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1098534 | Limit of S_n, where [imath]S_n = \int_{0}^{1}\frac{nx^{n-1}}{1+x} dx[/imath] for [imath]n\ge 1[/imath]
[imath]S_n = \int_{0}^{1}\frac{n x^{n-1}}{1+x} dx \quad n \ge 1[/imath] My attempt: [imath]|S_n| \le \int_{0}^{1}\left|\frac{n x^{n-1}}{1+x} dx\right| = \int_{0}^{1}\frac{n x^{n-1}}{1+x} dx \le \int_{0}^{1}n x^{n-1} dx[/imath] [imath]0 \le x \le 1[/imath], and [imath]n \ge 1[/imath]. So, [imath]|S_n|=[x^n]_{0}^{1}[/imath] and thus [imath]|Sn| = 1[/imath]. So, [imath]\lim_{n \to \infty} S_n = 1[/imath]. But the answer mentioned is [imath]0[/imath]. I don't know what I am doing wrong. Please help. | 292251 | Limit of [imath]s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx[/imath] as [imath]n \to \infty[/imath]
Let [imath]s_n[/imath] be a sequence defined as given below for [imath]n \geq 1[/imath]. Then find out [imath]\lim\limits_{n \to \infty} s_n[/imath]. \begin{align} s_n = \int\limits_0^1 \frac{nx^{n-1}}{1+x} dx \end{align} I have written a solution of my own, but I would like to know it is completely correct, and also I would like if people post more alternative solutions. |
1098969 | An Application: The Stone-Cech Compactification
If [imath]X[/imath] is completely regular, show that [imath]X[/imath] is open in [imath]βΧ[/imath] if and only if [imath]X[/imath] is locally compact. | 353413 | Stone-Čech compactification. A completely regular topological space is locally compact iff it is open in its Stone-Čech compactification.
I would like to show that a completely regular topological space is locally compact iff it is (weak-star) open in its Stone-Čech compactification. Does this hold in general? I.e given a compact subset [imath]X[/imath] of a normed space, then is it true that if [imath]\overline{U}= X[/imath] then, U is open iff U is locally compact? |
1099220 | Why is it called "characteristic [imath]0[/imath]" and not "characteristic [imath]\infty[/imath]?"
There may not be a good answer to this; it is perhaps more or less a matter of coincidence. But the thing is: For any ring, the characteristic of the ring is the order of the element [imath]1[/imath] in the ring --- except if this order is [imath]\infty[/imath], in which case the characteristic is zero. So wouldn't it be more natural to call it characteristic [imath]\infty[/imath]? Is there some smart reason behind this that I just haven't noticed yet, or is it just because mathematicians are weird? | 98605 | Why “characteristic zero” and not “infinite characteristic”?
The characteristic of a ring (with unity, say) is the smallest positive number [imath]n[/imath] such that [imath]\underbrace{1 + 1 + \cdots + 1}_{n \text{ times}} = 0,[/imath] provided such an [imath]n[/imath] exists. Otherwise, we define it to be [imath]0[/imath]. But why characteristic zero? Why do we not define it to be [imath]\infty[/imath] instead? Under this alternative definition, the characteristic of a ring is simply the “order” of the additive cyclic group generated by the unit element [imath]1[/imath]. My feeling is that there is a precise and convincing explanation for the common convention, but none comes to mind. I couldn't find the answer in the Wikipedia article either. |
1099316 | Extension of transcendental element
Let [imath]K[/imath] be a field, and let [imath]K\subset E\subset K(a)[/imath] be an extension, where [imath]a[/imath] is transcendental element over [imath]K[/imath]. How can I show that [imath]K(a)/E[/imath] is finite ? | 1095336 | [imath]X[/imath] is algebraic over [imath]E[/imath]
Let [imath]X[/imath] be transcendental over a field [imath]F[/imath], and let [imath]E[/imath] be a subfield of [imath]F(X)[/imath] properly containing [imath]F[/imath]. Prove that [imath]X[/imath] is algebraic over [imath]E[/imath]. Could we maybe use also the following?? Let [imath]f(x) \in E \Rightarrow f(x) \in F(X) \Rightarrow f(x)=\frac{h(X)}{g(X)}[/imath], where [imath]g(X) \neq 0, h,g \in F[x][/imath]. Or doesn't this help ?? |
1099364 | How to prove that [imath]2^x,3^x,5^x\in\mathbb N[/imath] implies [imath]x\in\mathbb N[/imath]?
Let [imath]x\in\mathbb R[/imath] and suppose that [imath]2^x,3^x[/imath] and [imath]5^x[/imath] are all integers. Does it imply that [imath]x[/imath] is also necessarily an integer? I read somewhere that the answer is "Yes" and a proof is known, but I haven't seen a proof and was not able to prove it myself. Could you please help me to find a proof? | 1087841 | Existence of [imath]x[/imath] such that [imath]2^x =a,3^x=b,5^x=c[/imath] for some integers [imath]a,b,c[/imath]
Conjecture: There does not exist a non-integer [imath]x[/imath] such that [imath]2^x=a[/imath] [imath]3^x=b[/imath] [imath]5^x=c[/imath] where [imath]a,b,c[/imath] are all integers. I'm aware that the similar question There does not exist a non-integer [imath]y[/imath] such that [imath]2^y=A[/imath] [imath]3^y=B[/imath] where [imath]A,B[/imath] are all integers. is a famous unsolved problem. (evidence in the corrolaries here or here) My idea was that the addition of the condition [imath]5^x[/imath] made the problem easier and therefore solvable. |
1099521 | Deriving formula for sum n(n+1)
Can you please describe how to derive a formula for first n members of [imath] S = 1\cdot 2 + 2\cdot 3 + 3\cdot 4 +\cdots +n(n+1)\mbox{?} [/imath] Thank you | 347585 | Formula for the [imath]1\cdot 2 + 2\cdot 3 + 3\cdot 4+\ldots + n\cdot (n+1)[/imath] sum
Is there a formula for the following sum? [imath]S_n = 1\cdot2 + 2\cdot 3 + 3\cdot 4 + 4\cdot 5 +\ldots + n\cdot (n+1)[/imath] |
255875 | Improper integration involving complex analytic arguments
I am trying to evaluate the following: [imath]\displaystyle \int_{0}^{\infty} \frac{1}{1+x^a}dx[/imath], where [imath]a>1[/imath] and [imath]a \in \mathbb{R}[/imath] Any help will be much appreciated. | 1577673 | how to integrate the definite integral using residue theorem?
How to evaluate [imath]\int_{0}^{\infty}\dfrac{1}{x^a+1}dx[/imath], where [imath]a>1[/imath]. I don't know where to start since [imath]x^a+1[/imath] could have infinitely many roots, then it is impossible(?) to evaluate its residues. And even we did find a way to do so, how are we gonna choose the contour at the first place? Any hint would be appreciated. |
1099155 | Logarithmic Series
I was doing a bit of math when I came across logarithmic series. I have no idea from where they come from. They seem so unrelated, that I have no intuition behind them at all. So, can anyone prove that [imath]\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^{n}}{n}=\ln(1+x)[/imath] [imath]\forall -1<x\leq1[/imath] or atleast point me in the right direction? | 878374 | taylor series of $\ln(1+x)$?
Compute the taylor series of [imath]$\ln(1+x)$[/imath] I've first computed derivatives (up to the 4th) of ln(1+x) [imath]f^{'}(x)[/imath] = [imath]\frac{1}{1+x}[/imath] [imath]f^{''}(x) = \frac{-1}{(1+x)^2}[/imath] [imath]f^{'''}(x) = \frac{2}{(1+x)^3}[/imath] [imath]f^{''''}(x) = \frac{-6}{(1+x)^4}[/imath] Therefore the series: [imath]$\ln(1+x) = f(a) + \frac{1}{1+a}\frac{x-a}{1!} - \frac{1}{(1+a)^2}\frac{(x-a)^2}{2!} + \frac{2}{(1+a)^3}\frac{(x-a)^3}{3!} - \frac{6}{(1+a)^4}\frac{(x-a)^4}{4!} + ...$[/imath] But this doesn't seem to be correct. Can anyone please explain why this doesn't work? The supposed correct answer is: As [imath]$\ln(1+x) = \int (\frac{1}{1+x})dx$[/imath] [imath]$\ln(1+x) = \Sigma_{k=0}^{\infty} \int (-x)^k dx$[/imath] |
1099104 | series convergence. [imath]\sum_{n=2}^\infty\frac{(-1)^n}{\sqrt[3]{n}+ (-1)^{\frac{n(n+1)}{2}}}[/imath]
[imath]\sum_{n=2}^\infty\frac{(-1)^n}{\sqrt[3]{n}+ (-1)^{\frac{n(n+1)}{2}}}[/imath] None of convergence tests I know (Leibnitz, Dirichlet, Abel) works becaues of the denominator. I know that sum of convergent series is convergent too and I tried using this information with no results. | 1089885 | Prove convergence of series
Let [imath]\displaystyle a_n=\sum_{n=2}^{\infty} \frac{(-1)^n}{n^{\frac{1}{3}}+(-1)^{\frac{n(n+1)}{2}}}[/imath] so I divide it into four series [imath]4k, 4k+1, 4k+2, 4k+3[/imath] and I pair for instance [imath]4k, 4k+1[/imath] and [imath]4k+2, 4k+3[/imath] and prove that these two series is convergent and conclude that since both series are convergent so the sum of it is also convergent but I'm not sure if it's legall second series I'm not sure is like this [imath]\displaystyle b_n=\frac{1}{1}+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+...[/imath] here I also split it into smaller parts namely [imath]\displaystyle \sum_{n=1}^{\infty}\frac{1}{4n-3}+\sum_{n=1}^{\infty}\frac{1}{4n-1}-\sum_{n=1}^{\infty}\frac{1}{2n}[/imath] and I prove that [imath]\displaystyle \frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}<\frac{1}{n^2} \cdot c [/imath] for some const c so it implies that origin series is convergent or not ? |
1099698 | Proof that [imath]a^b>b^a[/imath] if [imath]a are integers larger or equal to two and (a,b)\neq (2,3),(2,4)[/imath]
I would like a proof that if [imath]a<b[/imath] are integers with [imath]2\leq a,b[/imath] we have [imath]a^b>b^a[/imath] unless [imath]a=2,b=3[/imath] or [imath]a=2,b=4[/imath] . I would like to use as little calculus as possible. Here is my current solution: Case 1: [imath]a>2[/imath] Fix [imath]a[/imath] and start with [imath]b=a[/imath]. notice [imath]a^a=a^a[/imath]. Now increase [imath]b[/imath] by [imath]1[/imath] to now get [imath]a^{a+1}>(a+1)^a[/imath]. When [imath]a[/imath] is at least three this is true because the left side was multiplied by [imath]a[/imath] and the right side by [imath](1+\frac{1}{a})^a[/imath]. Since [imath](1+\frac{1}{a})^a=\sum\limits_{i=0}^a \binom{a}{i} \frac{1}{a^i}[/imath] is the sum of [imath]a+1[/imath] elements all of them smaller than [imath]1[/imath] and the last two terms add [imath]\binom{a}{a-1}\frac{1}{a^{a-1}}+\binom{a}{a}\frac{1}{a^a}=\frac{1}{a^{a-2}}+\frac{1}{a^a}<\frac{1}{3}+\frac{1}{27}<1[/imath] we conclude the sum is less than [imath]a[/imath]. Therefore the inequality holds. Notice that each time we add [imath]1[/imath] to [imath]b[/imath] the left side is multiplied by [imath]a[/imath] and the right by something smaller than [imath](1+\frac{1}{a})^a[/imath], hence the inequality holds for all [imath]b[/imath] larger than [imath]a[/imath]. Case [imath]a=2[/imath] is the same thing only we check [imath]b=3,4[/imath] by hand and then do something similar. I am not really happy with the current solution, can we find something simpler? If it can be more combinatorial it would be better. Notice I want as little calculus or inequalities as possible. | 410697 | Given [imath]a>b>2[/imath] both positive integers, which of [imath]a^b[/imath] and [imath]b^a[/imath] is larger?
Given [imath]a>b>2[/imath] both positive integers, which of [imath]a^b[/imath] and [imath]b^a[/imath] is larger? I tried an induction approach. First I showed that if [imath]b=3[/imath] then any [imath]a \geq4[/imath] satisfied [imath]a^b<b^a[/imath]. Then using that as my base case I tried to show that given any pair of positive integers [imath]a,b[/imath] satisfying [imath]a>b>2[/imath] and [imath]a^b<b^a[/imath], then [imath](a+1)^{b+1}<(b+1)^{a+1}[/imath] - but that is where I got stuck. Any help would be appreciated. |
1099905 | Calculating [imath] \lim_{n\to \infty} (1+\sin({1}/{n}))^{n}[/imath] without L'Hopital or series expansions
I am trying to calculate the following limit, without using the L'Hopital rule or series expansions: lim (1+sin(1/n))^(n), n->infinity I now that it is the same as: lim (1+sin(n))^(1/n), n->0 But that's about all I know... Any help? | 1099885 | What is the limit of
How can you calculate [imath]\lim_{x\rightarrow \infty}\left(1+\sin\frac{1}{x}\right)^x?[/imath] In general, what would be the strategy to solving a limit problem with a power? |
1099991 | Vandermonde determinants quotient
I have to prove that for any integers [imath]k_1<k_2<...<k_n[/imath] the quotient: [imath] \frac{V_n (k_1,k_2, ..., k_n)}{V_n (1, 2, ..., n)} [/imath] is an integer, where: [imath] V_n (k_1,k_2, ..., k_n) = \prod_{1 \le i < j \le n} (k_j - k_i)[/imath] is the Vandermonde determinant. I see that the denominator is equal to [imath]1^{n-1} \cdot 2^{n-2} \cdot 3^{n-3} \cdot ... \cdot (n-1)^1[/imath] but it doesn't seem to help. | 1085531 | How prove this [imath]\prod_{1\le i is integer[/imath]
let [imath]a_{i},i=1,2,\cdots,n[/imath] be postive integer ,show that [imath]1^{n-1}2^{n-2}\cdots (n-2)^2(n-1)|\prod_{1\le i<j\le n}(a_{i}-a_{j})[/imath] I know this [imath]\prod_{1\le i<j\le n}(a_{i}-a_{j})[/imath] is Vandermonde determinants,and I found [imath]1^{n-1}2^{n-2}\cdots (n-2)^2\cdot (n-1)=1!2!3!\cdots (n-1)!=\prod_{1\le i<j\le n}(j-i)[/imath] we only prove [imath]\prod_{1\le i<j\le n}\dfrac{a_{j}-a_{i}}{j-i}[/imath] is integer maybe consider Vandermonde determinants ? But I can't prove this |
1100189 | Proving by induction that [imath]\frac{n(n + 1)(2n + 1)}{6} = 0^2 + 1^2 + 2^2 + 3^2 + ... + n^2[/imath]
Note: I am asking this question as a simple introductory question to proofs by induction, to which I will give also my formal answer (which should be correct, if not, please comment) for future visitors of this website. I know that to prove something by induction, we need formally 3 steps (2, if we considere inductive hypothesis and inductive step together): Base case or basis: We determine what is the base case for our statement [imath]P(n)[/imath], and we immediately prove it. Inductive hypothesis: We assume our statement [imath]P(n)[/imath] is true for [imath]n[/imath]. Inductive step: We try to prove that our statement is also true for [imath]P(n + 1)[/imath]. If we can prove that, we can also prove that our statement is true for [imath]n + 2[/imath] (we don't need to do it), and so on. This step is usually the most difficult part, because it involves some algebraic manipulation, or some imagination in some way. For this last reason, it might be helpful watching other similar proofs by induction to solve the current one. For example, suppose our [imath]P(n)[/imath] is the following: [imath]\frac{n(n + 1)(2n + 1)}{6} = 0^2 + 1^2 + 2^2 + 3^2 + ... + n^2[/imath] which is basically saying that the sum of all squares of the natural numbers from [imath]0[/imath] to [imath]n[/imath] can be determined by the formula on the left side of the equation. How would we prove [imath]P(n)[/imath] is true for all [imath]n \in \mathbb{N}[/imath]? For more information about the proof by induction, see the Wikipedia page about induction called Mathematical induction. | 72636 | Compute [imath]1^2 + 3^2+ 5^2 + \cdots + (2n-1)^2[/imath] by mathematical induction
I am doing mathematical induction. I am stuck with the question below. The left hand side is not getting equal to the right hand side. Please guide me how to do it further. [imath]1^2 + 3^2+ 5^2 + \cdots + (2n-1)^2 = \frac{1}{3}n(2n-1)(2n+1)[/imath]. Sol: [imath]P(n):\ 1^2 + 3^2 + 5^2 + \cdots + (2n-1)^2 = \frac{1}{3}n(2n-1)(2n+1)[/imath]. For [imath]n=n_1 = 1[/imath] [imath]P(1) = \frac{1}{3}{3} = (1)^2.[/imath] Hence it is true for [imath]n=n_0 = 1[/imath]. Let it be true for [imath]n=k[/imath] [imath]P(k): 1^2 + 3^2 + 5^2 + \cdots + (2k-1)^2 = \frac{1}{3}k(2k-1)(2k+1).[/imath] We have to prove that it is true for [imath]P(k+1)[/imath]. [imath]P(k+1) = 1^1+3^2+5^2+\cdots+(2k+1)^2 = \frac{1}{3}(k+1)(2k+1)(2k+3)\tag{A}.[/imath] Taking LHS: [imath]\begin{align*} 1^2 + 3^2 + 5^2 + \cdots + (2k+1)^2 &= 1^2+3^2+5^2+\cdots + (2k+1)^2\\ &= 1^2 + 3^2 + 5^2 + \cdots + (2k-1)^2 + (2k+1)^2\\ &= \frac{1}{3}k(2k-1)(2k+1) + (2k+1)^2\\ &=\frac{k(2k-1)(2k+1)+3(2k+1)^2}{3}\\ &=\frac{(2k+1)}{3}\left[k(2k-1) + 3(2k+1)\right]\\ &=\frac{(2k+1)}{3}\left[2k^2 - k + 6k + 3\right]\\ &=\frac{1}{3}(2k+1)(2k^2 +5k + 3)\\ &=\frac{1}{3}(2k+1)(k+1)\left(k+\frac{3}{2}\right) \tag{B} \end{align*}[/imath] EDIT: Solving EQ (A): [imath]=(1/3)(2k^2+5K+3) (2K+1) \tag{C}[/imath] Comparing EQ(B) and EQ(C) Hence proved that it is true for [imath]n = k+1.[/imath] Thus the proposition is true for all [imath]n >= 1[/imath]. Thanks. |
1100184 | a problem in The Mean-Value Theorem
Let [imath]f(x)[/imath] be defined and continuous on the interval [imath][a,b][/imath] and differentiable on [imath](a,b)[/imath]. Prove that there is at least one number [imath]c[/imath] such that [imath]\frac{af(a)-bf(b)}{a-b}=f(c)+cf'(c).[/imath] | 1099141 | if f(x) if differentiable and continuous, prove [imath]\frac{af(a)-bf(b)}{a-b} = f(c) + cf'(c) [/imath]
Let [imath]f(x)[/imath] be differentiable at [imath](a,b)[/imath] and continuous at [imath][a,b][/imath] prove that there exist [imath] c \in (a,b)[/imath] such that: [imath] \frac{af(a)-bf(b)}{a-b} = f(c) + cf'(c) [/imath] I started with: let [imath]c \in (a,b) [/imath] [imath] f'(c) = \frac{f(b)-f(a)}{b-a} [/imath] But I have no clue how to continue from here, how can I transform it to this form? :S Any suggestions? |
732712 | My first proof employing strong induction / complete induction (very simple number theory). Please mark/grade.
What do you think about my first proof employing strong induction? What mark/grade would you give me? Theorem Every natural number greater than 1 is a product of one or more primes. Proof First, introducing a predicate [imath]P[/imath] over [imath]\mathbb{N}[/imath], we rephrase the theorem as follows. [imath]\forall n \in \{2, 3, \dots \}, \, P(n) \quad \text{where} \quad P(n) \, := \, n \text{ is a product of one or more primes}[/imath] We prove the theorem by strong induction; we induct on [imath]n[/imath]. Basis We have [imath]P(n)[/imath] for [imath]n = 2[/imath], since [imath]2[/imath] is a product of one prime, namely [imath]2[/imath]. Inductive step Below, we show that for all [imath]n \in \{2, 3, \dots \}[/imath], [imath]P(2) \land \dots \land P(n - 1) \land P(n) \Rightarrow P(n + 1)\text{.}[/imath] Let [imath]k \in \{2, 3, \dots \}[/imath]. We assume that [imath]P(2) \land \dots \land P(k - 1) \land P(k)[/imath] is true. In the following, we use this assumption to show that [imath]P(k + 1)[/imath] is true. We consider [imath]k + 1[/imath]. There are two cases: It is prime or not. Obviously, the cases are exhaustive. We consider the cases separately; in each consideration, we conclude that [imath]P(k + 1)[/imath] is true. In so doing, we conclude the inductive step for all cases. Case 1: Obviously, [imath]k + 1[/imath] is a product of one prime, namely [imath]k + 1[/imath]. Thus [imath]P(k + 1)[/imath] is true. Case 2: As [imath]k + 1[/imath] is not prime, by definition, it may be represented as a product of at least two natural numbers that are greater than [imath]1[/imath]. Thus there are [imath]i, j \in \{2, \dots, k - 1, k\}[/imath] such that [imath]ij = k + 1[/imath]. By the induction hypothesis, [imath]P(i)[/imath] and [imath]P(j)[/imath] are true. Hence, we can easily see that the product [imath]\,ij\,[/imath] is a product of primes. Thus [imath]P(ij)[/imath] is true. (Thus [imath]P(k + 1)[/imath] is true.) By the basis, the inductive step, and the method of strong induction; [imath]P(n)[/imath] is true for all [imath]n \in \{2, 3, \dots \}[/imath]. | 833746 | Complete induction proof that every [imath]n > 1[/imath] can be written as a product of primes
I'm trying to get through Spivak's Calculus on my own and even though I kinda understand induction I'm not so sure that's the case when it comes to complete induction. So I tried to do a starred problem which involves using it. But I'm not sure that my proof is valid, can someone check this for me? So the problem is that I have to prove that For every natural [imath]n>1[/imath], if [imath]n[/imath] is not prime we can write it as a product of primes. I'm starting with a base case. For [imath]n=2[/imath], [imath]n[/imath] is prime, so the assumption holds. Let's assume that for some [imath]n > 1[/imath] and also all numbers [imath]p <= n[/imath] the assumption holds. Then [imath]n+1 = cd[/imath]. If [imath]n+1[/imath] is not prime, then [imath]c < n+1[/imath] and [imath]d < n+1[/imath], but as they are natural numbers, we can also write that [imath]c <= n[/imath] and [imath]d <= n[/imath]. But we assumed that if [imath]p <= n[/imath], and [imath]p[/imath] is not prime, we can write it as a product of primes. So if [imath]c[/imath] or [imath]d[/imath] are not primes, we can write them as a product of primes. That means that [imath]n+1[/imath] can be written as a product of primes if it's not prime, which completes the proof. |
266391 | PDE Evans, 1st edition, Chapter 5, Problem 14
Let [imath]U[/imath] be bounded, with a [imath]C^1[/imath] boundary. Show that a ''typical'' function [imath]u\in L^p(U) (1\leq p < \infty)[/imath] does not have a trace on [imath]\partial U[/imath]. More precisely there does not exist a bounded linear operator [imath]T:L^p(U)\rightarrow L^p(\partial U) [/imath] such that [imath]Tu=u|_{\partial U}[/imath] whenever [imath]u\in C(\bar{U})\cap L^p(U)[/imath] | 332599 | A typical [imath]L^p[/imath] function does not have a well-defined trace on the boundary
This question is from PDE by Evans, 1st edition, Chapter 5, Problem 14. It has been posted here previously, however, I cannot quite put all the information together from the responses there. Hopefully you can help me now. The problem is as follows: Let [imath]U[/imath] be bounded with a [imath]C^1[/imath] boundary. Show that a ''typical'' function [imath]u \in L^p(U) \ (1 \leq p < \infty)[/imath] does not have a trace on [imath]\partial U[/imath]. More precisely, prove there does not exist a bounded linear operator \begin{equation} T:L^p(U) \to L^p(\partial U) \end{equation} such that [imath]Tu = \left. u \right|_{\partial U}[/imath] whenever [imath]u \in C(\overline{U}) \cap L^p(U)[/imath]. So what we're trying to find is a bounded sequence of functions that go to infinity at the boundary, correct? Because then [imath]Tu = \left. u \right|_{\partial U}[/imath] is undefined. I like the idea of using the [imath]\mathrm{dist}(x, \partial U)[/imath] function to do this. I was thinking about defining the function \begin{equation} u(x) = \frac{1}{\mathrm{dist}(x,\partial U)}, \end{equation} which must be in [imath]L^p(U)[/imath] since [imath]U[/imath] is bounded (hence integral of [imath]\varepsilon[/imath] over [imath]U[/imath] is finite, right?). But this function is not continuous at the boundary, and so [imath]u \notin C(\overline{U})[/imath]. Can this be modified somehow? Another option is to use that the boundary is [imath]C^1[/imath]. So for each [imath]x^0 \in \partial U[/imath] there exists an [imath]r > 0[/imath] and a [imath]C^1[/imath] function [imath]\gamma : \mathbb{R}^{n-1} \to \mathbb{R}[/imath] such that [imath]U \cap B(x^0,r) = \{ x \in B(x^0,r) \ | \ x_n > \gamma(x_1,\ldots, x_n) \}[/imath]. So around every point on the boundary we can find a ball where the n'th coordinate is greater than the [imath]C^1[/imath] function [imath]\gamma[/imath]. How can we use this? |
1100617 | If G is a p-group, the number of nonnormal subgroup is a multiple of p?
I want to show that If [imath]G[/imath] is a finite [imath]p[/imath]-group, then the number of nonnormal subgroups of [imath]G[/imath] is a multiple of [imath]p[/imath]. I think I need to consider a conjugation action. But then? | 279420 | Questions around the number of subgroups of a [imath]p[/imath]-group
Let [imath]G[/imath] pe a [imath]p[/imath] group. I have to show that the number of nonnormal subgroups is divisible by [imath]p[/imath] the number of subgroups differs from the number of normal subgroups by a power of [imath]p[/imath]. Are there any theorem that can help me to prove this ? We have discussed the Sylow theorems but I don't know how to apply them - if those are the theorems I need. (Does this theorem also hold for infinite groups [imath]G[/imath] ?) |
1093108 | Find sufficient condition for [imath]l[/imath] and [imath]m[/imath].
Find the sufficient condition for [imath]l[/imath] and [imath]m[/imath] for which the system of equations [imath]ab+bc+ca=l[/imath] [imath]abc=m[/imath] has positive integeral solutions for [imath]a,b,c[/imath]. This is not a standard question from some source but I need for some work. I have found some conditions but they are not sufficient, but necessary. | 1092832 | Find integer solution to the given system of equation.
find all positive integers [imath]a,b,c[/imath] such that [imath]abc=24[/imath] [imath]ab+bc+ca=38[/imath] If particular values are given then we can easily find the solution but I am searching for some short general method. Is there any sufficient condition for the constant values to have integer solution? |
259826 | Purely "algebraic" proof of Young's Inequality
Young's inequality states that if [imath]a, b \geq 0[/imath], [imath]p, q > 0[/imath], and [imath]\frac{1}{p} + \frac{1}{q} = 1[/imath], then [imath]ab\leq \frac{a^p}{p} + \frac{b^q}{q}[/imath] (with equality only when [imath]a^p = b^q[/imath]). Back when I was in my first course in real analysis, I was assigned this as homework, but I couldn't figure it out. I kept trying to manipulate the expressions algebraically, and I couldn't get anywhere. But every proof that I've seen since uses calculus in some way to prove this. For example, a common proof is based on this proof without words and integration. The proof on Wikipedia uses the fact that [imath]\log[/imath] is concave, which I believe requires the analytic definition of the logarithm to prove (correct me if I'm wrong). Can this be proven using just algebraic manipulations? I know that that is a somewhat vague question, because "algebraic" is not well-defined, but I'm not sure how to make it more rigorous. But for example, the proof when [imath]p = q = 2[/imath] is something I would consider to be "purely algebraic": [imath]0 \leq (a - b)^2 = a^2 + b^2 - 2ab,[/imath] so [imath]ab \leq \frac{a^2}{2} + \frac{b^2}{2}.[/imath] | 2600100 | algebrainc proof for [imath]\alpha\beta \leq \frac{\alpha^p}{p} + \frac{\beta^q}{q}[/imath] for conjugate exponents p and q
I found a very informal proof (geometric one, for by taking [imath]\alpha[/imath] as [imath]\beta[/imath] in the x-axis and y-axis and showing rectangle area is smaller than area under the plots). Is there a proper algebraic proof for the inequality ? [imath] \alpha\beta \leq \frac{\alpha^p}{p} + \frac{\beta^q}{q} [/imath] where [imath]\alpha, \beta > 0[/imath] [imath] \frac{1}{p} + \frac{1}{q} = 1 [/imath] |
1101372 | Solve [imath]z^3 + 5z^2 + (9 - 5i)z + 10 - 10i = 0[/imath]
Solve [imath]z^3 + 5z^2 + (9 - 5i)z + 10 - 10i = 0[/imath] I have never dealt with equations with complex numbers in them so this is interesting; first Ill expand. [imath] \implies z^3 + 5z^2- 5iz + 9z + 10 - 10i = 0[/imath] I am stuck again -_- any ideas? Thanks! | 867971 | Solving the complex polynomial
For the complex polynomial [imath]z^3 -5z^2 +(7-2i)z +6i-3 = 0 [/imath] [imath]1)[/imath] show that [imath]2+i [/imath] is a root. [imath]2)[/imath] solve the given equation. Attemp to solve: I'm not really sure how to solve this, but I considered using Vietes formulas to comstruct a system of equations with two unknowns, and from there to cacluate the roots.....I'm not sure how to do No. 1 though... help with this would be appricated... |
1101669 | [imath]f:[a,b]\to [c,d][/imath] is Riemann integrable bijective function imply [imath]f^{-1}[/imath] also Riemann integrable
Assume that [imath]f:[a,b]\to [c,d][/imath] is a Riemann integrable bijective function. Is [imath]f^{-1}[/imath] also Riemann integrable? I think it's true. I have tried to find a counterexample but failed since bijective is so strong. But I don't know where should I started with. | 19329 | the Riemann integrability of inverse function
If [imath]f \colon [a,b] \rightarrow [c,d][/imath] is a bijection, [imath]f\in \mathcal{R}[/imath] and [imath]f^{-1}[/imath] exists, then prove or disprove that [imath]f^{-1} \in \mathcal{R} [c,d][/imath]. Remark: I tried to use integration by parts to find [imath]\int_{c}^{d} f^{-1}[/imath] and to prove that was the right limit of Riemann sum, but failed. But I think this idea might be useful. |
1101701 | Closed form of product of complex numbers
I'm stuck in a proof where I want to get a closed form of something. This is the last thing I need to complete my proof: Apparently for small [imath]n\geq2[/imath], the product [imath]\prod\limits_{k=1}^{n-1} (1-\exp(\frac{2ki\pi}{n}))=n[/imath]. How can i show it for all [imath]n\geq2[/imath]? It does not seem too hard to show, so any hints are welcome. Thanks in advance | 986425 | Trying to evaluate [imath]\prod_{k=1}^{n-1}(1-e^{2k\pi i/n})[/imath] for my complex analysis homework
For my complex analysis homework, I am trying to show that the integral of the real function [imath]1/(1+x^n)[/imath], for integer [imath]n\ge2[/imath], along the positive real line is [imath]\int_0^{\infty}\frac{dx}{1+x^n} = \frac{\pi}{n\sin(\pi/n)}.[/imath] The suggested approach is to consider the contour integral [imath]\int_{\gamma}f(z)dz = \int_{\gamma}\frac{dz}{1+z^n},[/imath] where [imath]\gamma[/imath] is a contour construced such that it traverses all of the positive real line, but only a single simple pole is included. So I started out by factoring the polynomial in the denominator, which has (distinct) roots [imath]\omega_k=\exp(i[\pi+2k\pi]/n)[/imath], so that [imath]1+z^n=(z-\omega_0)(z-\omega_1)\cdots(z-\omega_{n-1})[/imath]. Using this, I attempt to find the residue at [imath]z=\omega_0[/imath] (which is the single pole included in [imath]\gamma[/imath] by construction) by [imath]\text{Res}_{\omega_0}f=\lim_{z\rightarrow\omega_0}\frac{1}{(z-\omega_0)\cdots(z-\omega_{n-1})}(z-\omega_0)= \frac{1}{(\omega_0-\omega_1)\cdots(\omega_0-\omega_{n-1})}.[/imath] I have found that for every [imath](\omega_0-\omega_k)=\exp(i\pi/n)-\exp(i\pi/n+2k\pi i/n)[/imath], I can extract an [imath]\omega_0[/imath] from the [imath]\omega_k[/imath], yielding [imath](\omega_0-\omega_k)=\exp(i\pi/n)[1-\exp(2k\pi i/n)][/imath] such that we are left with [imath]\text{Res}_{\omega_0}f=\frac{1}{\prod_{k=1}^{n-1} e^{i\pi/n}[1-e^{2k\pi i/n}]}=\frac{1}{e^{i\pi(n-1)/n}\prod_{k=1}^{n-1} \left(1 - e^{2k\pi i /n}\right)}.[/imath] Now, since I know the final answer, I strongly believe that [imath]\prod_{k=1}^{n-1}(1-e^{2k\pi i/n} )=n,[/imath] but I am having a very hard time trying to evaluate this product. Any pointers would be greatly appreciated. |
1101883 | Maximize area of rectangle fenced on three sides
A city decides to make a park by fencing off a section of riverfront property. Funds are allotted to provide [imath]80[/imath] meters of fence. The area enclosed will be a rectangle, but only [imath]3[/imath] sides will be enclosed by fence- the other side will be bound by the river. What is the maximum area that can be enclosed in this way? I know hoe to find the maximum area, I just can't decide how I will represent the dimensions. HELP! | 419571 | Maximize area of a rectangle
A rectangular lot is bordered on one side by a building and the other 3 sides by [imath]800[/imath]m of fencing. Determine the area of the largest lot possible. |
1102002 | Proving a set is uncountable.
I need to prove that the set of all functions [imath]\mathcal{F}:\mathbb{N}\rightarrow \left \{ 0,1 \right \}[/imath] is uncountable. I'm not too sure at all how to do this. My initial idea was to try and show that [imath]\mathcal{F} \rightarrow \mathbb{N}[/imath] was not a bijection but I'm not clear at all and need some big help. Thanks!!! | 304314 | The set of all functions from [imath]\mathbb{N} \to \{0, 1\}[/imath] is uncountable?
How can I prove that the set of all functions from [imath]\mathbb{N} \to \{0, 1\}[/imath] is uncountable? Edit: This answer came to mind. Is it correct? This answer just came to mind. By contradiction suppose the set is [imath]\{f_n\}_{n \in \mathbb{N}}[/imath]. Define the function [imath]f: \mathbb{N} \to \{0,1\}[/imath] by [imath]f(n) \ne f_n(n)[/imath]. Then [imath]f \notin\{f_n\}_{n \in \mathbb{N}}[/imath]. |
1102272 | If [imath]p \mid a^n[/imath] then does [imath]p^n \mid a^n[/imath]?
I'm trying to figure out if the statement is true or not and I need to prove it if so. Let [imath]p[/imath] be a prime and [imath]a[/imath] be an integer. If [imath]p\mid a^n[/imath] , is it true that [imath]p^n\mid a^n[/imath] ? I'm not sure how i would approach this problem. I started as [imath]a^n = p \cdot b[/imath] for some [imath]b \in \mathbb{Z}[/imath]. | 881779 | [imath]p[/imath] prime, [imath]p\mid a^k \Rightarrow p^k\mid a^k[/imath]
Suppose [imath]p[/imath] is a prime and [imath]a[/imath] and [imath]k[/imath] are positive integers. Prove that, if [imath]p \mid a^k[/imath], then [imath]p^k\mid a^k[/imath] also. I have already proven that if [imath]a,b,n\in\mathbb{N}[/imath] and if [imath]a^n\mid b^n[/imath], then [imath]a\mid b[/imath]. I tried using induction: [imath]k=1 \Rightarrow p\mid a^1 \Rightarrow p^1\mid a^1[/imath]. Assume the statement holds for some [imath]k[/imath]. Then if [imath]p\mid a^{k+1}\Rightarrow p\mid a^ka[/imath]. Since [imath]p[/imath] is prime, [imath]p\mid a[/imath] or [imath]p\mid a^k[/imath]. If [imath]p\mid a[/imath], we are done. If [imath]p\mid a^k[/imath] then by assumption, [imath]p^k\mid a^k[/imath] and so by the lemma, [imath]p\mid a[/imath] and so [imath]p^{k+1}\mid a^{k+1}[/imath]. Does this work? |
1102534 | Residue of [imath]f(z)[/imath] using Laurent Series at [imath]z=-2[/imath]
Calculate the residue of: [imath]f(z) = \frac{\psi(-z)}{(z+1)(z+2)^3} \space \text{at} \space z=-2[/imath] Where [imath]\psi(z)[/imath] is the digamma function, and [imath]\zeta(z)[/imath] is the Riemann-zeta function (below). The answer given is: [imath]\displaystyle \mathrm{Res}_{z=-2} f(z) = \frac{\pi^2}{6} +\zeta(3) - 3 +\gamma[/imath] Using the Laurent Series. Obviously we must find, the coefficient of [imath]\frac{1}{z+2}[/imath] in the series expansion. Using the series definition of [imath]\psi(-z)[/imath] zeta definition [imath]\psi(-z) = 1-\gamma + \sum_{k=1}^{\infty} (1-\zeta(k+1)) (z+2)^k[/imath] Or [imath]\psi(-z) = -\gamma + \sum_{n=0}^{\infty} \frac{z + 1}{(n+1)(z-n)}[/imath] How can I approach this? Thanks! | 1102571 | Find the residue at [imath]z=-2[/imath] for [imath]g(z) = \frac{\psi(-z)}{(z+1)(z+2)^3}[/imath]
Find the residue at [imath]z=-2[/imath] for [imath]g(z) = \frac{\psi(-z)}{(z+1)(z+2)^3}[/imath] I know that: [imath]\psi(z+1) = -\gamma - \sum_{k=1}^{\infty} (-1)^k\zeta(k+1)z^k[/imath] Let [imath]z \to -1 - z[/imath] to get: [imath]\psi(-z) = -\gamma - \sum_{k=1}^{\infty} \zeta(k+1)(z+1)^k[/imath] therefore we divide by the other part to get: [imath]\frac{\psi(-z)}{(z+1)(z+2)^3} = -\frac{\gamma}{(z+1)(z+2)^3} - \sum_{k=1}^{\infty} \frac{\zeta(k+1)(z+1)^{k-1}}{(z+2)^3}[/imath] I have to somehow get the coefficient of [imath]\frac{1}{z+2}[/imath] because I want to evaluate the residue of [imath]g(z)[/imath] at [imath]z=-2[/imath] The problem is I cant ever get a factor of [imath]\frac{1}{z+2}[/imath] what should I do? |
1103446 | Prove that function [imath]f[/imath] is continuous
Let [imath]f:\mathbb{R} \to \mathbb{R}[/imath] assume that for an arbitrary interval [imath](a,b) \subseteq \mathbb{R}[/imath] inverse image [imath]f^{-1}((a,b))[/imath] is union (not necessarily finite) of open intervals. Show that function [imath]f[/imath] is continuous. I'd be grateful for any hints, since I have problem with tackling this one | 105080 | Equivalence of continuity definitions
How to show that [imath](1)\Longleftrightarrow (2)[/imath] in metric spaces ? pre-image of open sets are open [imath]\delta[/imath]-[imath]\epsilon[/imath] definition of continuity |
1103488 | How to Evaluate [imath]\int _0^{\infty} \lfloor x \rfloor e^{-x} dx[/imath]?
How to Evaluate [imath]\int _0^{\infty} \lfloor x \rfloor e^{-x} dx[/imath] , where [imath]\lfloor x \rfloor[/imath] is the largest integer less than x? I am thinking of using something like [imath] \int_0^{\infty} (e^{-2}+2e^{-3}+....+ne^{-n-1}+.....)dx[/imath] and then use [imath]\int ne^{-n-1}= -e^{-n}[/imath] But the integral, AFAIK, is only finitely-additive, so now what? | 64067 | Evaluate the integral [imath]\int_0^{\infty} \lfloor x \rfloor e^{-x}\mathrm dx[/imath]
I'd like some help with the following integral: [imath]\int_0^\infty \lfloor x \rfloor e^{-x}\mathrm dx .[/imath] Thanks. |
1101460 | Prove that for an arbitrary (possibly infinite) language, that for a finite L-structure [imath]M[/imath], if [imath]M \equiv N[/imath] then [imath] M \cong N[/imath]
Prove that for an arbitrary (possibly infinite) language, that for a finite L-structure [imath]M[/imath], if [imath]M \equiv N[/imath] then [imath] M \cong N[/imath] I'm struggling to think of what to do, I presume the best thing is to keep it simple and assume the Language to only be relational. I'd start by assuming we have a finite Language and taking [imath]|M|[/imath] = k = [imath]|N|[/imath] and as I've assumed the language to only be relational and finite take {[imath]R_1,...,R_p[/imath]} to be the relation symbols. I'm guessing I need to show that there is some sentence [imath]\sigma[/imath] that is true in both [imath]M[/imath] and [imath]N[/imath]? I have no idea what that sentence would be could someone help me out? Thanks. Also how would I even begin to answer it in the infinite case? | 238909 | Complete first order theory with finite model is categorical
I am trying to prove that if [imath]T[/imath] is a complete first order theory that has a finite model then it has exactly one model up to isomorphism. To this end, I assumed that [imath]T[/imath] is complete with a finite model [imath]M_n[/imath]. Then I assumed that [imath]M_m[/imath] was another model of size [imath]m \geq n[/imath]. We know that if a theory has two finite models with different cardinalities then the theory is incomplete hence [imath]m = n[/imath]. Now two things remain to be shown: one is that any two models of finite size [imath]n[/imath] are isomorphic and the other is that every infinite model is also isomorphic to this finite model (is this even possible? but we clearly need to do something about the infinite case) Thanks for helping me finish this proof. For the record: this is an exercise in Just/Weese, page 84. Edit I'm looking for a sentence [imath]\varphi[/imath] such that if [imath]M_n \models \varphi[/imath] then [imath]M_n \cong M_m[/imath]. There are no assumptions on the language. But I think it's not possible to have a finite model for an infinite language so the language must be finite. Edit 2 After some more thinking, if there is a finite model [imath]M[/imath] of size [imath]n[/imath], let [imath] \varphi = \exists v_1, \dots , v_n ((v_1 \neq v_2) \land \dots \land (v_{n-1} \neq v_n)) \land \lnot \exists v_{n+1} ((v_{n+1} \neq v_1) \land \dots \land (v_{n+1} \neq v_n))[/imath] that is, [imath]\varphi[/imath] says that the model has exactly [imath]n[/imath] elements. Since [imath]T[/imath] is complete, either [imath]\varphi[/imath] or [imath]\lnot \varphi[/imath] is provable from [imath]T[/imath]. Since we have a model in which [imath]\varphi[/imath] is true we therefore know that [imath]T \vdash \varphi[/imath]. Hence any model of [imath]T[/imath] must have exactly [imath]n[/imath] elements. Now the question is, how do I show that any two [imath]n[/imath]-element models of [imath]T[/imath] must be isomorphic? |
1103740 | On infinitude of primes of certain form.
We know that there are infinite number of primes so as there are infinite number of primes of the form [imath]4n+3[/imath] where [imath]n\in Z^+[/imath]. A note on Burton's book (Elementary Number Theory) somehow says that it is of high chance to expect that there are also infinite number of primes of the form [imath]4n+1[/imath]. However it is not yet proven by the time the book was published. A quick search on net gives infinitude of primes of the form [imath]3n+1[/imath] and [imath]5n+1[/imath]. The question is: The problem: Are there infinitely many number of primes of the form [imath]4n+1[/imath] still an open problem or it was already proven? If so, where can I find the proof? Thanks a lot. | 244915 | Infinitely number of primes in the form $4n+1$ proof
Question: Are there infinitely many primes of the form [imath]4n+3[/imath] and [imath]4n+1[/imath]? My attempt: Suppose the contrary that there exist finitely many primes of the form [imath]4n+3[/imath], say [imath]k+1[/imath] of them: [imath]3,p_1,p_2,....,p_k[/imath] Consider [imath]N = 4p_1p_2p_3...p_k+3[/imath], [imath]N[/imath] cannot be a prime of this form. So suppose that [imath]N[/imath]=[imath]q_1...q_r[/imath], where [imath]q_i∈P[/imath] Claim: At least one of the [imath]q_i[/imath]'s is of the form [imath]4n+3[/imath]: Proof for my claim: [imath]N[/imath] is odd [imath]\Rightarrow q_1,...,q_r[/imath] are odd [imath]\Rightarrow q_i \equiv 1\ (\text{mod }4)[/imath] or [imath]q_i ≡ 3\ (\text{mod }4)[/imath] If all [imath]q_1,...q_r[/imath] are of the form [imath]4n+1[/imath], then [imath](4n+1)(4m+1)=16nm+4n+4m+1 = 4(\cdots) +1[/imath] Therefore, [imath]N=q_1...q_r = 4m+1[/imath]. But [imath]N=4p_1..p_k+3[/imath], i.e. [imath]N≡3\ (\text{mod }4)[/imath], [imath]N[/imath] is congruent to [imath]1\ \text{mod }4[/imath] which is a contradiction. Therefore, at least one of [imath]q_i \equiv 3\ (\text{mod }4)[/imath]. Suppose [imath]q_j\equiv 3\ (\text{mod }4)[/imath] [imath]\Rightarrow[/imath] [imath]q_j=p_i[/imath] for some [imath]1\leq i \leq k[/imath] or [imath]q_j =3[/imath] If [imath]q_j=p_i≠3[/imath] then [imath]q_j[/imath] | [imath]N = 4p_1...p_k + 3 \Rightarrow q_j=3[/imath] Contradiction! If [imath]q_j=3[/imath] ([imath]\neq p_i[/imath], [imath]1\leq i \leq k[/imath]) then [imath]q_j | N = 4 p_1...p_k + 3 \Rightarrow q_j=p_t[/imath] for some [imath]1 \leq i \leq k[/imath] Contradiction! In fact, there must be also infinitely many primes of the form [imath]4n+1[/imath] (according to my search), but the above method does not work for its proof. I could not understand why it does not work. Could you please show me? Regards |
1104064 | Euler path in cube
Suppose we have the cube [imath]3\times3\times3[/imath] divided by [imath]1\times1\times1[/imath] cubes. We want to prove that there isn't path from an edge cube to the cube in the center which passes through every cube and doesn't pass through an already passed cube. We can only go to adjacent cube if the two cubes have common side.I know the fact that we have Euler path if and only we have [imath]2[/imath] edges with odd degree but clearly in the [imath]3\times3\times3[/imath] cube if each cube is edge and each has vertices it is easy to see that there are more than [imath]2[/imath] cubes with odd vertices.Does this mean that there is no Euler path and we can't reach the middle cube? | 1102734 | A $3 \times 3 \times 3$ cube has no Hamiltonian path starting at the corner.
We have a [imath]$3\times3\times 3$[/imath] cube which has [imath]27[/imath] cubes each [imath]$1\times1\times1$[/imath] stuck together as usual. [imath]2[/imath] cubes are neighbours if they have a common face. The corner cubes are the [imath]8[/imath] cubes at the corners of the big cube. Can the worm start from a corner cube and eat its neighbour cube, continuing until at the end the last cube eaten is the one in the center of the big cube which has 6 neighbour cubes? |
1101340 | Non-negative integers to form a sum with restrictions?
How do I solve this? Number of non-negative solutions to [imath]x_1 + x_2 + x_3 + x_4 = 4[/imath] where [imath]0 \le x_i \le 3[/imath]? What's the general technique? I already know the technique for [imath]j \le x_i[/imath] but have no clue about upper bound restrictions. | 553960 | extended stars-and-bars problem(where the upper limit of the variable is bounded)
The problem of counting the solutions [imath](a_1,a_2,\ldots,a_n)[/imath] with integer [imath]a_i\geq0[/imath] for [imath]i\in\{1,2,\ldots,n\}[/imath] such that [imath]a_1+a_2+a_3+....a_n=N[/imath] can be solved with a stars-and-bars argument. What is the solution if one adds the constraint that [imath]a_i\leq r_i[/imath] for certain integers [imath]r_1,\ldots,r_n[/imath]? e.g. for [imath]n=3[/imath], [imath]N=6[/imath] and [imath](r_1,r_2,r_3)=(3,3,2)[/imath], the tuple [imath](a_1,a_2,a_3)=(2,3,1)[/imath] is a solution, but [imath](2,1,3)[/imath] is not a solution because [imath]a_3=3>2=r_3[/imath]. |
1104276 | The curve [imath]x^3-y^3=1[/imath] is asymptotic t the line [imath]x=y[/imath]. Find the point n the curve farthest from the line [imath]x=y[/imath].
The curve [imath]x^3-y^3=1[/imath] is asymptotic t the line [imath]x=y[/imath]. Find the point in the curve farthest from the line [imath]x=y[/imath] This is just need of further details in this question. Kindly give details about, if we are talking about asymtotic curve to the line [imath]x=y[/imath], what does it really means. And why we need the family of curves [imath]y=x+C[/imath] to look this problem(as in the answer given by Calvin). | 268927 | The curve [imath]x^3− y^3= 1[/imath] is asymptotic to the line [imath]x = y[/imath]. Find the point on the curve farthest from the line [imath]x = y[/imath]
(NBHM_2006_PhD Screening Test_Analysis) The curve [imath]x^3− y^3= 1[/imath] is asymptotic to the line [imath]x = y[/imath]. Find the point on the curve farthest from the line [imath]x = y[/imath] how should i solve this problem |
1104411 | Differentiable continuous function whose derivative is not continuous
Is there a function which is continuous and differentiable, but is not smooth function? By smooth I mean having continuous derivative. For example, the derivative of [imath]f(x)=x|x|/2[/imath] is [imath]f'(x)=|x|[/imath] which is continuous. So I consider this function smooth. | 724716 | Does a differentiable everywhere function have a continuous derivative?
If a function [imath]f[/imath] defined on [imath][a,b][/imath] and differentiable everywhere, does it mean that its derivative [imath]f'(x)[/imath] is continuous everywhere on [imath][a,b][/imath]? My understanding is 'yes'. Because if there were 'a gap' in [imath]f'(x)[/imath], we could integrate back to [imath]f(x)[/imath] and show that where the gap is, there would be a sharp "angle" where [imath]f(x)[/imath] is not differentiable. |
1104214 | Conformal map from the inside of the unit disk to the inside of an ellipse
I lack intuition when it comes to some conformal mappings and I'm presently looking for a conformal map taking the inside of a disk, let's say the unit disk and sending it to the inside of an ellipse. I know that a Joukowski transformation: [imath]z \rightarrow z + \frac{\eta}{z}[/imath] would take a circles and its exterior to an ellipse and its exterior but it is ill defined in its inside (singular in zero here). So it cannot be an inverse Joulkowski transform. I heard that may be some elliptic functions could do the job. Is it the direction I should look into? | 207934 | Conformal parametrization of an ellipse
I am looking to a formula for the conformal map from the unit disc in the interior of an ellipse centered in [imath]0[/imath] and with semiaxes [imath]a,b>0[/imath]. I know that depends on elliptic function, but I didn't find any references. Thanks in advance for any formula or reference. |
1006091 | Convergence in measure implies convergence almost everywhere of a subsequence
How can I prove that if a sequence of functions [imath]\{f_n\}[/imath] that converges to [imath]f[/imath] in measure on a space of finite measure, then there exists a subsequence of [imath]\{f_n\}[/imath] that converges to [imath]f[/imath] almost everywhere? | 1632288 | If [imath]f_n\to f[/imath] in measure, is there a subsequence s.t. [imath]f_{n_k}\to f[/imath] a.e.?
If [imath]f_n\to f[/imath] in measure, is there a subsequence s.t. [imath]f_{n_k}\to f[/imath] a.e. ? The convergence in measure is a little be abstract to me, I don't really see what it means (even if I know the definition). So I have difficulties to answer this question. |
1105409 | Transformation Matrix with respect to a basis.
I have a question regarding transformation matrices with regard to a basis. Lets say that there is a basis [imath]B = \{v_1,v_2,v_3\}[/imath] There is a formula that says that [imath][A]_B = C^{-1}AC[/imath] where [imath]C[/imath] is a matrix that has as column vectors the vectors that form the basis [imath]B[/imath] and [imath]A[/imath] is the transformation matrix with respect to the standard basis. My question is if I can think like this: Since [imath]A[/imath] is the transformation matrix with respect to the standard basis then every column vector has as its components its coordinates with respect to the standard basis. Can I use [imath]C^{-1}[/imath] (inverse of [imath]C[/imath]) to change the coordinates of every column vector of [imath]A[/imath] to their coordinates with respect to basis [imath]B[/imath] and use those coordinates (or vectors) as column vectors for [imath][A]_B[/imath]. Also I found a comment that states that: If [imath]B=(v_1,v_2,…,v_n)[/imath], then by definition the columns in [imath][A]_B[/imath] are [imath]Av_1,Av_2,…,Av_n[/imath] Can someone give me a link to read about the last statement. I have tried googling but came up with nothing. I am confused! | 1105253 | Intuitive understanding of the [imath]BAB^{-1}[/imath] formula for changing basis in linear transformations.
I would please like to have a better understanding on why we use the formula [imath]BAB^{-1}[/imath] when chaning basis on linear transformations. (I have not a complete understanding on this so there are probably errors in my question, please point them out so I can learn more...) For example: In [imath]\mathbb{R^3}[/imath] Lets say we have a basis [imath]e = {(e_1, e_2, e_3)}[/imath] and another basis [imath]f = {(f_1, f_2, f_3)}[/imath]. We have a change of basis matrix from [imath]f[/imath] to [imath]e[/imath] called [imath]B[/imath] that expresses what [imath]f_1, f_2, f_3[/imath] is in the basis [imath]e[/imath]. We have defined a linear transformation [imath]R_e[/imath] in the basis [imath]e[/imath] with the matrix [imath]A[/imath]. Let's say we have three vectors expressed in the base [imath]f[/imath]: [imath]P = \left( \begin{array}{c} s\\ t\\ u\\ \end{array} \right)[/imath] Now we want to apply the linear transformation [imath]R_e = A[/imath] to these vectors: First we have to change the basis of [imath]P[/imath] which gives us: [imath]BP[/imath]. [imath]BP[/imath] is now [imath]P[/imath] expressed in the basis [imath]e[/imath]. To apply the linear transformation we now only has to multiply [imath]BP[/imath] with [imath]A[/imath]: [imath]ABP[/imath]. But according to the formula it should be [imath]BAB^{-1}P[/imath]. Where is my thinking incorrect? |
1105530 | Expectation of the square of a process, martingales
Let [imath](X_n)_{n=1}^{\infty}[/imath] be a sequence of i.i.d. r.v. on [imath](\Omega,\mathcal{F},\mathbb{P})[/imath]. Assume that [imath]\mathbb{E}X_1=0[/imath] and [imath]\mathbb{E}X_1^2=1[/imath], consider [imath]\mathcal{F}_n=\sigma (X_1,...,X_n)[/imath], and define [imath]M_n=\sum\limits_{k=1}^n\frac{1}{k}X_1\cdot \cdot \cdot X_k[/imath]. Calculate [imath]\mathbb{E}(M_n^2)[/imath] My question is whether we can calculate [imath]\mathbb{E}(M_n^2)[/imath] without using the formula: [imath]Var(M_n)=\mathbb{E}(M_n^2)-(\mathbb{E}(M_n))^2[/imath]. Also we know that [imath](M_n)_{n=0}^{\infty}[/imath] is an [imath](\mathcal{F})_n)_{n=0}^{\infty}[/imath] martingale | 1105211 | Exercise on Martingales
I have been struggling with the following exercise and I was wondering whether my solution is correct or not. I am pretty sure about the second part of the question (the martingale part) but not so sure about the first part of the question. Please tell me what you think. Let [imath](X_n)_{n=1}^{\infty}[/imath] be a sequence of i.i.d. r.v. on [imath](\Omega,\mathcal{F},\mathbb{P})[/imath]. Assume that [imath]\mathbb{E}X_1=0[/imath] and [imath]\mathbb{E}X_1^2=1[/imath], consider [imath]\mathcal{F}_n=\sigma (X_1,...,X_n)[/imath], and define [imath]M_n=\sum\limits_{k=1}^n\frac{1}{k}X_1\cdot \cdot \cdot X_k[/imath]. Calculate [imath]\mathbb{E}(M_n^2)[/imath] and show that [imath](M_n)_{n=0}^{\infty}[/imath] is an [imath](\mathcal{F}_n)_{n=0}^{\infty}[/imath] martingale. Now my answer is the following: [imath]\mathrm{Var} (X_1)=(\mathbb{E}X_1^2)-(\mathbb{E}(X_1))^2=1[/imath] Now: [imath]\mathrm{Var}(M_n)=\mathrm{Var}(\sum_{k=1}^{n}\frac{1}{k}(X_1\cdot\cdot\cdot X_k))=\sum_{k=1}^{n}\frac{1}{k^2}\mathrm{Var}(X_1\cdot\cdot\cdot X_k)=\sum_{k=1}^{n}\frac{1}{k^2}[/imath], since [imath]\mathrm{Var}(X_1)=1[/imath] and [imath](X_n)[/imath] is an iid sequence. Also we have that: [imath]\mathbb{E}M_n=\mathbb{E}(\sum_{k=1}^{n}\frac{1}{k}(X_1\cdot\cdot\cdot X_k))=\sum_{k=1}^{n}\frac{1}{k}\mathbb{E}(X_1\cdot\cdot\cdot X_k)=0[/imath] since [imath]\mathbb{E}X_1=0[/imath] and [imath](X_n)[/imath] iid. Now using that: [imath]\mathbb{E}M_n^{2}=\mathrm{Var}(M_n)+(\mathbb{E}M_n)^2[/imath], we get that: [imath]\mathbb{E}M_n^2=0[/imath]. Now to prove that [imath](M_n)_{n=0}^{\infty}[/imath] is an [imath](\mathcal{F}_n){n=0}^{\infty}[/imath] martingale, I proceeded as follows: [imath]\mathbb{E}(M_{n+1}|\mathcal{F}_n)=\mathbb{E}(M_n+\frac{1}{n+1}(X_1\cdot\cdot\cdot X_{n+1})|\mathcal{F}_n)=\mathbb{E}(M_n|\mathcal{F}_n)+\frac{1}{n+1}\mathbb{E}(X_1\cdot\cdot\cdot X_n|\mathcal{F}_n)=M_n+\frac{1}{n+1}\mathbb{E}(X_1|\mathcal{F}_n)\cdot\cdot\cdot\mathbb{E}(X_{n+1}|\mathcal{F}_n)=M_n+\frac{1}{n+1}\mathbb{E}X_1\cdot\cdot\cdot \mathbb{E}X_{n+1}=M_n+0=M_n.[/imath] |
1105685 | Evaluating [imath]\lim_{x\to0} \frac{(1+x)^{1/x}-e}{x}[/imath]
Evaluate [imath]\lim_{x\to0} \frac{(1+x)^{1/x}-e}{x}[/imath] | 1101831 | Need hint for [imath]\lim_{x\to 0} \frac{(x+1)^\frac{1}{x}-e}{x}[/imath]
Please give me some hints or solution for this limit. I had it on my exam and I'm curious how to solve it. [imath] \lim_{x\to 0} \frac{(x+1)^\frac{1}{x}-e}{x} [/imath] |
1105930 | Prove that if [imath]a_{2n} \rightarrow g[/imath] and [imath]a_{2n+1} \rightarrow g[/imath] then [imath]a_n \rightarrow g[/imath]
The problem is in the question: Prove that if sequences [imath]a_{2n} \rightarrow g[/imath] and [imath]a_{2n+1} \rightarrow g[/imath] then [imath]a_n \rightarrow g[/imath]. I don't know how to prove that - it seems obvious when we look at the definition that for sufficiently large [imath]n[/imath] (let's say [imath]n>N[/imath]) we have [imath]|a_{2n}-g|[/imath] and [imath]|a_{2n+1}-g|[/imath] less than any given [imath]\epsilon[/imath] and these are all elements of [imath]a_n[/imath] when [imath]n>N[/imath] (even and odd terms). Does this need more formal proof? | 539964 | Convergence of a sequence whose even and odd subsequences converge
Suppose [imath]\{a_n\}[/imath] is a sequence such that the subsequences [imath]\{a_{2n−1}\}[/imath] and [imath]\{a_{2n}\}[/imath] converge to the same limit, say [imath]a[/imath]. Show that [imath]\{a_n\}[/imath] also converges to [imath]a[/imath]. |
1106030 | sum of a binomial coefficient
Trying without success to solve the following: what is the sum of [imath]\binom{80}{0}-\binom{80}{1}+\binom{80}{2}-\binom{80}{3}...-\binom{80}{79}+\binom{80}{80}[/imath] any help will be greatly appreciated | 1104744 | Calculating the sum of a binomial coefficient series
Calculate this: [imath]\bigl(\begin{smallmatrix} 80 \\0 \end {smallmatrix}\bigr)-\bigl(\begin{smallmatrix} 80 \\1 \end {smallmatrix}\bigr)+\bigl(\begin{smallmatrix} 80 \\2 \end {smallmatrix}\bigr)-\bigl(\begin{smallmatrix} 80 \\3 \end {smallmatrix}\bigr)+...-\bigl(\begin{smallmatrix} 80 \\79 \end {smallmatrix}\bigr)+\bigl(\begin{smallmatrix} 80 \\80 \end {smallmatrix}\bigr)[/imath] Can I get hints/suggestions for this? All I know is that: [imath]\bigl(\begin{smallmatrix} 80 \\0 \end {smallmatrix}\bigr)[/imath] and [imath]\bigl(\begin{smallmatrix} 80 \\80 \end {smallmatrix}\bigr)[/imath] must be equal to 1, right? I know there must be some kind of trick but I just can't find it. Thanks |
1106495 | How to prove that [imath]\cos(\pi÷11)+\cos(3\pi÷11)+\cos(5\pi÷11)+\cos(7\pi÷11)+\cos(9\pi÷11)=0.5[/imath]?
I need to prove that [imath]\cos\dfrac{\pi}{11}+\cos\dfrac{3\pi}{11}+\cos\dfrac{5\pi}{11}+\cos\dfrac{7\pi}{11}+\cos\dfrac{9\pi}{11}=\dfrac{1}{2}[/imath] How to do it? | 1105236 | Prove sum of [imath]\cos(\pi/11)+\cos(3\pi/11)+...+\cos(9\pi/11)=1/2[/imath] using Euler's formula
Prove that [imath]\cos(\pi/11)+\cos(3\pi/11)+\cos(5\pi/11)+\cos(7\pi/11)+\cos(9\pi/11)=1/2[/imath] using Euler's formula. Everything I tried has failed so far. Here is one thing I tried, but obviously didn't work. [imath]\Re e \{e^{\frac{\pi}{11}i}(1+e^{\frac{2\pi}{11}i}+e^{\frac{4\pi}{11}i}+e^{\frac{6\pi}{11}i}+e^{\frac{8\pi}{11}i}) \}=\frac{1}{2}[/imath] [imath]\Re e \{e^{\frac{\pi}{11}i}(1+\sqrt[11]{e^{2\pi i}}+\sqrt[11]{e^{4\pi i}}+\sqrt[11]{e^{6\pi i}}+\sqrt[11]{e^{8\pi i}}) \}=\frac{1}{2}[/imath] [imath]\Re e \{e^{\frac{\pi}{11}i}(1+\sqrt[11]{1}+\sqrt[11]{1}+\sqrt[11]{1}+\sqrt[11]{1}) \}=\frac{1}{2}[/imath] [imath]\Re e \{5e^{\frac{\pi}{11}i} \}=\frac{1}{2}[/imath] [imath]5\cos(\frac{\pi}{11})=\frac{1}{2}[/imath] Which isn't true :D Thanks in advance |
1106622 | closure of the unit ball
Is the closure of the unit ball of [imath]C^1[0,1][/imath] in [imath]C[0,1][/imath] compact? For this let us take a sequence [imath]x_n[/imath] in [imath]C^1[0,1][/imath] to show it has a convergent subseqence How to proceed with this.I am not so sure whether it is the way to ask a question here | 280546 | "The closure of the unit ball of [imath]C^1[0, 1][/imath] in [imath]C[0, 1][/imath]" and its compactness
[I really want to apologize if this problem looks a little too long.] The problem : This is taken from here [Question. 3.7 (c)] and it says... Prove or disprove the comapctness of the closure of the unit ball of [imath]C^1[0, 1][/imath] in [imath]C[0, 1][/imath]. What have I tried? I think it's pretty clear that the unit ball in [imath]C^1[0,1][/imath] is [imath]B= \{f \in C^{1}[0,1]: \lVert f \rVert \leq 1\}[/imath]. Let us denote [imath]K= \operatorname{cl}_{C[0,1]}(B)[/imath], the closure of [imath]B[/imath] in [imath]C[0,1][/imath]. I really didn't have much clue how to start working on this problem but finally thought about using the sequential approach. If [imath]a[/imath] and [imath]b[/imath] are two points such that [imath]0<a<b<1[/imath] then I think there can be functions [imath]f_a[/imath] and [imath]f_b[/imath] having the property that [imath]\lVert f_a-f_b \rVert = 1[/imath] which are sufficiently smooth to be in [imath]C^1[/imath]. Now as it is quite tedious to construct these functions explicitly, I scanned a hand-drawn picture of what I think they might look like. Here [imath]f_a[/imath] and [imath]f_b[/imath] take zero values in almost all of the interval [imath][0,1][/imath] and jumps up at [imath]a[/imath] and [imath]b[/imath] respectively. Also, as is clear from the figure, they don't assume non-zero values simultaneously. Now, my argument is clear: however close the two points [imath]a[/imath] and [imath]b[/imath] come to each other (remaining distinct) there will always be functions like [imath]f_a[/imath] and [imath]f_b[/imath]. They may get steeper and steeper but will never lose their [imath]C^1[/imath]-ness. So if I consider the sequence [imath](\frac{1}{n})[/imath] in [imath][0,1][/imath], I will get a sequence [imath](f_{\frac{1}{n}})[/imath] in [imath]B (\subseteq K)[/imath] where for [imath]m \neq n[/imath] we will have [imath]\lVert f_{\frac{1}{m}} - f_{\frac{1}{n}}\rVert = 1[/imath]. Hence the sequence [imath](f_{\frac{1}{n}})[/imath] cannot have a convergent subsequence, thereby proving that [imath]K[/imath] is not sequentally compact and hence not compact. But I'm not very sure about all these. A friend of mine told me that this same problem has a positive answer and that made me sufficiently confused. So, here comes my question... In the above argument, where have I gone wrong? What is the real answer? And how to prove it? Thanks a million for reading my extra-long question. And thanks a zillion for any help that you can offer. |
1106957 | Connection between field and the frobenius homomorphism
Let K be a field with char(K)>0. How do I prove that every algebraic extension of K is a separable extension if and only if [imath]\phi:x \rightarrow x^p[/imath] is surjective ? | 273124 | Separability by surjectivity of Frobenius Endomorphism
I'm trying to prove the following statement: Let [imath]F[/imath] be a finite field of prime characteristic [imath]p[/imath] and let [imath]E[/imath] be the field generated by [imath]F[/imath] and the elements [imath]\{t^{1/p},n \geq 1\}[/imath], where [imath]t[/imath] is an indeterminate. Then, any algebraic extension of [imath]E[/imath] is separable. I've already proved that the Frobenius Endomorphism on [imath]E[/imath] is surjective. I'm trying to use the following fact: Given [imath]\alpha[/imath] an element of an algebraic extension of [imath]E[/imath], then [imath]\alpha[/imath] is separable if, and only if, the derivate of [imath]Irr(\alpha,E)[/imath] is not zero. Then, I know that the irreducible polynomial can be expressed function of [imath]x^p[/imath], [imath]Irr(\alpha, E) = p(x^p)\in F[x].[/imath] However, I got stuck here and I don't know what else to do. Could someone give me and advice? |
746445 | Convergence and sum of geometric series [imath](e^{3-2n})[/imath] as [imath]n[/imath] goes from [imath]2[/imath] to [imath]\infty[/imath]
I have simplified the expression to: (e^3 / e^2n) This particular question asks to answer whether or not the series converges by virtue of |common ratio| < 1 alone, without using any other tests (besides checking if the limit DNE or does not equal 0), and then to find the sum using the formula first term/(1-r). The limit appears to go to 0, so the series does not diverge. As I cannot fit the the expression in the form a(r)^(n), I'm currently unable to either say whether or not the series converges and what the sum is. I just need help manipulating the expression to the form a(r)^(n), then I can answer the rest. Thanks! | 449113 | Geometric series sum [imath]\sum_2^\infty e^{3-2n}[/imath]
[imath]\sum_2^\infty e^{3-2n}[/imath] The formulas for these things are so ambiguous I really have no clue on how to use them. [imath]\frac {cr^M}{1-r}[/imath] [imath]\frac {1e^2}{1-\frac{1}{e}}[/imath] Is that a wrong application of the formula and why? |
1106632 | To prove triangle inequality for [imath]d : \mathbb C \times \mathbb C \to \mathbb R[/imath] ; [imath]d(x,y):=\frac {|x-y|} {\sqrt{1+|x|^2}+\sqrt{1+|y|^2}}[/imath]
Is the function [imath]d : \mathbb C \times \mathbb C \to \mathbb R[/imath] defined by [imath]d(x,y):=\dfrac {|x-y|} {\sqrt{1+|x|^2}+\sqrt{1+|y|^2}}[/imath] a metric ? I can easily prove it is symmetric and positive-definite ; but cant make any headway towards the triangle inequality . Please help | 1104145 | Proving that the triangle inequality holds for a metric on [imath]\mathbb{C}[/imath]
Show that [imath](X,d)[/imath] is a metric space where [imath]X =\Bbb C [/imath] and the distance function is defined as: [imath]d(x,y) = \frac {2|x-y|}{\sqrt {1+|x|^2} + \sqrt {1 + |y|^2}}, \text{ for } x,y \in \Bbb C.[/imath] I have done the proof of the first two propositions for being a metric, but I'm having a problem in proving the triangle inequality. |
1107246 | let [imath]x[/imath] be in finite group [imath]G[/imath] and let order of [imath]x[/imath] is [imath]p[/imath]. If [imath]h^{-1}xh = x^{10}[/imath] for a finite group show that [imath]p=3[/imath]
Let [imath]G[/imath] be a finite group, [imath]p[/imath] be the smallest prime divisor of [imath]|G|[/imath] and x [imath]\in[/imath] G an element of order [imath]p[/imath]. Suppose [imath] h \in G [/imath] is such that [imath]h^{-1}xh = x^{10}[/imath]. Show that [imath]p = 3[/imath]. I cant solve this problem i think it should be solved without using any special theorem please firstly give me a hint . | 696059 | Qualifying Exam Question On Elementary Group Theory
Question. Let [imath]G[/imath] be a finite group and [imath]p[/imath] be the smallest prime dividing [imath]|G|[/imath]. Let [imath]x[/imath] be n element of order [imath]p[/imath] in [imath]G[/imath]. Assume that there exists an element [imath]h\in G[/imath] such that [imath]hxh^{-1}=x^{10}[/imath]. Show that [imath]p=3[/imath]. What I have done so far is use the fact that [imath]o(hxh^{-1})=o(x)=p[/imath] and [imath]o(x^{10})=p/\gcd(p,10)[/imath]. By hypothesis we have [imath]p=p/\gcd(p,10)[/imath], giving [imath]\gcd(p,10)=1[/imath]. Here I am stuck. I have not been able to make use of the fact that [imath]p[/imath] is the smallest prime divisor of [imath]|G|[/imath]. Can somebody help? |
1106016 | If [imath]\gcd (a,b) = 1[/imath], what can be said about [imath]\gcd (a+b,a-b)[/imath]?
Suppose [imath]a, b \in \mathbb{Z}[/imath], [imath]a > b[/imath], and [imath]\gcd (a,b) = 1[/imath]. What can be said about [imath]\gcd (a+b,a-b)[/imath]? Is it true in general that [imath]\gcd (a+b,a-b) \leq 2[/imath]? | 1172907 | Greatest common divisor problem: If $\gcd(x, y) = 1$ then $\gcd(x + y, x - y) = 1$ or $2$
Prove that if [imath]\gcd(x, y) = 1[/imath] then [imath]\gcd(x + y, x - y) = 1[/imath] or [imath]$2$[/imath]. I know that any linear combination of [imath]x, y[/imath] is multiple of 1 since [imath]\gcd(x, y) = 1[/imath] then the set would be [imath]\{1, 2, 3, 4, 5, \ldots, x - y, \ldots, x + y, \ldots \}[/imath], [imath]x[/imath] and [imath]y[/imath] cannot be both even. But [imath]x[/imath] and [imath]y[/imath] can be both odd primes or even and odd for their GCD to be 1. If [imath]x[/imath] and [imath]y[/imath] are odd prime then [imath]x + y[/imath] and [imath]x - y[/imath] will give even integers, so their GCD could be 2. If [imath]x[/imath] and [imath]y[/imath] are even and odd then their GCD can be either 1, 3 or 5. This is what I think. Is it in the correct way? |
1107965 | Prove [imath]N\cap Z(P)\ne e[/imath] given a [imath]p[/imath]-group [imath]P[/imath] and a normal subgroup [imath]N[/imath].
Let [imath]P[/imath] be a [imath]p[/imath]-group and let [imath]N\triangleleft P[/imath]. Prove [imath]N\cap Z(P)\ne e[/imath] . Here's what I know so far: [imath]P[/imath], being a [imath]p[/imath]-group, is nilpotent and therefore is solvable. That means that it has its upper central series: [imath]e=Z_0\triangleleft Z_1\triangleleft Z_2....Z_n...[/imath] where [imath]Z_1=Z(P)[/imath] and that at some point, the series stabilizes at P forever... I also know that [imath]Z_{i+1}=[/imath]{ [imath]x\in P|\forall y\in P: [x,y]\in Z_i[/imath]}. (In my notebook it's been defined that way: [imath]Z_{i+1}=\pi^{-1}(Z(G[/imath]\ [imath]Z_i)[/imath] whichc confuses me a little. ) How can I show that the desired intersection is not empty? | 1036242 | Showing that [imath]|N \cap Z(G)| > 1[/imath] for normal subgroups of p-groups
I have a finite [imath]p[/imath]-group [imath]G[/imath] and a normal subgroup [imath]N[/imath] which is not the trivial subgroup. I am asked to show that [imath]|N \cap Z(G)| > 1[/imath]. There has been a similar question on MSE here: How to show that [imath]H \cap Z(G) \neq \{e\}[/imath] when [imath]H[/imath] is a normal subgroup of [imath]G[/imath] with [imath]\lvert H\rvert>1[/imath] Unfortunately, I cannot comment on other user's questions yet, so I made my own question. I know that the trick is to use conjugation but I can't follow the logic. Can someone elaborate? Especially on the last part of the top answer. |
1107986 | Induction - Prime Numbers
Prove that, for every natural number [imath]n > 2[/imath], there is a prime number between [imath]n[/imath] and [imath]n![/imath]. [Hint: There is a prime number that divides [imath]n! - 1[/imath].] | 1084296 | For all [imath]n>2[/imath] there exists a prime number between [imath]n[/imath] and [imath] n![/imath]
How to prove that there exists a prime number between [imath]n[/imath] and [imath] n![/imath], for all [imath] n> 2[/imath]? (Bertrand's postulate gives a much better bound, but this question is about obtaining a self-contained proof.) |
1107984 | Uniform continous functions?
Please if someone could tell me how to show that [imath]1\over x[/imath] is not uniformly continuous on [imath](0,1)[/imath]. I hope I've been clear enough, thanks. | 1069684 | Prove: [imath]\frac1x[/imath] is not a uniformly continuous function
I have to prove that [imath]\frac1x [/imath] with [imath]x \in (0,\infty)[/imath] is not uniformly continuous. I understand that the further you approach the [imath]0[/imath] the greater your [imath]e[/imath] and the smaller your [imath]\delta[/imath] becomes (in the limit even [imath]\delta = 0 [/imath] ?). But I need help with the "technical" proof, can someone give me an instruction to do this? |
1107942 | Fermat Numbers Proof
Fermat numbers are shown by: [imath]F_m = 2^{2^m} + 1[/imath]. How can I prove that for any [imath]m ≠ n[/imath], I can have [imath](F_m, F_n) = 1[/imath]? | 123524 | Fermat numbers are coprime
So today in my final for number theory I had to prove that the Fermat numbers ([imath]F_n=2^{2^n}+1[/imath]) are coprime. I know that the standard proof uses the following: [imath]F_n=F_1\cdots F_{n-1}+2[/imath] and then the [imath]\gcd[/imath] divides [imath]2[/imath], and it cannot be two, and hence the numbers are coprime. However, he asked us to use the hint: "Let [imath]l[/imath] be a prime dividing [imath]F_n[/imath]. What can you say about the order of [imath]2[/imath] in [imath](\mathbb{Z}/l\mathbb{Z})^\times[/imath]?" I have been thinking about it and I cannot figure out how to use his hint. Any ideas? |
1108023 | Separate numerator and denominator integral
I was given [imath] \frac{\int_0^{\pi/2}\sin^{\sqrt{2}+1}(x) \, dx}{ \int_0^{\pi/2}\sin^{\sqrt{2}-1}(x) \, dx} [/imath] How to evaluate this integral? Since denominator and numerator are different Integral substitution will fail , I tried with by part Integral but it lead to nowhere help? | 445808 | Integral [imath] \dfrac { \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx} { \int_0^{\pi/2} (\sin x)^{\sqrt 2 - 1} dx} [/imath]
I have this difficult integral to solve. [imath] \dfrac { \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx} { \int_0^{\pi/2} (\sin x)^{\sqrt 2 - 1} dx} [/imath] Now my approach is this: split [imath](\sin x)^{\sqrt 2 + 1}[/imath] and [imath](\sin x)^{\sqrt 2 - 1}[/imath] as [imath](\sin x)^{\sqrt 2}.(\sin x)[/imath] and [imath](\sin x)^{\sqrt 2 - 2}.(\sin x)[/imath] respectively, and then apply parts. But that doesn't seem to lead anywhere. Hints please! Edit: This is what I did (showing just for the numerator) [imath] \int_0^{\pi/2} (\sin x)^{\sqrt 2 + 1} dx [/imath] [imath] = \int_0^{\pi/2} (\sin x)^{\sqrt 2}.(\sin x) dx [/imath] [imath] = (-\cos x)(\sin x)^{\sqrt 2}\Bigg|_0^{\pi/2} + \int_0^{\pi/2}(\sin x)^{ \sqrt 2 - 1 }(\cos^2 x) dx [/imath] (taking [imath] v = \sin x [/imath] and [imath] u = (\sin x)^{\sqrt 2} [/imath] in the [imath] \int uv [/imath] formula) [imath] = \int_0^{\pi/2}\left( (\sin x)^{ \sqrt 2 - 1 } - (\sin x)^{ \sqrt 2 + 1 } \right) dx [/imath] Similarly for the denominator. This does give a reduction formula but then I don't see how to really use it for finding the answer. |
300531 | Integral representation of Euler's constant
Prove that : [imath] \gamma=-\int_0^{1}\ln \ln \left ( \frac{1}{x} \right) \ \mathrm{d}x.[/imath] where [imath]\gamma[/imath] is Euler's constant ([imath]\gamma \approx 0.57721[/imath]). This integral was mentioned in Wikipedia as in Mathworld , but the solutions I've got uses corollaries from this theorem. Can you give me a simple solution (not using much advanced theorems) or at least some hints. | 697383 | Integral from [imath]0[/imath] to [imath]\infty[/imath] of [imath]\ln(x)/e^x[/imath]
Show [imath]\int_0^\infty \frac{\ln(x)}{e^x} = -\gamma[/imath] (gamma is Euler-Mascheroni constant). Can anyone please prove this result? Also [imath] \int_0^\infty \frac{\left( \ln(x) \right)^2}{e^x}\mathrm dx. [/imath] |
1108182 | Show that [imath]<(1,2)(3,4),(1,2,3,4,5)> = A_5[/imath]?
I'm revising for my Group Theory exam and saw this in a past year paper question, and I'm not sure enumerating all 60 elements of [imath]A_5[/imath] and showing how I can get them using the given generating set is wise. So far, I've been trying to use Lagrange's Theorem. I know that [imath]<S>[/imath] is a subgroup of [imath]A_5[/imath] since elements of [imath]S[/imath] are even permutations. I also know that if [imath]R \subseteq S[/imath], then [imath]<R> \le <S>[/imath]. So if [imath]R=\{ (1,2,3,4,5)\}[/imath], then [imath]|<R>|=5[/imath] and so [imath]5 \mid <S>[/imath]. Knowing that [imath]|A_5| = 60[/imath], I think if I can find another subgroup of [imath]<S>[/imath] with 12 elements, or two subgroups with 3 and 4 elements respectively, then I can show that [imath]60 \mid <S>[/imath] and so [imath]<S> = A_5[/imath]. Any idea how I can find the other subgroup(s)? EDIT: I found this before I posted. I don't quite understand what was commented there, and I didn't know how to "hijack" the thread and ask for more responses. Sorry if it isn't good form to post the same question! | 1077664 | Is there a more elegant way of proving [imath]\langle (1,2)(3,4), (1,2,3,4,5) \rangle = A_5[/imath]
I'm trying to show the following [imath]\langle (1,2)(3,4), (1,2,3,4,5) \rangle = A_5[/imath] I managed to prove this but I think my solution is very inelegant. Here's my argument let [imath]J = \langle (1,2)(3,4), (1,2,3,4,5) \rangle,[/imath] then [imath]\lvert J \rvert \geq 8[/imath] and [imath]\lvert J \rvert[/imath] divides [imath]\lvert A_5 \rvert = 60,[/imath] so the possibilities are [imath]\lvert J \rvert = 10, 12, 15, 20, 30, 60[/imath]. Then I sat down and calculated 13 more elements of [imath]J[/imath], so now the possibilities are [imath]\lvert J \rvert = 30[/imath] or [imath]60[/imath]. But we can't have [imath]\lvert J \rvert = 30[/imath] because then [imath]J[/imath] would be normal (index 2 theorem) which would contradict [imath]A_5[/imath] being simple, so we must have [imath]J = A_5[/imath]. The calculation part make this proof quite long winded, is there a simpler way of getting the result? |
1103767 | Verify that the unbounded function belongs to [imath]W^{1,n}[/imath]
Verify that if [imath]n > 1[/imath], the unbounded function [imath]u = \log \log \left(1+\frac 1{|x|}\right)[/imath] belongs to [imath]W^{1,n}(U)[/imath], for [imath]U=B^0(0,1)[/imath]. This is PDE Evans, 2nd edition: Chapter 5, Exercise 14. In order to show that [imath]u \in W^{1,n}[/imath], where [imath]W^{1,n}[/imath] is a Sobolev space containing the function [imath]u[/imath] itself and its first-order derivative. Differentiating [imath]u=\log \log\left(1+\frac 1{|x|}\right)[/imath], I obtain (and please verify if you would like) [imath]u_{x_i}=\frac 1{\log \left(1+\frac 1{|x|}\right)}\frac 1{1+\frac 1{|x|}} \frac{x_i}{|x|^2}.[/imath] Given [imath]u'=v[/imath] in the weak sense, should I show, according to definition, that [imath]\int_{B(0,1)} u \phi' \, dx = -\int_{B(0,1)} v\phi dx[/imath] for all test functions [imath]\phi \in C_c^\infty(U)[/imath]? The integration by parts formula comes to my mind here. Or, should I be employing material from Section 5.6 (Sobolev inequalities) of the textbook? In particular, they discussed on page 280 the borderline case [imath]p=n[/imath], which mentions: The borderline case p=n. We assume next that [imath]p=n.[/imath] Owing to Theorem 2 (page 279) and the fact that [imath]p*=\frac{np}{n-p}\to+\infty[/imath] as [imath]p \to n[/imath], we might expect [imath]u \in L^\infty(U)[/imath], provided [imath]u \in W^{1,n}(U)[/imath]. This is however false if [imath]n>1[/imath]: for example, if [imath]U=B^0(0,1)[/imath], the function [imath]u=\log \log\left(1+\frac 1{|x|}\right)[/imath] belongs to [imath]W^{1,n}(I)[/imath] but not to [imath]L^\infty(U)[/imath]. | 1036001 | Show that [imath]u(x)=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right)[/imath] is in [imath]W^{1,n}(U)[/imath], where [imath]U=B(0,1)\subset\mathbb{R}^n[/imath].
The entire problem statement is: Let [imath]n>1[/imath] and let [imath]U=B(0,1)\subset\mathbb{R}^n[/imath]. Show that [imath]u:U\to\mathbb{R}[/imath] given by [imath]u(x)=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right)[/imath] is in [imath]W^{1,n}(U).[/imath] My attempt at the proof is as follows: To show that [imath]u\in W^{1,n}(U)[/imath] it suffices to show that [imath]|Du|\in L^{n}(U)[/imath]. Consider [imath]u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}|x|^{-2}\frac{x_i}{|x|},[/imath] which simplifies to, [imath]u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{x_i}{|x|^3}.[/imath] Thus, [imath]|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{1}{|x|^2}[/imath] which can be manipulated to, [imath]|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+|x|}\frac{1}{|x|}.[/imath] Moreover, since [imath]U=B(0,1)[/imath] we can bound the second term from above which gives, [imath]|Du|\leq\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}[/imath] Thus, we can instead show that [imath]\int_U\left(\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}\right)^n<\infty.[/imath] Converting this into polar coordinates we have with [imath]r=|x|[/imath], [imath]\int_0^R\int_{\partial B(0,r)}\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\,dSdr=\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\int_{\partial B(0,r)}dSdr[/imath] Note that [imath]\int_{\partial B(0,r)}\,dS=n\alpha(n)r^{n-1}[/imath], which is the surface area of [imath]B(0,r)[/imath]. Thus, we have then [imath]n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^nr^{n-1}\,dr=n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\right)^n\frac{1}{r}\,dr[/imath] And so this is where I get stuck in the problem, since I don't how I can evaluate that last integral to show it's finite. Thank you for any help or feedback! |
1108306 | Prove [imath]G[/imath] is cyclic if for every [imath]n\in \Bbb{N}[/imath], there are at most [imath]n[/imath] solutions to the equation [imath]x^n=e[/imath].
Let [imath]G[/imath] be a finite group. Prove G is cyclic if for every [imath]n\in \Bbb{N}[/imath], there are at most [imath]n[/imath] solutions to the equation [imath]x^n=e[/imath]. I am stuck with this :( Would appreciate your help. | 50791 | If [imath]x^m=e[/imath] has at most [imath]m[/imath] solutions for any [imath]m\in \mathbb{N}[/imath], then [imath]G[/imath] is cyclic
Fraleigh(7th ed) Sec10, Ex47. Let [imath]G[/imath] be a finite group. Show that if for each positive integer [imath]m[/imath] the number of solutions [imath]x[/imath] of the equation [imath]x^m=e[/imath] in [imath]G[/imath] is at most [imath]m[/imath], then [imath]G[/imath] is cyclic. I tried it a few hours but I couldn't solve it. So I read the solution. But I narrowly understood the solution, and it's still unclear to me. How can I solve it? |
1108448 | Zorn's lemma usage\problem.
Let [imath](A,\le)[/imath] be an ordered set. Show that if any chain has an upper bound then for any [imath]a\in A[/imath] there exist a maximal element such that [imath]a\le x[/imath]. I am stuck with this... Would appreciate any help... | 385082 | Zorn's Lemma and Posets
If [imath]A[/imath] is a poset in which every chain has an upperbound in [imath]A[/imath] ([imath]a[/imath] be any element in [imath]A[/imath]). There exists at least one maximal element [imath]m[/imath] in [imath]A[/imath] such that [imath]m \geq a[/imath]. Whats the difference between this and Zorn's Lemma? |
659302 | How to prove that [imath]\mathbb{Q}[/imath] ( the rationals) is a countable set
I want to prove that [imath]\mathbb{Q}[/imath] is countable. So basically, I could find a bijection from [imath]\mathbb{Q}[/imath] to [imath]\mathbb{N}[/imath]. But I have also recently proved that [imath]\mathbb{Z}[/imath] is countable, so is it equivalent to find a bijection from [imath]\mathbb{Q}[/imath] to [imath]\mathbb{Z}[/imath]? | 1136275 | Why is [imath]\mathbb Q [/imath] (rational numbers) countable?
By definition, a set [imath]S[/imath] is called countable if there exists an bijective function [imath]f[/imath] from [imath]S[/imath] to the natural numbers [imath]N[/imath]. If we take a function [imath]g\colon\mathbb{Z\times N\to Q}[/imath] given by [imath]g(m, n) = \frac{m}{ n} [/imath] to "construct" rational numbers, [imath]g[/imath] would only be a surjection from the countable set [imath]\mathbb{Z\times N}[/imath] to [imath]\mathbb Q[/imath]. It's not injective, or is it? |
1106367 | Prove that [imath]p[/imath] divides [imath]F_{p-1}+F_{p+1}-1[/imath]
Given the Fibonacci sequence [imath](F_n)[/imath], defined by [imath]F_0=0,F_1=1, F_{n+2}=F_{n+1}+F_n[/imath], and [imath]p[/imath] an odd prime number, how to prove that [imath]p[/imath] divides [imath]F_{p-1}+F_{p+1}-1[/imath]? Is induction a good idea here? | 1086066 | Prove Divisibility In Fibonacci Sequence Over A Prime Number
In The Fibonacci sequence which is defined as [imath] F_n=F_{n-1}+F_{n-2}, [/imath] lets say we have the number [imath]p[/imath] which is an odd prime. Prove that: [imath]F_{p-1} + F_{p+1} -1[/imath] Is divisible by [imath]p[/imath]. Prove that for any given [imath]n[/imath] real positive integer: [imath]F_{p^{n+1}-1} + F_{p^{n+1}+1} -(F_{p^{n}-1} + F_{p^{n}+1})[/imath] Is divisible by [imath]p^{n+1}[/imath] How to prove this? |
1109305 | Show that [imath]\lim_{n \to +\infty} \frac{\sum^n x_i}{\sum^n y_i}=a.[/imath]
Let [imath]\{y_n\}[/imath] a sequence so that [imath]\sum y_i=+\infty[/imath] and [imath]y_n>0, \forall n \in \mathbb{N}.[/imath] Show that if [imath]\lim_{n \to +\infty} \frac{x_n}{y_n}=a[/imath] then [imath]\lim_{n \to +\infty} \frac{\displaystyle \sum^n x_i}{\displaystyle \sum^n y_i}=a.[/imath] | 100338 | limit of quotient of two series
Let [imath]Y_{n} > 0[/imath] for all [imath] n\in \mathbb{N} [/imath], with [imath]\sum{Y_{n}}= +\infty[/imath]. If [imath]\displaystyle\lim\limits_{n\rightarrow \infty}\frac{X_{n}}{Y_{n}}= a[/imath] then [imath]\displaystyle\lim\limits_{n\rightarrow\infty}\frac{X_{1}+X_{2}+X_{3}+\dots+X_{n}}{Y_{1}+Y_{2}+Y_{3}+\dots+Y_{n}}= a[/imath] I need a idea for solution. can i supose that [imath]\lim\limits_{n\rightarrow\infty}Y_{n}[/imath] exist ? |
1109156 | Prove [imath]\lim_{n \to \infty}[/imath] [imath](1+\frac xn-o(\frac 1n))^n=e^x[/imath]
We know that [imath]\lim_{n \to \infty}[/imath] [imath](1+\frac xn)^n=e^x[/imath]. How to prove that [imath]\lim_{n \to \infty}[/imath] [imath](1+\frac xn-o(\frac 1n))^n=e^x[/imath]? Attempt of the proof: Let [imath]\epsilon>0[/imath] [imath]\exists n_0[/imath] such that [imath]\forall n\ge n_0[/imath] [imath]|o(\frac 1n)|<\epsilon\frac 1n[/imath] and [imath]|n*o(\frac 1n)|<\epsilon[/imath]. Thus [imath](1+\frac{x+\epsilon}{n})^n>(1+\frac xn-o(\frac 1n))^n>(1+\frac{x-\epsilon}{n})^n[/imath]. So [imath]e^x*e^\epsilon \ge \lim_{n \to \infty}[/imath] [imath](1+\frac xn-o(\frac 1n))^n \ge \frac{e^x}{e^\epsilon}[/imath] [imath]\forall \epsilon>0[/imath]. Say [imath]\lim_{n \to \infty}[/imath] [imath](1+\frac xn-o(\frac 1n))^n=y[/imath], it can not be that [imath]y<e^x[/imath], because if it is so, then [imath]\exists \epsilon>0[/imath] such that [imath]e^\epsilon y<e^x[/imath] and the second inequality above is violated. By the similar talking [imath]y>e^x[/imath] is impossible. This completes the proof. | 821399 | What is the proper way to handle the limit with little-[imath]o[/imath]?
I was hoping to show that [imath]\left(1-\frac{x}{n}+o\left(\frac{2x}{n}\right)\right)^n \xrightarrow{n\to\infty} e^{-x}[/imath] which would be just fine without the little-[imath]o[/imath]. Trying binomial formula: [imath]\left(1-\frac{x}{n}+o\left(\frac{2x}{n}\right)\right)^n = \sum_{k=0}^n \binom{n}{k} \left(1-\frac{x}{n}\right)^{n-k} \left( o\left( \frac{2x}{n} \right) \right)^k[/imath] Yet it doesn't look like it leads me somewhere where I could use [imath]\frac{o\left( \frac{2x}{n} \right)}{\frac{2x}{n}}\xrightarrow{n\to\infty} 0[/imath] What is the proper way to handle it? Edited after the comment by Claude Leibovici. |
1109169 | Verify stokes theorem example
Let [imath]c:I^2\rightarrow\mathbb{R}^3[/imath] be the singular [imath]2[/imath]-cube given by [imath]c(s,t)=(\frac{1}{2}s^2,st,\frac{1}{2}t^2)[/imath] Let [imath]\omega=xy^2dz[/imath] Questions: i) Compute [imath]c^*\omega[/imath] ii) Compute [imath]c^*d\omega[/imath] iii) Compute [imath]\int_cd\omega[/imath] iv) Without using Stokes theorem, compute [imath]\int_{\partial c}\omega[/imath] (Hence verifying Stokes theorem) Answers: (Parts (i)-(iii) checked and are correct) i) I have gotten: [imath]c^*\omega=\frac{1}{2}s^4t^2dt[/imath] ii) [imath]c^*d\omega=d(\frac{1}{2}s^4t^2dt)=2s^3t^2ds\wedge dt[/imath] iii) [imath]\int_cd\omega=2\int_0^1(\int_0^1s^3t^2ds)dt=2\frac{1}{4}\frac{1}{3}=\frac{1}{6}[/imath] iv) I am using the formula [imath]\int_{\partial c}\omega=\sum_{j=1}^2\sum_{\alpha=0,1}(-1)^{j+\alpha}\int_0^1c^*_{(j,\alpha)}\omega[/imath] I have computed that [imath]c_{(1,0)}(t)=(0,0,\frac{1}{2}t^2),c_{(1,1)}(t)=(\frac{1}{2},t,\frac{1}{2}t^2),c_{(2,0)}(t)=(\frac{1}{2}t^2,0,0),c_{(2,1)}(t)=(\frac{1}{2}t^2,t,\frac{1}{2})[/imath] So [imath]c_{(1,0)}^*\omega=0,c_{(2,0)}^*\omega=0,c_{(2,1)}^*\omega=0[/imath] Point of contention: My calculation: [imath]c_{(1,1)}^*\omega[/imath] is the only non-zero part for [imath]\partial c[/imath] And [imath]c_{(1,1)}^*\omega=\omega(\frac{1}{2},t,\frac{1}{2}t^2)=\frac{1}{2}\cdot t^2\cdot d(\frac{1}{2}t^2)[/imath] from the definition of [imath]\omega=xy^2dz[/imath] This gives [imath]\omega(\frac{1}{2},t,\frac{1}{2}t^2)=\frac{1}{2}t^2\cdot tdt=\frac{1}{2}t^2\cdot tdt[/imath] So I have gotten [imath]c_{(1,1)}^*\omega=\frac{1}{2}t^3dt[/imath] this is wrong since, Answer given: [imath]c_{(1,1)}^*\omega=\frac{1}{2}t^2dt[/imath] and the next step is [imath]\int_{\partial c}\omega=\int_0^1\frac{1}{2}t^2dt=\frac{1}{6}[/imath] Which must be correct since this verifies Stokes theorem from part (iii) And my answer must be wrong since it gives [imath]\int_{\partial c}\omega=\frac{1}{8}[/imath] The answer also has [imath]c_{(1,1)}(t)=(\frac{1}{2},t,\frac{1}{2}t^2)[/imath], but computes [imath]c_{(1,1)}^*\omega[/imath] as [imath]\frac{1}{2}t^2dt[/imath], whereas I get [imath]\frac{1}{2}t^3dt[/imath] Where am I going wrong with this calculation, it is probably a simple mistake, any help would be greatly appreciated. | 1097684 | Compute [imath]\int_cd\omega[/imath] and [imath]\int_{\partial c}\omega[/imath]
Question: Let [imath]c:I^2\rightarrow\mathbb{R}^3[/imath] be the singular [imath]2[/imath]-cube given by [imath]c(s,t)=\left(\frac{1}{2}s^2,st,\frac{1}{2}t^2\right)[/imath]Let [imath]x=(x,y,z)[/imath] denote the cartesian coordinates on [imath]\mathbb{R}^3[/imath] and let [imath]\omega[/imath] be the [imath]1[/imath]-form on [imath]\mathbb{R}^3[/imath] given by [imath]\omega=xy^2dz[/imath] i. Compute [imath]c^*\omega[/imath] ii. Compute [imath]c^*d\omega[/imath] iii. Compute [imath]\int_cd\omega[/imath] iv. Without using stokes theorem compute [imath]\int_{\partial c}\omega[/imath] Answer: i. [imath](c^*\omega)(s,t)=(\omega\circ c)(s,t)=\omega\left(\frac{1}{2}s^2,st,\frac{1}{2}t^2\right)=(\frac{1}{2}s^2)(st)(d(\frac{1}{2}t^2))=(\frac{1}{2}s^3t)(tdt)=\frac{1}{2}s^3t^2dt[/imath] ii. [imath]d\omega=d(xy^2dz)=y^2dx\wedge dz+2xydy\wedge dz+xy^2d(dz)=y^2dx\wedge dz+2xydy\wedge dz[/imath] [imath]c^*d\omega=(st)^2d(\frac{1}{2}s^2)\wedge d(\frac{1}{2}t^2)+s^3td(st)\wedge d(\frac{1}{2}t^2)[/imath] [imath]=(st)^3ds\wedge dt+d^3t^3ds\wedge dt=2(st)^3ds\wedge dt[/imath] I am stuck on parts iii. and iv. Is there a formula for these types of problems? I have not seen an example of one of these computed before What is the name of this type of problem so that I can look up methods for it? Any help would be greatly appreciated |
1109614 | Division of natural numbers
How can I prove that [imath]9|n[/imath] if and only if [imath]9[/imath] divides the sum of the digits of [imath]n[/imath], where [imath]n \in \mathbb{N}[/imath] | 830641 | Show that the number 9 divides the number [imath] m[/imath] if and only if the sum of the digits of the number [imath] m [/imath] is divisible by 9.
Show that the number 9 divides the number [imath] m[/imath] if and only if the sum of the digits of the number [imath] m [/imath] is divisible by 9. Show that the number 3 divides the number [imath] m[/imath] if and only if the sum of the digits of the number [imath] m [/imath] is divisible by 3. Are semigroup and monoid theory could be useful? |
1109806 | Calculate [imath]S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}[/imath].
Calculate [imath]S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}[/imath]. I know I posted this question already but I want a more detailed answer. For example, how you got from one step to another using the partial fraction formula. Thanks! | 1109391 | Calculate [imath] S =\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}. [/imath]
Calculate [imath]S =\displaystyle\sum_{k=1}^n\frac {1}{k(k+1)(k+2)}[/imath]. This sequence is neither arithmetic nor geometric. How can you solve this. Thanks! |
1109673 | [imath]\sum_{i=0}^{k} \binom{m}{i}\binom{n}{k-i} =\binom{m+n}{k}[/imath]
I'm trying to show that the equality [imath]\sum_{i=0}^{k} \binom{m}{i}\binom{n}{k-i} =\binom{m+n}{k}[/imath] Is true. I know it is since there is a good combinatorical argument for it. If we have a group of [imath]m[/imath] men and [imath]n[/imath] women, and we need to choose [imath]k[/imath] then we know its equal to [imath]\binom{m+n}{k}[/imath], but we can also say that we choose [imath]0[/imath] men and [imath]k[/imath] women, or we can choose [imath]1[/imath] man and [imath]k-1[/imath] women, or we can choose [imath]2[/imath] men and [imath]k-2[/imath] women, etc. So in theory they should be equal, but I can't work the math behind it. A hint in the right direction would be appreciated | 219928 | Inductive Proof for Vandermonde's Identity?
I am reading up on Vandermonde's Identity, and so far I have found proofs for the identity using combinatorics, sets, and other methods. However, I am trying to find a proof that utilizes mathematical induction. Does anyone know of such a proof? For those who don't know Vandermonde's Identity, here it is: For every [imath]m \ge 0[/imath], and every [imath]0 \le r \le m[/imath], if [imath]r \le n[/imath], then [imath] \binom{m+n}r = \sum_{k=0}^r \binom mk \binom n{r-k} [/imath] |
473180 | The Diophantine equation [imath]x^2 + 2 = y^3[/imath]
How to solve the Diophantine equation [imath]x^2 + 2 = y^3[/imath] with [imath]x,y>0[/imath] ? ([imath]x,y[/imath] are integers.) | 1218264 | find all integer solutions of [imath]y^2=x^3-2[/imath]
I’m blind about integer solutions of a polynomial. I have no number theory background, but I’m curious about how to figure out all integer solutions of a polynomial, for example this question. It is said there are only two solutions, but I can’t give a proof. What I can show is that there are no even solutions. I am actually curious about the general pattern to consider this kind of problem. Can anyone give me some ideas or references? Thanks! |
546368 | Smallest positive integer [imath]n[/imath] s.t. f(n) = [imath]n^2 + n + 41[/imath] is composite?
I'm pretty sure it's 40 but I'm not too sure if it's enough to show that: [imath]n(n+1) + 41[/imath] [imath]41(\frac{n(n+1)}{41} + 1)[/imath] and the smallest composite no# will be achieved when n+1 = 41, n =40? Am I missing anything here? --> Yes I still need to show that for n < 40, it's prime..ideas? Thanks | 706565 | Prove that [imath]n^2+n+41[/imath] is prime for [imath]n<40[/imath]
Here's a problem that showed up on an exam I took, I'm interested in seeing if there are other ways to approach it. Let [imath]n\in\{0,1,...,39\}[/imath]. Prove that [imath]n^2+n+41[/imath] is prime. I shall provide my own solution, though I am curious does anyone know how to do this without using PIDs? |
1110011 | Finite cyclic subgroups of [imath]GL_{2} (\mathbb{Z})[/imath]
How could we prove that any element of [imath]GL_2(\mathbb{Z})[/imath] of finite order has order 1, 2, 3, 4, or 6? I am aware of the proof supplied here at this link: https://www.maa.org/sites/default/files/George_Mackiw20823.pdf. But I am curious if there are any other proofs. | 318920 | Order of matrices in [imath]GL_2(\mathbb{Z})[/imath]
Let [imath]A\in GL_2\left(\mathbb{Z}\right)[/imath], the group of invertible matrices with integer coefficients, and denote by [imath]\omega(A)[/imath] the order of [imath]A[/imath]. How we prove that [imath]\left\{\omega(A);A\in GL_2\left(\mathbb{Z}\right)\right\}=\{1,2,3,4,6,\infty\}.[/imath] |
1110034 | "Expected value" of Thirteen card game !!
Thirteen cards numbered [imath]1[/imath] to [imath]13[/imath] are shuffled and dealt one at a time. "Match" occurs on deal [imath]k[/imath] if [imath]k[/imath]th card revealed is card number [imath]k[/imath] Let N be the total number of matches that occur in the thirteen cards. Determine [imath]E[N][/imath] Hi, the above is the problem that I have a trouble with. My first approach is that probability of first card with number [imath]1[/imath] is [imath]\frac{1}{13}[/imath] and probability of second card with number [imath]2[/imath] is [imath]\frac{1}{12}[/imath]. It keeps going until last card, so I think [imath]E[N][/imath] is [imath]\frac{1}{13}+\frac{1}{12}+\frac{1}{11}\cdots+1[/imath] I don't think this is right. There is a hint saying that "Write [imath]N= 1{A1}+1{A2}+....+1{A13}[/imath], where [imath]A_k[/imath] is the event that a match occurs on deal [imath]k[/imath]" Can anyone help me out? | 1109892 | Expected count of correct assignments
There are [imath]40[/imath] letters and [imath]40[/imath] envelopes with addresses. The letters are put in envelopes at random. What is the expected number of letters in their corresponding (correct) envelopes ? One envelope can hold only one letter. I start with [imath]E(X_i)=1\cdot\dfrac{1}{40}+0\cdot\dfrac{39}{40}=\dfrac{1}{40}[/imath]. [imath]X_i[/imath] being the [imath]i^{th}[/imath] letter placed correctly or not ([imath]1[/imath], if correctly placed and [imath]0[/imath], if not). I need to calculate the expected count of letters assigned correctly. Then add up all [imath]\implies[/imath] [imath]E(X_1+X_2+\ldots+X_{40})=\underbrace{\dfrac{1}{40}+\dfrac{1}{40}+\ldots+\dfrac{1}{40}}_{40\text{ times}}=1[/imath] Now my doubt is that if I put [imath]39[/imath] letters correctly, the [imath]40^{th}[/imath] one will be automatically assigned correctly . So should I add only [imath]39[/imath] times ? Any other mistakes I may be doing here ? Please advise. |
1110137 | Ideals of the quotient ring of A
Let A be a ring and b be an ideal of A. The quotient ring of A by b, denoted A/b is the ring of all equivalence classes A + b. Prove that the assignment [imath]c → c/b[/imath] induces a one-to-one correspondence between the ideals of A that contain b and the ideals of A/b. I am completely at a loss. I understand one-to-one to mean bijective so I assume it is necessary to show the mapping is both injective and surjective? I also don't know what c is supposed to be - is it an element of A? Please help! | 1109770 | Bijection between sets of ideals
Let [imath]A[/imath] be a ring and [imath]\mathfrak{b}[/imath] be an ideal of [imath]A[/imath]. Prove that the assignment [imath]\mathfrak{c} \mapsto \mathfrak{c}/\mathfrak{b}[/imath] induces a one-to-one correspondence between the ideals of [imath]A[/imath] that contain [imath]\mathfrak{b}[/imath] and the ideals of [imath]A/\mathfrak{b}[/imath]. This is a problem I've been given, typed exactly as I received it. I find it a bit unclear, but I think [imath]\mathfrak{c}[/imath] is meant to be an ideal of [imath]A[/imath] which contains [imath]\mathfrak{b}[/imath]. Then I must prove that the map given is bijective. I've seen quotient rings, but not between ideals before. Should I treat it as I would any other quotient ring? Assuming what I've said is correct, if the map above is [imath]f[/imath], then I must show that [imath]f(\mathfrak{c})=\mathfrak{c}/\mathfrak{b}=\lbrace x+y : x \in \mathfrak{c}, y \in \mathfrak{b} \rbrace[/imath] is a bijection? |
1110500 | give direct proof of the fact [imath]a^2 - 5a + 6[/imath] is even for any integer
I know this is true but I don't know how to prove it. I have worked it out for the integers from [imath]1[/imath] to [imath]10[/imath] but this is not direct proof, is there a formula I need? | 1098937 | Give a direct proof of the fact that [imath]a^2-5a+6[/imath] is even for any [imath]a \in \mathbb Z[/imath]
Give a direct proof of the fact that [imath]a^2-5a+6[/imath] is even for any [imath]a \in \mathbb Z.[/imath] I know this is true because any even number that is squared will be even, is it also true than any even number multiplied by 5 will be even?? is this direct proof enough? |
982344 | Fermat solved [imath]x^2+2=y^3[/imath] by infinite descent?
In a letter to Christiaan Huygens entitled "on problems in the theory of numbers: a letter to Christiaan Huygens", Fermat claism that he solved the diophantine [imath]x^2+2=y^3[/imath] using infinite descent. Here is the letter: " [1] For a long time I was unable to apply my method (i.e. infinite descent) to affirmative questions, because the twists and turns to get there are much more difficult than those which served me for the negative questions. [...] But finally a line of thought gone over many times showed me a light which did not fail, and affirmative questions surrendered to my method, with the help of some new principles which had to be joined with it of necessity. The progress in my thinking on these affirmative questions is this: if a prime number taken at one's discretion, which exceeds by one a multiple of four, is not a sum of two squares, there will be a prime of the same kind, less than the given one, and then a third still less etc., descending infinitely this way until you arrive at the number 5 which is the smallest of all those of this kind, which it follows cannot be the sum of two squares, which it is nonetheless. From which one must infer from that deduction of an impossibility that all those of this kind are consequently a sum of two squares. [...] [4] Finally I considered certain questions which, although negative, did not shrink from receiving very great difficulties, the way of applying the descent being quite as diverse as the preceding, as it will be easy to check. These are the following: No cube is a sum of two cubes. There is only one square in integers which, added to two, gives a cube. The said square is 25. There are only two squares in integers which, added to 4, give a cube. The said squares are 4 and 121. All the square powers of two, added to one, are prime numbers. This last question is of a very subtle and ingenious study and, even though it is posed affirmatively, it is negative; for to say that a number is prime is to say that it cannot be divided by any number." So the question is: Did Fermat have an infinite descent proof of [imath]x^2+2=y^3[/imath] (i.e. that the only integer solutions are (x,y)=(5,3), (-5,3) ) ? Has anyone ever found such a proof? | 677422 | Solve [imath]x^2+2=y^3[/imath] using infinite descent?
just so this doesn't get deleted, I want to make it clear that I already know how to solve this using the UFD [imath]\mathbb{Z}[\sqrt{-2}][/imath], and am in search for the infinite descent proof that Fermat claimed to have found. I've alaways been fascinated by this Diophantine equation [imath]x^2+2=y^3[/imath] in particular ever since I saw it, and I still have no clue how to attack it without [imath]\mathbb{Z}[\sqrt{-2}][/imath]. What's disappointing is that no one else seems interested in the hunt (an elementary proof using infinite descent). I know it's been studied extensively, and there have even been generalizations, such as Mordell's equation. However, I've never seen Fermat's original proof that [imath](x,y)=(\pm 5, 3)[/imath] is the only integer solution. Obviously, Fermat probably knew nothing of UFD's, which is why I believe there has to be an infinite descent proof like he claimed. Has anyone apart from him actually seen this proof? People mention it all the time, yet I can't find anything about it. As I said, I know that it involved infinite descent, but I've never seen it anywhere and no one seems to have any idea about it. Does anyone have ideas for this approach? I mean, infinite descent seems more effective for showing a contradiction, e.g. showing there are no solutions. But how could it work here? Also, why isn't it published anywhere in all this time? Could it really be that only Fermat knew his method of descent well-enough to make this problem submit to it? Thanks! |
1110247 | Is there a "smallest" divergent sum?
I'm having a look at analysis right now, and I just thought up this question after reading about the comparison test. Does there exist a "critical" infinite sum of real numbers (which is divergent) such that if [imath]S_n[/imath] is any infinite sum for which the terms in [imath]S_n[/imath] are all less than the terms in this critical sum for [imath]n > N_0[/imath] for some finite [imath]N_0[/imath], then [imath]S_n[/imath] converges? I know my terminology is somewhat incorrect. | 20378 | Which series converges the most slowly?
We say [imath]a_n[/imath] converges slower than [imath]b_n[/imath] if there exist an [imath]x[/imath] such that for all [imath]m>x[/imath], [imath]a_m>b_m[/imath] and both [imath]\sum a_n[/imath] and [imath]\sum b_n[/imath] converges. Ignoring constant factors, which type of function converges the slowest? |
1110730 | If A is positive semidefinite, then [imath]A+\alpha I[/imath] with [imath]\alpha\neq 0[/imath] is positive definite?
If [imath]A[/imath] is a symmetric positive semidefinite matrix, then [imath]A+\alpha I[/imath] with [imath]\alpha> 0[/imath] is positive definite? Or are there some conditions to [imath]\alpha[/imath] so that it verifies? [imath]I[/imath] is the identity matrix | 1110544 | How to prove the positive definite property of the semidefinite matrix with regularization?
If I have a positive semidefinite matrix [imath]A_{n\times n}[/imath], I want to transform it to positive definite matrix. I plug the following operation to the original matrix: [imath]B=A+\lambda I_{n}[/imath] where [imath]\lambda > 0[/imath]. I want to know whether B is positive definite matrix. Could anyone proof it? |
1110073 | How to prove the set [imath]\{\cos(n) \mid n \in \mathbb{N}\}[/imath] is dense in [imath][-1,1][/imath]
Prove that the set [imath]\{\cos(n) \mid n \in \mathbb{N}\}[/imath] is dense in [imath][-1,1][/imath]. | 656930 | Show that a set is dense in [imath][-1,1][/imath]
Show that [imath]\{\cos\ n:n \in \mathbb{N}\}[/imath] is dense in [imath][-1,1][/imath] by using the fact below: Suppose [imath]x[/imath] is irrational. Then there exists [imath]p_n,q_n \in \mathbb{Z}[/imath] such that [imath]\bigg|x -\frac{p_n}{q_n}\bigg| < \frac{1}{q_n^2}[/imath] I have no idea on how to apply the fact above to show the set is dense in [imath][-1,1][/imath]. Can anyone help me? |
938043 | Ito's integral from the definition
I am doing Oksendal's book exercises one by one. I got stuck in 3.2. I need to prove, from the definition that [imath]\int_{0}^{t}B_s^2\text{d}B_s=\frac{B_s^3}{3}-\int_{0}^{t}B_s\text{d}s,[/imath] where [imath]B_s[/imath] is Brownian motion and the integral is Ito's. Denote [imath]B_{t_i}[/imath] by [imath]B_i[/imath], for a partition [imath]\{t_i\}_{i=2}^n[/imath]. [imath]B_t^3=\sum_{0}^{n-1}\left(B_{i+1}^3-B_{i}^3\right)=\sum_{0}^{n-1}\left(B_{i+1}^2+B_{i+1}B_i+B_i^2\right)\left(B_{i+1}-B_i\right)[/imath] From the third terms we get [imath]\int_{0}^{t}B_s^2\text{d}B_s[/imath]. I could add and subtract [imath]B_i[/imath] to the [imath]B_{i+1}[/imath] to make appear sums like [imath]\sum(B_{i+1}-B_i)^2[/imath], which tend to the quadratic covariation ([imath]=t[/imath]), but there remain still some sums, like [imath]\sum (B_{i+1}-B_i)^3[/imath] that I don't know how to handle. Nevermind! I got the idea. I will keep the post so you can enjoy. I just looked again to the integral in the right-hand side and the [imath](t_{i+1}-t_i)[/imath] that appear in the sums. Replaced them conveniently. | 1110497 | Prove directly from the definition of the Ito's integral
I am trying to solve the exercises from the book Stochastic differential equations -An Introduction with applications by Bernt Oksendal and I am stuck on 1 question. Prove directly from the definition of Ito's integral that [imath]\int_0^t B_s^2 dB_s=\frac{1}{3}B_t^3-\int_0^t B_s ds[/imath] I have tried for a couple of hours but I simply cannot prove it . I would be grateful if somebody could help me out , even a hint would be very helpful |
1111069 | Deterministic integrals involving a Brownian motion
I am trying to work out the following two integrals involving a standard Brownian motion started at [imath]W_0 = 0[/imath]. The first expression is bewildering me a bit, since it seems like somehow the Itô isometry is involved and would help me work the expectation. [imath] \mathbb E\left[\left(\frac{1}{T}\int_0^T\,W_t\,dt \right)^2\right] [/imath] Secondly, it seems like to prove the following inequality it could be useful to reach an expression resulting in [imath]\frac{2}{\sqrt{\pi}}[/imath], but I am not sure how to do it. [imath] \mathbb P\left(\int_0^1\,W_t\,dt > \frac{2}{\sqrt{3}} \right) [/imath] Any help is welcome! | 250526 | Expected value of average of Brownian motion
For a standard one-dimensional Brownian motion [imath]W(t)[/imath], calculate: [imath]E\bigg[\Big(\frac{1}{T}\int\limits_0^TW_t\, dt\Big)^2\bigg][/imath] Note: I am not able to figure out how to approach this problem. All i can think of is that the term [imath]\frac{1}{T}\int\limits_0^TW_t\,dt[/imath] is like 'average'. But not sure how to proceed ahead. I'm relatively new to Brownian motion. I tried searching the forum for some hints..but could not find one. I will really appreciate if you could please guide me in the right direction. Thanks! |
1108981 | Let [imath]f \in O(D)[/imath] for some domain [imath]D[/imath]. Prove that if [imath]|f(z)|[/imath] is a constant, then [imath]f(z) =[/imath] const on [imath]D[/imath].
Let [imath]f \in O(D)[/imath] for some domain [imath]D[/imath]. Prove that if [imath]|f(z)|[/imath] is a constant, then [imath]f(z) =[/imath] const on [imath]D[/imath]. It seems to me it is the direct application of the following version of maximum modulus principle: Let f be a function holomorphic on some connected open subset [imath]D[/imath] of the complex plane [imath]\mathbb{C}[/imath] and taking complex values. If [imath]z_0[/imath] is a point in [imath]D[/imath] such that [imath]|f(z_0)|\ge |f(z)|[/imath] for all z in a neighborhood of [imath]z_0[/imath], then the function f is constant on D. | 244342 | If the absolute value of an analytic function [imath]f[/imath] is a constant, must [imath]f[/imath] be a constant?
I've been thinking how to prove that an analytic function [imath]f[/imath] is a constant if the absolute value of [imath]f[/imath] is a constant, but I haven't figured it out yet. What I was thinking is to use Cauchy-Riemann equations, but it didn't work well... If this is not true, I would like to know the counterexample... Here is what I tried: [imath]|f|=|u+iv|=\sqrt {u^2+v^2}[/imath] Thus [imath]u^2+v^2[/imath] is a constant. (1) [imath]\displaystyle u\frac {\delta u}{\delta x}+v\frac {\delta v}{\delta x}=0 [/imath] (2) [imath]\displaystyle u\frac {\delta u}{\delta y}+v\frac {\delta v}{\delta y}=0 [/imath] Plug Cauchy Riemann into (2). [imath]\displaystyle -u\frac {\delta v}{\delta x}+v\frac {\delta u}{\delta x}=0 [/imath] and I'm stuck here... |
1111301 | plane throw two points and paralell to another plane
I have [imath]\pi:4y−3z−4=0 \quad A=(2,4,4) \quad B=(2,−2,−4)[/imath]. I have to calculate the equation of the plane σ throw A and B and parallel (not orthogonal) to π. What is the solution? Thanks in advice! And if I have three points, what's the solution? | 1110831 | Find equation of a plane throw two given point and orthogonal to another space
I have [imath]\pi:4y-3z-4=0 \quad A=(2,4,4) \quad B=(2,-2,-4)[/imath]. I have to calculate the equation of the plane [imath]\sigma[/imath] throw [imath]A[/imath] and [imath]B[/imath] and orthogonal to [imath]\pi[/imath]. What is the solution? Thanks in advice! |
1111443 | Prove [imath]1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2[/imath] using Induction
Prove [imath]1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2[/imath] using Induction My proof so far: Let [imath]P(n)[/imath] be [imath]1^3 + 2^3 + \cdots + n^3 = (1 + 2 + \cdots + n)^2[/imath] Base Case [imath]P(1):[/imath] LHS = [imath]1^3 = 1[/imath] RHS = [imath](1)^2 = 1[/imath] Since LHS = RHS, therefore base case holds Induction Hypothesis Let [imath]n \in \mathbb{N}[/imath] be arbitrary Assume [imath]P(n)[/imath] holds Induction Step Prove [imath]P(n+1)[/imath] holds: [imath] \begin{align} & 1^3 + 2^3 + \cdots + \;n^3 + (n+1)^3 \\ = {} & (1 + 2 + \cdots + \; n)^2 + (n+1)^3 \text{ (by Induction Hypothesis)} \\ = {} & (1 + 2 + \cdots + \; n)^2 + (n^3 + 3n^2 + 3n + 1) \end{align} [/imath] This is where I get stuck. I don't know how to prove that my last step is equivalent to: [imath](1 + 2 + \cdots + \;n + (n+1))^2[/imath] | 996083 | Induction proof for a summation: [imath]\sum_{i=1}^n i^3 = \left[\sum_{i=1}^n i\right]^2[/imath]
Prove by induction: [imath]\sum_{i=1}^n i^3 = \left[\sum_{i=1}^n i\right]^2[/imath]. Hint: Use [imath]k(k+1)^2 = 2(k+1)\sum i[/imath]. Basis: [imath]n = 1[/imath] [imath]\sum_{i=1}^1 i^3 = \left[\sum_{i=1}^1 i\right]^2 \to 1^3 = 1^2 \to 1 = 1[/imath]. Hypothesis: Assume true for all [imath]n \le k[/imath]. So far I have the following: [imath]\sum_{i=1}^{k+1} i^3 = (k+1)^3 + \sum_{i=1}^k i^3[/imath] [imath](k+1)^3 + \left[\sum_{i=1}^k i\right]^2[/imath] |
1111602 | Compute the complex integration
Let, [imath]f(z)[/imath] be an analytic function. Then the value of [imath]\int_{0}^{2\pi}f\bigl(e^{it}\bigr)\cos t dt= ?[/imath] (a) 0 (b) [imath]2\pi f(0)[/imath] (c) [imath]2\pi f'(0)[/imath] (d) [imath]\pi f(0)[/imath]. [imath]\mathcal{My}{Attemt}:[/imath] First I put [imath]e^{it}=z[/imath]. Then integration becomes [imath]\int_{|z|=1}f(z)\frac{z+z^{-1}}{2}\frac{dz}{iz}[/imath] [imath]=\frac{1}{2i}\int_{|z|=1}\frac{f(z)(z^{2}+1)}{z^{2}}dz[/imath] [imath]\frac{1}{2i}2\pi if'(0)=\pi f'(0)[/imath] But the answer does not match any option. Where my mistake?? | 420415 | For [imath]f[/imath] an analytic function, what is
[imath]f[/imath] be analytic function, could any one tell me how to find the value of [imath]\int_{0}^{2\pi} f(e^{it})\cos t \,\mathrm dt[/imath] I am not able to apply any complex analysis result here, could any one give me hint? |
817062 | Proof that no polynomial with integer coefficients can only produce primes
Doing a discrete math review and am trying to solve problem 1.6 in the text found here: http://courses.csail.mit.edu/6.042/fall13/ch1-to-3.pdf - I believe I've gotten parts (a) and (b) correctly, but (c) is a bit tricky for me. Would appreciate a review of a/b and a hint for c if possible. Restatement here: For n = 40, the value of polynomial [imath]p(n)::=n^2 + n + 41[/imath] is not prime, as noted in Section 1.1. But we could have predicted based on general principles that no nonconstant polynomial can generate only prime numbers. In particular, let [imath]q(n)[/imath] be a polynomial with integer coefficients, and let [imath]c::=q(0)[/imath] be the constant term of q. (a) Verify that [imath]q(cm)[/imath] is a multiple of c for all [imath]m \in Z[/imath] Proof: Since q is a polynomial, it will be of the form: [imath]q(x) = a_nx^n+a_{n-1}x^{n-1} + ... + a_1x + c[/imath] If [imath]x = cm[/imath], then notice that cm will be inside every term of the polynomial, and since the final term is c, c can be factored out of every term. Hence, q(cm) is a multiple of c for all [imath]m \in Z[/imath]. b) Show that if q is nonconstant and c > 1, then as n ranges over the nonnegative integers, [imath]N[/imath], there are infinitely many [imath]q(n) \in Z[/imath] that are not primes. Proof: Continuing from the result found in (a), it is easy to see that for all [imath]m \in Z[/imath], there will be an infinite number of multiples of c that will be generated by q. Since c > 1, these multiples are guaranteed not to be prime. c) Conclude that for every nonconstant polynomial, q, there must be an [imath]n \in N[/imath] such that [imath]q(n)[/imath] is not prime. Hint: Only one easy case remains. c = 0 is trivially easy to prove using the (b) above. I assume the "easy" case is referring to c = 1, but in this case I'm not sure how to continue, since the result of (a) and (b) don't apply: I can't use them since adding 1 to any even number may make it prime. If the result of the terms not including c is odd, then adding 1 to that result makes it even and it is not prime. However, if the terms add up to an even number, I don't see a way of using the knowledge I have so far to prove conclusively that that number + 1 will NOT in fact be prime. | 2416348 | Prove that there is no non-constant polynomial [imath]p(n)[/imath] with integer coefficients that only takes prime values
Attempt (so far): Assume there exists a non-constant polynomial [imath]p: \mathbb Z \to \mathbb Z[/imath] with integer coeffecients that only takes on prime values. Let notate it as [imath]p(n)=d_j n^j + d_{j-1} n^{j-1}+ \dots +d_1 n + d_0 [/imath] where [imath]j\in \mathbb N[/imath]. Let [imath]k[/imath] be some composite number with [imath]r[/imath]-many factors. Then [imath]\begin{array} \ k &=& p(n_1)^{i_1} p(n_2)^{i_2} \dots p(n_r)^{i_r} \\ &=& (d_j {n_1}^j + d_{j-1} {n_1}^{j-1}+ \dots +d_1 {n_1}+d_0)^{i_1} \dots (d_j {n_r}^j + d_{j-1} {n_r}^{j-1}+ \dots +d_1 {n_r}+d_0)^{i_r} \end{array}[/imath] I don't really see how to progress from here unless I want to start doing ungodly amounts of computation. Could someone provide a hint of a path I should be taking? |
1112188 | A property P of morphisms of [imath]S[/imath]-schemes [imath]f : X \rightarrow Y[/imath] is local on [imath]X[/imath], or [imath]Y[/imath], or [imath]S[/imath] or
I have asked the same question on math.stackexhange here, but thought that is was a good idea to post it here also. I am learning schemes theory at school and I have for now only lectures notes that I am taking during the course. The professor is quite often using following expressions, without having defined them : if [imath]S[/imath] is a scheme and if [imath]f : X \rightarrow Y[/imath] is a morphism of [imath]S[/imath]-schemes he often says that some property P of [imath]f[/imath] is (1) local on [imath]X[/imath] (2) local on [imath]Y[/imath] (3) local on [imath]X[/imath] and [imath]Y[/imath] (4) local on [imath]X, Y[/imath] and [imath]S[/imath] and I have several question on this, and related to this. I went to the library to look at all Grothendieck's EGA's and found the same frequent use of the expression la question est locale sur in the same four cases - at least, maybe there are other than these four ? But I found in them no definition of the meaning of la question est locale sur, which means the question is local on in english. I guess that (1) means that [imath]f[/imath] has the property P if and only if for every open cover [imath](U_i)_{i\in I}[/imath] of [imath]X[/imath], all restrictions [imath]f_{|U_i} : U_i \rightarrow Y[/imath] have the property P, and that (2) means that [imath]f[/imath] has the property P if and only if for every open cover [imath](V_i)_{i\in I}[/imath] of [imath]Y[/imath], all corestrictions [imath]f^{-1} (V_i) \rightarrow V_i[/imath] have the property P. For (3) obviously it is equivalent to check (1) and (2), but I cannot state a more synthetic formulation. Does (3) means that [imath]f[/imath] has property P if and only if for every open cover [imath](U_i)_{i\in I}[/imath] of [imath]X[/imath] and for every open cover [imath](V_j)_{j\in J}[/imath] of [imath]Y[/imath] all morphisms [imath]U_i \cap f^{-1} (V_j) \rightarrow V_j[/imath] have the property P ? For (4) it is sufficient to define what means P is local on S. Does this means that [imath]f[/imath] has the property P if and only if for every open cover [imath](W_i)_{i\in I}[/imath] of [imath]S[/imath], every morphism [imath]p^{-1}(W_i) \cap f^{-1}(q^{-1} (W_i)) \rightarrow q^{-1} (W_i)[/imath] has the property P ? Or is there something more clever than this. Combining this with (3) is equivalent to (4), but here again, is there a more synthetic formulation for this ? In (1), (2), (3), (4) I used "local morphisms" (restrictions in (1), corestrictions in (2) etc). Is it possible to express these morphisms thanks to fiber products, and if so, how ? In (1), (2), (3), (4) I used the word every cover but sometimes in the course it suffice to have the fact for only one cover to have it for all. How does this work ? (I call Q this question.) Sometimes its not only open covers that are used, but open affine covers. For [imath]i\in\{1,2,3,4\}[/imath], is (i) equivalent to the assertion (i) with "open" replaced by "open affine", or even by "affine" ? Is answering to question Q easier with "open" replaced by "open affine", or only "affine" ? Of this note, what does the following sentence mean : the property P of the morphism of [imath]A[/imath]-algebras [imath]\varphi : B' \rightarrow B[/imath] is local on [imath]\textrm{Spec}(A)[/imath] (resp. on [imath]\textrm{Spec}(B')[/imath], resp. on [imath]\textrm{Spec}(B)[/imath], resp. on "combinations of the previous") ? I have a last question : all the previous questions are local for the topology of the schemes involved, which is a topology in the "classic" sense. Is all of this translatable to Grothendieck topologies ? If so, how ? And then, intuitively, in the case of the Zariski site, is the same ? I know that this was a lot of question, but all are intimately related to this "local on" stuff, so I preferred to ask all of them in one shot. | 1104790 | "This property is local on" : properties of morphisms of [imath]S[/imath]-schemes
I am learning schemes theory at school and I have for now only lectures notes that I am taking during the course. The professor is quite often using following expressions, without having defined them : if [imath]S[/imath] is a scheme and if [imath]f : X \rightarrow Y[/imath] is a morphism of [imath]S[/imath]-schemes he often says that some property P of [imath]f[/imath] is (1) local on [imath]X[/imath] (2) local on [imath]Y[/imath] (3) local on [imath]X[/imath] and [imath]Y[/imath] (4) local on [imath]X, Y[/imath] and [imath]S[/imath] and I have several question on this, and related to this. I went to the library to look at all Grothendieck's EGA's and found the same frequent use of the expression la question est locale sur in the same four cases - at least, maybe there are other than these four ? But I found in them no definition of the meaning of la question est locale sur, which means the question is local on in english. I guess that (1) means that [imath]f[/imath] has the property P if and only if for every open cover [imath](U_i)_{i\in I}[/imath] of [imath]X[/imath], all restrictions [imath]f_{|U_i} : U_i \rightarrow Y[/imath] have the property P, and that (2) means that [imath]f[/imath] has the property P if and only if for every open cover [imath](V_i)_{i\in I}[/imath] of [imath]Y[/imath], all corestrictions [imath]f^{-1} (V_i) \rightarrow V_i[/imath] have the property P. For (3) obviously it is equivalent to check (1) and (2), but I cannot state a more synthetic formulation. Does (3) means that [imath]f[/imath] has property P if and only if for every open cover [imath](U_i)_{i\in I}[/imath] of [imath]X[/imath] and for every open cover [imath](V_j)_{j\in J}[/imath] of [imath]Y[/imath] all morphisms [imath]U_i \cap f^{-1} (V_j) \rightarrow V_j[/imath] have the property P ? For (4) it is sufficient to define what means P is local on S. Does this means that [imath]f[/imath] has the property P if and only if for every open cover [imath](W_i)_{i\in I}[/imath] of [imath]S[/imath], every morphism [imath]p^{-1}(W_i) \cap f^{-1}(q^{-1} (W_i)) \rightarrow q^{-1} (W_i)[/imath] has the property P ? Or is there something more clever than this. Combining this with (3) is equivalent to (4), but here again, is there a more synthetic formulation for this ? In (1), (2), (3), (4) I used "local morphisms" (restrictions in (1), corestrictions in (2) etc). Is it possible to express these morphisms thanks to fiber products, and if so, how ? In (1), (2), (3), (4) I used the word every cover but sometimes in the course it suffice to have the fact for only one cover to have it for all. How does this work ? (I call Q this question.) Sometimes its not only open covers that are used, but open affine covers. For [imath]i\in\{1,2,3,4\}[/imath], is (i) equivalent to the assertion (i) with "open" replaced by "open affine", or even by "affine" ? Is answering to question Q easier with "open" replaced by "open affine", or only "affine" ? Of this note, what does the following sentence mean : the property P of the morphism of [imath]A[/imath]-algebras [imath]\varphi : B' \rightarrow B[/imath] is local on [imath]\textrm{Spec}(A)[/imath] (resp. on [imath]\textrm{Spec}(B')[/imath], resp. on [imath]\textrm{Spec}(B)[/imath], resp. on "combinations of the previous") ? I have a last question : all the previous questions are local for the topology of the schemes involved, which is a topology in the "classic" sense. Is all of this translatable to Grothendieck topologies ? If so, how ? And then, intuitively, in the case of the Zariski site, is the same ? I know that this was a lot of question, but all are intimately related to this "local on" stuff, so I preferred to ask all of them in one shot. |
534492 | space of schwartz, problem
my question is: Let [imath]f\in S(\mathbb{R})[/imath], with [imath]f(0)=0[/imath], then there exists [imath]g\in S(\mathbb{R})[/imath] such that: [imath] f(x)=xg(x)\;\text{ for all }\;x\in \mathbb{R}.[/imath] I need to prove this. | 104124 | Question on Schwartz function
I am seeing the Schwartz Space for the first time today and I have trouble understanding the following argument: Given that [imath]f \in S({\mathbb{R}})[/imath] with [imath]f(x_0) = 0[/imath] Taylor's theorem tells us that we can find a function [imath]g \in C^\infty(\mathbb{R})[/imath] such that \begin{equation} f(x) = g(x)(x - x_0) \end{equation} Hence, for [imath]x \neq x_0[/imath] we can write [imath]g(x)[/imath] as \begin{equation} g(x) = (x - x_0)^{-1} f(x) \end{equation} According to my notes, from this it should be clear that [imath]g \in S(\mathbb{R})[/imath], but how can I see this immediately ? In order to derive this I think would need to direvtly go via the definition, i.e. show for each [imath]k,m = 0,1,2, \dots[/imath] there exist constants [imath]c_{k,m}[/imath] such that \begin{equation} \sup_x |x^k g^{(m)}(x)| \leq c_{k,m} \end{equation} or is there a shorter way to infer this ? By "shorter" I mean for example a statement that somehow the constants that bound [imath]f[/imath] are also bounds on [imath]g[/imath], probably with some amendment. Edit: Does it maybe have to do with the fact that [imath]f[/imath], being smooth, has a Tayler expansion and so \begin{equation} g(x) = \sum^\infty_{n = 1} \frac{f^{(n)}(x_0)}{n!}(x - x_0)^{n-1} \quad (\text{note } f(x_0)) = 0 \end{equation} Thanks for your help! |
1113618 | Direct sum of ideals over Dedekind domain
I'm trying to show that Let [imath]\frak{a},\frak{b}[/imath] be two ideals of a Dedekind domain [imath]\cal{O}[/imath]. Show that there is an isomorphism \begin{equation*} \frak{a}\oplus\frak{b}\cong\cal{O}\oplus\frak{a}\frak{b}. \end{equation*} If [imath]\frak{a},\frak{b}[/imath] are relative prime, then the thing is easy. For the general case, there exists some [imath]c\in K[/imath] such that [imath]c\frak{a}[/imath] is relatively prime to [imath]\frak{b}[/imath]. Therefore \begin{equation*} c\frak{a}\oplus\frak{b}\cong\cal{O}\oplus c\frak{a}\frak{b}. \end{equation*} Now, how to prove [imath]\frak{a}\oplus\frak{b}\cong\cal{O}\oplus\frak{a}\frak{b}[/imath] from this? Eidt: I'm so fool to ignore that [imath]c\frak{a}\cong\frak{a}[/imath]. So there is no problem now. | 1074680 | Proof of Steinitz Theorem
I want a source containing the proof of Steinitz Isomorphism Theorem stating: For any Dedekind domain [imath]R[/imath] and any two nonzero ideals [imath]I[/imath] and [imath]J[/imath] of [imath]R[/imath] we have [imath]I⊕J≅R⊕IJ[/imath]. Thanks! |
1113761 | Number of elements of [imath]S_{10}[/imath] commuting with element (1 3 5 7 9)
Find Number of elements of [imath]S_{10}[/imath] commuting with element (1 3 5 7 9) I think we need to find order of centralizer of given permutation but how to find it? | 922195 | Calculating number of elements commmuting with [imath]\sigma\in S_{10}[/imath]
let [imath]S_{10}[/imath] denote the group of permutations on ten symbols [imath]{1,2,3,....,10}[/imath].Then how do we calculate number of elements of [imath]S_{10}[/imath] commuting with the element [imath]\sigma=(1\ 3\ 5\ 7\ 9)[/imath]? |
339036 | How to prove this inequality with factorials: [imath]n!>n^{\frac {n}{2}}[/imath]
This is what I am trying to solve: [imath]n!>n^{\frac {n}{2}}[/imath] I tried with induction and somehow it doesn`t work this way, I tried with logarithms and also didn´t find a way. Is there an elementary(if possible) way to prove this? Or is it the case that for some large [imath]n_0[/imath] this inequality reverses sign? | 785514 | Solving an inequality : [imath]n \geq 3[/imath] , [imath]n^{n} \lt (n!)^{2}[/imath].
I proved this inequality in the following way: Lemma: [imath]r \in \Bbb N, r \geq 3[/imath]. We have [imath]r^r \gt (r+1)^{r-1}[/imath]. Proof: We apply the AM-GM inequality to the [imath]r[/imath] positive integers where there are [imath]r-1[/imath] [imath](r+1)[/imath]'s and one [imath]1[/imath]. We obtain [imath]\frac {1+(r-1)(r+1)}{r} \gt ((r+1)^{r-1})^{\frac {1}{r}}[/imath] wherefrom we get (since the [imath]r^{{\rm th}}[/imath] power function is increasing), [imath]r^{r} \gt (r+1)^{r-1}[/imath]. Now, I used mathematical induction to prove the statement. We have from the lemma, [imath](k!)^{2} \gt k^k \implies (k!)^{2} \gt (k+1)^{k-1}[/imath] and multiplying this inequality by [imath](k+1)^{2}[/imath], we have [imath]((k+1)!)^{2} \gt (k+1)^{k+1}[/imath] and obviously [imath](3!)^{2} \gt 3^3[/imath]. Is there any direct proof of the statement that does not use induction and calculus? |
1114000 | Two iid random variables
Prove that for every two independent, identically distributed real random variables [imath]X[/imath] and [imath]Y[/imath], we have \begin{align*} P(|X-Y| \le 2) \le 3P(|X-Y| \le 1). \end{align*} | 244362 | To show that [imath]P(|X-Y| \leq 2) \leq 3P(|X-Y| \leq 1)[/imath]
I found this question while browsing through "The Probabilistic Method", by Noga Elon. Let X and Y be 2 independent and identically distributed real valued random variables. Prove that: [imath]P(|X-Y| \leq 2) \leq 3P(|X-Y| \leq 1)[/imath] So I tried the following: [imath]P\{|X-Y| \leq 2\} = P\{|X-Y| \leq 1\} + P\{X-Y \in (1,2]\cup[-2,-1)\}[/imath] [imath]= P\{|X-Y| \leq 1\} + P\{X-Y \in (1,2]\} + P\{X-Y \in [-2,-1)\}[/imath] [imath]= P\{|X-Y| \leq 1\} + 2P\{X-Y \in (1,2]\}[/imath] where the last step follows because [imath]X-Y[/imath] has a symmetric distribution. NB: A random variable Z has symmetric distribution if [imath]P(Z \leq z) = P(Z \geq -z) \quad \forall z \in \mathbb{R}[/imath] Thus the problem boils down to showing [imath]P\{X-Y \in (1,2]\} \leq P(|X-Y| \leq 1)[/imath] and I would be done. Unfortunately, I don't know how to proceed from here. I appreciate any help, hints, useful comments etc. I receive. |
992196 | Complexities of Recurrences in the form [imath]t(n) = t(\alpha n) + t(\beta n) + cn[/imath]
When considering recurrence relations, they are generally of a form similar to [imath]t(n) = t(\alpha n) + t(\beta n) + cn[/imath] and there are three cases to be considered for our values of [imath]\alpha[/imath] and [imath]\beta[/imath], being [imath]\alpha + \beta < 1[/imath] [imath]\alpha + \beta = 1[/imath] [imath]\alpha + \beta > 1[/imath] We are able to show that for our first case, we receive [imath]t(n) = O(n)[/imath], and in our second case [imath]t(n) = O(nlog n)[/imath], but are we able to analyze the big-Oh for the third case? I know we can consider the third case to be [imath]t(n) = \Omega(nlogn)[/imath] but is it possible to draw any further conclusion? | 780391 | Recursive Function - [imath]f(n)=f(an)+f(bn)+n[/imath]
I've got this recursive function [imath]f(n)=f(an)+f(bn)+n[/imath], and I need to find [imath]θ[/imath] on [imath]f(n)[/imath], as [imath]a+b>1[/imath]. Using a recursive tree, I managed to bound it by [imath]n\log(n)[/imath] from the bottom and by [imath]n^2[/imath] from the top, but I can't seen to find the [imath]θ[/imath] bound. I'd be glad for some help. Thanks in advance, Yaron. |
1114194 | Convexity over a line given a convex interval
Let [imath]f : \mathbb{R}^n \to \mathbb{R}_∞[/imath] be a function. I want to prove that [imath]f[/imath] is convex over the line [imath]L_{v,x_0}[/imath] iff [imath]\psi : \mathbb{R} \to \mathbb{R}_∞[/imath] [imath]\psi(t) := f (x_0 + tv)[/imath], is convex over [imath]I (x_0, v )[/imath]. where [imath]L_{v,x_0}:= \{x_0 + tv : t ∈ \mathbb{R}\}[/imath], and [imath]I(x_0,v):=\{t \in \mathbb{R} : x_0 + tv \in C\}[/imath] Any help is very much appreciated | 1109115 | Showing convexity of a function with the restriction over an arbitrary line
Let [imath]f : \mathbb{R}^n → \mathbb{R}_∞[/imath] be a function and let [imath]C ⊂ dom f[/imath] be a convex set. [imath]**Part I**[/imath] Prove that [imath]f[/imath] is a convex function if and only if [imath]f[/imath] is convex over every line [imath]L_{v,x_0}[/imath] where [imath]L_{v,x_0} := \{x_0 + tv : t \in \mathbb{R}\}[/imath]. i.e. f is convex iff when restricted to any line, it be convex on that certain line. So essentially to handle this proof I need to show that convexity for [imath]f[/imath] holds when its convex over any arbitrary line, and on the converse it does not hold when [imath]f[/imath] is not convex over that arbitrary line. If [imath]f[/imath] is convex then, [imath]\forall x,y \in C, a \in [0,1][/imath] it holds that [imath]f(ax+(1-a)y) \leq af(x) + (1-a)f(y) \space \space\space\space\space\space\space\space\space\space\space\space (1)[/imath] EDIT: [imath]**Part II**[/imath] Let [imath]f : \mathbb{R}^n \to \mathbb{R}_∞[/imath] be a function. I want to prove that [imath]f[/imath] is convex over the line [imath]L_{v,x_0}[/imath] iff [imath]\psi : \mathbb{R} \to \mathbb{R}_∞[/imath] [imath]\psi(t) := f (x_0 + tv)[/imath], is convex over [imath]I (x_0, v )[/imath]. where [imath]L_{v,x_0}:= \{x_0 + tv : t ∈ \mathbb{R}\}[/imath], and [imath]I(x_0,v):=\{t \in \mathbb{R} : x_0 + tv \in C\}[/imath] Any help is very much appreciated |
1114303 | For every continuous function [imath]f:[0,1]\to[0,1][/imath] there exists [imath]y\in [0,1][/imath] such that [imath]f(y)=y[/imath]
I want to prove that if we have a continuous function from the closed interval [0,1] to the closed interval [0,1], that there exists a value y in [0,1] such that [imath]f(y)=y[/imath]. I have an idea of a theorem that I want to use: Let f is a continuous real function on the interval [a,b]. If f(a) < f(b) and if c is a number such that f(a)< c < f(b), then there exists a point x in (a,b) such that f(x)=c. So I know that f(0) < f(1). So according to this theorem, for some number in between 0 and 1, there exists a point in the domain such that f(x)=this number. But I'm having trouble coming up with the argument | 1346543 | Prove that [imath]f(x) = x[/imath] has a solution on [0,1]
Let [imath]f:[0,1]\to [0,1][/imath] be a continuous function. Prove that [imath]f(x) = x[/imath] has a solution in [imath][0,1][/imath]. So, here, [imath]f(0) = 0[/imath] and [imath]f(1) = 1[/imath] and [imath]f'(x) = 1[/imath]. Now, can we conclude that there is at least one solution and that is... |
1114501 | Solve for [imath]m[/imath] in [imath]d^m = n[/imath]
I believe the answer is [imath]m = \lceil \sqrt[d]n \rceil[/imath] or [imath]\lfloor \sqrt[d]n \rfloor[/imath]. Can anyone help me? | 37036 | Solving for a variable that's an exponent
How would you figure out this? [imath]x^y = z[/imath] How do you find out [imath]y[/imath] if you know [imath]x[/imath] and [imath]z[/imath] ? |
1114819 | [imath]\sum_1^n 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} [/imath] converge or not?
how to check if this converge? [imath]\sum_{n=1}^\infty a_n[/imath] [imath]a_n = 2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1}[/imath] what i did is to show that: [imath]a_n =2\sqrt{n} - \sqrt{n-1} - \sqrt{n+1} > 2\sqrt{n} - \sqrt{n+1} - \sqrt{n+1} = 2\sqrt{n} - 2\sqrt{n+1} = -2({\sqrt{n+1} - \sqrt{n}}) = -2(({\sqrt{n+1} - \sqrt{n}})\frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}) = -2\frac{1}{\sqrt{n+1} + \sqrt{n}} > -2\frac{1}{2\sqrt{n+1}} = \frac{-1}{\sqrt{n+1}} = b_n [/imath] and we know that: [imath]\sum_{n=1}^\infty b_n[/imath] doesnt converge cause [imath]\sum_{n=1}^\infty \frac{1}{\sqrt{n}}[/imath] doesnt converge so from here my conclusion is that [imath]\sum_{n=1}^\infty a_n[/imath] doesnt converge but i know the final answer is that it does converge so what am i doing wrong? | 1109867 | Prove or disprove: [imath]\sum a_n[/imath] convergent, where [imath]a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}[/imath].
Let [imath]a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}[/imath]. Show that [imath]a_n>0\ \forall\ n\ge1[/imath]. Prove or disprove: [imath]\sum\limits_{n=1}^\infty a_n[/imath] is convergent. I can't show that [imath]a_n > 0\ \forall n\ge1[/imath]. I tried using induction but it wouldn't work. Attempt: [imath] \begin{align} 2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}&=(2\sqrt{n}-(\sqrt{n-1}+\sqrt{n+1}))\cdot {2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\\&={2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\end{align}[/imath] (The calculations are true for sure. No check is desired.) Denote [imath]a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}={2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\text{ and }b_n={1\over n^{2}}.[/imath] Then [imath]\begin{align*} \lim_{n\to \infty}{a_n\over b_n} & =\lim_{n\to \infty}n^{2}{2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})} \\ &=\lim_{n\to \infty}n^{2}\lim_{n\to \infty}{2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})} \\ &=0 \end{align*}[/imath] By the comparison test for series convergence, since [imath]\lim_{n\to \infty}{a_n\over b_n}=0[/imath], then if [imath]b_n[/imath] converges, which it does, so does [imath]a_n[/imath]. |
1114810 | How to compute this kind of limit?
Let [imath]x_0=a,x_1=b[/imath] [imath]x_{n+1}=\Big(1-\dfrac{1}{2n}\Big)x_n+\dfrac{ x_{n-1}}{2n}, n\ge1[/imath] Find [imath]\lim x_n.[/imath] If the limit exists, I can plug limit as [imath]l[/imath] to get a equation of [imath]l[/imath], whose root will be the limit. But in this case it becomes a identity. I am not familiar about how to compute such a limit. Please Help. | 1109254 | Limit of this recursive sequence: [imath]x_{n+1}=\bigl(1-\frac{1}{2n}\bigr)x_{n}+\frac{1}{2n}x_{n-1}.[/imath]
Consider the following sequence : [imath]x_{0}=a[/imath] , [imath]x_{1}=b[/imath] , [imath]x_{n+1}=\bigl(1-\frac{1}{2n}\bigr)x_{n}+\frac{1}{2n}x_{n-1}.[/imath] Find [imath]\displaystyle \lim_{n\to \infty}x_{n}.[/imath] I calculate [imath]x_{2}[/imath] , [imath]x_{3}[/imath] ,[imath]x_{4}[/imath] ,but I could not find any relationship between any two consecutive pair. But I found that the sum of the coefficients of [imath]a[/imath] & [imath]b[/imath] is equal to the term in denominator in each [imath]x_{i}[/imath]. How we find this limit...? |
1115149 | Show that [imath]q\equiv_8 1[/imath] when [imath]q[/imath] is an odd square number
Problem: Given: q is an odd squared number - show that: [imath]q\equiv_8 1[/imath] My assumption: [imath]\forall q\in N:\exists a \in Z: a =1\pmod{2}[/imath] and [imath]a^2=q[/imath]. Then I tried to show that it's only true satisfyingly if [imath]\mathrm{gcd}(q,8)\mid 1 \leftrightarrow x\cdot q+y\cdot 8=1[/imath]. But I don't know how to show that it is true for all numbers [imath]q[/imath]. All hints are welcome. | 268402 | Prove that odd perfect square is congruent to [imath]1[/imath] modulo [imath]8[/imath]
How can we prove that every odd perfect square is congruent to [imath]1[/imath] modulo [imath]8[/imath]? |
1115136 | Extracting differential equations
[imath]\frac{dx}{dy} = \frac{x(\alpha - \beta y)}{y(\delta x - \gamma)}[/imath] How do I extract two differential equations (y as a function of x and x as a function of y) from the equation above? I could separate the variables, but I don't see how that would help. | 1113587 | Division of differential equations
[imath]\frac{dx(t)}{dy(t)}=\frac{\alpha x(t) - \beta x(t) y(t)}{-\gamma y(t) + \delta x(t)y(t)}[/imath] How would one simplify this fraction? Maybe the chain rule could be of any use, but I don't see how. |
1042541 | How can we draw a line between two distant points using a finite-length ruler?
We want to join by a line two distinct points [imath]A[/imath] and [imath]B[/imath]. We have only a ruler of length [imath]l>0[/imath] and a pen. If [imath]AB>l[/imath] how can we do this? | 2341245 | How to join two points with only a small ruler.
Given two points [imath]A [/imath] and [imath]B [/imath] in the plane . is it possible to draw the segment [imath][A,B] [/imath] with only a ruler whose length is much smaller the the distance [imath]\| \vec {AB }\|.[/imath] I know the answer is yes and it uses the harmonic conjugate notion, but after many attempts, i still don't see how it is possible. Thanks a lot for any help. |
1032918 | [imath]Ax+By+Cz=D \text { has a solution iff } \gcd(\gcd(A,B),C)\mid D[/imath]
I read today that [imath]Ax+By+Cz=D \text { has a solution iff } \gcd(\gcd(A,B),C\mid D[/imath] but I can't find it again, I also can't find any Diophantine equations with 3 variables that doesn't have solutions so I'm starting to suspect that I'm remembering something wrong. My questions are: Are there Diophantine equations with 3 variables that has no solutions? Is the statement in the title correct? Note: [imath]A,B,C,D,x,y,z\in \mathbb Z[/imath] and [imath] A,B,C\neq0[/imath]. | 20906 | How to find an integer solution for general Diophantine equation [imath]ax + by + cz + dt... = N[/imath]
I know how to solve [imath]ax + by = c[/imath] using Extended Algorithm. But with more than variables, I'm lost :(. To verify if it has an integer solution is easy, since we only need to check for [imath]\gcd(a,b,c)|d[/imath]. Other than that, how can we find an integer solution for this equation? Thanks, Chan |
1116898 | What is the meaning of the notation [imath]]1, 1[[/imath]?
This may look like a silly question but I am struck in my work with this notation in one of the papers. What is meant by [imath]]1,1[[/imath] ? | 430851 | Notation for intervals
I have frequently encountered both [imath]\langle a,b \rangle[/imath] and [imath][a,b][/imath] as notation for closed intervals. I have mostly encountered [imath](a,b)[/imath] for open intervals, but I have also seen [imath]]a,b[[/imath]. I recall someone calling the notation with [imath][a,b][/imath] and [imath]]a,b[[/imath] as French notation. What are the origins of the two notations? Is the name French notation correct? Are they used frequently in France? Or were they perhaps prevalent in French mathematical community at some point? (In this MO answer Bourbaki is mentioned in connection with the notation [imath]]a,b[[/imath].) Since several answerers have already mentioned that they have never seen [imath]\langle a,b \rangle[/imath] to be used for closed intervals, I have tried to look for some occurrences for this. The best I can come up with is the article on Czech Wikipedia, where these notations are called Czech notation and French notation. Using [imath](a,b)[/imath] and [imath][a,b][/imath] is called English notation in that article. (I am from Central Europe, too, so it is perhaps not that surprising that I have seen this notation in lectures.) I also tried to google for interval langle or "closed interval" langle. Surprisingly, this lead me to a question on MSE where this notation is used for open interval. |
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