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2823907 | Showing [imath]\sum_{n=2}^{\infty} \frac{1}{n \ln(n^3)}[/imath] diverges
Two questions: Show whether [imath]\sum_{n=2}^{\infty} \frac{1}{n \ln(n^3)}[/imath] converges or diverges. According to wolfram, the series diverges by the comparison test, so I tried the following: for [imath]n[/imath] large, we get [imath]n < n \ln(n^3) [/imath] but then [imath]\frac{1}{n} > \frac{1}{n \ln(n^3)}[/imath], which gives me nothing, any tips? | 563258 | Does [imath]\sum_{n=2}^\infty (n\ln n)^{-1}[/imath] diverge?
Is [imath]\sum_{n=2}^\infty (n\ln n)^{-1}=\infty[/imath] ? This seems like elementary calculus, but I can't figure this out. Can anyone supply a hint? |
2812845 | Convergence or not of infinite series: [imath]\sum^{\infty}_{n=1}\frac{n}{1+n^2}[/imath]
How can we prove that the series [imath]\displaystyle \sum^{\infty}_{n=1}\frac{n}{1+n^2}[/imath] is convergent or divergent? Solution I try: [imath]\lim_{m\rightarrow \infty}\sum^{m}_{n=1}\frac{n}{1+n^2}<\lim_{m\rightarrow \infty}\sum^{m}_{n=1}\frac{n}{n^2}[/imath] Did not know how I can solve that problem from that point. | 2212504 | Prove that [imath]\sum_{n=1}^\infty\frac{n}{n^2+1}[/imath] is divergent
I'm trying to show that the series [imath]\sum_{n=1}^\infty\frac{n}{n^2+1}[/imath] is divergent. I'm trying to use a comparison test to do so. To do so, I first can see that [imath]\sum_{n=1}^\infty\frac{n}{n^2+1} \geq \sum_{n=1}^\infty\frac{n}{n^2+n} = \sum_{n=1}^\infty\frac{1}{n+1}[/imath] And I can use a change of variable [imath]k = n+1[/imath], so show [imath]\sum_{n=1}^\infty\frac{1}{n+1} = \sum_{k=2}^\infty\frac{1}{k} = \infty[/imath]. Then since [imath]\sum_{n=1}^\infty\frac{n}{n^2+1} \geq \sum_{k=2}^\infty\frac{1}{k}[/imath] it must be that [imath]\sum_{k=2}^\infty\frac{1}{k}[/imath] also diverges to infinity. Is this a valid way to show it? |
2770721 | Fundamental Group of [imath]GL_r(\mathbb{C})[/imath]
Let [imath]r \ge 2[/imath]. How to prove that the fundamental group [imath]\pi_1(GL_r(\mathbb{C}))[/imath] of the invertible matrices [imath]GL_r(\mathbb{C})[/imath] over [imath]\mathbb{C}[/imath] equals [imath]\mathbb{Z}[/imath]. What about [imath]\pi_1(GL_r(\mathbb{R}))[/imath]? | 3637 | Fundamental group of GL(n,C) is isomorphic to Z. How to learn to prove facts like this?
I know, fundamental group of [imath]GL(n,\mathbb{C})[/imath] is isomorphic to [imath]\mathbb{Z}[/imath]. It's written in Wikipedia. Actually, I've succeed in proving this, but my proof is two pages long and very technical. I want to find some better proofs of this fact (in order to compare to mine); to find some book or article, using which I can learn, how to calculate fundamental groups and, more generally, connectedness components of maps from one space to another; to find something for the reference, which I can use in order to learn, how to write this proofs nicely, using standard terminology. |
2824217 | Series of Natural Logarithms Raised to Powers
It's easy to see that for [imath]Re(s)>0[/imath], [imath]\zeta(s)=\displaystyle\sum_{n=1}^{\infty}\displaystyle\sum_{k=0}^{\infty}(-1)^k\frac{\ln^k(n)s^k}{k!}[/imath] from the series definition of the Riemann Zeta function and the Maclaurin series for [imath]n^{-s}[/imath] in terms of [imath]s[/imath]. This gives the series representation of the coefficient for the [imath]k[/imath]th term (i.e., the constant by which the term [imath]s^k[/imath] is scaled): [imath](-1)^k\frac{1}{k!}\displaystyle\sum_{n=1}^{\infty}\ln^k(n)[/imath] This begs the immediate question, [imath]\displaystyle\sum_{n=1}^{\infty}\ln^k(n)=?[/imath] It should be noted that this series is divergent, but any way to analytically find an answer that would make sense would be great, like something along the lines of the method by which [imath]\zeta(0)[/imath] or the p-adic definition of [imath]\sum_{n=0}^{\infty}2^n[/imath] is found. If you are not able to answer this question but are able to give me a representation of the Maclaurin series of the Riemann Zeta function without a double summation, that would be very nice too. | 2029777 | On the Maclaurin expansion of the Riemann zeta function and a related sequence.
I'm studying the Maclaurin series for the Riemann zeta function. I got that for [imath]\Re(s)\ge1[/imath] we have [imath]\zeta(s)=\lim\limits_{m\to\infty}\sum_{n=0}^\infty\left(\frac{(-1)^n}{n!}\sum_{i=1}^m{\ln(i)^n}\right)s^n[/imath] if we allow [imath]0^0=1[/imath] By looking at patterns, I formulated a guess for what the continuation would be for [imath]0\lt{\Re(s)}\lt1[/imath]. (Again, allow [imath]0^0=1[/imath]). [imath]\zeta(s)=\lim\limits_{m\to\infty}\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{a(k)}{(n-k)!}\ln(2)^k\sum_{i=1}^m(-1)^{(n-k+i)}\ln(i)^{(n-k)}\right)s^n[/imath] [imath]a(k)[/imath] is a sequence of rational numbers. Starting at [imath]k=0[/imath] they go [imath]\{1,2,3,\frac{13}{3},\frac{25}{4},\frac{541}{60},\frac{1561}{120}\frac{47293}{2520},\dots\}[/imath] I noticed that [imath]\frac{a(k-1)}{a(k)}[/imath] gets closer and closer to [imath]\ln2[/imath] for larger and larger [imath]k[/imath]. I took a guess for the sake of computation that [imath]a(k)=\prod_{i=0}^k\frac{\left[\frac{i!}{\ln(2)^{(i+1)}}\right]}{\left[\frac{i!}{\ln(2)^i}\right]}[/imath] where the brackets notate the nearest integer function. Please check me on my formulas and on this sequence and help me find a better formula to define the sequence. |
2824309 | Which kind of functions [imath]f(x, \ y)[/imath] can be written as [imath]g(x)\cdot h(y)[/imath]?
What are the properties of the function [imath]f(x, y)[/imath] that make it possible to be separated into a product of two one-variable functions, [imath]g(x)[/imath] and [imath]h(y)[/imath]? | 1764916 | Is that true that not every function [imath]f(x,y)[/imath] can be writen as [imath]h(x) g(y)[/imath]?
If not, why? Here [imath]h[/imath] and [imath]g[/imath] are two general function. |
2824849 | Evaluation of limit at negative infinity is wrong
[imath]\lim \limits_{x \to -\infty} \frac{x}{\sqrt{x^2+1}}[/imath] I tried to raise [imath]x[/imath] to the square and get: [imath]\lim \limits_{x \to -\infty} \sqrt\frac{x^2}{{x^2+1}}[/imath] afterwards, as I divide the numerator and denominator by [imath]x^2[/imath] I get [imath]\sqrt{\frac{1}{1+\frac1{x^2}}}[/imath] now after I set [imath]x[/imath] to minus infinity I get [imath]1[/imath]. The answer is wrong though.It should be [imath]-1[/imath] as written in the answers. Any guidance on how to solve it or directions to where I am mistaken will be appreciated. | 504750 | Simplifying radicals for limit leads to extra minus symbol
I am asked to calculate the limits of [imath]\dfrac{x}{\sqrt{1+x^2}}[/imath] for [imath]x\to\infty[/imath] and [imath]x\to-\infty[/imath] When I simplify the radical for [imath]x\to\infty[/imath], I generate: [imath]\dfrac{x}{\sqrt{1+x^2}} = \sqrt{\dfrac{x^2}{1+x^2}}[/imath], after which I can use l'hopital the two parts because they are [imath]\infty / \infty[/imath]. I do lim lim [imath]x\to-\infty[/imath] in similar fashion, but here an extra minus appears in the simplying radicals step that I can't quite place. I never actually use that [imath]x\to -\infty[/imath], so I'm kind of lost how that gets there? |
2824883 | INTEGRATION [imath]\int\frac{1}{\ln{x}}dx[/imath]?
How do we evaluate [imath]\int\frac{1}{\ln{x}}dx[/imath]? Is it a special integral, or some simple function, or what substitution will help? | 710808 | Integral of [imath]\int\frac{1}{\log x}dx[/imath]
What is the value of [imath]\int\frac{1}{\log x}dx[/imath] I have tried many times, but failed everytime. Can anyone help me out in solving this question. |
2824998 | ODE system of 2nd order
We have [imath]x''+bx=-a \quad \text{if} \: x'>0[/imath] and [imath]x''+bx=a \quad \text{if} \: x'<0.[/imath] With [imath]b>0[/imath] and [imath]x(0)=0[/imath] , Does this system have a solution? | 1201234 | solution of a ODE with a function of [imath]\dot{x}[/imath]
I have the equation: [imath]m\ddot{x}(t)+kx(t)=A[/imath] with m, k as constants and [imath]A = \left\{ \begin{array}{lr} a & : \dot{x}(t) <0\\ -a & : \dot{x}(t) >0 \end{array} \right. [/imath] a another constant. My question is how to solve for x(t) since A is a function of [imath]\dot{x}(t)[/imath] making it nonlinear. |
2824823 | If [imath]f : (a,b) \to \mathbb{R}[/imath] is differentiable, then [imath]f[/imath] is continuous: prove using [imath]\epsilon[/imath]-[imath]\delta[/imath]
Let [imath]f : (a,b) \rightarrow \mathbb{R}[/imath] be differentiable on [imath](a,b)[/imath]. Prove that [imath]f[/imath] is continuous by showing that [imath]\forall x_0 \in (a,b) \qquad \forall \epsilon>0 \qquad \exists \delta >0 [/imath] s.t.[imath] \; \;(x\in (a,b), |x-x_0| < \delta) \implies |f(x)-f(x_0)|<\epsilon[/imath] This is a 6 mark question on a practice Analysis 2 module. Frustratingly there are no solutions and I'm not sure where to start. Any help is greatly appreciated. | 2820625 | Proof of continuity implied by differentiability
I need to Let [imath]f : (a, b) → R[/imath] be differentiable on [imath](a, b)[/imath]. Prove that [imath]f[/imath] is continuous on [imath](a, b)[/imath] by showing that: [imath]∀x_0 ∈ (a, b)[/imath] [imath]∀ε > 0[/imath] [imath]∃δ > 0[/imath] s.t. [imath](x ∈ (a, b), |x − x_0| < δ) ⇒ |f(x) − f(x_0)| < ε[/imath] My understanding so far: When we say a function is differentiable at [imath]x_0[/imath], we mean that the limit: [imath]f^{\prime} (x) = \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0}[/imath] exists. When we say a function is continuous at [imath]x_0[/imath], we mean that: [imath]\lim_{x\to x_0} f(x) - f(x_0) = 0[/imath] My attempted proof: Let us suppose that [imath]f[/imath] is differentiable at [imath]x_0[/imath]. Then [imath] \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0} = f^{\prime} (x) [/imath] and hence [imath] \lim_{x\to x_0} f(x) - f(x_0) = \lim_{x\to x_0} \left[ \frac{f(x) - f(x_0)}{x-x_0} \right] \cdot \lim_{x\to x_0} (x-x_0) = 0[/imath] My issue: My issue may be a non issue, however for the 6 marks this question should be worth my proof seems shallow. Could someone please poke some holes in my proof, or add to it? Thank you |
2825459 | What is the maximum value of [imath]n[/imath] for which [imath]3^n[/imath] divides 1000!?
What is the maximum value of [imath]n[/imath] for which [imath]3^n[/imath] divides 1000! Am I supposed to look at it as a congruence... or apply euclid's algorithm somewhere? I'm lost I can't use simple arithmetics... Someone help. | 49670 | How to find maximum [imath]x[/imath] that [imath]k^x[/imath] divides [imath]n![/imath]
Given numbers [imath]k[/imath] and [imath]n[/imath] how can I find the maximum [imath]x[/imath] where: [imath]n! \equiv\ 0 \pmod{k^x}[/imath]? I tried to compute [imath]n![/imath] and then make binary search over some range [imath][0,1000][/imath] for example compute [imath]k^{500}[/imath] if [imath]n![/imath] mod [imath]k^{500}[/imath] is greater than [imath]0[/imath] then I compute [imath]k^{250}[/imath] and so on but I have to compute every time value [imath]n![/imath] (storing it in bigint and everytime manipulate with it is a little ridiculous) And time to compute [imath]n![/imath] is [imath]O(n)[/imath], so very bad. Is there any faster, math solution to this problem? Math friends?:) Cheers Chris |
2825116 | How many arrangements of k items from n items when there are duplicate items
Given the letters "BALL", how many arrangements of two letters exists? Using brute force, i.e. without using any counting technique, I get the following arrangements: AB AL BA BL LA LB LL Please let me know how I can get the correct answer using a counting formula. For example, if I wanted to know the number of arrangements that include all the letters, it would be [imath]\frac{4!}{2!}[/imath]; where the 2! accounts for the duplicate L's in BALL. | 125834 | A general formula for [imath]n[/imath] distinguishable arrangements of length [imath]\ell[/imath] of a sequence of a given size?
Is there a general formula for the number of distinguishable arrangements of length [imath]\ell[/imath] of a sequence of size [imath]s[/imath], given the repeated elements? So, For example, would there be a general formula to solve the problem: How many distinguishable 5-letter arrangements are there of the word "PROBABILITY" ? I'm looking for a general formula. |
2825703 | Prove that [imath]a_n[/imath] is bounded above and that it converges
Let [imath]{a_n}[/imath] be the sequence where [imath]a_1 =\sqrt{2}[/imath], and [imath]a _{n+1} = \sqrt{2+ \sqrt{a_n}}[/imath] for all n ∈ N. Prove that [imath]{a_n}[/imath] is bounded above by 2, and then prove that [imath]{a_n}[/imath] converges. My thought is to generally employ the monotonic convergence theorem: Proof. Because [imath]a_1= \sqrt{2}[/imath], and the sequence is increasing, we see that it is bound below by [imath]\sqrt{2}[/imath]. And, if we take the limit as n approaches infinity, we see the limit equals 2, so it is bounded above by 2. Further by the monotonic convergence theorem, as we see that the sequence is monotonically increasing and bounded above and below, it follows through the M.C.T. that it converges. QED My question really has to do with showing that it is monotonically increasing, which I believe I need to use induction for, however, I am not sure how exactly to do that, as my book glosses over the subject in a generic review of the concept, i.e. noting the notion that you do [imath]s(n)[/imath] and then show that [imath]s(n+1)[/imath] is also true for all n, but I'm fuzzy on the subject. Moreover, am I correct with my bound assumptions? | 2196026 | Using BMCT to prove the sequence [imath]a_{n+1} = \sqrt{2 + a_n}[/imath] is bounded and increasing.
My solution: Proof of (1) [imath]< \sqrt{2 + 2} \text{ by I. H.}[/imath] [imath]= 2[/imath] As required. Therefore by PMI [imath]\{a_n\}[/imath] is bounded above by 2 Proof of (2) [imath](a_n - 2)(a_n + 1) < 0[/imath] Therefore [imath]\{a_n \}[/imath] is strictly increasing. By BMCT it states that if a sequence is bounded above and strictly increasing, then it converges. Therefore this sequence by BMCT converges. Is this correct? Is there anything else I should say or add? |
2826805 | Simplifying [imath]\gcd(a^x + 1, a^y + 1)[/imath]
I am hoping those with more experience in number theory than me can help me out on this question. After learning that [imath]\gcd(a^m - 1, a^n - 1) = a^{\gcd(m, n)} - 1[/imath], I questioned what is [imath]\gcd(a^m + 1, a^n + 1)[/imath] ([imath]m[/imath] not equal to [imath]n[/imath]). I assumed WLOG that [imath]m > n[/imath] and then used the Euclidean Algorithm to get [imath]\gcd(a^m + 1, a^n + 1) = \gcd(a^{m-n} - 1, a^n + 1)[/imath]. However, I am now stuck. Please explain if the parity of [imath]x[/imath] and [imath]y[/imath] matter. QUESTION: Calculate [imath]\gcd(a^x + 1, a^y + 1)[/imath] | 2588518 | Find the GCD of [imath](p^n-1,p^m+1)[/imath].
Let [imath]p[/imath] be a prime number and [imath]n, m[/imath] positive integers. I want to find [imath]\gcd(p^n+1, p^m+1)[/imath] [imath]\gcd(p^n-1, p^m+1)[/imath] I know that [imath]\gcd(p^n-1, p^m-1)=p^{\gcd(n,m)}-1.[/imath] And It seems that if I can solve the first one, then the second one can be sovled too, since (assume that [imath]n>m[/imath]) [imath]\gcd(p^n+1, p^m+1)=\gcd(p^n-p^m, p^m+1)=\gcd(p^m(p^{n-m}-1), p^m+1).[/imath] At last I have [imath]\gcd(p^n+1, p^m+1)=\gcd(p^{n-m}-1, p^m+1).[/imath] So, my problem is that Find [imath]\gcd(p^n-1, p^m+1)[/imath]. It seems to me that it should have some formular which appears in some where, but I can't remember it. |
2821577 | prove that [imath]\{a_{n}\} [/imath] converges as [imath], a_{n+1}=\frac{a_{n}}{sin(a_{n})},\ \forall n\geq2[/imath] and [imath]0[/imath]
let [imath]\{a_{n}\}_{n=1}^{\infty} [/imath] be a sequence defined as [imath]0<a_{1}<\frac{\pi}{2}[/imath] and [imath] a_{n+1}=\frac{a_{n}}{\sin(a_{n})},\ \forall n\geq2[/imath] prove that [imath]\{a_{n}\} [/imath] converges, and find the limit of [imath]\{a_{n}\}[/imath]. | 2747256 | Given [imath]a_{n+1} = \frac{{a_n}}{\sin({a_n})}[/imath] with [imath]0 < a_1 < \frac{\pi}{2}[/imath]. Show that the limit [imath]\lim_{n\rightarrow\infty}{a_n}[/imath] exists and find it.
Given [imath]0 < a_1 < \frac{\pi}{2}[/imath], for [imath]n\geq 1[/imath] we define [imath]a_{n+1} = \dfrac{{a_n}}{\sin({a_n})}.[/imath] I need to show that [imath]\lim_{n\rightarrow\infty}{a_n} = L[/imath], and then I need to find [imath]L[/imath]. I can't find a way to bound [imath]\{a_n\}_n[/imath] so that [imath]0 <a_n < \frac{\pi}{2}[/imath] I've tried to use every convergence test I know, but I always have the same problem which I want to bound [imath]\{a_n\}_n[/imath] and can't to that without bounding [imath]\{\sin(a_n)\}_n[/imath]. |
2827331 | To prove: [imath]n!=\sum\limits_{r=0}^n (-1)^r \binom{n}{r} (n-r)^n[/imath]
How to algebraically prove that [imath]n!=\sum\limits_{r=0}^n (-1)^r \binom{n}{r} (n-r)^n[/imath] I was trying to find number of onto functions from [imath]A[/imath] to [imath]A[/imath] containing [imath]n[/imath] elements. Using the inclusion-exclusion principle I am getting [imath]\sum\limits_{r=0}^n (-1)^r \binom{n}{r} (n-r)^n.[/imath] We can also do it by simple combinatorics, as every element has to have a pre-image and number of elements in the domain equal to the number of elements in the codomain, the number of functions is [imath]n!.[/imath] Is there an algebraic way to prove these two are equal? | 1861402 | Proof of the summation [imath]n!=\sum_{k=0}^n \binom{n}{k}(n-k+1)^n(-1)^k[/imath]?
[imath]n!=\sum_{k=0}^n \binom{n}{k}(n-k+1)^n(-1)^k[/imath] Could anyone give the proof of the above equation? Thanks in advance! |
754023 | Show that there is no integer n with [imath]\phi(n)[/imath] = 14
I did the following proof and I was wondering if its valid. It feels wrong because I didn't actually test the case when purportedly [imath]n[/imath] is not prime, but please feel free to correct me. Assume there exists [imath]n[/imath] such that [imath]\phi(n) = 14[/imath]. Assume [imath]n[/imath] is prime. Then [imath]\phi(n) = n-1[/imath]. Then here, n-1 = 14, so n = 15. We know since Euler's totient function is multiplicative that [imath]\phi(xy)[/imath] = [imath]\phi(x)\phi(y)[/imath], so [imath]\phi(15) = \phi(3)\phi(5)[/imath], but alas [imath]\phi(3)\phi(5) = 2\cdot4 = 8 \ne 14[/imath]. If [imath]n[/imath] is not prime a similar argument follows since we know then that [imath]n[/imath] must be composed of prime numbers by the prime factorization theorem. | 3010321 | Find all number [imath]n\in \mathbb N[/imath] such that [imath]\varphi (n)=14[/imath]
Find all number [imath]n\in \mathbb N[/imath] such that [imath]\varphi (n)=14[/imath] [imath]\varphi (n)=p_1^{\alpha_1-1}\cdot p_2^{\alpha_2-1}\cdots p_n^{\alpha_n-1}(p_1-1)\cdots (p_n-1)=14[/imath] so number that divide 14 [imath]x|14[/imath] is [imath]x\in \{1,2,7,14\}[/imath] so [imath]p_i-1=1[/imath] or [imath]p_i-1=2[/imath] or [imath]p_i-1=7[/imath] or [imath]p_i-1=14[/imath] [imath]i\in \{1,2,\ldots n \}[/imath] so [imath]p_i=2, p_i=3, p_i=8, p_i=15[/imath] since [imath]8,15[/imath] is not prime number then [imath]p\in \{2,3\}[/imath]. If we say that [imath]n=2^{\alpha}[/imath] then [imath]\varphi (n)=2^{\alpha-1}=14[/imath] so there is not such [imath]\alpha \in \mathbb N[/imath] that [imath]\varphi(n)=14[/imath] so [imath]n\not=2^{\alpha}[/imath] The same is for [imath]n=3^{\beta}[/imath] and [imath] n=2^{\alpha}\cdot 3^{\beta}[/imath]. So there is not exist [imath]n\in \mathbb N[/imath] such that [imath]\varphi(n)=14[/imath] Is this ok? |
2828048 | characteristic function in probability theory
Suppose [imath]\mu[/imath] is a probability measure and [imath]f[/imath] is its characteristic function. Suppose [imath]f[/imath] has finite first order derivative at [imath]t=0[/imath], can you conclude that [imath]\mu[/imath] has finite expectation? I know that if [imath]f[/imath] has a finite derivative of even order [imath]k[/imath] at [imath]t=0[/imath], then [imath]\mu[/imath] has a finite moment of order [imath]k[/imath]. But I am not sure whether [imath]\mu[/imath] has finite expectation and how to prove it. Can anyone help me? | 793788 | Continuous probability distribution with no first moment but the characteristic function is differentiable
I am looking for an example of a continuous distribution function where the first moment does not exist but the characteristic function is differentiable everywhere. Cauchy distributions do not fulfill this, as their characteristic functions are not differentiable at [imath]0[/imath]. Does anybody here have an example? |
2828143 | One number is a multiple of other
Let [imath]S \subset N_{100}= \{1,2,3,4, \dots, 100 \}[/imath] with [imath]|S|=51[/imath]. Then prove for any such [imath]S[/imath] it is possible to select two numbers from [imath]S[/imath] such that one is a multiple of other. I wasn't able to proceed. How should I start. I think it needs Pigeon Hole Principle. | 2271176 | Choosing [imath]101[/imath] numbers from [imath]\{1, 2, \dots , 200\}[/imath].
I want to prove that if you choose [imath]101[/imath] numbers from the set [imath]\{1,2,3,4,\dots ,200\}[/imath], there are always two numbers such that one divides the other with no remainder. The prove should involve the "pigeonhole principle". I am not sure how to define the pigeonholes and how to define the pigeons. Any assistance with the proof will be most appreciated. Thank you. |
2826850 | [imath]x,y[/imath] and [imath]z[/imath] are consecutive integers, [imath]\frac {1}{x}+\frac {1}{y}+\frac {1}{z}[/imath]...
[imath]x,y[/imath] and [imath]z[/imath] are consecutive integers, such that [imath]\frac {1}{x}+ \frac {1}{y}+ \frac {1}{z} \gt \frac {1}{45} [/imath], what is the biggest value of [imath]x+y+z[/imath] ?. I assumed that [imath]x[/imath] was the smallest number so that I could express the other numbers as [imath]x+1[/imath] and [imath]x+2[/imath] and in the end I got to a cubic function but I didn't know how to find its roots. I probably didn't do anything important so I'd appreciate if you give me any hints or help. Thanks in advance. I have thought about using the AM-HM. | 2219688 | Finding Maximum value of consecutive positive integers with constraints
Let, [imath]x,y,z[/imath] be consecutive positive integers such that [imath]\frac{1}{x} + \frac{1}{y} + \frac{1}{z} > \frac{1}{10}[/imath] I need to find the maximum value of [imath]x+y+z[/imath] ? My attempt: I tried guessing first, To maximise [imath]x+y+z[/imath] it is necessary to have high value for [imath]x[/imath], But not too high which will contradict the constrain of the problem. As we can roughly split the [imath]0.1[/imath] into three parts so I started with numbers closer to [imath]30[/imath] with trial an error I got that for [imath]x=29[/imath] we get the optimum and final result would be 90 (=max. value of [imath]x+y+z[/imath]). Is there any way to arrive at this conclusion with using inequalities? Any hints will be appreciated... |
2827817 | Does [imath]dist(x, \partial A) \leq dist(x, \partial B)[/imath] if [imath]x \in A \subset B[/imath]?
Let [imath](M,d)[/imath] be a metric space, let [imath]A \subset B \subset M[/imath], and let [imath]x \in A[/imath]. Does the inequality below hold? [imath] dist(x, \partial A) \leq dist(x, \partial B) [/imath] Follow-up question: if not, does it hold if we add either/both of the following restrictions: [imath]A[/imath] and [imath]B[/imath] are open [imath](M,d)[/imath] is [imath]\mathbb{R}^n[/imath] with the usual distance NOTE: for any set [imath]E \subset M[/imath], we define [imath]dist(x, E) = \inf \{ d(x,z) : z \in E \}[/imath] | 1846840 | If [imath]A \subseteq B[/imath], does [imath]\mbox{dist}(x,\partial A) \le \mbox{dist}(x, \partial B)[/imath] hold for all [imath]x \in A[/imath]
Let [imath]A,B \subseteq \mathbb{R}^n[/imath] with euclidean metric. Furthermore let [imath] \mbox{dist}(X, Y) := \inf\{|x - y| : x \in X, y \in Y\}. [/imath] Does the following implication hold? [imath]A \subseteq B \implies \forall x \in A: \mbox{dist}(x,\partial A) \le \mbox{dist}(x, \partial B)[/imath] Problems could arrise if one of the boundaries [imath]\partial A[/imath], [imath]\partial B[/imath] is empty. But in this case [imath]A[/imath] or [imath]B[/imath] would be clopen, and hence equal to [imath]\mathbb{R}^n[/imath] or [imath]\emptyset[/imath]. I think I am better off assuming [imath]A \neq \emptyset[/imath] and [imath]B \neq \mathbb{R}^n[/imath] to avoid problems. Thanks to a suggestion by GEdgar I came up with the following proof: Let [imath]s \colon [0,1] \to \mathbb{R}^n[/imath] be the straight line with [imath]s(0) = x \in A[/imath] and [imath]s(1) = y \in \partial B[/imath]. I want to show, there exists [imath]t' \in [0,1][/imath] with [imath]s(t') \in \partial A[/imath]. It is safe to assume [imath]x \not \in \partial A[/imath], since then [imath]\mbox{dist}(x,\partial A) = 0[/imath] and the inequality holds trivially. Same goes for the assumption [imath]y \not \in \partial A[/imath]. Consider now the function [imath] f(x) := (1-2 \chi_A(x)) \cdot \mbox{dist}(x,\partial A). [/imath] One can check that this function is continuous on [imath]\mathbb{R}^n[/imath]: No problems on [imath]\mathbb{R}^n \setminus \partial A[/imath], since as a property of metric spaces [imath]\mbox{dist}(\cdot,\partial A)[/imath] is continuous and furthermore [imath](1-2 \chi_A(x))[/imath] is constant. Let [imath]\lim_{n \to \infty} x_n = x \in \partial A[/imath]. Then [imath] |(1-2 \chi_A(x_n)) \cdot \mbox{dist}(x_n,\partial A)| \leq \mbox{dist}(x_n,\partial A) \to 0 = \mbox{dist}(x,\partial A), [/imath] since the closedness of [imath]\partial A[/imath] implies [imath]\mbox{dist}(x,\partial A) = 0[/imath] iff [imath]x \in \partial A[/imath]. Now the claim follows by application of the intermediate value theorem. Is this proof correct? Does the claim also hold for more general metric spaces [imath](X,d)[/imath] which are only connected/path connected? Of course the definition of [imath]\mbox{dist}[/imath] would be modified accordingly [imath](| x - y | \to d(x,y))[/imath] |
2682399 | Comparing positive definite matrices and their inverses
Given two [imath]n\times n[/imath] symmetric real positive definite (p.d.) matrices [imath]A[/imath] and [imath]B[/imath], it is known that [imath]A \prec B \iff B^{-1} \prec A^{-1}[/imath] where [imath]A\prec B[/imath] means that [imath]B-A[/imath] is p.d.. Suppose [imath]A\prec B[/imath]. If I know some information about the "magnitude" of [imath]B-A[/imath], e.g., the minimum eigenvalue of [imath]B-A[/imath] (which is positive), can I infer information about the "magnitude" of [imath]A^{-1}-B^{-1}[/imath], e.g., its minimum eigenvalue? | 2482267 | How to prove positive semidefiniteness of two matrices through Schur Complement?
Let matrices [imath]X[/imath] and [imath]Y[/imath] be positive semidefinite. Show that [imath]X \succeq Y \succ 0[/imath] is equivalent to [imath]Y^{-1} \succeq X^{-1}[/imath]. The teacher tells me that it is easy to prove via the Schur complement. But I could not get his idea. Could anyone help me and give me some intructions? Thanks in advance! |
2829470 | Show that there is eigenvector that all composition is real number
If [imath]A[/imath] is a real matrix and [imath]\lambda[/imath] is a real eigenvalue of [imath]A[/imath], then there is a real eigenvector corresponding to [imath]\lambda[/imath]. How can I prove this? | 80064 | Must eigenvector be real?
If [imath]\lambda[/imath] is a real eigenvalue of a real matrix [imath]M[/imath], does there necessarily exist a real eigenvector of [imath]M[/imath] corresponding to [imath]\lambda[/imath]? Edit: Never mind. I figured it out.. If [imath]\lambda[/imath] is a real eigenvalue, then [imath]\det(\lambda I-M)=0[/imath], which means there exists a real vector [imath]x[/imath] such that [imath]Mx=\lambda x[/imath]. Right? |
2829140 | Convert equation system in to one ODE
I tried with a few algebraic manipulations with no success. for the system: [imath] \begin{pmatrix} x_1' \\ x_2' \end{pmatrix} = \begin{pmatrix} 4 & -5 \\ -7 & -3 \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} [/imath] Tried [imath]x_{1}'=x_{2}[/imath] thus from first equation I get [imath]x_{2}=4x_{1}-5x_{2}[/imath] and then replacing the [imath]x_{2}[/imath] in second equation with [imath]\frac{2}{3}x_{1}[/imath] and the result become [imath]y''=-9y[/imath]. This is not the final answer for some reason, They didn't do any example in class and my logic isn't working here... | 1242979 | How to turn a system of first order into a second order
So I have two equations [imath]X' = aX + bY[/imath] [imath]Y' = cX + dY[/imath] I want to convert it back to a second order equation with the form [imath]X'' + \alpha X' + \beta X[/imath] with [imath]\alpha,\beta[/imath] in terms of a,b,c,d. I have been racking my brain for hours trying to go backwards from a reduction of order, but just can't seem to figure it out. Any help would be much appreciated! |
2828529 | Find all function [imath]f:\Bbb{N}_0 \rightarrow \Bbb{N}_0 [/imath] satisfying the equation [imath]f(f(n) + f(n) = 2n +3k,[/imath] where [imath]k[/imath] is a fixed natural number
Find all function [imath]f: \Bbb{N}_0 \rightarrow \Bbb{N}_0 [/imath] satisfying the equation [imath]f(f(n) + f(n) = 2n +3k[/imath] for all [imath]n \in \Bbb{N}_0 [/imath] where [imath]k[/imath] is a fixed natural number. I have proceeded with solution to a point but I am not able to go further so lets say [imath]f(0)=m[/imath] [imath]\implies[/imath] [imath]f(m) =3k-m[/imath] [imath]f(3k-m)=f(f(m)=2m+3k-f(m)=3m[/imath] [imath]f(3m)=f(f(3k-m))=2(3k-m)+3k-f(3k-m)=9k-5m [/imath] after this I am not able to proceed through | 1535189 | Find all function satisfying [imath]f(f(n))+f(n)=2n+3k[/imath]
Find all functions [imath]f:\mathbb{N_{0}} \rightarrow \mathbb{N_{0}}[/imath] satisfying the equation [imath]f(f(n))+f(n)=2n+3k,[/imath] for all & [imath]n\in \mathbb{N_{0}}[/imath], where [imath]k[/imath] is a fixed natural number. A friend of mine suggested me to use recurrence relation. While it is unfortunate that I have problems understanding recurrence relation. Please refer information to understand difference equation, homogeneous equation & auxillary equation. Thanks in advance! Kindly don't down vote for not showing my attempt. I am seriously not able to understand the question and the process suggested to use. |
2829474 | Is [imath]A^2[/imath] is diagonalizable , then is [imath]A[/imath] diagonalizable when [imath]A[/imath] is non-singular?
I have a question in mind which is a slight generalization of a past question(famous). See here if [imath]A^2 \in M_{3}(\mathbb{R})[/imath] is diagonalizable then so is [imath]A[/imath] if you want the basic version. Now my question : If [imath]A^2[/imath] is diagonalizable , then is [imath]A[/imath] diagonalizable when [imath]A[/imath] is non-singular? Obviously, the previous counterexample of taking [imath]A[/imath] to be a nilpotent matrix of index will not work here as then [imath]|A|=0[/imath]. Any proofs? | 2349701 | If [imath]T^m[/imath] is diagonalizable for a [imath]m\in\mathbb N[/imath], then [imath]T[/imath] is diagonalizable.
Suppose that [imath]V[/imath] is a finite dimensional [imath]\mathbb C[/imath]-vector space, and suppose that [imath]T:V\rightarrow V[/imath] is injective. If there is a [imath]m\in\mathbb N[/imath] such [imath]T^m[/imath] is diagonalizable, then [imath]T[/imath] is diagonalizable. I've found the proof for the case that [imath]T^m=Id[/imath], but I can't adapt it to this case. |
2829693 | [imath]x-\frac{x^2}{2}<\ln(1+x)0[/imath]
Show that- [imath]x-\frac{x^2}{2}<\ln(1+x)<x-\frac{x^2}{2(1+x)}, x>0[/imath] I can prove this by observing that at [imath]x=0[/imath], the three functions in the above inequality (say [imath]f_1<f_2<f_3[/imath]) are zero. And for [imath]x>0[/imath], the derivative of the difference of [imath]f'_2-f'_1>0[/imath] and [imath]f'_3-f'_2>0[/imath] therefore the inequality is satisfied. I am looking for an elegant solution to this using some kind of Mean Value Theorem maybe, if possible. | 171659 | Proving inequality on functions [imath]x-\frac{x^2}{2}<\ln(1+x)[/imath]
To prove: [imath]x-\frac{x^2}{2}<\ln(1+x)<x-\frac{x^2}{2(1+x)},\quad\forall x>0[/imath] I have used Taylor series expansion at 0 for both the inequalites. The greater than by expanding [imath]\ln(1+x)[/imath] and the less than by expanding [imath]\int \ln(1+x)\,dx[/imath] at 0. Is there a cleaner / more elegant way of achieving the same? |
2829934 | Prove property of floor function
Prove that, for any [imath]a, b, c \in \mathbb{N}[/imath], [imath]\bigl\lfloor \frac{a}{bc} \bigr\rfloor = \Bigl\lfloor \frac{\bigl\lfloor \frac{a}{b} \bigr\rfloor}{c} \Bigr\rfloor[/imath]. I have no idea how to prove this. Can you help me, please? Thanks! | 1338558 | Proof for [imath]\left\lfloor\frac 1j\left\lfloor\frac nk\right\rfloor\right\rfloor=\left\lfloor\frac n{jk}\right\rfloor[/imath]
Problem: For positive integers [imath]n,j,k[/imath], prove that the following holds: [imath]\left\lfloor\frac 1j\left\lfloor\frac nk\right\rfloor\right\rfloor=\left\lfloor\frac n{jk}\right\rfloor[/imath] I simply need hints to start on this problem. I have verified that it's indeed true by checking a few examples and I can't seem to find a counter-example, so I'm almost sure that the problem is indeed correct. However, I can't seem to think of a proper (and rigorous) proof of this. Can someone please help me by giving me a few hints? I want to prove it myself, so I'd appreciate if people don't directly post the proof themselves. Also, please keep your hints as subtle as possible (there's no fun in proving something if the hint gives out most of the answer) :D |
2829976 | Proving [imath]\sum_{k=0}^{n-1} \text{cos}\left(\frac{2\pi k}{n} \right) = \sum_{k=0}^{n-1} \text{sin}\left(\frac{2\pi k}{n} \right) = 0[/imath]
I would like to prove the following, [imath]\sum_{k=0}^{n-1} \text{cos}\left(\frac{2\pi k}{n} \right) = \sum_{k=0}^{n-1} \text{sin}\left(\frac{2\pi k}{n} \right) = 0[/imath] This is equivalent to showing that if we have a regular n-gon, all of the vectors in [imath]\mathbb{R}^2[/imath] pointing to every vertex cancel eachother out to [imath]\textbf{0}[/imath]. It seems obvious enough and i've tried a number of examples for small [imath]n[/imath]. I am not sure how to start a proof though. | 1905189 | Proving [imath]\sum_{x=0}^{n-1} \cos\left(k +x{2\pi\over n}\right) =\sum_{x=0}^{n-1} \sin\left(k +x{2\pi\over n}\right) =0. [/imath]
Is there anybody who can help me show the following? [imath] \sum_{x=0}^{n-1} \cos\left(k +x{2\pi\over n}\right) =0 \qquad\hbox{and}\qquad \sum_{x=0}^{n-1} \sin\left(k +x{2\pi\over n}\right) =0 [/imath] I can only prove this claim for the sine expression when [imath]k=0[/imath], by grouping the output values such that they all cancel out (for both [imath]n[/imath] even and odd). I have thought about using complex numbers in exponential form to prove it, but I can't do it ... :P I would really appreciate an algerbraic proof. |
2829798 | A point and an ellipse
In particular, I have to find the exact value of the minimum distance from [imath]P(- \frac {15}{4},1)[/imath] to the ellipse [imath] \frac {x^2}{4} + \frac {y^2}{9} =1[/imath] Therefore, if [imath] \frac {x^2}{a^2} + \frac {y^2}{b^2} = 1[/imath] is an ellipse, with the parameterization [imath]x(θ)≔(a \cos θ,b \sin θ ),[/imath] I have to find the value of [imath]θ[/imath] giving the minimum distance from [imath]P(p,q)[/imath] (not on the ellipse) to the ellipse is given by a quartic in [imath]t= \tan( \frac {θ}{2}).[/imath] A necessary condition for [imath]x[/imath] to be the closest point to [imath]P[/imath] is that [imath]P-x[/imath] is perpendicular to the tangent vector in [imath]x ,[/imath] i.e. [imath](P-x(θ) ). x' (θ)=0[/imath] I can't handle the above condition to make an fuction (e.g. [imath]f(θ)[/imath]) to find the minimum value by calculating the derivative [imath]f'(θ)=0[/imath] for example. Then I have to prove that rational, non-zero values of [imath]a, b, p, q[/imath] can be found such that the quartic factorises as the product of two quadratics with rational coefficients. Any help? Thank you | 2829664 | Distance of point [imath]P[/imath] from an ellipse
If [imath] \frac {x^2}{a^2} + \frac {y^2}{b^2} = r^2[/imath] is an ellipse, with the parameterization [imath]x(θ)≔r(a \cos θ,b \sin θ ),[/imath] I have to find the value of [imath]θ[/imath] giving the minimum distance from [imath]P(p,q)[/imath] (not on the ellipse) to the ellipse is given by a quartic in [imath]t= \tan( \frac {θ}{2}).[/imath] A necessary condition for [imath]x[/imath] to be the closest point to [imath]P[/imath] is that [imath]P-x[/imath] is perpendicular to the tangent vector in [imath]x ,[/imath] i.e. [imath](P-x(θ) ). x' (θ)=0[/imath] I can't handle the above condition to make an fuction (e.g. [imath]f(θ)[/imath]) to find the minimum value by calculating the rerivative [imath]f'(θ)=0[/imath] for example. Then I have to prove that rational, non-zero values of [imath]a, b, p, q[/imath] can be found such that the quartic factorises as the product of two quadratics with rational coefficients. Any help? Thank you |
2830524 | If [imath]\{a_n\}[/imath] is a sequence such that [imath]0\leq a_{m+n}\leq a_m+a_n[/imath], show that [imath]\lim_{n\to\infty}\frac{a_n}{n}[/imath] exists
Let [imath]\{a_n\}[/imath] be a sequence satisfying [imath]\forall m,n\in\Bbb{N}:0\leq a_{m+n}\leq a_m+a_n[/imath]. Prove that [imath]\lim_{n\to\infty}\frac{a_n}{n}[/imath] exists. At first I tried to do it with some [imath]\epsilon-\delta[/imath]. We wish to prove that there exists some [imath]L[/imath] such that [imath]\forall\epsilon>0:\exists n_0\in \Bbb{N}:n>n_0\Rightarrow|\frac{a_n}{n}-L|<\epsilon[/imath]. Which then got me to something like this:[imath]-\epsilon\cdot n<a_n-Ln<\epsilon\cdot n[/imath] which got me nowhere and I didn't find a way to make use of the inequality given. The sequence is obviously bounded below but non-decreasing, which tells me nothing. I don't see any way to use this fact to prove the existence of given limit. Any useful tips, hints appreciated. | 1808404 | Prove [imath]\lim_{n\to\infty} \frac{a_n}{n}[/imath] exists for positive sequence where [imath]a_{n+m} \leq a_n + a_m[/imath]
Let [imath]\{a_n\}_{n=1}^\infty[/imath] be a positive sequence of real numbers such that [imath]a_{n+m} \leq a_n + a_m.[/imath] Prove that [imath]\lim_{n\to\infty} \frac{a_n}{n}[/imath] exists by showing that [imath]\lim_{n\to\infty} \frac{a_n}{n} = \inf_{n \geq 1} \frac{a_n}{n}.[/imath] A similar question has been posted previously, see Limit for sequence [imath]a_{m+n}\leq a_m+a_n[/imath] , and I have found that this is the general result of "Fekete's Subadditive Lemma" for sequences, but I am unhappy with the few proofs I've seen. Perhaps the clearest one I have found is http://web.mat.bham.ac.uk/R.W.Kaye/seqser/fekete.html , but this seems a bit tricky for what is expected on this qualifier question. We are given a hint "treat [imath]\liminf[/imath] and [imath]\limsup[/imath] separately". Does anyone know of a more direct way to show that this limit exists? |
2830647 | Adjoint mapping
I need to show the following: let [imath]\phi[/imath] be a linear function [imath]V \rightarrow V[/imath], [imath]V[/imath] with finite dimensions and with [imath]\phi^*[/imath] as adjoint function, than [imath]\phi[/imath] is normal if and only if there is a polynomial [imath]f[/imath] so that [imath]f(\phi)=\phi^*[/imath] can you help me out? Here is normal as [imath]\phi \circ \phi^*=\phi^*\circ\phi[/imath] to intend. Thank you | 2717863 | Suppose that [imath]T\in \mathcal L (V)[/imath]. prove that T is normal iff there exists a polynomal [imath]f(x) \in \Bbb C [x] [/imath] s.t [imath]T^* = f(T)[/imath]
Suppose that [imath]T\in \mathcal L (V)[/imath]. prove that T is normal iff there exists a polynomal [imath]f(x) \in \Bbb C [x] [/imath] s.t [imath]T^* = f(T)[/imath] [imath](\Leftarrow) [/imath] notice that [imath]T^* T= f(T) T = T f(T) = T T^* [/imath] as T commutes with a polynomail in T. [imath](\Rightarrow ) [/imath] Let [imath] v_1 , \cdots , v_n [/imath] be an orthonormal basis for T. [imath] p(x) = \sum_{i=0}^{k} a_i x^{i} [/imath] [imath] p(T) = a_0 I + a_1 T + \cdots + a_k T{k} [/imath] [imath]p(T)v_j = a_0 v_j + a_1 \lambda_{i} v_j + a_2 \lambda_{i}^2 v_j \cdots + a_k \lambda_{i}^{k} v_j [/imath] [imath] = p(\lambda_i) v_j [/imath] [imath] =\bar {\lambda_{i}} v_{j} [/imath] [imath]= T^{*} v_j [/imath] |
2831146 | About the proof of the inequality: [imath]\int_0^1 |f(x)|dx\le\text{max}\left(\int_0^1 |f’(x)|dx, \left\vert\int_0^1 f(x)dx\right\vert\right)[/imath]
设f(x)导数处处连续，证明： ∫|f(x)|dx≤MAX{∫|f'(x)|dx,|∫f(x)dx|}. 三个都为定积分 x from 0 to 1. 我用到了MAX{A,B}=(A+B+|A-B|)/2.加上|A-B|≥|A|-|B|. 但是和要证的不等号方向相反。 Translation: Let [imath]f[/imath] be a function such that [imath]f'[/imath] is continuous everywhere, prove [imath]\int_0^1 |f(x)|dx\le\text{max}\left(\int_0^1 |f’(x)|dx, \left\vert\int_0^1 f(x)dx\right\vert\right)[/imath] I have used [imath]\text{max}(A,B)=\frac{A+B+|A-B|}2[/imath] and [imath]|A-B|\ge|A|-|B|[/imath], however what I need to prove has the reversed inequality sign. | 59851 | An inequality of [imath]\int_0^1 |f(x)|dx[/imath]
Question: If [imath]f\in C^1[0,1][/imath], show that [imath]\int_0^1 |f(x)|dx\le\max\left\{\int_0^1 |f'(x)|\,dx,\;\bigg|\int_0^1 f(x)\,dx\bigg|\right\}.[/imath] I have tried to make connection between [imath]|f|[/imath] and [imath]|f'|[/imath] by using [imath](tf(t))'=f(t)+tf'(t),[/imath] integrate the equation with [imath]t[/imath] on [imath][0,1][/imath], it gets [imath]|f(1)|\leq \bigg|\int_0^1 f(t)\,dt\bigg|+\int_0^1 |f'(t)|\,dt,[/imath] but it is not the desired inequality. |
945678 | Killing form on some Lie algebra [imath]L[/imath] is zero. Is [imath]L[/imath] necessarily nilpotent?
I've solved an exercise in Humphreys that said: Show that if Lie algebra [imath]L[/imath] is nilpotent, then its Killing form is zero. I'm wondering is the opposite true? In Humphreys, we work mainly with Lie algebras over an algebraically closed field of characteristic zero, so let's assume that. This is what I've tried: Let's assume that Killing form is zero. That gives us solvability of [imath]L[/imath], so we know (as a corollary of Lie's theorem) that there is a basis of [imath]L[/imath] in which all matrices in [imath]ad \ L[/imath] are upper triangular. Then, [imath]\kappa(x,y) = tr(ad(x) \ ad(y))[/imath] is a scalar product of diagonals of [imath]ad(x)[/imath] and [imath]ad(y)[/imath] in that basis. If I could show that this somehow implies that for every [imath]x \in L[/imath] diagonal of [imath]ad(x)[/imath] is all zeros, it would follow all eigenvalues of [imath]ad(x)[/imath] are zero and this would imply that [imath]ad(x)[/imath] is nilpotent. Then, Engel's theorem would give us nilpotency. But I don't know how to do that. | 310272 | When is the Killing form null?
When is the Killing form [imath]\kappa[/imath] of a Lie algebra [imath]\mathfrak g[/imath] null, i.e. [imath]\kappa(\cdot,\cdot)=0[/imath]? Surely this is true for any Lie algebra with trivial bracket, but is this the only case? I can't seem to find any nontrivial examples. |
2831527 | Projection on a closed convex in a Banach space, without inner product.
It is a know fact that in any real Hilbert space [imath]E[/imath], given a point [imath]a \in E[/imath] and [imath]C[/imath] a closed convex and non-empty set in [imath]E[/imath], theres one and only one [imath]x \in C[/imath] such that [imath]d(a,x) = d(a,C)[/imath]. My question is: can it be extended to norms not coming from an inner product in a Banach space? It is easy to prove that: Uniqueness of such a [imath]x[/imath] is guaranteed when [imath](E, \| . \|)[/imath] is a strictly convex space. The existence always holds if [imath]E[/imath] is finite dimensional. So does the existence holds without any assumption on the dimension of [imath]E[/imath], even if it's a strictly convex Banach space? | 292529 | Least norm in convex set in Banach space
The following statement is true for Hilbert spaces [imath]H[/imath]: Every closed convex set [imath]{\cal C} \subset H[/imath] has a unique element [imath]x[/imath] such that for any [imath]y \in C[/imath], we have [imath]|x| \leq |y|[/imath]. Is this statement still true for Banach spaces? If not, what is a counterexample? |
2158435 | Are there arbitary large non-trivial powers NOT containing all digits?
Are there arbitary larger numbers of the form [imath]a^b[/imath] with positive integers [imath]a,b>1[/imath] and [imath]a\ne 0\mod 10[/imath] NOT containing all digits ? Here : Biggest powers NOT containing all digits. is either a very similar or the same question, but the question is so old that I decided to restate it. My current best example : The number [imath]1955^{39}[/imath] has [imath]129[/imath] digits, but does not contain the digit [imath]1[/imath]. If we assume that the digits are uniformly distributed in [imath][0,9][/imath] and considering the growth of the non-trivial powers, can we expect that we have enough chances to get , lets say , a [imath]1000[/imath]-digit example ? In other words, are there enough non-trivial powers with a given decimal-expansion-length, that we can expect that the answer to the question is yes ? | 627409 | Biggest powers NOT containing all digits.
Let [imath]m>1[/imath] be a natural number with [imath]m \not\equiv 0 \pmod{10}[/imath] Consider the powers [imath]m^n[/imath] , for which there is at least one digit not occurring in the decimal representation. Is there a largest [imath]n[/imath] with the desired property for any [imath]m[/imath]? If so, define [imath]n(m)[/imath] to be this number. Examples : [imath]m=2 \rightarrow 2^{168} = 374144419156711147060143317175368453031918731001856[/imath] does not contain the digit [imath]2[/imath]. All the powers above up to [imath]2^{10000}[/imath] conatin all the digits, so [imath]2^{168}[/imath] seems to be the biggest power with the desired property. [imath]m=3 \rightarrow 3^{106} = 375710212613636260325580163599137907799836383538729[/imath] does not contain the digit [imath]4[/imath]. All the powers above up to [imath]3^{10000}[/imath] contain all the digits, so [imath]3^{106}[/imath] seems to be the biggest power with the desired property. So, probably [imath]n(2)=168[/imath] and [imath]n(3)=106[/imath] hold. Is [imath]n(m)[/imath] defined for any [imath]m[/imath], and if yes, can reasonably sharp bounds be given? |
2831676 | When does Order of Second Partial Derivatives Matter?
My professor was saying that, for a function of multiple variables, usually the order in which you take the order of partial derivatives did not matter. (ex: [imath]f_{xy} = f_{yx}[/imath]). Under what circumstances is this not true? | 956095 | When does order of partial derivatives matter?
I've taken multivariate calculus and am wondering if I can see a specific function where the order of taking the partial derivative matters. I've been told that there are some exceptions where [imath] \dfrac{\partial ^2 f}{\partial x \partial y} \ne \dfrac{\partial ^2 f}{\partial y \partial x} [/imath], so I'm curious to see what this looks like. EDIT: And why would this true? |
1036761 | Product of two positive compact, self adjoint operators
If we have two positive compact , self adjoint operators; [imath]A[/imath], [imath]B[/imath]. Is the product [imath]AB[/imath] a positive operator? | 583921 | Product and sum of positive operators is positive
I want to show that for [imath]S,T\in B(H)[/imath] bounded operators on Hilbertspace with [imath]S\geq 0,T\geq 0[/imath] and [imath]ST=TS[/imath], we have [imath]S+T\geq 0[/imath], and [imath]ST\geq 0[/imath]. [imath]T\geq 0[/imath] means [imath](Tx,x)\geq 0[/imath]. To me it seems that [imath]((S+T)x,x) = (Sx,x)+(Tx,x)\geq 0[/imath]. But for [imath]ST[/imath] i like some help. There is a theorem in Rudin (12.32) that says that for every [imath]T\in B(H)[/imath] we have an equivalence: [imath]T[/imath] is positive [imath]\Leftrightarrow[/imath] [imath]T[/imath] is self adjoint, i.e. [imath]T=T^*[/imath] and [imath]\sigma(T)\subset [0,\infty)[/imath]. Since [imath]S, T[/imath] are self adjoint we know that [imath]ST=TS = S^*T^* = T^*S^*[/imath] ([imath]ST[/imath] is selfadjoint since [imath]S,T[/imath] commute). Also I read somewhere that if [imath]S,T [/imath] commute then [imath]\sigma(ST)\subset \sigma(S)\sigma(T)[/imath], but I am not sure if this is true. Clearly I cant prove it, but if it were true by this ''Theorem'' it would follow that [imath]ST \geq 0[/imath]. Is there a more straightforward way? |
2832944 | Gauss circle problem
I want to show that the number [imath]\mu(n)[/imath] of pairs of integers [imath](x,y)[/imath] with [imath]x^{2}+y^{2}<n[/imath] satisfies [imath]\mu(n)/n \rightarrow \pi [/imath] as [imath]n \rightarrow \infty[/imath] If we consider the lattice [imath]L=\mathbb{Z}<1,0>+\mathbb{Z}<0,1>[/imath] and [imath]X=\lbrace (x,y) \in \mathbb{Z}^{2}|x^{2}+y^{2}<n \rbrace[/imath] from Minkowski's theorm we know that [imath]X\cap L \neq \emptyset [/imath] i want to know that is there any formula for [imath]\mu(n)[/imath]? beacuse if we know the formula we can comput the limit. | 1889361 | Gauss circle problem : a simple asymptotic estimation.
Find the number of integer points lying in or inside a circle of radius [imath]n\in \mathbb N[/imath] centered at the origin. The problem asks for all [imath](a,b)\in \mathbb Z^2[/imath] such that [imath]a^2+b^2\leq n^2[/imath]. Looking at such points lying strictly in the North-East quadrant, pick an [imath]a\in \{1,\ldots,n\}[/imath] and let [imath]b[/imath] run through [imath]\{1,\ldots,\lfloor \sqrt{n^2-a^2}\rfloor\}[/imath] : there are exactly [imath]\displaystyle \sum_{k=1}^n \lfloor \sqrt{n^2-k^2} \rfloor[/imath] integer points strictly in the quadrant. Multiply that by [imath]4[/imath] to account for all [imath]4[/imath] quadrants, add the origin and points on the [imath]x[/imath] and [imath]y[/imath] axis and the total number is [imath]1+4\sum_{k=0}^n \lfloor \sqrt{n^2-k^2} \rfloor[/imath] I'm interested in asymptotics of this sum. A trivial estimate is the following:[imath]\begin{align}\sum_{k=0}^n \lfloor \sqrt{n^2-k^2} \rfloor &= \sum_{k=0}^n \sqrt{n^2-k^2} + O(n)\\ &= n^2\cdot \left(\frac 1n \sum_{k=0}^n \sqrt{1-\frac{k^2}{n^2}}\right) + O(n) \\ &=n^2 \left( \int_0^1 \sqrt{1-t^2}dt + o(1)\right) + O(n) \\ &= \frac \pi 4n^2 + o(n^2)\end{align} [/imath] Hence [imath]1+4\sum_{k=0}^n \lfloor \sqrt{n^2-k^2} \rfloor = \pi n^2 + o(n^2)[/imath] Is there a way to refine the estimate above and get [imath]\pi n^2 + O(n)[/imath] instead (or something better) ? |
2832744 | Find two non-singular matrices such that [imath]AB=-BA[/imath]
Question is precisely the title. I've been trying to come up with an example for an hour now and haven't found one. Do such matrices, [imath]A[/imath] and [imath]B[/imath] exist? | 2780319 | find nonsingular matrices [imath]B[/imath] and [imath]C[/imath] that satisfy [imath]BC+ CB= 0[/imath]
How to find two non singular matrices [imath]B[/imath] and [imath]C[/imath] such that [imath]BC+ CB= 0[/imath]. Clearly [imath]B[/imath] and [imath]C[/imath] must be square matrices of same order and the order is even.so it is possible we can find some example of order 2. But I cannot find any. I would be happy if someone helps me in finding some matrices.Thanks in advance. |
2833373 | Integral with log and sin
So I have the following integral. [imath]\int_0^{1} \frac{x^a-x^b}{\log x} \sin(\log x)\,\text{d}x. [/imath] I thought of a substitution like this [imath]\log x=t \implies (e^{a-1}-e^{b-1})\int_{-\infty}^{0}e^{-t}\frac{\sin t}{t}\,\text{d}t.[/imath] Is it correct? or am I missing something? | 2521153 | Integrating :[imath]\int_{0}^{1} \frac{x^{b}-x^{a}}{\log(x)}\sin(\log(x))dx[/imath]
Consider [imath]\int_{0}^{1} \frac{x^{b}-x^{a}}{\log(x)}\sin(\log(x))dx[/imath] I thought about making some substitution like : [imath]x^{y}[/imath](it's easy to calculate [imath]\int_{0}^{1}\frac{x^{b}-x^{a}}{\log(x)}dx = \int_{a}^{b}dy\int_{0}^{1}x^{y}dx=\log(b-1)-\log(a-1)[/imath]). But I couldn't find substitution like that. Any hints? |
2833316 | Showing the existence of an element [imath]g \in G[/imath], a finite abelian group, where [imath]g[/imath] has order [imath]LCM\{|x|: x \in G\}[/imath]
So I, more or less, need help understanding this problem. The first thing I have tried is to pick a specific group: [imath]\mathbf{Z}_5[/imath], under multiplication. To demonstrate, say we pick element [imath]2[/imath] and [imath]|2| = 4[/imath], and then the only other possible orders are [imath]\{1,4\}[/imath]. Indeed, [imath]4 = LCM(1,4)[/imath]. Not really quite sure, how to generalize this, yet. Any hints? What shall I think about? Definitely prime factorization of each element, hmm... I think! | 2002102 | Finite abelian group contains an element with order equal to the lcm of the orders of its elements
I will quote a question from my textbook, to prevent misinterpretation: Let [imath]G[/imath] be a finite abelian group and let [imath]m[/imath] be the least common multiple of the orders of its elements. Prove that [imath]G[/imath] contains an element of order [imath]m[/imath]. I figured that, if [imath]|G|=n[/imath], then I should interpret the part with the least common multiple as [imath]lcm(|x_1|,\dots,|x_n|)=m[/imath], where [imath]x_i\in G[/imath] for [imath]0\leq i\leq n[/imath], thus, for all such [imath]x_i[/imath], [imath]\exists a_i\in\mathbb{N}[/imath] such that [imath]m=|x_i|a_i[/imath]. I guess I should use the fact that [imath]|x_i|[/imath] divides [imath]|G|[/imath], so [imath]\exists k\in \mathbb{N}[/imath] such that [imath]|G|=k|x_i|[/imath] for all [imath]x_i\in G[/imath]. I'm not really sure how to go from here, in particular how I should use the fact that [imath]G[/imath] is abelian. |
2833618 | Prove that if [imath]a_n and \sum a_n and \sum c_n are convrgent then \sum b_n converge[/imath]
First of all argument bounded and monotone will not work because there is no restriction on sequences to be positive or negative, they can alternate irregularly. I tried to make a use of comparison test but there was too many cases to consider and it was just too messy. So I think the only way to approach the problem is directly from the definition. All I need to do is to prove that the sequence of partial sums [imath]B_n=\sum^n_{i=0} b_i[/imath] is convergent sequence. So I tried to prove that [imath]B_n[/imath] is Cauchy because this seemed easier. [imath]a_n<b_n<c_n \\ \therefore \ a_n+a_{n+1}+...+a_{n+m}<b_n+b_{n+1}+...+b_{n+m}<c_n+c_{n+1}+...+c_{n+m}\\ \therefore \ A_{n+m}-A_n<B_{n+m}-B_n<C_{n+m}-C_n[/imath] and by sandwich this goes to zero [imath]B_{n+m}-B_n[/imath] goes to zero as [imath]n[/imath] tend to infinity. The problem is that I need no this expressions but modulus of them and in this case it does not need to be true for example if [imath]A_{n+m}-A_n<0<B_{n+m}-B_n[/imath]. And here I am stuck | 549225 | Proof of sandwich/squeeze theorem for series.
I am interested in proving a theorem, which I suppose one may call a sandwich or squeeze theorem for series. Suppose we have three series: [imath]\sum^{\infty}_{n=1}a_{n}[/imath], [imath]\sum^{\infty}_{n=1}b_{n}[/imath] and [imath]\sum^{\infty}_{n=1}c_{n}[/imath]. We know that [imath]\sum^{\infty}_{n=1}a_{n}[/imath] and [imath]\sum^{\infty}_{n=1}c_{n}[/imath] converge; furthermore, let us assume that for all [imath]n\in\mathbb{N}[/imath], the following inequality holds: [imath]a_{n}<b_{n}<c_{n}[/imath]. Then, the series [imath]\sum^{\infty}_{n=1}b_{n}[/imath] will also converge. The only way that I can think of to approach the proof of the above would be via the Cauchy criterion, i.e. showing that [imath]\forall_{\varepsilon>0}\,\exists_{n_{\varepsilon}}\,\forall_{m>k>n_{\varepsilon}}\quad|b_{k+1}+...+b_{m}|<\varepsilon.[/imath] As I understand, in order to show that, we would have to somehow bound [imath]\left|\sum^{m}_{n=k+1}b_{n}\right|[/imath] by [imath]\left|\sum^{m}_{n=k+1}c_{n}\right|[/imath] and/or [imath]\left|\sum^{m}_{n=k+1}a_{n}\right|[/imath]. If we assume nonnegativity of the terms of [imath]\sum{}b_{n}[/imath] and [imath]\sum{}c_{n}[/imath], the task becomes trivial. However, without this assumption, I am having problems with finding the right bound. I would be thankful for some hints on how this could be done, or possibly advice on a different approach to the proof. Thank you in advance. |
2833248 | Find [imath]x[/imath] if [imath]\frac1{\sin1°\sin2°}+\frac1{\sin2°\sin3°}+\cdots+\frac1{\sin89°\sin90°} = \cot x\cdot\csc x[/imath]
If [imath]\dfrac1{\sin1°\sin2°}+\dfrac1{\sin2°\sin3°}+\cdots+\dfrac1{\sin89°\sin90°} = \cot x\cdot\csc x[/imath] and [imath]x\in(0°,90°)[/imath], find [imath]x[/imath]. I tried writing in [imath]\sec[/imath] form but nothing clicked. Any ideas? | 425966 | Finite Series - reciprocals of sines
Find the sum of the finite series [imath]\sum _{k=1}^{k=89} \frac{1}{\sin(k^{\circ})\sin((k+1)^{\circ})}[/imath] This problem was asked in a test in my school. The answer seems to be [imath]\dfrac{\cos1^{\circ}}{\sin^21^{\circ}}[/imath] but I do not know how. I have tried reducing it using sum to product formulae and found out the actual value and it agrees well. Haven't been successful in telescoping it. |
2551591 | [imath]X[/imath] and [imath]Y[/imath] are independent random variables,then density function of [imath]\frac{X+Y}{3} [/imath]?
[imath]X[/imath] and [imath]Y[/imath] are independent random variables each having the density [imath]f(t) = \frac{1}{\pi(1+t^2)} , -\infty < t < \infty[/imath]. Then the density function of [imath]\frac{X+Y}{3}[/imath] is? Let [imath]Y =U[/imath] and [imath]X+Y = V[/imath] then I was thinking of calculating the Joint distribution of [imath] U , V [/imath] and then calculating the marginal density function [imath]f_{V}[/imath]! So [imath]f(u,v) = |J| . f(x,y)[/imath] where the Jacobian is [imath]-3[/imath] hence [imath]|J| = 3.[/imath] So that [imath]f(u,v) = 3 . f(x,y) = 3. f(x) . f(y) = 3 . \frac{1}{\pi (1 + x^2)}\frac{1}{\pi (1 + y^2)} =3 . \frac{1}{\pi (1 + (3v - u)^2)}\frac{1}{\pi (1 + u^2)} [/imath] And finally [imath]f(v) = \int_{-\infty}^{\infty} \frac{3}{\pi^2 (1 + u^2)(1 + (3v - u)^2)} du[/imath] Now, are my above steps correct?. Also, it is becoming a bit complicated any other simpler method? EDIT It's a problem from previous year competition in our institute which had four options as answer - [imath]a) \frac{6}{\pi (4 + 9t^2)}[/imath] [imath]b) \frac{6}{\pi (9+4t^2)}[/imath] [imath]c) \frac{3}{\pi (1 + 9t^2)}[/imath] [imath]d) \frac{3}{\pi (9 + t^2)}[/imath] and usually there is a small amount of time for each question any reverse mathematics procedure or any tricks may be there apart from a few lengthy calculations, yes from answers I now know that characteristic functions are important to know about various properties of distributions but to a beginner in probability if unaware of Cauchy Random variable? | 2519667 | [imath]X[/imath] and [imath]Y[/imath] are independent rv having pdf [imath]f(t)=\frac{1}{\pi} \frac{1}{1+t^2}[/imath] determine pdf of [imath]Z=\frac{X+Y}{3}[/imath]
Suppose [imath]X[/imath] and [imath]Y[/imath] are independent random variables on [imath]\mathbb{R}[/imath] having pdf [imath]f(t)=\frac{1}{\pi} \frac{1}{1+t^2}[/imath]. Define [imath]Z=\frac{X+Y}{3}[/imath] determine the pdf of [imath]Z[/imath]. So the pdf of [imath](X,Y)[/imath] is [imath]f(x,y)=\frac{1}{\pi^2 (1+x^2)(1+y^2)}[/imath]. We shall find the CDF of [imath]Z[/imath] and then differentiate it. [imath]F(z)=P(Z \leq z) = \int_{-\infty}^{\infty} \int_{-\infty}^{3z-x} \ \frac{1}{\pi^2 (1+x^2)(1+y^2)} \ dy \ dx= \int_{-\infty}^{\infty} \frac{1}{\pi^2}\left(\frac{\pi}{2(1+x^2)}- \frac{\tan^{-1}(x-3z)}{1+x^2}\right) \ dx[/imath]. At this stage I am having trouble evaluating [imath]\int_{-\infty}^{\infty} \left(\frac{\tan^{-1}(x-3z)}{1+x^2}\right) \ dx[/imath]. How do I do this? Or otherwise is there different more elegant way to get the required pdf? |
2833717 | Show that [imath]C/c \cong (c/(C^{op}))^{op}[/imath]
I have trouble seeing that [imath]C/c \cong (c/(C^{op}))^{op}[/imath]. This is what I have managed so far: What am I doing wrong here? It seems to me that to get to [imath](c/(C^{op}))^{op}[/imath], would have to reverse the [imath]h[/imath], and then it no longer looks like [imath]C/c[/imath] (Edit) This is just to sum up what my mistake was, in case others fall in the same trap: What I missed was the fact, that (see diagram) in [imath]c/C^{op}[/imath], not only [imath]f[/imath] and [imath]g[/imath] get reversed (ie. [imath]f \rightarrow f^{op}[/imath] etc), but [imath]h[/imath] too becomes [imath]h^{op}[/imath]. Why did I not see that? Well, my thinking was along the lines "[imath]C = \cdot \xrightarrow{f} \cdot[/imath]", so "[imath]C^{op} = \cdot \xrightarrow{f^{op}} \cdot[/imath]", and to contruct the slice category, you create morphisms, let's call them [imath]h[/imath] ..., - but of course, those morphisms are between objects in the categories that are getting sliced (if that's the right term), so the same relationship applies here: [imath]h \rightarrow h^{op}[/imath]. Thanks for kindly helping me see this! | 2737731 | Show that [imath]\text{C}/c \cong (c/(\text{C}^{\text{op}}))^{\text{op}}[/imath]
I'm self studying from Category Theory in Context by Emily Riehl and have encountered the following question regarding slice categories: Let [imath]\text{C}[/imath] be a category and [imath]c[/imath] be an object in [imath]\text{C}[/imath]. Show that [imath]\text{C}/c \cong (c/(\text{C}^{\text{op}}))^{\text{op}}[/imath] The exercise appears in an chapter that introduces the notion of opposite categories and duality proofs. My intuition of why the statement is true comes from the following observation: Let [imath]x,y[/imath] be objects in [imath]\mathbf{C}[/imath] (or equivalently [imath]\mathbf{C}^{\text{op}}[/imath]). Let [imath]g^{\text{op}} : c \to y[/imath] and [imath]f^{\text{op}} : c \to x[/imath] be objects in [imath]c/(\mathbf{C}^{\text{op}})[/imath]. By the definition of slice category, there exists a morphism [imath]h^{\text{op}} : y \to x[/imath] in [imath]\mathbf{C}^{\text{op}}[/imath] such that the diagram below commutes: [imath]\hskip2in[/imath] Now the 'corresponding diagram' in the opposite category [imath](c/(\mathbf{C}^{\text{op}}))^{\text{op}}[/imath] would be: [imath]\hskip2in[/imath] which, as I understand it, would also be commutative. Now, to me, this indicates: that the objects in [imath](c/(\mathbf{C}^{\text{op}}))^{\text{op}}[/imath] are the same as the objects in [imath]\mathbf{C}/c[/imath], namely, a morphism in [imath]\mathbf{C}[/imath] from an object [imath]z \in \mathbf{C}[/imath] to [imath]c[/imath]. the commutativity of the seconds diagram describes morphisms [imath]h[/imath] in both [imath]\mathbf{C}/c[/imath] and [imath](c/(\mathbf{C}^{\text{op}}))^{\text{op}}[/imath] if you reverse the arrows again you return to the same diagram. This makes me think you could construct a map between morphisms by first making a map from [imath] \mathbf{C}/c \to (\mathbf{C}/c)^{\text{op}} ( \cong c/(\mathbf{C}^{\text{op}}) )[/imath] by [imath]h \mapsto h^{op}[/imath] Now, what I am really looking for is a way to structure a proof. I feel like I have some intuition but am very unsure of how to write it down in a concise way. I am new to category theory so I'm still very much learning how to construct these types of proofs using duality; I feel like what I have written is maybe overkill but I'm trying to be as explicit as possible to make sure I understand everything. In particular, I don't know how I would get a concrete correspondence between morphisms in the two categories that respects associativity of composition etc. Any help or pointers would be appreciated! |
2834402 | Interesting Recursive (Telescoping?) Inequality
Let [imath]x_0 = 5,[/imath] and [imath]x_{n+1} = x_n + \frac {1} {x_n}[/imath] for n = [imath]0, 1, 2, . . ..[/imath] Show that [imath]45 < x_{1000} < 45.1.[/imath] First noticing that [imath]x_{n+1}-x_n= \frac {1}{x_n},[/imath] we get [imath]\sum \limits_{n=1}^{1000} \frac{1}{x_n} = x_{1000} - 5[/imath] with some simple telescoping. Now we want to find some inequalities for [imath]\sum \limits_{n=1}^{1000} \frac{1}{x_n},[/imath] which I tried to do with more telescoping, but I've had no luck finding a good way to express [imath]\frac{1}{x_n}[/imath] between two bounds. | 2117757 | Show that [imath]45[/imath]
If [imath]x_0 = 5[/imath] and [imath]x_{n+1} = x_n + \frac {1}{x_n},[/imath] show that [imath]45<x_{1000}<45.1[/imath] This problem is taken from the list submitted for the [imath]1975[/imath] Canadian Mathematics Olympiad (but not used on the actual exam). SOURCE : CRUX(Page Number 3 ; Question Number 162) I tried writing out the first few terms : [imath]x_1 = 5+ \frac{1}{5} [/imath] [imath]x_2 = \big(5+\frac{1}{5}\big) + \big(5+\frac{1}{5}\big)^{-1} = \frac{x_0^2 + 1}{x_0} + \frac{x_0}{x_0^2 + 1} = \frac{(x_0^2 + 1)^2+x_0^2}{x_0(x_0^2+1)}[/imath] [imath]x_3 = \frac{(x_0^2 + 1)^2+x_0^2}{x_0(x_0^2+1)} + \frac{x_0(x_0^2+1)}{(x_0^2 + 1)^2+x_0^2} = Messy[/imath] I tried a lot but could not find any general formula for the [imath]n[/imath]th term. Does there even exist any? Also it is clear that [imath]\big(x_n + \frac{1}{x_n}\big)[/imath] is an increasing function. So I think sequence diverges, but how can the [imath]1000th[/imath] term be calculated or aprroximated? Any help would be gratefully acknowledged :). |
2834459 | [imath]AB[/imath] is invertible but [imath]A[/imath] or [imath]B[/imath] is not invertible.
Searching for an example: A Banach Space [imath]X[/imath], [imath]A,B[/imath] are operators on [imath]X[/imath] where [imath]AB[/imath] is invertible but [imath]A[/imath] or [imath]B[/imath] is not invertible. For finite dimensional case we know that result so the example will come from infinite dimension. Thank You. | 1302459 | [imath]A[/imath] and [imath]B[/imath] are bounded linear operators from the normed linear space [imath]X[/imath] to itself. If [imath]AB[/imath] is invertible are [imath]A[/imath] and [imath]B[/imath] invertible?
I think I understand the proof for square matrices, such that [imath](AB)^{-1}=B^{-1}A^{-1}[/imath], but I'm not sure if I can just say the same for the bounded linear operators A and B. Does the existence of [imath](AB)^{-1}:X\rightarrow X[/imath] imply the existence of [imath]B^{-1}[/imath] and [imath]A^{-1}[/imath]? |
1302793 | Why is an [imath]n[/imath]th power a norm in a Kummer extension?
Let [imath]F[/imath] be a [imath]p[/imath]-adic field containing the [imath]n[/imath]th roots of unity. Then by Kummer theory, [imath][F^{\ast} : F^{\ast n}][/imath] (which is finite) is equal to the cardinality of [imath]\textrm{Gal}(E/F)[/imath], where [imath]E[/imath] is [imath]F[/imath] adjoined with all the [imath]n[/imath]th roots of unity in [imath]F[/imath]. Also [imath]F^{\ast}/N_{E/F}(E^{\ast})[/imath] has cardinality [imath]|\textrm{Gal}(E/F)|[/imath] by local class field theory. This would give us [imath]N_{E/F}(E^{\ast}) = F^{\ast n}[/imath] provided we could show just one inclusion. Why is it that every element of [imath]F^{\ast n}[/imath] is a norm? | 1304415 | Why is [imath]K^{\ast n}[/imath] contained in the norm group?
http://www.bprim.org/cyclotomicfieldbook/rlmain.pdf In section 5, [imath]K[/imath] is a local [imath]p[/imath]-adic field containing the [imath]n[/imath]th roots of unity, and [imath]L = K(\sqrt[n]{x} : x \in K^{\ast})[/imath]. Kummer theory tells us that [imath][L : K] = [K^{\ast} : K^{\ast n}][/imath], and local class field theory tells us that [imath][L : K] = [K^{\ast} : N_{L/K}(L^{\ast})][/imath]. The author concludes that [imath]N_{L/K}(L^{\ast}) = K^{\ast n}[/imath], but I don't see how one of these things is contained in the other. Why are [imath]n[/imath]th powers norms from [imath]L[/imath], for example? |
2830201 | Prove that [imath]b\cdot\aleph_0=b[/imath], where b is an infinite set.
Prove that [imath]b\cdot\aleph_0=b[/imath], where [imath]b[/imath] is an infinite set. I tried by multiplying [imath]1\le\aleph_0\le b[/imath] with [imath]b[/imath] but then I have to prove that [imath]b\cdot b=b[/imath] and that seems harder than my original problem. | 690798 | How prove this equation [imath]\alpha=\aleph_{0}\alpha[/imath]
Question: let [imath]\alpha[/imath] be a infinite cardinal , show that: [imath]\alpha=\aleph_{0}\alpha[/imath] where [imath]\aleph_{0}[/imath] is the cardinality of the natural numbers is denoted aleph-null this equation is from a book,and I can't prove it,and the author say this is important equation,and can't prove it.Thank you for you help |
2834687 | Prove that a specific fixed point iteration is locally convergent
Let [imath]g : I \rightarrow \mathbb{R}[/imath] be a [imath]C^1[/imath] map such that [imath]g'(x) \ne 0[/imath] for any [imath]x[/imath] in [imath]I[/imath]. Assume that there exists [imath]r \in I[/imath] such that [imath]g(r)=0[/imath]. Prove that for [imath]\eta \in I[/imath] sufficiently close to [imath]r[/imath] then the fixed point iteration [imath]x_{k+1}=x_k-\frac{g(x_k)}{g'(x_k)},\,x_0=\eta[/imath] satisfies [imath]\lim_{k\rightarrow\infty}x_k=r[/imath]. I noticed that this is Newton's method for finding the roots of [imath]g[/imath], but I couldn't manage yet to prove the desired conclusion. | 786453 | Proof that Newton Raphson method has quadratic convergence
I've googled this and I've seen different types of proofs but they all use notations that I don't understand. But first of all, I need to understand what quadratic convergence means, I read that it has to do with the speed of an algorithm. Is this correct? Ok, so I know that this is the Newton-Raphson method: [imath]x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}[/imath] How do I prove that it converges? Thanks. |
1842693 | When is [imath]\sqrt{x/y^2}[/imath] equal to [imath]\sqrt{x}/y[/imath]?
The solution to the quadratics is given by [imath]r = -\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}[/imath], which is shortened to [imath]r = -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}[/imath], but I'm wondering how if this is justified, given that [imath]4a^2[/imath] can be negative if [imath]a \in \mathbb{C}[/imath], and [imath]\sqrt{\dfrac{x}{y}} \neq \dfrac{\sqrt{x}}{\sqrt{y}}[/imath] if [imath]x[/imath] and [imath]y[/imath] are negative, but given that we have [imath]a^2[/imath], is this justified? Is [imath]\sqrt{x/y^2} = \sqrt{x}/y[/imath]? Is [imath]r = -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}[/imath] always true? | 2831645 | Why doesn't the quadratic equation contain [imath]2|a|[/imath] in the denominator?
When deriving the quadratic equation as shown in the Wikipedia article about the quadratic equation (current revision) the main proof contains the step: [imath] \left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}} [/imath] the square root is taken from both sides, so why is [imath]\sqrt{4a^2} = 2a[/imath] in the denominator and not [imath] \sqrt{4a^2} = 2\left |a \right | [/imath] Could somebody explain this to me? Thank you very much |
2835469 | Solve [imath]a^b - b^a = 3[/imath]
For all positive integers [imath]a,b[/imath], given [imath]a^b - b^a = 3[/imath] Find all possible values of [imath]a[/imath] and [imath]b[/imath] I tried to simulate this using a program and found the one fit [imath]a = 4[/imath] and [imath]b = 1[/imath]. Would love to see a Mathematical solution for it. Please help. | 2065448 | Find integral solution of [imath]a^{b} - b^{a} = 3[/imath]
Find the integral solution of [imath]a^b - b^a = 3[/imath] I am a student of class 10 and got this question in a maths competetive exam. I have tried converting into a log equation in one variable, or solve it by using parity but couldn't solve it. Thanks in advance |
2522949 | Positive integers relatively prime to all the terms
Determine all positive integers relatively prime to all the terms of the infinite sequence [imath]a_n = 2^n + 3^n + 6^n − 1, n ≥ 1.[/imath] | 2803320 | IMO 2005 No 4. : All integers relatively prime to terms of infinite sequence
Here's No. 4 from the 2005 IMO. Q: Determine all positive integers relatively prime to all the terms of the infinite sequence [imath]a_n=2^n+3^n+6^n-1, n\geq1[/imath] I found the solution here, but I'm failing to understand it. It reads: Trivially, [imath]a_2 = 48[/imath] is divisible by 2 and 3. Okay, I get this, assuming they just picked n=2 since it's a small number. Let therefore p > 3 be a prime number. Wait what? What does p represent - does it represent n, or n+2? And from the previous step, how can we assume this? Then [imath]6a_{p−2} = 3·2^{p−1} + 2·3^{p−1} + 6^{p−1} −6 ≡ 3 + 2 + 1−6 ≡ 0 (\mathsf {mod \ p})[/imath]. Alone, I think I understand this line, although why multiply by 6? Is [imath]6a_{p-2}[/imath], or any term you would place instead, congruent to [imath]\mathsf {0(mod \ p)}[/imath] since that fulfills the definition of a prime number? And so the only such number is 1. Only such number refers to what number? Proved. So... what would be the answer? Thank you for helping! |
2834466 | Show that [imath]A[/imath] defined by a continuous function is open
Consider a continuous function defined all of [imath]R^n[/imath] i.e., [imath]h:R^n \to R[/imath]. Also obtain the infimum of [imath]h[/imath] [imath] a: = \inf_{x\in R^n} h(x) [/imath] Define the set [imath]A := \{ x \in R^n : h(x) > a \}[/imath]. Is [imath]A[/imath] open? I was trying to show that it is open. If [imath]A[/imath] is all of [imath]R^n[/imath] then it is open. We will consider the case when [imath]A \subset R^n[/imath]. If [imath]A[/imath] is a strict subset of [imath]R^n[/imath] then its complement exists. Denote it by [imath]A^c[/imath]. Also, since [imath]h(x)[/imath] is defined over all [imath]R^n[/imath] and [imath]a[/imath] is the infimum, we can obtain [imath]A^c[/imath] explicitly as [imath] A^c = \{ x \in R^n : h(x) = a \}[/imath] It remains to show that [imath]A^c[/imath] is closed. Suppose that [imath]A^c[/imath] is not closed. Then there is a sequence of points in [imath]A^c[/imath]: [imath]x_1, x_2, \dots[/imath] such that [imath]\lim_{n\to \infty} x_n \notin A^c[/imath]. This is a contradiction since [imath]h[/imath] is continuous and [imath] h(x_n) = a[/imath] for every [imath]n[/imath], and its limit is [imath]a[/imath]. Did I get this right? Is there a better proof? | 1671096 | Show that, for every real number a, the set {[imath]x ∈ S | f(x) ≤ a[/imath]} is a closed subset of [imath]\mathbb R^d[/imath]
Suppose that [imath]S[/imath] is a closed subset of [imath]\mathbb R^d[/imath] and that [imath]f[/imath] is a continuous real-valued function with domain [imath]S[/imath]. Show that, for every real number [imath]a[/imath], the set {[imath]x ∈ S | f(x) ≤ a[/imath]} is a closed subset of [imath]\mathbb R^d[/imath] I don't know how to start this. Do we use the fact that If [imath]S[/imath] is closed and bounded every sequence in [imath]S[/imath] has a subsequence that converges to a point in [imath]S[/imath]? |
2527871 | Is [imath]\mathbb{R}[/imath] isomorphic to [imath]\mathbb{R}(x)[/imath]?
Is [imath]\mathbb{R}[/imath] isomorphic (as a field) to [imath]\mathbb{R}(x)[/imath]? (where [imath]x[/imath] is an indeterminate on [imath]\mathbb{R}[/imath]) | 1897258 | Isomorphism between [imath]\Bbb R[/imath] and [imath]\Bbb R(X)[/imath]?
My questions are: [imath]1.[/imath] Is there a field morphism [imath]\Bbb R(X) \hookrightarrow \Bbb R[/imath] ? [imath]2.[/imath] If the answer to [imath]1.[/imath] is "yes", are [imath]\Bbb R[/imath] and [imath]\Bbb R(X)[/imath] isomorphic as fields? [imath] [/imath] [imath] [/imath] For [imath]1.[/imath], here are some comments: This is true for [imath]\Bbb C[/imath], see Can I embed [imath]\Bbb{C}(x)[/imath] into [imath]\Bbb{C}[/imath]?. The key point here is the fact that [imath]\Bbb C[/imath] is algebraically closed, of cardinality [imath]2^{\aleph_0}[/imath], and of characteristic [imath]0[/imath]. The same argument doesn't work for [imath]\Bbb R[/imath]. We can actually find a subfield [imath]F[/imath] of [imath]\Bbb R[/imath] such that [imath]F \cong F(X) \cong F(X,Y) \cong \cdots[/imath], see Is there a subfield [imath]F[/imath] of [imath]\Bbb R[/imath] such that there is an embedding [imath]F(x) \hookrightarrow F[/imath]?. Notice that [imath]\Bbb Q(X)[/imath] does not embed in [imath]\Bbb Q[/imath] (as a field). This is equivalent to: Is there an injective ring morphism [imath]\Bbb R[X] \hookrightarrow \Bbb R[/imath] ? (By taking the fraction fields). As for 2. : They are isomorphic as abelian groups, and actually as [imath]\Bbb Q[/imath]-vector spaces. They are not isomorphic as [imath]\Bbb R[/imath]-algebras, because they haven't the same transcendence degree over [imath]\Bbb R[/imath]. Let's try to see how to construct such a field isomorphism. We could try[imath]^{[1]}[/imath] to find a subset [imath]B[/imath] of [imath]\Bbb R[/imath] such that [imath]\Bbb R = \Bbb Q(B)[/imath] and [imath]C \subsetneq B \implies \Bbb R \neq \Bbb Q(C)[/imath]. The set [imath]B[/imath] must contain transcendental elements, pick such an [imath]x_0 \in B[/imath]. Then [imath]\Bbb R = \Bbb Q(B) = \Bbb Q(B \setminus \{x_0\})(x_0)[/imath] and we could try[imath]^{[2]}[/imath] to find a field isomorphism [imath]\Bbb Q(B \setminus \{x_0\}) \cong \Bbb Q(B)[/imath], so that [imath]\Bbb Q(B \setminus \{x_0\})(x_0) \cong \Bbb Q(B)(x_0) \cong \Bbb R(X)[/imath]. For [imath]{[1]}[/imath], I tried to use Zorn's lemma on [imath]\mathscr B = \{B \subset \Bbb R \mid \Bbb R = \Bbb Q(B)\}[/imath] and the partial order [imath]B_1 ≤ B_2 \iff B_1 \supset B_2[/imath]. But this is not clear to me why this should be an inductive set. For [imath]{[2]}[/imath], my idea was to define [imath]f : \Bbb Q(B \setminus \{x_0\}) \cong \Bbb Q(B)[/imath] by choosing a bijection [imath]b : B \setminus \{x_0\} \to B[/imath] (these sets should be uncountable), and then trying to extend [imath]f\vert_{B \setminus \{x_0\}} = b[/imath] to a field morphism. Edit for [imath]^{[1]}[/imath]: in this article (MINIMAL GENERATING SETS OF GROUPS, RINGS, AND FIELDS, by LORENZ HALBEISEN, MARTIN HAMILTON, AND PAVEL RUZICKA) the theorem 2.4. proves that [imath]\Bbb R[/imath] has no minimal generating set as field (or [imath]\Bbb Q[/imath]-algebra, i.e. as ring). The argument uses Artin-Schreier theorem. Minimal generating sets of modules are also discussed here. I think that [imath]B[/imath] is a minimal generating set of a subfield [imath]K \subset \Bbb R[/imath] over [imath]\Bbb Q[/imath] if and only if any element of [imath]B[/imath] cannot be written as a rational fraction of other elements of [imath]B[/imath]. Under these conditions and using (transfinite) induction, we can probably prove (and this was my first idea) that any function [imath]f: B \to \Bbb C[/imath] such that [imath]b[/imath] and [imath]f(b)[/imath] are conjugate over [imath]\Bbb Q[/imath] (i.e. have the same minimal polynomial) extends to a (unique) field morphism [imath]\sigma : K = \Bbb Q(B) \to \Bbb C[/imath]. Actually, the collection [imath]\mathscr B[/imath] defined above doesn't have to be an inductive set, at least if we replace [imath]\Bbb R[/imath] by [imath]K =\Bbb Q(\sqrt 2, \sqrt[4]{2},\sqrt[8]{2},\dots)[/imath], because the descending chain [imath]\left(E_n = \{\sqrt[2^m]{2} \mid m \geq n\}\right)_{n \geq 1}[/imath] satisfy [imath]K = \Bbb Q(E_n)[/imath], but [imath]K \neq \Bbb Q\left(\bigcap_{n≥1} E_n\right)=\Bbb Q[/imath], so the chain has no upper bound w.r.t. [imath]≤[/imath] in [imath]\mathscr B[/imath]. |
2834578 | Existence of biholomorphic function for two fixed points with [imath]f(z_1) = z_2[/imath]
Let [imath]U \subset C[/imath] be a simply connected open subspace and [imath]z_1, z_2 \in U[/imath]. I want to show that then there exists a biholomorphic function [imath]f: U \rightarrow U[/imath] with [imath]f(z_1) = z_2[/imath], but I have no idea how to do so. | 1582962 | Finding univalent function with [imath]f(z_1)=z_2[/imath]
Let [imath]\Omega[/imath] be a simply-connected domain. Let [imath]z_1,z_2\in\Omega[/imath]. Prove that exists an univalent function such that [imath]f(\Omega)=\Omega[/imath] and [imath]f(z_1)=z_2[/imath]. Since [imath]\Omega[/imath] is simply connected, one can apply Riemann mapping theorem to get [imath]g:\Omega\to\mathbb{D}[/imath] ([imath]\mathbb{D}[/imath] is the unit disk) with [imath]g(z_1)=0,g(\Omega)=\mathbb{D}[/imath]. I thought taking [imath]f(z)=g^{-1}(g(z)+g(z_2))[/imath] thus [imath]f(z_1)=g^{-1}(g(z_1)+g(z_2))=z_2[/imath]. On the other hand, [imath]f[/imath] does not map [imath]\Omega[/imath] to itself. How can I build a map which will satisfy the requirements above? |
2835430 | Wrong log reasoning?
From this question, we are given a system of equations and a restriction: Let [imath]a,b,c > 0[/imath] for [imath]\begin{cases} \log_a(b^x) & = & 2 \\ \log_b(c^x) & = & 2 \\ \log_c(a^x) & = & 5 \end{cases}[/imath] [imath]x =\,\,?[/imath] [imath]\log_a(b^x) = 2 \iff b^x=a^2[/imath] We now have [imath] b^x = a^2 \\ c^x = b^2 \\ a^x = c^5 [/imath] Adding these up and rearranging, we have [imath]a^x + b^x + c^x = a^2 + b^2 + c^5[/imath] Now [imath]x[/imath] does not seem to have a consistent solution. How is this possible? A comment suggested this might be due to an over-determined system, but this answer shows that it has the unique solution [imath]x^3 = 20[/imath]. From the original post, we have the following question (which was not answered) Looks to me like x is equal 2 different numbers at the same time which is strange, what am I doing wrong here? | 2827632 | A system of logarithmic equations with parameters
[imath]Let[/imath] [imath]a,b,c>0[/imath] [imath]\begin{cases}\log_a(b^x)=2\\\log_b(c^x)=2\\\log_c(a^x)=5\end{cases}[/imath] [imath]x=?[/imath] So my attempt is just to use the logarithmic definition: [imath]\log_a(b^x)=2\iff b^x=a^2[/imath] By similar logic, [imath]a^x=c^5[/imath] [imath]c^x=b^2[/imath] So, if add everything together, we should get: [imath]a^x+b^x+c^x=a^2+b^2+c^5[/imath] Looks to me like x is equal 2 different numbers at the same time which is strange, what am I doing wrong here? I'm going to be a maths student in the upcoming year, this is taken from the Tel-Aviv university preparation material - shouldn't be too complex. |
2834209 | Characterization of Topology by Boundary (solved)
Similar to my earlier question I try to understand the characterization of topology by different approaches. There is the "classical" one with open sets, the one with closed sets, the neighborhood characterization and one with closure and interior operator, respectively. (In addition, there is also a possibility to define a topology with respect to filter convergence but I did not dig into that yet. Are there further approaches besides the ones mentioned?). Since [imath]\mathrm{boundary}(A) = \mathrm{closure}(A)\setminus\mathrm{interior}(A)[/imath] it should be possible to characterize a topology by a boundary operator only. My thoughts so far: Theorem. Suppose that [imath]\mathcal T\subseteq2^X[/imath] is a family of sets and [imath]\partial:2^X\rightarrow 2^X[/imath]. Then [imath]\mathcal T[/imath] is a topology [imath]\partial A := \mathrm{closure}(A)\setminus\mathrm{interior}(A)[/imath] iff [imath]\partial[/imath] satisfies [imath]\partial\partial A\subseteq\partial A[/imath], [imath]\partial(A\cup B)\subseteq\partial A\cup\partial B[/imath], [imath]A\subseteq B[/imath] implies [imath]\partial A\subseteq \partial B\cup B[/imath], [imath]\partial A = \partial(X\setminus A)[/imath] and [imath]\mathcal T = \{U\subseteq\Omega : \partial U\cap U = \emptyset\}[/imath]. I managed to show "[imath]\Rightarrow[/imath]". I struggle with the reverse direction however. That [imath]\emptyset,X\in\mathcal T[/imath] is easy. But I could not manage to show the arbitrary unions and finite intersections. My attempt: Let [imath]A_i\in\mathcal T[/imath] for [imath]i\in I[/imath], [imath]I[/imath] finite. Then [imath]\partial\left(\bigcap_{i\in I}A_i\right) \cap \left(\bigcap_{i\in I}A_i\right) \subseteq \left(\bigcap_{i\in I}A_i\right)\cap(\partial A_i\cup A_i)[/imath] for each [imath]i\in I[/imath] by the third property. But Since [imath]A_i\in\mathcal T[/imath] it follows that [imath]\partial A_i\cap A_i =\emptyset[/imath]. So we have [imath]\partial\left(\bigcap_{i\in I}A_i\right) \cap \left(\bigcap_{i\in I}A_i\right) =\emptyset[/imath]. Is this correct? Let [imath]A_i\in\mathcal T[/imath] now for all [imath]i\in I[/imath] with arbitrary [imath]I[/imath]. Then [imath]\emptyset = A_i\cap\partial A_i[/imath] for all [imath]i[/imath] by assumption. Thus, [imath]\begin{align*}\emptyset &= \bigcup_{i\in I}(A_i\cap\partial A_i)\\&=\bigcup_{i\in I}(A_i\cap\partial A_i)\cap\partial\left(\bigcup_{i\in I}A_i\right)\\&\supseteq\bigcup_{i\in I}A_i\cap\partial\left(\bigcup_{i\in I}A_i\right)\end{align*}[/imath] because intersection with emptyset is emptyset. So, [imath]\bigcup_{i\in I}A_i\in\mathcal T[/imath]. Note In the other questions for which my question was flagged as duplicate answer this questions by showing that there exists a one-to-one relation between boundary and closure. I however wanted to prove that boundary characterizes a topology straight. | 1390194 | Axiomatization of a Topology using the Boundary
The axioms of a topological space are usually stated in the "open set" form: A topological space [imath]X[/imath] has a set of subsets [imath]\tau[/imath] whose members satisfy: [imath]\emptyset[/imath] and [imath]X[/imath] are in [imath]\tau[/imath]. [imath]\tau[/imath] is closed under arbitrary unions. [imath]\tau[/imath] is closed under finite intersections. We will call this the usual open set axiomatization of a topological space. The open set axiomatization is probably the most economical, but of course there are other ways as well, such as the Kuratowski closure axioms which axiomatizes the topology using the closure operator. In his note here, Pete L. Clark gives several alternative characterizations of a topology, most of which have details given. It was mentioned, I think originally by Willard, that it is "possible, but unrewarding to characterize a topology completely by its [boundary]". I could find no further references on this, and I am wondering how it might be done (unrewarding as it may be). So my question is this: How can a topological space by axiomatized using the boundary operation as the primitive notion? A complete answer should give a list of axioms required for the formulation, as well as a sketch of how they are equivalent to the usual open set formulation. |
2835693 | How Gelfond find his limit for [imath]\exp(\pi) [/imath]?
[imath] a_0 = \frac{1}{\sqrt 2} [/imath] [imath] a_{n+1} = \frac{( \sqrt {1 - a_n^2} -1)^2}{a_n^2} [/imath] [imath] \lim_{n \to \infty} \frac{4^{\frac{1}{2^n}}}{a_{n+1}^{\frac{1}{2^n}}} = \exp(\pi) [/imath] How did Gelfond find this nice result ? And how to prove it ? The 4 is trivial , but the rest is not. Is this related to trigonometry ? Is this related to continued fractions ? Are there analogues known for cube roots ? Notice a proof alone might not explain how he found the result. | 2489586 | Origin of rapidly converging sequence for [imath]e^\pi[/imath]
The following sequence converging to Gelfond's constant ([imath]e^\pi[/imath]) is apparently mentioned (and originates?) in the book "Mathematics by Experiment: Plausible Reasoning in the 21st Century.", which I unfortunately do not have access to. [imath]k_0 = 1/\sqrt{2},\quad k_{n+1}={\frac {1-{\sqrt {1-k_{n}^{2}}}} {1+{\sqrt {1-k_{n}^{2}}}}}[/imath] [imath]e^\pi =\lim_{n\to\infty} \left(\frac{4}{k_{n+1}}\right)^{2^{-n}}[/imath] How is this sequence and similar ones derived? |
2836308 | Problem in Arithmetic Mean - Geometric Mean inequality
Let a,b,c be positive real numbers, prove that [imath] \frac{a}{b} + \frac{b}{c} + \frac{c}{a} + \frac{3\sqrt[3]{abc}}{a+b+c} \geq 4[/imath] I am suppose to use AM-GM inequality, I tried [imath] \frac{a}{b} + \frac{b}{c} + \frac{c}{a} \geq 3 [/imath] and [imath] a + b + c\geq 3 \sqrt[3]{abc} [/imath] implying [imath] \frac{3\sqrt[3]{abc}}{a+b+c} \leq 1 [/imath] Now adding the two inequality could give me the desired result but the problem I face is with second inequality sign(it's less than or equal to 1 rather than greater | 189143 | Proving inequality [imath]\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4[/imath]
I started to study inequalities - I try to solve a lot of inequlites and read interesting .solutions . I have a good pdf, you can view from here . The inequality which I tried to solve and I didn't manage to find a solution can be found in that pdf but I will write here to be more explicitly. Exercise 1.3.4(a) Let [imath]a,b,c[/imath] be positive real numbers. Prove that [imath]\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{3\sqrt[3]{abc}}{a+b+c} \geq 4.[/imath] (b) For real numbers [imath]a,b,c \gt0[/imath] and [imath]n \leq3[/imath] prove that: [imath]\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+n\left(\frac{3\sqrt[3]{abc}}{a+b+c} \right)\geq 3+n.[/imath] |
2836538 | Proof that Darboux upper/lower sums always converge to the Darboux upper/lower integral as the partition gets thinner
I'm going through my calculus lecture notes, and at some point the following claim is made: Let [imath]f:\left[a,b\right]\to\mathbb{R}[/imath] be a bound function. For any [imath]\epsilon>0[/imath] there exists some [imath]\delta>0[/imath] s.t. for any partition [imath]P[/imath] of [imath]\left[a,b\right][/imath] for which [imath]\Delta P<\delta[/imath], it holds that [imath]\left|U\left(f,P\right)-\bar{\int}f\left(x\right)dx\right|<\epsilon[/imath] Where [imath]\Delta P[/imath] is the size of the biggest segment in the partition (The text, which is not in English, refers to [imath]\Delta P[/imath] as the “partition parameter”, I don't think this is the correct translation into English since I did not find any reference to it). [imath]U\left(f,P\right)[/imath] is the upper Darboux sum for said partition and [imath]\bar{\int}f\left(x\right)dx[/imath] is the infimum for the upper Darboux sums of all possible partitions. I have failed in proving this, and it seems important, could someone supply a source where this is proved? | 2045722 | Why is the upper Riemann integral the infimum of all upper sums?
I was reading the theory of Riemann integration when I cam across the following , If [imath]f[/imath] is bounded on [imath][a,b][/imath], and [imath]P = \{x_0,x_1,x_2.......x_n\}[/imath] is a partition of [imath][a,b][/imath], let [imath]M_j = \sup_{x_{j-1}\leq x\leq x_j}f(x)[/imath] The upper sum of f over P is [imath]S(P) = \sum_{j=1}^{n} M_j(x_j-x_{j-1})[/imath] and the upper integral of [imath]f[/imath] over [imath][a,b][/imath], denoted by [imath]\int_{a}^{b^-} f(x)dx[/imath] is the infimum of all upper sums. The theorem similarly goes on to state the result for lower sums. My doubt is : I do not understand how is [imath]\int_{a}^{b^-} f(x)dx[/imath] the infimum of all upper sums. I understand that if we refine the partition P, then the upper sum would decrease, so it may be a lower limit for all the upper sums computed on the refinements of P (but still being the lower limit does not prove that it is the infimum) and what about those partitions for which P itself is the refinement of? How do I know that it will be a lower limit for those, let alone a infimum? |
2837049 | What functions are in [imath]C_0(\Omega)[/imath] that are not in [imath]C_c(\Omega)[/imath]?
I use [imath]C_0(\Omega)[/imath] and [imath]C_c(\Omega)[/imath] pretty much interchangeably but I'm trying to be more precise now so I'd like to know what is the difference between these spaces? [imath]C_0[/imath] is commonly referred to as the set of all continuous functions on [imath]\Omega \in \mathbb{R}^n[/imath] that vanish at infinity (or on the boundary). [imath]C_c(\Omega)[/imath] is the space of all continuous functions on [imath]\Omega[/imath] with compact support. It holds that [imath]C_c(\Omega) \subset C_0(\Omega)[/imath]. So what functions are in [imath]C_0(\Omega)[/imath] that are not in [imath]C_c(\Omega)[/imath]? | 1378751 | Compact support vs. vanishing at infinity?
Consider the two sets [imath] C_0 = \{ f: \mathbb R \to \mathbb C \mid f \text{ is continuous and } \lim_{|x|\to \infty} f(x) = 0\}[/imath] [imath] C_c = \{ f: \mathbb R \to \mathbb C \mid f \text{ is continuous and } \operatorname{supp}{(f)} \text{ is bounded}\}[/imath] Aren't these two sets the same? What am I missing? |
2837360 | Simple ordinary differential equations, why is it not defined on this interval?
I'm wondering why this differential equation, [imath]xy'= 1[/imath] has no solution on the interval [imath]I:−1 < x < 1[/imath] per Ordinary Differential Equations by Morris Tenenbaum and Harry Pollard, page 29: example 4.22. From my understanding, this differential equation should have a solution as long as [imath]x ≠ 0[/imath]. It's "family" of solutions would then look like, [imath]y = log|x| + c_1[/imath] (which is given in the book), and seems to be defined for all [imath]x ≠ 0[/imath] as well. What am I missing here? | 2835140 | Why does the interval where the differential equation makes sense not include [imath]x = 0[/imath]?
Here is the ODE equation [imath]xy' = 2y[/imath]. A book states that the interval where it makes sense doesn't include [imath]0[/imath]. Can you explain why? As to me, I think it is defined everywhere. This problem is from the book "Ordinary Differential Equations - Harry Pollard, Morris Tenenbaum". The exact problem states: "Prove that the functions in the right-hand column below are solutions of the differential equations in the left-hand columns. (Be sure to state the common interval for which solution and differential equation make sense.)" [imath]xy' = 2y[/imath] and [imath]y = x^2[/imath] with the answer [imath]x \neq 0[/imath] |
2837498 | Functions of random variables problem
Random variable [imath]X[/imath] has uniform distribution on ([imath]0,1[/imath]). The question is to find the distribution of [imath]U=|X-1/3|[/imath]. So, by using the definition of continuous variables, I got to the equation [imath]Fx(u+1/3)-Fx(-u+1/3)[/imath], and that is correct. Now, I don't understand how to subtract these two functions, since one is non-zero on the interval [imath](-1/3,2/3)[/imath] and the other on the interval [imath](-2/3,1/3)[/imath]. The answer in the book is [imath]2u[/imath] on ([imath]0,1/3[/imath]), [imath]u+1/3[/imath] on ([imath]1/3,2/3[/imath]), and [imath]1[/imath] after that. | 2379178 | Random Variable absolute value distribution (PDF and CDF)
X has a Uniform distribution [imath]U(0,1)[/imath]. I have to find the distribution of the variable [imath]U =|X-1/3|[/imath]. This is what i can get so far : [imath]F(U) \;{= P\{U < u\} \\ = P\{| X-1/3 | < u\} \\ \ddots}[/imath] I know that i shall get something like [imath]P\{-(U-1/3) < X < U -1/3\}[/imath]. Could anyone explain how to get the PDF and the CDF for this problem and explain how can i easily change this absolute values for other problems ( other examples [imath]Z=1-|X|,~~G=|X|-1[/imath] etc..). |
2837403 | Let [imath]f[/imath] be a differentiable function such that it satisfies [imath]\int_1^{xy} f(t)dt=x\int_1^{y} f(t)dt+y\int_1^{x} f(t)dt[/imath] then find [imath]f(e)[/imath]
Let [imath]f:R^+ \to R [/imath] be a differentiable function such that [imath]f(1)=3[/imath] and satisfies [imath]\int_1^{xy} f(t)dt=x\int_1^{y} f(t)dt+y\int_1^{x} f(t)dt[/imath] [imath] \forall x,y\in R^+[/imath] then find [imath]f(e)[/imath] My try : Let [imath]F(a)=\int_1^{a} f(t)dt[/imath] Then our conditions give us [imath]F'(1)=3[/imath] and [imath]F(xy)=xF(y)+yF(x)[/imath] Now this form seems to be much similar to that of differentiation using product rule of two functions but this idea didn't help me much. I tried substituting [imath]x=y[/imath] in above equation to get [imath]F(x^2)=2xF(x)[/imath] But I don't seem to get anywhere with this. | 124049 | Computing [imath]f(x)[/imath] if [imath]\int_1^{xy} f(t) dt = y \int_1^x f(t) dt + x\int_1^y f(t) dt[/imath] and [imath]f(1) = 3[/imath]
This is a problem from Apostol, Calculus, Volume I, Chapter 6.9 (p. 238), that I was hoping someone could help with: A function [imath]f[/imath], continuous on the positive real axis, has the property that [imath]\int_1^{xy} f(t) dt = y \int_1^x f(t) dt + x \int_1^y f(t) dt[/imath] for all [imath]x > 0[/imath] and all [imath]y > 0[/imath]. If [imath]f(1) = 3[/imath], compute [imath]f(x)[/imath] for each [imath]x > 0[/imath]. My initial thought was to use properties of the integral to write: [imath]\begin{align*} \int_1^{xy} f(t) dt & = \int_1^x f(t) dt + \int_x^{xy} f(t) dt \\ & = \int_1^x f(t) dt + x \int_1^y f(xt) dt \end{align*}[/imath] and [imath]\begin{align*} \int_1^{xy} f(t) dt & = \int_1^y f(t) dt + \int_y^{xy} f(t) dt \\ & = \int_1^y f(t) dt + y \int_1^x f(yt) dt \end{align*}[/imath] Then each of these is also [imath]=y \int_1^x f(t)dt + x \int_1^ f(t)dt[/imath] from the given equation of the problem. I cannot seem to pull an equation for [imath]f(x)[/imath] out of this though (I might be missing some very simple manipulation...). I also thought in terms of letting [imath]A(x) = \int_1^x f(t)dt[/imath], so the problem is giving us a functional equation: [imath] A(xy) = yA(x) + xA(y)[/imath] This has obvious similarities to the functional equation of the logarithm. This problem is in the section of the text with the definition of the logarithm as an integral. I'm self-studying, so this isn't homework. Hints or full solutions are equally welcome. Thanks for the help. |
2832233 | Taylor coefficients of [imath]\exp(1/(1-z))[/imath]
I'm specifically interested in estimating the growth of the coefficients [imath](a_n)_{n\in\mathbb{N}}[/imath] of the Taylor series of [imath]\exp(1/(1-z))[/imath] centered in [imath]0[/imath]: knowing that [imath]\limsup_{n\rightarrow+\infty}|a_n|^{1/n}=1[/imath], what I'm trying to figure out is whether the growth of [imath](a_n)_{n\in\mathbb{N}}[/imath] is at most polynomial or super-polynomial... I tried to calculate explicitly the coefficients in order to find some regularity in the sequence of the coefficients, but I quickly got lost in the process. Thanks in advance for any answer or suggestion. | 126698 | Coefficient growth in the power series [imath]\sum u_n z^n = e^{1/(1-z)}[/imath]?
Let [imath]\sum u_n z^n[/imath] denote the power series of [imath]e^{1/(1-z)}[/imath]. As our radius of convergence is [imath]1[/imath], it follows that [imath]u_n[/imath] exhibits sub-exponential growth. On the other hand, [imath]\{u_n\}[/imath] must grow supra-polynomially, else transfer theorems like those found in Singularity Analysis of Generating Functions would then imply that the singularity at [imath]z=1[/imath] is regular. Heuristically, it would appear that [imath]u_n \sim \alpha n^{-3/4} e^{2\sqrt{n}},[/imath] for some [imath]\alpha \approx .162982[/imath]. This opinion is echoed, without support, as a comment on the OEIS page for A000262. Note: The sequence considered therein is the generating function of [imath]e^{z/(1-z)}[/imath], which has [imath]\mathbb{Z}[/imath] coefficients after scaling [imath]u_n[/imath] by [imath]n![/imath] How would one derive asymptotic results such as these? (Edited for spelling.) Note: I've posted my own solution in the answers below. In short, the saddle-point method applies. |
2837512 | What is the difference between these mathematical symbols
[imath]=[/imath] [imath]\cong[/imath] [imath]\equiv[/imath] Can anyone explain me "clearly" what's the difference between these [imath]3[/imath] symbols. The more I google about them, the more I get confused. Also, please tell me, how should I know which one to use? What are the hints? Thank you :) | 1269958 | The meaning of various equality symbols
I'm interested in knowing what is the meaning of the various equality symbols: [imath]=,\sim, \cong,\approx,\equiv[/imath]. For example, the speed of a car [imath]V[/imath] in m/s: what would be the meaning of each of these statements? [imath]V = 30\\ V\sim 30\\ V \cong 30\\ V \approx 30\\ V \equiv 30[/imath] |
2837692 | How do you prove that [imath]n^2-1[/imath] is composite when [imath]n>2[/imath]?
When [imath]n=2[/imath] we know that [imath]n^2-1[/imath] is prime but is this the case, when [imath]n>2[/imath]? | 1557483 | Prove that there exists only one prime number of the form [imath]p^2−1[/imath] where [imath]p≥2[/imath] is an integer
By factoring [imath]p^2 − 1[/imath], we have [imath](p + 1)(p - 1)[/imath]. I know that [imath]p = 2[/imath] which gives [imath]3[/imath] is the only solution. However, how do I prove that [imath]p = 2[/imath] is the only integer which gives a prime? |
2831130 | Cauchy's induction principle
Cauchy's induction principle states that: The set of propositions [imath]p(1),...,p(n),...[/imath] are all valid if: [imath]p(2)[/imath] is true. [imath]p(n)[/imath] implies [imath]p(n-1)[/imath] is true. [imath]p(n)[/imath] implies [imath]p(2n)[/imath] is true. How to prove Cauchy's induction principle? Can we use it to prove what we can prove with weak and strong induction? If yes how to prove using Cauchy's induction principle [imath] 1+2^1+2^2+...+2^n=2^{n+1}-1 [/imath] | 97350 | How does backwards induction work to prove a property for all naturals?
I was reading a blogpost here: http://mzargar.wordpress.com/2009/07/19/cauchys-method-of-induction/ One thing that threw me off was that after the first four large displayed equations, there is the statement "it is sufficient to prove that if the theorem holds for [imath]n=m+1[/imath], then it holds for [imath]n=m[/imath]." How is this type of induction valid? I browsed around for things like backward induction, reverse induction, and Cauchy induction, but couldn't find a justification for how this is valid. With the usual forward induction of verifying a base case and proving [imath]P(n)\implies P(n+1)[/imath], it's easy to intuitively understand how induction will show a property holds for all natural numbers (or at least starting at the base case). But with this reverse induction, it seems to me that if you prove [imath]P(m+1)\implies P(m)[/imath], then if you were able to verify a specific case like [imath]P(15)[/imath], then you would only know it's true for numbers up to [imath]15[/imath]. How does it actually prove the property for all naturals? |
2837354 | Proof of lemma about multiplication of ordinal number
Let [imath]\alpha[/imath], [imath]\beta[/imath] and [imath]\gamma[/imath] be ordinal numbers. (1) If $\alpha[imath]\prec[/imath]\beta$ and 0$\prec[imath]\gamma$, then $\alpha[/imath]\gamma[imath]\preccurlyeq[/imath]\beta[imath]\gamma$[/imath] (2) If $\gamma\alpha[imath]\prec[/imath]\gamma[imath]\beta$ and 0$\prec[/imath]\gamma$, then $\alpha[imath]\prec[/imath]\beta$ please let me know how to prove these things... | 912477 | Ordinal multiplication property: [imath]\alpha<\beta[/imath] implies [imath]\alpha\gamma\le\beta\gamma[/imath]
I'm having trouble proving the following two ordinal multiplication properties. If [imath]\alpha, \beta[/imath], and [imath]\gamma[/imath] are such that [imath]\alpha \lt \beta[/imath] and [imath]\gamma \gt 0[/imath], then [imath]\alpha\gamma \le \beta\gamma[/imath]. If [imath]\alpha, \beta[/imath], and [imath]\gamma[/imath] are such that [imath]\gamma\alpha \lt \gamma\beta[/imath] and [imath]\gamma \gt 0[/imath], then [imath]\alpha \lt \beta[/imath]. |
335590 | Derivative of sin(x) and cos(x) by definition of derivative and Taylor expansion
How to use the below to find the derivative of sin(x) (and cosine, too)? [imath]\sin(x)=\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n+1}}{(2n+1)!}\ , \ x\in \mathbb{R}[/imath] [imath]\cos(x)=\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n}}{(2n)!}\ , \ x\in \mathbb{R}[/imath] [imath]\lim_{h\to\infty}\frac{\sin(x+h)-\sin(x)}{h}[/imath] Thanks! | 335284 | Finding the derivatives of sin(x) and cos(x)
We all know that the following (hopefully): [imath]\sin(x)=\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n+1}}{(2n+1)!}\ , \ x\in \mathbb{R}[/imath] [imath]\cos(x)=\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n}}{(2n)!}\ , \ x\in \mathbb{R}[/imath] But how do we find the derivates of [imath]\sin(x)[/imath] and [imath]\cos(x)[/imath] by using the definition of a derivative and the those definition above? Like I should start by doing: [imath]\lim_{h\to\infty}\frac{\sin(x+h)-\sin(x)}{h}[/imath] But after that no clue at all. Help appreciated! |
2838063 | How to show in [imath]R^2[/imath], the [imath]f(x,y)=(x^4+y^4)^{1/4}[/imath] defines a norm on [imath]R^2[/imath].
How to show in [imath]R^2[/imath], the [imath]f(x,y)=(x^4+y^4)^{1/4}[/imath] defines a norm on [imath]R^2[/imath], i.e., how to prove the inequality: [imath]((x_1+x_2)^4+(y_1+y_2)^4)^{1/4}\le (x_1^4+y_1^4)^{1/4}+(x_2^4+y_2^4)^{1/4}[/imath] | 447747 | How to prove triangle inequality for [imath]p[/imath]-norm?
If [imath]\mathcal{M}=\{M_i : i\in I_n\}[/imath] is a collection of metric spaces, each with metric [imath]d_i[/imath], we can make [imath]M=\prod_{i\in I_n}M_i[/imath] a metric space using the [imath]p[/imath]-norm, we simply set [imath]d : M\times M\to \mathbb{R}[/imath] as: [imath]d((p_1,\dots,p_n),(q_1,\dots,q_n))=\left\|(d(p_1,q_1),\dots,d(p_n,q_n))\right\|_p[/imath] What I want to prove is that the [imath]p[/imath]-norm [imath]\left\|x\right\|_p=\left(\sum_{i=1}^{n}\left|x_i\right|^p\right)^{1/p}[/imath] is really a norm. Showing that [imath]\left\|x\right\|_p \geq 0[/imath] being zero if and only if [imath]x = 0[/imath] was easy. Showing that [imath]\left\|kx\right\|_p = \left|k\right|\left\|x\right\|_p[/imath] was also easy. The triangle inequality is the thing that is not being easy to show. Indeed, I want to show that: for every [imath]x,y \in \mathbb{R}^n[/imath] we have: [imath]\left(\sum_{i=1}^{n}\left|x_i+y_i\right|^p\right)^{1/p}\leq \left(\sum_{i=1}^{n}\left|x_i\right|^p\right)^{1/p}+\left(\sum_{i=1}^{n}\left|y_i\right|^p\right)^{1/p}.[/imath] I thought that it might not be as difficult as it seems, but after trying a little without sucess I've searched on the internet and the I found that we need measure theory to prove that. Is there any more elementary proof of this inequality? |
2838115 | What is wrong with this proof (Subgroup of Free group is a Free group)
Let [imath]G=\langle g_1, g_2,\dots\rangle[/imath] be a free group, with no relations. Let [imath]H[/imath] be a subgroup of [imath]G[/imath], with generators [imath]h_1,h_2,\dots\in G[/imath]. Suppose the generators have a relation, say [imath]h_{i_1}h_{i_2}\dots h_{i_m}=1[/imath]. Since each [imath]h_i[/imath] can be written as a product of [imath]g_j[/imath], then [imath]h_{i_1}h_{i_2}\dots h_{i_m}=1[/imath] implies that [imath]g_{j_1}\dots g_{j_k}=1[/imath] for some [imath]j_1,\dots,j_k[/imath]. This contradicts that [imath]G[/imath] has no relations for its generators. Hence [imath]H[/imath] is a free group. Intuitively, this is wrong (the proof is too simple). However I can't find the mistake. Thanks a lot for any help. | 2615108 | What is tricky about proving the Nielsen–Schreier theorem?
The Nielsen–Schreier theorem states (in part): Let [imath]F[/imath] be a free group, and [imath]H\le F[/imath] be any subgroup. Then [imath]H[/imath] is isomorphic to a free group. I have seen the topological proof of this theorem using the correspondence between coverings and subgroups of the fundamental group. This has always struck me as using rather strong theory for what it is being used to prove (though I do appreciate the beauty of the argument). In my head, I see the following (loose) argument: Let [imath]H[/imath] be a subgroup of [imath]F[/imath], and assume that [imath]H[/imath] is not free. Then there exists some nontrivial relation [imath]h_1h_2\dots h_n = 1[/imath]. But then this is also a nontrivial relation in [imath]F[/imath] implying that [imath]F[/imath] is not free, which is absurd. Thus [imath]H[/imath] is free. Clearly, there must be some problem with this. What are the stumbling blocks here? An issue I see is that the exact notation a relation has always seemed a little vague to me (some reduced word equal to the identity?), but that doesn't seem like it ought to be a large enough problem to invalidate the argument. |
2838524 | Given [imath]\lim a_n =a[/imath] and [imath]\lim b_n =b[/imath], show [imath]\lim\frac{1}{n}\sum_{k=1}^{n}a_k b_{n+1-k}=ab[/imath]
Given [imath]\lim a_n =a[/imath] and [imath]\lim b_n =b[/imath], show [imath]\lim\frac{1}{n}\sum_{k=1}^{n}a_k b_{n+1-k}=ab[/imath] As you can see, two limit are expanding in two different direction. I have no idea about how to do it. | 876520 | A sequence and convergence
Let [imath]\{x_n\}[/imath] and [imath]\{y_n\}[/imath] be sequences of real numbers which converge to [imath]\ell[/imath] and [imath]m[/imath] respectively. Show that [imath]\lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^{n}x_ky_{n+1-k}=\ell m[/imath] This is a problem I found in a book, it seemed to be routine, but now I can't solve it. What I tried : Let us take some [imath]\epsilon>0[/imath]. Now after a stage [imath]N[/imath], both [imath]|x_n-\ell|<\epsilon[/imath] and [imath]|y_n-m|<\epsilon[/imath]. Now I wanted to break the sum [imath]\left|\frac{1}{n}\sum_{k=1}^{n}x_ky_{n+1-k}-\ell m\right|[/imath]in two parts, one upto [imath]N[/imath], and other one for [imath]N+1[/imath] to [imath]n[/imath]. Now we need this to be arbitrarily small. But here I am lost. I don't know how to proceed anymore. Because when I take [imath]x_i's[/imath] to make up a finite sum, then the [imath]y_j's[/imath] are there too with [imath]j[/imath] very large. I think this is the main idea, but unable to do anything further. Can someone help me? Thanks. |
2838703 | Can you find the derivative of [imath]x![/imath]?
I need to find: [imath]\frac{d}{dx} x![/imath] Is there a way to find this? If not, is there a proof that shows we cannot find it? The graph of [imath]x![/imath] Function https://www.desmos.com/calculator/kwfazibe1r | 1633014 | Is there a way to evaluate the derivative of [imath]x[/imath]! without using Gamma function?
Taking the factorial function [imath]x![/imath] I wonder if there is a method to find the first derivative of this function without making any use of the Gamma function (or related integral representations of the factorial). Maybe something like that [imath] \begin{align} \frac{\text{d}}{\text{d} x}\ x! & = \frac{\text{d}}{\text{d} x}\ (x\cdot (x-1)\cdot (x-2)\cdot \cdots) \\\\ & = ((x-1)\cdot (x-2)\cdots) + x\cdot ((x-3)\cdot (x-4)\cdots) + x\cdot ((x-2)\cdot (x-4)\cdots) + \cdots \\\\ & = (x-1)! + x\cdot N(x) \end{align} [/imath] What I'm missing is a suitable [imath]N(x)[/imath] to express the remaining terms. Does such a function exist? I thought about this, observing term by term: [imath]N(x) = \left(\prod_{k = 3}^{n} (x-k)\right) + \left((x-2)\prod_{k = 4}^n (x-k)\right) + \left((x-2)\cdot(x-3)\prod_{k = 5}^n (x-k)\right) + \cdots[/imath] But still it seems quite messy, isn't it? |
2838750 | Irreducibility of a polynom
Any thoughts on how to show that [imath]p(x)=-1+\prod_{i=1}^n (x-i)[/imath] is irreducibile in [imath]\mathbb{Z}[x][/imath]? | 1845213 | Prove that the polynomial [imath]\prod\limits_{i=1}^n\,\left(x-a_i\right)-1[/imath] is irreducible in [imath]\mathbb{Z}[x][/imath].
Let [imath]n>1[/imath] be an integer. For [imath]a_1,a_2,\ldots,a_n\in\mathbb{Z}[/imath] with [imath]a_1< a_2< a_3 < \dots < a_n[/imath], prove that the polynomial [imath]f(x)=(x-a_1)(x-a_2)\cdots(x-a_n)-1\,.[/imath] is irreducible in [imath]\mathbb{Z}[x][/imath]. Please help! Thanks! |
2839280 | Why doesn't [imath]\sqrt{ab}=\sqrt{a}\sqrt{b}[/imath] work for [imath]a,b<0[/imath]?
It seems that [imath]\sqrt{ab}=\sqrt{a}\sqrt{b}[/imath] is only defined for [imath]a,b\geq 0[/imath], because it doesn't work for [imath]a,b<0[/imath]. I can see that it doesn't work. I would like to know why it doesn't work. Is there a less circular reason than "by definition"? This comes up in Khan Academy's i as the principal root of -1 and wikipedia's square root faulty proof: [imath] \begin{align*} -1&=ii\\ &=\sqrt{-1}\sqrt{-1}\\ &=\sqrt{(-1)(-1)}\\ &=\sqrt{1}\\ &=1 \end{align*} [/imath] They say [imath]ii=\sqrt{-1}\sqrt{-1}[/imath] is OK, and the faulty step is [imath]\sqrt{-1}\sqrt{-1 }=\sqrt{(-1)(-1)}[/imath], because [imath]\sqrt{ab}=\sqrt{a}\sqrt{b}[/imath] is only defined for [imath]a,b\geq 0[/imath]. | 2236790 | Why doesn't multiplying square roots of imaginary numbers follow [imath]\sqrt{a} \times \sqrt{b} = \sqrt{ab}[/imath]?
I've been following a series on understanding about how imaginary numbers came to be, and in the series, it mentions that imaginary numbers mostly follows the algebra rules for real numbers, such as adding or multiplying by real numbers. However, it specifically mentions this "inconsistency(?)" about multiplying square roots of imaginary numbers do not follow the rule for multiplying square roots of real numbers, namely [imath]\sqrt{a} \times \sqrt{b} = \sqrt{ab}[/imath]. For example, why can't I do this: [imath]\sqrt{-2} \times \sqrt{-3} = \sqrt{(-2)(-3)} = \sqrt{6}[/imath], which I know is the wrong answer. In the past, I've just memorized to factor out the imaginary parts first, and that [imath]i^2 = -1.[/imath] However, can anyone show me an explanation for this, or tell me where I could learn more about it? |
2839113 | Equality between the measure of a compact set and the limit of a series of open sets
Suppose [imath]E[/imath] is a given set, and [imath]O_n = \{x : d(x,E)<\frac1n\}[/imath]. Show that if [imath]E[/imath] is compact, then [imath]m(E) = \lim_{n \rightarrow \infty} m(O_n)[/imath] Also, show that this conclusion may be false for [imath]E[/imath] closed an unbounded, or [imath]E[/imath] open and bounded. I'm gonna need some help on this one... this measure theory stuff is so tricky! Thanks everyone. | 1669121 | Show that [imath]lim_{n \rightarrow \infty}m(O_n)=m(E)[/imath] when [imath]E[/imath] is compact.
Below is an attempt at a proof of the following problem. Any feedback would be greatly appreciated. Thx! Let [imath]E[/imath] be a set and [imath]O_n = \{x: d(x, E) < \frac{1}{n}\}[/imath]. Show If [imath]E[/imath] is compact then [imath]m(E) = lim_{n \rightarrow \infty} m(O_n)[/imath]. This is false for [imath]E[/imath] closed and unbounded or [imath]E[/imath] open and bounded. Note that all sets are real and measurable refers to Lebesgue measurable. Suppose [imath]E[/imath] is compact. Then [imath]m(E) < \infty[/imath]. Also, since each [imath]O_n[/imath] is an open set of [imath]\mathbb{R}^d[/imath], [imath]m(O_n) < \infty[/imath]. If either of the following is true, we have that [imath]lim_{n \rightarrow \infty}m(O_n)=m(E)[/imath]: [imath]O_n \nearrow E[/imath] [imath]O_n \searrow E[/imath] Now [imath]\bigcap_{n=1}^{\infty} O_n = \{ x:d(x,E) = 0\}[/imath]. That is, [imath]x \in \bigcap_{n=1}^{\infty}O_n \iff \exists n> \frac{1}{\epsilon},\ n \in \mathbb{N} \iff x\in \{x \mid d(x,E) =0\}[/imath]. Then we have that [imath]\bigcap_{n=1}^{\infty}O_n = E[/imath] and [imath]O_n \searrow E[/imath]. Thus [imath] \ m(E) = lim_{n \rightarrow \infty} m(O_n)[/imath]. Suppose [imath]E[/imath] is closed and unbounded. Let [imath]E = \{(x,y): y = 2 \}[/imath] and [imath]O_n = \{(x,y): d(x,E) < \frac{1}{n}\}[/imath]. Since [imath]E[/imath] is a line in [imath]\mathbb{R}^2[/imath], [imath]m(E) = 0[/imath]. Also, since the measure of a rectangle in [imath]\mathbb{R}^d[/imath] is its volume, [imath]m(O_n) = \left| O_n \right| = \infty[/imath], since the rectangle [imath]O_n[/imath] is unbounded. It is therefore apparent that [imath]m(E) \not = lim_{n \rightarrow \infty} m(O_n)[/imath]. Suppose [imath]E[/imath] is open and bounded. Let [imath]E=(0,1)[/imath] and [imath]O_n = \{x \mid d(x,E) < \frac{1}{n}\}[/imath]. [imath]\mathbb{R}-E[/imath] is closed and [imath]O_n \in \mathbb{R}-E[/imath]. Since [imath]\mathbb{R}-E[/imath] contains all its limit points [imath]lim_{n \rightarrow \infty}O_n = O \in \mathbb{R}-E[/imath]. Thus [imath]lim_{n \rightarrow \infty} m(O_n) \neq m(E)[/imath]. |
2840042 | Ideals in Commutative Rings Containing a Field
This is a question that always stumped me. It is from an undergrad book (Sharp, Steps in Commutative Algebra) and apparently concerns the Primary Decomposition/Prime Avoidance Theorem, so I expect the solution to be only a few lines, but I can't get it. Suppose that [imath]R[/imath] is a commutative ring containing an infinite field [imath]F[/imath]. Suppose that [imath]I, J_1, \dots, J_n[/imath] are ideals of [imath]R[/imath]. Suppose that [imath]I \subset \bigcup\limits_n J_n[/imath]. Then prove that [imath]I[/imath] is contained in some [imath]J_k[/imath]. I can't figure out how to use the presence of the field to turn it into a situation where the Prime Avoidance Theorem can be applied, but maybe this isn't the right method. | 1904561 | Exercise similar to prime avoidance theorem in R.Y. Sharp's textbook
I'm a beginner in commutative algebra, and I'm self-studying this subject via R.Y. Sharp's textbook. Let [imath]R[/imath] be a commutative ring with [imath]1[/imath] which contains an infinite field as a subring. Let [imath]I[/imath] and [imath]J_1,\ldots,J_n[/imath] where [imath]n\ge2 [/imath], be ideals of [imath]R[/imath] such that: [imath]I\subseteq J_1\cup J_2\cup\cdots\cup J_n.[/imath] Prove that [imath]I \subseteq J_j[/imath] for some [imath]j[/imath] with [imath]1\le j \le n[/imath]. It looks similar to the Prime Avoidance Theorem, but in this case each [imath]J_i[/imath] just to be ideals and I can't figure out what the role of "contains an infinite field as a subring" in this situation. I expect for some idea / suggestion. Thank in advance. |
2840119 | for [imath]f(x,y)=(x^2-y^2,2xy)[/imath] Sketch f(A) for [imath]x>0[/imath]
I want to solve part b) of the problem below. Another user has provided the picture below the problem which describes the map transformation. I am still uncertain about how to produce this image. I understand the picture depicting the domain, since [imath]A[/imath] is restricted by [imath]x>0[/imath], but how do we then get the image [imath]f(A)[/imath] to wrap around the way it does. Since the image consists of those points [imath](x^2-y^2,2xy)[/imath], and in complex form, [imath](x^2-y^2+i2xy)[/imath], why do we get that [imath]y\neq 0[/imath] for [imath]x\leq 0[/imath] in the image? it's only [imath]x[/imath] that's restricted in the image, not the real part [imath](x^2-y^2)[/imath], right? Thanks! | 2838005 | Show function [imath]f(x,y)=(x^2-y^2,2xy)[/imath] is [imath]1[/imath]-[imath]1[/imath] by Inverse Function Theorem
I'm trying to prove the problem below - which comes from Munkres' "Analysis on Manifolds" book in the section on the Inverse function theorem. Since its in the chapter on the Inverse Function Theorem I figured I'd start by showing that [imath]f[/imath] satisfies the conditions of the theorem. Writing the Jacobian shows that it's both [imath]C^r[/imath] and we get [imath]\det f'(x,y)=4x^2+4y^2\neq0[/imath] when [imath]x>0[/imath]. So we can apply the theorem but I'm unsure of how to proceed to show that [imath]f[/imath] is [imath]1[/imath]-[imath]1[/imath]. And I didn't see how to use the hint they provided. Thanks! Let [imath]f\colon \mathbf R^2\to \mathbf R^2[/imath] be defined by the equation [imath]f(x,y)=(x^2-y^2,2xy).[/imath] (a) Show that [imath]f[/imath] is one-to-one on the set of all [imath](x,y)[/imath] with [imath]x>0[/imath]. [Hint: If [imath]f(x,y)=f(a,b)[/imath], then [imath]\|f(x,y)\|=\|f(a,b)\|[/imath].] |
2840387 | Finding [imath] \lim_{x\rightarrow 0}\frac{(1+x)^{\frac{1}{x}}-e+\frac{ex}{2}}{x^2}[/imath] without using series expansion.
Finding [imath]\displaystyle \lim_{x\rightarrow 0}\frac{(1+x)^{\frac{1}{x}}-e+\frac{ex}{2}}{x^2}[/imath] without using series expansion. I am trying to solve it using D L Hopital Rule So [imath]\displaystyle \lim_{x\rightarrow 0}\frac{(1+x)^{\frac{1}{x}}\cdot \frac{1}{x^2}\bigg[\frac{x}{1+x}-\ln(1+x)\bigg]+\frac{e}{2}}{2x}[/imath] again i want to Diff with r to [imath]x,[/imath] but this is to complex please help me how to short my calculation , Thanks | 879430 | The limit of [imath]((1+x)^{1/x} - e+ ex/2)/x^2[/imath] as [imath]x\to 0[/imath]
[imath]\lim_{x\rightarrow 0}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}=\,?[/imath] by directly substituting [imath]x=0[/imath] i got [imath]\infty[/imath] by using L-H's rule, i got [imath]-1/8[/imath] the given options are [imath]a)\frac{24e}{11}[/imath] [imath]b)\frac{11e}{24}[/imath] [imath]c)\frac{e}{11}[/imath] [imath]d)\frac{e}{24}[/imath] may be the question would be [imath]\lim_{x\rightarrow 0}\frac{(1+x)^{1/x}-e+\frac{ex}{2}}{x^2}=\,?[/imath] |
2841054 | Show isomorphisms of abelian groups
Let [imath]d \in \mathbb{Z} \setminus \{0, 1\}[/imath] square free, [imath]K = \mathbb{Q}(\sqrt{d})[/imath], [imath]\mathcal{O}_K[/imath] the corresponding ring of integers, [imath]\mathfrak{a} \subseteq K[/imath] a fractional ideal [imath]n, n' \in \mathbb{N}[/imath] with [imath]n\mathfrak{a}, n'\mathfrak{a} \subseteq \mathcal{O}_K[/imath]. I'm currently trying to prove that the definition of the idealnorm for fractional ideals is well defined. Inside this proof I need the following result: [imath][n\mathfrak{a} : nn'\mathfrak{a}] = [\mathcal{O}_K : (n')][/imath] or more general [imath]n\mathfrak{a} / nn'\mathfrak{a} \cong \mathcal{O}_K / (n')[/imath] as abelian groups. I don't know how I could approach this step. Any help is welcome! Duplicate claim: My proof looks similar to the one posted here, but the linked post doesn't prove the question I asked here. | 1401876 | How to show that the norm of a fractional ideal is well-defined?
Sorry. This might probably be a really easy question, but I am only a beginner in algebraic number theory. So, please bear with me. Let [imath]K[/imath] be an algebraic number field and [imath]\mathcal{O}_K[/imath] its ring of integers. Preludium In my lecture we have defined the norm [imath]N_{K/\mathbb{Q}}(x)[/imath] for an element [imath]x \in K[/imath] in the usual way. In a following lecture we "extended" this to the norm of an ideal [imath]\mathfrak{a}[/imath] of [imath]\mathcal{O}_K[/imath] by setting [imath] \mathfrak{N}(\mathfrak{a}) := [\mathcal{O}_K : \mathfrak{a}] := \left| \mathcal{O}_K / \mathfrak{a} \right|[/imath] and showed that for the principal ideal [imath](a)[/imath] with [imath]a \in \mathcal{O}_K[/imath] we always have [imath]\mathfrak{N}((a)) = \left|N_{K/\mathbb{Q}}(a)\right|[/imath]. Later, we introduced the notion of a fractional ideal in [imath]K[/imath] being a non-zero, finitely generated [imath]\mathcal{O}_K[/imath]-submodule of [imath]K[/imath]. And we showed that a non-zero [imath]\mathcal{O}_K[/imath]-submodule [imath]\mathfrak{a}[/imath] of [imath]K[/imath] is a fractional ideal iff there is a [imath]c \in \mathcal{O}_K \setminus \{0\} [/imath] such that [imath]c\mathfrak{a} \subseteq \mathcal{O}_K[/imath] (thus [imath]c\mathfrak{a}[/imath] is an ideal of [imath]\mathcal{O}_K[/imath]). We now extended the norm [imath] \mathfrak{N}[/imath] by defining for each fractional ideal [imath]\mathfrak{a}[/imath] of [imath]K[/imath] the norm [imath] \mathfrak{N}(\mathfrak{a}) := \frac{[\mathcal{O}_K : c\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(c)\right|}, [/imath] [imath]c[/imath] being an element of [imath]\mathcal{O}_K \setminus \{0\}[/imath] such that [imath]c\mathfrak{a} \subseteq \mathcal{O}_K[/imath] (exists according to 2.). Question How can I show that this latter definition is independent of the chosen [imath]c[/imath]? Edit I changed the definition in (3) from [imath] \mathfrak{N}(\mathfrak{a}) := \frac{[\mathcal{O}_K : c\mathfrak{a}]}{N_{K/\mathbb{Q}}(c)} [/imath] to [imath]\mathfrak{N}(\mathfrak{a}) := \frac{[\mathcal{O}_K : c\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(c)\right|} [/imath] and added the absolute value everywhere else, too. I think that was a typo in my lecture notes. Thoughts (Is it better to add them as comments? I was not sure.) Let [imath]c, c' \in \mathcal{O}_K \setminus \{0\}[/imath] such that [imath]c \mathfrak{a} \subseteq \mathcal{O}_K [/imath] and [imath]c' \mathfrak{a} \subseteq \mathcal{O}_K[/imath]. At the moment I am contemplating if and in which way [imath]c[/imath] and [imath]c'[/imath] are related. I was wondering whether one always has [imath]c'=bc[/imath] (or vice versa) for an [imath]b \in \mathcal{O}_K[/imath]. Then we had [imath] \frac{[\mathcal{O}_K : c'\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(c')\right|} = \frac{[\mathcal{O}_K : bc\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(bc)\right|} = \frac{[\mathcal{O}_K : bc\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(b)\right| · \left|N_{K/\mathbb{Q}}(c)\right|}[/imath] and it remained to show that [imath][\mathcal{O}_K : bc\mathfrak{a}] = \left|N_{K/\mathbb{Q}}(b)\right| · [\mathcal{O}_K : c\mathfrak{a}][/imath] (maybe because [imath][\mathcal{O}_K : bc\mathfrak{a}] = [\mathcal{O}_K : (b)] · [\mathcal{O}_K : c\mathfrak{a}][/imath] ???). |
2837847 | locally injective holomorphic function
I know that if [imath]U[/imath] is an open subset of the complex plane [imath]\mathbb{C}[/imath] and [imath]f:U\to \mathbb{C}[/imath] is a holomorphic function and [imath]f[/imath] is one-to-one, then the derivative [imath]f'(z)[/imath] is different from zero for every [imath]z\in U[/imath]. But how to prove converse of this problem: If [imath]f'(z)\neq 0, \;\forall z\in U[/imath], then [imath]f(z)[/imath] is locally injective? | 582755 | If [imath]f:U\to V[/imath] is holomorphic and [imath]f'(z)\neq 0[/imath] for all [imath]z\in U[/imath], then[imath]f[/imath] is locally bijective.
I am trying to solve the following problem: Let [imath]f:U\to V[/imath] be a holomorphic function such that [imath]f'(z)\neq 0[/imath] for all [imath]z\in U[/imath]. Show that for all [imath]z_0\in U[/imath], there exists a disc [imath]D_\varepsilon(z_0)\subseteq U[/imath] such that [imath]f:D_\varepsilon(z_0)\to f(D_\varepsilon(z_0))[/imath] is bijective. Atempt: I am trying to use Rouche's Theorem, but I am stuck on a particular step. Let [imath]z_0\in U[/imath]. Since [imath]f'(z_0)\neq 0[/imath], there is a disc [imath]D_r(z_0)\subseteq U[/imath] such that [imath]f(z)=f(z_0)+(z-z_0)h(z),\quad\forall z\in D_r(z_0)[/imath] where [imath]h[/imath] is holomorphic on [imath]D_r(z_0)[/imath] and [imath]h(z)\neq 0[/imath] for all [imath]z\in D_r(z_0)[/imath]. Let [imath]0<\varepsilon<r[/imath] to be defined later and fix [imath]w\in D_\varepsilon(z_0)[/imath]. To show that [imath]f[/imath] is injective in [imath]D_\varepsilon(z_0)[/imath] is to show that the function [imath]f(z)-f(w)[/imath] has exactly one zero in [imath]D_\varepsilon(z_0)[/imath]. But [imath]F(z):=(z-z_0)h(z)[/imath] has exactly one zero in [imath]D_\varepsilon(z_0)[/imath], so we might want to apply Rouche's Theorem with [imath]F[/imath] and [imath]G(z):=f(z)-f(w)-(z-z_0)h(z)=f(z_0)-f(w)=-(w-z_0)h(w)[/imath] to conclude that [imath]F[/imath] and [imath]F+G[/imath] have the same number of zeros and hence [imath]f[/imath] is injective. But that means that we need to find [imath]0<\varepsilon<r[/imath] such that [imath]|G(z)|<|F(z)|[/imath] on [imath]\partial D_\varepsilon(z_0)[/imath]. That is, [imath]|w-z_0||h(w)|<|z-z_0||h(z)|[/imath] for all [imath]w\in D_\varepsilon(z_0)[/imath] and [imath]z\in\partial D_\varepsilon(z_0)[/imath]. Can we find such [imath]\varepsilon>0[/imath]? It seems intuitive since if we expand [imath]h[/imath] in a power series at [imath]z_0[/imath], [imath]h(z)=\sum_{n=0}^\infty a_n(z-z_0)^n[/imath], then [imath]|w-z_0||h(w)| = |w-z_0|\sum_{n=0}^\infty a_n|w-z_0|^n < |z-z_0|\sum_{n=0}^\infty a_n|z-z_0|^n.[/imath] for [imath]w\in D_\varepsilon(z_0)[/imath] and [imath]z\in\partial D_\varepsilon(z_0)[/imath]. But this is of course not a proof. |
2841533 | all functions that uphold [imath]f'(x)=f(1/x)[/imath]
Find all functions [imath]y:\mathbb{R}\to\mathbb{R}[/imath] such that [imath]y'(x)=y\Big(\frac{1}{x}\Big)[/imath] we can differentiate both sides to get a homogenous second order Euler equation [imath]y''(x)=-\frac{1}{x^2}y'\Big(\frac{1}{x}\Big)=-\frac{1}{x^2}y(x)\implies x^2y''+y=0[/imath] which is doable. My question is, am I missing something? From a short glance at WA it doesn't seem that all of the solutions of the Euler equation uphold the required criteria. | 31492 | Solving [imath]\forall x \in \mathbb{R}_+^*, f'(x) = f\left(\frac1{x}\right)[/imath]
I recently came across this equation : [imath]\forall x \in \mathbb{R}_+^*, f'(x) = f\left(\frac1{x}\right)[/imath]where [imath]f \in \mathcal{C}^1(\mathbb{R}, \mathbb{R})[/imath]. I've done the following, but I'm stuck at the end. Could you give me pointers? Thanks! Differentiating yields [imath]\forall x, f''(x) = -\frac1{x^2}f(x) \tag{$S_0$}[/imath]Solutions in the form [imath]x \mapsto \frac1{x^\phi}[/imath] work iff [imath]\phi(\phi+1) = -1 [/imath], ie. [imath]\phi = \frac{-1 \pm i \sqrt{3}}{2} =e^{\pm 2i\pi/3} = j, \overline{j}[/imath]. Elements of the vector space generated by the free pair [imath](x^j, x^\overline{j})[/imath] are therefore solutions of ([imath]S_0[/imath]). I then feed [imath]\lambda x^j + \mu x^\overline{j}[/imath] in the original equation, which yields [imath]-\lambda j\frac1{x^{j+1}}-\mu\overline{j}\frac1{x^{\overline{j} + 1}} = \frac{x^{j + \overline{j}}}{\lambda x^\overline{j} + \mu x^j}[/imath], then [imath](-\lambda j x^{\overline{j}+1} - \mu \overline{j} x^{j+1})(\lambda x^{\overline{j}} + \mu x^j) = x^{1+j+\overline{j}} = x^0 = 1[/imath], and [imath]-\lambda^2 j x^{2\overline{j} + 1} - \mu^2 \overline{j} x^{2j+1} - \lambda\mu(j + \overline{j}) = 0 [/imath]. Thus, [imath] \lambda^2 j x^{-2i\sin(2\pi/3)} + \mu^2 \overline{j} x^{2i\sin(2\pi/3)} = \lambda\mu[/imath] Does that mean that no solutions can be found to the original equation, except the trivial [imath]x \mapsto 0[/imath] one? Or that I didn't take the right approach? I can't figure out how to handle the last equality. |
2841740 | How to calculate the gradient of this function?
Let [imath]f(x) := \| y - B x \|_2^2[/imath] where [imath]y \in \mathbb R^k[/imath], [imath]x \in \mathbb R^n[/imath], and [imath]B \in \mathbb R^{k \times n}[/imath]. How to calculate the gradient of [imath]f[/imath] w.r.t. [imath]x[/imath]? | 369694 | Matrix Calculus in Least-Square method
In the proof of matrix solution of Least Square Method, I see some matrix calculus, which I have no clue. Can anyone explain to me or recommend me a good link to study this sort of matrix calculus? In Least-Square method, we want to find such a vector [imath]x[/imath] such that [imath]||Ax-b||[/imath] is minimized. Assume [imath]r=Ax-b[/imath] [imath]\Rightarrow\|r\|^2=x^TA^TAx-2b^TAx+b^Tb[/imath] [imath]\Rightarrow \nabla_x \|r\|^2=2A^TAx-2A^Tb[/imath] In the end we set the gradient to zero and find the minimized solution. I understand the whole idea, but I just don't know how exactly we did matrix calculus here, or say I don't know how to do the matrix calculus here. For example, can anyone tell me how we got those transpose in [imath]\|r\|^2[/imath](By what rule?) and how we got the gradient?(how do we take the gradient exactly in matrix format)? I'll really appreciate if you can help me out. Thanks! |
2841885 | Need help showing [imath]\left(\frac{n^2 - 2n + 1}{n^2-4n+2}\right)^n \to e^2[/imath]
I'm trying to work out this limit: [imath] \lim_{n \to \infty} \left(\frac{n^2 - 2n + 1}{n^2-4n+2}\right)^n = e^2 [/imath] I have managed to show that it is equal to: [imath]\lim_{n \to \infty}\left (1+\frac{2n-1}{n^2-4n+2}\right)^n[/imath] but I have no idea how to continue from here, help would be appreciated. | 2042972 | how does one solve [imath]\lim_{n\to \infty} (\frac{n^{2}-2n+1}{n^{2}-4n+2})^{n}[/imath]
I cant use lhopital's rule here is my attempt: Limit of [imath](\frac{n^{2}-2n+1}{n^{2}-4n+2})^{n}[/imath] =[imath]e^{\lim_{n\to \infty} n\times \ln(\frac{1 -2-n +\frac{1}{n^{2}}}{1 -\frac{4}{n} +\frac{2}{n^{2}}})}[/imath] I got stuck here |
2842445 | divisibility problem in entrance exam
[imath]\forall n \in \mathbb{N}-\{1\}[/imath], [imath]7^{2n}-48n-1[/imath] is divisible by, a) 25 b) 26 c) 1234 d) 2304 I tried using induction. But couldnt do. Any hint will be very helpful. | 1998537 | Prove by induction that [imath]7^{2n}-48n-1[/imath] is divisible by 2304
[imath]P(n):2304\mid7^{2n}-48n-1[/imath] I've done the base case; [imath]P(1)[/imath] is true because the expression then evaluates to zero, which is divisible by 2304. Now I'm stuck on the inductive step: proving [imath]P(m+1)[/imath] true if [imath]P(m)[/imath] is true. I do know this though: [imath]7^{2m+2}-48(m+1)-1 =49\cdot7^{2m}-48m-49[/imath] |
2841584 | Using differentials in integration by substitution
I am confused why we can introduce differentials into an integral when performing an integration by substitution. Consider the integral [imath]\int \frac{1}{ x \sqrt{1-x} } dx.[/imath] We can perform the substitutions [imath]x=\sin^2u,[/imath] [imath]dx=2\sin u \cos{u} du ,[/imath] on the integral to give [imath]\int \frac{1}{ \sin^2u \sqrt{1-\sin^2u} } 2\sin u \cos u du .[/imath] Why can you treat [imath]dx[/imath] as a differential? From my understanding the integration sign [imath]\int dx[/imath] works as as if it is an operator, just like how [imath]\frac {d}{dx}[/imath] works as an operator and not as a fraction. Which means [imath]\int[/imath] and [imath]dx[/imath] should not be interpreted separately. But at the same time treating [imath]dx[/imath] as a differential always work out fine so there must be some validness in treating it as an differential. | 402303 | Understanding the differential [imath]dx[/imath] when doing [imath]u[/imath]-substitution
I just finished taking my first year of calculus in college and I passed with an A. I don't think, however, that I ever really understood the entire [imath]\frac{dy}{dx}[/imath] notation (so I just focused on using [imath]u'[/imath]), and now that I'm going to be starting calculus 2 and learning newer integration techniques, I feel that understanding the differential in an integral is important. Take this problem for an example: [imath]\int 2x (x^2+4)^{100}dx[/imath] So solving this... [imath]u = x^2 \implies du = 2x dx \longleftarrow \text{why $dx$ here?}[/imath] And so now I'd have: [imath]\displaystyle \int (u)^{100}du[/imath] which is [imath]\displaystyle \frac{u^{101}}{101} + C[/imath] and then I'd just substitute back in my [imath]u[/imath]. I'm so confused by all of this. I know how to do it from practice, but I don't understand what is really happening here. What happens to the [imath]du[/imath] between rewriting the integral and taking the anti-derivative? Why is writing the [imath]dx[/imath] so important? How should I be viewing this in when seeing an integral? |
2842310 | show that for any irrational [imath]\alpha [/imath] the limit [imath]\lim _{n \to\infty} \sin n \alpha \pi [/imath] does not exist ..
1)show that for any irrational [imath]\alpha [/imath] the limit [imath]\lim _{n\to\infty} \sin n \alpha \pi [/imath] does not exist .. 2) show that for any rational [imath]\alpha [/imath] the limit [imath]\lim _{n \rightarrow \infty} \sin (n! \alpha \pi) [/imath] exist ? My attempts : For 1) I'm very confused ????? For 2) let [imath] \alpha =\frac {p}{q}[/imath] with [imath]p \in \mathbb{Z}[/imath] and [imath]q \in \mathbb{N}[/imath] . For [imath]n > q [/imath] the number [imath]n ! \alpha \pi[/imath] is a multiple of [imath]\pi [/imath], which means that the terms of the sequence , beginning with some Value [imath] n_{0} [/imath]of the index n, are all equal to [imath]0[/imath] as in 1) I'm very confused. How can I approach this kind of problems ?? Pliz help me Thanks un | 2185104 | Show that [imath]\lim_{n\to \infty}\sin{n\pi x} =0[/imath] if [imath]x\in \mathbb{Z},[/imath] but the limit fails to exist if [imath]x\notin \mathbb{Z}.[/imath]
Show that [imath]\lim_{n\to \infty}\sin{n\pi x} =0[/imath] if [imath]x\in \mathbb{Z},[/imath] but the limit fails to exist if [imath]x\notin \mathbb{Z}.[/imath] 1st part If [imath]x\in \mathbb{Z}[/imath] then [imath]\sin{n\pi x}=0[/imath] for all [imath]n,[/imath] giving the first part. Edit: 2nd part If [imath]x\notin \mathbb{Z},[/imath] I want to show that the limit doesn't exist. How to do that? |
2842505 | Evaluate [imath]\int_0^{\infty}\frac{\ln(x+\frac{1}{x})}{1+x^2}\cdot dx[/imath]
Evaluate: [imath]\int_0^{\infty}\dfrac{\ln(x+\frac{1}{x})}{1+x^2}\cdot dx[/imath] I tried substituting [imath]\ln(x+\frac{1}{x})=t[/imath] that to some extent created denominator but made it more tedious. Please give some hints to solve. | 1205543 | How to solve [imath]\int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx[/imath]?
Here is my question [imath]\int_0^{\infty} \frac{\log(x+\frac{1}{x})}{1+x^2}dx[/imath] I have tried it by substituting [imath]x[/imath] = [imath]\frac{1}{t}[/imath]. I got the answer [imath]0[/imath] but the correct answer is [imath]\pi log(2)[/imath]. Any suggestion would be appreciated. |
2632230 | Are there any subsets [imath]A, B[/imath] of [imath]\mathbb R[/imath] such that [imath]\sup(B)>\sup(A)[/imath] but none of the elements of [imath]B[/imath] are upperbounds of [imath]A[/imath]?
If not, then how would you prove it? I've thought of multiple examples that do not support this but cannot think of how to go about proving it. | 2842387 | If sup A [imath]\lt[/imath] sup B show that an element of [imath]B[/imath] is an upper bound of [imath]A[/imath]
(a) If sup A < sup B, show that there exists an element of [imath]b \in B[/imath] that is an upper bound for [imath]A[/imath]. I have argued that if sup A [imath]\lt[/imath] sup B, then choose an [imath]\epsilon>0[/imath] such that sup A +[imath]\epsilon \in B[/imath]. Since [imath]a \le [/imath] sup A for all [imath]a \in A[/imath], it follows [imath]a \lt [/imath] sup A + [imath]\epsilon[/imath] hence sup A + [imath]\epsilon [/imath] is an upper bound of A as well as an element of B. (b) Give an example to show that this is not always the case if we only assume sup A [imath]\le[/imath] sup B. I am having trouble extending my argument into an example which makes me think that it may not be correct. Question: Is my argument for (a) correct? If not how would one show this fact? Can you give an example asked for in (b)? Thanks in advance. I need to understand how I am assuming what I am supposed to prove in my approach. Can someone elaborate on that please? The other question doesn't really address this issue. |
2837733 | Can numbers with similar stopping times in the Collatz Conjecture be functionally described?
We slightly refine the Collatz sequence as follows: [imath]f(n) = \begin{cases} \text{Od}(n) & \text{if } n \equiv 0 \pmod 2\\ 3n + 1 & \text{if } n \equiv 1 \pmod 2\\ \end{cases} [/imath] Where Od(n) is the odd part of n. And [imath]a_i(n) = \begin{cases} n & \text{for } i = 0\\ f(a_{i-1}(n)) & \text{for } i > 0\\ \end{cases} [/imath] We wish to classify numbers by their refined Collatz stopping times, [imath]C(n) = s[/imath]. [imath]C(n) = s \iff a_s = 1[/imath] And we wish to describe sets of natural numbers grouped by stopping time class: [imath]C_s = \{n\in\mathbb{N} \big| C(n) = s\}[/imath] Note: \begin{align} C_0 &= \{1\}\\ C_1 &= \{n,a \in \mathbb{N} | n = 2^a\}\\ C_2 &= \bigg\{n,a \in \mathbb{N}, a>1 \bigg|n = \frac{4^a - 1}{3}\bigg\}\\ C_3 &= \{n\in\mathbb{N}, c_1\in C_1, c_2\in C_2\bigg|n=c_1 \times c_2\}\\ C_4 &= \bigg\{n\in \mathbb{N},a_1,a_2 \in \mathbb{Z}^*, r_1\in \{1,2\}, 3a_1+r_1\geq2\bigg|n=\frac{1}{3}\bigg(2^{3 - r_1}\times4^{a_2}\times\frac{4^{3a_1 + r_1} - 1}{3} - 1\bigg)\bigg\}\\ C_5 &= \{n\in\mathbb{N}, c_1\in C_1, c_4\in C_4\bigg|n=c_1 \times c_4\} \end{align} This seems to be quickly getting out hand. What is [imath]C_6[/imath]? What texts might have similar analysis? | 2711060 | Some details about 'Collatz Conjecture'?
Yes, there is no one who doesn't know this problem.My question is only about curiosity. [imath]C(n) = \begin{cases} n/2 &\text{if } n \equiv 0 \pmod{2}\\ 3n+1 & \text{if } n\equiv 1 \pmod{2} .\end{cases}[/imath] On this problem, I caught something like this.I'm sure, We all realized that. For example, [imath]n=19[/imath], we have [imath]6[/imath] odd steps. We know that, even steps are not important, because each even number is converted to an odd number. [imath]19\Longrightarrow 29 \Longrightarrow 11\Longrightarrow 17 \Longrightarrow13 \Longrightarrow 5 \Longrightarrow 1[/imath] Then, for [imath]n=77[/imath], We have also [imath]6[/imath] odd steps. [imath]77\Longrightarrow 29 \Longrightarrow 11\Longrightarrow 17 \Longrightarrow13 \Longrightarrow 5 \Longrightarrow 1[/imath] For [imath]n=9[/imath] [imath]9\Longrightarrow 7 \Longrightarrow 11 \Longrightarrow 17 \Longrightarrow 13\Longrightarrow 5 \Longrightarrow 1[/imath] Again we have [imath]k=6[/imath] odd steps. I want to know / learn / ask, for [imath]k=6[/imath], (Generalized: for any number [imath]k[/imath] ) can we produce a formula(s) to catch all such numbers, which gives the result [imath]1[/imath]? Thank you! |
2842567 | Counter-example: If [imath]U \subseteq X[/imath] is open and we have a S.E.S of Abelian sheaves, the corresponding S.E.S of rings/modules may not be exact
I'm looking for a counter-example for the problem in title when [imath]F_1[/imath] is not flabby. We know that if [imath]0\to F_1 \to F_2 \to F_3\to 0[/imath] is a short exact sequence of Abelian sheaves and [imath]F_1[/imath] is flabby then for any [imath]U \subseteq X[/imath] which is open we have the following exact sequence: [imath]0\to F_1(U) \to F_2(U) \to F_3(U)\to 0[/imath] Is there a counter-example that shows this may no longer be true if [imath]F_1[/imath] is not flabby? If yes, how a counter-example can be constructed? Edit: I'm looking for a solution without reference to manifolds or De Rham cohomology. Thanks. | 577545 | Exact sequence of sheaves with non exact sequence of global sections
Let [imath]X[/imath] be some topological space. By [imath]\mathcal{F}_i[/imath] we denote some sheaves of abelian groups on [imath]X[/imath]. The sequence of sheaves and morphisms [imath]\mathcal{F}_1\longrightarrow \mathcal{F}_2\longrightarrow \mathcal{F}_3\longrightarrow... [/imath] is said to be exact if for each [imath]x\in X[/imath] the corresponding sequence of stalks [imath](\mathcal{F}_1)_x\longrightarrow (\mathcal{F}_2)_x\longrightarrow (\mathcal{F}_3)_x\longrightarrow... [/imath] is exact. However if the sequence of sheaves is exact than the sequence of global sections is not necessarily exact! (The most famous example is the sequence of sheaves [imath]0\longrightarrow\mathbb{Z}\hookrightarrow \mathcal{O}\stackrel{\exp}{\longrightarrow} \mathcal{O}^*\longrightarrow0[/imath] considered as sheaves on [imath]\mathbb{C}-\{0\}[/imath], where [imath]\mathcal{O}[/imath] is a sheaf of holomorphic functions, [imath]\mathcal{O}^*[/imath] is a sheaf of holomorphic functions with no zeros). So, could you give me some easy examples of such phenomenon ? |
2844539 | If two fractions add up to 1, are their denominators the same?
If two fraction add up to 1, what is the relation between their denominators? Are their denominators equal? If [imath]\frac{a}{b}+\frac{c}{d}=1[/imath], then, is what is the relation between [imath]b[/imath] and [imath]d[/imath]? I was working on a problem. To solve the problem i need to know the relation between the denominators of fractions which add up to 1. I have tried to find out a relation by using the given equation but i could not obtain any useful result. I have searched the internet but could not find any help. | 687824 | If the sum of two irreducible fractions is an integer, then the denominators are equal
I have to show the following:"If the sum of two irreducible fractions with positive denominators is an integer, then the denominators are equal." [imath]\frac{a}{b}+\frac{c}{d}=k, \text{ where k an integer }[/imath] Since the fractions are irreducible, [imath](a,b)=1[/imath] and [imath](c,d)=1[/imath]. Right? But how can I continue?? |
2844247 | How to find [imath]\lim\limits_{x\to0}\frac{x^2+2\cos x-2}{x \sin^3x}[/imath]
How to evaluate [imath]\lim_{x\to0}\frac{x^2+2\cos x-2}{x \sin^3x}?[/imath] I tried using L'Hospital but it seemed that there is an infinite loop of 0/0 form | 1549506 | To prove the limit of [imath]\frac{x^2+2\cos x-2}{x\sin^3 x}[/imath] at zero is [imath]1/12[/imath]
To Prove [imath]\lim_{x \to 0}\frac{x^2+2\cos x-2}{x\sin^3 x}=\frac{1}{12}[/imath] I tried with L'Hospital rule but in vain. |
2844480 | If [imath]H[/imath] and [imath]K[/imath] are normal subgroups with [imath]H\cap K=\{e\}[/imath], then [imath]xy=yx[/imath] for [imath]x\in H, y\in K[/imath].
Claim : [imath]H[/imath] and [imath]K[/imath] are two normal subgroups of group [imath]G[/imath] such that [imath]H\cap K = \{e\}[/imath] then [imath] xy = yx [/imath] for [imath]x \in H[/imath] and [imath]y \in K[/imath]. Proof : Let [imath]x \in H[/imath] and [imath]y\in K[/imath] then I need to prove that [imath]xy =yx[/imath]. Let us assume that [imath]y^{-1}xy \neq x[/imath] [imath]y^{-1}xy = z[/imath] where [imath]z \in H[/imath] Question : I am not getting how to proceed further? | 253131 | [imath]H_1 ,H_2 \unlhd \, G[/imath] with [imath]H_1 \cap H_2 = \{1_G\} [/imath]. Prove every two elements in [imath]H_1, H_2[/imath] commute
This is the proof, which I mostly understand except for one bit: You have [imath]h_1 \in H_1[/imath] and [imath]h_2 \in H_2[/imath]. We also have [imath]h_1^{-1}(h_2^{-1}h_1h_2) \in H_1[/imath], because [imath]h_2^{-1}h_1h_2 \in h_2^{-1}H_1h_2 = H_1[/imath]. Similarly, we have [imath](h_1^{-1}h_2^{-1}h_1)h_2 \in H_2[/imath]. Therefore [imath] h_1^{-1}h_2^{-1}h_1h_2 \in H_1 \cap H_2 = \{1_G\} [/imath] and so [imath]h_1^{-1}h_2^{-1}h_1h_2 = \{1_G\}[/imath]. Let's first multiply everything on the left by [imath]h_1[/imath] [imath]h_1^{-1}h_2^{-1}h_1h_2 = \{1_G\}[/imath] [imath] h_1 h_1^{-1}h_2^{-1}h_1h_2 = h_1 \{1_G\}[/imath] [imath] h_2^{-1}h_1h_2 = h_1 \{1_G\}[/imath] Multiply both sides on the left by [imath]h_2[/imath] giving us [imath] h_1h_2 = h_2 h_1 [/imath] The bit I don't get is right at the beginning. Why is this correct: "We also have [imath]h_1^{-1}(h_2^{-1}h_1h_2) \in H_1[/imath], because [imath]h_2^{-1}h_1h_2 \in h_2^{-1}H_1h_2 = H_1[/imath]." |
2844821 | if [imath]{m/n}[/imath] is an approximation to [imath]\sqrt{2}[/imath], prove [imath]m/2n+n/m[/imath] is always a better approximation.
I noticed [imath](m/2n+n/m)^2 -2[/imath] = [imath](m/2n-n/m)^2[/imath]. So I set [imath](m/n)^2 - 2 = d[/imath] and substituted value for n in the above equation but can't show that resulting equation is less than d. | 1787700 | Approximation of [imath]\sqrt{2}[/imath]
I got the following problem in a chapter of approximations: If [imath]\frac{m}{n}[/imath] is an approximation to [imath]\sqrt{2}[/imath] then prove that [imath]\frac{m}{2n}+\frac{n}{m}[/imath] is a better approximation to [imath]\sqrt{2}.[/imath](where [imath]\frac{m}{n}[/imath] is a rational number) I am stuck in this problem. Any help will be appreciated. |
2844523 | Show that [imath]\sqrt b \in \mathbb Q(\sqrt{a^2-b})[/imath]
Let [imath]L[/imath] be the splitting field of [imath]f=X^4-2aX^2+b\in\mathbb Q[X][/imath]. Show that [imath]\sqrt b \in \mathbb Q(\sqrt{a^2-b})=:K[/imath] if [imath][L:\mathbb Q]=4[/imath]. EDIT: [imath]a,b[/imath] are chosen so that [imath]f[/imath] is irreducible. Can I argue like this? If we have [imath]\sqrt b\notin K[/imath], then [imath]K(\sqrt b)[/imath] is an extension of at least [imath]2[/imath] of [imath]K[/imath]. On the other hand, since [imath]f[/imath] is irreducible we have that [imath]\sqrt{a^2-b}\notin \mathbb Q[/imath] and therefor [imath]K/\mathbb Q[/imath] is an extension of degree [imath]2[/imath]. So, if [imath][L:K]=4[/imath] we obtain [imath]L=K(\sqrt b)=\mathbb Q(\sqrt b,\sqrt{a^2-b})[/imath] by degree comparison. But then [imath]L[/imath] is the splitting field of [imath]g=(X^2-b)(X^2-a^2+b)\neq f[/imath] which is in conflict with the uniqueness of splitting fields. | 2824792 | Galois extension with [imath]X^4-2aX^2+b\in\mathbb Q[X][/imath]
Considering the irreducible (by condition) polynomial [imath]f=X^4-2aX^2+b\in\mathbb Q[X][/imath] and the field extension [imath]K=\mathbb Q(\sqrt{a^2-b})[/imath] of [imath]\mathbb Q[/imath], I want to show the following result: For [imath][L:\mathbb Q]=4[/imath] we have [imath]\sqrt b\in K[/imath] where [imath]L[/imath] is the splitting field of [imath]f[/imath]. My thoughts so far: The four roots of [imath]f[/imath] are [imath]\pm\sqrt{a\pm\sqrt{a^2-b}}[/imath]. If [imath]L[/imath] is of degree [imath]4[/imath] it must be primitive since [imath]f[/imath] is irreducible, so [imath]L=\mathbb Q\left(\sqrt{a+\sqrt{a^2-b}}\right)[/imath]. Then, by multiplying the two roots [imath]\sqrt{a+\sqrt{a^{2}-b}}\cdot\sqrt{a-\sqrt{a^{2}-b}}=\sqrt{b}[/imath] we have that [imath]\sqrt{b}\in L[/imath]. Now I want to look at [imath]\mathbb Q\subset K\subset K(\sqrt b)\subset L[/imath] and differentiate some cases. The case where [imath][L:K]=1[/imath] is trivial and the case [imath][L:K]=4[/imath] can't happen because if we square any root and subtract [imath]a[/imath], we get [imath]\pm\sqrt{a^2-b}\in K[/imath] which means [imath][L:K]\leq 2[/imath]. So let's look at the case where [imath][L:K]=2[/imath]. As I see it, we get two subcases: For [imath][L:K(\sqrt b)]=2[/imath] we have [imath]K(\sqrt b)=K[/imath] which finishes the proof. For [imath][L:K(\sqrt b)]=1[/imath] we have [imath]K(\sqrt b)=L[/imath]. This means that [imath]L[/imath] is also the splitting field of [imath](x^2-b)(x^2-(a^2-b))[/imath]. Is this a contradiction? How do I finish this case? |
2844077 | Prove that [imath]-1[/imath] is an eigenvalue of an orthogonal matrix [imath]A \in M_{4 \times 4} (\Bbb R)[/imath] with [imath]\det(A)=-1[/imath]
Let [imath]A \in M_{4 \times 4} (\Bbb R)[/imath] be an orthogonal matrix with [imath]\det(A)=-1[/imath]. Prove that [imath]-1[/imath] is an eigenvalue of [imath]A[/imath]. I'm a bit lost. I know about all the basic orthogonal matrices' properties (including the ones about scalar product). I also know that orthogonal matrices' eigenvalues are [imath]\pm 1[/imath]. Any tips, please? | 335748 | If [imath]P[/imath] is real orthogonal matrix with [imath]\det P = -1[/imath], prove that [imath]-1[/imath] is an eigenvalue of [imath]P[/imath].
If [imath]P[/imath] is real orthogonal matrix with [imath]\det P = -1[/imath], prove that [imath]-1[/imath] is an eigenvalue of [imath]P[/imath]. can anyone help me please how can I solve this problem? |
555147 | Rational solutions of Pell's equation
1) [imath]D[/imath] is a positive integer, find all rational solutions of Pell's equation [imath]x^2-Dy^2=1[/imath] 2) What about [imath]D\in\Bbb Q[/imath] ? | 3044517 | How can we solve the diophantine equation?
How can we find all the primitive solutions of the diophantine equation [imath]x^2+3y^2=z^2[/imath] ? Some solutions are [imath](\pm n , 0 , \pm n)[/imath] for [imath]n\in \mathbb{N}[/imath]. How can we find also the other ones? |
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