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634770 | (AIME) number theory question
How many integers less than 1000 can be expressed in the form [imath]\frac{(x + y + z)^2}{xyz} [/imath] where [imath]x, y, z[/imath] are integers? So far, I've attempted substituting certain values of [imath]x, y, z[/imath]. For example, setting [imath]y = z = 1[/imath] gives [imath]f(x) = x + 4 + \frac{4}{x}[/imath]. I haven't been able to get this over 9 yet. | 627475 | How many integers less than [imath]1000[/imath] can be expressed in the form [imath]\frac{(x + y + z)^2}{xyz}[/imath]?
How many integers less than [imath]1000[/imath] can be expressed in the form [imath]\frac{(x + y + z)^2}{xyz}[/imath] where [imath]x, y, z[/imath] are positive integers? |
634828 | Two Perfect Squares--[imath](3n+1) \& (4n+1)[/imath].
Assume [imath]n[/imath] is a Natural Number which satisfies the following 2 properties simultaneously: [imath]01[/imath] . [imath](3n+1)[/imath]=[imath]a[/imath]12 for some Natural Number [imath]a[/imath]1. [imath]02[/imath] . [imath](4n+1)[/imath]=[imath]a[/imath]22 for some Natural Number [imath]a[/imath]2. Prove that [imath]n[/imath] is divisible by [imath]56[/imath]. For example, if we take [imath]n[/imath]=[imath]56[/imath], we have [imath](3n+1)[/imath]=[imath]169[/imath]=[imath]13[/imath]2 and [imath](4n+1)[/imath]=[imath]225[/imath]=[imath]15[/imath]2. I could prove that [imath]n[/imath] is a multiple of [imath]4[/imath] but coud not proceed any further.However my strategy was to prove that [imath]n[/imath] is divisible by 8 & then that [imath]n[/imath] is divisible by 7 because [imath]G.C.D (7,8)=1[/imath] and [imath]L.C.M.(7,8)=56[/imath]. | 575733 | If 4n+1 and 3n+1 are both perfect sqares, then 56|n. How can I prove this?
Prove that if [imath]n[/imath] is a natural number and [imath](3n+1)[/imath] & [imath](4n+1)[/imath] are both perfect squares, then [imath]56[/imath] will divide [imath]n[/imath]. Clearly we have to show that [imath]7[/imath] and [imath]8[/imath] both will divide [imath]n[/imath]. I considered first [imath]3n+1=a^2[/imath] and [imath]4n+1=b^2[/imath]. [imath]4n+1[/imath] is a odd perfect square. - so we have [imath]4n+1\equiv 1\pmod{8}[/imath]; from this [imath]2|n[/imath] so [imath]3n+1[/imath] is a odd perfect square. - so [imath]3n+1\equiv 1\pmod{8}[/imath] so [imath]8|n[/imath] but I can't show [imath]7|n[/imath]. How do I show this? Thanks for the help. |
634827 | Fourier transform involving a delta function
For ease let us suppose that I am trying to transform the function [imath]f(t)=\delta(a^2 -t^2)[/imath]. Therefore the Fourier transform is given by [imath]\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \delta(a^2 -t^2)\exp(-i\omega t)\,dt.[/imath] My question is how to evaluate [imath]\int_{-\infty}^{\infty}\exp(-i\omega t)\,dt[/imath], and how it relates to the delta function? I know that the Fourier transform of [imath]\delta(t)[/imath] is unity, but how does this change when considering [imath] \delta(a^2 -t^2)[/imath] ? Thanks very much. | 214763 | Calculating the Fourier transform of [imath]f(t)=Ae^{-i\omega_0 t}[/imath]
I am trying to calculate the Fourier transform of [imath]f(t)=Ae^{-i\omega_0 t}[/imath] I'm getting an infinity which is giving me problems. Here are my steps: [imath]F(\omega)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty f(t)e^{i\omega t}dt[/imath] [imath]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty Ae^{-i\omega_0 t}e^{i\omega t}dt[/imath] [imath]=\frac{A}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{t(i\omega - i\omega_0)}dt[/imath] [imath]=\frac{A}{\sqrt{2\pi}(i\omega - i\omega_0)} |e^{t(i\omega - i\omega_0)}|_{-\infty}^{\infty}[/imath] [imath]=\frac{A}{\sqrt{2\pi}(i\omega - i\omega_0)}(\infty - 0)[/imath] Where am I going wrong? Thanks. |
635136 | Proving summation identities
How would one go about proving the following identities? [imath]\sum_{i=1}^n \sum_{i\neq j}^n \frac{z_i}{z_i-z_j} = \frac{n(n-1)}{2}[/imath] [imath]\sum_{i=1}^n \sum_{i\neq j}^n \frac{z_i^2}{z_i-z_j} = (n-1)\sum_{i=1}^n z_i[/imath] [imath]\sum_{i=1}^n \sum_{i\neq j}^n \frac{z_i^3}{z_i-z_j} = (n-1)\sum_{i=j}^n z_i^2+\sum_{i<j}^n z_i z_j[/imath] [imath]\sum_{i=1}^n \sum_{i\neq j}^n \frac{z_i^4}{z_i-z_j} = (n-1)\sum_{i=j}^n z_i^3+\sum_{i<j}^n z_i z_j(z_i+z_j)[/imath] [imath]\sum_{i=1}^n \sum_{i\neq j}^n \frac{z_i^5}{z_i-z_j} = (n-1)\sum_{i=j}^n z_i^4+\sum_{i<j}^n z_i z_j(z_i^2+z_i z_j +z_j^2)[/imath] I see the obvious pattern here. The problem is that the algebra involving the summation is giving me some difficulty. | 628258 | General identity for a double summation theorem
I've been reading a research paper, and I'm interested in generalizing a certain theorem but I can't seem to understand how the following results are derived: [imath]\sum_{i=1}^{n}\sum_{j\neq i}^n \frac{1}{z_i-z_j}=0[/imath] [imath]\sum_{i=1}^{n}\sum_{j\neq i}^n \frac{z_i}{z_i-z_j}=\frac{1}{2}n(n-1)[/imath] [imath]\sum_{i=1}^{n}\sum_{j\neq i}^n \frac{z_i^2}{z_i-z_j}=(n-1)\sum_{i=1}^n z_i[/imath] [imath]\sum_{i=1}^{n}\sum_{j\neq i}^n \frac{z_i^3}{z_i-z_j}=(n-1)\sum_{i=1}^n z_i^2 + \sum_{i<j}^n z_i z_j[/imath] |
636216 | Prove that [imath]|f'(x)| \le \frac{A}2 \forall x \in [0,1] [/imath]
Let [imath]f[/imath] be twice differentiable in [imath][0,1][/imath] [imath]f(0) = f(1) = 0[/imath], [imath]|f''(x)|\le A[/imath]. Prove that [imath]|f'(x)| \le \frac{A}2, \forall x \in [0,1] [/imath]. Well this is what I came up with, [imath]f'(c_1) = \dfrac{f(x) -f(0)}{x-0} = \dfrac{f(x)}{x}[/imath] [imath]f'(c_2) = \dfrac{f(x) -f(1)}{x-1} = \dfrac{f(x)}{x-1}[/imath] for [imath]0\lt c_1,c_2 \lt 1[/imath] from rolle's theorem we know that there is : [imath]0\lt c_3 \lt 1[/imath] such that [imath]f'(c_3) =0[/imath] | 632850 | Proving that if [imath]|f''(x)| \le A[/imath] then [imath]|f'(x)| \le A/2[/imath]
Suppose that [imath]f(x)[/imath] is differentiable on [imath][0,1][/imath] and [imath]f(0) = f(1) = 0[/imath]. It is also known that [imath]|f''(x)| \le A[/imath] for every [imath]x \in (0,1)[/imath]. Prove that [imath]|f'(x)| \le A/2[/imath] for every [imath]x \in [0,1][/imath]. I'll explain what I did so far. First using Rolle's theorem, there is some point [imath]c \in [0,1][/imath] so [imath]f'(c) = 0[/imath]. EDIT: My first preliminary solution was wrong so I tried something else. EDIT2: Another revision :\ I define a Taylor series of a second order around the point [imath]1[/imath]: [imath] f(x) = f(1) + f'(1)(x-1) + \frac12 f''(d_1)(x-1)^2 [/imath] [imath] f(0) = f(1) + f'(1)(-1) + \frac12 f''(d_1)(-1)^2 [/imath] [imath] |f'(1)| = \frac12 |f''(d_1)| <= \frac12 A [/imath] Now I develop a Taylor series of a first order for [imath]f'(x)[/imath] around [imath]1[/imath]: [imath] f'(x) = f'(1) + f''(d_2)(x-1) [/imath] [imath] |f'(x)| = |f'(1)| + x*|f''(d_2)|-|f''(d_2)| \leq \frac{A}{2} + A - A = \frac{A}{2} [/imath] It looks correct to me, what do you guys think? Note: I cannot use integrals, because we have not covered them yet. |
549458 | Prove that the only prime triple is 3, 5, 7
Prove that the only prime triple is [imath]3,5,7[/imath]. I tried proving using this method: Multiplication of [imath]3[/imath] jumps back and forth between being an even and an odd number. Thus goes from odd to odd over an interval max size 6, and likewise from even to even. In either case these will have the following types: (even - odd - even - odd - even - odd) and (odd - even - odd - even - odd - even), with 3 being a multiple of the first instance in the combinations. In the both combinations, the first and fourth would be divisible by three, leaving only two possible primes left (two odd numbers). Thus the only triple prime can be a combination where three is considered a prime.... However, is this a formal proof? can anybody tell me how to do a proof for this questions? | 522585 | How can I prove that one of [imath]n[/imath], [imath]n+2[/imath], and [imath]n+4[/imath] must be divisible by three, for any [imath]n\in\mathbb{N}[/imath]
Intuitively it's true, but I just can't think of how to say it "properly". Take for example, my answer to the following question: Let [imath]p[/imath] denote an odd prime. It is conjectured that there are infinitely many twin primes [imath]p[/imath], [imath]p+2[/imath]. Prove that the only prime triple [imath]p[/imath], [imath]p+2[/imath], [imath]p+4[/imath] is the triple [imath]3,\ 5,\ 7[/imath]. And my solution: Given an odd integer [imath]n[/imath], between the three integers [imath]n[/imath], [imath]n+2[/imath] and [imath]n+4[/imath], one of them must be divisible by [imath]3[/imath]... Three possible cases are [imath]n=3k[/imath], [imath]n+2=3k[/imath], and [imath]n+4=3k[/imath]. The only such possible [imath]k[/imath] that makes [imath]n[/imath] prime is [imath]k=1[/imath]. In this case, given an odd prime [imath]p[/imath], either [imath]p=3[/imath], [imath]p+2=3[/imath], or [imath]p+4=3[/imath]. This would imply that [imath]p=3[/imath], [imath]p=1[/imath], or [imath]p=-1[/imath]. The only of these three that is prime is [imath]p=3[/imath], therefore the only three evenly distributed primes are [imath]3[/imath], [imath]5[/imath], and [imath]7[/imath]. Is there a "better" way that I can assert that one of the integers is divisible by 3? This feels too weak. |
636520 | Prove that the lebesgue integrals [imath]\lim_{n\to \infty} \int_{[0,1]^n} \frac{x_1^2+ \dots +x_n^2}{x_1+ \dots +x_n} \ d \mathrm{m}(\mathbf{x})=2/3[/imath]
I found this problem while studying for an analysis exam and have been puzzled ever since. If anybody has any hints on this, I would greatly appreciate it. Prove that the lebesgue integrals [imath]\lim_{n\to \infty} \int_{[0,1]^n} \frac{x_1^2+ \dots +x_n^2}{x_1+ \dots +x_n} \ d \mathrm{m}(\mathbf{x})=2/3[/imath] | 623433 | Asymptotic behaviour of a multiple integral on the unit hypercube
A few days ago I found an interesting limit on the "problems blackboard" of my University: [imath]\lim_{n\to +\infty}\int_{(0,1)^n}\frac{\sum_{j=1}^n x_j^2}{\sum_{j=1}^n x_j}d\mu = 1.[/imath] The correct claim, however, is: [imath]\lim_{n\to +\infty}\int_{(0,1)^n}\frac{\sum_{j=1}^n x_j^2}{\sum_{j=1}^n x_j}d\mu = \frac{2}{3}.[/imath] In fact, following @Tetis' approach we have: [imath]I_n = \int_{(0,1)^n}\frac{\sum_{j=1}^n y_j^2}{\sum_{j=1}^{n}y_j}d\mu = \int_{(-1/2,1/2)^n}\frac{\frac{1}{2}+\frac{2}{n}\sum_{j=1}^n x_j^2+\frac{2}{n}\sum_{j=1}^n x_j}{1+\frac{2}{n}\sum_{j=1}^n x_j}d\mu,[/imath] now setting [imath]x_j=-z_j[/imath] and summing the two integrals [imath]I_n = \int_{(-1/2,1/2)^n}\frac{\frac{1}{2}+\frac{2}{n}\sum_{j=1}^n x_j^2+\frac{4}{n^2}\left(\sum_{j=1}^n x_j\right)^2}{1-\frac{4}{n^2}\left(\sum_{j=1}^n x_j\right)^2}d\mu[/imath] follows, so: [imath] I_n-\frac{2}{3}=\int_{(-1/2,1/2)^n} \left(-\frac{1}{2}+\frac{2}{n}\sum_{j=1}^n x_j^2\right)\frac{\frac{4}{n^2}\left(\sum_{j=1}^n x_j\right)^2}{1-\frac{4}{n^2}\left(\sum_{j=1}^n x_j\right)^2}d\mu < 0,[/imath] [imath] \left| I_n-\frac{2}{3}\right|\leq\int_{\sum x_i^2\leq\frac{n}{4}}\left(\frac{1}{2}-\frac{2}{n}\sum_{j=1}^n x_j^2\right)\frac{\frac{4}{n^2}\left(\sum_{j=1}^n x_j\right)^2}{1-\frac{4}{n^2}\left(\sum_{j=1}^n x_j\right)^2}d\mu,[/imath] [imath] \frac{2}{3}-I_n\leq \frac{n^{n/2}}{2^{n+1}} \int_{\sum x_i^2\leq 1}\left(1-\sum_{j=1}^{n}x_j^2\right)\frac{x_1^2}{1-x_1^2}d\mu.[/imath] The last bound, anyway, is too crude, since the RHS is [imath] \Theta\left(\left(\sqrt{\frac{e\pi}{2}}\right)^n\frac{\log n}{n^{3/2}}\right).[/imath] My question now is: what is the asymptotic behaviour of [imath]I_n[/imath]? A second one is: can we prove [imath]I_n\geq\frac{2}{3}-\frac{C}{n}[/imath], for a suitable positive constant [imath]C[/imath], without the Central Limit Theorem? UPDATE: After a few I came out with a proof of my own. The challenge is now to give the first three terms of the asymptotics, and possibly a continued fraction expansion for [imath]I_n[/imath]. |
637160 | Integrating [imath]\int\sin^{-2}xdx[/imath]
I am trying to prove that [imath] \int\frac{1}{\sin^2(x)}dx = -\cot(x) + C [/imath] but I have difficulties, I don't know where to start, I can't substitute anything with [imath]sin(x)[/imath] because I don't have a [imath]cos(x)[/imath] to make it up for, I also know that: [imath] \sin^2x = \frac{1 - \cos(2x)}{2} [/imath] so I basically have: [imath] \int\frac{2}{1 - \cos(2x)}dx [/imath] and there I am stuck, how do I proceed? | 239808 | Integral of cosec squared ([imath]\operatorname{cosec}^2x[/imath], [imath]\csc^2x[/imath])
According to my sheet of standard integrals, [imath]\int \csc^2x \, dx = -\cot x + C[/imath]. I am interested in a proof for the integral of [imath]\operatorname{cosec}^2x[/imath] that does not require differentiating [imath]\cot x[/imath]. (I already know how to prove it using differentiation, but am interested in how one would calculate it without differentiation.) |
626254 | Give the characteristics of a function after a relationship
Let [imath]f\colon \mathbb{R}\to\mathbb{R}[/imath] be continuous, bounded function such that the space [imath]\operatorname{lin}(\{f_k\mid k \in \mathbb{N}\})[/imath] is finite-dimensional, where [imath]f_k(x)=f(x+k)[/imath]. What can we say about [imath]f[/imath]? Thank you in advance. | 627515 | study of subspace generated by [imath]f_k(x)=f(x+k)[/imath] with f continuous, bounded..
Let [imath]f:ℝ→ℝ[/imath] be continuous, bounded function such that the space [imath]\mathrm{lin}\{f_k(x)=f(x+k)∣k ∈\mathbb{N}\}[/imath] is finite-dimensional. Determine an expression of f. I started with: Let [imath]P_f(X) = \sum_{k=0}^n p_k X^k[/imath], the degree is n,and we define f on [0,n] but after I do not know how to continue. Thank you in advance. |
638499 | Image of homomorphism from ideal is ideal
Let [imath]A,B[/imath] be rings. If [imath]f:A\rightarrow B[/imath] is a homomorphism from [imath]A[/imath] onto [imath]B[/imath] with kernel [imath]K[/imath], and [imath]J[/imath] is an ideal of [imath]A[/imath] such that [imath]K\subseteq J[/imath], then [imath]f(J)[/imath] is an ideal of [imath]B[/imath]: My solution: Let [imath]a,b\in J[/imath]. Then [imath]a+b\in J[/imath], so [imath]f(a)+f(b)=f(a+b)\in f(J).[/imath] Also, [imath]-a\in J[/imath], so [imath]-f(a)=f(-a)\in f(J).[/imath] Finally, if [imath]c\in A[/imath], then [imath]f(a)f(c)=f(ac)\in f(J).[/imath] We don't need the fact that [imath]K\subseteq J[/imath], or am I mistaken somewhere? | 416709 | Must an ideal contain the kernel for its image to be an ideal?
I'm trying to learn some basic abstract algebra from Pinter's A Book of Abstract Algebra and I find myself puzzled by the following simple question about ring homomorphisms: Let [imath]A[/imath] and [imath]B[/imath] be rings. If [imath]f : A \to B[/imath] is a homomorphism from [imath]A[/imath] onto [imath]B[/imath] with kernel [imath]K[/imath], and [imath]J[/imath] is an ideal of [imath]A[/imath] such that [imath]K \subseteq J[/imath], then [imath]f(J)[/imath] is an ideal of [imath]B[/imath]. I'm clearly missing the obvious, but I don't see where the requirement [imath]K \subseteq J[/imath] comes into the proof. Since [imath]f[/imath] is also a homomorphism of additive groups, the image [imath]f(J)[/imath] must be closed under addition and negatives (correct?). Then for [imath]f(J)[/imath] to be an ideal we have to show that it is closed under multiplication by an arbitrary element [imath]b \in B[/imath]. Since [imath]f[/imath] is onto, there is some [imath]a \in A[/imath] such that [imath]b = f(a)[/imath]. Let [imath]j'[/imath] be any element of [imath]f(J)[/imath], so [imath]j' = f(j)[/imath] for some [imath]j \in J[/imath]. Then [imath]bj' = f(a)f(j) = f(aj)[/imath], and [imath]aj \in J[/imath] since [imath]J[/imath] is an ideal. Then it seems that [imath]f(J)[/imath] is closed under multiplication by [imath]B[/imath]. What mistaken assumption am I making? |
638540 | Show that if a prime number [imath]p|a^n[/imath] then [imath]p|a[/imath]
The title says it all, how can I prove the following: Show that if a prime number [imath]p|a^n[/imath] then [imath]p|a[/imath] | 637278 | Divisibility of prime numbers
I have this exercise in my worksheet in the discrete mathematics course.I don't understand the part that deals with prime numbers in integer-divisibility. "Show that for a prime number [imath]p[/imath], if a [imath]p\mid a^n[/imath] then [imath]p\mid a[/imath]" Can somebody show how ? |
638562 | Volume of an n-simplex (Without Probabilities)
Compute the volume of [imath] S_n=\{(x_1,x_2,...,x_n)\in\mathbb{R^n},x_i\geq 0,\displaystyle\sum_{k=0}^{n} x_i<1\} [/imath] I don't really have an idea how to solve it. My 'work': Perhaps I could use [imath] ||x||_1=|x_1|+...+|x_n| [/imath] So the set is [imath] S_n=\{(x_1,x_2,...,x_n)\in\mathbb{R^n},x_i\geq 0,B(0,1)\} [/imath] Where [imath] B(0,1)=\left\{M\in\mathbb{R^n} \,\mid\,d(M,0)<1\right\} [/imath] But I don't see where it leads me... Thank you in advance | 315298 | Volume of an n-simplex
It's rather tedious to show using Fubini's Theorem and induction on [imath]n[/imath] that the volume of the region [imath]x_1+x_2+...+x_n \leq 1[/imath] with [imath]x_1,...,x_n[/imath] nonnegative is [imath]\frac{1}{n!}[/imath]. Is there an easier way to see this? |
638741 | Center of mass in a straight rod
I got an assignment to prove that in a straight homogeneous rod, you can always choose a coordinate system in such a way that [imath]\int_S x_1dx_1dx_2=0 [/imath] [imath]\int_S x_2dx_1dx_2=0 [/imath] [imath]\int_S x_1x_2 dx_1dx_2=0[/imath] where S is the cross section of the rod Now, intuitively, I can understand that the first two expressions basically state that if you find the mass center of the rod, then all you have to do is to put the origin of the coordinate system in the mass center, and the integrals turn out to be zero. What I'm curious about is whether or not this is a good enough of an explanation, and also, a hint on how to obtain the third equality. | 638732 | Center of mass in a straight rod
I got an assignment to prove that in a straight homogeneous rod, you can always choose a coordinate system in such a way that [imath]\int_S x_1 \, dx_1 \, dx_2=0 [/imath] [imath]\int_S x_2 \, dx_1 \, dx_2=0 [/imath] [imath]\int_S x_1x_2 \, dx_1 \, dx_2=0[/imath] where [imath]S[/imath] is the cross section of the rod Now, intuitively, I can understand that the first two expressions basically state that if you find the mass center of the rod, then all you have to do is to put the origin of the coordinate system in the mass center, and the integrals turn out to be zero. What I'm curious about is whether or not this is a good enough of an explanation, and also, a hint on how to obtain the third equality. |
639093 | Definition: [imath]a \prec b[/imath] , [imath] a \npreceq b[/imath] , [imath]a,b \in \mathbb{R}[/imath]
let be [imath]a,b \in \mathbb{R}[/imath]: [imath]a \prec b[/imath] if [imath]a \preceq b \wedge a \neq b[/imath] [imath]a \npreceq b [/imath] if [imath]b \prec a[/imath] is correct? Thanks in advance! | 628529 | Does [imath]\neg(x > y)[/imath] imply that [imath]y \geq x[/imath]?
Given any arbitrary binary relation [imath]\geq[/imath] defined on some set [imath]S[/imath], we define a new binary relation [imath]>[/imath] on [imath]S[/imath] by: [imath] x > y \quad\text{iff}\quad (x \geq y) \wedge \neg(y \geq x) [/imath] In accordance with our usual intuition for inequalities, I would like to prove that: Claim: [imath] \neg(x > y) \quad\text{iff}\quad y \geq x [/imath] Unfortunately, all I could conclude was that: [imath] \neg(x > y) \quad\text{iff}\quad \neg(x \geq y) \vee (y \geq x) [/imath] Is my claim even true? If so, could somehow help me finish off my work? If not, what additional hypotheses should be added in order to salvage the claim? |
639181 | Infinite ring with nonzero characteristic
I was wondering as I read about characteristic of a ring: Is there an infinite ring with nonzero characteristic? We have [imath]1+1+\ldots+1=0[/imath], but that doesn't seem to imply that the number of elements in the ring is finite. | 356649 | Can a ring of positive characteristic have infinite number of elements?
For curiosity: can a ring of positive characteristic ever have infinite number of distinct elements? (For example, in [imath]\mathbb{Z}/7\mathbb{Z}[/imath], there are really only seven elements.) We know that any field/ring of characterisitc zero must have infinite elements, but I am not sure what happens above. |
530915 | "If [imath]1/a + 1/b = 1 /c[/imath] where [imath]a, b, c[/imath] are positive integers with no common factor, [imath](a + b)[/imath] is the square of an integer"
If [imath]1/a + 1/b = 1 /c[/imath] where [imath]a, b, c[/imath] are positive integers with no common factor, [imath](a + b)[/imath] is the square of an integer. I found this question in RMO 1992 paper ! Can anyone help me to prove this ? May be it's too easy and simple . But , I'm just novice in number theory ! | 828918 | Prove that [imath]a+b[/imath] is a perfect square
[imath]a, b, c[/imath] are natural numbers such that [imath]1/a + 1/b = 1/c[/imath] and [imath]gcd(a,b,c)=1[/imath]. Prove [imath]a+b[/imath] is a perfect square. |
639449 | Why is [imath]\det(e^X)=e^{\operatorname{tr}(X)}[/imath]?
I've seen on Wikipedia that for a complex matrix [imath]X[/imath], [imath]\det(e^X)=e^{\operatorname{tr}(X)}[/imath]. It is clearly true for a diagonal matrix. What about other matrices ? The series-based definition of exp is useless here. | 322640 | How to prove [imath]\det(e^A) = e^{\operatorname{tr}(A)}[/imath]?
Prove [imath]\det(e^A) = e^{\operatorname{tr}(A)}[/imath] for all matrices [imath]A \in \mathbb{C}_{n×n}[/imath]. |
639507 | Matrix to Matrix Power
Considering [imath]A^b[/imath], we have a nice definition and properties when [imath]A[/imath] is a matrix and [imath]b[/imath] is a real (or field) value. We also have a fairly useful definition for [imath]e^B[/imath], where [imath]B[/imath] is a matrix and [imath]e[/imath] is the Naperian base. I imagine this can be extended to an arbitrary base using logarithms. My question is whether there is a general definition for [imath]A^B[/imath] with [imath]A[/imath] and [imath]B[/imath] as matrices with some suitable properties. Some properties might be: [imath]A^b[/imath] and [imath]e^B[/imath] are special cases of the definition of [imath]A^B[/imath] [imath]A^I = A[/imath] [imath]A^{B + C} = A^BA^C \text{ under suitable conditions }[/imath] [imath](A^B)^C = A^{BC} \text{ under suitable conditions }[/imath] Is any such definition known? Can computations be done with it? What properties do we get and under what conditions? Thank you, this is for my own curiosity. Edit : Fly by Night found this previous question: Matrix raised to a matrix, but it doesn't really address the properties that I was asking about. The real usefulness in a definition is in what you can infer from it, so I want to keep this question open. | 164422 | Matrix raised to a matrix: [imath]M^N[/imath], is this possible? with [imath]M,N\in M_n(\Bbb K).[/imath]
I was wondering if there is such a valid operation as raising a matrix to the power of a matrix, e.g. vaguely, if [imath]M[/imath] is a matrix, is [imath] M^N [/imath] valid, or is there at least something similar? Would it be the components of the matrix raised to each component of the matrix it's raised to, resulting in again, another matrix? Thanks, |
639532 | Sup of holomorphic function [imath]f[/imath] is [imath]\geq 2|f'(0)|[/imath]
[imath]f:D \rightarrow \mathbb{C}[/imath] holomorphic and [imath]d:=\sup_{z,w\in D} |f(z)-f(w)|[/imath]. Prove that [imath]d\geq 2|f'(0)|[/imath] and is equal if and only if [imath]f(z)=a_0+a_1z[/imath] | 637943 | [imath] 2|f^{'}(0)| = \sup_{z, w \in D} |f(z)-f(w)|[/imath]
Let [imath]D = B(0,1) \subset \mathbb{C} [/imath] a disc, [imath]f[/imath] holomorphic on [imath]D[/imath]. I want to demonstrate that if [imath] 2|f^{'}(0)| = \sup_{z, w \in D} |f(z)-f(w)|[/imath] then [imath]f[/imath] is linear. I know this is a well-known result. Where can I find the proof ? |
557001 | Group classification generated by two elements
Classify groups that are generated by two elements [imath]x[/imath] and [imath]y[/imath] of order 2. Could someone help me please ? | 160168 | Prove a group generated by two involutions is dihedral
Prove a finite group generated by two involutions is dihedral Is my following argument correct? Let [imath]G=\langle x,y\rangle[/imath] be a group generated by involutions [imath]x,y[/imath]. Let [imath]n=\mathrm{ord}(xy)[/imath] to get a presentation [imath]G=\langle x,y\mid x^2=y^2=(xy)^n=1\rangle [/imath] so G is dihedral of order [imath]2n[/imath] ? Further note: I realise now my argument is not sufficient as it remains to show [imath]G[/imath] has no other relations. I just found an idea from a reference which claims "...So [imath]G[/imath] must have a presentation of the form [imath]G=\langle x,y\mid x^2=y^2=(xy)^m=1\rangle [/imath], then one has to show [imath]m=n[/imath]..." in which I do not understand why [imath]G[/imath] has exactly a presentation of such form (the presentation inovlves [imath]m[/imath])? That reference also showed [imath]|\langle x,y\rangle |=2n[/imath] which directly led to the conclusion: [imath]m=n[/imath] |
640493 | Why can infinite series be summed different ways to get different results?
[imath]S = 1 - \frac12 + \frac13 - \frac14 + \frac15 - \frac16 + \frac17 - \frac18 + \frac19 - \frac1{10} + \frac1{11} - \frac1{12}\ldots\text{(to infinity)}[/imath] Rearranged, this series looks like: [imath]S = \left(1 - \frac12\right) - \left(\frac14\right) + \left(\frac13 - \frac16\right) - \left(\frac18\right) + \left(\frac15 - \frac1{10}\right) - \left(\frac1{12}\right) + \left(\frac17 - \frac1{14}\right) \ldots\text{(to infinity)}\\ S = \left(\frac12\right) - \left(\frac14\right) + \left(\frac16\right) - \left(\frac18\right) + \left(\frac1{10}\right) - \left(\frac1{12}\right) + \left(\frac1{14}\right) \ldots\text{(to infinity)}[/imath] This rearranged infinite series contains every number that the original infinite series had. Further, [imath] 2S = 1 - \frac12 + \frac13 - \frac14 + \frac15 - \frac16 + \frac17 - \frac18 + \frac19 - \frac1{10} + \frac1{11} - \frac1{12} \ldots\text{(to infinity)}[/imath] Thus: [imath]2S = S[/imath] [imath]2 = 1[/imath] Mathematics disproven. Sorry. Jokes aside, I know that infinite series can be calculated in different ways to get different results. My question is: Why? While it makes sense with Grandi's and similar series, it doesn't make sense to me for a series whose final term is [imath]\frac1\infty = 0[/imath]. | 357224 | absolutely convergent series and conditionally convergent series rearrangement
I don't understand that why the terms of an absolutely convergent series can be rearranged in any order and all such rearranged series converge to the same sum. my textbook gives me an example that it is not so for conditionally convergent series [imath]1-\frac12+\frac13-\frac14+\ldots=\ln2[/imath] [imath]1+\frac13-\frac12+\frac15+\frac17-\frac14+\frac19+\frac1{11}-\frac16...=\frac32\ln2[/imath] which I also don't know how to prove |
640643 | existence of double covering
Let [imath]M[/imath] be a manifold , and [imath]\pi_1(M)=\mathbb{Z}[/imath]. then can we say, the double covering of [imath]M[/imath] exists and is unique? | 626087 | Double cover of [imath]GL(n,\mathbb{C})[/imath].
Is there any reference or simple proof to show that the group [imath]\frac{\mathbb{C}\times SL(n,\mathbb{C})}{2\mathbb{Z}}[/imath] is double cover of [imath]GL(n,\mathbb{C})[/imath]. |
640915 | prove that if [imath]k|m[/imath], [imath]\mathbb{Z}_m[/imath] has a subgroup of order [imath]k[/imath]
prove that if [imath]k|m[/imath], then [imath]\mathbb{Z}_m[/imath] has a subgroup of order [imath]k[/imath]. im not sure where to start with this. any help is appreciated. thanks! | 638883 | Prove that if [imath]k\mid m,[/imath] then [imath]Z_m[/imath] has a subgroup of order [imath]k.[/imath]
Prove that if [imath]k\mid m[/imath], then [imath]Z_m[/imath] has a subgroup of order [imath]k.[/imath] Ok, so this doesn't look like too hard of a problem. So do I just show that it is closed under multiplication and inverse? I just don't know how I would do that with a problem like this. Can anyone help me get started please? |
641089 | Prove or disprove [imath] lim_{n \to \infty} a_n = lim_{n \to \infty} a_{\pi(n)} [/imath], where [imath]\pi[/imath] is a bijective function
I have no idea how to solve the following task: Prove or disprove that: If [imath] (a_n)_{n \in \mathbb{N}} [/imath] converges and [imath] \pi : \mathbb{N} \mapsto \mathbb{N} [/imath] is a bijective function, then [imath] lim_{n \to \infty} a_n = lim_{n \to \infty} a_{\pi(n)}. [/imath] Since this is my homework, I'm not asking for a solution. Could you just give me some hints how to solve this? I think I should use the properties of a bijective function here, but I have no idea, how... Any help would be appreciated! Regards, Lena | 185778 | Rearrangement of sequences with limit [imath]0[/imath]
Is it true that every real sequence that converges to zero has the property that every rearrangement of it also converges to zero? I have a proof in mind, but I'm not 100% sure it's correct (although I'm pretty sure), so I just want a yes/no answer. |
642076 | A question about invertible
Let [imath]A[/imath] be a [imath]5\times5[/imath] matrix that can be written in the form [imath]A=BC[/imath], where [imath]B[/imath] is a [imath]5 \times 4[/imath] matrix and [imath]C[/imath] is a [imath]4 \times5[/imath] matrix. Prove that [imath]A[/imath] is not invertible. | 618837 | Product of matrices of different order is not invertible
If [imath]A[/imath] is an [imath]m \times n[/imath] and [imath]B[/imath] is an [imath]n \times m[/imath] matrix with [imath]n < m[/imath] then prove that [imath]C = AB[/imath] is not invertible. I am not getting any idea. May be I am missing vary basic point. [imath]C = AB[/imath] where both [imath]A[/imath] and [imath]B[/imath] are square matrices will be invertible iff both [imath]A[/imath] and [imath]B[/imath] be invertible. I know it and its proof. May be this question has been discussed earlier, but I am not seeing the link. Thank you for your help. |
641547 | Prove that this operator is continuous
Let [imath]X,Y,Z[/imath] be Banach spaces, and let [imath]T:X\to Y[/imath] be linear. Let [imath]J:Y\to Z[/imath] be linear, bounded and injective. If [imath]JT:X\to Z[/imath] is bounded, then T is bounded. | 179984 | Map bounded if composition is bounded
Let [imath]X,Y,Z[/imath] Banach spaces and [imath]A:X\rightarrow Y[/imath] and [imath]B:Y\rightarrow Z[/imath] linear maps with [imath]B[/imath] bounded and injective and [imath]BA[/imath] bounded. Prove that [imath]A[/imath] is bounded as well. If I knew that [imath]B(Y)[/imath] is closed I'd have a bounded linear map [imath]B^{-1}:B(Y)\rightarrow Y[/imath] by the bounded inverse theorem. Therefore [imath]A=B^{-1}BA[/imath] is bounded. How to prove the claim if [imath]B(Y)[/imath] is not closed. |
642182 | A question about rank
Let [imath]V[/imath] be a finite dimensional vector space and let [imath]T:V \to V[/imath] be a linear operator. If the rank of [imath]T^2[/imath] equals then rank of [imath]T[/imath], prove that [imath](\operatorname{range} T) \cap (\ker T)=\{0\}[/imath] I just want to know whether I'm correct or not. First [imath]T^2[/imath] implies that T is squared, then rank of [imath]T^2=[/imath] rank of [imath]T[/imath] means that [imath]T[/imath] is full rank is that correct? Then the range of [imath]T[/imath] is the space [imath]V[/imath].Is the [imath]\ker T[/imath] means that [imath]\ker T=\{v:T(v)=0\}[/imath]? If so range [imath]T \cap \ker T=\{0\}[/imath] right? | 520851 | Prove that if [imath]\operatorname{rank}(T) = \operatorname{rank}(T^2)[/imath] then [imath]R(T) \cap N(T) = \{0\}[/imath]
Let [imath]V[/imath] be a finite-dimensional vector space and let [imath]T:V\to V[/imath] be linear. Prove that if [imath]\operatorname{rank}(T) = \operatorname{rank}(T^2)[/imath] then [imath]R(T) \cap N(T) = \{0\}[/imath]. I don't see this implication, at all. Please give hints and explain conceptually. I suck at algebra. |
641946 | Proof of Reduction Formula for [imath]\int \cos^n (x) \ dx = \frac{1}{n}\cos^{n-1}x\sin x + \frac{n-1}{n}\int\cos^{n-2}x \ dx[/imath]
I ran into a question with proving the reduction formula: [imath] \int \cos^n x \ dx = \frac{1}{n}\cos^{n-1}x\sin x + \frac{n-1}{n}\int\cos^{n-2}x \ dx [/imath] I then attempted to prove by differentiation with respect to [imath]x[/imath], but something strange happened (have I just violated the Fundamental Theorem of Calculus?) After differentiation, the resulting expression is: [imath]\displaystyle\cos^n x=\bigg[\frac{1}{n}\bigg]\bigg[(n-1)\cos^{n-2}x(-\sin x)(\sin x)+\cos^{n-1}x\cos x+\frac{n-1}{n}\cos^{n-2}x\bigg][/imath] [imath]\displaystyle=\bigg[\frac{n-1}{n}\cos^{n-2}x\bigg]\bigg[\cos^2x+\cos^n x\bigg][/imath] [imath]\displaystyle \frac{n-1}{n}\bigg[\cos^nx+\cos^{2n-2}x\bigg][/imath] Edit: I corrected the derivative, but problem not solved. | 236543 | Proving a reduction formula for the antiderivative of [imath]\cos^n(x)[/imath]
I want to show that for all [imath]n\ge 2[/imath], it holds that [imath] \int \cos^n x\ dx = \frac{1}{n} \cos^{n-1} x \sin x + \frac{n-1}{n}\int \cos^{n-2} x\ dx. [/imath] I'm not even getting the result for the induction base [imath](n=2)[/imath]: Using integration by parts, I only get [imath] \int \cos^2 x\ dx = \cos x \sin x + \int \sin^2 x\ dx. [/imath] I'm suspecting that I need to use some trigonometric identity here. |
193435 | Prove [imath]\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx[/imath]
I am trying to prove [imath]\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx[/imath] This problem is a classic, but I seem to be missing one step or the understanding of two steps which I will outline below. [imath]I_n := \int\cos^n x \ dx = \int\cos^{n-1} x \cos x \ dx \tag{1}[/imath] First question: why rewrite the original instead of immediately integrating by parts of [imath]\int \cos^n x \ dx[/imath]? Integrate by parts with [imath]u = \cos^{n-1} x, dv = \cos x \ dx \implies du = (n-1)\cos^{n-2} x \cdot -\sin x, v = \sin x[/imath] which leads to [imath]I_n = \sin x \ \cos^{n-1} x +\int\sin^2 x (n-1) \ \cos^{n-2} x \ dx \tag{2}[/imath] Since [imath](n-1)[/imath] is a constant, we can throw it out front of the integral: [imath]I_n = \sin x \ \cos^{n-1} x +(n-1)\int\sin^2 x \ \cos^{n-2} x \ dx\tag{3}[/imath] I can transform the integral a bit because [imath]\sin^2 x + cos^2 x = 1 \implies \sin^2 x = 1-\cos^2 x[/imath] [imath]I_n = \sin x \ \cos^{n-1} x + (n-1)\int(1-\cos^2 x) \ \cos^{n-2} x \ dx \tag{4}[/imath] According to Wikipedia as noted here, this simplifies to: [imath]I_n = \sin x \ \cos^{n-1} x + (n-1) \int \cos^{n-2} x \ dx - (n-1)\int(\cos^n x) \ dx \tag{5}[/imath] Question 2: How did they simplify the integral of [imath]\int(1-\cos^2 x) \ dx[/imath] to [imath]\int(\cos^n x) \ dx[/imath]? Assuming knowledge of equation 5, I see how to rewrite it as [imath]I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x - (n-1) I_{n} \tag{6}[/imath] and solve for [imath]I_n[/imath]. I had tried exploiting the fact that [imath]\cos^2 x = \frac{1}{2} \cos(2x) + \frac{1}{2} [/imath] and trying to deal with [imath]\int 1 \ dx - \int \frac{1}{2} \cos (2x) + \frac{1}{2} \ dx[/imath] which left me with [imath]\frac{x}{2} - \frac{1}{4} \sin(2x)[/imath] after integrating those pieces. Putting it all together I have: [imath]I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x \left(-(n-1) (\frac{x}{2} - \frac{1}{4} \sin 2x) \right) \tag{7}[/imath] but I'm unsure how to write the last few terms as an expression of [imath]I_{something}[/imath] to get it to match the usual reduction formula of [imath]\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx[/imath] | 792070 | Reduction formula integral of [imath]\cos^n(x)[/imath]
I need to find the reduction formula for the integral of [imath]\cos^n(x)[/imath]. Ive split it into [imath]\cos(x)\cos^{(n-1)}x[/imath] in the hope of integrating by parts, but I'm unsure how to differentiate [imath]\cos^{(n-1)}[/imath], how should I proceed? |
642592 | If [imath]3n+1[/imath] is a perfect square where [imath]n[/imath] is a positive integer greater than [imath]1[/imath]. Then how to prove that [imath]n+1[/imath] would be sum of [imath]3[/imath] perfect squares?
If [imath]3n+1[/imath] is a perfect square where [imath]n[/imath] is a positive integer greater than [imath]1[/imath]. Then how to prove that [imath]n+1[/imath] would be sum of [imath]3[/imath] perfect squares? | 633651 | If [imath]n[/imath] is a positive integer greater than 1 such that [imath]3n+1[/imath] is perfect square, then show that [imath]n+1[/imath] is the sum of three perfect squares.
If [imath]n[/imath] is a positive integer greater than 1 such that [imath]3n+1[/imath] is perfect square, then show that [imath]n+1[/imath] is the sum of three perfect squares. My work: [imath]3n+1=x^2[/imath] [imath]3n+3=x^2+2[/imath] [imath]3(n+1)=x^2+2[/imath] [imath](n+1)=\dfrac{x^2+2}{3}[/imath] I have no clue what to do next. Please help! |
642699 | Explicit closed form [imath]\sum \frac{1}{an^2+bn+c}[/imath]
I know that [imath]\sum_{n\geq 1} \frac{1}{n^2}=\pi^2/6[/imath], is there a simple way to get an explicit closed form [imath]\sum \frac{1}{an^2+bn+c}[/imath], where [imath]a,b,c[/imath] are integers, [imath]a\neq 0[/imath]? | 161259 | What would be the value of [imath]\sum\limits_{n=0}^\infty \frac{1}{an^2+bn+c}[/imath]
I would like to evaluate the sum [imath]\sum_{n=0}^\infty \frac{1}{an^2+bn+c}[/imath] Here is my attempt: Letting [imath]f(z)=\frac{1}{az^2+bz+c}[/imath] The poles of [imath]f(z)[/imath] are located at [imath]z_0 = \frac{-b+\sqrt{b^2-4ac}}{2a}[/imath] and [imath]z_1 = \frac{-b-\sqrt{b^2-4ac}}{2a}[/imath] Then [imath] b_0=\operatorname*{Res}_{z=z_0}\,\pi \cot (\pi z)f(z)= \lim_{z \to z_0} \frac{(z-z_0)\pi\cot (\pi z)}{az^2+bz+c}= \lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b} [/imath] Using L'Hopital's rule. Continuing, we have the limit is [imath] \lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b}= \frac{\pi\cot (\pi z_0)}{2az_0+b} [/imath] For [imath]z_0 \ne 0[/imath] Similarly, we find [imath]b_1=\operatorname*{Res}_{z=z_1}\,\pi \cot (\pi z)f(z)=\frac{\pi\cot (\pi z_1)}{2az_1+b}[/imath] Then [imath]\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c} = -(b_0+b_1)=\\ -\pi\left( \frac{\cot (\pi z_0)}{2az_0+b} + \frac{\cot (\pi z_1)}{2az_1+b}\right)= -\pi\left( \frac{\cot (\pi z_0)}{\sqrt{b^2-4ac}} + \frac{\cot (\pi z_1)}{-\sqrt{b^2-4ac}}\right)= \frac{-\pi(\cot (\pi z_0)-\cot (\pi z_1))}{\sqrt{b^2-4ac}}= \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{\sqrt{b^2-4ac}} [/imath] Then we have [imath]\sum_{n=0}^\infty \frac{1}{an^2+bn+c} = \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{2\sqrt{b^2-4ac}}[/imath] Is this correct? I feel like I made a mistake somewhere. Could someone correct me? Is there an easier way to evaluate this sum? |
642676 | Average of [imath](1+X)^{-1}[/imath]
[imath]X[/imath] has the Poisson distribution with parameter [imath]λ[/imath]. I want to find the average of [imath](1+X)^{-1}[/imath] Some steps-of-thought would be appreciated. | 497643 | Compute the mean of [imath](1 + X)^{-1}[/imath] where [imath]X[/imath] is Poisson[imath](\lambda)[/imath]
Question Let [imath]X[/imath] be Poisson with parameter [imath]\lambda[/imath]. Compute the mean of [imath](1 + X)^{-1}[/imath]. (Introduction to Probability Theory, Hoel, pp. 104) Answer Key The answer key shows that the mean is [imath]\lambda^{-1}(1 - e^{-\lambda})[/imath] My Solution [imath]E(1 + X)^{-1} = \sum\limits_{j = 1}^\infty (1 + j)^{-1} \dfrac{\lambda^{(1 + j)^{-1}}e^{-\lambda}}{(1 + j)^{-1}!}[/imath] [imath]= e^{-\lambda} \sum\limits_{j = 1}^\infty \dfrac{\lambda^{(1 + j)^{-1}}}{((1 + j)^{-1} - 1)!}[/imath] [imath]= \lambda^{-1} e^{-\lambda} \sum\limits_{j = 1}^\infty \dfrac{\lambda^{(1 + j)^{-1} - 1}}{((1 + j)^{-1} - 1)!}[/imath] [imath]= \lambda^{-1} e^{-\lambda}(e^{\lambda} - 1)[/imath] [imath]= \lambda^{-1}(1 - e^{-\lambda})[/imath] EDIT: I found why I got the answer incorrect. Thanks for those who answer the question! :) |
642756 | Show a module is simple
Let [imath]M[/imath] be a non-zero module and suppose that every non-zero module homomorphism [imath]f:M\rightarrow N[/imath] is injective. Show that [imath]M[/imath] is simple. My idea is to let [imath]N[/imath] be a submodule; then pick a map to show it must be zero or [imath]M[/imath]. Thanks | 581997 | Simple module and homomorphisms
Are the following statements equivalent? i) [imath]M[/imath] is an [imath]R[/imath]-simple module. ii) Every [imath]R[/imath]-homomorphism (nonzero) from [imath]M[/imath] to an [imath]R[/imath]-module [imath]N[/imath] is a monomorphism. iii) Every [imath]R[/imath]-homomorphism (nonzero) from an [imath]R[/imath]-module [imath]N[/imath] to [imath]M[/imath] is an epimorphism. |
643257 | eigen value problem of the following matrix
[imath]A=\begin{bmatrix} 1 & -2 & 3 & -2 \\ 1 & 1 & 0 & 3 \\ -1 & 1 & 1 & -1 \\ 0 & -3 & 1 & 1 \\ \end{bmatrix}[/imath] 4.9 Pick out the smallest disc in the complex plane containing all the eigenvalues of [imath]A[/imath] from amongst the following: [imath]|z-1| \leq 7[/imath]; [imath]|z-1| \leq 6[/imath]; [imath]|z-1| \leq 4[/imath]. How to solve the problem 4.9? | 626035 | Find the smallest disk in complex plane containing the eigenvalues of a matrix (NBHM 2012)
Consider the given matrix [imath]\left[\begin{matrix}1& -2& 3& -2 \\1& 1& 0& 3\\-1& 1& 1& -1\\0& -3& 1& 1&\end{matrix}\right][/imath] Find out the smallest disk like ([imath]|z-1| < r[/imath] ) in the complex plane containing the eigenvalues of the given matrix. One easiest method is to find out the eigenvalues from the characteristic equation of the matrix. I can do it so no problem with it. I am looking for a method to solve the problem without the above method that can be evaluated in very short time. This type of method is possible, I believe. Thank you for your help |
283190 | If two CW complexes have isomorphic homotopy groups, are they homotopy equivalent?
I denote with [imath]\pi_{i}[/imath] i-homotopy group. If I have [imath]X,Y[/imath] CW-complex and [imath]\pi_{i}(X)=\pi_{i}(Y)[/imath] for all [imath]i[/imath]. Can I say that [imath]X[/imath] and [imath]Y[/imath] are homotopic equivalent? What type of equivalence is it? | 99302 | Spaces with equal homotopy groups but different homology groups?
Since it's fairly easy to come up with a two spaces that have different homotopy groups but the same homology groups ([imath]S^2\times S^4[/imath] and [imath]\mathbb{C}\textrm{P}^3[/imath]). Are there any nice examples of spaces going the other way around? Are there any obvious ways to approach a problem like this? |
643941 | A question of divisibility. (NBHM 2012)
The number [imath]18! + 1[/imath] is divisible by 437. How to prove it? I have [imath](19-1)! + 1 \equiv 0[/imath] (mod 19) and [imath]437 = 19 \times 23[/imath]. What to do next? Thank you for your help. | 247879 | how to prove [imath]437\,[/imath] divides [imath]18!+1[/imath]? (NBHM 2012)
I was solving some problems and I came across this problem. I didn't understand how to approach this problem. Can we solve this with out actually calculating [imath]18!\,\,?[/imath] |
376988 | Limit of [imath]\frac{\log(n!)}{n\log(n)}[/imath] as [imath]n\to\infty[/imath].
I can't seem to find a good way to solve this. I tried using L'Hopitals, but the derivative of [imath]\log(n!)[/imath] is really ugly. I know that the answer is 1, but I do not know why the answer is one. Any simple way to go about this? | 1608565 | Is [imath]\log(n!) \in\Theta(n \log n)[/imath]
Is [imath]\log(n!) \in\Theta(n \log n)[/imath]? I know it is [imath]O(n \log n)[/imath] because [imath]\log(n!) \leq \log(n^n)[/imath] which is the same as [imath]\log(n!) \leq n \log n[/imath]. But how can I show it is also [imath]\Omega(n \log n)[/imath]? |
420384 | the product space [imath]\mathbb{R}^I[/imath],where I denote [imath][0,1][/imath], has a countable dense subset.
show that the product space [imath]\mathbb{R}^I[/imath], where [imath]I[/imath] denote [imath][0,1][/imath], has a countable dense subset. | 488616 | Prove that the space has a countable dense subset
I'm doing this exercise in Munkres book and got no clue about the solution. Hope some one can help me solve this. Show that the product space [imath]R^{I}[/imath], where [imath]I = [0,1][/imath], has a countable dense subset. If [imath]J[/imath] has cardinality greater than [imath]\mathscr{P}(Z_{+})[/imath], then the product space [imath]R^{J}[/imath] does not have a countable dense subset With this problem, I even can't imagine the space [imath]R^{I}[/imath]. So if some one can talk something about this space, I really appreciate it (as I know, this space is rather important, right?) Thanks so much |
633691 | value of a complex integration along a curve
Let [imath]\gamma[/imath] be a closed and continuously differentiable path in the upper half plane [imath]\{z\in C: z= x+iy, x\in R, y\in R, y\gt 0 \}[/imath] not passing through the point [imath]i[/imath]. Describe the set of all possible values of the integral [imath]\frac{1}{2\pi i}\int_{\gamma} \frac{2i}{z^2+1} dz[/imath] If [imath]f[/imath] doesn't contain [imath]i[/imath] then the value would be zero. If [imath]f[/imath] contains [imath]i[/imath] then it would be [imath]1[/imath]. But is this all?? | 169068 | what are the possible values for integral
let [imath]\gamma[/imath] be a closed continuosly differentiable path in the upper half plane not passing through [imath]i[/imath]. What are the possible values of the integral [imath]\frac{1}{2\pi i}\int_{\gamma}\frac{2i}{z^2+1}dz[/imath] well the integral can be broken like [imath]\frac{1}{2\pi i}\int_{\gamma}\frac{2i}{(z+i)(z-i)}dz=[/imath] [imath]\frac{1}{2\pi i}\int_{\gamma}\frac{dz}{z-i}dz-\frac{1}{2\pi i}\int_{\gamma}\frac{dz}{z+i}dz=[/imath] by Cauchy Integral Formuale [imath]f(i)-f(-i)[/imath],so the second integral is [imath]0[/imath] as it is analytic in the upper half plane, but the first integral iss just [imath]n(\gamma,i)[/imath], which iss the windding number of [imath]\gamma[/imath] around [imath]i[/imath] what more I can say? Thank you. |
646624 | Summation of [imath]2^{(-2^{n})}[/imath]
By the ratio test, I know that this series convernges: [imath]\sum2^{(-2^{n})}[/imath], in the limit [imath]n[/imath] goes to infinity. Probably to something close to [imath].8[/imath] (if not equal to [imath].8[/imath]). The problem is, how do I demonstrate it? or to what does it really converge? Any hint is appreciated. | 583472 | Sum of Infinite Series [imath]1 + 1/2 + 1/4 + 1/16 + \cdots[/imath]
Everyone knows about the classic [imath] \sum_{i=1}^{\infty} \dfrac{1}{2^i} = 1 [/imath] However, is there any way to find [imath] \sum_{i=0}^{\infty} \dfrac{1}{2^{2^i}} = \dfrac12 + \dfrac14 + \dfrac{1}{16} + \dfrac{1}{256} + \cdots [/imath] |
646636 | What is the reason why we add 0.5 and subtract 0.5 in the normal probability to approximate the binomal probability?
For example, why does [imath]P(X \leq a) = P(Y \leq a + 0.5)[/imath] and [imath]P(X \geq a) = P(Y \geq a - 0.5)[/imath] where [imath]X[/imath] is the binomial random variable and [imath]Y[/imath] is the normal random variable? | 416150 | What is continuity correction in statistics
Can someone please explain to me the idea behind continuity correction and when is it necessary to add or subtract [imath]\dfrac{1}{2}[/imath] from the desired number (how do we tell whether we need to add or subtract), how do we tell when we need to use continuity correction? |
647032 | Show that [imath]H[/imath] is a normal subgroup of [imath]G[/imath]?
Let [imath]\mathbb M(n;\mathbb R)[/imath] denote the set of all real matrices (identified with [imath]\mathbb R^{n^2}[/imath] and endowed with its usual topology) and [imath]GL(n;\mathbb R)[/imath] denote the group of all invertible matrices. Let [imath]G[/imath] be a subgroup of [imath]GL(n;\mathbb R)[/imath]. Define [imath]H = \{ A \in G \mid \text{there exists } \phi : [0, 1] \rightarrow G \text{ continuous, such that } \phi (0)=A,\phi(1)=I\}.[/imath] Then is [imath]H[/imath] a normal subgroup? | 19070 | Showing that the path-connected component of the identity matrix in a subgroup of [imath]GL_n(\Bbb R)[/imath] is a normal subgroup
Let [imath]M(n;\mathbb{R})[/imath] denote the set of all [imath]n \times n[/imath] matrices with real entries (identified with [imath]\mathbb{R}^{n^{2}}[/imath] and endowed with its usual topology) and let [imath]GL(n;\mathbb{R})[/imath] denote the group of invertible matrices. Let [imath]G[/imath] be a subgroup of [imath]GL(n;\mathbb{R})[/imath]. Define [imath]H = \biggl\{ A \in G \ \biggl| \ \exists \ \varphi:[0,1] \to G \ \text{continuous such that} \ \varphi(0)=A , \ \varphi(1)=I\biggr\}[/imath] Then is [imath]H[/imath] normal in [imath]G[/imath]? For proving [imath]H[/imath] normal i should verify two things: First [imath]H[/imath] is a subgroup. For all [imath]A \in G, B \in H[/imath], we must have [imath]A \cdot B \cdot A^{-1} \in H[/imath]. That's it i am not able to proceed any further. |
647128 | Algebra: Unital Rings
I posted this earlier to no avail. It's an example of a topic i'm struggling with, I'm mostly looking for a sample answer that i can ask about and analyze to try and understand the "characteristic" portion. [imath]1[/imath]) Suppose [imath]Y^3 = Y[/imath] for each element [imath]Y[/imath] of a unital ring [imath]R[/imath]. Show that [imath]R[/imath] has a finite characteristic d that is a divisor of [imath]6[/imath]. | 646761 | Algebra struggles: 2 problems in 1
I'm struggling with some concepts in my algebra class. I've chosen these two completely unrelated examples that demonstrate my struggling topics (characteristics and function rings) the two are the following: [imath]1[/imath]) Let [imath]X[/imath] be a set, and let [imath]S[/imath] be a ring. Give a proof that the left distributive law holds for [imath](S^X, +, *)[/imath]. [imath]2[/imath]) Suppose [imath]Y^3 = Y[/imath] for each element [imath]Y[/imath] of a unital ring [imath]R[/imath]. Show that [imath]R[/imath] has a finite characteristic d that is a divisor of [imath]6[/imath]. Any example solutions for these proofs would be helpful. Thank you. |
647245 | If [imath]\sum n^2 a_n^2<\infty[/imath] then [imath]\sum a_n[/imath] is convergent.
Let [imath]\{a_n\}[/imath] be a sequence of positive integers, then prove/disprove: If [imath]\sum n^2 a_n^2<\infty[/imath] then [imath]\sum a_n[/imath] is convergent. | 296792 | If [imath]\sum a_n^2 n^2[/imath] converges then [imath]\sum |a_n|[/imath] converges
Let's suppose that [imath](a_n)[/imath] is a sequence so that [imath]\sum a_n^2 n^2[/imath] converges, so I have to prove that [imath]\sum |a_n|[/imath] converges. By the Cauchy criterion we have that, since [imath]\sum a_n^2 n^2 [/imath] converges if [imath]\varepsilon > 0[/imath] then there is a [imath]k\in\mathbb N[/imath] so for all [imath]m,n[/imath] naturals that satisfy [imath]m > n \geq k[/imath]: [imath] |n^2 a_n^2 + ... + m^2 a_m^2| < \varepsilon [/imath] but I can't get [imath]| |a_n| + ... + |a_m| | < \varepsilon[/imath] to show that [imath]\sum |a_n|[/imath] converges. Could you give me a hint? Is there any way to prove this without resorting Cauchy criterion. Thanks! |
647299 | Find the sum of the following infinite series:
Find the sum of the following infinite series: [imath]\frac{1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}- ...[/imath] stuck on this problem.seek your help | 635315 | Sum of a Hyper-geometric series. (NBHM 2011)
How to find the sum of the following series [imath]\frac{1}{5} - \frac{1\cdot 4}{5\cdot 10} + \frac{1\cdot 4\cdot 7}{5\cdot 10\cdot 15} - \dots\,.?[/imath] I have no idea. I have written the general term and tested its convergence by Gauss' test for convergence, but they are neither the question nor the answer. |
647036 | What are all the invertible elements in [imath]\mathbb{Z_m}[/imath] for [imath]m = 30[/imath]?
How can I know all of the invertible elements? Is it just all of the numbers that are relatively prime to 30? | 647219 | Finding invertible elements in [imath]\mathbb{Z}/m\mathbb{Z}[/imath]
Can anyone show me how to find invertible elements in [imath]\mathbb{Z}/m\mathbb{Z}[/imath], [imath]m[/imath] is say [imath]28[/imath]? Also I'm not very clear about what it mean by 'invertible element in [imath]\mathbb{Z}/m\mathbb{Z}[/imath]' |
647587 | Sum of a power series [imath]n x^n[/imath]
I would like to know: How come that [imath]\sum_{n=1}^\infty n x^n=\frac{x}{(x-1)^2}[/imath] Why isn't it infinity? | 919309 | Summing [imath]\sum_{n=0}^{\infty} n x^n[/imath]
I am trying to figure out how to do the infinite summation: [imath] \sum_{n=0}^{\infty} n x^n \qquad 0 \leq x < 1[/imath] The series converges so it seems to me that the limit must exist, but I'm having difficulties trying to find an exact answer for it. It is very much like an infinite geometric series, but the multiplication factor [imath]n[/imath] makes that you can't use the usual trick. The furthest I came was writing it out as: [imath] x + 2 x^2 + 3 x^3 + 4 x^4 + \dots[/imath] but that doesn't get me anywhere. Can someone give me a hint? |
647676 | How to check if [imath]n^2[/imath] will square root evenly without square rooting [imath]n^2[/imath]
How can I easily check if the square root of a number is a whole number and not a decimal without computing the square rood of said number. I could square root the number ((n^2)^1/2 or sqrt(n^2)) but I do not want to. | 131330 | Detecting perfect squares faster than by extracting square root
Given the radix-[imath]r[/imath] representation of a integer [imath]n[/imath], and a small integer constant [imath]k[/imath], there is an [imath]O(\log n)[/imath] algorithm for detecting whether [imath]n[/imath] is a multiple of [imath]k[/imath], namely, division, which produces as a byproduct the quotient [imath]\lfloor n/k\rfloor[/imath]. In general this is the best one can do. But for certain choices of [imath]r[/imath] and [imath]k[/imath], for example [imath]r=10[/imath] and [imath]k=2[/imath], there is an algorithm which answers the question much faster (constant time) without producing the quotient. Given the radix-[imath]r[/imath] representation of a integer [imath]n[/imath], we can extract the integer square root [imath]\lfloor\sqrt n\rfloor[/imath] in something like [imath]O(\log^3 n)[/imath] time by doing binary search, which Joriki notes below can be improved to [imath]O(\log^2 n)[/imath] with a sufficiently clever implementation. This gives an [imath]O(\log^2 n)[/imath] algorithm for determining whether [imath]n[/imath] is a perfect square. Is there a significantly faster algorithm which correctly decides whether [imath]n[/imath] is perfect square, without also producing the square root? I suspect not, but I would be interested to see a proof. |
647412 | Find the limit of the given function?
Let [imath]a>[/imath] 0 and [imath]k \in \mathbb N[/imath]. Evaluate [imath]\lim_{n\rightarrow \infty} a^{-nk}\prod_{j=1}^k(a+\frac{j}{n})^n.[/imath] How is this accomplished? | 643454 | Let [imath]a > 0[/imath] and let [imath]k \in \mathbb{N}[/imath]. Evaluate the limit
Let [imath]a > 0[/imath] and let [imath]k \in \mathbb{N}[/imath]. Evaluate: [imath]\lim_{n\to \infty}a^{-nk}\prod ^k_{j=1}\left(a+\frac{j}{n}\right)^n[/imath] Clueless on this problem. Seek your help. |
648005 | If [imath]P(B\text{ }|\text{ }A)=1-\epsilon[/imath] and [imath]P(C\text{ }|\text{ }B)=1[/imath] then [imath]P(C\text{ }|\text{ }A)\geq 1-\epsilon[/imath]
If [imath]1=P(C\text{ }|\text{ }B)=\frac{P(C\cap B)}{P(B)}[/imath] then we know that [imath]P(C\cap B)=P(B)[/imath]. If [imath]P(B\text{ }|\text{ }A)=1-\epsilon[/imath] for [imath]\epsilon\geq 0[/imath] then [imath]P(A)=\frac{P(B\cap A)}{1-\epsilon}[/imath]Using this I found that [imath]P(C\text{ }|\text{ }B)=(1-\epsilon)\frac{P(C\cap A)}{P(B\cap A)}[/imath] Now I have to show that [imath]\frac{P(C\cap A)}{P(B\cap A)}\geq 1[/imath] I don't know how to use the first assumption to show this. I need hints. Thanks. | 647948 | Given [imath]P(B\mid A)=1-\varepsilon[/imath] for some [imath]0<\varepsilon<1[/imath] and [imath]P(C\mid B)=1[/imath], prove that [imath]P(C\mid A)≥1-\varepsilon[/imath]
We need to show that, given [imath]P(B\mid A)=1-\varepsilon[/imath] for some [imath]0<\varepsilon<1[/imath] and [imath]P(C\mid B)=1[/imath], that [imath]P(C\mid A)≥1-\varepsilon[/imath]. Since we know that [imath]P(C\mid B)=1[/imath], it follows that [imath]P(B \cap C)=P(B)[/imath]. This however, doesn't really help much. I tried a lot of things, but I couldn't prove this question, although it seems pretty straight-forward. |
648221 | Compute [imath]\displaystyle\iint\limits_R\frac{y}{x+y^2}dA[/imath] where [imath]R=[0,1]\times[1,2][/imath]
Compute [imath]\displaystyle\iint\limits_R\frac{y}{x+y^2}dA[/imath] where [imath]R=[0,1]\times[1,2][/imath] [imath]\displaystyle \int_1^2\int_0^1\frac {y}{x+y^2}dxdy=\int_1^2y(\ln|x+y^2|)\Big|_0^1dy=\int_1^2(y\ln|1+y^2|-2y\ln|y|)dy=\Big(\frac {(y^2+1)\ln(y^2+1)-y^2-1}{2}-2(\frac {y^2\ln(y)}{2}-\frac {y^2}{4})\Big)\Big|_1^2=\frac {5\ln(5)-10\ln(2)}{2}[/imath] I got this answer with help from online integral calculator, but I don't really understand why [imath]\int_1^2(y\ln|1+y^2|-2y\ln|y|)dy=\Big(\frac {(y^2+1)\ln(y^2+1)-y^2-1}{2}-2(\frac {y^2\ln(y)}{2}-\frac {y^2}{4})\Big)\Big|_1^2[/imath] Can someone explain this please! Thank you. | 644396 | Compute [imath]\iint\limits_R\frac{y}{x+y^2}dA[/imath] where [imath]R=[0,1]\times[1,2][/imath]
Compute [imath]\displaystyle\iint\limits_R\frac{y}{x+y^2}dA[/imath] where [imath]R=[0,1]\times[1,2][/imath] [imath]\displaystyle\int_0^1\int_1^2\frac{y}{x+y^2}dydx=\int_0^1\int_1^2y(x+y^2)^{-1}dydx[/imath] How do I integrate the denominator if the superscript is [imath]-1[/imath]? Hints are appreciated! |
648557 | Why can't [imath]e^{x^2}[/imath] be integrated
My teacher told me that not only do we have to use the erf function to approximate error, but that it is proved impossible to integrate in real analysis (at least not Riemann-integrable). Is there a name for this proof, and can I have it? I am not a mathematician, but such a simple function has been around and it irks me terribly and need a detailed proof for closure. | 523824 | What is the antiderivative of [imath]e^{-x^2}[/imath]
I was wondering what the antiderivative of [imath]e^{-x^2}[/imath] was, and when I wolfram alpha'd it I got [imath]\displaystyle \int e^{-x^2} \textrm{d}x = \dfrac{1}{2} \sqrt{\pi} \space \text{erf} (x) + C[/imath] So, I of course didn't know what this [imath]\text{erf}[/imath] was and I looked it up on wikipedia, where it was defined as: [imath] \text{erf}(x) = \dfrac{2}{\sqrt{\pi}} \displaystyle \int_0^x e^{-t^2} \textrm{d}t [/imath] To my mathematically illiterate mind, this is a bit too circular to understand. Why can't we express [imath]\int e^{-x^2} \textrm{d}x[/imath] as a 'normal function'? Also, what is the use of the error function? |
648746 | Bounds on cardinality of sum-sets.
Let [imath]X\subset \mathbb{R}[/imath] and [imath]Y \subset \mathbb{R}[/imath] where X and Y have finite cardinalities. Let also, [imath]a,b \in \mathbb{R}/0[/imath]. What can we say about cardinality [imath]|aX+bY|=????[/imath] For example we can have lower bound and upper bound [imath]\max(|X|,|Y|) \le |aX+bY| \le |X||Y|[/imath]. What about more tight bounds? I am pretty sure one can show that [imath]|aX+bY|=|X||Y|[/imath] for almost all [imath]a,b[/imath], but I don't know how. Thanks, any help will be appreciated. Also, notation means [imath]aX+bY=\{ax+by: x\in X ,y\in Y \}[/imath]. | 647851 | Cardinality of sum-set
Let [imath]X\subset R[/imath] and [imath]Y \subset R[/imath] where X and Y have finite cardinalities. Let also, [imath]a,b \in R[/imath]. How to show that [imath]|aX+bY|=|X||Y|[/imath] almost everywhere (measure of [imath](a,b) \subset R^2[/imath] such that [imath]|aX+bY|<|X||Y|[/imath] is 0) ? Where we define [imath]|aX+bY|=\{ax+by:x \in X, y \in Y \}[/imath]. More, specifically let [imath]X=\{x=d_x*z: z \in \mathcal{Z}, d_x \in \mathcal{R}\}[/imath] (so, it's a lattice. Where each point is [imath]d_x[/imath] away from one another.) The above is easy if [imath]X=Y[/imath] then the proof of [imath]|aX+bX|=|X||X|[/imath] almost everywhere is fairly simple. Now what if [imath]Y[/imath] is lattice with [imath]d_y \neq d_x[/imath]. How to show it then? Thanks in advance. |
649144 | Bijection from a set of functions to a Cartesian product of sets
Let S be an arbitrary set. Let [imath]F=\{f:\{0,1\}\to S\}[/imath] be the set of functions from [imath]\{0,1\}[/imath] to S. Construct a bijection [imath]F→S \times S[/imath]. I think I would define the function [imath]a(f)=(f(0),f(1))[/imath] because we know that both [imath]f(0)[/imath] and [imath]f(1)[/imath] are in [imath]S[/imath], but I don't know where to go from there. | 640000 | Bijection Contruction
My knowledge of constructing bijections is very poor and I would really appreciate any clarification or assistance. The question I am having trouble with is the following: Let [imath]S[/imath] be an arbitrary set. Let [imath]F=\{\{0,1\} \xrightarrow{f} S \}[/imath] be the set of functions from [imath]\{0,1\}[/imath] to [imath]S[/imath]. Construct a bijection [imath]F \xrightarrow{\alpha} S \times S[/imath]. |
319963 | If [imath]2^n - 1[/imath] is prime from some integer [imath]n[/imath], prove that n must also be prime.
I understand the idea of the proof. I just want to make sure I wrote my proof well. Suppose [imath]n[/imath] is not prime. Then [imath]\exists x,y \in \mathbb{Z}[/imath] such that [imath]n = xy[/imath]. [imath]2^{xy} - 1 = (2^x)^y - 1[/imath] [imath] = (2^y - 1)(2^{y(x-1)} + 2^{y(x-2)} + ... + 2^{y} + 1)[/imath] Since [imath]2^{n} - 1[/imath] is divisible by [imath]2^y - 1[/imath] it must be that [imath]2^n - 1[/imath] is not prime. Contradiction. Thus [imath]n[/imath] must be prime. How does this look? | 2598242 | [imath]2^p -1[/imath] then [imath]p[/imath] is a prime
I have been trying to solve this following problem: If [imath]2^p-1[/imath] is a prime then prove that [imath]p[/imath] is a prime, where [imath]p \geq 2[/imath]. which way should I go to prove this? Using fermat's or Bezout's theorem? or is it something else? This is what I managed to come up with,I am confused about it and not sure if it is correct: If [imath]p[/imath] is a prime and [imath]2 \nmid p \rightarrow[/imath] gcd[imath](2,p)=1[/imath] so, [imath]2^{p-1} \equiv 1(p) [/imath] [imath]2^p\equiv2(p)[/imath] so, [imath]2^p\not\equiv 1(p) [/imath] and since [imath]2\not\equiv 1 (p)[/imath] conc: [imath]p\nmid2^p -1[/imath] Any hints are appreciated. |
649314 | [imath]\lim_{n\rightarrow +\infty} \frac{a_{n}}{a_{n+1}} = z_{0}[/imath] with [imath]z_{0}[/imath] pole
This is an exercise from Stein-Shakarchi. Suppose that [imath]f[/imath] is holomorphic in an open set containing the closed unit disc, except for a pole at [imath]z_{0}[/imath] on the unit circle. Show that if [imath]f(z) = \sum_{0}^{\infty}a_{n}z^{n}[/imath] in the open unit disc then[imath]\lim_{n\rightarrow +\infty} \frac{a_{n}}{a_{n+1}} = z_{0}[/imath] Any hint ? | 17932 | About the limit of the coefficient ratio for a power series over complex numbers
This is my first question in mathSE, hope that it is suitable here! I'm currently self-studying complex analysis using the book by Stein & Shakarchi, and this is one of the exercises (p.67, Q14) that I have no idea where to start. Suppose [imath]f[/imath] is holomorphic in an open set [imath]\Omega[/imath] that contains the closed unit disc, except for a pole at [imath]z_0[/imath] on the unit circle. Show that if [imath]f[/imath] has the power series expansion [imath]\sum_{n=0}^\infty a_n z^n[/imath] in the open unit disc, then [imath]\displaystyle \lim_{n \to \infty} \frac{a_n}{a_{n+1}} = z_0[/imath]. If the limit is taking on [imath]|\frac{a_n}{a_{n+1}}|[/imath] and assume the limit exists, by the radius of convergence we know that the answer is [imath]1[/imath]. But what can we say about the limit of the coefficient ratio, which is a pure complex number? I've tried to expand the limit directly by definition, with no luck. And I couldn't see how we can apply any of the standard theorems in complex analysis. I hope to get some initial directions about how we can start thinking on the problem, rather than a full answer. Thank you for the help! |
135471 | The direct sum of two closed subspace is closed? (Hilbert space)
I know that if [imath]X[/imath] is a Banach space, then, the direct sum of two closed subspace [imath]X_1[/imath] and [imath]X_2[/imath] is not necessarily closed. But what if [imath]X[/imath] is Hilbert? I assume there is something to do with the orthonormal basis, since this is something you can ask for a Hilbert space but not a Banach space. Moreover, is the projection to [imath]X_1[/imath] bounded? Namely, [imath]P[/imath] is defined on [imath]X_1\oplus X_2[/imath]: [imath]P(x)=x\quad \text{on $X_1$} \qquad P(x)=0\quad \text{on $X_2$}[/imath] is [imath]P[/imath] bounded? | 796159 | The subspace sum of closed subspaces is closed
Given an arbitrary Hilbert space [imath]\scr H[/imath] and closed subspaces [imath]A,B\subseteq\scr H[/imath] with trivial intersection, is it true that [imath]A+B=\{x+y:x\in A,y\in B\}[/imath] is closed? So far, I have the following: Let [imath]P_A(x)[/imath] be the unique [imath]z\in A[/imath] such that [imath]x-z\in B[/imath]. This is a well defined quantity for all [imath]x\in A+B[/imath] because [imath]A\cap B=0[/imath], and [imath]x=P_A(x)+P_B(x)[/imath] is a decomposition of [imath]x[/imath] into [imath]A[/imath] and [imath]B[/imath]. Given any [imath](x_n)_{n\in\Bbb N}\subseteq A+B[/imath] converging to [imath]x[/imath], we wish to show that [imath]x\in A+B[/imath]. Assuming that [imath](P_A(x_n))_{n\in\Bbb N}[/imath] and [imath](P_B(x_n))_{n\in\Bbb N}[/imath] converge, they must converge to points [imath]p_A\in A,p_B\in B[/imath] respectively, since [imath]A,B[/imath] are closed, and then [imath]x=p_A+p_B[/imath] because of the continuity of [imath]+[/imath]. It remains to show that the projected sequence is Cauchy, but I'm stuck at showing something along the lines of [imath]||P_A(x_n)-P_A(x_m)||\le C||x_n-x_m||[/imath], because there is no reason to believe (I think) that [imath]P_A[/imath] is bounded. |
649848 | Inner product spaces, examples for subspaces with certain properties
Let [imath]H[/imath] be a inner product space. Give examples for a subset [imath]U\subset H[/imath] so that (a) [imath]\overline{U}\neq U^{\bot\bot}[/imath] (b) [imath]\overline{U}\oplus U^{\bot}\neq H[/imath] I have thought about [imath]H=C[0,1][/imath] with the inner product [imath]\langle f,g\rangle = \int_{0}^{1}f(t)\overline{g(t)}dt[/imath], since this space is not complete and therefore pre-Hilbert, but I cannot find a subspace which satisfies satisfies neither (a) nor (b). | 647817 | Orthogonal complement examples
I am looking for an example such that in a pre-Hilbert space [imath]H[/imath] we have for a subspace [imath]U[/imath] that (i) [imath]\bar{U} \oplus U^\perp \neq H[/imath] (ii) [imath] \bar{U} \neq U^{\perp \perp}[/imath] Since finite and closed subspaces will probably not do it, I am not good at inventing nice examples. Does anybody here have a few at hand? |
649886 | How to find the solutions of a^2 = [29] mod 35?
How to find the solutions of: [imath]a^2 = [29]_{35}[/imath] ? (Brutal) Attempt: Since [imath]Z_{35}[/imath] contains only 35 elements, one could just brute force the calculation of [imath]a^2[/imath] for every element a [imath]\in Z_{35}[/imath] that GCD(a,35). When you find a number that verifies the equation, automatically you find another solution which is the inverse of that solution [imath]\in Z_{35}[/imath]. Do you know any non brutal way? NOTE: it doesn't have to require the quadratic congruence solving (I have only studied linear congruences) | 340484 | Find out all solutions of the congruence [imath]x^2 \equiv29 \mod 5[/imath].
Find out all solutions of the congruence [imath]x^2 \equiv29 \mod 5[/imath]. [Hint:Find the solutions of this congruence [imath]\mod 5[/imath] , [imath]\mod 7[/imath] , and [imath]\mod 7[/imath] , and then use the Chinese Remainder Theorem.] |
649970 | Does [imath]\frac12+\frac14+\frac18+\dots[/imath] equal [imath]1[/imath]?
Suppose I have something with length one unit. I divide it to two equal length [imath]0.5[/imath] unit and put left one in my left side and right one in my right side. I then do same for my right side and contact result's left one to my left side and replace my right side with it's right one. Even if I do these infinitely, there's something at my right side remaining. So it seems that [imath]\frac12+\frac14+\frac18+\dots[/imath] never taste value [imath]1[/imath]. Right? NOT DUPLICATE: Actually answers at here does not answer my question. because if I accept them then I have [imath]\frac12+\frac14+\frac18+\dots=\frac23+\frac29+\frac2{27}+\dots=1[/imath] Then I can define [imath]A_n = \frac12+\frac14+\frac18+\dots+\frac1{2^n}[/imath] And [imath]B_n = \frac23+\frac29+\frac2{27}+\dots+\frac2{3^n}[/imath] Then, it's quite can be seen that: [imath]A_n<B_n , n\in\mathbb N[/imath] Then how could I accept that both series can reach each other at value [imath]1[/imath]?! | 149373 | Does the fact that [imath]\sum_{n=1}^\infty 1/2^n[/imath] converges to [imath]1[/imath] mean that it equals [imath]1[/imath]?
I have a clueless friend who believes that [imath] \sum_{n=1}^\infty \frac{1}{2^n} [/imath] doesn't equal [imath]1[/imath] in the 'normal arithmetical sense'. He doesn't believe that this series flat out equals [imath]1[/imath] Is he correct? |
650891 | Does there always exists an injective homomorphism from [imath]G[/imath] into [imath]S_m[/imath] for [imath]m[/imath]
Does there always exists an injective homomorphism from [imath]G[/imath] into [imath]S_m[/imath] for some [imath]m<n[/imath] where [imath]|G|=n[/imath] I know that for [imath]|G|=n[/imath] there is always an injective homomorphism from [imath]G[/imath] into [imath]S_n[/imath] | 650841 | Group homomorphisms and which of the following statements are true (NBHM-[imath]2014[/imath])
Let [imath]G[/imath] be a finite group of order [imath]n\ge2[/imath]. Which of the following statements are true? a. There always exists an injective homomorphism from [imath]G[/imath] into [imath]S_n[/imath]. b. There always exists an injective homomorphism from [imath]G[/imath] into [imath]S_m[/imath] for some [imath]m\lt n[/imath]. c. There always exists an injective homomorphism from [imath]G[/imath] into [imath]GL_n(R)[/imath]. I think b. is right (don't know how to prove though). I also think that c. is false. But not sure. |
650163 | Find a uniformly continuous function such that [imath]a_{n+1}=f(a_n)[/imath]
[imath]a_{n+1} = a_n - a_n^2[/imath], [imath]a_1 = 2/3[/imath]. for [imath]n\ge1[/imath] a) Show the series converges and find its limit. b) find a uniformly continuous [imath]f:\mathbb{R}\rightarrow \mathbb{R}[/imath] such that: [imath]a_{n+1}=f(a_n) \forall n \in \mathbb{N}[/imath]. section (a) is very technically, I proved [imath]\{a_n\}[/imath] is monotonically decreasing and bounded bellow, And using limits arithmetic found [imath]L=0[/imath]. section (b) is quite tricky for me. I'd be glad for help. Thanks! | 836019 | Is there a uniformly continuous function [imath]f:\mathbb R \to \mathbb R[/imath] such that [imath]a_{n+1}=f(a_n)[/imath]?
Let [imath]a_1=\frac 2 3 , \ a_{n+1}=a_n-a^2_n[/imath] for [imath]n\ge 1[/imath], [imath]a_n[/imath] is monotonically decreasing and bounded: [imath]0\le a_{n+1} \le 1[/imath] and [imath]\displaystyle\lim_{n\to\infty} a_n=0[/imath]. Is there a uniformly continuous function [imath]f:\mathbb R \to \mathbb R[/imath] such that [imath]a_{n+1}=f(a_n), \ \forall n\in \mathbb N[/imath] ? I don't know how to start here but I'm pretty sure there is such function since [imath]a_n[/imath] is converging then from the definition of uniform continuity, we'll have [imath]\forall \epsilon >0 :\exists \delta >0: \forall n,m\in \mathbb N: |a_n-a_m|<\delta\Rightarrow |f(a_n)-f(a_m)|<\epsilon[/imath] [imath]a_{n+1}=f(a_n)=f(a_{n-1}-a_{n-1}^2)[/imath] so I need to find a function that will "push the index" by 1, like [imath]f(x)=x+1[/imath] and is uniformly continuous, but I have no idea how to find such function. |
651174 | Complex roots problem
I've got a complex equation with 4 roots that I am solving. In my calculations it seems like I am going through hell and back to find these roots (and I'm not even sure I am doing it right) but if I let a computer calculate it, it just seems like it finds the form and then multiplies by [imath]i[/imath] and negative [imath]i[/imath]. Have a look: http://www.wolframalpha.com/input/?i=%288*sqrt%283%29%29%2F%28z%5E4%2B8%29%3Di Here's me going bald: | 651214 | Complex root won't work
So I'm trying to get this: http://www.wolframalpha.com/input/?i=%288*sqrt%283%29%29%2F%28z%5E4%2B8%29%3Di And I've calculated [imath]z^4=16 \left( \cos (\frac{- \pi}{3})+ \sin ( \frac{- \pi}{3}) \right)[/imath] So I'm trying to find the roots using the formula: [imath] r^{1/n} = \left( \cos ( \frac { \theta + 2 \pi \cdot k}{n}) + i \cdot \sin ( \frac{ \theta + 2 \pi \cdot k}{n}) \right) [/imath] But my result does not equal. Take a look: http://www.wolframalpha.com/input/?i=16%5E%281%2F4%29*%28cos%28%28%28-pi%2F12%29%29%2F4%29%2Bi*sin%28%28%28-pi%2F12%29%29%2F4%29%29 This is for [imath]k=0[/imath]. What am I doing wrong? |
651339 | Open Nested Interval Question
Find a family [imath]\{ I_n \}[/imath] of open nested intervals such that no two [imath]I_n[/imath] are equal and the intersection is equal to [imath]\left[-2,2\right][/imath]. | 650998 | Family of Closed/Open Nested Intervals
Find a family [imath]\{I_n\}[/imath] of closed nested intervals, such that no two [imath]I_n[/imath]'s are equal and their intersection is [imath][-2,2][/imath]. An answer for the same question except for dealing with open nested intervals would also be appreciated. |
651605 | Is the interval [imath][0,1][/imath] equinumerous with [imath]\mathbb{R}[/imath]
I recall reading a proof that showed these two sets were equinumerous, but I'm having trouble finding it. Is there any intuitive method to show that they are in fact equinumerous? It seems like since [imath][0,1][/imath] is a subset of [imath]\mathbb{R}[/imath] that [imath]\mathbb{R}[/imath] should be larger. | 250867 | Bijection from finite (closed) segment of real line to whole real line
Is there a bijection from a finite (closed) segment of the real line to [imath]\mathbb{R}[/imath]? For example, is there a bijection from [imath][0,1][/imath] to [imath]\Bbb{R}[/imath]? If so, is there a straightforward example? If not, why? |
651509 | Show that if [imath]AB = BA[/imath], then [imath]\exp(A) \exp(B) = \exp(B) \exp(A)[/imath]
Show that if [imath]AB = BA[/imath], then [imath]\exp(A) \exp(B) = \exp(B) \exp(A)[/imath] here is what i got so far: [imath]\begin{align}\exp(A) \exp(B) &= (I + A + {\dfrac{1}{2}} A^2 + {\dfrac{1}{6}}A^3 + \ldots)(I + B + {\frac{1}{2}}B^2 + {\frac{1}{6}}B^3 + \ldots )\\ &= I + (A + B) + {\frac{1}{2}}(A^2 + 2AB + B^2) + {\frac{1}{6}}(A^3 + 3A^2 B + 3AB^2 + B^3 ) + \ldots\\ &= I + (A + B) + {\frac{1}{2}}(A + B)^2 + {\frac{1}{6}}(A + B)^3 + \ldots \end{align}[/imath] am i going the right direction so far...if so, how to get to show the rest until [imath]\exp(B)\exp(A)[/imath]. Thank you | 568450 | [imath]M,N\in \Bbb R ^{n\times n}[/imath], show that [imath]e^{(M+N)} = e^{M}e^N[/imath] given [imath]MN=NM[/imath]
I am working on the following problem. Let [imath]e^{Mt} = \sum\limits_{k=0}^{\infty} \frac{M^k t^k}{k!}[/imath] where [imath]M[/imath] is an [imath]n\times n[/imath] matrix. Now prove that [imath]e^{(M+N)} = e^{M}e^N[/imath] given that [imath]MN=NM[/imath], ie [imath]M[/imath] and [imath]N[/imath] commute. Now the left hand side of the desired equality is [imath]e^{(M+N)} = I+ (M+N) + \frac{(M+N)^2}{2!} + \frac{(M+N)^3}{3!} + \ldots [/imath] On the right hand side of the equation we have [imath]e^Me^N = \left(I + M + \frac{M^2}{2!} + \frac{M^3}{3!}\ldots\right) \left(I + N + \frac{N^2}{2!} + \frac{N^3}{3!} \ldots\right) [/imath] Now basically this is as far as I got... I am unsure on how to work out the product of the two infinite sums. Possibly I need to expand the powers on the left hand side expression but I am unsure how to do this in an infinite sum... If anyone could give me an answer or a hint that can help me forward I would greatly appreciate it. Thanks |
651886 | a Problem about Sequence
Let [imath]a_1[/imath] be an integer. Then we assume [imath] a_{n+1} = \begin{cases} 3a_n+1,&\text{$a_n$ is odd}\\ \frac{a_n}{2},&\text{$a_n$ is even} \end{cases} [/imath] Now we prove that for any [imath]a_1\in\mathbb N[/imath], there exists [imath]N[/imath] which satisfy: [imath]a_n=1,2[/imath] or [imath]4[/imath],[imath]n\geq{N}[/imath]. At first I want to give it a suitable category for the problem: analysis. And I want to use the basic method: evaluate the upper bound for [imath]a_n[/imath], however I find it's not easy because the iteration is rely on the odd or even property of [imath]a_n[/imath]. So I attempt the method of number theory. But I failed to find any way to go over it. Can anyone have idea? Thank you. | 2694 | What is the importance of the Collatz conjecture?
I have been fascinated by the Collatz problem since I first heard about it in high school. Take any natural number [imath]n[/imath]. If [imath]n[/imath] is even, divide it by [imath]2[/imath] to get [imath]n / 2[/imath], if [imath]n[/imath] is odd multiply it by [imath]3[/imath] and add [imath]1[/imath] to obtain [imath]3n + 1[/imath]. Repeat the process indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach [imath]1[/imath]. [...] Paul Erdős said about the Collatz conjecture: "Mathematics is not yet ready for such problems." He offered $500 USD for its solution. How important do you consider the answer to this question to be? Why? Would you speculate on what might have possessed Paul Erdős to make such an offer? EDIT: Is there any reason to think that a proof of the Collatz Conjecture would be complex (like the FLT) rather than simple (like PRIMES is in P)? And can this characterization of FLT vs. PRIMES is in P be made more specific than a bit-length comparison? |
651512 | Disjoint Union Topology universal property as instance of final topology's universal property
I'm studying the final topology on a set [imath]X[/imath] induced by a family [imath]\{f_\alpha:X_\alpha\rightarrow X\mid \alpha\in A\}[/imath] and know of its universal property, namely that given a function [imath]g:X\rightarrow Z[/imath], [imath]g[/imath] is continuous iff [imath]g\circ f_\alpha[/imath] is continuous for all [imath]\alpha\in A[/imath]. Showing this is not difficult, but it is not constructive. What bothers me about this is that my usual understanding of universal properties is that they refer to initial or terminal properties and morphisms; typically, they involve the existence of a unique morphism, as the definitions here suggest. As this answer points out, we may use the category of cones along with the forgetful functor from [imath]\mathbf{Top}[/imath] to [imath]\mathbf{Set}[/imath] to show that the above universal property is indeed as the above definitions give. To be honest, I don't know enough category theory to completely understand the process, which may be where my issue lies. The question I have is then: How (or if) the universal property of the disjoint union topology can be realized as a special case of the final topology's universal property, since the disjoint union topology is the final topology induced by the canonical injections of [imath]X_\alpha[/imath] into [imath]X=\amalg_{\alpha\in A}{X_\alpha}[/imath], which is constructive in nature, i.e. that it guarantees the existence of a unique morphism such that a certain diagram commutes. | 138360 | Universal property of initial topology
I'm learning some category theory and I thought I had understood universal objects but maybe I have not because I cannot write down the definition of initial topology in terms of categories. My understanding of universal objects so far: I thought universal objects are either initial or terminal objects in a comma category [imath](O \downarrow F)[/imath] where [imath]O[/imath] is a selection functor [imath]\textbf{1} \to O \in C[/imath] and [imath]F[/imath] is a forgetful functor into [imath]C[/imath]. In the example of a free group, [imath]F: \textbf{Group} \to \textbf{Set}[/imath] and [imath]O[/imath] is the set that generates the free group. The objects in this category are pairs [imath](f, G)[/imath] where [imath]f[/imath] is a (set-theoretic) map from [imath]S \to G[/imath]. Now I would like to do the same for the initial topology but I'm struggling to see what the objects are and also what [imath]F[/imath] should be. [imath]O[/imath] is probably going to be [imath]X[/imath], the space we would endow with the topology. So we probably want [imath]F[/imath] to be a functor [imath]\textbf{Top} \to \textbf{Set}[/imath]. The objects are probably [imath](f,Y)[/imath] where [imath]f[/imath] is a map [imath]X \to Y[/imath] (or [imath]Y \to X[/imath]?) such that there exists a continuous map [imath]f^\prime[/imath] with [imath]F(f^\prime) = f[/imath]. Is this right? If not: what are the objects? And is the initial topology an initial or a terminal object? Edit I think what I wrote above is wrong: I get one commutative diagram for each [imath]Y_i[/imath] in the family with respect to which the initial topology is defined. This means that the objects are [imath](g_i, Z)[/imath] where [imath]g_i : Z \to Y_i[/imath] and the comma category would be [imath](Y_i, F)[/imath]. Is this right? The diagram for [imath](Y_i \downarrow F)[/imath] would look like this: X --(f_i)--> Y_i ^ | | (g_i) | | Z Then the question is whether the unique continuous map [imath]\varphi[/imath] is [imath]Z \to X[/imath] or [imath]X \to Z[/imath]. |
192963 | If [imath]\lim_{x\to+\infty}[f(x+1)-f(x)]= \ell[/imath] then [imath]\lim\limits_{x\to+\infty}\frac{f(x)}x=\ell[/imath].
Let [imath]f:[o,+\infty)\rightarrow \Bbb R[/imath] a function bounded on each finite interval. i want to try that if [imath]\lim\limits_{x\rightarrow+\infty}[f(x+1)-f(x)]= L[/imath] then also [imath]\lim\limits_{x\rightarrow+\infty}\dfrac{f(x)}x = L[/imath] | 2579511 | If [imath]f[/imath] is continuous and [imath]\lim\limits_{x\to \infty}f(x+1)-f(x)=0[/imath], does this mean that [imath]\lim\limits_{x\to \infty}\frac{f(x)}{x}=0[/imath]?
Suppose that the function [imath]f[/imath] is continuous in [imath]\mathbb{R}[/imath] and [imath]\lim_{x\to \infty}\left(f(x+1)-f(x)\right)=0[/imath] then does this mean that [imath]\lim_{x\to \infty}\left(\frac{f(x)}{x}\right)=0[/imath] ? |
652341 | Count [imath] \lim_{n\rightarrow\infty}\frac{n!}{n^n}[/imath]
How to count this limit: [imath]\lim_{n\rightarrow\infty}\dfrac{n!}{n^n}[/imath] ? I have no idea how to even start. I will be glad for any tips or help. | 579825 | Compute the limit [imath]\lim_{n \to \infty} \frac{n!}{n^n}[/imath]
I am trying to calculate the following limit without Stirling's relation. \begin{equation} \lim_{n \to \infty} \dfrac{n!}{n^n} \end{equation} I tried every trick I know but nothing works. Thank you very much. |
652143 | Finding integer cubes that are [imath]2[/imath] greater than a square, [imath]x^3 = y^2 + 2[/imath]
I was given an example of a cube that is [imath]2[/imath] greater than a square number. The pair: [imath]27[/imath] and [imath]25[/imath]. What's the best way to find further pairs ? | 571473 | Solving the diophantine equation [imath]y^{2}=x^{3}-2[/imath]
It is known that the diophantine equation [imath]y^{2}=x^{3}-2[/imath] has only one positive integer solution [imath](x,y)=(3,5)[/imath]. The proof of it can be read from the book "About Indeterminate Equation" (in Chinese, by Ke Zhao and Sun Qi). But the method I have known of solving this problem is algebraic number theory. My question: Does there exists an elementary method for solving the diophantine equation [imath]y^{2}=x^{3}-2[/imath]? I have gotten an elementary method by the assumption that "[imath]x[/imath]" is a prime. That is, we can get a conclusion from the equation that [imath]3\mid x[/imath]. Since [imath]x[/imath] is a prime we get the only solution [imath](x,y)=(3,5)[/imath]. But if [imath]x[/imath] is not a prime, the answer seems to be difficult. Thank you for your help. |
652388 | Integral [imath]\int_{-1}^{1}\frac{\sqrt{(1-x^{2})}}{1+x^{2}}dx[/imath]
Consider [imath]\int_{-1}^{1}\frac{\sqrt{(1-x^{2})}}{1+x^{2}}dx[/imath] I have a problem with this integral; the method I know consists in calculating the complex integral of [imath]f(z) = \left( \frac{z-1}{z+1} \right)^{\frac{1}{2}} \frac{1+z}{1+z^{2}}[/imath] along the curve formed by the " shrinking dogbone contour" withe centres [imath]\{-1,1\}[/imath] and the circumference of radius [imath]R \rightarrow \infty[/imath]. But I obtain [imath]0[/imath], impossible. What's wrong with this method ? In particular, what are the residues of [imath]f(z) [/imath] in [imath]\{-i,i\} ? [/imath] | 428432 | Evaluating the integral [imath]\int_{-1}^{1} \frac{\sqrt{1-x^{2}}}{1+x^{2}} \, dx[/imath] using a dumbbell-shaped contour
I'm trying to evaluate the integral [imath] I =\int_{-1}^{1}\frac{\sqrt{1-x^{2}}}{1+x^{2}} \, dx [/imath] by using a dumbbell/dogbone contour, but I'm having difficulty determining the residue at infinity. I started by defining [imath]\sqrt{1-z^{2}}=\sqrt{(1-z)(1+z)} [/imath] so that it is a well-defined function if the line segment [imath][-1,1][/imath] is omitted. Similar to an example on Wikipedia, I choose the branches where [imath]0 < \arg(1-z) \le 2 \pi[/imath] and [imath]-\pi < \arg(1+z) \le \pi[/imath]. Then I integrated [imath] f(z) = \frac{\sqrt{1-z^{2}}}{1+z^{2}}[/imath] clockwise around a dumbbell contour. Just above the branch cut, [imath]\arg(1-z) = 2 \pi[/imath] and [imath]\arg(1+z) = 0[/imath]. And just below the branch cut, [imath]\arg(1-z) = 0[/imath] and [imath]\arg(1+z) = 0[/imath]. So the integral evaluates to [imath]-I[/imath] both above the cut and below the cut. Since the integrand is meromorphic outside the contour, I get [imath] - 2I = 2\pi i \left( \operatorname{Res}[f,i]+ \operatorname{Res}[f,-i] + \operatorname{Res}[f,\infty] \right),[/imath] where [imath]\begin{align} \operatorname{Res}[f,i] &=\lim_{z\to i }\frac{\sqrt{|1-z|e^{i\arg(1-z)}\ |1+z|e^{i\arg(1+z)}}}{z+i} =\frac{\sqrt{\sqrt{2}e^{\frac{7\pi i}{4}}\sqrt{2}e^{\frac{\pi i}{4}}}}{2i} \\ &=\frac{\sqrt{2}e^{\pi i}}{2i}=-\frac{\sqrt{2}}{2i}, \end{align}[/imath] [imath] \begin{align}\operatorname{Res}[f,-i] &=\lim_{z\to-i }\frac{\sqrt{|1-z|e^{i\arg(1-z)}\ |1+z|e^{i\arg(1+z)}}}{z-i}=\frac{\sqrt{\sqrt{2}e^{\frac{\pi i}{4}}\sqrt{2}e^{\frac{-\pi i}{4}}}}{-2i} \\ &=\frac{\sqrt{2}e^{ 0\pi i}}{-2i}=-\frac{\sqrt{2}}{2i}, \end{align}[/imath] and [imath] \begin{align}\operatorname{Res}[f,\infty] &=\operatorname{Res}\left[-\frac{1}{z^{2}}f\left(\frac{1}{z}\right), 0 \right]= \operatorname{Res}\left[-\frac{\sqrt{z^2-1}}{z(1+z^{2})},0\right]\\ &=-\lim_{z\to 0}\frac{\sqrt{z^2-1}}{1+z^{2}} = -(-1)^{\frac{1}{2}}. \end{align}[/imath] But what is [imath](-1)^{\frac{1}{2}}[/imath]? I assume it must be [imath]i[/imath], but I don't know how to argue that it can't be [imath]-i[/imath]. EDIT: There is also the issue that Ted Shifrin mentioned of assuming that [imath]\sqrt{z^{2}}=z[/imath]. |
653527 | Subgroups of index 3 are normal
[imath]G[/imath] has no subgroup with index 2. How can I show that all subgroups with index 3 are normal? (The index of the subgroup is the order of the quotient set.) | 2512896 | Proof verification on Group theory.
[imath]PROBLEM[/imath] :If [imath]G[/imath] is finite of odd order. And [imath]H[/imath] subgroup of [imath] G[/imath] with [imath][G:H]=3[/imath] then [imath]H[/imath] is normal. [imath]Proof[/imath]: I know that that there exists a homomorphism [imath]\phi : G \rightarrow S_3 [/imath] Such that [imath]ker(\phi)<H[/imath].I know that cause of i found it in the notes calling it Poincare's Trick.But i havent found a proof.(Id love to see a proof). Now that granted. I know [imath]\phi(G)<S_3[/imath] and since [imath]\phi[/imath] not an isomorphism since [imath]|G|=odd[/imath] and [imath]|S_3|=even[/imath]. So [imath]|\phi(G)|\mid |S_3|[/imath] so [imath]|\phi(G)|=2[/imath] or [imath]3[/imath]. Now i know [imath]G/ker\phi \simeq \phi(G) [/imath] meaning [imath]|G/ker\phi |= |\phi(G)| [/imath] And that must be an [imath]odd[/imath] since it is the [imath]index[/imath] of [imath]ker\phi [/imath] in [imath]G[/imath]. So [imath]|\phi(G)|=|G/ker\phi|=3[/imath] Now i know [imath]|G|=[G:H]|H|[/imath] and [imath]|G|=[G:Ker\phi]|Ker\phi|[/imath] substituting with [imath]3[/imath]. I get [imath]|H|=|Ker\phi| [/imath] and since [imath]Ker\phi < H[/imath] [imath]\Rightarrow[/imath] [imath]Ker\phi=H[/imath] .And i know [imath]Ker\phi \lhd G[/imath] So its [imath]H[/imath] too. Am i wrong?Afraid of some sort of circular reasoning.Also id love to see a proof of why there is such homomorphism. Im asking if my proof is correct and it is not a duplicate of the same problem.Since Im not asking for a proof but for a proof verification and a proof of a Trick mentioned!! |
653660 | Prove [imath]\{x \mid x^2 \equiv 1 \pmod p\}=\{1, -1\}[/imath] for all primes [imath]p[/imath]
One way to prove that [imath]\{x \mid x^2 \equiv 1 \pmod p\}=\{1, -1\}[/imath] is to use the fact that [imath]\{1, -1\}[/imath] is the only subgroup of the cyclic group of primitive residue classes modulo [imath]p[/imath] that has the order 2. Is there a more basic way to prove this? | 484165 | Number theory lemma - [imath]x^2 \equiv 1 \pmod p[/imath] implies [imath]x\equiv\pm 1\pmod p[/imath]
The proof is a follows: Let [imath]p[/imath] be a prime number and [imath]x[/imath] in Z. If [imath]x^2\equiv1\pmod p[/imath] then [imath]x\equiv\pm1\pmod p[/imath]. Proof: By assumption, [imath]p | x^2 - 1 = (x + 1)(x - 1)[/imath]. Thus [imath]p | x + 1[/imath] or [imath]p | x - 1[/imath]. This completes the proof. However if [imath]p | x + 1[/imath] then [imath]x[/imath] is congruent to [imath]-1 \mod p[/imath]. However why is [imath]p[/imath] congruent to [imath]1 \mod p[/imath] ?? I'm struggling seeing this, and been sitting for some hours now trying to figure it out - please help. |
654057 | a question about entire functions
Let [imath]f,g[/imath] be entire functions on [imath]\mathbb{C}[/imath]. If [imath]|f(z)|\leq |g(z)|[/imath] for all [imath]z\in\mathbb{C}[/imath], then prove there exists constant [imath]c[/imath] such that [imath]f=cg[/imath]. I have no idea... I intend to apply Liuville Theorem but this requires [imath]f/g[/imath] to be an entire function. Is there any version of improved Liuville Theorem? | 228514 | Entire function, Liouville and zeroes
Suppose [imath]f:\mathbb{C}\rightarrow\mathbb{C}[/imath] is an entire function. Let [imath]g:\mathbb{C}\rightarrow\mathbb{C}[/imath] be an entire function, which has no zeros. I have shown that [imath]\vert f(z) \vert \leq \vert g(z) \vert[/imath] for all [imath]z\in\mathbb{C}[/imath] implies [imath]f(z)=Cg(z)[/imath] for some constant [imath]C\in\mathbb{C}.[/imath] I have to decide if this also holds if [imath]g[/imath] is allowed to have zeros. Help? |
654288 | Let V be a complex vector space. If [imath]\langle T(v),v\rangle\in \mathbb R[/imath] for every [imath]v \in V[/imath], then is T self adjoint?
Let V be a complex vector space. If [imath]\langle T(v),v\rangle\in \mathbb R[/imath] for every [imath]v \in V[/imath], then is T self adjoint? I know how to prove reverse claim. Not sure about this one. Thanks! | 318440 | If [imath]\langle Ta, a\rangle \in \mathbb{R}[/imath] for all [imath]a[/imath] then [imath]T[/imath] is self-adjoint
I have having trouble with the following question: Let [imath]V[/imath] be a finite-dimensional complex inner product space, and let [imath]T[/imath] be a linear operator on [imath]V[/imath]. Prove that if [imath]\langle T\alpha, \alpha\rangle[/imath] is real for every [imath]\alpha[/imath] in [imath]V[/imath] then [imath]T[/imath] is self-adjoint. I have been focusing on trying to show that if [imath]\langle T\alpha, \alpha\rangle = \langle T^*\alpha, \alpha\rangle[/imath] for all [imath]\alpha[/imath] then [imath]T = T^*[/imath], but have been unable to do so. I also feel like I'm missing something really obvious. How should I proceed. |
580393 | Limit [imath]\lim_{n\to \infty} \frac{1}{2n} \log{2n\choose n}[/imath]
[imath]\lim_{n\to \infty} \frac{1}{2n} \log{2n\choose n}[/imath] I could not approach it beyond these simple steps, [imath]\lim_{n\to \infty} \frac{1}{2n} \log(\frac{2n!}{(n!)^2})[/imath] [imath]=\lim_{n\to \infty} \frac{1}{2n} [\log(2n)+\cdots +\log(n+1)-\log(n)-\cdots-\log1][/imath] [imath]=\lim_{n\to \infty} (\log(2n)^{1/2n}+\cdots+\log(n+1)^{1/2n}-\log(n)^{1/2n}-\cdots-\log1^{1/2n})[/imath] Now,I understand that I have to create a sum of limit and produce an integration or use the formula [imath]\lim_{n\to \infty} \log(1+\frac1x)^x=e[/imath] but I cannot do it. Please help! | 541232 | evaluation of [imath]\lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n}[/imath]
[imath]\displaystyle \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n} = [/imath] [imath]\underline{\bf{My\;\;Try}}::[/imath] Let [imath]\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \binom{2n}{n} = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \ln \left(\frac{(n+1)\cdot (n+2)\cdot (n+3)\cdot............\cdot (n+n)}{(1)\cdot (2)\cdot (3).........\cdot (n)}\right)[/imath] [imath]\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \left\{\ln \left(\frac{n+1}{1}\right)+\ln \left(\frac{n+2}{2}\right)+\ln \left(\frac{n+3}{3}\right)+..............+\ln \left(\frac{n+n}{n}\right)\right\}[/imath] [imath]\displaystyle y = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \sum_{r=1}^{n}\ln \left(\frac{n+r}{r}\right) = \lim_{n\rightarrow \infty}\frac{1}{2n}\cdot \sum_{r=1}^{n}\ln \left(\frac{1+\frac{r}{n}}{\frac{r}{n}}\right)[/imath] Now Using Reinman Sum [imath]\displaystyle y = \frac{1}{2}\int_{0}^{1}\ln \left(\frac{x+1}{x}\right)dx = \frac{1}{2}\int_{0}^{1}\ln (x+1)dx-\frac{1}{2}\int_{0}^{1}\ln (x)dx = \ln (2)[/imath] My Question is , Is there is any method other then that like Striling Approximation OR Stolz–Cesàro theorem OR Ratio Test If yes then please explain here Thanks |
654278 | Find the value of a and b if [imath]( x + 1 )[/imath] and [imath]( x -2 )[/imath] are factors of [imath] ax^3 - 4x^2 + bx - 12[/imath]
Find the value of [imath]a[/imath] and [imath]b[/imath] if [imath]( x + 1 )[/imath] and [imath]( x -2 )[/imath] are factors of [imath]ax^3 - 4x^2 + bx - 12[/imath]. This is regarding polynomials. Thank you | 654316 | Find the value of [imath]a[/imath] and [imath]b[/imath] if [imath]x+1[/imath] and [imath]x−2[/imath] are factors of [imath]ax^3−4x^2+bx−12[/imath]
Find the value of [imath]a[/imath] and [imath]b[/imath] if [imath]x+1[/imath] and [imath]x−2[/imath] are factors of [imath]ax^3−4x^2+bx−12[/imath] Attemption [imath] \begin{cases} f(-1) = -a - b - 16\\ f(2) = 8a + 2b - 28 \end{cases} [/imath] But then when I plug in [imath]b = 28 - 8a[/imath]. Then I get [imath]-a -(28- 8a) - 16=0[/imath][imath]a = 34/7[/imath] In the book it says [imath]a = 10[/imath] and [imath]b = -26[/imath]. So I am stumped. Please help |
654614 | An Euler-Mascheroni-like sequence
How does one compute the limit of the sequence: [imath]\sum_{k = 0}^{n}\frac{1}{3k+1} - \frac{\ln(n)}{3}[/imath] I would apreciate a hint. | 615695 | How to calculate the following sum:
How may one calculate [imath]\lim_{n\to\infty} \ \left(\left(\sum_{k=1}^{n} \frac{1}{3k-1}\right) - \frac{\ln n}{3}\right) \ ?[/imath] |
484813 | Inductive proof of the closed formula for the Fibonacci sequence
Fibonacci numbers [imath]f(n)[/imath] are defined recursively: [imath]f(n) = f(n-1) +f(n-2)[/imath] for [imath]n > 2[/imath] and [imath]f(1) = 1[/imath], [imath]f(2) = 1[/imath]. They also admit a simple closed form: [imath]\sqrt 5 f( n ) = \left(\dfrac{1+ \sqrt 5}{2}\right)^n- \left(\dfrac{1 - \sqrt 5}{2}\right)^n \tag1[/imath] How to prove (1) using induction? Remarks One could get (1) by the general method of solving recurrences: look for solutions of the form [imath]f(n)=r^n[/imath], then fit them to the initial values. But there should be a more concrete proof for this specific sequence, using the principle of mathematical induction. | 701823 | Proof by induction of a Fibonacci relation
We know: [imath]F_0 = 0[/imath] [imath]F_1 = 1[/imath] [imath]F_i = F_{i-1} + F_{i-2}[/imath] for [imath]i \geq 2[/imath] Prove by induction: [imath]F_i = \dfrac{\phi^i-{\phi^{*}}^i}{\sqrt{5}}[/imath] where [imath]\phi = (1+\sqrt{5}) / 2[/imath] and [imath]\phi = (1-\sqrt{5}) / 2[/imath]. My attempt: Base case: [imath]i = 0[/imath]: [imath]F_0 = 0[/imath] (easy to show) Assume true for [imath]i = k[/imath] and [imath]i = k + 1[/imath]. We have: [imath]F_k = \dfrac{\phi^k-{\phi^{*}}^k}{\sqrt{5}}[/imath] and [imath]F_{k+1} = \dfrac{\phi^{k+1}-{\phi^{*}}^{k+1}}{\sqrt{5}}[/imath] Show it holds for [imath]i = k + 2[/imath] to complete induction. This is where I'm stuck. I have tried: [imath]F_{k+2} = F_{k+1} + F_{k} = \dfrac{\phi^{k+1}-{\phi^{*}}^{k+1}}{\sqrt{5}} + \dfrac{\phi^k-{\phi^{*}}^k}{\sqrt{5}}[/imath] [imath]F_{k+2} = \dfrac{\phi^k(1+\phi)-{\phi^{*}}^{k}(1+{\phi^*})}{\sqrt{5}}[/imath] but I'm not sure where to go from here. Any help would be appreciated. |
655480 | Surjection Equivalence
I'm trying to show that if a function [imath]f:X\to Y[/imath] is surjective it's equivalent to saying that [imath]f \left(f^{-1}(B)\right) = B[/imath] for each [imath]B \subseteq Y[/imath]. The definition of surjective that I'm using is that [imath]f[/imath] is a surjective function from [imath]X[/imath] to [imath]Y[/imath] with [imath]f(X)=Y[/imath]. | 592793 | prove that [imath]f:X\rightarrow Y[/imath] is surjective if and only if [imath]f(f^{-1}(C))=C[/imath]
I need help with proving this: [imath]f:X\rightarrow Y[/imath] is surjective if and only if [imath]f(f^{-1}(C))=C[/imath] [imath]C\subseteq Y[/imath] Thanks. |
653507 | Is the space [imath]\mathbb N^ \mathbb N[/imath] metrisable?
Given the space [imath]\mathbb N^ \mathbb N[/imath] with the topology generated by basis sets of the form: [imath][V,n] = \{x \in \mathbb N^ \mathbb N ; V \text{ is an n prefix of x}\}[/imath] I can see that this space is separable. My question is: is it metrisable? or even completely metrisable? Thank you! | 250966 | Proof that [imath]\omega^\omega[/imath] is completely metrizable and second countable
I have almost solved the following problem but am stuck at the very end, can you help me finish it? Thank you for your help. Let [imath]n<\omega[/imath] and [imath]t\in {}^n\omega[/imath]. We define [imath]U_t=\{s\in {}^\omega\omega : t\subseteq s\}[/imath]. The family [imath]\mathcal B=\{U_t : t\in\bigcup {}^n\omega\}[/imath] is a basis for a topology [imath]\tau[/imath] on [imath]{}^\omega\omega[/imath]. This means that a set [imath]X[/imath] is in [imath]\tau[/imath] if and only if [imath]X[/imath] is a union of elements of [imath]\mathcal B[/imath]. EXERCISE 50 (PG): Show that [imath]\langle {}^\omega\omega,\tau \rangle[/imath] is a second countable completely metrizable space. Hint: For [imath]r,s\in {}^\omega\omega[/imath] such that [imath]r\ne s[/imath], the [imath]\varrho(r,s)=1/\min\{n : r(n)\ne s(n)\}[/imath]. Moreover, let [imath]\varrho(r,r)=0[/imath]. Show that [imath]\varrho[/imath] is a complete metric on [imath]{}^\omega\omega[/imath] that induces the topology [imath]\tau[/imath]. (i) Second countable: the set of [imath]n[/imath]-tuples is countable since it is a finite union of countable sets. (ii) complete with respect to [imath] d(f,g) = \frac{1}{\min \{n \mid f(n) \neq g(n) \}}[/imath] Let [imath]f_k[/imath] be a Cauchy sequence. This means that the index at which [imath]f_k, f_i[/imath] differ gets larger as [imath]i,k[/imath] get larger. Let [imath]f(n)[/imath] denote the pointwise limit, that is, [imath]\lim_{k \to \infty} f_k(n)[/imath]. Then [imath]f(n) \in \omega[/imath] for every [imath]n[/imath] since [imath]f_k(n)[/imath] is eventually constant. Hence [imath]f \in \omega^\omega[/imath]. (iii) [imath]d[/imath] induces the same topolgy as [imath]U_t[/imath]: [imath]\tau_t \subset \tau_d[/imath]: Let [imath]s \in U_t[/imath]. Then [imath]B(s, \frac{1}{n+1})[/imath] is an open ball contained in [imath]U_t[/imath] hence [imath]U_t[/imath] is open in [imath]\tau_d[/imath]. [imath]\tau_t \supset \tau_d[/imath]: Let [imath]B(s, \frac{1}{k})[/imath] be an open balls. If [imath]k < n[/imath], [imath]s \in U_s \subset B(s, \frac{1}{k})[/imath]. But if [imath]k \ge n[/imath] then I'm stuck. The ball [imath]B(s, \frac{1}{k})[/imath] consists of all sequences that agree with [imath]s[/imath] on the first [imath]k[/imath] coordinates. So it is quite small. How do I make the [imath]U_t[/imath] small enough to fit in this ball? I have thought of intersection but since I'm only allowed to do finite intersections I don't see what to do. |
655872 | Why does [imath]\int_1^\infty \sin(x \log x)\,\mathrm{d}x[/imath] converge?
Why does [imath]\int_1^\infty \sin(x\log x)\,dx[/imath] converge? I haven't got a clue about where to being with this, any help would be appreciated. | 651951 | Does the integral [imath]\int_{1}^{\infty} \sin(x\log x) \,\mathrm{d}x[/imath] converge?
I tried a couple of substitutions but have so far gotten nowhere. Can anyone guide me in the right direction? Thanks for your time. |
656114 | Show that [imath]2^n < n![/imath] for every positive integer [imath]n[/imath] with [imath]n\geq 4[/imath].
Using Mathematical induction prove the above proposition. Basis step can be verified easily. But how can i show that it is true for [imath]p(n+1)[/imath]. | 409609 | How to prove [imath]4(n!)>2^{n+2}[/imath] for [imath] n\geq 4[/imath] with induction
I've done the base step, but how do I prove it is true for [imath]n+1[/imath] without using a fallacy? [imath]4(n!)>2^{n+2}\quad \text{for } n\geq 4[/imath] Please help. |
247335 | Prove, formally that: [imath]\log_2 n! \ge n[/imath] , for all integers [imath]n>3[/imath].
Prove, formally that: [imath]\log_2 n! \ge n[/imath] for all integers [imath]n>3[/imath]. Hint: first prove that [imath]n! ≥2^n[/imath], for all integers [imath]n >3[/imath]. So far what I have: Base case, [imath]n = 4[/imath], [imath]4! = 24[/imath] [imath]2^4 = 16[/imath]. Therefore, it is true when [imath]n = 4[/imath]. So how do I proceed from here? I know I can log the whole equation which will lead to [imath]\log_2 n![/imath] and [imath]n[/imath]. But the linkage seems to be missing. Sorry for the untidiness about the math symbols. I am trying to input them correctly but apparently I am not doing it right. | 475891 | Proof by induction that [imath]n!\gt 2^{n}[/imath] for [imath]n \geq 4[/imath]
[imath]n!\gt 2^{n}[/imath] for [imath]n \geq 4[/imath]. Then there is an inductive proof: We assume that [imath]n! \gt 2^{n}[/imath] for every random [imath]n \in \mathbb{N}[/imath] with [imath]n\geq4[/imath], and we need to proof that [imath](n+1)!\gt 2^{n+1}[/imath]. Because [imath](n+1)\gt0[/imath] and [imath]n!\gt2^{n}[/imath], it follows that [imath](n+1)n!\gt(n+1)2^{n}[/imath]. In the left part of the equation we have [imath](n+1)![/imath], but in the right part we don't have [imath]2^{n+1}[/imath]. And here comes the part I don't understand: We want to replace[imath](n+1)2^{n}[/imath] with [imath]2^{n+1}[/imath], or [imath]n+1[/imath] by [imath]2[/imath]. We can do this without any problems, because it is sure that for [imath]n\geq4[/imath] that [imath]n+1\gt2[/imath], and thus that [imath](n+1)\cdot2^{n}\gt2\cdot2^{n}[/imath] How much I try, I just don't see the connection. So you multiply [imath]n+1\gt2[/imath] by [imath]2^{n+1}[/imath] on both sides, and that makes [imath](n+1)n!\gt(n+1)2^{n}[/imath] into [imath](n+1)!\gt2^{n+1}[/imath]? I don't see how you can get from [imath](n+1)2^{n}[/imath] to [imath]2^{n+1}[/imath] with what they're saying. To me, it seems like they're just saying "Hey, because [imath]n+1\gt2[/imath] is greater than [imath]2[/imath], we can just replace it with [imath]2^{n+1}[/imath]". |
656090 | How to evaluate the integral[imath] \frac{\log x}{1+x^2}[/imath]
How might I evaluate this integral: [imath]\int_0^\infty \frac{\log\ x}{1+x^2}dx[/imath] I tried substituing [imath]x = \tan\ v[/imath] but that didn't get me anywhere. Any suggestions as to a thought process for this one? | 652722 | Fast calculation for [imath]\int_{0}^{\infty}\frac{\log x}{x^2+1}dx=0[/imath]
I want to show that [imath]\int_{0}^{\infty}\frac{\log x}{x^2+1}dx=0[/imath], but is there a faster method than finding the contour and doing all computations? Otherwise my idea is to do the substitution [imath]x=e^t[/imath], integral than changes to [imath]\int _{-\infty}^{\infty}\frac{t e^t}{1+e^{2t}}dt[/imath]. Next step is to take the contour [imath]-r,r,r+i\pi,-r+i\pi[/imath] and integrate over it... |
656179 | Given f(x), a continous function on R such that [imath]\lim_{x\to \infty} f(x) = L[/imath], what is [imath]\int_0^\infty \frac{f(x)-f(2x)}{x}dx[/imath]?
I'm sure I'm missing something. Can anyone guide me in the right direction? Thanks for your time. | 646453 | A question related to a convergence of an integral
Let [imath]f[/imath] be [imath]f\in C([0,\infty ])[/imath], such that [imath]\lim_{x \to \infty} f(x) = L [/imath]. Calculate [imath] \int _{0}^\infty \frac{f(x)-f(2x)}{x}dx [/imath] Help? |
656198 | why the square root of x equals x to the one half power
Could someone explain how/why the square root of [imath]x[/imath] equals [imath]x[/imath] to the one half power? I know by definition it does, but is there any mathematical process we can go through to get from one to the other? In addition, I know the [imath]x[/imath] to the 2ed power is equal to [imath]x*x[/imath], but what is [imath]x^{1/2}[/imath] equal to? Is it one half of [imath]x[/imath] times one half of [imath]x[/imath]? No! I know all this stuff by definition, but I want to know why. Thanks | 537383 | Why is [imath]{x^{\frac{1}{2}}}[/imath] the same as [imath]\sqrt x [/imath]?
Why is [imath]{x^{\frac{1}{2}}}[/imath] the same as [imath]\sqrt x [/imath]? I'm currently studying indices/exponents, and this is a law that I was told to accept without much proof or explanation, could someone explain the reasoning behind this. Thank you. |
30949 | Expressing the maximum of several variables using elementary functions
It's well-known that [imath]\max(a,b)=\frac{a+b+|a-b|}{2}.[/imath] Is there a (good) generalization to several variables? Of course [imath]\max(a,b,c)=\max(a,\max(b,c))[/imath] and so [imath]\max(a,b,c)=\frac{a+\frac{b+c+|b-c|}{2}+|a-\frac{b+c+|b-c|}{2}|}{2}[/imath] [imath]=\frac{a+0.5b+0.5c+0.5\left|b-c|+|a-0.5b-0.5c-0.5|b-c|\right|}{2}[/imath] but I'd like a form that shows the natural symmetry better and which doesn't have so many operations. This is a practical problem working on a system which has an absolute value operator but no maximum and not much ability to execute conditional statements, but to be honest the real reason I'm interedted is an attempt to beautify something that is seemingly ugly. For the practical side I need 5-10 arguments and it's acceptable to assume that all arguments are at least 0, though of course it would be much more satisfying if this latter assumption was not needed. | 13253 | Nice expression for minimum of three variables?
As we saw here, the minimum of two quantities can be written using elementary functions and the absolute value function. [imath]\min(a,b)=\frac{a+b}{2} - \frac{|a-b|}{2}[/imath] There's even a nice intuitive explanation to go along with this: If we go to the point half way between two numbers, then going down by half their difference will take us to the smaller one. So my question is: "Is there a similar formula for three numbers?" Obviously [imath]\min(a,\min(b,c))[/imath] will work, but this gives us the expression: [imath]\frac{a+\left(\frac{b+c}{2} - \frac{|b-c|}{2}\right)}{2} - \frac{\left|a-\left(\frac{b+c}{2} - \frac{|b-c|}{2}\right)\right|}{2},[/imath] which isn't intuitively the minimum of three numbers, and isn't even symmetrical in the variables, even though its output is. Is there some nicer way of expressing this function? |
656602 | How do you show the sum of [imath]1/(n\log n)[/imath] diverges?
How do you show the sum [imath]\sum_{n=2}^\infty \frac{1}{n\log n}[/imath] diverges? I have tried to use the ratio test but the outcome was inconclusive as the limit was equal to [imath]1[/imath]. | 417036 | Is $\sum_{n=3}^\infty\frac{1}{n\log n}$ absolutely convergent, conditionally convergent or divergent?
Classify [imath]\sum_{n=3}^\infty \frac{1}{n\log(n)}[/imath] as absolutely convergent, conditionally convergent or divergent. Is it, [imath]\sum_{n=3}^\infty \frac{1}n[/imath] is a divergent [imath]p[/imath]-series as [imath]p=1[/imath], and [imath]\lim_{n\to\infty} \frac{1}{n\log(n)}{n} = 0[/imath] by comparison test. And this converges to [imath]0[/imath]. So, [imath]\sum_{n=3}^\infty \frac{1}{n\log (n)}[/imath] is conditionally convergence? I'm not sure if I'm doing right or not. Could you guide me? Thanks in advance! :) |
656468 | Prove that between every two rational numbers a/b and c/d that there is a rational number and there are an infinite number of rational numbers
So the full problem is Prove that between every two rational numbers [imath]a/b[/imath] and [imath]c/d[/imath] that: There is a rational number There are an infinite number of rational numbers I am having some trouble on which approach I should use, but I tried to solve part 1. Part 1) So I believe I proved that between every two rational numbers [imath]a/b[/imath] and [imath]c/d[/imath] that there is a rational number I have my base case as [imath]a,b,c,d[/imath] all being integers. My solution: For some number [imath]x[/imath], [imath] x = \frac{\dfrac a b + \dfrac c d}{2} = \frac{ \dfrac{ad}{bd} + \dfrac{cb}{bd}} 2 = \frac{ \dfrac{ad + cb}{bd} } 2 = \frac{ad + cb}{2bd} [/imath] I concluded that ad + cb must be an integers because addition/subtraction of integers leads to an integer. Same case for [imath]2db[/imath]. Not sure if this is correct though Part 2) Prove that there are an infinite number of rational numbers I am confused onto how to solve this part, where to start, and how to use the answer from part 1 into this problem. I've been toiling over this, so your input is greatly appreciated. :) | 421580 | Is there a rational number between any two irrationals?
Suppose [imath]i_1[/imath] and [imath]i_2[/imath] are distinct irrational numbers with [imath]i_1 < i_2[/imath]. Is it necessarily the case that there is a rational number [imath]r[/imath] in the interval [imath][i_1, i_2][/imath]? How would you construct such a rational number? [I posted this only so that the useful answers at https://math.stackexchange.com/questions/414036/rationals-and-irrationals-on-the-real-number-line/414048#414048 could be merged here before that question was deleted.] |
656771 | Showing that [imath]1+a>0 \implies (1+a)^n \ge 1 + na[/imath]
Establish the Bernoulli Inequality if [imath]1+a>0[/imath], then [imath](1+a)^n\ge 1+na[/imath] for [imath]n\ge 1.[/imath] So far I have [imath](1+a)^k+1 \ge 1+(k+1)a,[/imath] but I don't know what to do from here to make the left and right side equal. Step by step explanation please! | 46562 | Prove that in an ordered field [imath](1+x)^n \ge 1 + nx + \frac{n(n-1)}{2}x^2[/imath] for [imath]x \ge 0[/imath]
In an ordered field show that [imath]x \geq 0 \implies (1+x)^{n} \geq 1+nx+ \frac{1}{2}n(n-1)x^2[/imath] for every positive integer [imath]n[/imath]. I know that [imath](1+x)^{n} \geq 1+nx[/imath] (Bernoulli's inequality). To get the stronger inequality you can probably use induction also. But is there an easier way? The extra term on the right seems to be an "error" term and looks like a Taylor series expansion. |
421750 | A problem on continuity of a function on irrationals for [imath]f(x) = \sum_{r_n \leq x} 1/n^2[/imath]
Let [imath]\langle r_n\rangle[/imath] be an enumeration of the set [imath]\mathbb Q[/imath] of rational numbers such that [imath]r_n \neq r_m\,[/imath] if [imath]\,n\neq m.[/imath] [imath]\text{Define}\; f: \mathbb R \to \mathbb R\;\text{by}\;\displaystyle f(x) = \sum_{r_n \leq x} 1/n^{2},\;x\in \mathbb R.[/imath] Prove that [imath]f[/imath] is continuous at each point of [imath]\mathbb Q^c[/imath] and discontinuous at each point of [imath]\mathbb Q[/imath]. I find it difficult to prove especially the continuity on irrationals, I proved the discontinuity on a rational number in the following way is it correct? Let [imath] c \in \Bbb Q [/imath] then [imath] c=r_n [/imath] for some [imath]n \in \Bbb N [/imath] and Let [imath] \epsilon_0 = 1/n^{2} [/imath] Let [imath]\delta > 0 [/imath] be arbitrary and let [imath] x \in (c-\delta. c+\delta)[/imath] such that [imath]x<c [/imath] Then [imath]|f(x)-f(c)|>1/n^{2}=\epsilon_0 [/imath] How do prove that it is continuous on irrationals? | 253529 | Real Analysis Question: continuity of [imath]f(x) = \sum_{q_n[/imath]
Sorry, I was having trouble giving the question because I can't figure out how to type in mathematical symbols. Let me try again: [imath]\{q\}[/imath] is an enumeration of the rational numbers, and [imath]f(x)=\sum_{q_n< x}\frac1{n^2}[/imath] for [imath]x\in\Bbb R[/imath]. Or [imath]f(x)=\sum(1/n^2)[/imath], the index of summation [imath]q_n< x[/imath], and [imath]q[/imath] is an enumeration of the rational numbers. The goal is to prove that [imath]f[/imath] is continuous at each irrational and discontinuous at each irrational. I'm having trouble visualizing this series so I'm not sure why it is supposed to be different at irrational as opposed to a rational [imath]x[/imath] ... I think I need to use the epislon-delta definition of continuity but I'm not sure what epsilon to use. |
657000 | Order of an entire [imath]f [/imath] is [imath]\limsup_{r \rightarrow + \infty} \frac{\log \log M(r)}{\log r}[/imath]
An entire function is of finite order [imath]\rho[/imath] if [imath]\rho = \inf \{\lambda \geq 0 \ | \ \exists A, B > 0 \ s.t. \ |f(z)|\leq Ae^{B|z|^{\lambda}} \forall z \in \mathbb{C} \}[/imath] Write [imath]M(r) = \sup_{|z|=r}|f(z)|[/imath]. Then we have [imath]\rho = \limsup_{r \rightarrow + \infty} \frac{\log \log M(r)}{\log r}[/imath] Any hint ? | 475408 | Understanding the definition of the order of an entire function in Ahlfors's Complex Analysis
Let [imath]f: \mathbb C \to \mathbb C[/imath] be an entire function. The order of [imath]f[/imath] is defined by [imath]\lambda=\limsup_{r \to \infty} \frac{\log \log M(r)}{\log r}, [/imath] where [imath]M(r)=\max_{|z|=r} |f(z)| .[/imath] Ahlfors in his Complex Analysis claims that "According to this definition [imath]\lambda[/imath] is the smallest number such that [imath]M(r)\leq e^{r^{\lambda+\varepsilon}} [/imath] for any given [imath]\varepsilon > 0[/imath] as soon as [imath]r[/imath] is sufficiently large." Why is this true? My attempt: We know that [imath]\lambda=\lim_{\rho \to \infty} \sup_{r \geq \rho} \frac{\log \log M(r)}{\log r}. [/imath] From the definition of the limit we have that for any [imath]\varepsilon>0[/imath], there exists some [imath]\rho_0>0[/imath], such that [imath]\left\lvert \sup_{r \geq \rho} \frac{\log \log M(r)}{\log r}-\lambda \right\rvert \leq \varepsilon ,[/imath] for every [imath]\rho \geq \rho_0[/imath]. In other words [imath]\frac{\log \log M(r)}{\log r} \leq \lambda+\varepsilon [/imath] for every [imath]r \geq \rho_0[/imath]. From here it is easy to see that [imath]M(r)\leq e^{r^{\lambda+\varepsilon}}, [/imath] for all [imath]r \geq \rho_0[/imath]. I cannot see why [imath]\lambda[/imath] is the smallest number with this property. Thanks in advance. |
657170 | Biholomorphism between 2 non simply connected domains
Is the annulus of centre [imath]0[/imath] and radii [imath]1[/imath] and [imath]2[/imath] biholomorphic to the punctured disc [imath]\{ z \in \mathbb{C} \ | \ 0<|z| <1\}[/imath] ? Why ? I know the Riemann mapping theorem but here we have non simply connected domains. | 547514 | Is my proof correct? (Conformal equivalence of two circular annuli)
I want to show that the two annuli [imath]A=\{r<|z-z_0|<R\} [/imath] [imath]A'=\{r'<|z-z_0'|<R'\} [/imath] are conformally equivalent (i.e. there exists a biholomorphic map between the two) iff [imath]\frac{R}{r}=\frac{R'}{r'}. [/imath] Sufficiency: Suppose the ratios of the radii are the same. A suitable scaling (by [imath]\frac{r'}{r}[/imath]) will make [imath]A[/imath] congruent to [imath]A'[/imath], and a translation will map the scaled [imath]A[/imath] onto [imath]A'[/imath]. This is clearly a conformal homeomorphism. Necessity: The annulus [imath]A[/imath] has the following set of periods [imath]\left\{ \pm \frac{2 \pi}{\log (R/r)} \right\} [/imath] (this step required some computations which I will omit) Similarly [imath]A'[/imath] has [imath]\left\{ \pm \frac{2 \pi}{ \log(R'/r')} \right\}[/imath] as its set of periods. Since the set of periods is a conformal invariant. The ratios of the radii must indeed by the same. Is this proof correct? Thanks! |
657273 | If cardinality of every minimal cut is even then graph is Eulerian.
So I have a connected graph [imath]G[/imath], whose every minimal edge cut contains even number of edges. How to prove from this that the graph is Eulerian? If I knew from this, that every vertex has even degree I would be fine, but how do I deduce this? EDIT: I realise this question has been asked before, but why is the degree of every vertex even? The original answer doesn't elaborate on that. | 632149 | Minimal cut edges number in connected Eulerian graph.
I have to prove that Connected graph is Eulerian Graph if and only if the number of edges in minimal cut is even. Implication from left to right. Well if [imath]G[/imath] is Eulerian Graph then degree of each vertex must be even. It's oblivious if i try to disconnect graph by removing all edges connected to the vertex with minimal degree. But minimal cut can be done also by dividing graph into 2 connected graphs (each graph with more then one vertex) and i have no idea how to deal with it. Implication from right to left - completly no idea. |
657597 | [imath]\mathbb{R}^*[/imath] not cyclic
I need to show that the multiplicative group [imath]\mathbb{R}^*[/imath] of non-zero real numbers is not cyclic. This is what I'm thinking. I need to show that there exists no [imath]a[/imath] such that all elements of [imath]\mathbb{R}^*[/imath] are of the form [imath]a^n[/imath] with [imath]n\in \mathbb{Z}[/imath]. But I'm not sure how to do this. Should I provide a counter-example by choosing two real numbers and showing that they cannot be of the form [imath]a^n[/imath] with the same base? | 204579 | Prove that [imath]\mathbb{R^*}[/imath], the set of all real numbers except [imath]0[/imath], is not a cyclic group
Prove that [imath]\mathbb{R^*}[/imath] is not a cyclic group. (Here [imath]\mathbb{R^*}[/imath] means all the elements of [imath]\mathbb{R}[/imath] except [imath]0[/imath].) I know from the definition of a cyclic group that a group is cyclic if it is generated by a single element. I was thinking of doing a proof by contradiction but then that ended up nowhere. |
657624 | Does the number [imath]2.3\,5\,7\,11\,13\ldots[/imath] exist and, if so, is it rational or irrational &/or transcendental?
Does there exist a number which contains in its digits all of the prime numbers listed in order: [imath]2.3\,5\,7\,11\,13\ldots\ldots[/imath] if so, will it be rational or irrational &/or transcendental? | 197505 | Is [imath]0.23571113171923293137\dots[/imath] transcendental?
Is the following number transcendental? [imath]0.23571113171923293137\dots[/imath](Obtained by writing prime numbers consecutively from left to right, in the decimal expansion) |
657909 | The set of all finite subsets of [imath]\mathbb{N}[/imath] is countable
Can someone explain this in the most straightforward manner possible? I've looked at one other thread on here, but the mathematics was unfortunately too confusing for me. Is there some way I can understand why this is true without getting too much into mathematical bijections? | 263677 | How many subsets of [imath]\mathbb{N}[/imath] have the same cardinality as [imath]\mathbb{N}[/imath]?
How many subsets of [imath]\mathbb{N}[/imath] have the same cardinality as [imath]\mathbb{N}[/imath]? I realize that any of the class of functions [imath]f:x\to (n\cdot x)[/imath] gives a bijection between [imath]\mathbb{N}[/imath] and the subset of [imath]\mathbb{N}[/imath] whose members equal multiples of n. So, we have at least a countable infinity of sets which have the same cardinality of [imath]\mathbb{N}[/imath]. But, we could remove a single element from any countably infinity subset of the natural numbers and we still will end up with a countably infinite subset of [imath]\mathbb{N}[/imath]. So (the reasoning here doesn't seem quite right to me), do there exist uncountably many infinite subsets of [imath]\mathbb{N}[/imath] with the same cardinality of [imath]\mathbb{N}[/imath]? Also, is the class of all bijections [imath]f: \mathbb{N} \to \mathbb{N}[/imath] and a given countably infinite subset uncountably infinite or countably infinite? |
654697 | The Frobenius Coin Problem
I am asked to prove that: For integers [imath]n, x,y > 0[/imath], where [imath]x,y[/imath] are relatively prime, every [imath]n \ge (x-1) (y-1)[/imath] can be expressed as [imath]xa + yb[/imath], with nonnegative integers [imath]a,b \ge0[/imath]. How should I approach this? I have very limited knowledge in number theory. | 1870843 | Chicken nuggets come in boxes of [imath]389[/imath] and [imath]691[/imath]. What is the largest number of nuggests one cannot order?
Chicken nuggets come in boxes of [imath]389[/imath] and [imath]691[/imath]. What is the largest number of nuggets one cannot order? |
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