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766509 | Prove that the limit of the function a sequence is the same as the function of the limit of the sequence
Assume that [imath]\lim_{n\to \infty}a_n=a.[/imath] Suppose that the function [imath]f[/imath] is continuous everywhere including at [imath]a[/imath]. Form the sequence [imath](f(a_n))_{n=1}^{\infty}[/imath]. Prove that [imath]\lim_{n\to \infty}f(a_n)=f(\lim_{n\to \infty}a_n).[/imath] So show [imath]\lim_{n\to \infty}f(a_n)=f(a).[/imath] Help me, I am lost on how to even begin. | 588458 | Show that: lim[imath]_{n\rightarrow \infty}\,f(x_n)=f(\text{lim}_{n \rightarrow \infty} \, x_n)[/imath]
Let [imath]f:X \rightarrow Y[/imath] be a function. Prove that if [imath]f[/imath] is continuous, then for every convergent sequence [imath](x_n)[/imath] lim[imath]_{n\rightarrow \infty}\,f(x_n)=f(\text{lim}_{n \rightarrow \infty} \, x_n)[/imath] My attempt: Assume [imath]f[/imath] is continuous. Then [imath]\forall \epsilon >0, \exists \delta > 0[/imath] such that [imath]\forall p \in X[/imath] and [imath]q \in X: d(p,q) < \delta \implies d(f(q),f(q)) < \epsilon.[/imath] And also assume that [imath](x_n)[/imath] converges. Then [imath]\exists L[/imath] such that [imath]\forall \epsilon >0, \exists N[/imath] such that [imath]n>N \implies d(x_n, L) < \epsilon.[/imath] I need to use these definitions to show lim[imath]_{n\rightarrow \infty}\,f(x_n)=f(\text{lim}_{n \rightarrow \infty} \, x_n)[/imath] . Any hints? Edit: lim[imath]_{n \rightarrow \infty}(x_n) \in X[/imath] |
766665 | Computing [imath]\lim_{x\to 0^+} x^{\sin(x)}[/imath]
I'm trying to compute [imath] \lim_{x\to 0^+} x^{\sin(x)}. [/imath] I've been trying to get it into a form where I can apply L'Hopital's rule, but I haven't had any success. Namely, I haven't been able to massage it into a fraction form. Any advice? | 629459 | Why does [imath]\lim_{x\to 0^+} x^{\sin x}=1[/imath]?
I tried writing [imath]y=x^{\sin x}[/imath], so [imath]\ln y=\sin x\ln x[/imath]. I tried to rewrite the function as [imath]\ln(x)/\csc x[/imath] and apply l'Hopital to the last function, but it's a mess. Is there a by hand way to do it? |
766943 | Last two digit of number raised to exponents.
Find the last two digits of [imath]3^{3^{2014}}[/imath]. Attempt: First I try to work with [imath]3^{2014}[/imath]. So we can work on [imath]\text{mod 10}[/imath]. Then, [imath]\begin{align}&3^1 \equiv 3\pmod{10}\\ &3^2 \equiv 9 \pmod{10}\\ &3^3 \equiv 27 \equiv 7\pmod{10}\\ &3^4 \equiv 81 \equiv 1\pmod{10}\end{align}[/imath] So [imath]3^{2014} = 3^{(4\cdot503) + 2} \equiv 3^0 \cdot 3^2 \equiv 9 \pmod{10}[/imath]. So the last digit is 9. However I am ask to find the last two digit. Can anyone please help me? I would really appreciate all the help. | 765530 | Last digits, numbers
Can anyone please help me? 1) Find the last digit of [imath]7^{12345}[/imath] 2) Find the last 2 digits of [imath]3^{3^{2014}}[/imath]. Attempt: 1) By just setting the powers of [imath]7[/imath] we have [imath]7^1 = 7[/imath], [imath]7^2=49[/imath], [imath]7^3=343[/imath], [imath]7^4 = 2401[/imath], [imath]7^5 = 16807[/imath], [imath]7^6 = 117649[/imath], [imath]\dots[/imath] After the power of [imath]4[/imath], the last digits will repeat. Then by noticing the pattern the digits will end in [imath]7,9,3[/imath] and [imath]1[/imath]. Then we can divide the exponent [imath](12345)[/imath] by [imath]4[/imath] since this is the cycle that makes it repeat. Then [imath]12345 : 4[/imath] has remainder [imath]1[/imath]. So [imath]7^1 = 7[/imath] is the unit digit to [imath]7^{12345}[/imath]. So the last digit is [imath]7[/imath]. I know how to do it like this, the problem does not state how to find the last digit, but I know it has something do do with Euler's theorem. for part [imath]2[/imath]) I don't know how to start. Can anyone please help me? Thank you for the help. |
536770 | Prove if [imath]f(x) = g(x)[/imath] for each rational number x and [imath]f[/imath] and [imath]g[/imath] are continuous, then [imath]f = g[/imath]
[imath]f,g: \mathbb{R} \to \mathbb{R}[/imath] I'd like to see a sketch for this proof. [sorry for posting errors, I am on a cell phone] | 1244431 | Let [imath]f,g[/imath] be continuous from [imath]\mathbb R[/imath] to [imath]\mathbb R[/imath]
Let [imath]f, g[/imath] be continuous from [imath]\mathbb R[/imath] to [imath]\mathbb R[/imath], and suppose that [imath]f(r) = g(r)[/imath] for all rational numbers [imath]r[/imath]. Is it true that [imath]f(x) = g(x)[/imath] for all [imath]x \in \mathbb R[/imath]? |
589891 | Establish that 7 is a primitive root of any prime of the form [imath]p = 2^{4n} + 1[/imath].
Establish that 7 is a primitive root of any prime of the form [imath]p = 2^{4n} + 1[/imath]. [Hint: Because [imath]p ≡ 3[/imath] or [imath]5 \ (mod \ 7)[/imath], [imath](7/p) = (p/7)=-1[/imath]]. I get that [imath]p ≡ 2, 3, 5 \ (mod \ 7)[/imath]. Not only 3,5. How should i think? | 395652 | For every prime of the form [imath]2^{4n}+1[/imath], 7 is a primitive root.
What I want to show is the following statement. For every prime of the form [imath]2^{4n}+1[/imath], 7 is a primitive root. What I get is that [imath]7^{2^{k}}\equiv1\pmod{p}[/imath] [imath]7^{2^{k-1}}\equiv-1\equiv2^{4n}\pmod{p}[/imath] [imath]7^{2^{k-2}}\equiv(2^{n})^2\pmod{p}[/imath] Thus [imath](\frac{2^n}{p})=(\frac{7^{2^{k-2}}}{p})=1[/imath]. I think that [imath]7[/imath] is important because [imath]7[/imath] is a primitive root but I don't know how to use [imath]7[/imath]. |
767577 | Convergence problem in different norms
We remember the definition of strong convergence of a sequence in [imath]B(H)[/imath] for [imath]H[/imath] a Hilbert space: [imath]a_n\rightarrow a[/imath] iff [imath]\left\|a_nx-ax\right\|\rightarrow0[/imath] for all [imath]x\in H[/imath]. In general norm convergence implies strong convergence, but the converse isn't true in general. But it seems to be special cases: Let [imath](a_n)[/imath] a sequence in [imath]B(H)[/imath] which converges strongly to an operator [imath]a\in B(H)[/imath]. Furthermore assume that [imath]b[/imath] is a compact operator. The claim is that the sequence [imath](a_nb)[/imath] converges in norm (operator norm) to [imath]ab[/imath]. I can't find a good reason of estimation to conclude that this should be true. Moreover if we take the sequence [imath](ba_n)[/imath] then the conclusion that this converges to [imath]ba[/imath] isn't true. I think you have to use projections to find the counterexample, but more interesting is the question why [imath](a_nb)[/imath] well converges. Can someone help me with this problem? Thank you very much. | 528649 | Question about SOT and compact operators
I need some help with functional analysis / Hilbert space theory. If you have a favorite text to recommend, please let me know~ Here is my question: Given [imath]v_t[/imath] be the "squeeze operator" on [imath]H=L^2[0, 1][/imath], where [imath]v_t: L^2[0,1] \to L^2[0, \frac{2-t}{2}][/imath] acts on [imath]f \in L^2[0,1][/imath] by squeezing the domain of the function. We have that [imath]\{v_t\}[/imath] for [imath]t \in [0, 1][/imath] is a family of SOT continuous operators. I am wondering why given any [imath]p \in \mathbb{K}(H)[/imath], we have [imath]\{ v_tpv_t^* \}[/imath] is continuous in norm. I found the following facts on a reference suggested by Wikipedia (Hilbert Space Operators in Quantum Physics), but I am not sure how to prove them or how I may use them... [imath]T_n \to^{SOT} T[/imath] implies that for any [imath]p \in \mathbb{K}(H)[/imath], we have [imath]T_np \to Tp[/imath] in norm. Hints or suggestions would be greatly appreciated! |
765123 | How to use Fermat's little theorem to find [imath]50^{50}\pmod{13}[/imath]?
I don't understand how to use Fermat's little theorem to find remainders e.g if we are asked to find remainder of [imath]50^{50}[/imath] on division by [imath]13[/imath], what is a and what is [imath]p[/imath] in the formula? Also I wanted to check can we use both congruence classes as well as Fermat's little theorem to find such remainders? How would be write [imath]50^{50}[/imath] on division by [imath]13[/imath] in terms of congruence classes to find the remainder? I don't exactly understand what a congruence class is, what does it tell us and how is it related to this? | 1820027 | Find the remainder when [imath]11^{12}[/imath] is divided by [imath]13[/imath].
I am looking for an easier way than mine to solve the problem. Problem: Find the remainder when [imath]11^{12}[/imath] is divided by [imath]13[/imath]. Here is what I did. I simplify [imath]11^{12}[/imath] mod [imath]13[/imath] = [imath](–2)^{12}[/imath] mod [imath]13[/imath]. Then I don't know how to simply further. |
758538 | Equality of integrals: [imath] \int_{0}^{\infty} \frac {1}{1+x^2} \, \mathrm{d}x = 2 \cdot \int_{0}^{1} \frac {1}{1+x^2} \, \mathrm{d}x [/imath]
In Street-Fighting Mathematics (page 16), Prof. Sanjoy Mahajan states that [imath] \displaystyle\int_{0}^{\infty} \frac {1}{1+x^2} \, \mathrm{d}x = 2 \cdot \displaystyle\int_{0}^{1} \frac {1}{1+x^2} \, \mathrm{d}x [/imath] I realize this is equivalent to saying that [imath]\displaystyle\int_{0}^{1} \frac {1}{1+x^2} \, \mathrm{d}x = \displaystyle\int_{1}^{\infty} \frac {1}{1+x^2} \, \mathrm{d}x, [/imath] but why is this true? | 757569 | Showing that [imath]\int_{0}^{\infty} \frac{dx}{1 + x^2} = 2 \int_0^1 \frac{dx}{1 + x^2}[/imath]
I was reading an article in which it was stated that, with a change of variable, one could show that: [imath]\int_{0}^{\infty} \frac{dx}{1 + x^2} = 2 \int_0^1 \frac{dx}{1 + x^2}[/imath] I tried with [imath]t = 1 + \frac{1}{x}[/imath] but that doesn't work out, especially because the lower bound doesn't become [imath]0[/imath]. |
767161 | Concept of knots in B-splines
Given [imath]n+1[/imath] control points, B-spline blending functions are polynomials of degree [imath]d-1[/imath], [imath](1<d\leq n+1)[/imath]. This much is easy to comprehend. Now comes the part I am not able to make any sense of. Each polynomial function is defined over [imath]d[/imath] subintervals of the total range of [imath]u[/imath]. The selected set of subinterval endpoints [imath]u_j[/imath] is referred to as a knot vector. What is the purpose of knot vector? What is its physical significance? | 766330 | Physical significance of knot vector in B-spline.
A B-spline blending curve formulation is: [imath]P(u)=\sum_{k=0}^np_k B_{k,d}(u)[/imath] Given [imath]n+1[/imath] control points, B-spline blending functions are polynomials of degree [imath]d-1[/imath], [imath](1<d<=n+1)[/imath]. This much is easy to comprehend. Now comes the part I am not able to make any sense of. Each polynomial function is defined over [imath]d[/imath] subintervals of the total range of [imath]u[/imath]. The selected set of subinterval endpoints [imath]u_j[/imath] is referred to as a knot vector. What is the purpose of knot vector? What is its physical significance? |
251795 | Problem when integrating [imath]e^x / x[/imath].
I made up some integrals to do for fun, and I had a real problem with this one. I've since found out that there's no solution in terms of elementary functions, but when I attempt to integrate it, I end up with infinite values. Could somebody point out where I go wrong? So, I'm trying to determine: [imath] \int{\frac{e^x}{x}} \, dx [/imath] Integrate by parts, where [imath]u = 1/x[/imath], and [imath]v \, ' = e^x[/imath]. Then [imath]u \, ' = - 1/x^2[/imath], and [imath]v=e^x[/imath]. So, [imath]\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \int{\frac{e^x}{x^2}} \, dx[/imath] Integrate by parts again, [imath]u = 1/x^2[/imath], [imath]v \, ' = e^x[/imath], so that [imath]u \, ' = -2/x^3[/imath] and [imath]v=e^x[/imath]. So, [imath]\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + 2\int{\frac{e^x}{x^3}} \, dx[/imath] Repeat this process ad infinitum to get, [imath]\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + 2 \left( \frac{e^x}{x^3} + 3 \left( \frac{e^x}{x^4} + 4 \left( \frac{e^x}{x^5} + \, \cdots \right) \right) \right) [/imath] Expanding this gives, [imath]\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} + \frac{e^x}{x^2} + \frac{2e^x}{x^3} + \frac{6 e^x}{x^4} + \frac{24 e^x}{x^5} + \cdots [/imath] And factoring that gives, [imath]\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} \left( 1 + \frac{1}{x} + \frac{2}{x^2} + \frac{6}{x^3} + \frac{24}{x^4} + \cdots \right) [/imath] Now, considering the series itself, the ratio between the [imath]n^{th}[/imath] term and the [imath](n-1)^{th}[/imath] term = [imath]\Large \frac{n}{x}[/imath]. Eventually, [imath]n[/imath] will be larger than [imath]x[/imath], so the ratio between successive terms will be positive, so (assuming [imath]x[/imath] is positive), the series diverges, meaning (and I'm sure everybody will cringe upon seeing notation used like this), that: [imath]\int{\frac{e^x}{x}} \, dx = \frac{e^x}{x} \left( \infty \right) = \infty [/imath] | 2109927 | How to solve this integration question?
I have been trying to solve this question with the simple methods I knew but couldn't come to the solution. I would appreciate a bit of hint to solve this integral problem? [imath]\;\displaystyle\int\frac {e^x}x\,dx[/imath] Edit : Can't it be solved even if certain upper and lower limits are involved? |
768614 | The Proof of Wilson's Theorem using the auxiliary multiplicative modulous group
(self answered question, thanks for the hints Derek Holt provided:-)) problem 18,section 4 chapter 2 in Herstein's abstract algebra: Using the results of Problem 15 and 16,prove that if p is an positive odd prime number, then [imath](p-1)!\equiv -1 \pmod p[/imath]. (self answered question, thanks for the hints Derek Holt provided:-)) Problem 15: If [imath]x^2\equiv 1 \pmod p[/imath], then [imath]x\equiv 1\pmod p[/imath] and/or [imath]x\equiv -1\pmod p[/imath] Problem 16: If [imath]G=[/imath]{[imath]a_k|1\leq k\leq n[/imath]} is a finite abelian group, then [imath](\prod^n_1 a_k)^2=e.[/imath] | 307 | Prove that [imath](n-1)! \equiv -1 \pmod{n}[/imath] iff [imath]n[/imath] is prime [Wilson's Theorem]
How can I show that [imath](n-1)![/imath] is congruent to [imath]-1 \pmod{n}[/imath] if and only if [imath]n[/imath] is prime? Thanks. |
768550 | Legendre symbols with huge "p"s
I am in doubt with an exercise - I need to calculate [imath]\left(205\mid853\right)[/imath] I would use the fact Legendre's symbols are multiplicative, but then I would have something like [imath]\left(5\mid853\right)\left(41\mid853\right)[/imath] and it would not help much with Gauss lemma, since it would mean having sequences [imath](853-1)/2[/imath] longs to multiply, to operate [imath]\mod 853[/imath], and so on. Is there a faster trick I don't see? Thanks in advance. | 5424 | How can I find out whether a number is a quadratic residue in a large modulo?
Without strenuous arithmetic. Is there a program I can download to do so? What are the quadratic residues modulo [imath]5^4[/imath] or [imath]5^5[/imath]? Thanks! |
769174 | How to find the value of the infnite product [imath]\frac {7}{9}[/imath]x[imath]\frac {26} {28}[/imath]x [imath]\frac {63} {65}[/imath]x . . . x[imath]\frac {n^3-1} {n^3+1}[/imath]x . .
How to find the value of the infnite product [imath]\frac {7}{9}[/imath]x[imath]\frac {26} {28}[/imath]x [imath]\frac {63} {65}[/imath]x . . . x[imath]\frac {n^3-1} {n^3+1}[/imath]x . . Somebody please help me. | 462082 | How to compute [imath]\prod\limits^{\infty}_{n=2} \frac{n^3-1}{n^3+1}[/imath]
How to compute [imath]\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}\ ?[/imath] My Working : [imath]\prod^{\infty}_{n=2} \frac{n^3-1}{n^3+1}= 1 - \prod^{\infty}_{n=2}\frac{2}{n^3+1} = 1-0 = 1[/imath] Is it correct |
769331 | Showing definition for second derivative
Suppose [imath]f[/imath] is two-times differentiable at [imath]x[/imath]. Show that [imath] \lim_{h\to 0} \frac{f(x+h) - 2f(x) + f(x-h)}{h^2} = f''(x). [/imath] | 634885 | Proving [imath]\displaystyle\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)[/imath]
The function [imath]f[/imath] is differentiable twice at x. Prove that: [imath]\lim_{h\to0}\frac{f(x+h)-2f(x)+f(x-h)}{h^2}=f''(x)[/imath] Hint: use Peano's remainder (if [imath]f:I\to\mathbb R[/imath] is differentiable [imath]n[/imath] times on [imath]a\in I[/imath] then [imath]R_k(x)=o(|x-a|^k), \ x\to a[/imath]). I just don't see the connection here between the second derivative and the Taylor series which has to do with Peano's remainder... The remainder is defined to be the difference between the function and polynomial but what does it has to do in this case ? |
769364 | If a newborn baby is born a girl with probability [imath]p[/imath] and a boy with probability [imath]1-p[/imath]
Define the events [imath]A[/imath] = "both children are girls", [imath]B[/imath] = "at least one of the children is a girl" What is the probability [imath]Pr(A|B)[/imath]? Can someone explain how to do this? Both children are born independently | 762437 | Conditional Probability with boys and girls
Assume that a newborn baby is a girl with probability [imath]p[/imath] and a boy with probability [imath]1-p[/imath]. Also assume that the genders of dierent newborns are independent of each other. Consider a person who has two children. Define the following two events: A = "both children are girls" B = "at least one of the children is a girl" What is the conditional probability [imath]Pr(A \vert B)[/imath]? My attempt is to use the conditional probability formula [imath]P(A\vert B) = \frac{P(A\cap B)}{P(B)}[/imath] [imath]S = \{ BB, GG, GB, BG \}[/imath] [imath]P(A) = p\cdot p[/imath] since for both children to be girls [imath]P(B) = 2\cdot\{(1-p)\cdot p\} + (p)^2 = 2p - p^2[/imath] since the probability for at least one girl means the probability [imath]GG[/imath] plus [imath]GB[/imath] and [imath]BG[/imath], with NO [imath]BB[/imath] [imath]P(A\cap B) = ?[/imath] How do I get the intersection? I know the only intersection is [imath]GG[/imath] which is [imath]\frac{1}{4}[/imath] but I need to express it as a probability. My guess is that the intersection is [imath]p^2[/imath] since that is the [imath]P(A)[/imath], but I'm not sure, can some one convince me if I'm right? [imath]P(A\vert B) = \frac{p^2}{2p-p2} =\frac{p}{2-p}[/imath] Is this correct? |
770002 | How can the power law for sequences of real numbers be proved?
Let [imath](a_n)_n[/imath] and [imath](b_n)_n[/imath] be two sequences that converge to [imath]a[/imath], and respectively [imath]b[/imath] (with [imath] a_n \ge 0 , \forall n [/imath]). Prove that: [imath] \lim_{n \to \infty}{a_n}^{b_n} = a^b [/imath] How can you do that? An approach using the epsilon-convergence theorem seems kind of unpractical to me. Can you please help me? Edit: Please, stop marking my question as a duplicate! I'm intrested in a proof for sequences, not for limits! I began studying real analysis by learnin the sequences of real numbers first. Suppose that I don't know that: [imath]if \lim_{x \to k} {f(x)} ^{g(x)} = a^b, if\lim_{x \to k}f(x) =a, and\lim_{x \to k}g(x) =b[/imath] If I don'twant to use this, then how do I prove the statement? | 739133 | Limit Rule [imath]\lim f(x)^{g(x)}[/imath]
I want to know the following is true : If [imath] c,\ d\in {\bf R},\ \lim_{x\rightarrow 0} f(x)=c>0,\ \lim_{x\rightarrow 0} g(x) =d>0[/imath] then [imath] \lim_{x\rightarrow 0} f(x)^{g(x)} = c^d[/imath] In calculus book such formula cannot be found. Consider the problem : [imath]\lim_{x\rightarrow 0} (1+\sin\ 4x)^{\cot\ x} [/imath] To find a limit, we must use [imath]{\rm log}[/imath] and L'Hospital. But some student suggests that [imath] \lim_{x\rightarrow 0} (1+\sin\ 4x)^{\frac{1}{\sin\ 4x} \frac{\sin\ 4x}{\sin\ x}\cos\ x}=e^4[/imath] This argument is clear. But I know that it is informal. Is there a minus point in such way ? |
770517 | Algebraic structure question
Suppose there exist subgroups [imath]H,K\subseteq G[/imath] such that [imath]G = H \cup K[/imath]. Prove that [imath]H = G[/imath] or [imath]K = G[/imath]. I am not sure how to start? Thanks in advance. | 768653 | How can I Prove that H=G or K=G
Let subgroups [imath]H,K \subseteq G[/imath] such that [imath]G=H \cup K[/imath] Prove [imath]H=G[/imath] or [imath]K=G[/imath] |
770319 | Primes, Pseudo primes and poulet numbers
Prove the following statement! If [imath]a^2 -1[/imath] is nor divisible by prime [imath]p[/imath], then [imath]M[/imath] is Fermat pseudo prime, where [imath]M = AB[/imath] and [imath]A = (a^p -1)/(a-1)[/imath] and [imath]B = (a^p + 1)/(a+1)[/imath] | 694439 | Showing that [imath]n[/imath] is pseudoprime to the base [imath]a[/imath]
Show that if [imath]n=\frac{a^{2p}-1}{a^2-1},[/imath] where [imath]a[/imath] is an integer, [imath]a>1[/imath], and [imath]p[/imath] is an odd prime not dividing [imath]a(a^2-1)[/imath], then [imath]n[/imath] is pseudoprime to the base [imath]a[/imath]. Let [imath]n=\frac{a^{2p}-1}{a^2-1}=1+a^2+(a^2)^2+(a^2)^3+\cdots[/imath]. Then [imath]n-1=\frac{a^{2p}-1}{a^2-1}-1=\frac{a^{2p}-a^2}{a^2-1}=a^2\frac{(a^2)^{p-1}-1}{a^2-1}.[/imath] I have kind of hit a roadblock with this proof. I'm stuck trying to show that [imath]n-1[/imath] is even and that [imath]p | (a^{2p}-1)\equiv 0 \mod p[/imath]. Is there some obvious steps that I'm missing |
770446 | False proof? If [imath]f: G \rightarrow H[/imath] homomorphism, and [imath]H[/imath] abelian, then [imath]G[/imath] abelian?
Let [imath]f: G \rightarrow H[/imath] be homomorphism, and [imath]H[/imath] is abelian. So [imath]G \big/ \ker f \cong \operatorname{im}f[/imath]. Since [imath]\operatorname{im}f[/imath] is abelian, so is [imath]G \big/ \ker f[/imath]. So for every [imath]g_1,g_2 \in G[/imath]: [imath]g_1\ker f\cdot g_2\ker f=g_2\ker f \cdot g_1\ker f \Rightarrow g_1g_2\ker f=g_2g_1\ker f[/imath], hence [imath]g_1g_2=g_2g_1[/imath], so [imath]G[/imath] is abelian. | 761860 | proof verification: If [imath]f:G \rightarrow H[/imath] is group homomorphism, and [imath]H[/imath] is abelian, then [imath]G[/imath] is abelian
If [imath]f:G \rightarrow H[/imath] is group homomorphism, and [imath]H[/imath] is abelian, then [imath]G[/imath] is abelian. Is that statement correct? Here's my attempt of proof: Let [imath]a,b \in G[/imath], then: [imath]ab=f^{-1}f(ab)=f^{-1}f(a)f(b)=f^{-1}f(b)f(a)=f^{-1}f(ba)=ba[/imath] Is this correct? I relied on first on [imath]f[/imath] being group homomorphism, then [imath]H[/imath] being abelian. |
770999 | [imath]2[/imath] square matrices with equal rank
Suppose [imath]P[/imath] and [imath]Q[/imath] are [imath]n\times n[/imath] matrices of real numbers such that [imath]P^2=P[/imath] [imath]Q^2=Q[/imath] [imath]I−P−Q[/imath] is invertible, where [imath]I[/imath] is a [imath]n\times n[/imath] identity matrix. Show that [imath]P[/imath] and [imath]Q[/imath] have the same rank. | 630942 | Necessary condition for have same rank
Let [imath]P,Q[/imath] real [imath]n\times n[/imath] matrices such that [imath]P^2=P[/imath] , [imath]Q^2=Q[/imath] and [imath]I-P-Q[/imath] is an invertible matrix. Prove that [imath]P[/imath] and [imath]Q[/imath] have the same rank. Some help with this please , happy year and thanks. |
771320 | Evaluate [imath] \int_{-\pi/2}^{\pi/2} \frac{\cos(x)}{1+e^x} dx[/imath]
[imath] \int_{-\pi/2}^{\pi/2} \frac{\cos(x)}{1+e^x} dx[/imath] Using integration by parts [imath]\int u v = u \int v - \int u' \int v[/imath] with [imath]u(x) = \cos(x)[/imath] and [imath]v(x) = \frac{1}{1+e^x}[/imath], [imath] u \int_{-\pi/2}^{\pi/2} v dx = \left[ \cos(x) \int_{-\pi/2}^{\pi/2} \frac{1}{1+e^x} dx \right]_{-\pi/2}^{\pi/2} = \left[ \cos(x) ( x - log(1+e^x) \right]_{-\pi/2}^{\pi/2} [/imath] [imath] \int_{-\pi/2}^{\pi/2} u' \left( \int_{-\pi/2}^{\pi/2} v dx \right) dx = \int_{-\pi/2}^{\pi/2} -\sin(x) \int_{-\pi/2}^{\pi/2} \frac{1}{1+e^x} dx [/imath] [imath] = \int_{-\pi/2}^{\pi/2} -\sin(x) \left[ ( x - log(1+e^x) \right]_{-\pi/2}^{\pi/2} [/imath] Short of continuing ad nauseam, is there a better way to determine the answer (given as [imath]1[/imath]) ? | 60045 | Showing that [imath]\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx[/imath], when [imath]f[/imath] is even
I have a question: Suppose [imath]f[/imath] is continuous and even on [imath][-a,a][/imath], [imath]a>0[/imath] then prove that [imath]\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx[/imath] How can I do this? Don't know how to start. |
771078 | [imath]|G':G''| \le p^2[/imath] implies [imath]G'[/imath] is abelian
If [imath]G[/imath] is a finite [imath]p[/imath]-group, and [imath]|G':G''| \le p^2[/imath], then [imath]G'[/imath] is an abelian group. I'm reading its proof but I cannot understand a part: Suppose [imath]G''\neq1[/imath], then by a theorem, there exists a normal subgroup [imath]K[/imath] of [imath]G[/imath] contained as a subgroup of index [imath]p[/imath] in [imath]G''[/imath]. Then [imath](G/K)'=G'/K[/imath] and [imath](G/K)''=G''/K \neq 1[/imath]. So by another theorem, the center [imath]Z/K[/imath] of [imath]G'/K[/imath] is not cyclic, and so [imath]|Z:K| \ge p^2[/imath]. "" Hence [imath]|G':Z| \le p[/imath] "", and so [imath]G'/K/Z/K[/imath] is cyclic. Thus [imath]G'/K[/imath] is abelian, contradiction. Why is it true that [imath]|G':Z| \le p[/imath]? cf. The above proof looks simple but uses other theorems a lot. Does anyone know another good proof? | 757678 | If [imath][G' : G'']\leq p^2[/imath], then [imath]G'[/imath] is abelian.
Problem : Let [imath]G[/imath] be a p-group and [imath]G'[/imath] denote the commutator subgroup of [imath]G[/imath]. If [imath][G' : G'']\leq p^2[/imath], then [imath]G'[/imath] is abelian. It is easy to prove it for the case of [imath][G' : G'']=1[/imath] since G is solvable. But it is too hard to prove the other cases. Please give me any advice or solutions. Thanks in advance. |
407420 | Evaluating [imath]\int_{-1}^{1}\frac{\arctan{x}}{1+x}\ln{\left(\frac{1+x^2}{2}\right)}dx[/imath]
This is a nice problem. I am trying to use nice methods to solve this integral, But I failed. [imath]\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx, [/imath] where [imath]\arctan{x}=\tan^{-1}{x}[/imath] mark: this integral is my favorite one. Thanks to whoever has nice methods. I have proved the following: [imath]\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=\sum_{n=1}^{\infty}\dfrac{2^{n-1}H^2_{n-1}}{nC_{2n}^{n}}=\dfrac{\pi^3}{96}[/imath] where [imath]C_{m}^{n}=\dfrac{m}{(m-n)!n!},H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}[/imath] I also have got a few by-products [imath]\int_{-1}^{1}\dfrac{\arctan{x}}{1+x}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=-I_{1}-2I_{2}[/imath] where [imath]I_{1}=\int_{0}^{1}\dfrac{\ln{(1-x^2)}}{1+x^2}\ln{\left(\dfrac{1+x^2}{2}\right)}dx=\dfrac{\pi}{4}\ln^2{2}+\dfrac{\pi^3}{32}-2K\times\ln{2}[/imath] and [imath]I_{2}=\int_{0}^{1}\dfrac{x\arctan{x}}{1+x^2}\ln{(1-x^2)}dx=-\dfrac{\pi^3}{48}-\dfrac{\pi}{8}\ln^2{2}+K\times\ln{2}[/imath] and same methods,I have follow integral [imath]\int_{0}^{1}\dfrac{\ln{(1-x^4)}\ln{x}}{1+x^2}dx=\dfrac{\pi^3}{16}-3K\times\ln{2}[/imath] where [imath] K [/imath] denotes Catalan's Constant. | 2401903 | Hard Integral [imath]\int\limits_{-1}^{1}{\frac{\arctan x}{1+x}\ln \frac{1+x^{2}}{2}dx}=\frac{\pi ^{3}}{96}[/imath]
I am trying to calculate the integral [imath]\int\limits_{-1}^{1}{\frac{\arctan x}{1+x}\ln \frac{1+x^{2}}{2}dx}=\frac{\pi ^{3}}{96}[/imath] My idea is to use and try to calculate and prove that [imath]\int\limits_{-1}^{1}{\frac{\arctan x}{1+x}\ln \frac{1+x^{2}}{2}dx}=-\int\limits_{0}^{1}{\frac{\ln \left( 1-x^{2} \right)}{1+x^{2}}\ln \frac{1+x^{2}}{2}dx}-\int\limits_{0}^{1}{\frac{2x\arctan x}{1+x^{2}}\ln \left( 1-x^{2} \right)dx}[/imath] [imath]\frac{2x\arctan x}{1+x^{2}}=\operatorname{Re}\left( \frac{\ln \left( 1+ix \right)}{1+ix} \right)-\operatorname{Re}\left( \frac{\ln \left( 1-ix \right)}{1+ix} \right)[/imath] [imath]\int\limits_{0}^{1}{\frac{\ln \left( 1-x^{2} \right)}{1+x^{2}}\ln \frac{1+x^{2}}{2}dx}=\frac{\pi ^{3}}{32}+\frac{\pi }{4}\ln ^{2}2-2C\ln 2[/imath] [imath]\int\limits_{0}^{1}{\frac{\ln \left( 1\pm ix \right)\ln \left( 1\pm x \right)}{1+ix}dx}[/imath] |
772174 | solve [imath]x+\sin(x)=k[/imath] for [imath]x[/imath]
This question has been proposed to me and thus far it has baffled me: [imath] x + \sin(x) = k[/imath] solve for x. Another way of looking at it is find [imath]f^{-1}(x)[/imath] given that [imath]f(x)=x + \sin(x)[/imath]. Wolfram alpha doesn't even seem to know how to solve it, when you plug in a value for k, it tell you approximately the number but does not give an "exact" form. Any help? | 257873 | Why isn't the inverse of the function [imath]x\mapsto x+\sin(x)[/imath] expressible in terms of "the functions one finds on a calculator"?
The function [imath]f(x)=x+\sin(x)[/imath] is easily checked to be a bijection from the reals to itself, and so it has a unique inverse [imath]y\mapsto g(y)[/imath] such that [imath]f\circ g=g\circ f[/imath] are both the identity map. Now [imath]g[/imath] will almost certainly be a function which is not expressible using "the functions in a high-schooler's toolkit" (by which I guess I mean [imath]\exp[/imath], [imath]\log[/imath], and, if you like, the usual trigonometric functions and their friends like [imath]\sinh[/imath], although of course these can all be of course built from exponentials anyway). For purely recreational reasons (stemming from conversations I've had whilst teaching undergraduates) I'm interested in how one proves this sort of thing. A few years ago I was interested in a related question, and took the trouble to learn some differential Galois theory. My motivation at the time was learning how to prove things like why [imath]h(t):=\int_0^t e^{x^2} dx[/imath] is not expressible in terms of these calculator-button functions (I'm sure there's a better name for them but I'm afraid I don't know it). I've realised that since then I've forgotten most of what I knew, but furthermore I am also unclear about whether this is the way one is supposed to proceed. Is the idea that I come up with some linear differential equation satisfied by [imath]g[/imath] and then apply some differential Galois theory technique? In fact, one of the many things that I have forgotten is the following: if [imath]F[/imath] is a field equipped with a differential operator [imath]D[/imath], and [imath]E/F[/imath] is the field extension obtained by adding a non-zero root of [imath]Dh=ch[/imath], with [imath]c\in F[/imath], then the Galois group of [imath]E/F[/imath] is solvable, whereas the equation itself might not be, in terms of calculator-button functions, if I can't integrate [imath]c[/imath]. Can a more enlightened soul explain to me how one is supposed to proceed? I wonder whether I am somehow conflating two ideas and the differential Galois theory business is a red herring, but it seemed simpler to ask rather than continuing to flounder around. |
772176 | Limit as [imath]x[/imath] goes to [imath]0[/imath] of [imath]x^x[/imath]
[imath]\lim_{x \to 0}x^x[/imath] I know the answer is one but I have no idea how to get there. I tried taking a natural log and I think I need lhopitals rule but I keep going In circles. | 769644 | Evaluate [imath]\lim_{x \rightarrow 0} x^x[/imath]
I know when evaluated it gives one but I can't figure out how to prove it. Can anyone help? I believe it requires L'Hopitals rule and taking a natural log but i cannot figure out the exact math. |
772586 | How to prove [imath]\int_{-\pi}^\pi (f(x))^2dx\le \int_{-\pi}^\pi (f'(x))^2dx[/imath]
Let [imath]f[/imath] be [imath]C^1[/imath] in [imath][-\pi, \pi][/imath] and satisfies [imath]\int_{-\pi}^\pi f(x)dx=0[/imath], periodic boundary condition. Then, prove that [imath]\int_{-\pi}^\pi (f(x))^2dx\le \int_{-\pi}^\pi (f'(x))^2dx[/imath]. I try to prove it by Parseval's equality(with [imath]X_n=\sin nx[/imath]) and Schwartz inequality, but then some constants come out. Also why condition [imath]\int_{-\pi}^\pi f(x)dx=0[/imath] needs? Give some advice. | 619927 | Wirtinger Inequality: [imath]\int_0^1 f'^2 \geq \pi^2 \int_0^1 f^2 [/imath]
I have a basic question Let [imath]f[/imath] be a [imath]\mathcal{C}^1[/imath] map from [imath][0,1][/imath] into [imath]\mathbb{R}[/imath] vanishing in [imath]0[/imath] and [imath]1[/imath]. I want to prove that [imath]\int_0^1 f'^2 \geq \pi^2 \int_0^1 f^2 [/imath] and to see when we have equality. |
772914 | Real and Imaginary
[imath]Re\Big(({\frac{1+i\sqrt{3}}{1-i})^4\Big)} = 2[/imath] [imath]Im\Big(({\frac{1+i}{1-i})^5\Big)} = 1[/imath] I got that [imath]Re\Big(({\frac{1+i\sqrt{3}}{1-i})^4\Big)} = 1 \ne 2[/imath] And, that [imath]\Big(({\frac{1+i}{1-i})^5\Big)} = i [/imath] , which means that [imath]Im\Big(({\frac{1+i}{1-i})^5\Big)} = 1[/imath] Can you guys confirm that it's true? Thanks in advance! In this image I got to[imath](\frac{1−\sqrt3+i+i\sqrt3}{2})^4[/imath] and then, I'm not sure how to continue. | 772779 | Real and Imaginary
[imath]Re\Big(({\frac{1+i\sqrt{3}}{1-i})^4\Big)} = 2[/imath] [imath]Im\Big(({\frac{1+i}{1-i})^5\Big)} = 1[/imath] I got that [imath]Re\Big(({\frac{1+i\sqrt{3}}{1-i})^4\Big)} = 1 \ne 2[/imath] And, that [imath]\Big(({\frac{1+i}{1-i})^5\Big)} = i [/imath] , which means that [imath]Im\Big(({\frac{1+i}{1-i})^5\Big)} = 1[/imath] Can you guys confirm that it's true? Thanks in advance! |
773232 | Surjective continuous function [imath][0,1] \to [0,1]^2[/imath]
I was in class and the lecturer only mentioned that there exists such function. I didn't find anything about it on the internet. Is it simple? Maybe you could help me to understand that? thanks | 141958 | Why does the Hilbert curve fill the whole square?
I have never seen a formal definition of the Hilbert curve, much less a careful analysis of why it fills the whole square. The Wikipedia and Mathworld articles are typically handwavy. I suppose the idea is something like this: one defines a sequence of functions [imath]f_i(t) : [0,1]\to {\Bbb R}^2[/imath], and then considers the pointwise limit [imath]f(t) = \lim_{i\to\infty} f_i(t)[/imath]. But looking at the diagram, it is not clear that the sequence converges. I can imagine that we might think of following a single point in the range of [imath]f_i[/imath] as [imath]i[/imath] increases, and that point might move around, but only by [imath]2^{-i}[/imath] at each stage as we pass from [imath]f_i[/imath] to [imath]f_{i+1}[/imath], and so eventually approaches a limiting position. But I don't know how we'd get from there to showing that [imath]f_i(t)[/imath] converges for a particular value of the argument, especially for an argument other than a dyadic rational. And even if it does converge, every point in the range of each [imath]f_i[/imath] has at least one rational coordinate, so it is not at all clear why a point like [imath]({1\over\sqrt2}, {1\over\sqrt2})[/imath] should be in the range of [imath]f[/imath]. If there is an easy explanation, I would be glad to hear it, but I would also be glad to get a reference in English. |
773250 | Is there a sort of "two-sided semidirect product"?
Let [imath]G,H[/imath] be groups. Suppose we have both an action of [imath]G[/imath] on [imath]H[/imath], and an action of [imath]H[/imath] on [imath]G[/imath], both non-trivial. Let "[imath]\cdot[/imath]" define the former action, and [imath]\circ[/imath] define the latter. What can we say about the product: [imath] (g,h)\,(g',h') := \big(g\,(h\circ g'), h\,(g\cdot h')\big), [/imath] or the product: [imath] (g,h)\,(g',h') := \big(g\,(h\circ g'), (g'\cdot h)\,h'\big), [/imath] or anything similar where the other action is a right action? Do we have a group? Is this used somewhere? Is it useful, or too general? Thank you. | 107781 | Has this "generalized semidirect product" been studied?
If [imath]G[/imath] is a finite group with subgroups [imath]H[/imath] and [imath]K[/imath] such that [imath]HK = G[/imath] and [imath]H\cap K = \{1\}[/imath] we get that every element of [imath]G[/imath] can be written uniquely as [imath]hk[/imath] with [imath]h\in H[/imath] and [imath]k\in K[/imath]. This then gives a map from [imath]H\times K[/imath] to itself sending [imath](h,k)[/imath] to the unique [imath](h',k')[/imath] such that [imath]kh = h'k'[/imath]. Conversely, given a map [imath]\varphi[/imath] from [imath]H\times K[/imath] to itself with suitable properties, one can define a group structure [imath]G[/imath] on [imath]H\times K[/imath] such that [imath]H[/imath] and [imath]K[/imath] are subgroups with [imath]HK = G[/imath] and [imath]H\cap K = \{1\}[/imath] and such that the multiplication is given via the map as [imath](h,k)(h',k') = (h\varphi_1(h',k),\varphi_2(h',k)k')[/imath] where [imath]\varphi_1[/imath] and [imath]\varphi_2[/imath] are the projections on the first and second coordinate. Has this construction been studied before? It seems like it would be a natural generalization of a semidirect product, but I have never seen it mentioned anywhere. |
774166 | evaluating [imath] \sqrt {2+ \sqrt {3+ \sqrt{4+ \sqrt{5+\cdots}}} }[/imath]
How do we evaluate the infinite nested radical [imath] \sqrt {2+ \sqrt {3+ \sqrt{4+ \sqrt{5+\cdots}}} } [/imath] [imath]\space [/imath]? Please help N.B. :- It is not a duplicate | 654478 | Calculate Limit 0f nested square roots
It is an interesting task to try finding the limit of nested square root expressions. [imath]\lim_{n \to \infty}\left( 1 + \sqrt{2 + \sqrt{3+ ... + \sqrt {n + \sqrt{n+1}}}}\right)[/imath] How to solve this one? |
774082 | Particular form of states
On can prove that for a [imath]C^*[/imath]-algebra [imath]A[/imath] and an irreducible representation [imath](\psi,H)[/imath] of [imath]A[/imath] that for every unit vector of [imath]H[/imath] the state [imath]t_x(a)=\langle\psi(a)x,x\rangle[/imath] is pure. But now let [imath]H[/imath] be a separable Hilbert space of infinite dimension. I want to show that not all pure states of [imath]B(H)[/imath] are of the form [imath]A\mapsto\langle Ax,x\rangle[/imath] with [imath]x[/imath] a unit vector of [imath]H[/imath]. I think you have to go via the Calkin algebra [imath]B(H)/K(H)[/imath] to get a good answer, but i don't come further. Can someone give me some hints for this? Thanks. | 77820 | A question about pure state
For every unit vector [imath]x[/imath] in a Hilbert space [imath]H[/imath],let [imath]F_x[/imath] be the linear functional on [imath]\mathcal B(H)[/imath] (bounded linear operators) defined by [imath]F_x(T)=(Tx,x)[/imath]. Prove that each [imath]F_x[/imath] is pure state and [imath]\mathcal B(H)[/imath] has pure states which are not like this. |
69451 | How to show that [imath]\mathbb{Q}[/imath] is not [imath]G_\delta[/imath]
I read a section of a book and it made mention of the set of rationals not being a [imath]G_\delta[/imath]. However, it gave no proof. I read on wikipedia about using contradiction, but it made use of the Baire category theorem, which is unfamiliar to me. I was wondering if anyone could offer me a different proof; perhaps using the fact that the complement of [imath]G_\delta[/imath] is [imath]F_\sigma[/imath]. Thanks. | 1491919 | Baire's Theorem and Irrationals
I am asked to show that the irrational numbers are not a countable union of closed subsets of [imath]\mathbb{R}[/imath] given that if a complete metric space is the countable union of of closed subsets then at least one of them has a nonempty interior. So far: I assume that the irrationals are the countable union of closed subsets of [imath]\mathbb{R}[/imath]. I know that the irrationals are uncountable, so we cannot have this union consist of all singleton subsets. Therefore these closed sets must consist of intervals of irrational numbers. At this point I cannot see how these intervals must all have a nonempty interior. Since each of these sets is closed, its closure is itself, and and so the closure minus the boundary of the interval should be nonempty given that any interval no matter how small contains infinitely many rationals and irrationals. Can someone please point me in the right direction and/or point out the flaws in my reasoning? Thanks |
774130 | J. J. Rotman's proof that the cone is contractible
I was recently looking through J. J. Rotman's book: An Introduction to Algebraic Topology, where on page 23 he has the following result: Theorem 1.11. For every space [imath]X[/imath], the cone [imath]CX[/imath] is contractible. Proof. Define [imath]F:CX\times I\to CX[/imath] by [imath]F([x,t],s)=[x,(1-s)t+s][/imath]. [imath]\square[/imath] Now it is clear to me that if [imath]F[/imath] is continuous, then it is a homotopy from [imath]1_{CX}[/imath] to the constant map which sends each point of [imath]CX[/imath] to to the "top" point of the cone; hence, [imath]1_{CX}[/imath] is null-homotopic and so [imath]CX[/imath] is contractible as claimed. This is all well and good, but it is not immediately apparent to me which result Rotman is relying on to ensure the continuity of [imath]F[/imath]. Does anyone have any ideas about how to guarantee the continuity of [imath]F[/imath]? Thanks in advance. | 189978 | the cone is contractible
Let [imath]X[/imath] be a topological space. I want to show that the cone [imath]CX[/imath] is contractible. Here we construct a deformation retraction from [imath]CX[/imath] to the tip point of the cone [imath]H_t: CX\to CX;\; (x,t')\mapsto (x,t'(1-t))[/imath] is this correct? |
774279 | The center of the fundamental group of closed surface
[imath]S^g[/imath] is a closed surface with genus [imath]g[/imath], we know that the fundamental group [imath]\pi_1(S^g)=\{a_1,a_2,\dots ,a_g,b_1,\dots,b_g|a_1b_1a_1^{-1}b_1^{-1}\dots a_gb_ga_g^{-1}b_g^{-1}=1\}[/imath], how to calculate the center of [imath]\pi_1(S^g)[/imath]? I think the center of [imath]\pi_1(S^g)[/imath] is trivial, if [imath]g\ge 2[/imath], but I cannot prove it. | 231235 | Is the center of the fundamental group of the double torus trivial?
I know that the fundamental group of the double torus is [imath]\pi_1(M)=\langle a,b,c,d;a^{-1}b^{-1}abc^{-1}d^{-1}cd\rangle[/imath]. How can I calculate its center subgroup [imath]C[/imath]? Is [imath]C[/imath] trivial? Let [imath]p[/imath] be the quotient map from [imath]\pi_1(M)[/imath] to [imath]H_1(M)[/imath], maybe it's easy to prove that [imath]p(C)[/imath] is [imath]0[/imath] in [imath]H_1(M).[/imath] That will also solve my problem. Thanks, Yan |
773939 | Question Regarding Primitive Roots
Let m be a positive integer greater than 1. Prove that if r is a primitive root of m, then [imath]r^{φ(m)/2} ≡ -1[/imath] (mod m). | 773202 | [imath]g^\frac{p-1}{2} \equiv -1 \ (mod \ p)[/imath]
Let [imath]p[/imath] be an odd prime, and let [imath]g[/imath] be a primitive root modulo [imath]p[/imath]. Prove that [imath]g^\frac{p-1}{2} \equiv -1 \ (mod \ p)[/imath]. I've seen this solution on the math stack exchange but most of the solutions use quadratic reciprocity. Unfortunately, I skipped that section because it seemed very hard, but the proof that [imath]g^\frac{p-1}{2} \equiv -1 \ (mod \ p)[/imath] seems easy with it. Is there another way to solve this This question is proposed in the chapter of Finite Abelian Groups. DEF:In elementary number theory, an integer [imath]g[/imath] is called a primitive root for the modulus [imath]n[/imath] if [imath]\mathbb{Z}_n^\times[/imath] is a cyclic group and [imath][g]_n[/imath], is a generator for [imath]\mathbb{Z}_n^\times[/imath]. |
774691 | A more rigorous way to prove this?
I would like to prove the following statement [imath]x^n-a^n=(x-a)\sum^{n-1}_{k=0}x^ka^{n-k-1},\qquad\forall n\in\Bbb N_0[/imath] I can easily prove it by induction using polynomial long division or series expansion however I am unsure whether or not these are "rigorous enough". I have thought about this for a while but is there a way to prove this statement in a more rigorous way? I think that the use of ellipses and "do this for the remaining [imath]n[/imath]" gives me this uncertainty. Therefore I pose my question as follows: Is there a rigorous way to prove this statement, or is the use of ellipses rigorous enough? Thank you. | 592255 | Prove [imath]r^n - s^n = (r-s)\sum_{j=0}^{n-1} r^js^{n-j-1}[/imath] by induction
Prove [imath]r^n - s^n = (r-s)\sum_{j=0}^{n-1} r^js^{n-j-1} [/imath] [imath](1)[/imath]by induction. I've verified that [imath]n=1: r^1 - s^1 = (r-s)(r^0s{1-0-1}) = r-s[/imath] Assume [imath](1)[/imath] is true for [imath]n \le k[/imath]. That is [imath]r^k - s^k = (r-s)\sum_{j=0}^{k-1} r^js^{k-j-1} [/imath] is true. [imath] n = k + 1: \sum_{j=0}^{(k+1)-1} r^js^{(k+1)-j-1} =\sum_{j=0}^{k-1} r^js^{(k+1)-j-1} + r^k = s\sum_{j=0}^{k-1} r^js^{k-j-1} + r^k \Rightarrow (r-s)\sum_{j=0}^{(k+1)-1} r^js^{(k+1)-j-1} = (r-s)(s(r^k-s^k)+r^k)[/imath] [imath](2)[/imath] However I cannot write the last expression [imath](2)[/imath] as [imath]r^{k+1}-s^{k+1}[/imath]. Could someone help me out ? |
345480 | product of two uniformly continuous functions is uniformly continuous
Suppose that [imath]f[/imath] and [imath]g[/imath] are uniformly continuous functions defined on [imath](a,b)[/imath]. Prove that [imath]fg[/imath] is also uniformly continuous on [imath](a,b)[/imath]. My attempt: Since [imath]f[/imath] is uniformly continuous on [imath](a,b)[/imath], for all [imath]\epsilon>0[/imath], we have [imath]\delta_f(\epsilon)>0[/imath] such that for all [imath]x,y \in (a,b)[/imath], [imath]|x-y|<\delta_f[/imath], [imath]|f(x)-f(y)|<\epsilon[/imath] Since [imath]g[/imath] is uniformly continuous on [imath](a,b)[/imath], for all [imath]\epsilon>0[/imath], we have [imath]\delta_g(\epsilon)>0[/imath] such that for all [imath]x,y \in (a,b)[/imath], [imath]|x-y|<\delta_g[/imath], [imath]|g(x)-g(y)|<\epsilon[/imath] Notice that [imath]|f(x)g(x)-f(y)g(y)|=|f(x)g(x)-f(x)g(y)+f(x)g(y)-f(y)g(y)| \leq |f(x)||g(x)-g(y)| + |g(y)||f(x)-f(y)|[/imath] Here I don't know how to bound [imath]|f(x)|[/imath] and [imath]|g(y)|[/imath]. I have proven that uniformly continuous functions preserve boundedness of an interval , i.e. [imath]f[/imath] is bounded on [imath](a,b)[/imath]. Can anyone help me? | 2003847 | If f and g are uniformly continuous then fg is uniformly continuous
Given that [imath]f,g:(a,b) \rightarrow \mathbb{R}[/imath] and both uniformly continuous on [imath](a,b)[/imath] .... Can we say that [imath]fg[/imath] is uniformly continuous? .... If not please provide a counter-example and if true please provide me a hint Thank you |
775020 | For any integer n greater than 1, [imath]4^n+n^4[/imath] is never a prime number.
For any integer n greater than 1, show that [imath]4^n+n^4[/imath] is never a prime number. My approach: I tried to use mathematical induction. But somehow couldn't manage to prove it. Is it to be done by induction or can it be done by another method? | 489071 | Compositeness of [imath]n^4+4^n[/imath]
My coach said that for all positive integers [imath]n[/imath], [imath]n^4+4^n[/imath] is never a prime number. So we memorized this for future use in math competition. But I don't understand why is it? |
773759 | A Banach space [imath]X[/imath] is relfexive iff [imath]X^*[/imath] is reflexive.
I am trying to prove the following: A Banach space [imath]X[/imath] is reflexive iff [imath]X^*[/imath] is reflexive. Thus far, I have proven the forward direction: Let [imath]J_X:X\mapsto X^{**}[/imath] be the mapping defined by [imath]J_X(x)=x^{**}[/imath] and let [imath]J_{X^*}:X^*\mapsto X^{***}[/imath] be defined by [imath]J_{X^*}(x^*)=x^{***}[/imath]. The statement above is equivalent to showing that [imath]J_X(X)=X^{**}[/imath] if and only if [imath]J_{X^*}(X^*)=X^{***}[/imath]. Suppose that [imath]J_X(X)=X^{**}[/imath]. Let [imath]y\in X^{***}=B(X^{**},\mathbb{R})[/imath]. We show that there is some [imath]x^*\in X^*[/imath] such that [imath]J_{X^*}(x^*)=y[/imath], thus showing that [imath]J_{X^*}(X^*)=X^{***}[/imath]. Note that [imath]yJ_X:X\mapsto \mathbb{R}[/imath], and hence, [imath]yJ_X\in B(X,\mathbb{R})=X^*[/imath]. Since [imath]J_X(X)=X^{**}[/imath], if [imath]z\in X^{**}[/imath], then there is some [imath]x\in X[/imath] such that [imath]J_X(x)=z[/imath]. We have [imath] y(z)=y(J_X(x))=(J_X(x))(yJ_X)=z(yJ_X)=(J_{X^*}(yJ_X))(z). [/imath] Thus, [imath]y=J_{X^*}(yJ_X)[/imath], implying that [imath]X^*[/imath] is reflexive. I have no clue how to prove the other direction. I wanted to try a similar approach, but it's turning out to be kind of difficult (and ugly looking). I would greatly appreciate some help! Thanks! | 152343 | A Banach space is reflexive if and only if its dual is reflexive
How to show that a Banach space [imath]X[/imath] is reflexive if and only if its dual [imath]X'[/imath] is reflexive? |
765422 | the dual space of [imath]L^p[/imath]
I am reading some preliminary material to develop a good background in order to study PDE and I came across the following fact The dual space of [imath]L^p[/imath] is [imath]L^q[/imath] where [imath]q[/imath] is the Holder's Conjugate of [imath]p[/imath], so in other words [imath]\frac{1}{p}+\frac{1}{q}=1[/imath]. Can someone please explain to me how to prove this fact or indicate where can I read something about this that is understandable from somehow who doesn't have much knowledge of functional analysis. I know the definition of dual space and I was reading on wikipedia that there is a natural isomorphism from the space of linear functionals of [imath]L^p[/imath] to [imath]L^q[/imath], but I can't really reason on why so, and the intuition behind this. | 269402 | How should I prove the duality?
Rudin asked (Real Complex Analysis, First edition, Chapter 6, Problem 4): Suppose [imath]1\le p\le \infty[/imath], and [imath]q[/imath] is the exponent conjugate to [imath]p[/imath]. Suppose [imath]u[/imath] is a [imath]\sigma[/imath]-finite measure and [imath]g[/imath] is a measurable function such that [imath]fg\in L^{1}(\mu)[/imath] for every [imath]f\in L^{p}(\mu)[/imath]. Prove that then [imath]g\in L^{q}(\mu)[/imath]. I am being troubled with the fact that [imath]|g|_{q}[/imath] might be unbounded if we select wierd enough [imath]f[/imath]. Holder inequality only gives us [imath]|fg|_{1}\le |f|_{p}|g|_{q}\leftrightarrow |g|_{q}\ge \frac{|fg|_{1}}{|f|_{p}}[/imath]All the constructions I know proving [imath]g\in L^{q}[/imath] starts by assuming [imath]f\rightarrow fg[/imath] is a bounded linear operator，so I cannot use circular reasoning at here. It suffice to prove the statement for finite measure spaces and simple functions. So emulating Rudin we can assume [imath]|g|=\alpha g[/imath], where [imath]|\alpha|=1[/imath] and [imath]\alpha[/imath] is measurable. Let [imath]E_{n}=x:|g(x)|\le n[/imath] and let [imath]f=\chi_{E_{n}}\alpha g^{q-1}[/imath]. Then we have [imath]\int_{E_{n}}|g|^{q}d\mu=\int_{X}|fg|d\mu\le K_{n}[/imath] for some [imath]K_{n}<\infty[/imath]. But this constant obviously shift with the [imath]n[/imath] I choose, hence probably does not have a finite upper bound (for example [imath]K_{n}=n[/imath]). And I got stuck. In problem 6, Rudin now ask: Suppose [imath]1<p<\infty[/imath], and prove that [imath]L^{q}(\mu)[/imath] is the dual space of [imath]L^{p}(\mu)[/imath] even if [imath]\mu[/imath] is not [imath]\sigma[/imath]-finite. I keep thinking about it but do not know what is the best way to prove it. |
753243 | Some Continuity Question
Suppose [imath]f(x)[/imath] and [imath]g(x)[/imath] are continuous functions on [imath][a,b][/imath] with [imath]f[/imath] monotone increasing. Assume there exists a sequence [imath]x_n \in [a, b][/imath] such that for all [imath]n \in \mathbb{N}[/imath] , [imath]g(x_n) = f(x_{n+1})[/imath]. Show that there exists [imath]x_0 \in [a,b][/imath] such that [imath]g(x_0) = f(x_0)[/imath]. Just wondering if this question is wrong. My idea is that if I look at [imath]g(x) := x+1[/imath] and [imath]f(x) := x[/imath]. Choosing the sequence [imath]x_n = x[/imath]. Then I have [imath]g(x) = f(x+1)[/imath]. However, I don't think there is any [imath]x_0[/imath] such that [imath]x_0 + 1 = x_0[/imath]. What am I misunderstanding here? | 761299 | If there exist sequence such that [imath]g(x_n)=f(x_{n+1})[/imath], then we have [imath]g(x_0)=f(x_0)[/imath] for some [imath]x_0[/imath]
Suppose [imath]f(x)[/imath] and [imath]g(x)[/imath] are continuous functions on [imath][a,b][/imath] with [imath]f[/imath] monotone increasing. Assume there exists a sequence [imath]x_n \in [a, b][/imath] such that for all [imath]n \in N[/imath] , [imath]g(x_n) = f(x_{n+1})[/imath]. Show that there exists [imath]x_0 \in [a,b][/imath] such that [imath]g(x_0) = f(x_0)[/imath]. Can someone provide an example of functions that fulfills this condition? |
774726 | Minimal polynomial for sum of algebraic numbers.
If I have two algebraic numbers [imath]\alpha,\beta[/imath] such that [imath]A(\alpha) = 0[/imath] and [imath]B(\beta)=0[/imath] where [imath]A,B \in \mathbb{Q}[x][/imath] are the minimal polynomials of [imath]\alpha[/imath] and [imath]\beta[/imath] respectively. Knowing only the coefficients (and degree) of [imath]A[/imath] and [imath] B[/imath] is there an algorithm for generating the minimal polynomial of [imath]\alpha+\beta[/imath]? I asked a professor and he recommended computing powers of [imath]A(x) + B(x)[/imath] and claimed there was a way to reduce it to get the appropriate polynomial however I'm not seeing his method. | 155122 | How to prove that the sum and product of two algebraic numbers is algebraic?
Suppose [imath]E/F[/imath] is a field extension and [imath]\alpha, \beta \in E[/imath] are algebraic over [imath]F[/imath]. Then it is not too hard to see that when [imath]\alpha[/imath] is nonzero, [imath]1/\alpha[/imath] is also algebraic. If [imath]a_0 + a_1\alpha + \cdots + a_n \alpha^n = 0[/imath], then dividing by [imath]\alpha^{n}[/imath] gives [imath]a_0\frac{1}{\alpha^n} + a_1\frac{1}{\alpha^{n-1}} + \cdots + a_n = 0.[/imath] Is there a similar elementary way to show that [imath]\alpha + \beta[/imath] and [imath]\alpha \beta[/imath] are also algebraic (i.e. finding an explicit formula for a polynomial that has [imath]\alpha + \beta[/imath] or [imath]\alpha\beta[/imath] as its root)? The only proof I know for this fact is the one where you show that [imath]F(\alpha, \beta) / F[/imath] is a finite field extension and thus an algebraic extension. |
776446 | Is this sort of a representation possible in a group with these conditions?
Let [imath]G[/imath] be a finite group, and let [imath]T[/imath] be an automorphism of [imath]G[/imath] such that [imath]T(x) = x[/imath] for [imath]x \in G[/imath] if and only if [imath]x = e[/imath]. Then is it possible to represent every element [imath]g \in G[/imath] as follows? [imath]g = x^{-1} T(x)[/imath] for some [imath]x \in G[/imath]? | 41425 | Show that a finite group with certain automorphism is abelian
Let [imath]G[/imath] be a finite group and [imath]f:G\to G[/imath] an isomorphism. If [imath]f[/imath] has no fixed points (i.e., [imath]f(x)=x[/imath] implies [imath]x=e[/imath]) and if [imath]f\circ f[/imath] is the identity, then [imath]G[/imath] is abelian. (Hint: Prove that every element in [imath]G[/imath] has the form [imath]x^{-1}\cdot f(x)[/imath].) With the hint, I can see that for all [imath]t\in G[/imath], [imath]f(t)=t^{-1}[/imath], and since [imath]f[/imath] is an isomorphism, [imath]G[/imath] is abelian. But I can't see how to prove the hint or why it's evident. Source : Rotman J.J. Introduction to the theory of groups, exercise 1.50 |
776723 | Intermediate Extensions and Coding Sets by Ordinals
Lemma 15.43 in Jech's "Set Theory" states that if [imath]V \subseteq M \subseteq V[G][/imath] where [imath]G \subseteq \mathbb{P}[/imath] is some [imath]V[/imath]-generic filter and [imath]M[/imath] is a transitive models of ZFC, then [imath]M = V[D \cap G][/imath] for some complete subalgebra. In the proof, he states that since [imath]M[/imath] models choice, for every [imath]X \in M[/imath], there is a set of ordinals [imath]A_X[/imath] such that [imath]X \in V[A_X][/imath]. I do not see what this set [imath]A_X[/imath] should be and how to construct [imath]X[/imath] in [imath]V[A_X][/imath]. I believe the idea should be that the axiom of choice can be used to show that [imath]X[/imath] is in bijection with some ordinals. However, it is not clear why the coding of this set by ordinals is necessarily in [imath]V[/imath] or [imath]V[A_X][/imath] for an appropriately chosen set. Is this the correct approach? | 660275 | All models of [imath]\mathsf{ZFC}[/imath] between [imath]V[/imath] and [imath]V[G][/imath] are generic extensions of [imath]V[/imath]
I'm reading the proof of lemma 15.43 of Jech's Set Theory: Let [imath]G[/imath] be generic on a complete Boolean algebra [imath]B[/imath]. If [imath]M[/imath] is a model of [imath]\mathsf{ZFC}[/imath] such that [imath]V\subset M\subset V[G][/imath], then there exists a complete subalgebra [imath]D\subset B[/imath] such that [imath]M=V[D\cap G][/imath]. In the proof of this theorem the author states the following fact: "First we note that since [imath]M[/imath] satisfies the Axiom of Choice, there is for every [imath]X\in M[/imath] a set of ordinals [imath]A_X\in M[/imath] such that [imath]X\in V[A_X].[/imath]" My question is, why is this fact true? Thanks |
776054 | [imath]\langle x \rangle[/imath] is a direct summand of a finite abelian group where [imath]x[/imath] is maximal order
Let [imath]x[/imath] be an element of a finite abelian group [imath]G[/imath] where [imath]x[/imath] has maximal order. Then I want to show that [imath]\langle x\rangle[/imath] is a direct summand of [imath]G[/imath]. Note that I do not want to use finite abelian group classification theorem, because I want to use this fact to prove it. My strategy is to find a subgroup [imath]H\le G[/imath] where [imath]\langle x\rangle\cap H = \{e\}[/imath] and [imath]\langle x,H\rangle = G[/imath]. But I'm unsuccessful in finding such [imath]H[/imath]. | 85600 | Finite abelian groups - direct sum of cyclic subgroup
Let [imath]G[/imath] be a finite abelian [imath]p[/imath]-group. It is quite elementary to see that if [imath]g \in G[/imath] is an element of maximal order (and thus its span is a cyclic subgroup of [imath]G[/imath] of maximal order) then [imath]G[/imath] can be written as the direct sum [imath]G=\langle g \rangle \oplus H[/imath] for some [imath]H \leq G[/imath] (subgroup of [imath]G[/imath]). For a proof see this for example (page 2). My question: Do we need that [imath]G[/imath] is a [imath]p[/imath]-group or does it also work for arbitrary finite abelian groups? I think it is wrong for general groups because I looked around quite a bit and always only found the above theorem, but I could not find a counter-example. |
776954 | what will be the definite integration of the following equation
What will be the definite integration from 0 to inf of the cosine function [imath]\int_0^\infty\cos(r)\,\mathrm dr[/imath] | 776842 | Integral involving Cosine function
What will be the formula for the following: [imath]\int_0^\infty \cos(x)dx.[/imath] |
776477 | Trouble understanding Big O notation for a sum of n integers
This problem is an example in a Discrete Math textbook. How can big-O notation be used to estimate the sum of the first n positive integers? Solution: Because each of the integers in the sum of the first n positive integers does not exceed n, it follows that [imath]1+2+...+n \le n+n+...+n=n^2[/imath] From this inequality it follows that [imath]1+2+...+n[/imath] is [imath]O(n^2)[/imath]. I do not understand how the sum of [imath]n[/imath]'s is equal to [imath]n^2[/imath]. Can anyone explain to me what the book used to get this? | 747071 | Big-O notation and polynomials
In my text, I am given that the sum of the first n positive integers can be understood in terms of big-O notation. ''Since each of the integers in the sum of the first [imath]n[/imath] positive integers does not exceed [imath]n[/imath]'', we can write: [imath]1 + 2 + \cdots + n \leq n + n \cdots + n= n^2[/imath] Why does [imath]n + n +\cdots + n = n^2[/imath] ? |
777775 | Combinatorics - Counting in two ways?
What is the idea of proving a binomial identity by counting in two ways? Could you please illustrate this with this example? Thank you very much. [imath]\binom{2n}{n}= \sum_{k=0}^n {\binom nk}^2[/imath] (original screenshot) | 677307 | Counting two ways, [imath]\sum \binom{n}{k} \binom{m}{n-k} = \binom{n+m}{n}[/imath]
prove by counting two ways: I though to prove the right hand side I would say: Let n represent a number of boys and m a number of girls. We want to choose a group of n from boys and girls. But for the left hand side I want to keep the variables m and n girls and boys. But I don't know what k will be. Suggestions? |
778152 | Using Wilson's theorem, show that if [imath]p[/imath] is prime of the form [imath]p=4n+1[/imath] and if [imath]y=(\frac {p-1}2)![/imath], then [imath]y^2\equiv -1\mod p[/imath].
Using Wilson's theorem, show that if [imath]p[/imath] is prime of the form [imath]p=4n+1[/imath] and if [imath]y=(\frac {p-1}2)![/imath], then [imath]y^2\equiv -1\mod p[/imath]. My attempt: By Wilson, [imath](4n+1-1)!\equiv (4n+1-1) \mod p[/imath],[imath](4n)!\equiv4n\mod p[/imath]. The statement we want to show is equivalent to [imath]((2n)!)^2\equiv 4n\mod p[/imath], which is transformed into [imath]((2n)!)^2\equiv (4n)!\mod p[/imath] | 16206 | Show for prime numbers of the form [imath]p=4n+1[/imath], [imath]x=(2n)![/imath] solves the congruence [imath]x^2\equiv-1 \pmod p[/imath]. [imath]p[/imath] is therefore not a gaussian prime.
I need to show that for prime numbers of the form [imath]p=4n+1[/imath], [imath]x=(2n)![/imath] solves the congruence [imath]x^2 \equiv-1\pmod p[/imath]. I then need to show this implies p isn't a gaussian prime. I have started to solve this using Wilson's theorem that a number [imath]z[/imath] is prime iff [imath](z-1)!\equiv-1\pmod z[/imath]. Therefore the endpoint of my proof should be that [imath](p-1)!\equiv-1\pmod p[/imath]. As [imath]p[/imath] is of the form [imath]p=4n+1[/imath], I only need to prove that [imath]4n![/imath] is congruent to [imath]-1[/imath] modulo [imath]p[/imath]. Here is my working so far: Starting with the congruence [imath]x^2 \equiv -1\pmod p[/imath]: [imath]\eqalign{x^2 = -1\pmod p&\implies x^2 + 1 = kp \implies (x-i)(x+i) = kp\cr &\implies ((2n)!-i)((2n!)+i))=kp \implies 4n!- 1 =kp\cr}[/imath] This is where I start to run out of any ideas that seem to get me anywhere. Any tips would be greatly appreciated! |
131175 | Why does [imath](\frac{p-1}{2}!)^2 = (-1)^{\frac{p+1}{2}}[/imath] mod [imath]p[/imath]?
Assume [imath]p[/imath] is prime and [imath]p\ge 3[/imath]. Through experimentation, I can see that it's probably true. Using Wilson's theorem and Fermat's little theorem, it's equivalent to saying [imath]2^2 4^2 6^2 \cdots (p-1)^2 = (-1)^{\frac{p+1}{2}}[/imath] mod [imath]p[/imath], but I can't figure out any more than that. | 2611934 | Proof congruence identity modulo [imath]p[/imath]: [imath]2^2\cdot4^2\cdot\dots\cdot(p-3)^2\cdot(p-1)^2 \equiv (-1)^{\frac{1}{2}(p+1)}\mod{p}[/imath]
I want to show that for every odd prime [imath]p[/imath] the following congruence holds: [imath] 2^2\cdot4^2\cdot\dots\cdot(p-3)^2\cdot(p-1)^2 \equiv (-1)^{\frac{1}{2}(p+1)}\mod{p} [/imath] How would I approach this problem? |
778247 | Real Analysis Continuous Function Problem
Show that the only continuous function on [imath](-1,+1)[/imath], which is not identically zero and satisfies the equation [imath]f(x + y) = f(x)f(y)[/imath] for all [imath]x,y \in \mathbb{R}[/imath], is the exponential function [imath]f(x) = a^x[/imath] with [imath]a = f(1) > 0[/imath]. | 293371 | How do I prove that [imath]f(x)f(y)=f(x+y)[/imath] implies that [imath]f(x)=e^{cx}[/imath], assuming f is continuous and not zero?
This is part of a homework assignment for a real analysis course taught out of "Baby Rudin." Just looking for a push in the right direction, not a full-blown solution. We are to suppose that [imath]f(x)f(y)=f(x+y)[/imath] for all real x and y, and that f is continuous and not zero. The first part of this question let me assume differentiability as well, and I was able to compose it with the natural log and take the derivative to prove that [imath]f(x)=e^{cx}[/imath] where c is a real constant. I'm having a little more trouble only assuming continuity; I'm currently trying to prove that f is differentiable at zero, and hence all real numbers. Is this an approach worth taking? |
778301 | prove of [imath]17^n-12^n-24^n+19^n \equiv 0 \pmod{35} [/imath]
i came across this answer and i saw the given solution but i can not understand how it proves the given problem. Ok i get that [imath]lcm(5,7)= 35[/imath] and it is the same as the [imath](mod 35)[/imath]. Please can someone help me? | 755125 | how do i prove that [imath]17^n-12^n-24^n+19^n \equiv 0 \pmod{35}[/imath]
How do i prove that [imath]17^n−12^n−24^n+19^n≡0(\mod35)[/imath] for every possitive integer n. Can anyone give me a hint of how to start? |
779510 | How many solutions does the equation [imath]x+y+z=11[/imath] have?
How many solution does [imath]x+y+z=11[/imath] have where [imath]x, y, z[/imath] are non-negative integers. In light of the restrictions, its clear that [imath]x,y,z \in \{0,1,2,..11\}[/imath]. So, at face value I would assign a value for [imath]x[/imath] and determine the different combinations that [imath]y[/imath] and [imath]z[/imath] can hold. For example, For [imath]x=0[/imath], we have [imath]y+z=11[/imath]. With writing them out I found that there are [imath]12[/imath] different assigned combinations for [imath]y[/imath] and [imath]z[/imath] that satisfy the equation. For [imath]x=1[/imath], I got [imath]11[/imath]. Consequently, the pattern becomes clear whereby each one takes a value less by one. Hence, the number of solutions is [imath]1+2+3+4+5+6+7..+12=78[/imath]. I was wondering if there is an easier method perhaps with combinations equation [imath]C(a,b)[/imath]..? | 322170 | How many solutions does the equation [imath]x_1 + x_2 + x_3 = 11[/imath] have, where [imath]x_1, x_2, x_3[/imath] are nonnegative integers?
Help me understand problems of this type a bit more intuitively. The solution [imath]C(3+11−1,11)[/imath] seems simple enough, but I got stuck thinking about how many integers you are choosing from within [imath]x_1[/imath], [imath]x_2[/imath], and [imath]x_3[/imath] and I was trying to get an [imath]n[/imath] - as in [imath]C(n+r-1,r)[/imath] - this way. Also, I though the [imath]r[/imath] was 3 since you have [imath]x_1[/imath], [imath]x_2[/imath], [imath]x_3[/imath] choices. Basically, I went wrong at every step. Help! |
780360 | Solve for x when [imath]2222^{5555} + 5555^{2222} \equiv x \pmod{7}[/imath]
I need to find the remainder when [imath]2222^{5555} + 5555^{2222}[/imath] is divided by [imath]7[/imath]. I'm thinking that Fermat's Little Theorem might help. Any suggestions? | 279333 | What will be the one's digit of the remainder in: [imath]\left|5555^{2222} + 2222^{5555}\right|\div 7=?[/imath]
What will be the ones digit of the remainder in: [imath]\frac{\left|5555^{2222} + 2222^{5555}\right|} {7}[/imath] |
781035 | Proving that irreducibility of a matrix implies strong connectedness of the graph
I have tried to prove that if a matrix [imath]A\in\mathbb{C}^{n\times n}[/imath] is such that there are no two sets [imath]I,J\subseteq\{1,\dots,n\}[/imath] that are disjoint, complementary, nonempty, and such that for all [imath](i,j)\in I\times J[/imath] we have [imath]a_{ij}=0[/imath], [imath]a_{ij}[/imath] being the [imath]ij[/imath] entry of the matrix [imath]A[/imath], then the associate graph, as defined here, is strongly connected. The hypothesis will be called n-DISC (not Disjoint Index Set Condition) and the strong connectedness of the graph will be referred to as SCG in the proof. I have managed to give the following proof: I have proved it in special cases, and then assumed the same line of thought can be used for any size, but the proof is anyway long and very unelegant, so any suggestion of a more elegant proof is very welcome indeed. Instead of proving n-SCG implies DISC, I tried to prove that n-DISC implies SCG, because trying the former didn't seem to lead me anywhere, while the latter gave me something that seemed to be OK for any case. Let's start with [imath]n=2[/imath]. Then we have a [imath]2\times2[/imath] matrix and 2 nodes. Let's take one. it can't be cut off from the other one, otherwise we would have the DISC. So from this node, be it [imath]A[/imath], we can reach the other one, be it [imath]B[/imath]. The same argument on [imath]B[/imath] proves we can reach [imath]A[/imath] from [imath]B[/imath], and therefore the graph is SC. Let's go on with [imath]n=3[/imath]. We start at one node, be it [imath]A[/imath]. From it, as said above, we can reach at least one other node, be it [imath]B[/imath]. From [imath]B[/imath] we can reach at least another node. Now one is tempted to say it is the last node, [imath]C[/imath], but of course not. So we have to cases: either [imath]A\to B\to C[/imath], or [imath]A\leftrightarrow B[/imath]. In the former case, from [imath]C[/imath] we can reach at least one node. If it is [imath]A[/imath], we have concluded. If it is [imath]B[/imath], i.e. [imath]A\to B\leftrightarrow C[/imath] then from either [imath]B[/imath] or [imath]C[/imath] we can reach [imath]A[/imath], otherwise the DISC is violated by [imath]B,C[/imath] opposed to [imath]A[/imath]. If [imath]A\to B\leftrightarrow C\to A[/imath], we have concluded, and if [imath]A\leftrightarrow B\leftrightarrow C[/imath], we have also concluded. However, we still have the latter case: [imath]A\leftrightarrow B[/imath]. Of course, from one of those we must reach [imath]C[/imath]. So [imath]A\leftrightarrow B\to C[/imath] or [imath]C\leftarrow A\leftrightarrow B[/imath]. In the former case, we anyway have [imath]A\to B\to C[/imath], and we have shown before that this case leads to a positive conclusion. In the latter case, from [imath]C[/imath] we can reach either [imath]B[/imath] or [imath]A[/imath], otherwise we violate the supposed DISC. If [imath]C\to B[/imath], then [imath]A\to C\to B\to A[/imath], which concludes the proof. If [imath]C\to A[/imath], then [imath]C\leftrightarrow A\leftrightarrow B[/imath], which also concludes the proof. I tried it for [imath]n=4[/imath], but got bored in the process, concluded a positive case and left many uninvestigated cases. Now I fully well understand this proof is horribly complex and incredibly unelegant. This is why I'm asking if there is a simpler and more elegant proof for this statement. Naturally, this is only half of the theorem, which states the equivalence of n-DISC with SCG, i.e. the equivalence of the irreducibility of [imath]A[/imath] with the strong connectedness of its associated graph. Can someone help me here? | 750817 | Matrix graph and irreducibility
How do I prove that if [imath]A\in\mathbb C^{n\times n}[/imath] is a matrix then it is irreducible if and only if its associated graph (defined as at Graph of a matrix) is strongly connected? Update: Seeing as no-one answered for over a week, I tried to do it by myself. The first thing I did was try to show column or row permutation didn't change the strong connectedness of the graph. I didn't manage, and actually proved the opposite. On the way, though, I managed to show transposition doesn't. The argument is that transposition affects the graph in that it inverts all the arrows, but if there is a loop through all nodes then inverting the arrows means you go through it the other way round, so it's still there, and the graph stays strongly connectedness, and if there isn't, well, transposing can't make one appear, as otherwise transposing back would make it disappear, which we have proved impossible. This result may not be useful, but since I've done it I thought I might well write it down. Then I tried to think of what permuting both rows and columns does to the graph. Why that? Let's recall the notion of irreducible matrix: A matrix [imath]A\in\mathbb{C}^{n\times n}[/imath] is said to be reducible if there exists a permutation matrix [imath]\Pi[/imath] for which [imath]\Pi\cdot A\cdot > \Pi^T[/imath] is a block upper triangular matrix, i.e. has a block of zeroes in the bottom-left corner. So if this operation does not alter the graph's strong connectedness, then I can work on the reduced matrix to show its graph is not strongly connected and prove one implication. Now such multiplications as in the definition of a reducible matrix, with [imath]\Pi[/imath] a matrix that swaps line [imath]i[/imath] with line [imath]j[/imath] - what do they do to the graph? Swapping the lines makes all arrows that go out of [imath]i[/imath] go out of [imath]j[/imath] and viceversa; swapping the columns does the same for arrows leaving [imath]i[/imath] (or [imath]j[/imath]). So imagine we have a loop. Say it starts from a node other than [imath]i[/imath] and [imath]j[/imath]. At a certain point it reaches, say, [imath]i[/imath]. Before that, everything is unchanged. When the original loop reaches [imath]i[/imath], the new loop will reach [imath]j[/imath] and go out of it to the same node as it went out from [imath]i[/imath] to before the permutation, if that node wasn't [imath]j[/imath], in which case it will go to [imath]i[/imath]. When the original loop enters [imath]j[/imath], the new loop enters [imath]i[/imath], and same as before. So basically the result is just that [imath]i[/imath] and [imath]j[/imath] swap names, and the loop is the same as before taking the name swap into account. So this kind of operations do not alter the strong connectedness of the graph. Suppose [imath]A[/imath] is as follows: [imath]A=\left(\begin{array}{c|c}\Huge{A_{11}} & \Huge{A_{12}} \\\hline \Huge{0} & \Huge{A_{22}}\end{array}\right).[/imath] Suppose the [imath]\Huge{0}[/imath] is [imath]m\times m[/imath] with [imath]m\geq\frac{n}{2}[/imath]. Then we have [imath]m[/imath] nodes that are unconnected to other [imath]m[/imath] nodes, going out. But we don't have [imath]2m[/imath] nodes, or have exactly that many, so those [imath]m[/imath] nodes are cut off from all the other [imath]n-m[/imath], going out. So suppose there is a loop. If it starts at one of the [imath]m[/imath] nodes, it can never reach the other [imath]n-m[/imath], and if it starts at one of those, it can reach the [imath]m[/imath] nodes but never get back, so maybe we have a path through all the nodes, but it can't be a loop, i.e. a closed path. So the graph is not strongly connected. Now the definition doesn't say anything about the size of those blocks, so the problem I still have is that if [imath]m<\frac{n}{2}[/imath], the argument above fails because we have at least one node besides the [imath]m[/imath] nodes and the other [imath]m[/imath] nodes that can't be reached from the first [imath]m[/imath], and that node could be the missing link. Of course, when I said "can't be reached" up till now, I meant "be reached directly", i.e. not passing through other nodes. Of course, if the above is concluded, I have proved that reducibility implies non-strong-connectedness of the graph, so that a strongly connected graph implies irreducibility. But the converse I haven't even tried. So the questions are: how do I finish the above at points 3-4 and how do I prove the converse? Or maybe I'm missing something in the definition, in which case what is it? Update 2: I think I am missing something, as a [imath]3\times3[/imath] matrix with a 0 in the bottom-left corner and no other zeroes does have a strongly connected graph, since the only missing arrow is [imath]3\to1[/imath], but we have the loop [imath]1\to3\to2\to1[/imath]. So when is a matrix reducible? Update 3: Browsing the web, I have found some things. I bumped first into this link. Now if that is the definition of reducible matrix, then either I misunderstand the definition of the block triangular form, or the if and only if there doesn't hold, since a matrix with a square block of zeroes bottom-left definitely doesn't satisfy the disjoint set condition but definitely does satisfy the permutation condition, with no permutation at all. Maybe the first condition is equivalent to the non-strong-connection of the graph. Yes, because in that case there are [imath]\mu[/imath] nodes from which you can't reach the remaining [imath]\nu[/imath], so the graph is not strongly connected. So at least that condition implies the non-strong-connectedness of the graph. The converse seems a bit trickier. Looking for that definition, I bumped into this link. Note that no matrix there has a lone corner zero (which would be top-right as the link deals with lower triangular matrixes), and all of them satisfy the disjoint set condition in the link above. So what is the definition of block triangular matrix? If it is that there must be square blocks whose diagonal coincides with part of the original matrix's diagonal and below them there must only be zeroes, then I have finished, since the if and only if in the link above is valid, so reducibility implies non-strong-connectedness of the graph, and whoops, I'm not done yet, I still need the converse, so can someone finally come help me on that? And if it isn't, then what the bleep is it and how do I make this damn proof? |
686792 | Showing [imath]\rho (x,y)=\frac{d(x,y)}{1+d(x,y)}[/imath] is a metric
Show [imath]\rho (x,y)=\dfrac{d(x,y)}{1+d(x,y)}[/imath] is a metric on the metric space [imath]X[/imath], equipped with the Euclidean metric [imath]d[/imath]. I've already shown that the positivity [imath]\rho(x,y)\geq 0[/imath], the symmetry [imath]\rho(x,x)=\rho(y,y)=0[/imath]. I'm having trouble proving the last condition of a metric space though: the triangle inequality must be valid: [imath]\rho(x,y)\leq \rho(x,p)+\rho (p,y)[/imath], for [imath]p\in X[/imath]. So, [imath]\dfrac{d(x,y)}{1+d(x,y)}\leq \dfrac{d(p,x)}{1+d(p,x)}+\dfrac{d(p,y)}{1+d(p,y)}[/imath] Could someone provide a hint as to get started? No complete solutions please. Thanks in advance. | 1104633 | Showing [imath]d(x,y) = \frac{|x-y|}{1+|x-y|}[/imath] is a distance.
Show that [imath](\mathbb{N}, d)[/imath] is a metric space with [imath]d(x,y) = \frac{|x-y|}{1+|x-y|}[/imath] My attempt: let [imath]x,y \in \mathbb{N}[/imath], 1) [imath]d(x,y) = 0 \implies |x-y| = 0 \iff x = y[/imath] 2) [imath]d(x,y) = d(y,x)[/imath] 3) show for [imath]z \in \mathbb{N}[/imath], [imath]d(x,z) \leq d(x,y) + d(y,z)[/imath] Using the triangle ineuqality I managed to get: [imath]d(x,z) \leq \frac{|x-y|}{1+|x-y| - |y-z|} + \frac{|y - z|}{1 + |x-y| - |y-z|}[/imath] but i assume here I must use the fact that [imath]x,y,z \in \mathbb{N}[/imath] but I'm not sure how. Any hints please: edit: improvements on solution - Using fact that [imath]|x-y|[/imath] is a metric we have that [imath]d(x,y) \leq \frac{|x-y|}{1+|x-y| + |y-z|} + \frac{|y - z|}{1 + |x-y| + |y-z|}[/imath] (Since [imath]|x-z| \leq |x-y| + |y-z|[/imath]) Is this correct? From there I can use the fact that [imath]\frac{1}{1+|x-y|+|y-z|} \leq \frac{1}{1+|x-y|} [/imath] and [imath]\frac{1}{1+|x-y|+|y-z|} \leq \frac{1}{1+|y-z|}[/imath] to conclude? |
781286 | Constant function with maximum modulus
Suppose that [imath]f[/imath] is analytic on a domain [imath]D[/imath], which contains a simple closed curve [imath]\gamma[/imath] and the inside of [imath]\gamma[/imath]. If [imath]|f|[/imath] is constant on [imath]\gamma[/imath], then I want to prove that either [imath]f[/imath] is constant or [imath]f[/imath] has a zero inside [imath]\gamma[/imath] if [imath]f[/imath] is not constant, then the max/min modulus principle applies ... meaning [imath]|f|[/imath] can not have any local max/min on D I don't really know what to do next | 780539 | Minimum Modulus Principle for a constant fuction in a simple closed curve
Suppose that [imath]f[/imath] is analytic on a domain [imath]D[/imath], which contains a simple closed curve [imath]\gamma[/imath] and the inside of [imath]\gamma[/imath]. If [imath]|f|[/imath] is constant on [imath]\gamma[/imath], then I want to prove that either [imath]f[/imath] is constant or [imath]f[/imath] has a zero inside [imath]\gamma[/imath] Here is my take: if [imath]f[/imath] is constant, i dont see a reason why [imath]|f|[/imath] wouldnt be constant :) if [imath]f[/imath] is not constant, then the max/min modulus principle applies ... meaning [imath]|f|[/imath] can not have any local max/min on D now i am lost at this point ... |
781126 | Diophantine equation exercise
Prove that the diophantine equation [imath]x^4-2(y^2)=1[/imath] has only 2 solutions. Any hint on how to start and what to do .. I do not have a lot of experience on non linear diophantine equations and do not know how to approximate them. So far I can see that for [imath]y=0[/imath], [imath]x=1[/imath] is one solution. | 263647 | Solutions of Diophantine equations in Natural numbers
The one of solution of [imath]x^4 - 2y^2 = -1[/imath] is [imath]x = 1[/imath] and [imath]y = 1[/imath]. However, the solution [imath](1, 1)[/imath] of [imath]x^4 - 2y^2 = 1[/imath] is failed. We know [imath]x = 1[/imath] and [imath]y = 1[/imath] is small integers and we can check by trail method. In case more solutions are existing or not how to check? What are the solutions of [imath]x^4 - 2y^2 = 1[/imath]? |
782884 | Show the expected number of packets to buy to collect a whole set of minifigures
Minifigures are sold in packets, where each packet contains one minifigure, and from the outside of the packet it is impossible to tell which minifigure is inside. There are [imath]n[/imath] minigures to collect. Assuming that each packet that is bought is equally likely to contain any one of the figures, show that the expected number of packets that needs to be bought to collect the whole set is approximately [imath]n \log n[/imath]. | 782769 | calculating expected number of packets.
To go with the Lego Movie, Lego sell minifigures of the characters from the movie. They are sold in packets, where each packet contains one minifigure, and from the outside of the packet it is impossible to tell which minifigure is inside. There are [imath]n[/imath] minifigures to collect. Assuming that each packet that my son buys is equally likely to contain any one of the minifigures, show that the expected number of packets that he needs to buy to collect the whole set is approximately [imath]n \log n[/imath]. [Hint: express the random variable that is the total number of packets as a sum of simpler random variables, and use linearity of expectation.] |
783319 | Find [imath]det(xy^T)[/imath] where [imath]x[/imath], [imath]y[/imath] are vectors from [imath]R^n[/imath], [imath]n[/imath]>1
I represented [imath]x[/imath] as [imath][x_1\ x_2\ ... x_n]^T[/imath], and [imath]y^T[/imath] as [imath][y_1\ y_2\ ... y_n][/imath]. Multiplying them produces a matrix [imath]n[/imath]x[imath]n[/imath]: [imath] \begin{pmatrix}x_1y_1&x_1y_2&\dots& x_1y_n\\ x_2y_1&x_2y_2&\dots& x_2y_n\\ \\ \\ \\ \\ x_ny_1& x_ny_2&\dots& x_ny_n \end{pmatrix}[/imath] (I apologize, I don't know how to format a matrix using LaTeX). :) Obviously, for [imath]n=2[/imath], [imath]n=3[/imath] the determinant is [imath]0[/imath]. I just don't know how to prove it for [imath]n[/imath]. | 781444 | show that [imath]\det(A)=0[/imath] in this case
(a) Let [imath]x[/imath] and [imath]y[/imath] be [imath]n\times 1[/imath] matrices, [imath]n \ge 1[/imath], and let [imath]A=xy^T[/imath]. Show that [imath]\det(A)=0[/imath]. (b) Explain why the statment in part (a) is false if [imath]n=1[/imath]. |
783410 | Prove that [imath]|f(x)| \le \frac{3}{2}[/imath] when [imath]f(x)=ax^2+bx+c[/imath]
Suppose [imath]f(x) = ax^2+bx+c[/imath] where [imath]x \in [-1,1][/imath]. If [imath]f(-1),f(0),f(1)\in [-1,1][/imath] show that [imath]|f(x)| \le \frac{3}{2}[/imath] [imath]\forall x \in [-1,1][/imath]. This is how I tried: [imath]f(0)=c[/imath] [imath]f(1)=a+b+c[/imath] [imath]f(-1)=a-b+c[/imath] Putting [imath]f(0)=c[/imath] we get [imath]f(1)-f(0)=a+b[/imath], [imath]f(-1)-f(0)=a-b[/imath]. Solving for [imath]a[/imath] and [imath]b[/imath] respectively we get [imath]a=\frac{f(1)+f(-1)-2f(0)}{2}[/imath] and [imath]b=\frac{f(1)-f(-1)}{2}[/imath] Then [imath]|f(x)|=|(\frac{f(1)}{2})(x^2+x)+(\frac{f(-1)}{2})(x^2-x)+(f(0))(1-x^2)|[/imath] [imath]\le |\frac{x^2+x}{2}|+|\frac{x^2-x}{2}|+|1-x^2|[/imath]. Now individually the maximum of [imath]|x^2+x|[/imath] happens at [imath]x=1[/imath] , [imath]|x^2-x|[/imath] at [imath]x=-1[/imath] and [imath]|1-x^2|[/imath] at [imath]x=0[/imath]. So All I get [imath]|f(x)| \le 3[/imath]. But I need [imath]\frac{3}{2}[/imath]. Thanks for the help!! | 774069 | Let [imath]f(x)=ax^2+bx+c[/imath] where [imath]a,b,c[/imath] are real numbers. Suppose [imath]f(-1),f(0),f(1) \in [-1,1][/imath]. Prove that [imath]|f(x)|\le \frac{3}{2}[/imath] for all [imath]x \in [-1,1][/imath].
Let [imath]f(x)=ax^2+bx+c[/imath] where [imath]a,b,c[/imath] are real numbers. Suppose [imath]f(-1),f(0),f(1) \in [-1,1][/imath]. Prove that [imath]|f(x)|\le \frac{3}{2}[/imath] for all [imath]x \in [-1,1][/imath]. I made quite a few attempts but could not explicitly derive the required condition. I can understand that it happens, but cannot derive that. Please help. |
710220 | Prove: Any prime of the form [imath]3n+1[/imath] is also in the form [imath]6m+1[/imath]
Prove that any prime of the form [imath]3n+1[/imath] is also in the form [imath]6m+1[/imath]. So first I found an example... [imath]3(2)+1=7[/imath] and [imath]6(1)+1=7 [/imath]. How do I go about a formal proof though? | 78746 | Prove if [imath] p[/imath] is a prime of the form [imath]3n+1[/imath], then [imath]p[/imath] is also of the form [imath]6m+1[/imath]
Prove that if [imath]p[/imath] is a prime of the form [imath]3n+1[/imath], [imath]p[/imath] is also of the form [imath]6m+1[/imath]. I guess one could use the division algorithm for this; but, what's the best method? |
783624 | If [imath]y =\sqrt{5+\sqrt{5-\sqrt{5+ \cdots}}}[/imath], what is the value of [imath]y^2-y[/imath]?
If [imath]y =\sqrt{5+\sqrt{5-\sqrt{5 + \cdots}}},[/imath] what is the value of [imath]y^2-y[/imath] ? I am unable to get the clue due to these alternative signs of plus and minus please help on this thanks... | 588414 | [imath]\text{Let }y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} [/imath], what is the nearest value of [imath]y^2 - y[/imath]?
I found this question somewhere and have been unable to solve it. It is a modification of a very common algebra question. [imath]\text{Let }y=\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} [/imath], what is the nearest value of [imath]y^2 - y[/imath]? [imath]\textbf {(1) } 1 \qquad \textbf {(2) } \sqrt5 \qquad \textbf {(3) } 4 \qquad \textbf {(4) } 2\sqrt5[/imath] I worked the problem and got [imath]y^2 - y = 5+ \sqrt{5-\sqrt{5+\sqrt{5-...}}} -\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-...}}}} \\ = 5+ \sqrt{5-y} - y = 5-y + \sqrt{5-y}[/imath] What do I do next? |
783013 | Show that [imath]{\sin^4x\over a}+{\cos^4x\over b}={1\over a+b} \implies {\sin^6x\over a^2}+{\cos^6x\over b^2}={1\over (a+b)^2}[/imath]
If [imath]{\sin^4x\over a}+{\cos^4x\over b}={1\over a+b}[/imath] then show that [imath]{\sin^6x\over a^2}+{\cos^6x\over b^2}={1\over (a+b)^2}[/imath] | 639223 | If [imath]\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}[/imath], then show that [imath]\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}[/imath]
If [imath]\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}[/imath], then show that [imath]\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}[/imath] My work: [imath](\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b})=\frac{1}{a+b}[/imath] By squaring both sides, we get, [imath]\frac{\sin^8 x}{a^2}+\frac{\cos^8 x}{b^2}+2\frac{\sin^4 x \cos^4 x}{ab}=\frac{1}{(a+b)^2}[/imath] [imath]\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}-2\frac{\sin^4 x \cos^4 x}{ab}-\frac{\sin^6 x \cos^2 x}{a^2}-\frac{\sin^2 x \cos^6 x}{b^2}=\frac{1}{(a+b)^2}[/imath] So, now, we have to prove that, [imath]-2\frac{\sin^4 x \cos^4 x}{ab}-\frac{\sin^6 x \cos^2 x}{a^2}-\frac{\sin^2 x \cos^6 x}{b^2}=0[/imath] I cannot do this. Please help! |
784576 | If [imath]p~ \mid~ m^p + n^p[/imath], prove [imath]p~ \mid~ \frac{m^p + n^p}{m+n}[/imath].
If [imath]p \mid m^p + n^p[/imath], and [imath]p[/imath] is a prime greater than [imath]2[/imath], prove [imath]p \mid \frac{m^p + n^p}{m+n}.[/imath] No clue how to start. Clearly [imath]p \mid m + n[/imath], but then what. I feel very less information is given. Yet primes are full of surprises. Thanks in advance. | 783274 | If [imath]p \mid m^p+n^p[/imath] prove [imath]p^2 \mid m^p+n^p[/imath]
Prove that for a prime [imath]p>2[/imath], if [imath]p \mid m^p+n^p[/imath], prove [imath]p^2\mid m^p+n^p[/imath] From Fermats theorem I concluded [imath]p \mid m+n[/imath], so [imath]p^2\mid (m+n)^p[/imath]. How do I proceed next? Any hints are welcomed. |
776101 | Ideals and prime ideals in a commutative ring.
Let [imath]A_1[/imath] and [imath]A_2[/imath] be two ideals, and [imath]P_1[/imath] and [imath]P_2[/imath] be two prime ideals in a commutative ring [imath]R[/imath]. Assume that [imath]A_1 ∩ A_2 ⊆ P_1 ∩ P_2[/imath]. Is there at least an [imath]i[/imath] and [imath]j[/imath] such that [imath]A_i ⊆ P_j[/imath] is true? | 791651 | A problem involving ideals and prime ideals.
Please help me with a solution to this problem. Let [imath]R[/imath] be a commutative ring. Let [imath]A_1, A_2[/imath] be two ideals of [imath]R[/imath], and [imath]P_1, P_2[/imath] two prime ideals of [imath]R[/imath]. Assume that [imath]A_1 \cap A_2 \subseteq P_1 \cap P_2[/imath]. Show that there exist [imath]i, j \in \{1,2\}[/imath] such that [imath]A_i \subseteq P_j[/imath]. |
784993 | Proof of [imath]e^x - 1 \geq x[/imath] for [imath]{x: -1 \leq x < 0}[/imath]
Is this valid, and how can i prove that it holds. Proof of [imath]e^x - 1 \geq x \text{ for } {x:-1 \leq x < 0}[/imath] | 504663 | Simplest or nicest proof that [imath]1+x \le e^x[/imath]
The elementary but very useful inequality that [imath]1+x \le e^x[/imath] for all real [imath]x[/imath] has a number of different proofs, some of which can be found online. But is there a particularly slick, intuitive or canonical proof? I would ideally like a proof which fits into a few lines, is accessible to students with limited calculus experience, and does not involve too much analysis of different cases. |
785075 | [imath]a^2+ab+b^2=1[/imath], how many ordered integer solutions exist?
I found the answers to be a=-1, b=0; a=1,b=0; a=0,b=1 and a=0,b=-1. Only 4 pairs, however the answer is 6. Please tell me how to proceed. | 784821 | Number of integer solutions of the following equation
Consider the equation [imath]a^2+ab+b^2=1[/imath]. How many integer pairs are solutions to this? I found 4 pairs: [imath]a=-1, b=0[/imath]; [imath]a=0, b=-1[/imath]; [imath]a=1,b=0[/imath]; [imath]a=0, b=1[/imath]. But the solution says the answer is 6. Which other possibilities am I missing? Please post the answer with solution. |
785242 | Need help in proving that [imath]4[/imath] divides [imath]n[/imath].
I'm pretty sure this question has a duplicate, but since I can't seem to find it, I'm asking it again. If [imath]a_{1},a_{2}...a_{n}[/imath] are [imath]1[/imath] or [imath]-1[/imath] and if,[imath]a_{1}a_{2}a_{3}a_{4}+a_{2}a_{3}a_{4}a_{5}+...+a_{n}a_{1}a_{2}a_{3}=0[/imath]prove that [imath]n[/imath] is divisible by [imath]4[/imath]. What I've done so far is manage to prove that [imath]n[/imath] is even, which is pretty obvious. I haven't come up with any workable leads so far, so I'd be much obliged if you could give me some pointers. | 635462 | Let [imath]a_1,a_2,\cdots,a_n[/imath] be n numbers such that [imath]a_i[/imath] is either 1 or -1.If [imath]a_1a_2a_3a_4+\cdots+a_na_1a_2a_3=0[/imath] then prove that 4|n.
Let [imath]a_1,a_2,\cdots,a_n[/imath] be [imath]n[/imath] numbers such that [imath]a_i[/imath] is either [imath]1[/imath] or [imath]-1[/imath]. If [imath]a_1a_2a_3a_4+a_2a_3a_4a_5+\cdots+a_na_1a_2a_3=0[/imath] then prove that [imath]4 \mid n[/imath]. My work: By multiplying all the terms, we get, [imath]a_1^4a_2^4\ldots a_n^4=1.[/imath] I think that I will be able to represent [imath]4n[/imath] a power of [imath]1[/imath], but getting no clue. Please help! I also think that this problem can be done with invariance and extremal principal too. Please help with these approaches too! |
785945 | Prove there exists a countably infinite subset [imath]A[/imath] of [imath]P(\mathbb{N})[/imath] that satisfies given conditions
Prove that there exists a countably infinite set [imath]A \subseteq [/imath] [imath]P(\mathbb{N})[/imath] that satisfies all of the following conditions: [imath]i)[/imath] [imath]X \cap Y = \emptyset[/imath] for all [imath]X, Y \in A[/imath] such that [imath]X \neq Y[/imath] [imath]ii)[/imath] [imath]\mathbb{N} = \bigcup A [/imath] [imath]iii)[/imath] Every element in [imath]A[/imath] is countably infinite For all the sets I tried, the three conditions were satisfied but [imath]A[/imath] was not countably infinite, so now I'm stuck. Any help would be greatly appreciated. | 784533 | Help finding this set
Lets define the following: Let A be a set. A is innumerable if and only if there exists a bijective function from A to [imath]\mathbb{N}[/imath] Proof that there exists an innumerable set [imath]B \subseteq \mathcal P \left({\mathbb{N}}\right)[/imath] such that it has the following properties: 1) [imath]X\cap Y =\emptyset[/imath] for all [imath]X,Y \in B[/imath] such that [imath]X \neq Y[/imath] 2) [imath]\mathbb{N}= \bigcup B[/imath] 3) All elements of B are innumerable. Any help is appreciated. I don´t know how to find this set ... |
786289 | how to prove finite subset of a countable set is countable
Let A be a countable set. How to prove that the set [imath]\mathcal A = \{X \subseteq A : |X| = n [/imath] for some [imath]n \in \omega \}[/imath] of all finite subsets of A is countable?. I checked similar questions subset but they are not discussing about finite subsets. It seems the question is unclear because the finite subsets can be countable. | 361320 | Neatest proof that set of finite subsets is countable?
I am looking for a beautiful way of showing the following basic result in elementary set theory: If [imath]A[/imath] is a countable set then the set of finite subsets of [imath]A[/imath] is countable. I proved it as follows but my proof is somewhat fugly so I was wondering if there is a neat way of showing it: Let [imath]|A| \le \aleph_0[/imath]. If [imath]A[/imath] is finite then [imath]P(A)[/imath] is finite and hence countable. If [imath]|A| = \aleph_0[/imath] then there is a bijection [imath]A \to \omega[/imath] so that we may assume that we are talking about finite subsets of [imath]\omega[/imath] from now on. Define a map [imath]\varphi: [A]^{<\aleph_0} \to (0,1) \cap \mathbb Q[/imath] as [imath]B \mapsto \sum_{n \in \omega} \frac{\chi_B (n)}{2^n}[/imath]. Then [imath]\varphi[/imath] is injective hence the claim follows. (The proof of which is what is the core-fugly part of the proof and I omit it.). |
786435 | Does there exist an infinite field with characteristic [imath]p[/imath] for any prime [imath]p[/imath] that is not too big?
Does there exist an infinite field with characteristic [imath]p[/imath] for any prime [imath]p[/imath]? I haven't seen any such fields in my course yet. | 58424 | Example of infinite field of characteristic [imath]p\neq 0[/imath]
Can you give me an example of infinite field of characteristic [imath]p\neq0[/imath]? Thanks. |
786944 | Find the natural numbers [imath]n[/imath] in which [imath]n^2[/imath] divides [imath]584[/imath]?
I'm trying to find the natural numbers [imath]n[/imath] in which [imath]n^2[/imath] divides [imath]584[/imath] ? i tried all the ways i know but i get stuck. | 786718 | How to solve the equation [imath]n^2 \equiv 0 \pmod{584}[/imath]?
Well, I've confused when trying to solve this equation can anybody help me : [imath]n^2 \equiv 0 \pmod{584}[/imath] I tried to factorize the [imath]584[/imath] i got [imath]584=2^3\times73[/imath]. so [imath]n^2[/imath] has to be divisible by [imath]2^3[/imath] and [imath]73[/imath] in this same time. here i get stuck. |
786912 | If [imath]A^2=A[/imath], [imath]B^2=B[/imath] and [imath]I-(A+B)[/imath] is invertible, then [imath]R(A) = R(B)[/imath]
If [imath]I-(A+B)[/imath] is invertible and [imath]A[/imath] and [imath]B[/imath] are idempotent matrices, then how do I show that [imath]A[/imath] and [imath]B[/imath] have the same rank? | 782030 | Prove that matrices have equal rank.
If [imath]P[/imath] and [imath]Q[/imath] are [imath]n \times n[/imath] matrices of real numbers such that [imath]P^2=P[/imath] and [imath]Q^2=Q[/imath] and [imath]I-P-Q[/imath] is invertible where [imath]I[/imath] is an [imath]n \times n[/imath] identity matrix, Show that [imath]P[/imath] and [imath]Q[/imath] have the same rank. If [imath]P[/imath] is non-singular, then it can be shown that [imath]P=Q=I[/imath], so they have same rank. But I can't prove it when [imath]P[/imath] is singular. |
785225 | Is it possible to intuitively explain why conditional probability/expectation depends on sigma algebra
Is it possible to intuitively explain why conditional probability/expectation [imath](P(A|X)[/imath] or [imath]E[Y|X])[/imath] depends on sigma algebra on which it is conditioned not the value of the random variable ? Formally, it is due to Radon-Nykodin theorem. But in Breiman it says that the intuition behind this is that the relevant information contained in knowing [imath]X(\omega)[/imath] is the information regarding the location of [imath]\omega[/imath]. What is the meaning of it. | 780877 | One confusion over conditional expectation
Suppose for two random variables [imath]X_1[/imath] and [imath]X_2[/imath] [imath]\sigma(X_1) = \sigma(X_2)[/imath]. Why [imath]E[Y | X_1] = E[Y | X_2][/imath] a.e. ? the set where [imath]X_1= X_1(\omega)[/imath] can be different from the set [imath]X_2= X_2(\omega) [/imath]. How can then conditional expectation conditioned on the event can be equal ? |
582985 | Linear functional and transposition.
This is a question from Hoffman. Let [imath]V[/imath] be a finite-dimensional vector space over the field [imath]F[/imath] and let [imath]T[/imath] be a linear operator on [imath]V[/imath]. Let [imath]c[/imath] be a scalar and suppose there is a non-zero vector [imath]\alpha[/imath] in [imath]V[/imath] such that [imath]T\alpha=c\alpha[/imath]. Prove that there is a non-zero linear functional [imath]f[/imath] on [imath]V[/imath] such that [imath]T^tf=cf[/imath]. I know that essentially I need to prove that there exist an [imath]f[/imath] such that [imath]fT(\beta)=cf(\beta)[/imath] for all [imath]\beta[/imath] in [imath]V[/imath]. What puzzles me is that there doesn't seem to be anymore information on the linear transformation [imath]T[/imath] other than the fact that there is a non-zero vector [imath]\alpha[/imath] in [imath]V[/imath] such that [imath]T\alpha=c\alpha[/imath]; we have no idea of what we will get when we let [imath]T[/imath] operate on vectors other than [imath]\alpha[/imath] in [imath]V[/imath]. It therefore seems that there is a certain [imath]f[/imath] that can ignore the effect of [imath]T[/imath] somehow,but I can't find that [imath]f[/imath]. | 131005 | If [imath]T\alpha=c\alpha[/imath], then there is a non-zero linear functional [imath]f[/imath] on [imath]V[/imath] such that [imath]T^{t}f=cf[/imath]
I am self-studying Hoffman and Kunze's book Linear Algebra. This is exercise [imath]4[/imath] from page [imath]115[/imath]. It is in the section of The transpose of a Linear Transformation. Let [imath]V[/imath] be a finite-dimensional vector space over the field [imath]\mathbb{F}[/imath] and let [imath]T[/imath] be a linear operator on [imath]V[/imath]. Let [imath]c[/imath] be a scalar and suppose there is a non-zero vector [imath]\alpha[/imath] in [imath]V[/imath] such that [imath]T\alpha=c\alpha.[/imath] Prove that there is a non-zero linear functional [imath]f[/imath] on [imath]V[/imath] such that [imath]T^{t}f=cf.[/imath] ([imath]T^{t}[/imath] is the transpose of [imath]T[/imath].) I tried to solve this question by induction on [imath]\dim V[/imath]. I was able to show the base case, that is, when [imath]\dim V=1[/imath], but I got stuck in the inductive step. If [imath]\alpha [/imath] is a non-zero vector, then we can find a base [imath]\mathcal{B}=\{\alpha,\alpha_{1},\ldots\alpha_{m}\}[/imath] of [imath]V[/imath]. We can write [imath]V=W_{1}\oplus W_{2}[/imath], where [imath]W_{1}=\langle \alpha \rangle[/imath] and [imath]W_{2}=\langle \alpha_{1},\ldots,\alpha_{m}\rangle.[/imath] I can not show that [imath]T(W_{2})\subset W_{2}.[/imath] Anyway, [imath]\alpha \notin W_{2}.[/imath] EDIT: If [imath]T[/imath] is a linear transformation from [imath]V[/imath] to [imath]W[/imath], then [imath]T^{t}[/imath] is the linear transformation from [imath]W^{\star}[/imath] into [imath]V^{\star}[/imath] such that [imath](T^{t}f)(\alpha)=g(T\alpha)[/imath] for every [imath]f\in W^{\star}[/imath] and [imath]\alpha \in V[/imath]. |
788348 | can I show this set is closed?
Suppose [imath]X[/imath] is a normed vector space with norm [imath]||\cdot||[/imath]. Suppose [imath]S[/imath] is a linear subspace of [imath]X[/imath] and [imath]S[/imath] is closed. Let [imath]x\in X[/imath]. I was wondering whether the set [imath]\{||x+s||: s\in S\}[/imath] is closed in [imath]\mathbb{R}[/imath] | 296354 | Given a point [imath]x[/imath] and a closed subspace [imath]Y[/imath] of a normed space, must the distance from [imath]x[/imath] to [imath]Y[/imath] be achieved by some [imath]y\in Y[/imath]?
I think no. And I am looking for examples. I would like a sequence [imath]y_n[/imath] in [imath]Y[/imath] such that [imath]||y_n-x||\rightarrow d(x,Y)[/imath] while [imath]y_n[/imath] do not converge. Can anyone give a proof or an counterexample to this question? |
789103 | Number theory digit sum
How many natural numbers less than [imath]10^8[/imath] are there, with sum of digits equal to [imath]7[/imath]? My friend told me it is coefficient of [imath]x^7[/imath] in [imath]\frac{(x^{10} - 1)}{(x-1)^8}[/imath] How did he get this result? Can anyone please explain with logic. | 587783 | How many natural numbers less than [imath]10^8[/imath] are there, whose sum of digits equals [imath]7[/imath]?
How many natural numbers less than [imath]10^8[/imath] are there, whose sum of digits equals [imath]7[/imath]? I got it here.But is there any more effecient and easier way to solve than the link shows? |
789022 | Prove that if [imath]p: Y \to X[/imath] is a covering space and [imath]X[/imath] is path connected, then the cardinality of [imath]p^{-1}(X)[/imath] is constant.
Let [imath]p: Y \to X[/imath] be a covering space and [imath]X[/imath] is connected. I want to show that [imath]\forall x \in X[/imath] the cardinality of [imath]p^{-x}[/imath] is the same. [imath]\textbf{My Attempt:}[/imath] Let us first fix a point [imath]x_0 \in X[/imath] such that [imath]|p^{-1}(x_0)|=\kappa[/imath]. Then let [imath]H=\{x \in X \mid |p^{-1}(x)=\kappa| \}[/imath] Then let [imath]f: H \to \text{Cardinals}[/imath] be defined as [imath]f(x)=|p^{-1}(x)|[/imath]. Then [imath]f[/imath] is a constant function on [imath]H[/imath]. How do I show that [imath]H[/imath] is open? [imath]H[/imath] is the inverse image of [imath]\{\kappa\}[/imath] but this is a singleton set from the collection of cardinals. I am not really sure what to do with that. If I can show that [imath]H[/imath] is open, then I can prove by contradiction, using the assumption of connectedness, that all [imath]x \in X[/imath] are such that [imath]|p^{-1}(x)|= \kappa[/imath]. | 780920 | If $h : Y \to X$ is a covering map and $Y$ is connected, then the cardinality of the fiber $h^{-1}(x)$ is independent of $x \in X$.
In "Knots and Primes: An Introduction to Arithmetic Topology", the author uses the following proposition Let [imath]h: Y \to X[/imath] be a covering. For any path [imath]\gamma : [0,1] \to X[/imath] and any [imath]y \in h^{-1}(x) (x = \gamma(0))[/imath], there exists a unique lift [imath]\hat{\gamma} : [0,1] \to Y[/imath] of [imath]\gamma[/imath] with [imath]\hat{\gamma}(0) = y[/imath]. Furthermore, for any homotopy [imath]\gamma_t (t \in [0,1])[/imath] of [imath]\gamma[/imath] with [imath]\gamma_t = \gamma(0)[/imath] and [imath]\gamma_t(1) = \gamma(1)[/imath], there exists a unique lift of [imath]\hat{\gamma_t}[/imath] such that [imath]\hat{\gamma_t}[/imath] is the homotopy of [imath]\hat{\gamma}[/imath] with [imath]\hat{\gamma_t}(0) = \hat{\gamma}(0)[/imath] and [imath]\hat{\gamma_t}(1) = \hat{\gamma}(1)[/imath]. The author then follows with; "In the following, we assume that any covering space is connected. By the preceding proposition, the cardinality of the fiber [imath]h^{-1}(x)[/imath] is independent of [imath]x \in X[/imath]." I am not sure why this results is true. This is my attempt at explaining it to myself. Take two different [imath]x_1, x_2[/imath] and their fibers [imath]h^{-1}(x_1), h^{-1}(x_2)[/imath] such that [imath]y_1 \in h^{-1}(x_1)[/imath] and [imath]y_2 \in h^{-1}(x_2)[/imath]. Take two paths [imath]\gamma_1, \gamma_2[/imath] such that [imath]\gamma_1(0) = x_1[/imath] and [imath]\gamma_2(0) = x_2[/imath]. Then we get two lifts [imath]\hat{\gamma_1}, \hat{\gamma_2}[/imath] with [imath]\hat{\gamma_1}(0) = y_1[/imath] and [imath]\hat{\gamma_2}(0) = y_2[/imath]. Now, since our covering space is connected we can continuously deform [imath]\hat{\gamma_1}[/imath] into [imath]\hat{\gamma_2}[/imath] and conclude that every elements in [imath]h^{-1}(x_1)[/imath] is also in [imath]h^{-1}(x_2)[/imath] and vice versa. I feel that this is wrong but cannot figure out the right way to see this. Any help would be appreciated. |
788995 | Proving a metric with absolute value
I need to prove that function [imath]\mathbb R × \mathbb R → \mathbb R [/imath] : [imath]f(x,y) = \frac{|x-y|}{1 + |x-y|}[/imath] is a metric on [imath]\mathbb R[/imath]. First two axioms are trivial; it's the triangle inequality which is pain. [imath]\frac{|x-y|}{1 + |x-y|}[/imath] + [imath]\frac{|y-z|}{1 + |y-z|} ≥ \frac{|x-z|}{1 + |x-z|} ⇒ \frac{|x-y| + |y-z| + 2|(x-y)(y-z)|}{1 + |x-y| + |y-z| + |(x-y)(y-z)|}≥ \frac{|x-z|}{1 + |x-z|}[/imath], but then I am stuck. Can somebody show me way out of this? | 355493 | Defining a metric space
I'm studying for actuarial exams, but I always pick up mathematics books because I like to challenge myself and try to learn new branches. Recently I've bought Topology by D. Kahn and am finding it difficult. Here is a problem that I think I'm am answering sufficiently but any help would be great if I am off. If [imath]d[/imath] is a metric on a set [imath]S[/imath], show that [imath]d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}[/imath] is a metric on [imath]S[/imath]. The conditions for being a metric are [imath]d(X,Y)\ge{0}, d(X,Y)=0[/imath] iff [imath]X=Y[/imath], [imath]d(X,Y)=d(Y,X)[/imath], and [imath]d(X,Y)\le{d(X,Z)+d(Z,Y)}[/imath]. Thus, we simply go axiom by axiom. 1) Since both [imath]d(x,y)\ge{0}[/imath] and [imath]1+d(x,y)\ge{0},[/imath] it is clear that [imath]d_1(x,y)\ge{0}[/imath]. (Is this a sufficient analysis?) 2) [imath]d_1(x,x)=\frac{d(x,x)}{1+d(x,x)}=\frac{0}{1+0}=0[/imath]. 3) [imath]d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}=\frac{d(y,x)}{1+d(y,x)}=d_1(y,x).[/imath] 4) [imath]d_1(x,y)=\frac{d(x,y)}{1+d(x,y)}\le{\frac{d(x,z)+d(z,y)}{1+d(x,z)+d(z,y)}}=\frac{d(x,z)}{1+d(x,z)+d(z,y)}+\frac{d(z,y)}{1+d(x,z)+d(z,y)}\lt\frac{d(x,z)}{1+d(x,z)}+\frac{d(z,y)}{1+d(z,y)}=d_1(x,z)+d_1(z,y).[/imath] However, #4 is strictly less, not less than or equal to, according to my analysis, so where did I go wrong? |
789237 | How to integrate [imath]\int\frac{xe^{tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx[/imath]?
Original question: (updated question in section EDIT) How to evaluate the following integral? [imath]\int\frac{1}{(1+x^2)^\frac{3}{2}}\,dx[/imath] I tried substitution: [imath] t = 1+x^2 = \varphi \\ \frac{\,dt}{\,dx} = \varphi' = 2x \implies \,dt = 2x\,dx[/imath] ... this doesn't help me, then I thought, maybe this way: [imath] t = 1+x^2 \implies x = \sqrt{t - 1} = \varphi \\ \frac{\,dx}{\,dt} = \varphi' = \frac{1}{2\sqrt{t-1}} \implies \,dx = \frac{1}{2\sqrt{t-1}}\,dt[/imath] so I have: [imath]\int\frac{1}{(1+x^2)^\frac{3}{2}} \,dx= \frac{1}{2}\int\frac{\,dt}{t^{\frac{3}{2}}\sqrt{t-1}} = \frac{1}{2}\int\frac{\,dt}{\sqrt{t^4-t^3}}[/imath] which doesn't look very promising. Then I thought: [imath] t = x^2 \implies x = \sqrt{t} = \varphi \\ \frac{\,dx}{\,dt} = \varphi' = \frac{1}{2\sqrt{t}} \implies \,dx = \frac{\,dt}{2\sqrt{t}}[/imath] which gives me: [imath]\int\frac{1}{(1+x^2)^\frac{3}{2}} \,dx= \frac{1}{2}\int\frac{\,dt}{(1+t)^{\frac{3}{2}}\sqrt{t}}[/imath] The answer looks like someone could just "guess it": http://www.wolframalpha.com/share/clip?f=d41d8cd98f00b204e9800998ecf8427ea57hoiergo but I don't know how to "get there" by calculation. EDIT: The actual problem was the following integral: [imath]\int\frac{xe^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx[/imath] and my idea was to apply partial integration i.e. [imath]\int u\,dv = uv - \int v\,du[/imath] where [imath] u = e^{\tan^{-1}x} \implies \,du = e^{\tan^{-1}x} \cdot \frac{1}{1 + x^2} \,dx = \frac{e^{\tan^{-1}x}}{1+x^2} \,dx \\ \,dv = \frac{x}{(1+x^2)^\frac{3}{2}} \implies v = -\frac{1}{\sqrt{1+x^2}}[/imath] because [imath]t = 1 + x^2 = \varphi \implies \varphi' = 0 + 2x = 2x \\ \frac{\,dt}{\,dx} = \varphi' = 2x \implies \,dt = 2x\,dx \\ \int \frac{\frac{1}{2}\,dt}{t^\frac{3}{2}} = \frac{1}{2} \int\frac{\,dt}{t^\frac{3}{2}} = \frac{1}{2} \cdot \frac{t^{-\frac{1}{2}}}{-\frac{1}{2}} + C = -t^{-\frac{1}{2}} + C = -\frac{1}{\sqrt{1+x^2}} + C[/imath] and now we have: [imath]\int\frac{xe^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx = -\frac{e^{\tan^{-1}x}}{\sqrt{1+x^2}} - \int \bigg( -\frac{1}{\sqrt{1+x^2}} \bigg) \cdot \frac{e^{\tan^{-1}x}}{1+x^2}\,dx \\ = -\frac{e^{\tan^{-1}x}}{\sqrt{1+x^2}} + \int \frac{e^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx[/imath] now I want to solve [imath]\int \frac{e^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx[/imath] and my approach was the following: [imath] u = e^{\tan^{-1}x} \implies \,du = \frac{e^{\tan^{-1}x}}{1+x^2} \,dx \\ \,dv = \frac{1}{(1+x^2)^\frac{3}{2}}\,dx \implies v = ?[/imath] and this is the part I couldn't figure out i.e. [imath]\int\frac{1}{(1+x^2)^\frac{3}{2}}\,dx[/imath] but as you all pointed out, we can solve this by using trig (as shown here: Integrate Form [imath]du / (a^2 + u^2)^{3/2}[/imath]) or hyperbolic (as shown in the answer provided by georg) substitutions which gives us: [imath]\int\frac{1}{(1+x^2)^\frac{3}{2}}\,dx = \cdots = \frac{x}{\sqrt{1+x^2}} + C \implies v = \frac{x}{\sqrt{1+x^2}} [/imath] so we finally get: [imath]\int\frac{xe^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx = -\frac{e^{\tan^{-1}x}}{\sqrt{1+x^2}} + \int \frac{e^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx \\ = -\frac{e^{\tan^{-1}x}}{\sqrt{1+x^2}} + \frac{xe^{\tan^{-1}x}}{\sqrt{1+x^2}} - \int \frac{x}{\sqrt{1+x^2}}\cdot \frac{e^{\tan^{-1}x}}{1+x^2}\,dx \\ = -\frac{e^{\tan^{-1}x}}{\sqrt{1+x^2}} + \frac{xe^{\tan^{-1}x}}{\sqrt{1+x^2}} - \int\frac{xe^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx \\ \implies \\ 2 \int\frac{xe^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx = -\frac{e^{\tan^{-1}x}}{\sqrt{1+x^2}} + \frac{xe^{\tan^{-1}x}}{\sqrt{1+x^2}} \\ \int\frac{xe^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx = \frac{(x-1)e^{\tan^{-1}x}}{2\sqrt{1+x^2}} + C[/imath] which is the answer to the problem. I also noticed a different approach: [imath]\int\frac{xe^{\tan^{-1}x}}{(1+x^2)^\frac{3}{2}}\,dx = \int\frac{xe^{\tan^{-1}x}}{\sqrt{1+x^2}(1+x^2)}\,dx \\ t = \tan^{-1}x = \varphi \implies \varphi' = \frac{1}{1+x^2} \\ \frac{dt}{dx} = \varphi' \implies \,dt = \frac{\,dx}{1+x^2} \\ \int\frac{xe^{\tan^{-1}x}}{\sqrt{1+x^2}(1+x^2)}\,dx = \int\frac{\tan(t) e^t \,dt}{\sqrt{1 + \tan^2(t)}} = \int\frac{\tan(t) e^t \,dt}{\sqrt{\frac{1}{\cos^2(t)}}} = \int\sin(t) e^t \,dt \\ u = \sin t \implies \,du = \cos t\,dt \\ \,dv = e^t \,dt \implies v = e^t \\ \int\sin(t) e^t \,dt = \sin (t)e^t - \int e^t\cos(t)\,dt[/imath] now we solve [imath]\int e^t\cos(t)\,dt[/imath] : [imath] u = \cos t \implies \,du = -\sin t \,dt \\ \,dv = e^t\,dt \implies v = e^t \\ \int e^t\cos(t)\,dt = \cos(t)e^t - \int(-\sin(t))e^t\,dt[/imath] so we finally have: [imath] \int\sin(t) e^t \,dt = \sin(t)e^t - \bigg(\cos(t)e^t - \int(-\sin(t))e^t\,dt \bigg) \implies \\ 2\int\sin(t) e^t \,dt = \sin(t)e^t - \cos(t)e^t \implies \int\sin(t) e^t \,dt = \frac{(\sin(t) - \cos(t))e^t}{2} \\ \int\sin(\tan^{-1}x) e^{\tan^{-1}x} \,dx = \frac{(\sin(\tan^{-1}x) - \cos(\tan^{-1}x))e^{\tan^{-1}x}}{2} + C [/imath] and since: [imath]\sin(\tan^{-1}x) = \frac{x}{\sqrt{1+x^2}} ,\, \cos(\tan^{-1}x) = \frac{1}{\sqrt{1+x^2}}[/imath] we have: [imath] \int\sin(\tan^{-1}x) e^{\tan^{-1}x} \,dx = \frac{(x-1)e^{\tan^{-1}x}}{2\sqrt{1+x^2}} + C[/imath] and that's the answer. | 17666 | Integrate Form [imath]du / (a^2 + u^2)^{3/2}[/imath]
How does one integrate [imath]\int \dfrac{du}{(a^2 + u^2)^{3/2}}\ ?[/imath] The table of integrals here: http://teachers.sduhsd.k12.ca.us/abrown/classes/CalculusC/IntegralTablesStewart.pdf Gives it as: [imath]\frac{u}{a^2 ( a^2 + u^2)^{1/2}}\ .[/imath] I'm getting back into calculus and very rusty. I'd like to be comfortable with some of the proofs behind various fundamental "Table of Integrals" integrals. Looking at it, the substitution rule seems like the method of choice. What is the strategy here for choosing a substitution? It has a form similar to many trigonometric integrals, but the final result seems to suggest that they're not necessary in this case. |
789476 | Find the value of [imath]a^m+1/a^m+a^{2m}+1/a^{2m}[/imath]
If [imath]a+1/a+a^2+1/a^2+1=0[/imath], and [imath]m[/imath] a poitive integer, then find the value of: [imath]a^m+1/a^m+a^{2m}+1/a^{2m}[/imath] Only hints please.. | 785195 | Find the value of [imath]a^{2m}+a^m+{1\over a^m}+{1\over a^{2m}}[/imath]
Let [imath]a[/imath] be a complex number such that [imath]a^2+a+{1\over a}+{1\over a^2}+1=0[/imath] Let [imath]m[/imath] be a positive integer, find the value of [imath]a^{2m}+a^m+{1\over a^m}+{1\over a^{2m}}[/imath] My approach: I factorized equation 1 which yielded [imath](a+{1\over a})={-1^+_-(5^{1\over2})\over2}[/imath] I don't know if it is of any help. |
787645 | Partitions of Natural Numbers
This is a question from Complex analysis by Stein. The question is Prove that it is not possible to partition [imath]\mathbb N[/imath] into finitely many infinite AP's with distinct common differences.(other than the trivial [imath]a=d=1[/imath]) Infinite AP:=[imath]\{a,a+d,a+2d,......\}[/imath].I have no idea as to how to approach it. Any hints are welcome. Thanks. | 149040 | Prove: The positive integers cannot be partitioned into arithmetic sequences (using Complex Analysis)
An arithmetic sequence of step [imath]d[/imath] is a set of the form: {[imath]a, a+d, a+2d, a+3d, ...[/imath]} where [imath]a, d[/imath] are positive integers. Show that the positive integers cannot be partitioned into a finite number of arithmetic sequences s.t. each sequence has a distinct step [imath]d[/imath]. (Except the trivial sequence [imath]a=1, d=1[/imath].) Now the book gives me a hint: Write [imath]\sum_{n\in\mathbb{N}}z^n[/imath] as a sum of terms of the type [imath]\frac{z^a}{1-z^d}[/imath]. But it isn't clear that the power series can even be put such a form, and even if it can, why does that help me? |
493693 | Is convex set interval or ray?
Given an ordered set [imath]X[/imath], say a subset [imath]Y[/imath] of [imath]X[/imath] is convex in [imath]X[/imath] if for each pair of points [imath]a<b[/imath] of [imath]Y[/imath], the entire interval [imath](a,b)[/imath] of points of [imath]X[/imath] lies in [imath]Y[/imath]. An interval is of the form [imath](a,b),(a,b],[a,b),[a,b][/imath], and a ray is of the form [imath](a,+\infty),(-\infty,a),[a,+\infty),(-\infty,a][/imath]. Let [imath]X[/imath] be an ordered set. If [imath]Y[/imath] is a proper subset of [imath]X[/imath] that is convex in [imath]X[/imath], does it follow that [imath]Y[/imath] is an interval or a ray in [imath]X[/imath]? Well, suppose [imath]Y[/imath] has a smallest element [imath]a[/imath] and a largest element [imath]b[/imath]. Then [imath][a,b]\in Y[/imath], and so [imath]Y=[a,b][/imath] and [imath]Y[/imath] is an interval. What if [imath]Y[/imath] has no smallest element? So for each element [imath]a\in Y[/imath], there exists an element [imath]b\in Y[/imath] such that [imath]b<a[/imath]. What then? | 432027 | If nonempty, nonsingleton [imath]Y[/imath] is a proper convex subset of a simply ordered set [imath]X[/imath], then [imath]Y[/imath] is ray or interval?
This is from Exercise 7 in p. 92 in Munkres's Topology. Except for the trivial cases such as [imath]Y[/imath] is empty set or singleton, it seems if [imath]Y[/imath] is convex in an simply ordered set [imath]X[/imath] then [imath]Y[/imath] is interval or ray. But I cannot start my proof because I cannot use [imath]\sup[/imath] or [imath]\max[/imath] functions. What should I do? |
790122 | Prove: [imath]\lim\limits_{(x,y)\to (0,0)}\frac{xy}{(x^2+y^2)^\frac12}=0[/imath]
Prove: [imath]\lim_{(x,y)\to (0,0)}\frac{xy}{(x^2+y^2)^\frac12}=0[/imath] Thanks for your help. Thanks ahead:) | 101560 | Computing [imath]\lim_{(x,y)\to (0,0)}\frac{x+y}{\sqrt{x^2+y^2}}[/imath]
What is the result of [imath]\lim_{(x,y)\to (0,0)}\frac{x+y}{\sqrt{x^2+y^2}}[/imath] . I tried to do couple of algebraic manipulations, but I didn't reach to any conclusion. Thanks a lot. |
301800 | Find the minimum value of [imath]|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|[/imath] for real numbers [imath]x[/imath]
I'm tutoring for a college math class and we're doing putnam problems next week and this one stumped me: Find the minimum value of [imath]|\sin x+\cos x+\tan x+\cot x+\sec x+\csc x|[/imath] for real numbers [imath]x[/imath]. | 2234066 | Minimum value of : [imath]\left|\sin{x}+ \cos{x} + \sec{x} + \tan{x} + \cot{x} + \csc{x}\right|[/imath]
Find the minimum value of :[imath]\left|\sin{x}+ \cos{x} + \sec{x} + \tan{x} + \cot{x} + \csc{x}\right|.[/imath] What I have tried is trying to simplify the expression into a [imath]\sin{2x}[/imath] form but it did not go well. I guess there is much simpler way to solve the problem. I also thought of AM-GM inequality but for inequality to hold all the terms must be positive real numbers which in this case is not. |
790023 | Find the value of this combinatorial sum. [imath]\sum_{k=4}^{100}\binom{k-1}{3}[/imath]
How to compute this sum without laboring? [imath]\sum_{k=4}^{100}\dbinom{k-1}{3}.[/imath] Is it possible to reduce this to a single combinatorial term? | 583402 | Show that [imath] \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}[/imath]
I need a hand in showing that [imath] \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}[/imath] Thanks in advance for any help. |
789928 | Investigating the convergence of a series using the comparison limit test, Part II
I posted this question earlier, but as I don't know if a comment reply or edit will refresh this so people actually see, I'm going to post it again in hopes that someone knows what's going on. Here's the initial question: Replacing the sequence: [imath]x_{n}=1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}-2\sqrt{n},\,\,\,\, n=1,2,....[/imath] By the corresponding series, invesigate it's convergence. Hint: Take [imath]a_{1}=x_{1}[/imath] and [imath]a_{n}=x_{n}-x_{n-1}[/imath] for [imath]n>1[/imath]. Then [imath]x_{n}[/imath] is a sequence of partial sums for [imath]\sum_{k=1}^{\infty}a_{k}[/imath]. Use an expicit formula for [imath]a_{n}[/imath] and use the comparison in limit test to show that the series converges. Solution: Computing [imath]x_n - x_{n-1}[/imath], we get: [imath]x_{n}-x_{n-1} = \left(\sum_{k=1}^{n}\frac{1}{\sqrt{k}}-2\sqrt{n}\right)-\left(\sum_{k=1}^{n-1}\frac{1}{\sqrt{k}}-2\sqrt{n-1}\right)[/imath] [imath]a_{n} = \left(\sum_{k=1}^{n-1}\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{n}}-2\sqrt{n}\right)-\left(\sum_{k=1}^{n-1}\frac{1}{\sqrt{k}}-2\sqrt{n-1}\right)[/imath] [imath]a_{n} = \frac{1}{\sqrt{n}}-2\sqrt{n}+2\sqrt{n-1}[/imath] Now, the comparison test states that: If [imath]\lim_{n \to \infty} \frac{a_n}{b_n} = c[/imath] for some limit [imath]c[/imath], then either both [imath]a_n[/imath] and [imath]b_n[/imath] converge or diverge. Trick is, I'm really struggling to find a series [imath]b_n[/imath] that lets me show that [imath]a_n[/imath] diverges (i.e. that my fraction has a finite limit, hence [imath]a_n[/imath] diverges). | 788771 | Investigating the convergence of a series using the comparison limit test
Actually not sure how to approach this... but I may be missing something: Replacing the sequence: [imath]x_{n}=1+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{n}}-2\sqrt{n},\,\,\,\, n=1,2,....[/imath] By the corresponding series, invesigate it's convergence. Hint: Take [imath]a_{1}=x_{1}[/imath] and [imath]a_{n}=x_{n}-x_{n-1}[/imath] for [imath]n>1[/imath]. Then [imath]x_{n}[/imath] is a sequence of partial sums for [imath]\sum_{k=1}^{\infty}a_{k}[/imath]. Use an expicit formula for [imath]a_{n}[/imath] and use the comparison in limit test to show that the series converges. Note: This looks easy, but as I said... there's something not quite connecting. When writing out the summation, I'm not entirely sure if they're compatible, as one would go from k=1 to n, and the other would go from k=1 to n-1, yet still vary with k... so do I need to use another pronumeral for it? Or can I simply use k-1 instead of k? |
790259 | Help my homework: [imath]\int_0^1\int_{0}^1\frac{x^2-y^2}{(x^2+y^2)^2}dy\, dx[/imath]
I am trying to integrate [imath]\int_0^1\int_{0}^1\frac{x^2-y^2}{(x^2+y^2)^2}dy\, dx[/imath] In my book said that use tangent function but I don`t know how to evaluate it. Please help me. I want to know the process of evaluating this integral. One more question [imath]f(x,y)=1[/imath], if [imath]x[/imath] rational [imath]f(x,y)=2y[/imath], if [imath]x[/imath] irrational then why is [imath]f(x,y)[/imath] not integrable but the iterated integral exists over [imath][0,1]\times[0,1][/imath]? Thank you. | 787270 | How to Double integrals
I am trying to integrate [imath]\int_0^1\int_{0}^1\frac{x^2-y^2}{(x^2+y^2)^2}dy\, dx[/imath] In my book said that use tangent function but.. I don`t know.. help me. I want to know process of calculate this problem. and one more question f(x,y)=1, if x rational f(x,y)=2y, if x irrational then why f is not integrable? but iterated integral exists? over[0,1]x[0,1] |
790874 | [imath]\text{Gal}(L|K)[/imath] not abelian
[imath]f \in K[X][/imath] is irreducible and separable. [imath]L[/imath] the splitting field of [imath]f[/imath]. Show: [imath][L:K] > \text{deg}(f)[/imath] [imath]\Rightarrow[/imath] [imath]\text{Gal}(L|K)[/imath] is not abelian | 668357 | If Gal(K,Q) is abelian then |Gal(K,Q)|=n
Let [imath]f(x)\in \mathbb Q[x][/imath] irreducible of degree [imath]n[/imath] and [imath]K[/imath] its splitting field over [imath]\mathbb Q[/imath]. Prove that if [imath]\operatorname{Gal}(K/\mathbb Q)[/imath] is abelian, then [imath]|\operatorname{Gal}(K/\mathbb Q)|=n[/imath]. How can I prove this? |
791339 | understanding the difference between uniform continuity and continuity
We have defined continuity as [imath]\forall \epsilon, \exists \delta > 0 s.t. |x-c| < \delta, x \in D \implies |f(x) - f(c)| < \epsilon[/imath] and uniform continuity as [imath]\forall \epsilon, \exists \delta > 0 s.t. |x-y| < \delta, x,y \in D \implies |f(x) - f(y)| < \epsilon[/imath] The only difference I see is that we have [imath]x,y[/imath] in the domain rather than just [imath]x[/imath], but what does that really change? Furthermore, we have a theorem about the continuity on a closed domain implies uniform continuity, why does this work? I understand the proof but I'm just not sure why it works. | 293554 | difference of uniform continuity and continuity of map
[imath]f:X\rightarrow Y [/imath] is continuous iff every open ball in [imath]Y[/imath] has open pre-image. It is between topological spaces. [imath]f:X\rightarrow Y[/imath] is uniformly continuous if [imath]\forall \epsilon >0,\forall x \in X, |x-y|<\delta \implies |f(x)-f(y)|<\epsilon[/imath]. My question is what is the difference between uniform continuity and continuity of a map. |
791406 | Non-trivial solutions for [imath](x;\sin{x})\in\mathbb{Q}^2[/imath]?
Is it possible? For [imath]\cos{x}[/imath] were analogous solutions also okay. [imath](x;\sin{x})\in\mathbb{Q}^2[/imath] | 299124 | Is sin(x) necessarily irrational where x is rational?
My friend and I were discussing this and we couldn't figure out how to prove it one way or another. The only rational values I can figure out for [imath]\sin(x)[/imath] (or [imath]\cos(x)[/imath], etc...) come about when [imath]x[/imath] is some product of a fraction of [imath]\pi[/imath]. Is [imath]\sin(x) [/imath] (or other trigonometric function) necessarily irrational if [imath] x [/imath] is rational? Edit: Excluding the trivial solution of 0. |
416120 | [imath](a_n+a_{n+1})[/imath] is convergent implies [imath](a_n/n)[/imath] converges to [imath]0[/imath]
Let [imath](a_n)[/imath] be a sequence such that [imath](a_n+a_{n+1})[/imath] is convergent. Prove that [imath](a_n/n)[/imath] converges to [imath]0[/imath]. [imath](a_n+a_{n+1})[/imath] convergent to [imath]L[/imath] means that for all [imath]\epsilon[/imath], there exists [imath]N[/imath] such that for all [imath]n\geq N[/imath], [imath]|L-a_n-a_{n+1}|<\epsilon[/imath], or in other words [imath]L-\epsilon<a_n+a_{n+1}<L+\epsilon[/imath]. How to proceed from here? | 669790 | [imath]\lim_{n \rightarrow\infty} ~ x_{n+1} - x_n= c , c > 0[/imath] . Then, is [imath]\{x_n/n\}[/imath] convergent?
If [imath]\{x_n\}[/imath] is a sequence which satisfies [imath]\lim_{n \rightarrow\infty} ~ x_{n+1} - x_n= c[/imath] where [imath]c[/imath] is a real positive number. Then what can be said about the convergence/ divergence, boundedness/ unboundedness of [imath]\{x_n/n\}[/imath]. Attempt: [imath]\lim_{n \rightarrow\infty} ~ x_{n+1} - x_n= c[/imath] where [imath]c >0[/imath] => [imath]x_n[/imath] is unbounded and divergent. However, I am stuck on how to relate this to convergence/divergence of [imath]x_n/n[/imath] . Thanks for the help. |
791860 | Convergence of series [imath]\sum_{n=1}^{\infty} \frac{\ln^\beta(n)}{n^\alpha}[/imath]
I know if [imath]\beta = 0[/imath] it converges [imath]\iff[/imath] [imath]\alpha > 1[/imath]. Also, it doesn't converge if [imath]\alpha = 0[/imath]. I don't know what test to apply for the rest of cases. Any hints? | 267697 | Convergence of [imath]\sum\limits_{n=2}^\infty \frac{1}{n^\alpha \ln^\beta (n)} [/imath] for nonnegative [imath]\alpha[/imath] and [imath]\beta[/imath]
Study the convergence of the following series: [imath]\sum_{n=2}^\infty \frac{1}{n^\alpha \cdot\ln^\beta(n)} \text{ where }\alpha,\beta \geq 0 [/imath] Applying d'Alembert criterion I have that [imath] \lim_{n\to\infty} \frac{n^\alpha \ln^\beta(n)}{(n+1)^\alpha \ln^\beta (n+1)} = \lim_{n\to\infty} \left(\frac{n}{n+1}\right)^\alpha\left(\frac{\ln(n)}{\ln (n+1)}\right)^\beta = 1[/imath] so the nature of the series is inconclusive. If [imath]\alpha = \beta = 0[/imath], then the series diverges, since [imath]\sum_{n=2}^\infty 1 = \infty[/imath]. Should I study the rest of the cases (i.e. if [imath]\alpha = 0, \beta > 0[/imath] the root test and the ratio test are also inconclusive). What is the best form to study the series?. Thanks in advance |
766974 | The volume of a cone whose length of its side is [imath]R[/imath]
How can I compute the volume of a cone whose side length is [imath]R[/imath] and the vertex of the cone forms an angle [imath]2\theta[/imath]? | 792620 | Find the volume of a cone whose length of its side is [imath]R[/imath]
How can i compute the volume of a cone whose length of its side is [imath]R[/imath] and the vertex of the cone forms an angle [imath]2θ[/imath] . The top cone is a cap of a sphere of radius [imath]R[/imath]. I tried to solve first in 2 dimension and then use solid of revolution: Some help please to find the volume of this cone. |
792655 | Is the "set" of all algebraic extensions a set?
Consider a field [imath]K[/imath]. Now, consider the class of all algebraic extensions of [imath]K[/imath]. Is this a set? Since I think it isn't, how to prove it isn't? If the class were of all extensions of [imath]K[/imath], I think I could argue with Zorn's Lemma to get a maximal element and enlarge it by taking the field of fractions of the polynomials with coefficients in that maximal element (proving, thus, that it is a not a set), but I can't repeat this argument for algebraic extensions. So, what to do ? | 621944 | Proof of Existence of Algebraic Closure: Too simple to be true?
Having read the classical proof of the existence of an Algebraic Closure (originally due to Artin), I wondered what is wrong with the following simplification (it must be wrong, otherwise why would we bother with the extra complications in Artin's proof?): Theorem: Let [imath]K[/imath] be a field, then [imath]K[/imath] has an algebraic closure [imath]\bar{K}[/imath] (i.e an algebraic extension that is algebraically closed). "Proof": Define [imath]A=\{ F \supset K | F \text{ is an algebraic extension of } K\}[/imath] and inherit this with the usual partial order of inclusion. One can check that Zorn's lemma applies (union of a nested chain of algebraic extensions is itself algebraic). Thus take [imath]\overline{K}[/imath] to be a maximal element. It must be algebraically closed for otherwise there is an irreducible polynomial with root in some strictly bigger field. [imath]\blacksquare[/imath] Now here is what I suspect is false about this proof: The definition of [imath]A[/imath] smells like your usual set theory paradoxes like Russell's paradox. In fact one could just as well use the same technique to prove that there exists a "largest set" which of course there does not. I am however under the impression that "most" working mathematicians ignore set theory foundations and just "do" mathematics, so is there a safe way of doing this (i.e: do "concrete everyday mathematics" by avoiding set theory) without getting burnt? |
793159 | Derivatives and functions
Let [imath]f(x)[/imath] and [imath]g(x)[/imath] be continuous non decreasing functions defined on [imath]\Bbb{R}[/imath] such that [imath]f''(x)= g(x)[/imath] and [imath]g''(x)= f(x)[/imath]. Suppose [imath]f(x)g(x)[/imath] is linear in [imath]x[/imath] in [imath](a,b)[/imath]. Then prove that [imath]f(x)=g(x)=0 \quad\forall x\in (a,b)[/imath] | 792089 | A calculus problem with functions such that [imath]f''(x) = g(x)[/imath] and [imath]g''(x) = f(x)[/imath]
Let: [imath]f(x)[/imath] and [imath]g(x)[/imath] be twice differentiable, non-decreasing functions. [imath]f''(x) = g(x)[/imath] and [imath]g''(x) = f(x)[/imath]. [imath]f(x) \cdot g(x)[/imath] is a linear function. Then we have to show that [imath]f(x) = g(x) = 0[/imath]. I am really not able to do anything useful, any help would be appreciated :) |
793370 | What is the minimum value of [imath]\operatorname{trace}(AA^T)[/imath]
[imath]A_{n\times n}[/imath] be a non singular matrix, could any one tell me what is the minimum value of [imath]\operatorname{trace}(AA^T)[/imath] | 791140 | Find [imath]\min(\operatorname{trace}(AA^T))[/imath] for invertible [imath]A_{n\times n}[/imath]
For invertible [imath]A_{n\times n}[/imath] find [imath]\min(\operatorname{trace}(AA^T))[/imath] (a) [imath]0[/imath] (b) [imath]1[/imath] (c) [imath]n[/imath] (d) [imath]n^2[/imath] Clearly for [imath]A=I[/imath], it is [imath]n[/imath], and I am unable to get any lower values, but how do I prove it. |
793409 | Does every group have a maximal normal subgroup?
Define [imath]P=\{M\subset G: M\unlhd G, M\neq G\}[/imath]. Let [imath]\mathscr{C}[/imath] be a chain of [imath]P[/imath]. I have shown that [imath]\bigcup \mathscr{C}[/imath] is a normal subgroup of [imath]G[/imath], but i don't know how to prove [imath]\bigcup\mathscr{C}\neq G[/imath]. Moreover, i'm not sure whether this is true. Is it true? If so, how do i prove it is not [imath]G[/imath]? | 419091 | Does every infinite group have a maximal subgroup?
[imath]G[/imath] is an infinite group. Is it necessary true that there exists a subgroup [imath]H[/imath] of [imath]G[/imath] and [imath]H[/imath] is maximal ? Is it possible that there exists such series [imath]H_1 < H_2 < H_3 <\cdots <G [/imath] with the property that for every [imath]H_i[/imath] there exists [imath]H_{i+1}[/imath] such that [imath]H_i < H_{i+1}[/imath]? |
793475 | Show that [imath]F(B)=B[/imath]and [imath]F(C)=C[/imath] if [imath]F: \mathcal{P}A \to \mathcal{P}A[/imath] and that [imath]F[/imath] has the monotonicity property
Assume that [imath]F: \mathcal{P}A \to \mathcal{P}A[/imath] and that [imath]F[/imath] has the monotonicity property: [imath]X \subseteq Y \subseteq A \implies F(X) \subseteq F(Y)[/imath] Define [imath]B = \bigcap \{ X \subseteq A | F(X) \subseteq X \} \; \mathtt{and} \; C = \bigcup \{ X \subseteq A | X \subseteq F(X) \}[/imath] Show that [imath]F(B)=B[/imath] and [imath]F(C)=C[/imath]. So far I can get: 1. [imath]F(A) \subseteq A \;\& \; \varnothing \subseteq F(\varnothing)[/imath] 2. given any [imath]X_1 \subseteq F(X_1)[/imath] and [imath]F(X_2) \subseteq X_2[/imath] [imath](X_1 ,X_2 \subseteq A)[/imath] I have [imath]B \subseteq X_2 ,F(B) \subseteq F(X_2) \subseteq X_2[/imath] and [imath]X_1 \subseteq C, X_1 \subseteq F(X_1) \subseteq F(C)[/imath] But: I cannot find a way to show the relation between [imath]F(B)[/imath] and [imath]B[/imath].What should I do? | 723181 | On Tarski-Knaster theorem
Let [imath]P(X)[/imath] denote the power set of [imath]X[/imath] (the set of all subsets of [imath]X[/imath]) with a partial order given by inclusion. If [imath]F: P(X) \to P(X)[/imath] is monotone (order preserving), then [imath]F[/imath] has a fixed point. How are we going to prove that without using the term (complete lattice)?! |
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