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1844998
|
The set of natural numbers that don't belong to a set
Describe which natural numbers do not belong to the set [imath]E = \left\{\left[n+\sqrt{n}+\frac{1}{2}\right] \mid n \in \mathbb{N}\right\}.[/imath] The answer is the set of positive perfect squares. I am not sure how to prove it. We know that [imath]\left[n+\sqrt{n}+\frac{1}{2}\right] = n+\left[\sqrt{n}+\frac{1}{2} \right][/imath], but what do I do from here?
|
850483
|
[imath]m[/imath] doesn't come in the sequence [imath]a_n=[n+\sqrt{n}+\frac{1}{2}][/imath] iff m is a perfect square
Let [imath](a_n)[/imath] be a sequence of positive integers defined by [imath]a_n=[n+\sqrt{n}+\frac{1}{2}][/imath] Show that a natural number [imath]m[/imath] doesn't occur in the sequence iff it is a square.
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1846139
|
Natural log operation on [imath]e^{i \tau}[/imath]
I have a layman's question: [imath]e^{i\pi} = -1[/imath] ... euler's identity and [imath]e^{i\tau}[/imath] = 1 where [imath]\tau = 2\pi[/imath] so if I assume that [imath]e^x=y[/imath] implies [imath]ln(y)=x[/imath], which it should, then [imath]ln(1) = i\tau[/imath] but [imath]ln(1) = 0 \ne i\tau[/imath] What am I doing wrong?
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1319940
|
What's wrong with this?
What's wrong with this : [imath]e^{i\pi} = -1[/imath] [imath]\therefore e^{2i\pi} = 1[/imath] [imath]\therefore log \left( e^{2i\pi} \right ) = log(1) = 0[/imath] [imath]\therefore 2i\pi = 0[/imath]
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1846376
|
Helping understanding finding a limit without L'hopital's rule?
[imath]\lim \limits_{x \to 0}x\sin\left(\frac{1}{x}\right)[/imath] I need to find and prove this limit. I can easily plug it into wolfram alpha, but I want to make sure I learn something. It's been 3 years since my Calculus 2 course and I just cannot remember how to tackle this one. Hints?
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1066434
|
Calculate: [imath] \lim_{x \to 0 } = x \cdot \sin(\frac{1}{x}) [/imath]
Evaluate the limit: [imath] \lim_{x \to 0 } = x \cdot \sin\left(\frac{1}{x}\right) [/imath] So far I did: [imath] \lim_{x \to 0 } = x\frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x}\cdot x} [/imath] [imath] \lim_{x \to 0 } = 1 \cdot \frac{x}{x} [/imath] [imath] \lim_{x \to 0 } = 1 [/imath] Now of course I've looked around and I know I'm wrong, but I couldn't understand why. Can someone please show me how to evaluate this limit correctly? And tell me what I was doing was wrong.
|
1845818
|
An example of subset [imath]A[/imath] such that [imath]A \cap K[/imath] is open in [imath]K[/imath] for each compact set [imath]K[/imath], but [imath]A[/imath] is not open.
Let [imath]X[/imath] be a topological space. For any [imath]A \subseteq X[/imath], consider two possible conditions on [imath]A[/imath]: 1) [imath]A[/imath] is open in [imath]X[/imath]; 2) [imath]A \cap K[/imath] is open in [imath]K[/imath], for each compact set [imath]K \subseteq X[/imath]. Then [imath](1)[/imath] implies [imath](2)[/imath]. but, necessarily [imath](2)[/imath] doesn't imply [imath](1)[/imath]. Give an example of [imath]A[/imath] such that [imath]A[/imath] satisfies in [imath](2)[/imath] but [imath]A[/imath] is not open in [imath]X[/imath].
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1241768
|
Spaces in which "[imath]A \cap K[/imath] is closed for all compact [imath]K[/imath]" implies "[imath]A[/imath] is closed."
Let [imath]X[/imath] denote a topological space. For any [imath]A \subseteq X[/imath], consider two possible conditions on [imath]A[/imath]. [imath]A[/imath] is closed [imath]A \cap K[/imath] is closed, for all compact [imath]K \subseteq X[/imath]. If [imath]X[/imath] is Hausdorff, then clearly (1) implies (2). If [imath]X[/imath] is metrizable, then the converse holds; see here. Questions. What are some Hausdorff topological spaces in which (2) does not imply (1)? For Hausdorff topological spaces: what weaker conditions than metrizability are known to imply that (2) implies (1)? Are there any interesting characterizations of those Hausdorff spaces in which (2) implies (1)?
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1846384
|
Interesting Trigonometry Identity
Prelude While studying trigonometry, I came across this very interesting problem. It wasn't very difficult to solve, however it's result was quite interesting. I have given the solution below. Try to solve it yourself before looking. Problem The diagram below shows three equal squares, with angels [imath]\alpha[/imath], [imath]\beta[/imath], [imath]\gamma[/imath] as marked. Prove that [imath] \alpha + \beta = \gamma [/imath]. Hint If you are not in the midst of a trigonometry course, chances are you will need to see this hint, as it is an uncommon identity. Hover below to see a hint. [imath] \arctan a + \arctan b = \arctan{\frac{a + b}{1 - ab}} [/imath] This identity was solved earlier in the book I found this problem in (see Postscript). Try to prove this identity for yourself, too! Postscript This problem came from the textbook: "Trigonometry" by I. M. Gelfand.
|
1605964
|
Show that the angles satisfy [imath]x+y=z[/imath]
How can I show that [imath]x+y=z[/imath] in the figure without using trigonometry? I have tried to solve it with analytic geometry, but it doesn't work out for me.
|
1843658
|
How would I find the nth derivative of a function, where n is imaginary? What about where n is not a constant?
Forgive me if this question has already been asked. I was unable to find anything relevant to this question. The [imath]n[/imath]th derivative of a function, [imath]f^n(x)[/imath] is well-defined for [imath]n\in\mathbb{Z}^+[/imath]. As well, it can be defined for the case where [imath]n\in\mathbb{R}^+[/imath], (i.e. fractional derivatives). What about in the case where [imath]n\in\mathbb{C}[/imath]? What would it mean to take the imaginary derivative of a function? Similarly, let [imath]n[/imath] be a function of some variable, [imath]x[/imath], (i.e. [imath]n=\cos(x)[/imath]). Could you define [imath]f^n(x)=f^{\cos(x)}(x)[/imath]? And again, what would it even mean to take the [imath]\cos(x)[/imath]th derivative of a function? As well, if [imath]F'(x)=f(x)[/imath], does [imath]f^{-1}(x)=\int\,f(x)\,dx=F(x)[/imath], where [imath]f^{-1}[/imath] is the negative derivative, and not the inverse? Edit: Duplicates do not answer the final part of this question (i.e. the [imath]\cos(x)[/imath]th derivative of a function). Edit 2: It's interesting because, let's say that [imath]f^{-1}(x)=\int\,f(x)\,dx=F(x)[/imath]. Operator theory tells us that when we apply two half derivatives in succession to a function (i.e. [imath]D^{\frac{1}{2}}D^{\frac{1}{2}}f(x)[/imath], that we obtain the first derivative of that function [imath]f(x)[/imath]. This is consistent with the idea that the antiderivative is nothing more than [imath]D^{-1}f(x)=D^{-1}F'(x)[/imath]. But consider the complex derivative [imath]f^i(x)[/imath]. Following operator theory, one would claim that the antiderivative of this function could be written as [imath]f^{-i}(x)[/imath]. However, this is obviously not the case as this implies that [imath]f(x)=f^0(x)=f'{x}[/imath], recalling that [imath]i^2=-1[/imath], such that [imath](i)(-i)=1[/imath].
|
1331819
|
Non-integer order derivative
I do not know much about fractional calculus, except what I have read in a few short posts at MSE and https://en.wikipedia.org/wiki/Fractional_calculus. I know that order of a derivative can be extended from rational values to real values, but all I know is what is written here. So my question is, what is the simplest way to understand and/or define [imath]\dfrac{d^n}{dx^n}f(x),[/imath] for [imath]n\in\mathbb{R}[/imath]? For example what is [imath]\dfrac{d^\pi}{dx^\pi}f(x)[/imath]? Also, what about complex values? What might we say about [imath]\dfrac{d^{s}}{dz^s}f(z)[/imath] for [imath]s\in\mathbb{C}[/imath]? And what are some uses for such derivatives?
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1846729
|
Synthetic division for: [imath]\frac{60 x^{3}+43x^{2}-34x-24}{3x+2}[/imath]
If I have a polynomial to which the solutions are integers, in this case, I know how to perform the synthetic division. Also, I know how to perform the present division using long division. But I don't know how to do it with synthetic division. I watched videos on youtube, but the polynomials presented in the shows have integer solutions. So I would greatly appreciate If you could walk me through the solution using synthetic division. I'm sure I'll learn something new. Thank you. [imath]\frac{60 x^{3}+43x^{2}-34x-24}{3x+2}[/imath] Edit: The answer you're referring to has two part: One part giving the rule and I don't know how to translate it into the question in the hand; and the second part is using long division. If the long division is the only way, please let me know and I'll accept that as an answer.
|
825470
|
Synthetic division of polynomials by factor of the form [imath](ax+b)[/imath]
Find the quotient and remainder when [imath]6x^4-11x^3+5x^2-7x+9[/imath] is divided by [imath](2x-3)[/imath]. I expressed the divisor [imath](2x-3)[/imath] as [imath]2(x-\frac{3}{2})[/imath] and conducted synthetic division by [imath]\frac{3}{2}[/imath] and obtained the coefficients of the quotients as [imath]6, -2, 2[/imath] and [imath]-4[/imath] and the remainder as [imath]3[/imath]. Then I divided the quotients by [imath]2[/imath]. Why should I do this? and why should I not divide the remainder by [imath]2[/imath]?
|
1846869
|
Simplifying $\frac{2\sqrt{19}}{3}\cos{\left(\frac{1}{3}\arccos{\frac{7}{\sqrt{76}}}\right)}-\frac{1}{3}$
Using Vieta's Trignometric Solutions, I am getting [imath]\dfrac{2\sqrt{19}}{3}\cos{\left(\dfrac{1}{3}\arccos{\dfrac{7}{\sqrt{76}}}\right)}-\dfrac{1}{3}\tag{1}[/imath] And it is also equal to [imath]2\left(\cos\left(\frac {4\pi}{19}\right)+\cos\left(\frac {6\pi}{19}\right)+\cos\left(\frac {10\pi}{19}\right)\right)\tag{2}[/imath] My Question: How would you get from [imath](1)[/imath] to [imath](2)[/imath], and is there a way to generalize it? Anything helps!
|
1837628
|
Roots of [imath]y=x^3+x^2-6x-7[/imath]
I'm wondering if there is a mathematical way of finding the roots of [imath]y=x^3+x^2-6x-7[/imath]? Supposedly, the roots are [imath]2\cos\left(\frac {4\pi}{19}\right)+2\cos\left(\frac {6\pi}{19}\right)+2\cos\left(\frac {10\pi}{19}\right)[/imath], [imath]2\cos\left(\frac {2\pi}{19}\right)+2\cos\left(\frac {14\pi}{19}\right)+2\cos\left(\frac {16\pi}{19}\right)[/imath] and [imath]2\cos\left(\frac {8\pi}{19}\right)+2\cos\left(\frac {12\pi}{19}\right)+2\cos\left(\frac {18\pi}{19}\right)[/imath]. Anything helps. I don't think substituting [imath]x[/imath] with something like [imath]t+t^{-1}[/imath] will work.
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1847800
|
Types of simple Lie algebra
What exactly is Simple algebra of type [imath]A_2[/imath]? I found that it has something to do with root systems, which I also don't really know what those are. Any idea? Thanks!
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1652408
|
The root system of [imath]sl(3,\mathbb C)[/imath]
I want to determine the root-system of the lie algebra [imath]sl(3,\mathbb C)[/imath]. Does someone know a good (and complete) reference for this problem? I know that the root-system is [imath]A_2[/imath] but I want to see a complete proof (calculation) for this. Thanks in advance.
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1847343
|
Is [imath]\mathbb{R}[/imath] a finite field extension?
Is there a field [imath]K \subseteq \mathbb{R}[/imath] with [imath][\mathbb{R} : K] < \infty[/imath]? My intuition tells me no: I would imagine that [imath]K(\alpha)[/imath] would be missing some [imath]n[/imath]th root of [imath]\alpha[/imath] for all [imath]x \in \mathbb{R} \setminus K[/imath]. But I'm not sure how to approach this: it's certainly not true of all finite field extensions (e.g. [imath]\mathbb{R} \subseteq \mathbb{C}[/imath]), and there are irrationals [imath]\beta[/imath] s.t. [imath]\beta[/imath] has arbitrarily (but finitely) many roots in [imath]\mathbb{Q}(\beta)[/imath]: e.g. [imath]n[/imath]th powers of numbers satisfying [imath]x^n - x - 1[/imath].
|
211582
|
Is there a proper subfield [imath]K\subset \mathbb R[/imath] such that [imath][\mathbb R:K][/imath] is finite?
Is there a proper subfield [imath]K\subset \mathbb R[/imath] such that [imath][\mathbb R:K][/imath] is finite? Here [imath][\mathbb R:K][/imath] means the dimension of [imath]\mathbb R[/imath] as a [imath]K[/imath]-vector space. What I have tried: If we can find a finite subgroup [imath]G\subset Gal (\mathbb C/\mathbb Q)[/imath] such that [imath]G[/imath] contains the complex conjugation, it will be done by letting [imath]K[/imath] be the fixed field of [imath]G[/imath]. But I don't know whether such a group exists. Maybe we can start with finding a suitable subgroup of [imath]Gal(\bar{\mathbb Q}/\mathbb Q)[/imath] and then lift it to [imath]Gal(\mathbb C/\mathbb Q)[/imath], where [imath]\bar{\mathbb Q}[/imath] denotes the algebraic closure of [imath]\mathbb Q[/imath]. By isomorphism extension theorem, we can find many automorphisms of [imath]\mathbb C[/imath], none of them carries [imath]\mathbb R[/imath] to itself except for the identity. This is because [imath]Gal(\mathbb R/\mathbb Q)[/imath] is the trivial group. For example, now suppose [imath]\{x_\alpha\}\subset \mathbb R[/imath] is a transcendence basis over [imath]\mathbb Q[/imath]. Let [imath]\sigma[/imath] be a permutation of [imath]\{x_\alpha\}[/imath], then by the isomorphism extension theorem, [imath]\sigma[/imath] extends to an automorphism of [imath]\mathbb C[/imath], which we still denote by [imath]\sigma[/imath]. Then [imath]L=\sigma(\mathbb R)[/imath] is a copy of [imath]\mathbb R[/imath] and [imath]\mathbb R[/imath] is algebraic over [imath]K=\mathbb R\cap L[/imath]. How large can [imath]K[/imath] be? Is it possible that [imath][\mathbb R:K][/imath] is finite? Does anyone has some ideas? Thanks!
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1847557
|
Find all the algebraic operations
How many algebraic operations can be defined on a set S with n elements and how many of them are comutative. Ok so I know that a algebraic operation is a law (let's say *) such that [imath]*:S\times S \rightarrow S[/imath]. So for each pair of the cartesian product associate a element from S. I was thinking like this: Let [imath]A = \{i \in \mathbb{N}|1\leq i \leq |S \times S|\}[/imath], and let [imath]A' = \{\overline{a_1a_2...a_n}\|n = A \ and\ a_i \in A, \forall i = \overline{1,n}\}[/imath]. So the cardinal of A' is the answare.
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375529
|
Counting binary operations on a set with [imath]n[/imath] elements
I am trying to solve following problem but not able to find any way to proceed. Let [imath]S[/imath] be a set having [imath]n[/imath] elements. Can we count about number of binary operations that can be defined on a set? Can we also count number of commutative binary operations defined on [imath]S[/imath]? Thanks for the help and suggestions
|
1848567
|
Finding number of ways of selecting 6 gloves each of different colour from 9 pair of gloves?
There are nine pairs of gloves each of different colors in how many ways can we arrange six gloves such that each is of different color? I tried like this : First number of ways in which we can select 6 colors out of nine it is [imath]^9C_6[/imath] ways. Then we can arrange it in 6! ways and also we can select either of two gloves out of the 6 in [imath]2^6[/imath] ways. Hence required number of ways should be [imath]^9C_6[/imath] .[imath]2^6[/imath].6!
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1848331
|
Finding number of ways of selecting 6 gloves each of different colour from 18 gloves?
There are nine pairs of gloves each of different colors in how many ways can we select six gloves such that each is of different color? I tried like this : First number of ways in which we can select 6 colors out of nine it is [imath]^9C_6[/imath] ways. Then we can arrange it in 6! ways and also we can select either of two gloves out of the 6 in [imath]2^6[/imath] ways. Hence required number of ways should be [imath]^9C_6[/imath] .[imath]2^6[/imath].6!
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1848605
|
Square matrix A is singular [imath]\iff \operatorname{rank}(A)?[/imath]
I know that if a matrix is not of a full rank than it is singular, but is it always true that singular matrices are of non-full ranks?
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592562
|
A square matrix [imath]n \times n[/imath] is an invertible matrix iff the rank of the matrix is [imath]n[/imath].
Why does the following sentence is true? A square matrix [imath]n \times n[/imath] is an invertible matrix iff the rank of the matrix is [imath]n[/imath].
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1804135
|
Find [imath]\sum\limits_{r=0}^n(-1)^r\binom{n}{r}^{-1}[/imath] for [imath]n[/imath] even
If [imath]n[/imath] is an even natural number, then find [imath]\sum_{r=0}^n \left(\frac{(-1)^r}{\binom{n}{r}}\right)[/imath] I tried to solve the question using conventional method, by trying to use calculus, but I don't think that would be applicable here, because no binomial expansion (as far I as know) can give a coefficient in denominator hence I got stuck at it.
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1402886
|
What is [imath]\sum_{r=0}^n \frac{(-1)^r}{\binom{n}{r}}[/imath]?
Find a closed form expression for [imath]\sum_{r=0}^n \dfrac{(-1)^r}{\dbinom{n}{r}}[/imath] where [imath]n[/imath] is an even positive integer. I tried using binomial identities, but since the binomial coefficient is in the denominator, I can't do much. Also, I'm seeking a non-calculus, elementary solution, if possible. Thanks.
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1849011
|
What is the actual meaning of power which is less than 1?
[imath]3^4 = 3 \cdot 3 \cdot 3 \cdot 3[/imath] [imath]3^2 = 3 \cdot 3[/imath] [imath]3^1 = 3[/imath] But my problem is for below 1: Example: [imath]3^{0.7} = \ ?[/imath] Can you please explain it to clear my doubt?
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932496
|
Why is the expression [imath]\underbrace{n\cdot n\cdot \ldots n}_{k \text{ times}}[/imath] bad?
For writing a (german) article about the power with natural degree I have the following question: In school one defines the power with natural degree via [imath]n^k = \underbrace{n\cdot n\cdot \ldots \cdot n}_{k \text{ times}}[/imath] In calculus normally recursion is used to introduce the power: [imath]\begin{align} n^0 & := 1 \\ n^{k+1} & := n \cdot n^k \end{align}[/imath] Question: What are the disadvantages of the expression [imath]\underbrace{n\cdot n\cdot \ldots \cdot n}_{k \text{ times}}[/imath] and why it is not used to define the power in advanced mathematics (although it is intuitive)? Note: I'm sorry, I had an ambiguous question before. I do not want to know, why we define the power. I want to know, why we use recursion and not the expression [imath]\underbrace{n\cdot n\cdot \ldots \cdot n}_{k \text{ times}}[/imath] to define the power in advanced mathematics... My ideas so far: [imath]\underbrace{\ldots}_{k \text{ times}}[/imath] is no operator, which was introduced before. definitions with recursion lead naturally to a scheme, how properties of these concepts can be proved via induction
|
1846538
|
Set-builder notation for [imath]\{1 , 3 , 5 , -7 , -9 , -11 , 13 , 15 , 17 , \ldots\}[/imath]
I need a set-builder notation for [imath]\{1 , 3 , 5 , -7 , -9 , -11 , 13 , 15 , 17 , \ldots\}[/imath]. It is an infinite set of odd positive integers, which are multiplied by [imath]-1[/imath] three by three. One of the terms of this question is not to use floor function.
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1839378
|
Three negative and three positive [imath]1[/imath]s in a serie
I want an alternating series formula that generates three negatives and three positives together.I know that we can do the same except that it is two negative and three positive can performed by triangle numbers.(See the link below.But I cannot find anything that generates three negative and three positive.For this we should do three odd and three even series and then put it in the power of [imath]-1[/imath]. Recommended:How to create alternating series with happening every two terms update1:The series:[imath]{1,1,1,-1,-1,-1,1,1,1,-1,-1,-1,...}[/imath] update2:We didn't learned floor function or trigonometry yet.
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1848849
|
Where is the logical error resulting in my negative variance?
I have three random variables [imath]X,Y,Z[/imath] that all have a mean of [imath]\mu=10[/imath] and a standard deviation of [imath]\sigma=2[/imath]. Now, I would like to say that they are negatively correlated with [imath]\rho_{X,Y}=-1, \rho_{Y,Z}=-1, \rho_{X,Z}=-1[/imath]. My question would be, what is the variance of [imath]X+Y+Z[/imath]? Normally, we would say: [imath]Var(X+Y+Z)=Var(X)+Var(X)+Var(Z)+2Cov(X,Y)+2Cov(Y,Z)+2Cov(X,Z)[/imath] However, in our simple example, this leads to [imath]3\cdot2^2+3\cdot2\cdot(-1)\sqrt{2^2\cdot2^2}=-12[/imath] which cannot make sense. Where is my logical error? I presume because of some transitivity, [imath]\rho_{X,Z}[/imath] cannot be [imath]-1[/imath] if the other two [imath]\rho_i[/imath] are -1 but which principle is behind this?
|
284877
|
Correlation between three variables question
I was asked this question regarding correlation recently, and although it seems intuitive, I still haven't worked out the answer satisfactorily. I hope you can help me out with this seemingly simple question. Suppose I have three random variables [imath]A[/imath], [imath]B[/imath], [imath]C[/imath]. Is it possible to have these three relationships satisfied? [imath] \mathrm{corr}[A,B] = 0.9 [/imath] [imath] \mathrm{corr}[B,C] = 0.8 [/imath] [imath] \mathrm{corr}[A,C] = 0.1 [/imath] My intuition is that it is not possible, although I can't see right now how I can prove this conclusively.
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1849281
|
Average distance between two points on a unit square.
Consider the unit square [imath]S =[0,1]\times[0,1][/imath]. I'm interested in the average distance between random points in the square. Let [imath] \mathbf{a} = \left< x_1,y_1 \right>[/imath] and [imath] \mathbf{b} = \left< x_2,y_2 \right>[/imath] be random points in the unit square. By random, I mean that [imath]x_i[/imath] and [imath]y_i[/imath] are uniformly distributed on [imath][0,1][/imath]. The normal approach is to use multiple integration to determine the average value of the distance between [imath]\mathbf{b}[/imath] and [imath]\mathbf{a}[/imath]. I would like to try another approach. [imath]\mathbf{a}[/imath] and [imath]\mathbf{b}[/imath] are random vectors, and each element has known distribution. So, the vector between them also has known distribution. The difference between two uniformly random variables has triangular distribution. So [imath]\mathbf{c} = \mathbf{b} - \mathbf{a}[/imath]. Then, the average distance is the expectation of [imath]\lVert \mathbf{c} \rVert[/imath]. Perhaps it would be easier to calculate the expectation of [imath]\lVert \mathbf{c} \rVert^2[/imath]. In any case, I am not sure how to calculate the expectation for [imath]\lVert \mathbf{c} \rVert^2[/imath]. Can someone guide me in the right direction?
|
1254129
|
Average distance between two random points in a square
A square with side [imath]a[/imath] is given. What is the average distance between two uniformly-distributed random points inside the square? For more general "rectangle" case, see here. The proof found there is fairly complex, and I am looking for a simpler proof for this special case. I expect it could be significantly simpler. See also "line" case.
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1849907
|
Proving if [imath]\lim_{n\rightarrow\infty}a_n=L [/imath] then [imath]\lim_{n\rightarrow\infty} \frac{a_1+a_2+\cdots+a_n}n=L [/imath]
Good morning, i have a big problem with this proof. I haven't idea about this proof. Problem: Suppose [imath]\lim_{n\rightarrow\infty}a_n =L [/imath] then [imath]\lim_{n\rightarrow\infty}\ \frac{a_1+a_2+\cdots+a_n}n=L [/imath] I've tried this: [imath]| a_n-L|<\varepsilon\Rightarrow L-\varepsilon<a_n<L+\varepsilon[/imath] Please help!!
|
1439987
|
Suppose [imath]\lim \limits_{n \to ∞} a_n=L[/imath]. Prove that [imath]\lim\limits_{n \to ∞} \frac{a_1+a_2+\cdots+a_n}{n}=L[/imath]
I haven't been able to solve this by using the definition of limit. I think maybe trying induction will work, though. Can someone help? Suppose [imath]\lim \limits_{n \to ∞} a_n=L[/imath]. Prove that [imath]\lim\limits_{n \to ∞} \dfrac{a_1+a_2+\cdots+a_n}{n}=L[/imath]
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1568696
|
How to prove that if [imath]\lim_{n \rightarrow \infty}a_n=A[/imath], then [imath]\lim_{n \rightarrow \infty}\frac{a_1+...+a_n}{n}=A[/imath]
I'm wondering if I have a sufficient proof of the following: If [imath](a_n)[/imath] is a sequence such that [imath]\lim_{n \rightarrow \infty}a_n=A[/imath], then [imath]\lim_{n \rightarrow \infty}\frac{a_1+...+a_n}{n}=A[/imath]. My approach: For all [imath]\varepsilon > 0[/imath], there exists [imath]N[/imath] such that for all [imath]k>N[/imath], [imath]|a_k -A|<\varepsilon[/imath]. So we can break the limit up as follows [imath]\lim_{n\rightarrow \infty} \frac{a_1+...+a_k}{n} + \lim_{n \rightarrow \infty}\frac{a_{k+1}+...+a_n}{n} \overset{\epsilon \rightarrow 0}{=} 0 + \lim_{n \rightarrow \infty}\frac{nA}{n}=A[/imath] Is this on the right track, or am I missing something about breaking up the limit in the way I have?
|
1933042
|
Prove that, if [imath]\lim_{j\rightarrow\infty}a_j= \ell[/imath], then [imath]\lim_{j\rightarrow\infty}m_j= \ell[/imath].
Let [imath]a_j[/imath] be a sequence of real numbers. Define [imath]m_j=\frac{a_1+a_2+\dots +a_j}{j}[/imath]. Prove that, if [imath]\lim_{j\rightarrow\infty}a_j= \ell[/imath], then [imath]\lim_{j\rightarrow\infty}m_j= \ell[/imath]. Give an example to show that the converse is not true. I have been thinking about this question a lot. I know I could rewrite this as [imath]m_j=\frac{\sum_{j=1}^\infty a_j}{j}[/imath] But I am having a really hard time after that. Thanks for any help you can give.
|
1850353
|
Why is the characteristic function of a probability distribution function uniformly continuous?
Why is the characteristic function of a probability distribution function uniformly continuous? This is from page151 of Chung's A Course in Probability Theory. Specifically, why is the last inequality true? [imath]|\phi(t+u)-\phi(t)|=\left|\int (e^{i(t+u)x}-e^{itx})dF(x)\right|\leq\int |e^{iux}-1|dF(x)[/imath].
|
108657
|
Continuity of the Characteristic Function of a RV
Defining the Characteristic Function [imath] \quad \phi(t) := \mathbb{E} \left[ e^{itx} \right] [/imath] for a random variable with distribution function [imath]F(x)[/imath] in order to show it is uniformly continuous I say [imath] |\phi(t+u) - \phi(t)| = \left |\int e^{itx}(e^{iux} - 1) dF(x) \right| \le \\ \int 1 \cdot|e^{iux} -1|dF(x) \to 0 \quad as \quad u\to0 [/imath] Now my question is, does the convergence I state in the last line follow directly, or do I need to be a little carful before I conclude it is true ? (i.e. can I directly use that [imath]|e^{itu} -1| \to 0 ? )[/imath]
|
1712938
|
Prove that [imath]\limsup s_n + \liminf t_n \leq \limsup (s_n + t_n)[/imath]
Prove that [imath]\limsup s_n + \liminf t_n \leq \limsup (s_n + t_n)[/imath]. [imath]s_n[/imath] and [imath]t_n[/imath] are bounded. I don't know how to start this, can anyone help and give a formal proof of this?
|
70478
|
Properties of [imath]\liminf[/imath] and [imath]\limsup[/imath] of sum of sequences: [imath]\limsup s_n + \liminf t_n \leq \limsup (s_n + t_n) \leq \limsup s_n + \limsup t_n[/imath]
Let [imath]\{s_n\}[/imath] and [imath]\{t_n\}[/imath] be sequences. I've noticed this inequality in a few analysis textbooks that I have come across, so I've started to think this can't be a typo: [imath]\limsup\limits_{n \rightarrow \infty} s_n + \liminf\limits_{n \rightarrow \infty} t_n \leq \limsup\limits_{n \rightarrow \infty} (s_n + t_n) \leq \limsup\limits_{n \rightarrow \infty} s_n + \limsup\limits_{n \rightarrow \infty} t_n[/imath]. I understand the right two-thirds of the equation, and can prove it. I'm stuck on the far-left third. I don't understand how that is true.
|
1850735
|
Prove: [imath]\mid\sum_{i=1}^{n} x_i\mid\leq \sum_{i=1}^{n}\mid x_i\mid[/imath]
[imath]\mid\sum_{i=1}^{n} x_i\mid\leq \sum_{i=1}^{n}\mid x_i\mid[/imath] If [imath]n[/imath] is even we will divide the sum into groups of [imath]2[/imath] [imath]x[/imath]'s namely [imath]\mid x+x \mid \leq \mid x\mid+\mid x \mid[/imath] and will repeat the process to get [imath]\mid\sum_{i=1}^{n} x_i\mid\leq \sum_{i=1}^{n}\mid x_i\mid[/imath] If [imath]n[/imath] is odd, we will divide the sum into an even number of [imath]x[/imath]'s called [imath]a[/imath] and the leftover [imath]x[/imath] called [imath]b[/imath], by using the proof for even [imath]n[/imath] and the triangle inequality [imath]\mid a+b\mid\leq \mid a \mid +\mid b \mid= \mid\sum_{i=1}^{n} x_i\mid\leq \sum_{i=1}^{n}\mid x_i\mid[/imath] Is the proof valid?
|
195582
|
General Proof for the triangle inequality
I am trying to prove: [imath]P(n): |x_1| + \cdots + |x_n| \leq |x_1 + \cdots +x_n|[/imath] for all natural numbers [imath]n[/imath]. The [imath]x_i[/imath] are real numbers. Base: Let [imath]n =1[/imath]: we have [imath]|x_1| \leq |x_1|[/imath] which is clearly true Step: Let [imath]k[/imath] exist in the integers such that [imath]k \geq 1[/imath] and assume [imath]P(k)[/imath] is true. This is where I am lost. I do not see how to leverage the induction hypothesis. Here is my latest approach: Can you do the following in the induction step: Let [imath]Y[/imath] = |[imath]x_1[/imath] +...+[imath]x_n[/imath]| and Let [imath]Z[/imath] = |[imath]x_1[/imath]| +...+ |[imath]x_n[/imath]| Then we have: |[imath]Y[/imath] + [imath]x_n+1[/imath]| [imath]\leq[/imath] [imath]Z[/imath] + |[imath]x_n+1[/imath]|. [imath]Y[/imath] [imath]\leq[/imath] [imath]Z[/imath] from the induction step, and then from the base case this is just another triangle inequality. End of proof.
|
1846323
|
Upper limits for [imath]s_n[/imath] [imath]\leq[/imath] [imath]t_n[/imath]
Let [imath](s_n)[/imath] and [imath](t_n)[/imath] be two sequences of real numbers. Suppose there exists [imath]N_0[/imath] such that for all [imath]n>N_0[/imath], [imath]s_n \leq t_n[/imath]. Two Questions: Suppose that [imath]\lim s_n[/imath] and [imath]\lim t_n[/imath] both exist. Show that lim [imath] s_n \leq [/imath]lim [imath] t_n[/imath]. Is this a reasonable answer: Proof: By assumption since there exists [imath]N_0[/imath] such that for all [imath]n>N_0[/imath] then [imath]s_n \leq t_n[/imath] then sup{[imath]t_n: n>N_o[/imath]} [imath]\geq[/imath] sup{[imath]s_n: n>N_o[/imath]} and thus lim sup{[imath]t_n: n>N_o[/imath]} [imath]\geq[/imath] lim sup{[imath]s_n: n>N_o[/imath]} and since both of these limits exist and are equal to lim [imath]t_n[/imath] and lim [imath]s_n[/imath] respectively then the desired result follows. Second question where I am stumped. show that lim inf [imath]s_n[/imath] [imath]\leq[/imath] lim inf [imath]t_n[/imath] and that lim sup [imath]s_n[/imath] [imath]\leq[/imath] lim sup [imath]t_n[/imath]
|
176093
|
Proof that if [imath]s_n \leq t_n[/imath] for [imath]n \geq N[/imath], then [imath]\liminf_{n \rightarrow \infty} s_n \leq \liminf_{n \rightarrow \infty} t_n[/imath]
This is half of Theorem 3.19 from Baby Rudin. Rudin claims the proof is trivial. What I've come up with so far doesn't seem trivial, however, and is probably also wrong (my problem with it is pointed out below). This makes me wonder whether I'm overlooking some useful fact and/or using an unprofitable approach. Theorem. If [imath]s_n \leq t_n[/imath] for [imath]n \geq N[/imath], where [imath]N[/imath] is fixed, then [imath] \liminf_{n \rightarrow \infty} s_n \leq \liminf_{n \rightarrow \infty} t_n. [/imath] Proof. Suppose that [imath]s_n \leq t_n[/imath] if [imath]n \geq N[/imath], but that [imath] \liminf_{n \rightarrow \infty} s_n > \liminf_{n \rightarrow \infty} t_n. [/imath] Let [imath]E_s[/imath] denote the set of all subsequential limits of [imath]\{s_n\}[/imath], and let [imath]E_t[/imath] denote the set of all subsequential limits of [imath]\{t_n\}[/imath]. Then [imath] \inf E_s > \inf E_t. [/imath] This implies that there exists some [imath]x \in E_t[/imath] such that [imath]\inf E_s > x > \inf E_t[/imath], since otherwise [imath]\inf E_t[/imath] would not be the greatest lower bound of [imath]E_t[/imath]. Hence some subsequence of [imath]\{t_n\}[/imath], say [imath]\{t_{n_i}\}[/imath], converges to [imath]x < \inf E_s[/imath]. Lemma. (from Rudin) If [imath]x < \liminf_{n \rightarrow \infty} s_n[/imath], then there exists an integer [imath]N[/imath] such that if [imath]n \geq N[/imath], then [imath]s_n > x[/imath]. By the lemma, there exists an integer [imath]N_0[/imath] such that if [imath]n \geq N_0[/imath], then [imath]s_n > x[/imath]. Now, let [imath]\epsilon = \inf_{n \geq N_0} \{s_n - x\}[/imath]. This is where I think my proof breaks down. Can't [imath]\epsilon[/imath] be zero? Then, since [imath]\{t_{n_i}\}[/imath] converges to [imath]x[/imath], there exists an integer [imath]N_1[/imath] such that if [imath]n_i \geq N_1[/imath], then [imath]|t_{n_i} - x| < \epsilon[/imath]. But this means that, if [imath]n_i \geq \max\{N, N_0, N_1\}[/imath], we have both [imath] s_{n_i} > t_{n_i}, [/imath] as well as [imath]s_{n_i} \geq t_{n_i}[/imath], a contradiction.
|
1850772
|
Towers of coins puzzle
Let [imath]n[/imath] be a natural number. You are given [imath]\frac{n(n+1)}{2}[/imath] coins which are arranged in towers. Every turn you pick up the top coin of each tower and gather all these into a new tower. Prove that eventually (after finitely many turns) you get towers of the following heights: [imath]\{{ 1,2,3,\dots,n }\}[/imath]. For example: for [imath]n = 6[/imath] (number of coins [imath]21[/imath]): and for the initial state of towers: [imath]\{{17, 2, 1, 1}\}[/imath] Inital state: [imath]\{{17, 2, 1, 1}\}[/imath] Turn 1: [imath]\{{16, 1, 4}\}[/imath] Turn 2: [imath]\{{15, 3, 3}\}[/imath] Turn 3: [imath]\{{14, 2, 2, 3}\}[/imath] Turn 4: [imath]\{{13, 1, 1, 2, 4}\}[/imath] Turn 5: [imath]\{{12, 1, 3, 5}\}[/imath] Turn 6: [imath]\{{11, 2, 4, 4}\}[/imath] Turn 7: [imath]\{{10, 1, 3, 3, 4}\}[/imath] Turn 8: [imath]\{{9, 2, 2, 3, 5}\}[/imath] Turn 9: [imath]\{{8, 1, 1, 2, 4, 5}\}[/imath] Turn 10: [imath]\{{7, 1, 3, 4, 6}\}[/imath] Turn 11: [imath]\{{6, 2, 3, 5, 5}\}[/imath] Turn 12: [imath]\{{5, 1, 2, 4, 4, 5}\}[/imath] Turn 13: [imath]\{{4, 1, 3, 3, 4, 6}\}[/imath] Turn 14: [imath]\{{3, 2, 2, 3, 5, 6}\}[/imath] Turn 15: [imath]\{{2, 1, 1, 2, 4, 5, 6}\}[/imath] Turn 16: [imath]\{{1, 1, 3, 4, 5, 7}\}[/imath] Turn 17: [imath]\{{2, 3, 4, 6, 6}\}[/imath] Turn 18: [imath]\{{1, 2, 3, 5, 5, 5}\}[/imath] Turn 19: [imath]\{{1, 2, 4, 4, 4, 6}\}[/imath] Turn 20: [imath]\{{1, 3, 3, 3, 5, 6}\}[/imath] Turn 21: [imath]\{{2, 2, 2, 4, 5, 6}\}[/imath] Turn 22: [imath]\{{1, 1, 1, 3, 4, 5, 6}\}[/imath] Turn 23: [imath]\{{2, 3, 4, 5, 7}\}[/imath] Turn 24: [imath]\{{1, 2, 3, 4, 6, 5}\}[/imath]
|
438970
|
Problem about a process with bins of balls
A friend of mine give me this problem for fun: Given [imath]\frac {n(n+1)}{2}[/imath] balls, first we divide arbitrarily these balls in baskets, after that we make another basket with one ball of each basket e do this procedure infinitely. I want to prove that one time this stabilizes with 1 ball in one basket, 2 balls in another basket, ..., n balls in another basket. It seems easy to solve, he says we can use some concept of energy (???), I'm trying with some concepts of combinatorics without any success. Thanks in advance.
|
1851056
|
Give the smallest positive value of [imath]x[/imath] for which [imath]\tan x[/imath] is undefined.
Give the smallest positive value of [imath]x[/imath] for which [imath]\tan x[/imath] is undefined. An answer and explanation on how to solve would be great.
|
632996
|
Why is [imath]\tan((1/2)\pi)[/imath] undefined?
Why is the trigonometric function [imath]\tan((1/2)\pi)[/imath] undefined?
|
1850880
|
If [imath](a,b)=1[/imath] then there exist positive integers [imath]x[/imath] and [imath]y[/imath] s.t [imath]ax+by=1[/imath].
How can i prove that if [imath]\gcd(a,b)=1[/imath] there exist [imath]x>0[/imath] and [imath]y>0[/imath] such that [imath]ax-by=1[/imath]?
|
72374
|
How you get from [imath]\gcd(a,b)=1 \Longleftrightarrow ax+by=1 \Longrightarrow a \nmid b[/imath]?
How you get from [imath]\gcd(a,b)= 1 \Longleftrightarrow ax+by=1 \Longrightarrow a \nmid b[/imath]?
|
1851229
|
The rank of a square 2 by 2 block matrix with singular square blocks
I have tried to prove or disprove the following statement using Penrose-Moore pseudoinverses, but all of the literature on that subject seems to focus on the case [imath]m \not= n[/imath], i.e. rectangular matrices, rather than singular square matrices. Let [imath]M = \begin{bmatrix} M_1 & M_2 \\ M_3 & M_4 \end{bmatrix}[/imath], where [imath]M[/imath] is a [imath]2m \times 2m[/imath] matrix, and [imath]M_1 , M_2, M_3, M_4[/imath] are [imath]m \times m[/imath] square matrices, all with determinant zero. Moreover, no two adjacent blocks of the [imath]M_1, M_2, M_3, M_4[/imath] are symmetric (hence the answers given here are non-examples, because all of the blocks are symmetric). Show that [imath]M[/imath] has determinant zero, or find a counterexample where [imath]M[/imath] is not singular. Any help or insights with this problem would be greatly appreciated; I know this is probably a simple problem, but I am not sure how to apply the Laplace expansion in the case that the block matrices just have determinant zero, rather than being all zeros. This answer here says that the answer is affirmative and trivial, but I am not seeing it unfortunately. The condition regarding symmetry may seem random, but it is motivated by the fact that in such cases we can replace the blocks by their characteristic polynomial and still achieve desirable results. See, for example, here: http://www.ee.iisc.ac.in/new/people/faculty/prasantg/downloads/blocks.pdf
|
532328
|
If A, B, C, D are non-invertible [imath]n \times n[/imath] matrices, is it true that their [imath]2n \times 2n[/imath] block matrix is non-invertible?
Is it true that [imath] \left( \begin{array}{ccc} A & B \\ C & D \\\end{array} \right)[/imath] is non-invertible? Assume that the matrix is over a field.
|
1850709
|
What is the rank of the matrix consisting of all permutations of one vector?
Let [imath]a=(a_1,...,a_n)^\top\in\mathbb{R}^n[/imath] be a column vector and let [imath]M_1,...,M_{n!}[/imath] denote all [imath]n\times n[/imath] permutation matrices. When is the rank of the matrix that consists of all possible permutations of [imath]a[/imath]: [imath] A=[M_1 a \,|\; ... \; |\, M_{n!} a]\in\mathbb{R}^{n\times n!} [/imath] equal to [imath]n[/imath]? Obviously, [imath]rank(A)\le n[/imath] and if all entries of [imath]a[/imath] are identical, then [imath]rank(A)=1[/imath]. Moreover, if [imath]A[/imath] has rank [imath]n[/imath], then there exist two entries [imath]i,j[/imath] s.t. [imath]a_i\not=a_j[/imath]. Is the converse statement also true?
|
75427
|
Subspace generated by permutations of a vector in a vector space
Let [imath]K[/imath] be a field. Consider the vector space [imath]K^n[/imath] over the field [imath]K[/imath]. Suppose [imath](a_1,a_2, ... ,a_n) \in K^n[/imath]. What is the dimension of the subspace generated by all the permutations of [imath](a_1,a_2,...,a_n)[/imath]? I think there are 4 different cases [imath]a_1=a_2=...=a_n=0[/imath] [imath]a_1=a_2=a_3=...=a_n \ne 0[/imath] [imath]a_1+a_2+...+a_n=0,[/imath] [imath] a_1 \ne a_2[/imath] [imath]a_1+a_2+...+a_n \ne 0[/imath] , [imath]a_1 \ne a_2[/imath]
|
253860
|
The pebble sequence
Let we have [imath]n\cdot(n+1)/2[/imath] stones grouped by piles. We can pick up 1 stone from each pile and put them as a new pile. Show that after doing it some times we will get the following piles: [imath]1, 2, \ldots n[/imath] stones. Example: [imath]n = 3[/imath] Let we have 2 piles with 3 stones of each. [imath]3 3 \to 2 2 2 \to 1 1 1 3 \to 2 4 \to 1 3 2[/imath]
|
2003121
|
Moving pebbles and invariable state
Suppose there are [imath]\frac{n(n+1)}{2}[/imath] pebbles divided into small heaps. For each step, one pebble from each heap is taken to form a new heap. I need to prove that no matter what state one start with, eventually the number of pebbles from each heap is [imath]1, 2, ..., n.[/imath] For example, when [imath]n=5[/imath] [imath](7,8) \rightarrow (6,7,2) \rightarrow (5,6,1,3) \rightarrow (4,5,2,4) \rightarrow (3,4,1,3,4) \rightarrow (2,3,2,3,5) \rightarrow (1,2,1,2,4,5) \rightarrow (1,1,3,4,6) \rightarrow (2,3,5,5) \rightarrow (1,2,4,4,4) \rightarrow (1,3,3,3,5) \rightarrow (2,2,2,4,5) \rightarrow (1,1,1,3,4,5) \rightarrow (2,3,4,6) \rightarrow (1,2,3,5,4)[/imath] I have try to construct a state function that decreases for each operation. I also draw a digraph for case [imath]n=3[/imath] but I did not recognized any particular pattern.
|
742625
|
Invertible e converges series.
If T is a linear transformation on [imath]R^n[/imath] with [imath]||T - I||<1[/imath], prove that [imath]T[/imath] is invertible and that the series [imath]\sum_{k=0}^\infty(I-T)^k[/imath] converges absolutely to [imath]T^{-1}.[/imath] (Use the geometric series)
|
98562
|
Proof of Neumann Lemma
Prove that if [imath]\|A\| < 1[/imath], then [imath]I-A[/imath] is invertible. Here, [imath]\|\cdot\|[/imath] is a matrix norm induced by a vector norm. This lemma is referred to as Neumann Lemma. Any ideas on how to go ahead with this? Thanks.
|
1851600
|
Closure of [imath]\frac {1} {n} [/imath]
I have as a definition of the closure of a set [imath]E[/imath] in a metric space [imath](S,d)[/imath] that [imath]E^-[/imath] is the intersection of all closed sets containing E (Elementary analysis the theory of calculus by Kenneth Ross). I have as a hunch that the closure of [imath]E = \{\frac{1}{n}: n \in \Bbb N\} [/imath] is [imath][0,1][/imath] since this set seems to contain every element in [imath]E[/imath]. How can I prove this to be true?
|
1399295
|
Closure of [imath]\{\frac{1}{n}\}:=M[/imath]
In [imath]\mathbb{R}[/imath] with standard topology the closure of the above set [imath]M[/imath] is [imath]M \cup 0[/imath]. I fail showing this by purely topological means. I.e. not using the fact that this holds in the reals as a metric space and deduce it for the induced topology. I assumed that there exist a closed subset [imath]A[/imath] containing [imath]M[/imath] but not 0. Hence, the complement of [imath]A[/imath] is an open subset containing 0. Does this already yield a contradiction or do I need a different approach? Could you help me on this?
|
1805915
|
Basic question on second covariant derivative
I am having a question on the wikipedia article https://en.wikipedia.org/wiki/Second_covariant_derivative Using the notation therein I don't get why [imath](\nabla_{u}\nabla_{v}w )^a=u^c\nabla_{c}v^b\nabla_{b}w^a[/imath]. It is clear that [imath](\nabla_{u}\nabla_{v}w )^a=(u^c\nabla_{c}v^b\nabla_{b}w)^a[/imath]. What do I miss?
|
1102581
|
Covariant derivative ambiguity
I'm studying general relativity and am running into an ambiguity with the covariant derivative. The covariant derivative acting on a scalar is, in a co-ordinate basis, simply [imath]\nabla_X f = X^a \nabla_a f = X^a \partial_a f [/imath] whilst when acting on vectors or tensors, the covariant derivative includes extra terms involving the connection coefficients [imath]\Gamma[/imath]. My problem is that I'm not sure what happens when the covariant derivative acts on the components of a vector. So the [imath]X^a[/imath] in the above expression are each scalar functions, and so when we write [imath]\nabla_Y X^a[/imath], do we assume the covariant derivative acts on [imath]X^a[/imath] as scalars, or do we include connection coefficient terms? To clarify, consider [imath] \nabla_X Y = X^a \nabla_a(Y^b e_b) = X^a (\nabla_a Y^b) e_b + X^a Y^b (\nabla_a e_b) = X^a (\partial_a Y^b) e_b + X^a Y^b \Gamma^c{}_{ba} e_c [/imath] As you can see, when the covariant derivative acts on the [imath]Y^b[/imath], it acts as a simple partial derivative, since the [imath]Y^b[/imath] are here considered functions premultiplying the vectors [imath]e_b[/imath]. However, in this equation from this page: [imath](\nabla_X \nabla_Y Z)^a = X^c \nabla_c Y^b \nabla_b Z^a = X^c Y^b \nabla_c \nabla_b Z^a + (X^c \nabla_c Y^b) \nabla_b Z^a = (\nabla^2_{X,Y} Z)^a + (\nabla_{\nabla_X Y} Z)^a[/imath] When the covariant derivative acts on the [imath]Y^b[/imath] term it treats it like a vector. What's going on here? Thanks.
|
432319
|
Inequality between two sequences preserved in the limit?
Let [imath](a_n)_{n\in \mathbb{N}}[/imath] and [imath](b_n)_{n\in \mathbb{N}}[/imath] be two real sequences that satisfy [imath]a_n\geq b_n, \forall n \in \mathbb{N}[/imath] and converge to some [imath]a,b[/imath], respectively. Is it always true that [imath]a \geq b[/imath]?
|
1850471
|
Suppose that [imath](s_n)[/imath] converges to [imath]s[/imath], [imath](t_n)[/imath] converges to [imath]t[/imath], and [imath]s_n \leq t_n \: \forall \: n[/imath]. Prove that [imath]s \leq t[/imath].
I'm stuck with the proof of the following: Suppose that [imath](s_n)[/imath] converges to [imath]s[/imath], [imath](t_n)[/imath] converges to [imath]t[/imath], and [imath]s_n \leq t_n \: \forall \: n[/imath]. Prove that [imath]s \leq t[/imath]. I've tried starting with [imath]s_n \leq t_n \: \forall : n[/imath] and the definitions of each limit (i.e. [imath]|s_n - s| \leq \epsilon \: \forall \: n > N_1[/imath]), but I'm not really getting very far. Any help is appreciated!
|
1852008
|
Why there is no value for [imath]x[/imath] if [imath]|x| = -1[/imath]?
According to the definition of absolute value negative values are forbidden. But what if I tried to solve a equation and the final result came like this: [imath]|x|=-1[/imath] One can say there is no value for [imath]x[/imath], or this result is forbidden. That reminds me that same thinking in the past when mathematical scientists did not accept the square root of [imath]-1[/imath] saying that it is forbidden. Now the question is :"is it possible for the community of math to accept this term like they accept imaginary number. For example, they may give it a sign like [imath]j[/imath] and call it unreal absolute number then a complex number can be expanded like this: [imath]x = 5 +3i+2j[/imath] , where [imath]j[/imath] is unreal absolute number [imath]|x|=-1[/imath] An other example, if [imath]|x| = -5[/imath], then [imath]x=5j[/imath] The above examples are just simple thinking of how complex number may expanded You may ask me what is the use of this strange new term? or what are the benefits of that? I am sure this question has been raised before in the past when mathematical scientists decided to accept [imath]\sqrt{-1}[/imath] as imaginary number. After that they knew the importance of imaginary number.
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1345191
|
Is there a number whose absolute value is negative?
I've recently started to think about this, and I'm sure a couple of you out there have, too. In Algebra, we learned that [imath]|x|\geq0[/imath], no matter what number you plug in for [imath]x[/imath]. For example: [imath]|-5|=5\geq0[/imath] We also learned that [imath]x^2\geq0[/imath]. For example: [imath](-5)^2=25\geq0[/imath] The exception for the [imath]x^2[/imath] rule is imaginary numbers (which we learn later on in Algebra II). Imaginary numbers are unique, in that their square is a negative number. For example: [imath]4i^2=-4[/imath] These imaginary numbers can be used when finding the "missing" roots of a polynomial equation. My question to you is this: Is there any number whose absolute value is negative, and how could it be used?
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1852590
|
Compact convergence [imath]\sum_{n = 1}^{\infty} f(z^n)[/imath]
Let [imath]f \colon \mathbb{E} \to \mathbb{C}[/imath] holomorphic with [imath]f(0) = 0[/imath]. Ist there a way to prove the compact convergence of the series [imath]\sum_{n = 1}^{\infty} f(z^n)[/imath] without using the Taylor series and then applying some theorem for double serieses?
|
1022157
|
show that [imath]f(z)+f(z^2)+\cdots + f(z^n)+\cdots[/imath] converges locally uniformly to an analytic function in the unit disk.
If [imath]f(z)[/imath] is analytic in the unit disk and [imath]f(0)=0[/imath], show that [imath]f(z)+f(z^2)+\cdots+ f(z^n)+\cdots[/imath] converges locally uniformly to an analytic function in the unit disk. I am thinking to apply Weierstrass [imath]M[/imath] test for convergence or since all the [imath]f(z^n)[/imath] are analytic in the unit disk, so each of them can be expressed as the taylor series. Then rewriting the series again might work. But I am not sure. Any suggestions!
|
1852222
|
Why is a curve parameterized by arc length necessarily a unit speed curve?
I apologize if this is trivial but I have not been able to figure it out. For a curve [imath]\sigma(t)[/imath], I have a definition for arc length: [imath]s(t)=\int_{t_0}^t |\sigma'(t)|dt[/imath] We reparameterize a curve [imath]\sigma[/imath] by observing that [imath]s[/imath] has an inverse, [imath]t(s)[/imath], and the resulting reparameterized curve has underlying assignment [imath]s \mapsto t \mapsto \sigma(t)[/imath]. A curve is a unit speed curve if [imath]\forall t\qquad |\sigma'(t)|=1[/imath] Thanks
|
1634974
|
Prove that for any piecewise smooth curve it is possible to find the parametrisation
Prove that for any piecewise smooth curve it is possible to find the parametrisation [imath]\phi[/imath] that is consistent with its length, ie. length of a curve segment between [imath]\phi(a)[/imath] and [imath]\phi(b)[/imath] is equal to [imath]|b-a|[/imath]. It seems intuitive, but I don't have any idea how can I show this in the formal way.
|
774434
|
Generating function of [imath]\binom{3n}{n}[/imath]
Wolfram alpha tells me the ordinary generating function of the sequence [imath]\{\binom{3n}{n}\}[/imath] is given by [imath]\sum_{n} \binom{3n}{n} x^n = \frac{2\cos[\frac{1}{3}\sin^{-1}(\frac{3\sqrt{3}\sqrt{x}}{2})]}{\sqrt{4-27x}}[/imath] How do I prove this?
|
557230
|
Ordinary generating function for [imath]\binom{3n}{n}[/imath]
The ordinary generating function for the central binomial coefficients, that is, [imath]\displaystyle \sum_{n=0}^{\infty} \binom{2n}{n} x^{n} = \frac{1}{\sqrt{1-4x}}[/imath] follows from the generalized binomial theorem and an application of the duplication formula for the gamma function. But what about the ordinary generating function for [imath] \displaystyle \binom{3n}{n}[/imath]? According to Wolfram Alpha, [imath] \sum_{n=0}^{\infty} \binom{3n}{n} x^{n} = \frac{2\cos \big(\frac{1}{3} \arcsin (\frac{3 \sqrt{3x}}{2})\big)}{\sqrt{4-27x}} [/imath] Any suggestions on how to prove this? EDIT: Approaching this problem using the fact that [imath] \text{Res} \Big[ \frac{(1+z)^{3}}{z^{n+1}},0 \Big] = \binom{3n}{n}[/imath] you get that [imath] \sum_{n=0}^{\infty} \binom{3n}{n} x^{n} = \frac{-1}{2 \pi i x} \int_{C} \frac{dz}{z^{3}+3z^{2}+3z - \frac{z}{x}+1}[/imath] where [imath]C[/imath] is a circle centered at [imath]z=0[/imath] such that every point on the circle satisfies [imath] \displaystyle\Big|\frac{x(1+z)^{3}}{z} \Big| < 1[/imath]. Evaluating that contour integral would appear to be quite tedious.
|
1853086
|
Find all the numbers [imath]n[/imath] such that [imath]\frac{6n-8}{2n-5}[/imath] can't be reduced.
Find all the numbers [imath]n[/imath] such that [imath]\frac{6n-8}{2n-5}[/imath] can't be reduced. Attempt: It can't be reduced when [imath]\gcd(6n-8,2n-5)=\color{red}1[/imath] [imath]1 = \gcd(6n-8,2n-5)=\gcd(4n-3,2n-5)=\gcd(2n+2,2n-5)=\gcd(7,2n-5)[/imath] that's equale to one when [imath]2n-5\not\equiv 0 \pmod 7[/imath] i.e [imath]n\not\equiv 6 \pmod 7[/imath] Is my attempt correct?
|
1852984
|
Find all the numbers [imath]n[/imath] such that [imath]\frac{4n-5}{60-12n}[/imath] can't be reduced.
Find all the numbers [imath]n[/imath] such that [imath]\frac{4n-5}{60-12n}[/imath] can't be reduced. Attempt: [imath]\gcd(4n-5,60-12n)=(4n-5,-8n+55)=(4n-5,-4n+50)=(4n-5,45)[/imath] [imath]n=1: (4-5,45)=1\quad \checkmark\\ n=2: (3,45)=3\quad \times\\ n=3: (7,45)=1\quad \checkmark\\ n=4: (11,45)=1\quad \checkmark\\ n=5: (15,45)=15\quad \times\\ n=6: (19,45)=1\quad \checkmark\\ n=7: (23,45)=1\quad \checkmark\\ n=8: (27,45)=9\quad \times\\ \vdots[/imath] So the answer is that it can't be reduced for [imath]n=1,3,4,6,7,..[/imath] i.e [imath]\bigg\{n\bigg|n\notin \begin{cases}a_1=2\\a_n=a_{n-1}+3\end{cases}\bigg\}[/imath] I want to verify that my solution is correct
|
1852940
|
Vector as directional derivate?
The Poor Man’s Introduction to Tensors - by Justin C. Feng. https://web2.ph.utexas.edu/~jcfeng/notes/Tensors_Poor_Man.pdf On page 3, "Before I can tell you what a tensor is, I must tell you what a vector really is; in fact, you will later see that a vector is a type of tensor. A vector is simply a directional derivative. Before you write me off as a nut, take a look at the directional derivative of some scalar function [imath]f(x^i)[/imath]" [imath]\nu.\nabla(f(x^j))=\nu^i\dfrac{\partial}{\partial{x^i}}f(x^j)[/imath] To me this is still ok, the author later removes the function itself and claim, [imath]\nu.\nabla=\nu^i\dfrac{\partial}{\partial{x^i}}[/imath] currently i see [imath]\nu.\nabla[/imath] as some form of operator. But my trouble arises when on page 4, equation 19, author then gives equation, stating the following "Some people get rid of the function f altogether, and write the following:" [imath]\nu=\nu^i\dfrac{\partial}{\partial{x^i}}[/imath] and [imath]\nu^i = \nu(x^i)[/imath] My problem is that seem that this definition of vector seems very un-intuitive to me,maybe because i am acquainted with all the old definition. It is as if vector were a operator. Can anyone give me the intuition behind this? Or can someone help me make its sense in term of old definitions of vector i have like in simple Euclidean geometry. Edit: Just to be exact from the possible duplicate, My problem with intuition is exactly this new operator kind of definition As compared to n-tuple that i have been seeing all along. Exactly how do they connect if they do so, Or if they are different or if one generalizes the other.
|
75510
|
Motivation behind the definition of tangent vectors
I've been reading the book Gauge, Fields, Knots and Gravity by Baez. A tangent vector at [imath]p \in M[/imath] is defined as function [imath]V[/imath] from [imath]C^{\infty}(M) [/imath] to [imath]\mathbb R[/imath] satisfying the following properties: [imath]V(f+g)=V(f) + V(g)[/imath]. [imath]V(\alpha f)= \alpha V(F)[/imath]. [imath]V(fg) = V(f)g(p) + V(g)f(p)[/imath]. Can someone explain me what is the physical interpretation of tangent vectors and the above definition?
|
1853395
|
Defining binary operations of isomorphisms
The question given is: The map [imath]f : \Bbb Z \rightarrow \Bbb Z[/imath] defined by [imath]f(n) = n + 1[/imath] for [imath]n \in \Bbb Z[/imath] is one to one and onto Z. Give the definition of a binary operation [imath]*[/imath] on [imath]\Bbb Z[/imath] such that [imath]f[/imath] is an isomorphism mapping [imath](\Bbb Z, +)[/imath] onto [imath](\Bbb Z, * )[/imath]. So I know this means I need to equate [imath]f(a + b) = f(a) * f(b)[/imath] with a defined [imath]*[/imath] of my choosing. If I define [imath]f(a) * f(b)[/imath] as [imath]f(a) + f(b) - 1[/imath] is this an acceptable definition? Because then [imath]f(a+b) = (a+b) + 1[/imath] and [imath]f(a) * f(b) = (a+1) + (b + 1) - 1 = (a + b) + 1[/imath], and the identity element would be [imath]1[/imath] since [imath]f(0) = 1[/imath]?
|
210827
|
Basic Isomorphism Question
The map [imath]\phi: \mathbb{Z} \to \mathbb{Z}[/imath] defined by [imath]\phi (n) = n + 1[/imath] for [imath]n \in \mathbb{Z}[/imath] is one to one and onto [imath]\mathbb{Z}[/imath]. Give the definition of a binary operation [imath]*[/imath] on [imath]\mathbb{Z}[/imath] such that [imath]\phi[/imath] is an isomorphism mapping a) [imath]\langle\mathbb{Z},+\rangle[/imath] onto [imath]\langle\mathbb{Z}, *\rangle[/imath] I absolutely do not understand this. What is the solution doing? Why are we starting out with [imath]m*n[/imath]? And why do we have [imath]\phi(m-1) * \phi(n-1)[/imath]?
|
1853783
|
Find the derivative of [imath]y = \frac{\sqrt[3]{4x^3+8}}{(x+2)^5}[/imath]
Can anyone solve 1b? Find the derivatives of each of these functions. b. [imath]y = \frac{\sqrt[3]{4x^3+8}}{(x+2)^5}[/imath] I am really confused, I used the quotient rule for differentiation of the chain rule and keep getting stuck.After the [imath]y′=\frac{f′(x)g(x)−g′(x)f(x)}{g(x)^2}[/imath]. so far this is what I have done
|
1853770
|
Find the derivative of [imath]y= \frac{(4x^3 +8)^{\frac{1}{3}}}{(x+2)^5}[/imath]
How can we find the derivative of [imath]y= \frac{(4x^3 +8)^{\frac{1}{3}}}{(x+2)^5}[/imath]? so far this is what I have done, and am confused about what to do after? sorry for the messy handwriting
|
1853915
|
Let [imath]f[/imath] be a function [imath]f:\mathbb{R}\rightarrow \mathbb{R}[/imath] such that [imath]f(x)f''(x)\leq 0[/imath] for all [imath]x\in \mathbb{R}[/imath].
Let [imath]f[/imath] be a function [imath]f:\mathbb{R}\rightarrow \mathbb{R}[/imath] such that the derivative [imath]f'(x)[/imath] and all its successive derivatives exist everywhere and also [imath]f(x)f''(x)\leq 0[/imath] for all [imath]x\in \mathbb{R}[/imath]. Let [imath]\alpha[/imath] and [imath]\beta[/imath] be two consecutive roots of [imath]f(x)=0[/imath]. Prove that [imath]f'''(\gamma)=0[/imath] for some [imath]\gamma \in(\alpha,\beta)[/imath] I feel that the problem may not be true I have been able to draw the curve of such a function..
|
1853675
|
Application of Derivatives rigorous proof
Let [imath]f:R\rightarrow R[/imath] be a function such that all its successive derivatives exist in all [imath]R[/imath] and also [imath]f(x)f''(x)\leq 0[/imath] everywhere. If [imath]\alpha[/imath] and [imath]\beta[/imath] be two successive roots of [imath]f(x)=0[/imath]. Then prove that [imath]f'''(x)=0[/imath] for atleast one [imath]\gamma \in(\alpha,\beta)[/imath]. My Attempt: I began by taking an example [imath]f(x)=(x-1)(2-x)[/imath] and let [imath]x=1[/imath] and [imath]x=2[/imath] be its two successive roots.The statement is trivially true. Then I took the function [imath]f(x)=e^{-x}(x-1)(2-x)[/imath].The statement is true here also so on and so forth. But what would be the exact proof I wonder. It appears to be a question of Rolle's Theorem.
|
1854088
|
How to show that [imath]\emptyset\in\tau_{\beta}[/imath]?
Problem. Let [imath]X[/imath] be a non-empty set and [imath]\beta[/imath] be a basis set (in the linked definition of basis for a topology replace the term by basis set). Then the set [imath]\tau_\beta:=\left\{\displaystyle\bigcup_{B\in \beta_0} B: \beta_0\subseteq \beta\right\}[/imath] forms a topology on [imath]X[/imath]. I have already shown that [imath]\tau_{\beta}[/imath] is a topology assuming that [imath]\emptyset\in\tau_{\beta}[/imath] but I can't show that [imath]\emptyset\in \tau_\beta[/imath]. I think that it holds vacuously but I am not sure how to prove it. Can anyone help?
|
1826226
|
Why is this the basis for the indiscrete topology?
Indiscrete topology only has [imath]\{\varnothing, X\}[/imath] where [imath]X[/imath] is the entire space. I know this is probably a dumb question. Why is the basis (according to wikipedia) for this topology just [imath]\{X\}[/imath] instead of [imath]\{\varnothing, X\}[/imath]? I understand that [imath]\{X\}[/imath] satisfies the definition. Suppose [imath]\mathcal{B} = \{X\}[/imath] Then [imath]\mathcal{B}[/imath] covers [imath]X[/imath] - Check! For all [imath]B_1, B_2 \in \mathcal{B}, \ldots[/imath] - Trivially satsifed since we don't have [imath]B_1, B_2[/imath] But how does this basis generate [imath]\varnothing[/imath]?
|
1853933
|
On a property of a holomorphic function on a unit disk
Let [imath]f[/imath] be a holomorhic function on the open unit disk [imath]D[/imath]. Suppose that there is an [imath]r\in[0,1)[/imath] such that restriction of [imath]f[/imath] to the annulus [imath]U=\left\{z\in C:r<\left|z\right|<1\right\}[/imath] is one-to-one. Prove that [imath]f[/imath] is one-to-one on [imath]D[/imath]. Firstly, [imath]f[/imath] is one-to-one if and only if derivative of [imath]f(x)[/imath] is not zero. So my aim is to use the boundary property of this function to prove the result. I tried Cauchy integral theorem. But it doesn't seem to work very well. Neither do the maximum and minimum principle.
|
1844776
|
Prove that a holomorphic function injective in an annulus is injective in the whole ball
Let [imath]f: B(0,1) \rightarrow \mathbb{C}[/imath] be holomorphic and suppose [imath]\exists\ r \in (0,1)[/imath] such that [imath]f[/imath] is injective in [imath]A = \{z \in \mathbb{C} : r < |z| < 1\}[/imath]. Prove that [imath]f[/imath] is injective. I tried using Rouché theorem, or the identity theorem, but I don't know what to do. Any hints? :)
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1186161
|
Equivalent form of biconditional
I'm reading How to Prove It: A Structured Approach (Velleman) Second Ed. Doing all the end of chapter exercises for chapter 1 and having trouble on problem 5a which reads Show that [imath]P \leftrightarrow Q[/imath] is equivalent to [imath](P \wedge Q) \vee (\neg P \wedge \neg Q)[/imath] Clearly they're equivalent to each other based on the truth tables. But is that really the best way to 'show' it? I was able to derive the second form from the first in all the other questions that asked to show two forms are equivalent so far. Here's what happens when I try to derive it: [imath]P \leftrightarrow Q[/imath] [imath] \text{Form of biconditional} [/imath] [imath](P \rightarrow Q) \wedge (Q \rightarrow P)[/imath] [imath] \text{Form of conditional}[/imath] [imath] (\neg P \vee Q) \wedge (\neg Q \vee P)[/imath] From here it appears to match the form of a distributive law but with mismatched negations. I just don't know where to go from here...
|
459494
|
proving logical equivalence [imath](P \leftrightarrow Q) \equiv (P \wedge Q) \vee (\neg P \wedge \neg Q)[/imath]
I am currently working through Velleman's book How To Prove It and was asked to prove the following [imath](P \leftrightarrow Q) \equiv (P \wedge Q) \vee (\neg P \wedge \neg Q)[/imath] This is my work thus far [imath](P \to Q) \wedge (Q \to P)[/imath] [imath](\neg P \vee Q) \wedge (\neg Q \vee P)[/imath] (since [imath](P \to Q) \equiv (\neg P \vee Q)[/imath]) [imath]\neg[\neg(\neg P \vee Q) \vee \neg (\neg Q \vee P)][/imath] (Demorgan's Law) [imath]\neg [(P \wedge \neg Q) \vee (Q \wedge \neg P)][/imath] (Demorgan's Law) At this point I am little unsure how to proceed. Here are a few things I've tried or considered thus far: I thought that I could perhaps switch some of the terms in step 3 using the law of associativity however the [imath]\neg[/imath] on the outside of the two terms prevents me from doing so (constructing a truth table for [imath]\neg (\neg P \vee Q) \vee (\neg Q \vee P)[/imath] and [imath]\neg (\neg P \vee \neg Q) \vee \neg (P \vee Q)[/imath] for sanity purposes) I can't seem to apply the law of distribution since the corresponding terms are negated. Applying demorgans law to one of the terms individually on step 2 or 3 doesnt seem to get me very far either. Did I perhaps skip something? Am I even on the right track? Any help is appreciated
|
1854507
|
Let [imath]k[/imath] be a field of characteristic [imath]\neq 2[/imath]. Then [imath]x^6-xy+y^6[/imath] is irreducible in [imath]k[x,y][/imath].
There is no obvious way to apply Eisenstein's criterion; and if I assume by contradiction that [imath]x^6-xy+y^6=f(x,y)g(x,y)[/imath], f with homogeneous degree [imath]\leq 3[/imath]. Then I have that [imath]f(x,y)=\sum_{i,j} a_{ij}x^i y^j[/imath], [imath]g(x,y)=\sum_{i,j} b_{ij}x^i y^j[/imath], so [imath]f(x,y)g(x,y)=\sum_{i,j,k,l}a_{ij} b_{kl} x^{i+k} y^{j+l}=\sum_{i,j} x^i y^j\left(\sum_{i_0+i_1=i}\sum_{j_0+j_1=j}a_{i_0 j_0} b_{i_1 j_1} \right) =x^6-xy+y^6[/imath] so that [imath]\sum_{i_0+i_1=i}\sum_{j_0+j_1=j}a_{i_0 j_0} b_{i_1 j_1}= \left\{\begin{array}{ll} 1 & \quad i=6,j=0\\ -1 & \quad i=j=1\\ 1 & \quad i=0,j=6\\ 0 & \quad \mathrm{else} \end{array}\right.[/imath] which doesn't seem like a particularly fruitful approach. Other than painstakingly writing out the 10 possible terms for [imath]f[/imath] and the 28 possible terms for [imath]g[/imath] to get a horrendous 38-variable quadratic system of equations, I was wondering if there was some easier way to do this.
|
303681
|
Irreducibility of Polynomials in [imath]k[x,y][/imath]
I'm working through some Hartshorne problems and have noticed that in order to do certain problems properly one must prove a given polynomial [imath]f\in k[x,y][/imath] is irreducible. For example, in problem I.5.1(b) we are studying the polynomial [imath]f=xy-x^6-y^6[/imath]. How can I show this is irreducible? The standard tricks (Eisenstein, linearity in one variable, etc...) don't seem to work here. Edit: Some of the other polynomials in the problem are also giving me trouble. They are [imath]g=x^3-y^2-x^4-y^4[/imath] and [imath]h=x^2y+xy^2-x^4-y^4[/imath]. Any ideas for these?
|
1854521
|
Proof about prime factors: every prime factor of [imath]4n^2+1[/imath] is congruent to [imath]1 \pmod 4[/imath].
Show that if [imath]n[/imath] is an integer, then every prime factor of [imath]4n^2+1[/imath] is congruent to [imath]1 \pmod 4[/imath]. (Hint: if [imath]p\mid 4n^2+1[/imath]), then what can you say about [imath](-1\mid p)[/imath]? Approach: I went over all the cases for n and concluded that [imath]4n^2+1 \equiv 1\pmod 4[/imath]. We know that [imath]4n^2+1[/imath] can be represented as the product of prime numbers, so let [imath]4n^2+1=q_1\cdots q_\ell[/imath]. This implies that [imath]q_i \mid 4n^2+1[/imath] b) Show that there are infinitely many primes congruent to 1(mod 4) Approach The common way to do this is by contradiction. Assume there are finitely many primes congruent to 1 mod 4. Call them [imath]p_1,....,p_k[/imath]. Now let's build a number n congruent to 1 mod 4 with these primes. n=[imath]4p_1*....*p_{k}+1 \equiv 1(mod\text{ } 4)[/imath]. By definition we can represent n as a product of primes. Let [imath]n=q_1,...,q_l[/imath]. If [imath]q_j=p_j[/imath] then [imath]q_j | 4p_1*...*p_k[/imath] and [imath]q_j | 1[/imath] which can't be possible, so contradiction. I am just wondering if there is a way to do the latter by using what was just done.
|
1854445
|
Is it possible that [imath]n^2+1[/imath] has some divisor of the form [imath]4k+3[/imath]?
Given an integer [imath]n[/imath], we are asked to investigate about the existence of integer divisors of [imath]n^2+1[/imath] of the form [imath]4k+3[/imath]. Can you provide some insights about it?
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1851511
|
Factorial Proof by Induction Question: [imath] \frac1{2!} + \frac2{3!} + \dots+ \frac{n}{(n+1)!} = 1 - \frac1{(n+1)!} [/imath]?
[imath]\text{Use the PMI to prove the following for all natural numbers n.}[/imath] [imath] \frac{1}{2!} + \frac{2}{3!} + \cdot \cdot \cdot + \frac{n}{(n+1)!} = 1 - \frac{1}{(n+1)!} [/imath] So for this question I get stuck because I cannot seem to make the left side and right side be equivalent. [imath]\color{Green}{Proof} :[/imath] [imath]\mathbf (I) \; Basis \; Step :[/imath] Show [imath]p(1).[/imath] For [imath]n=1[/imath] [imath]\frac{1}{(1+1)!} = 1- \frac{1}{(1+1)!} \Rightarrow \frac12 = \frac12[/imath] [imath]\mathbf (II) \; Inductive \; Step : [/imath] Assume [imath]p(k)[/imath] for a PAC [imath]k \in â„• ,\; \; \;\; n=k[/imath] [imath] \frac{1}{2!} + \frac{2}{3!} +\cdot\cdot\cdot + \frac{k}{(k+1)!} = 1-\frac{1}{(k+1)!} \; \; \text{Induction Hypothesis}[/imath] Left Hand Side : We need to show : [imath]p(k+1)[/imath] i.e. [imath]n = k+1[/imath] [imath]\frac{k}{(k+1)!} + \frac{k+1}{(k+1+1)!} \Rightarrow \ [/imath] [imath]\frac{(k+1)}{(k+1+1)!} +1 - \frac{1}{(k+1)!} [/imath] Right Hand Side: [imath] n =k+1[/imath] [imath]\left( 1- \frac{1}{(k+1+1)!}\right) = 1- \frac{1}{(k+2)!}[/imath] This is where I have made my mistake because I think I have made an arithmetic mistake on the left side because I cannot get them to be verifiable. Any hints on how to correct the error would be appreciated. Have a good one.
|
1821557
|
Prove if [imath]n \in \mathbb N[/imath], then [imath]\frac{1}{2!}+\cdots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}[/imath]
Prove if [imath]n \in \mathbb N[/imath], then [imath]\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{n}{\left(n+1\right)!} = 1-\frac{1}{\left(n+1\right)!}[/imath] So I proved the base case where [imath]n=1[/imath] and got [imath]\frac{1}{2}[/imath] Then since [imath]n=k[/imath] implies [imath]n=k+1[/imath] I setup the problem like so: [imath]\frac{k}{(k+1)!}+\frac{(k+1)}{(k+2)!}=1-\frac{1}{(k+2)!}[/imath] After trying to simplify it I got the following: [imath]\frac{k(k+2)!+(k+1)(k+1)!}{(k+1)!(k+2)!}=1-\frac{1}{(k+2)!}[/imath] However, I'm having trouble simplifying it to match the RHS. Hints?
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221926
|
Entire function constant, where [imath]f(z)=f(z+1)[/imath] and [imath]|f(z)|< e^{|z|}[/imath].
I came across this old exam problem. Suppose [imath]f(z)[/imath] is entire and [imath]|f(z)|< e^{|z|}[/imath], and also [imath]f(z)=f(z+1)[/imath]. Show [imath]f(z)[/imath] is a constant. I am able to show the singularity at infinity is not a pole. But I can't rule out it being an essential singularity.
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2837870
|
Prove that this entire function [imath]f[/imath] is constant
I need help for this problem: If [imath]f:\mathbb{C}\to\mathbb{C}[/imath] is entire function such that [imath]f(z+1)=f(z)[/imath] and [imath]|f(z)|<e^{|z|}[/imath], [imath]\forall z\in\mathbb{C}[/imath], prove that [imath]f[/imath] is constant function. [Hint: Look function [imath]g(z)=\frac{f(z)-f(0)}{\sin \pi z}[/imath]]
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1854353
|
How to Show that this relation is not well defined.
We represent an element of the domain as an equivalence class [imath]\bar x[/imath], and use the notation [imath] \left[ x \right][/imath] for equivalence classes in the codomain. Show that this is not well defined. [imath] f: \Bbb Z_{4} \rightarrow \Bbb Z_{6} \; \text{ given by, } f(\bar x) = \left[ 2x+1\right][/imath] My question is how do does one determine/ show that this is not a function? How does one show that this not one to one? [imath]f(\bar 0) = 2(0) +1 = [1][/imath] [imath]f(\bar 1) = [3][/imath] [imath]f(\bar 2) = [5][/imath] [imath]f(\bar 3) = [7][/imath] [imath]f(\bar 4) = [9][/imath] [imath]f(\bar 5) = [11][/imath] [imath]f(\bar 6) = [13][/imath] I also guessing here that if a number is a multiple of another then that may play a role in determining it if is one-one.
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1852755
|
The map [imath]f:\mathbb{Z}_3 \to \mathbb{Z}_6[/imath] given by [imath]f(x + 3\mathbb{Z}) = x + 6\mathbb{Z}[/imath] is not well-defined
By naming an equivalence class in the domain that is assigned at least two different values prove that the following is not a well defined function. [imath]f : \Bbb Z_{3} \to \Bbb Z_{6} \;\;\;\text{ given by } f(\overline x) = [x] [/imath] In this case we represent an element of the domain as an [imath]\bar x[/imath] and use the notation [imath][x][/imath] for equivalence classes in the co-domain. [imath]f(\overline0) = [0] \;,[/imath] [imath] \Bbb Z_{3} \quad (3x+0)\;\; \overline 0 = \{ ...-6,-3,0,3,6... \}, \; \Bbb Z_{6}\; (6x+0)\; \overline0 =\{ ...-12,-6,0,6,12...\}[/imath] [imath]f(\overline1) = [1], [/imath] [imath]\qquad \; (3x+1) \; \;\;\;\overline 1 = \{ ...-5,-2,1,4,7 ... \},\; \; (6x+1)\;\overline1 =\{...-11,-5,1,7,13.. \}[/imath] [imath]f(\overline2) = [2], [/imath] [imath]\qquad \qquad \qquad \;\overline 2 = \{ ...-4,-3,2,5,8 ... \},\;\;\overline 2 = \{ ...-10,-4,2,8,14 ...\},\;[/imath] [imath]f(\overline3) = [3] ,[/imath] [imath]\qquad \qquad \qquad \qquad \qquad \qquad,\; \quad \quad \quad \; \; \; \;\overline 3 = \{ ...-9,-3,3,9,15 ... \},[/imath] [imath]f(\overline4) = [4],\qquad \qquad \qquad\qquad \qquad\qquad \; \quad \quad \quad \quad \; \;\overline 4 = \{ ...-8,-2,4,10,16... \}, [/imath] [imath]f(\overline5) = [5], \qquad \qquad \qquad\qquad \qquad\qquad \; \quad \quad \quad \quad \;\;\overline 5 = \{ ...-7,-1,5,11,17... \}, [/imath] [imath]f(\overline6) = [6] ,[/imath] So my main question for this problem is how to find out if this question is not a function. From the information I have gathered here I still cannot see why this is not a function any help on showing how this is not function would be much appreciated. The set of equivalence classes for the relation [imath]\cong_{m}[/imath] is denoted [imath]\Bbb Z_{m}[/imath] The [imath] 3x+0 \text{ and } 6x+0[/imath] are just showing how I got [imath]\overline 0 [/imath]
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1854420
|
Unable to prove that a statement is an equivalent form of the Axiom of Choice
The exercise 22 page 158 of Elements of Set Theory by B. Enderton is the following: Show that the following statement is another equivalent version of the axiom of choice: For any set [imath]A[/imath] there is a function [imath]F[/imath] with [imath]\text{dom }F=\bigcup A[/imath] and such that [imath]x\in F(x)\in A[/imath] for all [imath]x\in\bigcup A[/imath]. The versions of the axiom of choice that I know for the moment are: (although the Cardinal comparability and Zorn's lemma were stated as well, the author didn't prove yet that they're equivalent to the Axiom of Choice) and the following form: (Exercise 18 in the same page) I proved that AC III implies the statement of the exercise as follows: Let [imath]A[/imath] be a set According to AC III there exists a function [imath]f:\mathcal{P}A\setminus\{\emptyset\}\to A[/imath] such as [imath]\forall X\in\mathcal{P}A\setminus\{\emptyset\},\,f(X)\in X[/imath]. Let \begin{align*} g:&\bigcup A\,\,\to \mathcal{P}A\setminus\{\emptyset\}\\ &x \quad\quad\mapsto\{a\in A\mid x\in a\} \end{align*} [imath]g[/imath] is a well defined function/ Let [imath]F=f\circ g[/imath]. Then [imath]\text{dom }F=\text{dom }g=\bigcup A[/imath]. Let [imath]x\in\bigcup A[/imath]. [imath]F(x)=f(g(x))=f\left(\{a\in A\mid x\in a\}\right)[/imath]. Thus [imath]F(x)\in\{a\in A\mid x\in a\}[/imath]. Thus [imath]x\in F(x)\in A[/imath].[imath]\tag*{$\square$}[/imath] I couldn't prove the other way. I had two ideas:( [imath]F:\bigcup A\to A[/imath] such that [imath]\forall a\in\bigcup A,\,a\in F(a)\in A[/imath]) 1) We could try to construct an "inverse" of [imath]F[/imath]. But [imath]F[/imath] isn't necessarily injective nor surjective. Thus such a construction would require the axiom of choice. 2) We could consider [imath]f=F\circ g[/imath] where [imath]g:\,A\to \bigcup\bigcup A[/imath] and [imath]F:\,\bigcup\bigcup A\to\bigcup A[/imath] but I couldn't find a suitable [imath]g[/imath]. Could you please help me? Thanks you in advance!
|
37799
|
Does a "backwards" choice function imply the Axiom of Choice?
One of my favorite formulations of the Axiom of Choice is that for any nonempty family [imath]A[/imath] of nonempty sets, there is a choice function [imath]F\colon A\to\cup A[/imath] such that [imath]F(X)\in X[/imath] for each [imath]X\in A[/imath]. Using this, I was able to prove that there is also a function [imath]f\colon\cup A\to A[/imath] such that [imath]x\in f(x)\in A[/imath] for all [imath]x\in\cup A[/imath]. I did this by considering any [imath]x\in\cup A[/imath], and letting [imath]B_x=\{X\in A\mid x\in X\}[/imath], and then letting [imath]C=\{t\in\mathscr{P}(A)\mid \exists_{x\in\cup A}t=B_x\}[/imath], i.e. [imath]C[/imath] is the set of all [imath]B_x[/imath]. Now each [imath]B_x[/imath] is nonempty, so there is a choice function [imath]F[/imath] such that [imath]F(B_x)\in B_x[/imath] for each [imath]B_x[/imath]. Defining [imath]f(x)=F(B_x)[/imath], I have [imath]x\in f(x)\in A[/imath] for every [imath]x\in\cup A[/imath]. I suppose this is what I mean by a "backwards" choice function, even though it's not really one. Is this equivalent to AC? That is, if for any set [imath]A[/imath], there exists a function [imath]f\colon\cup A\to A[/imath] such that [imath]x\in f(x)\in A[/imath] for all [imath]x\in\cup A[/imath], then AC holds? I couldn't immediately see a way to imply the Axiom of Choice, in any equivalent formulation assuming that the above is true for any set [imath]A[/imath]. Is it just a one way implication? Thanks.
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521861
|
Proof by induction that [imath]\sum_{i=1}^n \frac{i}{(i+1)!}=1- \frac{1}{(n+1)!}[/imath]
Prove via induction that [imath]\sum_{i=1}^n \frac{i}{(i+1)!}=1- \frac{1}{(n+1)!}[/imath] Having a very difficult time with this proof, have done pages of work but I keep ending up with 1/(k+2). Not sure when to apply the induction hypothesis and how to get the result [imath]1- \frac{1}{(n+2)!}[/imath]. Please help! thanks guys, youre the greatest!
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1858566
|
Prove [imath]\sum_{i=1}^{n}\frac{i}{(i+1)!}= 1-\frac1{(n+1)!}[/imath]
Required to prove: [imath]1-\frac1{(k+2)!}[/imath] [imath]\begin{align*} \sum_{i=1}^{k+1}\frac{i}{(i+1)!}&=\sum_{i=1}^k\frac{i}{(i+1)!}+\frac{k+1}{(k+2)!}\\ &= 1-\frac1{(k+1)!}+\frac{k+1}{(k+2)!}\tag{induction hyp.}\\[1em] \end{align*}[/imath] At my inductive solution i got stuck at the above section where i add the [imath]k+1[/imath] value to the inductive solution. Please help.
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1855008
|
what is asymptotic behavior of [imath]\sum \frac {1}{\sqrt[\alpha] k}[/imath]
Asymptotic behavior of [imath]\sum \frac {1}{\sqrt[\alpha] k}[/imath] for [imath]\alpha=1[/imath]? is [imath]\ln k[/imath] what about [imath]\alpha > 1[/imath] ? the suggested link is for [imath]\alpha > \frac{1}{2}[/imath] my question is about [imath] 0< \alpha < \frac{1}{2}[/imath]
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988710
|
Asymptotic behavior of [imath]\sum\limits_{k=1}^n \frac{1}{k^{\alpha}}[/imath] for [imath]\alpha > \frac{1}{2}[/imath]
As the title states, I'm interested in the asymptotic behavior of [imath]\sum\limits_{k=1}^n \frac{1}{k^{\alpha}} , \alpha > \frac{1}{2}[/imath] for [imath]n \to \infty [/imath]. Any hints/ideas?
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1855465
|
Help determining if the series [imath]\sum_{n=0}^{\infty }\frac{n^2}{3^n}[/imath] converges or diverges
[imath]\sum_{n=0}^{\infty }\frac{n^2}{3^n}[/imath] I've tried using the ratio test but I can't seem to figure out what I'm doing wrong: [imath]\lim_{n \rightarrow \infty}\left|\frac{(n+1)^2}{3^{n+1}}\cdot \frac{3^n}{n^2}\right|=\lim_{n \rightarrow \infty}\left|\frac{(n+1)^2}{3n^2}\right|=\frac{1}{3}\lim_{n \rightarrow \infty}\left|\frac{(n+1)^2}{n^2}\right|[/imath] Where do I go from here? It's been a few years since I've done any sequences and series.
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593996
|
How to prove [imath]\sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6[/imath]?
I'd like to find out why \begin{align} \sum_{n=0}^{\infty} \frac{n^2}{2^n} = 6 \end{align} I tried to rewrite it into a geometric series \begin{align} \sum_{n=0}^{\infty} \frac{n^2}{2^n} = \sum_{n=0}^{\infty} \Big(\frac{1}{2}\Big)^nn^2 \end{align} But I don't know what to do with the [imath]n^2[/imath].
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1855566
|
Can this be done using Sylow theorems?
Let [imath]p[/imath] and [imath]q[/imath] be distinct primes.Suppose that [imath]H[/imath] is a proper subset of the integers and [imath]H [/imath]is a group under addition that contains exactly three elements of the set {[imath]p,p+q,pq,p^q,q^p[/imath]}.Determine which of the following are the three elements in [imath]H[/imath]. a. [imath]pq,p^q,q^p[/imath] b. [imath]p,p+q,q^p[/imath] c. [imath]p,p^q,q^p[/imath] d. [imath]p,pq,p^q[/imath]
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529517
|
Additive group contains exactly three elements of the set [imath]\{p,p+q,pq,p^q,q^p\}[/imath]
Let [imath]p[/imath] and [imath]q[/imath] be distinct primes. There is a proper subgroup [imath]J[/imath] of the additive group of integers which contains exactly three elements of the set [imath]\{p,p+q,pq,p^q,q^p\}[/imath]. Which three elements are in [imath]J[/imath]? I know the solution is [imath]\{p,pq,p^q\}[/imath]. However I do not understand why. I thought [imath]J[/imath] can not even be a finite group. Maybe there is serious misunderstanding of the problem. Can anyone give me a hand? Thanks.
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1856128
|
Why does a = b (mod n) iff a - b is divisible by n?
I am specifically asking why the statement [imath]a \equiv b \;(\bmod\; n)[/imath] is equivalent to the statement [imath]a = b + kn[/imath], where k is some positive integer. Why is it that the difference of a and b has to be a multiple of n in order for them to have the same remainder when divided by n?
|
235343
|
Why [imath]a\equiv b\pmod n[/imath] implies [imath]a[/imath] and [imath]b[/imath] have equal remainder when divided by [imath]n[/imath]?
How does [imath]a\equiv b\pmod n[/imath] implies [imath]a[/imath] and [imath]b[/imath] have the same remainder when divided by [imath]n[/imath]? I don't understand the huge jump from modular equivalence to having equal remainders. I see by definition [imath]a\equiv b\pmod n\,[/imath] implies [imath]n\mid (a-b)\,[/imath] so [imath]\,a-b=nk\,[/imath] so [imath]\,a=b+nk[/imath]. But I do not see how this implies that [imath]a[/imath] and [imath]b[/imath] have the same remainder when divided by [imath]n[/imath].
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1856394
|
If two real matrices are conjugated over [imath]\mathbb{C}[/imath], are they then also conjugated over [imath]\mathbb{R}[/imath]?
As in the title: If two real (square) matrices are conjugated over [imath]\mathbb{C}[/imath], are they then also conjugated over [imath]\mathbb{R}[/imath]?
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1907255
|
Conjugate in [imath]\mathbb{C}[/imath] implies conjugate in [imath]\mathbb{R}[/imath]?
If [imath]A,B\in M(n,\mathbb{R})[/imath] and there exists [imath]P\in GL(n,\mathbb{C})[/imath] such that [imath]A=PBP^{-1}[/imath], does that imply that there exists [imath]Q\in GL(n,\mathbb{R})[/imath] such that [imath]A=QBQ^{-1}[/imath]?
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312098
|
Cartesian products and families
I have some trouble trying to understand Halmos' explanations about generalized Cartesian products. I've read these entries already: Cartesian products of families in Halmos' book. This also has the page I'm stuck on. One-to-one correspondence between a Cartesian product and a set of families set theoretic function, products of sets (product versus Cartesian product) Is his generalization the same as a Cartesian product of several sets? As here: Multiple products What puzzles me is [imath]\{a,b\}[/imath]. Where does it come from, what does it do? I know that it is supposed to be arbitrary. How does it identify distinct elements of [imath]X \times Y[/imath]? Does some function map each [imath]a \in A[/imath] to an element in [imath]X[/imath]? Or is it always the same [imath]a[/imath]? What does he mean by the set [imath]Z[/imath] of all families [imath]z[/imath], indexed by [imath]\{a,b\}[/imath], such that [imath]z_a \in X[/imath] and [imath]z_b \in Y[/imath] ? What does does an element of [imath]Z[/imath], namely one family [imath]z[/imath], consist of? By family, does he mean an indexing function or the indexed elements? This comes to a very important question for me: Does family mean a function or the set of all indexed elements? I will try to explain how I understand it thus far: [imath]Z[/imath] contains elements of the form [imath]\{(a,x),(b,y)\}[/imath]. The choice of [imath]x[/imath] depends on [imath]a[/imath], so an [imath]x[/imath] in a [imath]z \in Z[/imath] equals the [imath]z_a[/imath] of [imath]X[/imath]. Now, we want to map these pairs to [imath]X \times Y[/imath], so we have a function [imath]f[/imath] that takes [imath]z \in Z[/imath] and connects it to the corresponding ordered pair [imath](z_a, z_b)[/imath], where [imath]z_a \in X[/imath]. [imath]X \times Y[/imath] is not the same as [imath]Z[/imath]; they only, effectively, contain equal elements. He goes on: If [imath]\{X_i\}[/imath] is a family of sets ([imath]i \in I[/imath]), the Cartesian product of the family is, by definition, the set of all families [imath]\{x_i\}[/imath] with [imath]x_i \in X_i[/imath] for each [imath]i[/imath] in [imath]I[/imath]. How I understand it: [imath]\{X_i\}[/imath] contains, as elements, several sets, each single one a set [imath]X[/imath] identified by an index/element of [imath]I[/imath]. Then, we go over all indeces [imath]i[/imath] in [imath]I[/imath], take from each set [imath]X[/imath], namely [imath]X_i[/imath], an element [imath]x[/imath] and add it to current set of the newly created sets/families. So, the lower index [imath]i[/imath] does not refer to an element of [imath]X_i[/imath] indexed by [imath]i[/imath], but rather calls all elements of [imath]X_i[/imath] [imath]x_i[/imath]. We end up with a set, defined as Cartesian product, of sets, in which each element (set) contains ordered elements, each from a different [imath]X_i[/imath]. How does the ordering occur? [imath]\prod_i X_i[/imath], the Cartesian product, becomes equal to [imath]X^I[/imath], if all [imath]X_i[/imath] are the same set [imath]X[/imath]. [imath]X^I[/imath] means all possible combinations of mapping from [imath]I[/imath] to [imath]X[/imath]. We randomly take an elements [imath]x[/imath] from some [imath]X_i[/imath], put it in a set/family with other randomly obtained [imath]x[/imath], put the entire set into a set denoted [imath]X^I[/imath], and continue doing so, till we have no more options. Then, each set in the new [imath]X^I[/imath] contains a mapping of some [imath]i[/imath] to some [imath]x[/imath]. Since we exhausted the options, they are indeed all combinations. Where did I go wrong? Someone needs to spoonfeed my the answer; I am completely confused by now.
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2101462
|
Elaborate definition of [imath]\prod_{i} X_i[/imath] where [imath]X_i \in \left \{X_i \right \}_{i \in I}[/imath]
Can you elaborate below definition ideally with an example or two? I am starting to read the online book "Algebra and Topology" by Schapira Sierre, but got stuck at the very beginning. He gives the product of a family of sets indexed by [imath]I[/imath], [imath]\left \{ X_i \right \}_{i \in I}[/imath], on page 8 as follows: [imath]\prod_i X_i := \left \{ \left \{ x_i \right \}_{i \in I} \mid \forall i \in I: x_i \in X_i \right \}[/imath]
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1856403
|
[imath]R[/imath]-module [imath]M[/imath] which is artinian, but not noetherian
As I' m currently dealing with artinian and noetherian modules, I' m asking myself whether there is an artinian module which is not noetherian. I think so, but even after I thought a lot about this, I didn't find an example. Does anybody find one?
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173230
|
Example of Artinian module that is not Noetherian
I've just learned the definitions of Artinian and Noetherian module and I'm now trying to think of examples. Can you tell me if the following example is correct: An example of a [imath]\mathbb Z[/imath]-module [imath]M[/imath] that is not Noetherian: Let [imath]G_{1/2}[/imath] be the additive subgroup of [imath]\mathbb Q[/imath] generated by [imath]\frac12[/imath]. Then [imath]G_{1/2} \subset G_{1/4} \subset G_{1/8} \subset \dotsb[/imath] is a chain with no upper bound hence [imath]M = G_{1/2}[/imath] as a [imath]\mathbb Z[/imath]-module is not Noetherian. But [imath]M[/imath] is Artinian: [imath]G_{1/2^n}[/imath] are the only subgroups of [imath]G_{1/2}[/imath]. So every decreasing chain of submodules [imath]G_i[/imath] is bounded from below by [imath]G_{1/2^{\min i}}[/imath]. Edit In Atiyah-MacDonald they give the following example: Let [imath]G[/imath] be the subgroups of [imath]\mathbb{Q}/\mathbb{Z}[/imath] consisting of all elements whose order is a power of [imath]p[/imath], where [imath]p[/imath] is a fixed prime. Then [imath]G[/imath] has exactly one subgroup [imath]G_n[/imath] of order [imath]p^n[/imath] for each [imath]n \geq 0[/imath], and [imath]G_0 \subset G_1 \subset \dotsb \subset G_n \subset \dotsb[/imath] (strict inclusions) so that [imath]G[/imath] does not satisfy the a.c.c. On the other hand the only proper subgroups of [imath]G[/imath] are the [imath]G_n[/imath], so that [imath]G[/imath] does satisfy d.c.c. (Original images here and here.) Does one have to take the quotient [imath]\mathbb{Q}/\mathbb{Z}[/imath]?
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1855647
|
What is the number of possible [imath] x[/imath] values in [imath]\frac{x}{100}= \sin(x)[/imath]
Problem [imath]\frac{x}{100}= \sin(x)[/imath] We are asked to find the number of possible values of [imath]x[/imath] in this scenario and I had tried to figure it out by the use of trigonometric identities but then i had realized that there are no trig identities that can help me... or is there? steps that i had tried: I first multiplied 100 on both sides to get [imath]x[/imath] by itself to get: [imath]100(\sin x)=x[/imath] Then i determined that [imath]\sin x \le 1[/imath] therefore I had determined that [imath]\frac{x}{100}[/imath] is [imath]\le 1[/imath] But when i look at the answer choices all of them are less than [imath]100[/imath] meaning that all of them are going to make [imath]\frac{x}{100}[/imath] less than 100 when [imath]x[/imath] is substituted.
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1814707
|
How many solutions has the equation [imath]\sin x= \frac{x}{100}[/imath] ?
How many solutions has the equation [imath]\sin x= \frac{x}{100}[/imath] ? Usually when I was asked to solve this type of problem, I would solve it graphically but this one seems to be trickier. It doesn't seem wise to put [imath]f(x)=\sin x[/imath] and [imath]g(x)=\frac{x}{100}[/imath] in the same graph and then counting all the intersection points. What would be some algebraic methods to solve this?
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1856753
|
S is saturated if and only if R\S, the complement of S in R, is the union of some (possibly empty) family of prime ideals of R
I got problem which is the same here but I can not solve follow the way 2 of them pointing out Show [imath]R \setminus S[/imath] is a union of prime ideals Please help me explain more. And one more thing I wanna ask that the converse is still right. Could you give the solution of the converse?
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1418829
|
Show [imath]R \setminus S[/imath] is a union of prime ideals
I'm stuck on the following question: Let [imath]R[/imath] be a commutative ring with [imath]1[/imath], and [imath]S \subseteq R[/imath] a saturated multiplicative set (that is, [imath]1 \in S[/imath] and [imath]x, y \in S[/imath] if and only if [imath]xy \in S[/imath]). Show that [imath]R \setminus S[/imath] is a union of prime ideals. Here's my attempt so far. If [imath]a \in R \setminus S[/imath], then the ideal [imath]Ra[/imath] does not intersect [imath]S[/imath], because if [imath]ra[/imath] were in [imath]S[/imath] for some [imath]r \in R[/imath], then [imath]a[/imath] would have to be in [imath]S[/imath]. Thus, it follows by Zorn's lemma that there exist ideals [imath]I[/imath] of [imath]R[/imath] which are maximal with respect to the property that [imath]I \cap S = \emptyset[/imath]. Let [imath]I[/imath] be any such ideal. If I can show that [imath]I[/imath] is prime, the conclusion will follow. Let [imath]xy \in I[/imath]. I want to show that [imath]x[/imath] or [imath]y[/imath] is in [imath]I[/imath]. Suppose [imath]x \not\in I[/imath]. Then [imath]I + Rx[/imath] intersects [imath]S[/imath], so there exists [imath]s \in S, a \in I, r \in R[/imath] such that [imath]s = a + rx[/imath]. Then [imath]sy = ay + rxy \in I[/imath], so [imath]sy[/imath], and hence [imath]y[/imath], cannot be in [imath]S[/imath]. Similarly if [imath]y[/imath] is not in [imath]I[/imath], then [imath]x \not\in S[/imath]. So we are reduced to proving the following is impossible: [imath]xy \in I[/imath], [imath]x[/imath] and [imath]y[/imath] are not in [imath]S[/imath], but they aren't in [imath]I[/imath] either.
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1856897
|
A **proof** for [imath]\sum_{i=0}^{t-2}{\frac{1}{t+3i}} \leq \frac{1}{2}[/imath]
I need a proof for the inequality: [imath]\sum_{i=0}^{t-2}{\frac{1}{t+3i}} \leq \frac{1}{2}[/imath] for all natural numbers [imath]t \geq 2[/imath]. For [imath]t=2[/imath] both sides are equal. Can someone find a proof for all [imath]t[/imath]? maybe by proving monotonicity (non-increasing) in [imath]t[/imath]?
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1856855
|
A proof for the inequality [imath]\sum_{i=0}^{t-2}{\frac{1}{t+3i}} \leq \frac{1}{2} [/imath] for all [imath]t \geq 2[/imath]
I'm struggling with proving the following inequality: [imath]\sum_{i=0}^{t-2}{\frac{1}{t+3i}} \leq \frac{1}{2}[/imath] for all [imath]t \geq 2[/imath]. I think it is monotonic non-increasing in [imath]t[/imath], which would suffice. Thanks in advance!
|
918631
|
Prove that [imath]\dim(U_{\perp}) = \dim(V ) − \dim(U)[/imath].
Let [imath]V[/imath] be a finite-dimensional inner product space over field [imath]F[/imath], and let U be a subspace of [imath]V[/imath] . Prove that the orthogonal complement [imath]U_{\perp}[/imath] of [imath]U[/imath] with respect to the inner product [imath]\langle \cdot , \cdot\rangle[/imath] on [imath]V[/imath] satisfies [imath]\dim(U_{\perp}) = \dim(V ) − \dim(U)[/imath].
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1467135
|
Why [imath]\dim U+\dim U^\perp=\dim V[/imath]?
Let [imath]V[/imath] be a vector space. Given [imath]U\le V[/imath], define [imath]U^\perp=\{f\in V^\ast\mid f(u)=0\,,\;\forall u\in U\}[/imath] and given [imath]W\le V^\ast[/imath], define [imath]W^\perp=\{u\in V\mid f(u)=0\;,\;\forall f\in W\}[/imath]. This defines a Galois connection between subspaces of [imath]V[/imath] and subspaces of [imath]V^\ast[/imath]. If [imath]V[/imath] is finite dimensional, why is [imath]\dim U+\dim U^\perp=\dim V[/imath]? Unfortunately, I am not very comfortable with linear algebra. Any hint will be appreciated.
|
1857165
|
How do I see that [imath]a = bu[/imath] for some unit [imath]u[/imath]?
Suppose elements [imath]a[/imath] and [imath]b[/imath] in a domain satisfy [imath]a \mid b[/imath] and [imath]b \mid a[/imath]. How do I see that [imath]a = bu[/imath] for some unit [imath]u[/imath]?
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246591
|
Associates in Integral Domain
Let [imath]x[/imath] and [imath]y[/imath] be nonzero elements of an integral domain [imath]D[/imath]. Then [imath]x[/imath] and [imath]y[/imath] are associates if and only if [imath]x = yd[/imath] for some unit [imath]d \in D[/imath]. I am done proving the [imath]\Leftarrow[/imath] part. For [imath]\Rightarrow[/imath] what I did was: If [imath]x[/imath] and [imath]y[/imath] are associates then [imath]x \mid y[/imath] and [imath]y \mid x[/imath], so [imath]x \mid y[/imath] implies that [imath]xs = y[/imath] for some [imath]s \in D[/imath] and [imath]y \mid x[/imath] implies that [imath]yt = x[/imath] for some [imath]t \in D[/imath]. That is, [imath]yts=y[/imath] which implies [imath]ts=1[/imath]. Therefore [imath]s[/imath] is a unit and [imath]t[/imath] is a unit. Am I on the right track? Can I say that the proof is complete?
|
957274
|
[imath]f(a+b)=f(a)+f(b)[/imath] but [imath]f[/imath] is not linear
Can you show me a continuous function [imath]f \colon \mathbb{R}^n\to\mathbb{R}^m[/imath] that satisfies [imath]f(a+b)=f(a)+f(b)[/imath] but is not linear? We have that [imath]f(0)=f(0+0)=2f(0)\implies f(0)=0\\ f(x-x)=f(0)=f(x)+f(-x)=0\implies f(-x)=-f(x)\\ f(nx)=f(x+x+\dots+x)=f(x)+\dots+f(x)=nf(x)\quad \forall n \in \mathbb{N}[/imath] But [imath] f(-nx)=-f(nx)=-nf(x) [/imath] So: [imath] f(ax)=af(x) \quad \forall a \in \mathbb{Z} [/imath]
|
2198432
|
Real analysis, continuity of f
Dose there a function [imath]f: \mathbb{R} \to \mathbb{R}[/imath] satisfying [imath]f(a+b) = f(a)+f(b)\quad\forall a,b\in\mathbb{R}[/imath] But not of the form [imath]f(x)= \lambda x[/imath] ? Such [imath]f[/imath] must necessarily be discontinuous
|
9770
|
Understanding imaginary exponents
Greetings! I am trying to understand what it means to have an imaginary number in an exponent. What does [imath]x^{i}[/imath] where [imath]x[/imath] is real mean? I've read a few pages on this issue, and they all seem to boil down to the same thing: Any real number [imath]x[/imath] can be written as [imath]e^{\ln{x}}[/imath] (seems obvious enough.) Mumble mumble mumble This is equivalent to [imath]e^{\cos{x} + i\sin{x}}[/imath] Clearly I'm missing something in step 2. I understand (at least I think I do) how the complex number [imath]\cos{x} + i\sin{x}[/imath] maps to a point on the unit circle in a complex plane. What I am missing, I suppose, is how this point is related to the natural log of [imath]x[/imath]. Moreover, I don't understand what complex exponentiation is. I can understand integer exponentiation as simple repeated multiplication, and I can understand other things (like fractional or negative exponents) by analogy with the operations that undo them. But what does it mean to repeat something [imath]i[/imath] times?
|
1859276
|
Complex numbers as exponents
Is there any formula to calculate [imath]2^i[/imath] for example? What about [imath]x^z[/imath]? I was surfing through different pages and I couldn't seem to find a formula like de Moivre's with [imath]z^x[/imath].
|
1857342
|
Difference between norm and distance.
I was wondering the difference between norm and distance. My teacher told me that a norm always induce a distance, but that the reciprocal is not true. So, let [imath](E,\|\cdot \|)[/imath] a normed space. I agree that we can give a structure of metric space by setting [imath]d(x,y)=\|x-y\|[/imath]. Now let [imath](E,d)[/imath] a metric space. Why [imath]\|x\|=d(x,0)[/imath] would not be a norm over [imath]E[/imath] ?
|
172028
|
Difference between Norm and Distance
I'm now studying metric space. Here, I don't understand why definitions of distance and norm in euclidean space are repectively given in my book. I understand the difference between two concepts when i'm working on non-euclidean space, but is there any even slight difference between these two concepts when it is [imath]\mathbb{R}^k[/imath]?
|
648183
|
Is there a Relationship between sum and sum of reciprocals?
Is there any relationship between the sum of positive numbers and the sum of the reciprocals of the same numbers? For example: [imath]A=1+2+3.5[/imath] and [imath]B=1+1/2+ 10/35[/imath] Thanks!
|
1857424
|
Is there any known relationship between [imath]\sum_{i=1}^{n} f(i)[/imath] and [imath]\sum_{i=1}^{n} \dfrac {1}{f(i)}[/imath]
If we knew [imath]\sum_{i=1}^{n} f(i)=S[/imath] ([imath]n[/imath] can be [imath]\infty[/imath]), is there anything we could say about [imath]\sum_{i=1}^{n} \dfrac {1}{f(i)}[/imath] in terms of [imath]S[/imath]? The only thing I was able to find on the web is link to a dead question. If there is no known relationship, is it possible for there to exist one that we just haven't discovered?
|
1857218
|
[imath] z = 1 + 2i [/imath] - Prove that [imath] z^n \notin \mathbb{R} [/imath]
[imath] z = 1 + 2i \ (complex \ number) \\ z^n = a_n + b_ni \ (a_n, b_n \in \mathbb{Z}, n \in \mathbb{N}^*) \\ We \ know \ that: \ b_{n+2} - 2b_{n+1} + 5b_n = 0 \\ a_{n+1}=a_n-2b_n \\ b_{n+1}=b_n+2a_n [/imath] Prove that [imath] z^n \notin \mathbb{R} [/imath] What I tried: We know that [imath] b_{n+2} - 2b_{n+1} + 5b_n = 0[/imath] If [imath] z^n \in \mathbb{R} [/imath] it means that [imath] b_n = 0 [/imath] So, [imath] b_{n+2} - 2b_{n+1} = 0 [/imath] and I tried to get a contradiction but I can't because I get [imath] -3b_n = 0 [/imath]. How can I solve this problem? Thank you!
|
1856963
|
[imath] z^n = a_n + b_ni [/imath] Show that [imath] b_{n+2} - 2b_{n+1} + 5b_n = 0 [/imath] (complex numbers)
[imath] z = 1+2i \ (complex \ number) \\ z^n = a_n + b_ni \ \ \ (a_n, b_n \in \mathbb{Z}, n \in \mathbb{N}^*) [/imath] Prove that [imath] b_{n+2} - 2b_{n+1} + 5b_n = 0[/imath] How can I solve this? Thank you! EDIT: Or please tell me your ideas.
|
1858382
|
When is orthogonal projection compact?
Let [imath]M[/imath] be a closed subspace of a Hilbert space [imath]H[/imath]. Let [imath]P[/imath] be the orthogonal projection on [imath]M[/imath]. I was told to find the eigenvalues and eigenvectors of [imath]P[/imath] and moreover say when it is compact. Since [imath]M[/imath] is closed we have [imath]H\cong M\oplus M^\perp[/imath] and this is an eigenspace decomposition so [imath]P[/imath] has eigenvalues [imath]1,0[/imath] and eigenvectors the elements of [imath]M,M^\perp[/imath] respectively. Is there any formal explanation I emphasize in the infinite dimensional case? If [imath]M[/imath] is finite dimensional [imath]P[/imath] is finite rank so compact. I don't see anything but that...
|
241318
|
Showing that the orthogonal projection in a Hilbert space is compact iff the subspace is finite dimensional
Suppose that we have a Hilbert Space [imath]H[/imath] and [imath]M[/imath] is a closed subspace of [imath]H[/imath]. Let [imath]T\colon H\rightarrow M[/imath] be the orthogonal projection onto [imath]M[/imath]. I have to show that [imath]T[/imath] is compact iff [imath]M[/imath] is finite dimensional. So if we assume that [imath]M[/imath] is finite dimensional then [imath]\overline{T(B(0,1))}[/imath] is a closed bounded set in a finite dim vector normed space and so it is compact. Which gives that [imath]T[/imath] is compact. But I am unsure how to prove that if [imath]T[/imath] is compact then [imath]M[/imath] is finite dimensional? Thanks for any help
|
1827086
|
Does there exist [imath]A[/imath] of infinite order in [imath]\{ A \in GL_2(\mathbb{R}) : A^T = A^{-1} \}[/imath]?
Does there or does there not exist [imath]A[/imath] of infinite order in [imath]\{ A \in GL_2(\mathbb{R}) : A^T = A^{-1} \}[/imath]? I know that elements of the form [imath]A=\begin{bmatrix}\cos(\theta) & \sin(\theta)\\\pm\sin(\theta) & \pm\cos(\theta)\end{bmatrix}[/imath] are in this group, but do not know how to calculate their order.
|
319210
|
Matrix Group induction proof and order of elements question
[imath]H[/imath] is the set of the matrices [imath]A[/imath] of the form [imath]A= \begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}[/imath] where [imath]0\leq \theta < 2\pi[/imath] is a group with respect to matrix multiplication. Question part (1) Prove by induction that for any [imath]\theta[/imath] and any positive integer [imath]n[/imath], [imath]\begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}^{n} = \begin{pmatrix} \cos(n\theta) & \sin (n\theta) \\ -\sin (n\theta) & \cos (n\theta)\end{pmatrix}[/imath] When [imath]n =1[/imath] we have [imath]\begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}^{1} = \begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}[/imath] and clearly this holds for the base case. Now I'm going to assume true when [imath]n=k[/imath] and try and prove for [imath]k+1[/imath]. So we have [imath]\begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}^{k}\begin{pmatrix} \cos(\theta) & \sin (\theta) \\ -\sin (\theta) & \cos (\theta)\end{pmatrix}^{1} = \begin{pmatrix} \cos(k\theta + \theta) & \sin (k\theta + \theta) \\ -\sin (k\theta + \theta) & \cos (k\theta + \theta)\end{pmatrix}[/imath] Now I'm sure I can cheese this, but is there some clever way to pull this apart? I want to just be able to say something like the determinant of the left side is [imath]1^{k+1}[/imath] and its the same on the right ... Question part (2) Find an element of [imath]H[/imath] of order 5. Here I'm totally lost, I seem to think that every element is of order 1 cause the determinant is 1. Question part (3) Find an element of [imath]H[/imath] of infinite order. Again I'm lost; I want to somehow make the determinant 0 for this but it can never be 0 for any [imath]\theta[/imath].
|
1858733
|
On the relationship between [imath]\text{SL}_2(5)[/imath] and [imath]A_5[/imath]
I have two questions. What is the quickest way to see from scratch that [imath]\text{SL}_2(5)/\{\pm I\}[/imath] is isomorphic to the alternating group [imath]A_5[/imath]? Does [imath]\text{SL}_2(5)[/imath] have any subgroups isomorphic to [imath]A_5[/imath]?
|
93762
|
Elementary isomorphism between [imath]\operatorname{PSL}(2,5)[/imath] and [imath]A_5[/imath]
At this Wikipedia page it is claimed that to construct an isomorphism between [imath]\operatorname{PSL}(2,5)[/imath] and [imath]A_5[/imath], "one needs to consider" [imath]\operatorname{PSL}(2,5)[/imath] as a Galois group of a Galois cover of modular curves and consider the action on the twelve ramified points. While this is a beautiful construction, I wonder if this really is necessary. Is there a construction of a map that takes a representative matrix of a class in [imath]\operatorname{PSL}(2,5)[/imath] and uses some relatively simple computation to produce a permutation in [imath]S_5[/imath] that can be shown to be even? I don't mind if describing the construction and providing the verification that it is well-defined and does what it should takes several pages. I'd just like to think that it's possible.
|
1858245
|
Prove that [imath]\sin x + 2x \ge \frac{3x.(x+1)}{\pi} [/imath]
Prove that [imath]\sin x + 2x \ge \frac{3x.(x+1)}{\pi}\quad\forall x \in \left[0,\frac{\pi}{2}\right].[/imath] My work: [imath] 3x^2 + (3-2\pi )x - \pi \sin x \le 0 [/imath] [imath] f(x) = 3x^2 + (3-2\pi )x -\pi \sin x [/imath] [imath]f(0)=0 [/imath] [imath]f\left({\pi\over 2 }\right) ={\pi\over 2 }\left(1 -{\pi\over 2 } \right)[/imath] What should I do next ?
|
1424145
|
Prove that [imath]\sin x+2x\ge\frac{3x(x+1)}{\pi}[/imath] for all [imath]x\in [0,\frac{\pi}{2}][/imath]
Question: Prove that [imath]\sin x+2x\ge\frac{3x(x+1)}{\pi}[/imath] for all [imath]x\in [0,\frac{\pi}{2}][/imath] How did [imath]\pi[/imath] come in the expression? This is how I tried to solve. [imath]\sin x\le x[/imath] so [imath]\sin 2x\le 2x[/imath] [imath]\sin x+2x\le 3x[/imath]. Applying AM GM inequality: [imath]\sin x+2x\ge 2\sqrt{2x\sin x}[/imath] [imath]3x\ge 2\sqrt{2x\sin x}[/imath] This how far I can go.
|
1859458
|
Solution of differential equation [imath]\frac{dy}{dx}=\frac{1}{xy(x^2 \sin y^2+1)}[/imath]
Solve the given differential equation. [imath]\frac{dy}{dx}=\frac{1}{xy(x^2 \sin y^2+1)}[/imath] I have been trying to solve given differential equation using elementary approaches but no manipulation is a leading to a solvable form. Could someone help me with this?
|
1742373
|
Solution of [imath]\frac{dy}{dx}=\frac{1}{xy(x^2 \sin y^2+1)}[/imath]
Find the solution of following differential equation: [imath]\frac{dy}{dx}=\frac{1}{xy(x^2 \sin (y^2)+1)}[/imath] Could someone hint me something to get through this problem?
|
1859987
|
Relation between HCF, LCM and product of multiple numbers
It is well known that for two numbers [imath]a [/imath] and [imath]b[/imath], [imath]\text {lcm} (a,b)\times \text {hcf} (a,b)=ab[/imath] Does there exist a similar equality/ inequality between HCF, LCM and product of multiple numbers? (i.e. some relation between [imath]\text {hcf} (a_1,a_2,...,a_n)[/imath], [imath]\text{lcm} (a_1,a_2,...,a_n)[/imath] and [imath]a_1\times a_2 \times ...\times a_n[/imath])
|
319297
|
GCD to LCM of multiple numbers
If I know the GCD of 20 numbers, and know the 20 numbers, is there a formula such that I input the 20 numbers and input their GCD, and it outputs their LCM? I know that [imath]\frac{\left| a\cdot b\right|}{\gcd(a,b)} = \text{lcm}(a,b).[/imath] So is it[imath]\frac{\left| a\cdot b\cdot c\cdot d\cdot e\cdot f\right|}{\gcd(a,b,c,d,e,f)}?[/imath]If not, what is it?
|
1860330
|
Largest positive integer [imath]n[/imath] that satisfies a given inequality
Largest positive integer value of [imath]n[/imath] in [imath]n \cdot \left(\frac{abc}{a+b+c}\right)\leq (a+b)^2+(a+b+4c)^2[/imath] where [imath]a, b, c \in \mathbb{R^{+}}[/imath]. [imath]\bf{My\; Try::}[/imath] We can write [imath](a+b)^2+(a+b+4c)^2= (a+b)^2+[(a+2c)+(b+2c)]^2[/imath] Now Using [imath]\bf{A.M}\geq G.M,[/imath] We get [imath]\frac{(a+2b)+(b+2c)}{2}\geq \left[(a+2b)(b+2c)\right]^{\frac{1}{2}}[/imath] So [imath](a+b+4c)^2\geq 4(a+2b)(b+2c)[/imath] So [imath](a+b+4c)^2+(a+b)^2\geq 4(a+2b)(b+2c)+(a+b)^2[/imath] [imath] = 4(ab+2b^2+2ac+4bc)+(a^2+b^2+2ab)[/imath] Now How can i solve after that, Help required, Thanks
|
1075570
|
What is the largest [imath]k[/imath] such that [imath] \frac { k(abc) }{ a+b+c } \le \left( a+b \right) ^{ 2 }+\left( a+b+4c \right) ^{ 2 } [/imath]?
Find the largest value of [imath] k [/imath] such that [imath]k \cdot \left(\frac { abc }{ a+b+c }\right) \le \left( a+b \right) ^{ 2 }+\left( a+b+4c \right) ^{ 2 }[/imath]
|
1860415
|
Prove [imath]\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}=\sec \theta\csc\theta+1[/imath]
I did it by converting every trigonometry stuff into [imath]{\sin}[/imath] and [imath]\cos[/imath]. But I want to know if there is a shortcut (without converting everything to [imath]\sin[/imath] and [imath]\cos[/imath]) to do this. Please help.
|
429047
|
Trigonometric identity: [imath]\frac {\tan\theta}{1-\cot\theta}+\frac {\cot\theta}{1-\tan\theta} =1+\sec\theta\cdot\csc\theta[/imath]
I have to prove the following result : [imath]\frac {\tan\theta}{1-\cot\theta}+\frac {\cot\theta}{1-\tan\theta} =1+\sec\theta\cdot\csc\theta[/imath] I tried converting [imath]\tan\theta[/imath] & [imath]\cot\theta[/imath] into [imath]\cos\theta[/imath] and [imath]\sin\theta[/imath]. That led to a huge expression which I wasn't able to simplify. Please help!!!
|
1860541
|
Value of [imath]f^2(4)+g^2(4)[/imath]
If [imath]f(x)=g'(x),g(x)=-f'(x)[/imath] for all real [imath]x[/imath] and [imath]f(2)=4=f'(2)[/imath] then value of [imath]f^2(4)+g^2(4)[/imath] is ? Now the above is true when we have a constant function with constant [imath]0[/imath]. But then that would not satisfy third condition ?what can this function be? Thanks !
|
1857054
|
Sum of square of function
If [imath]f'(x) = g(x)[/imath] and [imath]g'(x) = - f(x)[/imath] for all real [imath]x[/imath] and [imath]f(5) =2 =f'(5)[/imath] then we have to find [imath]f^2[/imath][imath](10) + g^2(10)[/imath] I tried but got stuck
|
1412056
|
Let [imath]A_1 A_2 \dotsb A_{11}[/imath] be a regular 11-gon inscribed in a circle of radius 2.
Let [imath]A_1 A_2 \dotsb A_{11}[/imath] be a regular 11-gon inscribed in a circle of radius 2. Let [imath]P[/imath] be a point, such that the distance from [imath]P[/imath] to the center of the circle is 3. Find [imath]PA_1^2 + PA_2^2 + \dots + PA_{11}^2.[/imath]
|
1361517
|
Find [imath]LK_1^2 + LK_2^2 + \dots + LK_{11}^2[/imath].
[imath]K_1 K_2 \dotsb K_{11}[/imath] is a regular [imath]11[/imath]-gon inscribed in a circle, which has a radius of [imath]2[/imath]. Let [imath]L[/imath] be a point, where the distance from [imath]L[/imath] to the circle's center is [imath]3[/imath]. Find [imath]LK_1^2 + LK_2^2 + \dots + LK_{11}^2[/imath]. Any suggestions as to how to solve this problem? I'm unsure what method to use.
|
1861085
|
Normal subgroup.
Let [imath]N[/imath] be subgroup of a group [imath]G[/imath]. Suppose that, for each [imath]a\in G[/imath], there exists [imath]b\in G[/imath] such that [imath]Na=bN[/imath]. Prove that [imath]N[/imath] is a normal subgroup. Please guide me with a proof. Thank you for your kindness. This is Exercise, Hungerford.
|
381462
|
Normal subgroups and cosets
Let [imath]N[/imath] be a subgroup of a group [imath]G[/imath]. Suppose that, for each [imath]a[/imath] in [imath]G[/imath], there exists [imath]a, b[/imath] in [imath]G[/imath] such that [imath]Na=bN[/imath]. Prove that [imath]N[/imath] is a normal subgroup. Attack: I found [imath]b^{-1}N[/imath] = [imath]Na^{-1}[/imath] but I am stuck! Any help will be appreciated
|
1860187
|
Proof of the square root inequality [imath]2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}[/imath]
I stumbled on the following inequality: For all [imath]n\geq 1,[/imath] [imath]2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}.[/imath] However I cannot find the proof of this anywhere. Any ideas how to proceed? Edit: I posted a follow-up question about generalizations of this inequality here: Square root inequality revisited
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263115
|
Proof of inequality [imath]2(\sqrt{n+1}-\sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1})[/imath] using induction
Prove that [imath]2(\sqrt{n+1}-\sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1})[/imath] if [imath]n \ge 1[/imath] using induction. Can someone help me with this problem please. Base case is easily shown, and for the inductive step I know I have to use the inequality for [imath]n\gt1[/imath] to show that it is true for [imath]n+1[/imath], but I am not quite sure how to. Thanks.
|
1861152
|
The total number of solutions (real) of equation: [imath]2^x+3^x+4^x-5^x=0 ?[/imath]
The total number of solutions (real) of equation: [imath]2^x+3^x+4^x-5^x=0 ?[/imath] I have no idea how to solve this problem. Can someone point me in the right direction?
|
886972
|
Solving an equation in [imath]x[/imath], in which [imath]x[/imath] occurs as exponent four times
Find the number of solutions to the equation [imath]2011^x+2012^x+2013^x-2014^x=0[/imath] The answer seems to be zero, but I have no idea why. Please avoid considering complex solutions and other scary things.
|
1861726
|
Integration by Parts and the Constant of Integration
The constant of integration only seems to be used at the very end of integration by parts despite the use of integrals beforehand. An example of this would be: [imath]$$∫ x\sin \left(x\right)\ dx = x∫ \sin \left(x\right)\ dx - ∫ x'\left(∫ \sin \left(x\right)\ dx\right)\ dx$$[/imath] Ordinarily, the right side of the equation would be simplified to: [imath]$$x\left(-\cos \left(x\right)\right) - ∫-\cos \left(x\right)\ dx$$[/imath] And further to: [imath]$$-x\left(\cos \left(x\right)\right) + \sin \left(x\right)$$[/imath] Then finally arranged and given the constant of integration: [imath]$$\sin \left(x\right) - x\cos \left(x\right)+ C$$[/imath] What I am confused about is why [imath]C[/imath] is only added at the very end of this instead of at each integral. I would be more inclined to use try something more like this: [imath]$$x\left(-\cos \left(x\right) +C_1\right) - ∫ -\cos \left(x\right) + C_2\ dx$$[/imath] Which would simplify to: [imath]$$-x\left(\cos \left(x\right) -C_1\right) - ∫-\cos \left(x\right)\ dx + ∫ C_2\ dx $$[/imath] And further to: [imath]$$-x\cos \left(x\right) +C_1x +\sin \left(x\right)+ C_3 + C_2x + C_4 $$[/imath] Which finally arranges itself as: [imath]$$\sin \left(x\right) - x\cos \left(x\right) + C_5x + C_6$$[/imath] Where [imath]C_5=C_1 + C_2[/imath] and [imath]C_6=C_3 + C_4[/imath] I also feel I should probably mention I am a bit of an oblivious idiot so if the answer is completely obvious or my math is full of errors, I apologize.
|
26869
|
Constants of integration in integration by parts
After finishing a first calculus course, I know how to integrate by parts, for example, [imath]\int x \ln x dx[/imath], letting [imath]u = \ln x[/imath], [imath]dv = x dx[/imath]: [imath]\int x \ln x dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2x} dx.[/imath] However, what I could not figure out is why we assume from [imath]dv = x dx[/imath] that [imath]v = \frac{x^2}{2}[/imath], when it could be [imath]v = \frac{x^2}{2} + C[/imath] for any constant [imath]C[/imath]. The second integral would be quite different, and not only by a constant, so I would like to understand why we "forget" this constant of integration. Thanks.
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743360
|
Showing [imath]a_n=\sin(n)[/imath] does not converge
Show that [imath]a_n=\sin(n)[/imath] does not converge My idea: Take two subsequences: [imath]a_{n_k}=\sin(\frac {\pi k} 2)[/imath] , [imath]a_{n_l}=\sin(\frac {2\pi l} 3)[/imath] So: [imath]\forall n[/imath] : [imath]\lim_{n\to\infty} a_{n_k}=1[/imath], [imath]\lim_{n\to\infty} a_{n_l}=-1[/imath] So two inifinite subsequence converge to different limits thus the sequence [imath]a_n[/imath] doesn't converge. Is that correct ? Edit: The contrapositive of the defintion of a limit of a sequence: [imath]\exists\epsilon>0 : \forall n\in N : \exists n>N \Rightarrow |x_n-L|>\epsilon[/imath] Take [imath]\epsilon =1[/imath] and we know that [imath]\sin(n)[/imath] is bounded so lets take it's supermum: 1 [imath]|1-1|>\epsilon=1\Rightarrow 0>1 \Rightarrow[/imath] Contradiction.
|
967548
|
How to show that [imath]\sin(n)[/imath] does not converge?
I am supposed to show that [imath]\sin(n)[/imath] does not converge by constructing two subsequences: one subsequence contains terms of [imath]\sin(n)[/imath] that are between [imath]1/2[/imath] and [imath]1[/imath], and the other subsequence contains terms of [imath]\sin(n)[/imath] that are between [imath]-1/2[/imath] and [imath]-1[/imath]. But how can I do this? Which [imath]k(n)[/imath]--for subsequence [imath]x_{k(n)}[/imath]--should I choose?
|
1861390
|
For which [imath]\alpha[/imath], [imath]\beta[/imath] does [imath]\int\limits_1^{\infty} x^{\alpha} \cdot (\ln x)^\beta dx[/imath] converge?
For which [imath]\alpha[/imath] and [imath]\beta[/imath] does the following integral converge ?: [imath] \int_{1}^{\infty}x^{\alpha}\,\ln^{\beta}\left(x\right)\,\mathrm{d}x [/imath] Here is my analysis: I noticed that the function under the integral is of the following type: [imath]\,\mathrm{f}\left(x\right)\,\mathrm{g}\left(x\right)[/imath]. If we had [imath]\,\mathrm{g}\left(x\right) = 1[/imath], so [imath]\displaystyle{\int_{1}^{\infty}x^{\alpha}\,\mathrm{d}x}[/imath] converges when [imath]\alpha < -1[/imath]. When [imath]\beta > 0[/imath] the power [imath]\alpha[/imath] should somehow overcome that value. If [imath]\,\mathrm{g}\left(x\right)[/imath] was simply [imath]\,\mathrm{g}\left(x\right) = x^{\beta}[/imath], then the relationship between [imath]\alpha[/imath] and [imath]\beta[/imath] would have been linear, so [imath]\alpha + \beta < -1[/imath]. I cannot come up with the relationship in case when [imath]\,\mathrm{g}\left(x\right) = \ln^{\beta}\left(x\right)[/imath].
|
1574329
|
For which values of [imath]\alpha[/imath] and [imath]\beta[/imath] does the integral $\int\limits_2^{\infty}\frac{dx}{x^{\alpha}\ln^{\beta}x}$ converge?
I'm trying to find out for which values of [imath]\alpha[/imath] and [imath]\beta[/imath] the integral [imath]$\int\limits_2^{\infty}\frac{dx}{x^{\alpha}\ln^{\beta}x}$[/imath] does converge. I know that when [imath]\alpha=1[/imath] then [imath]\beta[/imath] must be greater than [imath]1[/imath]. I tried to use integration by parts but It didn't work, so I would appreciate some hints. Thanks in advance.
|
1861573
|
About [imath]\frac{\partial^n f}{\partial x^n}[/imath] ,for [imath]f(x)[/imath],What should I think, when [imath]n[\in (\mathbb R\backslash \mathbb Q)^+][/imath] or [imath](\in \mathbb C)[/imath]?
About [imath]\frac{\partial^n f}{\partial x^n}[/imath] ,for [imath]f(x)[/imath],What should I think, when [imath]n[\in (\mathbb R\backslash \mathbb Q)^+][/imath] ,pozitive irrational? or [imath](\in \mathbb C)[/imath] For example; [imath]f(x)=x^2+3x\quad[/imath] and [imath]\quad n=\sqrt2\quad\to\quad \dfrac{\partial^{(\sqrt 2)} f}{\partial x^{(\sqrt2)}}=?[/imath] and, [imath]f(x)=x^2+3x\quad[/imath] and [imath]\quad n=\sqrt i\quad\to\quad\dfrac{\partial^{(\sqrt i)} f}{\partial x^{(\sqrt i)}}=?[/imath] And I think,I did grammar mistakes in the tittle, sorry about this.
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878707
|
Non integer derivative of [imath]1/p(x)[/imath]
I need to find the [imath]k[/imath]'th derivative of [imath]1/p(x)[/imath], where [imath]p(x)[/imath] is a polynomial and [imath]k\in\mathbb{R}[/imath] It dosen't have to be an explicit formula, an algorithm which finds a formula for some [imath]k[/imath] is fine.
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1862379
|
Prove that [imath]\operatorname{trace}(A) = 0[/imath] if and only if [imath]A^2 = 0[/imath].
Let [imath]A\in M_{n \times n}[/imath] such that rank of [imath]A[/imath] is [imath]1[/imath]. Prove that [imath]\operatorname{trace}(A) = 0[/imath] if and only if [imath]A^2 = 0[/imath].
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1090421
|
Questions about matrix rank, trace, and invertibility
(a) Prove that a square matrix [imath]T[/imath] of rank one has [imath]\text{tr}(T)=0[/imath] if and only if [imath]T^2=0[/imath]. (b) Consider a matrix [imath]A[/imath] of the form [imath]A=aI+T[/imath], where [imath]a\ne0[/imath], [imath]I[/imath] is the identity matrix, and [imath]T[/imath] has rank one and zero trace. Find the inverse and the determinant of [imath]A[/imath]. (c) Find the inverse of [imath]A[/imath] as above when [imath]T[/imath] has rank one but nonzero trace [imath]\text{tr}(T)=b[/imath]. For which value of [imath]b[/imath] is [imath]A[/imath] not invertible? I'm still stuck on part (a), but campus buildings are closing soon, so I'll be working from home but would love to get some hints / comments on this question. I'll have limited access to this site - on my phone. For part (a), I've been trying to look at the SVD of matrix [imath]A[/imath], since one can read off the rank very easily - by looking at the number of non-zero singular values of [imath]A[/imath]. Then I am trying some block matrix multiplication to see whether [imath]T^2 = 0[/imath], from assuming that [imath]\text{tr}(T) =0[/imath]. So far, no luck. Do you think I should stick with this SVD approach, or is it better to play around with the definition and properties of nilpotent operators? Any other hints for the other parts of the question would be greatly appreciated. Thanks!
|
1862587
|
Arbitrary distributive law for Sets
How can we interchange union and intersection in [imath]\bigcap_{\gamma < \kappa}[\bigcup_{n<\omega}I(\gamma,n)][/imath]
|
471744
|
Proof that [imath]\bigcap_{a\in A} \bigg(\, \bigcup_{b\in B} F_{a,b} \, \bigg) = \bigcup_{f\in ^AB} \bigg(\, \bigcap_{a \in A} F_{a,f(a)}\,\bigg) [/imath]
I found an interesting exercise of other amazing book (Jech's) that I'm not totally sure about how to do it. Prove the following form of the distributive law: [imath]\bigcap_{a\in A} \bigg(\, \bigcup_{b\in B} F_{a,b} \, \bigg) = \bigcup_{f\in ^AB} \bigg(\, \bigcap_{a \in A} F_{a,f(a)}\,\bigg) [/imath] Assuming that [imath]F_{a,b_1} \cap F_{a,b_2} = \emptyset [/imath] for each [imath]a\in A[/imath] and [imath]b_1,b_2\in B[/imath] and [imath]b_1\not=b_2[/imath]. First of all, I'm not completely sure about what [imath] F_{a,b}[/imath] really means. I suppose that is the range of a family with a domain [imath]A \times B[/imath]. And second, I cannot figure out how to do the converse (assuming that the first part it is correct, I used as normal in this kind of proof element chasing) Proof: ([imath]\Rightarrow[/imath]) ... ([imath]\Leftarrow[/imath]) Suppose that [imath]z\in \bigcup_{f\in ^AB} \big(\, \bigcap_{a \in A} F_{a,f(a)}\,\big)[/imath]; then there is [imath]f\in \,^AB[/imath] such that [imath]z\in \bigcap_{a \in A} F_{a,f(a)}\, [/imath]. Let [imath]a \in A[/imath] be arbitrary. Claim 1 [imath]\, \bigcap_{a \in A} F_{a,f(a)}\subseteq F_{a,f(a)} \subseteq \bigcup_{b\in B} F_{a,b}[/imath]. Proof of the Claim 1: For the first inclusion: Suppose [imath]z\in\bigcap_{a \in A} F_{a,f(a)}[/imath]; then for each [imath]a\in A[/imath] we have that [imath]z\in F_{a,f(a)}[/imath]. Then [imath]\bigcap_{a \in A} F_{a,f(a)}\subseteq F_{a,f(a)}[/imath]. For the second inclusion: Now suppose [imath]z\in F_{a,f(a)}[/imath]. Clearly [imath]f(a)\in B[/imath] because [imath]f\in ^AB[/imath]. Then there is some [imath]b\in B[/imath] such that [imath]z\in F_{a,b}[/imath] and hence that [imath]z\in \bigcup_{b\in B} F_{a,b}[/imath]. [imath]\square[/imath] Since [imath]a[/imath] was arbitrary it follows that [imath]z\in \bigcup_{b\in B} F_{a,b}[/imath] (by claim 1) for each [imath]a\in A[/imath] and hence that [imath]z\in \bigcap_{a\in A} \big(\, \bigcup_{b\in B} F_{a,b} \, \big)[/imath] as desired. I really, really would very much appreciate some help with that. Thanks in advance as usual.
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1862617
|
Why [imath]\lim_{z\to 0}\frac{(\bar{z})^2}{z^2}[/imath] doesn't exist?
Why this limit does not exist? [imath]\lim_{z\to 0}\frac{(\bar{z})^2}{z^2}[/imath] I know that the limit exists and equals to [imath]1[/imath] is [imath]z[/imath] is real. When [imath]z[/imath] is complex, I tried [imath]\frac{(\bar{z})^2}{z^2}=\frac{e^{-2i\theta}}{e^{2i\theta}}=e^{-4i\theta}[/imath] Then put the limit back, [imath]\lim_{z\to 0} \frac{(\bar{z})^2}{z^2}=\lim_{z\to0} e^{-4i\theta}[/imath] Then I got a little bit confused because I wasn't sure if [imath]\lim_{z\to 0}[/imath] is equivalent to [imath]\lim_{r\to 0}[/imath]? I was thinking something like "[imath]z[/imath] gets small iif [imath]\|z\|[/imath] gets small iif [imath]r[/imath] gets small".
|
222172
|
Does [imath]\lim_{z\rightarrow0}\frac{\bar{z}^2}{z^2}[/imath] exist?
For[imath]\lim_{z\rightarrow0}\frac{\bar{z}^2}{z^2}[/imath] Does [imath]\lim_{z\rightarrow0}\frac{\bar{z}^2}{z^2}[/imath] exist? If the answer is no, why? Does [imath]\bar{z}[/imath] represents [imath]a-bi[/imath]?
|
451001
|
Proof of the greatest integer theorem: for every real number [imath]x[/imath] there exists a unique greatest integer less than or equal to [imath]x[/imath]
To define the function [imath]f(x)=|[x]|[/imath] where [imath]|[x]|[/imath] is the greatest integer that is less or equal to [imath]x[/imath], we need to prove that indeed such an integer exists. In other words, [imath]\forall x\in \mathbb{R} \;\;\exists !\, n\in\mathbb{Z}\;\;:\;\;n\leq x<n+1.[/imath] My first attempt is by defining the set [imath]A=\{y\in\mathbb{Z}| y\leq x\}[/imath]. Since [imath]\mathbb{N}[/imath] is not bounded above then [imath]\exists n\in \mathbb{N}(n>-x)[/imath]. Then [imath]\exists n'\in \mathbb{Z}(n'<x)[/imath], namely [imath]n'=-n[/imath]. Therefore [imath]A\neq\emptyset[/imath]. Also [imath]A[/imath] is bounded above by [imath]x[/imath]. Then [imath]\sup(A)[/imath] exists. **Here my problem is how to prove that [imath]\sup(A)\in A[/imath]. Also my question is that I'm not sure if this is enough because [imath]sup(A)[/imath] is unique or do I still need to prove uniqueness. EDIT: After all the hints in the answers, chich I appreciate so much, I have arrived to this attempts of solution, please feel free to comment and help me saying if something is wrong: First approach: considering tree cases ([imath]x<0, x=0, x>0[/imath]). 1) If [imath]x=0[/imath] then I can think of [imath]n=0[/imath]. 2) If [imath]x>0[/imath] then I can think of the set [imath]B=\{y\in \mathbb{N}|x\leq y\}[/imath]. By using the well-ordering principle then [imath]\min(B)-1<x\leq \min(B)[/imath] and then if [imath]x=\min(B) [/imath] I can make [imath]n=\min(B)[/imath]; if [imath]\min(B)-1<x<\min(B)[/imath] then [imath]n=\min(B)-1[/imath]. 3) if [imath]x<0[/imath] then [imath]-x>0[/imath] and therefore [imath]\exists m \in \mathbb{N}(m\leq -x < m+1)[/imath]. Then [imath]\exists n'\in \mathbb{Z}(-n'<x\leq-n'-1)[/imath], namely [imath]n'=m+1[/imath]. Again, if [imath]x=-n'-1[/imath] then I can make [imath]n=-n'-1[/imath]. If [imath]-n'<x\leq-n'-1[/imath] then [imath]n=-n'[/imath]. Second approach. Since [imath]\mathbb{N}[/imath] is not bounded above then let [imath]x_{0}\in \mathbb{N}[/imath] be such that [imath]x_{0}>\sup(A)[/imath]. Let's consider the set [imath]B=\{x_{0}-y: y\in A\}[/imath]. Since [imath]B\subseteq \mathbb{N}[/imath] then, by the well ordering principle, [imath]\min(B)[/imath] exists. Moreover [imath]\exists y_{0}\in A\forall y\in A(\min(B)=x_{0}-y_{0}\leq x_{0}-y)[/imath]. This means that [imath]\exists y_{0}\in A\forall y\in A (y\leq y_{0})[/imath]. Therefore [imath]\max(A)=y_{0}[/imath]. In this case I can make [imath]n=\max(A)[/imath].
|
117734
|
Proof of greatest integer theorem: floor function is well-defined
I have to prove that [imath]\forall x \in \mathbb{R},\exists\text{ exactly ONE }n \in \mathbb{Z} \text{ s.t. }n \leq x < n+1\;.[/imath] I'm done with proving that there are at least one integers for the solution. I couldn't prove the "uniqueness" of the solution, and so I looked up the internet, and here's what I found: Let [imath]\hspace{2mm}n,m \in \mathbb{Z} \text{ s.t. }n \leq x < n+1[/imath] and [imath]m \leq x < m+1[/imath]. Since [imath]n \leq x \text{ and } -(m+1) < -x[/imath], by adding both, [imath]n + (-m-1) < (x-x) = 0[/imath]. And (some steps here) likewise, [imath](x-x) < n+m+1[/imath]. Now, can I add up inequalities like that, even when the book is about real analysis (and in which assumptions are supposed to be really minimal)? Or should I also prove those addition of inequalities? Thank you :D
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1862822
|
b) How many onto functions are there from A to C?
Let [imath]A=\{1,2,3,4\}[/imath] Let [imath]B= \{a,b\}[/imath] Let [imath]C= \{ \text{hiking, baseball, hockey} \}[/imath] a) How many onto functions from A to B [imath](1,a) (1,b) (2,a) (2,b) (3,a) (3,b) (4,a) (4,b)[/imath] Thus 8. b) How many onto functions are there from A to C? This is where I am lost. Would I just count the non-onto functions???
|
334420
|
Number of onto functions
What are the number of onto functions from a set [imath]\Bbb A [/imath] containing m elements to a set [imath]\Bbb B[/imath] containing n elements. I found that if m = 4 and n = 2 the number of onto functions is 14. But is there a way to generalise this using a formula? If yes, what is this formula and how is it derived? reference I referred to this question but my doubt was not cleared: How many one to one and onto functions are there between two finite sets? If not, Then what is the standard way of doing it? If you explain this to me with an example please explain with the example of m = 5 and n = 3.
|
1862419
|
Generic Element of Compositum of Two Fields
I'm interested in understanding compositum of general fields better. Assume we have [imath]\Omega/K/F[/imath] and [imath]\Omega/L/F[/imath] field extensions, and consider the composite [imath]KL[/imath]. It seems to me that every [imath]a \in KL[/imath], exists [imath]n_1,n_2 \in \mathbb{N}[/imath], [imath] \left \lbrace k_i \right \rbrace_{i=1}^{n_1}, \left \lbrace k'_i \right \rbrace_{i=1}^{n_2} \subseteq K[/imath] and [imath] \left \lbrace l_i \right \rbrace_{i=1}^{n_1}, \left \lbrace l'_i \right \rbrace_{i=1}^{n_2} \subseteq L[/imath] such that: [imath] a = \frac{\sum_{i=1}^{n_1} k_i l_i}{\sum_{j=1}^{n_2} k'_j l'_j} [/imath] (That is since the family of all such sums is closed under addition, multiplication, and distribution because of the abelian nature of fields, therefore all numbers of this form are subfields that contain both K and L) Is it a correct way to refer to a compositum of fields? When is it correct that the general item of an extension is [imath]a = \sum_{i=1}^{n} k_i l_i[/imath]?
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1226078
|
The elements in the composite field [imath]FK[/imath]
Where [imath]F[/imath],[imath]K[/imath] are two fields. What does the element in the composite field [imath]FK[/imath] look like? All the elements are generated by the elements of [imath]F[/imath] and [imath]K[/imath]? (combination of the elements of [imath]F[/imath] and [imath]K[/imath]) I think [imath]FK[/imath] is pretty close to the free product of [imath]F[/imath] and [imath]K[/imath]; is [imath]FK \subsetneq F*K[/imath]? Thank you!
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1862770
|
Prove [imath]f_n(x)=\frac{x^n}{\sqrt{3n}}[/imath] for [imath]x \in [0,1][/imath] is uniformly convergent
I'm completely confused by uniform convergence, but I put together the following proof just based on my other questions here and examples I read online. Discussion: Let [imath]\epsilon \gt 0[/imath] We want to find [imath]N[/imath] so that for all [imath]x[/imath] in [imath][0,1][/imath] and [imath]n \gt N[/imath], [imath]\left|f_n(x)-f(x)\right|=\left|\frac{x^n}{\sqrt{3n}}-0\right|\lt \epsilon[/imath] [imath]\left|\frac{x^n}{\sqrt{3n}}\right|\le\frac{1}{\sqrt{3n}}\le\frac{1}{\sqrt{3N}}\lt\epsilon[/imath] if [imath]\frac{1}{3\epsilon^2}\lt N[/imath] Proof: Give [imath]\epsilon \gt 0[/imath], let [imath]N=\frac{1}{3\epsilon^2}[/imath]. Then for all [imath]x[/imath] in [imath][0,1][/imath] and [imath]n \gt N[/imath], [imath]\left|f_n(x)-f(x)\right|=\left|\frac{x^n}{\sqrt{3n}}-0\right|\le\frac{1}{\sqrt{3n}}\le\frac{1}{\sqrt{3N}}=\frac{1}{\sqrt{\frac{1}{3\epsilon^2}}}=\epsilon[/imath] I'm completely confused by this entire subject, I've been reading the definition over for the last 3 hours and I just do not understand what is going on.
|
1862353
|
Determine whether [imath]f_n(x)=\frac{x^n}{\sqrt{3n}}[/imath] for [imath]x \in [0,1][/imath] is uniformly convergent
This is a follow up to my below question: Pointwise convergence: State [imath]f(x) = \lim f_n(x)[/imath] I'm trying to determine whether [imath]f_n(x)=\frac{x^n}{\sqrt{3n}}[/imath] for [imath]x \in [0,1][/imath] is uniformly convergent. I know if it is then it must satisfy [imath]\left|f_n(x)-f(x)\right|\lt \epsilon[/imath] Since I know [imath]f(x)=0[/imath] I know it must be [imath]\left|f_n(x)-f(x)\right|\lt \epsilon = \left|\frac{x^n}{\sqrt{3n}}-0\right|\lt \epsilon =\left|\frac{x^n}{\sqrt{3n}}\right|\lt \epsilon[/imath] After doing some algebra I get: [imath]\frac{1}{\epsilon}\lt \frac{\sqrt[n]{3n}}{x}[/imath] Where do I go from here and have I made any mistakes? Does this means that [imath]f[/imath] is uniformly convergent?
|
1863869
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Prove that if [imath]A_1,A_2,...,A_n[/imath] are finite sets, then [imath]\bigcup_{i=1}^{n} A_{i}[/imath] is finite.
Let, S be a set defined such that [imath] S =\{n \in \mathbb{N}:[/imath] if [imath]A_1,A_2,...,A_n[/imath] are finite sets, then [imath]\bigcup_{i=1}^{n} A_{i}[/imath] is finite[imath] \} [/imath]. Let n = 1 and assume [imath]A_n = A_1 [/imath] is finite. Then, [imath]\bigcup_{i=1}^{n} A_{i} =A_1[/imath],which means [imath]\bigcup_{i=1}^{n} A_{i} =A_1[/imath] is finite. Therefore, [imath]1\in S[/imath]. Suppose for some [imath]n \in \mathbb{N}[/imath] that [imath]n \in S[/imath]. Then, the predicate if [imath]A_1,A_2,...,A_n[/imath] are finite sets, then [imath]\bigcup_{i=1}^{n} A_{i}[/imath] is finite is true for [imath]n[/imath]. Suppose, [imath]A_{n+1} [/imath] is finite. Then, [imath]\bigcup_{i=1}^{n+1} A_{i} = \bigcup_{i=1}^{n} \cup A_{n+1}[/imath]. Since, [imath]\bigcup_{i=1}^{n}[/imath] is finite by our induction hypothesis and since [imath]A_{n+1}[/imath] is finite, it follows that [imath]\bigcup_{i=1}^{n+1} A_{i}[/imath] is finite. Therefore, [imath]n+1 \in S[/imath]. By the principle of induction it follows that [imath]S= \mathbb{N}[/imath] please critique my proof and give any advice. Thank you!
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1862603
|
Proving [imath] \bigcup_{i=1}^n A_{i} \text{ is finite.} [/imath] by Induction.
Prove : If $A_{1},A_{2},...,A_{n} \text{ are finite sets, then } [imath]$ \bigcup_{i=1}^n A_{i} \text{ is finite.} [/imath] Proof: (I) Basis Step : $p(1)[imath] is true because it is true because it is finite. There is a first one and a last one.[/imath] (II) Induction Step: Suppose p(k)$ is ture. [imath] \bigcup_{i=1}^k A_{i} \text{ is finite.} [/imath] Show : $P(k+1)$ is true LHS. [imath] \bigcup_{i=1}^{k+1} A_{i} \text{ is finite.} [/imath] Here is where I get stuck because I do not know what the right hand side of this problem is. I also do not know if you are to multiply or add to the original left hand side. How does one prove this is true for all $n \in \Bbb N ?$
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1863891
|
To solve the radius of convergence of a power series in complex analysis
Let [imath]f(z)=z+\sum_{n=2}^\infty a_nz^n[/imath] have a positive radius of convergence. Does there exist a series [imath]g(z)=z+\sum_{n=2}^\infty b_nz^n[/imath], satisfying [imath] f(g(z))=z\text{?}\tag{49} [/imath] Does this series have a positive radius of convergence? Are the coefficients [imath]\left\{b_n\right\}[/imath] uniquely determined? I don't think completely expand all the terms of f (g(z)) is a good idea, can you give me some other methods to solve it? Thank you so much. Original image.
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465423
|
Power series [imath] f(g(z)) = z [/imath]
Let [imath] f(z) = z + \sum\limits_{n=2}^{\infty}a_nz^n [/imath] have a positive radius of convergence. Does there exist a series [imath] g(z) = z + \sum\limits_{n=2}^{\infty}b_nz^n [/imath], satisfying \begin{equation} f(g(z))=z? \end{equation} Does this series have a positive radius of convergence? Are the coefficients [imath] \{b_n\} [/imath] uniquely determined?
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1864163
|
Separable but not reduced?
Say a [imath]\Bbbk[/imath]-algebra is separable if [imath]L\otimes _\Bbbk A[/imath] is reduced for every field extension [imath]L/\Bbbk[/imath], and reduced if its underlying ring is reduced. Separable always implied reduced, and I found a result saying the converse is true over perfect fields. However, I would like an instructive counterexample of a reduced [imath]\Bbbk[/imath]-algebra which is not separable. How can such a thing happen?
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308015
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Tensor product of reduced [imath]k[/imath]-algebras must be reduced?
Let [imath]A[/imath], [imath]B[/imath] be two reduced [imath]k[/imath]-algebras. Then if an element of the form [imath]\sum a_{i}\otimes b_{j}[/imath] is nilpotent, we can compose it with any [imath]k[/imath]-homomorphism [imath]f[/imath] from [imath]A[/imath] to [imath]k[/imath] to get a homomorphism from [imath]A\otimes B[/imath] to [imath]B[/imath]. This map's image is [imath]\sum f(a_{i})\otimes b_{j}[/imath]must be nilpotent as well. So if we assume [imath]b_{j}[/imath] is linearly independent in the first place, we may reason that [imath]f(a_{i})=0[/imath]. This showed that [imath]a_{i}[/imath] are in the Jacobson radical of [imath]A[/imath]. However, we only know [imath]\mathrm{Nil}(A)=0[/imath], and it is not clear to me how to pass from the Jacobson radical to the nilradical without using some commutative algebra machinery. Here is what I think that might work. If [imath]A[/imath] is of finite type, then by Nullstellensatz we can assert the above is true. But if [imath]A[/imath] is not of finite type, then I can write [imath]A=k[x_{1},...,x_{n}][X][/imath], where [imath]n[/imath] is the transcendence degree of [imath]k(A)[/imath] and [imath]k[X][/imath] is finite over [imath]k[/imath]. Then I should be able to pass from [imath]k[X][/imath] to [imath]A[/imath] by inductive argument on the transcendental degree of [imath]k(A)[/imath], since prime ideals are maximal in a UFD (I am working with [imath]k[x][/imath] by introducing [imath][x][/imath]). The proof I have for the case of finite type is not satisfying. Given [imath]I\subset k[X][/imath], using Nullstellensatz one can show that [imath]I_{X}V_{X}[I]=\sqrt{I}[/imath] where [imath]V_{X}[I][/imath] is the set of all points in [imath]X[/imath] on which [imath]I[/imath] vanishes, and [imath]I_{X}(V_{X}[I])[/imath] be the ideal in [imath]k[X][/imath] which vanishes on [imath]V_{X}[I][/imath]. Granting this, let [imath]I=(a_{i})[/imath], then we want to show [imath]I=0[/imath], which is [imath]V_{X}[I]=X[/imath]. This follows since if [imath]U\subset W[/imath], then [imath]V_{X}[U]\supset V_{X}[W][/imath]. Since [imath]V_{X}[I]\supset V_{X}[m_{x}][/imath] for all maximal ideal corresponding to points, it must be [imath]X[/imath]. This proof is unnecessarily winding by using the ideal-variety correspondence. Ideally I should be able to construct a pure algebraic proof. The thing not clear to me is whether the above strategy really works. Namely I am not sure if I can really induct on the transcendental degree of [imath]k(A)[/imath]. To make matters simple assume we showed the statement holds for [imath]A[/imath] with transdental degree [imath]<n[/imath]. For [imath]A[/imath] with transdental degree [imath]n[/imath] we can thus write it as [imath]k[A'][x][/imath]. Then any maximal ideal [imath]m'[/imath] of [imath]k[A'][x][/imath] must correspond to a maximal ideal [imath]m=m'\cap k[A'][/imath] in [imath]k[A'][/imath]. But it seems not possible to show that any prime ideal is maximal. Indeed considering the simple example of [imath]k[x,y][/imath], where [imath](x)[/imath] is a prime ideal but not maximal since [imath]k[x,y]/(x)=k[y][/imath] is not a field. The problem is in the inducting process the base ring is not neccessarily a field, and so this strategy cannot work. I am thinking maybe I can show [imath]m'=m[f(x)][/imath] with [imath]f(x)[/imath] some irreducible polynomial in [imath]k[A'][x][/imath]. Because we know [imath]k[A'][x][/imath] is noetherian by Hilbert's basis theorem, so in particular [imath]m'[/imath] must have a finite generating set. And we can reduce the number of irreducible polynomial to 1 since [imath]m'[/imath] is a maximal ideal. But I still do not know how to use this to finish the proof.
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1864472
|
Infinitely pairs of positive integers such that [imath]1+2+\cdots+k = (k+1)+(k+2)+\cdots+N[/imath]
Prove that there are infinitely many pairs [imath](k,N)[/imath] of positive integers such that [imath]1+2+\cdots+k = (k+1)+(k+2)+\cdots+N[/imath]. I thought of transforming this into a Pell equation, but I didn't see a way of doing that. Should we show that if the equation is true for some [imath](k,N)[/imath] it is also true for another?
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1864263
|
Find general solution for the equation [imath]1 + 2 + \cdots + (n − 1) = (n + 1) + (n + 2) + \cdots + (n + r) [/imath]
A positive integer [imath]n[/imath] is called a balancing number if [imath]1 + 2 + \cdots + (n − 1) = (n + 1) + (n + 2) + \cdots + (n + r) \tag{1}[/imath] for some positive integer [imath]r[/imath]. Problem: Find the general solution (closed-form expression) of equation [imath](1)[/imath], in other words, find [imath]f_1(m)=n, f_2(m)=r[/imath] where [imath]f_1, f_2[/imath] are solution of [imath](1)[/imath] for any integer [imath]m \in \mathbb Z[/imath]. I have asked a similar question with a constrain involving Pell's equation.
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1864520
|
Is there a function [imath]y=f(x)[/imath] such that [imath]\frac{d^3y}{dx^3} = f(x)[/imath]?
The exponential function [imath]y=e^x[/imath] is its own derivative, the hyperbolic functions [imath]y=\frac{e^x+e^{-x}}{2}, y= \frac{e^x-e^{-x}}{2}[/imath] are equal to their own second derivatives, and the trigonometric functions [imath]y=sin(x)[/imath], [imath]y=cos(x)[/imath] are equal to their own fourth derivatives. Is there a function that is equal to its own third derivative?
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1080016
|
Function whose third derivative is itself.
I'm looking for a function [imath]f[/imath], whose third derivative is [imath]f[/imath] itself, while the first derivative isn't. Is there any such function? Which one(s)? If not, how can we prove that there is none? Notes: [imath]x\longmapsto c\cdot e^x, c \in R[/imath] are the functions whose derivative is itself. [imath]x\longmapsto \cosh(x)={e^x+e^{-x}\over 2}[/imath] and [imath]x\longmapsto \sinh(x)={e^x-e^{-x}\over 2}[/imath] have their second derivatives equal to themselves. [imath]x\longmapsto f(x)[/imath], has its third derivative equal to itself. [imath]x\longmapsto \cos(x)[/imath] and [imath]x\longmapsto \sin(x)[/imath] have their fourth derivatives equal to themselves.
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536232
|
Value of tan(pi/2)
I understand that this is a very stupid question but I'm not getting the answer. At [imath]x=\pi/2[/imath], what is the value of [imath]tan(x)[/imath]? Should it be [imath]-\infty[/imath] or [imath]+\infty[/imath]? Text tells it to be [imath]+\infty[/imath]. But why? Geometrically thinking, it comes out to be [imath]+\infty[/imath]. But how to explain the graph which has both the values at [imath]\pi/2[/imath]?
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189621
|
Is [imath]\tan(\pi/2)[/imath] undefined or infinity?
The way I have understood, [imath]0/0[/imath] is undefined or indeterminate because, if [imath]c=0/0[/imath] then [imath]c\cdot 0=0[/imath], where [imath]c[/imath] can be any finite number including [imath]0[/imath] itself. If we also observe a fraction [imath]F=a/b[/imath] where [imath]a,b[/imath] are positive real numbers, the value of [imath]F[/imath] increases with the decrements of [imath]b[/imath]. Being [imath]0[/imath] the least non-negative integer, if [imath]b[/imath] tends [imath]0[/imath] then [imath]F[/imath] tends to [imath]\infty[/imath] which is greater than all finite numbers. I have also heard that no number is equal to infinity, a variable can tend to infinity. Any ratio is also a variable or is resolved to a number. Now, [imath]\tan(\pi/2) = \frac{\sin(\pi/2)}{\cos(\pi/2)}=\frac{1}{0},[/imath] is it undefined or infinity? Also we know [imath]\tan2x= \frac{2\tan x}{1-\tan^2x}[/imath] Is this formula valid for [imath]x=\pi/2[/imath]? Any rectification is more than welcome. Some references: tan(PI/2) = ? Infinity is not a number Woah! I thought tan(pi/2) was undefined?
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