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6695 | Ways to evaluate [imath]\int \sec \theta \, \mathrm d \theta[/imath]
The standard approach for showing [imath]\int \sec \theta \, \mathrm d \theta = \ln|\sec \theta + \tan \theta| + C[/imath] is to multiply by [imath]\frac{\sec \theta + \tan \theta}{\sec \theta + \tan \theta}[/imath] and then do a substitution with [imath]u = \sec \theta + \tan \theta[/imath]. I like the fact that this trick leads to a fast and clean derivation, but I also find it unsatisfying: It's not very intuitive, nor does it seem to have applicability to any integration problem other than [imath]\int \csc \theta \,\mathrm d \theta[/imath]. Does anyone know of another way to evaluate [imath]\int \sec \theta \, \mathrm d \theta[/imath]? | 717528 | Why is [imath]\int \sec x\,dx[/imath] equal to [imath]\ln | \sec x + \tan x |[/imath]?
Everyone knows that [imath]\int \sec x\,dx = \ln | \sec x + \tan x |[/imath]? But how to reach it through a conscious deduction, through a clear and objective way? |
1057 | Group With an Endomorphism That is "Almost" Abelian is Abelian.
Suppose a finite group has the property that for every [imath]x, y[/imath], it follows that \begin{equation*} (xy)^3 = x^3 y^3. \end{equation*} How do you prove that it is abelian? Edit: I recall that the correct exercise needed in addition that the order of the group is not divisible by 3. | 1782335 | Abelian finite group
This is a (maybe be simple) problem from Group Theory, but being a beginner, I am unable to take even a first step forward. Let [imath]G[/imath] be a finite group whose order is not divisible by [imath]3[/imath].Suppose that [imath](ab)^3=a^3b^3\ \ [/imath] [imath]\forall\ \ a,b\in G[/imath]. I am to prove that [imath]G[/imath] must be an abelian group. Please help. |
44406 | How do I get the square root of a complex number?
If I'm given a complex number (say [imath]9 + 4i[/imath]), how do I calculate its square root? | 970081 | Find square roots of [imath]8 - 15i[/imath]
Find the square roots of: [imath]8-15i.[/imath] Could I get some working out to solve it? Also what are different methods of doing it? |
40149 | Intuition of the meaning of homology groups
I am studying homology groups and I am looking to try and develop, if possible, a little more intuition about what they actually mean. I've only been studying homology for a short while, so if possible I would prefer it if this could be kept relatively simple, but I imagine it is entirely possible there is no real answer to my query anyway. As I said above, I want to gain a little deeper understanding of what the n-th homology group actually means: I can happily calculate away using Mayer-Vietoris but it doesn't really give me a great deal of intuition about what the n-th homology group actually means. For example, with homotopy groups, the fundamental group is in some sense a description of how loops behave on the object in question, and it is obvious to me why that is what it is for say, the torus or the circle. However, I have no idea what, if anything, I am actually saying about a triangulable object when I talk about it having 0-th homology group this or 1st homology group that. The best I have been able to find online or in my limited book selection is the brief description "intuitively, the zeroth homology group counts how many disjoint pieces make up the shape and gives that many copies of [imath]\Bbb Z[/imath], while the other homology groups count different types of holes". What 'different types of holes' are there, roughly speaking? I appreciate that it may often be completely non-obvious what the low-order homology groups are for some complicated construction, but perhaps in simpler examples it might be more explicable. Are there (simple) cases where I could say, just from looking something like e.g. the torus, what its zero-th or first or second etc. homology group was based on the nature of the object? I guess in the zero-th case it is, as my source (http://teamikaria.com/hddb/wiki/Homology_groups) above says, related to the number of disjoint pieces. Can we delve deeper than this for the other homology groups? Any book/website suggestions would be welcomed (preferably websites as I am nowhere near a library!) - I have Hatcher but not a great deal else, and I haven't gleaned as much as I wish to from that alone. Of course I know that there is a great deal we don't know about homology groups even today, so I don't expect some magical all-encompassing answer, but any thoughts you could provide would be appreciated. I hope this question is appropriate for SE Mathematics, apologies if not! -M | 2292374 | Understanding the geometric meaning of homology and cohomology group?
I'm learning some algebraic topology books which introduce me to singular cubial homology and cohomology ( not simplicial as usual so sometimes it's hard for me . I think the difference between these theory is triangulation ) . I will go straight to main problem , if you have studied Poicare duality theorem you may know the way to define a orientable manifold through group homology . In higher dimension of manifold , it can't be imagined . My question is the intuition of homology and cohomology group in higher dimension . How these groups affect to orientation or somethings like this . For instance , rank of [imath]H_{0}[/imath] is the number of component , [imath]H_{1}[/imath] is abelianization of fundamental group . I try to find some ideas on Internet but the only I get is like " the number of [imath]n[/imath] dimension holes the space has " . I can't happily calculate when I don't really understand what they actually mean . |
283 | Is [imath]0[/imath] a natural number?
Is there a consensus in the mathematical community, or some accepted authority, to determine whether zero should be classified as a natural number? It seems as though formerly [imath]0[/imath] was considered in the set of natural numbers, but now it seems more common to see definitions saying that the natural numbers are precisely the positive integers. | 468587 | Isn't zero natural enough to be included in the set of natural numbers?
I always define [imath]\mathbb{N}[/imath] to include [imath]0[/imath] but some authors don't. Since the elements of [imath]\mathbb{N}[/imath] are used for counting, shouldn't [imath]0\in\mathbb{N}[/imath]? [imath]0[/imath] is the number of cows in a classroom for example. Moreover, [imath]0\in\mathbb{N}[/imath] is a consequence of the Peano axioms and in fact the digit [imath]0[/imath] is used in writing integers ([imath]10[/imath], [imath]205[/imath]) so why on earth would anyone define [imath]\mathbb{N}=\{1;2;3;\cdots\}[/imath] in the twentieth and 21th century? Is he a babylonian? |
99913 | Easy way to show that [imath]\mathbb{Z}[\sqrt[3]{2}][/imath] is the ring of integers of [imath]\mathbb{Q}[\sqrt[3]{2}][/imath]
This seems to be one of those tricky examples. I only know one proof which is quite complicated and follows by localizing [imath]\mathbb{Z}[\sqrt[3]{2}][/imath] at different primes and then showing it's a DVR. Does anyone know any simple quick proof? | 979906 | Find the ring of algebraic integers.
Find the ring of algebraic integers in [imath]K=\mathbb Q(\sqrt[3]{2})[/imath]. So, I know that [imath]K=\{a+b\sqrt[3]{2}+c\sqrt[3]{2}^2 \mid a,b,c \in \mathbb Q\}[/imath]. My professor has done very little on this topic. I know that an algebraic integer is the root of a monic polynomial with integer coefficients. I have no clue on where to start with this question. Any help is much appreciated! |
75246 | Surjectivity of Composition of Surjective Functions
Suppose we have two functions, [imath]f:X\rightarrow Y[/imath] and [imath]g:Y\rightarrow Z[/imath]. If both of these functions are onto, how can we show that [imath]g\circ f:X\rightarrow Z[/imath] is also onto? | 1141670 | How to prove that the composition of two surjective functions is surjective
I know that the map [imath]f:A\to B[/imath] is a surjective function (onto) if for all [imath]b[/imath] in [imath]B[/imath], there exists an [imath]a[/imath] in [imath]A[/imath] such that [imath]f(a)=b[/imath] But I am having trouble getting started with this proof since it involves the composition of two surjective functions. |
71851 | Prove that any rational can be expressed in the form [imath]\sum\limits_{k=1}^n{\frac{1}{a_k}}[/imath], [imath]a_k\in\mathbb N^*[/imath]
Let [imath]x\in\mathbb{Q}[/imath] with [imath]x>0[/imath]. Prove that we can find [imath]n\in\mathbb{N}^*[/imath] and distinct [imath]a_1,...,a_n \in \mathbb{N}^*[/imath] such that [imath]x=\sum_{k=1}^n{\frac{1}{a_k}}[/imath] | 1768472 | Let [imath]r[/imath] be a rational no. with [imath]0. Then r can be expressed as a sum of reciprocals of finitely many distinct positive integers[/imath]
Let [imath]r[/imath] be a rational no. with [imath]0<r<1[/imath]. Then [imath]r[/imath] can be expressed as a sum of reciprocals of finitely many distinct positive integers . Show that this is true even for [imath]r>1[/imath]. If there is a particular theorem or an article on this please share.. |
5119 | Example where union of increasing sigma algebras is not a sigma algebra
If [imath]\mathcal{F}_1 \subset \mathcal{F}_2 \subset \dotsb[/imath] are sigma algebras, what is wrong with claiming that [imath]\cup_i\mathcal{F}_i[/imath] is a sigma algebra? It seems closed under complement since for all [imath]x[/imath] in the union, [imath]x[/imath] has to belong to some [imath]\mathcal{F}_i[/imath], and so must its complement. It seems closed under countable union, since for any countable unions of [imath]x_i[/imath] within it, each of the [imath]x_i[/imath] must be in some [imath]\mathcal{F}_j[/imath], and so we can stop the sequence at any point and take the highest [imath]j[/imath] and we know that all the [imath]x_i[/imath]'s up to that point are in [imath]\mathcal{F}_j[/imath], and thus so must be their union. There must be some counterexample, but I don't see it. | 1122526 | Showing that if [imath]A_{1},A_{2},...[/imath] are all algebras then the union of all of them is an algebra
I am not sure how to show this. It seems obvious but maybe its not. The help would be greatly appreciated! |
26888 | The union of a strictly increasing sequence of [imath]\sigma[/imath]-algebras is not a [imath]\sigma[/imath]-algebra
The union of a sequence of [imath]\sigma[/imath]-algebras need not be a [imath]\sigma[/imath]-algebra, but how do I prove the stronger statement below? Let [imath]\mathcal{F}_n[/imath] be a sequence of [imath]\sigma[/imath]-algebras. If the inclusion [imath]\mathcal{F}_n \subsetneqq \mathcal{F}_{n+1} [/imath] is strict, then the union [imath]\bigcup_{n=1}^{\infty}\mathcal{F}_n[/imath] is not a [imath]\sigma[/imath]-algebra. | 535749 | Union of increasing [imath]\sigma[/imath]-algebras which is no [imath]\sigma[/imath]-Algebra
Give an example for an increasing series of [imath]\sigma[/imath] algebras [imath] \mathcal{A}_1\subset\mathcal{A}_2\subset\ldots [/imath] so that [imath]\bigcup_{i=1}^{\infty}\mathcal{A}_i [/imath] is no [imath]\sigma[/imath]-algebra. Could you pls give me a hint how to find such an example? |
51788 | Compactness of Multiplication Operator on [imath]L^2[/imath]
Suppose we have an bounded linear operator A that operates from [imath]L^2([a,b]) \mapsto L^2([a,b])[/imath]. Now suppose that [imath]A(f)(t) = tf(t)[/imath]. Is A compact? Edit: I know [imath]A = A^*[/imath] but I'm not really sure how to start on this. It's not homework, just summer fun :D | 236669 | No Nonzero multiplication operator is compact
Let [imath]f,g \in L^2[0,1][/imath], multiplication operator [imath]M_g:L^2[0,1] \rightarrow L^2[0,1][/imath] is defined by [imath]M_g(f(x))=g(x)f(x)[/imath]. Would you help me to prove that no nonzero multiplication operator on [imath]L^2[0,1][/imath] is compact. Thanks. |
30774 | Prove that [imath]a_{n}=0[/imath] for all [imath]n[/imath], if [imath]\sum a_{kn}=0[/imath] for all [imath]k\geq 1[/imath]
Let [imath]\sum a_{n}[/imath] be an absolutely convergent series such that [imath]\sum a_{kn}=0[/imath] for all [imath]k\geq 1[/imath]. Help me prove that [imath]a_{n}=0[/imath] for all [imath]n[/imath]. Thank you! | 1321310 | [imath]\sum_{n=1}^{\infty} a_n[/imath] converges absolutely and [imath]\sum _{n=1}^\infty a_{kn}=0 ,\forall k \ge 1 [/imath] ; then [imath]a_n=0 , \forall n \in \mathbb N [/imath]?
Suppose that the series [imath]\sum_{n=1}^{\infty} a_n[/imath] of real terms converges absolutely and [imath]\sum _{n=1}^\infty a_{kn}=0 ,\forall k \in \mathbb N [/imath] , then how to prove that [imath]a_n=0 , \forall n \in \mathbb N [/imath] ? |
159659 | Which sets are removable for holomorphic functions?
Let [imath]\Omega[/imath] be a domain in [imath]\mathbb C[/imath], and let [imath]\mathscr X[/imath] be some class of functions from [imath]\Omega[/imath] to [imath]\mathbb C[/imath]. A set [imath]E\subset \Omega[/imath] is called removable for holomorphic functions of class [imath]\mathscr X[/imath] if the following holds: every function [imath]f\in\mathscr X[/imath] that is holomorphic on [imath]\Omega\setminus E[/imath] is actually holomorphic on [imath]\Omega[/imath], possibly, after being redefined on [imath]E[/imath]. (An example of the above: [imath]E[/imath] is a line interval, [imath]\mathscr X[/imath] consists of continuous functions. In this case [imath]E[/imath] is removable, which is shown in the answer.) It is clear that the larger [imath]\mathscr X[/imath] is, the smaller is the class of removable sets. In the extreme case, if [imath]\mathscr X[/imath] contains all functions [imath]\Omega\to\mathbb C[/imath], there are no nonempty removable sets. Indeed, if [imath]a\in E[/imath], then [imath]f(z)=\frac{1}{z-a}[/imath] (arbitrarily defined at [imath]z=a[/imath]) is holomorphic on [imath]\Omega\setminus E[/imath] but has no holomorphic extension to [imath]\Omega[/imath]. The problem of describing removable sets is nontrivial in many classes [imath]\mathscr X[/imath] such as [imath]L^{\infty}(\Omega)[/imath], bounded functions [imath]C(\Omega)[/imath], continuous functions [imath]C^{\alpha}(\Omega)[/imath], Hölder continuous functions [imath]\mathrm{Lip}(\Omega)[/imath], Lipschitz functions Which sets are removable for holomorphic functions in these classes? | 1073421 | Need help with holomorphic functions on a domain interval removed.
I want to prove that for a region [imath]\Omega[/imath] with interval [imath]I=[a,b]\subset\Omega[/imath], if [imath]f[/imath] is continuous in [imath]\Omega[/imath] and [imath]f\in H(\Omega-I)[/imath], then actually [imath]f\in H(\Omega)[/imath]. Is this problem related to the removable singularity? and are there some other sets that could draw same conclusion? |
11477 | Fibonacci addition law [imath]F_{n+m} = F_{n-1}F_m + F_n F_{m+1}[/imath]
Question: Let [imath]F_n[/imath] the sequence of Fibonacci numbers, given by [imath]F_0 = 0, F_1 = 1[/imath] and [imath]F_n = F_{n-1} + F_{n-2}[/imath] for [imath]n \geq 2[/imath]. Show for [imath]n, m \in \mathbb{N}[/imath]: [imath]F_{n+m} = F_{n-1}F_m + F_n F_{m+1}[/imath] My (very limited) attempt so far: after creating a small list of the values [imath]F_0=0, F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, F_6=8, F_7=13, F_8=21, F_9=34, F_{10}=55[/imath] i can see that yes it does seem to work for instance [imath]F_{6+3}=F_5 F_3 +F_6 F_4 = 10 +24 = 34 = F_9[/imath]. However, I really don't know where to begin as showing that this must hold in general terms. Should I be looking to use limits? Or perhaps induction? What is the best way to solve this? | 1761673 | Prove equality for Fibonacci sequence
I have to show that [imath]F_{n+k} = F_{k}F_{n+1} + F_{k-1}F_{n}[/imath], where [imath]F_{n}[/imath] is nth Fibonacci element. I was trying with mathematical induction applied to n and saying k is constant. step for [imath]n=1[/imath] [imath]F_{k+1} = F_{k}F_{2} + F_{k-1}F_{1}[/imath] which is true, because [imath]F_1=F_2=1[/imath] step let say our theorem is true for some n step I was trying to do it like this: [imath]F_{n+1+k} = F_{n+k}+F_{n+k-1} = F_{k}F_{n+1} + F_{k-1}F_{n} + F_{n+k-1}[/imath] but there is [imath]F_{n+k-1}[/imath]and I do not know what to do with it. Thanks! |
48850 | Continuity of the function [imath]x\mapsto d(x,A)[/imath] on a metric space
Let [imath](X,d)[/imath] be a metric space. How to prove that for any closed [imath]A[/imath] a function [imath]d(x,A)[/imath] is continuous - I know that it is even Lipschitz continuous, but I have a problem with the proof: [imath] |d(x,a) - d(y,a)| \leq d(x,y) [/imath] for any [imath]a\in A[/imath] - but we cannot just replace it by [imath]|d(x,A) - d(y,A)|\leq d(x,y)[/imath] since the minimum (or infimum in general) can be attained in different points [imath]a\in A[/imath] for [imath]x[/imath] and [imath]y[/imath], so we only have that [imath] |d(x,A)-d(y,A)|\leq d(x,y)+\sup\limits_{a,b\in A}d(a,b) [/imath] which does not mean continuity. | 418042 | Show that the function [imath]f: X \to \Bbb R[/imath] given by [imath]f(x) = d(x, A)[/imath] is a continuous function.
I'm studying for my Topology exam and I am trying to brush up on my metric spaces. Suppose [imath](X, d)[/imath] is a metric space and [imath]A[/imath] is a proper subset of [imath]X[/imath]. Show that the function [imath]f: X \to \Bbb R[/imath] given by [imath]f(x) = d(x, A)[/imath] is a continuous function. I know that showing the pre-image of an open set is open in [imath]X[/imath] is an option for continuity. Yet, I would like to know how to show continuity with open balls or neighborhoods given the context of the problem. By the way, is this the Euclidean metric? Or am I jumping the gun a bit there? |
206851 | Generalisation of Dominated Convergence Theorem
Wikipedia claims, if [imath]\sigma[/imath]-finite the Dominated convergence theorem is still true when pointwise convergence is replaced by convergence in measure, does anyone know where to find a proof of this? Many thanks! Statement of the theorem: Let [imath]\mu[/imath] be [imath]\sigma[/imath]-finite, [imath]|f_n|\leq g[/imath] and [imath]f_n\rightarrow f[/imath] in measure, then we must have [imath]\int f_n \rightarrow \int f[/imath] and [imath]\int|f_n-f| \rightarrow 0[/imath] | 358161 | Convergence of functions
Assume that [imath](X,M,\mu)[/imath] is a [imath]\sigma[/imath]-finite space. Suppose that [imath]|f_n|\leq g\in L^+[/imath] and [imath]f_n\rightarrow f[/imath] in measure. Show that [imath]\int f=\lim_{n\rightarrow\infty}\int f_n[/imath]. I tried taking a subsequence of [imath]f_n[/imath], call it [imath]f_{n_j}[/imath] and [imath]f_{n_j}\rightarrow f[/imath] almost everywhere. Also, [imath]f_{n_j}\leq g\in L^+[/imath]. So by Dominated Convergence Theorem, [imath]f=\lim_{j\rightarrow\infty}\int f_{n_j}[/imath] And then I say that by [imath]\lim_{n\rightarrow\infty}\int f_n=\lim_{j\rightarrow\infty}\int f_{n_j}[/imath], I get the equality. However, I received a comment on my solution saying that the final statement only holds if [imath]\lim_{n\rightarrow\infty}\int f_n[/imath] exists. Can anybody provide a clearer explanation why this is the case and how I can fix my proof? |
130471 | Convergence/divergence of [imath]\sum\frac{a_n}{1+na_n}[/imath] when [imath]a_n\geq0[/imath] and [imath]\sum a_n[/imath] diverges
A question from Rudin (Principles) Chapter 3: Let [imath]a_n\geq0[/imath] and [imath]\sum a_n[/imath] diverges. What can be said about convergence/divergence of [imath]\sum\frac{a_n}{1+na_n}[/imath]? This one is being recalcitrant. Given that [imath]x>y[/imath] implies [imath]\frac{x}{1+nx}>\frac{y}{1+ny}[/imath] and when [imath]a_n=1/n\log n[/imath] the sum in question diverges, it seems plausible that in general the sum will always diverge, but I can't get a proof out. If it does diverge, it does so pretty slowly as [imath]\frac{a_n}{1+na_n}=\frac{1}{n}-\frac{1}{n+n^2a_n}\leq\frac{1}{n}.[/imath] | 1505389 | Prove that if [imath]\sum a_n[/imath] is divergent then [imath]\sum a_n/(1 + na_n)[/imath] diverges too
I had this question where it was given [imath]a_n > 0[/imath] and [imath]\sum a_n[/imath] is divergent and I was to prove that [imath]\sum\frac{a_n}{1 + na_n}[/imath] diverges too, to be frank I could not make any worthy headway on this problem, so any ideas ? |
220410 | A characterization of functions from [imath]\mathbb R^n[/imath] to [imath]\mathbb R^m[/imath] which are continuous
Greets I came up the other day with the following question: Is it true that [imath]f:\mathbb{R}^n\longrightarrow{\mathbb{R}^m}[/imath] is continuous if and only if [imath]f[/imath] maps compact sets onto compact sets and maps connected sets onto connected sets? I'm having trouble showing the "if", and I haven't found a counterexample to this part, so I would appreciate ideas to prove or disprove this. Thanks | 1212136 | A map of two Euclidean spaces preserving connectedness and compactness is continuous
Let [imath] f:\mathbb R^n \to \mathbb R^m[/imath]. If [imath]f[/imath] preserves connectedness and compactness then [imath]f[/imath] is continuous. How can this be proven? I don't really know where to start. |
239825 | Can anyone explain why [imath]a^{b^c} = a^{(b^c)} \neq (a^b)^c = a^{(bc)}[/imath]
I'm so puzzled about this: [imath]a^{b^c} = a^{(b^c)} \neq (a^b)^c = a^{(bc)}.[/imath] Why isn't [imath]a^{b^c}[/imath] equal to [imath]a^{(bc)}[/imath]? Why is [imath]a^{b^c}[/imath] instead equal to [imath]a^{(b^c)}[/imath]? And how is it possible that [imath](a^b)^c = a^{(bc)}[/imath]? My mind is pretty much exploding from trying to understand this. | 2029904 | Notation: Precedence in multiple levels of exponentiation
When people write [imath]a^{b^c}[/imath], does it refer to [imath](a^b)^c=a^{bc}[/imath] or [imath]a^{(b^c)}[/imath]? What about the notation a^b^c (without superscripts)? Is there an agreed-upon convention? |
75130 | How to prove that [imath]\lim\limits_{x\to0}\frac{\sin x}x=1[/imath]?
How can one prove the statement [imath]\lim_{x\to 0}\frac{\sin x}x=1[/imath] without using the Taylor series of [imath]\sin[/imath], [imath]\cos[/imath] and [imath]\tan[/imath]? Best would be a geometrical solution. This is homework. In my math class, we are about to prove that [imath]\sin[/imath] is continuous. We found out, that proving the above statement is enough for proving the continuity of [imath]\sin[/imath], but I can't find out how. Any help is appreciated. | 400971 | Finding [imath]\lim_{x\to 0}\frac{\sin x}{x} [/imath]
How to find [imath]\lim_{x\to 0}\frac{\sin x}{x} [/imath] |
271138 | convergence of a series involving [imath]x^\sqrt{n}[/imath]
I was trying to prove the convergence of the series [imath]\sum_{n=1}^{\infty}x^{\sqrt{n}}[/imath], for [imath]0<x<1[/imath]. Unfortunately, I could not make one of the standard convergence tests give me an answer. Does anybody of you have a suggestion? any help is much appreciated! many thanks! | 382109 | Finding the supremum of the following set
I am stuck on the following problem: Let [imath]P=\{x \in \Bbb R: x\ge 0,\sum_{n=1}^{\infty}x^{\sqrt n}< \infty\}[/imath].Then what is the supremum of [imath]P[/imath]? Can someone help me out by providing some explanation? Thanks in advance. |
64432 | What is so special about [imath]\alpha=-1[/imath] in the integral of [imath]x^\alpha[/imath]?
Of course, it is easy to see, that the integral (or the antiderivative) of [imath]f(x) = 1/x[/imath] is [imath]\log(|x|)[/imath] and of course for [imath]\alpha\neq - 1[/imath] the antiderivative of [imath]f(x) = x^\alpha[/imath] is [imath]x^{\alpha+1}/(\alpha+1)[/imath]. I was wondering if there is an intuitive (probably geometric) explanation why the case [imath]\alpha=-1[/imath] is so different and why the logarithm appears? Some answers which I thought of but which are not convincing: Taking the limit [imath]\alpha=-1[/imath] either from above or below lead to diverging functions. Some speciality of the case [imath]\alpha=-1[/imath] are that both asymptotes are non-integrable. However, the antidrivative is a local thing, and hence, shouldn't care about the behavior at infinity. | 848143 | Why, intuitively, does [imath]log(x)[/imath] come in as the integral of [imath]1/x[/imath], wheras the integral of other powers of [imath]x[/imath] are powers of [imath]x[/imath]?
Question in title really, something I always found strange when I was learning calculus. I can see that [imath]\int \frac{1}{x} dx[/imath] can't be [imath]\frac{x^0}{0}[/imath] since this is not defined, and then the definite integral [imath]\int_1^t \frac{1}{x} dx[/imath] comes down to [imath]\lim_{\delta \rightarrow 0} \frac{t^{\delta}-1}{\delta} = \log (t).[/imath] But this understanding just comes from l'Hopital's rule, and also it still just seems really bizaare that the log function should fit into the set of power functions like this. Can anyone de-mystify this at all? |
244409 | If [imath]f^2[/imath] is Riemann Integrable is [imath]f[/imath] always Riemann Integrable?
Problem: Suppose that [imath]f[/imath] is a bounded, real-valued function on [imath][a,b][/imath] such that [imath]f^2\in R[/imath] (i.e. it is Riemann-Integrable). Must it be the case that [imath]f\in R[/imath] ? Thoughts: I think that this is not necessarily true, but I am having trouble refuting or even proving the above. Of course, the simplest way to prove that it is not necessarily true would be to give an example, but I am unable to think of one! I also have tried using [imath]\phi(y)=\sqrt y[/imath] and composing this with [imath]f^2[/imath] (to try show [imath]f[/imath] is continuous); however, the interval [imath][a,b][/imath] may contain negative numbers so I can't utilise [imath]\phi[/imath] in that case. Question: Does there exist a function [imath]f[/imath] such that [imath]f^2\in R[/imath] but [imath]f[/imath] [imath]\not\in R[/imath] ? Or conversely, if [imath]f^2\in R[/imath] does this always imply [imath]f[/imath] [imath]\in R[/imath] ? (If so, could you provide a way of proving this). | 2064203 | If [imath]f[/imath] is bounded and if [imath]f^2[/imath] is Riemann integrable, then is f Riemann integrable?
This was on a past exam paper, and it was asking if [imath]f[/imath] is bounded and if [imath]f^2[/imath] is Riemann integrable, then is f Riemann integrable? If I had to guess, I'd say no. I tried creating an [imath]\epsilon[/imath] argument, but I feel like I'm grasping at straws, so I'd appreciate a hint towards a slightly more concrete proof. |
56307 | Simultaneous diagonalization of commuting linear transformations
Let [imath]V[/imath] be a vector space of finite dimension and let [imath]T,S[/imath] linear diagonalizable transformations from [imath]V[/imath] to itself. I need to prove that if [imath]TS=ST[/imath] every eigenspace [imath]V_\lambda[/imath] of [imath]S[/imath] is [imath]T[/imath]-invariant and the restriction of [imath]T[/imath] to [imath]V_\lambda[/imath] ([imath]T:{V_{\lambda }}\rightarrow V_{\lambda }[/imath]) is diagonalizable. In addition, I need to show that there's a base [imath]B[/imath] of [imath]V[/imath] such that [imath][S]_{B}^{B}[/imath], [imath][T]_{B}^{B}[/imath] are diagonalizable if and only if [imath]TS=ST[/imath]. Ok, so first let [imath]v\in V_\lambda[/imath]. From [imath]TS=ST[/imath] we get that [imath]\lambda T(v)= S(T(v))[/imath] so [imath]T(v)[/imath] is eigenvector of [imath]S[/imath] and we get what we want. I want to use that in order to get the following claim, I just don't know how. One direction of the "iff" is obvious, the other one is more tricky to me. | 1841311 | Problem about linear algebra
Suppose we have two [imath]n \times n[/imath] square matrices A and B such that [imath]AB=BA[/imath]. It is known that A, B and AB all have n distinct eigenvectors that is a basis of [imath]\mathbb{C}^n[/imath]. Can we then show that there is a basis of [imath]\mathbb{R}^n[/imath] that comprises entirely of vectors that are eigenvectors of both A and B? I have no idea on solving this problem. Any form of help is appreciated. Thanks! |
207910 | Prove convergence of the sequence [imath](z_1+z_2+\cdots + z_n)/n[/imath] of Cesaro means
Prove that if [imath]\lim_{n \to \infty}z_{n}=A[/imath] then: [imath]\lim_{n \to \infty}\frac{z_{1}+z_{2}+\cdots + z_{n}}{n}=A[/imath] I was thinking spliting it in: [imath](z_{1}+z_{2}+\cdots+z_{N-1})+(z_{N}+z_{N+1}+\cdots+z_{n})[/imath] where [imath]N[/imath] is value of [imath]n[/imath] for which [imath]|A-z_{n}|<\epsilon[/imath] then taking the limit of this sum devided by [imath]n[/imath] , and noting that the second sum is as close as you wish to [imath]nA[/imath] while the first is as close as you wish to [imath]0[/imath]. Not sure if this helps.... | 443770 | Proving a statmenet about convergence of complex sequence
Let [imath]x_k \in \mathbb C[/imath] for [imath]k \in \mathbb N \cup {0}[/imath] and let [imath]y_k = \frac{(x_0 + x_1 + ... + x_k)}{k+1}[/imath]. We want to prove that if [imath]x_k[/imath] converges to [imath]x[/imath] ([imath]x \in \mathbb C[/imath]) as [imath]k \rightarrow \infty[/imath] then [imath]y_k[/imath] also converges to [imath]x[/imath] as [imath]k \rightarrow \infty[/imath]. This seems like a simple interesting exercise but having a little trouble. If [imath]x_k \rightarrow x[/imath] then from this I tried to make conclusions about convergence of [imath]y_k[/imath] since it depends of [imath]x_k[/imath]. As [imath]k \rightarrow \infty[/imath] then [imath]y_k \rightarrow \frac{(x_0 + x_1 + ... + x)}{k+1}[/imath] but how can we conclude [imath]\frac{(x_0 + x_1 + ... + x)}{k+1} = x[/imath]? Intuitively we also notice that all the finite terms over k+1 will go to 0 as [imath]k \rightarrow \infty[/imath]. |
50316 | [imath]x^x=y[/imath]. How to solve for [imath]x[/imath]?
I tried looking for ways to solve this equation and came across something like Lambert's W function, which, by the way, I did not understand a bit, because I've never learned it nor do I have a decent mathematical background. I also came across one more method called the Newton's method, but never used it either. Can the procedure to solve this equation be made a little bit clearer? I just plugged in the value on my calculator and it solved it for me, I guess it must have used some trial and error. Please advise. | 1281246 | How to solve for [imath]y[/imath] the equation [imath]x= y^y[/imath]?
I need an equation where I receive a number that when raised to itself equals the input. Formally: in [imath]x=y^y[/imath] solve for [imath]y[/imath]. Intro to Calculus level knowledge. If the Lambert function is necessary, please explain it to me. |
198787 | Prove that if c is a common divisor of a and b then c divides the gcd of a and b..
If [imath]c[/imath] is a common divisor of [imath]a[/imath] and [imath]b[/imath] then [imath]c[/imath] divides the greatest common divisor of [imath]a[/imath] and [imath]b[/imath]. What can we use to prove this? | 1753209 | Let [imath]a^n, a^m \in (a^k)[/imath] for some positive integer [imath]k[/imath]. Then [imath]k \mid n, m.[/imath] Hence [imath]k \mid \operatorname{gcd(n, m)}?[/imath]
Let [imath]a^n, a^m \in (a^k)[/imath] for some positive integer [imath]k[/imath]. Then [imath]k \mid n, m.[/imath] Hence [imath]k \mid \operatorname{gcd(n, m)}[/imath]. How is it possible? Let [imath]k = 6, n = 12, m =24.[/imath] Context: Let [imath]H[/imath] be the smallest subgroup of [imath](a)[/imath] s.t. [imath]a^n, a^m \in H[/imath]. Prove that [imath]H = (a^{\operatorname{gcd(n, m)}}).[/imath] Proof: Since [imath]G[/imath] is cyclic, [imath]H[/imath] is cyclic. Hence [imath]H = (a^k)[/imath] for some [imath]k \in \mathbb N.[/imath] Since [imath]a^n, a^m \in H,[/imath] then [imath]k \mid n,m.[/imath] Hence [imath]k \mid \operatorname{gcd(n, m)}.[/imath] Thus [imath]a^{\operatorname{gcd(n, m)}} \in H.[/imath] Hence [imath](a^{\operatorname{gcd(n, m)}}) \subset H.[/imath] Also, since [imath]a^{\operatorname{gcd(n, m)}} \mid n, m[/imath] then [imath]a^n, a^m \in (a^{\operatorname{gcd(n, m)}})[/imath]. Since [imath]H[/imath] is the smallest subgroup of [imath](a)[/imath] containing [imath]a^n, a^m[/imath] and [imath]a^n, a^m \in (a^{\operatorname{gcd(n, m)}})[/imath], we conclude that [imath]H = (a^{\operatorname{gcd(n, m)}}).[/imath] |
36364 | What does [imath]\ll[/imath] mean?
I saw two less than signs on this Wikipedia article and I was wonder what they meant mathematically. http://en.wikipedia.org/wiki/German_tank_problem EDIT: It looks like this can use TeX commands. So I think this is the symbol: [imath]\ll[/imath] | 625763 | What is the difference between [imath]\gg[/imath] and [imath]\gt[/imath]?
What is the difference between [imath]\gg[/imath] and [imath]\gt[/imath] ? Thank you [imath]\infty[/imath] times. |
305954 | Can you someone help me to find the indefinite integral, step by step. I did my self, and getting wrong answer.
can you please someone tell me how to do this indefinite integral in steps [imath]\int[/imath][imath]cos(\sqrt{6x})\over\sqrt{6x}[/imath] dx | 305351 | what are the possible answers we can get for the below intergral?
Could you please tell me what are the possible answers (if there is more than one) for the following indefinite integral? [imath]\int \dfrac{\cos(\sqrt{6x})}{\sqrt{6x}}dx[/imath] |
184609 | Why is the last digit of [imath]n^5[/imath] equal to the last digit of [imath]n[/imath]?
I was wondering why the last digit of [imath]n^5[/imath] is that of [imath]n[/imath]? What's the proof and logic behind the statement? I have no idea where to start. Can someone please provide a simple proof or some general ideas about how I can figure out the proof myself? Thanks. | 827467 | How do you prove that [imath] n^5[/imath] is congruent to [imath] n[/imath] mod 10?
How do you prove that [imath]n^5 \equiv n\pmod {10}[/imath] Hint given was- Fermat little theorem. Kindly help me out. This is applicable to all positive integers [imath]n[/imath] |
37327 | [imath]\infty = -1 [/imath] paradox
I puzzled two high school Pre-calc math teachers today with a little proof (maybe not) I found a couple years ago that infinity is equal to -1: Let x equal the geometric series: [imath]1 + 2 + 4 + 8 + 16 \ldots[/imath] [imath]x = 1 + 2 + 4 + 8 + 16 \ldots[/imath] Multiply each side by 2: [imath]2x = 2 + 4 + 8 + 16 + 32 \ldots[/imath] Again from the equation in step 1, move the [imath]1[/imath] term to the left hand of the equation: [imath]x - 1 = 2 + 4 + 8 + 16 + 32 \ldots[/imath] So the following appears to be true: [imath]2x = x - 1 \implies x = -1[/imath] This is obviously illogical. The teachers told me the problem has to do with adding the two infinite geometric series, but they weren't positive. I'm currently in Pre-calc, so I have extremely little knowledge on calculus, but a little help with this paradox would be appreciated. | 352650 | Paradox of Infinity?
If a series such as '[imath]a[/imath]' below adds to infinity: [imath]a = 1 + 2 + 4 + 8 + 16 + \cdots\to \infty[/imath] Multiplying '[imath]a[/imath]' by [imath]2[/imath] yields: [imath]2a = 2 + 4 + 8 + 16 + \cdots\to \infty[/imath] However when I subtract these two series, I find a paradoxical answer. Series '[imath]a[/imath]', which supposedly adds to infinity also equals -1. [imath]2a - a = (2 + 4 + 8 + 16 + \cdots) - (1 + 2 + 4 + 8 + 16 +\cdots)[/imath] [imath]2a - a = -1 + (2 - 2) + (4 - 4) + (8 - 8) + \cdots[/imath] [imath]a = -1 + 0 + 0 + 0 + 0 + \cdots[/imath] [imath]a = -1[/imath] Does this mean that infinity equals [imath]-1[/imath]? |
304480 | A Combinatorial Question to Solve a System of Equations
Suppose we have [imath]N[/imath] integer-valued variables [imath]i_1[/imath], [imath]i_2[/imath], [imath]\cdot\cdot\cdot[/imath], [imath]i_N[/imath], such that each variable can take integer values from 0 to [imath]k[/imath], and the sum of these [imath]N[/imath] variables is also equal to [imath]k[/imath]. Formally, we want to solve the constrained linear equation: \begin{cases} i_1+i_2+\cdot\cdot\cdot + i_N = k \\ 0 \leq i_j \leq k & \text{where [imath]1\leq j \leq N[/imath]} \end{cases} Is there anyone know why the number of solutions is equal to [imath]\binom{k+N-1}{N-1}[/imath]? Thanks in advance very much. | 686 | Combinations of selecting [imath]n[/imath] objects with [imath]k[/imath] different types
Suppose that I am buying cakes for a party. There are [imath]k[/imath] different types and I intend to buy a total of [imath]n[/imath] cakes. How many different combinations of cakes could I possibly bring to the party? |
78533 | prove that [imath]\frac{(2n)!}{(n!)^2}[/imath] is even if [imath]n[/imath] is a positive integer
Prove that [imath]\frac{(2n)!}{(n!)^2}[/imath] is even if [imath]n[/imath] is a positive integer. For clarity: the denominator is the only part being squared. My thought process: The numerator is the product of the first [imath]n[/imath] even numbers and the product of the first [imath]n[/imath] odd numbers. That is, [imath](2n!) = (2n)(2n-2)(2n-4)\cdots(2n-1)(2n-3)(2n-5).[/imath] In effect, the product of even numbers can be cancelled out with [imath]n![/imath] resulting in the following quotient: [imath] \frac{(2^n)(2n-1)(2n-3)}{(n!)}\;.[/imath] To me this looks even thanks to the powers of [imath]2[/imath]. But I am not convinced for some reason. Was this approach to naive? Sorry for the poor notation, I don't know any coding languages, my apologies. | 667404 | show that [imath]2n\choose n[/imath] is divisible by 2
I tried using induction, but in the inductive step, I get: If [imath]2n\choose n[/imath] is divisible then I want to see that [imath]2n +2\choose n +1[/imath] [imath]{2n +2\choose n +1} = (2n + 2)!/(n+1)!(n +1)! = {2n\choose n} (2n+1)(2n+2)/(n+1)^2[/imath] [imath]= {2n\choose n} 2(2n+1)/(n+1)[/imath] Then [imath]{2n\choose n}[/imath] is divisible by [imath]2[/imath] by hypothesis, but [imath]2(2n+1)/(n+1)[/imath] is not even an integer. What can I do? |
295466 | Algebraic sets in [imath]\mathbb{A}^2[/imath]
Deduce that if [imath]Z[/imath] is an algebraic set in [imath]\mathbb{A}^2[/imath] and [imath]c\in\mathbb{C}[/imath] then [imath]Y = \{ a\in\mathbb{C} : (a,c) \in Z \}[/imath] is either finite or all of [imath]\mathbb{A}^1[/imath]. Deduce that [imath]\{ ( z,w) \in \mathbb{A}^2 : |z|^2 + |w|^2 = 1 \}[/imath] is not an algebraic set in [imath]\mathbb{A}^2[/imath]. | 295445 | Proving algebraic sets
i) Let [imath]Z[/imath] be an algebraic set in [imath]\mathbb{A}^n[/imath]. Fix [imath]c\in \mathbb{C}[/imath]. Show that [imath]Y=\{b=(b_1,\dots,b_{n-1})\in \mathbb{A}^{n-1}|(b_1,\dots,b_{n-1},c)\in Z\}[/imath] is an algebraic set in [imath]\mathbb{A}^{n-1}[/imath]. ii) Deduce that if [imath]Z[/imath] is an algebraic set in [imath]\mathbb{A}^2[/imath] and [imath]c\in \mathbb{C}[/imath] then [imath]Y=\{a\in \mathbb{C}|(a,c)\in Z\}[/imath] is either finite or all of [imath]\mathbb{A}^1[/imath]. Deduce that [imath]\{(z,w)\in \mathbb{A}^2 :|z|^2 +|w|^2 =1\}[/imath] is not an algebraic set in [imath]\mathbb{A}^2[/imath]. |
28568 | Bijection between an open and a closed interval
Recently, I answered to this problem: Given [imath]a<b\in \mathbb{R}[/imath], find explicitly a bijection [imath]f(x)[/imath] from [imath]]a,b[[/imath] to [imath][a,b][/imath]. using an "iterative construction" (see below the rule). My question is: is it possible to solve the problem finding a less exotic function? I mean: I know such a bijection cannot be monotone, nor globally continuous; but my [imath]f(x)[/imath] has a lot of jumps... Hence, can one do without so many discontinuities? W.l.o.g. assume [imath]a=-1[/imath] and [imath]b=1[/imath] (the general case can be handled by translation and rescaling). Let: (1) [imath]X_0:=]-1,-\frac{1}{2}] \cup [\frac{1}{2} ,1[[/imath], and (2) [imath]$f_0(x):=\begin{cases} -x-\frac{3}{2} &\text{, if } -1,[/imath] so that the graph of [imath]f_0(x)[/imath] is made of two segments (parallel to the line [imath]y=x[/imath]) and one segment laying on the [imath]x[/imath] axis; then define by induction: (3) [imath]X_{n+1}:=\frac{1}{2} X_n[/imath], and (4) [imath]f_{n+1}(x):= \frac{1}{2} f_n(2 x)[/imath] for [imath]n\in \mathbb{N}[/imath] (hence [imath]X_n=\frac{1}{2^n} X_0[/imath] and [imath]f_n=\frac{1}{2^n} f_0(2^n x)[/imath]). Then the function [imath]f:]-1,1[\to \mathbb{R}[/imath]: (5) [imath]f(x):=\sum_{n=0}^{+\infty} f_n(x)[/imath] is a bijection from [imath]]-1,1[[/imath] to [imath][-1,1][/imath]. Proof: i. First of all, note that [imath]\{ X_n\}_{n\in \mathbb{N}}[/imath] is a pairwise disjoint covering of [imath]]-1,1[\setminus \{ 0\}[/imath]. Moreover the range of each [imath]f_n(x)[/imath] is [imath]f_n(]-1,1[)=[-\frac{1}{2^n}, -\frac{1}{2^{n+1}}[\cup \{ 0\} \cup ]\frac{1}{2^{n+1}}, \frac{1}{2^n}][/imath]. ii. Let [imath]x\in ]-1,1[[/imath]. If [imath]x=0[/imath], then [imath]f(x)=0[/imath] by (5). If [imath]x\neq 0[/imath], then there exists only one [imath]\nu\in \mathbb{N}[/imath] s.t. [imath]x\in X_\nu[/imath], hence [imath]f(x)=f_\nu (x)[/imath]. Therefore [imath]f(x)[/imath] is well defined. iii. By i and ii, [imath]f(x)\lesseqgtr 0[/imath] for [imath]x\lesseqgtr 0[/imath] and the range of [imath]f(x)[/imath] is: [imath]f(]-1,1[)=\bigcup_{n\in \mathbb{N}} f(]-1,1[) =[-1,1][/imath], therefore [imath]f(x)[/imath] is surjective. iv. On the other hand, if [imath]x\neq y \in ]-1,1[[/imath], then: if there exists [imath]\nu \in \mathbb{N}[/imath] s.t. [imath]x,y\in X_\nu[/imath], then [imath]f(x)=f_\nu (x)\neq f_\nu (y)=f(y)[/imath] (for [imath]f_\nu (x)[/imath] restrited to [imath]X_\nu[/imath] is injective); if [imath]x\in X_\nu[/imath] and [imath]y\in X_\mu[/imath], then [imath]f(x)=f_\nu (x)\neq f_\mu(y)=f(y)[/imath] (for the restriction of [imath]f_\nu (x)[/imath] to [imath]X_\nu[/imath] and of [imath]f_\mu(x)[/imath] to [imath]X_\mu[/imath] have disjoint ranges); finally if [imath]x=0\neq y[/imath], then [imath]f(x)=0\neq f(y)[/imath] (because of ii). Therefore [imath]f(x)[/imath] is injective, hence a bijection between [imath]]-1,1[[/imath] and [imath][-1,1][/imath]. [imath]\square[/imath] | 1006445 | Proving [imath](0,1)[/imath] and [imath][0,1][/imath] have the same cardinality
Prove [imath](0,1)[/imath] and [imath][0,1][/imath] have the same cardinality. I've seen questions similar to this but I'm still having trouble. I know that for [imath]2[/imath] sets to have the same cardinality there must exist a bijection function from one set to the other. I think I can create a bijection function from [imath](0,1)[/imath] to [imath][0,1][/imath], but I'm not sure how the opposite. I'm having trouble creating a function that makes [imath][0,1][/imath] to [imath](0,1)[/imath]. Best I can think of would be something like [imath]x \over 2[/imath]. Help would be great. |
160738 | How to define a bijection between [imath](0,1)[/imath] and [imath](0,1][/imath]?
How to define a bijection between [imath](0,1)[/imath] and [imath](0,1][/imath]? Or any other open and closed intervals? If the intervals are both open like [imath](-1,2)\text{ and }(-5,4)[/imath] I do a cheap trick (don't know if that's how you're supposed to do it): I make a function [imath]f : (-1, 2)\rightarrow (-5, 4)[/imath] of the form [imath]f(x)=mx+b[/imath] by \begin{align*} -5 = f(-1) &= m(-1)+b \\ 4 = f(2) &= m(2) + b \end{align*} Solving for [imath]m[/imath] and [imath]b[/imath] I find [imath]m=3\text{ and }b=-2[/imath] so then [imath]f(x)=3x-2.[/imath] Then I show that [imath]f[/imath] is a bijection by showing that it is injective and surjective. | 299006 | What is the most effective way to implement Hilbert's hotel?
Assuming I need to find an onto and 1-to-1 function from [imath](a,b)[/imath] to [imath](0,1)[/imath], well that's not a hard job. But things are getting bit more complicated when I'm asked to do the exact same but from [imath][a,b)[/imath] to [imath](0,1)[/imath] or from [imath](a,b)[/imath] to [imath](0,1][/imath] and so on. What is the most effective way to find those required functions? because I have the feeling that there is a scheme that I can work by to handle those kind of problems handling with the cardinality of the continuum, [imath]\aleph[/imath]. |
29441 | Prove that [imath]n[/imath] is a sum of two squares?
Problem Let [imath]n = p_1.p_2.p_3 \cdots p_k.m^2[/imath], where [imath]p_1, p_2, p_3 \cdots p_k[/imath] are distinct primes. Prove that n is sum of two squares if and only if [imath]p_i[/imath] is either 2 or [imath]p_i \equiv 1 \pmod{4}[/imath] For [imath]p_i = 2[/imath] , this is trivial case since [imath]2m^2 = m^2 + m^2[/imath]. For [imath]p_i \equiv 1 \pmod{4}[/imath], I tried to use the fact that the product of a number of the form [imath]4k + 1[/imath] is also in this form. So I come up with: [imath]n = (4k + 1) \cdot m^2 = 4km^2 + m^2[/imath] Apparently, [imath]m^2[/imath] is a square, but I could not figure out how to prove [imath]4km^2[/imath] is a square, since k is in unknown form. Any idea? Thanks, | 1678756 | solutions for the diophantine equation [imath]x^2+y^2=n[/imath]
Are there any solutions for the diophantine equation [imath]x^2+y^2=n[/imath] ? For [imath]n \in \mathbb{P} \wedge n \equiv1\pmod4[/imath] solutions are widely known. Can we generalize a bit? |
7938 | [imath]n!+1[/imath] being a perfect square
One observes that \begin{equation*} 4!+1 =25=5^{2},~5!+1=121=11^{2} \end{equation*} is a perfect square. Similarly for [imath]n=7[/imath] also we see that [imath]n!+1[/imath] is a perfect square. So one can ask the truth of this question: Is [imath]n!+1[/imath] a perfect square for infinitely many [imath]n[/imath]? If yes, then how to prove. | 767510 | [imath]1+n!=m^{2}[/imath] for some n,m[imath]\in\mathbb{N}[/imath]
I have no idea whether this is known or not and I couldn't find anything related on Google. While I was studying , I come up with this idea [imath]1+n!=m^{2} [/imath] for some [imath]n,m\in\mathbb{N}[/imath] [imath]1+4!=5^{2}[/imath] [imath]1+5!=11^{2}[/imath] [imath]1+?!=?^{2}[/imath] and the question is what is the next number? Wolfram Alpha gave me this interesting graph: Thanks in advance for your interest. |
93409 | Does every Abelian group admit a ring structure?
Given some Abelian group [imath](G, +)[/imath], does there always exist a binary operation [imath]*[/imath] such that [imath](G, +, *)[/imath] is a ring? That is, [imath]*[/imath] is associative and distributive: \begin{align*} &a * (b * c) = (a*b) * c \\ &a * (b + c) = a * b + a * c \\ &(a + b) * c = a * c + b * c \\ \end{align*} We also might have multiplicative identity [imath]1 \in G[/imath], with [imath]a * 1 = 1 * a = a[/imath] for any [imath]a \in G[/imath]. Multiplication may or may not be commutative. Depending on the definition, the answer could be no in the case of the group with one element: then [imath]1 = 0[/imath]. But the trivial ring is not a very interesting case. For cyclic groups the statement is certainly true, since [imath](\mathbb{Z}_n, +, \cdot)[/imath] and [imath](\mathbb{Z}, +, \cdot)[/imath] are both rings. What about in general? Is there some procedure to give arbitrary abelian groups ring structure? | 1006716 | Ring structures on abelian groups
My question is: given an abelian group [imath]G[/imath] with addition [imath]+[/imath], is there some natural multiplicative structure that arises so that we can define a ring [imath](G, +, \cdot)[/imath]. For instance, multiplication on [imath]\mathbb{Z}[/imath] and [imath]\mathbb{Z}_n[/imath] are entirely determined by addition, since it must be that [imath]ma = a + \ldots + a[/imath], where the addition is [imath]m[/imath] times. For finitely generated abelian groups [imath]G[/imath], we know that its representation according to the Fundamental Theorem of Finitely Generated Abelian Groups is [imath]G = \mathbb{Z}_{p_1^{r_1}} \times \ldots \times \mathbb{Z}_{p_n^{r_n}} \times \mathbb{Z} \times \ldots \times \mathbb{Z}[/imath] Since each of those factors has a natural ring structure, we can define a ring structure on [imath]G[/imath] as the product of these ring structures. That leaves the question: can we define a "natural" ring structure on infinitely generated abelian groups [imath]G[/imath]? |
80453 | How to prove that [imath]\lim\limits_{n \to \infty} \frac{k^n}{n!} = 0[/imath]
It recently came to my mind, how to prove that the factorial grows faster than the exponential, or that the linear grows faster than the logarithmic, etc... I thought about writing: [imath] a(n) = \frac{k^n}{n!} = \frac{ k \times k \times \dots \times k}{1\times 2\times\dots\times n} = \frac k1 \times \frac k2 \times \dots \times \frac kn = \frac k1 \times \frac k2 \times \dots \times \frac kk \times \frac k{k+1} \times \dots \times \frac kn [/imath] It's obvious that after k/k, every factor is smaller than 1, and by increasing n, k/n gets closer to 0, like if we had [imath]\lim_{n \to \infty} (k/n) = 0[/imath], for any constant [imath]k[/imath]. But, I think this is not a clear proof... so any hint is accepted. Thank you for consideration. | 1042129 | A limit of a sequence
I'm trying to prove the following limit [imath](\frac{2^n}{n!}) \to 0[/imath] But it seems difficault to me. How can I prove it? Thanks. |
24456 | Matrix multiplication: interpreting and understanding the process
I have just watched the first half of the 3rd lecture of Gilbert Strang on the open course ware with link: http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/ It seems that with a matrix multiplication [imath]AB=C[/imath], that the entries as scalars, are formed from the dot product computations of the rows of [imath]A[/imath] with the columns of [imath]B[/imath]. Visual interpretations from mechanics of overlpaing forces come to mind immediately because that is the source for the dot product (inner product). I see the rows of [imath]C[/imath] as being the dot product of the rows of [imath]B[/imath], with the dot product of a particular row of [imath]A[/imath]. Similar to the above and it is easy to see this from the individual entries in the matrix [imath]C[/imath] as to which elements change to give which dot products. For understanding matrix multiplication there is the geometrical interpretation, that the matrix multiplication is a change in the reference system since matrix [imath]B[/imath] can be seen as a transormation operator for rotation, scalling, reflection and skew. It is easy to see this by constructing example [imath]B[/imath] matrices with these effects on [imath]A[/imath]. This decomposition is a strong argument and is strongly convincing of its generality. This interpreation is strong but not smooth because I would find smoother an explanation which would be an interpretation begining from the dot product of vectors and using this to explain the process and the interpretation of the results (one which is a bit easier to see without many examples of the putting numbers in and seeing what comes out which students go through). I can hope that sticking to dot products throughout the explanation and THEN seeing how these can be seen to produce scalings, rotations, and skewings would be better. But, after some simple graphical examples I saw this doesn't work as the order of the columns in matrix [imath]B[/imath] are important and don't show in the graphical representation. The best explanation I can find is at Yahoo Answers. It is convincing but a bit disappointing (explains why this approach preserves the "composition of linear transformations"; thanks @Arturo Magidin). So the question is: Why does matrix multiplication happen as it does, and are there good practical examples to support it? Preferably not via rotations/scalings/skews (thanks @lhf). | 1698783 | Understanding matrix multiplication
I have a hard time understanding, on an intuitive level, what matrix multiplication actually does. I have used it a lot, but I do not really know what it does. I know that [imath]Ax = y[/imath], where [imath]A[/imath] is a matrix and [imath]x[/imath] is an [imath]n[/imath]-tuple, is just another way writing a system of equations. Seeing it this way, matrix addition is quite intuitive. Since every elementary matrix can be seen as representing an elementary row operation, I can understand what matrix multiplication does with invertible matrices. However, that still does not explain anything when we are working with matrices that are not invertible. What is a good way of thinking about matrix multiplication? |
117194 | Fastest way to try all passwords
Suppose you have a computer with a password of length [imath]k[/imath] in an alphabet of [imath]n[/imath] letters. You can write an arbitrarly long word and the computer will try all the subwords of [imath]k[/imath] consecutive letters. What is the smallest word that contains all combinations of [imath]k[/imath] letters as subword? (i.e. the fastest way to hack the computer :) ) The smallest word that contains [imath]n^k[/imath] subwords of size [imath]k[/imath] has length [imath]k-1+n^k[/imath] and based on some easy cases, we would like to prove that it is in fact possible to find a word of such length that contains all possible passwords. The problem can be translated into a problem in graph theory, by taking as vertices all words of length [imath]k[/imath]. We tried [imath]k=2[/imath], where you can prove the conjecture by induction. For [imath]n=2[/imath] and small [imath]k[/imath] it also works. | 1320278 | Puzzle: Cracking the safe
A safe is protected by a four-digit [imath](0-9)[/imath] combination. The safe only considers the last four digits entered when deciding whether an input matches the passcode. For instance, if I enter the stream [imath]012345[/imath], I am trying each of the combinations [imath]0123[/imath], [imath]1234[/imath], and [imath]2345[/imath]. Clearly, a 40000-length string [imath]000000010002...9999[/imath] is guaranteed to crack the safe. Can we try each of the 10000 combinations using a shorter string? What's the shortest string we can devise to try every combination? |
166553 | How is the Integral of [imath]\int_a^bf(x)~dx=\int_a^bf(a+b-x)~dx[/imath]
Can Some one tell me what this method is called and how it works With a detailed proof [imath]\int_a^bf(x)~dx=\int_a^bf(a+b-x)~dx[/imath] I've been using this a lot in definite integration but haven't seemed to have realized why it is true. But whatever it is it always seems to work. Basically a proof of how it is always true. | 1064896 | Show that [imath]\int_0^a f(x)dx=\int_0^a f(a-x)dx[/imath]
I don't really know where to start with this one. Can you just ignore the [imath]f(..)[/imath] and deal exclusively with what's inside the brackets? |
45008 | Equivalent conditions for a preabelian category to be abelian
Let's fix some terminology first. A category [imath]\mathcal{C}[/imath] is preabelian if: 1) [imath]Hom_{\mathcal{C}}(A,B)[/imath] is an abelian group for every [imath]A,B[/imath] such that composition is biadditive, 2) [imath]\mathcal{C}[/imath] has a zero object, 3) [imath]\mathcal{C}[/imath] has binary products, 4) [imath]\mathcal{C}[/imath] has kernels and cokernels. A category [imath]\mathcal{C}[/imath] is abelian if it is preabelian and satisfies: 5) every monomorphism is a kernel and every epimorphism is a cokernel. Define the coimage of a map to be the cokernel of its kernel, and the image to be the kernel of its cokernel. We have the following commutative diagram: where [imath]\overline{f}[/imath] is the only existing map (because of universality of kernel and cokernel). I'm having trouble proving the following: A preabelian category [imath]\mathcal{C}[/imath] is abelian iff [imath]\overline{f}[/imath] is an isomorphism. The converse is easily shown, I'm having trouble proving [imath]\Rightarrow[/imath]... | 220051 | Image in abelian categories
[imath]\def\im{\operatorname{im}}\def\coker{\operatorname{coker}}[/imath]For a morphism [imath] f: A\to B[/imath] in an abelian category, we let [imath]\im f:=\ker(\coker f)[/imath]. Then the morphism [imath]A\to \im f[/imath] is an epimorphism and [imath]\coker(\ker f\to A).[/imath] May I have their proofs? |
20170 | Is there a measurable set [imath]A[/imath] such that [imath]m(A \cap B) = \frac12 m(B)[/imath] for every open set [imath]B[/imath]?
Is there a measurable set [imath]A[/imath] such that [imath]m(A \cap B)= \frac12 m(B)[/imath] for every open set [imath]B[/imath]? Edit: (t.b.) See also A Lebesgue measure question for further answers. | 1647113 | Existence of a subset [imath]S\subset\mathbb R[/imath] s.t. [imath]\forall a, S\cap [a,b] has Lebesgue measure (b-a)/2?[/imath]
I am trying to either find an example of such a set, or prove that no such set exists. I know of examples of dense sets with measure [imath]1/2[/imath] on specific intervals, such as [imath][0,1][/imath], but I haven't been able to find any set that satisfies this more general property. |
625 | Why is the derivative of a circle's area its perimeter (and similarly for spheres)?
When differentiated with respect to [imath]r[/imath], the derivative of [imath]\pi r^2[/imath] is [imath]2 \pi r[/imath], which is the circumference of a circle. Similarly, when the formula for a sphere's volume [imath]\frac{4}{3} \pi r^3[/imath] is differentiated with respect to [imath]r[/imath], we get [imath]4 \pi r^2[/imath]. Is this just a coincidence, or is there some deep explanation for why we should expect this? | 766642 | Validity of proof for surface area of a sphere
On a geometry test I forgot the formula for the surface area of a sphere so I derived it and ended up being right. But it seems like my derivation is wrong. I got the surface area formula by taking the derivative of the volume formula, [imath]\frac{4}{3}\pi r^3[/imath]. My reasoning was that taking the derivative is equivalent taking the volume of a sphere with radius [imath]r+h[/imath] minus the volume of a sphere with radius [imath]r[/imath] and then dividing by [imath]h[/imath] as [imath]h[/imath] goes to [imath]0[/imath]. I figure that if you took the part of the larger sphere that is not in the smaller sphere and laid it flat it would be a prism with height [imath]h[/imath], so its base area (i.e. surface area of the sphere) would be the volume divided by [imath]h[/imath]. So my questions are: First, why must [imath]h[/imath] go to [imath]0[/imath]? First I tried it with [imath]h = 1[/imath] thinking, since the height is [imath]1[/imath], the base area (i.e. surface area of sphere) would equal the volume. But when [imath]h=1[/imath] you get the wrong formula. Secondly, is the rest of the reasoning correct? Thanks, Elliot |
16244 | If [imath]\int_0^x f \ dm[/imath] is zero everywhere then [imath]f[/imath] is zero almost everywhere
I have been thinking on and off about a problem for some time now. It is inspired by an exam problem which I solved but I wanted to find an alternative solution. The object was to prove that some sequence of functions converges weakly to zero in [imath]L^2[/imath]. I managed to show (with some help) that the limit [imath]f[/imath] (of a subsequence) satisfies [imath]\int_0^x f \ dm=0[/imath] for all [imath]x>0 [/imath]. From this I want to conclude that [imath]f=0[/imath] a.e. I can do this with the fundamental theorem of calculus in its Lebesgue version but there ought to be a more elementary proof. Can someone here help me out? | 1837775 | Zero integral implies zero function almost everywhere
Assume [imath]f[/imath] is Riemann integrable and further assume that [imath]\int_a^x f=0[/imath] for all [imath]x[/imath]. How would I go about showing that [imath]f[/imath] itself is [imath]0[/imath] almost everywhere? I am new to Lebesgue's measure theory so I am hoping for a somewhat elementary proof if possible? I know that almost everywhere means all except a set of measure zero. I was wondering if I could get a point to start on? We are told [imath]f[/imath] is Riemann integrable, so that means by Lebesgue's criterion for Riemann integration that there are at most countably infinitely many discontinuities. Thank you! |
114462 | A map is continuous if and only if for every set, the image of closure is contained in the closure of image
As a part of self study, I am trying to prove the following statement: Suppose [imath]X[/imath] and [imath]Y[/imath] are topological spaces and [imath]f: X \rightarrow Y[/imath] is a map. Then [imath]f[/imath] is continuous if and only if [imath]f(\overline{A})\subseteq \overline{f(A)}[/imath], where [imath]\overline{A}[/imath] denotes the closure of an arbitrary set [imath]A[/imath]. Assuming [imath]f[/imath] is continuous, the result is almost immediate. Perhaps I am missing something obvious, but I have not been able to make progress on the other direction. Could anyone give me a hint which might illuminate the problem for me? | 740588 | Proving that [imath]f(\bar Z)\subset\overline {f(Z)}[/imath] when [imath]f[/imath] is a continuous map
I'm trying to solve this question from my textbook: Let [imath]f:X\rightarrow Y[/imath] be a continuous map and let [imath]Z \subset X[/imath]. Prove the inclusion [imath]f(\bar Z)\subset\overline {f(Z)}[/imath]. Thanks in advance for any help! |
93816 | Proving that an additive function [imath]f[/imath] is continuous if it is continuous at a single point
Suppose that [imath]f[/imath] is continuous at [imath]x_0[/imath] and [imath]f[/imath] satisfies [imath]f(x)+f(y)=f(x+y)[/imath]. Then how can we prove that [imath]f[/imath] is continuous at [imath]x[/imath] for all [imath]x[/imath]? I seems to have problem doing anything with it. Thanks in advance. | 2445643 | Conditions about continuous functions
Say we have [imath]f(x+y) = f(x) + f(y) \quad \forall x,y \in \mathbb R[/imath] and [imath]f[/imath] is continuous at one point at least. I wish to show there must be some [imath]c[/imath] such that [imath]f(x)=cx[/imath] for all [imath]x[/imath]. Think I can do so by first showing [imath]f[/imath] is continuous everywhere I'm not sure how then let [imath]f(q) = 1[/imath] somehow and show that [imath]f(q) = cq[/imath] where [imath]q[/imath] is rational. But then the aim is to show for all real [imath]x[/imath] so I am not sure~ |
10400 | Comaximal ideals in a commutative ring
Let [imath]R[/imath] be a commutative ring and [imath]I_1, \dots, I_n[/imath] pairwise comaximal ideals in [imath]R[/imath], i.e., [imath]I_i + I_j = R[/imath] for [imath]i \neq j[/imath]. Why are the ideals [imath]I_1^{n_1}, ... , I_r^{n_r}[/imath] (for any [imath]n_1,...,n_r \in\mathbb N[/imath]) also comaximal? | 328890 | If a finite set of ideals generates a ring, then so does any set of arbitrary powers of those ideals.
In Lang's Algebra, pg 95 (3rd Revised Ed.), he concludes a proof on the Chinese Remainder Theorem with: In the same vein as above, we observe that if [imath]\mathfrak{a_1},\dots,\mathfrak{a_n}[/imath] are ideals of a ring [imath]A[/imath] such that [imath] \mathfrak{a_1}+ \dots + \mathfrak{a_n}=A,[/imath] and if [imath]\mathit{v_1},\dots,\mathit{v_n}[/imath] are positive integers, then [imath]\mathfrak{a_1}^\mathit{v_1}+\dots+\mathfrak{a_n}^\mathit{v_n}=A.[/imath] He states the proof is a trivial consequence of the CRT. However, I have been unable to find one that is satisfactory. I can see that there is a set of elements, one from each ideal which sum to 1. I am at a loss to see how we can find such a set that satisfies the above claim. |
125065 | Partitioning a natural number [imath]n[/imath] in order to get the maximum product sequence of its addends
Suppose we have a natural number [imath]n \ge 0[/imath]. Given natural numbers [imath]\alpha_1,\ldots,\alpha_k[/imath] such that [imath]k\le n[/imath] [imath]\sum_i \alpha_i = n[/imath] what is the maximum value that [imath]\Pi_i \alpha_i[/imath] can take? I'm quite sure that there is a theorem telling me the result, but I cannot find it. For sure an upper bound is [imath]n^k[/imath] but I'm searching for a real upper bound. I'm pretty sure that upper the bound should be [imath]n^2[/imath], but I don't know I could prove it. | 1860096 | Maximize the product of the partitions of an integer
Let [imath]n>0[/imath] be an integer. Consider all partitions of [imath]n[/imath], i.e. all possible ways of writing [imath]n[/imath] as a finite sum of positive integers, [imath]n=n_1+n_2+\cdots+n_k.[/imath] What partition maximizes the product [imath]n_1n_2\cdots n_k[/imath]? Examples: [imath]2=2[/imath] [imath]3=3[/imath] [imath]4=2+2[/imath] [imath]5=3+2[/imath] [imath]6=3+3[/imath] [imath]7=3+2+2[/imath] [imath]8=3+3+2[/imath] [imath]9=3+3+3[/imath] [imath]10=3+2+2[/imath] [imath]20=3+3+3+3+3+3+2[/imath] [imath]30=3+3+3+3+3+3+3+3+3+3[/imath] [imath]40=3+3+3+3+3+3+3+3+3+3+3+3+2+2[/imath] [imath]50=3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+3+2[/imath] Conjecture: It is the partition that has the maximum number of [imath]3[/imath]'s among those consisting only of [imath]3[/imath]'s and [imath]2[/imath]'s. |
52373 | Proof that [imath]\gcd(ax+by,cx+dy)=\gcd(x,y)[/imath] if [imath]ad-bc= \pm 1[/imath]
I'm having problems with an exercise from Apostol's Introduction to Analytic Number Theory. Given [imath]x[/imath] and [imath]y[/imath], let [imath]m=ax+by[/imath], [imath]n=cx+dy[/imath], where [imath]ad-bc= \pm 1[/imath]. Prove that [imath](m,n)=(x,y)[/imath]. I've tried to give a proof, but I suspect it's wrong (or at least not very good). I would be very thankful for any hints/help/advice! My proof: We observe that since [imath]ad-bc= \pm 1[/imath], [imath]ad=bc \pm 1[/imath], and [imath](ad,bc)=1[/imath]. Now, [imath](a,b) \mid m[/imath] and [imath](c,d) \mid n[/imath] but [imath](a,b) = ((d,1)a,(c,1)b)=(ad,a,bc,b)=(ad,bc,a,b)=((ad,bc),a,b)=(1,a,b)=1.[/imath] Similarly, we determine [imath](c,d)=1[/imath]. So, [imath]1=(a,b) \mid m[/imath] and [imath]1=(c,d) > \mid n[/imath]. But [imath](x,y)[/imath] also divide [imath]m[/imath] and [imath]n[/imath]. Since [imath](x,y) \geq (a,b)=(c,d)=1[/imath], this implies that [imath](x,y)=m,n[/imath]. Hence [imath](m,n)=((x,y),(x,y))=(x,y)[/imath]. | 2947523 | [imath]h\mid (3a + 5b)[/imath], prove [imath]h\mid a[/imath] and [imath]h\mid b[/imath]
I have this homework question. "For any integer [imath]a[/imath] and [imath]b[/imath], prove that [imath]\gcd(a,b) = \gcd(3a+5b,11a+18b)[/imath]." I know that if [imath] g = \gcd(a,b)[/imath] and [imath]h = \gcd(3a+5b,11a+18b)[/imath] then [imath]g = h[/imath] iff [imath]g \leq h [/imath] and [imath]h \leq g[/imath]. I successfully proved that [imath]g \leq h[/imath]. Now, to prove that [imath]h \leq g[/imath], I need to prove that [imath]h\mid (a,b)[/imath], but I can't seem to find how I should prove this. [imath]\because h = gcd(3a+5b,11a+18b) \Rightarrow h \mid (3a+5b)[/imath] From here I'm stuck on how to get [imath]a[/imath] and [imath]b[/imath] seperate. Any hint would be very helpful. Edit: Since this question was marked duplicate and I was given these 1,2,3 links to check, I did check them and didn't find my answer because all of these questions have given that [imath]gcd = 1[/imath], whereas my question doesn't tell if [imath]gcd = 1[/imath] and furthermore these questions are a bit complex for me to understand since I'm a new learner of number theory. |
115228 | Solution(s) to [imath]f(x + y) = f(x) + f(y)[/imath] (and miscellaneous questions...)
My lecturer was talking today (in the context of probability, more specifically Kolmogorov's axioms) about the additive property of functions, namely that: [imath]f(x+y) = f(x) + f(y)[/imath] I've been trying to find what functions satisfy this. Intuition says that, for functions over [imath]\mathbb{R}[/imath], the only functions should be of the form [imath]f(x) = ax[/imath] for some real a. Unfortunately I've only shown this is true when the domain of the function is the rational multiples of a given real number. My question is if it is possible to extend this result (that [imath]f(x) = ax[/imath] given additivity) to the real numbers, possibly without assuming the continuity of f. It seems to me that additivity introduces so many constrains on a function that nothing but the trivial case would be able to sneak through. The following is a summary of my thoughts to date, though they're obviously long and not 'compulsory reading'. :) When x is rational - Preliminary Investigation It is not hard to see that: [imath]f(x + x + x) = 3f(x)[/imath] and more generally, for [imath]a\in \mathbb{N}[/imath], [imath]f(ax) = af(x)[/imath] It is not too hard to prove (well, it took half a bus trip ... ) that this also applies first for [imath]a\in \mathbb{Z}[/imath] and then for [imath]a\in \mathbb{Q}[/imath], (for the latter you just need to consider [imath]a=m/n[/imath] and then note that: [imath]f\Big(\frac{m}{n}x\Big)=mf\Big(\frac{x}{n}\Big)=\frac{m}{n}\cdot nf\Big(\frac{x}{n}\Big)=\frac{m}{n}\cdot f\Big(n\frac{x}{n}\Big)=\frac{m}{n}\cdot f(x)[/imath] The reason this little equation is cool is that we can set [imath]x = 1[/imath] and get: [imath]f(a)=a\cdot f(1)[/imath] which is equivalent to what was expected intuitively, namely (after changing [imath]a[/imath] to [imath]y[/imath] and [imath]f(1)[/imath] to [imath]a[/imath]) [imath]f(y) = a\cdot y[/imath] as long as y is rational [imath]y[/imath] is a rational multiple of a real number But we can do a bit better than that. If we substitute in [imath]x = \sqrt{2}[/imath] or any other real number in [imath]f(ax) = af(x)[/imath] (which we know for rational [imath]a[/imath]), you can conduct the exact same argument above and show that, for instance [imath]f(y) = \Big(\frac{f(\sqrt{2})}{\sqrt{2}}\Big)\cdot y=a\cdot y[/imath] Whenever [imath]y = \frac{m}{n}\sqrt{2}[/imath] i.e. whenever [imath]y[/imath] is a rational multiple of [imath]\sqrt{2}[/imath]. Note however, that the value of the coefficient [imath]a[/imath] (i.e. the slope of the line) is apparently completely unrelated to the value taken in the case where [imath]y[/imath] is purely rational. What I'm actually asking We still haven't shown that [imath]f(x) = ax[/imath] for all [imath]x \in \mathbb{R}[/imath], as the slope of the line may change depending on what real number we are taking rational multiples of. As far as I've shown now, we might have [imath]f(x) = x[/imath] when [imath]x[/imath] is rational, [imath]f(x) = 3x[/imath] when [imath]x[/imath] is a rational multiple of [imath]\sqrt{2}[/imath], etc. I still feel that [imath]f(x) = ax[/imath] for all [imath]x \in \mathbb{R}[/imath]. One reason for thinking this comes from noting that [imath]f(2) = f(2-\sqrt{2})+f(\sqrt{2})[/imath] [imath]2[/imath], [imath]2-\sqrt{2}[/imath] and [imath]\sqrt{2}[/imath] are not rational multiples of each other, however the equation above gives a restraint on the slopes of the lines formed by their rational multiples (which we'll call [imath]a_1, a_2[/imath] and [imath]a_3[/imath] for the slopes on the rational multiples of [imath]2, 2-\sqrt{2}[/imath] and [imath]\sqrt{2}[/imath] respectively). We have [imath]2a_1 = (2-\sqrt{2}) a_2 + \sqrt{2} a_3[/imath] There's so many constraints here - all the rational multipes have the same coefficient, whenever 2 (or more) numbers which aren't rational multiples of each other are added together we get another constraint on their coefficients. The trivial solution is just that[imath]f(x) = ax[/imath] over [imath]x \in \mathbb{R}[/imath] and I really struggle to see how any other solution could possible squeeze through all these constraints. Is there an additive function on [imath]\mathbb{R}[/imath] not of the form [imath]f(x) = ax[/imath]? | 536735 | Find all [imath]f: \mathbb{Q} \rightarrow \mathbb{R}[/imath] such that [imath]f(x+y) = f(x)+f(y)[/imath]
i have to find all functions [imath]f: \mathbb{Q} \rightarrow \mathbb{R}[/imath], such that [imath]f(x+y)=f(x)+f(y)[/imath]. So functions of the form [imath]f(x) := ax, a \in \mathbb{R}[/imath] satisfy the above condition: [imath] f(x+y)=a(x+y)=ax+ay=f(x)+f(y) [/imath] But how do i proove that all functions that satisfy the above condition have the form [imath]f(x) := ax, a \in \mathbb{R}[/imath]? Thanks in advance! |
59738 | Probability for the length of the longest run in [imath]n[/imath] Bernoulli trials
Suppose a biased coin (probability of head being [imath]p[/imath]) was flipped [imath]n[/imath] times. I would like to find the probability that the length of the longest run of heads, say [imath]\ell_n[/imath], exceeds a given number [imath]m[/imath], i.e. [imath]\mathbb{P}(\ell_n > m)[/imath]. It suffices to find the probability that length of any run of heads exceeds [imath]m[/imath]. I was trying to approach the problem by fixing a run of [imath]m+1[/imath] heads, and counting the number of such configurations, but did not get anywhere. It is easy to simulate it: I would appreciate any advice on how to analytically solve this problem, i.e. express an answer in terms of a sum or an integral. Thank you. | 513808 | Probability of Runs of Heads of Length N
For example: [imath]“THHTHTTHHHTHTHTTHHTHT”[/imath] contains 1 run of heads of length 3, 2 runs of length 2, and 4 runs of length 1. Assuming [imath]P(H) = p[/imath] and [imath]P(T) = (1-p)[/imath], calculate (using properties such as conditional probability and Bayes’ Rule) the probability of runs of [imath]H[/imath]eads of length [imath]n[/imath]. Where: [imath]P(H)[/imath] is the probability of heads [imath](p)[/imath] [imath]P(T)[/imath] is the probability of tails [imath](1-p)[/imath] The implication that conditional probability and Bayes' Rule can be used would imply that this solution is much more simple than I am making it. That being said, numerous searches have returned complex solutions containing recursive functions, often the Fibonacci series. Furthermore, none of the solutions I have seen are consistent with one another. From the derived equation, I know that I can then use the linearity of expectation to quickly solve for the probability of runs of length [imath]k = 1, 2, ... 10[/imath] for a total number of events (flips) [imath]N >> 1[/imath] What I could use assistance with is calculating runs of [imath]H[/imath]eads of length n given the probabilities for heads [imath](p)[/imath] and tails [imath](p-1)[/imath]. |
244644 | Question about set notation: what does [imath]]a,b[[/imath] mean?
In the following question here the notation [imath]c\in ]a,b[[/imath] is used. What does this mean? I have never seen it before. | 1077112 | What is the meaning of the notation [imath]]a,b[[/imath]?
I've seen the notation [imath]]a,b[[/imath] in several questions on this site, but I am not familiar with it. Can someone clue me in? |
43032 | How to obtain tail bounds for a sum of dependent and bounded random variables?
Note: I divide this question to two separated question not to be duplicate version. I am looking for tail bounds (preferably exponential) for the sum of dependent and bounded random variables. Consider [imath]K_{ij}=\sum_{r=1}^N\sum_{c=1}^N W_{ir}W_{jc}[/imath] where [imath]i \neq j[/imath], [imath]W\in \{+1, -1\}[/imath] and [imath]W[/imath] are i.i.d. random variables [imath]\operatorname{Bernoulli}(0.5)[/imath]. How can I obtain an exponential bound over [imath]Pr[K_{ij} \geq \epsilon] \leq \exp(?)[/imath] where [imath]\epsilon[/imath] is a positive value. Answer: [imath]K_{ij}[/imath] can be considered as the multiplication of two independent Binomial random variables, i.e., [imath]K_{ij}=\left(\sum_{r=1}^N W_{ir}\right)\left(\sum_{c=1}^N W_{jc}\right)[/imath], then the moment generating function can be evaluated and then by using the Chernoff we can have a tail bound for [imath]K_{ij}[/imath]. Thanks a lot in advance. | 42997 | How to obtain tail bounds for a linear combination of dependent and bounded random variables?
I am looking for tail bounds (preferably exponential) for a linear combination of dependent and bounded random variables. consider [imath]K_{ij}=\sum_{r=1}^N\sum_{c=1}^N W_{ir}C_{rc}W_{jc}[/imath] where [imath]i \neq j[/imath], [imath]W\in \{+1, -1\}[/imath] and [imath]W[/imath] follows [imath]\operatorname{Bernoulli}(0.5)[/imath], and [imath]C=\operatorname{Toeplitz}(1, \rho, \rho^2, \ldots, \rho^{N-1})[/imath], [imath]0 \leq \rho < 1[/imath]. I will be to happy if you give me any pointer to how I can evaluate the moment generating function of [imath]K_{ij}[/imath] to have bound for [imath]Pr\{K_{ij} \geq \epsilon\}\leq \min_s\exp(-s\epsilon)E[\exp(K_{ij}s)][/imath] based on chernoff bound ans [imath]s \geq 0[/imath]. For a hint you can look at my another question entitled as ``How to obtain tail bounds for a sum of dependent and bounded random variables?'' which is a special case of this problem where [imath]C_{rc}=1, 1 \leq c \leq N, 1 \leq r \leq N[/imath]. Thanks a lot in advance. |
33970 | Finding the limit of [imath]\frac{Q(n)}{P(n)}[/imath] where [imath]Q,P[/imath] are polynomials
Suppose that [imath]Q(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots+a_{1}x+a_{0} [/imath]and [imath]P(x)=b_{m}x^{m}+b_{m-1}x^{m-1}+\cdots+b_{1}x+b_{0}.[/imath] How do I find [imath]\lim_{x\rightarrow\infty}\frac{Q(x)}{P(x)}[/imath] and what does the sequence [imath]\frac{Q(k)}{P(k)}[/imath] converge to? For example, how would I find what the sequence [imath]\frac{8k^2+2k-100}{3k^2+2k+1}[/imath] converges to? Or what is [imath]\lim_{x\rightarrow\infty}\frac{3x+5}{-2x+9}?[/imath] This is being asked in an effort to cut down on duplicates, see here: Coping with abstract duplicate questions. and here: List of abstract duplicates. | 1477946 | How to solve this limit: [imath]\lim_{n \to \infty} \frac{(2n+2) (2n+1) }{ (n+1)^2}[/imath]
[imath] \lim_{n\to\infty}\frac{(2n+2)(2n+1)}{(n+1)^{2}} [/imath] When I expand it gives: [imath] \lim_{n\to\infty} \dfrac{4n^{2} + 6n + 2}{n^{2} + 2n + 1} [/imath] How can this equal [imath]4[/imath]? Because if I replace [imath]n[/imath] with infinity it goes [imath]\dfrac{\infty}{\infty}[/imath] only. |
66052 | Prove that the center of a group is a normal subgroup
Let [imath]G[/imath] be a group. We define [imath]H=\{h\in G\mid \forall g\in G: hg=gh\},[/imath] the center of [imath]G[/imath]. Prove that [imath]H[/imath] is a (normal) subgroup of [imath]G[/imath]. | 330109 | Elements which commute with a given element form a subgroup
Let [imath](G,∗)[/imath] be a group and [imath]a\in G[/imath] then so that set of elements [imath]x[/imath] of [imath]G[/imath] such that [imath]a∗x = x∗a[/imath] is a subgroup of [imath]G[/imath]. I have tried by using theorem that [imath]H[/imath] is as subgroup of [imath]G[/imath] if and only if for any [imath]a,b\in H[/imath] , [imath]a∗b^{-1}\in H[/imath] but didn’t get the result Please help me |
301005 | Is there an example of a function [imath]f: \mathbb{Z} \to \{\text{finite subsets of }\mathbb{Z}\}[/imath]?
In my last question, I asked for a proof of "Are the set of all finite subsets in [imath]\mathbb{Z}[/imath] countable?" . I had a good answer that showed me that it is an [imath]f: \mathbb{N} \to \{\text{finite subsets of }\mathbb{Z}\}[/imath]. So knowing that there exists a bijection [imath]\mathbb{N} \leftrightarrow \mathbb{Z}[/imath], then it is proved. But I am curious about an example (if it exists) of a function [imath]f: \mathbb{Z} \to \{\text{finite subsets of }\mathbb{Z}\}[/imath] Does such an example exist? | 200389 | Show that the set of all finite subsets of [imath]\mathbb{N}[/imath] is countable.
Show that the set of all finite subsets of [imath]\mathbb{N}[/imath] is countable. I'm not sure how to do this problem. I keep trying to think of an explicit formula for 1-1 correspondence like adding all the elements in each subset and sending that sum to itself in the natural numbers, but that wouldn't be 1-1 because, for example, the set {1,2,3} would send to 6 and so would the set {2,4}. Multiplying all the elements in each subset and sending that product to itself in the natural numbers wouldn't work either since, for example, {2,3} would send to 5 and so would the set {1,5}. Any ideas? |
96826 | The Monty Hall problem
I was watching the movie 21 yesterday, and in the first 15 minutes or so the main character is in a classroom, being asked a "trick" question (in the sense that the teacher believes that he'll get the wrong answer) which revolves around theoretical probability. The question goes a little something like this (I'm paraphrasing, but the numbers are all exact): You're on a game show, and you're given three doors. Behind one of the doors is a brand new car, behind the other two are donkeys. With each door you have a [imath]1/3[/imath] chance of winning. Which door would you pick? The character picks A, as the odds are all equally in his favor. The teacher then opens door C, revealing a donkey to be behind there, and asks him if he would like to change his choice. At this point he also explains that most people change their choices out of fear; paranoia; emotion and such. The character does change his answer to B, but because (according to the movie), the odds are now in favor of door B with a [imath]1/3[/imath] chance of winning if door A is picked and [imath]2/3[/imath] if door B is picked. What I don't understand is how removing the final door increases the odds of winning if door B is picked only. Surely the split should be 50/50 now, as removal of the final door tells you nothing about the first two? I assume that I'm wrong; as I'd really like to think that they wouldn't make a movie that's so mathematically incorrect, but I just can't seem to understand why this is the case. So, if anyone could tell me whether I'm right; or if not explain why, I would be extremely grateful. | 1141915 | Conditional probability problem of three choices
I have the following problem where I have difficulties grasping the intuition: Lets say we have three boxes, with two of them empty and one containing a gold price. Lets say we randomly select one of the boxes. After our selection, we are given which one of the remaining two boxes does not contain the price. Now the question is: Should I stick with my original selection or select another box from the two possible alternatives left. What are the probabilities? I empirically tried this problem by making a computer program to repeat this experiment 1,000,000 times with first staying with the original choice and then always changing the selection. I got the probabilities to be: [imath]P(golden\; price\;with\;original\;selection)\approx33\%[/imath] [imath]P(golden\; price\;with\;changing\;selection)\approx 66\%[/imath] Intuitively the probabilities seem at first to be 50% for both of these choices, but it seems it's not the case. I can't grasp on why?... P.S. please let me know if my question is unclear |
257392 | If [imath]\gcd(a,b)=1[/imath], then [imath]\gcd(a+b,a^2 -ab+b^2)=1[/imath] or [imath]3[/imath].
Hint: [imath]a^2 -ab +b^2 = (a+b)^2 -3ab.[/imath] I know we can say that there exists an [imath]x,y[/imath] such that [imath]ax + by = 1[/imath]. So in this case, [imath](a+b)x + ((a+b)^2 -3ab)y =1.[/imath] I thought setting [imath]x = (a+b)[/imath] and [imath]y = -1[/imath] would help but that gives me [imath]3ab =1.[/imath] Any suggestions? | 522289 | Given that [imath]gcd(a,b)=1[/imath], prove that [imath]gcd(a+b,a^2-ab+b^2)=1[/imath] or [imath]3[/imath], also when will it equal [imath]1[/imath]?
It is an exercise on the lecture that i am unable to prove. Given that [imath]gcd(a,b)=1[/imath], prove that [imath]gcd(a+b,a^2-ab+b^2)=1[/imath] or [imath]3[/imath], also when will it equal [imath]1[/imath]? |
18179 | Finding Value of the Infinite Product [imath]\prod \Bigl(1-\frac{1}{n^{2}}\Bigr)[/imath]
While trying some problems along with my friends we had difficulty in this question. True or False: The value of the infinite product [imath]\prod\limits_{n=2}^{\infty} \biggl(1-\frac{1}{n^{2}}\biggr)[/imath] is [imath]1[/imath]. I couldn't do it and my friend justified it by saying that since the terms in the product have values less than [imath]1[/imath], so the value of the product can never be [imath]1[/imath]. I don't know whether this justification is correct or not. But i referred to Tom Apostol's Mathematical Analysis book and found a theorem which states, that The infinite product [imath]\prod(1-a_{n})[/imath] converges if the series [imath]\sum a_{n}[/imath] converges. This assures that the above product converges. Could anyone help me in finding out where it converges to? And, Does there exist a function [imath]f[/imath] in [imath]\mathbb{N}[/imath] ( like [imath]n^{2}[/imath], [imath]n^{3}[/imath]) such that [imath]\displaystyle \prod\limits_{n=1}^{\infty} \Bigl(1-\frac{1}{f(n)}\Bigr)[/imath] has the value [imath]1[/imath]? | 513053 | Is [imath]\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1[/imath]
Question is to check if [imath]\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=1[/imath] we have [imath]\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=\prod \limits_{n=2}^{\infty}(\frac{n^2-1}{n^2})=\prod \limits_{n=2}^{\infty}\frac{n+1}{n}\frac{n-1}{n}=(\frac{3}{2}.\frac{1}{2})(\frac{4}{3}.\frac{2}{3})(\frac{5}{4}.\frac{3}{4})...[/imath] In above product we have for each term [imath]\frac{a}{b}[/imath] a term [imath]\frac{b}{a}[/imath] except for [imath]\frac{1}{2}[/imath].. So, all other terms gets cancelled and we left with [imath]\frac{1}{2}[/imath]. So, [imath]\prod \limits_{n=2}^{\infty}(1-\frac{1}{n^2})=\frac{1}{2}[/imath]. I would be thankful if some one can assure that this explanation is correct/wrong?? I am solving this kind of problems for the first time so, it would be helpful if some one can tell if there are any other ways to do this.. Thank you |
31097 | A lady and a monster
A famous problem: a lady is in the center of the circular lake and a monster is on the boundary of the lake. The speed of the monster is [imath]v_m[/imath], and the speed of the swimming lady is [imath]v_l[/imath]. The goal of the lady is to come to the ground without meeting the monster, and the goal of the monster is to meet the lady. Under some conditions on [imath]v_m,v_l[/imath] the lady can always win. What if these conditions are not satisfied? Edited: the monster cannot swim. If the conditions are not satisfied, then monster can always perform a strategy such that the lady will not escape the lake. On the other hand this strategy is not desirable for both of them because they do not reach their goals. As there was mentioned, this deals with undecidability of the problem. On the other hand, if you imagine yourself to be this lady/monster, you can be interested in the strategy which is not optimal. What is it? If there are such strategies in the game theory? Edited2: My question is more general in fact. If we have a game with one parameter [imath]v[/imath] when two players [imath]P_1, P_2[/imath] are enemies and if [imath]v>0[/imath] then for any strategy of [imath]P_2[/imath] the player [imath]P_1[/imath] wins. If [imath]v\leq 0[/imath] then for any strategy of [imath]P_2[/imath] there is a strategy of [imath]P_1[/imath] such that [imath]P_2[/imath] does not win and vice versa. I am interested in this case. From the mathematical point of view as I have understood the problem is undecidable since there is no an ultimate strategy neither for [imath]P_1[/imath] nor for [imath]P_2[/imath]. But we are solving somehow these problem IRL. Imagine that you are a lady in this game - then you would like to win anyway even while knowing that your strategy can be covered by the strategy of the monster. On the other hand, the monster knows that if he will cover all strategies of the lady she will never reach the shore and he will never catch her. I mean they have to develop some non-optimal strategies. I hope now it's more clear. | 2646032 | What it the best constant in the riddle of the man and the tiger?
[imath]\newcommand{\man}{\text{man}}[/imath] [imath]\newcommand{\tiger}{\text{tiger}}[/imath] Consider the following situation: A man is standing in the center of a circle of radius [imath]r[/imath]. On the circle there is a tiger. The man move in arbitrary piecewise smooth paths with a fixed speed, and the tiger can move only along the circle, where [imath]V_{\tiger}=cV_{\man} [/imath], [imath]c>1[/imath]. I model the man and the tiger as "points" - they have no actual volume. Question: For which values of [imath]c[/imath] can the man escape from the enclosing circle without meeting the tiger? (assuming the tiger does its best to catch him). If [imath]c < \pi[/imath] then its trivial. The man just goes in a straight line in the opposite direction from the tiger: The man reaches the circle after a time of [imath]\frac{r}{V_{\man}}[/imath] and the tiger reaches that same point after [imath]\frac{\pi r}{V_{\tiger}}[/imath]. In fact, there is a more sophisticated idea, which works for any [imath]c < \pi +1[/imath], as I explain below. Is [imath]\pi+1[/imath] the highest speeds ration when the game is solvable? The man starts by going to an intermediate circle with sufficiently small radius [imath]r_1[/imath], in the sense that [imath] \frac{r_1}{V_{\man}} < \frac{r}{V_{\tiger}}= \frac{r}{cV_{\man}} \Rightarrow r_1 < \frac{r}{c}. \tag{1}[/imath] The man than starts circling the small circle until it lies on the "opposite" point of the tiger, so the distance between them is [imath]r+r_1[/imath]. If [imath]r_1[/imath] satisfies [imath] \frac{r-r_1}{V_{\man}} < \frac{\pi r}{V_{\tiger}}=\frac{\pi r}{cV_{\man}}, \tag{2}[/imath] then the man can win. Combining [imath](1),(2)[/imath] we see that [imath]r_1[/imath] must satisfy [imath] r(1-\frac{\pi}{c}) < r_1 < \frac{r}{c},[/imath] so this technique can work if and only if [imath] 1-\frac{\pi}{c} < \frac{1}{c} [/imath], i.e. [imath]c < \pi +1[/imath]. I am also not entirely sure how to formalize this question mathematically. We are looking for a winning "strategy"-but what is a strategy? In my solution with the intermediate circle, what the man does (e.g. direction of movement) is affected by the tiger's moves. In that case one could say that there is a periodic path of the man such that no matter what the tiger does, there exist a "breaking point" where the man can run free. But is this really the shape of the best solution? |
28476 | Finding the limit of [imath]\frac {n}{\sqrt[n]{n!}}[/imath]
I'm trying to find [imath]\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}} .[/imath] I tried couple of methods: Stolz, Squeeze, D'Alambert Thanks! Edit: I can't use Stirling. | 516578 | What is the limit [imath] \underset{n\to\infty}{\lim} \frac {{n!}^{1/n}}{n} [/imath]
What is the following limit equal to and how do I prove it? [imath] \underset{n\to\infty}{\lim} \frac {{n!}^{1/n}}{n}. [/imath] I've been trying for a while and I can't seem to get it. |
22472 | Nested sequences of balls in a Banach space
This seems to be a fairly easy question but I'm looking for new points of view on it and was wondering if anyone might be able to help. (By the way- this question does come from home-work, but I've already solved and handed it, and I'm posting this out of interest, so no HW tag.) Let [imath]B_n=B(x_n,r_n)[/imath] be a sequence of nested closed balls in a Banach space [imath]X[/imath]. Prove that [imath]\bigcap_{n=1}^\infty B_n\neq\varnothing[/imath]. As I said before, it should be rather simple. When the radii decrease to 0, it's just a matter of selecting any sequence of points in [imath]B_n[/imath], and it must be Cauchy- and the limit is in the intersection. My question is what to do when the radii do not decrease to 0? I got some tips about multiplying the balls by a sequence of decreasing scalars, or reducing the radii so that they decrease to 0, but found too many pathological cases for both methods. Finally- I used a geometric arguemnt (which i've shown to work in any normed space) that if [imath]B(x_1,r_1)\subset B(x_2,r_2)[/imath] then [imath]\| x_1-x_2\|\leq|r_1-r_2|[/imath]. This turned out to be some kind of technical catastrophe, but it worked... Still, if anyone knows of a more elegant solution, I'd love to hear about it. | 1478376 | Let [imath]X[/imath] be a Banach space, and [imath]B_1\supseteq B_2 \supseteq\cdots [/imath] . Show that [imath]\bigcap\limits_{i=1}^\infty B_i\neq\emptyset[/imath]
Let [imath]X[/imath] be a Banach space, and [imath]B_1\supseteq B_2 \supseteq \cdots [/imath] a sequence of closed balls with radius [imath]r_i[/imath] and center [imath]x_i[/imath]. Show that [imath]\bigcap_{i=1}^\infty B_i\neq\emptyset[/imath] I proved that [imath]r_i\leq r_j[/imath] for [imath]B_i\subseteq B_j[/imath] and [imath]r_i\rightarrow\alpha\geq0[/imath] so, for [imath]\varepsilon>0\ \exists N\in\mathbb{N}[/imath] such that [imath]|r_i-r_j|<\varepsilon[/imath] for [imath]i,j>N[/imath] and just left to show that [imath]\|x_i-x_j\|<\varepsilon[/imath] so the centers is a chauchy sequence and it converges, so my question is how can I prove that [imath]\|x_i-x_j\|<\varepsilon[/imath]? |
37971 | Identity for convolution of central binomial coefficients: [imath]\sum\limits_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}[/imath]
It's not difficult to show that [imath](1-z^2)^{-1/2}=\sum_{n=0}^\infty \binom{2n}{n}2^{-2n}z^{2n}[/imath] On the other hand, we have [imath](1-z^2)^{-1}=\sum z^{2n}[/imath]. Squaring the first power series and comparing terms gives us [imath]\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}2^{-2n}=1[/imath] that is, [imath]\sum_{k=0}^n \binom{2k}{k}\binom{2(n-k)}{n-k}=2^{2n}[/imath] My question: is there a more direct, combinatorial proof of this identity? I've been racking my brains trying to come up with one but I'm not having much success. | 360053 | Finding a closed form expression for this sum
For non-negative [imath]n[/imath], find [imath] \sum_{k=0}^n \binom{2k}{k}\binom{2n-2k}{n-k}. [/imath] I can't figure this out. Any ideas? |
387 | Sum of reciprocals of numbers with certain terms omitted
I know that the harmonic series [imath]1 + \frac12 + \frac13 + \frac14 + \cdots[/imath] diverges. I also know that the sum of the inverse of prime numbers [imath]\frac12 + \frac13 + \frac15 + \frac17 + \frac1{11} + \cdots[/imath] diverges too, even if really slowly since it's [imath]O(\log \log n)[/imath]. But I think I read that if we consider the numbers whose decimal representation does not have a certain digit (say, 7) and sum the inverse of these numbers, the sum is finite (usually between 19 and 20, it depends from the missing digit). Does anybody know the result, and some way to prove that the sum is finite? | 329807 | Convergence of [imath]\sum^n_{k=1}\frac1k[/imath] after removing terms containing the digit [imath]p[/imath]
We know that [imath]\sum^n_{k=1}\frac1k[/imath] diverges. But if I were to pick a digit [imath]p[/imath] so that [imath]p[/imath] is an integer between [imath]0[/imath] and [imath]9[/imath] inclusive, and then I removed all terms in the sum [imath]\sum^n_{k=1}\frac1k[/imath] containing the digit [imath]p[/imath], what does the sum converge to, if it converges at all? After a few hours considering this problem, I have decided that the problem is beyond me and I need help. |
82489 | How to compute [imath]\int_0^{\pi/2}\frac{\sin^3 t}{\sin^3 t+\cos^3 t}dt[/imath]?
Calculating with Mathematica, one can have [imath]\int_0^{\pi/2}\frac{\sin^3 t}{\sin^3 t+\cos^3 t}\,\mathrm dt=\frac{\pi}{4}.[/imath] How can I get this formula by hand? Is there any simpler idea than using [imath]u = \sin t[/imath]? Is there a simple way to calculate [imath] \int_0^{\pi/2}\frac{\sin^n t}{\sin^n t+\cos^n t}\,\mathrm dt [/imath] for [imath]n>3[/imath]? Could anyone come up with a reference for this exercise? | 439851 | Evaluate the integral [imath]\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\,\mathrm dx[/imath].
Evaluate the integral [imath]\int^{\frac{\pi}{2}}_0 \frac{\sin^3x}{\sin^3x+\cos^3x}\, \mathrm dx.[/imath] How can i evaluate this one? Didn't find any clever substitute and integration by parts doesn't lead anywhere (I think). Any guidelines please? |
128657 | How to prove that a simple graph having 11 or more vertices or its complement is not planar?
It is an exercise on a book again. If a simple graph [imath]$G$[/imath] has 11 or more vertices,then either G or its complement [imath]$\overline { G } $[/imath] is not planar. How to begin with this? Induction? Thanks for your help! | 3026767 | Show that [imath]K_n[/imath] is not the union of two planar graphs
Show that [imath]K_n[/imath] is not the union of two planar graphs for [imath]n\ge 11[/imath] I know that a graph [imath]G[/imath] is planar iff it does not have [imath]K_5[/imath] or [imath]K_{3,3}[/imath] as its induced subgraphs But how to use it in the above problem. Please help. |
105633 | Prove that [imath]a=b[/imath], where [imath]a[/imath] and [imath]b[/imath] are elements of the integral domain [imath]D[/imath]
Let [imath]D[/imath] be an integral domain and [imath]a,~b \in D[/imath]. Suppose that [imath]a^n=b^n[/imath] and [imath]a^m=b^m[/imath] for any two some [imath]m,~n[/imath] such that [imath](m,n)=1[/imath]. Prove that [imath]a=b[/imath]. I know that [imath]ab≠0[/imath] since [imath]D[/imath] contains no divisors of zero, but I don’t have an idea as to how to prove this. | 82678 | [imath]a^m=b^m[/imath] and [imath]a^n=b^n[/imath] imply [imath]a=b[/imath]
Let [imath]D[/imath] be an integral domain and let [imath]a^m=b^m[/imath] and [imath]a^n=b^n[/imath] where [imath]m[/imath] and [imath]n[/imath] are relatively prime integers, [imath]a,b \in D[/imath]. How do I show [imath]a=b[/imath]? |
34271 | Order of general- and special linear groups over finite fields.
Let [imath]\mathbb{F}_3[/imath] be the field with three elements. Let [imath]n\geq 1[/imath]. How many elements do the following groups have? [imath]\text{GL}_n(\mathbb{F}_3)[/imath] [imath]\text{SL}_n(\mathbb{F}_3)[/imath] Here GL is the general linear group, the group of invertible n×n matrices, and SL is the special linear group, the group of n×n matrices with determinant 1. | 1069116 | Order of [imath]\mathrm{GL}_n(\mathbb F_p)[/imath] for [imath]p[/imath] prime
While proving some facts about matrix group operations on finite fields, I stumbled across the following question: What is the order of the group of invertible [imath]n\times n[/imath] matrices over a finite field of prime order [imath]p[/imath]? The answer seems to be [imath]\left|\mathrm{GL}_n(\mathbb F_p)\right|=\prod_{k=1}^n\left(p^n-p^{k-1}\right)\text,[/imath] but I have not yet seen a satisfying proof: I took a look at this one, but found it to be somewhat sloppy. Is there a nice formal proof of this fact? |
67148 | If [imath]a^3 =a[/imath] for all [imath]a[/imath] in a ring [imath]R[/imath], then [imath]R[/imath] is commutative.
Let [imath]R[/imath] be a ring, where [imath]a^{3} = a[/imath] for all [imath]a\in R[/imath]. Prove that [imath]R[/imath] must be a commutative ring. | 360958 | Prove that [imath]R[/imath] is a commutative ring if [imath]x^3=x[/imath]
Let [imath]R[/imath] be a ring satisfying: [imath]\forall x\in R, \; x^3=x[/imath]. Prove that [imath]R[/imath] is a commutative ring. |
48080 | Sum of First [imath]n[/imath] Squares Equals [imath]\frac{n(n+1)(2n+1)}{6}[/imath]
I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals: [imath]\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}[/imath] I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other? | 387664 | Proving [imath]\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}[/imath] without induction
I was looking at: [imath]\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}[/imath] It's pretty easy proving the above using induction, but I was wondering what is the actual way of getting this equation? |
65861 | Summation of natural number set with power of [imath]m[/imath]
Who knows about the summation of this series: [imath]\sum\limits_{i=1}^{n}i^m [/imath] where [imath]m[/imath] is constant and [imath]m\in \mathbb{N}[/imath]? thanks | 811693 | Calculate [imath]1*1+2*2+3*3+4*4+....+r*r[/imath]
If [imath]1+2+3+...+r[/imath] is in otherwords [imath]\frac{r(r+1)}{2}[/imath], then what is the answer with squares? Thank you for all the help. |
11464 | How to compute the formula [imath]\sum \limits_{r=1}^d r \cdot 2^r[/imath]?
Given [imath]1\cdot 2^1 + 2\cdot 2^2 + 3\cdot 2^3 + 4\cdot 2^4 + \cdots + d \cdot 2^d = \sum_{r=1}^d r \cdot 2^r,[/imath] how can we infer to the following solution? [imath]2 (d-1) \cdot 2^d + 2. [/imath] Thank you | 667191 | Difficulty with understanding summations
I am in advance sorry if this question is too easy for this site, but I am having real problem understanding how to solve this summation: [imath]\sum_{i=1}^n{i*2^i}[/imath] I understand basics of summations but i don't know where to start, please help. |
122274 | Why [imath]x^{p^n}-x+1[/imath] is irreducible in [imath]{\mathbb{F}_p}[/imath] only when [imath]n=1[/imath] or [imath]n=p=2[/imath]
I have a question, I think it concerns with field theory. Why the polynomial [imath]x^{p^n}-x+1[/imath] is irreducible in [imath]{\mathbb{F}_p}[/imath] only when [imath]n=1[/imath] or [imath]n=p=2[/imath]? Thanks in advance. It bothers me for several days. | 509277 | Dummit exercise 14.3.11: extension degree of finite fields
Dummit and Foote's exercise 14.3.11 asks to prove that [imath]f(x) = x^{p^{n}}-x+1[/imath] is irreducible over [imath]\mathbb{F}_{p}[/imath] iff [imath]n=1[/imath] or [imath]n=p=2[/imath]. To prove the 'only if' part, the exercise suggest to prove that if [imath]\alpha[/imath] is a root of [imath]f(x)[/imath], then so is [imath]\alpha + a[/imath] for every [imath]a \in \mathbb{F}_{p^{n}}[/imath]. This implies that [imath]\mathbb{F}_{p}(\alpha)[/imath] contains [imath]\mathbb{F}_{p^{n}}[/imath] and that [imath][\mathbb{F}_{p}(\alpha):\mathbb{F}_{p^{n}}] = p[/imath]. I've proven everything except that [imath][\mathbb{F}_{p}(\alpha):\mathbb{F}_{p^{n}}] = p[/imath]. Can anyone give me some ideas for this? Thanks. |
76946 | Prove the inequality [imath]n! \geq 2^n[/imath] by induction
I'm having difficulty solving an exercise in my course. The question is: Prove that [imath]n!\geq 2^n[/imath]. We have to do this by induction. I started like this: The lowest natural number where the assumption is correct is [imath]4[/imath] as: [imath]4! \geq 2^4 \iff 24 \ge 16[/imath]. The assumption is: [imath]n! \ge 2^n[/imath]. Now proof for [imath](n+1)[/imath] which brings me to: [imath](n+1)! \ge 2^{(n+1)}[/imath] I think I can rewrite it somehow like this: [imath] {(n+1)} \times {n!} \stackrel{\text{(definition of factorial)}}{\ge} 2^n \times 2 [/imath] [imath] (n+1) \times 2^n \ge 2^n \times 2 [/imath] Then I think I can eliminate the [imath]2^n[/imath] and have something like this: [imath]n+1 \ge 2[/imath], or [imath]n \ge 1[/imath]. But I think I'm wrong here some where and was hoping somebody has some advice on this. How can I prove the above assumption? Any help would be appreciated, kind regards. | 296790 | Prove [imath]n!>2^n[/imath] for [imath]n\geq4[/imath] using induction.
I just want to know if my proof to this question is correct. First, I proved it was true for [imath]n = 4[/imath]. [imath]4!>2^4[/imath] [imath]24>16[/imath] Then, I assumed that it was true for [imath]n=k[/imath]. [imath]k!>2^k[/imath] Afterwards, I tried to prove it for [imath]n=k+1[/imath]. [imath](k+1)!>2^{k+1}[/imath] [imath](k+1)k!>(2)(2^k)[/imath] Since [imath]k!>2^k[/imath], am I allowed to simply do [imath](k+1)(2^k)>(2)(2^k)[/imath] [imath](k+1)>2[/imath] and this is true for all [imath]k\geq3[/imath]. Would the last two steps be illegal because I only assumed that [imath]k!>2^k[/imath] is true? |
88300 | If [imath]f(x)[/imath] is continuous on [imath][a,b][/imath] and [imath]M=\max \; |f(x)|[/imath], is [imath]M=\lim \limits_{n\to\infty} \left(\int_a^b|f(x)|^n\,\mathrm dx\right)^{1/n}[/imath]?
Let [imath]f(x)[/imath] be a continuous real-valued function on [imath][a,b][/imath] and [imath]M=\max\{|f(x)| \; :\; x \in [a,b]\}[/imath]. Is it true that: [imath] M= \lim_{n\to\infty}\left(\int_a^b|f(x)|^n\,\mathrm dx\right)^{1/n} ? [/imath] Thanks! | 318634 | Show that [imath]\lim((\int_{a}^{b}f^{n})^\frac{1}{n})=\sup\{f(x):x\in[a,b]\}[/imath]
Another exercise (this one is 7.2.18) from "Introduction to Real Analysis" by Bartle and Sherbert that I'm struggling with: Let [imath]f[/imath] be continuous on [imath][a,b][/imath], let [imath]f(x)>=0[/imath] for [imath]x\in[a,b][/imath], and let [imath]M_{n}:=(\int_{a}^{b}f^{n})^\frac{1}{n}[/imath]. Show that [imath]\lim(M_{n})=\sup\{f(x):x\in[a,b]\}[/imath]. I think that as [imath]f[/imath] is continuous, thus bounded, on given interval, one should start with representing [imath]M_{n}[/imath] as [imath](M^{n}\int_{a}^{b}(\frac{f}{M})^{n})^\frac{1}{n}[/imath], where [imath]M[/imath] is essential supremum of [imath]f[/imath] on given interval, and then proceed from there; however, I'm not sure how to formalize further steps... Note that this particular exercise is following exercises on MVT for integrals. So another approach I tried was to follow from the fact that there exists [imath]c_{1},\ldots,c_{n}\in[a,b][/imath] such that [imath]M_{n}=(f(c_{1})\int_{a}^{b}f^{n-1})^\frac{1}{n}=\ldots=(f(c_{1})\ldots f(c_{n})(b-a))^\frac{1}{n}[/imath], but to no avail. |
49383 | How does [imath] \sum_{p grow asymptotically for \text{Re}(s) < 1 ?[/imath]
Note the [imath] p < x [/imath] in the sum stands for all primes less than [imath] x [/imath]. I know that for [imath] s=1 [/imath], [imath] \sum_{p<x} \frac{1}{p} \sim \ln \ln x , [/imath] and for [imath] \mathrm{Re}(s) > 1 [/imath], the partial sums actually converge to a finite limit called the prime zeta function, which has an analytic continuation to the whole right-half plane but the actual series diverges in the critical strip. So anyway, I'm wondering what the asymptotic behavior of the partial sums are in the limit as [imath] x \to \infty [/imath] for a given value of [imath] s [/imath] with [imath] \mathrm{Re}(s) < 1 [/imath]. At first I intuitively conjectured it might be something vaguely like the following [imath] \sum_{p<x} \frac1{p^s} \sim f(s) \pi(x)^{1-s} , \quad f(s) = \lim_{n\to\infty} \int_0^1 g_n(u) u^{-s} du [/imath] but after some thought I'm not sure if this kind of formula will work after all. Any ideas? Note again: I'm asking about asymptotics when [imath] \mathrm{Re}(s) < 1 [/imath]. | 679950 | prime zeta function when [imath]0[/imath]
I would like to know if there is a good estimate for the sum which concerns all primes not exceeding [imath]x[/imath]: [imath]\sum\limits_{p\leq x}\frac{1}{p^s}[/imath][imath]0<s<1[/imath]. Only this. Thanks in advance! |
58943 | The simply connected coverings of two homotopy equivalent spaces are homotopy equivalent
This is exercise 1.3.8 in Hatcher: Let [imath]\tilde{X}[/imath] and [imath]\tilde{Y}[/imath] be simply-connected covering spaces of path connected, locally path-connected spaces [imath]X[/imath] and [imath]Y[/imath]. Show that if [imath]X\simeq Y[/imath] then [imath]\tilde{X}\simeq \tilde{Y}[/imath]. I tried applying the lifting criterion, but I seem to be hitting a dead end. Any help would be much appreciated. | 136405 | Homotopy equivalence of universal cover
As part of am exam question (Q21F here), I'm trying to prove that if [imath]X[/imath] and [imath]Y[/imath] are path-connected, locally path-connected spaces with universal covers [imath]\widetilde{X}[/imath] and [imath]\widetilde{Y}[/imath], respectively, then if [imath]X \simeq Y[/imath] then [imath]\widetilde{X} \simeq \widetilde{Y}[/imath]. My attempt might be correct, but it seems too complicated and amongst the jumble I may have made some incorrect assumptions. So what I'm looking for is (a) verification or correction; and (b) ideas as to how I could simplify my argument. My argument: Let [imath]p:\widetilde{X} \to X[/imath] and [imath]q:\widetilde{Y} \to Y[/imath] be the covering maps, and let [imath]X \overset{f}{\underset{g}{\leftrightarrows}} Y[/imath] be a homotopy equivalence. Fix a point [imath]\widetilde{x} \in \widetilde{X}[/imath], let [imath]x=p(x) \in X[/imath] and [imath]y=f(x) \in Y[/imath], and fix [imath]\widetilde{y} \in q^{-1}(\{y\}) \subseteq \widetilde{Y}[/imath]. Also let [imath]x' = g(y) \in X[/imath] and fix [imath]\widetilde{x'} \in p^{-1}(\{x'\}) \subseteq \widetilde{X}[/imath]. Define a map [imath]\widetilde{f} : \widetilde{X} \to \widetilde{Y}[/imath] as follows. For [imath]\widetilde{z} \in \widetilde{X}[/imath] let [imath]\widetilde{u}:[0,1] \to \widetilde{X}[/imath] be a path from [imath]\widetilde{x}[/imath] to [imath]\widetilde{z}[/imath]. Let [imath]z=p(z)[/imath] so that [imath]u=p\widetilde{u}[/imath] is a path in [imath]X[/imath] from [imath]x[/imath] to [imath]z[/imath]. Let [imath]v=fu[/imath], so that [imath]v[/imath] is a path in [imath]Y[/imath] from [imath]y[/imath] to [imath]f(z)[/imath]. Lift [imath]v[/imath] to a path [imath]\widetilde{v}[/imath] in [imath]\widetilde{Y}[/imath] with [imath]\widetilde{v}(0) = \widetilde{y}[/imath]. Define [imath]\widetilde{f}(\widetilde{z}) = \widetilde{v}(1)[/imath]. Notice that [imath]q\widetilde{v}=v[/imath] so that [imath]q\widetilde{f} = fp[/imath]. Define [imath]\widetilde{g}:Y \to X[/imath] analogously: For [imath]\widetilde{z} \in \widetilde{Y}[/imath] let [imath]\widetilde{a}:[0,1] \to \widetilde{Y}[/imath] be a path from [imath]\widetilde{y}[/imath] to [imath]\widetilde{z}[/imath]. Let [imath]z=q(z)[/imath] so that [imath]a=q\widetilde{a}[/imath] is a path in [imath]Y[/imath] from [imath]y[/imath] to [imath]z[/imath]. Let [imath]b=fa[/imath], so that [imath]b[/imath] is a path in [imath]X[/imath] from [imath]x'[/imath] to [imath]g(z)[/imath]. Lift [imath]b[/imath] to a path [imath]\widetilde{b}[/imath] in [imath]\widetilde{X}[/imath] with [imath]\widetilde{b}(0) = \widetilde{x'}[/imath]. Define [imath]\widetilde{g}(\widetilde{z}) = \widetilde{b}(1)[/imath]. Likewise, notice that [imath]p\widetilde{g}=gq[/imath]. Claim: [imath]\widetilde{X} \overset{\widetilde{f}}{\underset{\widetilde{g}}{\leftrightarrows}} \widetilde{Y}[/imath] is a homotopy equivalence. We have [imath]p\widetilde{g}\widetilde{f} = gq\widetilde{f}=gfp \simeq p[/imath] and [imath]q\widetilde{f}\widetilde{g} = fp\widetilde{g} = fgq \simeq q[/imath]. This shows that [imath]\widetilde{g}\widetilde{f}[/imath] and [imath]\widetilde{f}\widetilde{g}[/imath] are covering translations (deck transformations) and are therefore homotopic to the respective identity maps. So we have a homotopy equivalence. Any comments would be appreciated. |
4764 | Sine function dense in [imath][-1,1][/imath]
We know that the sine function takes it values between [imath][-1,1][/imath]. So is the set [imath]A = \{ \sin{n} \ : \ n \in \mathbb{N}\}[/imath] dense in [imath][-1,1][/imath]. Generally, for showing the set is dense, one proceeds, by finding out what is [imath]\overline{A}[/imath] of this given set. And if [imath]\overline{A} = [-1,1][/imath], we are through with the proof, but i having trouble here! Similarly can one do this with cosine function also, that is proving [imath]B= \{ \cos{n} \ : \ n \in \mathbb{N}\}[/imath] being dense in [imath][-1,1][/imath] | 313943 | Closure of the set [imath]\{\sin(n): n\in\mathbb{N}, n > 0\}[/imath] is the interval of reals [imath][-1,1][/imath]?
Please prove or disprove. Any help is appreciated Thanks in advance |
3408 | Characterizing non-constant entire functions with modulus $1$ on the unit circle
Is there a characterization of the nonconstant entire functions [imath]f[/imath] that satisfy [imath]|f(z)|=1[/imath] for all [imath]|z|=1[/imath]? Clearly, [imath]f(z)=z^n[/imath] works for all [imath]n[/imath]. Also, it's not difficult to show that if [imath]f[/imath] is such an entire function, then [imath]f[/imath] must vanish somewhere inside the unit disk. What else can be said about those functions? Thank you. | 1387347 | Characterize all entire functions [imath]f[/imath] such that [imath]|f(z)|=1[/imath], whenever [imath]|z|=1[/imath]
Characterize all entire functions [imath]f[/imath] such that [imath]|f(z)|=1[/imath], whenever [imath]|z|=1[/imath]. I think either [imath]f(z)=c[/imath] or [imath]f(z)=cz[/imath] where [imath]|c|=1[/imath]. But I am not sure, any hint would be great, thanks. |
14429 | What's the cardinality of all sequences with coefficients in an infinite set?
My motivation for asking this question is that a classmate of mine asked me some kind of question that made me think of this one. I can't recall his exact question because he is kind of messy (both when talking about math and when thinking about math). I'm kind of stuck though. I feel like the set [imath]A^{\mathbb{N}} = \{f: \mathbb{N} \rightarrow A, f \text{ is a function} \}[/imath] should have the same cardinality as the power set of A, if A is infinite. On the other hand, in this post, it is stated that the sequences with real coefficients have the same cardinality as the reals. It's easy to see that [imath]A^{\mathbb{N}} \subseteq P(A)[/imath], but (obviously) I got stuck on the other inclusion. Is there any general result that says anything else? References would be appreciated. EDIT To clarify the intetion of this question: I want to know if there are any general results on the cardinality of [imath]A^{\mathbb{N}}[/imath] other that it is strictly less than that of the power set of A. Also, I was aware that the other inclusion isn't true in general (as the post on here I linked to gave a counterexample), but thanks for pointing out why too. :) | 591465 | Cardinality of the set of all natural sequences is [imath]2^{\aleph_0}[/imath]
I was wondering how can you prove that [imath]\mathbb{N}^\mathbb{N} \sim 2^\mathbb{N}[/imath] (where [imath]\mathbb{N}^\mathbb{N}[/imath] is the set of all functon [imath]f:\mathbb{N}\rightarrow \mathbb{N}[/imath]). I think I can show that [imath]2^\mathbb{N} \sim n^\mathbb{N}[/imath] for some [imath]n\in\mathbb{N}[/imath] with a tedious bijection, but I'm not sure what to do with [imath]\mathbb{N}^\mathbb{N}[/imath]. I apologize if the question lacks details/notation, I have only studied elementary set theory and we haven't mentioned [imath]\mathbb{N}^\mathbb{N}[/imath]'s cardinality at all (I usually ended up finding injections/bijections to/from [imath]2^\mathbb{N}[/imath], but now I'm curious as to why the above correct). Thanks in advance. |
39802 | Why does [imath]1+2+3+\cdots = -\frac{1}{12}[/imath]?
[imath]\displaystyle\sum_{n=1}^\infty \frac{1}{n^s}[/imath] only converges to [imath]\zeta(s)[/imath] if [imath]\text{Re}(s) > 1[/imath]. Why should analytically continuing to [imath]\zeta(-1)[/imath] give the right answer? | 354265 | 1+2+3+4+... = -1/12
Consider the zeta function [imath]\zeta(s)= \sum \limits_{n=1}^{\infty} \frac{1}{n^s}[/imath]. It is established that [imath] \zeta(-1) = -\frac{1}{12}[/imath]. Reference (Equation 90) Then we have [imath] \zeta(-1) = \sum \limits_{n=1}^{\infty} \frac{1}{n^{-1}}= 1+2+3+4 + ... = -\frac{1}{12}[/imath]. But of course this series is divergent. So what is the problem here? |
305708 | Cyclotomic polynomials, irreducibility
I need to decide if certain cyclotomic polynomials are irreducibles over the [imath]\mathbb{F}_q[/imath]. For example, if [imath]\Phi_{12}(x)[/imath] is irreducible over [imath]\mathbb{F}_9[/imath]. Anyone can help me? Ok, i think i should aclare something: this question is not a duplicate of the question i made before... if you take the time to read my other question you will see that this is in fact a question that come up from that, because there was mention a criterion to decide if a cyclotomic polynomial was irreducible over [imath]\mathbb{F}_q[/imath], and i get an answer, but i note that the criterion requieres that [imath](n,q)=1[/imath] so i made this question...please someone could give an idea? | 305111 | Irreducible cyclotomic polynomial
I want to know if there is a way to decide if a cyclotomic polynomial is irreducible over a field [imath]\mathbb{F}_q[/imath]? |
13344 | Proof of [imath]\int_0^\infty \left(\frac{\sin x}{x}\right)^2 \mathrm dx=\frac{\pi}{2}.[/imath]
I am looking for a short proof that [imath]\int_0^\infty \left(\frac{\sin x}{x}\right)^2 \mathrm dx=\frac{\pi}{2}.[/imath] What do you think? It is kind of amazing that [imath]\int_0^\infty \frac{\sin x}{x} \mathrm dx[/imath] is also [imath]\frac{\pi}{2}.[/imath] Many proofs of this latter one are already in this post. | 484907 | Integration of [imath]\int_0^\infty\frac{\sin^2x}{x^2}dx[/imath]
I have been trying very hard to find the answer to the following integral[imath]\int_0^\infty\frac{\sin^2x}{x^2}dx[/imath] given that [imath]\int_0^\infty\frac{\sin x\cos x}{x}dx = \frac{\pi}{4}[/imath] |
79287 | Conditions Equivalent to Injectivity
Let [imath]A[/imath] and [imath]B[/imath] be sets, where [imath]f : A \rightarrow B[/imath] is a function. Show that the following properties are valid equivalent*: [imath]f[/imath] is injective. For all [imath]X, Y \subset A[/imath] is valid: [imath]f(X \cap Y)=f(X)\cap f(Y)[/imath] For all [imath]X \subset Y \subset A[/imath] is valid: [imath]f(X \setminus Y)=f(X) \setminus f(Y)[/imath]. I do know what injective is, but I thought number (2.) and (3.) were valid for any kind of function. Just to see if I understood this right: [imath]f(X\cap Y)[/imath] means, first, make the intersection from [imath]X[/imath] and [imath]Y[/imath] and then map it to [imath]B[/imath] via [imath]f[/imath]. [imath]f(X)\cap f(Y)[/imath] actually means, map all [imath]f(X)[/imath] and [imath]f(Y)[/imath] and intersect both. Aren't those both properties valid for all functions? I can't think of a counter example. Thanks in advance guys! *edited by SN. | 1105696 | If [imath]f[/imath] is an injection, [imath]f(S_1 \cap S_2) = f(S_1) \cap f(S_2)[/imath]
I need to prove that if [imath]S_1[/imath] and [imath]S_2[/imath] are subsets of a set [imath]X[/imath], and if [imath]f: X \to Y[/imath] is an injection, prove that [imath]f(S_1 \cap S_2) = f(S_1) \cap f(S_2)[/imath]. I know I need to show inclusion in both directions but not sure how to begin. |
9321 | Understanding proof of completeness of [imath]L^{\infty}[/imath]
I'm reading page number 4 here. In particular the section where it deals with the case [imath]p=\infty[/imath], that is , showing that [imath]L^{\infty}[/imath] is complete. http://www.core.org.cn/NR/rdonlyres/Mathematics/18-125Fall2003/5E3917E2-C212-463B-9EDB-671486133388/0/18125_lec15.pdf Two questions: 1) Why is the convergence uniform? where it says "for [imath]x \in N^{c}[/imath] , [imath]f_{n}[/imath] is a Cauchy sequence of complex numbers. Thus [imath]f_{n} \rightarrow f[/imath] uniformly. Clearly we have pointwise convergence but why is it uniform? 2) I don't see why [imath]||f_{n} - f||_{\infty} \rightarrow 0[/imath]. Can you please explain this step? Thanks. (3/2015) Edit: The original link appears to be broken. This document seems to provide a similar (maybe even identical) proof to the one the OP talks about with slight notational differences. | 742046 | Proving the Rietz-Fischer Theorem for [imath]p = \infty[/imath]
Rietz-Fischer Theorem: Let [imath]E[/imath] be a measurable set and [imath]1 \le p \le \infty[/imath]. Then every rapidly Cauchy sequence in [imath]L^p(E)[/imath] converges both with respect to the [imath]p[/imath]-norm and pointwise almost everyone on [imath]E[/imath] to a function in [imath]L^p(E)[/imath]. Question: How does one prove this in the case that [imath]p = \infty[/imath]? Every proof I have seen thus far seems to assume that [imath]p \in [1,\infty)[/imath]. For example, the wikipedia proof makes this assumption and then concludes by remarking "the case [imath]p = \infty[/imath] reduces to a simple question about uniform convergence". Upon thinking about this for a bit it's not obvious how to proceed. |
7757 | How to prove this binomial identity [imath]\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}[/imath]?
I am trying to prove this binomial identity [imath]\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}[/imath] but am not able to think something except induction,which is of-course not necessary (I think) here, so I am inquisitive to prove this in a more general way. The left side of an identity occurs while solving another problem (concerning binomial theorem) so I am more interested in deriving the right side from the left side, else I have to remember it now onward. EDIT: I am more interested in an algebraic proof rather than combinatorial argument or something involving calculus (however I liked svenkatr and Bill Dubuque solution), hence I am removing the combinatorics tag. | 478859 | Prove that [imath]\sum_{k=1}^{n}k\binom{n}{k}=n\cdot2^{n-1}[/imath]
I want to prove the following: [imath]\sum_{k=1}^{n}k\binom{n}{k}=n\cdot2^{n-1}[/imath] what I did is(use binominal): [imath]\binom{n}{k}X^k\cdot 1^{n-k} = (X+1)^n[/imath] [imath]k\binom{n}{k}X^k\cdot 1^{n-k} = k(X+1)^k-1[/imath] now I replace [imath]k[/imath] by [imath]n[/imath] and insert [imath]X=1[/imath] , I will get now [imath]n\cdot2^{n-1}[/imath] what you are suggesting? there is another way to do that? thanks! |
82309 | Slowing down divergence 2
Let [imath]f(x)[/imath] and [imath]g(x)[/imath] be positive nondecreasing functions such that [imath] \sum_{n>1} \frac1{f(n)} \text{ and } \sum_{n>1} \frac1{g(n)} [/imath] diverges. (Why) must the series [imath]\sum_{n>1} \frac1{g(n)+f(n)}[/imath] diverge? | 858339 | For [imath]a_n,b_n\uparrow[/imath] and [imath]\sum \frac{1}{a_n}[/imath], [imath]\sum \frac{1}{b_n}[/imath] divergent is the series [imath]\sum \frac{1}{a_n+b_n}[/imath] also divergent?
Let [imath]a_n[/imath] and [imath]b_n[/imath] are strictly increasing to [imath]+\infty[/imath] sequences such that the series [imath]\sum \frac{1}{a_n}[/imath] and [imath]\sum \frac{1}{b_n}[/imath] are divergent. Is it true that the series [imath]\sum \frac{1}{a_n+b_n}[/imath] is also divergent? At first sight it looks true, so I tried to prove that, but after some failed attempts I start to believe now that there might exist a counterexample. Any thoughts? |
255614 | Prove that [imath]\sum_{n=1}^{\infty}\ a_n^2[/imath] is convergent if [imath]\sum_{n=1}^{\infty}\ a_n[/imath] is absolutely convergent
Suppose that [imath]\displaystyle\sum_{n=1}^{\infty}\ a_n[/imath] is absolutely convergent. How can we prove that [imath]\displaystyle\sum_{n=1}^{\infty}\ a_n^2[/imath] is convergent? | 36429 | If [imath]\sum_{n=1}^{\infty} a_n[/imath] is absolutely convergent, then [imath]\sum_{n=1}^{\infty} (a_n)^2[/imath] is convergent
Let [imath]\sum_{n=1}^{\infty} a_n[/imath] be a series in R. Prove that if [imath]\sum_{n=1}^{\infty} a_n[/imath] is absolutely convergent, then [imath]\sum_{n=1}^{\infty} (a_n)^2[/imath] is convergent. |
40998 | The sum of [imath](-1)^n \frac{\ln n}{n}[/imath]
I'm stuck trying to show that [imath]\sum_{n=2}^{\infty} (-1)^n \frac{\ln n}{n}=\gamma \ln 2- \frac{1}{2}(\ln 2)^2[/imath] This is a problem in Calculus by Simmons. It's in the end of chapter review and it's associated with the section about the alternating series test. There's a hint: refer to an equation from a previous section on the integral test. Specifically: [imath]L=\lim_{n\to\infty} F(n)=\lim_{n\to\infty}\left[a_1+a_2+\cdots+a_n-\int_1^n\! f(x)\,\mathrm{d}x\right][/imath] Here, [imath]\{a_n\}[/imath] is a decreasing sequence of positive numbers and [imath]f(x)[/imath] is a decreasing function such that [imath]f(n)=a_n[/imath], and [imath]\gamma[/imath] is this limit in the case that [imath]a_n=\frac{ 1}{n}[/imath]. New users can't answer their own questions inside of 8 hours, so I'm editing my question to reflect the answer. Ok, I got it. Following the hint in the book [imath]L=\lim_{n\to\infty}\left[\frac{\ln 2}{2}+\frac{\ln 3}{3}+\cdots+\frac{\ln n}{n}-\int_2^n\! \frac{\ln x}{x}\,\mathrm{d}x\right][/imath] [imath]=\lim\left[\frac{ \ln 2}{2}+\cdots+\frac{ \ln n}{n}-\left.\frac{ \ln^2x}{2}\right|_2^n\right][/imath] The partial sum for the positive series is: [imath]\left(\frac{\ln^2n}{2}-\frac{\ln^2}{2}\right)+L+o(1)[/imath] Returning to the original, alternating series: [imath]-S_{2n}=\frac{ -\ln 2}{2}+\frac{ \ln 3}{3}-\frac{\ln 4}{4}+\frac{\ln 5}{5}-\cdots[/imath] [imath]=\frac{\ln 2}{2}+\frac{\ln 3}{3}+\cdots+\frac{\ln 2n}{2n}-2\left(\frac{\ln 2}{2}+\frac{\ln 4}{4}+\cdots+\frac{\ln 2n}{2n}\right)[/imath] Consider the partial sum in parentheses [imath] \frac{\ln 2}{2}+\frac{\ln 4}{4}+\cdots+\frac{\ln 2n}{2n}=\frac{\ln 2}{2}+\frac{\ln 2 +\ln 2}{4}+\frac{\ln 2+\ln 3}{6}+\cdots+\frac{\ln 2+\ln n}{2n}[/imath] [imath]=\frac{1}{2}\left(\ln 2\left(1+\frac{ 1}{2}+\cdots+\frac{ 1}{n}\right)+\left(\frac{ \ln 2}{2}+\frac{ \ln 3}{3}+\cdots+\frac{ \ln n}{n}\right)\right)[/imath] Now, plug that back in [imath]-S_{2n}=\left(\frac{\ln 2}{2}+\cdots+\frac{ \ln 2n}{2n}\right)-\ln 2\left(1+\frac{ 1}{2}+\cdots+\frac{ 1}{n}\right)-\left(\frac{ \ln 2}{2}+\frac{ \ln 3}{3}+\cdots+\frac{ \ln n}{n}\right)[/imath] [imath]=\frac{ \ln^2(2n)}{2}-\frac{ \ln^2 2}{2}+L+o(1)-\ln 2\left(\ln n +\gamma+o(1)\right)-\left(\frac{ \ln^2 n}{2}-\frac{ \ln^2 2}{2}+L+o(1)\right)[/imath] [imath]=\frac{ (\ln 2 +\ln n)^2}{2}-(\ln 2)(\ln n)-\gamma\ln 2-\frac{ \ln^2 n}{2}+o(1)[/imath] [imath]=\frac{ \ln^2 2}{2}+(\ln 2)(\ln n)+\frac{ \ln^2 n}{2}-(\ln 2)(\ln n)-\gamma \ln 2 - \frac{ \ln^2 n}{2}+o(1)[/imath] [imath]-S_{2n}\to\frac{ \ln^2}{2}-\gamma\ln 2[/imath] Which gives the desired result [imath]\sum_2^{\infty}(-1)^n \frac{ \ln n}{n}=\gamma\ln 2 -\frac{ \ln^2 2}{2}[/imath] | 1627615 | How to prove this series: [imath]\displaystyle \sum_{n=1}^{\infty }\frac{\left ( -1 \right )^{n}\ln n}{n}=\gamma \ln 2-\frac{1}{2}\ln^22[/imath]
How to prove this series [imath]\sum_{n=1}^{\infty }\frac{\left ( -1 \right )^{n}\ln n}{n}=\gamma \ln 2-\frac{1}{2}\ln^22[/imath] and \begin{align*} \sum_{n=1}^{\infty }\frac{\left ( -1 \right )^{n}\ln \left ( n+1 \right )}{n+1}&=\frac{1}{2}\ln^22-\gamma \ln 2\\ \sum_{n=1}^{\infty }\frac{\left ( -1 \right )^{n}\ln \left ( n+2 \right )}{n+2}&=\gamma \ln 2-\frac{1}{2}\ln^22-\frac{1}{2}\ln 2 \end{align*} So I want to know, is there a closed form for [imath]\sum_{n=1}^{\infty }\frac{\left ( -1 \right )^{n}\ln \left ( n+k \right )}{n+k}~~~\left ( k>0 \right )[/imath] |
20661 | The sum of an uncountable number of positive numbers
Claim:If [imath](x_\alpha)_{\alpha\in A}[/imath] is a collection of real numbers [imath]x_\alpha\in [0,\infty][/imath] such that [imath]\sum_{\alpha\in A}x_\alpha<\infty[/imath], then [imath]x_\alpha=0[/imath] for all but at most countably many [imath]\alpha\in A[/imath] ([imath]A[/imath] need not be countable). Proof: Let [imath]\sum_{\alpha\in A}x_\alpha=M<\infty[/imath]. Consider [imath]S_n=\{\alpha\in A \mid x_\alpha>1/n\}[/imath]. Then [imath]M\geq\sum_{\alpha\in S_n}x_\alpha>\sum_{\alpha\in S_n}1/n=\frac{N}{n}[/imath], where [imath]N\in\mathbb{N}\cup\{\infty\}[/imath] is the number of elements in [imath]S_n[/imath]. Thus [imath]S_n[/imath] has at most [imath]Mn[/imath] elements. Hence [imath]\{\alpha\in A \mid x_\alpha>0\}=\bigcup_{n\in\mathbb{N}}S_n[/imath] is countable as the countable union of finite sets. [imath]\square[/imath] First, is my proof correct? Second, are there more concise/elegant proofs? | 419739 | All finite sums bounded implies countable
Let [imath]X[/imath] be a set of positive real numbers, and let [imath]S[/imath] be the set of all finite sums of members of [imath]X[/imath]. Suppose that [imath]S[/imath] is bounded. Prove that [imath]X[/imath] is countable. Not much idea on how to start here. A contrapositive would require proving that if [imath]X[/imath] is uncountable, then [imath]S[/imath] is unbounded. A big ask, it seems. |
26363 | Square roots -- positive and negative
It is perhaps a bit embarrassing that while doing higher-level math, I have forgot some more fundamental concepts. I would like to ask whether the square root of a number includes both the positive and the negative square roots. I know that for an equation [imath]x^2 = 9[/imath], the solution is [imath]x = \pm 3[/imath]. But if simply given [imath]\sqrt{9}[/imath], does one assume that to mean only the positive root? And when simply talking about the square root of a number in general, would one be referring to both roots or just the positive one, when neither is specified? | 380359 | Why is [imath]\sqrt{4} = 2[/imath] and Not [imath]\pm 2[/imath]?
I've always been told that if [imath]\ x^2 = 4,[/imath] [imath] =>x = \pm2[/imath] But recently, Prof. mentioned that if [imath]x = \sqrt{4}[/imath], Then [imath]x = +2(only)[/imath] I am very skeptical about this because they both mean the same thing and still yield different results! So how is the above statement justified? |
93463 | Is [imath]\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})[/imath]?
Is [imath]\mathbf{Q}(\sqrt{2}, \sqrt{3}) = \mathbf{Q}(\sqrt{2}+\sqrt{3})[/imath] ? [imath]\mathbf{Q}(\sqrt{2},\sqrt{3})=\{a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6} \mid a,b,c,d\in\mathbf{Q}\}[/imath] [imath]\mathbf{Q}(\sqrt{2}+\sqrt{3}) = \lbrace a+b(\sqrt{2}+\sqrt{3}) \mid a,b \in \mathbf{Q} \rbrace [/imath] So if an element is in [imath]\mathbf Q(\sqrt{2},\sqrt{3})[/imath], then it is in [imath]\mathbf{Q}(\sqrt{2}+\sqrt{3})[/imath], because [imath]\sqrt{6} = \sqrt{2}\sqrt{3}[/imath]. How to conclude from there? | 396276 | How can I show this field extension equality?
How can I show this field extension equality [imath]\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})[/imath]? |
304487 | Unit Tangent and Unit Normal Vectors -- Calculus III Question
Consider the following vector function. [imath]r(t) = \left\langle 2t \cdot \sqrt{2}, e^{2t}, e^{-2t}\right\rangle[/imath] (a) Find the unit tangent and unit normal vectors [imath]T(t)[/imath] and [imath]N(t)[/imath]. [imath]T(t) =[/imath] [imath]N(t) =[/imath] (b) Use this formula to find the curvature. [imath]κ(t) =[/imath] I am getting bogged down in the math. I know how to calculate the three things but I am having trouble getting the derivative of [imath]T(t)[/imath] after solving for it. I have gotten [imath]T(t)[/imath] to equal [imath]\frac{1}{2 e^{2t} + 2 e^{-2t}} \left\langle 2 \cdot \sqrt{2}, 2 e^{2t},-2 e^{-2t}\right\rangle[/imath]. Thank you! | 305007 | Normal tangent vector and normal vector problem
Consider the following vector function. [imath]r(t) = \left\langle 2t \cdot \sqrt{2}, e^{2t}, e^{-2t}\right\rangle[/imath] (a) Find the unit tangent and unit normal vectors [imath]T(t)[/imath] and [imath]N(t)[/imath]. [imath]T(t) =[/imath] [imath]N(t) =[/imath] (b) Use this formula to find the curvature. [imath]κ(t) =[/imath] I am getting bogged down in the math. I know how to calculate the three things but I am having trouble getting the derivative of [imath]T(t)[/imath] after solving for it. I have gotten [imath]T(t)[/imath] to equal [imath]\frac{1}{2 e^{2t} + 2 e^{-2t}} \left\langle 2 \cdot \sqrt{2}, 2 e^{2t},-2 e^{-2t}\right\rangle[/imath]. Thank you! |
296686 | Union of Countable Sets is Countable
Let [imath]\{E_n\}[/imath], [imath]n = 1, 2, 3, \ldots[/imath], be a sequence of countable sets, and put [imath]S = \displaystyle \bigcup_{n=1}^{\infty} E_n[/imath]. Prove that [imath]S[/imath] is countable. | 55181 | countably infinite union of countably infinite sets is countable
How do you prove that any collection of sets [imath]\{X_n : n \in \mathbb{N}\}[/imath] such that for every [imath]n \in \mathbb{N}[/imath] the set [imath]X_n[/imath] is equinumerous to the set of natural numbers, then the union of all these sets, [imath]\bigcup_{i\in \mathbb{N}}[/imath] [imath]X_i[/imath] is also equinumerous to the set of natural numbers? (By equinumerous I mean there exists a one-to-one and onto function [imath]f:X_n \to \mathbb{N}[/imath].) Is this statement false as it stands? |
180073 | Approximating [imath]\pi[/imath] with least digits
Do you a digit efficient way to approximate [imath]\pi[/imath]? I mean representing many digits of [imath]\pi[/imath] using only a few numeric digits and some sort of equation. Maybe mathematical operations also count as penalty. For example the well known [imath]\frac{355}{113}[/imath] is an approximation, but it gives only 7 correct digits by using 6 digits (113355) in the approximation itself. Can you make a better digit ratio? EDIT: to clarify the "game" let's assume that each mathematical operation (+, sqrt, power, ...) also counts as one digit. Otherwise one could of course make artifical infinitely nested structures of operations only. And preferably let's stick to basic arithmetics and powers/roots only. EDIT: true. logarithm of imaginary numbers provides an easy way. let's not use complex numbers since that's what I had in mind. something you can present to non-mathematicians :) | 1990401 | Find two rational numbers [imath]\frac ab[/imath] satisfying [imath]\mid \pi - \frac ab\mid < \frac{1}{\sqrt 5 b^2}.[/imath]
Find two rational numbers [imath]\frac ab[/imath] satisfying [imath]\mid \pi - \frac ab\mid < \frac{1}{\sqrt 5 b^2}.[/imath] I dont know how to find such rationals. IS there a method? or trial? |
66145 | [imath]\mathbb{Q}/\mathbb{Z}[/imath] has a unique subgroup of order [imath]n[/imath] for any positive integer [imath]n[/imath]?
Viewing [imath]\mathbb{Z}[/imath] and [imath]\mathbb{Q}[/imath] as additive groups, I have an idea to show that [imath]\mathbb{Q}/\mathbb{Z}[/imath] has a unique subgroup of order [imath]n[/imath] for any positive integer [imath]n[/imath]. You can take [imath]a/n+\mathbb{Z}[/imath] where [imath](a,n)=1[/imath], and this element has order [imath]n[/imath]. Why would such an element exist in any subgroup [imath]H[/imath] of order [imath]n[/imath]? If not, you could reduce every representative, and then every element would have order less than [imath]n[/imath], but where is the contradiction? | 391327 | [imath]\mathbb{Q}/\mathbb{Z}[/imath] has cyclic subgroup of every positive integer [imath]n[/imath]?
I would like to know whether [imath](\mathbb{Q}/\mathbb{Z},+)[/imath] has [imath]1[/imath]. Cyclic subgroup of every positive integer [imath]n[/imath]? [imath]2[/imath]. Yes, unique one. [imath]3[/imath]. Yes, but not necessarily unique one. [imath]4[/imath]. Does not have cyclic subgroup of every positive integer [imath]n[/imath]. What I know about the given group is infinite group but every element has finite order. Please help how to proceed. |
143173 | Showing the inequality [imath]|\alpha + \beta|^p \leq 2^{p-1}(|\alpha|^p + |\beta|^p)[/imath]
I have a small question that I think is very basic but I am unsure how to tackle since my background in computing inequalities is embarrassingly weak - I would like to show that, for a real number [imath]p \geq 1[/imath] and complex numbers [imath]\alpha, \beta[/imath], I have [imath]\begin{equation} |\alpha + \beta|^p \leq 2^{p-1}(|\alpha|^p + |\beta|^p) \end{equation}[/imath] I thought it would be best to rewrite this as [imath]\begin{equation} \left|\frac{\alpha + \beta}{2}\right|^p \leq \frac{|\alpha|^p + |\beta|^p}{2} \end{equation}[/imath] but then I am unsure what to do next - is this a sensible start anyways ? Any help would be great ! (P.S. this is not a homework question - I am currently trying to brush up my knowledge of [imath]L^p[/imath] spaces, and this inequality came up as a statement. I thought it might be worthwhile to make sure I can fill in the gaps to improve my skills in computing inequalities.) | 2174788 | Common inequality
I was trying to prove that for arbitrary [imath]A,B\in\mathbb{R}[/imath] and any [imath]p\in\mathbb{N}[/imath] it holds [imath](A+B)^p \leq 2^{p-1}(A^p+B^p)[/imath] I'd appreciate some advice. Thanks a lot. |