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966078 | sequence[imath]\frac{A}{B}\rightarrow \alpha\neq 0[/imath], show if A is summable, B is summable.
If sequence[imath]\frac{A}{B}\rightarrow \alpha\neq 0[/imath], show if A is summable, B is summable. I was trying to use contrapositive of the statement suppsoe A converges to L and B to M.[imath]If M\neq 0 then \frac{A}{B}\rightarrow \frac{L}{M}[/imath] But it didn't work out well since we do not know if B has a limit or not. How do I show B converges? | 269837 | Limit comparison test proof
We had the following theorem in class: Let [imath](a_n)[/imath] and [imath](b_n)[/imath] be sequences and [imath]b_n>0[/imath] and [imath]\lim_{n\rightarrow\infty}\frac{a_n}{b_n}=L[/imath] with [imath]L\in\mathbb R\backslash\{0\}[/imath]. Then [imath]\sum a_n[/imath] converges if and only if [imath]\sum b_n[/imath] converges. So by recapitulating the lecture I've tried to prove it but I didn't get it. It's obvious that you have to use the comparison test, but how? Can anybody help? Thanks a lot! |
1021705 | Inequality for positive definite matrices
Let [imath]A[/imath], [imath]B[/imath] be positive definite symmetric matrices. such that [imath]A \leq B[/imath], i.e., [imath](A - B)[/imath] is a negative semi-definite matrix. Is it true to conclude that [imath]B^{-1} \leq A^{-1}[/imath]? (I need to prove it in order to use in the proof of a theorem in stability of control systems) For trivial examples [imath]A=aI[/imath] and [imath]B=bI[/imath], where [imath]a \leq b[/imath], the above statement is true. In addition, I think it can simply be proven for the case that [imath]A[/imath], [imath]B[/imath] commute, i.e., [imath](A B = BA)[/imath]. Thanks in advance. | 473879 | Prove that [imath]A\ge0, B\ge0[/imath] and [imath]A\ge B[/imath] implies [imath]B^{-1}\ge A^{-1}[/imath]
Does anyone know how to prove the following: Suppose [imath]A[/imath] and [imath]B[/imath] are both positive definite and [imath]A - B[/imath] is positive semi-definite. Show that [imath]B^{-1} - A^{-1}[/imath] is also positive semi-definite. I really appreciate any comments! |
1022180 | What does the anitpodal map do to the homology groups of a sphere?
Let [imath]A: S^n \to S^n[/imath] denote the map sending a point on a sphere to the exact opposite point. Let [imath]A_*: H_n(S^n) \to H_n(S^n)[/imath] denote the action of [imath]A[/imath] on the homology group. Then [imath]A_*[/imath] is an isomorphism from [imath]\mathbb{Z}[/imath] to [imath]\mathbb{Z}[/imath], but the generator might change. For instance, on [imath]S^1[/imath], if [imath]x_0[/imath] is the base point on a clockwise loop and [imath]x_1[/imath] is a point a little clockwise of [imath]x_0[/imath], then their image under [imath]A[/imath] will have the same orientation, so [imath]1[/imath] will map to [imath]1[/imath] by [imath]A_*[/imath]. On [imath]S^2[/imath], however, a clockwise loop on the top of the sphere will map to a counterclockwise loop on the bottom, so here [imath]1[/imath] maps to [imath]-1[/imath]. What can I say about higher dimensional spheres? | 452962 | The degree of antipodal map.
I am trying to solve the problem A map without fixed points - two wrong approaches. But I am not certain about the degree of antipodal map. I my thought, since the preimage of a point [imath]y \in S^k[/imath] is just [imath]-y[/imath], the degree is just [imath]+1[/imath] or [imath]−1[/imath], depending on the orientation of [imath]-y[/imath]? |
247224 | Derangements with repetitive numbers
I have a very simple problem. Lets assume that I have a well shuffled deck of 52 cards. I start drawing the top card always and when the card matches its rank I lose. J=11 Q=12 K=13. If there were only 13 cards I could easily use the [imath]\ \frac{!n}{n!}[/imath] for derangements in order to solve this. The problem is that there are 52 cards so when I pass 13 I start from 1 again so I don't know what is the probability to win. Example of the game 1st card: 4 - Continue 2nd Card: A - continue 3rd Card: K - Continue 4th Card: K - Continue 5th Card: 6 - Continue 6th Card: 9 - Continue 7th Card: 10 - Continue 8th Card: A - Continue 9th Card: J - Continue 10th Card: 3 - Continue 11th Card: 2 - Continue 12th Card: 8 - Continue 13th Card: A - Continue 1st card: 5 - Continue 2nd Card 2 LOSE So actually I have to count from 1 to 13 4 times and if I draw all 52 cards then I win. What's the probability? | 1456597 | The probability of a match in a card game
I found the following problem in the book Understanding Probability: An ordinary deck of 52 cards is thoroughly shuffled. The dealer turns over the cards one at a time, counting as he goes ``ace, two, three, [imath]\ldots[/imath] , king, ace, two, [imath]\ldots[/imath]," and so on, so that the dealer ends up calling out the thirteen ranks four times each. A match occurs if the card that comes up matches the rank called out by the dealer as he turns it over. The author states that the inclusion-exclusion rule can be used to find the value 0.9838 for the probability of at least one match. I could not reproduce this result. As pointed by Joriki, a derivation of this result is given in Derangements with repetitive numbers This quite complicated derivation is based on an integral representation involving a Laguerre polynomial of degree four. However, for the specific example, you would expect that a a direct application of the inclusion-exclusion formula must be possible. |
1022197 | In what sense does the Maclaurin series for [imath]\ln(x+1)[/imath] converge on [imath][-1,1][/imath]?
This is regarding Modes of Convergence. The question is as follows. Given [imath]f(x) = \ln(x+1)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}[/imath]. Let [imath]f_N(x)=\sum_{n=1}^{N}(-1)^{n-1}\frac{x^n}{n}[/imath] Determine whether [imath]f_N[/imath] converges pointwise, uniformly, or in the [imath]L^2[/imath] sense to [imath]f[/imath] on [imath]-1\leq x\leq1[/imath]. I am having a hard time determining whether or not [imath]f_N(x)[/imath] does converge pointwise, uniformly, or in the [imath]L^2[/imath] sense to [imath]f(x)[/imath] I know [imath]|f(x)-f_N(x)|\rightarrow 0[/imath] as [imath]N\rightarrow \infty[/imath] for every [imath]x[/imath] in [imath]-1<x<1[/imath]; however, I am not sure whether it converges pointwise or uniformly on the closed interval [imath][-1,1][/imath]. The end point [imath]-1[/imath] is what causes the complications here. I want to say on the close interval, it does not converge uniformly because for all [imath]\epsilon>0[/imath], there does not exist a [imath]N[/imath] such that for all [imath]n \geq N[/imath], [imath]|f(x)-f_N(x)|\leq \epsilon[/imath] for all [imath]x \in [-1,1].[/imath] Any suggestions? Thanks in advance for your time. | 1021299 | Error Analysis and Modes of Convergences
I have the following question regarding Modes of Convergence Recall that the Taylor series of the function [imath]f(x) = \ln(1+x)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}[/imath]. Let [imath]f_{N}(x)=\sum_{n=1}^{N}(-1)^{n-1}\frac{x^n}{n}[/imath] be the partial sum. Determine whether [imath]f_N[/imath] converges pointwise, uniformly, or in the [imath]L^2[/imath] sense to [imath]f[/imath] on [imath]-1\leq x\leq1[/imath]. I am kinda sketchy on whether it converges pointwise - well in general actually - for [imath]-1[/imath] since [imath]\ln(1+-1) = \ln(0)[/imath] is undefined. If I disregard this, let's take a look at what we know. [imath]\ln(x+1)[/imath] = [imath]\sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}[/imath] converges on [imath]|x|<1[/imath] by the ratio test. The ends would be of concern, but [imath]1[/imath] is okay. Not so much [imath]-1[/imath]. Anywho, this is what I have. Pointwise Converges: Consider [imath]f(x) = \sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}[/imath] and its [imath]N[/imath]th partial sum [imath]f_{N}(x)=\sum_{n=1}^{N}(-1)^{n-1}\frac{x^n}{n}[/imath]. The infinite series converges pointwise to [imath]f(x)[/imath] on [imath]-1\leq x \leq 1[/imath] if [imath] \begin{eqnarray} | \sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^n}{n}-\sum_{n=1}^{N}(-1)^{n-1}\frac{x^n}{n}|\\ |(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+...)-(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\frac{x^5}{5}+...)|\\ |(x-x)+(-\frac{x^2}{2}+\frac{x^2}{2})+(\frac{x^3}{3}-\frac{x^3}{3})+(-\frac{x^4}{4}+\frac{x^4}{4})+(\frac{x^5}{5}-\frac{x^5}{5})+...| \longrightarrow 0\\ \end{eqnarray} [/imath] as [imath]N \rightarrow \infty[/imath], for every [imath]x\in[-1,1][/imath]. However, for some reasoning I am doubting it actually converges on the entire interval. I could just be over thinking it. The point is to show the error between [imath]f(x)[/imath] and [imath]f_N(x)[/imath] tends to zero as [imath]N\rightarrow \infty[/imath] for every [imath]x[/imath] on the interval and it seems to do just that. Uniformly Convergence is similar; however, it focuses on uniform error between [imath]f(x)[/imath] and [imath]S_N(x)[/imath] tends to zero as [imath]N\rightarrow \infty[/imath]. If want to conclude it also tends to zero. Same goes for [imath]L^2[/imath] sense. Any suggestions? Thank you for your time and Thanks in advance for any feedback. |
1022730 | In a formal language, how does one show that [imath]\neg \neg \bot \neq( \phi \wedge \psi) [/imath]
In a formal language, how does one show that [imath]\neg \neg \bot \neq( \phi \wedge \psi) [/imath] Or how do one go about showing that the former is not a proposition. I've just started reading Dalen's Logic and structure and there the set of propositions is defined as the smallest set [imath]X [/imath] satisfying, i) [imath]p _i \in X(i \in N ) [/imath], [imath]\bot \in X [/imath] ii) [imath]\phi , \psi \in X \implies ( \phi \wedge \psi),( \phi \vee \psi), ( \phi \to \psi),( \phi \leftrightarrow \psi) \in X[/imath] iii) [imath]\phi \in X \implies (\neg \phi) \in X [/imath] To show that [imath]\neg \neg \bot [/imath] is not a proposition one can go about to show that assuming [imath]X [/imath] and then setting [imath]Y=X \setminus \{\neg \neg \bot \} [/imath] again satisfies i),ii) and iii) and is smaller that [imath]X [/imath]. But this hinges on that one can show that [imath]\neg \neg \bot \neq( \phi \wedge \psi) [/imath], where nothing has been said about syntax. Dalen states that one should "look at the brackets", can one determine difference from this? Thanks in advance! | 748821 | Why [imath]¬¬\bot \not\in PROP[/imath]?
I'm reading Dalen's Logic and Structure. He gives the definition of the set [imath]PROP[/imath]: Definition [imath]\bf2.1.2[/imath] The set [imath]PROP[/imath] of propositions is the smallest set [imath]X[/imath] with the properties [imath]\begin{array}{rl} \rm(i)&p_i\in X(i\in N),\bot\in X,\\ \rm(ii)&\varphi,\psi\in X\Rightarrow(\varphi\wedge\psi),(\varphi\vee\psi),(\varphi\to\psi),(\varphi\leftrightarrow\psi)\in X,\\ \rm(iii)&\varphi\in X\Rightarrow(\neg\varphi)\in X.\\ \end{array}[/imath] Then there are some examples: And he says that [imath]¬¬\bot \not\in PROP[/imath], but I don't understand it. Isn't [imath]¬¬ \bot=\bot[/imath] due to the property of double negation? Considering this is true, [imath]\bot\in PROP[/imath]. Of course, in this question, I'm pressuposing some kind of evaluation, that [imath]¬¬ \bot[/imath] becomes [imath]\bot[/imath] due to the law of the double negation, but perhaps it should be hold in an unevaluated form. Ex:. [imath]1+1=2[/imath] because we presupose the [imath]1[/imath] merges with [imath]1[/imath] and then becomes two. If the merge (evaluation) does not happen, then I guess: [imath]1+1=1+1[/imath] [imath]2=2[/imath] [imath]2\neq 1+1[/imath] [imath]1+1\neq 2[/imath] Perhaps it's something like that, I took the idea from Mathematica. Sometimes it does treats things as different in the absence of full evaluation. |
1023676 | Prove that if a fraction is broken up into two the resulting two fractions cannot both have a larger value
I am working on research involving probability tables. I simplified the problem to the following. Say we have the following: [imath]C_1, C_2, C_3, C_4[/imath] [imath]\forall i, C_i > 0[/imath] [imath]x = \frac{C_1 + C_2}{C_1 + C_2 + C_3 + C_4}[/imath] [imath]y = \frac{C_1}{C_1 + C_3}[/imath] [imath]z = \frac{C_2}{C_2 + C_4}[/imath] I am almost positive (by just filling in numbers) that both [imath]y[/imath] and [imath]z[/imath] cannot [imath]> x[/imath] or [imath]< x[/imath]. Meaning that unless [imath]y = z[/imath], then either [imath]y < x[/imath] and [imath]z > x[/imath] or [imath]y > x[/imath] and [imath]z < x[/imath]. Does anyone have suggestions for an approach to prove this? ==EDIT== Basically, I am trying to prove that either of these two situations are impossible: [imath]z \geq y > x[/imath] [imath]x > y \geq z[/imath] | 205654 | Given 4 integers, [imath]a, b, c, d > 0[/imath], does [imath]\frac{a}{b} < \frac{c}{d}[/imath] imply [imath]\frac{a}{b} < \frac{a+c}{b+d} < \frac{c}{d}[/imath]?
We were trying to come up with an easy way to generate a rational number in between two existing rational numbers with a fairly low numerator and denominator (the way we were doing this earlier was to find the average of the two rationals, but that results in a denominator of up to [imath]c * d[/imath]. Does this inequality hold for all values of [imath]a, b, c, d[/imath]? |
1023160 | Congruence for the sums of odd powers of integers
Does someone know how to prove ***EDIT by induction**** that for all integers [imath]n\ge1[/imath], [imath]k\gt0[/imath] [imath]\sum\limits_{i=1}^{n} {i^{2k+1}}\equiv 0\ \ \ \pmod{\frac{n(n+1)}{2}}[/imath] I thought this should be a quick outcome from known polynomial expressions for sums of powers, or easy by induction. But I have not been able to write down such a proof. Thanks for help.. I make the above EDIT to 'deduplicate' (somehow) the question. This is where I am: Let [imath]P_{m+1}[/imath] be the [imath]m+1[/imath] degree polynomial such that [imath]P_{m+1}(n) =\sum\limits_{i=1}^{n} {i^{m}}[/imath] From [imath](n+\frac{1}{2})^{m+1}-(\frac{1}{2})^{m+1}=\sum\limits_{i=1}^{n}((i+\frac{1}{2})^{m+1}-(i-\frac{1}{2})^{m+1})[/imath] after expansion of the binomes on the RHS and rearrangement: [imath](n+\frac{1}{2})^{m+1}-(\frac{1}{2})^{m+1}=\sum\limits_{j=0}^{m+1}\frac{\binom{m+1}{j}}{2^{j}}(1-(-1)^j)P_{m+2-j}(n)[/imath] I obtain a recurrence relationship for the [imath]P_{2k}[/imath] of even degrees [imath]2^{2k+2}(2k+2)P_{2k+2}(n)=(2n+1)^{2k+2}-1-\sum\limits_{j=1}^{k}2^{2j}\binom{2k+2}{2j-1}P_{2j}(n)[/imath] that is [imath]2^{2k}(k+1)P_{2k+2}(n)=\binom{n+1}{2}\sum\limits_{j=0}^{k}(2n+1)^{2j}-\sum\limits_{j=1}^{k}2^{2j-3}\binom{2k+2}{2j-1}P_{2j}(n)[/imath] Now if I suppose that, for all [imath]1\le j \le k[/imath], for all [imath]n[/imath], all the [imath]P_{2j}(n)\equiv 0\ \ \pmod{\frac{n(n+1)}{2}}[/imath] , which is true for [imath]j=1[/imath], and since all the [imath]2^{2j-3}\binom{2k+2}{2j-1}[/imath] are integers even when [imath]j=1[/imath], then I have [imath]2^{2k}(k+1)P_{2k+2}(n)\equiv 0 \pmod{\frac{n(n+1)}{2}}[/imath] which is almost (but not quite :-() enough to complete the proof by induction | 427744 | Showing that [imath]1^k+2^k + \dots + n^k[/imath] is divisible by [imath]n(n+1)\over 2[/imath]
For any odd positive integer [imath]k\geq1[/imath], the sum [imath]1^k+2^k + \dots + n^k[/imath] is divisible by [imath]n(n+1)\over 2[/imath]. I used induction principle for the solution but cannot prove it. I took [imath]P(k) = 1^k+2^k+\dots+n^k[/imath]. For [imath]P(1)[/imath] it is true. For [imath]P(n)[/imath] let it be true. But for [imath]P(n+1)[/imath] I cannot solve it. |
1025271 | Why is the naive notion of a product ideal not necessarily additively closed?
Considering the product ideal [imath]IJ = \{ \sum_{i=1}^n a_ib_i | a_i \in I, b_i \in J \forall i\}[/imath], I've always seen it written that the more naive notion [imath]IJ = \{ ij | i \in I, j \in J\}[/imath] is not an ideal because in general it is not additively closed and is thus not a subring. I am unable to come up with a proof or any examples. Any ideas? | 21440 | Defining the Product of Ideals
Where does the "naive definition" of the product of two ideals [imath]I[/imath] and [imath]J[/imath], [imath]IJ = \{ ij \mid i \in I, j \in J \}[/imath] fall apart? (The product of two ideals [imath]I[/imath] and [imath]J[/imath] is defined to be [imath]IJ := \sum_i a_ib_i[/imath], where each [imath]a_i[/imath] is in [imath]I[/imath] and each [imath]b_i[/imath] is in [imath]J[/imath].) Note: This is a question from a problem set I'm working on. I am not expecting a full solution for the answers, but I'd like some help knowing where to look. It seems to work fine in [imath]\mathbb{Z}[/imath] and [imath]\mathbb{C}[x][/imath]. |
1025317 | Calculation of [imath]\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{m^2n}{3^n\left(m\cdot 3^n+n\cdot 3^m\right)}[/imath]
Calculation of [imath]\displaystyle \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{m^2n}{3^n\left(m\cdot 3^n+n\cdot 3^m\right)}[/imath]. [imath]\bf{My\; Try::}[/imath] Let [imath]\displaystyle S = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{m^2n}{3^n\left(m\cdot 3^n+n\cdot 3^m\right)}=\sum_{m=1}^{\infty}\left[\frac{1}{3}\cdot \frac{m^2}{ \left(3m+3^m\right)}+\frac{1}{3^2}\cdot \frac{2m^2}{(3^2m+2\cdot 3^m)}+.........\right][/imath] Actually I did not understand how can i solve it, Help me Thanks | 847928 | I need help with a double sum
I need help with the following exercise: determining the value of the next sum: [imath]\sum _{ m=1 }^{ \infty }{ \sum _{ n=1 }^{ \infty }{ \frac { { m }^{ 2 }n }{ { 3 }^{ m }(n{ 3 }^{ m }+m{ 3 }^{ n }) } } } [/imath] First time I face a double sum. I have not the slightest idea how to begin to work it |
1025494 | Solve the equation [imath]((x+y i)-\frac{1}{x+y i})/{(2 i)} = 2[/imath]
Solve the equation [imath]\frac{\left((x+y i)-\frac{1}{x+y i}\right)}{2 i} = 2[/imath] So far, I got [imath](0, 2-\sqrt{3}i)[/imath] and [imath](0, 2+ \sqrt{3}i)[/imath] as solutions for [imath]x[/imath] and [imath]y[/imath]. Do I require [imath]2[/imath] more solutions? | 1023390 | Find all the values of [imath]w[/imath] ∈ C that satisfy the equation.
Find all values of [imath]\omega \in \mathbb{C}[/imath] such that [imath]\frac{\omega - \frac1\omega}{2i} = 2.[/imath] So what I have is: Let [imath]\omega[/imath] be represented as [imath]x + yi[/imath] (by definition of complex numbers) Then [imath]\frac1\omega = \frac{x-yi}{x^2 + y^2}[/imath] (by properties of complex numbers) So, [imath]\frac{x+yi - \frac{x-yi}{x^2 + y^2}}{2i} = 2[/imath] Eventually, I get, via expansion and simplification: [imath]\frac{x(x^2 + y^2 - 1) + yi(x^2 + y^2 + 1)}{2i(x^2 + y^2)}[/imath] Any hints on how I can continue? I feel like I'm close but missing something. |
1019329 | Lemma about a prime times a unit
I came across this Lemma: "Let [imath]R[/imath] be an integral domain, and let [imath]a,u\in R[/imath] such that [imath]u[/imath] is invertible. Then [imath]a[/imath] is a prime if and only if [imath]au[/imath] is a prime. I tried to prove it unsuccessfully, but would appreciate your help with a formal proof. thanks :) | 1024123 | [imath]R[/imath] integral domain : [imath]u\in R^*, a \text{ is prime} \iff au \text{ is prime}[/imath]
[imath]R[/imath] integral domain : [imath]u\in R^*,\; a \text{ is prime} \iff au \text{ is prime}[/imath] I started by looking at [imath]auu^{-1}[/imath]. What should I do next? I'd be glad for help. Note: [imath]u \in R^*[/imath] meaning is [imath]u[/imath] is invertible. |
1026349 | What values will make this expression real
For what values of [imath]z[/imath] , where [imath]z ∈ C[/imath], will make the expression, $(iz - 1)[imath]/[/imath](z-i)$, real? So far, I've noted that if the expression, let's name it [imath]w[/imath], is real, then [imath]w[/imath] is equal to its conjugate. Then I equated them and solved, getting [imath]z = ±1[/imath]. I want to know if this method is correct, and are the answers that I got correct, or should I be using a different methodology? Thanks. | 1025518 | Describe all the complex numbers [imath]z[/imath] for which [imath](iz − 1 )/(z − i)[/imath] is real.
Describe all the complex numbers [imath]z[/imath] for which [imath](iz − 1 )/(z − i)[/imath] is real. Your answer should be expressed as a set of the form [imath]S = \{z \in\mathbb C : \text{conditions satisfied by }z\}[/imath]. I started solving for [imath]((iz − 1 )/(z − i)) = \overline{ ((iz − 1 )/(z − i))}[/imath]. I got stuck. Is this the right way to go about this? |
1026388 | If [imath]\lim_{x \to \infty}f(x) = l[/imath], then [imath]\lim_{x \to 0} f(\frac1{x^2}) = l[/imath]
Suppose that [imath]\lim_{x \to \infty}f(x) = l[/imath], [imath]l \in \mathbb R[/imath]. Prove that [imath]\lim_{x \to 0} f(\frac1{x^2}) = l[/imath]. Is it enough for me to say: Let [imath]g(x) = \frac1{x^2}[/imath]. So, [imath]\lim_{x \to 0} g(x) = \infty[/imath]. Then [imath]\lim_{x \to 0} f(\frac1{x^2}) = \lim_{g(x) \to \infty}f(g(x)) = l[/imath] | 1025329 | limit of reciprocal of function
Suppose that [imath]\displaystyle \lim_{x \to \infty} f(x) = \ell[/imath] where [imath]\ell \in \mathbb R[/imath]. Prove that [imath]\displaystyle \lim_{x \to 0} f(1/x^2) = \ell[/imath]. Does this involve reciprocal of functions? I am guessing that if [imath]f(x) = \ell[/imath] as [imath]x \to \infty[/imath], then [imath]f(1/x^2)[/imath] as [imath]x[/imath] goes to 0 will be same as [imath]f(x)[/imath] as [imath]x \to \infty[/imath]. Therefore the limit will be same. Am I right? Thanks. |
936014 | Existence of finite indexed normal subgroup for a given finite indexed subgroup.
Prove that if H has finite index n then there is a normal subgroup N of G with [imath]N \subset H [/imath] and [G:N][imath] \le [/imath]n!. I tried to solve the problem but could not done exactly. Since [G:H]=n , Let A={[imath]a_i H[/imath]:i=1,2,..,n}. Consider the left multiplication group action on A .Then kernal of the action is intersection N= [imath]\cap a_iHa_i^{-1}[/imath] i=1,2..,n. Then N is the largest normal subgroup of G contained in H. Now it is clear that [imath][G:a_iHa_i^{-1}][/imath] =n for all i.{by orbit stabliser theorem}.Now how to find [G:N]? yes I got it. By Caley's theorem G/N is isomorphic to a subgroup of [imath] S_{A}[/imath] .Therefone [G:N] divides n!. | 1756953 | Every subgroup of finite index contained in an infinite group [imath]G[/imath] contains a normal subgroup of [imath]G[/imath].
Let [imath]H<G[/imath] such that [imath]H[/imath] has finite index in the infinite group [imath]G[/imath]. Then [imath]G[/imath] contains a normal subgroup of finite index contained in [imath]H[/imath]. Can I create a subgroup of index [imath]2[/imath] in [imath]G[/imath] using elements of [imath]H[/imath]? |
1026620 | limit of a sequence of iterated logarithms
I was playing around with the family of sequences [imath]s(x)[/imath] defined for [imath]x > 0[/imath] as [imath]s(x)_0 = x[/imath], [imath]s(x)_{n+1} = \log(1+s(x)_n)[/imath] and I noticed a strange behavior. Specifically, regardless of the choice of [imath]x[/imath], [imath]s(x)_{10^n} \approx 2\cdot 10^{-n}[/imath] as [imath]n[/imath] grows: > let logs x = x : logs (log (1 + x)) > logs 1000 !! 1000 2.0066131337990703e-3 > logs 1000 !! 10000 2.0008131530487204e-4 > logs 1000 !! 100000 2.0000966401761077e-5 > logs 1000 !! 1000000 2.000011198587142e-6 Based on this observation, I am wondering whether [imath]n s(x)_n \rightarrow 2[/imath] for all [imath]x > 1[/imath] and if so, why. | 5408 | Limit of the sequence [imath]nx_{n}[/imath] where [imath]x_{n+1} = \log (1 +x_{n})[/imath]
Suppose [imath]x_{1}>0[/imath], and consider the sequence, [imath]\{x_{n}\}[/imath] defined as follows: [imath]x_{n+1}=\log(1+x_{n}) \quad n\geq 1 [/imath] Find the value of [imath]\displaystyle \lim_{n \to \infty} nx_{n}[/imath] I am having trouble solving it. One thing is clear, that since [imath]x_{n}>0[/imath] and [imath]x_{n+1} < x_{n}[/imath], we can have a sequence which converges an [imath]f[/imath] which satisfies [imath]f=\log(1+f)[/imath], so that [imath]f=0[/imath]. Any way as to how we can proceed from here. |
1026990 | Essential part to undestand a proof .
In the proof of the the lemma suppose [imath]G[/imath] is a finite abelian [imath]p[/imath]-group, and let [imath]C[/imath] a cyclic subgroup of maximal order, then [imath]G=C\oplus H[/imath] for some subgroup [imath]H[/imath] at http://torus.math.uiuc.edu/jms/m317/handouts/finabel.pdf, They say that since [imath]H\cap(C+K)= K[/imath], we have [imath]H\cap C=\{e\}[/imath]. But how do they have that [imath]H\cap(C+K)= K[/imath]? Can someone prove that please is because I don't see why is that true with the arguments presented in that proof. thanks a lot in advance :) I am sorry for this but my doubt was in the part that I pst here and in the other question the answer was to other thing :) | 1026772 | A doubt with a part of a certain proof.
Well, in the proof of the following lemma suppose [imath]G[/imath] is a finite abelian [imath]p[/imath]-group, and let [imath]C[/imath] a cyclic subgroup of maximal order, then [imath]G=C\oplus H[/imath] for some subgroup [imath]H[/imath] at http://torus.math.uiuc.edu/jms/m317/handouts/finabel.pdf, They say that since [imath]H\cap(C+K)= K[/imath], we have [imath]H\cap C=\{e\}[/imath]. But how do they have that this part [imath]H\cap(C+K)= K[/imath]? Can someone prove that please is because I don't see why is that true with the arguments presented in that proof. I have the following attempt : We know that [imath]G/K=(C+K)/K \cong H' \Rightarrow H \cap (C+K)=K[/imath] If there is [imath]x \in H \cap C, x \neq e,[/imath] then [imath]H \cap (C + K) = K \Rightarrow x \in K.[/imath] But by your assumption [imath]C\cap K = \{e\}[/imath] more precisely : I want to prove (convincing myself), why is this rght. In the proof of the lemma suppose [imath]G[/imath] is a finite abelian [imath]p[/imath]-group, and let [imath]C[/imath] a cyclic subgroup of maximal order, then [imath]G=C\oplus H[/imath] for some subgroup [imath]H[/imath] at http://torus.math.uiuc.edu/jms/m317/handouts/finabel.pdf they have that since [imath]H\cap(C+K)= K[/imath], we have [imath]H\cap C=\{e\}[/imath]. But how do they have the part of [imath]H\cap(C+K)= K[/imath]? is because they say that [imath]H'[/imath] is the preimage of a certain map, then they can do the following: [imath]G/K= (C+K)/K \oplus H/K[/imath] but I do not know if that is enough to justificate that step, or how can they have the conclusion that [imath]H\cap(C+K)= K[/imath],and well the part that [imath]H\cap C=\{e\}[/imath] is obvious from here, and in the proof of the same lemma they say that because [imath]K[/imath] has prime order then [imath]K\cap C=\{e\}[/imath], is that the only reason? Can someone prove those things please is because I don't see why is that true with the arguments presented in that proof. Thanks in advance. I am back in the discussion and I have edited my post to be clearler :) |
1026506 | Integration [imath]I_n=\int _0^{\pi }\:sin^{2n}\theta \:d\theta [/imath]
If [imath]I_n=\int _0^{\pi }\:sin^{2n}\theta \:d\theta [/imath], show that [imath]I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}[/imath], and hence [imath]I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi [/imath] Hence calculate [imath]\int _0^{\pi }\:\:sin^4tcos^6t\:dt[/imath] I knew how to prove that [imath]I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}[/imath] ,, but I am not very good at English, what does it mean Hence [imath]I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi [/imath] do we need to prove this part as well or is it just a hint to use? and for the other calculation to find [imath]\int _0^{\pi }\:\:sin^4tcos^6t\:dt[/imath] is there something in the first part that I can do to help me solve this question because otherwise it becomes very long and it's part of the question. | 1026939 | Integration [imath]I_n=\int _0^{\pi }\sin^{2n}\theta d\theta [/imath]
If [imath]I_n=\int _0^{\pi }\sin^{2n}\theta \:d\theta [/imath], show that [imath]I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}[/imath], and hence [imath]I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi [/imath] Hence calculate [imath]\int _0^{\pi }\:\sin^4t\cos^6t\:dt[/imath] I knew how to prove that [imath]I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}[/imath], but I am not very good at English, what does it mean Hence [imath]I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi [/imath] do we need to prove this part as well or is it just a hint to use? and for the other calculation to find [imath]\int _0^{\pi }\:\sin^4t\cos^6t\:dt[/imath] is there something in the first part that I can do to help me solve this question because otherwise it becomes very long and it's part of the question. |
1026854 | A hard question on surjective group homomorphism
Say [imath]G[/imath] and [imath]H[/imath] are finite groups, and there exists a surjective group homomorphism from [imath]G × G[/imath] to [imath]H × H[/imath]. Must there exist a surjective group homomorphism from [imath]G[/imath] to [imath]H[/imath]? I have no idea how to do this. Help please | 221152 | Can we ascertain that there exists an epimorphism [imath]G\rightarrow H[/imath]?
Let [imath]G,H[/imath] be finite groups. Suppose we have an epimorphism [imath]G\times G\rightarrow H\times H[/imath] Can we find an epimorphism [imath]G\rightarrow H[/imath]? |
1027393 | If [imath]H[/imath] and [imath]K[/imath] are subgroups of a group G, and [imath]a, b \in G[/imath]. How do I prove the following relations hold?
Either [imath]Ha \cap Kb = \emptyset[/imath], or [imath]Ha \cap Kb = (H \cap K)c[/imath], for some [imath]c \in G[/imath]. I tried supposing [imath]Ha \cap Kb \not= \emptyset[/imath] but I can't see how it follows that [imath]Ha \cap Kb = (H \cap K)c[/imath]. | 338650 | Intersection of cosets
I'm not really sure how to go about proving the following. Any help will be appreciated. Question: Let [imath]H[/imath] and [imath]K[/imath] be subgroups of a group [imath]G[/imath]. Prove that the intersection [imath]xH \cap yK[/imath] of two cosets of [imath]H[/imath] and [imath]K[/imath] is either empty or else is a coset of the subgroup [imath]H \cap K[/imath]. |
1027300 | Proof of the First Isomorphism Theorem for Rings
The statement is the First Isomorphism Theorem for Rings from Abstract Algebra by Dummit and Foote. I'd like to check if all is ok. In particular I'm a bit worried about the (*) line. It looks a bit awkward, but is it too bad? Any suggestions? Theorem: 1) If [imath]\varphi:R \rightarrow S[/imath] is a homomorphism of rings, then the kernel of [imath]\varphi[/imath] is an ideal of [imath]R[/imath], the image of [imath]\varphi[/imath] is a subring of [imath]S[/imath] and [imath]R/ker \varphi[/imath] is isomorphic as a ring to [imath]\varphi(R)[/imath]. 2) If [imath]I[/imath] is any ideal of [imath]R[/imath], then the map [imath]R \rightarrow R/I[/imath] defined by [imath]r \rightarrow r+I[/imath] is a surjective ring homomorphism with kernel [imath]I[/imath]. Thus every ideal is the kernel of a ring homomorphism and vice versa. Proof: Let [imath]\varphi: R \rightarrow S[/imath] be a ring homomorfism. If [imath]r \in R[/imath] and [imath]r' \in ker \varphi[/imath], then we have [imath]rr',r'r \in ker \varphi[/imath] (so that it is closed under multiplication by elements of [imath]R[/imath]) since [imath]\varphi (rr')=\varphi (r) \varphi (r')=\varphi (r)0 =0=0 \varphi (r)=\varphi (r') \varphi (r)=\varphi (r'r);[/imath] since [imath]ker \varphi[/imath] is also a subring of [imath]R[/imath], it is an ideal of [imath]R[/imath]. It's clear that [imath]\varphi (R)[/imath] is a subring of [imath]S[/imath]. Now, let [imath]I[/imath] be an ideal of [imath]R[/imath], so that [imath]R/I[/imath] is also a ring, and define [imath]\pi:R \rightarrow R/I[/imath] by [imath]\pi (r)=r+I[/imath]. We know [imath]\pi[/imath] is a group homomorphism with kernel [imath]I[/imath], and for [imath]r,s \in R[/imath], we have [imath]\pi (rs)=(rs)+I=(r+I)(s+I)=\pi (r) \pi (s),[/imath] so that [imath]\pi[/imath] is in fact a ring homomorphism. Define then [imath]\phi: R/ker \varphi \rightarrow \varphi(R)[/imath] by [imath]\phi (r+(ker \phi))= \varphi (r),[/imath] for each [imath](r+(ker \phi)) \in R/ker \varphi[/imath], for some [imath]r \in R[/imath]. This is well defined because if [imath]r' \in (r+(ker \varphi)),[/imath] then [imath]\phi (r'+(ker \varphi))=\varphi(r')=\varphi(r)=\phi (r+(ker \varphi)).[/imath] Also, this is a ring isomorphism because for each [imath]\varphi (s) \in \varphi (R)[/imath] for some [imath]s \in R[/imath], we have [imath](*) \ \phi^{-1}\{\varphi (s)\}= \phi^{-1} \varphi[r+(ker \varphi)]= \phi^{-1} \varphi[\pi^{-1} \{r+(ker \varphi)\}]= \{r+(ker \varphi)\},[/imath] a set with a single element of [imath]R/ker \varphi[/imath], so that it is a bijection, and for every [imath]r+(ker \varphi), r'+(ker \varphi) \in R/ker \varphi[/imath], for some [imath]r,r' \in R[/imath], we have [imath]\phi [(r+(ker \varphi))+(r'+(ker \varphi))]=\phi [(r+r')+(ker \varphi)]=\varphi(r+r')=\varphi(r)+ \varphi(r')=\phi [r+(ker \varphi)]+ \phi [r'+(ker \varphi)],[/imath] and [imath]\phi [(r+(ker \varphi)) (r'+(ker \varphi))]=\phi [(rr')+(ker \varphi)]=\varphi(rr')=\varphi(r) \varphi(r')=\phi [r+(ker \varphi)] \phi [r'+(ker \varphi)],[/imath] so that it is a ring homomorphism. | 1027298 | Proof of the First Isomorphism Theorem for Groups
The statement is the First Isomorphism Theorem for Groups from Abstract Algebra by Dummit and Foote. This proof was left as a exercise, so I'd like to check if all is ok. In particular, I'm a bit worried about the (*) line. It looks a bit awkward, but is it too bad? Any suggestions? Theorem: If [imath]\varphi : G \rightarrow H[/imath] is a homomorphism of groups, then [imath]ker \varphi \trianglelefteq G[/imath] and [imath]G/ker \varphi \cong \varphi (G)[/imath]. Proof: Let [imath]\varphi:G \rightarrow H[/imath] be a group homomorphism. If [imath]g \in G[/imath] and [imath]h \in ker \varphi[/imath], then we have that [imath]ghg^{-1} \in ker \varphi[/imath] because [imath]\varphi (ghg^{-1})= \varphi (g) \varphi (h) \varphi (g^{-1})= \varphi (g) \varphi (g^{-1}) = \varphi (gg^{-1}) = \varphi (1) =1,[/imath] so that [imath]ker \varphi \trianglelefteq G[/imath]. Now, define [imath]\phi: G/ker \varphi \rightarrow \varphi (G)[/imath] by [imath]\phi (g(ker \varphi))=\varphi(g),[/imath] for each [imath]g(ker \varphi) \in G/ker \varphi[/imath], for some [imath]g \in G[/imath]. This is well defined because if [imath]g' \in g(ker \varphi),[/imath] then [imath]\phi (g'(ker \varphi))= \varphi(g')= \varphi(g)= \phi (g(ker \varphi)).[/imath] Also, this is a group isomorphism because for each [imath]\varphi (h) \in \varphi (G),[/imath] for some [imath]h \in G[/imath], we have [imath](*) \ \phi^{-1}\{\varphi(h)\}= \phi^{-1} \varphi[g(ker \varphi)]= \phi^{-1} \varphi(\pi^{-1}\{g(ker \varphi)\})= \{g(ker \varphi)\},[/imath] a set with a single element of [imath]G/ker \varphi[/imath], so that it is a bijection, and for every [imath]g(ker \varphi), g'(ker \varphi) \in G/ker \varphi[/imath], for some [imath]g,g' \in G[/imath], we have [imath]\phi [g(ker \varphi) g'(ker \varphi)]=\phi [(gg')(ker \varphi)]=\varphi(gg')=\varphi(g) \varphi(g')=\phi [g(ker \varphi)] \phi [g'(ker \varphi)],[/imath] so that it is a group homomorphism. |
1027607 | Cardinality of the basis for [imath]\mathbb{R}[/imath] over [imath]\mathbb{Q}[/imath]?
This question came up as a discussion I had with a friend. Clearly, the basis is not of finite cardinality since that would imply the set [imath]\mathbb{R}[/imath] has the cardinality [imath]\aleph_{0}[/imath] which is false. Therefore the basis clearly has to be infinite. However, the question that came up was that is the basis countably infinite or uncountably infinite. So I'd like to know the cardinality of the basis and also appreciate some pointers to the proof. Thanks in advance. :) | 583601 | Take [imath]\mathbb{R}[/imath] as a vector space over [imath]\mathbb{Q}[/imath], then a basis for [imath]\mathbb{R}[/imath], then [imath]|B|=2^{\aleph_0}[/imath]
I am trying to prove that Given [imath]\mathbb{R}[/imath] as a vector space over [imath]\mathbb{Q}[/imath], and a basis for [imath]\mathbb{R}[/imath], then [imath]|B|=2^{\aleph_0}[/imath]. Proof: Suppse that [imath]|B|<2^{\aleph_0}[/imath], then [imath]\mathbb R = span\{ b_i ; b_i \in B \}[/imath] So [imath] \mathbb R = \{ \Sigma_{i=1}^{\infty}\alpha_i b_i ; \alpha_i \in \mathbb Q, b_i \in B, i \in \mathbb N \} = \bigcup_{n=1}^{\infty}\{\Sigma_{i=1}^n \alpha_ib_i\}[/imath]. Which is a contradiction since a countable union of countable union of countable sets is countable and [imath]\mathbb R[/imath] is not. What do you think? Thank you, Shir |
1010637 | Properties possessed by [imath]H , G/H[/imath] but not G
i) Does there exist a group [imath]G[/imath] with a normal subgroup [imath]H[/imath] such that [imath]H , G/H[/imath] are abelian but [imath]G[/imath] is not ? ii) Does there exist a group [imath]G[/imath] with a normal subgroup [imath]H[/imath] such that [imath]H , G/H[/imath] are cyclic but [imath]G[/imath] is not ? | 2551974 | If G/H is a cyclic group, then G is a cyclic group
I know "If [imath]G[/imath] is cyclic, then [imath]G/H[/imath] is cyclic" is true, but I was wondering if it true if, "If [imath]G/H[/imath] is a cyclic group, then [imath]G[/imath] is a cyclic group"? I think this would be true since [imath]H[/imath] is a subgroup of [imath]G[/imath] then [imath]G/H \in G[/imath] and since it's cyclic then [imath]G[/imath] must be cyclic right? Do I have the right reasoning/is this this a true statement? |
1026797 | How to find the integer solutions of [imath]\frac{2^m-1}{2^{m+x}-3^x}=2a+1[/imath]?
Is there a way to find all integer triplets of [imath](x, m, a)[/imath] for the following equation. [imath]\frac{2^m-1}{2^{m+x}-3^x}=2a+1[/imath] | 669827 | Divisibility of [imath]2^n - 1[/imath] by [imath]2^{m+n} - 3^m[/imath].
For what values of [imath]m,n[/imath] natural, do [imath]2^n - 1[/imath] is divisible by [imath]2^{m+n} - 3^m[/imath]? Thank you very much. |
725592 | How to evaluate [imath]\int_0^1\frac{1+x^4}{1+x^6}\,dx[/imath]
[imath]\int_0^1\frac{1+x^4}{1+x^6}\,dx[/imath] Can anyone help me solve the question? I am struggling with this. | 101049 | Compute: [imath]\int_{0}^{1}\frac{x^4+1}{x^6+1} dx[/imath]
I'm trying to compute: [imath]\int_{0}^{1}\frac{x^4+1}{x^6+1}dx.[/imath] I tried to change [imath]x^4[/imath] into [imath]t^2[/imath] or [imath]t[/imath], but it didn't work for me. Any suggestions? Thanks! |
1029198 | Prove that [imath]p(x)=(x-1)(x-2) \cdots (x-n) + 1[/imath] is irreducible over [imath]\mathbb{Z}[/imath] for all [imath]n \geq 1[/imath], [imath]n \neq 4[/imath].
Prove that [imath]p(x)=(x-1)(x-2) \cdots (x-n) + 1[/imath] is irreducible over [imath]\mathbb{Z}[/imath] for all [imath]n \geq 1[/imath], [imath]n \neq 4[/imath]. I do not clearly see how to solve this problem and what is so special about the integer [imath]4[/imath] here? Thanks for your help in advance | 503765 | Prove that the polynomial [imath](x-1)(x-2)\cdots(x-n) + 1[/imath], [imath] n\ge1 [/imath], [imath] n\ne4 [/imath] is irreducible over [imath]\mathbb Z[/imath]
I try to solve this problem. I seems to come close to the end but I can't get the conclusion. Can someone help me complete my proof. Thanks Show that the polynomial [imath]h(x) = (x-1)(x-2)\cdots(x-n) + 1[/imath] is irreducible over [imath]\mathbb Z[/imath] for all [imath]n\ge1[/imath] and [imath] n\ne4[/imath]. Suppose [imath]h(x) = f(x) g(x)[/imath], then we must have [imath]f(i)g(i) = 1[/imath] for all [imath]i = 1,2,...n[/imath]. So both [imath]f(i)[/imath] and [imath]g(i)[/imath] are [imath]1[/imath] or [imath]-1[/imath]. In either case, [imath]m(x) = f(x) - g(x)[/imath] has degree smaller than [imath]n[/imath] and have [imath]n[/imath] different roots ([imath]1,2,...,n[/imath]). So we must have [imath]m(x) = 0[/imath]. Then [imath]h(x) = f(x)^{2}[/imath]. So [imath]n[/imath] must be even. Let [imath]n = 2k[/imath]. Because [imath]f(x)[/imath] has degree [imath]k[/imath], there are [imath]k[/imath] values from [imath]\{1,2,...,2k\}[/imath] at which [imath]f(x)[/imath] is [imath]1[/imath] and [imath]k[/imath] values at which [imath]f(x)[/imath] is [imath]-1[/imath]. This is where I got stuck. Hope some one can help me solve this. Thanks. |
1029887 | Evaluate the integral (using partial fractions maybe?)
Evaluate the following integral [imath]\int{\frac{1}{(x+a)(x+b)}}[/imath] (this might involve partial fraction decomposition, [imath]\int{\frac{1}{x^2+x(a+b)+ab}}[/imath] this is what my first step was) | 1027798 | Evaluate the following integral?
Evaluate the following integral. [imath]\int\frac1{(x+a)(x+b)}~dx[/imath] [imath]{a}\neq{b}[/imath] I do not know where to being solving this integral. |
1029958 | Pigeon hole drunken mailman
Two letters need to be delivered to each of [imath]n[/imath] houses. How many ways can a postman deliver two letters to each house such that each house receives at least one incorrect letter? Right now I have the total number to deliver two letters to [imath]n[/imath] houses as [imath]N = \frac{(2n)!}{2^n}[/imath]. I know derangement is equal to [imath]N - N_1 + N_2 - N_3 + ... + (-1)^kN_k[/imath]. I have [imath]N_1 =\displaystyle{n \choose 1}\bigg[\frac{(2n-2)!}{2^{(n-1)}}\bigg][/imath] but I can't seem to find a pattern or find a solution. | 1029634 | Applying derangement principle to drunken postman problem.
Two letters need to be delivered to each of n houses. How many ways can a postman deliver two letters to each house such that each house receives at least one incorrect letter? I got stuck and don't now how to progress. Can someone provide hint? 1) [imath](2n)!/2^n[/imath] is the number of all possible onto functions from our domain of houses to the codomain of letters. 2) Let [imath]A_i[/imath] be property that two delivered letters are correct. [imath] |A_1' \cup A_2' \cup... \cup A_n'|=(2n)!/2^n-|A_1 \cap A_2 \cap... \cap A_n|[/imath] I tried everything I could come up with but I cannot find [imath]|A_1 \cap A_2 \cap... \cap A_n|[/imath]. I feel like I need to find union first but in order to do that I should find intersection. Really confused. Any hint would be appreciated. Edit: Corrected mistake in the equation. |
1029337 | Alpha and Omega Limit Sets in Polar Coordinates
I guess here I am not sure how to get started, I know the definitions: The [imath]ω[/imath]-limit sets of points are the set of points that the system of equation approach as time goes to infinity, and the [imath]α[/imath]-limit sets of points are the points approached as t goes to negative infinity. But I am not sure how the eigenvalues or vectors help me determine the limit sets. But I don't understand how to apply them. My conclusion for both of these question is that [imath]α(0)=ω(0)={0}[/imath] because in both cases the origin is a fixed point. Is this correct? Are the more limit sets than these? Consider the systems of equations and find all the [imath]α[/imath]- and [imath]ω[/imath]-limit sets of points in the plane. a. [imath]\dot {x}=y+x(x^2+y^2-1)[/imath] [imath]\dot {y}=-x+y(x^2+y^2-1)[/imath] which in polar coordinates, is given by: [imath]\dot {r}=r(r^2-1)[/imath] [imath]\dot {ø}=-1[/imath] [imath]r^2=x^2+y^2[/imath] [imath]r \dot {r}=x \dot {x}+y \dot {y}[/imath] [imath]=r^2(r^2-1)[/imath] [imath]r \dot=r(r-1)(r+1)[/imath] So I understand how the got the value for [imath]\dot {r}[/imath], but I don't understand how to get the value for [imath]\dot {ø}[/imath] The fixed points will be when [imath]\dot {r}=0[/imath] so when r=1, -1, 0. From my book: "Because [imath]\dot {ø}[/imath]=-1, thus the solution goes clockwise around the origin at unit angular speed." Why?? b. [imath]\dot {x}=y+x(1-x^2-y^2)(x^2+y^2-4)[/imath] [imath]\dot {y}=-x+y(1-x^2-y^2)(x^2+y^2-4)[/imath] which in polar coordinates, is given by: [imath]\dot {r}=r(1-r^2)(r^2-4)[/imath] [imath]\dot {ø}=-1[/imath] | 1015264 | System of equations, limit points
This is a worked out example in my book, but I am having a little trouble understanding it: Consider the system of equations: [imath]x'=y+x(1-x^2-y^2)[/imath] [imath]y'=-x+y(1-x^2-y^2)[/imath] The orbits and limit sets of this example can be easily determined by using polar coordinates. (My question: what is the motivation for that thinking? What should clue me in to thinking that I should use polar coordinates?) The polar coordinate satisfies [imath]r^2=x^2+y^2[/imath] so by differentiating with respect to t and using the differential equations we get: [imath]r\cdot r'=x\cdot x'+y\cdot y'[/imath] (I am unclear about how the book even got this first equation from [imath]r^2=x^2+y^2[/imath]) [imath]=x\cdot y+x^2(1-r^2)-x\cdot y+y^2(1-r^2)[/imath] Substitute in [imath]x'[/imath] and [imath]y'[/imath] and then multiple out and replace with [imath]r[/imath], I get this step [imath]=r^2(1-r^2)[/imath] cancel terms, I get this step too [imath]r'=r(1-r)[/imath] similarly, the angle variable [imath]\theta[/imath] satisfies [imath]\tan\theta=\frac yx[/imath], so the derivative with respect to [imath]t[/imath] yields [imath]\sec^2(\theta)\theta'=x^{-2}[x^2+xy(1-r^2)-y^2-xy(1-r^2)]=-\frac{r^2}{x^2}[/imath] so [imath]\theta=1[/imath] Thus the solution goes clockwise around the origin at unit angular speed. I don't understand the [imath]\theta[/imath] step at all or how they reached the conclusion of clockwise around the origin with unit angular speed.. But then it just jumps to saying "the origin is a fixed point, so α(0)=ω(0)={0} but I have no idea how they reached this conclusion.. |
1030335 | Proving binomial coefficients identity: [imath]\binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r} = \binom{n+1}{r+1}[/imath]
Let [imath]n[/imath] and [imath]r[/imath] be positive integers with [imath]n \ge r[/imath]. Prove that: [imath]\binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r} = \binom{n+1}{r+1}.[/imath] Tried proving it by induction but got stuck. Any help with proving it by induction or any other proof technique is appreciated. | 1025139 | Proving a combinatorics equality: [imath]\binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r} = \binom{n+1}{r+1}[/imath]
How to prove the following? Should I use induction or something else? Let [imath]n[/imath] and [imath]r[/imath] be positive integers with [imath]n \ge r[/imath]. Prove that [imath]\binom{r}{r} + \binom{r+1}{r} + \cdots + \binom{n}{r} = \binom{n+1}{r+1}.[/imath] Attempted start: Basis step: [imath]{\binom{1}{1}} = {\binom{2}{2}}[/imath] true. Where do I go from here? |
1031124 | integration, anti- derivative, complex
Let [imath]\gamma(w,R)[/imath] denote the circular contour [imath]t\mapsto w+Re^{it}[/imath] where [imath]0\lt t\lt2\pi[/imath]. Evaluate [imath]\int_\gamma\dfrac1{1+z^2}dz[/imath] when [imath]\gamma[/imath] is: [imath]\gamma(1,1),\quad\gamma(i,1),\quad\gamma(0,2),\quad\gamma(3i,\pi).[/imath] I have solved these using logarithms as an anti-derivative and then using the complex logarithm rule, however, I seem to be getting [imath]- \pi k[/imath] as a solution to all 4 problems. Any help as to where I'm going wrong would be helpful. | 1029158 | Circular contour integration.
solving one of the 5 options would be much appreciated as this will give me an idea on how to solve the rest. Let [imath]\gamma(w,R)[/imath] denote the circular contour [imath]t\mapsto w+Re^{it}[/imath] where [imath]0\lt t\lt2\pi[/imath]. Evaluate [imath]\int_\gamma\dfrac1{1+z^2}dz[/imath] when [imath]\gamma[/imath] is: [imath]\gamma(1,1),\quad\gamma(i,1),\quad\gamma(-i,1),\quad\gamma(0,2),\quad\gamma(3i,\pi).[/imath] |
1029697 | Let [imath]\lambda>1[/imath] and show the equation [imath]\lambda - z -e^{-z} = 0[/imath] has exacly one solution in the half plane [imath]\{z:Re(z)>0\}[/imath]
Let [imath]\lambda>1[/imath] and show the equation [imath]\lambda - z -e^{-z} = 0[/imath] has exacly one solution in the half plane [imath]\{z:Re(z)>0\}[/imath]. Show that this solution must be real. What happens to the solution as [imath]\lambda \rightarrow 1 [/imath]? | 320055 | [imath]\lambda-z-e^{-z}=0[/imath] has one solution in the right half plane
Let [imath]\lambda > 1[/imath] , want to show that the equation [imath]\lambda-z-e^{-z}=0[/imath] has exactly one solution in the right half plane [imath]\{z:Re(z)>0\}[/imath]. Moreover, the solution must be real.I tried to use Rouche's theorem on [imath]g(z)=\lambda - z[/imath] and [imath]f(z)=e^{-z}[/imath] to get that the number of zeros of [imath]f+g[/imath] and the number of zeros of [imath]g(z)[/imath] is the same, and since [imath]g(z)[/imath] has only one solution then the equation about must also have one solution the problem is I don't know how to choose the correct curve [imath]\gamma[/imath] such that this will work.for the second part I used the IVT to show that [imath]\lambda -x-e^{-x}[/imath] has a zero in [imath](0,\lambda)[/imath] to conclude that the solution is real. is this acceptable? Thank you for your help. |
1031887 | Do I use induction or is there another way to prove [imath]\binom{r}{r}+\binom{r+1}{r}+\cdots+\binom{n}{r}=\binom{n+1}{r+1}[/imath]?
Prove the following statement is true: [imath]\binom{r}{r}+\binom{r+1}{r}+\cdots+\binom{n}{r}=\binom{n+1}{r+1}[/imath]. Since [imath]\binom{r}{r}=\binom{n}{r}=\dfrac{n!}{r!(n-r)!}[/imath], is that to form a basis step? If so, how do I induce k+1 for n+1 and r+1 (where n ≥ r and both positive integers)? At the same time, or in two steps? | 1031651 | How to prove this statement: [imath]\binom{r}{r}+\binom{r+1}{r}+\cdots+\binom{n}{r}=\binom{n+1}{r+1}[/imath]
Let [imath]n[/imath] and [imath]r[/imath] be positive integers with [imath]n \ge r[/imath]. Prove that Still a beginner here. Need to learn formatting. I am guessing by induction? Not sure what or how to go forward with this. Need help with the proof. |
1031815 | Any algorithm or theorem to decide whether two functions are equivalent?
Any algorithm or theorem to decide whether two functions that are polynomials,rationals and analytic over [imath]\mathbb{N}[/imath] or [imath]\mathbb{Q}[/imath] or [imath]\mathbb{R}[/imath] or [imath]\mathbb{C}[/imath] are equivalent ? | 584230 | How do you validate that two math expressions are equal?
Let's say you have a few expressions like the following: [imath]\begin{array}((x+17)^2 \\ x^2 + 34x + 289 \end{array} \\ 288 + \frac{x^2}{2} + \frac{x^2}{2} + 34x + 1 \\ [...] [/imath] You get the idea: there's an infinite number of ways of representing the same expression. The challenge is that my application is validating answers to math problems, input by math students practicing a certain concept. I'd like to make sure that whenever they submit an answer, the checker is completely insensitive to the issues of ordering, simplification, formatting etc. which would be plain frustrating. I'd also ideally like to support more than one variable such as x, y, z in the same expression. What's a reliable way to check that two expressions are the same? The most brute force way I can think of would be to just plug a number into each variable of each expression and see what the result is. I have a hunch however that there might be ways of getting false positives this way. That, and I might be either missing something really obvious about this, or there might be a better way of doing the comparison. Would love to hear what you think. |
1031089 | element in field, not a square
I am doing a specific exercise where the quaternion group is realised as a Galois group of some field extension. It goes like this: let [imath]K = \mathbb Q(\sqrt 2,\sqrt 3)[/imath] and [imath]\alpha = (2 +\sqrt 2)(3 +\sqrt 6)[/imath]. I want to show that [imath]\alpha[/imath] is not a square in [imath]K[/imath]. I tried to solve equations explicitly to derive some contradiction but I didn't succeed. Any suggestions? Thanks in advance. | 248224 | Galois group and the Quaternion group
Let [imath]E=\mathbb{Q}(\sqrt{2},\sqrt{3})[/imath] and [imath] L = E \left( \sqrt{ ( \sqrt{2}+2 ) ( \sqrt{3} + 3)} \right) \ . [/imath] I want to show that [imath]L/\mathbb{Q}[/imath] is a Galois extension with the Quaternion group as its Galois group. I know [imath]E/\mathbb{Q}[/imath] is Galois and [imath]L/E[/imath] is also Galois, but it is not true in general that if [imath]K_1/K_0[/imath] is Galois and [imath]K_2/K_1[/imath] is Galois then [imath]K_2/K_0[/imath] is Galois (take [imath]K_0 = \mathbb{Q}[/imath], [imath]K_1 = \mathbb{Q}(\sqrt{2})[/imath] and [imath]K_2 = \mathbb{Q}(\sqrt[4]{2})[/imath] as a counter-example). |
1032049 | Proving quadratic inequalities.
I am trying to prove that [imath]e^{k+1} ≥ 3 + 3k + k^2[/imath] with, [imath]k>2[/imath] WhatI have done so far: What we are trying to prove is that [imath]e^n≥1+n+n^2[/imath] is a true statement. Since [imath]n=3[/imath] holds, this is our base case. Then, the inductive hypothesis is: [imath]e^k≥1+k+k^2[/imath] Hence, I need to show [imath]e^{k+1}≥1+(k+1)+(k+1)^2=3+3k+3k^2[/imath] is a true statement. | 1031730 | Proving quadratic inequalities?
I am trying to prove that [imath]e^{k+1} ≥ 3 + 3k + k^2[/imath] with, [imath]k>2[/imath] WhatI have done so far: What we are trying to prove is that [imath]e^n≥1+n+n^2[/imath] is a true statement. Since [imath]n=3[/imath] holds, this is our base case. Then, the inductive hypothesis is: [imath]e^k≥1+k+k^2[/imath] Hence, I need to show [imath]e^{k+1}≥1+(k+1)+(k+1)^2=3+3k+3k^2[/imath] is a true statement. |
1032281 | Inequalities textbook request
At university I have got a problem set with lots of inequalities. Unfortunately there are no explanations given how to do them. In Highschool we only did very easy inequalities. Therefore I am looking for a resource for inequalities. Especially for more difficult inequalities like [imath]1 \leq z \overline {z} \leq 4 , |\Im(z)|<\Re (z),[/imath] where [imath]z[/imath] is a complex number. I would be glad at any recommendations. | 1032245 | How to solve this Complex inequality system
[imath]1 \leq z \overline {z} \leq 4[/imath] and [imath]|\Im(z)|<\Re (z)[/imath] How can I solve this system of inequalities? ([imath]\Im[/imath] is the imaginary part and [imath]\Re[/imath] is the real part of a complex number). I have tried so far: [imath]z=x+yi[/imath] and then I get [imath]1 \leq x^2+y^2 \leq 4[/imath] and [imath] |y|<x[/imath]. How can I solve this? |
1032430 | [imath]\sum a_n = \sum{(\sqrt{n+1} - \sqrt{n})}[/imath] Prove divergence
[imath]\sum a_n = \sum{(\sqrt{n+1} - \sqrt{n})}[/imath]. I want to show that this diverges. I think the only way I can do it, is by using the comparison test, and finding a series less than it that diverges, but I can't think of a series that is used often/ easy to prove divergence for that is less than this one for all [imath]n[/imath]. I was origionally thinking something like [imath]\frac{1}{\sqrt{n}}[/imath] or [imath]\frac{1}{n}[/imath], but those don't work | 962301 | Does this series diverge: [imath](\sqrt 2-\sqrt 1)+(\sqrt 3-\sqrt 2)+(\sqrt 4-\sqrt 3)+(\sqrt 5-\sqrt 4)+\dots[/imath]?
[imath](\sqrt 2-\sqrt 1)+(\sqrt 3-\sqrt 2)+(\sqrt 4-\sqrt 3)+(\sqrt 5-\sqrt 4)…[/imath] I have found partial sums equal to natural numbers. The first 3 addends sum to 1. The first 8 sum to 2. The first 15 sum to 3. When the minuend in an addend is the square root of a perfect square, the partial sum is a natural number. So I believe this series to be divergent. Am I right? Have I used correct terminology? How would this be expressed using sigma notation? Is there a proof that this series diverges? |
785 | Is there a general formula for solving 4th degree equations (quartic)?
There is a general formula for solving quadratic equations, namely the Quadratic Formula. For third degree equations of the form [imath]ax^3+bx^2+cx+d=0[/imath], there is a set of three equations: one for each root. Is there a general formula for solving equations of the form [imath]ax^4+bx^3+cx^2+dx+e=0[/imath] ? How about for higher degrees? If not, why not? | 2678815 | Quartic Formula existance
There is a formula for cubic equations and quadratic equations. Is there a formula for the equations in the form [imath]ax^4+bx^3+cx^2+dx+f=0[/imath] I also skipped [imath]e[/imath] because of Euler's constant. |
1031845 | Eigenvectors on complex space
Suppose [imath]U[/imath] is k-dimensional vector space on complex field, and [imath]S\in L(U)[/imath]. Assume [imath]S[/imath] has at least [imath]m[/imath] distinct nonzero eigenvalues. Show that [imath]\def\null{\operatorname{null}} \null (S^{k-m})= \null (S^{k-m+1})[/imath] Here is my approach: Suppose [imath] S \in L(U) [/imath] and k is non-negative integer and [imath]S[/imath] has at least [imath]m[/imath] distinct non-zero eigenvalues. If [imath]S^{k-m} v = 0 [/imath] for all [imath]v \in U[/imath], then [imath]S^{k-m+1} v = S (S^{k-m})=S(0)=0[/imath]. Thus, [imath] \null (S^{k-m}) = \null (S^{k-m+1})[/imath]. Please give me feedback on my answer. If it is wrong, direct me to right track. Thanks in advance! | 1032375 | Null Space of Transformation
I am given that [imath]V[/imath] is n-dimensional vector space over [imath]\mathbb{C}[/imath] and [imath]T \in L(V)[/imath]. And [imath]T[/imath] has least [imath]m[/imath] distinct nonzero eigenvalues. How do I show that [imath]\text{null}(T^{n-m}) = \text{null}(T^{n-m+1})[/imath]? I'm really not sure how to start here so just the first couple steps would be very helpful. Thanks!! |
1033359 | Using [imath]\log[/imath] and [imath]\ln[/imath] in Integration
I found in some integral equations where they use [imath]\log(n)[/imath] and in some other with [imath]\ln(n)[/imath]. Suppose [imath] \int_{n_0}^{\large\frac{n_0}{2}} \frac{1}{n}dn [/imath] Which formula should I use ? [imath] \log(n)\ \mbox{or}\ \ln(n) [/imath] | 987493 | Textbook clarification: [imath]\log = \ln[/imath]
Textbook reads: All logarithms are natural logarithms: [imath]\log = \ln[/imath]. Does this mean [imath]n\log(n) = n\ln(n)[/imath]? |
1033604 | Why is [imath]\sqrt{x}[/imath] a function?
How [imath]\sqrt{x}[/imath] can be a function when [imath]\sqrt{4}[/imath] is equal to [imath]-2[/imath] and [imath]2[/imath]? | 284179 | Why is [imath]y = \sqrt{x-4}[/imath] a function? and [imath]y = \sqrt{4 - x^2}[/imath] should be a circle
So why is it a function, even though for example [imath]x = 8[/imath]; you'll have [imath]y = +2[/imath] and [imath]y = -2[/imath]. It'll fail the vertical line test. But every textbook considers it as a function. Did I misunderstand something? Edit: Wait how come [imath]y = \sqrt{4 - x^2}[/imath] is a function too when it can be transformed into [imath] y = \sqrt{4-x^2} [/imath] [imath] y^2 = 4- x^2[/imath] [imath] x^2 + y^2 = 4 [/imath] [imath] \frac{x^2}{4} + \frac{y^2}{4} = 1[/imath] which is an equation of a circle and not a function |
1034213 | Inverse matrix as a sum of matrix powers
I have matrix [imath] A\in \mathbb{C}^{n x n}[/imath] and [imath]A[/imath] is invertible. How can I show that coefficients [imath]c_0,...,c_{n-1}[/imath] exist : [imath]A^{-1} = c_0I+c_1A+...+c_{n-1}A^{n-1}[/imath] I tried to solve it first by multiplying the equation above with [imath]A[/imath] which gives: [imath]I = c_0A + c_1A^2 + ... +c_{n-1}A^n[/imath] And I continued then by knowing that every square matrix can be written with Jordan canonical form ([imath]A = VJ_aV^-1[/imath]): [imath]I = c_0VJ_aV^{-1} + c_1VJ_a^2V^{-1}+...+c_{n-1}VJ_a^nV^{-1}[/imath] Which can be written in form: [imath]I = V(c_0J_a+c_1J_a^2+....+c_{n-1}J_a^n)V^{-1}[/imath] After this point my task is to show that the identity matrix can be presented with sum of coefficients and Jordan canonical forms, but I don't have any idea how to do that or even if my approach to this problem is right. | 744378 | Inverse of a matrix is expressible as a polynomial?
Let [imath]A[/imath] be an [imath]n \times n[/imath] matrix. Prove that if A is invertible, then there exists a polynomial [imath]p[/imath], such that [imath]A^{-1}=p(A)[/imath] Thus far: Let [imath]W[/imath] denote the [imath]k[/imath] dimensional A-cyclic subspace spanned by a vector [imath]v[/imath]. Then, [imath]I_n=\sum_{i=0}^{k} a_{i}A^i[/imath] for some scalar [imath]a_i[/imath]. What I want to do is multiply both sides by [imath]A^{-1}[/imath], so that way it just falls into my hands, but Im not sure that is 'legal'. |
1034023 | If [imath]f(z)g(z)=0[/imath] then [imath]f(z)=0[/imath] or [imath]g(z)=0[/imath]
Let [imath]D[/imath] be a domain and let [imath]f,g[/imath] be analytic in [imath]D[/imath]. I need to prove that if [imath]f(z)g(z)=0[/imath] for all [imath]z\in D[/imath], then [imath]f(z)=0[/imath] for all [imath]z\in D[/imath] or [imath]g(z)=0[/imath] for all [imath]z\in D[/imath].This is my answer. Assume there exists [imath]z_0 \in D[/imath] such that [imath]f(z_0)\neq 0[/imath] and [imath]g(z_0)\neq 0[/imath]. Since [imath]f,g[/imath] are continuous in [imath]D[/imath] there exists [imath]r_1,r_2>0[/imath] such that [imath]f(z)\neq 0[/imath] for all [imath]z\in D_{r_1}(z_0)[/imath] and [imath]g(z)\neq 0[/imath] for all [imath]z\in D_{r_2}(z_0)[/imath]. Ler [imath]r=min[/imath]{[imath]r_1,r_2[/imath]}. Then, [imath]f(z)g(z)\neq0,z\in D_r(z_0)[/imath] This is a contradiction. Thus [imath]f(z)=0[/imath] for all [imath]z\in D[/imath] or [imath]g(z)=0[/imath] for all [imath]z\in D[/imath]. Is this correct? Any improvements?Better answers? Thanks | 912950 | If [imath]f,g[/imath] are entire functions and[imath]\ fg\equiv 0[/imath] then either [imath]f \equiv 0[/imath] or [imath]g\equiv0. [/imath]
Let [imath]f,g[/imath] be entire functions such that [imath]g \not\equiv 0.[/imath] If [imath]fg\equiv0[/imath] in [imath]\mathbb{C},[/imath] could anyone advise me how to show [imath]f \equiv0[/imath] in [imath]\mathbb{C} \ ?[/imath] Thank you. |
1034437 | Question about logical implication [imath]P\to Q[/imath]
Having come across mathematical logic, a question suddenly came into my mind. We commonly know that the truth value of [imath]P\to Q[/imath] given as: [imath]\begin{matrix} P&Q&P \Rightarrow Q \\ T&T&T\\ T&F&F\\ F&T&T\\ F&F&T \end{matrix}[/imath] I do not understand how [imath]P\to Q[/imath] holds when P is false. For example, let me propose a statement: Let [imath]n[/imath] be a nonzero real number. P: [imath]n[/imath] is a rational number Q: [imath]n\cdot0=k[/imath], where [imath]k[/imath] is a nonzero real number. We obviously know that Q is a false statement. Hence, I omit the case when Q is true. [imath]\begin{matrix} P&Q&P\Rightarrow Q \\ T&F&F\\ F&F&T \end{matrix}[/imath] P True: [imath]n[/imath] is a rational number; P False: [imath]n[/imath] is not a rational number; [imath]n[/imath] is an irrational number. How is it that [imath]P\to Q[/imath] holds true when P is false? | 70736 | In classical logic, why is [imath](p\Rightarrow q)[/imath] True if [imath]p[/imath] is False and [imath]q[/imath] is True?
Provided we have this truth table where "[imath]p\implies q[/imath]" means "if [imath]p[/imath] then [imath]q[/imath]": [imath]\begin{array}{|c|c|c|} \hline p&q&p\implies q\\ \hline T&T&T\\ T&F&F\\ F&T&T\\ F&F&T\\\hline \end{array}[/imath] My understanding is that "[imath]p\implies q[/imath]" means "when there is [imath]p[/imath], there is q". The second row in the truth table where [imath]p[/imath] is true and [imath]q[/imath] is false would then contradict "[imath]p\implies q[/imath]" because there is no [imath]q[/imath] when [imath]p[/imath] is present. Why then, does the third row of the truth table not contradict "[imath]p\implies q[/imath]"? If [imath]q[/imath] is true when [imath]p[/imath] is false, then [imath]p[/imath] is not a condition of [imath]q[/imath]. I have not taken any logic class so please explain it in layman's terms. Administrative note. You may experience being directed here even though your question was actually about line 4 of the truth table instead. In that case, see the companion question In classical logic, why is [imath](p\Rightarrow q)[/imath] True if both [imath]p[/imath] and [imath]q[/imath] are False? And even if your original worry was about line 3, it might be useful to skim the other question anyway; many of the answers to either question attempt to explain both lines. |
1034842 | proof of divisibility of n(n+1)(2n+1) by 6
How can I prove that [imath]n(n+1)(2n+1)[/imath] (where [imath]n[/imath] is a positive integer) is divisible by 6? As the product is even it is divisible by 2. But I do not know how to prove that it is divisible by 3 | 878430 | How to show [imath]n(n+1)(2n+1) \equiv 0 \pmod 6[/imath]?
I've been asked to show that: [imath]n(n+1)(2n+1) \equiv 0 \pmod 6[/imath] I found in a previous question that: [imath]n(n+1)[/imath] was divisible by [imath]2[/imath] and resulted in an even number e.g [imath]n(n+1) \equiv 0 \pmod 2[/imath] so I figured I needed to find: [imath](2n+1) \equiv 0 \pmod 3[/imath] in order to complete [imath]n(n+1)(2n+1) \equiv 0 \pmod 6[/imath] but I am unsure on how to find [imath](2n+1) \equiv 0 \pmod 3[/imath] Is this the right way to find the mod 6 and if so could you tell me how I could find [imath](2n+1) \equiv 0 \pmod 3?[/imath] |
1035213 | Does [imath]\sum |\sin n| / n[/imath] converge?
How can I prove if the following series converges? [imath]\sum_{n\geqslant1} \frac{|\sin n|}{n}[/imath] I can't use differential or integral calculus. I've tried using Dirichlet and Cauchy tests, but they didn't get me anywhere. | 1034059 | Covergence test of [imath]\sum_{n\geq 1}{\frac{|\sin n|}{n}}[/imath]
I need to prove that [imath]\sum_{n\geq 1}{\frac{|\sin n|}{n}}[/imath] is convergent. How should I do it? |
1035179 | Find the leading order uniform approximation to the boundary value problem [imath]\epsilon y''+y'\sin x+y\sin 2x = 0[/imath]?
[imath]\epsilon y''+y'\sin x+y\sin 2x = 0[/imath] with boundary conditions [imath]y(0)=\pi[/imath] and [imath]y(\pi)=0[/imath] as [imath]\epsilon \rightarrow 0[/imath]. I don't know how to find out where the boundary layer is? I thought initially it was at x=0 but this just leads to the outer solution of [imath]y=Ae^{-2\sin x}[/imath] if i've done it correctly and applying the boundary conditions means A=0 which means its not a boundary layer. I'm so lost. Any hints? Thanks! | 1035335 | Find the leading order uniform approximation when the conditions are not [imath]0[/imath]
[imath]\epsilon y''+y'\sin x+y\sin 2x = 0[/imath] with boundary conditions [imath]y(0)=\pi[/imath] and [imath]y(\pi)=0[/imath] as [imath]\epsilon \rightarrow 0[/imath]. I don't know how to find out where the boundary layer is? I thought initially it was at [imath]x=0[/imath], but this just leads to the outer solution of [imath]y=Ae^{-2\sin x}[/imath] if I've done it correctly and applying the boundary conditions means [imath]A=0[/imath] or [imath]A=\pi[/imath]. Am I right so far? What do I do now? I'm so lost. Any hints? Thanks! |
1035473 | Nth Derivative of the function
Find the [imath]n^{th}[/imath] derivative of [imath]f(x) = e^x\cdot x^m[/imath] If i am not wrong i have following [imath]1^{st}[/imath] Derivative: [imath]e^x\cdot m \cdot x^{m-1} + x^m\cdot e^x[/imath] [imath]2^\text{nd}[/imath] Derivative: [imath]e^x\cdot m \cdot (m-1)\cdot x^{m-2} + 2 \cdot e^x\cdot m \cdot x^{m-1} + x^m\cdot e^x[/imath] [imath]3^\text{rd}[/imath] Derivative: [imath]e^x\cdot m \cdot (m-1) \cdot (m-2)\cdot x^{m-3} + 3 \cdot e^x \cdot m \cdot (m-1) \cdot x^{m-2} + 3 \cdot e^x \cdot m \cdot x^{m-1} + \cdot x^m \cdot e^x[/imath] From here how do I calculate the [imath]n^{th}[/imath] derivative? Thanks. :) | 1035189 | Nth Derivative of a function
Find the [imath]n^{th}[/imath] derivative of [imath]f(x) = e^x\cdot x^n[/imath] If i am not wrong i have following [imath]1^{st}[/imath] Derivative: [imath]e^x\cdot n\cdot x^{n-1} + x^n\cdot e^x[/imath] [imath]2^\text{nd}[/imath] Derivative: [imath]e^x\cdot n\cdot (n-1)\cdot x^{n-2} + 2 \cdot e^x\cdot n\cdot x^{n-1} + x^n\cdot e^x[/imath] [imath]3^\text{rd}[/imath] Derivative: [imath]e^x\cdot n\cdot (n-1)\cdot (n-2)\cdot x^{n-3} + 3 \cdot e^x\cdot n \cdot (n-1)\cdot x^{n-2} + 4 \cdot e^x \cdot n \cdot x^{n-1} + 2 \cdot x^n\cdot e^x[/imath] From here how do I calculate the [imath]n^{th}[/imath] derivative? Thanks. :) |
991885 | Show that [imath]\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}[/imath] for positive [imath]a,b,c[/imath]
Show that [imath]\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}[/imath], if [imath]a,b,c[/imath] are positive. Well, I got that [imath]bc(a+b+c)+ac(a+b+c)+ab(a+b+c)\geq9abc[/imath]. | 193771 | Prove that [imath]\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0[/imath]
Please help me for prove this inequality: [imath]\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0[/imath] |
1035763 | Indefinite integrals with natural logs
I know the integral of [imath]\frac{1}{x}[/imath] is [imath]\log(x)[/imath] but I'm not sure how to solve this problem, any help would be appreciated: [imath] \int^{3}_{2} \frac{1}{x \ln x} [/imath] I think I need to substitute [imath]x\ln x[/imath] but from there I get stuck. | 634276 | Evaluating [imath]\int \frac{\operatorname dx}{x\log x}[/imath]
How to integrate [imath]\frac{1}{x\log x}[/imath]? Could you give me some ideas on how to integrate this? thanks. i've tried setting [imath]u=(\log x)^{-1}[/imath]. [imath]\dfrac{\mathrm du}{\mathrm dx} = x^{-1}[/imath] But it didnt work... |
1035877 | I need someone who is willing to actually answer this volume of revolution question!
I'd really appreciate if someone could help me so I could get going on these problems, but this is confusing me... and it's been holding me up for the last couple hours. How can I find the volume of the solid when revolving the region bounded by [imath]y=1-\frac{1}{2}x[/imath], [imath]y=0[/imath], and [imath]x=0[/imath] about the line [imath] x=-1[/imath]? How could I set it up? I'd REALLY appreciate if someone could take the time to answer this so I don't spend all night on one problem. | 1035761 | Volume of Revolution. Am I doing it right?
How do I find the volume when revolving the region bounded by [imath]y=1-\frac{1}{2}x[/imath], [imath]y=0[/imath], and [imath]x=0[/imath] about the line [imath] x=-2[/imath]? Would it be [imath]x=2-2y[/imath] so radius [imath]r(y) = 2-2y -(-2) [/imath] => [imath]r(y)= 4-2y[/imath] [imath]π\int (4-2y)^2 dy[/imath] ? What would be my limits of integration? Would it be from 0 to 2?? yes? |
1035984 | Show that the [imath]n[/imath]th derivative of [imath]f(x)[/imath] is zero for all [imath]n \geq 0[/imath].
Let [imath]f:\mathbb{R} \rightarrow \mathbb{R}[/imath] be defined by [imath]f(x)=e^{-\frac{1}{x^2}}[/imath] for [imath]x \neq 0[/imath] and [imath]f(x)=0[/imath] for [imath]x=0[/imath]. I want to show that [imath]f^{(n)}(0)=0[/imath] for all [imath]n \ge 0[/imath]. That is, the [imath]n[/imath]th derivative at [imath]x=0[/imath] is zero for all [imath]n \ge 0[/imath]. Any hints on how to get started? Thanks. | 629156 | Showing that this function is infinitely differentiable
Show, that the function [imath] \mathcal E: \mathbb R \to \mathbb R: x \mapsto \begin{cases} \exp(-\frac{1}{x^2}), & \text{if x $\neq$ 0}, \\ 0, & \text{otherwise}, \end{cases} [/imath] is infinitely differentiable and that [imath]\frac{d^k\mathcal E}{dx}(0) = 0[/imath] for all [imath]k \in \mathbb N[/imath]. We just introduced differentiation, so the solution should not contain very advanced techniques to solve this. We also got the tip it should be done by induction. @fgp I found that it is [imath]f^{(n)}(x) = P_n \left(\frac1x\right)e^{-\frac1{x^2}}[/imath] where [imath]P_n[/imath] is a polynomial with integer coefficients. I could also do the initial step for the induction: For [imath]n=1:[/imath] [imath]f'(x) = \frac2{x^2}e^{-\frac1{x^2}} = P_1 \left(\frac1t\right) e^{-\frac1{x^2}}[/imath] where [imath]P_1(x)=2x^2.[/imath] I am stuck at the induction step: [imath]f^{(n+1)}(x) = [/imath]..... |
1036004 | If [imath]a \equiv b \bmod n[/imath], then [imath]\gcd(a, n)= \gcd(b,n)[/imath]
Again, I have been stuck in a problem of modular arithmetic. Given that [imath]a,b, n \in \mathbb Z [/imath] and [imath]n>0[/imath] and [imath]a \equiv b \bmod n[/imath]. Show that [imath]\gcd(a, n)= \gcd(b,n)[/imath]. | 25648 | Show that if [imath]a \equiv b \pmod n[/imath], [imath]\gcd(a,n)=\gcd(b,n)[/imath]
My problem is how to somehow relate the the gcd and congruence. I know that [imath](a,b) = ax + by[/imath]. I also know that [imath]a \equiv b \pmod n[/imath] means [imath]n\mid a-b[/imath]. Any hints? Thanks! |
576689 | Problem involving ideals contained in prime ideals
So far I have used the hint and have done the following: Let [imath]i=1,2[/imath]. Suppose that [imath]I\subseteq P_1\cup P_2[/imath], but that [imath]I[/imath] is not properly contained in [imath]P_i[/imath] for any [imath]i[/imath]. Then there is an [imath]a_i[/imath] such that [imath]a_i \in I\setminus P_i[/imath]. Then [imath]a_1+a_2 \in I\setminus (P_1\cup P_2)= \emptyset[/imath]. This is not possible, so [imath]I\subseteq P_1[/imath] or [imath]I\subseteq P_2[/imath]. Now I want to extend this result for [imath]i=n[/imath], where [imath]n[/imath] is a positive integer greater than or equal to [imath]2[/imath]. I want to try a similar approach where I let [imath]a_j \in I\setminus \left( \displaystyle \bigcup_{i=1, i\neq j}^n P_i \right)[/imath] so then [imath]a_j\in P_j[/imath], but I have absolutely no idea what to do from there. Please help me! Thank you! | 1016551 | Ideal contained in a finite union of prime ideals
Let [imath]I \subset R[/imath] be an ideal and [imath]P_i[/imath] [imath](i=\{1,...,n\})[/imath] prime ideals with [imath]I\subseteq\bigcup_{i=1}^nP_i[/imath]. Prove that then [imath]I[/imath] is contained in one [imath]P_i[/imath]. I don't know how to show this because I don't have any approach. So I am looking for something to start with or something like a sequence of tips (Since I know this is a large proof) Thanks in advance! |
1036717 | Showing a Group [imath]G[/imath] is not Simple
Let [imath]G[/imath] be a finite group of order [imath]pq[/imath], where [imath]p,q[/imath] are distinct prime numbers. Show that [imath]G[/imath] is not simple. Here is my attempt: [imath]|G|=pq[/imath]. If [imath]G[/imath] is not simple, then it has non-trivial subgroups, i.e., subgroups other than the identity and itself. Let [imath]H[/imath] be a subgroup of order [imath]p[/imath]. By Lagrange's Theorem, [imath]|H| \mid |G| \Rightarrow p \mid pq[/imath]. Similarly, let [imath]K[/imath] be a subgroup of order [imath]q[/imath] so that [imath]|K| \mid |G| \Rightarrow q \mid pq[/imath]. Hence, [imath]G[/imath] is not simple. I was hoping somebody could verify my proof and point out any errors. As always, any help or advice is greatly appreciated. | 222853 | Group of order pq is not simple
Is the following correct way of showing that there is no simple group of order [imath]pq[/imath] where [imath]p[/imath] and [imath]q[/imath] are distinct primes? If [imath]|G|=n=pq[/imath] then the only two Sylow subgroups are of order [imath]p[/imath] and [imath]q[/imath]. From Sylow's third theorem we know that [imath]n_p | q[/imath] which means that [imath]n_p=1[/imath] or [imath]n_p=q[/imath]. If [imath]n_p=1[/imath] then we are done (by a corollary of Sylow's theorem) If [imath]n_p=q[/imath] then we have accounted for [imath]q(p-1)=pq-q[/imath] elements of [imath]G[/imath] and so there is only one group of order [imath]q[/imath] and again we are done. Is that correct? |
1036613 | Faithful representation of a finite group over reals
https://mathoverflow.net/questions/77325/finite-groups-with-faithful-real-two-dimensional-representation Suppose [imath]\pi :G \rightarrow SL_2(\mathbb{R})[/imath] is a faithful representation of a finite group [imath]G[/imath], then I want to prove that [imath]G[/imath] must be cyclic. I knew that every finite subgroup of [imath]GL_2(\mathbb{R})[/imath] is conjugate to a subgroup of the orthogonal group [imath]O_2(\mathbb{R})[/imath] and that [imath]G \cong \pi (G) \leq SL_2(\mathbb{R})[/imath] but I don't know how to proceed. | 200354 | Faithful representation implies group is cyclic
Suppose there is a faithful representation [imath]\rho:G\to SL_2(\mathbb{R})[/imath]. Prove that [imath]G[/imath] is cyclic. I know there has to be something special about its representation being special (no pun intended) because e.g. the Klein 4 group has a non-special representation. Also it has to be important that it's in two dimensions, because [imath]SO(3)[/imath] contains non-cyclic groups. Apart from that, I haven't really made any progress. Any hints? |
1036879 | Inequality about squareroots
If [imath]a,b\geq 0[/imath] show that [imath]\left| \sqrt{a}-\sqrt{b}\right|\leq\sqrt{\left|a-b\right|}[/imath]. WLOG we can assume that [imath]a\geq b[/imath]. If one of them is [imath]0[/imath] this is trivial. So assume none of them is [imath]0[/imath]. Now, [imath]\left| \sqrt{a}-\sqrt{b}\right|\leq\sqrt{\left|a-b\right|} \iff a-2\sqrt{ab}+b\leq a-b \iff -2\sqrt{ab}\leq -2b \iff \sqrt{ab}\geq b \iff ab\geq b^2\iff a\geq b[/imath] Is this proof correct? | 1035430 | Inequality [imath] \vert \sqrt{a}-\sqrt{b} \vert \leq \sqrt{ \vert a -b \vert } [/imath]
I have the following inequality on my class notes that I haven't been able to prove, I was even wondering if it is actually true: [imath] \forall a,b \in \mathbb{R}^{\ge0} \left( \left| \sqrt{a}-\sqrt{b} \right| \leq \sqrt{ \vert a -b \vert}\right) [/imath] Thanks |
1035069 | Definite integral of general polylogarithm
[imath]\int_{0}^{1} Li_k(x) dx[/imath] [imath]Li_k(x) = \sum_{n=1}^{\infty} \frac{x^n}{n^k}[/imath] [imath]\int_{0}^{1} Li_k(x) \,dx = \sum_{n=1}^{\infty} \int_{0}^{1} \frac{x^n}{n^k} \,dx[/imath] From Fubini's theorem, I suppose we were allowed to interchange. [imath]\int_{0}^{1} Li_k(x) \,dx = \sum_{n=1}^{\infty} [\frac{x^{n+1}}{(n+1)n^k}]_{0}^{1}[/imath] [imath]\int_{0}^{1} Li_k(x) \,dx = \sum_{n=1}^{\infty} \frac{1}{(n+1)n^k}[/imath] Let [imath]S = \displaystyle \sum_{n=1}^{\infty} \frac{1}{(n+1)n^k}[/imath] I am having the big trouble in evaluating the sum [imath]S[/imath]. | 981650 | Closed-form of [imath]\int_0^1 \operatorname{Li}_p(x) \, dx[/imath]
While I've studied integrals involving polylogarithm functions I've observed that [imath]\int_0^1 \operatorname{Li}_p(x) \, dx \stackrel{?}{=} \sum_{k=2}^p(-1)^{p+k}\zeta(k)+(-1)^{p+1},\tag{1}[/imath] for any integer [imath]p\geq2[/imath]. Here [imath]\zeta[/imath] is the Riemann zeta function. After that I have three questions. [imath]1^\text{st}[/imath] Question. Is [imath](1)[/imath] true? If it is, how could we prove it? [imath]2^\text{nd}[/imath] Question. If it's a well-known result could you give any reference? [imath]3^\text{rd}[/imath] Question. I think there is also a similar closed-form of [imath]\int_0^b \operatorname{Li}_p(x) \, dx[/imath], for any integer [imath]b \geq 1[/imath]. What is the closed-form of this integral? |
1037918 | Suppose that [imath]N_1[/imath] is a normal subgroup of [imath]G_1[/imath]. Is the image [imath]f(N_1)[/imath] of [imath]N_1[/imath] a normal subgroup of [imath]G_2[/imath]?
Let [imath]f:G_1 \to G_2[/imath] be a homomorphism between multiplicative groups. Suppose that [imath]N_1[/imath] is a normal subgroup of [imath]G_1[/imath]. Is the image [imath]f(N_1)[/imath] of [imath]N_1[/imath] a normal subgroup of [imath]G_2[/imath]? | 532094 | the image of normal subgroups
I want to find an example of a homomorphism [imath]f:G\to H[/imath] such that [imath]A[/imath] is a normal subgroup of [imath]G,[/imath] but [imath]f(A)[/imath] is not so in [imath]H.[/imath] I know that if [imath]f[/imath] was onto [imath]f(A )[/imath] must be normal , but otherwise i want to find an example! |
1038037 | Is [imath]\mathbb{C}\bigotimes_\mathbb{R}\mathbb{C}\simeq \mathbb{C}\bigotimes_{\mathbb{C}}\mathbb{C}[/imath]?
I'm trying to see if for several cases changing the ring in a tensor product affects the result or doesn't. Now I'm trying to prove [imath]\mathbb{C}\bigotimes_\mathbb{R}\mathbb{C}\simeq \mathbb{C}\bigotimes_{\mathbb{C}}\mathbb{C}[/imath] if it's true, or to show why it isn't. I've been unable to find an isomporphism between those two, but I don't know how would I proceed in order to show that there is no possible function that could define one. | 431657 | Are [imath]\mathbb{C} \otimes _\mathbb{R} \mathbb{C}[/imath] and [imath]\mathbb{C} \otimes _\mathbb{C} \mathbb{C}[/imath] isomorphic as [imath]\mathbb{R}[/imath]-vector spaces?
Are [imath]\mathbb{C} \otimes _\mathbb{R} \mathbb{C}[/imath] and [imath]\mathbb{C} \otimes _\mathbb{C} \mathbb{C}[/imath] isomorphic as [imath]\mathbb{R}[/imath]-vector spaces? I am having a very hard time at digesting tensor products and I do not know how to "compare" tensor products over different rings. My hunch is that they are not isomorphic. It is easy to see that [imath]\mathbb{C} \otimes _\mathbb{R} \mathbb{C}[/imath] is an [imath]\mathbb{R}[/imath]-vector space of dimension [imath]4[/imath]. I suspect that [imath]\mathbb{C} \otimes _\mathbb{C} \mathbb{C}[/imath] is also an [imath]\mathbb{R}[/imath]-vector space but of lower dimension, but I have no idea how to show this or if indeed this intuition is correct. Thank you. |
1038172 | Compute limit [imath]\lim_{n\rightarrow\infty}\frac{1*3*5*...*(2n-1) }{ 2*4*6*...*(2n)}=0[/imath]
[imath]\lim_{n\rightarrow\infty}\left( \frac{1*3*5*...*(2n-1) }{ 2*4*6*...*(2n)}\right)^3=0[/imath] Having products at fractions I cant figure out how to calculate this limit. | 1025630 | To show for following sequence [imath]\lim_{n \to \infty} a_n = 0[/imath] where [imath]a_n[/imath] = [imath]1.3.5 ... (2n-1)\over 2.4.6...(2n)[/imath]
How can I show [imath]\lim_{n \to \infty} a_n = 0[/imath] [imath]a_n = {1.3.5 ... (2n-1)\over 2.4.6...(2n)}[/imath] I have shown that [imath]a_n[/imath] is monotonically decreasing. I thought to shown sequence is bounded from below then it automatically would converge and hence my question will be solve. But I'm unable to show its boundedness... Or there maybe another method to prove this. Thanks |
1038784 | If [imath]\sum{a_k}[/imath] converges, then [imath]\lim ka_k=0[/imath].
I want to prove the following statement: Suppose that [imath]\displaystyle\sum_{k=1}^{\infty}a_k[/imath] converges, where [imath](a_k)_{k\in\mathbb{N}}\subseteq\mathbb{R}[/imath] is monotone. Then [imath]\displaystyle\lim_{k\to\infty}ka_k=0[/imath]. I believe we have several cases. For example, if [imath](a_k)_{k\in\mathbb{N}}[/imath] is monotone increasing and there exists [imath]k[/imath] such that [imath]a_k>0[/imath], then obviously [imath]\displaystyle\sum_{k=1}^{\infty}a_k[/imath] is not convergent. Then, we could conclude that if some [imath]a_k>0[/imath] then we can suppose that [imath](a_k)_{k\in\mathbb{N}}[/imath] is monotone decreasing. By the same argument, we can conclude that if some [imath]a_k<0[/imath] then [imath](a_k)_{k\in\mathbb{N}}[/imath] must be monotone increasing. So, I believe we only need to take care of the case where [imath]a_k\ge 0[/imath] for each [imath]k\in\mathbb{N}[/imath] and [imath](a_k)_{k\in\mathbb{N}}[/imath] is monotone decreasing (the other case would be symmetric). Any hint to prove this? I have been thinking a lot ot time... Thanks. | 383769 | If [imath]x_{n}[/imath] is decreasing and [imath]\sum x_{n}[/imath] converges, then prove that [imath]\lim nx_{n} = 0[/imath]
I need a little hint in a proof on convergence of a sequence. The problem I have is: Suppose [imath](x_n)[/imath] is a monotone decreasing sequence of real numbers such that [imath]\sum x_n[/imath] converges, prove that: [imath]\lim_{n \to \infty} nx_n=0[/imath] My idea is: proving this fact aims to prove that given [imath]\epsilon >0[/imath] there's some natural [imath]n_0 \in \mathbb{N}[/imath] such that if [imath]n >n_0[/imath] we have [imath]|nx_n|<\epsilon[/imath]. Now, I know that [imath]\sum x_n[/imath] converges, so that given [imath]\epsilon'>0[/imath] there's some [imath]k_0 \in \mathbb{N}[/imath] such that if [imath]k > k_0[/imath] we have: [imath]\left|\sum_{i=1}^{k}x_i - S\right|<\epsilon'[/imath] Where [imath]S = \sum x_n[/imath]. Now, since [imath](x_n)[/imath] is monotone decreasing we know that we must have [imath]x_1 > \cdots > x_k[/imath] so that the sum of all [imath]x_i[/imath] should be less or equal to [imath]k x_k[/imath]. So I know that I have: [imath]\left|\sum_{i=1}^{k}x_i - S\right|\leq\left|\sum_{i=1}^{k}x_i\right|+|S|\leq|kx_k|+|S|[/imath] I feel that the proof will come from this, however I'm stuck at this point. Can someone give just a little hint on how to proceed from here? Thanks very much in advance for your help. |
1038440 | Finding the roots of a polynomial on a complex plane
I use an online calculator in order to calculate [imath]x^5-1=0[/imath] I get the results x1=1 x2=0.30902+0.95106∗i x3=0.30902−0.95106∗i x4=−0.80902+0.58779∗i x5=−0.80902−0.58779∗i I know that this is the correct answer because my roots have to be on the complex plane but I do not understand how can I get the results! | 39202 | How to find roots of [imath]X^5 - 1[/imath]?
How to find roots of [imath]X^5 - 1[/imath]? (Or any polynomial of that form where [imath]X[/imath] has an odd power.) |
1038851 | Deducing if the series converges.
[imath]\displaystyle \sum\limits_{}^{} \dfrac{1}{k(ln(k)^2)}[/imath] Integral test [imath]\int_{} \frac {1}{u^2} du = \int u^{-2} du = \frac {-1}{u} = \frac{-1}{\ln k} +c[/imath] | 240166 | Proving the convergence of [imath]\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2}[/imath]
I am attempting to prove if the below expression converges or diverges: [imath]\sum_{n=2}^{\infty} \frac{1}{n (\ln n)^2}[/imath] I decided to try using the limit convergence test. So then, I set [imath]f(n) = \frac{1}{n (\ln n)^2}[/imath] and [imath]g(n) = \frac{1}{n}[/imath] and did the following: [imath] \lim_{n \to \infty} \frac{f'(n)}{g'(n)} = \lim_{n \to \infty} \frac{-\frac{\ln(n) + 2}{n^2 (\ln n)^3}}{-\frac{1}{n^2}} = \lim_{n \to \infty} \frac{\ln(n) + 2}{n^2 (\ln n)^3} \cdot n^2 = \lim_{n \to \infty} \frac{1}{(\ln n)^2} + \frac{2}{(\ln n)^3} = 0 [/imath] I know that [imath]\sum \frac{1}{n}[/imath] diverges due to properties of harmonic series, and so concluded that my first expression [imath]\frac{1}{n (\ln n)^2}[/imath] also must diverge. However, according to Wolfram Alpha, and as suggested by this math.SE question, the expression actually converges. What did I do wrong? |
1039007 | Proving uniqueness of [imath]e[/imath]
Let's define [imath]e[/imath] as the number [imath]a[/imath] such that [imath]\frac {d}{dx} a^x = a^x[/imath]. I'm trying to prove that this [imath]a[/imath] has to be [imath]e[/imath]. I don't see any way of proceeding from here except by the limit definition (I'm not assuming I know what the [imath]\ln[/imath] function is, or else there'd be a much easier definition of [imath]e[/imath] to be had). [imath]\frac {d}{dx} a^x=\lim_{h\to 0} \frac {a^{x+h}-a^x}{h}=\lim_{h\to 0} a^x\frac{a^h-1}{h}=a^x\left(\lim_{h\to 0} \frac{a^h-1}{h}\right)[/imath] So clearly [imath]e[/imath] must be the number that makes that limit on the far right equal to [imath]1[/imath]. I'm not sure how to evaluate this. Using L'Hopital's rule, I just get [imath]\lim_{h\to 0} \frac {d}{dh} a^h[/imath], which doesn't particularly help. So my question: How can I prove that there is a unique number such that [imath]\lim_{h\to 0} \frac {a^h-1}{h}=1[/imath]? | 925454 | Proving that a definition of e is unique
We can define [imath]e[/imath] as the number such that [imath]\lim_{h \to 0} \frac{e^h-1}{h}=1[/imath]. However, of course we can only define [imath]e[/imath] this way if it is unique, i.e., there is no other value [imath]c[/imath] for which that is a true statement. Could someone prove this uniqueness for me? I am tutoring a calculus student and he asked me something to this effect and I couldn't see why. Thanks! |
1038823 | Steps to construct the Field of fractions of Gaussian Integers [imath]\mathbb{Z}[i][/imath]
i don't know how to construct such field [imath]\mathbb{Q[i]} [/imath] from [imath]\mathbb{Z[i]}[/imath]. I know the following: [imath](a+bi,c+di)\sim (m+ni,r+si)[/imath] iff [imath](a+bi)(r+si)=(c+di)(m+ni)[/imath] is the equivalence relation and if [imath]\frac ab+\frac cdi\in\Bbb Q(i)[/imath], then: [imath]\frac ab+\frac cdi = \frac {ad+bci}{bd},[/imath] where both numerator and denominator are Gaussian Integers. On the other hand, I know that i need to use the universality of field of fractions, but I'm not clear how to construct such field. | 1037686 | Quotient field of gaussian integers
Let [imath]D[/imath] be the set of all gaussian integers of the form [imath]m+ni[/imath] where [imath]m,n \in Z[/imath]. Carry out the construction of the quotient field [imath]Q[/imath] for this integral domain. Show that this quotient field is isomorphic to the set of all complex numbers of the form [imath]a + bi[/imath] where [imath]a,b[/imath] are rational numbers. Well I know how to show that the quotient field is isomorphic, I just don't know how to construct such a quotient field. In the case of integers, we have the quotient field [imath][a,b] = [c,d][/imath] if and only if [imath]ab = bc[/imath], so I am guessing maybe the quotient field for the gaussian integers should follow that, [imath][m+ni,r+si] = [m_1 + n_1i,r_1+s_1i][/imath] if and only if [imath](m+ni)(r_1+s_1i)=(r+si)(m_1+n_1i)[/imath], is that right? |
1039126 | Show that [imath]x\otimes x+2\otimes2[/imath] is not an elementary tensor in [imath]I\otimes_{A}I[/imath]
Let [imath]A=\mathbb{Z}[x][/imath] and [imath]I=(2,x)\lhd A.[/imath] Show that [imath]x\otimes x+2\otimes2[/imath] is not an elementary tensor in [imath]I\otimes_{A}I[/imath]. I have [imath]I\otimes_A I = \frac{L_A(I\times I)}{T}[/imath] where [imath]T[/imath] is the submodule generated by the elementary tensors [imath]\{(am,n)-a(m,n);(m,an)-a(m,n);(m,n)+(m,n')-(m,n+n');(m,n)+(m',n)-(m+m',n)\}[/imath] I haven't been able to to much work here. I have a couple of ideas: showing that [imath]x\otimes x+2\otimes2[/imath] can't be written in one of the form above to prove is not elementary. A possible alternative would show that the tensor given is can be written as a linaer combination of elementary tensors (and is this always true? question arise because not every tensor consists only of elementary tensors). | 823642 | [imath]2 \otimes_{R} 2 + x \otimes_{R} x[/imath] is not a simple tensor in [imath]I \otimes_R I[/imath]
This question is related to Properties of the element [imath]2 \otimes_{R} x - x \otimes_{R} 2[/imath]. Another exercise of Dummit-Foote is to show that Let [imath]I = (2, x)[/imath] be the ideal generated by [imath]2[/imath] and [imath]x[/imath] in the ring [imath]R = \mathbb{Z}[x][/imath] . Show that the element [imath]2 \otimes_R 2 + x \otimes_R x[/imath] in [imath]I \otimes_R I[/imath] is not a simple tensor, i.e., cannot be written as [imath]a \otimes_R b[/imath] for some [imath]a, b \in I [/imath]. I know how to show that a tensor is not simple in tensor products like [imath]M \otimes_R N[/imath] where [imath]M[/imath] and [imath]N[/imath] are free [imath]R[/imath]-modules, but here we can't use this fact; so how to do it ? |
1039254 | Let [imath]X = \mathbb{R}[/imath] and [imath]Y = \left \{ x \in \mathbb{R}\mid x ≥ 1 \right \}[/imath]. Define [imath]G : X → Y[/imath] by [imath]G(x) = e^{x^2}[/imath]. Prove that [imath]G[/imath] is onto.
Let [imath]X = \mathbb{R}[/imath] and [imath]Y = \left \{ x \in \mathbb{R}\mid x ≥ 1 \right \}[/imath]. Define [imath]G : X → Y[/imath] by [imath]G(x) = e^{x^2}[/imath]. Prove that [imath]G[/imath] is onto. | 1039141 | The function [imath]G: x \mapsto 2^{x^2}[/imath] maps [imath]\mathbb{R}[/imath] onto [imath]\{ x \in \mathbb{R} : x \geq 1 \}[/imath]
Let [imath]X = \mathbb{R}[/imath] and [imath]Y = \{x \in \mathbb{R} :x ≥ 1\}[/imath], and define [imath]G : X → Y[/imath] by [imath]G(x) = e^{x^2}.[/imath] Prove that [imath]G[/imath] is onto. Is this going along the right path and if so how do get the function to equal [imath]y[/imath]? [imath]G: \mathbb{R} \to\mathbb{N}_1[/imath]. Let $y[imath]\in [/imath]\mathbb{N_1}$. claim: [imath]\sqrt{\ln y}[/imath] maps to [imath]y[/imath]. Does [imath]\sqrt{\ln y}[/imath] belong to [imath]\mathbb{N_1}[/imath]? Yes because [imath]y \in \mathbb{N_1}[/imath], [imath]G( \sqrt{\ln y})=e^{(\sqrt{\ln y})^2}[/imath]. |
1038704 | Product Measures Counter Example
I am aware that the product of two Lebesgue [imath]\sigma[/imath] algebra is not a Lebesgue [imath]\sigma[/imath] algebra. Could someone illustrate this with an example. | 679883 | Is the n-dim lebesgue measure the product of the lebesgue measure?
Let [imath]\mathfrak{M}[/imath] be the sigma-algebra of the n-dimensional Lebesgue measurable sets (Completion of translation invariant measure having 1 for [imath][0,1]^n[/imath]) Let [imath]\Sigma[/imath] be the sigma-algebra of the Lebesgue-measurable sets in [imath]\mathbb{R}[/imath]. Is [imath]\mathfrak{M}[/imath] the sigma-algebra generated by [imath]\{\prod_{i=1}^n p(i) : p\in \Sigma^n\}[/imath]? |
1039443 | How to use Cauchy sequence on [imath]|a_{n+2}-a_{n+1}| \le q|a_{n+1}-a_n|[/imath]
I want to apply Cauchy prinicpal on the following question: If exists constant [imath]0<q<1[/imath] such that [imath]|a_{n+2}-a_{n+1}| \le q|a_{n+1}-a_n|[/imath] for any [imath]n[/imath] then [imath]a_n[/imath] converges. Now if I just think about it what I see is that as [imath]n[/imath] goes up the distance between his members becoming smaller and smaller since even if i take the last 2 elements and multiple them by a small number ([imath]<1[/imath]) the distance is still smaller. My conclusion is that the sequence will probably converge because of that, but my method is not math is just some logic which may be wrong. I feel like Cauchy can help me here but I have no idea how to apply it | 602887 | Prove that [imath] \left(a_{n}\right)_{n=1}^{\infty} [/imath] converges when [imath]|a_{n+1}-a_{n}| for 0[/imath]
I'm stuck on a homework question, and could really use some help. Here is said question: "Assume that for every [imath]n[/imath] the following occurs: [imath]|a_{n+1}-a_{n}|<q|a_{n}-a_{n-1}|[/imath] when [imath] 0<q<1 [/imath] Prove that the series [imath] \left(a_{n}\right)_{n=1}^{\infty} [/imath] converges. Hint: use Cauchy sequence" OK I'm really stuck on this one, I don't even have a clue where to begin. Any help is appreciated. Thank you! |
1039472 | Combinatorics-graph colouring
Show that if [imath]K_9 [/imath]is coloured red and blue and contains no red triangle and no blue [imath]K_4[/imath], then every vertex must have red degree [imath]3[/imath] and blue degree [imath]5[/imath]. I have absolutely no idea how to proceed :( | 1035927 | Can the complete graph [imath]K_9[/imath], be 2-coloured with no blue [imath]K_4[/imath] or red triangles?
I am working on the following problem on 2-coloured complete graphs: [imath]K_9[/imath] is coloured red and blue and contains no red triangle and no blue [imath]K_4[/imath] then every vertex must have red degree 3 and blue degree 5. Is this possible? I am pretty sure the answer is 'no' but I am not sure how to explain my answer. Any help would be greatly appreciated! |
1039935 | CDF of maximum of iid rvs
I am having a small doubt regarding maximum of random variables. I have [imath]Z= \max\{ X_1, X_2,\dots X_p, \dots X_N\}[/imath] where all [imath]X_i[/imath] are independent, identically distributed. Now, If for sure, I know that [imath]X_p>X_{p+1}>\dots>X_N[/imath] the can i say that the CDF of [imath]Z[/imath], [imath]\begin{align*}P(Z<z)&= P\left(\max\{ X_1, X_2,\dots X_p, \dots X_N\}<z\right)\\&=P\left(\max\{ X_1, X_2,\dots X_p\}<z\right)\\&=\left[P(X_1<z)\right]^{N-p-1}\end{align*}[/imath] How to find CDF of [imath]Z[/imath] in this case ? Is there any theorem that reduce the size of total variables for certain condition ? | 1035113 | Maximum of RVS independent and identically distributed
I am having a small doubt regarding maximum of random variables. I have [imath]Z= \max\{ X_1, X_2,\dots X_p, \dots X_N\}[/imath] where all [imath]X_i[/imath] are independent, identically distributed. Now, If for sure, I know that [imath]\{X_1,\dots,X_p\}>X_{p+1}>\dots>X_N[/imath] the can i say that the CDF of [imath]Z[/imath], [imath]\begin{align*}P(Z<z)&= P\left(\max\{ X_1, X_2,\dots X_p, \dots X_N\}<z\right)\\&=P\left(\max\{ X_1, X_2,\dots X_p\}<z\right)\\&=\left[P(X_1<z)\right]^{N-p-1}\end{align*}[/imath] How to find CDF of [imath]Z[/imath] in this case ? Is there any theorem that reduce the size of total variables for certain condition ? |
1040061 | Proving a Relation that is a Function by Division Algorithm
Let A=B=[imath]\mathbb{N}[/imath] R is: (a,b)[imath]\in[/imath]R iff for some q[imath]\in[/imath]Integers a=5q+b WHERE 0[imath]\leq[/imath]b<5 Given a relation, show that it's a function. To Show: 1) [imath]\forall[/imath]a[imath]\in[/imath]A[imath]\exists[/imath]b[imath]\in[/imath]B((a,b)[imath]\in[/imath]R) 2) [imath]\forall[/imath]a[imath]\in[/imath]A, there is a unique b[imath]\in[/imath]B((a,b)[imath]\in[/imath]R) Proof: Let a[imath]\in[/imath]A. Assume (a,b)[imath]\in[/imath]R and (a,c)[imath]\in[/imath]R where b,c[imath]\in[/imath]B. Do MATH [imath]\therefore[/imath] b=c. QED What math should I do to get the full proof. Thanks! | 1040001 | Is the relation on integers, defined by [imath](a,b)\in R\iff a=5q+b[/imath], a function?
Let [imath]A=B=\mathbb N[/imath]. Relation [imath]R[/imath] is: [imath](a,b)\in R[/imath] iff for some [imath]q \in \mathbb Z[/imath] we have [imath]a=5q+b[/imath] Given a relation, show that it's a function. To Show: [imath]\forall a \in A \ \exists b \in B[/imath] such that [imath](a,b)\in R[/imath] [imath]\forall a \in A [/imath] there is a unique [imath]b\in B[/imath] such that [imath](a,b) \in R[/imath]. Proof Outline: Let [imath]a \in A[/imath]. Assume [imath](a,b)\in R[/imath] and [imath](a,c) \in R[/imath] where [imath]b,c \in B[/imath]. then some steps leading to.... [imath]\therefore[/imath] b=c. QED What math should I do to complete the proof? |
1040106 | Open connected subset of [imath] \mathbb R^2 [/imath]is path connected
Is open connected subset of [imath] \mathbb{R^2} [/imath] is path connected? | 319103 | Show that if [imath]U[/imath] is an open connected subspace of [imath]\mathbb{R}^2[/imath], then [imath]U[/imath] is path connected
Show that if [imath]U[/imath] is an open connected subspace of [imath]\mathbb{R}^2[/imath], then [imath]U[/imath] is path connected. (Hint:Show that given [imath]x_0 \in U[/imath], the set of points can be joined to [imath]x_0[/imath] by a path in [imath]U[/imath] is both open and closed in [imath]U[/imath].) This should not be too difficult, but I am stuck at some point. Let [imath]x_0 \in U[/imath] Let [imath]A = \{ x \in U | \text{there is a path connecting [/imath]x[imath] and [/imath]x_0[imath]} \}[/imath] [imath]\subset U[/imath]. I want to show that [imath]A[/imath] is open, so let [imath]x \in A[/imath]. By openness of [imath]U[/imath], we can find a basic open set [imath]\prod_{i=1}^2 (a_i,b_i)[/imath] such that [imath]\prod_{i=1}^2 (a_i,b_i) \subset U[/imath]. But I am stuck in showing that [imath]\prod_{i=1}^2 (a_i,b_i)[/imath] lies entirely in [imath]A[/imath]. |
1040458 | Closed set on topological space
This is a problem on topological spaces and continuous functions. If [imath]f,g \to\mathbb{R}[/imath] are continuous functions, then [imath]T=\{x\in X: f(x)=g(x)\}[/imath] is closed on X | 199617 | The set of points where two maps agree is closed?
Let [imath]f,g\colon X \to Y[/imath] be continuous maps. Let [imath]Y[/imath] be Hausdorff. Is the set [imath]A := \{x\in X \, : \, f(x)=g(x) \}[/imath] necessarily closed ? |
1039697 | Questions about the Moduli Space of Vector Bundles of rank 2 with trivial determinant over a curve of genus [imath]g[/imath]
Let [imath]M_0[/imath] be the moduli space of rank 2 semi-stable vector bundles over X with trivial determinant which is a singular projective variety of dimension [imath]3g-3[/imath]. [imath]M_0[/imath] is constructed as the [imath]GIT[/imath] quotient of the [imath]R(2,p)[/imath] by [imath]SL(p)[/imath]. Let [imath]M_0^{s}[/imath] be the points of moduli space corresponding to the semi-stable points of [imath]R(2,p)[/imath]. Then the singular locus of [imath]M_0[/imath] is the Kummer Variety [imath]\mathbb{K}[/imath] or the complement of [imath]M_0^{s}[/imath]. The elements of [imath]\mathbb{K}[/imath] are of the form [imath]L\oplus L^{-1}[/imath], where L is a line bundle of degree 0. [imath]\mathbb{K}_0[/imath] be bundles of the form [imath]L\oplus L^{-1}[/imath] such that [imath]L=L^{-1}[/imath], i.e., [imath]L^2[/imath] is trivial. Question 1- What is meant by a Kummer Variety? Why [imath]\mathbb{K}_0[/imath] is the Kummer Variety of dimension g? [imath]\mathbb{Z_2}[/imath] acts on [imath]Jac_0[/imath] by involution i.e., [imath](0,L)\longmapsto L[/imath] and [imath](1,L)\longmapsto L^{-1}[/imath] [imath]\mathbb{K}[/imath] [imath]\cong[/imath] [imath]Jac_0//\mathbb{Z_2}[/imath]. There are [imath]2^{2g}[/imath] number of fixed points [imath]\mathbb{Z_2^{2g}}=\{[L\oplus L^{-1}:L\cong L^{-1}]\}[/imath]of the action. Question2.- Is this isomorphism is isomorphism as varieties? Question3- Why are there are [imath]2^{2g}[/imath] number of fixed points of the action. Thus [imath]M_0[/imath] has a stratification [imath]M_0=M_0^{s}\coprod(\mathbb{K}-\mathbb{Z_2^{2g}})\coprod Z_2^{2g}[/imath]. Over each point in the deepest strata [imath]\mathbb{Z_2^{2g}}[/imath] there is unique closed orbit in [imath]R(2,p)^{ss}[/imath]. By deformation theory, the normal space of the orbit at a point h, which represents [imath]L\oplus L^{-1}[/imath] where [imath]L\cong L^{-1}[/imath] is, [imath]H^{1}(End_0(L\oplus L))\cong H^{1}(\mathcal(O))\oplus \mathcal{sl}(2)[/imath] Question4- Why is the last statement true? It will be helpful if someone provides some reference where I can find my answers. | 1040498 | Questions about the Moduli Space of Vector Bundles of rank 2 with trivial determinant
Let [imath]M_0[/imath] be the moduli space of rank 2 semi-stable vector bundles over X with trivial determinant which is a singular projective variety of dimension [imath]3g-3[/imath]. [imath]M_0[/imath] is constructed as the [imath]GIT[/imath] quotient of the [imath]R(2,p)[/imath] by [imath]SL(p)[/imath]. Let [imath]M_0^{s}[/imath] be the points of moduli space corresponding to the semi-stable points of [imath]R(2,p)[/imath]. Then the singular locus of [imath]M_0[/imath] is the Kummer Variety [imath]\mathbb{K}[/imath] or the complement of [imath]M_0^{s}[/imath]. The elements of [imath]\mathbb{K}[/imath] are of the form [imath]L\oplus L^{-1}[/imath], where L is a line bundle of degree 0. [imath]\mathbb{K}_0[/imath] be bundles of the form [imath]L\oplus L^{-1}[/imath] such that [imath]L=L^{-1}[/imath], i.e., [imath]L^2[/imath] is trivial. Question 1- What is meant by a Kummer Variety? Why [imath]\mathbb{K}_0[/imath] is the Kummer Variety of dimension g? [imath]\mathbb{Z_2}[/imath] acts on [imath]Jac_0[/imath] by involution i.e., [imath](0,L)\longmapsto L[/imath] and [imath](1,L)\longmapsto L^{-1}[/imath] [imath]\mathbb{K}[/imath] [imath]\cong[/imath] [imath]Jac_0//\mathbb{Z_2}[/imath]. There are [imath]2^{2g}[/imath] number of fixed points [imath]\mathbb{Z_2^{2g}}=\{[L\oplus L^{-1}:L\cong L^{-1}]\}[/imath]of the action. Question2.- Is this isomorphism is isomorphism as varieties? Question3- Why are there are [imath]2^{2g}[/imath] number of fixed points of the action. Thus [imath]M_0[/imath] has a stratification [imath]M_0=M_0^{s}\coprod(\mathbb{K}-\mathbb{Z_2^{2g}})\coprod Z_2^{2g}[/imath]. Over each point in the deepest strata [imath]\mathbb{Z_2^{2g}}[/imath] there is unique closed orbit in [imath]R(2,p)^{ss}[/imath]. By deformation theory, the normal space of the orbit at a point h, which represents [imath]L\oplus L^{-1}[/imath] where [imath]L\cong L^{-1}[/imath] is, [imath]H^{1}(End_0(L\oplus L))\cong H^{1}(\mathcal(O))\oplus \mathcal{sl}(2)[/imath] Question4- Why is the last statement true? |
742341 | Find the last two digits of the number [imath]9^{9^9}[/imath]
Find the last two digits of the number [imath]9^{9^9}[/imath] . [Hint: [imath]9^9 \equiv 9 \pmod {10} [/imath]; hence, [imath]9^{9^9}[/imath] = [imath]9^9+10k[/imath] ;now use the fact that [imath]9^9 \equiv 89 \pmod {100}[/imath]] | 842643 | Find the last two digits of [imath]9^{{9}^{9}}[/imath]
I have to find the last two decimal digits of the number [imath]9^{{9}^{9}}[/imath]. That's what I did: [imath]m=100 , \phi(m)=40, a=9[/imath] [imath](100,9)=1, \text{ so from Euler's theorem,we have :} 9^{40} \equiv 1 \pmod{100}[/imath] [imath]9^{{9}^{9}} \equiv 9^{9^{2+2+2+2+1}} \equiv 9^{9^2 \cdot 9^2 \cdot 9^2 \cdot 9^2 \cdot 9} \equiv 9^{81 \cdot 81 \cdot 81 \cdot 81 \cdot 9} \equiv 9^{(40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot 9} \equiv (9^{(40+40+1)})^{(40+40+1) \cdot (40+40+1) \cdot (40+40+1) \cdot 9} \equiv 9^9 \equiv 9^4 \cdot 9^4 \cdot 9 \equiv 6561 \cdot 6561 \cdot 9 \equiv 3721 \cdot 9 \\ \equiv 21 \cdot 9 \equiv 89 \pmod{100}[/imath] So,the last two digits are [imath]8 \text{ and } 9[/imath]. [imath]$[/imath][imath]But,is there also an other way two calculate the last two digits of [/imath]9^{{9}^{9}}$ or is the above the only one? |
871748 | Prove that one of the sides of a Pythagorean triangle is a multiple of 3.
Prove that if [imath]\{a,b,c\}[/imath] forms a Pythagorean triple, then at least one of the three numbers must be divisible by [imath]3[/imath]. Need some help with this one. | 611347 | In any Pythagorean triplet at least one of them is divisible by [imath]2[/imath], [imath]3[/imath] and [imath]5[/imath].
Show that if [imath]x[/imath], [imath]y[/imath], [imath]z[/imath] are integers such that [imath]x^2 + y^2 = z^2[/imath], then at least one of them is divisible by [imath]2[/imath], at least one is divisible by [imath]3[/imath], and at least one is divisible by [imath]5[/imath]. I know that [imath]x,y,z[/imath] are sides of a right-angled triangles. If I start putting values, then I can show, but have no clue about mathematical approach. How do I solve it mathematically, using modular arithmetic? |
1040031 | Show that both [imath]S[/imath] and [imath]T[/imath] are bounded.
Let [imath]S[/imath] and [imath]T[/imath] be linear map defined on hilbert space [imath]H[/imath] s.t. [imath]\langle Su,v \rangle=\langle u,Tv\rangle [/imath] [imath]\forall u,v \in H[/imath]. Then show that both [imath]S[/imath] and [imath]T[/imath] are bounded. [imath]\textbf{TRY-}[/imath] If I can show [imath]S[/imath] is bounded than [imath]T[/imath] is bounded but I am unable to show [imath]S[/imath] is bounded. I was working on it but getting nowhere. Any help is appreciated. Thanks! | 1034278 | If a linear operator has an adjoint operator, it is bounded
This is a question I'm struggling with for a while: Let [imath]H[/imath] be a Hilber space. Let [imath]T,S: H\rightarrow H[/imath] be linear operators (not neccessarily bounded) such that for every [imath]x,y\in H[/imath]: [imath]\langle Tx,y\rangle=\langle x,Sy \rangle[/imath]. Prove [imath]T,S[/imath] are bounded. What I did so far: First attempt: [imath]||Tx||^2=\langle Tx,Tx\rangle=\langle x,STx\rangle\le||x||*||STx||\le||x||*||S||*||Tx||[/imath] I'm not entirly sure I'm allowed to do the last inequallity, since [imath]||S||[/imath] might be [imath]\infty[/imath]. It follows [imath]||Tx||\le ||x||*||S||[/imath] and analogly [imath]||Sx||\le ||x||*||T||[/imath] From the first I'd get [imath]||T||\le ||S||[/imath] and from the second [imath]||S||\le||T||[/imath]. Therefore [imath]||S||=||T||[/imath]. I don't know where to go from here. Second Attempt: assume [imath]T[/imath] isn't bounded, therefore so is [imath]S[/imath], therefore there are series [imath](x_n),(y_n)\subset H[/imath] such that [imath]||x_n||=||y_n||=1[/imath] and [imath]\lim ||Tx_n||=\infty,\lim ||Sy_n||=\infty[/imath]. I'll keep it going soon, but it didn't go so well.. I'd love some guide. Not neccessarily the whole solution, but something that would tell me what i'm missing. Thanks :) |
1040945 | Irreducible representations of [imath]S_n[/imath]
I want to prove that [imath]S_n[/imath] has an irreducible representation of dimension [imath]n-1[/imath]. Intuitively, I know that the [imath]\forall n[/imath], the trivial representation is irreducible, and there should be some complimentary subspace which is irreducible. However, I am new to representation theory, and I am unable to formalize the proof. Any hints and ideas on how to go about this are welcome. | 212025 | Permutation module of [imath]S_n[/imath]
Let [imath]G=S_n[/imath] and let [imath]V[/imath] be the permutation module of [imath]G[/imath] with basis [imath]\{x_1,\ldots,x_n\}.[/imath] Let [imath]\lambda, \mu \in \mathbb{C}[/imath] to allow one to define a [imath]\mathbb{C}G[/imath]-homomorphism [imath]\rho:V \to V[/imath] by [imath]\rho(x_j):=\lambda x_j+\mu\sum_{i \neq j}x_i.[/imath] By using the above fact or otherwise, how can we prove that [imath]V[/imath] is the direct sum of two non-isomorphic irreducible [imath]\mathbb{C}G[/imath] -submodules? I tried to prove this by construction. A familiar irreducible submodule in this case is the [imath]1[/imath]-dimensional space [imath]U:=\operatorname{span}\{x_1+\cdots+x_n\}[/imath]. I intend to find another [imath](n-1)[/imath]-dimensional submodule [imath]W[/imath] which makes [imath]V=U\oplus W[/imath] hold, but it's hard to do so. Is there a way to use the fact instead of a random construction? |
1007566 | Calculate the leading-order asymptotic behaviour (Laplace's Method)
thanks in advance! Calculate the leading-order asymptotic behaviour of the integral [imath]I(x) = \int_{0}^{2\pi} (1+t^2) e^{x \cos t} dt \mbox{ as } x \mbox { tends to infinity}[/imath] So far I know there are two maximas at [imath]0 \mbox{ and }2\pi[/imath] so it can be assumed to be the integral [imath]\int_{0}^{\epsilon} (1+t^2) e^{x \cos t} dt + \int_{2\pi-\epsilon}^{2\pi} (1+t^2) e^{x \cos t} dt[/imath] then using taylor expansions this can be transformed to: [imath]\int_{0}^{\epsilon} (1+t^2) e^{x (1-\frac{t^2}{2})} dt + \int_{2\pi-\epsilon}^{2\pi} (1+t^2) e^{x (1-\frac{1}{2}(t-2\pi)^2)} dt[/imath] but where do I go from here? How do I find the leading order term? | 1009076 | Calculate the leading order asymptotic behaviour (with two maxima)
thanks in advance! Calculate the leading-order asymptotic behaviour of the integral [imath]I(x) = \int_{0}^{2\pi} (1+t^2) e^{x \cos t} dt \mbox{ as } x \mbox { tends to infinity}[/imath] So far I know there are two maximas at [imath]0 \mbox{ and }2\pi[/imath] so it can be assumed to be the integral [imath]\int_{0}^{\epsilon} (1+t^2) e^{x \cos t} dt + \int_{2\pi-\epsilon}^{2\pi} (1+t^2) e^{x \cos t} dt[/imath] then using taylor expansions this can be transformed to: [imath]\int_{0}^{\epsilon} (1+t^2) e^{x (1-\frac{t^2}{2})} dt + \int_{2\pi-\epsilon}^{2\pi} (1+t^2) e^{x (1-\frac{1}{2}(t-2\pi)^2)} dt[/imath] but where do I go from here? How do I find the leading order term? |
1041377 | a question about compact set, how to prove there exists f(y)=y
Let (X,p) be a compact metric space.Suppose that f X->X is a function such that, for all [imath]x_1[/imath],[imath]x_2[/imath] [imath]\in[/imath]X, if [imath]x_1\neq x_2[/imath] then p(f([imath]x_1[/imath]),f([imath]x_2[/imath]))<[imath]p([/imath]x1[imath],[/imath]x2[imath])[/imath]. Prove that there exists a unique point y which in X such that f(y)=y. (there is a hint below the question:Consider the function g:X->R defined by setting g(x)=p(x,f(x)).) this is my thought: I know g(X) should be a closed and bounded set in the real numbers, since X is compact. so m<=g(X)<=M, if m=0, it's obvious that there exists a x in X such that f(x)=x, but if m>0, I want to prove if m>0, then X cannot be a compact set, I am trying to find an open covering which there are no finite open sets which can cover X. If I can find such an open covering, then I think I will finish the question, but I don't know how to find it. Can someone help me to find it. Or, can someone have other methods to solve this question? I am struggling with it, and have no idea how to continue? Hope someone can help me solve this question! | 409822 | Contraction Map on Compact Normed Space has a Fixed Point
Let [imath]K[/imath] be a compact normed space and [imath]f:K\rightarrow K[/imath] such that [imath]\|f(x)-f(y)\|<\|x-y\|\quad\quad\forall\,\, x, y\in K, x\neq y.[/imath] Prove that [imath]f[/imath] has a fixed point. |
1041805 | finding line integral with potential function
Evaluate the line integral [imath] \int_C (\ln y) e^{-x} \,dx - \dfrac{e^{-x}}{y}\,dy + z\,dz [/imath] where C is the curve parametrized by [imath]r(t)=(t-1)i+e^{t^4}j+(t^2+1)k[/imath] for [imath]0\leq t\leq 1[/imath] I know that the potential function is [imath]f(x, y, z) = -e^{-x}\ln y + \frac{z^2}{2} + C[/imath] but exactly how would I use that to evaluate the line integral? | 1041391 | line integral (multivariable calculus)
Evaluate the line integral [imath] \int_C (\ln y) e^{-x} \,dx - \dfrac{e^{-x}}{y}\,dy + z\,dz [/imath] where C is the curve parametrized by [imath]r(t)=(t-1)i+e^{t^4}j+(t^2+1)k[/imath] for [imath]0\leq t\leq 1[/imath] |
1042102 | The following series is an irrational number
[imath](*)[/imath] Let [imath]\{ x_n \}[/imath] be a sequence where [imath]x_n[/imath] is either [imath]1[/imath] or [imath]-1[/imath]. Then [imath] \sum_{n=0}^{\infty} \frac{ x_n}{n!} \; \; \; \text{is irrational}[/imath] This problem arise when I was trying to prove that [imath]e[/imath] is irrational. If the result above is true, then we can put [imath]x_n = (-1)^n [/imath] and hence [imath] \frac{1}{e} = \sum \frac{(-1)^n}{n!} \; \; \; \text{is irrational} [/imath] which would imply that [imath]e[/imath] is irrational. So, my question is: Is [imath](*)[/imath] true? | 367183 | Prove the series [imath] \sum_{n=1}^\infty \frac{1}{(n!)^2}[/imath] converges to an irrational number
How can one prove that the series [imath]\displaystyle \sum\limits_{n=1}^\infty \frac{1}{(n!)^2}[/imath] converges to an irrational number? There's no need to use Taylor expansion, integrals or any advanced/professional techniques. It can be proved using only basic techniques. My first attempt was to assume that the series does converge to a rational number [imath] \frac{p}{q}[/imath] and then it follows by definition that for every positive epsilon there exists an integer k such that [imath] {\sum_{n=1}^{k} \frac{1}{(n!)^2} -\frac{p}{q}}<=\epsilon[/imath] I tried to get a contradiction but I failed. |
1042823 | Convergence test of the series [imath]\sum\sin100n[/imath]
I need to prove that [imath]\sum_{n=1}^{\infty} {\sin{100n}} \; \text{diverges}[/imath] I think the best way to do it is to show that [imath]\lim_{n\to \infty}{\sin{100n}}\not=0[/imath]. But how do I prove it? | 1042726 | Proof that [imath]\lim_{n\to\infty}{\sin{100n}}[/imath] does not exist
How to prove that [imath]\lim_{n\to\infty}{\sin{100n}}[/imath] doesn't exist? Some possible approaches: It would be enough to find two subsequences [imath]n_{k}[/imath] that converge to two different numbers. But it's not clear how to find [imath]n_k[/imath] so that [imath]\sin 100n_k[/imath] converge. Show that [imath]\sin (100(n+1))-\sin 100n[/imath] does not approach [imath]0[/imath]. This is not obvious, either. |
1043525 | Order of [imath]GL(n, \mathbb Z_p)[/imath]
Let [imath]G[/imath] be the group of all [imath]n\times n[/imath] matrices with entries of the matrices from the field [imath]\mathbb Z_p[/imath] , [imath]p[/imath] prime, such that determinant of every matrix is not [imath][0][/imath] , w.r.t. matrix multiplication ; then what is the order of [imath]G[/imath] ? | 1603389 | What is order of general linear group GL(3,Zp)?
I have tried for [imath]GL(2,\mathbb{Z}_p)[/imath] its [imath]= (p^2-1)(p^2-p)[/imath] Since as number of ways in which first row can be filled is [imath]p^2-1[/imath] by omitting the way in which all entries r zero and second row can be filled in [imath]p^2-p[/imath] ways by omitting the ways in which d entries are multiples of first row. I have also tried for for [imath]GL(3,\mathbb{Z}_p)[/imath] its found as [imath]=(p^3-1)(p^3-p)(p^3-2p)[/imath]. Is it right? |
1043611 | Prove that [imath](\Bbb{R},+)[/imath] is isomorphic to [imath](\Bbb{C},+)[/imath].
Prove that [imath](\Bbb{R},+)[/imath] is isomorphic to [imath](\Bbb{C},+)[/imath]. I don't know how to proceed in this question. On doing a customary internet search, I found that this can be proven using Axiom of Choice. Does there not exist an explicit isomorphism? | 302514 | Are [imath](\mathbb{R},+)[/imath] and [imath](\mathbb{C},+)[/imath] isomorphic as additive groups?
Are [imath](\mathbb{R},+)[/imath] and [imath](\mathbb{C},+)[/imath] isomorphic as additive groups? I know that there is a bijection between [imath]\mathbb{R}[/imath] and [imath]\mathbb{C}[/imath], and this question asks whether they are isomorphic as abelian groups, are they referring to the additive abelian group? If so is there any simple isomorphism I can find? I know nothing about Hamel basis. Thanks. |
1043716 | How to evaluate the following limit? [imath]\lim\limits_{x\to1}\frac{x^n+x^{n-1}+x^{n-2}+\cdots+x-n}{x-1}.[/imath]
How to evaluate the following limit? [imath]\lim_{x\to1}\dfrac{x^n+x^{n-1}+x^{n-2}+\cdots+x-n}{x-1}.[/imath] I could simplify but couldn't proceed further | 451799 | How to show [imath]\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}[/imath]?
I am able to evaluate the limit [imath]\lim_{x \to 1} \frac{x + x^2 + \dots + x^n - n}{x - 1} = \frac{n(n + 1)}{2}[/imath] for a given [imath]n[/imath] using l'Hôspital's (Bernoulli's) rule. The problem is I don't quite like the solution, as it depends on such a heavy weaponry. A limit this simple, should easily be evaluable using some clever idea. Here is a list of what I tried: Substitute [imath]y = x - 1[/imath]. This leads nowhere, I think. Find the Taylor polynomial. Makes no sense, it is a polynomial. Divide by major term. Dividing by [imath]x[/imath] got me nowhere. Find the value [imath]f(x)[/imath] at [imath]x = 1[/imath] directly. I cannot as the function is not defined at [imath]x = 1[/imath]. Simplify the expression. I do not see how I could. Using l'Hôspital's (Bernoulli's) rule. Works, but I do not quite like it. If somebody sees a simple way, please do let me know. Added later: The approach proposed by Sami Ben Romdhane is universal as asmeurer pointed out. Examples of another limits that can be easily solved this way: [imath]\lim_{x \to 0} \frac{\sqrt[m]{1 + ax} - \sqrt[n]{1 + bx}}{x}[/imath] where [imath]m, n \in \mathbb{N}[/imath] and [imath]a, b \in \mathbb{R}[/imath] are given, or [imath]\lim_{x \to 0} \frac{\arctan(1 + x) - \arctan(1 - x)}{x}[/imath]. It sems that all limits in the form [imath]\lim_{x \to a} \frac{f(x)}{x - a}[/imath] where [imath]a \in \mathbb{R}[/imath], [imath]f(a) = 0[/imath] and for which [imath]\exists f'(a)[/imath], can be evaluated this way, which is as fast as finding [imath]f'[/imath] and calculating [imath]f'(a)[/imath]. This adds a very useful tool into my calculus toolbox: Some limits can be evaluated easily using derivatives if one looks for [imath]f(a) = 0[/imath], without the l'Hôspital's rule. I have not seen this in widespread use; I propose we call this Sami's rule :). |
1043872 | Show that this is the number of moves that have to be done to solve the problem
Prove that [imath]2^n − 1[/imath] moves are necessary and sufficient to solve the Towers of Hanoi problem. Could you give me some hints how I could do that?? | 302165 | Basic proof by Mathematical Induction (Towers of Hanoi)
I am new to proofs and I am trying to learn mathematical induction. I started working out a sample problem, but I am not sure if I am on the right track. I was wondering if someone would be kind enough to comment on my work so far, and give me some hints as to how I should proceed.Many thanks in advance! [imath]S_n[/imath] is the minimum number of moves it takes to solve towers of Hanoi where [imath]n[/imath] is a positive integer. [imath]S_n = 2^n-1[/imath] Base Case: [imath]\begin{align*}S_1 &= 2^1-1 \\&= 1 \end{align*}[/imath] Assume true for [imath]k[/imath]: [imath]S_k = 2^k-1[/imath] Hence, [imath]\begin{align*}S_{k+1} &= (2^1-1) , (2^2-1) , (2^3-1) , \ldots , (2^k-1) , (2^{k+1}-1) \\&= (2^k-1) + 2^{k+1}-1\end{align*}[/imath] This is where I am stuck. |
1044880 | Proof about holomorphic function
I have to prove that if [imath]\Omega[/imath] is convex, [imath]f[/imath] is holomorphic on [imath]\Omega[/imath] and [imath]\text{Re}(f')>0 [/imath] then [imath]f[/imath] is injection. Any ideas? | 709866 | f is differentiable on convex domain D and Re(f')>0 implies f is injective on D
Suppose that [imath]f[/imath] is differentiable on a convex domain (open and connected) D and Re([imath]f')>0[/imath] in D. How can we prove that f is injective on D? I found some answers using the mean value theorem for holomorphic functions, but I am not familiar with that theorem. Actually I want to use the mean value theorem for integrals, which tells us that the value of an differentiable function [imath]f[/imath] at the center [imath]z_{0}[/imath] of the disk [imath]|z-z_{0}|<r[/imath] is equal to the integral mean of its values on the boundary of that disk: [imath]f(z_{0})=\int^{2\pi}_{0}{f(z_{0}+re^{it})dt}[/imath]. Can you help me proving it by the mean value theorem for integrals? |
1045100 | How is it possible that you get two digit numbers multiplied by three at the beginning if you divide 1 by 97, etc.?
I divided [imath]1[/imath] by [imath]97[/imath] and I got this: [imath]0.0103092783\dots[/imath]; I got two digit numbers being multiplied by three at the beginning. I divided [imath]1[/imath] by [imath]997[/imath] and I got a similar thing: [imath]0.001003009027081243731\dots[/imath]; I got three-digit numbers being multiplied by three at the beginning. How does this happen? It's something very remarkable, isn't it? | 1029279 | Why do divisions like 1/98 and 1/998 give us numbers continuously being multiplied by two each time in decimal form?
For example, when I divided [imath]1[/imath] by [imath]98[/imath], I got an amazing result of [imath]0.0102040816326530612244897...[/imath] where so many numbers get multiplied by two every time in the right pattern with some carrying. Also, when I divided [imath]1[/imath] by [imath]998[/imath], I got an amazing result of [imath]0.0010020040080160320641282...[/imath]. I have a longer result from my cool memory, which is [imath]0.001002004008016032064128256513026052104208517034068136272545090180360721442885...[/imath]What explains why this doubling thing is happening? I love it when I see it! Also I don't know if it repeats (I'm talking about 1/998). I think it does because I looked it up and it said 498 numbers repeat every time, so it's rational. I mean seriously, how does this doubling happen? I hope I can receive good information from you! |
1044713 | Find all the triangles satisfying [imath]\cos(A)\cos(B)+\sin(A)\sin(B)\sin(C)=1[/imath]
I am trying to solve the problem of finding all triangles with angles [imath]A[/imath], [imath]B[/imath] and [imath]C[/imath] (in [imath][0,\pi][/imath]) such that [imath]\cos A\cos B+\sin A\sin B\sin C=1[/imath]. In the case where the triangle has a right angle, it can be proved that [imath]C=\frac{\pi}{2}[/imath] (thus [imath]C[/imath] is the right angle and [imath]A=B=\frac{\pi}{4}[/imath]. If we assume that the triangle is isosceles, then the triangle is flat or has a right angle (hence is the above solution). My intuition says that these are the only solutions but I struggle to prove it. Any idea? | 469599 | If [imath]\sin(a)\sin(b)\sin(c)+\cos(a)\cos(b)=1[/imath] then find the value of [imath]\sin(c)[/imath]
If [imath]\sin(a)\sin(b)\sin(c)+\cos(a)\cos(b)=1,[/imath];where abc are the angles of the triangle.! then find the value of [imath]\sin(c)[/imath]. By trial and error put this triangle as right angled isosceles and got the answer..!! But i want a complete proof! |
1045565 | Proof [imath] e^{x-1} \geq{x} [/imath] by writting [imath]e^x[/imath] as a infinite series
By writing [imath]e^x[/imath] as a infinite series proof that for every [imath]x\geq{0}[/imath] is true: [imath] e^{x-1} \geq{x} [/imath] Also when is true [imath]e^{x-1} = x[/imath]? | 1045269 | By expanding [imath]e^x[/imath] into a series prove the following inequality
By expanding [imath]e^x[/imath] into a series [imath]\sum e^x[/imath] prove that [imath]\forall x \in \mathbb{R}, x \ge 0 \implies e^{x-1} \ge x[/imath] Also show when this inequality becomes equality. I'm not really sure how to attack this problem at all, any tips would be very welcome. |
1044332 | Proving convergence of series
Prove whether the following series converge or diverge. [imath]\sum \limits_{n=1}^{\infty} \frac{(2n)!}{(4^n)(n!)^2(n^2)}[/imath] I think this series converge and I tried to justify using the ratio test but I got [imath]\lim \limits_{n\to\infty} \frac{a_{n+1}}{a_n}=1[/imath] which means the ratio test cannot be applied. Please give me some ideas or hints on how to solve this question, thanks to anybody who helps. | 1045156 | Using Ratio test/Comparison test
I have [imath]\displaystyle\sum_{n=1}^{\infty}{\frac{(2n)!}{(4^n)(n!)^2(n^2)}}[/imath] and need to show whether it diverges or converges. I attempted to use the ratio test, but derived that the limit of [imath]\dfrac{a_{n+1}}{a_n}=1[/imath] and hence the test is inconclusive. So I now must attempt to use the comparison test, but I am struggling to find bounds to compare it to to show either divergence or convergence. |
1045707 | Tetrations of non-integers?
A Tetration is defined as [imath]{^{n}a} = \underbrace{a^{a^{\cdot^{\cdot^{a}}}}}_n[/imath] or, by a recursion function, [imath]{^{n}a} := \begin{cases} 1 &\text{if }n=0 \\ a^{\left[^{(n-1)}a\right]} &\text{if }n>0 \end{cases} [/imath] That is, iterated exponentiation (like exponentiation is iterated multiplication). Exponentiation can be expressed as [imath]e^{n\ln a}[/imath] [1] Is there a like way to express Tetrations? That is, a way to calculate for both negative and decimal Tetrations? | 56663 | Is There a Natural Way to Extend Repeated Exponentiation Beyond Integers?
This question has been in my mind since high school. We can get multiplication of natural numbers by repeated addition; equivalently, if we define [imath]f[/imath] recursively by [imath]f(1)=m[/imath] and [imath]f(n+1)=f(n)+m[/imath], then [imath]f(n) = m \times n[/imath]. Likewise, we get exponentiation by repeated multiplication. If [imath]g(1)=m[/imath] and [imath]g(n+1)=mg(n)[/imath], then [imath]g(n) = m^n[/imath]. In my high school mind it was natural to imagine a new function defined by repeated exponentiation: [imath]h(1)=m[/imath] and [imath]h(n+1)=m^{h(n)}[/imath]. These definitions only make sense for [imath]n[/imath] a natural number, but of course there are standard very mathematically satisfying ways to define multiplication and exponentiation by any real number. My question is this: Can the function [imath]h[/imath] defined above also be extended in a natural way to [imath]\mathbb{R}^{>0}[/imath]? The question is in the spirit of seeking an extension of [imath]f(n)=n![/imath] to [imath]\mathbb{R}[/imath] and arriving at [imath]\Gamma(x)[/imath]. Let me focus the question, and attempt to make precise what I mean by "in a natural way." Take [imath]h(1)=2[/imath] and [imath]h(n+1)=2^{h(n)}[/imath]. [imath]h[/imath] is now defined on [imath]\mathbb{N}[/imath], and [imath]h(2)=4[/imath], [imath]h(3)=16[/imath], [imath]h(4)=2^{16}=65,536[/imath] etc. Is it possible to extend the domain of definition of [imath]h[/imath] to all positive reals in such a way that a) The functional equation [imath]h(x+1)=2^{h(x)}[/imath] continues to be satisfied for all [imath]x[/imath] in the domain. b) [imath]h[/imath] is [imath]C^\infty[/imath]. (Analytic would be even better but this seems maybe too much to hope for?) c) All [imath]h[/imath]'s derivatives are monotone. These requirements are my attempt to codify what would count as "natural." I am open to suggestions about what would be a better list of requirements. If such a function exists, I would like to know how to construct it; if it doesn't, I would like to know why (i.e. outline of proof), and if relaxing some of the requirements (e.g. just the first derivative monotone) would make it possible. (If the function exists, I am also interested in the questions, "is it unique?" "Could we add some natural requirements to make it unique?" But my main query is about existence.) |
238970 | How to evaluate fractional tetrations?
Recently I've come across 'tetration' in my studies of math, and I've become intrigued how they can be evaluated when the "tetration number" is not whole. For those who do not know, tetrations are the next in sequence of iteration functions. (The first three being addition, multiplication, and exponentiation, while the proceeding iteration function is pentation) As an example, 2 with a tetration number of 2 is equal to [imath]2^2[/imath] 3 with a tetration number of 3 is equal to [imath]3^{3^3}[/imath] and so forth. My question is simply, or maybe not so simply, what is the value of a number "raised" to a fractional tetration number. What would the value of 3 with a tetration number of 4/3 be? Thanks for anyone's insight | 1666659 | How would you define non-integer tetration?
Tetration is defined for all [imath]n\in \Bbb{N}[/imath] by [imath] {^1}a = a \\ {^{n+1}}a = a^{\left({^n}a\right)} [/imath] Thus [imath]{^3}a[/imath] means [imath]a^{a^a}[/imath]. Here [imath]a[/imath] could be any real (or indeed even complex) value, but only integer [imath]n[/imath] in the tetration is defined by this definition. But I've seen questions here dealing with real valued tetration ([imath]{^x}a[/imath] for real [imath]x[/imath]) and even one dealing with complex tetration ([imath]{^z}a[/imath] for complex [imath]z[/imath]). let's take one step at a time: How does one define tetration with a rational hyperpower [imath]{^q}a : q \in \Bbb{Q}[/imath] (From there, the move to real tetration is the same sort of "easy" as defining real exponentiation.) And what does [imath]{^z}a[/imath] mean for complex [imath]z[/imath]? |
1045635 | Why is the area of the circle [imath]πr^2[/imath]?
I searched many times about the cause of the circle area formula but I did not know anything so ... Why is the area of the circle [imath]\pi r^2[/imath]? Thanks for all here. | 466762 | Why is [imath]\pi r^2[/imath] the surface of a circle
Why is [imath]\pi r^2[/imath] the surface of a circle? I have learned this formula ages ago and I'm just using it like most people do, but I don't think I truly understand how circles work until I understand why this formula works. So I want to understand why it works and not just how. Please don't use complicated symbols. |
1046328 | Show that T is nilpotent if and only if all of [imath]tr(T^n)[/imath], where n is a positive integer are zero.
Let [imath]V[/imath] be a finite dimensional vector space over [imath]\mathbb{C}[/imath] and let [imath]T: V \to V[/imath] be a linear operator. We say that T is nilpotent if [imath]T^p = 0_V[/imath] for some positive integer p. Show that T is nilpotent if and only if all of [imath]tr(T^n)[/imath], where n is a positive integer are zero. I can prove T is nilpotent implies all of [imath]tr(T^n)[/imath] are zero, but I can't prove the other direction. | 191879 | Nilpotent matrices over field of characteristic zero
From this Wikipedia link. Let [imath]K[/imath] be a field of characteristic zero and let [imath]A[/imath] be an [imath]n \times n[/imath] matrix over [imath]K[/imath]. Prove the following: (a) [imath]N[/imath] is nilpotent iff [imath]$\mathrm{tr}(N^m)=0$[/imath] for all [imath]0<m \leq n[/imath] iff [imath]$\mathrm{tr}(N^m)=0$[/imath] for all [imath]m \in \mathbb{N_{+}}[/imath]. (b) If [imath]N[/imath] is nilpotent, then [imath]N[/imath] is similar to a strictly upper triangular matrix. Additional Query: Is above true if [imath]K[/imath] is not assumed to be algebraically closed? (Then one can't apply Jordan normal form anymore.) |
1046360 | Complex numbers property proof.
I eas given this quesstion in one of my Linear Algebra course with the excercises regarding minimal polynomialsm eigenvalues and diagonalizable matrix: Show that for any two numbers [imath]a,b \in \mathbb{C}[/imath] exist [imath]c,d \in \mathbb{C}[/imath] such that [imath]c+d=a[/imath] and [imath]cd=b[/imath]. I have no idea if and how it's related to the material we just covered, and have no clue how to attack it. I tried prooving it straight forward algebrically, but it's just got messy. Any hints? clues? ideas? Thanks :) | 1042574 | Prove that [imath]\forall a,b\in\mathbb{C} \exists c,d \in \mathbb{C}: c+d = a \land cd = b[/imath]
Prove that: [imath]\forall a,b\in\mathbb{C} \exists c,d \in \mathbb{C}: c+d = a \land cd = b[/imath] We just learned about the characteristic/minimal polynomial and diagonalization but I am not sure if it has something to do with this question. I'd be glad for guidance! |
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