(4,2)+C(7,3)C(4,3)+C(7,4)C(4,4)[/imath] |
2640368 | A combinatorial argument for IMO 1972, Problem 3.
I'm trying to prove the following question from IMO 1972 Prove that for any positive integers [imath]m,n[/imath], [imath]\frac{(2m)!(2n)!}{m!n!(m+n)!}[/imath] is always an integer. I tried thinking of a combinatorial argument, and came up with this: Imagine that there are two bags with [imath]2m[/imath] and [imath]2n[/imath] objects respectively. This expression is the number of ways of selecting [imath]m[/imath] objects from the first bag and [imath]n[/imath] objects from the second bag, such that the order in which you select these objects is not important. For instance, you can keep switching between bags, and also changing the order in which you select the objects. However, I am not convinced that this is a correct argument. Where am I going wrong? Thanks | 1072416 | An example where [imath]\frac{(2m)!(2n)!}{m!n!(m + n)!}[/imath] is the number of ways of counting something?
Prove that for all non-negative integers [imath]m,n[/imath], [imath]\frac{(2m)!(2n)!}{m!n!(m + n)!}[/imath] is an integer. There is a answer given here to this question here. I've seen how it can be proven using recurrence equations as well. When I was trying to solve this problem I was looking to represent it as the number of ways of counting something. Can anyone give such an example, thus proving the question another and very natural way? |
1683440 | Prove that [imath](fg)^{(n)} = \sum_{k=0}^n \binom{n}{k}f^{(k)}(x)g^{(n-k)}(x)[/imath]
Assume [imath]f[/imath] and [imath]g[/imath] are differentiable at [imath]x[/imath]. Prove that [imath](fg)^{(n)} = \sum_{k=0}^n \binom{n}{k}f^{(k)}(x)g^{(n-k)}(x)[/imath] I am assuming here [imath]fg = f(x) g(x)[/imath]. Then we can prove this via induction. If [imath]n = 0[/imath] we have [imath]1 = 1[/imath] which is true. Now assume it is true for some [imath]m[/imath] we need to show it is possible for [imath]m+1[/imath]. We have [imath](fg)^m = \sum_{k=0}^m f^{(k)}(x)g^{m-k}(x)[/imath]. How do we show that? | 2451243 | Proving Leibniz theorem using induction
Let [imath]f[/imath] and [imath]g[/imath] be [imath]n[/imath] times differentiable functions then prove that [imath]\frac {d^n}{dx^n}(fg)=\sum _{i=0}^n \binom{n}{ i} f^{(i)}g^{(n-i)} [/imath] where [imath]f^{(k)} [/imath] is [imath]k[/imath]-th derivative with respect to [imath]x [/imath]. Now I started with Mathematical Induction. I know its true for [imath]n=1[/imath] so skipped it. Let it be true for [imath]m<n [/imath] thus [imath]\frac {d^m}{dx^m}(fg)=\sum_{i=0}^{m} \binom {m}{i}f^{(m)}g^{(m-i)} =s .[/imath] We need to prove this for [imath]m+1[/imath] . Note that [imath]m+1 <n [/imath] is also true. So we see that [imath]\frac {d^{m+1}}{dx^{m+1}}(fg)=\frac {d}{dx}(s).[/imath] But now problem here is that I don't know how to differentiate this whole [imath]s[/imath] as we cant use Leibniz rule to prove Leibniz rule. |
2641117 | Prove that the polygon of [imath]n[/imath] sides with the biggest area, inscribed in a circumference, is always a regular polygon
Prove that the polygon of [imath]n[/imath] sides with the bigger area, inscribed in a circumference, is always a regular polygon. Ex (because the way in which I wrote the statement maybe was not clear and produces confusion): For [imath]n=4[/imath]: Prove that the inscribed quadrilateral with bigger area is always a square. I thought that this problem was easy, but i don't see the way to prove this Any hints? Edit This problem is mean to be solved without trigonometry, so the post linked as duplicated doesn't help me | 1919440 | Area of a cyclic polygon maximum when it is a regular polygon
My question: Let [imath]n[/imath] points [imath]A_1, A_2,\ldots,A_n[/imath] lie on given circle then show that [imath]\operatorname{Area}(A_1A_2\cdots A_n)[/imath] maximum when [imath]A_1A_2\cdots A_n[/imath] is an [imath]n[/imath]-regular polygon. |
2641731 | Can a meromorphic function have removable singularities?
Authors usually give the following definition (or some variation) of meromorphic function: Let [imath]G[/imath] be a region (i.e. open connected subset of [imath]\mathbb{C}[/imath]). A function [imath]f[/imath] is said to be meromorphic in [imath]G[/imath] if it's holomorphic in [imath]G[/imath] except in a discrete set of poles. However I think we should also consider removable singularities in the definition of meromorphic function. There is a theorem that says: If [imath]f[/imath] and [imath]g[/imath] are holomorphic in [imath]G[/imath] then [imath]f/g[/imath] is meromorphic in [imath]G[/imath]. This theorem is no longer true if we take the usual definition. Take for example [imath]f,g:\mathbb{C}\to \mathbb{C}, f(z)=g(z)=z[/imath] and define [imath]h:\mathbb{C}-\{0\}\to\mathbb{C}, h(z)=f(z)/g(z)[/imath]. Then [imath]h=1[/imath] in [imath]\mathbb{C}-\{0\}[/imath], however [imath]h[/imath] is not meromorphic in [imath]\mathbb{C}[/imath] in the sense of the previous definition because [imath]0[/imath] is not a pole, is a removable singularity. Most authors only mention poles in the definition of meromorphic functions, but I think that's not strictly correct. If I'm right, I would like to see an author considering removable singularities in the definition of meromorphic functions. There is a similar question here definition of meromorphic function : it may have removable singularities but I don't like the phrasing of the question and the answer. I would like to have a reference text considering these details. For example Lang, Conway and Ahlfors all give the definition without removable singularities. | 2540475 | definition of meromorphic function : it may have removable singularities
I'd like to know about the definition of meromorphic function. Usually I see the definition of meromorphic function as follows: Let [imath]D\subset\mathbb C[/imath] be a connected open set, a function [imath]f[/imath] defined on a subset [imath]U[/imath] of [imath]D[/imath] and with value in [imath]\mathbb C[/imath] is meromorphic on [imath]D[/imath] if the following conditions are satisfied: [imath]P(f)=D\setminus U[/imath] is a set of poles [imath]P(f)[/imath] is discrete in [imath]D[/imath] [imath]f[/imath] is holomorphic on [imath]U[/imath]. However, in the book " Complex anlysis for mathematics and engineering" by John H. Mathews and Russel W. Howeell, [imath]P(f)=D\setminus U[/imath] is a set of poles and removable singularities. I think removable singularities are not real singularities, since we can extend the function to the holomorphic function. Thus, two definitions may be almost same. I'd like to know how other people think about this question. Would you give any comments about this question? Thanks in advance! |
2641407 | Proof of inequality involving positive number
Let [imath]x[/imath] be a real number such that [imath]x> 0[/imath]. Prove that [imath]x + \frac{1}{4x} \geq 1[/imath]. I tried to do this but cannot prove it. | 2638803 | Prove that [imath]x+\frac1{4x} ≥ 1[/imath] for [imath]x>0[/imath]
Let [imath]x[/imath] be a real number such that [imath]x > 0[/imath]. Prove that x +[imath]\frac {1} {4x} ≥ 1[/imath]. Not really sure on the correct way to approach it/is valid and could use some help. Answer: Proof Strategy: Proof by cases: [imath]x = 1[/imath] [imath]x > 1[/imath] [imath]x < 1[/imath] -- Case 1: [imath]x = 1[/imath] [imath]x+\frac1{4x}[/imath] [imath]= (1) + 1/(4(1))[/imath] [imath]= 1.25 \ge 1[/imath] Case is true Case 2: [imath]x > 1[/imath], in this case [imath]x = 2[/imath]. [imath]x+\frac1{4x} [/imath] [imath]= (2) + 1/(4(2))[/imath] [imath]= 2.125 \ge 1[/imath] Case is true Case 3: [imath]x < 1[/imath], in this case [imath]x = 0.5[/imath]. [imath]= (0.5) + 1/(4(0.5))[/imath] [imath]= 1 \ge 1[/imath] Case is true Since all possible cases were satisfied therefore [imath]x+\frac1{4x} \ge 1 [/imath] when [imath]x > 0[/imath]. |
2641710 | Conjugation in [imath]S_8[/imath]
I am reviewing for a test and i find the following good exercice. Would you please help me to prove the second part of the question b) here for question a) i find [imath]|G|=32[/imath] since [imath]G=C_8\ltimes C_2\times C_2[/imath] Thanks for your help in advance. | 2639383 | Subgroup of [imath]S_8[/imath] isomorphic to [imath]G[/imath]
Please help me to prove the following result: I recall that the holomorph [imath]Hol(\Gamma)[/imath] of a group [imath]\Gamma[/imath] is the semidirect product [imath]Hol(\Gamma)=\Gamma\ltimes_{\phi} Aut(\Gamma)[/imath] where [imath]\phi: Aut(\Gamma)\to Aut(\Gamma)[/imath] is the identity map. Let [imath]G=Hol(C_8)[/imath]. In addition let [imath]f(x)\in \mathbb{Q}[x][/imath] be an irreducible polynomial of degree [imath]8[/imath], [imath]L[/imath] a splitting field of [imath]f(x)[/imath] over [imath]\mathbb{Q}[/imath] and suppose that [imath]Gal(L/\mathbb{Q})\cong G[/imath] Show that there is a subgroup of [imath]S_8[/imath] isomorphic to [imath]G[/imath] Thanks in advance. |
2642456 | What does it mean to localize a ring at a prime ideal?
I am not that well acquainted with commutative algebra, so that I am kind of struggling to understand the concept of "localizing a ring at a prime ideal." For one thing I can say is that localizing alludes to a geometric meaning. Can somebody clarify this concept from both perspectives: algebraic and geometric ? Since a ring ( for example [imath]\mathbb{Z}[/imath] ) can contain many prime ideals, I presume that the introduced localization is not unique. Is this correct ? What are the consequences ? How then do you make a choice on picking up the most suitable prime ideal for the purpose of the localization ? How do you procede in constructing such "localization" ? Many thanks. | 349957 | Geometric meaning of completion and localization
Let [imath]R[/imath] be a commutative ring with unit, [imath]I[/imath] an ideal of [imath]R[/imath] and consider the following three constructions. The localization [imath]R_I[/imath] of [imath]R[/imath] at [imath]I[/imath] (i.e. the localization of [imath]R[/imath] at the multiplicative system [imath]R\setminus I[/imath]) gives a morphism [imath] f_1:\operatorname{Spec}(R_I)\to \operatorname{Spec}(R)=:X [/imath] The completion [imath]\widehat{R}_I[/imath] of [imath]R[/imath] at [imath]I[/imath] gives a morphism [imath] f_2:\operatorname{Spec}(\widehat{R}_I)\to X [/imath] For the special case [imath]I=(a)[/imath], the localization [imath]R_a[/imath] (i.e. the localization of [imath]R[/imath] at the multiplicative system [imath]\{1,a,a^2,\ldots\}[/imath]) gives a morphism [imath] f_3:\operatorname{Spec}(R_a)\to X [/imath] My question is: What is the geometric meaning of all three constructions and how are they related? This is what I ''know'' already or describes at least the style of answer I would appreciate. As remarked in the comments, [imath]I[/imath] has to be a prime ideal. [imath]R\to R_I[/imath] is injective iff [imath]R\setminus I[/imath] contains no zero divisors which is the case if [imath]I[/imath] is a prime ideal. The scheme [imath]\operatorname{Spec}(R_I)[/imath] is the intersection of all the neightbourhoods of [imath]I[/imath] in [imath]X[/imath]. This is a little contra-intuitive for me since the last statement sounds like ''[imath]f_1[/imath] is injective'' but the [imath]\operatorname{Spec}[/imath]-operator should turn 'injective'' and ''surjective'' around somehow (I know this is literally not true but I only want to get a feeling like ''is contained'', ''is bigger'', ''is smaller'', etc.). [imath]\operatorname{Spec}(R_a)[/imath] is somehow the opposite of [imath]\operatorname{Spec}(R_{(a)})[/imath] (= the intersection of all the neightbourhoods of [imath](a)[/imath] in [imath]X[/imath]) because [imath]\operatorname{Spec}(R_a)[/imath] seems to be something like the union of all the open sets of [imath]X[/imath] not containing the point [imath](a)[/imath]. [imath]R\to\widehat{R}_I[/imath] is injective iff [imath]\cap I^n=(0)[/imath] and this holds very often (e.g. for [imath]R[/imath] noetherian and either an integral domain or a local ring). Hence (this is probably false intuition as remarked before), [imath]f_2[/imath] should be something like a ''projection'' (from something ''big'' into something ''small''). But what is geometrically the difference between localization and completion. I don't have a geometric idea of completion at all. As remarked by Qiaochu Yuan in the comments below, one should not think of [imath]\operatorname{Spec}[/imath] as injective-surjective switching. |
2136793 | Trigonometric equations involving [imath]\prod_{r=1}^{n}\cos\left(\frac{x}{2^n}\right)[/imath]
It is a well known fact in trigonometry that, for [imath]0 < x < \pi[/imath], we have [imath]\displaystyle \prod_{r=1}^{n}\cos\left(\frac{x}{2^r}\right) = \frac{\sin x}{2^n \sin\left(\frac{x}{2^n}\right)}[/imath]. This result proves useful when dealing with limits of certain functions, or summing trigonometric series. I started to wonder if there are any other useful applications of this result. Can anyone suggest any trigonometric equations that this result makes easier to solve? A pretty obvious one is [imath]\frac{1}{8}\csc\left(\frac{x}{8}\right)=\prod_{r=1}^{3}\cos\left(\frac{x}{2^r}\right)[/imath], but I am looking to see if there are more creative examples. | 2143974 | Evaluating limit [imath]\lim_{k\to \infty}\prod_{r=1}^k\cos{\left(\frac {x}{2^r}\right)}[/imath]
I stumbled across the following question which asked to evaluate... [imath]\lim_{k\to \infty}\prod_{r=1}^k\cos{\left(\frac {x}{2^r}\right)}[/imath] I at first tried writing few terms [imath]\cos{\left(\frac {x}{2}\right)}\cos{\left(\frac {x}{4}\right)}\cos{\left(\frac {x}{8}\right)}...[/imath] I used the Half-angle formula to write[imath]\cos{\left(\frac {x}{2}\right)}=\pm\sqrt{\frac{1+\cos(x)}{2}}[/imath] Therefore, [imath]\sqrt{\frac{1+\cos(x)}{2}}\sqrt{\frac{1+\sqrt{\frac{1+\cos(x)}{2}}}{2}}...[/imath] As there are infinitely many two's in the denominator, the denominator goes to [imath]\infty[/imath] which means [imath]\lim_{k\to \infty}\prod_{r=1}^k\cos{\left(\frac {x}{2^r}\right)}=0[/imath] So..My question is ...Am I correct?... If not, Could you please give me some hint to how should I proceed ? |
2642649 | Suppose there are 15 people consisting of 8 women and 7 men.
How many ways are there to form a committee with at least 3 women and at least 2 men? My answer is [imath]2^{15}- 2^7\binom{8}{0}-2^7\binom{8}{1}-2^7\binom{8}{2}-2^8\binom{7}{0}-2^8\binom{7}{1}[/imath] = 25984 but the answer given is 26280. Can someone tell me where I got wrong? Thank you! | 1024795 | Combinations - 17 women and 21 men to form a committee of size 7
How many committees are possible if a committee must have [imath]3[/imath] women and [imath]4[/imath] men? [imath]_{38}C_3+_{38}C_4[/imath] or [imath]\frac{38!}{3!35!}+\frac{38!}{4!34!} = 8,435+73,815 = 82,251[/imath] How many committees are possible if a committee must consist of all women or all men? [imath]_{17}C_7+_{21}C_7[/imath] or [imath]\frac{17!}{7!10!}+\frac{21!}{7!14!} = 19,448+116,280 = 135,728[/imath] How many committees are possible if a committee must have at least 3 women? [imath]_{38}C_3+_{35}C_4[/imath] or [imath]\frac{38!}{3!35!}+\frac{35!}{4!31!} = 8,435+52,360 = 60,795[/imath] How many committees are possible if a committee must have exactly one woman? [imath]_{38}C_1+_{37}C_6[/imath] or [imath]\frac{38!}{1!37!}+\frac{37!}{6!31!} = 38+2,324,821 = 2,324,821[/imath] Am I going about these the correct way? |
2642360 | How do I prove this identity involving polar coordinates and [imath]\nabla[/imath]?
Given that [imath]\begin{aligned} &x=r\cos(\theta),\\ &y=r\sin(\theta),\\ & \qquad \text{and}\\ &x^2+y^2=r^2 \end{aligned}[/imath] Use the chain rule to show that [imath]\nabla=\mathbf{\hat{r}}\frac\partial{\partial r}+\mathbf{\hat{\theta}}\frac1r\frac\partial{\partial\theta}.[/imath] I derived that [imath]\,\mathbf{\hat{r}}=\cos(\theta)\mathbf{\hat{\text{i}}}+\sin(\theta)\mathbf{\hat{\text{j}}}\,[/imath] and that [imath]\,\mathbf{\hat{\theta}}=-\sin(\theta)\mathbf{\hat{\text{i}}}+\cos(\theta)\mathbf{\hat{\text{j}}}.[/imath] But I can't seem to gro from here. | 586848 | How to obtain the gradient in polar coordinates
I'm not sure on how to find the gradient in polar coordinates. The thing that troubles me the most is how to find the unit vectors [imath]\hat{r}[/imath] and [imath]\hat{\theta}[/imath]. My approach for the rest is expressing the partial derivatives in respect of [imath]r[/imath] and [imath]\theta[/imath] using the chain rule. How can I get around solving this problem? |
2642944 | Probability of no two consecutive heads (or tails) after n trials?
Please help! H/T, [imath]50\%[/imath]-[imath]50\%[/imath] independent events. I pick Heads. What is the probability that after [imath]29[/imath] attempts of a fair coin, there's no consecutive Heads (...HH...) at all? I would appreciate seeing how it's worked out... Thanks a bunch. | 1355511 | Probability that no two consecutive heads occur?
A fair coin is tossed [imath]10[/imath] times. What is the probability that no two consecutive tosses are heads? Possibilities are (dont mind the number of terms): [imath]H TTTTTTH[/imath], [imath]HTHTHTHTHTHTHT[/imath]. But except for those, let [imath]y(n)[/imath] be the number of sequences that start with [imath]T[/imath] [imath]T _[/imath], there are two options, [imath]T[/imath] and [imath]H[/imath] so, [imath]y(n) = y(n - 1) + x(n-1) = y(n - 1) + y(n - 2)[/imath] Let [imath]x(n)[/imath] be the number of sequences that start with [imath]H[/imath], [imath]H _[/imath] the next option is only [imath]T[/imath] hence, [imath]x(n) = y(n - 1)[/imath] The total number of sequences [imath]F(n)[/imath] is: [imath]F(n) = y(n) + x(n) = 2y(n - 1) + y(n - 2)[/imath] We are after [imath]F(10)[/imath], [imath]F(10) = 2y(9) + y(8)[/imath] [imath]F(3) = 2y(2) + y(1) = 1 + 4 = 5[/imath] [imath]F(4) = 2y(3) + y(2) = 6 + 2 = 8[/imath] But I'm not quite sure where this will take me though. |
2643189 | Limit of probability question 2
Assume for all finite [imath]r>r_0[/imath] that [imath]\lim_{n\to\infty}\mathbb{P}(\mathcal{A}_{r,n})=0,[/imath] where [imath]\mathcal{A}_{r,n}[/imath] is an event on some random variables, and [imath]n[/imath] and [imath]r[/imath] are deterministic variables. How to prove the following? [imath]\lim_{n\to\infty}\mathbb{P}(\lim_{r\to\infty} \mathcal{A}_{r,n})=0.[/imath] This is a part of a bigger problem. I need to prove the second equation but I have proved the first one. Is the second equation always true? If yes, how can I prove that? | 2643173 | Limit of probability question
Assume for all finite [imath]r>r_0[/imath] that [imath]\lim_{n\to\infty}\mathbb{P}(\mathcal{A}_{r,n})=0,[/imath] where [imath]\mathcal{A}_{r,n}[/imath] is an event on some random variables, and [imath]n[/imath] and [imath]r[/imath] are deterministic variables. How to prove the following? [imath]\lim_{n,r\to\infty}\mathbb{P}(\mathcal{A}_{r,n})=0.[/imath] This is a part of a bigger problem. I need to prove the second equation but I have proved the first one. Is the second equation always true? If yes, how can I prove that? |
2643734 | Properties of logs. Why does the log of a power, does the exponent become a coefficient?
I'm having trouble seeing number 3 clearly. Or rather, I'm having trouble seeing a proof. I've tested a series of numbers and it works, but I'm having trouble seeing why. Can anyone clarify? Here is where I get confused: I start with this log: [imath] \log_2 4^1 = 2[/imath] now if I raise both sides to the 2nd power then: [imath] \log_2 4^{1^2} = 2^2[/imath] [imath] = \log_2 4^2 = 4[/imath] [imath] = 2 \cdot \log_2 4 = 4[/imath] but if I raise both sides to the power of 1.5 then: [imath] \log_2 4^{2 \cdot 1.5} \ne 4^{1.5} \ne 8[/imath] | 2646444 | Why does this log rule make sense? How to solve this ln derivative?
Why does this log rule make sense: [imath] \log(x)^{1/2}= \frac{1}{2} \log x?[/imath] I can see why this makes sense: [imath]\log x^4=4\log(x)[/imath] But I can't really make sense of why this works. Like why does: [imath]log_24^3=3 \log_24 [/imath] This is a problem I am having trouble finding the derivative for: [imath]H(z) = \ln \sqrt{\frac{a^2-z^2}{a^2 + z^2}}=\frac{1}{2}\ln \frac{a^2-z^2}{a^2 + z^2}=\frac{1}{2} \ln(a^2 - z^2) - \ln(a^2 + z^2)[/imath] right? so: [imath]H'(z) = \frac{1}{2} \cdot \frac{1}{a^2 - z^2} \cdot -2z - \frac{-1}{a^2 + z^2} \cdot 2z[/imath] [imath] = \frac{-2z}{2(a^2 - z^2)} - \frac{2z}{a^2 + z^2}[/imath] Is this right? |
2643622 | Solve the equation [imath] { \sqrt{4+\sqrt{4+\sqrt{4-x}}}}=x [/imath]
Prove that one root of the Equation [imath]{ \sqrt{4+\sqrt{4+\sqrt{4-x}}}}=x [/imath] is [imath]x=2\left(\cos\frac{4\pi}{19}+\cos\frac{6\pi}{19}+\cos\frac{10\pi}{19}\right) [/imath] My progress: After simplyfying the given Equation I got [imath] x^8-16x^6+88x^4-192x^2+x+140=0[/imath] Then I tried to factorise it, got this [imath] x^8-16x^6+88x^4-192x^2+x+140=(x^3+x^2-6x-7)(x^5-x^4-9x^3+10x^2+17x-20)[/imath] [imath] x^8-16x^6+88x^4-192x^2+x+140=(x^3+x^2-6x-7)(x^3-2x^2-3x+5)(x^2+x-4)[/imath] Should I now solve cubic? | 1065862 | Something strange about [imath]\sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x[/imath] and its friends
This post can be generalized to, [imath]\begin{align} \sqrt{ 2+ \sqrt{ 2 + \sqrt{ 2-x}}}=x&,\quad\quad x = -2\cos\left(\frac{8\pi}{9}\right)=1.8793\dots\quad\quad\quad \\ \\ \sqrt{ 4+ \sqrt{ 4 + \sqrt{ 4-x}}}=x&,\quad\quad x = 2\sum_{n=1}^3\cos\left(\frac{2\pi\, s_1(n)}{19}\right)=2.5070\dots\quad\\ \\ \sqrt{ 8+ \sqrt{ 8 + \sqrt{ 8-x}}}=x&,\quad\quad x = -1-2\sum_{n=1}^6\cos\left(\frac{2\pi\, s_2(n)}{37}\right)=3.3447\dots\\ \\ \sqrt{ 14+ \sqrt{ 14 + \sqrt{14-x}}}=x&,\quad\quad x =\, \color{red}? \pm 2\sum_{n=1}^{\color{red}{7\,?}}\cos\left(\frac{2\pi\, s_3(n)}{\color{red}{63\,?}}\right)=\dots\quad\quad\quad\quad\\ \\ \sqrt{ 22+ \sqrt{ 22 + \sqrt{22-x}}}=x&,\quad\quad x = -1+2\sum_{n=1}^{16}\cos\left(\frac{2\pi\, s_4(n)}{97}\right)=5.2065\dots\\ \\ \sqrt{ 32+ \sqrt{ 32 + \sqrt{ 32-x}}}=x&,\quad\quad x = -2-2\sum_{n=1}^{23}\cos\left(\frac{2\pi\, s_5(n)}{139}\right)=6.1716\dots \end{align}[/imath] and so on, where the sequences [imath]s_k(n)[/imath] are, [imath]\begin{aligned} s_1(n) &= 2, 3, 5.\\ s_2(n) &= 2, 9, 12, 15, 16, 17.\\ s_3(n) &=\, \color{red}?\\ s_4(n) &= 4, 6, 9, 10, 11, 14, 15, 17, 21, 23, 25, 26, 32, 35, 39, 48. \end{aligned}[/imath] etc. Note that by repeated squaring, we get, [imath]((x^2 - a)^2 - a)^2 - a + x = 0[/imath] which has two cubic factors when [imath]a=k^2+k+2[/imath], with the [imath]x_i[/imath] above as sums of [imath]\cos(z)[/imath], and a root of the same cubic family, [imath]x^3 + k x^2 - (k^2 + 2k + 3)x - ((k + 1)^3 - k^2)=0[/imath] This has a negative discriminant, [imath]D =-(4k^2+6k+9)^2[/imath], implying all roots are real. [imath]\begin{array}{|c|c|c|} k&a&\sqrt{D}\\ 0&2&9\,i\\ 1&4&19\,i\\ 2&8&37\,i\\ 3&14&\color{red}{63}\,i\\ 4&22&97\,i\\ 5&32&139\,i\\ \end{array}[/imath] Questions: Can anybody find the sequence [imath]s_3(n)[/imath]? (I used Mathematica's PowerMod, but it only works for primes.) Why do the others have a constant (namely [imath]-1,-1,-2[/imath]) added to the sum of cosines? Can we predict its value, or can it be found only by trial and error? |
2643401 | number of non-zero, distinct, eigen values
Let [imath]V^T[/imath] denote the transpose of a vector [imath]V[/imath]. Consider two non-zero [imath]p[/imath]-dimensional column vectors [imath]a[/imath] and [imath]b[/imath], for [imath]p\ge 2[/imath]. How many non-zero distinct eigen-values does the [imath]p\times p[/imath] matrix [imath]ab^T + b a^T[/imath] have? | 1637362 | Finding the null space of symmetric matrix generated by outer product
Let [imath]p, q \in \mathbb{R}^n[/imath] such that [imath]\|p\| = \|q\| = 1[/imath] and define [imath]A = pq^T + qp^T[/imath] Find the null space of [imath]A[/imath]. I am not having very much luck. I have managed to show that [imath]p + q[/imath] and [imath]p - q[/imath] are eigenvectors, but I can't seem to connect those two. I know that since [imath]A[/imath] is generated by an outer product, it is of rank one. That implies that the rank of the null space is [imath]n - 1[/imath], but I can't seem to get any further. Can anyone help? |
2643898 | Let [imath]a, b, c, d[/imath] be positive integers satisfying [imath]ab=cd[/imath]. Prove that [imath]a+b+c+d[/imath] ݀ is composite.
I started with an even odd arguement, where I tried to show that no matter which numbers you pick, the sum would always be even. However, this doesn't work for the arrangment of 3 evens and one odd. Where do I go from here? | 383394 | Prove [imath]a+b+c+d [/imath] is composite
Let [imath]a,b,c,d[/imath] be natural numbers with [imath]ab=cd[/imath]. Prove that [imath]a+b+c+d[/imath] is composite. I have my own solution for this (As posted) and i want to see if there is any other good proofs. |
2644490 | Ring Isomomorphism?
[imath]R=\left\lbrace\left.\begin{bmatrix}a & -b\\b&a\end{bmatrix}\,\right| a, b\in \mathbb{R}\right\rbrace[/imath] [imath](R,+,\cdot)[/imath] is ring with binary operation [imath]+[/imath] and [imath]\cdot[/imath] , Give me a hint to show that [imath]R[/imath] is isomorphic to [imath]\mathbb{C}[/imath] where [imath](\mathbb{C},+,\cdot)[/imath] is the ring of complex numbers with binary operation [imath]+[/imath] and [imath]\cdot[/imath]. | 1028371 | Complex number isomorphic to certain [imath]2\times 2[/imath] matrices?
I have been trying to prove this, but I am having trouble understanding how to prove the following mapping I found is injective and surjective. Just as a side note, I am trying to show the complex ring is isomorphic to special [imath]2\times2[/imath] matrices in regard to matrix multiplication and addition. Showing these hold is simple enough. [imath]\phi:a+bi \rightarrow \begin{pmatrix} a & -b \\ b & a \end{pmatrix}[/imath] This is what I have so far: Injective: I am also confused over the fact that there are two operations, and in turn two neutral elements (1 and 0). Showing that the kernel is trivial is usually the way I go about proving whether a mapping is injective, but I just can't grasp this. [imath] \phi(z_1) = \phi(z_2) \implies \phi(z_1)\phi(z_2)^{-1} = I = \phi(z_1)\phi(z_2^{-1}) = \phi(z_2)\phi(z_2^{-1}).[/imath] So if we can just show that the kernel of [imath]\phi[/imath] is trivial, then it also shows that [imath]z_1 = z_2[/imath]. The only complex number that maps to the identity matrix is one where [imath]a = 1[/imath] and [imath]b = 0[/imath], [imath]a + bi = 1 + 0i = 1[/imath]. Using a similar argument for addition we can just say that the only complex number [imath]z[/imath] such that [imath]\phi(z) = 0\text{-matrix}[/imath], is one where [imath]a=0[/imath] and [imath]b=0[/imath], [imath]a+bi=0+0i=0[/imath]. Surjective: I forgot to add this before I posted, but I honestly don't really understand how to prove this because it just seems so obvious. All possible [imath]2\times2[/imath] matrices of that form have a complex representation because the complex number can always be identified by its real parts and since the elements of the [imath]2\times2[/imath] matrix are real then the mapping is obviously onto. I have always had trouble understanding when I can say that I have "rigorously" proved something, so any help would be appreciated! |
1044732 | An entire function [imath]f(z)[/imath] is real iff [imath]z[/imath] is real
Let [imath]f[/imath] be an entire function s.t. [imath]\forall z \in \mathbb{C}, f(z)\text{ is real}\iff z\text{ is real}[/imath]. Prove that [imath]\exists a,b \in \mathbb{R}[/imath] s.t. [imath]a\neq 0[/imath] and [imath]\forall z \in \mathbb{C}, f(z)=az+b[/imath]. I think I have a proof for this using some advanced theorems, but I want to find one that can be done by elementary means (by a student in undergraduate complex analysis). | 2634606 | Holomorphic function satisfying [imath]f^{-1}(\Bbb R)=\Bbb R[/imath] is of the form [imath]f(z)=az+b[/imath]
Let [imath]f[/imath] be a holomorphic function defined on [imath]\Bbb C[/imath] such that [imath]f^{-1}(\Bbb R)=\Bbb R[/imath]. Prove that there exists [imath]a,b \in \Bbb R, \; a \neq 0 \;[/imath] such that [imath]f(z)=az+b \;[/imath] for every [imath]z \in \Bbb C[/imath]. I was given a hint: define the function [imath]g(z)=\frac{f(z)-f(0)}{z}[/imath]. such [imath]g[/imath] is also holomorphic in [imath]\Bbb C[/imath] since the only singularity is in [imath]0[/imath], and it is a removable singularity since we can define [imath]g(0)=f'(0)[/imath]. Next I thought about proving [imath]g[/imath] is bounded, and thus constant (using Liouville's theorem), but pretty much out of ideas. How can I finish the proof/solve in another way? |
861082 | consider a square of side length [imath]x[/imath], find the area of the region which contains the points which are closer to its centre than the sides.
Any ideas how to start. I am having trouble figuring out the region itself All ideas are appreciated thanks | 363992 | Probablility of a dart landing closer to the center than the edge of a square dartboard?
The question is: There is a square dart board with a side length of [imath]2[/imath]m, and a dart has equal probability to land anywhere on the board. What is the probability that the dart will land closer to the center than to the edges? So I know that if I center the square at the origin, I have to find [imath]f(x)[/imath] such that every point on [imath]f(x)[/imath] is equidistant between the origin and the line [imath]x=1[/imath] or [imath]y=1[/imath] (I'm only working with the first quadrant since by symmetry the probability should be the same). Another reasonable assumption that I made is that [imath]f(x)[/imath] is reflected over [imath]y=x[/imath], so I only have to work with the first [imath]45[/imath] degrees of the first quadrant. So to find [imath]f(x)[/imath] I did the following: [imath]\sqrt{x^2+[f(x)]^2}=1-x[/imath], so [imath]f(x)=\sqrt{1-2x}[/imath]. Now I know that I just find the area under this curve and divide it by [imath]1/8[/imath] the area of the square to find the probability. I just wanted to know if my [imath]f(x)[/imath] was okay? Thanks! |
2620671 | For a non-diagonalizable [imath]n \times n[/imath] matrix, how to prove that [imath]\exists (A_m)[/imath] is diagonalizable matrix such that [imath]A_m[/imath] converges to [imath]A[/imath].
For a non-diagonalizable [imath]n \times n[/imath] matrix, how to prove that [imath]\exists (A_m)[/imath] is diagonalizable matrix such that [imath]A_m[/imath] converges to [imath]A[/imath]. Here we define norm on [imath]M(n, \Bbb C)[/imath] s.t [imath]||X||=(\sum _{i=1}^\infty\sum_{j=1}^\infty|x_{ij}|^2)^{1/2}[/imath] and [imath]A_m[/imath] converges to [imath]A[/imath] iff [imath]||A_m - A|| \to 0[/imath] as [imath]m\to \infty[/imath] I was thinking to use Jordon Canonical form but can't get it as the calculation are getting twisted. If someone has some detailed proof please help. | 107945 | Diagonalizable matrices with complex values are dense in set of [imath]n\times n[/imath] complex matrices.
My linear algebra professor proved that Diagonalizable matrices with complex values are dense in set of [imath]n \times n[/imath] complex matrices. He defined a metric (I believe) that was somehow related to the usual metric on [imath]\mathbb{R}^{n^2}[/imath]. Then somehow proved that diagonalizable matrices were dense because for any matrix [imath]A[/imath] if [imath]\det(A - \lambda I) = 0[/imath] on an open subset, then [imath]\det(A - \lambda I)[/imath] was the zero polynomial. I Googled around a bit and found some stuff talking about the Zariski topology, and I am not sure this is what I want. Most of what I have found on this topology is much more general than what he was doing. Does anyone know any resources that would explain how to go about proving this statement? Edit: I found out how to prove this the way my professor did. We define the norm of a matrix by [imath] |A| = \max\{|Ax| : |x| = 1 \}.[/imath] Now we have a distance [imath]d(A, B) < \epsilon[/imath]. You can prove that if [imath](A - B)_{ij} < \epsilon/n[/imath], then [imath]|A - B| < \epsilon[/imath]. Let [imath]p(x)[/imath] be the characteristic polynomial of [imath]A[/imath], an [imath]n \times n[/imath] matrix. [imath]p(x) = \prod_1^n (x - x_i) = x^n + (-1)\sigma_1 x^{n - 1} + \cdots + (-1)^n\sigma_n[/imath] where [imath]\sigma_1, \ldots, \sigma_n[/imath] are the elementary symmetric polynomials of [imath]x_1, \ldots, x_n[/imath] which are the eigenvalues. The discriminant of the characteristic polynomial is a symmetric polynomial, therefore it can be written in terms of the elementary symmetric polynomials, which in turn can be written in terms of the entries of the matrix. But since the discriminant is a polynomial, it only has finitely many roots. Therefore for [imath]\epsilon > 0[/imath], by changing the entries of the matrix less than [imath]\epsilon / n[/imath] we can find a new matrix [imath]B[/imath] such that [imath]|B - A| < \epsilon[/imath] and the discriminant is not zero. The discriminant being not zero means [imath]B[/imath] has distinct eigenvalues, thus has a basis of eigenvectors. Therefore, it is diagonalizable. Thus the set of diagonalizable matrices is dense in the set of matrices with respect to that metric. |
2645590 | [imath]\sum_{k=n}^{2n} \binom{k}{n} 2^{-k}[/imath]
I don't have any idea to solve this problem. Find the value of [imath]\sum_{k=n}^{2n} \binom{k}{n} 2^{-k}[/imath] For every integers [imath]n\geq1[/imath] Could you give me a hint to solve it? Once I get a hint, I will try to solve it again. Thanks | 2642884 | Determine the value of summation form.
Determine the value of [imath]\displaystyle \sum_{k=n}^{2n} \binom{k} {n} 2^{-k}[/imath] for [imath]n \geq 1[/imath] My attempt: I have asked this question before in another math forum but it leads no solution. My friend suggested me to search for something like series-k. I also looked at my textbook in combinatoric section (which apparently leads to Binomial Theorem), but really I don't have any idea for this one. Could you help me? |
2645983 | Finding the Laplace Transform
I am trying to find the Laplace Transform of [imath]\frac{1-\cos(ax)}{x}.[/imath] I have tried to directly compute the integral and also tried to see if I could use the convolution theorem in some way but both lead nowhere. Any help would be appreciated! | 1904748 | How do I calculate [imath]u(w)=\int_0^\infty \frac{1-\cos(wt)}{t}\,e^{-t}\,dt[/imath]?
How do I calculate [imath]u(w)=\int_0^\infty \frac{1-\cos(wt)}{t}\,e^{-t}\,dt[/imath] I tried to do it, I use partial integration but I get lost. Is there any nice simple way to calculate it? |
2646122 | Proof if [imath]AB = BA[/imath] is not verified, then [imath]AB[/imath] is not symmetric.
They ask me the following question: Let [imath]A[/imath] and [imath]B[/imath] be two symmetric matrices of the same order, show that if [imath]AB = BA[/imath] is not verified, then [imath]AB[/imath] is not symmetric. I have tried the following but I do not know if it is correct: if [imath]AB \neq BA \Rightarrow (AB)^t = B^tA^t = BA \neq AB[/imath] | 2490623 | If A and B are symmetric is AB=BA true?
Question if we have matrices [imath]A[/imath] and [imath]B[/imath] which are symmetric. Both of these satisfy: [imath] A=A^T[/imath] Is [imath]AB=BA[/imath] true ? Attempt to solve Now if [imath]A[/imath] and [imath]B[/imath] are symmetric then [imath]AB[/imath] most be symmetric too. [imath] AB=(AB)^T=B^TA^T=BA [/imath] problem is how do you proof that if [imath]A[/imath] and [imath]B[/imath] are symmetric also [imath]AB[/imath] is symmetric ? If somone could provide some insight on this that would ge highly appreciated. Thanks, Tuki |
2646385 | Evaluate the Finite Sum with Binomial Coefficient
Evaluate the following sum: [imath]\sum_{k=0}^n\binom{n}{k}\frac{k}{n}x^k(1-x)^{n-k}[/imath] This almost looks like the Binomial Theorem, except there's that pesky little [imath]\frac{k}{n}[/imath] term in there that changes things up a bit. I'm not quite sure what steps to take from here. How do you evaluate this sum? | 629874 | Evaluate: [imath] \sum _ {k=1}^{n}{\frac{k}{n}\binom{n}{k}t^k(1-t)^{n-k}}[/imath]
Evaluate: [imath] \sum _ {k=1}^{n}{\frac{k}{n}\dbinom{n}{k}t^k(1-t)^{n-k}} [/imath] [imath]\dbinom{n}{k}[/imath]stands for the usual binomial coefficient giving the number of ways of choosing [imath]k[/imath] objects from n objects. Totally stuck.How can I able to solve this? |
2646601 | Prove that R is an equivalence relation on S.
let [imath]S[/imath] be the union of disjoint sets [imath]A_1, \dots , A_k[/imath]. Let [imath]R[/imath] be the relation consisting of pairs [imath](x,y) \in S \times S[/imath] such that [imath]x,y[/imath] belong to the same member of [imath]\{A_1, \dots, A_k\}[/imath]. Prove that [imath]R[/imath] is an equivalence relation on [imath]S[/imath]. To answer the question I did the following, I am not sure if it is correct. Reflexive For any [imath]x \in A_i[/imath], [imath](x,x) \in R[/imath] and this also hold for any [imath]y \in A_i[/imath], so therefore reflexivity holds symmetry For any [imath]x,y \in A_i[/imath] where [imath]x = y[/imath], then [imath](x,y)=(y,x) \ in R[/imath] therefore symmetry holds Transitivity given [imath](x,y),(y,z) \in R[/imath] then [imath]x=y[/imath] and [imath]y=z[/imath] so therefore [imath]x=z[/imath] and Transitivity holds I don't know if the above counts as a valid proof but that is what i was able to come up with. | 2642523 | Union of disjoint sets : Equivalence relations
I'm not exactly sure how to go about doing this question. I've attempted it but I'm not exactly sure if it's correct. Question: Let [imath]S[/imath] be the union of disjoint sets [imath]A_1, \cdots, A_k[/imath]. Let [imath]R[/imath] be the relation consisting of pairs [imath](x, y) \in S \times S[/imath] such that [imath]x, y[/imath] belong to the same member of [imath]\{A1, \cdots, A_k\}[/imath]. Prove that [imath]R[/imath] is an equivalence relation on [imath]S[/imath]. The three axioms are: reflexivity, symmetry, transitivity, I have a brief idea of how to do the first 2, however, for transitivity, I don't have any ideas. Could anyone help out on this please? |
2646478 | [imath]\left\{e\right\}[/imath] Closed Implies [imath]T_0[/imath] and Regular Topological Group
Let [imath]G[/imath] be a topological group with identity [imath]e[/imath] such that [imath]\left\{e\right\}[/imath] is closed. Then [imath]G[/imath] is [imath]T_0[/imath] and regular. [imath]G[/imath] is trivially [imath]T_0[/imath]. Let [imath]C\subset G[/imath] be closed and [imath]C\not\ni g\in G[/imath]. For every [imath]h\in C[/imath], I was able to find an open neighborhood [imath]U_h\not\ni g[/imath] of [imath]h[/imath]. I am stuck trying to find an open neighborhood [imath]U[/imath] of [imath]g[/imath] disjoint from [imath]\bigcup_{h\in C}U_h[/imath]. | 2039061 | Help understanding proof of every topological group is regular
I found a proof that any topological group is regular here, but I got lost in the last part. The whole argument goes like this: Consider the map [imath]f:G \times G \to G[/imath] defined by [imath]f(a,b)=ab^{-1}[/imath]. This map is always continuous in a topological group. Now take [imath]x \in U[/imath], where [imath]U[/imath] is an open set in [imath]G[/imath]. Then [imath]f^{-1}(U)[/imath] contains [imath](x,e)[/imath], so we have [imath](x,e)\in V \times W \subseteq f^{-1}(U)[/imath] for some open subsets [imath]V[/imath] and [imath]W[/imath] such that [imath]x\in V[/imath] and [imath]e\in W[/imath]. Hence, [imath]x\in V[/imath]. Furthermore, [imath]V\cap (X-U)W=\emptyset[/imath], since any element in the intersection corresponds to [imath]a\in V[/imath], [imath]b\in W[/imath], such that [imath]ab^{-1}\notin U[/imath], which is a contradiction. Since [imath](X-U)W[/imath] is an open set containing [imath]X-U[/imath], [imath]X[/imath] is regular. The boldface argument is what I don't understand. How can I see that [imath]V\cap (X-U)W=\emptyset[/imath]? |
1828775 | Prove [imath]\forall n \in \mathbb{N}: \int_{0}^{\frac{\pi}{2}} |\frac{\sin(nx)}{x}|dx \geq \frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k}[/imath]
As my further preparation to Putnam competition, I came across such inequality to prove: [imath]\forall n \in \mathbb{N}: \int_{0}^{\pi} \left|\frac{\sin(nx)}{x}\right|dx \geq \frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k}[/imath] The problem is that I got stuck in spite of spending by far 6 days on this problem! Because I got really stuck, I am very determined to see how to prove such inequality. Help very, very appreciated! | 683576 | Show that,[imath]\int_0^\pi \left|\frac{\sin nx}{x}\right|\mathrm{d}x \ge \frac{2}{\pi}\left(1+\frac12+\cdots+\frac{1}{n}\right)[/imath]
Show that,[imath]\int_0^\pi \bigg|\dfrac{\sin nx}{x}\bigg|\mathrm{d}x \ge \dfrac{2}{\pi}\bigg(1+\dfrac12+\cdots+\dfrac{1}{n}\bigg)[/imath] I could not approach the problem at all. Please help. |
202924 | Finding [imath]\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!} +\frac{5}{3!+4!+5!}+\cdots+\frac{2008}{2006!+2007!+2008!}[/imath]
How to find : [imath]\frac{3}{1!+2!+3!}+\frac{4}{2!+3!+4!} +\frac{5}{3!+4!+5!}+\cdots+\frac{2008}{2006!+2007!+2008!}[/imath] | 2647385 | A sum of series problem: [imath]\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}[/imath]
I have a question regarding the sum of this series: [imath]\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}[/imath] My approach: I found that this sum is equal to: [imath]\sum_{n=3}^{2008}\frac{n}{(n-2)!+(n-1)!+(n)!}[/imath] I reduced it to : [imath]\sum_{n=3}^{2008}\frac{1}{n(n-2)!}[/imath] Please suggest how to proceed further. |
2552775 | Show that [imath]\sin x[/imath] lies between [imath]x-x^3/6[/imath] and [imath]x \;[/imath] [imath]\forall x \in R[/imath]
Show that [imath]\sin x[/imath] lies between [imath]x-x^3/6[/imath] and [imath]x \;[/imath] [imath]\forall x \in R[/imath] I am getting: [imath]\sin(x) = x - \frac{x^3}{3!} + R_4(x)[/imath] where [imath]R_4(x) = \frac{\cos(c)x^5}{5!}[/imath] for some [imath]c[/imath] between [imath]0[/imath] and [imath]x[/imath] I want to prove [imath]R_4(x)\geq 0[/imath] to arrive at the result [imath]x-x^3/3 \leq \sin(x)[/imath]. for:[imath]0 \leq x \leq \pi/2 \Rightarrow 0<c<\pi/2 \Rightarrow R_4(x)\geq 0[/imath] but for: [imath]-\pi/2 \leq x < 0 \Rightarrow -\pi/2 < c < 0 \Rightarrow R_4(x) < 0[/imath] How can I proceed with this ? There are many cases that I need to check | 390899 | Proving that [imath]x - \frac{x^3}{3!} < \sin x < x[/imath] for all [imath]x>0[/imath]
Prove that [imath]x - \frac{x^3}{3!} < \sin(x) < x[/imath] for all [imath]x>0[/imath] This should be fairly straightforward but the proof seems to be alluding me. I want to show [imath]x - \frac{x^3}{3!} < \sin(x) < x[/imath] for all [imath]x>0[/imath]. I recognize this shouldn't be too difficult but perhaps finals have fried my brain. |
1711275 | Limit involving Zeta function
Evaluate [imath]\displaystyle \lim_{x \to 1} (x-1) \zeta (x)[/imath] I'm not much familiar with the properties of zeta function. An elementary solution is appreciated. Thanks. | 412616 | Showing that [imath]\displaystyle\lim_{s \to{1+}}{(s-1)\zeta(s)}=1[/imath]
I need prove the following: ([imath]\zeta(s)[/imath] is the Riemann zeta function) [imath]\displaystyle\lim_{s \to{1+}}{(s-1)\zeta(s)}=1[/imath] I really don't know, I have tried hard, but have obtained nothing for now. |
2647938 | Find all the subgroups of this group
I found the following subgroups: [imath]H_1=\{I_2,A_\pi\}[/imath] [imath], H_2=\{I_2,F_v\}[/imath] [imath]H_3=\{I_2\ ,F_h\}[/imath] [imath]H_4=\{I_2\}[/imath] [imath]H_5=\{I_2,A_\pi\,F_v,F_h\}[/imath] Is my answer correct? Are there more subgroups? | 697722 | Subgroups of the Klein-4 Group
Can anyone explain to me the subgroups of the Klein-4 group? I'm trying to view it this way: I want some groups that are not empty and [imath]ab^{-1} \in H[/imath], where [imath]H[/imath] denotes the subgroups I am looking for. So the Klein-4 group is as follows: [imath]V_4 = \langle a,b : a^2 = b^2 = (ab)^2 = 1\rangle [/imath]. I'm a bit confused by the [imath]\langle [/imath] and [imath]\rangle [/imath] to denote a cyclic group but I sort of follow what's going on. I guess the group itself and the trivial subgroup are there...But can you explain to me how to find the rest? Possible answer: Trivial subgoups [imath] \{\ e \}\ [/imath] has order one. Order 2: [imath]\{\ e,a \}\ , \{\ e,b \}\ , \{\ e,ab \}\ [/imath] Order 4: The group itself. As the order of the subgroups must divide the order of the group, these must be all the subgroups. |
2648201 | Prove that for any fixed non-negative random variable [imath]X[/imath], [imath]\lim\limits_{n→∞}\frac1nE\left(\frac1X I_{\{X>1/n\}}\right) =0[/imath]
Show [imath]\lim_{n\to\infty} \frac{1}{n}E\left(\frac{1}{X} I_{\{X>1/n\}}\right) = 0.[/imath] I am totally lost on this problem. Any hint or reference to study will be greatly appreciated. | 2633866 | Durrett Ex.1.3.16
Let [imath]X \ge 0[/imath]. Show \begin{align} \lim_{y\to\infty} y\,\Bbb{E}\left[ \frac1X \cdot1_{X>y} \right] &= 0 \quad\text{and} \\ \lim_{y \downarrow 0} y\,\Bbb{E}\left[ \frac1X \cdot1_{X>y} \right] &= 0. \end{align} N.B.: The author doesn't assume [imath]\Bbb{E}[1/X] < \infty[/imath]. My try: Since the expectation appears on the RHS of Markov's inequality, I don't think this inequality can be applied to solve this question. [imath] \begin{aligned} y\,\Bbb{E}\left[ \frac1X \cdot1_{X>y} \right] &= \Bbb{E}\left[ \frac{y}{X} \cdot1_{X>y} \right] \\ &< \Bbb{E}\left[ 1 \cdot1_{X>y} \right] \\ &= \Bbb{P}(X > y) \\ &\xrightarrow[y\to0]{} \Bbb{P}(X \ge 0) = 1 \end{aligned} \tag1 \label1 [/imath] For the first part [imath]y \to \infty[/imath], I apply Markov's inequality at the row [imath]\Bbb{P}(X > y)[/imath] in \eqref{1}, so that [imath]\Bbb{P}(X > y) \le \frac{\Bbb{E}[X]}{y} \xrightarrow[y\to\infty]{} 0?[/imath] But I don't have [imath]\Bbb{E}[X] < \infty[/imath]. Thanks for any help. |
2648521 | Trigonometric Inequality [imath]\sin{A}+\sin{B}-\cos{C}\le\frac32[/imath]
To prove [imath]\sin{A}+\sin{B}-\cos{C}\le\frac32[/imath] Given [imath]A+B+C=\pi[/imath] and [imath]A,B,C>0[/imath] I have managed to convert LHS to [imath]1-4\cos{\frac C2}\sin{\frac{A+C-B}2}\cos{\frac{A-B-C}2}[/imath] but that clearly isn't very helpful One other conversion was [imath]\sin A+\sin B+\cos A\cos B-\sin A\sin B[/imath] but not sure how to proceed. Any hints will be appreciated | 646584 | If [imath]A,B,C[/imath] are the angle of a triangle, then show that [imath]\sin A+\sin B-\cos C\le \dfrac3 2[/imath]
If [imath]A,B,C[/imath] are the angle of a triangle, then show that [imath]\sin A+\sin B-\cos C\le \dfrac32[/imath] I tried substituting [imath]C=180^\circ-(A+B)[/imath] and got stuck. I also tried using the formula [imath]\sin A+\sin B=2\sin \frac{A+B}{2}\cos \frac{A-B}{2}[/imath] without any progress. Please help! |
2644242 | Evaluate a sum which almost looks telescoping but not quite:[imath]\sum_{k=2}^n \frac{1}{k(k+2)}[/imath]
Suppose I need to evaluate the following sum: [imath]\sum_{k=2}^n \frac{1}{k(k+2)}[/imath] With partial fraction decomposition, I can get it into the following form: [imath]\sum_{k=2}^n \left[\frac{1}{2k}-\frac{1}{2(k+2)}\right][/imath] This almost looks telescoping, but not quite... so at this point I am unsure of how to proceed. How can I evaluate the sum from here? | 747822 | Help with telescoping sum [imath]\sum_{i=3}^n \frac{1}{i(i+3)} [/imath]
How can I use the telescoping technique to compute the following sum? I'm having issues getting started. I know the basic steps but I don't know how to perform them. I know I have to separate the fraction into A and B. After that I have to perform the sum but I'm not sure what comes next. [imath]\sum_{i=3}^n \frac{1}{i(i+3)} [/imath] Thanks for any help! |
2648611 | Show that [imath]d(x,y)= \left|\frac{1}{x} - \frac{1}{y}\right|[/imath] defines a metric on [imath](0,\infty)[/imath].
Exercise: Show that [imath]d(x,y)= \left|\frac{1}{x} - \frac{1}{y}\right|[/imath] defines a metric on [imath](0,\infty)[/imath]. What I've tried: I need to check the following properties i) [imath]0\leq d(x,y)<\infty,[/imath] for all [imath]x,y\in(0,\infty)[/imath]. ii) [imath]d(x,y)=0[/imath] if and only if [imath]x=y[/imath] iii) [imath]d(x,y)=d(y,x),[/imath] for all [imath]x,y\in(0,\infty)[/imath]. iv) [imath]d(x,z)\leq d(x,y) + d(y,z)[/imath] for all [imath]x,y,z\in(0,\infty)[/imath]. I think I was able to solve the first three properties i): If [imath]\dfrac{1}{x} - \dfrac{1}{y}[/imath] is positive then [imath]d(x,y)[/imath] is positive, if [imath]\dfrac{1}{x} - \dfrac{1}{y}[/imath] is negative then [imath]d(x,y) = \left|\dfrac{1}{x} - \dfrac{1}{y}\right|[/imath] is positive, so [imath]d(x,y)\geq 0[/imath]. Since [imath]x,y\in(0,\infty)[/imath] we have that [imath]\left|\dfrac{1}{x} - \dfrac{1}{y}\right|\in(0,\infty)[/imath]. ii): If [imath]x = y[/imath] then [imath]d(x,y) = 0[/imath], if [imath]x\neq y[/imath] then [imath]\dfrac{1}{x} \neq \dfrac{1}{y}[/imath] and [imath]d(x,y) \neq 0[/imath]. iii): If [imath]\left|\dfrac{1}{x} - \dfrac{1}{y}\right|^2 = \left|\dfrac{1}{y} - \dfrac{1}{x}\right|^2[/imath] then [imath]\left|\dfrac{1}{x} - \dfrac{1}{y}\right| = \left|\dfrac{1}{y} - \dfrac{1}{x}\right|[/imath]. I know that [imath]\left|\dfrac{1}{x} - \dfrac{1}{y}\right|^2 = 1/x^2 - 2/xy + 1/y^2 = \left|\dfrac{1}{y} - \dfrac{1}{x}\right|^2,[/imath] hence [imath]\left|\dfrac{1}{x} - \dfrac{1}{y}\right| = \left|\dfrac{1}{y} - \dfrac{1}{x}\right|^2[/imath] and [imath]d(x,y) = d(y,x)[/imath]. I don't know how I should show that the fourth property holds! Question: How do I show that [imath]\left|\dfrac{1}{x} - \dfrac{1}{y}\right|\leq \left|\dfrac{1}{x} - \dfrac{1}{z}\right| + \left|\dfrac{1}{z} - \dfrac{1}{y}\right|[/imath], or equivalently, that [imath]d(x,y) \leq d(x,z) + d(z,y)[/imath]? Thanks! | 1461723 | Show that [imath]\left| \frac{1}{x}-\frac{1}{z} \right| \le \left|\frac{1}{x} - \frac{1}{y} \right| + \left|\frac{1}{y} - \frac{1}{z} \right|[/imath].
I'm trying to prove that [imath] \left| \frac{1}{x}-\frac{1}{y} \right| [/imath] is a metric on [imath]X = \mathbb R^+[/imath]. The properties of non-negativity and symmetry are obvious (do I actually even need to prove them?), but the tricky one is, as usual the inequality property, i.e. [imath] d(x,z) \le d(x,y) + d(y,z) \text{ for any } z \in X. [/imath] So, ok, I have that [imath] d(x,z) \le d(x,y) + d(y,z) \\ \iff \left| \frac{1}{x}-\frac{1}{z} \right| \le \left|\frac{1}{x} - \frac{1}{y} \right| + \left|\frac{1}{y} - \frac{1}{z} \right| \\ \iff \left| \frac{1}{x} \right|- \left| \frac{1}{z} \right| \le \left| \frac{1}{x}-\frac{1}{z} \right| \le \left|\frac{1}{x} - \frac{1}{y} \right| + \left|\frac{1}{y} - \frac{1}{z} \right| [/imath] ...and that's about as far as I've gotten–which is to say, not very far. I also tried multiplying both sides by [imath] \left| \frac{1}{x}-\frac{1}{z} \right| [/imath], but that wasn't very fruitful. If someone could point me in the right direction, it'd be greatly appreciated. Thanks. |
2648642 | X is r.v. on ([imath]\Omega[/imath],F,P) where [imath]E[\left\lvert X \right\rvert^a] < \infty[/imath]. Prove [imath]E[\left\lvert X \right\rvert^b] < \infty[/imath] for all [imath]0[/imath]
X is r.v. on ([imath]\Omega[/imath],F,P) where [imath]E[\left\lvert X \right\rvert^a] < \infty[/imath]. Prove [imath]E[\left\lvert X \right\rvert^b] < \infty[/imath] for all [imath]0<b\le a[/imath] How would I go about proving this using something in the realm of Lebesgue Integration or Domination thm. If I am off base with this please let me know. Having trouble even visualizing how to prove, would appreciate any help or hints. edit: added abs value. sorry for that oversight! | 580542 | Suppose [imath]\mu(X)<\infty[/imath]. Show that if [imath]\ f\in\ L^q(X,\mu)[/imath] for some [imath]1\le q<\infty[/imath] then [imath]\ f\in\ L^r(X,\mu)[/imath] for [imath]1\le r[/imath]
Suppose [imath]\mu(X)<\infty[/imath]. Show that if [imath]\ f\in\ L^q(X,\mu)[/imath] for some [imath]1\le q<\infty[/imath] then [imath]\ f\in\ L^r(X,\mu)[/imath] for [imath]1\le r<q[/imath] In the solution the assistant wrote this: If [imath]1\le r<q[/imath] then [imath]q/r\ge 1[/imath]. Let p be the dual exponent to [imath]q/r[/imath] (i.e. [imath]\frac{1}{q/r}+\frac{1}{p}=1)[/imath] then by Hölder Inequality [imath]\int_{X}|f|^rd\mu\le\underbrace{(\int_{X}|f|^qd\mu)^{r/q}}_{??????}(\int_X1^pd\mu)^{1/p}<\infty[/imath] for me the middle part doesn't make sense (it is not the hölder inequality) can somebody show it ? instead of this can't we say (only in this case) that [imath]\int_{X}|f|^rd\mu\le (\int_{X}|f|d\mu)^r[/imath] then [imath](\int_{X}|f|d\mu)^r\le ((\int_{X}|f|^qd\mu)^{1/q}(\int_X1^pd\mu)^{1/p})^r<\infty[/imath] ? |
2615932 | Integral of [imath]\int_{-\infty}^\infty x^2e^{-\frac {(x-\mu)^2}{2\sigma^2}}dx[/imath]
Calculate this integral [imath]\int_{-\infty}^\infty x^2e^{-\frac {(x-\mu)^2}{2\sigma^2}}dx[/imath] I know the trick can be using [imath]E[x^2]=var[x]+E^2[x][/imath], but how to solve it in an analytic way? | 1596354 | Solve the integral [imath]\frac 1 {\sqrt {2 \pi t}}\int_{-\infty}^{\infty} x^2 e^{-\frac {x^2} {2t}}dx[/imath]
To find the Variance of a Wiener Process, [imath]Var[W(t)][/imath], I have to compute the integral [imath] Var[W(t)]=\dots=\frac 1 {\sqrt {2 \pi t}}\int_{-\infty}^{\infty} x^2 e^{-\frac {x^2} {2t}}dx=\dots=t. [/imath] I've tried integration by parts to solve the integral but end up with [imath] \dots=\frac 1 {\sqrt {2 \pi t}} \left(0 - \int_{-\infty}^{\infty} -\frac {x} {t} \cdot e^{-\frac {x^2} {2t}}\cdot \frac {x^3} 3 dx\right), [/imath] which is even worse and probably wrong. Can anyone please help me compute the first integral and show how it becomes equal to [imath]t[/imath]? (I know that the Variance for a Wiener Process (Standard Brownian motion)is defined as [imath]t[/imath] but want to prove it with the integral above.) |
2649625 | How to prove a Linear Transformation is an identity mapping
Firstly, what's the definition of a identity map? I've thought of one but don't know if it's correct :" [imath]P[/imath] is a identity map if [imath]P(x)=x[/imath] for all [imath]x[/imath] on its domain." Second, how to prove a projection transformation is an identity map on a specific domain. More specifically, Let [imath]P: V \to V[/imath] be a projection. Show that the restriction of [imath]P[/imath] to [imath]\operatorname{range}(P)[/imath] is the identity map on [imath]\operatorname{range}(P)[/imath]. | 552850 | Proof of an elementary property of Projection Operators
I'm asked to show the following: Let [imath]X[/imath] be a linear space, and let [imath]P : X \rightarrow X[/imath] be a projection operator. Restricted to the linear space [imath]range(P)[/imath], the projection [imath]P[/imath] is the identity operator, that is, [imath]Px = x[/imath] for all [imath]x \in range(P)[/imath]. If anyone could offer a hint for how to show this. I'm rusty with linear algebra so don't really know what I'm supposed to bring in outside of what's specified in the lemma/definition of projection. What is so special about the range of P that would make this so? |
2649426 | The convergence of [imath] \lim\limits_{x\to\infty}\frac{f(x)}x=0[/imath]
Let [imath]f[/imath] be continuous such that [imath]\lim _{x\to \infty }\left(f\left(x+1\right)-f\left(x\right)\right)=0[/imath] show that [imath]\lim _{x\to \infty }\left(\frac{f\left(x\right)}{x}\right)=0[/imath] My idea was applying the sequential criterion for continuity: that if [imath]f[/imath] is continuous at [imath]c[/imath] if and only if for every sequence [imath]x_n[/imath] that converges to [imath]c[/imath], [imath]f(x)[/imath] converges to [imath]f(c)[/imath]. Also, I have the idea that since [imath]\lim_{x\to \infty }\left(f\left(x\right)\right)=\lim_{x\to\infty}\left(f\left(x+1\right)\right),[/imath] I can use it somehow in the demonstration, or am I wrong? | 2119937 | Limit of ratio [imath]{f(x)\over x}[/imath] equal to limit of difference [imath]f(x+1)-f(x)[/imath].
Using my right to ask for help once again. I can't solve the following problem for quite a long time. Just have no idea how to do that. Here it is. Let [imath]f(x)[/imath] be bounded on any interval [imath](1,b), b>1[/imath]. Then [imath]\lim_{x\to+\infty}{f(x) \over x}=\lim_{x\to+\infty}(f(x+1)-f(x))[/imath]. It is from the Russian book "Lekcii po matematicheskomu analizu" by G.I.Arkhipov, V.A.Sadovnichy and V.N.Chubarikov, 2nd ed., Moscow, 2000, page 676. Please, keep in mind, that times, when I did homework had passed long time ago. Now I'm solving problems for pleasure. |
2649051 | Inverse trigonometic functions
Question: Prove that [imath]\tan^{-1}(\frac{1}{2} \tan2A)+\tan^{-1}(\cot A)+\tan^{-1}(\cot^{3} A)=0[/imath] MyProblem We can just use the formula of [imath]\tan^{-1} A +\tan^{-1} B[/imath] but I think it would be a waste of time. Is there any other shorter and simpler method to solve it. | 1114007 | Simplify [imath]\arctan (\frac{1}{2}\tan (2A)) + \arctan (\cot (A)) + \arctan (\cot ^{3}(A)) [/imath]
How to simplify [imath]\arctan \left(\frac{1}{2}\tan (2A)\right) + \arctan (\cot (A)) + \arctan (\cot ^{3}(A)) [/imath] for [imath]0< A< \pi /4[/imath]? This is one of the problems in a book I'm using. It is actually an objective question , with 4 options given , so i just put [imath]A=\pi /4[/imath] (even though technically its disallowed as [imath]0< A< \pi /4[/imath]) and got the answer as [imath]\pi [/imath] which was one of the options , so that must be the answer (and it is weirdly written in options as [imath]4 \arctan (1) [/imath] ). Still , I'm not able to actually solve this problem. I know the formula for sum of three arctans , but it gets just too messy and looks hard to simplify and it is not obvious that the answer will be constant for all [imath]0< A< \pi /4[/imath]. And I don't know of any other way to approach such problems. |
2649900 | Number of nonzero Fourier coefficients for any [imath]x[/imath] can be at most countably many
I have been trying to prove Lemma 3.5 in "Introduction to Functional Analysis" by Kreyszig, which says that: Any [imath]x[/imath] in an inner product space [imath]X[/imath] can have at most countably many nonzero Fourier coefficients [imath]\langle x,e_k\rangle[/imath] with respect to an orthonormal family [imath](e_k), k\in I,[/imath] in X. This is my attempt at proof using the hint provided in the book: Let [imath]n_m[/imath] be the number of Fourier coefficients which are greater than [imath]1/m[/imath]. By Bessel inequality: [imath]\|x\|^2\geq\sum_{k=1}^{\infty}\big|\langle x,e_k\rangle\big|^2> n_m\frac{1}{m^2}.[/imath] [imath]\implies n_m< m^2\|x\|^2. [/imath] This last inequality is clearly true for any [imath]m=1,2,....[/imath]. Which implies that maximum number of nonzero Fourier coefficients [imath]n_{\frac{1}{\infty}}<\infty.[/imath] But I am stuck here. It is clear that number of nonzero coefficients might be infinite, but how does it imply [imath]n_m[/imath] is still countable. The same question has been asked at: Fourier coefficients [imath]\langle x,e_k\rangle[/imath] are at countable. But I don't understand the last sentence of the answer: Therefore there are only countably many i with [imath]|\langle x,e_i\rangle|>0.[/imath] | 2274906 | Fourier coefficients [imath]\langle x,e_k\rangle[/imath] are at countable
Say we have an Inner Product space [imath](X, \langle\cdot ,\cdot \rangle)[/imath]. Let [imath](e_k)[/imath] be an orthonormal sequence, where [imath]k \in I[/imath]. Then we know by Bessel's inequality that [imath] \sum_1^\infty |\langle x,e_k\rangle|^2 \le \|x\|^2[/imath] The [imath]\langle x,e_k\rangle[/imath] are called Fourier coefficients of x (with respect to [imath]e_k[/imath]). How do we then show that any [imath]x \in X[/imath] can have at most countably many non-zero Fourier coefficients? First we suppose that the orthonormal family [imath](e_k)[/imath] is uncountable, otherwise it's trivial. So my first question is, how can there exist uncountably many [imath]e_k[/imath]? Is there an example? What I thought could help is the following: form the coefficients [imath]\langle x,e_k\rangle[/imath]. We know that for every [imath]n=1,2,3,\ldots[/imath] [imath] \sum_1^n |\langle x,e_k\rangle|^2 \le \|x\|^2[/imath] But I couldn't make any progress. |
2650770 | Riemann Sum look-alike
[imath]\lim_{n \rightarrow \infty} \sum_{k=1}^n \sqrt{n^4+k} \ sin\frac{2k\pi}{n}[/imath] I have found this Riemann Sum look-alike, but I have no idea how to approach it. Can you perhaps help me solve this? The answer should be [imath]-\frac{1}{4\pi} [/imath]. | 2638504 | How to calculate the limit [imath]\lim _{n\to \infty }\sum _{k=1}^n\sqrt{n^4+k}\sin(\frac{2k\pi }{n})[/imath]? Is it a Riemann sum?
I just came across this limit and I suppose it can be computed using a Riemann sum but I can't get it right. [imath]\lim _{n\to \infty }\sum _{k=1}^n\sqrt{n^4+k}\sin\left(\frac{2k\pi }{n}\right)[/imath] Any ideas? |
2650262 | Solving a pair of quadratic equations for two unknowns efficiently
I have a pair of quadratic equations in the form [imath]ax^2 + bx + cy^2 + dy + e = 0[/imath] [imath]fx^2 + gx + hy^2 + ky + l = 0[/imath] and there should be a solution in the range [imath]-2 \leq x,y \leq 2[/imath] which I need to find programmatically. I can eliminate one variable by squaring to get a quartic equation which I could solve numerically and substitute back, but I'm not sure about numeric stability and spurious roots. Is there a better way of solving this? Edit: This question Solution af a system of 2 quadratic equations has answers with comments "In order to solve this degree 4 equation you need either some luck, or some patience " and "How would one programatically find the lines that constitute a degenerate conic?". That does not seem to demonstrate a better/simpler solution. | 175380 | Solution af a system of 2 quadratic equations
I have a system of two quadratic equations with unknowns [imath]x[/imath] and [imath]y[/imath]: [imath]a_{1 1} x y + a_{1 2} x^2 + a_{1 3} y^2 + a_{1 4} x + a_{1 5} y + a_{1 6} = 0,\\ a_{2 1} x y + a_{2 2} x^2 + a_{2 3} y^2 + a_{2 4} x + a_{2 5} y + a_{2 6} = 0,[/imath] where [imath]a_{i j}[/imath] are arbitrary scalars. Is there an algebraic solution of the above system? |
2645750 | Existence of totally real number fields of any degree
Are there simple/canonical examples for totally real number fields of degree [imath]n[/imath] for any natural number [imath]n \in \mathbb{N}[/imath]? Of course, the question can be reformulated by asking if there exists an irreducible polynomial of degree [imath]n[/imath] with real roots only, i.e. all roots in the algebraic closure [imath]\mathbb{C}[/imath] are real. Currently, I just started to read about field theory, but in the standard books (and soucres) I could not find an answer to my question. [imath]\mathbb{Q}(\cos(2\pi/n))[/imath] is an example of degree [imath]\varphi(n)/2[/imath]. In the Paper On 2-Rank of the Ideal Class Groups of Totally Real Number Fields of Ichimura one can find a construction of such polynomials (with some additional properties), using Hilbert's Irreducibility Theorem. My question arose from Geometry of Numbers/Analytic Number Theory: Any totally real number field gives rise for an 'admissible lattice' [imath]\Lambda[/imath], that means [imath]\inf_{(m_1,\ldots,m_n) \in \Lambda \setminus \{0\}} |m_1 \cdots m_d| >0[/imath] in standard coordinates, as follows. Under the map [imath]\mathbb{K} \rightarrow \mathbb{R}^n[/imath], [imath]x \mapsto (\sigma_1(x),\ldots,\sigma_n(n))[/imath], where [imath]\sigma_1,\ldots,\sigma_n[/imath] denote all embeddings (which are real since [imath]\mathbb{K}[/imath] is totally real) of the number field [imath]\mathbb{K}[/imath], the image of the ring of integers [imath]\mathcal{O}_{\mathbb{K}}[/imath] is an admissible lattice. Skriganov has shown in Constructions of Uniform Distributions in Terms of Geometry of Numbers, 1994, that 'admissible lattices' are very uniformly distributed in parallelepipeds. (The remainder term is only of type [imath]\log^{d-1}{r}[/imath].) | 1867852 | Totally real Galois extension of given degree
Let [imath]n≥1[/imath] be an integer. I would like to prove (or disprove) the existence of a subfield [imath]K \subset \Bbb R[/imath] such that [imath]K/\Bbb Q[/imath] is Galois and has degree [imath]n[/imath]. It is easy to construct such a subfield for [imath]n=2^k[/imath]: one can take [imath]K=\Bbb Q(\sqrt 2, \sqrt 3, \sqrt 5, ..., \sqrt{p_k})\;[/imath] where [imath]p_k[/imath] is the [imath]k[/imath]-th prime number. I found that the irreducible polynomial [imath]x^3-6x+2 \in \Bbb Q[X][/imath] (Eisenstein's criterion with [imath]p=2[/imath]) has three real roots, so its splitting field [imath]K[/imath] satisfies my conditions for [imath]n=3[/imath]. Then I'm done with [imath]n=3\cdot 2^k[/imath] thanks to compositum of fields. Apparently, a Galois extension of odd degree must be totally real. In all cases, I don't know how to construct, in general or in particular cases ([imath]n[/imath] prime for instance), such a totally real Galois extension [imath]K/\Bbb Q[/imath] of degree [imath]n[/imath]. Thank you for your help! |
2651126 | How do I calculate the ratio of a distribution of random variables
I'm quite lost with regards to solving this problem. Any help is appreciated. Thanks. Suppose [imath]X_1, X_2, . . .[/imath] are independent discrete random variables, having the same distribution, and [imath]E[X_i]>0[/imath], for each [imath]i[/imath]. Is it true that for any two positive integers [imath]m[/imath] and [imath]n[/imath] [imath]E[\frac{X_1+...+X_m}{X_1+...+X_n}]=\frac{m}{n}[/imath] ? | 852890 | Expectation of random variables ratio
Let [imath]X_1, X_2, \dots, X_n[/imath] be [imath]n[/imath] positive iid random variables. Then show that [imath]E\left(\frac{\sum_{j=1}^k X_j}{\sum_{i=1}^{n} X_i}\right) = \frac{k}{n}.[/imath] Because of the linearlity of the expectation I known that [imath]E\left(\frac{\sum_{j=1}^k X_j}{\sum_{i=1}^{n} X_i}\right) = \sum_{j=1}^k E\left(\frac{X_j}{\sum_{i=1}^{n} X_i}\right)[/imath], so it's enougth to show [imath]E\left(\frac{X_1}{\sum_{i=1}^{n} X_i}\right) = \frac{1}{n}[/imath]. But I'm unable to deal with the [imath]X_i[/imath] in the denominator. |
2651133 | Maclaurin Series of [imath]e^{t^2/2}[/imath]
How do we find Maclaurin Series of [imath]e^{t^2/2}[/imath]? I see one way on one book to do it is applying the fact that [imath]e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!}[/imath], then [imath]e^{t^2/2} = \sum_{n=0}^{\infty}\frac{(\frac{1}{2}t^2)^n}{n!}[/imath] But I cannot really understand why can we directly subsitute [imath]\frac{1}{2}t^2[/imath] into [imath]e^x[/imath] equation since it is not a constant. May someone explain to me? I am really confused. | 106649 | Multiplying Taylor series and composition
I have two questions: A. I know the taylor series for [imath]\arctan(x)[/imath] and for [imath]e^x[/imath]. How do I find the series for [imath]\arctan(x)\cdot e^x[/imath] ? B. Say I want to find the series for [imath]\arctan(g(x))[/imath], do I just subtitue [imath]x[/imath] with [imath]g(x)[/imath], Or do I have to start the process of finding the series from the beginning? Thanks! |
2650168 | Does any [imath]n[/imath]-th order ODE have to have [imath]n[/imath] arbitrary contants in its general solution?
In the lecture, my proffesor stated that Any [imath]n[/imath]-th order ODE has [imath]n[/imath] arbitrary constants in its general solution. However, we have later seen that some ODEs have some singularities, so the so called "the general solution" is not so general at all in many cases, and this lead me to question that is there any execeptions to the rule that I have quoted. For any ordinary DE, does the general solution of this ODE has to have [imath]n[/imath] arbitrary constants ? If so, how ? if not, can you provide an example ? Edit: As I have pointed out, I'm asking this question for ODEs, and not PDEs. | 2554972 | How many "constants of integration" will a PDE have?
For smooth cases, we expect the solution of an [imath]n[/imath]th-order ODE to have [imath]n[/imath] constants of integration. Is there a name to the theorem which more precisely makes this claim? Is there an analogous theorem for PDEs? That is, do we expect [imath]n[/imath] "integration functions" for an [imath]n[/imath]th-order PDE? |
2651626 | If [imath]V[/imath] over [imath]\mathbb{C}[/imath], [imath]\dim V\ge 2[/imath], let [imath]q: V \rightarrow \Bbb{C}[/imath] be a Quadratic form. s.t. exists [imath]v \ne 0[/imath] such that [imath]q(v)=0[/imath]
If [imath]V[/imath] is a vector space over [imath]\mathbb{C}[/imath] and [imath]\dim V\ge 2[/imath], prove that if [imath]q: V \rightarrow \Bbb{C}[/imath] is a Quadratic form then exists [imath]v \ne 0[/imath] such that [imath]q(v)=0[/imath] I'm not sure I even understand why this is true. Any tips? | 81798 | Quadratic form under [imath]\mathbb{C}[/imath] finite-dimension vector space always have a nonzero vector that yields zero?
Let [imath] V [/imath] be a vector space of finite dimension on [imath]\mathbb{C} [/imath] and [imath] \dim(V) \gt 1 [/imath]. Show that for every quadratic form [imath] q : V \to \mathbb{C} [/imath] there exists [imath] 0 \neq v \in V [/imath] such that [imath] q(v) = 0 [/imath]. It certainly has something to do with the fact that we are in [imath]\mathbb{C} [/imath] and not [imath] \mathbb{R} [/imath], but I'm since [imath] q [/imath] can be represented by any non-singular complex matrix, I don't know how this is even possible. (edit: V is of finite dimension) |
2651565 | Show that cardinality of [imath]\mathbb{Z}^n/T(\mathbb{Z}^n)[/imath] equals the absolute value of the determinant of the matrix representing the homomorphism T
I am self teaching module theory and in the section on modules over PID, I have ran into the following problem. Show that the cardinality of M= [imath]\mathbb{Z}^n/T(\mathbb{Z}^n)[/imath] equals the absolute value of the determinant of the matrix representing the homomorphism T. Note: Assuming determinant of the matrix is not 0. I am thinking let [imath]e_1,...e_n[/imath] be the standard basis for [imath]Z^n[/imath]. Then [imath]T(e_1)= c_11 e_1 + ... c_1n e_n[/imath], etc.. so if I find the invariant factors of M by applying column and row operations to the matrix that represents this transformation until its a diagonal matrix, then I will have 1's and invariant factors on the diagonal, and hence M will be isomorphic to [imath]Z^n/(a_1) \bigoplus ... \bigoplus Z^n/(a_n)[/imath] where the [imath]a_i[/imath] are the invariant factors and hence the cardinality will just be the product of the invariant factors, which I guess should be the absolute value of the determinant representing the matrix? I am not sure if this approach would work or not. Could someone help me? | 317065 | Question on determinants of matrices changing between integer matrices
The following problem came up from a though I had while reading: Let's say we have [imath]M=\mathbb{Z}^n[/imath] and we have another free [imath]\mathbb{Z}[/imath]-module, [imath]N[/imath], inside of [imath]M[/imath] also with rank [imath]n[/imath]. We know we can make a matrix [imath]A[/imath] that changes [imath]N[/imath] to its representation in [imath]M[/imath] (ie has as columns the basis [imath]N[/imath] expressed in coordinates coming from the basis of [imath]M[/imath]). If the index of [imath]N[/imath] in [imath]M[/imath] is [imath]m[/imath], I am pretty sure just from tinkering that the determinant of [imath]A[/imath] should also be [imath]m[/imath], however, I have not been able to show this. It is unclear to me how to proceed. I though maybe I could do something related to the points in [imath]\mathbb{Z}^n[/imath] that are within the unit box of [imath]N[/imath], but I cannot really make sense of this. Thank you for any help or direction. |
2652010 | Prove that for all rational number [imath]a[/imath] and [imath]b[/imath], [imath]\frac{a+b}{2}[/imath] ≥ [imath]\sqrt{ab}[/imath]
Prove that for all rational numbers (can be integer) a y b, [imath]\frac{a+b}{2}[/imath] ≥ [imath]\sqrt{ab}[/imath] | 1444157 | Let a and b be real numbers such that 0 < a < b. Prove [imath]\frac{a+b}2 > \sqrt{ab} > \frac{2ab}{a + b}[/imath]
Let a and b be real numbers such that 0 < a < b. Prove [imath]\frac{a+b}2 > \sqrt{ab} > \frac{2ab}{a + b}[/imath] How can I prove this? Been working for hours and got nowhere. I see [imath]\frac{a+b}{2}[/imath] and [imath]\frac{2ab}{a + b}[/imath] are almost reciprocals. Is this important? Please explain step by step. |
2648754 | Does the scheme-theoretic definition of a manifold include Hausdorffness and second-countability?
Wikipedia has that: Scheme-theoretically, a manifold is a locally ringed space, whose structure sheaf is locally isomorphic to the sheaf of continuous (or differentiable, or complex-analytic, etc.) functions on Euclidean space. I wonder if the locality assumption on the structure sheaf (every stalk is a local ring) implies Hausdorffness. If we let the line with two origins have the structure sheaf inherited from the quotient, won't the stalks at each of the origins be [imath]\Bbb R(y)[x]_{(x)}[/imath] and [imath]\Bbb R(x)[y]_{(x)}[/imath], which are local rings? What about second countability? | 107319 | Sanity check about Wikipedia definition of differentiable manifold as a locally ringed space
Most textbooks introduce differentiable manifolds via atlases and charts. This has the advantage of being concrete, but the disadvantage that the local coordinates are usually completely irrelevant- the choice of atlas and chart is arbitrary, and rarely if ever seems to play any role in differential geometry/topology. There is a much better definition of differentiable manifolds, which I don't know a good textbook reference for, via sheaves of local rings. This definition does not involve any strange arbitrary choices, and is coordinate free. Paragraph 3 in Wikipedia (which is the actual definition) states: A differentiable manifold (of class [imath]C_k[/imath]) consists of a pair [imath](M, \mathcal{O}_M)[/imath] where [imath]M[/imath] is a topological space, and [imath]\mathcal{O}_M[/imath] is a sheaf of local [imath]R[/imath]-algebras defined on [imath]M[/imath], such that the locally ringed space [imath](M,\mathcal{O}_M)[/imath] is locally isomorphic to [imath](\mathbb{R}^n, \mathcal{O})[/imath]. This confuses me, because I don't see why such a sheaf should be acyclic, or where conditions like "paracompact" or "complete metric space" or "second countable Hausdorff" are implicit. So either: The wikipedia entry has a mistake (I would want to sanity-check this before editing the entry, because this is such a fundamental definition which thousands must have read). Somewhere in that definition, the condition that [imath]M[/imath] be paracompact is implicit. Question: Should the definition above indeed require that [imath]M[/imath] be second-countable Hausdorff or paracompact or whatever? Or is it somehow implicit somewhere, and if so, where? Also, is this definition given carefully in any textbook? Update: I have editted the Wikipedia article to require that [imath]M[/imath] be second-countable Hausdorff. But I'm still wondering if there is a textbook covering this stuff, and whether requiring the sheaf to be acyclic might have worked instead as an alternative. |
2650976 | Restriction of Cartesian product measurable?
Let [imath](X,\mathcal{X})[/imath] and [imath](Y,\mathcal{Y})[/imath] be measurable spaces. We consider the product of the two spaces, denoted [imath](X \times Y,\mathcal{X} \overline{\times} \mathcal{Y})[/imath], where [imath]\mathcal{X} \overline{\times} \mathcal{Y}[/imath] is the smallest [imath]\sigma[/imath]-algebra generated by [imath]\{ A \times B \mid A \in \mathcal{X},B \in \mathcal{Y} \}[/imath]. For a measurable set [imath]S \in \mathcal{X} \overline{\times} \mathcal{Y}[/imath] and an element [imath]x \in X[/imath], wikipedia states that the following set is measurable: [imath] S_x := \{ y \in Y \mid (x,y) \in S \} [/imath] How can I prove this? | 1153161 | Every section of a measurable set is measurable? (in the product sigma-algebra)
I'm studying measure theory, and I came across a theorem that says that, given two [imath]\sigma[/imath]-finite measure spaces [imath]\left(X,\mathcal M,\mu\right)[/imath] and [imath]\left(Y,\mathcal N,\nu\right)[/imath], and a set [imath]E\in \mathcal M\otimes\mathcal N[/imath], the function [imath]x\mapsto \nu\left(E_x\right)[/imath] is measurable. The definition of the product-[imath]\sigma[/imath]-algebra [imath]\mathcal M\otimes\mathcal N[/imath] is the [imath]\sigma[/imath]-algebra generated by the rectangles [imath]A\times B[/imath] with [imath]A\in\mathcal M[/imath] and [imath]B\in\mathcal N[/imath]. Also, the [imath]x[/imath]-section [imath]E_x[/imath] is defined as [imath]E_x=\left\{y\in Y:\left(x,y\right)\in E\right\}[/imath]. My problem is that I can't even convince myself that, given [imath]E\in\mathcal M\otimes \mathcal N[/imath], the sections [imath]E_x[/imath] are in [imath]\mathcal N[/imath] for all [imath]x[/imath]. It looks like it has something to do with unitary sets being measurable (which shouldn't be necessary, I think), but I can't prove it even assuming that [imath]\left\{x\right\}\in\mathcal M[/imath] for all [imath]x[/imath]. |
2651370 | Prove that there exist discontinuous functions [imath]f: \mathbb{R} \to \mathbb{R}[/imath] such that [imath]f(x+y) = f(x) + f(y)[/imath].
The hint given is that we can assume the set of real numbers has a basis. I've seen several other questions of this form, but all of them use terminology which I am unfamiliar with ([imath]Q[/imath]-linear, extend linearly to [imath]\mathbb{R}[/imath] etc). Edit : I know what a Hamel basis is, and I don't have to use the axiom of choice to show that it exists or anything. I'm essentially told that I can assume it exists. | 2377638 | Additive function [imath]T: \mathbb{R} \rightarrow \mathbb{R}[/imath] that is not linear.
A function [imath]T:V \rightarrow W[/imath] is additive if [imath]T(x+y) = T(x) + T(y)[/imath] for every [imath]x, y \in V[/imath]. Prove that there exists an additive function [imath]T: \mathbb{R} \rightarrow \mathbb{R}[/imath] that is not linear. My attempt: Let [imath]T[/imath] be the function [imath]T: \mathbb{R}[/imath] (over the field [imath]\mathbb{Q}[/imath]) [imath]\rightarrow \mathbb{R}[/imath] (over the field [imath]\mathbb{R}[/imath]). The set [imath]\{1, \sqrt{2}\} \subseteq \mathbb{R}[/imath] is linearly independent for the vector space [imath]\mathbb{R}[/imath] over [imath]\mathbb{Q}[/imath]. Then, there must exist a linearly independent set [imath]W \subseteq \mathbb{R}[/imath] (over [imath]\mathbb{Q}[/imath]) such that [imath]\{1, \sqrt{2}\} \subseteq W \subseteq \text{span}(W) = \mathbb{R}[/imath] (over [imath]\mathbb{Q}[/imath]). I have been told that the function defined as [imath]T(1) = 1[/imath] and [imath]T(w) = 0[/imath] for all [imath]w \in \text{span}(W) \setminus \{1\}[/imath] is additive but not linear, but I cannot see why this is? I can see why it is not linear, clearly [imath]T(\sqrt{2} \cdot 1) = 0[/imath] but [imath]\sqrt{2} T(1) = \sqrt{2}[/imath]. But, why is [imath]T[/imath] additive? For example, [imath]T(1+1) = T(2) =0[/imath] but [imath]T(1) + T(1) = 1+1 = 2[/imath]? Is there a mistake somewhere? |
2652262 | Calculate the limit of [imath]a_n = n(2^{1/n}-1)[/imath]
On an exam we were asked to prove the sequence converges and find the value of: [imath]\lim_{n\to \infty}a_n = n(2^{1/n}-1)[/imath] On the test I tried every common convergence test we learned, to no avail. Monotone and bounded, pinch theorem, n-th root, ratio test, and so on. Does anyone have any idea how to solve this? Thanks | 151558 | Proving [imath]\lim_{n\rightarrow\infty} n(2^{1/n} - 1) = \log 2[/imath]
How do you prove [imath]\lim_{n\rightarrow\infty} n(2^{1/n} - 1) = \log 2[/imath] ? Background: in computer science, if you allocate CPU time to [imath]n[/imath] processes by rate-monotonic scheduling, all the processes get sufficient amount of time when the quantity [imath]U[/imath] called CPU utilization is less than or equal to [imath]n(2^{1/n} - 1)[/imath]. It is monotonously decreasing and tends to [imath]\log 2[/imath] when [imath]n\rightarrow\infty[/imath], so if [imath]U \le \log 2[/imath], you can be sure all the processes will be given sufficient CPU time. |
2650518 | Exact form of distance preserving map [imath]T:\mathbb R^2 \to \mathbb R^2.[/imath]
Let [imath]T:\mathbb R^2 \to \mathbb R^2[/imath] be distance preserving, i.e., [imath]|T(a)-T(b)|=|a-b|[/imath] for all [imath]a,b\in \mathbb R^2.[/imath] From here can we say the exact form of [imath]T[/imath]? If [imath]T(x)=Ax+b,[/imath] where [imath]A^TA=I[/imath] and [imath]x,b\in \mathbb R^2[/imath] and [imath]b[/imath] is fixed, then [imath]T[/imath] is distance preserving. But my question is that 'Can we say that the above mentioned mappings are only distance preserving map?' | 866445 | Rigid motion on [imath]\mathbb{R}^2[/imath] which fixes the origin is linear
Let [imath]V=\mathbb{R}^2[/imath] be an inner product space with the standard inner product, and let [imath]T[/imath] be a rigid motion of [imath]V[/imath]. Suppose [imath]T(0)=0[/imath], prove that [imath]T[/imath] is linear. (A rigid motion of an inner product space is a function (not necessarily linear) [imath]f:V \to V[/imath], such that [imath]||f\alpha-f\beta||=||\alpha-\beta||[/imath].) It is obvious that [imath]T[/imath] preserves the inner product (T is unitary), but I don't know how to prove that [imath]T[/imath] is linear. Thanks in advance |
2652927 | Why isn't minimal sigma-algebra simply the set containing all countable unions and complements
For [imath]X[/imath] a set and [imath]A\subset X[/imath] the sigma-algebra generated by [imath]A[/imath] is [imath]\sigma(A)=\bigcap\{\Sigma: \text{$\Sigma$ $\sigma$-algebra s.t. $A\subset\Sigma$}\}.[/imath] However why can' t we simply say [imath]\sigma(A)=\{\text{all countable unions and complements of elements of $A$}\}\cup \{\emptyset,X\}?[/imath] What would be a good counterexample? | 211233 | Generate the smallest [imath]\sigma[/imath]-algebra containing a given family of sets
My teacher gave me an example of performing the subject: Example Let [imath]\Omega = \Bbb R[/imath] and [imath]\mathcal R = \{(-\infty,-1),(1,+\infty)\}[/imath]. Then [imath]\sigma(\mathcal R) = \{\emptyset, \Bbb R, (-\infty,-1), (1,+\infty), [-1, \infty), (-\infty,1], (-\infty,-1)\cup(1,+\infty),[-1,1]\}[/imath]. There was a different example, where she also generated the smallest [imath]\sigma[/imath]-algebra for the family of sets [imath]\mathcal A = \{A,B\} \subset 2^\Omega[/imath] in the same way: [imath]\sigma(\mathcal R) = \{\emptyset, \Omega, A, B, A^C,B^C,A\cup B, (A\cup B)^C\}[/imath]. I certainly understand why [imath]\emptyset[/imath], [imath]\Omega [/imath] and a family of sets itself are there, and I certainly know that [imath]\sigma[/imath]-algebra is closed under the operations of complement and union. What I don't understand about the other elements of generated [imath]\sigma[/imath]-algebras: why we are taking exactly them? Does this work in a general case? |
2654664 | For a non-zero principal ideal [imath]I=(x)[/imath] of a ring of integers of an algebraic number field, [imath]|A/I|=| N_{L|\mathbb Q } (x)|[/imath]
Let [imath]L[/imath] be an extension field of [imath]\mathbb Q[/imath] with [imath][L:\mathbb Q]=n[/imath]. Let [imath]A[/imath] be the ring of integers of [imath]L[/imath] i.e. the Integral closure of [imath]\mathbb Z[/imath] in [imath]L[/imath]. Then [imath]A[/imath] is a Dedekind domain. If [imath]I[/imath] is a non-zero ideal of [imath]A[/imath], then I can show that [imath]I[/imath] is a free [imath]\mathbb Z[/imath]-module of rank [imath]n[/imath] and [imath]aA \subseteq I[/imath] for some [imath]0 \ne a \mathbb Z[/imath] . Moreover [imath]A/I[/imath] is finite. My question is : If [imath]I=(x)[/imath] is a non-zero principal ideal, then how to show that [imath]|A/I|=| N_{L|\mathbb Q } (x)|[/imath] ? where [imath]N_{L|\mathbb Q}[/imath] denotes the field Norm function . | 1735863 | Norm of element [imath]\alpha[/imath] equal to absolute norm of principal ideal [imath](\alpha)[/imath]
Let [imath]K[/imath] be a number field, [imath]A[/imath] its ring of integers, [imath]N_{K / \mathbf{Q}}[/imath] the usual field norm, and [imath]N[/imath] the absolute norm of the ideals in [imath]A[/imath]. In some textbooks on algebraic number theory I have seen the fact: [imath]\vert N_{K / \mathbf{Q}}(\alpha) \vert = N(\alpha A)[/imath] for any [imath]\alpha \in A[/imath]. However, I wasn't able to find a proof (neither in books nor by myself). Can someone explain to me, why this is true? Thanks! |
2233442 | Find all primes [imath]p[/imath] such that [imath]x^3+x+1\equiv0\pmod p[/imath] has [imath]3[/imath] incongruent solutions.
Find all primes [imath]p[/imath] such that [imath]x^3+x+1\equiv0\pmod p[/imath] has [imath]3[/imath] incongruent solutions. By a standard result in number theory, it has [imath]3[/imath] incongruent solutions iff there exists [imath]q(x),r(x)\in\mathbb{Z}[x][/imath] with [imath]\deg r(x)<3[/imath] such that [imath]x^p-x=(x^3+x+1)\,q(x)+p\cdot r(x).[/imath] I am unable to proceed from here. I tried to write out the coefficients of [imath]q(x)[/imath] and multiply out, but it was in vain. How do I solve this problem? Thanks in advance! I am unclear about what tags to add, so please fix this. | 378331 | Solve [imath]x^3+x \equiv 1 \pmod p[/imath]
Find all the primes [imath]p[/imath] so that [imath]x^3+x \equiv 1 \pmod p \tag{1}[/imath] has integer solutions. We consider [imath]x[/imath] and [imath]y[/imath] are the same solutions iff [imath]x\equiv y \pmod p.[/imath] We can prove that [imath](1)[/imath] cannot have exactly [imath]2[/imath] solutions unless [imath]p=31[/imath]. When [imath]p=31[/imath], the solutions of [imath](1)[/imath] are [imath]x \equiv {17\text{ (double), } 28} \pmod {31}.[/imath] My question is: when does [imath](1)[/imath] have no solution, and when does [imath](1)[/imath] have [imath]1[/imath] solution and when does [imath](1)[/imath] have [imath]3[/imath] solutions? (As Ma Ming has pointed out, an equation of degree [imath]3[/imath] has no more than [imath]3[/imath] solutions.) I have proved that if [imath]p \not = 31[/imath], and [imath]a[/imath] is a solution of [imath](1)[/imath], then [imath](1)[/imath] has [imath]3[/imath] solutions iff [imath]\left(\frac{a-1}{p} \right)=\left(\frac{a+3}{p} \right),[/imath] here [imath]\left( \frac{a}{p} \right)[/imath] is the Jacobi symbol. Thanks in advance! |
2655046 | if [imath]p[/imath] and [imath]q[/imath] are successive odd primes and [imath]p + q = 2r[/imath] then [imath]r[/imath] is composite,
As [imath]p[/imath] and [imath]q[/imath] are successive odd primes, for example if [imath]p = 3[/imath], [imath]q = 5[/imath] then [imath]p + q = 8 = 2 \times 4[/imath] here 4 is a composite number. But how to prove it generally in all conditions? | 1103458 | Odd Primes Problem Proof
Given the odd prime numbers, Prove that if [imath]x[/imath] and [imath]y[/imath] are adjacent odd primes in this list, then [imath]x + y[/imath] has [imath]3[/imath] prime factors. The factors need not be distinct. Here is an example I have provided: [imath]3 + 5 = 8 = 2 \cdot 2 \cdot 2[/imath]. Therefore, [imath]8[/imath] has [imath]3[/imath] repeated factors of [imath]2[/imath]. |
2654812 | Let [imath]U =\{ p\in \mathbb{P_4}(\mathbb{R}): p(6) = 0 \}[/imath]. Find a Basis of [imath]U[/imath].
LADR 3rd ed. S.Axler|Section 2.C|problem 4. page.48 Let [imath]U = \{ p\in \mathbb{P_4}(\mathbb{R}): p(6) = 0 \}[/imath]. I take the general form of a polynomial of degree 4 : [imath]p(x)=e+dx^1+cx^2+bx^3+ax^4[/imath] and then I tried to narrow the form down to one that represents polynomials of degree 4 with the property that p(6)=Zero [imath]0=e+d6^1+c6^2+b6^3+a6^4[/imath] [imath]e=-d6^1-c6^2-b6^3-a6^4[/imath] [imath]p(x)=e+dx^1+cx^2+bx^3+ax^4[/imath] turns into: [imath]p(x)=d6^1-c6^2-b6^3-a6^4+dx^1+cx^2+bx^3+ax^4[/imath] [imath]p(x)=d(x-6)+c(x^2-6^2)+b(x^3-6^3)+a(x^4-6^4)[/imath] According to a worked-out solution I found on the net, [imath]\{(x-6),(x^2-6^2),(x^3-6^3),(x^4-6^4)\}[/imath] is indeed supposed to be a basis for the subspace described above. What I fail to understand is how [imath]\{(x-6),(x^2-6^2),(x^3-6^3),(x^4-6^4)\}[/imath] can be linearly independent and thus a basis for the subspace(assuming the answer is correct). [imath]0=p(\textbf{6})=d(\textbf{6}-6)+c(\textbf{6}^2-6^2)+b(\textbf{6}^3-6^3)+a(\textbf{6}^4-6^4)[/imath] If x=6, then p(x) is zero, but the coefficients do not have to be zero. Thank you. | 1621816 | Is my basis for [imath]U=\{p \in \mathcal P_4(\mathbb R) \mid p(6)=0\}[/imath] correct?
[imath]\mathcal P_4(\mathbb R)[/imath] is the set of polynomials with degree at most [imath]4[/imath] with real coefficients. [imath]U=\{p \in \mathcal P_4(\mathbb R) \mid p(6)=0\}[/imath]. Would [imath]\langle(x-6),(x-6)^2, (x-6)^3, (x-6)^4\rangle[/imath] form a basis? Suppose [imath]a(x-6)+b(x-6)^2+c(x-6)^3+d(x-6)^4=0[/imath]. Then we know [imath]a=b=c=d=0[/imath] since the RHS has no [imath]x, x^2, x^3, x^4[/imath] terms. So the set of vectors is linearly independent. Also, since [imath]\dim U < 5[/imath], we cannot extend a basis of [imath]U[/imath] beyond length [imath]4[/imath]. Hence [imath]\dim U=4[/imath]. So, [imath]\langle(x-6),(x-6)^2, (x-6)^3, (x-6)^4\rangle[/imath] is a basis of [imath]U[/imath]. |
2655413 | Prove the inequality [imath]\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}\le\frac{3\sqrt3}{4}[/imath]
Let [imath]a,b,c[/imath] are nonnegative real numbers such that [imath]a^2+b^2+c^2=1[/imath]. Prove the inequality [imath]\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}\le\frac{3\sqrt3}{4}[/imath] I tried the method of Lagrange multipliers and Jensen's inequality but I have not been proved this inequality | 106473 | Prove that [imath]x_1^2+x_2^2+x_3^2=1[/imath] yields [imath] \sum_{i=1}^{3}\frac{x_i}{1+x_i^2} \le \frac{3\sqrt{3}}{4} [/imath]
Prove this inequality, if [imath]x_1^2+x_2^2+x_3^2=1[/imath]: [imath] \sum_{i=1}^{3}\frac{x_i}{1+x_i^2} \le \frac{3\sqrt{3}}{4} [/imath] So far I got to [imath]x_1^4+x_2^4+x_3^4\ge\frac{1}3[/imath] by using QM-AM for [imath](2x_1^2+x_2^2, 2x_2^2+x_3^2, 2x_3^2+x_1^2)[/imath], but to be honest I'm not sure if that's helpful at all. (You might want to "rename" those to [imath]x,y,z[/imath] to make writing easier): [imath] \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2} \le \frac{3\sqrt3}4[/imath] |
2656276 | Prove if [imath]C=C_1+\cdots+C_k[/imath] then [imath]\dim(C)=\dim(C_1)+\cdots+\dim(C_k)[/imath]
Let [imath]\gamma:V\rightarrow V[/imath] linear. with [imath]\lambda_1,\ldots,\lambda_k[/imath] his eigenvalues. and [imath]C_i=\ker(\lambda_i I_d - \gamma)[/imath] for all [imath]1\leq i\leq k[/imath]. Prove if [imath]C=C_1+\cdots+C_k[/imath] then [imath]\dim(C)=\dim(C_1)+\cdots+\dim(C_k)[/imath] Definition: [imath]\gamma:V\rightarrow V[/imath] is diagonalizable if the matrix associated to [imath]\gamma[/imath] is diagonal. I have a idea of how prove this. Basically i need show [imath]\;\sum_{i=1}^k C_i=\bigoplus_{i=1}^k C_i[/imath] I'm trying to generalize the idea with this: Suppose that for [imath]\;1\le i\neq j\le n\;[/imath] , we have [imath]x\in\ker (\lambda_iI-\gamma)\cap\ker(\lambda_jI-\gamma)\implies \gamma x=\lambda_ix=\lambda_jx[/imath] and since [imath]\;\lambda_i\neq\lambda_j\;[/imath] , we get that [imath]\;x=0\;[/imath] . Moreover, as [imath]C=C_1+C_2[/imath] then [imath]C=C_1\oplus C_2[/imath] but i have serious problem trying to generalize that. can someone help me? | 2654863 | Prove about direct sum
Prove if [imath]W=W_1+\cdots+W_k[/imath] then [imath]\dim(W)=\dim(W_1)+\cdots+\dim(W_k)[/imath] I'm trying to prove this but is a little complicated for me. Can someone help me? |
1140521 | Multiplication of adjacency matrix of two graphs
Consider two [imath]N \times N[/imath] adjacency matrices [imath]A[/imath] and [imath]B[/imath] belonging to different graphs(unweighted and undirected). What does their product [imath]AB[/imath] represent? I know that the powers of the adjacency matrix gives the number of paths between any two vertices of length equal to the power. But I am having trouble extendeding this logic to the above problem | 114334 | Product of adjacency matrices
I was wondering if there was any meaningful interpertation of the product of two [imath]n\times n[/imath] adjacency matrices of two distinct graphs. |
2655203 | Given [imath]x, y[/imath] are acute angles such that [imath]\sin y= 3\cos(x+y)\sin x[/imath]. Find the maximum value of [imath]\tan y[/imath].
Given [imath]x, y[/imath] are acute angles such that [imath]\sin y= 3\cos(x+y)\sin x[/imath]. Find the maximum value of [imath]\tan y[/imath]. Attempt at a solution: [imath](\sin y = 3(\cos x \cos y - \sin x \sin y) \sin x) \frac{1}{\cos y}[/imath] [imath]\tan y = (3 \cos x - 3 \sin x \tan y) \sin x[/imath] [imath]\tan y + 3 \sin^2 x + \tan y = 3 \sin x \cos x[/imath] [imath]\tan y = \frac{3 \sin x \cos x}{1 + 3 \sin^2 x}[/imath] I have also tried substituting [imath]0[/imath], [imath]30[/imath], [imath]45[/imath], [imath]60[/imath], [imath]90[/imath] to the values of [imath]x[/imath]. | 2647273 | Trigonometry Olympiad Question
Given [imath]x, y[/imath] are acute angles such that [imath]\sin y= 3\cos(x+y)\sin x[/imath]. Find the maximum value of [imath]\tan y[/imath]. Attempt at a solution: [imath][sin y = 3(cos x cos y - sin x sin y) sin x] 1/cos y[/imath] [imath]tan y = (3 cos x - 3 sin x tan y) sin x[/imath] [imath]tan y + 3 sin^2 x + tan y = 3 sin x cos x[/imath] [imath]tan y = (3 sin x cos x) / (1 + 3 sin x)[/imath] I'm stuck here... |
2649338 | Probabilities regarding [imath]\lim \sup[/imath] and [imath]\lim \inf[/imath] of sequence of independent random variables.
Exercise : Let [imath]X_n, n=1,2,3,\dots[/imath] be a sequence of independent continuous random variables. It's also given that all of [imath]X_n[/imath] have the same probability density function [imath]f(x)[/imath], for which it is [imath]f(x) > 0 \space \forall \space x \in \mathbb R[/imath]. For a given real number [imath]b[/imath], consider the following events : [imath]A=\lim \sup_n \{X_n >b\}, \space B=\lim \inf_n \{X_n>b\}[/imath] [imath]G = \lim \sup_n \{X_n \leq b\}, \space D = \lim \inf_n\{X_n\leq b\}[/imath] Explain with simple words what each of the events above means and calculate their probabilities of happening. Discussion : It seems something that may have to do like Borrel - Cantelli Theorems, but I'm not even close to sure on how to proceed and make conclusions. Any help will be greatly appreciated as this is a past exam problem. | 1625796 | Probability of lim sup, lim inf for sequence of random variables.
Maybe this is extremely simple, but i havent found a specific answer for this online. For a sequence of independent continuous random variables [imath]X_n[/imath] ,[imath]n=1,2,3,...[/imath] , all with the same probability density function [imath]f(x)>0[/imath] for all real [imath]x[/imath], and a specific real number [imath]b[/imath], what do the following events mean in simple terms and what is the probability of each? a)[imath]\lim_n\sup\{X_n \gt b\}[/imath] b)[imath]\lim_n\inf\{X_n \gt b\}[/imath] c)[imath]\lim_n\sup\{X_n \le b\}[/imath] d)[imath]\lim_n\inf\{X_n \le b\}[/imath] |
2656474 | Norm Equals Supremum of Inner Product when ||w||=1?
Let [imath]V[/imath] be an inner product space, and [imath]v\in V[/imath]. I want to show that [imath]||v||=\sup\{\langle\,v,w\rangle:||w||=1\}[/imath] What I know: Since this is a generic inner product, [imath]||v||=\sqrt{\langle\,v,v\rangle}[/imath]. So, if [imath]||v||=0[/imath] then [imath]v=0[/imath]. The equality is trivial if [imath]v=0[/imath], so I know I need to consider when [imath]v\neq 0[/imath]. Also, we have that [imath]||w||=1[/imath] this implies that [imath]w=\frac{v}{||v||}[/imath] if we use it in terms of [imath]v[/imath]. I am not sure if I am pulling the correct information, but I am confused on how to proceed to prove the equality statement. | 1295407 | Norm is sup of inner products (proof).
Let [imath]V[/imath] be a vector space with an inner product [imath]\langle.,. \rangle[/imath] and associated norm [imath]|| . ||[/imath] Then: Could I have a proof of this fact? |
2643604 | Find an Unbiased Estimator of a Function of a Parameter
Suppose that [imath]Y_1,...,Y_n[/imath] is an IID sample from a uniform distribution [imath]U(\theta,1)[/imath] The method of moments estimator is [imath]\hat \theta=2\bar Y-1[/imath]. Find an unbiased estimator, call it [imath]\hat \beta[/imath], of the quantity: [imath]\frac{1-\theta}{\sqrt{3n}}[/imath] I noticed that if I take the expected value of this arbitrarily chosen estimator: [imath]\frac{1-\hat \theta}{\sqrt{3n}}[/imath] I end up with [imath]E\left( \frac{1-\hat \theta}{\sqrt{3n}} \right)=E\left( \frac{1-(2\bar Y-1)}{\sqrt{3n}} \right) =\frac{1-\theta}{\sqrt{3n}}[/imath] So I don't quite understand if the goal here is for the expected value of my estimator to equal EXACTLY [imath]\theta[/imath] or if it's supposed to equal the original function of [imath]\theta[/imath]. So this abitrarily chosen estimator of mine; is an an unbiased estimate of the given quantity; is it the [imath]\hat \beta[/imath] I seek? | 2642733 | Unbiased Estimator of the Standard Error of another Estimator
Suppose that [imath]Y_1,...,Y_n[/imath] is an IID sample from a uniform [imath]U(\theta, 1)[/imath] distribution. The method of moments estimator for [imath]\theta[/imath] is [imath]\tilde \theta=2\bar Y-1[/imath]. The standard error of [imath]\tilde \theta[/imath] is [imath]\sigma_{\tilde \theta}=\frac{1-\theta}{\sqrt{3n}}[/imath] Find an unbiased estimator of [imath]\sigma_{\tilde \theta}[/imath] and show that it is unbiased. So normally I know how to find the estimator of a parameter of a distribution, but I don't quite understand how I'm supposed to find an estimator of the standard deviation of another estimator of a parameter of a distribution... so I really don't know how to go about this. What do I do? |
2657579 | Calculating [imath]\lim_{x\to 0} \left[\frac{1}{\sin^2x}-\frac{1}{x^2}\right][/imath]
How can I calculate this limit? [imath]\lim_{x\to 0} \left[\frac{1}{\sin^2x}-\frac{1}{x^2}\right] = \,?[/imath] I don't have any idea how to do it. | 629257 | Evaluate [imath]\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {{{\sin }^2}x}} - {1 \over {{x^2}}}} \right)[/imath]
I tried l'Hospital but that will require a lot (and I mean A LOT!!!) of differentiating Is there a shortcut? [imath]\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {{{\sin }^2}x}} - {1 \over {{x^2}}}} \right)[/imath] Thanks in advance |
2657765 | Exercise IV.17 in Kunen (1980): Absolute in all transitive models of a finite set of ZF axioms implies [imath]\Delta_1[/imath]
Exercise 17, chapter IV in Kunen's 1980 "Set Theory" reads: "Show that for any formula [imath]\phi[/imath], the following are equivalent: [imath]\phi[/imath] is [imath]\Delta_1^{ZF}[/imath]. There is a finite set [imath]S[/imath] of axioms in [imath]ZF[/imath] such that: [imath]ZF \vdash \forall M\left(M \text{ transitive } \wedge \left(\bigwedge_{\sigma \in S}\sigma^M \right) \to \phi \text{ is absolute for } M \right) [/imath] Hint: For [imath](2)\implies (1)[/imath], use reflection." I can prove [imath](1)\implies (2)[/imath] by proving [imath]\phi[/imath] is [imath]\Delta_1^M[/imath] and hence both downward and upward absolute. But I cannot figure out the other direction. I can use Reflection Theorem to prove there are transitive [imath]M[/imath] satisfying [imath]\bigwedge_{\sigma \in M}\sigma^M[/imath] but that doesn't seem to help. One of the bottleneck is I have no idea how to prove a formula [imath]\phi[/imath] is [imath]\Delta_1^{ZF}[/imath] without knowing exactly what [imath]\phi[/imath] is. Any suggestion is appreciated. | 2648707 | Show that [imath]\textbf{ZF}[/imath] shows [imath]\phi[/imath] is [imath]\Delta_1[/imath]
Fix a [imath]\phi[/imath] Suppose for some finite [imath]S :=\ [/imath]{[imath]\psi_1,....\psi_n[/imath]},axioms of [imath]\textbf{ZF},[/imath] where $\textbf{ZF}\vdash\ \forall M[imath][(M$ is transitive and $(\psi_1\wedge .....\wedge\psi_n)^{M}[/imath])\rightarrow \phi$ is absolute for [imath]M[/imath] [imath]][/imath] [imath](\psi_1\wedge .....\wedge\psi_n)^{M}[/imath] means the conjunction relativise to [imath]M[/imath] Show that [imath]\textbf{ZF}[/imath] shows [imath]\phi[/imath] is [imath]\Delta_1[/imath] Attempt: The idea is to use reflection theorem follow by mostowski collapse. We first reflect [imath]S[/imath] onto some [imath]R(\alpha)[/imath] and apply mostowski collapse to make it transitive,call it [imath]M[/imath], thus by assumption we conclude that [imath]\phi[/imath] is absolute for [imath]M[/imath]. But I'm stuck from here on. Can we conclude absoluteness for [imath]M[/imath] implies [imath]\phi[/imath] must either be [imath]\Delta_0[/imath] or [imath]\Delta_1[/imath] ? Any help or insight is appreciated. Cheers |
2656283 | Determine eigenvalue pairs.
Assume that [imath]A[/imath] is an [imath]n\times n[/imath] matrix. Assume that [imath]u[/imath] and [imath]v[/imath] are eigenvectors corresponding to eigenvalues [imath]5[/imath] and [imath]9[/imath]. Determine all pairs of real numbers [imath]α[/imath] and [imath]β[/imath] such that [imath]αu+βv[/imath] is also an eigenvector of [imath]A[/imath]. Could someone please advise if this works! Since u and v are eigenvectors corresponding to eigenvalues 5 and 9, We have Au = 5u …(1) Av = 9v …(2) Now we want to find all pairs of α and β lets call them α_i and β_i such that (α_i u+β_i v) is eigen vector of A i.e. [imath] A(α_i u+β_i v)=e_i (α_i u+β_i v) < => Aα_{i} u+Aβ_{i} v=e_{i} (α_{i} u+β_{i} v) < => α_{i} Au+β_{i} Av=e_{i} (α_{i} u+β_{i} v) < => α_{i} 5u+β_{i} 9v=e_{i} (α_{i} u+β_{i} v) < => e_{i}=((α_{i} 5u+β_{i} 9v))/((α_{i} u+β_{i} v) ) …(3)[/imath] e_{i} will be=k when [imath] α_{i}[/imath] happens to be the multiple of 9 i.e.[imath]α_{i}=9*k, β_{i} [/imath] happens to be multiple of 5 i.e.5*k except when k = 0 Thus for[imath] (α_{i} u+β_{i} v) [/imath] to be eigenvector of A [imath] (α_{i},β_{i} )=[(9*k,5*k)] ∀ k = ±1,±2,±3[/imath]…and k!=0 So my question is different from the possible duplicate highlighted below - as it has a specific linear combination to deal with | 2256942 | If the sum of eigenvectors is an eigenvector, then they all correspond to the same eigenvalue
I believe I am very close to finishing this proof, but I cannot figure out the last part. If anybody could check my work and maybe give me a little hint, it would be greatly appreciated! Let [imath]V[/imath] be a finite-dimensional vector space and [imath]T \in \mathcal{L}(V)[/imath], and let [imath]\mathbf{u,v} \in V[/imath] be eigenvectors of [imath]T[/imath]. Claim: If [imath]\mathbf{u} + \mathbf{v}[/imath] is an eigenvector of [imath]T[/imath], then [imath]\mathbf{u}, \mathbf{v}[/imath], and [imath]\mathbf{u+v}[/imath] all correspond to the same eigenvalue. Proof (So far!): Suppose [imath]T(\mathbf{u}) = \lambda_1 \mathbf{u}[/imath] and [imath]T(\mathbf{v}) = \lambda_2 \mathbf{v}[/imath] with [imath]\mathbf{u}, \mathbf{v} \neq \mathbf{0}[/imath]. Now suppose [imath]T(\mathbf{u+v}) = \lambda_3(\mathbf{u+v})[/imath] with [imath]\mathbf{u+v}\neq \mathbf{0}[/imath]. Then, [imath] T(\mathbf{u}) + T(\mathbf{v}) = \lambda_3 \mathbf{u} + \lambda_3 \mathbf{v}\\ \lambda_1\mathbf{u} + \lambda_2\mathbf{v} = \lambda_3 \mathbf{u} + \lambda_3\mathbf{v}\\ \lambda_1\mathbf{u} + \lambda_2\mathbf{v} - \lambda_3 \mathbf{u} -\lambda_3\mathbf{v} = \mathbf{0}\\ (\lambda_1 - \lambda_3) \mathbf{u} + (\lambda_2 - \lambda_3)\mathbf{v} = \mathbf{0} [/imath] Now I know in order to show that [imath]\lambda_1 = \lambda_2 = \lambda_3[/imath], I must show that the only solution to the last line is the trivial one. This would imply that [imath]\mathbf{u}[/imath] and [imath]\mathbf{v}[/imath] are linearly independent which I am unconvinced of! The only information I have to my advantage I haven't used yet is the fact that [imath]\mathbf{u}, \mathbf{v}, \mathbf{u+v} \neq \mathbf{0}[/imath]. I really cannot see how this information can help me though. Perhaps I am going about this wrong, but that's why I want to ask! Thanks for your help. |
2640503 | If [imath]P \in M_7(\mathbb{R})[/imath] has rank 4, rank of [imath]P + aa^T[/imath], where [imath]a[/imath] is a column vector?
Let [imath]P[/imath] be a [imath]7 \times 7[/imath] matrix of rank [imath]4[/imath] with real entries and let [imath]a \in \mathbb{R}^7[/imath] be a column vector. Then the rank of [imath]P + aa^T[/imath] is at least ______. This question was in the 2017 IIT JAM paper. Any ideas? | 2145647 | Least rank of a matrix
Let [imath]A[/imath] be a [imath]7\times 7[/imath] matrix such that it has rank [imath]3[/imath] and [imath]a[/imath] be [imath]7\times 1[/imath] column vector. Then least possible rank of [imath]A+(a a^T)[/imath] is? ([imath]a^T[/imath] is transpose of the vector) Intuitively I think it's [imath]2[/imath] as [imath]a a^T[/imath] has rank [imath]1[/imath], and the most it could affect is one row of [imath]A[/imath]. |
2657136 | Given 3 points, how can I find a quadratic equation that intersects all of these points?
Given 3 points, [imath](x_1, y_1), (x_2, y_2), (x_3, y_3)[/imath], how might I find an equation intersecting all of these points? Given just 2 points, to find a linear equation, this is the formula: [imath]y\ =\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}x+\frac{\left(y_1+y_2-\left(x_1\left(\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\right)+x_2\left(\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}\right)\right)\right)}{2}[/imath] In the same form of [imath]y=...[/imath], what is the formula for a quadratic equation with 3 points? | 88881 | Getting a standard form quadratic from a set of points ([imath]3[/imath] points)
I came home from school today, pulled out my homework, now I'm stumped. I don't want the answer, I just want to know how to do it. Here is the question that I'm reading: Determine a quadratic function in standard form for the set of points [imath](x_1,y_2)(x_2,y_2)(x_3,y_3)[/imath] Where the [imath]x[/imath]'s and [imath]y[/imath]'s represent the points. Here are the points I am working with (remember, I don't want the answer, I just want to know how to do it, there are lots of questions on this so I want to be able to know how to do it). [imath](-1, 2), (0, 1), (1, -4)[/imath] Thanks! |
2658569 | Help with Combinatorial argument Proof
Show by a Combinatorics argument [imath]{2n\choose {2}} = 2{n\choose 2} + n^2[/imath] I guess I'm having trouble with the "Combinatorial argument" part. Can anyone help out? | 1456022 | How can you prove [imath]\binom {2n}{2} = 2\binom{n}{2} + n^2[/imath]?
Is there a way one can prove that this is true: [imath]{2n \choose 2} = 2{n\choose 2} + n^2[/imath] ? I am thinking it may involve binomial theorem. |
2658614 | Prove [imath]\sum_{i=0}^{2n}{(-1)^i \binom{2n}{i}^3} = (-1)^n \frac{(3n)!}{(n!)^3}[/imath]
I've tried to look at it combinatorically but I couldn't find a good model for [imath]\binom{2n}{k}^3[/imath], I've also tried to solve it using induction but the changes in steps are too much to keep track of Can any one give a hint on any kind of proof for this? | 497470 | Proving [imath]\sum_{k=0}^{2m}(-1)^k{\binom{2m}{k}}^3=(-1)^m\binom{2m}{m}\binom{3m}{m}[/imath] (Dixon's identity)
I found the following formula in a book without any proof: [imath]\sum_{k=0}^{2m}(-1)^k{\binom{2m}{k}}^3=(-1)^m\binom{2m}{m}\binom{3m}{m}.[/imath] I don't know how to prove this at all. Could you show me how to prove this? Or If you have any helpful information, please teach me. I need your help. Update : I crossposted to MO. |
2658479 | How can I prove this inequality with powers of radicals?
The inequality goes like this: [imath]\sqrt7^{\sqrt5}>\sqrt5^{\sqrt7}[/imath] I have to prove this without using approximate numbers and I just cannot find out how. | 1705690 | Prove that [imath]x^y < y^x[/imath]
Assuming that [imath]e<y<x[/imath], prove that [imath] x^y < y^x[/imath]. I think this must be easy, but I can't work it out. Thanks in advance for any kind of help. |
2659165 | How many ways are there to distribute [imath]7[/imath] (identical) apples, [imath]6[/imath] oranges and [imath]7[/imath] pears among [imath]3[/imath] without restrictions?
How many ways are there to distribute [imath]7[/imath] (identical) apples, [imath]6[/imath] oranges and [imath]7[/imath] pears among [imath]3[/imath] without restrictions? Here is my solution below, is this the correct method? If not how can I fix it? thanks! Using stars and bars on each fruit, Apples, [imath]\dbinom{7+3-1}{3-1} = \dbinom{9}{2} = 36[/imath] Oranges, [imath]\dbinom{6+3-1}{3-1} = \dbinom{8}{2} = 28 [/imath] Pears, [imath]\dbinom{7+3-1}{3-1} = \dbinom{9}{2} = 36[/imath] Then adding these cases together would give the total ways. | 2659101 | ways are there to distribute [imath]7[/imath] (identical) apples, [imath]6[/imath] oranges and [imath]7[/imath] pears among [imath]3[/imath] different people with each person getting at least [imath]1[/imath] pear?
How many ways are there to distribute [imath]7[/imath] (identical) apples, [imath]6[/imath] oranges and [imath]7[/imath] pears among [imath]3[/imath] different people with each person getting at least [imath]1[/imath] pear? Below are my workings but I am not sure if they are correct, looking for some help, thanks! Using the inclusion/exclusion principle: Include the number of ways to distribute them such that at most [imath]\color\red3[/imath] people have pears: [imath]3^7\cdot\binom{3}{\color\red3}\cdot\color\red3^6=1594323[/imath] Exclude the number of ways to distribute them such that at most [imath]\color\red2[/imath] people have pears: [imath]3^7\cdot\binom{3}{\color\red2}\cdot\color\red2^6=419904[/imath] Include the number of ways to distribute them such that at most [imath]\color\red1[/imath] person has pears: [imath]3^7\cdot\binom{3}{\color\red1}\cdot\color\red1^6=6561[/imath] Hence the number of ways to distribute them such that each person has pears is: [imath]1594323-419904+6561=1180980[/imath] |
2658448 | Number of solution of exponential equation modulo p (prime)
I have to find the number of solution of this congruential equation [imath]x^k\equiv 1 (mod\space p)[/imath]. Surely I know that [imath]x^{p-1}\equiv 1 (mod\space p)[/imath] so, [imath]x^d \equiv 1[/imath] where [imath]d[/imath] is the MCD between [imath](p-1,k)[/imath]. Now I have to count how many divisor of [imath]d[/imath] there are, but I don't know how. | 2549305 | Number of solutions to [imath]x^n \equiv 1 \mod p[/imath]
I wish to find the number of solutions to the congruence [imath]x^n \equiv 1 \mod p[/imath] where [imath]n[/imath] is any natural number, [imath]p[/imath] prime. I tried using primitive roots and Fermat but I don't know where to go from here. |
2658980 | Prove that [imath]\mathbb{Q}(\sqrt[3]{p}, \sqrt{q}) = \mathbb{Q}(\sqrt[3]{p}\cdot\sqrt{q})[/imath]
Let [imath]p, q[/imath] be distinct primes and [imath]L := \mathbb{Q}(\sqrt[3]{p}, \sqrt{q})[/imath]. I want to show that [imath]L = \mathbb{Q}(\sqrt[3]{p}\cdot\sqrt{q})[/imath]. Of course one inclusion is clear, but how do I get the other one? | 638561 | [imath]\mathbb{Q}(\sqrt{p},\sqrt[3]{q})= \mathbb{Q}(\sqrt{p}\cdot \sqrt[3]{q})[/imath] ??
Let [imath]p,q[/imath] be primes, [imath]p≠q[/imath], then I have to show that [imath]\mathbb{Q}(\sqrt{p},\sqrt[3]{q})= \mathbb{Q}(\sqrt{p}\cdot \sqrt[3]{q})[/imath] So far I've tried a lot of things with minimal polynomials and bases, but I'm far from any argument that I find sufficient. This is an example from my algebra course, first chapter about field extension. So far we know about basic definitions, minimal polynomials and their properties, algebraic elements and algebraic fields. Could someone maybe give me a hint about how to proceed? Thanks a lot!!! |
2658570 | Solving minimization problem [imath]L_2[/imath] IRLS
In the article ''' Chartrand, Rick, and Wotao Yin. "Iteratively reweighted algorithms for compressive sensing." Acoustics, speech and signal processing, 2008. ICASSP 2008. IEEE international conference on. IEEE, 2008. ''' It is given that one iteration of iterative reweighed least squares problem can be written as [imath]\min \sum_{i=1}^N w_i u_i^2 [/imath] subject to [imath]\Phi u = b [/imath] which has a closed form solution given as [imath]u^{(n)} = Q_n \Phi^T (\Phi Q_n \Phi^T)^{-1} b[/imath]. where [imath]Q_n[/imath] is the diagonal matrix containing entries [imath]1/w_i[/imath] And the closed form equation can be derived from the Euler Lagrange equation. Can someone please help me to get the derivation of the iteration | 2658169 | Solving minimization problem [imath]L_2[/imath] IRLS (Iteration derivation)
In the article ''' Chartrand, Rick, and Wotao Yin. "Iteratively reweighted algorithms for compressive sensing." Acoustics, speech and signal processing, 2008. ICASSP 2008. IEEE international conference on. IEEE, 2008. ''' It is given that one iteration of iterative reweighed least squares problem can be written as [imath]\min \sum_{i=1}^N w_i u_i^2 [/imath] subject to [imath]\Phi u = b [/imath] which has a closed form solution given as [imath]u^{(n)} = Q_n \Phi^T (\Phi Q_n \Phi^T)^{-1} b[/imath]. where [imath]Q_n[/imath] is the diagonal matrix containing entries [imath]1/(|u_i^{(n-1)}|^2 + \epsilon )^{1/2} [/imath] And the closed form equation can be derived from the Euler Lagrange equation. Can someone please help me to get the derivation of the iteration?\ I understood till [imath]\min_u u^TWu+λ|| \Phi u−b||_2^2 [/imath] which can be written as [imath](\bigtriangledown=0)[/imath] which gives [imath] W u + \lambda (\Phi^T \Phi u - \Phi^T b) =0[/imath] |
2659464 | Let R be a PID, B a torsion R module and p a prime in R. Prove that if [imath]pb=0[/imath] for some non zero b in B, then [imath]\text{Ann}(B)[/imath] is a subset of (p)
I have tried to show it but to no avail. Its an exercise in chapter 12.1 of Dummit and Foote which I am learning by myself. Thanks in advance for any help. | 2179743 | Torsion Modules Annihilators
Let [imath]R[/imath] be a PID, let [imath]B[/imath] be a torsion [imath]R[/imath]-module, and let [imath]p[/imath] be a prime in [imath]R[/imath]. Show that if [imath]pb = 0[/imath] for some nonzero [imath]b \in B[/imath], then [imath]\operatorname{Ann}(B)[/imath] is a subset of [imath](p)[/imath]. Well we can let [imath]\operatorname{Ann}(B) = (t)[/imath] for some [imath]t \in R[/imath] since it is an ideal in [imath]R[/imath]. How to make use of the fact that [imath]B[/imath] is torsion (i.e. there is r in R such that rm=0 for each m in M) to prove the desired result? |
2659142 | Why is [imath]e^x[/imath] the only function that is its own derivative?
I've heard that [imath]f(x) = Ae^x[/imath] is only function (both elementary and non-elementary) that satisfies the property [imath]f(x)=\frac{df(x)}{dx}[/imath]. Is this true (and if it's true, is there a definitive way to prove it)? | 1291511 | Are the any non-trivial functions where [imath]f(x)=f'(x)[/imath] not of the form [imath]Ae^x[/imath]
This may seem like a silly question, but I just wanted to check. I know there are proofs that if [imath]f(x)=f'(x)[/imath] then [imath]f(x)=Ae^x[/imath]. But can we 'invent' another function that obeys [imath]f(x)=f'(x)[/imath] which is non-trivial? |
2659949 | Prove that [imath]\frac1{1+a_1}+\frac1{1+a_2}+...+\frac1{1+a_k}\ge\frac k{1+\sqrt[k]{a_1a_2...a_k}}[/imath]
Let [imath]a_1, a_2, ..., a_k[/imath] are real numbers greater thah [imath]1[/imath]. Prove that [imath]\frac1{1+a_1}+\frac1{1+a_2}+...+\frac1{1+a_k}\ge\frac k{1+\sqrt[k]{a_1a_2...a_k}}[/imath] I have this inequality proved by Sturm's method. Is there another way? May be Jensen's inequality. | 1600051 | Prove that [imath]\frac{1}{1+x_1}+\frac{1}{1+x_2}+\cdots+\frac{1}{1+x_n} \geq \frac{n}{\sqrt[n]{x_1x_2\cdots x_n}+1}[/imath]
If [imath]x_1,x_2,\ldots,x_n[/imath] are real numbers larger than [imath]1[/imath], prove that [imath]\dfrac{1}{1+x_1}+\dfrac{1}{1+x_2}+\cdots+\dfrac{1}{1+x_n} \geq \dfrac{n}{\sqrt[n]{x_1x_2\cdots x_n}+1}[/imath] Attempt AM-GM doesn't work here since we will get an upper bound. I don't see Cauchy-Schwarz working either. Thus, I think a substitution might work, but I am unsure of which one to use. |
2659723 | Binomial coefficient: Search for multiples of 13
Question: The coefficients of how many terms in the expansion of (1+x)2018 are multiples of 13? So, we've to investigate the powers of 13 in [imath]{2018 \choose r}[/imath], where 0 ≤ r ≤ 2018 I tried using the following: [imath]s_p(N!) = \left \lfloor \frac{N}{p} \right \rfloor + \left \lfloor \frac{N}{p^2} \right \rfloor + \left \lfloor \frac{N}{p^3} \right \rfloor + \cdots[/imath] where, [imath]s_p(N!)[/imath] denotes the highest exponent of prime p in n!, where n is a natural number. [imath]{2018 \choose r}[/imath], may be written as [imath]\frac{2018!}{r!(2018-r)!}[/imath]. Now, I tried to figure out the exponent of 13 in numerator for several values of r, but couldn't find a pattern that would lead me to the desired answer. Could someone please give a detailed solution to this problem, and also explain how to approach such problems? Is there any generalisation for the multiples of a prime number p in [imath]{n \choose r}[/imath]? P.S. The answer, to the best of my recollection, is 1395. | 2636627 | Divisiblity of binomial coefficients
Find the number of binomial coefficients in the expansion of [imath](1+x)^{2018}[/imath] that are divisible by [imath]13[/imath] On searching the internet for a long time I found that these type of questions are relevantly solved using either Kummer 's theorem or the Luca's theorem. But I haven't studied them yet. So can someone please explain some method to solve these problem without using those theorems. |
2659640 | Radical Ideal in the Coordinate Ring is Intersection of Maximal Ideals
I need to prove that a radical ideal in the coordinate ring [imath]\mathbb{C}[V][/imath] (where V is an algebraic variety) is the intersection of all maximal ideals containing it. The hint I was given was to consider the correspondence between ideals in a quotient ring [imath]R/I[/imath] and the ideals of [imath]R[/imath] containing [imath]I[/imath]. Any suggestions for where to begin? I don't even know where to start. | 732907 | Intersection of all maximal ideals containing a given ideal
Let [imath]I[/imath] be a proper ideal in [imath]k[x_1,....,x_n][/imath], where [imath]k[/imath] is an algebraically closed field. Show that [imath]\sqrt{I}= \cap M[/imath], where [imath]M[/imath] runs through all maximal ideals containing [imath]I[/imath]. I am confused by what we can say is in the intersection of all maximal ideals. For the proof, since k is algebraically closed, I think I would want to use Nullstellensatz, but once I have that [imath]\sqrt{I}=I(V(I))[/imath], I am not sure what to do next. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.