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1888470
|
Inverse of a function defined as an integral
I am trying to get the inverse of: [imath]g(x) = \int _a ^x h(t) dt[/imath] That is, I want to find a general expression for [imath]g^{-1}(x)[/imath] and was wondering whether there are theorems I can apply. Thanks
|
1467784
|
Inverse of a function's integral
The function [imath]g[/imath] is strictly positive. Let the function [imath]f[/imath] be defined as [imath]f(x) = \int_0^x g(u) du[/imath] Is there a way to express [imath]f^{-1}(x)[/imath] in terms of [imath]g[/imath]?
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1581242
|
One of Hermitian metric's properties?
We now define a Hermitian manifold is a complex manifold in which unmixed components of metric tensor vanish [imath]g_{ij}=g_{\bar{i}\bar{j}}=0[/imath]. Is this a propert of a Hermitian manifold? Or is it an extra imposed condition on the metric - this case Kaehler metric?
|
1573599
|
Why do those terms vanish if the metric is Hermitian?
On this page, the author says: We now define a Hermitian manifold as a complex manifold where there is a preferred class of coordinate systems in which unmixed components of metric tensor vanish ([imath]g_{\alpha\beta}=g_{\bar{\alpha}\bar{\beta}}=0[/imath]). My question is why do those vanish if the metric is Hermitian?
|
1888261
|
How do we prove that [imath]r_1^2+r_2^2+r_3^2+r_4^2=16R^2-a^2-b^2-c^2[/imath]
Prove that [imath]r_1^2+r_2^2+r_3^2+r^2=16R^2-a^2-b^2-c^2[/imath] Where r denotes the respective in-radius and R the circum radius a,b,c are the sides of the triangle I have tried using the [imath]\frac{\Delta}{(s-a)\dots}[/imath] But I get a fraction with a irreducible [imath]\frac{denominator}{numerator}[/imath].
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1352532
|
Proving a relation between inradius ,circumradius and exradii in a triangle
Prove that in a triangle [imath]r^2+r_1^2+r_2^2+r_3^2=16R^2-(a^2+b^2+c^2)[/imath] where the symbols have their usual meanings. I am looking for a smaller or elegant proof using trigonometry. A geometric proof would be super cool. I did prove it but it took me ages to do the calculation. Thanks.
|
1888473
|
Calculating Fibonacci numbers
In my book they calculated the [imath]28[/imath]th Fibonacci number and said [imath]F_{28} = 3 \times 13 \times 19 \times 281 = 317811[/imath]. This made me wonder if there was an easier way to find the [imath]28[/imath]th Fibonacci number than by doing it by hand.
|
8
|
How are we able to calculate specific numbers in the Fibonacci Sequence?
I was reading up on the Fibonacci Sequence, [imath]\text {{1,1,2,3,5,8,13,....}}[/imath] when I've noticed some were able to calculate specific numbers. So far I've only figured out creating an array and counting to the value, which is incredibly simple, but I reckon I can't find any formula for calculating a Fibonacci number based on it's position. Is there a way to do this? If so, how are we able to apply these formulas to arrays?
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409752
|
prove that the order of [imath]2[/imath] mod p is [imath]2^{n+1}[/imath]
I'm trying to prove this: Let [imath]p[/imath] be a prime factor of [imath]F_n=2^{2^n}+1[/imath], prove that order of [imath]2[/imath] mod p is [imath]2^{n+1}[/imath]. I know that [imath]2^{2^{n+1}} \equiv 1 \mod p[/imath]. But it only means that [imath]ord(2)_p[/imath] its a power of [imath]2[/imath] and divides [imath]2^{n+1}[/imath]. how I can prove that there is no other power of 2, less than [imath]2^{n+1}[/imath]?
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380774
|
Show that ord[imath]_{p}2 = 2^{n + 1}[/imath].
Let [imath]p[/imath] be a prime divisor of the Fermat number [imath]F_{n} = 2^{2^{n}} + 1[/imath]. Show that ord[imath]_{p}2 = 2^{n + 1}[/imath] The order of the element modulo some integer is the least positive integer such that it divides the "maximum" order. Some attempt I made: Assume that Fermat number is prime. By Fermat Theorem, we have: [imath]2^{2^{2^{n}}} \equiv 1 \bmod (2^{2^{n}} + 1)[/imath] I want to show that ord[imath]_{p}2 = 2^{n + 1}[/imath] i.e. [imath]2^{2^{n + 1}} \equiv 1 \bmod (2^{2^{n}} + 1)[/imath]. The question I have is: How do you know that [imath]2^{n + 1}[/imath] is a least positive integer such that we obtain 1 mod Fermat number? Can this be done by Modular Arithmetic, or is there a theorem that I need to use? Well, I found the orders of an element that divide the order [imath]F_{n} - 1[/imath], which are basically in the set [imath]\{2^{k} | k \in \mathbb{N}\}[/imath], but I don't know where to go from here. Any comments or suggestions?
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1889110
|
Prove the following statement regarding geometry by induction
Prove by induction that [imath]n[/imath] straight lines that lie in a plane and are as such that none are parallel and there doesn't exist a point where some three lines cross, divide the plane on [imath]1+\frac{n(n+1)}{2}[/imath] parts.
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1437160
|
Induction proof: n lines in a plane
Assume that there are [imath]n[/imath] infinitely long straight lines lying on a plane in such a way that no two lines are parallel, and no three lines intersect at a single point. Prove that these lines divide the plane into [imath]\frac{n^2+n+2}{2}[/imath] regions. (Hint: Any two non-parallel straight lines on a plane must intersect at exactly one point.) Can someone tell me where to start here? I know to prove the base where [imath]n=0[/imath], but don't know how to proceed. Any help is greatly appreciated!
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1888999
|
Square root of 2
I was watching some videos on algebra and found someone said that [imath] \frac{1}{\sqrt 2} [/imath] [imath]= \frac {\sqrt 2}{2} [/imath] Can someone explain this to me please?
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1082664
|
How [imath]\frac{1}{\sqrt{2}}[/imath] can be equal to [imath]\frac{\sqrt{2}}{2}[/imath]?
How [imath]\frac{1}{\sqrt{2}}[/imath] can be equal to [imath]\frac{\sqrt{2}}{2}[/imath]? I got answer [imath]\frac{1}{\sqrt{2}}[/imath], but the real answer is [imath]\frac{\sqrt{2}}{2}[/imath]. Anyway, calculator for both answers return same numbers.
|
1889171
|
Show that [imath]x^8 \equiv 1 \pmod{32}[/imath]
Prove that if [imath]x[/imath] is odd, then [imath]x^8 \equiv 1 \pmod{32}[/imath]. I tried using the expansion [imath](2k+1)^8 = 256 k^8+1024 k^7+1792 k^6+1792 k^5+1120 k^4+448 k^3+112 k^2+16 k+1,[/imath] but I don't see how that helps. Is there an easier way? How would we prove in general that [imath]x^{{2^k}} \equiv 1 \pmod{2^{k+2}}[/imath] if [imath]x[/imath] is odd?
|
313501
|
Prove that [imath]a^{2^n}=1 \mod 2^{n+2}[/imath]
I would like to prove that [imath]a^{2^n}\equiv 1 \pmod {2^{n+2}}[/imath] I tried induction but could not get it. Thank you very much!
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1887632
|
Why is [imath]\lim_{x\to 1}\sqrt{1-x}\sum_{n=0}^\infty x^{n^2}=\sqrt{\pi}/2\,\,[/imath]?
It is well-known that [imath]\sum_{n=0}^\infty x^n=1/(1-x)[/imath]. However, it seems difficult to find [imath]\sum_{n=0}^\infty x^{n^2}[/imath]. A natural problem is study its behavior around [imath]1[/imath]. A problem states that [imath]\lim_{x\to 1}\sqrt{1-x}\sum_{n=0}^\infty x^{n^2}=\sqrt{\pi}/2.[/imath] How can we prove this? Abel summation formula? Any others? I have no idea.
|
179249
|
Compute the limit of [imath]\sqrt{1-a}\sum\limits_{n=0}^{+\infty} a^{n^2}[/imath] when [imath]a\to1^-[/imath]
I need some suggestions, hints for the limit when [imath]a \to 1^{-}[/imath] of [imath]\sqrt{\,1 - a\,}\,\sum_{n = 0}^{\infty}a^{n^{2}}.[/imath]
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1889338
|
Probability question involving quadratic equation
Each coefficient in the equation [imath] ax^2 +bx+c = 0 [/imath] is determined by throwing an ordinary die. Find the probability that the equation will have no real root. The only thing I know is [imath]a,b,c\in\{1,2,3,4,5,6\}[/imath]
|
322939
|
Probability of Obtaining the Roots in a Quadratic Equation by Throwing a Die Three Times
Question : The coefficients a,b,c of the quadratic equation [imath]ax^2+bx+c=0[/imath] are determined by throwing a die three times and reading off the value shown on the uppermost face of each die. i.e, if you throw a 1, 5 and 3 respectively, the equation is [imath]1x^2+5x+3=0[/imath]. Find the probabilities that the quadratic equation you obtain : has real roots; has complex roots; has equal roots. Thank you for your attention.
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1188685
|
Correct notation for "slice" of Integers
Say I want to have the set of all integers between any two given integers [imath]a[/imath] and [imath]b[/imath]. For example, if I want the set of all integers between [imath]-3[/imath] and [imath]7[/imath] I would get: [imath] \{-3,-2,-1,0,1,2,3,4,5,6,7\} [/imath] How would I go about writing this for any given [imath]a,b[/imath]? Note: If I wanted the real numbers between [imath]a[/imath] and [imath]b[/imath], I could use [imath][a,b][/imath], therefore since the Integers are a subset of the Reals, I could write [imath][a,b]\cap\mathbb Z[/imath], but that just seems inelegant. Is there any other way?
|
1889436
|
Notation for discrete intervals
The well-known notation for continuous intervals is [imath][a,b][/imath]. But what's the case for discrete intervals? Actually they are sets of finite elements [imath]\left\{a, a+1, ..., b-1, b\right\}[/imath] or infinite elements [imath]\left\{0, 1, 2, ...\right\}[/imath]. Is there any special notation or common practice for discrete intervals?
|
1889758
|
Prove [imath]f_{1} + f_{3} + f_{5} +...+f_{2n-1} = f_{2n}[/imath]?
Prove [imath]f_{1} + f_{3} + f_{5} +...+f_{2n-1} = f_{2n}[/imath] when [imath]f_{n}[/imath] is the [imath]n[/imath]-th Fibonacci number. Before I try to prove this, I am trying to understand this in English. What is happening in this equation? How can [imath]2n-1 = 2n[/imath]? What is the English interpretation of this?
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1250262
|
Sum of odd Fibonacci Numbers
Trying to prove that the sum of odd-index consecutive Fibonacci numbers is the next even-index Fibonacci number. I have a gap in my proof that I cannot figure out. I know that induction would be easier and I have already done it that way, I am looking to use alpha and beta. [imath]\sum_{i=1}^n F_{2i-1} = F_{2n}[/imath] [imath]\begin{align*} \sum_{i=1}^n F_{2i-1}&=\sum_{i=1}^1 \frac{1}{\sqrt{5}}(\alpha^{2i-1}-\beta^{2i-1})\\ &=\frac{1}{\sqrt{5}}\left(\sum_{i=1}^1\alpha^{2i-1}-\sum_{i=1}^1\beta^{2i-1}\right)\\ &\;\;\vdots\\ &=\frac{1}{\sqrt{5}}[(\alpha^n-\beta^n)(\alpha^n+\beta^n)]\\ &=\frac{1}{\sqrt{5}}(\alpha^{2n}-\beta^{2n})\\ &=F_{2n} \end{align*}[/imath]
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1889993
|
Why study Gal([imath]\overline{\mathbb{Q}}|\mathbb{Q})[/imath]?
I was told by one of my teachers that Galois Representation is a major tool to study Gal([imath]\overline{\mathbb{Q}}|\mathbb{Q}[/imath]). If we know this Galois group, what knid of questions can we answer ? In short, what motivated people to study this group ?
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805088
|
Why there is much interest in the study of [imath]\operatorname{Gal}\left(\overline{\mathbb Q}/\mathbb Q\right)[/imath]?
Let's start for a simple quote from wikipedia: "No direct description is known for the absolute Galois group of the rational numbers. In this case, it follows from Belyi's theorem that the absolute Galois group has a faithful action on the dessins d'enfants of Grothendieck (maps on surfaces), enabling us to "see" the Galois theory of algebraic number fields." What does wikipedia mean exactly by "a direct description" of [imath]\operatorname{Gal}\left(\overline{\mathbb Q}/\mathbb Q\right)[/imath]? It seems that the absolute group of rationals is very important in mathematics, in fact there are several tools from algebraic geometry whereby we try to study it (I'm measuring the importance of an object simply estimating the amount of forces invested on investigations about its nature). But why to give this importance to this particular object? For example why [imath]\operatorname{Gal}\left({\overline K}/K\right)[/imath], where [imath]K[/imath] is a generic field, is not "beautiful" as [imath]\operatorname{Gal}\left(\overline{\mathbb Q}/\mathbb Q\right)[/imath]? Thanks in advance.
|
449623
|
Every infinite subset of a countable set is countable.
Here is the proof I tried to weave while trying to prove this theorem: Theorem. Every infinite subset of a countable set is countable. Proof. Let [imath]A[/imath] be a countable set and [imath]E\subset A[/imath] be infinite. Then [imath]A\thicksim\mathbb{N}[/imath]. This implies that there is a sequence [imath]\left\{x_{n}\right\}_{n\in\mathbb{N}}[/imath] where [imath]x_{n}\in A[/imath]. Construct a sequence [imath]\left\{n_{k}\right\}_{k\in\mathbb{N}}[/imath] where [imath]x_{n_{k}}\in A[/imath], [imath]n_{1}[/imath] is the smallest positive integer such that [imath]x_{n_{1}}\in A[/imath], and [imath]n_{k+1}>n_{k}[/imath]. Then [imath] E=\bigcup_{k=1}^{\infty}x_{n_{k}}, [/imath] and [imath]E[/imath] is countable, because [imath]E\thicksim\mathbb{N}[/imath]. [imath]\blacksquare[/imath] Is it convincing?
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2153510
|
Showing that [imath]\exists f: \mathbb {N} \rightarrow A[/imath] that is injection
Suppose [imath]A[/imath] is infinite. Also suppose [imath]\exists g:A \rightarrow \mathbb {N}[/imath] that is 1-1. This is all the info I have. I need to show the result in the title because then I would have some function [imath]h[/imath] that is a bijection from [imath]A [/imath] to [imath]\mathbb{N} [/imath] and that would mean that [imath]A [/imath] is countable. But I just do not see how to obtain that function [imath]f [/imath] in title...
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1878068
|
[imath]f(x)[/imath] if [imath]f(xy)=f(x) +f(y) +\frac{x+y-1}{xy}[/imath]
Let [imath]f[/imath] be a differentiable function satisfying the functional time [imath]f(xy)=f(x) +f(y) +\frac{x+y-1}{xy} \forall x,y \gt 0 [/imath] and [imath]f'(1)=2[/imath] My work Putting [imath]y=1[/imath] [imath]f(1)=-1[/imath] [imath]f'(x)=\lim_{h\to 0}\frac{f(x+h) - f(x)}{h}[/imath] But I don't know anything about [imath]f(x+h)[/imath] so what to do in this problem ?
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2847864
|
The value of [imath][f(e^{100})][/imath]
Let [imath]f[/imath] be a differentiable function satisfying the functional rule [imath]f(xy) = f(x) +f(y) + \frac{x+y-1}{xy}[/imath] such that [imath]x,y>0[/imath] and [imath]f'(1) = 2[/imath]. Then what is the value of [imath][f(e^{100})][/imath]. Where [imath]\lfloor k \rfloor[/imath] denotes the greatest integer less than or equal to [imath]k[/imath]. My try :
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1889701
|
Proving [imath]S^3[/imath] and [imath]\mathbb{R}^3[/imath] are not homeomorphic
Prove [imath]S^3[/imath] and [imath]\mathbb{R}^3[/imath] are not homeomorphic. I've encountered this question on a PhD exam in topology. This is at a level where we are expected to understand cohomology already, so there are already a lot of obvious one line proofs I could give (e.g. they don't have the same homology groups). But this appears among the "give a detailed answer" questions, as opposed to the more computational questions in the latter half. (The heading says to show all work and support all statements to the best of my ability.) So, I am confused about the level of detail I would have to include here. Is there an obvious choice for how to prove this directly without relying on any one liners that assume higher level stuff? I realize this question is a little opinion based, but maybe the answer will be unambiguous to those with more experience in topology. How would you answer this question?
|
382211
|
Compactness of a Sphere
This fact is mentioned liberally in literature along with subsequent mention of Heine-Borel and I am trying to get my head around it. What would be a formal proof of this if we take for example, a unit sphere in [imath]\mathbb R^3 [/imath]? In addition, how could it be shown, non-trivially of course, that this sphere is not homeomorphic with [imath]\mathbb R^2 [/imath]? As I understand this involves finding a bijective function with continuous inverse?
|
1890544
|
Finding a Closed Form
Find a closed form for [imath]S_n = 1 \cdot 1! + 2 \cdot 2! + \ldots + n \cdot n!.[/imath] for integer [imath]n \geq 1.[/imath] Your response should have a factorial. Another induction problem, I tried some examples but they didn't really tell me anything.
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976943
|
Closed Form for Factorial Sum
I came across this question in some extracurricular problem sets my professor gave me: what is the closed form notation for the following sum: [imath]S_n = 1\cdot1!+2\cdot2!+ ...+n \cdot n![/imath] I tried computing some terms, and the only "vague" thing I noticed was that maybe I should be subtracting a term, but I'm really not sure. I went around looking on StackExchange's archives for a closed form of [imath]S_n = 1!+2!+ ...+ n![/imath] but that didn't help me with my problem much. Any pointers?
|
1890512
|
Algebraic Proof Using Induction
Let [imath]n[/imath] be a positive integer, and let [imath]x \ge -1[/imath]. Prove, using induction, that [imath](1 + x)^n \ge 1 + nx.[/imath] I don't know what to do, I can't expand the left side. I'm not familiar with induction, so can someone please provide an answer? Thanks!
|
181702
|
Proof by induction of Bernoulli's inequality [imath] (1+x)^n \ge 1+nx[/imath]
I am working on getting the hang of proofs by induction, and I was hoping the community could give me feedback on how to format a proof of this nature: Let [imath]x > -1[/imath] and [imath]n[/imath] be a positive integer. Prove Bernoulli's inequality: [imath] (1+x)^n \ge 1+nx[/imath] Proof: Base Case: For [imath]n=1[/imath], [imath]1+x = 1+x[/imath] so the inequality holds. Induction Assumption: Assume that for some integer [imath]k\ge1[/imath], [imath](1+x)^k \ge 1+kx[/imath]. Inductive Step: We must show that [imath](1+x)^{k+1} \ge 1+(k+1)x[/imath] Proof of Inductive Step: [imath]\begin{align*} (1+x)^k &\ge 1+kx \\ (1+x)(1+x)^k &\ge (1+x)(1+kx)\\ (1+x)^{k+1} &\ge 1 + (k+1)x + kx^2 \\ 1 + (k+1)x + kx^2 &> 1+(k+1)x \quad (kx^2 >0) \\ \Rightarrow (1+x)^{k+1} &\ge 1 + (k+1)x \qquad \qquad \qquad \square \end{align*}[/imath]
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1890678
|
Tightest convex set containing curves of length 1
What is the smallest (either in terms of perimeter or area) convex shape in [imath]R^2[/imath] such that all curves of length 1 are properly contained within it ? I'm thinking a right triangle with sides of [imath]1/\sqrt(2)[/imath], but I'm not sure how to reason about this rigorously. Any pointers would be appreciated.
|
93808
|
What is the smallest convex set includes all smooth unit curves?
I try to understand: is there a smallest in area convex set that every smooth curve with length 1 can be placed inside it by translation and rotation? I only have a upper bound [imath]S \leq \frac14+\frac{\pi}{16}[/imath] because of convex hull of two circles radius [imath]\frac14[/imath] and simple lower bound [imath]S\geq\frac1{4\pi}[/imath]. Does this set exist and what is its length?
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1890539
|
Why isn't [imath]\mathbb{Z}[\sqrt{5}][/imath] integrally closed?
The book I'm reading casually mention that [imath]\mathbb{Z}[\sqrt{5}][/imath] isn't integrally closed without explaining why. Could someone show me why this is the case?
|
100639
|
What are the integers [imath]n[/imath] such that [imath]\mathbb{Z}[\sqrt{n}][/imath] is integrally closed?
I was recently reading about integral ring extensions. One of the first examples given is that [imath]\mathbb{Z}[/imath] is integrally closed in its quotient field [imath]\mathbb{Q}[/imath]. Another is that [imath]\mathbb{Z}[\sqrt{5}][/imath] is not integrally closed in [imath]\mathbb{Q}(\sqrt{5})[/imath] since for example [imath](1+\sqrt{5})/2\in\mathbb{Q}[\sqrt{5}][/imath] is integral over [imath]\mathbb{Z}[/imath] as a root of [imath]X^2-X-1[/imath], but [imath](1+\sqrt{5})/2\notin\mathbb{Z}[\sqrt{5}][/imath]. Now I'm curious, can we find what are all integers [imath]n[/imath] such that [imath]\mathbb{Z}[\sqrt{n}][/imath] is integrally closed (equal to its integral closure in its quotient field)? One thing I do know is that unique factorization domains are integrally closed, so I think rings like [imath]\mathbb{Z}[\sqrt{-1}][/imath], [imath]\mathbb{Z}[\sqrt{-2}][/imath], [imath]\mathbb{Z}[\sqrt{2}][/imath] and [imath]\mathbb{Z}[\sqrt{3}][/imath] are integrally closed, as they are Euclidean domains, and thus are UFDs. But can we say what all integers [imath]n[/imath] are such that [imath]\mathbb{Z}[\sqrt{n}][/imath] is integrally closed? Thanks!
|
1890875
|
How to calculate [imath]\int_{0}^1 \sqrt{\frac{1-x^2}{1+x^2}}dx[/imath]?
I encounter the following question: [imath]\int_{0}^1 \sqrt{\frac{1-x^2}{1+x^2}}dx=?[/imath] It looks simple. But I doubt there exists analytic solution. Could anyone help? Thank you! Icarus 369 has pointed out Page [ Evaluating [imath]\int\sqrt{\frac{1-x^2}{1+x^2}}\mathrm dx[/imath] ]. The question indeed has no elementary solution. Thanks Icarus 369. The elliptic integration turns out to be very interesting. FYI: http://web.mst.edu/~lmhall/SPFNS/sfch3.pdf
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1410439
|
Evaluating [imath]\int\sqrt{\frac{1-x^2}{1+x^2}}\mathrm dx[/imath]
Evaluating [imath]\int\sqrt{\frac{1-x^2}{1+x^2}}\mathrm dx[/imath] I had read the similar problem, but it doesn't work.
|
1608420
|
Proof for conjugate cycles
Let [imath]\alpha =(a_1,a_2,...,a_s)[/imath] be a cycle and [imath]\pi[/imath] a permutation in [imath]S_n[/imath]. Then [imath]\pi\alpha\pi^{-1}[/imath] is the cycle [imath](\pi(a_1),...,\pi(a_s))[/imath]. I'm having trouble understanding the proof of this: Take [imath]\pi(a_i)[/imath]. Then [imath]\pi\alpha\pi^{-1}\pi(a_i)=\pi\alpha(a_i)=\pi(a_{i+1\; (\mathrm{mod}\;\mathrm{s})})[/imath], then follows on from here. But I'm not understanding how this proves the statement, since it feels like it's proving that [imath]\pi\alpha=(\pi(a_1),...,\pi(a_s))[/imath].
|
712501
|
Conjugate cycles
Prove the following in [imath]S_n[/imath]: Let [imath]\alpha = (a_1,a_2,\ldots,a_s)[/imath] be a cycle and let [imath]\pi[/imath] be a permutation in [imath]S_n[/imath]. Then [imath]\pi\alpha\pi^{-1}[/imath] is the cycle [imath](\pi(a_1),\ldots,\pi(a_s)).[/imath] I'm not sure where to even start.
|
1890811
|
a coding theory question about inequality
Let [imath]F[/imath] be a finite collection of binary strings of finite lengths, and assume that no two distinct concatenations of two finite sequences of codewords result in the same binary sequence. Let [imath]N_i[/imath] denote the number of strings of length [imath]i[/imath] in [imath]F[/imath]. Prove that[imath]\sum_{i}{\frac{N_i}{2^i}}\le{1}[/imath]
|
383857
|
Probabilistic method Kraft-McMillan inequality
I am trying to solve as many problems as I can but I am a little confused on this one. It is in chapter 1, problem 9 (Probabilistic method, Alon & Spencer). The problem says Let [imath]F[/imath] be a finite collection of binary string of finite lengths and assume that no two distinct concatenations of two finite sequences of codewords result in the same binary sequence. Let [imath]N_i[/imath] denote the number of string of length [imath]i[/imath] in [imath]F[/imath]. Prove [imath]\sum_i \frac{N_i}{2^i}\leq 1[/imath] The problem that I have is that I understand the problem as if [imath]w_1w_2=w_3w_4[/imath] then we must have [imath]w_1=w_3[/imath] and [imath]w_2=w_4[/imath], but if this is what they mean then let [imath]F=\{1,0,100\}[/imath] then the possible contatinations are: [imath]11,00,10,01,1001,1100,0100,1000,100100[/imath], so no two are the same, but then we would have [imath]N_1=2,N_3=1[/imath], so [imath]N_1/2+N_3/2^3=2/2+1/8>1[/imath]. I assume I must be understanding the condition incorrectly.
|
1890958
|
Why is the formula for the sum of infinite terms of a geometric progression only defined for [imath]r < 1[/imath]?
The formula for the the sum of infinite terms of a geometric progression is [imath]S_{\infty} = \dfrac{a}{1 - r}[/imath], where [imath]r < 1[/imath]. Why is this only defined for [imath]r < 1[/imath]? My teacher doesn't know why.
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542120
|
When is the formula for the infinite geometric series valid
When is the formula [imath]S_{\infty} = \dfrac{a}{1-r}[/imath] valid? Does |[imath]r| <1[/imath]?
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1891226
|
Function expressed the sum of an odd and even function
Show that a function defined on a domain [imath]D[/imath] symmetric about the origin can be expressed in a unique way as the sum of an even and an odd function. Sorry but I have no idea about how to start.
|
121775
|
Deriving even odd function expressions
What is the logic/thinking process behind deriving an expression for even and odd functions in terms of [imath]f(x)[/imath] and [imath]f(-x)[/imath]? I've been pondering about it for a few hours now, and I'm still not sure how one proceeds from the properties of even and odd functions to derive: [imath]\begin{align*} E(x) &= \frac{f(x) + f(-x)}{2}\\ O(x) &= \frac{f(x) - f(-x)}{2} \end{align*}[/imath] What is the logic and thought process from using the respective even and odd properties, [imath]\begin{align*} f(-x) &= f(x)\\ f(-x) &= -f(x) \end{align*}[/imath] to derive [imath]E(x)[/imath] and [imath]O(x)[/imath]? The best I get to is: For even: [imath]f(x)-f(-x)=0[/imath] and for odd: [imath]f(x)+f(-x)=0[/imath] Given the definition of [imath]E(x)[/imath] and [imath]O(x)[/imath], it makes a lot of sense (hindsight usually is) but starting from just the properties. Wow, I feel I'm missing something crucial.
|
1161248
|
What is the correct integral of [imath]\frac{1}{x}[/imath]?
I understand that the graphs of [imath]\log(x)[/imath] and [imath]\ln(x)[/imath] both have derivatives (changes in slope) that follow the pattern of: [imath]\frac{d}{dx}\log_{b}x= \frac{1}{(x\ln(b))}[/imath] However, depending on the source, I have seen different definitions for the integral of [imath]\frac{1}{x}[/imath]: [imath]\int\frac{1}{x}dx=\ln| x |+C[/imath] [imath]\int\frac{1}{x}dx=\log| x |+C[/imath] [imath]\int\frac{1}{x}dx=\ln x+C[/imath] [imath]\int\frac{1}{x}dx=\log x +C[/imath] I believe that only the top two definitions are close to being valid, and I also think that [imath]\ln| x |+C[/imath] is the only correct answer, based on the formula for the derivative given above, and the fact that [imath]ln(e) = 1[/imath]. Is that incorrect? Can the answer to this be shown graphically as well as algebraically?
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234624
|
Is the integral of [imath]\frac{1}{x}[/imath] equal to [imath]\ln(x)[/imath] or [imath]\ln(|x|)[/imath]?
The inconsistency I see between mathematical subjects is really confusing me. I understand that it isn't possible for [imath]e^x[/imath] to be less than zero for real [imath]x[/imath], which is probably why they say that the integral is [imath]\ln(|x|)[/imath]. Before I ramble on too much, I just want to ask: Is there a set of guidelines to follow to help me choose whether to let the integral of [imath]\frac{1}{x}[/imath] equal to [imath]\ln(x)[/imath] or [imath]\ln(|x|)[/imath]? Thanks, Aralox
|
1891781
|
To find all functions [imath]h:\mathbb{R}\to\mathbb{R}[/imath] such that [imath]h(x+y) = h(x)+h(y)[/imath] and [imath]h(xy)=h(x)h(y)[/imath]
Problem: Find all functions [imath]h:\mathbb{R}\to\mathbb{R}[/imath] such that [imath]h(x+y) = h(x)+h(y)[/imath][imath]h(xy)=h(x)h(y)[/imath] for all [imath]x,y\in\mathbb{R}[/imath]. My attempt: From [imath]h(x+y) = h(x) +h(y)[/imath], we can derive that [imath]h(x) = k x[/imath] for all rational [imath]x[/imath] and some rational [imath]k[/imath]. Since [imath]h(xy)=h(x)h(y)[/imath], we get [imath]h(x)=x[/imath] for all rational [imath]x[/imath]'s. In fact, it is true for all real algebraic numbers, if I argue in a similar manner. However, I cannot extend this to other irrational numbers (Although I have a feeling that I may need to use completeness theorem for [imath]\mathbb{R}[/imath]). Any hints as on how to extend this to all irrational numbers?
|
684078
|
Functional equations [imath]f(x+y)= f(x) + f(y)[/imath] and [imath]f(xy)= f(x)f(y)[/imath]
Let [imath]f:\mathbb{R}\to \mathbb{R}[/imath] is a function such that for all real [imath]x[/imath] and [imath]y[/imath], [imath]f(x+y)= f(x) + f(y)[/imath] and [imath]f(xy)= f(x)f(y)[/imath], then prove that [imath]f[/imath] must be one of the two following functions: [imath]f:\mathbb{R}\to \mathbb{R}[/imath] defined by [imath]f(x)=0[/imath] for all real [imath]x[/imath] OR [imath]f:\mathbb{R}\to\mathbb{R}[/imath] defined by [imath]f(x)=x[/imath] for all real [imath]x[/imath] I got to the point where putting the two equations together, you get [imath]f(x+y)f(x)= f(xy) + f(x)^2[/imath] and plugging in [imath]f(x)=x[/imath] checks with it. So am I going in the right direction or am I just doing some guess work? Is there a more elegant way of doing it? Thanks
|
353028
|
Is there an idempotent element in a finite semigroup?
Let [imath](G,\cdot)[/imath] be a finite semigroup. Is there any [imath]a\in G[/imath] such that: [imath]a^2=a[/imath] It seems to be true in view of theorem 2.2.1 page 97 of this book (I'm not sure). But is there an elementary proof? Theorem 2.2.1. [R. Ellis] Let [imath]S[/imath] be a compact right topological semigroup. Then there exists an idempotent in it. This theorem is also known as Ellis–Numakura lemma.
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1974341
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Existence of an idempotent element in a finite monoid.
I'm trying to show that if [imath]E[/imath] is a finite monoid, then there exists [imath]s[/imath] such that [imath]s²=s[/imath], i.e, there exists an idempotent element in [imath]E[/imath]. What I did is this: Take an element [imath]a[/imath] of [imath]E[/imath], and consider the application [imath]F_a :E\to E[/imath] such that for [imath]x[/imath] in [imath]E[/imath], [imath]F_a(x)=ax[/imath]. Now, if [imath]E[/imath] was a group, it's easy to see that [imath]F_a[/imath] is a permutation of [imath]E[/imath] (consider [imath]F_{sym}(a)[/imath]) since [imath]E[/imath] is finite. Then there exists at least one element [imath]x[/imath] such that [imath]F_a(x)=x[/imath]. Take [imath]x=a[/imath] we'll have [imath]a²=a[/imath]. But what about the initial case, where [imath]E[/imath] is just a finite monoid? Thanks.
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1892580
|
Two reflections generating the Dihedral group [imath]D_n[/imath]
Let [imath]l_1[/imath] and [imath]l_2[/imath] be two lines intersecting at an angle of [imath]\pi/n[/imath] on [imath]\mathbb{R}^2[/imath]. Let [imath]r_1[/imath] and [imath]r_2[/imath] denote the reflection by [imath]l_1[/imath] and [imath]l_2[/imath] respectively. I am to show that [imath]r_1[/imath] and [imath]r_2[/imath] generate a Dihedral group [imath]D_n[/imath]. First I claim that [imath]D_n[/imath] can be generated by a reflection [imath]r[/imath] about a line and rotation by angle [imath]2\pi/n[/imath], [imath]\rho_{2\pi/n}[/imath], because reflection by any other line intersecting at an angle of [imath]2\pi m/n[/imath] can be obtained by first applying [imath]\rho_{2\pi/n}[/imath] [imath]m[/imath] times, followed by [imath]r[/imath], and then followed by [imath]\rho_{2\pi/n}^{-1}[/imath] [imath]m[/imath] times. Then I deduce from drawing and basic geometry that [imath]r_1 r_2 = \rho_{2\pi/n}[/imath] Therefore [imath]r_1[/imath] and [imath]r_2[/imath] generates a Dihedral group [imath]D_n[/imath]. I am not sure if my answer is correct or rigorous enough.
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589439
|
Prove two reflections of lines through the origin generate a dihedral group.
Let [imath]l_1[/imath] and [imath]l_2[/imath] be the lines through the origin in [imath]R^2[/imath] that intersect in an angle π/n and let [imath]r_i[/imath] be the reflection about [imath]l_i[/imath]. Prove the [imath]r_1[/imath] and [imath]r_2[/imath] generate a dihedral group [imath]D_n[/imath]. Attempt: So [imath]D_n[/imath] is the dihedral group of order 2n generated by [imath]ρ_θ[/imath], where θ = 2π/n, and a reflection r' about a line l through the origin. The product [imath]r_1r_2[/imath] of these reflections preserves orientation and is a rotation about the origin. Its angle of rotation is ±2θ. Could I just say that by the definition of the dihedral group, we need to find a reflection and a rotation which fit the descriptions?
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564417
|
Joint cdf and pdf of the max and min of independent exponential RVs
Let [imath]X[/imath] and [imath]Y[/imath] be independent random variables. Each has an exponential distribution with parameter [imath]\lambda[/imath]. Define two new random variables by [imath]W = \min({X,Y}) [/imath] [imath]Z = \max({X,Y})[/imath] Find the joint cdf of [imath]W[/imath] and [imath]Z[/imath], and use it to find their joint pdf. I know I need to start out by splitting up the event [imath]\{W \le w, Z \le z\}[/imath] as the union of [imath]\{W \le w, Z \le z, X \le Y \}[/imath] and [imath]\{W \le w, Z \le z, X > Y \} [/imath]
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565624
|
The joint density of the max and min of two independent exponentials
Let [imath]X=\min(S,T)[/imath] and [imath]Y=\max(S,T)[/imath] for independent exponential variables [imath]S[/imath] and [imath]T[/imath]. Find the joint density of [imath]X[/imath] and [imath]Y[/imath]. Are [imath]X[/imath] and [imath]Y[/imath] independent? How would you suggest I approach this?
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1892862
|
Is there a formula to calculate the following sum?
I tried googling, but I may not be searching with correct terms. [imath]\sum_{k=1}^{n} k^k = 1^1 + 2^2 +3^3 + 4^4 + \cdots + n^n[/imath] So, is there a formula to calculate this?
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45066
|
Closed form formula for [imath]\sum\limits_{k=1}^n k^k[/imath]
Is there a way of finding a formula for [imath]\sum\limits_{k=1}^n k^k[/imath]? Maybe I'm missing something really obvious, but I've looked around a bit on the Internet and I haven't been able to find anything. So, what I'm looking for is a formula in closed form to generate the sequence [imath]1,5,32,288,3413,\dots[/imath]
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1892090
|
Solving an Ito Diffusion and finding the Expectation/Variance
I have an example in which [imath]Y_t[/imath] is the Ito diffusion with generator [imath]A(f )(x) = αxf '(x) + 2x^2f ''(x).[/imath] Assume [imath]Y_0 = y ∈ R^+.[/imath] I would like to find [imath]Y_t[/imath] and from that would like to find [imath]E [Y_t ][/imath] and [imath]Var [Y_t ][/imath]. Can someone please walk me through this example?
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1890461
|
Finding [imath]X_t[/imath] of an Itô Diffusion
Someone please help me with this: I have that [imath]X_t[/imath] is the Ito diffusion with genertator [imath]A(f)(x)=\alpha xf'(x)+f''(x).[/imath] Then, if [imath]X_0=x \in \mathbb{R}^+[/imath], how do I find [imath]X_t[/imath]?
|
1893156
|
[imath]a_1=1[/imath], [imath]a_{n+1}=\frac{1}{2}\left(a_n+\frac{2}{a_n}\right)[/imath]. Show that the sequence is decreasing
[imath]a_1=1[/imath], [imath]a_{n+1}=\frac{1}{2}\left(a_n+\frac{2}{a_n}\right)[/imath]. Show that the sequence is decreasing. [imath]a_{n+1}[/imath] is the Arithmetic mean of [imath]a_n[/imath] and [imath]\frac{2}{a_n}[/imath]. I can say that if [imath]a_n > \frac{2}{a_n}[/imath], then [imath]a_{n+1}<a_n[/imath]. But thats not always true. What should be my approach?
|
768354
|
[imath]a_1=3[/imath] and [imath]a_{n+1}=\frac{a_n}{2} + \frac{1}{a_n}[/imath]. Show that it monotonically decreases and find the limit.
What I've done so far: I have proved that this sequence is bounded below by 0, which is a very rough estimate. I know that the infimum is [imath]\sqrt2[/imath]. Anyway, the question first asks me to prove that the sequence decreases monotonically. And I've tried the following: Suppose [imath]a_{n+1} \le a_n[/imath], then we would get [imath]\frac{1}{a_n} \le \frac{a_n}{2}[/imath]. After some algebra, I ended up with [imath]\frac{1}{a_n^2 +2}\le\frac{a_n^2 +2}{2}[/imath]. Now, [imath]a_{n+2} - a_{n+1}=\frac{2a_n}{a_n^2 +2}-\frac{a_n^2 +2}{4a_n}[/imath]. I was looking for some similarities between these two expressions but it doesn't seem to work out. A more general question is, if the sequence is defined recursively, what are some common strategies to find the limit? I just started studying analysis and this kind of questions kind of troubles me. Thank you guys.
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1893271
|
A Counting Proof
For fixed positive integers [imath]m[/imath] and [imath]n[/imath], let [imath]S=\sum_{k=0}^m (-1)^k\binom{m}{k}(m-k)^n.[/imath]Show that [imath]S = 0[/imath] if [imath]n < m[/imath] and [imath]S = n![/imath] if [imath]n = m[/imath]. I have asked this question before: Proof Involving Sigmas, but have not received any answer that I could relate to. I know that I have to use PIE(principle of inclusion and exclusion) and a counting argument. Here's my work so far: There are [imath]m[/imath] people, [imath]\binom{m}{k}[/imath] represents all the ways to pick [imath]k[/imath] people out of [imath]m[/imath] people. [imath](m-k)^n[/imath] represents all the ways of giving the people that weren't chosen [imath]n[/imath] distinguishable objects. What I don't get is how to apply PIE, why the signs are alternating, and how does [imath]S = 0[/imath] if [imath]n < m[/imath] and [imath]S = n![/imath] if [imath]n = m[/imath]. Could anyone offer an answer by the counting argument I started? Thanks!
|
1886100
|
Proof Involving Sigmas
For fixed positive integers [imath]m[/imath] and [imath]n[/imath], let [imath]S=\sum_{k=0}^m (-1)^k\binom{m}{k}(m-k)^n.[/imath]Show that [imath]S = 0[/imath] if [imath]n < m[/imath] and [imath]S = n![/imath] if [imath]n = m[/imath]. I know that I have to use PIE(principle of inclusion and exclusion) and a counting argument. Here's my work so far: There are [imath]m[/imath] people, [imath]\binom{m}{k}[/imath] represents all the ways to pick [imath]k[/imath] people out of [imath]m[/imath] people. [imath](m-k)^n[/imath] represents all the ways of giving the people that weren't chosen nn distinguishable objects. What I don't get is how to apply PIE, why the signs are alternating, and how does [imath]S = 0[/imath] if [imath]n < m[/imath] and [imath]S = n![/imath] if [imath]n = m[/imath]. Could anyone offer an answer by the counting argument I started? Thanks!
|
1893421
|
Does [imath]G\!-\!v_1 \simeq G\!-\!v_2[/imath] mean an automorphism of [imath]G[/imath] maps [imath]v_1[/imath] to [imath]v_2[/imath]?
For a simple undirected graph [imath]G[/imath], suppose we have two vertices [imath]v_1[/imath] and [imath]v_2[/imath] such that [imath]G-v_1 \simeq G-v_2[/imath]. Does this necessarily mean that there is an automorphism of [imath]G[/imath] that maps [imath]v_1[/imath] to [imath]v_2[/imath]? This is just a condition that I've assumed to be true for awhile now without thinking too hard about whether or not it's actually true. Using the condition that [imath]G-v_1 \simeq G-v_2[/imath] has been a useful way to characterize two vertices as being "the same," but I was wondering if this actually corresponds to automorphisms of the graph.
|
310893
|
Find a graph on eight vertices with a pair of pseudosimilar vertices
Can somebody give me an example of a graph on eight vertices that has a pair of pseudosimilar vertices? Or, is it not possible? If it is not possible, please do not give me a proof, as I would like to figure out how to prove it myself. Definition: vertices [imath]u[/imath] and [imath]v[/imath] in a graph [imath]X[/imath] are similar if there is an automorphism of [imath]X[/imath] that maps [imath]u[/imath] to [imath]v[/imath]. If [imath]u[/imath] and [imath]v[/imath] are similar, then the vertex-deleted subgraphs [imath]X-u[/imath] and [imath]X-v[/imath] are isomorphic. If [imath]X-u[/imath] and [imath]X-v[/imath] are isomorphic but [imath]u[/imath] and [imath]v[/imath] are not similar, we say that they are pseudosimilar.
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602781
|
Is it a Fermat polynomial?
A Fermat polynomial is a polynomial which can be written as the sum of squares of two polynomials with integer coefficients. Let [imath]f(x)[/imath] be a Fermat polynomial such that [imath]f(0)=1000[/imath]. Prove that [imath]f(x)+2x[/imath] is not a Fermat polynomial.
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595435
|
A polynomial is called a Fermat's polynomial...
A polynomial is called a Fermat polynomial if it can be written as the sum of the squares of two polynomials with integer coefficients. Suppose that [imath]f(x)[/imath] is a Fermat polynomial such that [imath]f(0) = 1000[/imath]. Prove that [imath]f(x) + 2x[/imath] is not a Fermat polynomial. What I have done: Let [imath]p(x)[/imath] be a Fermat polynomial such that [imath]p(0)[/imath] is divisible by [imath]4[/imath]. Suppose that [imath]p(x) = g(x)^2+h(x)^2[/imath] where [imath]g(x)[/imath] and [imath]h(x)[/imath] are polynomials with integer coefficients. Therefore [imath]g(0)^2 + h(0)^2[/imath] is divisible by [imath]4[/imath]. Since [imath]g(0)[/imath] and [imath]h(0)[/imath] are integers, their squares are either [imath]1 (mod 4)[/imath] or [imath]0 (mod 4)[/imath]. It therefore follows that [imath]g(0)[/imath] and [imath]h(0)[/imath] are even. From here, I can not proceed any further. Please help
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939360
|
[imath]\mathbb{R}[/imath] with the finite complement topology
Let [imath]X=\mathbb{R}[/imath] be given with the collection [imath]\tau[/imath] where [imath] \tau = \{U\subset X: |X\setminus U|<\aleph_0\}\cup \{\emptyset\} [/imath] I was sitting at my computer, when I suddenly asked myself: "How is [imath]\tau[/imath] actually a topology on [imath]\mathbb{R}[/imath]?". It is supposed to satisfy the arbitrary union condition, but it doesn't. Indeed [imath] \bigcup_{i\in \mathbb{N}} \mathbb{R}\setminus \{i\} \notin \tau [/imath] but each of [imath]\mathbb{R}\setminus \{i\} \in \tau[/imath]. Can someone explain to me how [imath]\tau[/imath] is a topology? edit as Thomas Andrews points out, I was confusing unions with intersections and, in general, just being an idiot.
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1833828
|
Show finite complement topology is, in fact, a topology
My attempt to prove the following is below: Let X be an infinite set. Show that [imath]\mathscr{T}_1=\{U \subseteq X : U = \emptyset [/imath] or [imath] X\setminus U [/imath] is finite [imath] \}[/imath] My book calls this set the "finite complement topology" I just want to know if this proof is correct. (I intuitively understand why its a topology, but I'm not sure if I'm showing it properly) Using the definition of topology: [imath]\mathscr{T}[/imath] is a topology on X if and only if the following are true: (i) X and Ø are elements of [imath]\mathscr{T}[/imath]. (ii) [imath]\mathscr{T}[/imath] is closed under finite intersections. (iii) [imath]\mathscr{T}[/imath] is closed under arbitrary unions. Let:[imath]V_k \subseteq X[/imath] be any finite set. (ie: [imath]V_1, V_2...[/imath] are all finite sets) So Then, [imath]U_k = X \setminus V_k[/imath], would be the subsets of [imath]\mathscr{T}[/imath]. Now to verify the definition. (i) X and Ø are elements of [imath]\mathscr{T}[/imath]. Ø is given in the definition of [imath]\mathscr{T}_1[/imath]. Also, the empty set is considered finite, with cardinality zero. Thus [imath]X[/imath] is in [imath]\mathscr{T}_1[/imath] because its complement is finite. (ii) [imath]\mathscr{T}[/imath] is closed under finite intersections. [imath]U_k \cap U_j = X \setminus (V_k \cup V_j) [/imath], due to De Morgan's law. Since the Union of two finite sets is also finite, this set is still in [imath]\mathscr{T}_1[/imath]. Or Symbolically (using bars || for cardinality): If [imath]|V_k \cup V_j| < |\mathbb{N}| [/imath], then [imath](U_k \cap U_j) \in \mathscr{T}_1[/imath]. This reasoning can be extended to any finite amount of intersections. (iii) [imath]\mathscr{T}[/imath] is closed under arbitrary unions. [imath]U_k \cup U_j = X \setminus (V_k \cap V_j) [/imath], due to De Morgan's law.Since the Intersection of two finite sets is also finite, this set is still in [imath]\mathscr{T}[/imath]. Or Symbolically: If [imath]|V_k \cap V_j| < |\mathbb{N}| [/imath], then [imath](U_k \cup U_j) \in \mathscr{T}_1[/imath]. This reasoning can be extended to any finite amount of Unions. Now let's consider the infinite case: [imath]V_k \cap ... \cap V_j[/imath] will only contain points that are in EVERY set being intersected. So if [imath]V_k \cap ... [/imath] was infinite, that would imply the every [imath]V_k[/imath] being intersected is also infinite, but [imath]V_k[/imath] is specifically defined as being finite. Thus even with infinite Unions, [imath]\mathscr{T}_1[/imath] is still closed.
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1893224
|
If [imath]1\leq p < r \leq \infty[/imath] , then [imath]\| x \|_p \geq \| x \|_r [/imath]
For [imath]1\leq p \leq \infty[/imath] , let [imath]\| \cdot \|_p [/imath] be the [imath]\mathcal{\ell_p}[/imath] norm on [imath]\mathbb{R}^n[/imath]. I need to show that if [imath]1\leq p < r \leq \infty[/imath] , then [imath]\| x \|_p \geq \| x \|_r [/imath]. I have tried Holder's inequality and Minkowski inequality here but they didn't solve my purpose. I am completely stuck at this point.
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218046
|
Relations between p norms
The [imath]$p$[/imath]-norm on [imath]$\mathbb R^n$[/imath] is given by [imath]$\|x\|_{p}=\big(\sum_{n=1}^\infty |x_{n}|^p\big)^{1/p}$[/imath]. For [imath]0 < p < q[/imath] it can be shown that [imath]$\|x\|_p\geq\|x\|_q$[/imath] (1, 2). It appears that in [imath]$\mathbb{R}^n$[/imath] a number of opposite inequalities can also be obtained. In fact, since all norms in a finite-dimensional vector space are equivalent, this must be the case. So far, I only found the following: [imath]$\|x\|_{1} \leq\sqrt n\,\|x\|_{2}$[/imath](3), [imath]$\|x\|_{2} \leq \sqrt n\,\|x\|_\infty$[/imath] (4). Geometrically, it is easy to see that opposite inequalities must hold in [imath]$\mathbb R^n$[/imath]. For instance, for [imath]$n=2$[/imath] and [imath]$n=3$[/imath] one can see that for [imath]0 < p < q[/imath], the spheres with radius [imath]\sqrt n[/imath] with [imath]$\|\cdot\|_p$[/imath] inscribe spheres with radius [imath]$1$[/imath] with [imath]$\|\cdot\|_q$[/imath]. It is not hard to prove inequality (4). According to Wikipedia, inequality (3) follows directly from Cauchy-Schwarz, but I don't see how. For [imath]$n=2$[/imath] it is easily proven (see below), but not for [imath]$n>2$[/imath]. So my questions are: How can relation (3) be proven for arbitrary [imath]$n\,$[/imath]? Can this be generalized into something of the form [imath]$\|x\|_{p} \leq C \|x\|_{q}$[/imath] for arbitrary [imath]$0?[/imath] Do any of the relations also hold for infinite-dimensional spaces, i.e. in [imath]$l^p$[/imath] spaces? Notes: [imath]$\|x\|_{1}^{2} = |x_{1}|^2 + |x_{2}|^2 + 2|x_{1}||x_{2}| \leq |x_{1}|^2 + |x_{2}|^2 + \big(|x_{1}|^2 + |x_{2}|^2\big) = 2|x_{1}|^2 + 2|x_{2}|^2$[/imath], hence [imath]$=2\|x\|_{2}^{2}$[/imath] [imath]$\|x\|_{1} \leq \sqrt 2 \|x\|_{2}$[/imath]. This works because [imath]$|x_{1}|^2 + |x_{2}|^2 \leq 2|x_{1}\|x_{2}|$[/imath], but only because [imath](|x_{1}| - |x_{2}|)^2 \geq 0[/imath], while for more than two terms [imath]$\big(|x_{1}| \pm |x_{2}| \pm \dotsb \pm |x_{n}|\big)^2 \geq 0$[/imath] gives an inequality that never gives the right signs for the cross terms.
|
743055
|
Find a basis of the [imath]k[/imath] vector space [imath]k(x)[/imath]
Suppose [imath]x[/imath] is a transcendental over field [imath]k[/imath] and [imath]k(x)[/imath] is the field of fractions of [imath]k[x][/imath]. Can we explicitly express a basis of the [imath]k[/imath] vector space [imath]k(x)[/imath]？
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126747
|
A basis for [imath]k(X)[/imath] regarded as a vector space over [imath]k[/imath]
Can anyone give an explicit basis of the [imath]k[/imath]-vector space [imath]k(X) = \operatorname{Quot}(k[X])[/imath] of rational functions over [imath]k[/imath]? The dimension is given by [imath]\dim_k k(X) = \max(|k|, |\mathbb N|).[/imath] If [imath]k[/imath] is infinite, this follows from [imath]|k| \leqslant |k(X)| \leqslant |k[X] \times k[X]| = |k \times k| = |k|[/imath] and the linear independence of [imath]\lbrace \frac{1}{X - \alpha} \mid \alpha \in k\rbrace.[/imath] If [imath]k[/imath] is finite, the result can be obtained similarly from [imath]|k(X)|=|\mathbb N|[/imath] and the linear independence of the monomials [imath]X^n[/imath], [imath]n \geqslant 0[/imath].
|
1892005
|
Is there such thing as a "smallest positive number that isn't zero"?
My brother and I have been discussing whether it would be possible to have a "smallest positive number" or not and we have concluded that it's impossible. Here's our reasoning: firstly, my brother discussed how you can always halve something, [imath]$(1, 0.5, 0.25, \dotsc)$[/imath]. I myself believe that it is impossible because of something I managed to come up with. You can put an infinite amount of zeroes in the decimal place before a number, [imath]$(0.1, 0.01, 0.001, \dotsc)$[/imath]. I am not entirely sure if our reasoning is correct though. I have been told that there is a smallest number possible but I decided to see for myself.
|
455639
|
What is the meaning of infinitesimal?
I have read that an infinitesimal is very small, it is unthinkably small but I am not quite comfortable with with its applications. My first question is that is an infinitesimal a stationary value? It cannot be a stationary value because if so then a smaller value on real number line exist, so it must be a moving value. Moving value towards [imath]0[/imath] so in most places we use its magnitude equal to zero but at the same time we also know that infinitesimal is not equal so in all those places were we use value of infinitesimal equal to [imath]0[/imath] we are making an infinitesimal error and are not [imath]100\%[/imath] accurate, maybe [imath]99.9999\dots\%[/imath] accurate, but no [imath]100\%[/imath]! So please explain infinitesimal and its applications and methodology in context to the above paragraph or elsewise intuitively please.
|
1892703
|
Parameter on [imath]\mathrm{SU}(4)[/imath] and [imath]\mathrm{SU}(2)[/imath]
From my reading, [imath]\mathrm{SU}(4)[/imath] has 15 parameters and [imath]\mathrm{SU}(2)[/imath] has 3 paramaters that range differently with certain parameter (rotation angle). And all the parameter is linearly independent to each other. My question are: What the characteristic of each of the parameter? If I choose two of them, what is the reason behind the theory? Can anyone explain to me or suggest any book/paper to me to understand the parameter itself.
|
1892705
|
Parameters on [imath]SU(4)[/imath] and [imath]SU(2)[/imath]
From my reading, [imath]SU(4)[/imath] has 15 parameters and [imath]SU(2)[/imath] has 3 parameters that range differently with certain parameters (rotation angle). And all the parameters are linearly independent to each other. My question are: What the characteristic of each of the parameters? If I choose two of them, what is the reason behind the theory? Can anyone explain to me or suggest any book/paper to me to understand the parameters themselves?
|
1893469
|
Product symbol, order of multiplication if factors are not commutative?
Regarding the product notation... if we have non-commutative things (for example matrices): [imath]\prod_{i=1}^N {\bf A}_i[/imath] Is it by default assumed to mean [imath]{\bf A}_1{\bf A}_2\cdots {\bf A}_N[/imath] or [imath]{\bf A}_N{\bf A}_{N-1} \cdots {\bf A}_1[/imath] or maybe even an arbitrary order?
|
399365
|
Big Greeks and commutation
Does a sum or product symbol, [imath]\Sigma[/imath] or [imath]\Pi[/imath], imply an ordering? Clearly if [imath]\mathbf{x}_i[/imath] is a matrix then: [imath]\prod_{i=0}^{n} \mathbf{x}_i[/imath] depends on the order of the multiplication. But, even if one accepts that it has a sequence, it is not clear if it should mean [imath]\mathbf{x}_0\mathbf{x}_1 \cdots \mathbf{x}_{n-1}\mathbf{x}_n[/imath] or [imath]\mathbf{x}_n\mathbf{x}_{n-1} \cdots \mathbf{x}_{1}\mathbf{x}_0[/imath]. A similar question, is there a "big" wedge product convention? [imath]\overset{n}{\underset{i=0}{\Huge\wedge}} \;{}^{\Large{\mathbf{x}_i} \;=\; \mathbf{x}_0 \wedge \mathbf{x}_1 \;\cdots \mathbf{x}_{n-1}\; \wedge \mathbf{x}_{n}} [/imath]
|
1893559
|
Why aren't all square roots irrational?
The more known proof of square root of [imath]2[/imath] is by contradiction when we assume it can be expressed as an irreducible fraction and later finding that it isn't irreducible, but... if we assume the same conditions for example: square root of [imath]9[/imath] we find that it isn't irreducible either. So, what's the trouble here?
|
457291
|
The contradiction method used to prove that the square root of a prime is irrational
The contradiction method given in certain books to prove that sqare root of a prime is irrational also shows that sqare root of [imath]4[/imath] is irrational, so how is it acceptable? e.g. Suppose [imath]\sqrt{4}[/imath] is rational, [imath]\begin{align} \sqrt{4} &=p/q \qquad\text{where pand q are coprimes} \\ 4 &=p^2/q^2\\ 4q^2&=p^2 \tag{1} \\ 4&\mid p^2\\ 4&\mid p\\ \text {let }p&=4m \qquad\text{for some natural no. m} \\ p^2&=16m^2\\ 4q^2&=16m^2 \qquad\text{(from (1) )}\\ q^2&=4m^2\\ 4& \mid q^2\\ 4&\mid q \end{align} [/imath] but this contradicts our assumption that [imath]p[/imath] and [imath]q[/imath] are coprime since they have a common factor [imath]p[/imath]. Hence [imath]\sqrt{4}[/imath] is not rational. But we know that it is a rational. Why?
|
1894383
|
Why is there no equivalence in the definition of injection?
Everywhere I found that the definition of an injection only uses an implication. However, I can't think a case where if [imath]f[/imath] is an injective function, and if [imath]a = b[/imath], then [imath]f(a) \neq f(b)[/imath]. Why do we use an implication ?
|
76186
|
Definition of injective function
From wikipedia I obtain the following definition of an injective function : Let [imath]f[/imath] be a function whose domain is a set [imath]A[/imath]. The function [imath]f[/imath] is injective if for all [imath]a[/imath] and [imath]b[/imath] in [imath]A[/imath], if [imath]f(a) = f(b)[/imath], then [imath]a = b[/imath]; that is, [imath]f(a) = f(b)[/imath] implies [imath]a = b[/imath]. From this I conclude that a function [imath]f[/imath] is injective if the below statement is true for all [imath]a,b \in A[/imath]: [imath]f(a)=f(b) \implies a=b[/imath] My question is: Can I re-formulate the above statement as [imath]f(a)=f(b) \iff a=b[/imath] ?
|
1894688
|
If [imath]xy+yz+zx=1[/imath] , show that [imath]\dfrac x{1-x^2}+\dfrac y{1-y^2}+\dfrac z{1-z^2}=\dfrac{4xyz}{(1-x^2)(1-y^2)(1-z^2)}[/imath].
It is to be solved using trigonometry. I tried taking [imath]x[/imath], [imath]y[/imath] and [imath]z[/imath] as [imath]\sin(a)[/imath], [imath]\sin(b)[/imath] and [imath]\sin(c)[/imath] respectively, but could get no further.
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1834114
|
If xy + yz + zx = 1, ...........
If [imath]xy + yz + zx = 1[/imath], then show that [imath]\dfrac{x}{1-x^2} + \dfrac{y}{1-y^2} + \dfrac{z}{1-z^2} = \dfrac{4xyz}{(1-x^2)(1-y^2)(1-z^2)}[/imath] I tried doing the sum algebraically, that is, by solving the LHS but, that method isn't really getting us anywhere. Also, I found this sum in an a book on trigonometry. So how can I solve it with trigonometry?
|
106508
|
Existence of a Riemannian metric inducing a given distance.
Let [imath]M[/imath] be a smooth, finite-dimensional manifold. Suppose [imath]M[/imath] is also a metric space, with a given distance function [imath]d: M \times M \rightarrow \mathbb{R}_{+}[/imath], which is compatible with the original (manifold) topology on [imath]M[/imath]. Question: is there a Riemannian metric [imath]g[/imath] on [imath]M[/imath] such that the distance [imath]d_g(p, q) = \inf_{\gamma \in \Omega(p, q)} L_g(\gamma)[/imath] coincides with [imath]d[/imath]? I believe that this setting is very standard, but for the sake of completeness: [imath]L_g[/imath] denotes the Riemannian length of the curve [imath]\gamma[/imath], and [imath]\Omega(p, q)[/imath] the set of all piecewise smooth curves [imath]\gamma : [a, b] \rightarrow M[/imath] s.t. [imath]\gamma(a) = p[/imath] and [imath]\gamma(b) = q[/imath]. Thanks in advance.
|
2304085
|
Riemannian metric from metric
Let [imath]M[/imath] be a manifold (finite dimension) and a [imath]d:M\times M\mapsto \mathbb{R}^+[/imath] metric on [imath]M[/imath]. Does this define a Riemannian manifold [imath](M, g)[/imath]? Is it possible to derive a metric tensor [imath]g[/imath] from the metric [imath]d[/imath]? I don't care about the pathologies, let's say that you have geodesic completeness or any kind of regularities you want on [imath]M[/imath] and [imath]d[/imath]. Just for motivational background: it is well known how to get a metric on a Riemannian manifold. My question is the other way around.
|
1894725
|
Proof about floor function: [imath]\lfloor x\rfloor+\lfloor x + \frac{1}{n} \rfloor+\cdots + \lfloor x + \frac{(n-1)}{n} \rfloor = \lfloor nx \rfloor[/imath]
How can we prove that [imath]\left\lfloor x\right\rfloor + \left\lfloor x + \frac{1}{n} \right\rfloor + \left\lfloor x + \frac{2}{n} \right\rfloor + \cdots + \left \lfloor x + \frac{(n-1)}{n} \right\rfloor = \left\lfloor nx \right \rfloor[/imath] Where [imath]\lfloor x\rfloor[/imath] denotes greatest integer less than or equal to [imath]x[/imath]. I was able to prove it when [imath]n = 2[/imath] or [imath]3[/imath]. Please see this link. But I can't prove it generally. I tried it using Principle of mathematical Induction but couldn't. Can someone prove it using some other way using properties of greatest integer function?
|
5650
|
Proving [imath]\sum\limits_{k=0}^{n-1} \Bigl[x + \frac{k}{n}\Bigr] = [nx][/imath]
Right, this is an exercise in Apostol, which I am not being able to solve. I was able to prove this result for a small case, that is the case when [imath]n=2[/imath], [imath][x] + \Bigl[x + \frac{1}{2}\Bigr]=[2x][/imath], but I am struggling with the generalization. Prove that [imath]\sum\limits_{k=0}^{n-1} \Biggl[x + \frac{k}{n}\Biggr] = [nx],[/imath] where [imath][ \ ][/imath] denotes the greatest integer function. Can this be proved via induction, I ask this because I have shown it to be already true for [imath]n=2[/imath].
|
691458
|
Subfield of [imath]\mathbb R[/imath] with algebraic closure as [imath]\mathbb C[/imath].
Does there exist a proper subfield of [imath]\mathbb{R}[/imath] whose algebraic closure is [imath]\mathbb{C}[/imath] ? A weaker question: Does there exist a proper subfield [imath]F[/imath] of [imath]\mathbb{R}[/imath] such that [imath]\mathbb{R}[/imath] is algebraic over [imath]F[/imath]?
|
1140259
|
Is [imath]\mathbb{R}[/imath] an algebraic extension of some proper subfield?
Is there a (proper) subfield [imath]K[/imath] of [imath]\mathbb{R}[/imath] such that [imath]\mathbb{R}[/imath] is an algebraic extension of [imath]K[/imath]? From this question, Is there a proper subfield [imath]K\subset \mathbb R[/imath] such that [imath][\mathbb R:K][/imath] is finite?, it is clear that for such an [imath]K[/imath], [imath][\mathbb{R}:K]=\infty[/imath]. This question seems to be non-trivial, and I suspect any existence result will be non-constructive. Any references will be appreciated.
|
1894173
|
Sum of reciprocal of primes
If [imath]p_n[/imath] is the nth prime, then prove that:[imath]\frac1p_1+\frac1p_2+....+\frac1p_n[/imath] is not an integer. Since [imath]p_1 ,p_2,..., p_n[/imath] are all prime their reciprocal is not an integer so the sum should not be an integer. But I am not sure.
|
66642
|
A finite sum of prime reciprocals
How can you prove that [imath]\sum\limits_k \frac1{p_k}[/imath], where [imath]p_k[/imath] is the [imath]k[/imath]-th prime, does not result in an integer?
|
223043
|
What is the cardinality of a transcendence basis of [imath]\mathbb{C}[/imath] over [imath]\mathbb{Q}[/imath]?
What is the cardinality of a transcendence basis of [imath]\mathbb{C}[/imath] over [imath]\mathbb{Q}[/imath]? Is it that of the continuum? Proof?
|
103177
|
Why does [imath]\mathbb{C}[/imath] have transcendence degree [imath]\mathfrak{c}[/imath] over [imath]\mathbb{Q}[/imath]?
It's pretty well known that [imath]\text{trdeg}(\mathbb{C}/\mathbb{Q})=\mathfrak{c}=|\mathbb{C}|[/imath]. As a subset of [imath]\mathbb{C}[/imath], of course the degree cannot be any greater than [imath]\mathfrak{c}[/imath]. I'm trying to understand the justification why it cannot be any smaller. The explanation in my book says that if [imath]\mathbb{C}[/imath] has an at most countable (i.e. finite or countable) transcendence basis [imath]z_1,z_2,\dots[/imath] over [imath]\mathbb{Q}[/imath], then [imath]\mathbb{C}[/imath] is algebraic over [imath]\mathbb{Q}(z_1,z_2,\dots)[/imath]. Since a polynomial over [imath]\mathbb{Q}[/imath] can be identified as a finite sequence of rationals, it follows that [imath]|\mathbb{C}|=|\mathbb{Q}|[/imath], a contradiction. I don't see why the polynomial part comes in? I'm know things like a countable unions/products of countable sets is countable, but could someone please explain in more detail this part about the polynomial approach? Since [imath]\mathbb{C}[/imath] is algebraic over [imath]\mathbb{Q}(z_1,z_2,\dots)[/imath], does that just mean that any complex number can be written as a polynomial in the [imath]z_i[/imath] with coefficients in [imath]\mathbb{Q}[/imath]? For example, [imath] \alpha=q_1z_1^3z_4z_6^5+q_2z_{11}+q_3z^{12}_{19}+\cdots+q_nz_6z_8z^4_{51}? [/imath] Is the point just that the set of all such polynomials are countable? Thanks,
|
1895934
|
Solving limit [imath]\lim_{n\rightarrow \infty} (\frac{1}{n+1}+ ... +\frac{1}{2n})[/imath]
Could anyone help me solving the following limit? [imath]\lim_{n\rightarrow \infty} (\frac{1}{n+1}+ ... +\frac{1}{2n})[/imath] I think it should just be 0 since we can distribute the limite inside the sum, but I am not sure of this answer. Thanks so much for your help!
|
179205
|
Evaluate [imath]\lim\limits_{n\to \infty}\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{6n}[/imath]
Show that [imath]\lim_{n\to \infty}\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{6n}\right)=\log 6[/imath] Here I need to use the definition of integral but I faced problem in range . Please help.
|
1895809
|
Showing a set involving complement is a topological space.
Question: Let [imath]X[/imath] be a set. Let [imath]\tau[/imath] consists of [imath]\varnothing[/imath] , and any set [imath]U \subseteq X[/imath] such that [imath]X\setminus U[/imath] is either finite or countable. Show that [imath]\tau[/imath] is a topology on [imath]X[/imath]. Looking only for [imath]\mathbf{hints}[/imath]. Let [imath]\tau=\left \{ \varnothing,U \right \}[/imath]. Obviously, [imath]X \subseteq X[/imath] so [imath]X \in \tau[/imath] Now, [imath]\tau=\left \{ \varnothing, \left \{ U_{i} \right \}_{i \in I} \right \}[/imath] with [imath]X \in \left \{ U_{i} \right \}_{i \in I}: \exists i \in I[/imath] By hypothesis, [imath]X\setminus U[/imath] is countable so it is expected that [imath]\exists[/imath] a bijection [imath]f:\mathbb{Z}^{+}\rightarrow X\setminus U[/imath] I would like to take this further. My intuition says that if I can show that the elements in [imath]U[/imath] are open sets, then the elements in [imath]U[/imath] are elements of [imath]\tau[/imath] which then satisfies one of the axiom for topological space. However, I am unable to do so. Any Hint is appreciated. Thanks in advance.
|
1837814
|
Show "countable complement topology" is a topology
I'm teaching my self topology with the aid of a book. I'm trying to prove the following is a topology: Let X be an infinite set. Show that [imath]\mathscr{T}_2=\{U \subseteq X : U = \emptyset [/imath] or [imath] X\setminus U [/imath] is countable[imath] \} [/imath] is a topology. My book calls this set the "countable complement topology" Please let me know if my proof is valid. corollary [A]: The union of countable sets are countable. proof: I'm going to take as a given that that the union of finite set are finite, and just prove the case of countable infinite sets. A set is countable infinite if there is a bijection between the set and the natural numbers. Let [imath]A[/imath] and [imath]B[/imath] be a countable infinite sets. Now let [imath]f[/imath] be the bijection between [imath]A[/imath] and [imath]\mathbb{N}[/imath] and [imath]g[/imath] be [imath]B[/imath]'s bijection. [imath]f:\mathbb{N} \rightarrow A[/imath] [imath]g: \mathbb{N} \rightarrow B[/imath] There can always exist a bijection [imath]h[/imath] that maps [imath]A \cup B [/imath] to [imath]\mathbb{N}[/imath]. To account for non disjoint sets, let [imath]B' = B \setminus A[/imath]; (note: [imath]A \cup B = A \cup B'[/imath]) [imath]h: \mathbb{N} \rightarrow A \cup B'[/imath] [imath] h(x) = \left\{ \begin{array}{ll} f(\frac{1}{2}x); & \quad x \in 2\mathbb{N} \\ g(\frac{x+1}{2}); & \quad x \in 2\mathbb{N}+1 \end{array} \right. [/imath] [imath]h^{-1}(x)[/imath] would map [imath]A[/imath] to the even numbers, and [imath]B'[/imath] to the odd numbers. Therefore, The union of countable sets are countable. (QED) Now to verify the definition. (i) X and Ø are elements of [imath]\mathscr{T}[/imath]. Ø is given in the definition of [imath]\mathscr{T}_2[/imath]. Also, the empty set is considered finite, with cardinality zero. Thus [imath]X[/imath] is in [imath]\mathscr{T}_1[/imath] because its complement is finite, and finite numbers are countable. (ii) [imath]\mathscr{T}[/imath] is closed under finite intersections. Let:[imath]V_k \subseteq X[/imath] be any countable set. (ie: [imath]V_1, V_2...[/imath] are all countable sets) So Then, [imath]U_k = X \setminus V_k[/imath], would be the subsets of [imath]\mathscr{T}_2[/imath]. [imath]U_k \cap U_j = X \setminus (V_k \cup V_j) [/imath], due to De Morgan's law. Thus, [imath]U_k \cap U_j \in \mathscr{T}_2[/imath], since the Union of two countable sets is also countable[A]. And This reasoning can be extended to any finite amount of intersections. (iii) [imath]\mathscr{T}[/imath] is closed under arbitrary unions. Let [imath]A=\bigcup_{i \in I} U_i[/imath] be an arbitrary union of sets whose complements are are countable.Then by De Morgan's law, [imath]A=X \setminus (\bigcap_{i \in I}V_i)[/imath], for some countable sets [imath]V_i[/imath]. By the definition of intersection, [imath]\bigcap_{i \in I}V_i \subseteq V_i[/imath] for all [imath]i \in I[/imath]. Since [imath]V_i[/imath] is countable so is any of its subsets. Hence the complement of [imath]A[/imath] is countable.
|
1895958
|
Brownian motion first hitting time negative barrier
for the standard cases the answer is well known, but somehow I couldn't derive the answer for the following setting: Suppose that [imath]X_t = \sigma B_t + ct[/imath], where [imath]B[/imath] is a Brownian motion, [imath]c>0[/imath] is a constant and [imath]X_0=0[/imath]. We define [imath]H_a = \inf \{ t: X_t =a \}[/imath] for [imath]\underline{a <0}[/imath] as the first hitting time. What is the density of [imath]H_a[/imath] in this case? It is different from the standard case because now [imath][H_a \leq t] = \left[ \inf_{s \leq t} X_s \leq a \right] \tag{1}[/imath] Thanks!
|
1053294
|
Density of first hitting time of Brownian motion with drift
I just started learning about Brownian motion and I am struggling with this question: Suppose that [imath]X_t = B_t + ct[/imath], where [imath]B[/imath] is a Brownian motion, [imath]c[/imath] is a constant. Set [imath]H_a = \inf \{ t: X_t =a \}[/imath] for [imath] a >0[/imath]. Show that for [imath]c \in \mathbb{R}[/imath], the density of [imath]H_a[/imath] is \begin{equation} f_{H_a} (t) = \frac{ a \exp \Big\{ \frac{- (a-ct)^2}{2t} \Big\} }{\sqrt{2 \pi t^3}}. \end{equation}
|
1831305
|
Polynomial ring with isomorphic quotients
If [imath]R[/imath] is a commutative ring and [imath]f(x), g(x) \in R[x][/imath] two polynomials such that [imath]R[x]/f(x)\cong R[x]/g(x)[/imath] as [imath]R[/imath]-algebras, what can we say about [imath]f[/imath] and [imath]g[/imath]? Or given [imath]f(x)\in R[x][/imath], what can we say about the set [imath]\{g \in R[x]\mid R[x]/f(x) \cong R[x]/g(x)\}[/imath]? Can we conclude that [imath]\mathrm{deg}(f) = \mathrm{deg}(g)[/imath]? Is there an automorphism of [imath]R[x][/imath] taking [imath]f[/imath] to [imath]g[/imath]? I'm most concerned about the cases where [imath]R=k[/imath] is a field, [imath]R=\mathbb{Z}[/imath], or [imath]R=\mathcal{O}_k[/imath] is the ring of integers of a number field, but I'd be interested in the most general results possible.
|
869335
|
What is necessary and/or sufficient for polynomials to provide isomorphic quotient rings?
Let [imath]R[/imath] be a commutative ring (with identity). Let [imath]f,g\in R\left[x\right][/imath] both be monic polynomials of degree [imath]d[/imath]. Then the underlying abelian groups of the rings [imath]R\left[x\right]/\left(f\left(x\right)\right)[/imath] and [imath]R\left(x\right)/\left(g\left(x\right)\right)[/imath] are isomorphic. Actually both are isomorphic to the direct sum [imath]R^{\oplus d}[/imath]. Questions: What sort of relation between [imath]f[/imath] and [imath]g[/imath] will ensure that [imath]R\left[x\right]/\left(f\left(x\right)\right)[/imath] and [imath]R\left(x\right)/\left(g\left(x\right)\right)[/imath] are also isomorphic as rings (with identities)? What sort of relation between [imath]f[/imath] and [imath]g[/imath] will ensure that [imath]R\left[x\right]/\left(f\left(x\right)\right)[/imath] and [imath]R\left(x\right)/\left(g\left(x\right)\right)[/imath] are also isomorphic as [imath]R[/imath]-algebras? Maybe my questions are too broad. Partial answers (e.g. with extra conditions on [imath]R[/imath]) or references are also very welcome.
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1896352
|
Prove that [imath]D_4[/imath] cannot be expressed as the internal direct product of two proper subgroups.
Prove that [imath]D_4[/imath] cannot be expressed as the internal direct product of two proper subgroups. I know that it can't be an internal direct product of two rotational subgroups because no rotation combination gives a reflection, but for the other cases I can't seem to make any progress.
|
170012
|
Prove that the dihedral group [imath]D_4[/imath] can not be written as a direct product of two groups
I like to know why the dihedral group [imath]D_4[/imath] can't be written as a direct product of two groups. It is a school assignment that I've been trying to solve all day and now I'm more confused then ever, even thinking that the teacher might have missed writing out that he means normal subgroups. On another thread it was stated (as the answer to this question) that the direct product of two abelian groups is again abelian. If we consider the direct product of abelian subgroups [imath]H[/imath],[imath]K\in G[/imath] where [imath]HK=G[/imath] (for all [imath]g \in G[/imath], [imath]g=hk[/imath] [imath]h \in H[/imath], [imath]k \in K[/imath].) I can't understand why this would imply [imath]g=kh[/imath]? It is not stated anywhere that [imath]H[/imath],[imath]K[/imath] has to be normal! But if this implication is correct I do understand why [imath]D_4[/imath] (that is non-abelian) can't be written as a direct product of two groups. But if it's not, as I suspect, I need some help! We know that all groups of order 4 and 2 are abelian, (since [imath]4=p^2[/imath]), but only 4 of the subgroups of [imath]D_4[/imath] are normal: Therefor I can easily show that [imath]D_4[/imath] can't be a direct product of normal subgroups: The only normal subgroups of [imath]D_4[/imath] is the three subgroups of order 4, (index 2 theorem): [imath]\{e,a^2,b,a^2b\}[/imath], [imath]\langle a\rangle[/imath], [imath]\{e,a^2,ab,a^3b\}[/imath] and the center of [imath]D_4=\{e,a^2\}[/imath] We can see that these are not disjoint. So [imath]D_4[/imath] can't be a direct product of normal subgroups. The reason for this being that the center is non-trivial. But why can't [imath]D_4[/imath] be a direct product of any two groups? If we write the elements of [imath]D_4[/imath] as generated by [imath]a[/imath] and [imath]b[/imath], [imath]a^4=e[/imath], [imath]b^2=e[/imath], [imath]ba=a^3b[/imath] why isn't [imath]D_4=\langle b\rangle\langle a\rangle[/imath] ? I calculated the products of the elements of these two groups according to the rule given above and ended up with [imath]D_4[/imath], and also [imath]\langle a\rangle[/imath], [imath]\langle b\rangle[/imath] is disjoint...? Why is this wrong? Very thankful for an answer!
|
1896081
|
Trigonometric Indefinite Integration
Integrate: [imath]\int\frac{\cos5x+\cos4x}{1-2\cos 3x}\; dx[/imath] I tried using sums and products formula but couldn't make it. How to approach this problem?
|
1866647
|
Another way to evaluate [imath]\int\frac{\cos5x+\cos4x}{1-2\cos3x}{dx}[/imath]?
What I've done is this:[imath]\int\dfrac{\cos5x+\cos4x}{1-2\cos3x}{dx}[/imath] [imath]\int \dfrac{\sin 3x}{\sin 3x}\left[\dfrac{\cos5x+\cos4x}{1-2\cos3x}\right]{dx}[/imath] [imath]\dfrac {1}{2}\int\dfrac{\sin 8x -\sin 2x +\sin 7x -\sin x}{\sin 3x - \sin 6x}[/imath] [imath]-\dfrac {1}{2}\int\dfrac{ \sin \frac{7x}{2} +\sin \frac{5x}{2} } {\sin \frac{3x}{2} }[/imath] [imath]-\int\dfrac{ \sin {3x}\cos \frac{x}{2} } {\sin \frac{3x}{2} }[/imath] [imath]-\int\dfrac{2\sin \frac{3x}{2} \cos \frac{3x}{2}\cos \frac{x}{2} } {\sin \frac{3x}{2} }[/imath] [imath]-\int {2\cos \frac{3x}{2}\cos \frac{x}{2} }[/imath] [imath] -\left(\frac{\sin 2x}{2} +\sin x \right) +c [/imath] Is there any other way to do so ? Is it possible to do it by substitution ?
|
1226110
|
Is the sequence defined by the recurrence [imath]a_{n+2}=\frac{1}{a_{n+1}}+\frac{1}{a_n}[/imath] convergent?
Let [imath]a_0=1[/imath], [imath]a_1=1[/imath] and [imath]a_{n+2}=\frac1{a_{n+1}}+\frac1{a_n}[/imath] for every natural number [imath]n[/imath]. How can I prove that this sequence is convergent? I know that if it's convergent, it converges to [imath]\sqrt2[/imath] since if [imath]\lim\limits_{n\to\infty}a_n=a[/imath] then: [imath]\lim_{n\to\infty}\left(a_{n+2}-\frac1{a_{n+1}}-\frac1{a_n}\right)=a-\frac2a=0[/imath] [imath]\therefore\quad a^2=2[/imath] Now it's easy to see that every [imath]a_n[/imath] is positive, so [imath]a\ge0[/imath] and thus [imath]a=\sqrt2[/imath]. Assuming the sequence is convergent, I can calculate an estimation of the rate of convergence too. Let [imath]\epsilon_n:=a_n-\sqrt2[/imath]. We have: [imath]\epsilon_{n+2}=\frac1{a_{n+1}}-\frac1{\sqrt2}+\frac1{a_n}-\frac1{\sqrt2}=-\frac{a_{n+1}-\sqrt2}{\sqrt2a_{n+1}}-\frac{a_n-\sqrt2}{\sqrt2a_n}=-\frac{\epsilon_{n+1}}{\sqrt2a_{n+1}}-\frac{\epsilon_n}{\sqrt2a_n}[/imath] Now because [imath]a_n\sim\sqrt2+\epsilon_n[/imath] and [imath]\lim\limits_{n\to\infty}\epsilon_n=0[/imath], therefore from the above equation: [imath]\epsilon_{n+2}\lesssim-\frac{\epsilon_{n+1}+\epsilon_n}2[/imath] which yields [imath]\epsilon_n\lesssim\alpha\left(\frac{-1-\sqrt7i}4\right)^n+\beta\left(\frac{-1+\sqrt7i}4\right)^n[/imath] for some constants [imath]\alpha[/imath] and [imath]\beta[/imath], using induction on [imath]n[/imath]. Hence [imath]\frac1{\sqrt2}[/imath] is a good estimation for the rate of convergence since [imath]\left|\frac{-1\pm\sqrt7i}4\right|=\frac1{\sqrt2}[/imath]. (Edit: Thanks to @AlexRavsky for his confirming graphs.) Edit (some more of my thoughts): Let [imath]b_n:=\min\left\{a_n,a_{n+1},\frac2{a_n},\frac2{a_{n+1}}\right\}[/imath]. It's easy to see that [imath]b_n\le a_n\le\frac2{b_n}[/imath] and [imath]b_n\le a_{n+1}\le\frac2{b_n}[/imath]. Now using induction we can prove that [imath]b_n\le a_{n+m}\le\frac2{b_n}[/imath]. Especially, [imath]a_{n+2}\ge b_n[/imath] and [imath]\frac2{a_{n+2}}\ge b_n[/imath] which yields [imath]b_{n+1}\ge b_n[/imath]. The problem can be solved if I show that the sequence [imath]\left(b_n\right)_{n=0}^\infty[/imath] increases to [imath]\sqrt2[/imath].
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1890531
|
Proof of a limit for a recursively-defined sequence
I have a sequence defined by [imath] a_1=1,\quad a_2=1,\quad a_n=\frac{1}{a_{n-1}}+\frac{1}{a_{n-2}}\text{ for } n\ge2. [/imath] Now, if [imath]\lim_{n\to \infty}a_n=g[/imath] then [imath]\lim_{n\to \infty}a_n=\lim_{n\to \infty}\Bigl(\frac{1}{a_{n-1}}+\frac{1}{a_{n-2}}\Bigr)=\frac{2}{g}[/imath] so [imath]g=\sqrt{2}[/imath] or [imath]g=-\sqrt{2}[/imath] but [imath]a_n>0[/imath] so [imath]g=\sqrt{2}[/imath]. Now, how do I prove that it has an actual limit? Also, it can be proven that [imath]1\le a_n\le2[/imath], and it's not monotonic because [imath]a_4 \gt a_5 \lt a_6[/imath]. Also, it's not monotonic after some [imath]N\in\mathbb N[/imath]
|
1897210
|
Continuous, absolutely integrable, unbounded function
For my course in Fourier analysis I need an example of a continuous unbounded function that is absolutely integrable. I can't seem to think of one and so need some help with this. I also need an continuous bounded function that isn't absolutely integrable and I was thinking [imath]f(x)=\begin{cases} \frac{\sin(x)}{x}, \text{if } x\neq 0 \\ 1, \text{ if } x=0\end{cases}[/imath]. Is this correct?
|
482934
|
Continuous unbounded but integrable functions
Many tricky exercises concern the quest for functions that satisfy particular conditions. For example, let us consider the spaces [imath]C_p( \mathbb R), 1 \leq p < \infty[/imath], of continuous functions on [imath]\mathbb R[/imath] such that [imath]\int_{\mathbb R} \lvert f(x) \rvert^p \mbox{d}x < \infty[/imath]. I found the following exercises rather demanding: Find a function [imath]f\in C_1(\mathbb R)[/imath] such that [imath]f[/imath] is unbounded; Find a function [imath]f \in C_1(\mathbb R) \backslash C_2(\mathbb R)[/imath]. so one can deduce that [imath]C_p \not\subset C_q[/imath] if [imath]p < q[/imath]. I'm quite sure that a function defined to be zero everywhere except for triangular peaks of height [imath]k[/imath] and base [imath]1/k^3[/imath] centered on positive integers [imath]k[/imath] is a solution for both the exercises. (it's a simple matter to give to this function an explicit form, but I think it would be rather unclear.) Is it correct? Can anyone give other examples? (for the one or the other exercise, not necessary a solution of both of them!)
|
239078
|
Problem related polynomial ring over finite field of intergers
if [imath]f(x)[/imath] is in [imath]F[x][/imath]. [imath]F[/imath] is field of integer mod [imath]p[/imath]. [imath]p[/imath] is prime and [imath]f(x)[/imath] is irreducible over [imath]F[/imath] of degree [imath]n[/imath] . prove that [imath]F[x]/(f(x))[/imath] is a field with [imath]p^n[/imath] elements.
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542210
|
Proving [imath]\mathbb{F}_p/\langle f(x)\rangle[/imath] with [imath]f(x)[/imath] irreducible of degree [imath]n[/imath] is a field with [imath]p^n[/imath] elements
I am attempting to prove what the title says, that [imath]\mathbb{F}_p/\langle f(x)\rangle[/imath] with [imath]f(x)[/imath] irreducible of degree [imath]n[/imath] is a field with [imath]p^n[/imath] elements. I have already proven that for any field [imath]K[/imath] and polynomial [imath]f(x)[/imath] in [imath]K[x][/imath], [imath]K[x]/\langle f(x)\rangle[/imath] is a field if and only if [imath]f(x)[/imath] is irreducible in [imath]K[x][/imath]. So I know that [imath]\mathbb{F}_p/\langle f(x)\rangle[/imath] is a field for sure. Where I'm lost is figuring out how to know for sure that there are [imath]p^n[/imath] elements. It's fairly clear if [imath]n\in\{ 0 ,1\}[/imath], but I don't think induction is going to help. I know from a proposition in class that since [imath]F_p = \mathbb{Z}/\langle p\rangle[/imath] and [imath]F_p = \mathbb{Z}/\langle p\rangle[/imath] is an integral domain, [imath]F_p[x] = (\mathbb{Z}/\langle p\rangle)[x] = (\mathbb{Z}[x])/(\langle p\rangle[x])[/imath], but I'm not sure whether or not that's helpful information.
|
1897715
|
Propagating uncertainties with trigonometric functions
I have [imath]\tan (\frac{2.05 × 10^{-4}± 2.89 × 10^{-6}}{2})=\frac{x}{667.6 × 10^6}[/imath] [imath]\therefore x = (667.6 × 10^6) \cdot \tan(\frac{2.05 × 10^{-4}± 2.89 × 10^{-6}}{2})[/imath] As described in the title, I am trying to find the absolute uncertainty. Please include working out.
|
1897678
|
Trigonometric Uncertainty Propagation
I have [imath]tan (\frac{2.05 × 10^{-4}± 2.89 × 10^{-6}}{2})=\frac{x}{667.6 × 10^6}[/imath] [imath]\therefore x = (667.6 × 10^6) \cdot tan(\frac{2.05 × 10^{-4}± 2.89 × 10^{-6}}{2})[/imath] I want to find [imath]x[/imath] with the absolute uncertainty. I don't know how to start, as I am unsure how I would propagate uncertainties with [imath]tan[/imath]. Please include your working out when you explain the method of propagating uncertainties in this case.
|
1897928
|
Suppose [imath]G[/imath] is a group with order [imath]n[/imath]. Assume [imath](n,\phi(n)) = 1[/imath]. Prove that [imath]G[/imath] is Abelian group?
I have thought about this question alot. Unfortunately couldn't picture any good path to the desirable result. Suppose [imath]G[/imath] is a group with order [imath]n[/imath]. Assume [imath](n,\phi(n)) = 1[/imath], So they're relatively prime. Prove that [imath]G[/imath] is Abelian group. I am looking forward to hearing from you. Any hints, tips, or creative solutions are welcome.
|
588522
|
Any group of order [imath]n[/imath] satisfying [imath]\gcd (n, \varphi(n)) =1[/imath] is cyclic
Sorry for the last mistaken problem I just posted. Now I know that only having the order being odd square free is not enough for a group to be cyclic. Here's the complete problem which the main goal is to show that any group of order [imath]n[/imath] satisfies [imath]\gcd (n,\varphi(n))[/imath]=1 is cyclic. It asks me to do it in the following way: (a) If [imath]n > 2[/imath], [imath]n[/imath] is an odd squarefree integer. (b) Show that there is a [imath]H = G/N[/imath] of prime order. (c) Show that [imath]G \cong N \times H[/imath] and [imath]G[/imath] is abelian. (d) [imath]G[/imath] is cyclic. Any ideas is appreciated.
|
1897572
|
Calculating the integral of an infinite product
I came across the following question which asked Prove that [imath]\int_{0}^{1} \prod_{k=1}^{\infty} (1-x^k)=\frac{4\pi\sqrt{3}\sinh{\frac{\pi\sqrt{23}}{3}}}{{\sqrt{23}\cosh{\frac{\pi\sqrt{23}}{2}}}}[/imath] Note: This is the one of the first integrals I have seen like this. I have been looking up other questions (similar to mine) and am still unable to figure out a method to try. Attempt: I first examined different cases for [imath]k[/imath] and I am able to compute those integrals fine, since they are just polynomials when expanded. However, I am still stuck with where to begin. [imath]\int_{0}^{1} \prod_{k=1}^{\infty} (1-x^k)=(1-x)(1-x^2)(1-x^3) \cdots (1-x^n) \cdots[/imath] If I were to factor a [imath]-1[/imath] from each term I can see that each term, after the first, contains a factor of [imath](x-1)[/imath] which might help with some kind of substitution, but that is about as far as I got. If someone could please offer a hint that will help me get going with this problem. Thank you.
|
1883140
|
An integration of product [imath](1-x^n)[/imath]
Prove [imath]\int_0^1\prod_{n=1}^\infty(1-x^n)dx=\frac{4\pi\sqrt3}{\sqrt{23}}\frac{\sinh\frac{\pi\sqrt{23}}3}{\cosh\frac{\pi\sqrt{23}}2}.[/imath] In fact, product [imath](1-x^n)[/imath] is difficult to compute, I hope you can show me some ideas, thank you!
|
1898282
|
An infinite series [imath]1+\frac{1}{2+\frac{1}{3+\frac{1}{4+.......}}}[/imath]
How can we find the value to which the following series converges, if it converges to a finite number? If else, how can we prove that it is divergent?[imath]1+\frac{1}{2+\frac{1}{3+......}}[/imath]
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1871733
|
Convergence of a Harmonic Continued Fraction
Does this continued fraction converge? [imath]\cfrac { 1 }{ 1+\cfrac { 1 }{ 2+\cfrac { 1 }{ 3+\cfrac { 1 }{ 4+\ddots } } } } [/imath] ([imath][0;1,2,3,4, \dots][/imath]) I tried approximating a few values but I couldn't make out whether it converges or diverges. Can anyone provide a proof whether this converges or diverges. If it converges, please add what it converges to.
|
1898126
|
Coefficient of [imath]x^{-2}[/imath] in expression [imath](x-1/x)^{12}[/imath]
Find the coefficient of [imath]x^{-2}[/imath] in [imath](x-1/x)^{12}[/imath] Can anyone help me with this by providing the process to find the answer? I know the answer is [imath]-792[/imath] but I can't find it.
|
1885813
|
Find the coefficient of the term [imath]x^2[/imath] in [imath]\left(x+\frac 2x\right)^{4}[/imath]
Find the coefficient of the term [imath]x^2[/imath] in [imath]\left(x+\frac 2x\right)^{4}[/imath] This is what I have done so far. We know [imath]a[/imath] = [imath]x[/imath] and [imath]b[/imath] = [imath]\frac2x[/imath] and [imath]n[/imath] = 4 So what I have done is [imath]4c2[/imath] x [imath](x)^2[/imath] x ([imath]\frac2x[/imath])[imath]^2[/imath] As we know that [imath]^2[/imath] + [imath]^2[/imath] = 4, and 4 is the [imath]n[/imath] value. The answer is one though? What have I done wrong? P.S. this was the tutorial that helped me.
|
1898286
|
The uniform limit of holomorphic functions is also a holomorphic function
Let [imath]f_{n}[/imath] be a sequence of holomorphic functions on an open set [imath]U \in \mathbb{C}[/imath], which tends to f uniformly on [imath]U[/imath]. Prove that f is holomorphic on U I have proved a simple case which is if [imath]U[/imath] is a simply connected domain, then a continuous function [imath]g[/imath] on [imath]U[/imath] is holomorphic if and only if [imath]\int_{\gamma}g=0[/imath] for any closed curve [imath]\gamma[/imath] in [imath]U[/imath]. But I don't know how to apply to the general case [imath]U[/imath] is an open set. Thanks.
|
368664
|
Uniform limit of holomorphic functions
Let [imath]\{f_n\}[/imath] be a sequence of holomorphic functions defined in a generic domain [imath]D \subset \mathbb{C}[/imath]. Suppose that there exist [imath]f[/imath] such that [imath]f_n \to f[/imath] uniformly. My question is: is it true that [imath]f[/imath] is holomorphic too?
|
534024
|
Definition of [imath] 1 + \frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{\ddots}}}}[/imath]
Is there a definition of [imath] 1 + \frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{\ddots}}}}[/imath]? I am somewhat familiar with continued fractions; that is, I am aware that their convergence depends on whether our input is rational or not. This is not a homework problem, I am just curious about this topic which has gotten little attention in my current studies. If this expression is defined, what is it and does it converge?
|
1917265
|
Finding 1/(2+1/(3+1/(4+...)))
I was wondering how to find the value of the following series: [imath]\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{4+\cfrac{1}{5+\cfrac{1}{6+\ddots}}}}}[/imath] How can I solve this?
|
133361
|
Cyclotomic extension over Q special case of Kronecker Weber theorem.
For [imath]d\in\mathbb{Q}[/imath], how could one show that [imath]Q(\sqrt{d})[/imath] lies in a cyclotomic extension of [imath]\mathbb{Q}[/imath] without using the Kronecker-Weber theorem?
|
282757
|
Square roots of integers and cyclotomic fields
For every [imath] N \in \mathbb Z[/imath] there exists an integer [imath]n[/imath] such that [imath] \sqrt N \in \mathbb Q(\zeta_n)[/imath]. I am struggling where to start this question, please suggest me few hints.
|
1898960
|
Do there exist arbitrary large [imath]n,n+1[/imath] so that they are both prime powers?
Hello I am rather new to number theory and only know some very basics. This question has been bugging me for some time. Is there an upper bound to [imath]\{n,n+1\}[/imath] for which [imath]\begin{array}{rr}n=&{p_1}^{k_1}\\n+1=&{p_2}^{k_2}\end{array}[/imath] Where [imath]p_1,p_2[/imath] prime. Preferrably with [imath]k_1>k_0[/imath] and [imath]k_2>k_0[/imath] for some smallest exponent [imath]k_0[/imath].
|
735653
|
consecutive prime power
I'm interesting on consecutive prime power numbers. I see that there is the Mersenne primes and the Fermat Primes that give solutions and [imath](8,9)[/imath]. In Sloane collection it is referred on A006549 and it is written: David W. Wilson and Eric Rains found a simple proof that in this case of Catalan's conjecture either n or n+1 must be a power of 2 and the other number must be a prime, except for n=8. But I can't find this proof on the web (and I don't find it myself...). I tried the first case. I get a proof for [imath]2^k+1=3^q[/imath], but found it complicated. If someone has something simple? Simple: I see that it's enough to consider the case [imath](2^k, q^n)[/imath] because... ok it's clear I see that it's enough to show that [imath]2^k=-1[/imath] mod [imath]q[/imath] or [imath]q^2[/imath] is impossible. But for exemple that give nothing for [imath]2^k+1=5^q[/imath] because [imath]2[/imath] is primite modulo [imath]5[/imath] and [imath]25[/imath] and so modulo all the [imath]5^q[/imath].
|
1898829
|
How to solve this indefinite real integral?
I have to solve the following real integral [imath] \int_{0}^{\infty} \frac{x^3} {{e^x -1}{}} dx[/imath] Using complex analysis I split it in 4 different integral, according to big Circle lemma the integral along the big circle is zero that heads to: [imath] P.V. \int_{0}^{\infty} \frac{x^3} {{e^x -1}{}} dx = \int_{\gamma_\epsilon}^{} \frac{x^3} {{e^x -1}{}} [/imath] Using little circle lemma the second integral gives [imath] 8\pi^4 [/imath] It's this result correct?
|
99843
|
Contour integral for [imath]x^3/(e^x-1)[/imath]?
What contour and integrand do we use to evaluate [imath] \int_0^\infty \frac{x^3}{e^x-1} dx [/imath] Or is this going to need some other technique?
|
1899155
|
Conditional Trigonometric Identities on [imath]A+B+C=π[/imath].
If [imath]A+B+C=\pi[/imath] then prove that: [imath]\tan A\tan B+\tan B\tan C+\tan C \tan A=1+\sec A\sec B\sec C[/imath] My Approach: Given, [imath]A+B+C=\pi[/imath] [imath]A+B=\pi-C[/imath] [imath]\sin(A+B)=\sin(\pi-C)=\sin C[/imath] [imath]\cos(A+B)=\cos(\pi-C)=-\cos C[/imath] Now, [imath]\text{L.H.S.}=\tan A \tan B+\tan B\tan C+\tan C\tan A[/imath] [imath]=\frac {\sin A\sin B}{\cos A\cos B}+\frac {\sin B\sin C}{\cos B\cos C} + \frac {\sin C\sin A}{\cos C\cos A}[/imath] I got struck at here. Please help me to move on.
|
1849467
|
Given [imath]A+B+C=180^{\circ}[/imath], find value of [imath]\tan A\cdot\tan B+\tan B\cdot\tan C+\tan A\cdot \tan C-\sec A\cdot\sec B\cdot\sec C[/imath]
Given [imath]A+B+C=180^{\circ}[/imath], find value of [imath]\tan A\cdot\tan B+\tan B\cdot\tan C+\tan A\cdot \tan C-\sec A\cdot\sec B\cdot\sec C[/imath] I know about some basic conditional identities but don't know how to use them here.
|
1899217
|
Characteristic of a non unital integral domain
Theorem: the characteristic of a non unital integral domain must be [imath]0[/imath] or a prime. Assume that [imath]mn[/imath] is the characteristic of a the integral domain for every [imath]a[/imath] in the integral domain we have [imath](mn)a^2=(ma)(na)=0[/imath] then [imath]ma[/imath] or [imath]na[/imath] must be zero. So for every [imath]a[/imath] in the integral domain [imath]na[/imath] OR [imath]ma[/imath] is zero but I can't say one of [imath]m[/imath] and [imath]n[/imath] works for every [imath]a[/imath] !
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1077466
|
Integral domain without unity has prime characteristic?
By an integral domain, we mean here, a ring (not necessarily with unity) in which [imath]ab=0[/imath] implies [imath]a=0[/imath] or [imath]b=0[/imath]. Question: If an integral domain without unity has positive characteristic, is it necessarily prime? An integral domain [imath]D[/imath] is said to be of finite characteristic if there exists a positive integer [imath]m[/imath] such that [imath]ma=0[/imath] for all [imath]a\in D[/imath]. [cf. Topics in Algebra- I. N. Herstein, 2nd Ed., p. 129] My question is a slight modification of Problem 6 in [Topics in Algebra- I. N. Herstein, 2nd Ed., p. 130] Topics in Algebra- Herstein, 2nd Ed.
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1899387
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Where Is The Problem In Proving [imath]1=2[/imath] Through Continued Fractions
I can't seem to figure out why the proof is false. Everyone knows that [imath]1 \ne 2[/imath] and that [imath]1=2[/imath] is only proven through mathematical fallacies. However I can't figure out where the problem is in the following fallacy: Observe that [imath]1=\frac{2}{3-1}[/imath] Substitute [imath]\frac{2}{3-1}[/imath] for the [imath]1[/imath] in the denominator [imath]1=\frac{2}{3-\frac{2}{3-1}}[/imath] Substitute again [imath]1=\frac{2}{3-\frac{2}{3-\frac{2}{3-1}}}[/imath] Therefore [imath]1[/imath] can be written as the infinite fraction: [imath]1=\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots}}}[/imath] Now notice that [imath]2=\frac{2}{3-2}[/imath] Substitute 2 similar to how 1 was substituted above [imath]2=\frac{2}{3-\frac{2}{3-2}}[/imath] And through repeated substitution we get [imath]2=\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots}}}[/imath] Notice that [imath]1=\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots}}}[/imath] and [imath]2=\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots}}}[/imath] Therefore, [imath]1=2[/imath] Just looking at this I know that it cannot be true. But they both seem to share the same infinite fraction, so what is it that I am missing?
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417280
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Continued fraction fallacy: [imath]1=2[/imath]
It's easy to check that for any natural [imath]n[/imath] [imath]\frac{n+1}{n}=\cfrac{1}{2-\cfrac{n+2}{n+1}}.[/imath] Now, [imath]1=\frac{1}{2-1}=\frac{1}{2-\cfrac{1}{2-1}}=\frac{1}{2-\cfrac{1}{2-\cfrac{1}{2-1}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-1}}}}=\ldots =\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\dots}}}}},[/imath] [imath]2=\cfrac{1}{2-\cfrac{3}{2}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{4}{3}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{5}{4}}}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{6}{5}}}}}=\ldots =\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\ldots}}}}}.[/imath] Since the right hand sides are the same, hence [imath]1=2[/imath].
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1899864
|
Commuting matrix exponentials: necessary condition.
In related questions (here, here and here), it was shown that if [imath]A[/imath] and [imath]B[/imath] commute, then [imath]e^A[/imath] and [imath]e^B[/imath] also commute (and incidentally [imath]e^A e^B = e^{(A+B)}[/imath]). Here, the commutative property of A and B is a sufficient condition. Is it also a necessary condition? If no, is there another necessary condition for [imath]e^A[/imath] and [imath]e^B[/imath] to commute?
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349180
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If [imath]e^A[/imath] and [imath]e^B[/imath] commute, do [imath]A[/imath] and [imath]B[/imath] commute?
It is known that if two matrices [imath]A,B \in M_n(\mathbb{C})[/imath] commute, then [imath]e^A[/imath] and [imath]e^B[/imath] commute. Is the converse true? If [imath]e^A[/imath] and [imath]e^B[/imath] commute, do [imath]A[/imath] and [imath]B[/imath] commute? Edit: Addionally, what happens in [imath]M_n(\mathbb{R})[/imath]? Nota Bene: As a corollary of the counterexamples below, we deduce that if [imath]A[/imath] is not diagonal then [imath]e^A[/imath] may be diagonal.
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1899814
|
Proof [imath]\mathbb{Z}[i]/(2- i) \cong \mathbb{Z_5}[/imath]
My worked example sheet states Prove [imath]\mathbb{Z}[i]/(2- i) \cong \mathbb{Z_5}[/imath] using the following steps: Let [imath]G = \mathbb{Z}[i]/(2- i)[/imath] [imath]G = \{ \bar{x} + \bar{y}\bar{i} \mid x,y \in \mathbb{Z} \}[/imath] [imath]G = \{ \bar{x} \mid x \in \mathbb{Z} \}[/imath], because [imath]\bar{2} = \bar{i}[/imath] Show that [imath]G = \{ \bar{0},\ldots,\bar{4}\}[/imath] Finally, prove [imath]G \cong \mathbb{Z_5}[/imath] The proof is clear to me but I do not know how to rigorously show [imath]G = \{ \bar{x} + \bar{y}\bar{i} \mid x,y \in \mathbb{Z} \}[/imath] and [imath]G = \{ \bar{0},\ldots,\bar{4}\}[/imath]. For the first one, I think this is just the definition of a quotient ring but is that enough for rigour? While for the second one, we have the relation [imath]\bar{5} = \bar{0}[/imath] in [imath]R[/imath] but how can I show ( rather write ) [imath]G[/imath] has the mentioned elements. I guess, I am asking for a formal way of writing this. PS: The proof for similar problem is available on the forum, but my problem is how much detail is enough. ( For me the context is an examination. ) Pardon my elementary question.
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1320365
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[imath]\mathbb Z[i]/ \langle 1+2i \rangle \cong \mathbb Z_5[/imath]
I am trying to prove that [imath]\mathbb Z[i]/ \langle 1+2i \rangle[/imath] is isomorphic to [imath]\mathbb Z_5[/imath]. The only thing that came to my mind was trying to apply the first isomorphism theorem using an appropiate function. If I consider the euclidean function [imath]N: Z[i] \setminus \{0 \} \to \mathbb N[/imath] defined as [imath]N(a+bi)=a^2+b^2[/imath], then I can express any element [imath]z[/imath] in [imath]\mathbb Z[i][/imath] as [imath]z=(1+2i)q+r[/imath] with [imath]N(r)=1,2,3,4[/imath] or [imath]r=0[/imath]. If I define [imath]f:\mathbb Z[i] \to \mathbb Z_5[/imath] as [imath]f(x=(1+2i)q_x+r_x)=\overline{r_x}[/imath], then it is clear that [imath]f(x)=0[/imath] if and only if [imath]x \in \langle 1+2i \rangle[/imath]. The problem with this function is that it doesn't satisfy [imath]f(x+y)=f(x)+f(y)[/imath] and it is not surjective for if [imath]r_x=f(x)=3 (5)[/imath], then if [imath]r_x=a+bi[/imath], we have [imath]a^2+b^2=3[/imath], which is absurd. I don't know what else to do, any suggestions would be appreciated.
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1899852
|
Evaluating [imath]\prod\limits_{n\ge2}\left(1-\frac2{n(n+1)}\right)^2[/imath]
Let [imath]x_n=\left(1-\frac13\right)^2\left(1-\frac16\right)^2\left(1-\frac1{10}\right)^2\cdots\left(1-\frac2{n(n+1)}\right)^2,n\ge2[/imath] Then the value of [imath]\lim_{n\to\infty}x_n[/imath] equals? I tried to take log both sides: [imath]\log x_n=2\sum\limits_{k=2}^{n}\log(1-\frac{2}{k(k+1)}) [/imath] Despite taking limit both sides, I did not get it of the standard form of [imath]1^{\infty}[/imath].
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1897569
|
Finding the limit [imath]\lim_{n\rightarrow\infty}(1-\frac{1}{3})^2(1-\frac{1}{6})^2(1-\frac{1}{10})^2...(1-\frac{1}{n(n+1)/2})^2[/imath]
Any hints on how to find the following limit ?. I haven't been able to figure it out still. [imath] \lim_{n \to \infty}\left(\, 1 - \frac{1}{3}\, \right)^{2} \left(\, 1 - \frac{1}{6}\, \right)^{2}\left(\, 1 - \frac{1}{10}\, \right)^{2}\ldots\left[\, 1 - \frac{1}{n\left(\, n + 1\, \right)/2}\, \right]^{2} [/imath] Pardon me if it's a silly question. Edit: Solved upto this much [imath] \prod_{n = 2}^{\infty}\left[\, 1-\frac{2}{n\left(\, n + 1\, \right)}\, \right] = \prod_{n = 2}^{\infty}\left(\, \frac{n + 2}{n + 1}\,\frac{n - 1}{n}\, \right) [/imath] I did'nt notice most of the terms cancel out.
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257382
|
Fermat's equation with real exponents
Just out of curiosity : has the equation [imath] x^r+y^r=z^r,\qquad(x,y,z)\in\Bbb Z^3,\quad r\in\Bbb R, [/imath] been studied? Any non trivial result for [imath]r\in\Bbb R\setminus\Bbb N[/imath] ?
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394548
|
Fermat Last Theorem for non Integer Exponents
We now that Fermat's last theorem is true so there are not positive integer solutions to [imath]x^n+y^n=z^n[/imath] for [imath]n\in\mathbb{N}[/imath] and [imath]n>2[/imath]. But what about if [imath]n\in\mathbb{R}[/imath] or [imath]n\in\mathbb{R}^+[/imath]?
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1900269
|
It has been shown that every open set in ℝ is the union of disjoint open intervals. Is this true for [imath]ℝ^n[/imath] in general?
It has been shown that every open set in [imath]\mathbb R[/imath] is the union of disjoint open intervals. Is this true for [imath]\mathbb R^n[/imath] in general? It seems to me that it is not, but I cannot think of an example.
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1341759
|
Why an open set in [imath]R^n[/imath] might not be written as countable union of disjoint open intervals?
There is a theorem saying that every open set in [imath]R[/imath] can be written as a countable union of disjoint open intervals. And we can also show any open set in [imath]R^n[/imath] can be written as the union of countable many bounded open intervals (see here). I understand the proofs of the above two statements, but it feels strange that [imath]R^n[/imath] does not share the same property as [imath]R[/imath]. Is there any explanation that why an open set in [imath]R^n[/imath] might not be written as countable union of disjoint open intervals?
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1900618
|
A question from my real analysis qualifying exam
Let [imath]f[/imath] be in [imath]L_1(\Bbb R^N)[/imath] and [imath]g[/imath] be in [imath]L_q(\Bbb R^N)[/imath] for [imath]q \in (1, \infty)[/imath]. Show that [imath](f*g)(x)= \int_{\Bbb R^N}f(x-y)g(y)dy[/imath] is in [imath]L_q(\Bbb R^N)[/imath] and that [imath]||(f*g)(x) ||_q \le ||f||_1||g||_q[/imath] I had trouble getting past the following part. [imath]||(f*g)(x) ||_q^q= \int_{\Bbb R^N} \big |\int_{\Bbb R^N}f(x-y)g(y)dy \big|^q dx \le \int_{\Bbb R^N} \left(\int_{\Bbb R^N}|f(x-y)||g(y)|dy \right)^q dx [/imath] Any help is appreciated
|
722689
|
Inequality for the p norm of a convolution
Let [imath]1\le p < \infty[/imath]. Let [imath]g \in L^p(\Bbb R^n)[/imath], and [imath]f \in L^1(\Bbb R^n)[/imath]. Show: [imath]||f*g||_p \le ||f||_1||g||_p[/imath], where [imath]*[/imath] indicates the convolution defined by [imath] f*g (x) := \int f(x-t) g(t)dt.[/imath] Here is a relevant hint, apparently: Let [imath]q[/imath] be the conjugate of [imath]p[/imath]. Note that: [imath]|f*g| \le \int [|g(t)||f(x-t)|^{1/p}]|f(x-t)|^{1/q}dt,[/imath] then apply Holder's inequality. I did exactly as the hint suggested, and ended up with what looked like a meaningless inequality compared to the final answer. I even took the [imath]p[/imath] power of both sides. Any help?
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1895482
|
Showing that [imath]\frac{a + c}{b + d} \leq \max(\frac{a}{b},\frac{c}{d})[/imath]
I need help in proving (or disproving) the following assumption: [imath]\frac{a + c}{b + d} \leq \max(\frac{a}{b},\frac{c}{d})[/imath] where [imath]a,b,c,d \geq 0[/imath] are positive integers. Both fractions [imath]\frac{a}{b}[/imath] and [imath]\frac{c}{d}[/imath] are between 0 and 1 and therefore the conditions [imath]a \leq b[/imath] and [imath]c \leq d[/imath] hold. Any help or ideas are appreciated, thank you!
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2541142
|
Prove that [imath]\min{\left(\frac{a_1}{b_1},\frac{a_2}{b_2}\right)}\leq\frac{a_1+a_2}{b_1+b_2}\leq\max{\left(\frac{a_1}{b_1},\frac{a_2}{b_2}\right)}[/imath]
How can i prove the following inequality [imath]\ \min{\left(\frac{a_1}{b_1},\frac{a_2}{b_2}\right)}\leq\frac{a_1+a_2}{b_1+b_2}\leq\max{\left(\frac{a_1}{b_1},\frac{a_2}{b_2}\right)} [/imath] for any real numbers [imath]a_1,a_2[/imath] and positive numbers [imath]b_1,b_2[/imath]. Can anyone give me some hint or reference for the proof of this inequality?
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1900365
|
Help proving the product of any four consecutive integers is one less than a perfect square
Apparently this is a true statement, but I cannot figure out how to prove this. I have tried setting [imath](m)(m + 1)(m + 2)(m + 3) = (m + 4)^2 - 1 [/imath] but to no avail. Could someone point me in the right direction?
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155040
|
Prove that the product of four consecutive positive integers plus one is a perfect square
I need to prove the following, but I am not able to do it. This is not homework, nor something related to research, but rather something that came up in preparation for an exam. If [imath]n = 1 + m[/imath], where [imath]m[/imath] is the product of four consecutive positive integers, prove that [imath]n[/imath] is a perfect square. Now since [imath]m = p(p+1)(p+2)(p+3)[/imath]; [imath]p = 0, n = 1[/imath] - Perfect Square [imath]p = 1, n = 25[/imath] - Perfect Square [imath]p = 2, n = 121[/imath] - Perfect Square Is there any way to prove the above without induction? My approach was to expand [imath]m = p(p+1)(p+2)(p+3)[/imath] into a 4th degree equation, and then try proving that [imath]n = m + 1[/imath] is a perfect square, but I wasn't able to do it. Any idea if it is possible?
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150554
|
UFDs are integrally closed
Let [imath]A[/imath] be a UFD, [imath]K[/imath] its field of fractions, and [imath]f[/imath] an element of [imath]A[T][/imath] a monic polynomial. I'm trying to prove that if [imath]f[/imath] has a root [imath]\alpha \in K[/imath], then in fact [imath]\alpha \in A[/imath]. I'm trying to exploit the fact of something about irreducibility, will it help? I havent done anything with splitting fields, but this is something i can look for.
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2435585
|
If the rational number [imath]r[/imath] is also an algebraic integer, show that [imath]r[/imath] is an integer
If the rational number [imath]r[/imath] is also an algebraic integer, show that [imath]r[/imath] is an integer So [imath]r[/imath] is a rational number that is a root of a polynomial [imath]p(x)[/imath] over the integers, right? I need to show that [imath]r[/imath] is an integer. I know that for [imath]p(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_0[/imath] we have that [imath]a_0[/imath] is the product of all roots. I don't think it helps because we could have two roots [imath]a/b[/imath] and [imath]b/a[/imath] and their product is an integer, so the fact that the polynomial is over the integers won't do anything in this case. Somebody have an idea?
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1901736
|
Manifold with largest symmetry group
Is there any [imath]n[/imath]-dimensional (smooth) manifold [imath]M_{0}[/imath] which has largest symmetry group, i.e. for any [imath]n[/imath]-dimensional manifold [imath]M[/imath], [imath]Sym(M)[/imath] can be embedded in [imath]Sym(M_{0})[/imath]? Intuitively, if it exists, I think sphere or whole euclidean space will give answer (because it looks highly symmetric), but I don't know how to show or even such manifold exists. Is there any known results with more specific conditions? (specific [imath]n[/imath], compact manifold, complex manifold, Riemannian manifold (with isometry group), ...) Thanks in advance.
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587189
|
Riemannian Manifolds with [imath]n(n+1)/2[/imath] dimensional symmetry group
Given a [imath]n[/imath]-dimensional connected Riemannian manifold [imath](M,g)[/imath], its symmetry group [imath]G[/imath] can be considered as a subbundle of orthonormal frame bundle of [imath]M[/imath] (which I call [imath]F_OM[/imath]), yielding: [imath]\dim G\le \dim F_OM=\dim M +\dim O(n)={n+n(n-1)\over 2}=\frac{n(n+1)}{2}[/imath] (Here the embedding [imath]\phi:G \hookrightarrow F_OM[/imath] is defined by singling out an arbitrary point [imath]p \in M[/imath] and orthonormal frame [imath](v_1,...,v_n) \in F_{O,p} M[/imath] and defining [imath]\phi: g \mapsto (gp,g_*v_1,...,g_*v_n)[/imath]. However, it takes some work to verify that this map is indeed injective.) These considerations made me wonder which manifolds have a maximally large symmetry group, i.e. which [imath]M[/imath] do satisfy [imath]\dim G= \frac{n(n+1)}{2}[/imath]. Of, course [imath]\Bbb R^n,\Bbb S^n[/imath] and [imath]\Bbb H^n[/imath] have got this property, but are there any other exotic examples? (Some thoughts of mine on this problem: One can see that [imath]M[/imath] is homogenuous and isotropic. Each stabiliser of a point has to be isomorphic to either [imath]SO(n)[/imath] or [imath]O(n)[/imath]. In particular, [imath]M[/imath] has constant curvature, which implies that [imath]M[/imath] has to be one of [imath]\Bbb R^n,\Bbb S^n[/imath] and [imath]\Bbb H^n[/imath] if [imath]M[/imath] is complete and simply-connected.)
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1381245
|
Eigenvectors of the companion matrix
Suppose one has an Hermitian square matrix [imath]A[/imath] with [imath]p[/imath] is the characteristic polynomial [imath] p(x)= a_0 + a_1 x + \cdots + a_{n-1}x^{n-1} + x^n ~, [/imath] and define the companion matrix of [imath]p[/imath] as [imath] \tilde{A}=\begin{bmatrix} 0 & 0 & \dots & 0 & -a_0 \\ 1 & 0 & \dots & 0 & -a_1 \\ 0 & 1 & \dots & 0 & -a_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \dots & 1 & -a_{n-1} \end{bmatrix}.[/imath] For the definition of companion matrix, the eigenvalues of a the hermitian matrix [imath]A[/imath] coincide with the eigenvalues of the non-hermitian matrix [imath]\tilde{A}[/imath] (and with the roots of the polynomial [imath]p[/imath]). The question is: What about eigenvectors? Is there a relationship between the eigenvectors of the original hermitian matrix [imath]A[/imath] and the eigenvectors of the companion matrix [imath]\tilde{A}[/imath]? Note that there is no unitary transformation between the two matrices (one is hermitian and the other is not). Therefore one cannot find a unitary matrix [imath]U[/imath] such that [imath]\tilde{A}=U A U^\dagger[/imath] so that the eigenvectors transforms simply as [imath]\tilde{v}=U v[/imath]. Maybe in this case one can still define a non-unitary transformation which does the job. This is a follow up of this question.
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223577
|
Eigenspace of the companion matrix of a monic polynomial
How do I prove that the eigenspace of an [imath]n\times n[/imath] companion matrix [imath] C_p=\begin{bmatrix} 0 & 1 & 0 &\cdots & 0\\ 0 & 0 & 1 &\cdots & 0 \\ \vdots&\vdots &\vdots&\ddots&\vdots\\ 0 & 0 & 0 &\cdots &1 \\ -\alpha_0 &-\alpha_1 &-\alpha_2 &\cdots&-\alpha_{n-1} \end{bmatrix} [/imath] equals [imath]\operatorname{Span}\{v_{\lambda} \} [/imath] where [imath]v_{\lambda}[/imath] is an eigenvector of the companion matrix w.r.t. the eigenvalue [imath]\lambda[/imath]: [imath] v_{\lambda} = \begin{bmatrix} 1 \\ \lambda\\ \lambda^{2} \\ \vdots\\ \lambda^{n-1} \end{bmatrix}. [/imath]
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186337
|
[imath]\operatorname{Aut}(\mathbb Z_n)[/imath] is isomorphic to [imath]U_n[/imath].
I've tried, but I can't solve the question. Please help me prove that: [imath]\operatorname{Aut}(\mathbb Z_n)[/imath] is isomorphic to [imath]U_n[/imath].
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1983490
|
Prove [imath]Aut(G) \cong \mathbb{Z}_n^{\times}[/imath] where [imath]G=(\mathbb{Z}_n , +)[/imath].
Prove [imath]Aut(G) \cong \mathbb{Z}_n^{\times}[/imath] where [imath]G=(\mathbb{Z}_n , +)[/imath]. [imath]Aut(G)[/imath] denotes the set of all automorphisms with respect to composition. [imath]\mathbb{Z}_n^{\times}[/imath] denotes the set of all invertible elements of [imath]\mathbb{Z}_n[/imath] with respect to multiplication. Please do not offer a full proof as an answer (where is the fun in that?). I am interested only in finding a specific map that works. My thoughts: Define [imath]\psi : Aut(G) \rightarrow \mathbb{Z}_n^{\times}[/imath] as [imath]\psi ([m])=\phi_m([x])=[mx][/imath] (a function which is defined in a previous exercise). I am not confident that this works. Should I continue looking for isomorphisms in this direction? Is the opposite direction more fruitful?
|
1902679
|
Hypothesis Test of Calculating P-Value for Average Height
Question: In 2003-2006 the average height of an American aged 20-29 was reported to be 67.25. Many believe that college students are shorter than average. A random sample of students produced the following data: n=140, mean = 65.39, s = 4.64 Type in the test statistic rounded to TWO decimal places Solution Attempt: I attempted to tackle this problem by choosing its sampling distribution as 1 sample T-test. I'm given no probabilities, so I don't think it's a Z-test. Would the setup for the formula for solving for a t-value then be: [imath]t = \frac{65.39 - 67.25}{\frac{4.64}{\sqrt{140}}} = -4.74[/imath] Taking this value then and putting into R-Studio as [imath]p value = pt(-4.74, df = 139, lower.tail = FALSE) = .999999[/imath] This means if I choose a 95% confidence interval ... I fail to reject right? Because .9999 >> ..05
|
1902619
|
The Average Height of an American is (Hypothesis Test)
Problem The average height of an American was reported to be 70. Many believe that college students are shorter than average. A random sample of students produced the following data: [imath]n=140[/imath], [imath]mean = 65.39[/imath], [imath]s = 4.64[/imath] Find the test statistics Steps Taken to Solve I decided that this type of problem was a 1 sample T-test. These are my steps to the Hypothesis Test. Null Hypothesis = No difference, so [imath]M = M_o = 0[/imath] Alternative Hypothesis = [imath]M < 50[/imath] [imath]t = \frac{50 - 0}{\frac{4.64}{\sqrt{140}}} = 166.7467[/imath] Is this approach correct so far?
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1903507
|
How to formalize "It doesn't matter how places the brackets"
Let [imath]X[/imath] be a set and [imath]X\times X\to X: (a, b)\mapsto a\cdot b[/imath] an associative operation on [imath]X[/imath]. Now one can prove by induction that it doesn't matter how one places the brackets in a product [imath]x_1\cdot x_2\cdot\text{ }\dots\text{ } \cdot x_n[/imath], the product always evaluates to the same value. This result justifies the notation of a product [imath]x_1\cdot x_2\cdot\text{ }\dots\text{ } \cdot x_n[/imath] without any brackets. But how can one formalize it?
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709196
|
Definition of General Associativity for binary operations
Let's say we are talking about addition defined in the real numbers. Then, by induction we define [imath]\sum_{i=0}^{0}a_i=a_0[/imath] and [imath]\sum_{i=0}^{n}a_i=\sum_{i=0}^{n-1}a_i+a_n[/imath] for [imath]n> 1.\:[/imath] Now, how do you define general associativity? I know that this has something to do with the fact that [imath]\sum_{i=0}^{n}=\sum_{i=0}^{k}+\sum_{i=k+1}^{n}[/imath] for [imath]0\leq k<n[/imath], being [imath]\sum_{i=k}^{n}a_i=\sum_{i=0}^{n-k}a_{i+k}[/imath] by definition. But the thing is that this doesn't quite define the notion of different ways of arranging brackets, like for example [imath](a_0+(a_1+a_2))+((a_3+a_4)+(a_5+a_6))[/imath]. So my question is how they formally define this process of bracketing. Think of the case when someone just tell you to prove the general associativity for the real numbers. How do they actually define this property in order to prove it? is it necessary to have one? Look at for example this proof, specifically at the point where professor M. Zuker says: "Let us now assume that any bracketing of [imath]a_1, a_2,...,a_k[/imath] equals the standard form for [imath]1\leq k\leq n-1[/imath], where [imath]n>3[/imath]". But, then again my question: what is the definition of bracketing? is it actually necessary to have a definition of bracketing or this proof works whatever the definition of bracketing is? Also I have found this paper by William P. Wardlow - A generalized general associative law- that contains different proofs of this general associativity law. The first one, the one that he suggests as his favorite, is done by Nathan Jacobson in his book "Lectures of Abstract Algebra" Vol. 1 page 20. Looking at this proof there is one point where he says "Consider now any product associated with [imath](a_1, a_2,..., a_n)[/imath]...", which means "any bracketing associated with...". Then again the same question. I hope you understand my point. If not, please fell free of asking anything related to my question. Edit: For clarification let's say we are talking about sum in the real numbers. Then, 1.- [imath](...(((a_0+a_1)+a_2)+a_3)...+a_n)[/imath] is a representation of the formal definition by recursion, meaning [imath]\sum_{i=0}^{n}a_i[/imath] just as defined above: [imath]\sum_{i=0}^{0}a_i=a_0[/imath] and [imath]\sum_{i=0}^{n}a_i=\sum_{i=0}^{n-1}a_i+a_n[/imath] for [imath]n> 1[/imath]. 2.- What is the formal definition of [imath]a_0+(a_1+(a_2+...+(a_{n-1}+a_n)...)[/imath] ? 3.- What is the formal definition of something like [imath](a_0+((a_1+a_2)+a_3))+(((a_4+a_5)+a_6)+....+(a_{n-1}+a_n))[/imath]?
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1903580
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[imath]f[/imath] is continuous [imath]\iff[/imath] [imath]\forall S \subseteq X, f(\text{cl}(S)) \subseteq \text{cl}(f(S))[/imath]
Let [imath]X,Y[/imath] be topological spaces and [imath]f:X \to Y[/imath]. Prove that [imath]f[/imath] is continuous [imath]\iff[/imath] [imath]\forall S \subseteq X, f(\text{cl}(S)) \subseteq \text{cl}(f(S))[/imath] Here's my attempt: [imath](\Rightarrow)[/imath] Suppose [imath]f[/imath] is continous. Since [imath]\text{cl}(f(S))[/imath] is closed, then [imath]f^{-1} ( \text{cl}(f(S)))[/imath] is closed. Since [imath]f(S) \subseteq \text{cl} (f(S))[/imath], then [imath]S \subseteq f^{-1} ( \text{cl}(f(S)))[/imath]. A subset of a closed set implies the closure is also a subset. Hence [imath]\text{cl}(S) \subseteq f^{-1} ( \text{cl}(f(S)))[/imath], which means [imath]f(\text{cl}(S)) \subseteq \text{cl}(f(S))[/imath]. [imath](\Leftarrow)[/imath] Suppose [imath]f(\text{cl}(S)) \subseteq \text{cl}(f(S)) \forall S \subseteq X[/imath] [imath]\text{cl}(f(S))[/imath] is closed, then if I prove that [imath]f^{-1}\text{cl}(f(S))[/imath] I would have [imath]f[/imath] continuous, but I can't do it. If the condition were [imath]\text{cl} f^{-1}(A)\subseteq f^{-1}(\text{cl}A）[/imath] as I found in other posts, I can do it. Maybe this condition I have is mispelled? Thanks
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594979
|
for any [imath]A\subseteq X[/imath], [imath]f(\overline{A})\subseteq\overline{f(A)}[/imath] , if and only if [imath]f: X \to Y[/imath] is continuous.
Let [imath]X[/imath] and [imath]Y[/imath] be two topological spaces. Prove that for any [imath]A\subseteq X[/imath], [imath]f(\overline{A})\subseteq\overline{f(A)}[/imath] , if and only if [imath]f: X \to Y[/imath] is continuous. I am stuck on the converse. Suppose [imath]f: X \to Y[/imath] is continuous. Then for every closed set C in Y, [imath]f^{-1}(C)[/imath] is closed in X. WTS for any [imath]A\subseteq X[/imath], [imath]f(\overline{A})\subseteq\overline{f(A)}[/imath]. Since [imath]\overline{f(A)}[/imath] is a closed set in Y, [imath]f^{-1} \circ \overline{f(A)}[/imath] = [imath] \overline{ f^{-1} \circ \overline{f(A)} }[/imath] Also, [imath]f(\overline{A}) \subseteq \overline{ f(\overline{A})}[/imath]
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1903674
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Expected value inequality for a monotone function
A "simple" expected value inequality confuse me. If [imath]f(x)[/imath] is a monotone increasing function defined on the interval [imath][a,b][/imath], and [imath]c \in [a,b][/imath], is it possible to prove the following inequality? [imath]\mathrm E\left(\sum_{t=a}^{c}f(t)\right)\leq \mathrm E \left(\sum_{t=a}^{b}f(t)\right)[/imath]
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1903699
|
Expectation inequality for a set and its subset.
Given that a function [imath]g(x)[/imath] is a monotone increasing function. Its domain is an interval [c,d]. There are two sets [imath]a_j \in A, j= 1,...,J[/imath] and [imath]b_k \in B,k=1,...,K[/imath] on this interval, which satisfy [imath]c_z \in C = B[/imath]\A and [imath]c_z \geq a_j[/imath] for all [imath]a_j \in A[/imath]. For example, the interval [imath][c,d]=[0,8][/imath],[imath]A=\{0,1,2,3,4,5\}[/imath] and [imath]B=\{0,1,2,3,4,5,6,7\}[/imath]. In this case, does the following inequality hold? If yes, could you give me some hints to prove that? [imath]\frac{1}{K} \sum_{k=1}^{K}g(b_k) \geq \frac{1}{J} \sum_{j=1}^{J}g(a_j)[/imath]
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1898807
|
Find all real solutions to [imath][x]+[2x]+[4x]+[8x]+[16x]+[32x]=12345[/imath]
Find all real solutions of the equation: [imath][x]+[2x]+[4x]+[8x]+[16x]+[32x]=12345[/imath] Where [imath][.][/imath] is greatest integer fuction.
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1851042
|
Find the number of solutions to [imath] \lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \ldots + \lfloor 32x \rfloor =12345[/imath]
Find the number of solutions of the equation [imath]\lfloor x \rfloor + \lfloor 2x \rfloor + \lfloor 4x \rfloor + \lfloor 8x \rfloor + \lfloor 16x \rfloor + \lfloor 32x \rfloor =12345,[/imath] where [imath]\lfloor\,\cdot\,\rfloor[/imath] represents the floor function. My work: I use the fact that [imath]\lfloor nx \rfloor =\sum_{k=0}^{n-1} \left\lfloor x +\frac kn \right\rfloor.[/imath] So the equation becomes [imath]\lfloor x \rfloor +\sum_{k=0}^{1} \left\lfloor x +\frac k2 \right\rfloor +\sum_{k=0}^{3} \left\lfloor x +\frac k4 \right\rfloor +\sum_{k=0}^{7} \left\lfloor x +\frac k8 \right\rfloor \\ \qquad {}+\sum_{k=0}^{15} \left\lfloor x +\frac k{16} \right\rfloor +\sum_{k=0}^{31} \left\lfloor x+\frac k{32} \right\rfloor \\ = \lfloor x \rfloor + \left\lfloor x+\frac 12 \right\rfloor + \left\lfloor x+\frac 64 \right\rfloor + \left\lfloor x+\frac{28}{8} \right\rfloor \\ \qquad {}+ \left\lfloor x+\frac{120}{16} \right\rfloor + \left\lfloor x+\frac{496}{32} \right\rfloor[/imath] What should I do next?
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1904683
|
Probability of having an even number of a specific face value of a die
Here is the question: Given a fair die rolled [imath]n[/imath] times, what is the probability of having an even number of [imath]6[/imath]s? My approach: [imath]\Omega = \{(x_{1}, x_{2}, ..., x_{n}) :1 \le x_{i} \le 6\}[/imath] So [imath]|\Omega| = 6^{n}[/imath] Since the probability is space is a finite uniform one, [imath]P(A) = \frac{|A|}{|\Omega|}[/imath] Let [imath]A_{i}[/imath] be the event that there are exactly [imath]i[/imath] sixes in the [imath]n[/imath]-tuple. Then, [imath]A = \bigcup\limits_{i=0}^{\lceil{(n-1)/2}\rceil} A_{2*i}[/imath] Since these events are disjoint, [imath]P(A) = P(\bigcup\limits_{i=0}^{\lceil{(n-1)/2}\rceil} A_{2*i}) = \sum\limits_{i=0}^{\lceil{(n-1)/2}\rceil} P(A_{2*i})[/imath] Now, [imath]|A_{i}| = \binom{n}{i} * 5^{n-i}[/imath]. Hence, [imath]P(A) = \frac{1}{6^n} \sum\limits_{i=0}^{\lceil{(n-1)/2}\rceil} \binom{n}{2i} * 5^{n-2i}[/imath] As a side note, when I fed this sum into wolfram alpha engine, it couldn't compute it. The book however says the anwer is [imath]\frac{1}{2}(1 + (\frac{2}{3})^{n})[/imath] Any suggestions?
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1648601
|
Probability of an even number of sixes
We throw a fair die [imath]n[/imath] times, show that the probability that there are an even number of sixes is [imath]\frac{1}{2}[1+(\frac{2}{3})^n][/imath]. For the purpose of this question, 0 is even. I tried doing this problem with induction, but I have problem with induction so I was wondering if my solution was correct: The base case: For [imath]n=0[/imath], our formula gives us [imath]\frac{1}{2}[1+(\frac{2}{3})^0] =1[/imath]. This is true, because if we throw the die zero times, we always get zero sixes. Suppose it's true for [imath]n=k[/imath]. Then the odds of an even number of sixes is [imath]\frac{1}{2}[1+(\frac{2}{3})^n][/imath], and thus the odds of an odd number of sixes is [imath]1 - \frac{1}{2}[1+(\frac{2}{3})^n][/imath]. For [imath]n=k+1[/imath], there are two ways the number of sixes are even: a. The number of sixes for [imath]n=k[/imath] was even, and we do not throw a six for [imath]n=k+1[/imath]: [imath] \frac{5}{6} \cdot \frac{1}{2}[1+(\frac{2}{3})^n][/imath] b. The number of sixes for [imath]n=k[/imath] was odd, and we throw a six for [imath]n=k+1[/imath]: [imath]\frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])[/imath] So the probability [imath]p[/imath] for an even number of sixes at [imath]n=k+1[/imath] is [imath] \frac{5}{6} \cdot \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])[/imath] I have two questions How do I get from [imath] \frac{5}{6} \cdot \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])[/imath] to [imath]\frac{1}{2}[1+(\frac{2}{3})^n][/imath]? I seem to have done something wrong, I can't get the algebra correct, I get [imath]p = \frac{1}{3}[1+(\frac{2}{3})^n] + \dfrac{1}{6}[/imath] Other than that, is my use of induction correct? Is it rigorous enough to prove the formula?
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369735
|
Does a symmetric matrix with main diagonal zero is classified into a separate type of its own? And does it have a particular name?
I have a symmetric matrix as shown below [imath]\begin{pmatrix} 0&2&1&4&3 \\ 2&0&1&2&1 \\ 1&1&0&3&2 \\4&2&3&0&1 \\ 3&1&2&1&0\end{pmatrix}[/imath] Does this matrix belong to a particular type? I am CS student and not familiar with types of matrices. I am researching to know the particular matrix type since I have a huge collection of matrices similar to this one. By knowing the type of matrix, maybe I can go through its properties and work around easier ways to process data efficiently. I am working on a research project in Data Mining. Please help. P.S.: Only the diagonal elements are zero. Non diagonal elements are positive.
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79779
|
Properties of zero-diagonal symmetric matrices
I'm interested in the properties of zero-diagonal symmetric (or hermitian) matrices, also known as hollow symmetric (or hermitian) matrices. The only thing I can come up with is that it cannot be positive definite (if it's not the zero matrix): The Cholesky decomposition provides for positive definite matrices [imath]A[/imath] the existence of a lower triangular [imath]L[/imath] with [imath]A=LL^*[/imath]. However the diagonal of [imath]LL^*[/imath] is the inner product of each of the rows of [imath]L[/imath] with itself. Since the diagonal of [imath]A[/imath] consists of zeros, so [imath]L[/imath] (and thus [imath]A[/imath]) must be the zero matrix. The sorts of questions that interest me are: which symmetric matrices can be transformed orthogonally into a zero-diagonal matrix? what can we say about the eigen-values of a zero-diagonal symmetric matrix?. any other interesting known properties??
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1906276
|
Is there a vector space with a non-trivial zero vector?
Usually, the zero vector in a vector space can be found quite easily. Often it's just the intuitive sense of 'zero' in the set that the vector space is defined over. For example, if we consider the vector space of real continuous functions in an interval, then the zero vector is simply the zero function. If we consider the vector space of complex numbers, the zero vector is simply [imath]0+0i[/imath]. What is an example of a vector space where the zero vector is non-trivial? Obviously the idea of non-triviality is somewhat subjective, but that is very much unavoidable for a question of this nature.
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1714461
|
Zero vector of a vector space
I know that every vector space needs to contain a zero vector. But all the vector spaces I've seen have the zero vector actually being zero (e.g. [imath]\mathbf{0}=\langle0,0,\ldots,0\rangle[/imath]). Can't the "zero vector" not involve zero, as long as it acts as the additive identity? If that's the case then are there any graphical representations of a vector space that does not contain the origin?
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1904113
|
Limit [imath]c^n n!/n^n[/imath] as [imath]n[/imath] goes to infinity
Let [imath]c>0[/imath] be a real number. I would like to study the convergence of [imath]a_n := c^n n!/n^n[/imath], when [imath]n \to \infty[/imath], in function of [imath]c[/imath]. I know (from this question) that [imath]n!>(n/e)^n[/imath], so that [imath]c^n n!/n^n>1[/imath] for [imath]c ≥e[/imath]. But this doesn't imply that the sequence goes to infinity. And I'm not sure what to do for [imath]c<e[/imath]. I tried usual tests (D'Alembert...), without any success. I would like to avoid using Stirling approximation. Thank you for your help!
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1971533
|
Limit of a Sequence (limit at infinity)
Who can help me to find the following limit? [imath]\lim _{n\to \infty}\frac{e^n\times n!}{n^n}=?[/imath]
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1906664
|
How are random variables random?
Given a probability space [imath](\Omega, \mathcal F, \mu)[/imath], a random variable is a function [imath]X:\Omega\to\Bbb R[/imath]. That's the definition I was given of what a random variable is. I don't see what's random about them, or how they capture the idea of the 'uncertainty' of some events. Could anyone provide some intuition about random variables?
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687467
|
Why is the function [imath]\Omega\rightarrow\mathbb{R}[/imath] called a random variable?
I do not understand the relation of a normal variable "x", which is to me just a placeholder for an element of a set, and a random variable, which is a mapping from the set of all possible events to [imath]\mathbb{R}[/imath]. To make the question more concrete, some parts I am struggling with: Normal functions with variables can be evaluated, e.g. [imath]f(x)=2x[/imath] will plot to a line, can I do this with a function of a random variable? Furthermore the randomness part seems to be missing, why is it not neccessary to define how the random variable obtains its random value? Edit: now that I feel I understood the definition, I cannot say what exacly what was missing for understanding. Nevertheless reading the following links, the comments and answer here, did the trick. The concept of random variable Definition of a real-valued random variable Understanding the definition of a random variable Random Variable Definition of random variable, Borel [imath]\sigma[/imath]-algebra What exactly is a random variable?
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1906409
|
Let [imath]A,B[/imath] and [imath]U[/imath] be sets so that [imath]A\subseteq U[/imath] and [imath]B \subseteq U[/imath]. Prove that [imath]A \subseteq B[/imath] iff [imath](U\setminus B) \subseteq (U\setminus A).[/imath]
Let [imath]A,B[/imath] and [imath]U[/imath] be sets so that [imath]A\subseteq U[/imath] and [imath]B \subseteq U[/imath]. Prove that [imath]A \subseteq B[/imath] iff [imath](U\setminus B) \subseteq (U\setminus A).[/imath] I've this question as my graded assignment. But for the entire module, I am completely lost and would really appreciate if anybody can help me with this question. Thank you so much!
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1905506
|
Let [imath]A,B[/imath] and [imath]U[/imath] be sets so that [imath]A\subseteq U[/imath] and [imath]B\subseteq U[/imath]. Prove that [imath]A\subseteq B[/imath] iff [imath](U\setminus B)\subseteq(U\setminus A)[/imath].
Let [imath]A,B[/imath] and [imath]U[/imath] be sets so that [imath]A\subseteq U[/imath] and [imath]B\subseteq U[/imath]. Prove that [imath]A\subseteq B[/imath] iff [imath](U\setminus B)\subseteq(U\setminus A)[/imath]. The forward implication is easy to prove but i got stuck at the backward implication, i.e. [imath](U\setminus B)\subseteq(U\setminus A)\Rightarrow A\subseteq B[/imath]. Can someone help me out with it? Thank you.
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1897773
|
Utility to demand function
Suppose that Sally’s preferences over baskets containing food (good x), and clothing (good y), are described by the utility function U(x,y)=√x+y Sally’s corresponding marginal utilities are, Ux=[imath]1\over2√x[/imath] and Uy=[imath]1[/imath] Use Px to represent the price of food, Py to represent the price of clothing, and I to represent Sally’s income. Question 1: Find Sally's food demand function, and Sally's clothing demand function. For the purposes of this question you should assume that I/Py greater than or equal Py/4Px. Even by using the I=Py(y)+Px(x) formula, I tried using [imath]MUx\over MUy[/imath]=[imath]Px\over Py[/imath] and letting one of the price be 1. But I am unable to substitute into the income function as I could find y value.
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1896159
|
Microeconomics (quasi-linear utility function)
Suppose that Sally’s preferences over baskets containing food (good x), and clothing (good y), are described by the utility function [imath]U(x,y)=\sqrt{x}+y[/imath]. Sally’s corresponding marginal utilities are, [imath]U^\prime_x=\frac{1}{2\sqrt{x}}[/imath] and [imath]U^\prime_y=1[/imath]. Use Px to represent the price of food, Py to represent the price of clothing, and I to represent Sally’s income. Question 1: Find Sally's food demand function, and Sally's clothing demand function. For the purposes of this question you should assume that I/Py greater than or equal Py/4Px. Clothing demand function : y= I/py - Py/4Px Food demand function : x = P^2y/4P^2x After deriving demand function for both Clothing and Food, the following question confused me. Question 3: Now assume that I/Py less than Py/4Px. Find Sally's food demand function, and Sally's clothing demand function. How do i get about doing this following question and getting the NEW Demand function for Food and Clothing? I do know it might possible be a corner solution.. But how do i explain it in terms of the function as i am not given the price of food and price of clothing in numerical form?
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1905056
|
If [imath]3\mid a[/imath], then [imath]3\mid a^2[/imath]
I want to prove for all integers [imath]a[/imath] that if [imath]3\mid a[/imath] then [imath]3\mid a^2[/imath]. Suppose that [imath]a \in \mathbb{Z}[/imath] and [imath]3\mid a[/imath]. Then [imath]a = 3k[/imath] for any integer k. Hence [imath]a^2 = (3k)(3k) = 9k^2 = 3(3k^2)[/imath] Since [imath]k[/imath] is a integer, [imath]3k^2[/imath] is an integer. Thus [imath]a^2 = 3b[/imath] for some integer [imath]b[/imath]. Therefore [imath]3\mid a^2[/imath]. Is this proof correct?
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310758
|
Prove that for all integers [imath] a [/imath] and [imath] b [/imath], if [imath] a [/imath] divides [imath] b [/imath], then [imath] a^{2} [/imath] divides [imath] b^{2} [/imath].
I just need to know that if [imath] a [/imath] divides [imath] b [/imath], where [imath] a [/imath] and [imath] b [/imath] are integers, does [imath] a^{2} [/imath] divide [imath] b^{2} [/imath]?
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1906852
|
Proving [imath]-(-z) = z[/imath]
I've managed to prove this: [imath]0 = (-1+1)z = -1\cdot z + z \rightarrow [/imath] [imath]-1\cdot z = -z[/imath] but how to prove that [imath]-(-z) = z[/imath]?
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1904039
|
Proof Verification : Prove -(-a)=a using only ordered field axioms
I need to prove for all real numbers [imath]a[/imath], [imath]-(-a) = a[/imath] using only the following axioms: Thanks to many members of the Mathematics Stackexchange Community, I have the following proof worked out: Theorem: The Additive Inverse Identity is Unique [imath] ( \forall a,b,c \in \mathbb{R} )(a+b=0) \land (a+c=0) \\ ( \forall a,b,c \in \mathbb{R} )(a+b=a+c) \\ ( \forall a,b,c \in \mathbb{R} )(b=c) \\ \text{QED} \\ [/imath] Theorem [imath]\textbf{a} \cdot \textbf{0 = 0}[/imath] [imath] \begin{align} a \cdot 0 &= a \cdot 0 \\ a \cdot (0 + 0) &= a \cdot 0 \\ a \cdot 0 + a \cdot 0 &= a \cdot 0 \\ a \cdot 0 + a \cdot 0 + (-a) \cdot 0 &= a \cdot 0 + (-a) \cdot 0 \\ a \cdot 0 &= 0 \\ &\text{QED} \\ \end{align} [/imath] Theorem: [imath]-\textbf{1} \cdot \textbf{(a) = (}-\textbf{a)}[/imath] [imath] \begin{align} a \cdot 0 &= 0 \\ a \cdot \left[ 1 + (-1) \right] &= a + (-a) \\ 1 \cdot a + (-1) \cdot a &= a + (-a) \\ -1 \cdot (a) &= (-a) \\ &\text{QED} \\ \end{align} [/imath] Theorem: [imath]-\textbf{(}-\textbf{a) = a}[/imath] [imath] \begin{align} 0 &= 0 \\ -a \cdot 0 &= 0 \\ -a \cdot \left[ 1 + (-1) \right] &= 0 \\ -a \cdot 1 + -a \cdot (-1) &= 0 \\ -a + \left[ -(-a) \right] &= 0 \\ -(-a) &= a \\ \text{QED} \\ \end{align} [/imath] Does everything look alright? Have I missed anything? Also--why is it necessary to show that the additive inverse is unique? Many thanks in advance.
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1906473
|
Solve [imath]\frac{xdx-ydy}{xdy-ydx}=\sqrt{\frac{1+x^2-y^2}{x^2-y^2}}[/imath]
Solve [imath]\dfrac{xdx-ydy}{xdy-ydx}=\sqrt{\dfrac{1+x^2-y^2}{x^2-y^2}}[/imath] I tried doing this- [imath]\sqrt{x^2-y^2}d(x^2-y^2)=x^2\sqrt{1+x^2-y^2}d(\frac{x}{y})[/imath] but it does not get better from here. I even tried putting [imath]\sqrt{x^2-y^2}[/imath] as [imath]u[/imath] but that does not simplify either.
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1015021
|
solve[imath]\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}[/imath]
solve the differential equation. [imath]\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}[/imath] The question is from IIT entrance exam paper. I have tried substituting [imath]x^2=t\ and \ y^2=u[/imath] but was not a worth try. Thanks in advance.
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1906878
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Prove [imath]\left|\int_a^b \phi(t) \ dt \right|\le \int_a^b |\phi(t)| \ dt[/imath]
Let [imath]\phi(t) = \phi_1(t)+i\phi_2(t)[/imath] be a complex function in a real interval [imath]t[/imath] In order to prove: [imath]\left|\int_a^b \phi(t) \ dt \right|\le \int_a^b |\phi(t)| \ dt[/imath] I tried: [imath]\int_a^b \phi(t) \ dt = \int_a^b \phi_1(t) + i\phi_2(t) \ dt = \int_a^b \phi_1(t) \ dt + \int_a^b \phi_2(t) \ dt \rightarrow[/imath] [imath]\left|\int_a^b \phi(t) \ dt \right| = \sqrt{\left(\int_a^b \phi_1(t) \ dt\right)^2+\left(\int_a^b \phi_2(t) \ dt\right)^2}\le \sqrt{\left(\int_a^b \phi_1(t) \ dt\right)^2}+\sqrt{\left(\int_a^b \phi_2(t) \ dt\right)^2} = [/imath] [imath]\left|\int_a^b \phi_1(t) \ dt\right|+\left|\int_a^b \phi_2(t) \ dt\right|\le \int_ a^b |\phi_1(t)| \ dt + \int_a^b |\phi_2(t)| \ dt = \int_a^b |\phi_1(t)|+|\phi_2(t)| \ dt[/imath] But how to arrive at [imath]\int_a^b |\phi(t)| \ dt[/imath]?
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1772406
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Prove [imath]\left| \int_a^b f(t) dt \right| \leq \int_a^b \left| f(t) \right| dt[/imath]
I've been given a proof that shows the following: If [imath]f:[a,b]\to \mathbb C[/imath] is a continuous function and [imath]f(t)=u(t)+iv(t)[/imath] then [imath]\left| \int_a^b f(t) dt \right| \leq \int_a^b \left| f(t) \right| dt[/imath] The proof begins by letting [imath]\theta[/imath] be the principle argument of the complex number [imath]\int_a^b f(t)dt[/imath] and there is one step in the proof I don't understand; can anyone explain to me why [imath]\int_a^b e^{-i\theta}f(t) dt=\Re\bigg(\int_a^be^{-i\theta}f(t)dt\bigg)[/imath] Would this not imply that the imaginary part of the left hand side is equal to [imath]0[/imath]? I'm not sure why the LHS and RHS would be equal here.
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