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2107710
|
Show [imath]0< \frac{2x}{\pi} < \sin x[/imath]
How do I go about showing [imath]0 \leq \frac{2x}{\pi} \leq \sin x [/imath] for [imath]x\in[0,\pi/2][/imath]? I am completely stuck where to start. Many thanks. (I see it is a step in the proof of Jordan's lemma, but I'm not interested in this, and the proofs I find do not explain this actual step, ta).
|
980551
|
How to show [imath] \sin x \geq \frac{2x}{\pi}, x \in [0, \frac{\pi}{2}][/imath]?
I have tried the following: [imath] f(x) = \sin x-\frac{2x}{\pi} \\ f'(x)= \cos x-\frac{2}{\pi} \\ f''(x) = -\sin x \leq 0 [/imath] But this doesn't seem to be heading in the right direction as it would appear [imath]f'(x)[/imath] is decreasing and has a zero within the domain, so I'm not sure how this might be used to prove the inequality.
|
2108234
|
[imath]n[/imath] is a prime number [imath]\iff [/imath] [imath]\frac{Tn(x)}{x}[/imath] is irreducible on [imath]Q[x][/imath]
Let [imath]n[/imath] be an odd number. Prove that [imath]n[/imath] is a prime number [imath] \iff[/imath] [imath]\frac{T_n(x)}{x}[/imath] is irreducible on [imath]Q[x][/imath] We have that [imath] [/imath] [imath]T_n(x)[/imath] is Chebyshev polynominal please help me :((
|
109214
|
Chebyshev Polynomials and Primality Testing
Let [imath]T(n,x)[/imath] be the nth Chebyshev polynomial of the first kind and let [imath]U(n-1,x)[/imath] the [imath](n-1)[/imath]th Chebyshev polynomial of the second kind. Would any one kindly help show that 1) [imath]n[/imath] is prime iff [imath]T(n,x)[/imath] is irreducible in [imath]\mathbb{Z}[x][/imath]. 2) [imath]n[/imath] is prime iff [imath]U(n-1,x)[/imath], expressed in powers of [imath](x^2-1)[/imath], is irreducible in [imath]\mathbb{Z}[x][/imath]. Many Thanks!! No "ordering" is implied here. The wording of the question was done in similar fashion as any question in any math/research question. I do not see how a person who is asking for help would be ordering people to help. Any way back to the topic, For the first part of the question, I noticed that if n is prime, then T(n,x) satisfies Eisenstein's Irreducibility Criterion. But I am not sure how to show if T(n,x) is irreducible then n is prime.
|
945033
|
Helix around helix parametric equation?
I know the parametric equation for a [imath]3D[/imath] helix is: [imath]x = R \cos t[/imath] [imath]y = R \sin t[/imath] [imath]z = h t[/imath] Can somebody explain to me this parametric equation (image and equation from Wolfram) for a "Helix around helix" / Slinky: [imath]x = [R + a \cos(\omega t)] \cos t[/imath] [imath]y = [R + a \cos(\omega t)] \sin t[/imath] [imath]z = h t + a \sin(\omega t)[/imath]. I don't understand what are the variables '[imath]a[/imath]' and '[imath]\omega[/imath]' supposed to represent. I assume '[imath]h[/imath]' is the height? How could I expand those equations for an [imath]n[/imath]-number of helices? (say this is [imath]2[/imath] helices, what would change in the formula for [imath]3[/imath] helices?) I would like to be able to generate such structures in Solid Works. I can get the simple [imath]3D[/imath] helix from the equations, but if I try with the "Slinky" one it only makes strange (yet beautiful) shapes. Any help? I know this is way over my head, but the coil-made of a coil-made of a coil structure has obsessed me for quite some time now (I would like to sculpt it in real-life, but first model it in [imath]3D[/imath]) Thanks!
|
470538
|
Helix in a helix
I am trying to work out a "helix in a helix" mathematically. Intuitively I think of this as a steel cable, which is made up of a number of smaller steel cables all bound together in spiral. If I wanted to find the length of one of the individual cables, it would be bound in a spiral in the smaller cable and then those cables bound in a larger spiral cable. I know that if I wanted to do a helix whose ends meet, I would use the parametrization [imath]((a+b\cos(\omega{t}))\cos{t},(a+b\cos(\omega{t})\sin{t},b\sin(\omega{t})),t=0..2\pi[/imath] I've been trying to map out in my head how to, instead of curling the helix, making the helix travel in the path of a helix. I've achieved it partially with [imath]((a+b\cos(\omega{t}))\cos{t},(a+b\cos(\omega{t})\sin{t},t),t=0..\infty[/imath] But this doesn't keep the smaller helix in tact, and turns it into a sine wave helix. I've also tried [imath]((a+b\cos(\omega{t}))\cos{t},(a+b\cos(\omega{t})\sin{t},tb\sin(\omega{t})),t=0..2\pi[/imath] But this gives me sort of a nautilus shape where the helix curls into itself and increases in size and curls around into itself. What am I missing? EDIT: Also, what if we wanted to do this again, like a 'helix in a helix in a ... in a helix'?
|
2107995
|
test convergence: [imath]\sum \frac{1}{a_n}[/imath]
[imath]a_n [/imath]- sequence of all natural numbers which in notation doesn't use the digit 6 [imath]\sum \frac{1}{a_n}[/imath] I have already tried to compare [imath]\sum(\frac{1}{6}+\frac{1}{16}+\frac{1}{26}+...)\ge\sum \frac{1}{10n}\to \infty[/imath] So this gives me nothing.
|
583218
|
The series of reciprocals of the integers that do not contain 9 in their decimal representation
Does the following series converge or diverge? [imath]\sum_{n=1}^{\infty} a_n[/imath] where [imath]a_n = \frac 1 b_n[/imath], and [imath](b_n)_n[/imath] is the subsequence of [imath](n)_n[/imath] whose terms do not have a [imath]9[/imath] in their decimal notation. I have no idea how the sequence [imath]b_n[/imath] even looks like. If I have an idea of how to prove that a series converges or not with criteria but I don't if I don't know what [imath]b_n[/imath] is.
|
2107809
|
Solving the functional equation f(x+y)=f(x)+f(y)+y√f(x)
The question is to solve the functional equation [imath]f(x+y)=f(x)+f(y)+y\sqrt{f(x)}[/imath] [imath]\forall x,y \in \mathbb R[/imath] I tried to put x=y=0,y=x and y=-x in the given functional equation.I ended up getting [imath]f(2x)=3f(x)+f(-x)[/imath] and [imath]f(x)+f(-x)=x\sqrt{f(x})[/imath] from where I am unable to proceed.Thanks.
|
1762184
|
Solution of functional equation [imath]f(x+y)=f(x)+f(y)+y\sqrt{f(x)}[/imath]
If [imath]x,y\in \mathbb{R}[/imath] and [imath]f(x+y)=f(x)+f(y)+y\sqrt{f(x)}[/imath] and [imath]f'(0)=0\;,[/imath] Then [imath]f(x)[/imath] is [imath]\bf{My\; Try::}[/imath] Using [imath]f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\rightarrow 0}\frac{f(x)+f(h)+h\sqrt{f(x)}-f(x)}{h}[/imath] Now Put [imath]x=y=0[/imath] in [imath]f(x+y)=f(x)+f(y)+y\sqrt{f(x)}\;,[/imath] We get [imath]f(0)=0[/imath] So we get [imath]f(0)=0[/imath] So [imath]f'(x) = \sqrt{f(x)}+\lim_{h\rightarrow 0}\frac{f(h)}{h}=\sqrt{f(x)}[/imath] So [imath]\int\frac{f'(x)}{\sqrt{f(x)}}dx = 1\int dx\Rightarrow 2\sqrt{f(x)}=x+c[/imath] Now Put [imath]x=\;,[/imath] We get [imath]c=0[/imath] So we get [imath]2\sqrt{f(x)}=x\Rightarrow 4f(x)=x^2\Rightarrow \displaystyle f(x)=\frac{x^2}{4}[/imath] Can we solve it some short way, If yes then please explain here, Thanks
|
2108036
|
Big O and equality
On this problem, I'm not sure what Big O definition they are referring. How would the big o definition help show this? I get that [imath]|y-y_h|[/imath]< [imath]M\beta (h)[/imath], where [imath]y_h[/imath] is an approximate value to y at some point h, but don't really understand the question Use the definition of [imath]O[/imath] to show that if [imath]y = y_h + O(h^p)[/imath], then [imath]hy = hy_h + O(h^{p+1})[/imath].
|
2107971
|
Big O definition and showing equality
On this problem, I'm not sure what Big O definition they are referring. How would the big o definition help show this? Use the definition of [imath]O[/imath] to show that if [imath]y = y_h + O(h^p)[/imath], then [imath]hy = hy_h + O(h^{p+1})[/imath].
|
2108426
|
A simple algebra that is not semisimple
I was said that can exists simple algebras [imath]A[/imath] that are not semi-simples in the following sense: [imath]A[/imath] is simple, i.e. doesn't have non trivial ideals; [imath]A[/imath] is not semi-simple, i.e. is not a semi-simple module over itself. Does anybody has an example? Thanks in advance
|
1340293
|
why is a simple ring not semisimple?
A simple module is a semisimple module . A module [imath]M[/imath] is called semisimple if every submodule is a direct summand of [imath]M[/imath] Since a simple module has [imath]\{0\}[/imath] and [imath]M[/imath] as its submodules so it is semisimple. But why is a simple ring not semisimple because a A ring is semisimple if it is semisimple as a module over itself How to construct such a ring?When will the result hold?
|
2108070
|
Would Legendre's Conjecture imply Riemann's Hypothesis?
Riemann's hypothesis implies that the gap [imath]g_n[/imath] between primes is [imath]g_n=O(\sqrt{p_n}\log{p_n})[/imath] And Legendre's conjecture is equivalent to [imath]g_n=O(\sqrt{p_n})[/imath] Then, would proving the second conjecture imply the Riemann's Hypothesis to be true?
|
482726
|
Would proof of Legendre's conjecture also prove Riemann's hypothesis?
Legendre's conjecture is that there exists a prime number between [imath]n^2[/imath] and [imath](n+1)^2[/imath]. This has been shown to be very likely using computers, but this is merely a heuristic. I have read that if this conjecture is true, the biggest gap between two consecutive primes is [imath]O(\sqrt{p})[/imath]; the Riemann Hypothesis, on the other hand, implies that this gap is [imath]O(\sqrt{p}\log{p})[/imath], which is a wider gap for sufficiently large inputs. This leaves me asking the following question, which I am aware may be a bit naive, but I want to be sure: Would a proof of Legendre's conjecture also be considered a proof of Riemann's hypothesis? EDIT: I am aware that the way the asymptotic above were expressed was in terms of prime numbers, and that the Riemann Hypothesis is concerned about everything in between as well; would a proof of this sort need to show this upper bound for all inputs, or would the prime numbers be sufficient? EDIT 2: It seems to me that, if a proof of Legendre's conjecture could result in a proof of the Riemann hypothesis, that this would come from the [imath]\log{x}[/imath] term of the asymptotic, showing that this term is never smaller than what results from the distance between primes. In other words, this would have to be an inductive proof, showing an initial case and, as a result of that initial case and the fact that the gap is [imath]O(\sqrt{p})[/imath] for that case, all future cases must therefore be [imath]O(\sqrt{p} \log{p})[/imath]. I hate to add to my question once again, but please tell me if this is completely off.
|
2087705
|
Evaluating [imath]\sum^{n-1}_{j=0} \binom{n}{j}\binom{n}{j+1}[/imath]
Is there a simpler expression for the following sum? [imath](n\in\mathbb{N})[/imath] [imath]S_n =\binom{n}{0}\binom{n}{1} + \binom{n}{1}\binom{n}{2} + \dots + \binom{n}{n-1}\binom{n}{n}[/imath] It seems like [imath]S_n = \binom{2n}{n-1}[/imath], however I have no clue as to how I can prove that relation. I also tried re-writing the sum as [imath]S_n = \binom{n}{0}\binom{n}{n-1} + \binom{n}{1}\binom{n}{n-2} + \dots + \binom{n}{n-1}\binom{n}{0} = \sum^{n-1}_{j=0}\binom{n}{j}\binom{n}{n-j-1}[/imath] Which resembles a special case of Vandemonde's Identity. Is there a connection between the two?
|
351363
|
Find a simple formula for [imath]\binom{n}{0}\binom{n}{1}+\binom{n}{1}\binom{n}{2}+...+\binom{n}{n-1}\binom{n}{n}[/imath]
[imath]\binom{n}{0}\binom{n}{1}+\binom{n}{1}\binom{n}{2}+...+\binom{n}{n-1}\binom{n}{n}[/imath] All I could think of so far is to turn this expression into a sum. But that does not necessarily simplify the expression. Please, I need your help.
|
2108463
|
determinant of 4x4-matrix occuring in Zarhin's trick
What's the easiest/fastest way of calculating the determinant of [imath]\begin{pmatrix}a & b & c & d\\ -b & a & -d & c\\ -c & d & a & -b \\ -d & -c & b & a\end{pmatrix}[/imath]? The result is [imath](a^2+b^2+c^2+d^2)^2[/imath]. This determinant occurs in Zarhin's trick (if [imath]A[/imath] is an Abelian variety, [imath](A \times A^t)^4[/imath] is principally polarised).
|
706513
|
Calculating the determinant gives [imath](a^2+b^2+c^2+d^2)^2[/imath]?
I need to calculate the following determinant in order to prove the following equality: [imath]\det\begin{pmatrix} a & b & c & d \\ -b & a & -d & c \\ -c & d & a & -b \\ -d & -c & b & a \end{pmatrix} = (a^2+b^2+c^2+d^2)^2.[/imath] I tried using Gauss-algorithm to get an easier matrix, but I'm not sure if I did it correctly. Calling the [imath]4[/imath] lines [imath]I[/imath], [imath]II[/imath], [imath]III[/imath] and [imath]IV[/imath], I did: (1) [imath]II \cdot a[/imath] (2) [imath]III \cdot a[/imath] (3) [imath]IV \cdot a[/imath] After this I did: (4) [imath]II' + I \cdot b[/imath] (5) [imath]III' + I \cdot c[/imath] (6) [imath]IV' + I \cdot d[/imath] So finally I got the following matrix: [imath]\begin{pmatrix} a & b & c & d \\ 0 & a^2+b^2 & bc-ad & ac+bd \\ 0 & ad+bc & a^2+c^2 & cd-ab \\ 0 & bd-ac & ab-cd & a^2+d^2 \end{pmatrix}.[/imath] I thought this would make the determinant a bit easier, unfortunately I must have done something wrong. Is multiplication with single lines allowed as I have done it? Thank you very much.
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2108605
|
[imath]\lim_{x\to \infty} f(x)= \text{Does not exist}[/imath]
let [imath]f[/imath] Periodic function (Except Constant functions) then : [imath]\lim_{x\to \infty} f(x)= \text{Does not exist}[/imath] is it right ?? such that : [imath]\lim_{x\to \infty} \sin x= \text{Does not exist}[/imath] [imath]\lim_{x\to \infty} \cos x= \text{Does not exist}[/imath] [imath]\lim_{x\to \infty} \tan x= \text{Does not exist}[/imath] [imath]\lim_{x\to \infty} \cot x= \text{Does not exist}[/imath] [imath]\lim_{x\to \infty} \lfloor x\rfloor +\lfloor -x \rfloor= \text{Does not exist}[/imath] is it right ??
|
1037185
|
Why do non-constant periodic functions have no limit at infinity?
Why do periodic functions (like [imath]\cos[/imath] or [imath]\sin[/imath] or [imath]\tan[/imath]) have no limit at infinity? I can guess that it is because their values don't converge but repeat over and over, but I would like to know what is the formal analytical (that is with [imath]\epsilon[/imath]'s and [imath]\delta[/imath]'s) proof of the statement.
|
1433279
|
Find solution to system of differential equations with initial conditions
I have a system [imath] \begin{cases} x_1'(t) = -x_2(t) \\ x_2'(t) = -x_1(t) \end{cases} [/imath] I have found the linearly independent solutions [imath] \begin{split} \vec{x}(t) &= e^{\lambda_1 t} \vec{v}_1 = e^t \begin{pmatrix}-1 \\ 1\end{pmatrix}, \\ \vec{x}(t) &= e^{\lambda_2 t} \vec{v}_2 = e^{-t} \begin{pmatrix}1 \\ 1\end{pmatrix}. \end{split} [/imath] and the general solution [imath] \vec{x}(t) = c_1 e^t \begin{pmatrix}-1 \\ 1\end{pmatrix} + c_2 e^{-t} \begin{pmatrix}1 \\ 1\end{pmatrix} [/imath] Is it correct that both all the linearly independent solutions and the general solution are all named [imath]\vec{x}(t)[/imath] or should the linearly independent solutions be named [imath]\vec{x}_1(t)[/imath] and [imath]\vec{x}_2(t)[/imath]? Now I have to find the solution to which [imath]x_1(0) = -1[/imath] and [imath]x_2(0) = 1[/imath]. How can I do this? I know I have to choose the correct values of [imath]c_1[/imath] and [imath]c_2[/imath] in order to fulfil the conditions. But I don't know where to substitute [imath]x_1(0) = -1[/imath] and [imath]x_2(0) = 1[/imath]. In my system, I have [imath]x_1'(t)[/imath] and [imath]x_2'(t)[/imath] but these are not vector function although I only have found vector functions as solutions. Is it because my general solution [imath]\vec{x}(t)[/imath] consists of the functions [imath]x_1(t) = -c_1 e^t + c_2 e^{-t}[/imath] and [imath]x_2(t) = c_1 e^t + c_2 e^{-t}[/imath], so I just have to find [imath]c_1[/imath] and [imath]c_2[/imath] such that [imath]x_1(0) = -c_1 e^0 + c_2 e^0 = -c_1 + c_2 = -1[/imath] and [imath]x_2(0) = c_1 e^0 + c_2 e^0 = c_1 + c_2 = 1[/imath], resulting in [imath]c_1 = 1[/imath] and [imath]c_2 = 0[/imath]?
|
930914
|
Find solutions for an differential equation system
I have a differential equation system [imath]x_1'(t) = -x_2(t)[/imath] [imath]x_2'(t) = -x_1(t)[/imath] I see that I can write [imath]\dot{x} = Ax[/imath] where [imath]A = \begin{pmatrix}0 & -1 \\ -1 & 0\end{pmatrix}[/imath] The complete real solution will be [imath]x(t) = c_1 e^{-t} \begin{pmatrix}1 \\ 1\end{pmatrix} + c_2 e^t \begin{pmatrix}-1 \\ 1\end{pmatrix}[/imath] where [imath]c_1,c_2 \in \mathbb{R}[/imath]. It this correct? Now I need to find the solution such that [imath]x_1(0) = -1[/imath] and [imath]x_2(0) = 1[/imath]. How can I do this? I think I could split my complete solution in [imath]x_1(t) = c_1 e^t - c_2 e^{-t}[/imath] and [imath]x_2(t) = c_1 e^t + c_2 e^{-t}[/imath] and then find [imath]c_1[/imath] and [imath]c_2[/imath] by solving [imath]x_1(0) = -1[/imath] and [imath]x_2(0) = 1[/imath] with respect to [imath]c_1[/imath] and [imath]c_2[/imath] and finally substitute these values in the complete solution [imath]x(t)[/imath]. Is this correct?
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2108179
|
[imath]i^2 = (-i)^2[/imath] apparently contradiction
I got this question in a class and can't figure out the answer: [imath]i^2 = (-i)^2 = -1[/imath] [imath](e^{πi(2n + 1/2)})^2 = (e^{πi(2m - 1/2)})^2[/imath] [imath]e^{πi(4n + 1)} = e^{πi(4m - 1)}[/imath] [imath]\ln(e^{πi(4n + 1)}) = \ln(e^{πi(4m - 1)})[/imath] [imath]πi(4n + 1) = πi(4m - 1)[/imath] [imath]m = n + 1/2[/imath] Which isn't possible for m and n both being integers. But I can't figure out what exactly is wrong with what is written here.
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366603
|
Incoherence using Euler's formula
Using the relation [imath]\ e^{ix} = \cos(x) + i\sin(x)[/imath] and substituting for [imath]\ x = \pi[/imath], we have the well-known Euler identity, [imath] e^{i\pi} = -1[/imath]. Substitute also for [imath] x = -\pi [/imath], we have [imath] e^{-i\pi} = -1[/imath]. So we can say [imath] e^{i\pi} = e^{-i\pi}[/imath] and taking [imath]\ln[/imath] on both sides, [imath]i\pi=-i\pi[/imath]. There is clearly something wrong here. Can some help me to figure out what's wrong? EDIT : I actually had a confusing output from my calculator while trying to solve the problem :
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2109614
|
If [imath]0 Prove that q[/imath]
If [imath]0<p<q<1[/imath] Prove that [imath]q<p+q-pq<1[/imath] I'm not sure how to set out a proof for this
|
1039666
|
how to prove [imath]a+b-ab \le 1[/imath] if [imath]a,b \in [0,1][/imath]?
Given: [imath]0 \le a \le 1[/imath] [imath]0 \le b \le 1[/imath] Prove: [imath]a + b - ab \le 1[/imath]
|
2108981
|
Find all the functions for which [imath]f(x^{2})-f(y^{2})=(x+y)(f(x)-f(y)):\forall x, y\in\mathbb{R}[/imath]
Hey guys I need help on solving the following problem Find all the functions for which : [imath]f(x^{2})-f(y^{2})=(x+y)(f(x)-f(y)):\forall x, y\in\mathbb{R}[/imath] I started with taking [imath]x=y[/imath], and there you see that the both sides are equal to [imath]0[/imath], therefore the solution is the function [imath]y=x[/imath], but after that I can't prove that that is the only solution (if it is the only). Would appreciate some help, thanks.
|
379978
|
Find all functions [imath]f[/imath] such that [imath](x+y)f(x)+f(y^2)=(x+y)f(y)+f(x^2)[/imath]
Find all functions [imath]f[/imath] that assign a real number [imath]f(x)[/imath] to every real number [imath]x[/imath] such that [imath](x+y)f(x)+f(y^2)=(x+y)f(y)+f(x^2)[/imath] I've tried subbing in heaps of values but I keep getting things like [imath]f(0)=f(0)[/imath] and other such useless results. Any help would be hugely appreciated.
|
2107915
|
Prove that [imath] \frac{n(n-1)(n-k+1)}{k(k-1)\cdots1} = \frac{n!}{(n-1)!k!}[/imath].
Prove that [imath] \frac{n(n-1)(n-k+1)}{k(k-1)\cdots1} = \frac{n!}{(n-k)!k!}[/imath]. [imath]= \frac{n(n-1)...(n-k+1)(n-k)!}{(n-k)!(k(k-1)...1)(3 \times 2 \times 1)}[/imath] [imath]= \frac{(n(n-1)(n-2)...(n-k+1)}{(k(k-1)....1)}[/imath] [imath]= \frac{n(n-1)..(n-k+1)}{k!}[/imath] My question is how is it that [imath](n-k)![/imath] is able to appear in the numerator and denominator on the right side of the equation? This is not obvious to me. What is (n-k)! in expanded form?
|
2104993
|
Prove that [imath] \frac{n(n-1)(n-k+1)}{k(k-1)\cdots1} = \frac{n!}{(n-1)!k!}[/imath]
Prove that [imath] \frac{n(n-1)(n-k+1)}{k(k-1)\cdots1} = \frac{n!}{(n-1)!k!}[/imath]. Proof: This is what I have tried: [imath]\frac{n!}{k!(n-k)!} \times (n-k)! = \frac{n(n-1)\cdots(n-k+1)(n-k)!}{k(k-1)\cdots1(n-k)!}[/imath] [imath]\frac{n!}{k!} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k(k-1)\cdots1}[/imath] This is where my attempt ends. Are my steps correct? If not how can I improve them? New Questions: How is it that n-k magically appears on the right side? I get that n-k and 3 times times 2 times 1 cancels out in the numerator and the denominator which leaves [imath]\frac{n(n-1)(n-2)...(n-k+1)}{k(k-1)....1}[/imath] which is equivalent to [imath]{n \choose k}[/imath] or \frac{n!}{(n-k)!k!}....is that correct?
|
1283151
|
Let [imath]f(x)=7x^{32}+5x^{22}+3x^{12}+x^2[/imath]. Find the remainder when [imath]x^2+1[/imath] divides [imath]f(x)[/imath] and [imath]xf(x)[/imath].
Let [imath]f(x)=7x^{32}+5x^{22}+3x^{12}+x^2[/imath]. Find the remainder when [imath]x^2+1[/imath] divides [imath]f(x)[/imath] and [imath]xf(x)[/imath]. I tried this problem two ways, substituting [imath]x=1,-1[/imath] in [imath]f(x)[/imath] to find the remainder, and by long division, but, that's not getting me to the answer. I think there is a shorter and elegant technique for solving this question. Please help. Thank you.r
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1279417
|
Let [imath]f(x) =7x^{32}+5x^{22}+3x^{12}+x^2[/imath]. Then find its remainder in the following cases.
Let [imath]f(x) =7x^{32}+5x^{22}+3x^{12}+x^2[/imath]. (i) Then find the remainder when [imath]f(x)[/imath] is divided by [imath][x^2+1][/imath]. (ii) Also find the remainder when [imath]xf(x)[/imath] is divided by [imath][x^2+1][/imath]. Given both the remainders will be of the form [imath]4(ax+b)[/imath]. The 'polynomial long division method' can be applied, but that won't be logical enough, because its too long a process. Can you tell me any other shorter & easier process? Thanks.
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2109667
|
Prove that if [imath]a, b, c \geq 0[/imath] then [imath](a + b)(b + c)( a + c) \geq 8abc[/imath]
Prove that if [imath]a, b, c\ge 0[/imath], then [imath](a + b)(b + c)(a + c)\ge 8abc[/imath].
|
1942608
|
AM-GM inequality basic
[imath]a, b, c[/imath] are positive real numbers. Prove that [imath]8abc\le (a+b)(b+c)(c+a)[/imath] I haven't really gotten anywhere, just tried to open up [imath](a+b)(b+c)(c+a)[/imath] and got null.
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2109939
|
a bounded analytic function on both real and imaginary axis
Does there exist an analytic function [imath]f:\mathbb{C}\rightarrow\mathbb{C}[/imath] such that [imath]f[/imath] is bounded on both real and imaginary axes? According to Liouville's theorem the function must be constant but I am confused that any bounded function [imath]f[/imath] on [imath]\mathbb{C}[/imath] must be like [imath]f(z)=c[/imath] where c is a constant. I am convinced that it must be bounded with respect to one axis but how about other axis. For example, [imath]f(z)=1[/imath] then it is bounded w.r.t real axis but how about imaginary axis? Question may be easy for other but I am unable to visualize it .
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1342366
|
Does there exists an entire function [imath]f: \mathbb C \to \mathbb C[/imath] which is bounded on real line and imaginary line?
Does there exists a nonconstant entire function [imath]f: \mathbb C \to \mathbb C[/imath] which is bounded on real line and imaginary line? Clearly,[imath] f(z)=sin(z)[/imath] is an example of an entire function which is bounded on real line and [imath] f(z)= e^z[/imath] is example of a function which is bounded on imaginary line.But I'm unable to find a function which is bounded on both the lines.Any ideas?
|
2108067
|
Find angle between two matrices
For the inner product space [imath]M_{2\times2} (\Bbb{C})[/imath], with inner product [imath]\langle A,B\rangle = \operatorname{tr}(AB^*)[/imath], find the angle between [imath]A=\begin{bmatrix}2+i&1\\-1&0\end{bmatrix} [/imath] [imath]B=\begin{bmatrix}3-2i&1+i\\0&-1\end{bmatrix}[/imath] I tried to use [imath]\cos(z)=\frac{\langle A,B\rangle}{\|A\|\|B\|}=\frac{5+6i}{4\sqrt{7}}[/imath] but how can I find the angle? is it supposed to be a real number?
|
2079945
|
What is the angle between those two matrices over [imath]\mathbb{C}[/imath]?
Let [imath]\left<\;,\;\right>[/imath] be an inner product defined by [imath]\left<A,B\right>=[/imath] tr[imath]\left(A\overline{B}^{T}\right)[/imath] over [imath]M_n\left(\mathbb{C}\right).[/imath] I am looking for the angle between: [imath]A=\begin{pmatrix}2+i & 1\\-1&0\end{pmatrix}\quad B=\begin{pmatrix}3-2i & 1+i\\0 & -i\end{pmatrix}[/imath] First we can see that:[imath]\overline{A}=\begin{pmatrix}2-i & 1\\-1&0\end{pmatrix}\quad \overline{B}=\begin{pmatrix}3+2i & 1-i\\0 & i\end{pmatrix}[/imath] [imath]\Downarrow[/imath] [imath]\overline{A}^T=\begin{pmatrix}2-i & -1\\1&0\end{pmatrix}\quad \overline{B}^T=\begin{pmatrix}3+2i & 0\\1-i & i\end{pmatrix}[/imath] And therefore: [imath]\left<A,B\right>=[/imath] tr[imath]\left(A\overline{B}^{T}\right)=\left(2+i\right)\left(3+2i\right)+1-i=5+6i[/imath] [imath]\left<A,A\right>=\left(2+i\right)\left(2-i\right)+1+1=7[/imath] [imath]\left<B,B\right>=\left(3-2i\right)\left(3+2i\right)+\left(1+i\right)\left(1-i\right)+1=16[/imath] Now as the angle between any two vectors is defined we get that: [imath]\cos{\alpha}=\frac{\left<A,B\right>}{\Vert{A}\Vert\Vert{B}\Vert}=\frac{5+6i}{\sqrt{112}}[/imath] [imath]\Downarrow[/imath] [imath]\alpha = \;?[/imath] I didn't even know a complex number could be an image for the cosine function. Thanks for any help.
|
2109576
|
Find basis and dimension for [imath]V[/imath]
Let [imath]S= \{ 1,2,\ldots,n \}[/imath]. and [imath]V[/imath] be the set of all functions [imath]f: S \to \mathbb{R}[/imath]. [imath]V[/imath] is a vector space defined by: [imath](f+ g)(x) = f(x) + g(x) \text{ and } (cf)(x)=cf(x). [/imath] Find a basis for [imath]V[/imath] and [imath]\dim V[/imath]. I'm not sure how to answer this question, it seems different than how I learned solving a basis, which was through a set of vectors, anyone know how I can tackle this?
|
2105344
|
Basis and dimension of vector space
Let [imath]V=\{f : S \to \mathbb{R} \}[/imath] be the set of all the functions [imath]f : S \to \mathbb{R}[/imath]. Let [imath]S=\{ 1,2,...,n[/imath]}, now my train of thought was that the basis of V must be a linearly independent set that Span V so somehow show that there is a set of functions on S that can compose as a linear combination and map to any number in [imath]\mathbb{R}[/imath]. Any idea how to get a general answer for basis for dimension of V?
|
2099578
|
number of solutions for a system of integer equations
I want to find the number of integer solutions for the following system of linear equations without computing the actual solutions ([imath]c_1,\ldots,c_n \in \mathbb{Z}[/imath] and [imath]r\in \mathbb{N}[/imath] ) : [imath]x_1+x_2=c_1[/imath] [imath]x_3+x_4=c_2[/imath] ... [imath]x_{2n-1}+x_{2n}=c_n[/imath] [imath]|x_1|+|x_2|+\ldots+|x_{2n}|\leq r[/imath] [imath]x_1,\ldots,x_{2n} \in \mathbb{Z}[/imath]
|
2086807
|
Finding all solutions for a system of equations with constraint on the sum of absolute values
I am looking for a fast algorithm to find all integer solutions for the following system of linear equations ([imath]c_1,\ldots,c_n \in \mathbb{Z}[/imath] and [imath]r\in \mathbb{N}[/imath] ): [imath]x_1+x_2=c_1[/imath] [imath]x_3+x_4=c_2[/imath] ... [imath]x_{2n-1}+x_{2n}=c_n[/imath] [imath]|x_1|+|x_2|+\ldots+|x_{2n}|=r[/imath] [imath]x_1,\ldots,x_{2n} \in \mathbb{Z}[/imath] Also can we find the number of solutions before solving the system?
|
2110479
|
Show that [imath]f[/imath] has an essential singularity at o
Let [imath]f[/imath] be a holomorphic, non-constant function on [imath]D=\{0<|z|<10\}[/imath]. It is given that for every [imath]n \in \mathbb{N}[/imath], [imath]|f(\frac{1}{n})| \le \frac{1}{n!}[/imath]. Show that [imath]f[/imath] has an essential singularity at 0. I have no idea how to even approach this. Why is it not possible for [imath]f[/imath] to have a pole at 0? Note. In the original question it was written [imath]n \in \mathbb{C}[/imath] but I considered this a typo. Thanks in advance!
|
2104353
|
holomorphic function on punctured disk satisfying [imath]\left|f\left(\frac{1}{n}\right)\right|\leq\frac{1}{n!}[/imath] has an essential singularity at [imath]0[/imath]
Let [imath]f[/imath] be holomorphic non-constant on [imath]D=\left\{ 0<\left|z\right|<10\right\}[/imath] . Given that for all [imath]n\in\mathbb{N}[/imath]: [imath]\left|f\left(\frac{1}{n}\right)\right|\leq\frac{1}{n!}[/imath] prove that [imath]f[/imath] has an essential singularity at [imath]0[/imath]. Find an example of [imath]f[/imath] satisfying the condition. My idea was to assume for contradiction the singularity is not essential, which means that either the limit [imath]z\to0[/imath] of [imath]f[/imath] exists or of [imath]1/f[/imath], and from [imath]\left|f\left(\frac{1}{n}\right)\right|\leq\frac{1}{n!}[/imath] we get that the limit of [imath]f[/imath] exists and is [imath]0[/imath]. I then define [imath]h(z)=f(z)[/imath] for [imath]z\neq0[/imath] and [imath]h(0)=0[/imath], which is holomorphic on the entire disk, and derive a contradiction from there and the uniqueness theorem. But then the suitable holomorphic function is the Gamma function, which we have not really discussed in class and I'm not sure about it's properties. Any ways of solving this which avoids using the gamma function?
|
2110432
|
Prove [imath]\|v\|_{H^1(\Omega )}\leq C(\|f\|_{L^2(\Omega )}+\|v\|_{H^{1/2}(\partial \Omega )}+\|\partial _\nu v\|_{H^{-1/2}(\partial \Omega )})[/imath]
Let [imath]\Omega \subset \mathbb R^n[/imath] be a smooth, bounded and open. Assume that [imath]u=0[/imath] if [imath]u\in H^1(\Omega )[/imath] and satisfies [imath]-\Delta u-u=0\quad \text{on }\Omega \quad \text{and}\quad \partial _\nu u=u=0\quad \text{on }\partial \Omega .[/imath] This property is know as unique continuation principle. Let [imath]f\in L^2(\Omega )[/imath] and assume that [imath]v[/imath] is a solution to [imath]-\Delta v-v=f[/imath] in [imath]\Omega [/imath]. Prove [imath]\|v\|_{H^1(\Omega )}\leq C(\|f\|_{L^2(\Omega )}+\|v\|_{H^{1/2}(\partial \Omega )}+\|\partial _\nu v\|_{H^{-1/2}(\partial \Omega )}).[/imath] There is just things that I don't understand in the proof. Proof : Assume by contradiction there exist sequences [imath]v_n\in H^1(\Omega)[/imath] and [imath]f_n\in L^2(\Omega)[/imath] such that [imath]\|v_n\|_{H^1(\Omega)}=1\quad\text{ and}\quad \|f_n\| _ {L^2(\Omega )}+\|v_n\| _ {H^{1/2}(\Omega )}+\|\partial _\nu v_n\| _ {H^{-1/2}(\partial \Omega )}\to 0.[/imath] By the Reileigh-Kondrashov compactness theorem there exists a weakly convergent subsequence of [imath]\{v_n\}[/imath] in [imath]H^1(\Omega)[/imath] such that [imath]\{v_n\}[/imath] converges strongly in [imath]L^2(\Omega),[/imath] i.e., [imath]\nabla v_n\rightharpoonup \nabla v[/imath] in [imath]L^2(\Omega)[/imath] and [imath]v_n\to v[/imath] in [imath]L^2(\Omega)[/imath] (after relabeling the subsequence). Multiplying the equality [imath]-\Delta v_n-v_n=f_n [/imath] by any function [imath]\varphi\in H^1(\Omega)[/imath] and integrating by parts we get [imath]\int_\Omega \nabla\varphi\nabla v_ndx=\int_\Omega \varphi v_n+ \int_\Omega \varphi f_n+\int_{\partial\Omega}\partial_\nu v_n\varphi \ \ \ (1).[/imath] Sending now [imath]n[/imath] to infinity, the second and the third summand will go to zero by our contradiction assumption, thus we get by the weak convergence [imath]v_n\rightharpoonup v,[/imath] [imath]\int_\Omega\nabla\varphi\nabla v=\int_\Omega\varphi v \ \ \ \ (2).[/imath] On the on the other hand if we take [imath]\varphi=v_n[/imath] in (1) and let [imath]n[/imath] go to infinity we get [imath]\|\nabla v_n\|^2\leq 2\|v_n\|^2[/imath] for [imath]n[/imath] big enough. Now, (2) implies that [imath]-\Delta v-v=0[/imath] in the weak sense. Also, by the definition of [imath]H^{1/2}[/imath] and [imath]H^{-1/2}[/imath] norms and the strong convergence [imath]v_n\to v[/imath] we get [imath]\|v\| _ {H^{1/2}(\partial \Omega)}=\|\partial_\nu v \| _ {H^{-1/2}(\partial\Omega)}=0,[/imath] Question 1 : I don't understand how we get [imath]\|v\| _ {H^{1/2}(\partial \Omega)}=\|\partial_\nu v \| _ {H^{-1/2}(\partial\Omega)}=0.[/imath] The thing I know is that [imath]\|v\|_{H^{1/2}(\partial \Omega) }=\int_{\partial \Omega } \hat u(x)(1+|x|)^2dx[/imath] and [imath]\|\partial_\nu v\|_{H^{-1/2}(\partial \Omega )}=\sup_{\|\varphi \|=1}|\left<\partial _\nu v,\varphi\right>|[/imath]. So how do I get that those norms are [imath]0[/imath] ? thus from the given unique continuation principle we get [imath]v=0[/imath] in [imath]\Omega.[/imath] From the strong convergence [imath]v_n\to v[/imath] we get [imath]\|v_n\| _ {L^2(\Omega)}\to 0[/imath], thus the inequality [imath]\|\nabla v_n\|^2\leq 2\|v_n\|^2[/imath] implies [imath]\|v_n\| _ {H^1(\Omega)}\to 0[/imath] which is a contradiction. Question 2 : Why the inequality [imath]\|\nabla v_n\|^2\leq 2\|v_n\|^2[/imath] implies [imath]\|v_n\| _ {H^1(\Omega)}\to 0[/imath] ?
|
2088646
|
Prove that [imath]\|v\|_{H^1(\Omega )}\leq C(\|f\|_{L^2(\Omega )}+\|v\|_{H^{1/2}(\partial \Omega )}+\|\partial _\nu v\|_{H^{1/2}(\partial \Omega )}))[/imath]
Let [imath]\Omega [/imath] be a smooth bounded subset of [imath]\mathbb R^d[/imath]. Assume that [imath]u=0[/imath] if [imath]u\in H^1(\Omega )[/imath] and [imath]u[/imath] satisfies [imath]-\Delta u-u=0\ \ in\ \ \Omega \quad \text{and}\quad \partial _\nu u=u=0\ \ on \ \ \partial \Omega .[/imath] Let [imath]f\in L^2(\Omega )[/imath] and assume that [imath]v[/imath] is solution of [imath]-\Delta v-v=f[/imath] in [imath]\Omega [/imath]. Prove that [imath]\|v\|_{H^1(\Omega )}\leq C(\|f\|_{L^2(\Omega )}+\|v\|_{H^{1/2}(\partial \Omega )}+\|\partial _\nu v\|_{H^{-1/2}(\partial \Omega )})),[/imath] where [imath]C[/imath] is independent of [imath]v[/imath] and [imath]f[/imath]. My attempts My attempts are unfortunately very few... I'm stuck on this problem since yesterday. I tried to make the substitution [imath]w=v-u[/imath], then [imath]w[/imath] is a solution of [imath]\nabla v-v=f[/imath], but it doesn't look to help a lot.
|
2110237
|
What is the relation between the Laplacian Operator and the Laplacian Matrix?
In what way are the Laplacian operator (defined on functions from [imath]\Bbb{R}^n[/imath] to [imath]\Bbb{R}[/imath] which are twice-differentiable) and the Laplacian matrix (defined on simple graphs) similar or related?
|
1657788
|
Discrete Laplacian
I was wondering if anybody would explain (or please point me to a nice reference) as to why the ``discrete Laplacian" on a graph is actually called a Laplacian. Namely, how is it related to the standard Laplacian on [imath]\mathbb{R}^n[/imath] ? Is there a sense in which the discrete Laplacian on say, a square lattice, would converge to the standard Laplacian as the lattice spacing tends to [imath]0[/imath] ?
|
2111218
|
Solving [imath]au_x + bu_y = f(x,y)[/imath]
The prompt is: Solve [imath]au_x + bu_y = f(x,y)[/imath], where [imath]f(x,y)[/imath] is a given function. If [imath]a \neq 0[/imath], write the solution in the form [imath]u(x,y) = (a^2+b^2)^{-1/2}\int_L fds + g(bx-ay)[/imath], where [imath]g[/imath] is an arbitrary function of one variable, L is the characteristic line segment from the y axis to the point (x,y), and the integral is a line integral. (Hint: Use the coordinate method.) My attempt at a solution: I changed variables to [imath]w = ax + by[/imath] [imath]e = bx - ay[/imath] Now [imath] u_x = au_{w}+ bu_{e}[/imath] and [imath]u_y = bu_w-au_e[/imath] so the PDE becomes [imath](a^2+b^2)u_w = f(x,y)[/imath]. Integrating both sides gives you [imath]u = g(e) + \dfrac{\int{f}dw}{(a^2+b^2)}[/imath] I am just having trouble getting from [imath]\int{f}(a^2+b^2)^{-1}dw[/imath] to [imath](a^2+b^2)^{-1/2}\int_L fds[/imath] If I understand correctly, converting the integral to a line integral over the characteristic curve will introduce a factor of [imath](a^2+b^2)[/imath], but I am new to PDE and I do not really know how to change the integral in [imath]w[/imath] to a line integral over the characteristic curve in [imath]ds[/imath].
|
304321
|
Show [imath]au_x+bu_y=f(x,y)[/imath] gives [imath]u(x,y)=(a^2+b^2)^{-\frac{1}{2}}\int_{L}fds +g(bx-ay)[/imath] if [imath]a\neq 0[/imath].
For my homework I am asked to do the following: Solve [imath]au_x+bu_y=f(x,y)[/imath], where [imath]f(x,y)[/imath] is a given function. If [imath]a\neq 0[/imath] write the solution in the form [imath]u(x,y)=(a^2+b^2)^{-\frac{1}{2}}\int_{L}fds +g(bx-ay)[/imath] where the integral is a line integral and [imath]L[/imath] is the characteristic line segment from the [imath]y[/imath]-axis to the point [imath](x,y)[/imath] and [imath]g[/imath] is an arbitrary function of one variable. A hint to use the coordinata method (change of coordinates) is given. For the [imath]g(bx-ay)[/imath] part we have [imath]g_x(bx-ay)=bg'[/imath] and [imath]g_y=-ag'[/imath] so this satisfies [imath]ag_x+bg_y=0[/imath] and therefore is the homogeneous solution. For the rest I realized that [imath]au_x+bu_y[/imath] is the directional derivative of [imath]u[/imath] along the characteristic line [imath]c=bx-ay[/imath] and therefore integrating along this line to solve seems reasonable. However I am unclear about the particulars. If anyone could help me out I would be very thankful. Also, isn't it important that besides specifying [imath]a\neq 0[/imath] we also have [imath]b\neq 0[/imath]?
|
2111697
|
Induction proof for an inequality with a recursive formula [imath]a_n = a_{\lfloor{n/2}\rfloor} + a_{\lceil{n/2}\rceil} + 3n + 1[/imath]
We have this recursive formula defined [imath]a_1=3[/imath] & [imath]a_n = a_{\lfloor{n/2}\rfloor} + a_{\lceil{n/2}\rceil} + 3n + 1[/imath] for [imath]n \geq 2[/imath]. And we need to prove [imath]n \leq 2^k \Longrightarrow a_n \leq 3 * k2^k+4*2^k-1[/imath] for all [imath]n,k \in \mathbb{Z}^+[/imath]. I think we wanna do the induction over k For our basic case we choose [imath]n = 2 , k = 1[/imath] [imath]a_2 = a_{\lfloor{2/2}\rfloor}+a_{\lceil{2/2}\rceil}+3*2+1 =[/imath] [imath] 3 + 3 + (6+1) = 13[/imath] [imath]2\leq2^1 \Longrightarrow 13 \leq 3*1*2^1+4*2^1-1[/imath] [imath]2\leq2 \Longrightarrow 13 \leq 13[/imath] Our basic case is proven, so we can move on to the induction step So we will assume that the expression is true for [imath]n \leq 2^k[/imath] and we will now show it is true for [imath]n \leq 2^{k+1}[/imath] This is where I have kinda reached a dead end. My guess is that we want to move from [imath]n \leq 2^k \Longrightarrow a_n \leq 3 * k2^k+4*2^k-1[/imath] to [imath]n \leq 2^{k+1} \Longrightarrow a_n \leq 3 * (k+1)2^{k+1}+4*2^{k+1}-1[/imath] And we could get closer to the P(k+1) expression by multiplying by 2 on both sides of P(k) like this: [imath]n*2 \leq 2^k*2 \Longrightarrow a_n*2 \leq (3 * k2^k+4*2^k-1)*2[/imath] But I don't know
|
2083461
|
Induction with floor, ceiling [imath]n\le2^k\implies a_n\le3\cdot k2^k+4\cdot2^k-1[/imath] for [imath]a_n=a_{\lfloor\frac{n}2\rfloor}+a_{\lceil\frac{n}2\rceil}+3n+1[/imath]
Via induction I need to prove an expression is true. the expression is: [imath]n \leq 2^k \longrightarrow a_n \leq 3 \cdot k2^k + 4 \cdot 2^k-1[/imath] for all [imath]n,k \in \mathbb{Z^+}[/imath] [imath]a_n[/imath] is a recursive function where [imath]a_1 = 3[/imath] [imath]a_n = a_{\lfloor \frac{n}{2} \rfloor} + a_{\lceil \frac{n}{2} \rceil} + 3n+1[/imath] I am stuck at the point where I need to prove that [imath]P(n+1)[/imath] is true [imath]a_{\lfloor \frac{n+1}{2} \rfloor} + a_{\lceil \frac{n+1}{2} \rceil} + 3(n+1) + 1 \leq 3 \cdot k 2^k + 4 \cdot 2^k -1 [/imath] It is the fact that I don't know how to rewrite the floor and ceiling functions to something else. Can someone give me some ideas how to proceed with this?
|
2111499
|
How to find a function where [imath]f[/imath] is differentiable and [imath]f'[/imath] is unbounded
The problem reads as follows: Find a map [imath]f: [0,1] →ℝ[/imath] such that [imath]f[/imath] is differentiable everywhere but [imath]f'[/imath] is unbounded. Obviously [imath]f[/imath] is continuous, and [imath]f'[/imath] is unbounded if [imath]∀M \geq 0 \ ∃x[/imath] such that [imath]|f'(x)|>M[/imath] How would you go about finding this function?
|
1324892
|
Derivable function on an interval [imath] [a,b][/imath] with derivative unbounded on [imath][a,b][/imath]
I'm searching for a derivable function defined on an interval [imath] [a,b][/imath] whose derivative is not bounded on [imath][a,b][/imath].
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2112057
|
Cardinal Arithmetic [imath]a^ba^c = a^{b+c}[/imath]
Prove that for any cardinal numer [imath]a,b,c[/imath] that [imath]a^b a^c = a^{b+c}[/imath]. Start with three non-empty sets [imath]A,B[/imath] and [imath]C[/imath] such that [imath]A∩B=\emptyset[/imath],and show that [imath]|A|^{|B|} \dot |A|^{|C|} = A^{|B| + |C|}[/imath] Workings: This would be done by establishing a bijection between [imath]A^B \times A^C[/imath] and [imath]A^{B \cup C}[/imath] But I'm not sure what the bijection could possible be. Any help will be appreciated. Note: I know of this Notation on proving injectivity of a function [imath]f:A^{B\;\cup\; C}\to A^B\times A^C[/imath]. But that uses notation and concepts I am not familiar with.
|
842444
|
Let [imath]A,B,C[/imath] be sets, and [imath]B \cap C=\emptyset[/imath]. Show [imath]|A^{B \cup C}|=|A^B \times A^C|[/imath]
Let [imath]A,B,C[/imath] be sets, and [imath]B \cap C=\emptyset[/imath]. Show [imath]|A^{B \cup C}|=|A^B \times A^C|[/imath] by defining a bijection [imath]f:A^{B \cup C} \rightarrow A^B \times A^C[/imath]. Any hints on this one? Thank you!
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413477
|
How can be possible [imath]\cap R_M=R[/imath]?
I'm doing this question in Hungerford's book: I didn't understand how can be possible this intersection be equal to [imath]R[/imath], because [imath]R_M[/imath] is [imath]S^{-1}R[/imath] with [imath]S=R-M[/imath], maybe the author means they are isomorphic? I need a hint to begin to solve this question. Thanks a lot
|
380686
|
Show that [imath]M=\bigcap_{\mathfrak{p}\in\operatorname{Spec}(R)}M_\mathfrak{p}=\bigcap_{\mathfrak{m}\in\text{Max}(R)}M_\mathfrak{m}[/imath] for certain [imath]M[/imath].
[imath]\newcommand{\Spec}{\operatorname{Spec}}[/imath] [imath]\newcommand{\mSpec}{\operatorname{Max}}[/imath] This is a homework from my algebra course. I am in a situation where I think I have found a solution, though somehow there's a condition in the question that I don't need. Important: I don't want help on the problem itself, I just want to know what's wrong with my proof! Exercise: Let [imath]R[/imath] be an integral domain with quotient field [imath]K[/imath]. Let [imath]M \subset K[/imath] be an [imath]R[/imath]-submodule of [imath]K[/imath]. Then for each prime ideal [imath]\mathfrak{p} \subset R[/imath] we can regard [imath]M_\mathfrak{p} \subset K[/imath]. Show that [imath]M = \bigcap_{\mathfrak{p} \in \Spec(R)} M_\mathfrak{p} = \bigcap_{\mathfrak{m} \in \mSpec(R)} M_\mathfrak{m}[/imath] as [imath]R[/imath]-submodules of [imath]K[/imath]. My proof: the inclusions from left to right are obvious (since [imath]R[/imath] is an integral domain and [imath]M[/imath] is torsionfree the inclusion [imath]M \rightarrow M_\mathfrak{a}[/imath], [imath]a \mapsto \frac{a}{1}[/imath] is injective, so [imath]M[/imath] can be seen as an [imath]R[/imath]-submodule of any of the [imath]M_\mathfrak{p}[/imath]. The inclusion [imath]\bigcap_{\mathfrak{p} \in \Spec(R)} M_\mathfrak{p} \subset \bigcap_{\mathfrak{m} \in \mSpec(R)} M_\mathfrak{m}[/imath] is always trivial. Now for the inclusion [imath]\bigcap_{\mathfrak{m} \in \mSpec(R)} M_\mathfrak{m} \subset M[/imath]: Let [imath]\frac{1}{s} \cdot m \in \bigcap_{\mathfrak{m} \in \mSpec(R)} M_\mathfrak{m} \Rightarrow \forall \mathfrak{m} \in \mSpec(R): \frac{1}{s} \cdot m \in M_\mathfrak{m} \Rightarrow \forall \mathfrak{m} \in \max(R): s \notin \mathfrak{m}[/imath]. Therefore [imath]s[/imath] is a unit in [imath]R[/imath] and [imath]\frac{1}{s} \cdot m = s^{-1} \cdot m \in M[/imath]. However, I don't see where my proof uses the fact that [imath]M[/imath] is a submodule of [imath]K[/imath] (torsionfree and integral domain would be enough). This confuses me a bit, so I am afraid to have made a mistake. It would be nice if someone could check this solution and tell me what I have done wrong, because right now I seem to be blind. Thanks a lot. Edit: I've just realized that all the [imath]M_\mathfrak{p}[/imath] must be submodules of some module [imath]P[/imath] for the [imath]\bigcap[/imath] to be defined. But is this really all the problem?
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927291
|
A Commutative Ring Having a Unique Prime Ideal (Dummit and Foote, Prob 7.4.40(i))
I am trying to solve Problem 7.4.40 from Dummit and Foote, a part of which states: Let [imath]R[/imath] be a commutative ring with [imath]1\neq 0[/imath] such that [imath]R[/imath] has exactly one prime ideal. Then every element of [imath]R[/imath] is either nilpotent or a unit. ATTEMPT: Let [imath]P[/imath] be the unique prime ideal of [imath]R[/imath]. We will show that every element of [imath]R\setminus P[/imath] is a unit. Let [imath]r\in R\setminus P[/imath]. Then the ideal [imath](r)[/imath] is either equal to [imath]R[/imath] or is a proper ideal of [imath]R[/imath]. If [imath](r)=R[/imath] then [imath]r[/imath] is a unit. So assume [imath](r)\subsetneq R[/imath]. But then [imath](r)[/imath] is contained in a maximal ideal [imath]M[/imath] of [imath]R[/imath]. Clearly [imath]M\neq P[/imath]. But [imath]M[/imath] is also a prime ideal of [imath]R[/imath] since it is a maximal ideal of [imath]R[/imath], giving a contradiction. So we have shown that [imath]x\in R\setminus P\Rightarrow r[/imath] is a unit. Also, since [imath]P[/imath] is prime, we have [imath]\mathfrak N(R)\subseteq P[/imath]. What I am struggling with is showing that [imath]\mathfrak N(R)=P[/imath]. NOTE: The result that [imath]\mathfrak N(R)[/imath] is the intersection of all the prime ideal of [imath]R[/imath], where [imath]R[/imath] is a commutative ring with [imath]1\neq 0[/imath], is not yet discussed in the textbook (See the comment in Problem 7.4.26 in D&F). So I think there must be an elementary argument which shows [imath]\mathfrak N(R)=P[/imath] in the question at hand. Thanks.
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595552
|
Each element of a ring is either a unit or a nilpotent element iff the ring has a unique prime ideal
Let [imath]R[/imath] be a ring. Prove that each element of [imath]R[/imath] is either a unit or a nilpotent element iff the ring [imath]R[/imath] has a unique prime ideal. Help me some hints.
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2112555
|
Proof using Prime Decomposition
My question reads: If c^2=ab and (a,b)=1, prove that a and b are perfect squares. I began my proof by [imath]a=p_1 p_2 \cdots p_n[/imath] and [imath]b=q_1 q_2\cdots q_m[/imath]. Then I gave [imath]c[/imath] its own decomposition as well and said [imath]c=s_1 s_2\cdots s_t[/imath]. From there I squares the c and just got the same except now with 2 in the exponent for each s. I am not too sure how to continue on from there. Would I need to re-index? Also, I am not too sure when to bring in the fact that their gcd is 1.
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284636
|
If a and b are relatively prime and ab is a square, then a and b are squares.
If [imath]a[/imath] and [imath]b[/imath] are two relatively prime positive integers such that [imath]ab[/imath] is a square, then [imath]a[/imath] and [imath]b[/imath] are squares. I need to prove this statement, so I would like someone to critique my proof. Thanks Since [imath]ab[/imath] is a square, the exponent of every prime in the prime factorization of [imath]ab[/imath] must be even. Since [imath]a[/imath] and [imath]b[/imath] are coprime, they share no prime factors. Therefore, the exponent of every prime in the factorization of [imath]a[/imath] (and [imath]b[/imath]) are even, which means [imath]a[/imath] and [imath]b[/imath] are squares.
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2112966
|
Minimum value of [imath]\,\left(a^3 + b^3 + c^3 -3abc\right)[/imath]
If [imath] a, b , c [/imath] are distinct positive integers satisfying the inequality [imath] ab + bc + ca \ge 107[/imath]. Then what would be the least value of [imath]\left(a^3 + b^3 + c^3 -3abc\right)[/imath] So I basically tried doing two things, first one being applying AM > GM, but I could not because some terms would be negative. The second try was that I simplified [imath]a^3 + b^3 + c^3 -3abc [/imath] into [imath] \left(a+b+c\right) \left(a^2 + b^2 + c^2 -ab- bc- ca\right)[/imath] but was stuck in finding the minimum values of [imath] (a+b+c )[/imath]. Is there any specific way to solve this problem?
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1448104
|
Minimum value of the expression given below.
Is [imath]a,b,c[/imath] are three different positive integers such that : [imath] ab+bc+ca\geq 107 [/imath] Then what is the minimum value of [imath]a^3+b^3+c^3-3abc[/imath]. I expanded this expression to [imath](a+b+c)((a+b+c)^2-3(ab+bc+ca))[/imath] and tried to find the minimum value of [imath](a+b+c)[/imath] by AM-GM inequalities but the problem is that the minimum value occurs at scenario where [imath]a=b=c[/imath], which is ommitted by assumptions of this question. I'd appreciate some help.
|
2112837
|
How do I prove that [imath]6 \mid n^3- n[/imath] for all [imath]n ≥ 1[/imath]?
How do I prove that [imath]6 \mid n^3 - n[/imath] for all [imath]n \ge 1[/imath]? I'm having difficulty understanding this problem. All help is greatly appreciated.
|
1202868
|
Proving divisibility of [imath]a^3 - a[/imath] by [imath]6[/imath]
As part of a larger proof, I need to show why [imath]a^3-a[/imath] is always divisible by [imath]6[/imath]. I'm having trouble getting started.
|
2113032
|
Evaluate [imath]1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}+\dots[/imath]
[imath]1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}+\frac{1\cdot 3\cdot 5\cdot 7}{3\cdot 6\cdot 9\cdot 12}+\dots[/imath] How do I proceed on evaluating this sum.
|
1692678
|
Prove that [imath]1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6 \cdot 9}+.........=\sqrt{3}[/imath]
Prove that [imath]1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6 \cdot 9}+.........=\sqrt{3}[/imath] [imath]\bf{My\; Try::}[/imath] Using Binomial expansion of [imath](1-x)^{-n} = 1+nx+\frac{n(n+1)}{2}x^2+\frac{n(n+1)(n+2)}{6}x^3+.......[/imath] So we get [imath]nx=\frac{1}{3}[/imath] and [imath]\frac{nx(nx+x)}{2}=\frac{1}{3}\cdot \frac{3}{6}[/imath] We get [imath]\frac{1}{3}\left(\frac{1}{3}+x\right)=\frac{1}{3}\Rightarrow x=\frac{2}{3}[/imath] So we get [imath]n=\frac{1}{2}[/imath] So our series sum is [imath](1-x)^{-n} = \left(1-\frac{2}{3}\right)^{-\frac{1}{2}} = \sqrt{3}[/imath] Although I know that this is the simplest proof, can we solve it any other way someting like defining [imath]a_{n}[/imath] and then use Telescopic sum. Thanks.
|
2113262
|
Need help with a question about analytic functions on a simply connected domain.
Let [imath]V\subseteq\mathbb{C}[/imath] be a simply connected domain such that [imath]V\neq\mathbb{C}[/imath], with [imath]a,b\in V, a\neq b.[/imath] Prove that if [imath]g:V\rightarrow V[/imath] is analytic with fixpoints [imath]a,b[/imath], then [imath]\forall z\in V, g(z) = z[/imath]. Here's my attempt at a solution. Let [imath]f(z) = g(z) - z.[/imath] As [imath]V[/imath] is simply connected, for every cycle [imath]\gamma\in V[/imath], [imath]0 = \int_{\gamma}{f(z)}dz = \int_{\gamma}{(g(z)-z)}dz = \int_{\gamma}{g(z)}dz - \int_{\gamma}{z}dz,[/imath] so for every cycle [imath]\gamma\in V[/imath], [imath]\int_{\gamma}{g(z)}dz = \int_{\gamma}{z}dz.[/imath] This can only be the case if [imath]\forall z\in V, g(z) = z[/imath], so we are done. I'm not sure whether my last line is correct or not, but I have a feeling that it isn't. Could anyone either confirm that this is a correct proof or point me in the right direction if it's incorrect? Thanks!
|
1582821
|
Analytic map with two fixed points on a simply connected domain is the identity
Let [imath]f:U\rightarrow U[/imath] be analytic, where [imath]U \subsetneqq \mathbb{C}[/imath] is a simply connected domain. Suppose that [imath]f[/imath] has at least two fixed points. Prove that [imath]f[/imath] is the indentity map. I am able to prove the theorem in the case where [imath]U = \mathbb{D}[/imath] is the unit disc (the proof uses Schwarz lemma, and is discussed in this question). The proof however skips the (rigorous) details of how the Riemann Mapping Theorem allows us to reduce from [imath]U[/imath] to [imath]\mathbb{D}[/imath]. Any details are appreciated.
|
2112967
|
The square of a number is even if and only if the number itself is even
I wish to prove that square of a number is even if and only if the number itself is even. Let's call that number [imath]a[/imath], we need to prove that [imath] a^2\text{ is even}\Leftrightarrow a\text{ is even} [/imath] We need to prove two things: If [imath]a[/imath] is even, then [imath]a^2[/imath] is also even. I think that I can do this part. If [imath]a[/imath] is even, then we can write [imath]a=2k[/imath] where [imath]k[/imath] is some integer. Now [imath] a^2=(2k)^2=4k^2=2(2k^2) [/imath] which shows that [imath]a^2[/imath] is also even. If [imath]a^2[/imath] is even, then [imath]a[/imath] is also even. I am having difficulty here. If [imath]a^2[/imath] is even, then I can write [imath]a^2=2m[/imath] . for some integer [imath]m[/imath] but taking square root gives [imath]a=\sqrt{2m}[/imath] which can be written as [imath]a=2\sqrt{\frac{m}{2}}[/imath] but the problem is that will [imath]\sqrt{\frac{m}{2}}[/imath] always end up as some integer? I do not think so!
|
2106275
|
Prove that if n is even then [imath]n^2[/imath] is even and if n is odd then [imath]n^2[/imath] is odd?
How can I prove that if [imath]n[/imath] is even then [imath]n^2[/imath] is even and if [imath]n[/imath] is odd then [imath]n^2[/imath] is odd?
|
2113416
|
Which is this largest integer?
The largest integer say [imath]n[/imath] for which [imath](n+5)[/imath] divides [imath]n^5+5[/imath] is [imath]3115[/imath] [imath]3120[/imath] [imath]3125[/imath] [imath]3130[/imath] I just want a hint not the solution.
|
672854
|
How can I find all integers [imath]x≠3[/imath] such that [imath]x−3|x^3−3[/imath]
How can I find all integers [imath]x≠3[/imath] such that [imath]x−3|x^3−3[/imath]? I tried expand [imath]x^3−3[/imath] as a sum but I couldn't find a way after that.
|
2113496
|
Integration Problem
how can I prove this identity: [imath] n \in N_0, m \in N [/imath] [imath] \int_0^1 x^{n}(1-x)^{m}dx\ = \frac{n!m!}{(n+m+1)!} [/imath] I thought about the geometric series but I am not sure Thanks in advance!
|
164719
|
Prove that [imath]\int_0^1t^{p-1}(1-t)^{q-1}\,dt=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}[/imath] for positive [imath]p[/imath] and [imath]q[/imath]
I'm trying to prove that for [imath]p,q>0[/imath], we have [imath]\int_0^1t^{p-1}(1-t)^{q-1}\,dt=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}.[/imath] The hint given suggests that we express [imath]\Gamma(p)\Gamma(q)[/imath] as a double integral, then do a change of variables, but I've been unable thus far to express it as a double integral. Can anyone get me started or suggest an alternate approach? Note: This wasn't actually given to me as the [imath]\Gamma[/imath] function, just as a function [imath]f[/imath] satisfying [imath]f(p)=\int_0^\infty e^{-t}t^{p-1}\,dt[/imath] for all [imath]p>0[/imath], but I recognized that. This is in the context of an advanced calculus practice exam.
|
2112996
|
Let [imath]f(z) = f(z^2)[/imath], for all [imath]z\in\mathbb{D}[/imath] with [imath]f(1) = 1[/imath], then which of the following is incorrect?
Let [imath]f(z)[/imath] be analytic on [imath]\mathbb{D}[/imath] = {[imath]{z\in\mathbb{C}:|z-1|<1}[/imath]} such that [imath]f(1) = 1[/imath], if [imath]f(z) = f(z^2)[/imath] for all [imath]z\in\mathbb{D}[/imath], then which of the following statements is NOT correct? 1) [imath]f(z) = [f(z)]^2[/imath] 2) [imath]f(\frac{z}{2}) = \frac{1}{2}f(z)[/imath] 3) [imath]f(z^3) = [f(z)]^3[/imath] 4) [imath]f'(1) = 0[/imath] This question is also in mathstack. If we will assume [imath]f(z)=1[/imath] for all [imath]z\in\mathbb{D}[/imath] then this function clearly satisfies all the hypothesis so (2) is incorrect. I strongly believe that the only function which satisfy the above hypothesis is [imath]f(z)=1[/imath]. But how will I prove this?
|
825865
|
Let [imath]f(z)[/imath] be analytic on [imath]\mathbb{D}[/imath] = {[imath]{z\in\mathbb{C}:|z-1|<1}[/imath]} such that [imath]f(1) = 1[/imath], if [imath]f(z) = f(z^2)[/imath] for all [imath]z\in\mathbb{D}[/imath],
Let [imath]f(z)[/imath] be analytic on [imath]\mathbb{D}[/imath] = {[imath]{z\in\mathbb{C}:|z-1|<1}[/imath]} such that [imath]f(1) = 1[/imath], if [imath]f(z) = f(z^2)[/imath] for all [imath]z\in\mathbb{D}[/imath], then which of the following statements are correct? 1) [imath]f(z) = [f(z)]^2[/imath] 2) [imath]f(\frac{z}{2}) = \frac{1}{2}f(z)[/imath] 3) [imath]f(z^3) = [f(z)]^3[/imath] 4) [imath]f'(1) = 0[/imath] I can't understand how to do solving this problem. please anyone give me some hints.
|
2113505
|
Proving [imath]1 \geq \frac{\sin(\theta)}{\theta} \geq \frac{2}{\pi}[/imath] for [imath]\theta \in [0,\frac{\pi}{2}][/imath]
In proving Jordan's lemma, the inequality [imath]1 \geq \frac{\sin(\theta)}{\theta} \geq \frac{2}{\pi}[/imath] is used, for [imath]\theta \in [0,\frac{\pi}{2}][/imath]. I got as far as establishing [imath]1 \geq \sin(\theta) \geq 0[/imath] and [imath]\theta^{-1} \in [\frac{2}{\pi}, \infty)[/imath]. I'm looking for a modest hint.
|
407517
|
The sine inequality [imath]\frac2\pi x \le \sin x \le x[/imath] for [imath]0[/imath]
There is an exercise on [imath]\sin x[/imath]. How could I see that for any [imath]0<x< \frac \pi 2[/imath], [imath]\frac 2 \pi x \le \sin x\le x[/imath]? Thanks for your help.
|
777213
|
Field extensions - if [imath](E : F) = n[/imath] then [imath](E(x) : F(x)) = n[/imath]
Well, pretty much everything is in the title - I'm looking for the proof of the following statement: if we have a field extension [imath]F \subset E[/imath] then the degree of the extension [imath]F(x) \subset E(x)[/imath] [fields of rational functions] is the same as that of [imath]E[/imath] over [imath]F[/imath]. Will be grateful for any hints!
|
1587229
|
If a field [imath]F[/imath] is an algebraic extension of a field [imath]K[/imath] then [imath](F:K)=(F(x):K(x))[/imath]
Suppose [imath]K[/imath] is a field and [imath]F[/imath] is an algebraic extension of some degree [imath]n=(F:K)[/imath]. It is stated that the field of rational functions [imath]F(x)[/imath] is in fact an algebraic extension of the field [imath]K(x)[/imath] and moreover [imath](F(x):K(x))=n[/imath]. How do I approach this exercise? I'm new in this area so any help would be greatly appreciated!
|
2114559
|
Let [imath]R[/imath] be a ring, and let [imath]C[/imath]={[imath]x\in R: xy=yx[/imath] for all y in [imath]R[/imath]}
Let [imath]R[/imath] be a ring, and let [imath]C[/imath]={[imath]x\in R: xy=yx[/imath] for all y in [imath]R[/imath]} Prove that if [imath]x^2-x \in C[/imath] for all [imath]x[/imath] in [imath]R[/imath], then [imath]R[/imath] is commutative. There's a hint given in the book that says "Show that [imath]xy+yx \in C[/imath] by considering [imath]x + y[/imath], and then show that [imath]x^2 \in C[/imath]". I manage to show [imath]xy+yx \in C[/imath], but don't really know how to show [imath]x^2 \in C[/imath]. Any help is appreciated, thanks!
|
2032689
|
Let [imath]R[/imath] be a ring such that for all [imath]a,b[/imath] in [imath]R[/imath], [imath](a^2-a)b=b(a^2-a)[/imath]. Then [imath]R[/imath] is commutative
Let [imath]R[/imath] be a ring such that for all [imath]a,b[/imath] in [imath]R[/imath], [imath](a^2-a)b=b(a^2-a)[/imath]. Then [imath]R[/imath] is commutative. I am having difficulties making the connection. Any help would be appreciated.
|
2114387
|
Sum of telescoping series
How can you change this series into a telescoping series so then you can find its sum? [imath]\sum_{n=1}^{\infty} \frac{1}{\sqrt n + \sqrt{n+1}}[/imath]
|
2104451
|
Value of: [imath] \frac {1} {\sqrt {4} + \sqrt {5}} + \frac {1} {\sqrt {5} + \sqrt {6}} + ... + \frac {1} {\sqrt {80} + \sqrt {81}}[/imath]
I have the following question: Find the value of: [imath] \frac {1} {\sqrt {4} + \sqrt {5}} + \frac {1} {\sqrt {5} + \sqrt {6}} + ... + \frac {1} {\sqrt {80} + \sqrt {81}}[/imath] The book already provide the answer for this question but it didn't include the ways. Hint or anything that will help is appreciated
|
2115125
|
Prove that the system of congruences has a solution
I'm doing the following exercise and I don't know how to prove it. Suppose [imath]a,b \in \mathbb{Z}[/imath], and [imath]d=\gcd(a,b)[/imath]. I have to prove that if [imath]x\equiv y \bmod d[/imath], then the system [imath]X\equiv x \mod a[/imath] [imath]X\equiv y \mod b[/imath] has a solution. I don't even how to start... Thanks for your help :)
|
2060713
|
Deriving Chinese Remainder Theorem from gcd Bezout identity
We want to find the solution [imath]x[/imath] to the congruence system [imath]$$\begin{align} x &\equiv r_1 \!\!\!\pmod{\!m_1}\\ x &\equiv r_2 \!\!\!\pmod{\!m_2}\end{align},\ \ {\rm where}\ \ \gcd(m_1, m_2) = 1$$[/imath] These can be rewritten as [imath]$$\begin{align} x &= r_1 + m_1 j\\ x &= r_2 - m_2k\end{align}$$[/imath] for unknown integers [imath]j, k[/imath]. So we set those two equations as equal and rearrange them to [imath]m_1j + m_2k = r_2 - r_1[/imath] Now assume we perform [imath]\text{egcd}(m_1, m_2)[/imath], which gives us the solution to [imath]m_1u + m_2v = \gcd(m_1, m_2)[/imath] Let the integer [imath]$\,h\,$[/imath] make [imath]$\,h \gcd(m_1, m_2) = r_2 - r_1$[/imath]. Then [imath]$\qquadhm_1u + hm_2v = h\gcd(m_1, m_2)= r_2 - r_1 = m_1j + m_2k$[/imath] At this point I am lost because I don't think we can just assume [imath]hm_1u = m_1j[/imath] and extract [imath]j = hu[/imath]. How do I get the values of [imath]j[/imath] or [imath]k[/imath] so I can get the value of [imath]x[/imath]?
|
2115550
|
How many solutions does [imath]1=x^π[/imath] have?
I was wondering how many solutions there are to [imath]1 = x^\text{irrational number}[/imath], since the cube root of 1 has 3 solutions and the 4th root has 4 etc and since the number of solutions to [imath]x = x^{a/b}[/imath] is b (where [imath]a[/imath] and [imath]b[/imath] share no factors), how many would [imath]x^π=1[/imath] have? Infinity, none or something else?
|
930666
|
How many roots have a complex number with irrational exponent?
If a rational exponent on a complex number [imath]z^q[/imath] is the representation of a finite number of roots, then if the exponent is irrational this mean that there is infinite countable roots? If this is the case... the cardinality of the number of roots is the same for any irrational exponent? Thanks in advance. NOTE: there are different questions about irrational exponents but no one answer what Im searching so please dont mark this as repeated.
|
2115536
|
Example of quadratic extension
I am looking for an example of a quadratic extension K/F that is not given by [imath]K=F(\sqrt{\beta})[/imath] with [imath]\beta \in F[/imath] and [imath]\beta[/imath] not a square in [imath]F[/imath]. I Know that every quadratic extension (with char(F) [imath]\neq 2[/imath]) are of the form [imath]F(\sqrt{\alpha})[/imath], with [imath]\alpha[/imath] not a square in F. Therefore I assumed [imath]F= \mathbb{Z}/2\mathbb{Z}[/imath]. I am not sure what K could be.
|
290003
|
Quadratic extensions in characteristic [imath]2[/imath]
I recently saw in class that the degree [imath]2[/imath] extensions of a field of characteristic [imath]\neq 2[/imath] are given by square roots of non-squares in the base field. I wonder what happens in the case of characteristic [imath]2[/imath] fields. For finite fields of characteristic [imath]2[/imath], it's clear what's going on, but what about others like [imath](\mathbb{Z}/2\mathbb{Z})(X)[/imath]?
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2115119
|
Proving that [imath]\det (M) = \det (A) \det(D-CA^{-1}B)[/imath]
So the task is the following: Be [imath]M\in M_{m\times n}(K)[/imath] a block matrix [imath]M = \bigg( \begin{array}{cc} A & B \\ C & D \\ \end{array} \bigg) [/imath] with [imath]A\in M_m (K)[/imath], [imath]B,C^t\in M_{m\times n}(K)[/imath] and [imath]D\in M_n (K)[/imath] and let [imath]A[/imath] be invertable. Show that: (a) [imath]\det (M)=\det (A) \det (D-C(A^{-1})B[/imath]. (b) If [imath]m=n[/imath] and [imath]AB=BA[/imath], [imath]\det (M)=\det (DA-CB)[/imath]. Note for (a): Multiplicate [imath]M[/imath] with a proper block matrix. I am really lost on this one... and would really appreciate any help.
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1811433
|
Let [imath]M=\big( \begin{smallmatrix} A & B \\ C & D \end{smallmatrix} \big)[/imath]. Prove [imath]\det(M)=\det(A)\cdot \det(D-C·A^{-1}·B)[/imath].
Let (shown in matrix blocks) [imath]M=\big( \begin{smallmatrix} A & B \\ C & D \end{smallmatrix} \big)[/imath] a square matrix such that [imath]A[/imath] is invertible and [imath]D[/imath] is a square matrix. I have to prove that [imath]\det(M)=\det(A)\cdot \det(D-C·A^{-1}·B)[/imath] I also have an indication to use it: Consider previously these cases: \begin{align*} A_1 &= \begin{pmatrix} A & 0 \\ 0 & I \end{pmatrix},\\ A_2 &= \begin{pmatrix} I & 0 \\ 0 & D \end{pmatrix},\\ A_3 &= A_1·A_2,\\ A_4 &= \begin{pmatrix} I & A^{-1}·B \\ 0 & I \end{pmatrix},\text{ and}\\ A_5 &= A_3\cdot A_4. \end{align*} The problem is difficult in general for me and I don't even know how to use the indication.
|
2115789
|
Prove there is no nonzero integer solution to [imath]x^2 - 2y^2 = 0[/imath]
By a contradiction. Then prove that there is also no nonzero rational solution. I'm lost in my first proof course. I appreciate any help.
|
1699080
|
Why must [imath]a[/imath] and [imath]b[/imath] both be coprime when proving that the square root of two is irrational?
Suppose we wish to prove that the square root of two is irrational. We begin by assuming that it is rational. Namely, where both [imath]a[/imath] and [imath]b[/imath] are integers [imath]\frac{a}{b} = \sqrt 2 % MathType!MTEF!2!1!+- % faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0x % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa % aiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGHbaabaGaamOyaaaacq % GH9aqpdaGcaaqaaiaaikdaaSqabaaaaa!31D3! [/imath] Why is it so important that both [imath]a[/imath] and [imath]b[/imath] be coprime and the fraction be irreducible? That is, if [imath]a[/imath] and [imath]b[/imath] are not coprime and the fraction is indeed reducible, then[imath]\frac{a}{b} \ne \sqrt 2 % MathType!MTEF!2!1!+- % faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbqedmvETj % 2BSbqefm0B1jxALjharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0x % bbL8FesqqrFfpeea0xe9Lq-Jc9vqaqpepm0xbba9pwe9Q8fs0-yqaq % pepae9pg0FirpepeKkFr0xfr-xfr-xb9Gqpi0dc9adbaqaaeGaciGa % aiaabeqaamaabaabaaGcbaWaaSaaaeaacaWGHbaabaGaamOyaaaacq % GHGjsUdaGcaaqaaiaaikdaaSqabaaaaa!3294! [/imath]
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2115175
|
If [imath]f \in {\mathcal{C}}_c({\mathbb{R}}^n)[/imath], then [imath]f[/imath] is uniformly continuous
If [imath]f \in {\mathcal{C}}_c({\mathbb{R}}^n)[/imath], prove that [imath]f[/imath] is uniformly continuous on [imath]{\mathbb{R}}^n[/imath]. I think I have to consider [imath]g = f\big|_{\mbox{supp} f}[/imath] because [imath]g[/imath] is uniformly continuous on [imath]\mbox{supp} f[/imath], by Heine-Cantor theorem: [imath]\mbox{supp} f[/imath] is a compact set in [imath]{\mathbb{R}}^n[/imath] and [imath]g[/imath] is continuous on [imath]\mbox{supp} f[/imath], because [imath]f[/imath] is continuous on [imath]{\mathbb{R}}^n[/imath]. But what about [imath]f[/imath] on [imath]{\mathbb{R}}^n[/imath]? Mi thought is the next: fix [imath]\varepsilon > 0[/imath]. By the continuity on [imath]x \in {\mathbb{R}}^n[/imath], exists [imath]{\delta}_x > 0[/imath] such that if [imath]y \in B(x , {\delta}_x)[/imath], then [imath]f(y) \in (f(x) - \varepsilon , f(x) + \varepsilon)[/imath]. Then [imath]{\{B(x , {\delta}_x)\}}_{x \in \mbox{supp} f}[/imath] is an open cover of [imath]\mbox{supp} f[/imath]. By the compactness of [imath]\mbox{supp } f[/imath], exists a finitely cover [imath]{\{{\mathcal{O}}_j\}}_{j = 1}^m[/imath] such that [imath] \mbox{supp } f \subset \bigcup_{j = 1}^m {\mathcal{O}}_j \subset \bigcup_{x \in \mbox{supp } f} B(x , {\delta}_x)\mbox{.} [/imath] Now I can't finish my argument. Thank you very much.
|
445735
|
Compactly supported continuous function is uniformly continuous
Let [imath]f:\mathbb R \rightarrow \mathbb R[/imath] be continuous and compactly supported. How can I prove that [imath]f[/imath] is uniformly continuous ? I was trying to prove it by contradiction but get stuck. My attempt was as follows: Let [imath]E[/imath] be the compact support of [imath]f[/imath]. On [imath]E[/imath] we know that [imath]f[/imath] is uniformly continuous. If we assume that [imath]f[/imath] is not uniformly continuous we know [imath] \exists \epsilon > 0 \forall \delta > 0 \exists x,y \in \mathbb R: |x-y|<\delta \wedge |f(x)-f(y)| \geq \epsilon [/imath] Let [imath]\delta_n := \frac 1n[/imath] and fix this [imath]\epsilon > 0[/imath]. Compute a [imath]\delta > 0[/imath] s.t. [imath]\forall x,y \in E:|x-y| < \delta \rightarrow |f(x)-f(y)| <\epsilon[/imath]. Let [imath]n \geq N[/imath] s.t. [imath]\frac 1N < \delta[/imath]. Then we may assume wlog that [imath]x_n \in E[/imath] and [imath]y_n \in \mathbb R \setminus E[/imath] wehere [imath]x_n,y_n[/imath] are the points corresponding with [imath]\delta_n[/imath]. This gives a sequence of points where [imath]|x_n-y_n| \rightarrow 0[/imath] and [imath]x_n \in E[/imath] and [imath]y_n \in \mathbb R \setminus E[/imath]. I now want to use somehow the continuitiy of [imath]f[/imath]. How can I do this ? Is this approach a good one ? Can it be more simple ? New idea: I know that [imath]E[/imath] is compact so [imath](y_n)_{n=0}^\infty[/imath] has a convergent subsequence [imath](y_{n_j})_{j=0}^\infty[/imath] whit limit say [imath]y \in E[/imath]. Now [imath] |x_{n_j}-y| \leq |x_{n_j}-y_{n_j}|+|y_{n_j}+y| \rightarrow 0 [/imath] So I can take [imath]x_{n_j}[/imath] close to [imath]y[/imath] to get [imath]|f(x_{n_j})-f(y)| = |f(y)| <\frac \epsilon 2[/imath] by the continuity. I can also take [imath]y_{n_j}[/imath] close to [imath]y[/imath] to get [imath]|f(y_{n_j})-f(y)| <\frac \epsilon 2[/imath]. We further have [imath] |f(y_{n_j})| \leq |f(y_{n_j})-f(y)| + |f(y)| [/imath] s.t. [imath] |f(y)| \geq |f(y_{n_j})| - |f(y_{n_j})-f(y)| \geq \frac \epsilon 2 [/imath] because [imath]|f(y_{n_j})| \geq \epsilon[/imath] per construction. So we have [imath]|f(y)| \leq \frac \epsilon 2[/imath] and [imath]|f(y)| > \frac \epsilon 2[/imath] which is a contradiction. New solution: Let [imath]\epsilon > 0[/imath]. Let [imath]\delta_1[/imath] for the uniform continuity on [imath]E[/imath]. Further [imath] \forall x \in E,\ \exists \delta_x > 0\ \forall y \in \mathbb{R}: |x-y|< \delta_x \rightarrow |f(x)-f(y)| < \epsilon [/imath] Compute an open finite over of [imath]E[/imath] [imath] E \subseteq \bigcup_{i=1}^N B\left(x_i,\frac{\delta_{x_i}}2\right) [/imath] Write [imath]\delta_i := \delta_{x_i}[/imath]. Let [imath]\delta_2 := \min_{i=1,\cdots,N} \frac {\delta_i} 2[/imath] and [imath]\delta := \min(\delta_1,\delta_2)[/imath]. Assume [imath]|x-y|< \delta[/imath]. If [imath]x,y \in E[/imath] or [imath]x,y \notin E[/imath] we are done. Otherwise assume [imath]x \in E[/imath] and [imath]y \notin E[/imath]. Then [imath]x \in B\left(x_i,\frac{\delta_i}{2}\right)[/imath] for some [imath]i[/imath]. Further [imath] |y-x_i| \leq |y-x|+|x-x_i|\leq \delta_2 + \frac{\delta_i}2 \leq \delta_i [/imath] thus [imath]y \in B(x_i,\delta_i)[/imath] which proves the claim.
|
2116972
|
Prove that [imath] (x+y)^n \le x^n +y^n[/imath] with [imath]x,y \ge 0 [/imath] and [imath]0 \lt n \le 1[/imath]
I have no clue on this one so I hope you can help me out on this one. Let [imath]x,y \ge 0 [/imath] and [imath]0 \lt n \le 1[/imath]. Then [imath] (x+y)^n \le x^n +y^n[/imath]
|
1990936
|
Is this inequality true? [imath] (x + y)^{\alpha} < x^{\alpha} + y^{\alpha} [/imath], for positive [imath]x[/imath] & [imath]y[/imath], and for [imath]0 < \alpha < 1[/imath]
If [imath]0 < \alpha < 1[/imath], then [imath] (x + y)^{\alpha} < x^{\alpha} + y^{\alpha} [/imath] for [imath]x[/imath], [imath]y[/imath] positive. Is this inequality true in general? I tried using Young's Inequality: For [imath]z,t > 0[/imath], and for [imath]n[/imath], [imath]m[/imath] such that [imath]n+m=1[/imath], then [imath] z^n + t^m \leq nz + mt [/imath] So, using this we have [imath] (x+y)^{\alpha} \cdot 1^{1 - \alpha} \leq \alpha(x+y) + (1-\alpha) = \alpha x + \alpha y + (1-\alpha)[/imath] which is not as tight as I want.
|
2116127
|
[imath]\sum\limits_{k=0}^{\lfloor{n/2}\rfloor}(-1)^k\binom{n-k}{k}=\frac{2}{3}\left(\cos\frac{\pi(n-1)}{3}+\cos\frac{\pi n}{3}\right)[/imath]
Be [imath]n\in\mathbb{N}_0[/imath] . This is a simple formula but not obviously : [imath]\sum\limits_{k=0}^{\lfloor{n/2}\rfloor}(-1)^k\binom{n-k}{k}=\frac{2}{3}\left(\cos\frac{\pi(n-1)}{3}+\cos\frac{\pi n}{3}\right)[/imath] Can be found a short proof for that ?
|
1960944
|
Sum of the series [imath]\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........[/imath]
The sum of the series [imath]\binom{n}{0}-\binom{n-1}{1}+\binom{n-2}{2}-\binom{n-3}{3}+..........[/imath] [imath]\bf{My\; Try::}[/imath] We can write it as [imath]\displaystyle \binom{n}{0} = [/imath] Coefficient of [imath]x^0[/imath] in [imath](1+x)^n[/imath] Similarly [imath]\displaystyle \binom{n-1}{1} = [/imath] Coefficient of [imath]x^1[/imath] in [imath](1+x)^{n-1}[/imath] Similarly [imath]\displaystyle \binom{n-2}{2} = [/imath] Coefficient of [imath]x^2[/imath] in [imath](1+x)^{n-2}[/imath] Now, how can I solve it after that, Help Required, Thanks
|
2115010
|
Prove [imath]s^\lambda t^{(1-\lambda)}\leq \lambda s +(1-\lambda)t[/imath]
Prove [imath]s^\lambda t^{(1-\lambda)}\leq \lambda s +(1-\lambda)t[/imath], for all [imath]s,t\geq 0[/imath] and [imath]\lambda \in (0,1)[/imath]
|
81432
|
An application of Jensen's Inequality
Given that [imath]\{\phi_n\}[/imath] is a sequence of non-negative numbers whose sum is [imath]1[/imath] and [imath]\{\psi_n\}[/imath] is a sequence of positive numbers, how can I show that [imath] \prod_{n=1}^{\infty}~\psi_n^{\phi_n}~\leq~\sum_{n=1}^{\infty}~\phi_n\psi_n~? [/imath] Thanks. PS: I'm not too sure about the title. Perhaps, someone could give it a better title?
|
2117468
|
If [imath]\alpha [/imath] and [imath]\beta[/imath] are roots of equation [imath]a\tan\theta +b \sec\theta=c[/imath]. Prove that [imath]\tan(\alpha + \beta)=\frac{2ac}{a^2-c^2}[/imath]
If [imath]\alpha [/imath] and [imath]\beta[/imath] are roots of equation [imath]a\tan\theta +b \sec\theta=c[/imath]. Prove that [imath]\tan(\alpha + \beta)=\frac{2ac}{a^2-c^2}[/imath] i have converted tan to sin and cos and reached to [imath]\sin^2\theta(a^2-c^2+2ac) + c^2-b^2-ac=0[/imath]. how do i proceed thanks
|
1490652
|
prove that [imath]\tan(\alpha+\beta) = 2ac/(a^2-c^2)[/imath]
if [imath]\alpha[/imath] , [imath]\beta[/imath] are the values of [imath]\theta[/imath] satisfying [imath]\displaystyle a\tan(\theta)+b\sec(\theta) = c[/imath] prove that [imath]\displaystyle\tan(\alpha+\beta) = \dfrac{2ac}{a^2-c^2}[/imath] How do i solve this, suggestions about how to go about doing this
|
2116392
|
Prove that [imath]\ln{n} \lt \sqrt{n}[/imath] for [imath]n \in \mathbb{N}^{*}[/imath] without derivative
Prove that [imath]\ln{n} \lt \sqrt{n}[/imath] for [imath]n \in \mathbb{N}^{*}[/imath]. I have tried to prove that using induction but I really don't know how to do it. I know how to solve it by creating a function [imath]f(x) = \sqrt{x}-\ln{x}[/imath] and then making the derivative, and then show that is bigger than 0. But I need it solved without math analysis (without derrivative, maybe with induction). How can I do this? Thank you very much!
|
65793
|
How to prove [imath]\log n \leq \sqrt n[/imath] over natural numbers?
It seems like [imath]\log n \leq \sqrt n \quad \forall n \in \mathbb{N} .[/imath] I've tried to prove this by induction where I use [imath] \log p + \log q \leq \sqrt p \sqrt q [/imath] when [imath]n=pq[/imath], but this fails for prime numbers. Does anyone know a proof?
|
2117787
|
Find [imath]x[/imath] positive integer such that [imath]x^4+x^3+x^2+x+1[/imath] is a perfect square
Find the number of positive integers [imath]x[/imath] for which [imath]x^4+x^3+x^2+x+1[/imath] is a perfect square. My attempts: Let [imath]x^4+x^3+x^2+x+1=k^2\implies (x+1)^2(x^2-x+1)=(k-x)(k+x)[/imath] I analysed this a bit found [imath]x=0[/imath] as one which satisfy all condition, how do I find other, please help, try to continue this further, if any other elegant method then add that too.
|
1835990
|
Integer squares of the form [imath]x^4+x^3+x^2+x+1[/imath]
I want to solve [imath]{y^2}=x^4+x^3+x^2+x+1[/imath] in [imath]\mathbb{Z}[/imath]. I can find four solutions. Is there another solution? I know that there are six solutions. [imath](-1, \pm 1)\;,\;(0, \pm 1).[/imath] My try: \begin{align} & x^4+x^3+x^2+x+1=0\quad \Rightarrow \underbrace{\left( x+\frac{1}{x} \right)^2}_{t^2}+\underbrace{\left( x+\frac{1}{x} \right)}_{t}-1=0\,\Rightarrow t\notin \mathbb{Z} \\ & x^4+x^3+x^2+x+1={\pm 1}^2\quad\Rightarrow \,x=-1\,\,,\,x=0.\\ \end{align} Thanks.
|
2117874
|
Is it true that [imath](I\cap J)^n=I^n\cap J^n[/imath]
Let [imath]R[/imath] be a ring, and let [imath]I, J\subset R[/imath] be two ideals of [imath]R[/imath]. Given [imath]n\in \mathbb{N}[/imath], is it true that [imath](I\cap J)^n=I^n\cap J^n[/imath] ? I've shown that the inclusion [imath](I\cap J)^n\subset I^n\cap J^n[/imath] is true, and I'm trying to find a counterexample to the opposite inclusion. I've tried with some rings but I did not find a counterexample.
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447660
|
Example of two prime ideals whose intersection of the squares not equal to the square of the intersection
In this topic the OP raised the following question: Let [imath]R[/imath] be a commutative noetherian ring and [imath]\mathfrak p,\mathfrak q \in \operatorname{Spec}(R)[/imath]. Is it true that [imath](\mathfrak p\cap \mathfrak q)^2=\mathfrak p^2 \cap \mathfrak q^2[/imath]? Obviously, we always have [imath](\mathfrak p\cap \mathfrak q)^2 \subseteq \mathfrak p^2 \cap \mathfrak q^2[/imath] and there is no reason to think that, in general, the converse holds. What remained unsolved in that topic is to Give an example of prime ideals [imath]\mathfrak p,\mathfrak q[/imath] (in a noetherian ring) such that [imath](\mathfrak p\cap \mathfrak q)^2 \neq \mathfrak p^2 \cap \mathfrak q^2.[/imath] Edit. It would be nice to have such an example for [imath]R[/imath] a noetherian integral domain.
|
2118024
|
Do isometries preserve the cross-product?
Suppose I have an isometry [imath]f:\Bbb R^3 \rightarrow \Bbb R^3[/imath]. In an exercise given in my geometry notes it asks to determine whether isometries preserve the cross product, so that If I have two vectors [imath]a = (a_1,a_2,a_3)[/imath] and [imath]b=(b_1,b_2,b_3)[/imath] that it is true that [imath]u \times v = f(u) \times f(v)[/imath]. Then I know that [imath]a \times b = [ (a_2b_3-a_3b_2), - (a_1b_3-a_3b_1), a_1b_2-a_2b_1)[/imath] but I am having difficulty preceeding. Edit: It is also pointed out in the comments that [imath]a \times b = ||a||b|| sin \theta n[/imath] where [imath]\theta[/imath] is the angle between [imath]a,b[/imath] and [imath]n[/imath] is a unit vector perpendicular to the plane containing [imath]a[/imath] and [imath]b[/imath]. So If I have points in [imath]P,Q,R, \in \Bbb R^3[/imath] and let [imath]u= Q-P, v= R-P, u'=f(Q)-f(P), v' = f(R)-f(P)[/imath] then [imath]|u|=|P-Q|=|f(P)-f(Q)|=|u'|[/imath] and similarly [imath]|v| = |v'|[/imath], and since isometries preserve angles then [imath]|u||v|sin \theta = |u'||v'|sin \theta [/imath] but what about the unit vector perpendicular to the plane spanned by [imath]u,v[/imath] and [imath]u',v'[/imath]? Any hints or insights much appreciated.
|
296858
|
Rotations preserve cross products?
In relation to this question here, I'm trying to prove that [imath]\bf R(a \times b) = R(a) \times R(b)[/imath] for any given rotation [imath]\bf R[/imath] in [imath]\mathbb{R}^3[/imath]. Edit: If one can prove that this holds for the unit vectors (e.g. [imath]\bf R(\hat j) \times R(\hat k) = R(\hat i)[/imath]) then that's all there is to it, given the distributivity of cross product and linearity of [imath]\bf R[/imath].
|
2118177
|
Summation of [imath]1 + \frac{1\cdot3}{6} + \frac{1\cdot3\cdot5}{6\cdot8} \cdots[/imath]
My attempt is as follows [imath] \prod_{r=1}^n (2r-1)\cdot\prod_{r=1}^{n} (2r)= (2r)![/imath] [imath]\prod_{r=1}^{n} (2r)= 2^n.r![/imath] The general term for the denominator or [imath]2^{r-1}\frac{(r+1)!}{2}[/imath] [imath]T_r = \frac{\prod_{r=1}^n (2r-1)}{2^{r-2}(r+1)!} = \frac{(2r)!}{2^{2r-2}(r+1)!.r!} = \frac{\binom{2r}{r}}{(r+1)2^{2r-2}}[/imath] How can I proceed with the summation?
|
479610
|
Sum of the series [imath]1+\frac{1\cdot 3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots[/imath]
Decide if the sum of the series [imath]1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+ \cdots[/imath] is: (i) [imath]\infty[/imath], (ii) [imath]1[/imath], (iii) [imath]2[/imath], (iv) [imath]4[/imath].
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1854319
|
Tensor product [imath]Z[1/2]\otimes Z/3[/imath]
Compute [imath]\mathbb{Z}[1/2]\otimes_{\mathbb{Z}} \mathbb{Z}/3[/imath] My initial instinct was that it is equal to [imath]\mathbb{Z}/3[1/2][/imath], but I couldn't show this is correct by the universal propert. Hence, I think that my instinct misled me. However, I would like to know what is the right way to think about it, without unnecessary guesses. Thank you!
|
1357331
|
Is [imath]\mathbb{Z}/3\otimes \mathbb{Z}[1/2]=\mathbb{Z}/3[/imath]?
The title says all. Is it true that [imath]\mathbb{Z}/3\otimes_{\mathbb{Z}} \mathbb{Z}[1/2]=\mathbb{Z}/3[/imath]? I know that [imath]\mathbb{Z}/2\otimes_{\mathbb{Z}}\mathbb{Z}[1/2]=0[/imath] because [imath]1\otimes 1=2\otimes \frac{1}{2}=0\otimes \frac{1}{2}=0[/imath].
|
587838
|
Showing that a certain map is not flat by explicit counterexample
I wish to show that the injection [imath]k[y^2, y^3] \rightarrow k[y][/imath] is not flat. I know of geometric ways to see this, but I wish to see explicitly [imath]k[y^2, y^3][/imath]-modules (or localizations thereof) [imath]0 \rightarrow M' \rightarrow M[/imath] which does not remain injective upon tensoring with [imath]k[y][/imath]. Is such a search futile?
|
669818
|
Direct proof of non-flatness
Consider [imath]k[/imath] a field and the rings [imath]A=k[X^2,X^3]\subset B=k[X][/imath]. How to prove that [imath]B[/imath] is not flat over [imath]A[/imath] by using only the definition of flatness that it maintains exact sequences after making tensor products?
|
2118683
|
If the polynomial [imath]x^4-6x^3+16x^2-25x+10[/imath] is divided by another polynomial [imath]x^2-2x+k[/imath], the
If the polynomial [imath]x^4-6x^3+16x^2-25x+10[/imath] is divided by another polynomial [imath]x^2-2x+k[/imath], theremainder is [imath]x+a[/imath], find [imath]k[/imath] and [imath]a[/imath]. My Attempt, [imath]f(x)=x^4-6x^3+16x^2-25x+10[/imath] [imath]g(x)=x^2-2x+k[/imath] [imath]R=x+a[/imath] Here, the divisor is in the quadratic form. so how do I use the synthetic division
|
170581
|
Theorem for Dividing Polynomials
When a polynomial [imath]P(x)=x^4- 6x^3 +16x^2 -25x + 10[/imath] is divided by another polynomial [imath]Q(x)=x^2 - 2x +k,[/imath] then the remainder is [imath]x+a.[/imath] I have to find the values of [imath]a[/imath] and [imath]k[/imath]. Can somebody tell me shortest way to get these values? Which theorem should be applied here?
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2118543
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Let [imath]A[/imath] and [imath]B[/imath] be sets. Suppose that there is some set [imath]X[/imath] such that [imath]A \cap X = B \cap X[/imath] and [imath]A \cup X = B \cup X[/imath]. Show that [imath]A = B[/imath].
Let [imath]A[/imath] and [imath]B[/imath] be sets. Suppose that there is some set [imath]C[/imath] such that [imath]A \cap C = B \cap C[/imath] and [imath]A \cup C = B \cup C[/imath]. Show that [imath]A = B[/imath]. My sketch is the following. Suppose first that sets are non-empty and instead that [imath]A[/imath] were not equal to [imath]B[/imath]. Then there must exist an element in either [imath]A[/imath] or [imath]B[/imath] that is not in the other. In particular, choose a in [imath]A\setminus B[/imath]. Then this a must be in [imath]A \cap C = B \cap C[/imath]. But this contradicts the fact that a was not in [imath]B[/imath]. In fact, I am interested in the logic of my steps. Any comment would be helpful.
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1951669
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Let [imath] A [/imath], [imath] B [/imath] and [imath] C [/imath] be sets. If [imath] A \cup B = A \cup C [/imath] and [imath] A \cap B = A \cap C [/imath], then show that [imath] B = C [/imath].
I’m stuck on this one problem in my textbook regarding proofs in set theory. I’ve done the following so far: Let [imath] x \in B [/imath]. As [imath] B \subseteq A \cup B [/imath], we have [imath] x \in A \cup B [/imath], so [imath] x \in A \cup C [/imath] because [imath] A \cup B = A \cup C [/imath]. Hence, [imath] x \in (A \cup B) \cap (A \cup C) [/imath], which yields [imath] x \in A \cup (B \cap C) [/imath]. Now, either [imath] x \in A [/imath] or [imath] x \in B \cap C [/imath]. If [imath] x \in B \cap C [/imath], then [imath] x \in C [/imath] because [imath] B \cap C \subseteq C [/imath], so [imath] B \subseteq C [/imath]. This is only half of the proof, though, as I have to prove that [imath] C \subseteq B [/imath] as well; and I also haven’t even considered the case [imath] x \in A [/imath] for the last step in what I have so far. I’ve drawn many different Venn diagrams, and although I see why the statement is true, I just can’t formalize it. Any pointers, help or guidance is very much appreciated. Thanks!
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838082
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[imath]\alpha[/imath] and [imath]\beta[/imath] are solution of [imath]a \cdot \tan\theta + b \cdot \sec\theta = c[/imath] show [imath] \tan(\alpha + \beta) = \frac{2ac}{a^2 - c^2}[/imath]
If [imath]\alpha[/imath] and [imath]\beta[/imath] are the solution of [imath]a \cdot \tan\theta + b \cdot \sec\theta = c[/imath], then show that [imath] \tan(\alpha + \beta) = \frac{2ac}{a^2 - c^2}[/imath] I did the following: [imath]a \cdot \tan\theta + b \cdot \sec\theta = c[/imath] Squaring both sides, [imath]a^2 \cdot \tan^2\theta + b^2 \cdot \sec \theta + 2 \cdot a \cdot b \cdot \tan \theta \cdot \sec \theta = c^2[/imath] This implies [imath](a^2 + b^2) \cdot \sin^2 \theta + 2 \cdot a \cdot b \cdot \sin \theta + (b^2 - c^2) = 0[/imath] Hence I get [imath]\sin \alpha + \sin \beta = \frac{-2ab}{a^2 + c^2}[/imath] and [imath]\sin \alpha \cdot \sin \beta = \frac{b^2 - c^2}{a^2 - c^2}[/imath] But I am not able to solve it further.
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2383208
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Solving equations involving both algebra, trigonometry (and possible use of inequalities)
If [imath]\alpha[/imath] and [imath]\beta[/imath] are solutions [imath]a\tan\theta+ b\sec\theta = c[/imath] then show that: [imath]\tan(\alpha+\beta)= \dfrac{2ac}{a^2-c^2}[/imath] This is the first time I am attempting such a question so I am confused. Nevertheless, I gave it a try: [imath]\dfrac{a\sin\theta}{\cos\theta}+\dfrac{b}{\cos\theta}=c [/imath] [imath]\implies a\sin\theta -c\cos\theta= -b[/imath] Now, [imath]-\sqrt{a^2+c^2}\le a\sin\theta -c\cos\theta\le\sqrt{a^2+c^2}[/imath] or, [imath]-\sqrt{a^2+c^2}\le -b\le\sqrt{a^2+c^2}[/imath] I don't know how to proceed from here. Also, is this a valid approach to the question?
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2118849
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If [imath]S = x_1 + x_2 + .. + x_n[/imath], Prove that [imath] (1+x_1)(1+x_2)..(1+x_n) \le 1 + S + \frac{S^2}{2!} + .. + \frac{S^n}{n!}[/imath]
Let [imath]x_1, x_2, \ldots ,x_n[/imath] be positive real numbers, and let [imath] S = x_1 + x_2 + \cdots + x_n.[/imath] Prove that [imath] (1+x_1)(1+x_2)\ldots(1+x_n) \le 1 + S + \frac{S^2}{2!} + \cdots + \frac{S^n}{n!}[/imath] Here's my attempt: Case 1 : Let [imath]x_1 = x_2= x[/imath] (When all terms are equal) [imath]LHS = (1+x)(1+x) = 1+2x+x^2[/imath] [imath]RHS = 1+\frac{x+x}{1!}+\frac{(x+x)^2}{2!}=1+2x+\color{blue}{2}x^2[/imath] Hence when all terms are equal, [imath]LHS<RHS[/imath]. Case 2 : [imath]x_1\ne x_2[/imath] [imath]LHS=(1+x_1)(1+x_2)=1+{x_1}{x_2}+(x_1+x_2)[/imath] [imath]RHS=1+ (x_1+x_2)+\frac{(x_1+x_2)^2}{2!} =1+{x_1}{x_2}+(x_1+x_2)+\color{blue}{\left(\frac{x_1^2+x_2^2}{2}\right)}[/imath] Hence when even one term is not equal, [imath]LHS\lt RHS[/imath]. From the two cases, it's clear that under no circumstance [imath]LHS \gt RHS[/imath]. And hence the (partial) proof. Is there any way I can improve my proof? If you look again I've considered only two terms and my intuition tells me that it would apply to the entire range but how do I state it mathematically? Or can you think of a better or more rigorous proof than this? Thanks again!
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326721
|
Prove that: [imath](1+a_1)(1+a_2)...(1+a_n) \le 1 + S_n + (S_n)^2/2! + ... + (S_n)^n/n![/imath]
I am currently working on this problem from Hardy's Course of Pure Mathematics and have gotten stuck near the end. I was wondering if someone could help me determine what to go next. Question If [imath]a_1, a_2, ...,a_n[/imath] are all positive and [imath]S_n=a_1+a_2+...+a_n[/imath] then: [imath](1+a_1)(1+a_2)...(1+a_n) \le 1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!}[/imath] My Attempt Proof by induction: I had first shown that it was true for [imath]n=1[/imath] and [imath]n=2[/imath]. Now suppose that it is true for n. Then it must also be true for [imath]n+1[/imath] [imath](1+a_1)(1+a_2)...(1+a_n)(1+a_{n+1}) \le 1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!} + \dfrac{(S_{n})^{n+1}}{(n+1)!}[/imath] Define: [imath](1+a_1)(1+a_2)...(1+a_n)=x[/imath] and [imath]1 + S_n + \dfrac{(S_n)^2}{2!} + ... + \dfrac{(S_n)^n}{n!}=y[/imath] Then: [imath]x+xa_{n+1} \le y+\dfrac{(S_{n})^{n+1}}{(n+1)!}[/imath] By the induction assumption, we know that [imath]x \le y[/imath]. What I can't figure out From this point, what I think the next natural step would be is to show that [imath]xa_{n+1} \le \dfrac{(S_{n})^{n+1}}{(n+1)!}[/imath] I know this rearranges to: [imath]xa_{n+1} \le \dfrac{(S_{n})^{n}}{n!} \times \dfrac{S_n}{(n+1)} [/imath] And feel there may be something I can do here, but haven't been able to figure anything out.
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2119276
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Proof that [imath]2^{10}+5^{12}[/imath] is a composite number
A hint or full answer will help a lot, because I have no idea what to do.
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978222
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Prove that [imath]2^{10}+5^{12}[/imath] is composite
Prove that [imath]2^{10}+5^{12}[/imath] is composite I need to solve this using only high school mathematics. Any ideas?
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2119547
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How can I prove that [imath](n^2+11n)/6[/imath] has no remainder?
I must prove, that when dividing the following by 6 it yields no remainder: [imath]n^2+11n [/imath] Is this done with mathematical induction method? and what other technique can I use? Update: some may find this question as a potential duplicate of the correct case when we have [imath]n^3[/imath] instead of [imath]n^2[/imath]. I want to clarify that this is not true.
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1204306
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Prove by induction that [imath]n^3 + 11n[/imath] is divisible by [imath]6[/imath] for every positive integer [imath]n[/imath].
Prove by induction that [imath]n^3 + 11n[/imath] is divisible by [imath]6[/imath] for every positive integer [imath]n[/imath]. I've started by letting [imath]P(n) = n^3+11n[/imath] [imath]P(1)=12[/imath] (divisible by 6, so [imath]P(1)[/imath] is true.) Assume [imath]P(k)=k^3+11k[/imath] is divisible by 6. [imath]P(k+1)=(k+1)^3+11(k+1)=k^3+3k^2+3k+1+11k+11=(k^3+11k)+(3k^2+3k+12)[/imath] Since [imath]P(k)[/imath] is true, [imath](k^3+11k)[/imath] is divisible by 6 but I can't show that [imath](3k^2+3k+12)[/imath] is divisible by 6
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2108698
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If [imath]R=F[x,y,z]/(x^2-y^2z)[/imath] why [imath]\operatorname{trdeg}_F R=2[/imath]?
Let [imath]F[/imath] be a field. I know that [imath]F[x,y,z][/imath] is a UFD, that [imath]x^2-y^2z\in F[x,y,z][/imath] is prime. Let [imath]R:=F[x,y,z]/(x^2-y^2z)[/imath]. Q1) Why [imath]R[/imath] is an integral domain ? It's clear that [imath]R[/imath] is a field, and thus it's a domain, but why is it integral ? (maybe a field is always integral). Let [imath]S[/imath] be the integral closure of [imath]R[/imath]. We have that [imath]\bar x^2=\bar y^2\bar z\implies \left(\frac{\bar x}{\bar y}\right)^2=\bar z[/imath], we have that [imath]\frac{\bar x}{\bar y}[/imath] is integral over [imath]R[/imath]. Q2) Why now we have that [imath]S\supset F[\frac{\bar x}{\bar y},\bar y][/imath] ? After, they say that [imath]\frac{\bar x}{\bar y}[/imath] and [imath]\bar y[/imath] are algebraically independent, otherwise, [imath]2>\operatorname{trdeg}_F \operatorname{Frac} F\left[\frac{\bar x}{\bar y},\bar y\right]\geq \operatorname{trdeg}_F R=2,[/imath] but I don't understand each inequality. Q3) I mean, why [imath]2>\operatorname{trdeg}_F \operatorname{Frac} F\left[\frac{\bar x}{\bar y},\bar y\right]\geq \operatorname{trdeg}_F R=2[/imath], why [imath]\operatorname{trdeg}_F \text{Frac}( F\left[\frac{\bar x}{\bar y},\bar y\right])\geq\operatorname{trdeg}_F R[/imath], and why [imath]\operatorname{trdeg}_F R=2[/imath] ?
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2119160
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Compute integral closure of [imath]F[x,y,z]/(x^2-y^2z)[/imath].
I want to compute integral closure of [imath]R:=F[x,y,z]/(x^2-y^2z)[/imath]. Let [imath]S[/imath] the integral closure. I have proved that [imath]\frac{\bar x}{\bar y}[/imath] and [imath]\bar y[/imath] are integral over [imath]R[/imath] and that [imath]\{\bar y,\frac{\bar x}{\bar y}\}[/imath] is a transcendence basis of [imath]\text{Frac}(R)[/imath]. My questions are the following one : Q1) Why [imath]S\supset F[\bar y,\frac{\bar x}{\bar y}][/imath] ? I agree that if [imath]S\supset R[\bar y,\frac{\bar x}{\bar y}]\subset S[/imath], but why [imath]F[\bar y,\frac{\bar x}{\bar y}]\subset S[/imath] ? Q2) Why [imath]\text{trdeg}_F(\text{Frac}(R))\leq \text{trdeg}_F(\text{Frac}(F[\bar y,\frac{\bar x}{\bar y}])) \ \ ?[/imath] With those answer, I can conclude.
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2119982
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Automorphism group of the direct product of cyclic groups
I was wondering what the order of the group [imath]Aut(Z_{p^a} \times Z_{p^b})[/imath] is, where [imath]p[/imath] is prime and [imath]a [/imath] and [imath]b[/imath] are natural numbers, eventually equal. I had found a way on internet that was using matrices, but it looks like I cannot find that topic anymore, and I think I hadn't really understood that explanation, could someone please explain how to count them? By the way, it seems that using matrices is the best option (as I clearly remember the explanation I had found wasn't very long and looked like it didn't require many calculations), but couldn't I just calculate how many "good" possibilities there are? I mean, I would send the generator of [imath]Z_{p^b}[/imath] in another element of order [imath]p^b[/imath], then the generator of [imath]Z_{p^a}[/imath] in another element of order [imath]p^a[/imath]. The problem is that the cyclic subroups generated by the images of the generators must have a trivial intersection. So I was hoping to find how many choices I had for the second generator, but here it gets tricky and I don't know how to proceed, how to count how many "good" choices I have. Thanks for the help
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27200
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Number of automorphisms of a direct product of two cyclic [imath]p[/imath]-groups
Suppose I have [imath]G = Z_{p^m} \times Z_{p^n}[/imath] for [imath]m, n[/imath] distinct natural numbers and [imath]p[/imath] a prime. Is there a combinatorial way to determine the number of automorphisms of [imath]G[/imath]?
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2119600
|
Find the sum of the series [imath]\frac{1}{1\cdot 2\cdot 3}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{7\cdot 8\cdot 9}+...=[/imath]
Given : [imath]\frac{1}{1\cdot 2\cdot 3}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{7\cdot 8\cdot 9}+ ...[/imath] What is the sum of this series, how can one rewrite it to look simpler ? EDIT : Actually I found how to rewrite it : [imath]\sum _{n=1}^{\infty }\:\frac{1}{\left(3n-2\right)\left(3n-1\right)3n}[/imath]
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573119
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How find this sum [imath]\sum\limits_{n=0}^{\infty}\frac{1}{(3n+1)(3n+2)(3n+3)}[/imath]
Find this sum [imath]I=\sum_{n=0}^{\infty}\dfrac{1}{(3n+1)(3n+2)(3n+3)}[/imath] My try: let [imath]f(x)=\sum_{n=0}^{\infty}\dfrac{x^{3n+3}}{(3n+1)(3n+2)(3n+3)},|x|\le 1[/imath] then we have [imath]f^{(3)}(x)=\sum_{n=0}^{\infty}x^{3n}=\dfrac{1}{1-x^3}[/imath] then we find the [imath]f(x)[/imath],Following is very ugly(can you someone can post your follow solution,) have other simple methods? Thank you very much.
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2118125
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constant to the [imath]\log(n)[/imath] equals [imath]n[/imath] to [imath]\log(\text{constant})[/imath]
Could someone explain to me why [imath]x^{\log(n)} = n^{\log(x)}[/imath] in simple terms? I tried to simply take the [imath]log[/imath] of both sides but it doesn't work out or simplify.
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320116
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Logarithm proof problem: [imath]a^{\log_b c} = c^{\log_b a}[/imath]
I have been hit with a homework problem that I just have no idea how to approach. Any help from you all is very much appreciated. Here is the problem Prove the equation: [imath]a^{\log_b c} = c^{\log_b a}[/imath] Any ideas?
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2120317
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Show that the sequence of functions [imath]{f_n}(x)=\frac{x}{1+nx^2}[/imath] converges uniformly
For all [imath]n\geq 1[/imath], let [imath]f_n: \mathbb{R}\rightarrow \mathbb{R}[/imath] be defined by [imath]f_n(x) = \frac{x}{1+nx^2}[/imath]. Show that the sequence of functions [imath]{f_n}[/imath] converges uniformly to some function [imath]f:\mathbb{R}\rightarrow\mathbb{R}[/imath]. My attempt: I think as [imath]n\rightarrow\infty[/imath], [imath]f_n(x) = \frac{\frac{1}{n}x}{\frac{1}{n}+x^2}[/imath] goes to [imath]\frac{1}{nx}[/imath], which goes to [imath]0[/imath] (is this true? I'm having trouble rigorously justifying it). Then I want to show that for all x, [imath]\sup_{x\in\mathbb{R}}|f_n(x)|\rightarrow 0[/imath] as [imath]n\rightarrow\infty[/imath]. This I'm having trouble with, as the reasoning seems similar to above, but I'm not positive. Any help appreciated!
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578337
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Uniform convergence of a function
For [imath]n=1,2,3,\dots,[/imath] and [imath]|x| < 1[/imath] I need to prove that [imath]\frac{x}{1+nx^2}[/imath] converges uniformly to zero function. How ?. For [imath]|x| > 1[/imath] it is easy.
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1829075
|
Let [imath]f_1[/imath] and [imath]f_2[/imath] be bounded functions on [imath][a, b][/imath]. Prove that [imath]L(f_1) + L(f_2) \leq L(f_1 + f_2)[/imath] (Darboux intergral)
Let [imath]f_1[/imath] and [imath]f_2[/imath] be bounded functions on [imath][a, b][/imath]. Prove that [imath]L(f_1) + L(f_2) \leq L(f_1 + f_2)[/imath] For this question specifically, I know that I am supposed to show that the [imath]\sup L(f_1,p) + \sup L(f_2,p) \leq \sup L(f_1+f_2, p)[/imath] But I get confused when considering the sup of these three functions since the partition is any arbitrary partition. I have also thought about trying to show [imath]\inf f_1 + \inf f_2 \leq \inf (f_1+f_2)[/imath] first in every interval, but I got stuck in terms of how to prove the above statement. New to real analysis, would appreciate any insights.
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1290105
|
Let [imath]f_1 , f_2: I\mapsto \mathbb{R}[/imath] bounded functions. Show that [imath]L(f_1)+L(f_2)\leq L(f_1+f_2)[/imath] (Riemann integral)
Let [imath]f_1 , f_2: I\mapsto \mathbb{R}[/imath] bounded functions. Show that [imath]L(f_1)+L(f_2)\leq L(f_1+f_2)[/imath] where [imath]L(F)[/imath] is the supremum of the lower sums of the Riemann integral. I tried to by contradicction [imath]L(f_1)+L(f_2)> L(f_1+f_2)[/imath] and isoleting [imath]L(f_1)> L(f_1+f_2)-L(f_2)[/imath] and usig the definition of sup. I also tried to use [imath]\sup(A)+\sup(B)=\sup(A+B)[/imath] but it din't work. Thanks.
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2119360
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Prove that [imath]\sum\limits_{cyc} a^7 \geq \sum\limits_{cyc}a^4b^3[/imath]
Prove that [imath]a^7+b^7+c^7\ge a^4b^3+b^4c^3+c^4a^3[/imath] SOURCE : "A Brief Introduction to Olympiad Inequalities" by Evan Chen It was one of the practice problems. Equality case is easy. I tried AM-GM and Muirhead, but could not seem to find a suitable proof for the inequality. Any hint would be really helpful. Thanks in advance ! ^__^
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1325652
|
Prove that [imath]a^7 + b^7 + c^7 \geq a^4b^3 + b^4c^3 + c^4a^3[/imath]
Prove that [imath]a^7 + b^7 + c^7 \geq a^4b^3 + b^4c^3 + c^4a^3[/imath] Values [imath]a,b,c[/imath] are all positive reals. I tried Muirhead and a few AM [imath]\geq[/imath] GM. This problem is equivalent to proving [imath]a^4b^3 + b^4c^3 + c^4a^3 \geq c^2a^5 + b^2c^5+a^2b^5[/imath].
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2119784
|
Need help for proving: [imath]f(f^{−1}(A)) ⊆ A[/imath].
Informations: [imath]f:X \longrightarrow X[/imath] and [imath]A \subseteq X[/imath]. How can i prove this statement: [imath]f(f^{-1}(A)) \subseteq A[/imath] This is my thoughts until now: [imath]f^1(A)=\{x\in X |f(x)\in A\} \subseteq X[/imath]. [imath]f(A)=\{f(x)|x\in A\}[/imath] [imath]f(f^1(A))=\{y\in A:\exists \in f^1(A):y=f(x)\} \in A[/imath]
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2119866
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Need help for proving that: [imath]f(f^{-1}(A)) ⊆ A[/imath]
I realised I needed to show more information, which I now did: [imath]f: X \rightarrow X \,\,\,\rm{and}\,\,\,\, A \subseteq X[/imath] Proof that: [imath]f(f^{-1}(A)) \subseteq A[/imath] This is my proof: By defintion: [imath]f^{-1}(A)=\{x \in X\mid f(x) \in A\}[/imath] and [imath]f(A)=\{f(x) \mid x \in A\} = \{y \in X \mid \exists x \in A: y=f(x)\} \subseteq X[/imath] Therefore we can end the proof by a final definition:\ [imath]f(f^{-1}(A))=\{y \in A: \exists x \in f^{-1}(A):y=f(x)\} \subseteq A[/imath] Is this a legit "proof"? And is it even a proof, when i only use definitions?
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2120893
|
Method of finding all non-trivial homomorphism [imath]\mathbb{Z_{14}}\rightarrow \mathbb{Z_{21}}[/imath]
I have the following question : Find all non-trivial homomorphism [imath]\mathbb{Z_{14}}\rightarrow \mathbb{Z_{21}}[/imath] I wonder how to solve such problems when the groups are pretty big to check the definition of each possibility. Any ideas? Thank you.
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617497
|
Finding all homomorphisms between two groups - couple of questions
Consider [imath]\mathbb{Z}_{15}[/imath], and [imath]\mathbb{Z}_{18}[/imath]. Let's say I want to find all homomorphisms [imath]f:\mathbb{Z}_{15}\rightarrow \mathbb{Z}_{18}[/imath]. I'm not interested in the answer in particular, mostly I'm concerned about understanding the properties of homomorphism, so I can answer these kind of questions myself. So, first of all, I know that homomorphism of cyclic group is completely determined by it's generator. But, will any mapping do? For example, the easiest one to find is [imath]f(1)=0[/imath], where [imath]Imf=\left\{0\right\}[/imath], and [imath]Ker=G[/imath] (correct me if I'm wrong, this kind of [imath]f[/imath] can be defined between any two groups). Now, what things do I need to consider, when trying to find another one (if it exist)? Can I decide that [imath]f(1)=1[/imath]? (It is not onto, but that shouldn't bother me) And what about [imath]f(1)=2[/imath]? and so on... My second question is: what about non-cyclic groups? Consider [imath]D_{10}[/imath] and [imath]\mathbb{Z}_{18}[/imath], for example. Do I need to go and define [imath]f[/imath] for each and every [imath]g\in D_{10}[/imath]? (it doesn't have a generator) A link to a useful (and simple) summary regarding homomorphisms properties will also be great. Thank you in advance.
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2120468
|
How do I integrate [imath]\sqrt{a^2 - x^2}[/imath]?
How do I integrate [imath]\displaystyle \int \sqrt{a^2 - x^2}dx [/imath] It can be solved using trig substitution, but don't know how to solve. Thank you.
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694853
|
Different methods of evaluating [imath]\int\sqrt{a^2-x^2}dx[/imath]:
Is there a simple and nice way to solve [imath]\int\sqrt{a^2-x^2}dx[/imath]: PS:I am not looking for a substitution like [imath]x=a\sin p[/imath],
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2121155
|
I was solving limits and it said [imath]\lim_{x→0}[x^2/\tan(x)\sin(x)]=0[/imath]
I tried using [imath][x][y]=[xy][/imath] but clearly, that is wrong I know that [imath]\lim_{x→0}[\sin(x)x]=0,\lim_{x→0}[\sin(x)x]=0[/imath] and [imath]\lim_{x→0}[\tan(x)x]=1[/imath].
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2121119
|
Does [imath][x]\cdot [y]=[xy][/imath]? Here [imath][.][/imath] is the GIF
I was solving limits and it said [imath]\lim_{x\to 0} \left[\frac{x^2}{\tan(x)\sin(x)}\right]=0[/imath] I tried thinking about it but since I am not really good at math I couldn't proceed. I know that [imath]\lim_{x\to 0} \left[\frac{\sin(x)}{x}\right] = 0[/imath] and [imath]\lim_{x\to 0} \left[\frac{\tan(x)}{x}\right] = 1[/imath]
|
437859
|
Compute [imath]\mathrm{Aut}(S)[/imath] of the ring [imath]S=\mathbb{Q}[x]/(x^2)[/imath]
This problem once again is from a previous exam. The problem is to compute the group of automorphisms of the ring [imath]S[/imath] where [imath]S=\mathbb{Q}[x]/(x^2)[/imath] My thoughts: Well [imath]x^2[/imath] is reducible over [imath]\mathbb{Q}[/imath] so I am completely stuck. Also what exactly is the theory I need to know to solve this problem?. I don't even know where to look in the text book to find the relevant theory for this. In particular I have not learned Galois Theory. In fact we don't need to know Galois' theory for my exam. So I'm hoping this doesn't use any of that heavy machinery. Can you guys help? Thanks for your help and answers.
|
270637
|
Determine [imath]\text{Aut}(S)[/imath] where [imath]S = \mathbb{Q}[x]/(x^2)[/imath]
Here's the problem: if [imath]S = \mathbb{Q}[x]/(x^2)[/imath], compute the group of automorphisms [imath]\mbox{Aut}(S)[/imath]. First off, I have the solution and I know it's isomorphic to the multiplicative group [imath]\mathbb{Q}\backslash\{0\}[/imath], so I'm trying to understand it. It uses what is called the substitution principle and the fact that the restriction of any endomorphism [imath]\phi: \mathbb{Q}[x]\to \mathbb{Q}[x][/imath] to [imath]\mathbb{Q}[/imath] is the identity. It says that these two facts implies that the endomorphisms of [imath]\mathbb{Q}[x][/imath] are in 1-1 correspondence with [imath]h \in \mathbb{Q}[x][/imath], where the endomorphism corresponding to [imath]h[/imath] is [imath]\phi_h : \mathbb{Q}[x] \to \mathbb{Q}[x][/imath] given by [imath]\phi_h(f) = f(h)[/imath] for [imath]f \in \mathbb{Q}[x][/imath]. I guess I just don't understand how it applied the substitution principle? It's probably obvious, but for some reason I am not getting it. An explanation would be greatly appreciated! Or if you have a different solution, that'd be cool too. Thanks!!
|
1932358
|
Find the value of [imath]\frac{\alpha^{2014}+\beta^{2014}+\alpha^{2016}+\beta^{2016}}{\alpha^{2015}+\beta^{2015}}[/imath] for the roots of [imath]x^2-3x+1[/imath]
I got this problem from my textbook, not school. I tried various methods but was unable to solve the problem. Let [imath]\alpha[/imath] and [imath]\beta[/imath] be the roots of the polynomial [imath]x^2-3x+1[/imath]. I need to find the value of [imath] \frac{\alpha^{2014}+\beta^{2014}+\alpha^{2016}+\beta^{2016}}{\alpha^{2015}+\beta^{2015}}[/imath]
|
789901
|
Find [imath]\alpha^{2016} + \beta^{2016} + \alpha^{2014} + \beta^{2014} \over \alpha^{2015} + \beta^{2015}[/imath] for zeroes of the polynomial [imath]x^2+3x +1[/imath]
We have a polynomial [imath]x^2 + 3x + 1[/imath]. There are 2 zeroes of it, [imath]\alpha[/imath] and [imath]\beta[/imath] Now, what we need to find out is(What I couldn't) is as follows: [imath] \alpha^{2016} + \beta^{2016} + \alpha^{2014} + \beta^{2014} \over \alpha^{2015} + \beta^{2015}[/imath] Any ideas on how to do it? Thanks a lot! -bone
|
2121712
|
Can a continuous function have an indefinite integral that is discontinuous at a point?
I was asked to evaluate the indefinite integral [imath]\int \frac{dx}{2 + \sin x}[/imath]. I got the result as [imath] \int \frac{dx}{2 + \sin x } = \frac{2}{\sqrt{3}} \arctan \bigg( \frac{2\tan(\frac{x}{2}) + 1}{\sqrt{3}} \bigg) + C [/imath] [imath]f(x) = 2 + \sin x[/imath] is continuous for all [imath]x \in \mathbb{R}[/imath], though its anti-derivative is discontinuous at [imath]x = (2k + 1) \pi[/imath] and thus non differentiable at [imath]x = (2k + 1)\pi[/imath] for [imath]k \in \mathbb{Z}[/imath]. Does this mean the indefinite integral is invalid at [imath]x = \pi[/imath] and its multiples?
|
1653406
|
Anti-derivative of continuous function [imath]\frac{1}{2+\sin x}[/imath]
I use tangent half-angle substitution to calculate this indefinite integral: [imath] \int \frac{1}{2+\sin x}\,dx = \frac{2}{\sqrt{3}}\tan^{-1}\frac{2\tan \frac{x}{2}+1}{\sqrt{3}}+\text{constant}. [/imath] Wolfram Alpha also give the same answer. However, [imath]\frac{2}{\sqrt{3}}\tan^{-1}\frac{2\tan \frac{x}{2}+1}{\sqrt{3}}[/imath] is discontinuous on [imath](n+1)\pi[/imath] where [imath]n[/imath] is any integer. Why is an anti-derivative of a continuous function discontinuous?
|
2121872
|
Powers of distinct primes in PID are relatively prime
Let [imath]R[/imath] be a principal ideal domain containing distinct prime elements [imath]p, q[/imath]. Let [imath]n, m \geq 1[/imath] be integers. Is there a simple proof for the claim that the elements [imath]p^n[/imath] and [imath]q^m[/imath] of [imath]R[/imath] are relatively prime, i.e. that [imath](p^n) + (q^m) = R[/imath], or equivalently that the greatest common divisor of [imath]p^n[/imath] and [imath]q^m[/imath] is [imath]1[/imath]?
|
166839
|
If [imath]\gcd(a,b)=1[/imath], then [imath]\gcd(a^n,b^n)=1[/imath]
If [imath]\gcd(a,b)=1[/imath], then [imath]\gcd(a^n,b^n)=1[/imath] This seems clear, but I don't know how to prove this.. I was trying to show this by induction such that if [imath]a^{n+1}[/imath] = [imath]rs[/imath] and [imath]b^{n+1}[/imath] = [imath]rt[/imath], then [imath]s,t[/imath] are divisible by [imath]a,b[/imath] respectively, but i think this is a wrong way..
|
2122018
|
[imath]|f(z)|\leq 1[/imath], then possible value of [imath](e^f)''(0)[/imath]
Let [imath]f[/imath] be an analytic function on [imath]\bar{D}=\{z\in \mathbb{C}:|z|\leq1\}[/imath]. Assume that [imath]|f(z)|\leq1~ \forall z\in \bar{D}[/imath]. Then, which of the following is NOT a possible value of [imath](e^f)''(0)[/imath]. (i)6 (ii)2 (iii)[imath]\frac{7}{9}e^{\frac{1}{9}}[/imath] (iv)[imath]\sqrt{2}+i\sqrt{2}[/imath]. I calculated [imath](e^f)'=f'e^f[/imath]. Here if [imath](e^f)'(0)[/imath] were [imath]0[/imath] then I could have used Schwarz Lemma to find the estimate. But here it is not [imath]0[/imath]. Don't know how to proceed.
|
1632871
|
which of the following is NOT a possible value of [imath](e^{f})''(0)[/imath]??
Let [imath]f[/imath] be an analytic function on [imath]\bar{} = \{z \in \mathbb{C}: |z| \le 1\}[/imath]. Assume that [imath]|()| ≤ 1[/imath] for each [imath]z\in \bar{D}[/imath]. Then, which of the following is NOT a possible value of [imath](e^{f})''(0)[/imath]?? [imath](A) 2[/imath] [imath](B) 6[/imath] [imath](C) \frac{7}{9}e^{\frac{1}{9}}[/imath] [imath](D) √2 + √2[/imath] So I take a look at Cauchy Integral Formula and then [imath](e^{f})''(0)=\frac{1}{\pi}\int_{|z|=1}\frac{e^{f(z)}}{z^3}[/imath] Taking modulus on both sides we have [imath]|(e^{f})''(0)| \le \frac{e}{\pi}2\pi=2e[/imath] Hence (b) cannot be the choice. Is this alright!! Thanks for the help!
|
2122164
|
Can there exist a holomorphic function such that [imath]f(\frac1n) = \frac{1}{2^n}[/imath] where [imath]n \in \Bbb Z[/imath]
Can there exist a holomorphic function defined on the unit disk [imath]D = \{ z \mid |z| <1\}[/imath] such that [imath]f(\frac1n) = \frac{1}{2^n}[/imath] where [imath]n \in \Bbb Z[/imath]? I thought of using CR equations to find a contradiction but unable to do.
|
392682
|
There does not exist an entire function which satisfies [imath]f({1\over n})={1\over 2^n}[/imath]?
There does not exist an entire function which satisfies [imath]f({1\over n})={1\over 2^n}[/imath], what I tried is if possible then define [imath]g(z)=f(z)-{1\over 2^{1\over z}}[/imath] Then [imath]g({1\over n})=0[/imath] and so [imath]g(z)[/imath] is entire and its [imath]0[/imath] set has limit point in it and so [imath]f(z)={1\over 2^{1\over z}}[/imath] which is not analytic at [imath]0[/imath]? Please help! Edit: OOps! the way I defined [imath]g(z)[/imath] that is not entire! could any one give me hint?
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2121039
|
Proving equality of index when [imath]G=AB[/imath]
Let [imath]G[/imath] be a group and [imath]A,B \leq G[/imath]. If [imath]B[/imath] is of finite index in [imath]G[/imath], then [imath]A\cap B[/imath] is of finite index in [imath]A[/imath] and we have [imath][A:A\cap B]\leq[G:B][/imath] It is fairly easy to show that the inequality holds, but what bothers me more is the statement that equality holds if and only if [imath]G=AB[/imath]. Has someone a hint for me how to prove this? Edit. The answer of Dietrich Burde is really nice! I learned alot from it. Also I learned how to alter my proof to yield the desired result. Consider the mapping [imath]\{x(A \cap B) : x \in A\} \to \{xB : x \in G\}\\ x(A \cap B) \mapsto xB[/imath] Assume [imath]xB = yB[/imath]. Then [imath]y^{-1}x \in B[/imath] but also [imath]y^{-1}x \in A[/imath] so [imath]y^{-1}x \in A \cap B[/imath] which implies [imath]x(A \cap B) = y(A \cap B)[/imath]. Hence the mapping is injective and thus the inequality follows. Now if [imath]G = AB[/imath] we have that [imath]\{xB : x \in G\} = \{xB : x \in AB\} =\{xB : x \in A\}[/imath] and thus the mapping is clearly surjective.
|
1873474
|
If [imath][G:K][/imath] is finite, then [imath][H:H \cap K] = [G:K][/imath] iff [imath]G = HK[/imath] (Hungerford Proposition 4.8, Proof)
The following is from Hungerford's graduate algebra book, Chapter 1, Section 4, Proposition 8. The proposition states that: If [imath]H[/imath] and [imath]K[/imath] are subgroups of a group [imath]G[/imath], then [imath][H:H \cap K] \leq [G:K][/imath]. If [imath][G:K][/imath] is finite, then [imath][H:H \cap K] = [G:K][/imath] if and only if [imath]G = HK[/imath]. I am trying to work my way through the proof (the book only provides an outline). I have show that [imath][H:H \cap K] \leq [G:K][/imath]. I am now trying to show the second part, that [imath][H:H \cap K] = [G:K][/imath] if and only if [imath]G = HK[/imath]. The outline says that to show this we prove two things: 1.) [imath][H:H \cap K] = [G:K][/imath] if and only if [imath]f[/imath] is surjective, where [imath]f[/imath] is the map given by [imath](H \cap K)h \rightarrow Kh[/imath] (I have shown this part). 2.) [imath]f[/imath] is surjective if and only if [imath]G = HK[/imath]. I am stuck on showing the second part. Thank you in advance for any help you can give.
|
2122797
|
Find limit of [imath]\lim_{x\to \infty} \sqrt{x^2+x}-\sqrt{x^2-x}[/imath]
I need to find the limit of the following sequence: [imath]\lim_{x\to \infty} \sqrt{x^2+x}-\sqrt{x^2-x}[/imath] I can transform the above to: [imath]\frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}}[/imath] But I can't seem to prove that the term is diminishing and that its superlum is [imath]1[/imath] (which would prove the limit). Am I going about this completely the wrong way?
|
597177
|
Find the limit: [imath]\displaystyle\lim_{x\to \infty} (\sqrt{x^2+x}-\sqrt{x^2-x})[/imath].
Find the following limit: [imath] \lim_{x\to \infty} (\sqrt{x^2+x}-\sqrt{x^2-x} )[/imath] I tried to simplify using conjugation. This gave me the following: [imath] \lim_{x\to \infty} \frac{2x}{\sqrt{x^2+x}+\sqrt{x^2-x}} [/imath] When I plug in the [imath]\infty[/imath], I'm left with [imath] \frac{\infty}{\infty} [/imath]. Did I mess up somewhere, or does the limit not exist?
|
2122159
|
Is the next claim about automorphism group [imath]{\rm \mathit Aut} _\mathbb{Q} (\mathbb{C})[/imath] correct?
Let [imath]G={\rm \mathit Aut} _\mathbb{Q} (\mathbb{C})[/imath] be the automorphism group. [imath]K \subset \mathbb{C}[/imath] is defined as [imath]K =\ \bigr\{ \alpha \in \mathbb{C} \ |\ \sigma(\alpha) = \alpha \ ( \forall \sigma \in G)\bigl\}[/imath] . Then [imath]K= \mathbb{Q} \ [/imath] is true??
|
41296
|
What is [imath]\mathbb{C}^{Aut(\mathbb{C}/\mathbb{Q})}[/imath]?
Let [imath]Aut(\mathbb{C}/\mathbb{Q})[/imath] be the field automorphisms of [imath]\mathbb{C}[/imath], and [imath]\mathbb{C}^{Aut(\mathbb{C}/\mathbb{Q})}[/imath] the subfield of [imath]\mathbb{C}[/imath] fixed by this group. I supsect that it is equal to [imath]\mathbb{Q}[/imath] but I have difficulty proving it. Here is what I do : let [imath]x \notin \mathbb{Q}[/imath]. We have to find an automorphism not fixing [imath]x[/imath]. It is easy to find an embeding [imath]\mathbb{Q}(x) \rightarrow \mathbb{C}[/imath] not sending [imath]x[/imath] on [imath]x[/imath], and use Zorn's lemma to extend it to a maximal subfield (of [imath]\mathbb{C}[/imath]) [imath]K \supset \mathbb{Q}(x)[/imath]. But it is not true that [imath]K=\mathbb{C}[/imath]. So how to handle it ?
|
2122974
|
Is there an inner product on [imath]\mathcal{C}(\Bbb R,\Bbb R)[/imath]?
Is there an inner product on the vector space of continuous real functions [imath]\mathcal{C}(\Bbb R,\Bbb R)[/imath] ? If so what is it and if not why ?
|
814754
|
Inner product on [imath]C(\mathbb R)[/imath]
With Axiom of choice it is possible to construct an inner product on [imath]C(\mathbb R)[/imath]. My question is, is it possible to explicitly construct an inner product on [imath]C(\mathbb R)[/imath]? I.e. to give a closed formular to calculate the inner product? I know it is straight-forward to write down a scalar product using a Hamel basis. This is not the answer I am looking for. This question came to me, when a student asked me in the lecture today 'whether there are vector spaces without inner products'. So I tried to find scalar produces for function spaces. I think I managed to write one down for [imath]L^1((0,1))[/imath]. But I failed to construct one for [imath]C(\mathbb R)[/imath].
|
2123035
|
Solving system of linear equations ( 4 variables, 3 equations)
Finding the solution(s) for: [imath] \begin{align*} \begin{cases} 2x_1+x_2+3x_3+2x_4 &=5\\ x_1+x_2+x_3+2x_4&=3\\ -x_2+x_3+6x_4&=3 \end{cases} \end{align*} [/imath] I tried using elimination to rewrite the system in row echelon form, then back substituting. I have no idea what I did wrong, or if I was doing the elimination process correctly.. but I kept getting stuck. I am confused and not really sure how to proceed, I must be awful because I've been staring at this problem for hours.
|
602927
|
Solve a linear system with more variables than equations
Suppose that, after a series of elementary row operations the augmented matrix of a linear system with variables [imath]x_1[/imath], [imath]x_2[/imath], [imath]x_3[/imath], [imath]x_4[/imath] is transformed into reduced row echelon form as follows: [imath]\left(\begin{array}{cccc|c}1 & 0 & 0 & 1 & 0\\0 & 1 & 0 & 2 & 1 \\0 & 0 & 1 & 3 & 0 \end{array}\right)[/imath]. Can I solve the linear system as below? Let [imath]t[/imath] be an arbitrary real number. Then solving each linear equation corresponding to the augmented matrix for leading variable and setting [imath]x_4=t[/imath], we get [imath]x_1=-t, x_2=1-2t[/imath], and [imath]x_3=-3t[/imath]. Thus the general solution of the linear system is \begin{align} x_1=-t\\ x_2=1-2t\\ x_3=-3t\\ \end{align} where t is an arbitrary real number.
|
1333944
|
How to evaluate [imath]\lim_{ n\to \infty }\frac{a_n}{2^{n-1}}[/imath], if [imath]a_0=0[/imath] and [imath]a_{n+1}=a_n+\sqrt{a_n^2+1}[/imath]?
Let [imath]a_1,a_2,..,a_n[/imath] be sequence of real numbers such that [imath]a_{n+1}=a_{n}+\sqrt{1+a_n^2}[/imath] and [imath]a_0=0[/imath]. How to evaluate [imath]\lim_{ n\to \infty }\frac{a_n}{2^{n-1}}[/imath] ?
|
1175041
|
Convergence of sequence given by [imath]x_1=1[/imath] and [imath]x_{n+1}=x_n+\sqrt{x_n^2+1}[/imath]
Be [imath](x_n)_n\ge 1 [/imath] such that [imath]x_1=1[/imath] and [imath]x_{n+1}=x_n+\sqrt{x_n^2+1}[/imath] for every [imath]n\ge 1[/imath]. Prove that the sequence [imath]y_n=(2^n/x_n)_n \ge 1[/imath] is convergent and find it s limit. Being positive and decreasing, [imath]y_n[/imath] is clearly convergent. But finding it's limit really put me intro trouble. Any help? Thank you!
|
2123325
|
Let [imath]\omega[/imath] be a complex number such that [imath]\omega^5 = 1[/imath] and [imath]\omega\neq 1[/imath], find ...
Let [imath]\omega[/imath] be a complex number such that [imath]\omega^5 = 1[/imath] and [imath]\omega \neq 1[/imath]. Find [imath]\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3}[/imath] I've been having trouble with this unit, need help on solving this problem.
|
1345022
|
Complex numbers - roots of unity
Let [imath]\omega[/imath] be a complex number such that [imath]\omega^5 = 1[/imath] and [imath]\omega \neq 1[/imath]. Find [imath]\frac{\omega}{1 - \omega^2} + \frac{\omega^2}{1 - \omega^4} + \frac{\omega^3}{1 - \omega} + \frac{\omega^4}{1 - \omega^3}.[/imath] I have tried adding the first two and the second two separately, then adding those sums but how do I get a numerical value as the answer? Thanks
|
2123996
|
A closed form of [imath]\sum_{n=0}^\infty\frac{n!!}{n!}[/imath]
It's fairly easy to see, using the ratio test, that the series [imath]\sum_{n=0}^\infty\frac{n!!}{n!}[/imath] converges, where [imath]n!![/imath] is the double factorial. However, I'd be interested if there is a way of calculating its value analytically. I've tried around a little with WolframAlpha and I know that [imath]\sum_{n=0}^\infty\frac{n!!}{n!}\approx 4.05941[/imath] Wolfram also suggests a couple of possible closed forms, which look very different. Since this sum is related to [imath]e[/imath] somehow, the following two are what I'd find most likely: [imath]\sum_{n=0}^\infty\frac{n!!}{n!}=-8-e+2e^2[/imath] [imath]\sum_{n=0}^\infty\frac{n!!}{n!}=\Gamma(e+1)+e-2-\frac{5}{2e}[/imath] However, I have no clue if either of the two is actually correct, and if it were, how one would go about proving this result to be true - none of the approaches to series that I know seem to be getting me anywhere. Is there a way to find a closed form for this series, and if so, how would one go about deriving it?
|
1482211
|
Double factorial series
My question is pretty simple. Since [imath]n! \gt n!![/imath], it's clear by the comparison test that [imath]\sum_{n=0}^\infty \frac {1}{n!!}[/imath] converges. But what value does the sum converge to? How does one go about determining its value (if possible)?
|
2123998
|
Why infinity minus infinity, 0 multiply infinity, infinity divided by infinity and 0/0 are not defined?
Suppose that the arithmetic operation on extended complex plane are defined via arithmetic operations on the corresponding sequences, why [imath]∞-∞，0*∞，∞/∞，0/0 [/imath]are not defined? Can anyone give me some example about that? I know that ∞ + ∞ is undefined because if we have two sequences: [imath]1, 3, 3, 5, 5, 7, 7, 7, 9,....[/imath] and [imath]-1,-2,-3,-4,-5,-6,-7,-8,....[/imath] the sum of the above sequences will be: [imath]0,1,0,1,0,1,0,1,...[/imath] which is not converging or diverging. so ∞+∞ is meaningless. Can I use it in [imath] ∞-∞，0*∞，∞/∞，0/0[/imath]? Can anyone give me some example about that?
|
1647795
|
Give examples showing why [imath]0\cdot \infty[/imath], [imath]\infty/\infty[/imath], and [imath]0/0[/imath] are meaningless
Assuming arithmetic operations on [imath]\overline{\mathbb{C}}[/imath] (that's the extended complex plane) are defined via arithmetic operations on the corresponding sequences, I need to give examples showing why [imath]0\cdot \infty[/imath], [imath]\infty/\infty[/imath], and [imath]0/0[/imath] are meaningless. I also had to show the same thing for [imath]\infty - \infty[/imath], but that was easy: Consider the sequences [imath]\{1,3,3,5,5,7,7,9,\cdots \}[/imath] and [imath]\{1,2,3,4,5,6,7,8,\cdots \}[/imath], both of which belong to the class [imath]\infty[/imath] (i.e., the equivalence class of all sequences converging to [imath]\infty[/imath]). Then, subtracting the second sequence from the first, term-by-term, we obtain the resulting sequence [imath]\{0,1,0,1,0,1,\cdots\}[/imath], which does not belong to any convergence class, since it has no limit. However, I can't approach the three cases I'm asking on here about in the same way. I must admit, I'm a little confused as to what is being asked. For reference, it is problems 2-4 on p. 79 of Markushevich's Theory of Functions of a Complex Variable, Vol I. If someone could furnish me with an example, at least for [imath]0 \cdot \infty[/imath], and maybe either [imath]0/0[/imath] or [imath]\infty/\infty[/imath], perhaps I could use that to help me figure out the other case on my own. Thank you.
|
2083311
|
Proving that the function [imath]f(N)=\sum_{n=1}^N \cos(n^2)[/imath] is unbounded
I am trying to prove that the function [imath]f(N)=\sum_{n=1}^N \cos(n^2)[/imath] is unbounded. Admittedly, I don't know for certain that it is, but someone told me, and I would like to see it for myself. First, I thought it might be because the square numbers mod [imath]2\pi[/imath] was not uniformly distributed, but numerics seem to imply that they are. My next thought is then that it has to do with the fact that we might have positive or negative subsequences in the sum of arbitrary length. However, what is a good strategy for proving this kind of statement?
|
1209178
|
Is [imath]\sum_{k=1}^{n} \sin(k^2)[/imath] bounded by a constant [imath]M[/imath]?
I know [imath]\sum_{k=1}^{n} \sin(k)[/imath] is bounded by a constant. How about [imath]\sum_{k=1}^{n} \sin(k^2)[/imath]?
|
1489366
|
Proving that [imath]a \mathbb{Z} \cap b\mathbb{Z} = \operatorname{lcm}(a,b)\mathbb Z[/imath]
Given [imath]m \in \mathbb{Z}[/imath], let [imath]m\mathbb{Z}[/imath] denote the set of integer multiples of [imath]m[/imath], i.e. [imath]m\mathbb{Z} := \{mk\mid k \in \mathbb{Z}\}[/imath]. Now let [imath]a,b \in \mathbb{Z}[/imath] with [imath]a,b[/imath] not both [imath]0[/imath]. Prove that [imath]a\mathbb{Z} \cap b\mathbb{Z} = \operatorname{lcm}(a,b)\mathbb{Z}[/imath]. I am trying to write a proof for this, but I am unsure of what method to use. Also I am confused by [imath]mk\mid k[/imath], because wouldn't [imath]m=1[/imath] for this to be true.
|
447862
|
Proving [imath]a\mathbb{Z}\cap b\mathbb{Z}=[a,b]\mathbb{Z}[/imath]
The question is Prove that [imath]a\mathbb{Z}\cap b\mathbb{Z}=[a,b]\mathbb{Z}[/imath]. Hint:First prove [imath]b\mid a \Leftrightarrow a\mathbb Z\le b\mathbb Z[/imath] and then prove [imath]a\mathbb Z+b\mathbb Z=(a,b)\mathbb Z[/imath] I managed to prove [imath]b\mid a\Leftrightarrow a\mathbb Z\le b\mathbb Z[/imath]. About the next claim: From the first statement ([imath]b\mid a...[/imath]) it follows that since [imath](a,b)\mid a[/imath] and [imath](a,b)\mid b, a\mathbb Z+b\mathbb Z\le(a,b)\mathbb Z[/imath]. My problem is proving the opposite side (a.k.a [imath]a\mathbb Z+b\mathbb Z\ge(a,b)\mathbb Z[/imath]). How can I do so and how can I proceed from [imath](a,b)[/imath] to [imath][a,b][/imath]?
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33688
|
ZFC + [imath]\exists[/imath] Standard model [imath]\rightarrow[/imath] Con(ZFC + [imath]\exists \omega[/imath]-model)
[imath]ZFC + \exists V_\alpha[/imath] model of [imath]ZFC \vdash Con(ZFC + \exists[/imath] transitive standard model of [imath]ZFC)[/imath] and then [imath]ZFC + \exists[/imath] transitive standard model of [imath]ZFC \vdash Con(ZFC + \exists \omega-model[/imath] of [imath]ZFC)[/imath] For the first one : We can always find a countable extentional [imath]M \subset V_\alpha[/imath] elementary equivalent to [imath]V_\alpha[/imath]. Let [imath]M'[/imath] be the mostowski collapse of [imath]M[/imath]. [imath]M' \approx M[/imath] so [imath]M'[/imath] is model of ZFC. And because [imath]M'[/imath] is countable and transitive then [imath]M' \in V_\alpha[/imath] (since [imath]H_{\omega_1} \subset V_{\omega_1}[/imath] and [imath]\alpha[/imath] is surely far larger than [imath]\omega_1[/imath]). So [imath]V_\alpha[/imath] is the model of '[imath]\exists[/imath] a standard transitive model of ZFC'. For the second one : I don't really know how to do it... Does anyone have an idea ?
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2126493
|
existence of transitive standard models of ZFC independent of existence of models of ZFC.
If we consider the sentences in ZFC: EM that express that exist a model of ZFC. SM that express that exist a standard transitive model of ZFC. OM that express that exist a [imath]\omega[/imath]-model of ZFC. By Godel second incompleteness theorem we know that [imath]\mathsf{ZFC}\nvdash \mathsf{EM}[/imath] What is known about: [imath]\mathsf{ZFC+EM}\nvdash \mathsf{OM}[/imath] ? or [imath]\mathsf{ZFC+OM}\nvdash \mathsf{SM}[/imath] ? or [imath]\mathsf{ZFC+EM}\nvdash \mathsf{SM}[/imath] ?
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2124929
|
Divergent but bounded partial sums
I am trying to show that the series [imath]\sum _{n=1}^{\infty} \sin n[/imath] diverges but has bounded partial sums. Plugging in some terms we see that, [imath]\sum _{n=1}^{\infty} \sin n = \sin1 +\sin2 +\sin3+...+\sin n[/imath] My idea is to try and use [imath]e^{i\theta} = \cos\theta+i\sin\theta[/imath] [imath]e^{in}=\cos n+\sin n[/imath] [imath]\sin n = Im(e^{in})[/imath] But how can I use this to show that a finite geometric series won't converge to anything, therefore diverge, but is bounded?
|
59589
|
Show that [imath]A_n=\sum\limits_{k=1}^n \sin k [/imath] is bounded?
Let [imath]A_n=\sum\limits_{k=1}^n \sin k [/imath] , show that there exists [imath]M>0[/imath] , [imath]|A_n|<M [/imath] for every [imath]n[/imath] .
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2108287
|
Is the weak derivative of any function of the form [imath]f(\lfloor g(x) \rfloor)[/imath] [imath]0[/imath] everywhere?
I recently found this thing called a weak derivative. I think it is what I have been trying to express for quite some time, but it's definition makes it mildly hard for me to prove the following (which is probably quite trivial... or not): Let [imath]f[/imath] and [imath]g[/imath] be continuous functions and [imath]x'[/imath] denote the weak derivative, then [imath]\forall_{x \in R} f(\lfloor g(x) \rfloor)'=0[/imath]. I am not real sure how to prove it if it is true. The form I gave is my definition of piecewise constant. Mostly, I am concerned on whether or not it is what has this property as I've been trying for quite some time to formally define such a 'derivative'. If it does, then my next thing to ask is whether or not there is an associated weak integral as that is more what I care about.
|
849353
|
Weak Derivative Heaviside function
I have to prove that the Heaviside function [imath] H(x):=\begin{cases} 1 &\mbox{if } x \in [0,+\infty) \\ 0 &\mbox{otherwise}\end{cases} [/imath] doesn't admit weak derivative in [imath]L^1_{loc}(\mathbb{R})[/imath]. Is it correct the following solution? Let's suppose by contradiction that [imath]\exists w\in L^1_{loc}(\mathbb{R})[/imath] such that [imath]\int_{\mathbb{R}}H(x)v'(x)dx=\int_{\mathbb{R}}w(x)v(x)dx\qquad\forall v\in C^{\infty}_c(\mathbb{R}) [/imath] By the Fundamental Theorem of Calculus we have that [imath]\int_0^{+\infty}w(x)v(x)dx=v(0)\qquad\forall v\in C^{\infty}_c(\mathbb{R}) [/imath] By the Lebesgue Dominated Convergence Theorem we have that [imath]\lim_{r\to 0}\int_{-r}^{r}|w(x)|dx=0 [/imath] Then there exists [imath]\delta>0[/imath] such that [imath]\int_{-\delta}^{\delta}|w(x)|dx\le 1/2[/imath]. Let [imath]v\in C^{\infty}_c(\mathbb{R})[/imath] such that [imath]\operatorname{supp}(v)\subset [-\delta,\delta][/imath] and [imath]\max (v)=v(0)=1[/imath]. Then [imath]1=v(0)=\int_0^{+\infty}w(x)v(x)dx=\int_{-\delta}^{\delta}w(x)v(x)dx \le \max(v)\int_{-\delta}^{\delta}|w(x)|dx\le 1/2[/imath] Contradiction!!!
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2124782
|
Continuing logarithm [imath]\log(\log(\dots\log(z)))[/imath]
I don't know the best way to describe it in technical terms, but what is the result of a continuing logarithm of [imath]z[/imath], for example: [imath]\log(\log(\dots\log(z)))[/imath] Where it is taking the logarithm of the logarithm and so on, for an infinite amount of times? How would this type of thing behave? Does it converge, go off to infinity or an infinitesimal? Does the resulting behavior depend on whether [imath]z[/imath] is imaginary or real? Positive or negative?
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1587350
|
Does logging infinitely converge?
Trying to evaluate [imath]\ln(\ln(\ln(\ln(\cdots\ln(x)\cdots))))[/imath]For some fixed [imath]x[/imath] produces a complex answer that appears to converge, at least sometimes. So I want a proof that this converges for either some [imath]x[/imath], no [imath]x[/imath], or all [imath]x[/imath]. If it converges for all [imath]x[/imath] or some [imath]x[/imath], what does it converge to? If it diverges, is there a way we can evaluate it like we evaluate diverging sums? And after all of that, does it appear to converge to the same value, no matter what [imath]x[/imath] value we start with? I know [imath]\ln(z)=\ln(|z|)+i\arg(z)[/imath], but I can't repeat this process without a given [imath]z[/imath]. (where [imath]z[/imath] is complex). A similar post of mine found here does not answer my question and focuses more on the limits, calculus, and infinites. This question asks for consideration from a complex-analysis point of view, considering convergence of value in the complex plane.
|
2124657
|
Find all permutations of [imath]S_4[/imath] that commutes with (123)
This is what I have. Not sure if this is right. [imath]{4\times3\times2\over3}=8 [/imath] I think this is too easy to be true
|
2118144
|
Prove/Disprove: Find [imath]a\in S_4[/imath] that is not a power of [imath]b[/imath] such that [imath]ab=ba[/imath] while [imath]b=(123)[/imath]
Prove/Disprove: Find [imath]a\in S_4[/imath] that is not a power of [imath]b[/imath] such that [imath]ab=ba[/imath] while [imath]b=(123)[/imath] What I did: I think this statement is false. So we need to find [imath]aba^{-1}=b[/imath], we know that [imath]id\in S_4[/imath] will surely [imath]id*b*id^{-1}=b[/imath] but we get [imath]id^k=b[/imath], I wonder if there are more elements such that [imath]ab=ba[/imath], how to find them? Any help will be appreciated.
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