qid
stringlengths 1
7
| Q
stringlengths 87
7.22k
| dup_qid
stringlengths 1
7
| Q_dup
stringlengths 97
10.5k
|
|---|---|---|---|
2911404
|
Prove [imath]|xax^{-1}| = |a|[/imath]
So, I have a question that says for any group elements [imath]a[/imath] and [imath]x[/imath], prove that [imath]|xax^{-1}| =|a|[/imath], where [imath]|\cdot|[/imath] is the order of an element. I know that [imath]xx^{-1} = e[/imath] and [imath]ae=a[/imath], but I'm not sure how to get the [imath]x[/imath]'s on the same side. Currently, for my proof, which is not much, I have: Let [imath]n = |xax^{-1}|[/imath], then [imath](xax^{-1})^n = e[/imath]. Honestly, I have no idea where to go from here, or if what I've got is even right. I would really appreciate just a point in the right direction.
|
734317
|
Prove that any conjugate of [imath]a[/imath] has the same order as [imath]a[/imath].
I am trying to prove that any conjugate of [imath]a[/imath] has the same order as [imath]a[/imath]. Let [imath]G[/imath] be a group and let [imath]a \in G[/imath]. An element [imath]b \in G[/imath] is called a conjugate of [imath]a[/imath] if [imath]b=xax^{-1}[/imath]. My professor gave us a hint. Prove [imath]a^k=e[/imath] iff [imath]b^k=e[/imath] for all [imath]k \in \mathbb{Z}[/imath] implies [imath]o(a) = o(b)[/imath]. My first move is to [imath]b^k = (xax^{-1})^k = (x^{-1})^ka^kx^k = (x^{-1})^kex^k = (x^{-1})^kx^k=x^{-k+k}=x^0[/imath] I am not sure that I am doing this correctly? I am trying to get [imath]b^k=e[/imath]. However when I raise [imath](xax^{-1})^k[/imath], I am not sure if I what I have done is correct? I am kind of confused here.
|
2912077
|
Show that a matrix is a multiple of the identity
Let [imath]A[/imath], [imath]B[/imath], [imath]C[/imath] three complex 2x2 matrices such that [imath]A^2=B^3=I\;(A\neq I\neq B),\quad ABA=B^{-1},\quad AC=CA,\quad BC=CB.[/imath] Show that [imath]C=rI[/imath] for some [imath]r\in\mathbb{C}[/imath]. I got the minimal polynomials of [imath]A[/imath] and [imath]B[/imath], but I can't go further to find the minimal polynomial of [imath]C[/imath]. Any suggestion?
|
327223
|
If [imath]C[/imath] commutes with certain matrices [imath]A[/imath] and [imath]B[/imath], why is [imath]C[/imath] a scalar multiple of the identity?
I'm self studying Steven Roman's Advanced Linear Algebra, and this is problem 10 of Chapter 8. Let [imath]A,B\in M_2(\mathbb{C})[/imath], [imath]A^2=B^3=I[/imath], [imath]ABA=B^{-1}[/imath], but [imath]A\neq I[/imath] and [imath]B\neq I[/imath]. If [imath]C\in M_2(\mathbb{C})[/imath] commutes with [imath]A[/imath] and [imath]B[/imath], then [imath]C=rI[/imath] for some [imath]r\in\mathbb{C}[/imath]. Is there a way to solve this without writing out arbitrary matrices and attempting to solve a huge system of equations? The only thing I observe is that [imath]A=A^{-1}[/imath], so [imath]B\sim B^{-1}[/imath], so [imath]B[/imath] and [imath]B^{-1}[/imath] have the same characteristic polynomial. I'm stymied trying to show [imath]C[/imath] is diagonal, let alone a multiple of [imath]I[/imath]. Thanks for any ideas. I should add that I know that the center [imath]Z(M_n(\mathbb{C}))[/imath] consists of scalar multiples of [imath]I[/imath], but I don't see any reason to assume or prove [imath]C[/imath] commutes with everything.
|
2911865
|
Closed form of [imath]\sum^\infty_{n=1} \frac{\sin n}n[/imath]
What is the closed form of [imath]\sum^\infty_{n=1}\frac{\sin n}n[/imath]? My approach: [imath]\sum^\infty_{n=1}\frac{\sin n}n=\Im\left[\sum^\infty_{n=1}\frac{e^{in}}n\right]=\Im\left[-\text{Log}(1-e^{i})\right][/imath] where [imath]\text{Log}(1):=0[/imath]. [imath]\Im\left[-\text{Log}(1-e^i)\right]=-\arg(1-e^i)=-\frac{1-\pi}2[/imath] Thus, [imath]\sum^\infty_{n=1}\frac{\sin n}n=\frac{\pi-1}2[/imath] And as a bonus, [imath]\sum^\infty_{n=1}\frac{\cos n}n=\ln(2\sec1)[/imath] Is this correct?
|
879726
|
Evaluation of Sum of [imath] \sum_{n=1}^{\infty}\frac{\sin (n)}{n}[/imath].
If [imath]\displaystyle S = \sum_{n=1}^{\infty}\frac{\sin (n)}{n}.[/imath] Then value of [imath]2S+1 = [/imath] Using Fourier Series Transformation I am Getting [imath]2S+1=\pi[/imath] But I want to solve it Using Euler Method and Then Use Logarithmic Series. [imath]\bf{My\; Try::}[/imath] Using [imath]\displaystyle \sin (n) = \left(\frac{e^{in}-e^{-in}}{2i}\right)[/imath]. So [imath]\displaystyle S = \sum_{n=1}^{n}\frac{\sin (n)}{n} = \frac{1}{2i}\sum_{n=1}^{\infty}\frac{e^{in}}{n}-\frac{1}{2i}\sum_{n=1}^{\infty}\frac{e^{-in}}{n}[/imath] Now Using [imath]\displaystyle \ln(1-x) = -x-\frac{x^2}{2}-\frac{x^3}{3}...............\infty[/imath] So Let [imath]\displaystyle S = -\frac{1}{2i}\ln(1-e^{i})+\frac{1}{2i}\ln(1-e^{-i})[/imath] Now How can I solve after that Help me Thanks
|
2912165
|
Lower bound for complex polynomial beyond circle or radius R
If we have a polynomial with [imath]c_i[/imath] a complex number [imath]c_nz^n + c_{n-1}z^{n-1} + \cdots + c_1 z + c_0[/imath] then [imath]|P(z)| > \frac{|c_n|R^n}{2}[/imath] When |z| > R for some R I have tried using the triangle inequality where I obtain, [imath]|P(Z)| \leq |c_n||z|^n + \cdots + |c_0|[/imath]. But I seem to keep getting stuck. Any hints on how to move forward? Thank you!
|
39567
|
For a polynomial [imath]p(z)[/imath], prove there exist [imath]R>0[/imath], such that if [imath]|z|=R[/imath], then [imath]|p(z)|\geq |a_n|R^n/2[/imath]
For a polynomial [imath]p(z)=a_0+\ldots+a_nz^n[/imath], prove there exist [imath]R>0[/imath], such that if [imath]|z|=R[/imath], then [imath]|p(z)|\geq |a_n|R^n/2[/imath]. I get [imath]|p(z)|\geq |a_n|R^n[/imath], which makes that factor of [imath]1/2[/imath] useless. So I must have done something wrong. [imath]|\frac{\partial^n{f}}{\partial z^n}(P)|\leq \frac{\sup_{z\in \overline{D}(P,r)} |f(z)|k!}{r^k}[/imath]. This is the Cauchy estimate. [imath]|\frac{\partial^n{p}}{\partial z^n}(z)| = |a_n n!|[/imath] [imath]|\frac{\partial^n{p}}{\partial z^n}(0)|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|n!/R^n[/imath] [imath]|a_n n!|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|n!/R^n[/imath] [imath]R^n |a_n|\leq \sup_{\zeta\in \overline{D}(0,|z|)} |p(\zeta)|[/imath] There exist a [imath]z_0[/imath], such that [imath]|z_0|>0[/imath] and [imath]\sup_{\zeta\in \overline{D}(0,|z_0|)} |p(\zeta)| = |p(z_0)|[/imath]. This is saying there exist a disk centered at the origin, such that [imath]|p|[/imath] is maximum on some point on the boundary. Let [imath]R=|z_0|[/imath], we have [imath]R^n |a_n|\leq |p(z_0)|[/imath] Edit: I know what I got wrong. [imath]z_0[/imath] is just one point on the circle [imath]|z| = R[/imath]. One can chose another point [imath]z_1[/imath] on the circle, such that [imath]p(z_1)<p(z_0)[/imath].
|
2760005
|
[imath]f(x,y) := \begin{cases} \frac{xy}{x^2+y^2}, &\mbox{if } (x,y) \neq (0,0)\\ 0, & \mbox{if } (x,y)=(0,0) \end{cases}[/imath] not differentiable in [imath](0,0)^T[/imath]
[imath]f(x,y) := \begin{cases} \frac{xy}{x^2+y^2}, &\mbox{if } (x,y) \neq (0,0)\\ 0, & \mbox{if } (x,y)=(0,0) \end{cases}[/imath] I need to show that all partial derivatives in [imath](0,0)^T[/imath] exist which I already did, but how can I show that [imath]f[/imath] is not differentiable in [imath](0,0)^T[/imath]?
|
2496400
|
If [imath]F(0,0)=0[/imath] and [imath]F(x,y)= \frac{xy}{x^2+y^2}[/imath] for [imath](x,y)\neq (0,0)[/imath] then [imath]F[/imath] is differentiable at [imath](0,0)[/imath]?
Let [imath]F:\mathbb{R^2}\rightarrow \mathbb{R}[/imath] be a function, such as : [imath]F(x,y)=\begin{cases} 0 & (x,y)=(0,0) \\ \frac{xy}{x^2+y^2} & (x,y)\neq (0,0) \end{cases}[/imath] Is [imath]F[/imath] differentiable at [imath](0,0)[/imath]? My attempt : Consider the limit, [imath]\lim_{(x,y)\rightarrow (0,0)} F(x,y)[/imath] along the line [imath]y=x[/imath]. The limit equals to [imath]\frac{1}{2}[/imath]. However, [imath]F(0,0)=0[/imath]. Hence [imath]F[/imath] is not continuous at [imath](0,0)[/imath] which implies that [imath]F[/imath] isn't differentiable as well. Is this approach okay ?
|
2912249
|
show [imath]\lim\limits_{(x,y)\to(0,0)}\frac{|x|^a|y|^b}{|x|^c+|y|^d}[/imath] exists and equals [imath]0[/imath]?
suppose [imath]f(x,y)=\frac{|x|^a|y|^b}{|x|^c+|y|^d}[/imath] and [imath]a,b,c,d >0[/imath] how would you show that if [imath]\frac{a}{c} + \frac{b}{d} >1[/imath] then [imath]\lim\limits_{(x,y)\to(0,0)}\frac{|x|^a|y|^b}{|x|^c+|y|^d}[/imath] exists and equals [imath]0[/imath]? I've been trying to use the squeeze theorem and set up an inequality but im really struggling.
|
451867
|
Show that [imath] \lim_{(x,y) \to 0} \frac {|x|^ \alpha |y|^ \beta} {|x|^ \gamma + |y|^ \delta} \text {exists} \iff \alpha/\gamma + \beta/\delta > 1.[/imath]
Ted Shifrin on this site posed an interesting problem to me: show that [imath] \lim_{(x,y) \to (0,0)} \frac {|x|^ \alpha |y|^ \beta} {|x|^ \gamma + |y|^ \delta} \text {exists} \iff \frac\alpha\gamma + \frac\beta\delta > 1, \,\,\,\,\text{where } \alpha, \beta, \gamma, \delta >0[/imath] I think I've got the [imath](\Leftarrow)[/imath] direction as follows: assume WLOG that [imath]\gamma \leq \delta[/imath]. Then switch to polar coordinates and get [imath]\frac {r^ {\alpha + \beta} (*) } {r^\gamma (1(*) + r^{\delta - \gamma}(*))},[/imath] where [imath](*)[/imath] represents some trig stuff that is bounded near zero. Edit: I need to make sure that [imath](1(*) + r^{\delta - \gamma}(*))[/imath] is bounded below here. The other direction is giving me some trouble. It seems we need a lower bound for the fraction (something to force to zero), and I'm not sure where to find one. In particular, I'm not sure what to do with the denominator. Any ideas?
|
2912816
|
f is analytic on [imath]D(0,1)\setminus\{0\}[/imath] with an essential singularity at [imath]0,[/imath] show f is not one-to-one
I am working on following Question; f is analytic on [imath]D(0,1)\setminus\{0\}[/imath] with an essential singularity at 0, show f is not one-to-one. I am a bit stuck. All I have been able to figure out is the following; [imath]\lim_{z \rightarrow 0} f(z)[/imath] does not exist since the singularity is not removable. [imath]f(D(0,1)\setminus\{0\})[/imath] is open by the Open Mapping Theorem [imath]f(D(0,1)\setminus\{0\})[/imath] is dense in [imath]\mathbb{C}[/imath], by Casorati-Weierstrass. Now I suspect that we should try the proof by contradiction route and suppose that [imath]f[/imath] is one-to-one and show that then the singularity is maybe not essential, potentially removable. Or maybe we should construct two sequence converging to different limits while the images of the sequences converge to the same thing. Cheers for the help in advance; If possible please do not give the whole answer but rather give hints and when I feel I have a solution I will post it as an answer (Of course I will accept the most useful hint as the answer to this question)
|
53303
|
Why can't an analytic function be injective in a neighborhood of an essential singularity?
Let [imath]D \subset \mathbb{C}[/imath] be a domain and let [imath]a \in D[/imath]. Suppose [imath]f: D \smallsetminus \{a\} \to \mathbb{C}[/imath] is analytic and that [imath]a[/imath] is an essential singularity of [imath]f[/imath]. Show that [imath]f[/imath] cannot be univalent (= injective) in any neighborhood of [imath]a[/imath]. This is a trivial consequence of the Picard theorem. But I don't know if there is any elementary approach.
|
2912939
|
Lucas Numbers have the same prime divisors?
Let [imath]L_n[/imath] be the [imath]n[/imath] th Lucas number, with [imath]L_0 = 2, L_1=1, L_{n+1} = L_n + L_{n-1}[/imath] Is it true that for every even number [imath]n \geq 4[/imath], if the number [imath]L_n - 2[/imath] has a prime divisor [imath]p > 2[/imath], then the number [imath]L_{n+1} - 1[/imath] is also divided by [imath]p[/imath]? If not, then is the statement still correct with [imath]n = 2018[/imath]? If not, how many number [imath]n[/imath] satisfy that if the number [imath]L_n - 2[/imath] has a prime divisor [imath]p > 2[/imath], then the number [imath]L_{n+1} - 1[/imath] is also divided by [imath]p[/imath]? (Sorry, English is my second language)
|
2472354
|
Prove that if prime [imath]p[/imath] divide [imath]a_{2k}-2[/imath], then [imath]p[/imath] divide also [imath]a_{2k+1}-1[/imath].
Sequence [imath]a_0,a_1,a_2,...[/imath] satisfies that [imath]a_0=2,a_1=1,a_{n+1}=a_n+a_{n-1}[/imath] Prove that if [imath]p[/imath] is a prime divisor of [imath]a_{2k}-2[/imath],then [imath]p[/imath] is also a prime divisor of [imath]a_{2k+1}-1[/imath] If [imath]x_{1,2}={1\pm\sqrt{5}\over 2}[/imath] then [imath]a_k = x_1^k+x_2^k[/imath] and [imath]a_{2k}-2(-1)^k = (x_1^k-x_2^k)^2[/imath].
|
2912710
|
Proof of the open interval [imath](-1,1)[/imath] with respect to [imath]*[/imath] is a group
Hi,so I just can't get my head through the solution for 4th problem.I successfully proved associative law and when I wrote an equation for finding identity element I did get the same thing,but then I don't understand how they get [imath]e(1 - x^2) = 0[/imath]. I always get [imath]x + e = x^2e + x[/imath]. I'd appreciate if someone could explain. I understand that logically speaking, it should be [imath]0[/imath].
|
785470
|
Proving a set is an abelian group.
I am trying to prove that [imath](G, *)[/imath] is an abelian group with [imath]G=(-1,1)[/imath] and [imath]a*b=[/imath][imath]\frac{x+y}{1+xy}[/imath]. Thus far I have found that the identity element [imath]e=0[/imath]. From here, I set [imath]a*b=0[/imath] and found [imath]a^{-1}[/imath] to be [imath]-a[/imath]. My work for trying to prove closure and that the set is abelian is: Let [imath]a,b \in G[/imath], where [imath]a=e[/imath] and [imath]b=a^{-1}[/imath]. Does [imath]a*b=b*a[/imath]? Evaluating, I see that [imath]a^{-1}=a^{-1}[/imath] which then proves that G is closed and G is an abelian group. Did I go about this proof correctly and efficiently?
|
2913073
|
Will this series converge? [imath]\sum \frac {1/2 + (-1)^n}{n}[/imath]
Will this series converge? [imath]\sum \frac {1/2 + (-1)^n}{n}[/imath] MY Try: Dirichilet , Abel , libnitz rules can not be used. [imath]\sum \frac {1/2 + (-1)^n}{n} = \sum \frac {1}{2n} + \sum\frac {(-1)^n}{2n} [/imath]. Is it possible to write in that way? If yes the how? Can anyone please help me out?
|
2891645
|
Convergence of [imath]\sum_{n=1}^{\infty} \frac{\frac{1}{2}+(-1)^n}{n}[/imath]
Does the following series converge?: [imath]\sum_{n=1}^{\infty} \frac{\frac{1}{2}+(-1)^n}{n}[/imath] Liebnitz test shouldn't work here...Also I try to find out the partial sum sequence but it was not so fruitful ....Please help.
|
2913105
|
Does there exist a continuous bijection from real line onto the unit circle?
Does there exist a continuous bijection from [imath]\mathbb R[/imath] onto [imath]S^1[/imath] ? I know that there isn't any continuous bijection from [imath]S^1[/imath] onto [imath]\mathbb R[/imath] because such a continuous bijection would be a homeomorphism , but [imath]S^1[/imath] is compact whereas [imath]\mathbb R[/imath] is not, so impossible. But I'm not sure about this other direction. Please help.
|
1731407
|
Continuous bijection from [imath](a,b) \to S^1[/imath]?
This question started bothering me after working on an exercise. I know that there cannot be a contiuous bijection [imath]S^1 \to (a,b)[/imath] because if there was it would be a homeomorphism but [imath]S^1[/imath] and [imath](a,b)[/imath] are not homeomorphic. But the theorem that implies this is that a continuous bijection from a compact into a Hausdorff space is a homeomorphism. Hence it cannot be applied to the opposite direction. I still suspect that the answer will turn out to be negative, i.e. there is no continuous bijection from [imath](a,b) \to S^1[/imath] but I don't see how to prove it because the inverse is not required to be continuous. So somehow there still remains a faint possibility for such a map. Please could someone help me resolve my confusion and tell me whether there is or is not a continuous bijection [imath](a,b) \to S^1[/imath]?
|
622704
|
Transcendental numbers involving primes?
Is the prime zeta function value [imath] P(2)=\sum_{p \in \mathrm{primes}} \frac{1}{p^2} = 0.452247420041065498506543364832247934173231343\ldots [/imath] a transcendental number ? What about the following sum ? [imath]\sum_{p \in \mathrm{primes}} \frac{1}{p^p} = 0.2873582513062241797364180458789322069559088\ldots [/imath] For the last sum, Liouville's criterion might help because of the fast converging series. The first sum might have been checked, as it is natural to come to such a sum. Perhaps, it is at least known if it is irrational.
|
1529996
|
Value of [imath]\sum 1/p^p[/imath]
A very simple question, but I can't seem to find anything relating to it : Is there any research, are there any results that have focused on or given insight on [imath]\sum 1/p^p[/imath], [imath]{p \in \mathbb P}[/imath] ? A very basic series, converges extremely fast, its value is around .29. What more can there be said about it ? From what little I know about more advanced number theory, similar sequences (I can think of a few similar ones that I can't find any relevant research or results about) can be very non-trivial to compute or to analyse.
|
2913323
|
about limit of exponential function
Maybe the answer is obvious. I'm sorry for this I know for all [imath]x \in \mathbb{R}[/imath] that [imath] \lim_\limits{n \to \infty}\left(1 + \frac{x}{n} \right)^{n} = \exp(x). [/imath] Now suppose I have a sequence [imath]\{x_{n}\}_{n \in \mathbb{N}}[/imath] such that [imath] \lim_\limits{n \to \infty} x_{n} = x \in \mathbb{R}. [/imath] Can I also conclude that [imath] \lim_\limits{n \to \infty}\left(1 + \frac{x_{n}}{n} \right)^{n} = \exp(x)? [/imath]
|
891271
|
Limit of [imath]\lim\limits_{n\to\infty} (1 + \frac{x_n}{n})^n[/imath]
Many websites and calculus books give this well known result \begin{equation} \lim\limits_{n\to\infty} \left(1 + \frac{x}{n}\right)^n = e^x \end{equation} However, a textbook I was reading casually mentioned that if [imath]x_n \rightarrow x[/imath] then \begin{equation} \lim\limits_{n\to\infty} \left(1 + \frac{x_n}{n}\right)^n = e^x \end{equation} Why is this true? It seems very intuitive but I feel some explanation is missing. Thank you!
|
2913495
|
Problem with complex equation [imath]\sin z=\cos z[/imath]
I attempted to solve [imath]\sin{z} = \cos{z}[/imath] I used Eulero formulas for complex sin and cos. Then, in order to eliminate \i from sin's denominator, I raised to square, but then I get an impossible equation. Is it the right way? Edit: I saw another discussion, but I want understand if my method is correct
|
170192
|
Solving the complex equation [imath]\sin(z) = \cos(z)[/imath]
To find the complex numbers z satisfying [imath]\sin(z) = \cos(z)[/imath], can I say: [imath]\sin(z) = \frac{(e^{iz}-e^{-iz})}{2i}=\frac{(e^{iz}+e^{-iz})}{2}[/imath] and solve for z? So we then reduce this to [imath]-e^{-iz} = e^{-iz}[/imath] but this doesn't look right
|
2911517
|
An equality with factorials
Prove that [imath]2\times 2! + 3\times 3!+4\times 4! +....+n\times n!=(n+1)!-2[/imath] I know that it can be proved by mathematical induction, but I want to prove it without using the mathematical induction. I tied the equation [imath]C^n_0 + C^n_1 + C^n_2 +...+C^n_n=2^n[/imath] But I did not get any thing useful.
|
2421440
|
How to simplify the summation kk! without using induction?
[imath]\sum_{k=1}^nk(k!)[/imath] I know the answer is (n+1)!-1..I can solve this question using principle of mathematical induction...but I would like to know if there is any other alternative approach
|
2913101
|
Explanation of composition of two onto functions?
My book says that if functions [imath]f[/imath] and [imath]g[/imath] are both onto then [imath]f\circ g[/imath] and [imath]g\circ f[/imath] may or may not be onto. Why is this so? Would someone please help me understand this, maybe with an example or diagrammatically? My book states that[imath] f\circ g[/imath] and [imath]g\circ f[/imath] may or may not be onto. I think this might be related to the domains and codomains of g and f which may not be equal. I am aware that there is a similar looking question and therefore I'm clarifying mine. With respect to the domains and codomains would someone please explain why the composite functions may or may not be onto?
|
1463025
|
Prove composition of two surjections is surjection
Suppose [imath]f: A \to B[/imath] and [imath]g: B \to C[/imath] are both surjections. Since [imath]f[/imath] is surjective, then for every [imath]b \in B[/imath] there exists [imath]a \in A[/imath] such that [imath]f(a)= b[/imath]. Since [imath]g[/imath] is surjective, for every [imath]c \in C[/imath], there's [imath]b \in B[/imath] such that [imath]g(b) = c[/imath]. Then [imath]g(f(a)) = g(b) = c[/imath]. Would this work? edit; I didn't realize there are similar questions on this site. Don't mind if this post is deleted.
|
2909492
|
How to construct an example for the entropy equation: [imath]H(Z)=H(X)+H(Y)[/imath] where [imath]Z=X+Y[/imath]
Given [imath]Z=X+Y[/imath] where X and Y are two random variables, under what conditions does [imath]H(Z)=H(X)+H(Y)[/imath]? Notice [imath]Z[/imath] is a function of [imath](X,Y)[/imath], therefore [imath]H(Z)\leq H(X,Y)[/imath], and [imath]H(X,Y)\leq H(X)+H(Y)-I(X;Y)[/imath]. Therefore when [imath]Z[/imath] and [imath](X,Y)[/imath] is a one-to-one function and [imath]X[/imath] is independent of [imath]Y[/imath]. My question is, how to find an example that satisfy this problem ? I have a simple example that let [imath]X=0[/imath] so that [imath]Z=Y[/imath], which satisfy the conditions. However, can anyone gives a more non-trivial example ? Thanks.
|
809383
|
Entropy of sum is sum of entropies
Having [imath]X[/imath] and [imath]Y[/imath] discrete random variables above finite set. Z is defined as [imath]Z=X+Y[/imath] when does the following happen: [imath]H(Z)=H(X)+H(Y)[/imath]
|
2914686
|
Prove that if every linear operator which commutes with [imath]T[/imath] is a polynomial in [imath]T[/imath], then [imath]T[/imath] admits a cyclic vector [imath]v[/imath] in [imath]V[/imath] (finite dimensional).
Let [imath]T:V→V[/imath] be a linear map over a finite dimensional vector space V with [imath]dim[/imath] [imath]V=n[/imath]. Prove that if every linear operator which commutes with [imath]T[/imath] is a polynomial in [imath]T[/imath], then T has a cyclic vector [imath]v[/imath] in [imath]V[/imath]. Let [imath]S[/imath] = the set of all polynomials in [imath]T[/imath]. Then [imath]S=C(T)[/imath] as every polynomial in [imath]T[/imath] commutes with [imath]T[/imath] and we are told the converse is also true. Suppose we do not have a cyclic vector [imath]v[/imath] i.e the set [imath]A[/imath] = {[imath]v,Tv,...,T^{n−1}v[/imath]} is not a basis for V. Then [imath]A[/imath] is either linearly dependent, or it does not span [imath]V[/imath]. Suppose A is linearly dependent but it spans V. Then we can get rid of one element in A and have a spanning set of size [imath]< n[/imath] which is impossible. Suppose A is linearly independent and it does not span [imath]V[/imath]. This is impossible as [imath]dim[/imath] [imath]V =n[/imath]. Suppose [imath]A[/imath] is linearly dependent and does not span V. Now in order to arrive at a contradiction, I need to somehow construct a linear operator which commutes with [imath]A[/imath] and is not a polynomial in [imath]T[/imath]. Any help would be great. Edit: The question really has not been answered in the other link. The answer only refers to a different website in which the text is incredibly difficult to read. Having another answer on stackexchange in clear text would make it much more accessible for the average reader.
|
414012
|
proof about commutative operators and T-cyclic vectors
Let [imath]V[/imath] be a finite dimensional vector space over [imath]F[/imath]. Let [imath]T:V \to V[/imath] be a linear operator. Prove that if every linear operator [imath]U[/imath] which commutes with [imath]T[/imath] is a polynomial of [imath]T[/imath], than [imath]T[/imath] has a [imath]T[/imath]-cyclic vector. I don't really know where to start... can someone please point me in the right direction?
|
2916029
|
Interior points.
Assume that {[imath]{x_n}[/imath]} is a weakly convergent sequence in an infinite dimensional Banach space [imath]X[/imath]. Show that [imath] \overline c\overline o \overline n \overline v[/imath]{[imath]x_n[/imath]} has no interior point. Can anyone help me with the proof of this one. I've no clue how to start, nor how to continue it.
|
1051666
|
[imath]\overline{\text{conv}}\{x_n: n\in \mathbb{N}\}[/imath] has empty interior for a w-conv seqeunce [imath](x_n)_n[/imath].
Given [imath]x_n \to x[/imath] weakly in a Banach space X, [imath]\dim X = \infty[/imath]. Show that [imath]\text{int}\left\{ \overline{\text{conv}}\{x_n:n\in\mathbb N \}\right\}[/imath] is empty. Can anyone help me with this problem? Any hints? Prob. Krein-Millmann is needed but I don't see how to use it.
|
2915798
|
Values of this integeration
The values of [imath]I=\int_{0}^{1} {x^4} ({1-x})^4\dfrac1{1+x^2}dx[/imath] My Failed Attempt I tried applying F(x) =F(a+b-x) in hopes the denominator will cancel out but it's not. I tried to further simplify the expression but with no luck [imath]I=\int_{0}^{1} {x^4} ({1-x})^4\dfrac1{1+({1-x})^2}dx[/imath] Adding both integral isn't helping me much
|
129625
|
Finding [imath]\int_0^1{\frac{x^4(1-x)^4}{1+x^2}}dx[/imath]
The question I am working on: Evaluate [imath]\frac{1}{2} \int^1_0{x^4 (1-x)^4 } dx \le \int^1_0{\frac{x^4 (1-x)^4}{1+x^2}} dx \le \int^1_0{x^4 (1-x)^4 } dx[/imath] So using integration by parts to solve: (letting [imath]u=(1-x)^4[/imath] and [imath]dv=x^4[/imath]) [imath]\int{x^4 (1-x)^ 4} dx = \frac{4}{5}x^5(x-1) - \frac{4}{5}\left(\frac{x^7}{7} - \frac{x^6}{6}\right)+c[/imath] Is it correct so far? If so ... [imath]\int^1_0{ x^4 (1-x)^4 } dx = \frac{4}{5}\left(\frac{1}{6}-\frac{1}{7}\right)=\frac2{105}[/imath] [imath]\frac{1}{2} \int^1_0{ x^4 (1-x)^4 } dx = \frac{2}{5}\left(\frac{1}{6}-\frac{1}{7}\right)=\frac1{105}[/imath] But [imath]\int\frac{{x^4(1-x)^4}}{1+x^2} dx = ??[/imath] Since I found [imath]\int{{x^4(1-x)^4}} dx[/imath]. I thought of integration by parts, letting [imath]u=\frac{1}{1+x^2}[/imath], [imath]dv = x^4(1-x)^4[/imath]. But I will get a very complicated [imath]v[/imath] to integrate later? Same if I did it the other way around?
|
2916789
|
Complex Integration doubt in the solution
I'm trying to find the value of [imath]\int_0^\infty e^{-ax}\cos(bx)\,dx,\quad a>0.[/imath] by integrating [imath]e^{−Az} ,A=\sqrt{a^2+b^2}[/imath], over an appropriate sector with angle [imath]ω[/imath], with [imath]cos ω=\dfrac{a}{A}[/imath]. I'm seeing some solutions and I have the following questions: 1) It seems [imath]ω= cos^{−1}(a/A)[/imath] is strictly between 0 and [imath]\pi/2[/imath]? 2) Why is it true that [imath]\cos(\theta) \geq 1 - \dfrac{2\theta}{\pi} [/imath]? how to prove? Can someone answer both questions with details?
|
300019
|
Complex Integration of [imath]\int_0^\infty e^{-ax}\cos(bx)\,dx[/imath]
Out of Stein's book, we're asked to show find a formula for [imath]\int_0^\infty e^{-ax}\cos(bx)\,dx,\quad a>0.[/imath]While this is very doable via integration by parts, I'm asked to use contour integration, where we're suggested to integrate over a sector with angle [imath]\omega[/imath] such that [imath]\cos(\omega)=a/\sqrt{a^2+b^2}.[/imath] I've attempted this multiple times, and I keep having trouble with some integrals. I've set the contour up so that on the first segment, it's on the real axis, so we have the integral [imath]\int_0^R e^{-az}\cos(bz)\,dz.[/imath] Then I parameterize the arc as [imath]z(\theta)=Re^{i\theta}[/imath] for [imath]0\leq\theta\leq \omega[/imath], so the second integral becomes [imath]\int_0^\omega e^{-a(Re^{i\theta})}\cos(b(Re^{i\theta}))\left(iRe^{i\theta}\right)\,d\theta.[/imath]The final segment I parameterized as [imath]z(t)=Re^{i\omega}(1-t)[/imath] and set up the final integral as [imath]\int_0^1e^{-a(Re^{i\omega}(1-t))}\cos\big(b(Re^{i\omega}(1-t))\big)(-Re^{i\omega})\,dt.[/imath]I've tried finding some way to bound one of the last two integrals so that I can show one of them goes to [imath]0[/imath] as [imath]R\to\infty[/imath], but I've not had any luck. Will someone make a suggestion if my approach and parameterizations are correct? Thanks! Update: My thoughts are really that the integral which goes to zero is the arc. I keep working it down in the following way; we know that it is \begin{align} &\leq R\int_0^\omega\left|e^{-aR(\cos\theta+i\sin\theta)}\cdot\left(\frac{e^{ibRe^{i\theta}}+e^{-ibRe^{i\theta}}}{2}\right)\right|\,d\theta\\ &\leq\frac{R}{2}\int_0^\omega\left|e^{-aR\cos\theta}\cdot\left(e^{ibR(\cos\theta+i\sin\theta)}+e^{-bR(\cos\theta+i\sin\theta)}\right)\right|\,d\theta\\ &\leq\frac{R}{2}\int_0^\omega\left|e^{-aR\cos\theta-bR\sin\theta}\right|+\left|e^{-aR\cos\theta+bR\sin\theta}\right|\,d\theta. \end{align} At this point, it is easy to show that the first term tends to zero, since [imath](-aR\cos\theta)<0[/imath] and [imath]bR\sin\theta>0[/imath] (since [imath]b[/imath] and [imath]\sin\theta[/imath] have the same sign). The second term, however is what causes me trouble. I just finished working it out again, and I get that it only goes to zero if [imath]a^2>b^2[/imath], which isn't necessary in the general formula when achieved by integration by parts. I am really at a loss... Added Solution: See the solution I've posted and please leave comments on your thoughts about it. Thanks!
|
832810
|
if A is a countable dense subset of [imath]\mathbb{R}[/imath] then there exists [imath]B\subset A[/imath] dense and countable in [imath]\mathbb{R}[/imath]
I found this result in so many books and I tried to demonstrate it but I'm not convinced with my proof, if A is a countable dense subset of [imath]\mathbb{R}[/imath] then there exists [imath]B\subset A[/imath] dense and countable in [imath]\mathbb{R}[/imath].
|
60916
|
Smallest dense subset of [imath]\mathbb{R}[/imath]
I am not sure if what I am looking for even makes sense (or) exists. Anyway I would be happy if someone can clear my confusion. The set of real numbers [imath]\mathbb{R}[/imath] is obtained as completion of [imath]\mathbb{Q}[/imath]. However, [imath]\mathbb{Q}[/imath] is not the only set which is dense in [imath]\mathbb{R}[/imath]. [imath]\mathbb{Q} \backslash \mathbb{Z}[/imath] is also a dense subset of [imath]\mathbb{R}[/imath]. I am wondering if it makes sense to talk of the "smallest" dense subset of [imath]\mathbb{R}[/imath]. To phrase what I am looking for precisely and what I mean by smallest, I am looking for a dense set [imath]A[/imath] of [imath]\mathbb{R}[/imath] such that if [imath]B[/imath] is a proper subset of [imath]A[/imath] then [imath]B[/imath] is not dense in [imath]\mathbb{R}[/imath]. I am able to "see" that the set [imath]A[/imath], I am looking for doesn't exist since any open interval consists of infinite rationals. But I am unable to precisely argue out to myself and convince why [imath]A[/imath] doesn't exist. Could some one throw more light on this?
|
2916771
|
Functions satisfying [imath]f(x)+f(\frac{1}{1-x})=x[/imath] with [imath]x\in\mathbb{R}\setminus\{0,1\}.[/imath]
I have used this identity: if [imath]g(x)=1/(1-x),[/imath] then [imath]g^{-1}(x)=1-\frac{1}{x},[/imath] to get all functions satisfying: [imath]f(x)+f(\frac{1}{1-x})=x[/imath] with [imath]x\in\mathbb{R}\setminus\{0,1\},[/imath] but I didn't get a general form of its solution. My question here is: Is there any simple method to solve the titled functional equation?
|
1541303
|
Find [imath]f[/imath] if [imath] f(x)+f\left(\frac{1}{1-x}\right)=x [/imath]
Find [imath]f[/imath] if [imath] f(x)+f\left(\frac{1}{1-x}\right)=x [/imath] I think, that I have to find x that [imath]f(x) = f\left(\frac{1}{1-x}\right)[/imath] I've tried to put x which make [imath]x = \frac{1}{1 - x}[/imath], but this equation has no roots in real numbers.
|
2916817
|
Functional analysis orthogonal sequence.
Let [imath]\{{e_n}\}\ n=1,2,3,...[/imath]be an orthonormal sequence in a Hilbert space [imath]H[/imath] and let [imath]x\not=0 \in H[/imath] then [imath]<x, e_n> \to 0[/imath] as [imath]\ n \to \infty[/imath]. By Bessel's inequality we have [imath]\sum_{n=1}^{\infty}|<x,e_n>|^2\leq||x||^2[/imath] and therefore [imath]<x, e_n> \to 0[/imath] as [imath]\ n \to \infty[/imath]. Is my argument correct? But the answer says the limit is [imath]1[/imath]. Could you please explain to me.
|
470368
|
[imath]3.\lim_{n\to\infty}\langle x,e_n\rangle=?[/imath]
given that [imath]\{e_i\}_{n=1}^{\infty}[/imath] is an orthonormal sequence in a hilbert space [imath]H[/imath], and [imath]x\ne 0\in H[/imath]. Then could any one tell me which of the following is true? [imath]1.\lim_{n\to\infty}\langle x,e_n\rangle[/imath] does not exist [imath]2.\lim_{n\to\infty}\langle x,e_n\rangle=\|x\|[/imath] [imath]3.\lim_{n\to\infty}\langle x,e_n\rangle=1[/imath] [imath]4.\lim_{n\to\infty}\langle x,e_n\rangle=0[/imath] I just know one relations about orthonormal sequence in Hilbert space [imath]\sum_{n=1}^{\infty}|\langle x,e_i\rangle|^2\le \|x\|^2[/imath]. Thank you for help.
|
2917277
|
Infinite series involving e
Consider: [imath]f(x) = \frac{1}{e}\sum_{n=0}^{\infty}\frac{n^x}{n!}[/imath] When [imath]x[/imath] is a whole number f(x) is also a whole number. Is there a way to resolve this summation?
|
1109420
|
Evaluting sum [imath]\sum \limits_{n=0}^\infty\frac{n^k}{n!}[/imath]
Inspired by this question,I was interested if the following sum has a closed form.Looking for [imath]k[/imath] integer I found the Dobinski's formula so that the sum when [imath]k[/imath] is natural number is [imath]e\cdot B_k[/imath] where [imath]B_k[/imath] is the [imath]k[/imath]-th Bell number.I am interested if it's known whether the sum [imath]\sum_{n=0}^\infty\frac{n^k}{n!}[/imath] has a closed form for some other values of [imath]k[/imath].
|
2917006
|
Prove that *every* subset of a metric space [imath]M[/imath] can be written as the intersection of open sets.
Prove that every subset of a metric space [imath]M[/imath] can be written as the intersection of open sets. My attempt: If [imath]A\subset M[/imath] is open, [imath]A[/imath] can be written as [imath]A\cap M[/imath], which is the intersection of 2 open sets. If [imath]A\subset M[/imath] is closed, it can presumably be written as the intersection of infinitely many open sets. But I'm not sure how. And what if [imath]A\subset M[/imath] is neither open nor closed?
|
2369777
|
Show that any set in a metric space can be written as the intersection of open sets
Show that any set contained in the metric space [imath](X, d)[/imath] can be written as the intersection of open sets. Definitions: A set [imath]A \subseteq X[/imath] is open if [imath]\forall x \in A[/imath], [imath]\exists \varepsilon>0[/imath] such that [imath]B_{\varepsilon}(x) \subseteq A[/imath]. A set [imath]C \subseteq X[/imath] is closed if and only if [imath]X \setminus C[/imath] is open. I also know that the intersection of a finite collection of open sets is open and that any open ball contained in [imath]X[/imath] is open. How can I prove this question?
|
2916872
|
Bijection Partition identity
Hello I have a doubt about the next partition identities [imath]P(n: even \; number \; of \; odd \; parts)=P(n: distinct\; parts, \; number \; of \; odd\; parts\; is\; even)[/imath] Also for the case when both instances of "even" are changed to odd. I need to show a bijection of but I can't figure out how. I have been trying with one example (n=6) to see any particular characteristic however I can not have the equality. I trying to solve the problem by using a bijection between the partitions of each side because I am not allowed to use function now. So, any hint would be appreciated!
|
447485
|
Prove : [imath]p[/imath](n│even number of ODD parts)=[imath]p[/imath](n│distinct parts ,number of ODD parts is even )
I'm trying to prove the following Integer Partition claim : [imath]p[/imath](n│even number of ODD parts) = [imath]p[/imath](n│distinct parts ,number of ODD parts is even) . So I tried to prove a stronger claim : [imath]p[/imath](n│number of odd parts) = [imath]p[/imath](n│number of distinct parts) : Using generating functions : DISTINCT PARTS = [imath](1+x)\cdot(1+x^2 )\cdot(1+x^3 )\cdot(1+x^4 )\cdot\cdot\cdot\cdot[/imath] = [imath]\frac {1-x^2} {1-x} \cdot \frac {1-x^4} {1-x^2} \cdot \frac {1-x^6} {1-x^3} \cdot \frac {1-x^8} {1-x^4}\cdot\cdot\cdot\cdot[/imath] [imath] \frac {1} {1-x} \cdot \frac {1} {1-x^3} \cdot \frac {1} {1-x^5} \cdot \frac {1} {1-x^7} \cdot\cdot\cdot\cdot [/imath] = ODD PARTS But I'm not sure regarding the proof , does it really prove that even number of odd parts equals distinct parts ,where number of ODD parts is even ? Thanks
|
2809087
|
Integral [imath]\int_{-2}^0 \frac{x}{\sqrt{e^x+(x+2)^2}}dx[/imath]
I am trying to evaluate [imath]\int_{-2}^0 \frac{x}{\sqrt{e^x+(x+2)^2}}dx[/imath] So far I had no succes using trig substitution or integration by parts, also some random substitution like [imath]x=2t[/imath] and moved the exponential to the numerator, but I am stuck. Could you perhaps give me an idea? (this is a college admission problem)
|
2699140
|
Calculate [imath] \int_{-2}^{0} \frac{x}{\sqrt{e^x+(x+2)^2}} dx [/imath]
[imath] \int_{-2}^{0} \frac{x}{\sqrt{e^x+(x+2)^2}} dx [/imath] = -[imath] 2.887270... [/imath] The function has no antiderivate and there's no symmetry here to help, if you solve this I would be thankful if you would for a highschooler ^^
|
2778196
|
Why is the nontrivial left ideal of [imath]M_ 2(D)[/imath] minimal?
I'm totally confused by this question. I appreciate any help or answer. thanks in advance why is the nontrivial left ideal of M_2(D) minimal? that D is a division ring
|
2838341
|
why each left ideal of this is minimal?
please first read these then answer my question Let [imath]D[/imath] be a division ring and r, n be two positive integer numbers such that [imath]0 ≤ r ≤ n[/imath]. Then [imath]H_r(D)[/imath] denotes the left ideal of [imath]M_n(D)[/imath] containing all matrices whose [imath]j[/imath]th column is zero, for every [imath]r < j ≤ n[/imath]. Theorem A: Let [imath]D[/imath] be a division ring and [imath]n[/imath] be a positive integer number. Then for every left ideal [imath]I[/imath] of [imath]M_n(D)[/imath], there exist an invertible matrix [imath]P[/imath] and integer [imath]r[/imath], [imath]0 ≤ r ≤ n[/imath], such that [imath]I = PH_r(D)P^{−1}[/imath]. Now from theorem A how I can conclude that each left ideal of [imath]M_2(D)[/imath] is minimal? I'm totally confused with this
|
2918052
|
Proof that if [imath]R_{1};R_{2}=R_{2};R_{1}[/imath] is a relation on a set [imath]S[/imath], then [imath]R_{1}[/imath] and [imath]R_{2}[/imath] are bijective maps
For the surjectivity part, I believe I have managed to prove the claim. And for Injectivity, I've tried to show that if [imath]aR_{1};R_{2}c[/imath] and [imath]bR_{2};R_{1}c[/imath], then [imath]a=b[/imath] but I am not entirely sure what [imath]a=b[/imath] means in this context. Do I have to show that there is an equivalence relation containing (a,b)?
|
2918206
|
If [imath]R_{1}[/imath] and [imath]R_{2}[/imath] are relations on a set S with [imath]R_{1};R_{2}=R_{2};R_{1}[/imath]. Then [imath]R_{1}[/imath] and [imath]R_{2}[/imath] are bijective maps
For the surjectivity part, I showed that if [imath](a,b)\in I[/imath] then [imath](a,c)\in R_{1}[/imath] and [imath](c,b)\in R_{1}[/imath] and [imath](a,c)\in R_{2}[/imath] and [imath](c,b)\in R_{2}[/imath] for some [imath]c\in S[/imath]. Now for arbituary [imath](x,y)\in R_{1}[/imath] we have [imath](x,z)\in I[/imath] which implies that [imath](y,z)\in R_{2}[/imath]. But also [imath](y,z)\in R_{1}[/imath], so [imath]R_{1}[/imath] is surjective. Similar argument for [imath]R_{2}[/imath] And for Injectivity, I've tried to show that if [imath]aR_{1};R_{2}c[/imath] and [imath]bR_{2};R_{1}c[/imath], then [imath]a=b[/imath] but I am not entirely sure what [imath]a=b[/imath] means in this context. Do I have to show that there is an equivalence relation containing (a,b)? edit: [imath]R_{1};R_{2}[/imath] is a composition of two relations and [imath]I=R_{1};R_{2}[/imath]
|
2918998
|
Resolving nested dependent summation [imath] \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} j [/imath]
I have the following summation: [imath] \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} j [/imath] In which the inner sum depends on the outer one, in the index on the bottom. Is there a way to change the indexes in order to remove the dependence? What is the easiest way to get a closed form from it?
|
2918464
|
Getting a closed form from [imath]\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1[/imath]
I need to get a closed form from [imath] \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1 [/imath] Starting from the most outer summation, I got [imath] \sum_{k=1}^{j} 1 = j [/imath] But now I don't know how to proceed with: [imath] \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} j [/imath] Could you guys please help me? Thanks in advance.
|
2912613
|
Prove that for each [imath]\alpha \in [0,1][/imath], there exists [imath]E \in M[/imath] with [imath]\mu(E)=\alpha[/imath]
Let [imath](X,M,\mu)[/imath] be a measure space. Assume that [imath]\mu[/imath] is a non-atomic finite measure, namely for every [imath]E \in M[/imath] with [imath]\mu(E)>0[/imath], there exists [imath]F \subset E[/imath] such that [imath]0<\mu(F)<\mu(E)[/imath]. Suppose that [imath]\mu(X)=1[/imath]. Show that for each [imath]\alpha \in [0,1][/imath], there exists [imath]E \in M[/imath] with [imath]\mu(E)=\alpha[/imath]. Thoughts on Question: I've thought about this question for a very long time but I can't seem to figure it out. Here is my initial partial attempt. Proof (first following a hint given to me): Let [imath]E \in M[/imath] with [imath]\mu(E)>0[/imath]. Also let [imath]\epsilon>0[/imath]. Since [imath]\mu[/imath] is a non-atomic finite measure, there exists [imath]F\subset E[/imath] so that [imath]0<\mu(F)<\mu(E)[/imath]. We know that [imath]\mu(F)+\mu(E \cap F^c)=\mu(E)[/imath] by additivity. We know that one of [imath]F[/imath], [imath]E\cap F^c[/imath] must satisfy [imath]0<\mu(F_1)\leq 1/2*\mu(E)[/imath]. Let [imath]F_1[/imath] represent the element that does. Let's continue this process again. Let [imath]F_2 \subset F_1[/imath]. Then we have: [imath]\mu(F_2) \leq 1/2* \mu(F1) \leq 1/4*\mu(E)[/imath]. For every n, there exists [imath]F_n[/imath] so that [imath]0<F_n\leq 1/2^n*\mu(E)[/imath]. Take n big enough so that [imath]1/(2^n) \leq \epsilon[/imath]. Thus we know that for [imath]\epsilon>0[/imath] we can find [imath]G\subset E[/imath] so that [imath]0<\mu(G)<\epsilon[/imath]. Concerns: So I have now proved that for [imath]\epsilon>0[/imath] there exists [imath]0<\mu(G)<\epsilon[/imath],as per a hint but how do I use this to prove the the result the question is asking:for each [imath]\alpha \in [0,1][/imath], there exists [imath]E \in M[/imath] with [imath]\mu(E)=\alpha[/imath]? I have seen some things online but I can't seem to follow any of them. Also is proof correct for the part above? Did I say anything, even something small inaccurately? Could someone, if possible, help me clean up my proof above because I want to precise as possible? Thank you very much. I just need help utilizing this to prove the result.
|
254728
|
Simpler proof - Non atomic measures
Suppose that [imath](X,\mathcal{E},\mu)[/imath] is a non-atomic finite measure space (i.e. for every [imath]E \in \mathcal{E}[/imath] with [imath]\mu(E)>0[/imath] there exists [imath]F \subset E[/imath] measurable such that [imath]0<\mu(F) <\mu(E)[/imath].) a) Prove that for every [imath] \varepsilon >0[/imath] there is a finite partition of [imath]X[/imath] in measurable subsets [imath]X_1,..,X_n[/imath] such that [imath]\mu(X_i)\leq \varepsilon[/imath]. b) Prove that for every [imath]\alpha \in [0,\mu(X)][/imath] there exists [imath]E \in \mathcal{E}[/imath] with [imath]\mu(E)=\alpha[/imath]. I guess that a) is given to prove b) more easily. I have the following idea of solution for a) (inspired from this Wikipedia post, which proves b) ) Denote [imath] \Gamma = \{ (X_1,..,X_n) : n >> \mu(X)/\varepsilon, X_i \text{ are disjoint }, \mu(X_i)\leq \varepsilon \}[/imath] ordered by componentwise inclusion. Totally ordered parts [imath](Y_\alpha)[/imath] of [imath]\Gamma[/imath] have an upper bound element [imath](\bigcup Y_\alpha^i)_{i=1}^n[/imath] which is still in [imath]\Gamma[/imath]. By Zorn's lemma [imath]\Gamma[/imath] has maximal elements. If a maximal [imath](X_1..X_n)[/imath] element is not a partition, then we can replace it with something better, of the form [imath](X_1,..,X_n\cup A)[/imath] where [imath]A \subset X \setminus (X_1 \cup..\cup X_n)[/imath] and [imath]\mu(A)>0[/imath]. My questions are: 1) Is my solution of a) correct? I feel that the part with the upper bound element may not work, since there might be noncountable unions. 2) What is a simpler solution of a)?
|
1307370
|
Closedness of a mapping [imath]p : [0,1] \cup [2,3] \to [0,2][/imath]
Below is an example (Example 1, p.137) from Munkres' book Topology. Let [imath]X[/imath] be the subspace [imath][0,1] \cup [2,3][/imath] of [imath]\mathbb{R}[/imath], and let [imath]Y[/imath] be the subspace [imath][0,2][/imath] of [imath]\mathbb{R}[/imath]. The map [imath]p:X \to Y [/imath] defined by [imath]p(x) = \begin{cases} x, & \text{for }x\in [0,1] \\ x-1, & \text{for }x\in [2,3] \end{cases}[/imath] is readily seen to be surjective, continuous and closed. It is clear to me that [imath]p[/imath] is surjective and continuous, but I cannot not see "readily" why [imath]p[/imath] is closed. To show [imath]p[/imath] is closed, I let [imath]U[/imath] be a closed subset of [imath]X[/imath] and tried to show [imath]p(U)[/imath] is closed in [imath]Y[/imath], but I am not sure how to proceed. How can I see/prove that [imath]p[/imath] is closed?
|
1252991
|
Example 1, Sec. 22 in Munkres' TOPOLOGY, 2nd edition: How to verify that this map is closed?
Let [imath]X[/imath] be the subspace [imath][0,1] \cup [2,3][/imath] of [imath]\mathbb{R}[/imath], and let [imath]Y[/imath] be the subspace [imath][0,2][/imath] of [imath]\mathbb{R}[/imath]. The map [imath]p \colon X \to Y[/imath] defined by [imath] p(x) \colon= \begin{cases} x \ &\mbox{ for } \ x \in [0,1], \\ x-1 &\mbox{ for } \ x \in [2,3] \end{cases} [/imath] is surjective and continuous. This much is clear to me. But Munkres asserts that this map is also a closed (and hence a quotient) map. How to verify this? Now if [imath]A[/imath] is the subspace [imath][0,1) \cup [2,3][/imath] of [imath]X[/imath], then the map [imath]q \colon A \to Y[/imath] obtained by restricting [imath]p[/imath] is continuous and surjective. Alright! Now what stumps me is the following assertion by Munkres: The set [imath][2,3][/imath] is saturated with respect to [imath]q[/imath]. How to show this?
|
2920005
|
Unable use countability argument in following problem
Suppose f is analytic function defined everywhere in [imath]\mathbb C[/imath] and such [imath]z_0\in \mathbb C[/imath] at least one coefficent in expansion [imath]f(z)=\sum^\infty_{n=0}c_n(z-z_0)^n[/imath] is equal to [imath]0[/imath]. Then prove that [imath]f[/imath] is polynomial I understood that for each [imath]z\in \mathbb C [/imath] there exist some [imath]n\in N[/imath] such that [imath]f^{(n)}(z)=0[/imath] There is hint available for this problem says that use countability argument . I know that zero of holomorphic function is at most countable by Identity theorem. Here we get each [imath]z[/imath] some [imath]n[/imath]. It does not say about for all [imath]n[/imath] some [imath]n[/imath] exist . I am not able to use hint. Any Help will be appreciated
|
943952
|
Prove that [imath]f[/imath] is a polynomial if one of the coefficients in its Taylor expansion is 0
Suppose [imath]f[/imath] is an analytic function defined everywhere in [imath]\mathbb{C}[/imath] and such that for each [imath]z_0 \in \mathbb{C}[/imath] at least one coefficient in the expansion [imath]f(z) = \sum_{n = 0}^{\infty}c_n(z-z_0)^n[/imath] is equal to [imath]0[/imath]. Prove that [imath]f[/imath] is a polynomial. The problem hints to use the fact that [imath]c_nn! = f^{(n)}(z_0)[/imath] and use a countability argument. My attempt at a solution The only thing I can think of in this case is that the coefficients [imath]c_n[/imath] of the above Taylor expansion are defined as: [imath]c_n = \frac{f^{(n)}(z_0)}{n!}[/imath] The only way one of these [imath]c_n[/imath] could be zero is if the derivative of order [imath]n[/imath] vanishes everywhere. Thus, this would mean that [imath]f[/imath] is a polynomial of order [imath]k < n[/imath]. Does this suffice as a proof? How would I use a countability argument to prove this instead? Thanks.
|
2921056
|
A difficulty in understanding the definition for limit superior of [imath]{x_{n}}[/imath]
The definition is given by the following formula: [imath]\limsup_{n \rightarrow \infty} x_{n} := \lim_{n \rightarrow \infty}(\sup_{m\geq n} x_{m}) [/imath] I could not understand the meaning of [imath](\sup_{m\geq n} x_{m})[/imath] and why we have this number m and why it must be [imath]\geq n[/imath], could anyone explain this for me please?
|
2294183
|
How to explain the following definition of [imath]\lim\sup s_n[/imath] intuitively?
How to explain the following definition of [imath]\lim\sup s_n[/imath] intuitively? let [imath]\lim\sup s_n[/imath]=[imath]l[/imath] The definition is ([imath]\forall t<l,\forall N,\exists n>N, s_n>t[/imath]) and ([imath]\forall t>l,\exists N,\forall n>N, s_n<t[/imath]) For the first one, if we have a increasing sequence from very negative to [imath]0[/imath], how is it still true. i.e. [imath]\{-1, -0.5, ... , -0.00001,...\}[/imath], here [imath]\lim\sup s_n=0[/imath]. let [imath]t=-1,[/imath] so [imath]t<0[/imath], but not all terms are greater than [imath]t[/imath]. For the second one, I am also confused. I think It would be great if someone could show some valid examples. Also when we prove the such limit exists, is it necessary to show both characteristics?
|
2920934
|
Condition when any union of subgroups is subgroup
Let [imath]G[/imath] be a group and [imath]\{H_i\}_{i\in \mathcal{I}}[/imath] for some (not necessarily countable) indexing set [imath]\mathcal{I}[/imath] be a family of subgroups of [imath]G[/imath] none of which contain all the others. State and prove a general condition that will imply [imath]\bigcup \limits_{i\in\mathcal{I}}H_i[/imath] is a subgroup of [imath]G[/imath]. After thinking for a day two I ran into the following idea: What if I put the following condition: For any pairs [imath]i,j\in \mathcal{I}[/imath] there is [imath]k\in \mathcal{I}[/imath] such that [imath]H_i\subset H_k[/imath] and [imath]H_j\subset H_k[/imath]. And it's quite easy to show that in this case the union of [imath]\{H_i\}_{i\in \mathcal{I}}[/imath] is a subgroup! But I don't know what to do with the condition "none of which contain all the others"?
|
367983
|
When can I say that [imath]\cup H_i[/imath] is a group?
Let [imath]G[/imath] be a group and [imath]\{H_i\}_{i\in I}[/imath] a family of subgroups. I would like to find a condition that will imply that [imath]\cup_{i\in I} H_i[/imath] is a subgroup. I know that it's not true in general, I need help to find this condition. Any help is welcome. Thanks a lot.
|
2921179
|
Prove the module of a holomorphic function cannot be [imath]\frac{K}{\cosh x}[/imath]
Prove that there does not exist any holomorphic function [imath]f(z)[/imath], such that [imath]|f(z)| =\frac{K}{\cosh x}[/imath] where [imath]K>0[/imath] is a constant, and [imath]x=Re(z)[/imath]. My idea is to use the maximum modulus principle. [imath]|f(z)| =\frac{K}{\cosh x} = \frac{2K}{e^x+e^{-x}} \leq K[/imath]. An entire function with bounded modulus must be a constant. This leads to a contradiction. My question is, in my attempt, I assume [imath]f[/imath] to be holomorphic on [imath]\mathbb{C}[/imath]. Is it possible to prove the same result when [imath]f[/imath] is holomorphic on [imath]\Omega[/imath] where [imath]\Omega[/imath] is a domain? And is it possible not to use maximum modulus principle? Thank you for any help!
|
2921051
|
No analytic function [imath]f[/imath] has modulus [imath]|f(z)|=1/\cosh(\Re z)[/imath]
An analytic function [imath]f(z) = f(x+iy)[/imath] in [imath]\mathbb{C}[/imath] cannot have modulus [imath]\frac{A}{\cosh(x)}[/imath] for some constant [imath]A \neq 0[/imath]. Can we do so simply using the Cauchy-Riemann Equations? I tried working by contradiction: Say there is such a [imath]f(z)[/imath]. Given [imath]f(z) = u(x,y) + i v(x,y)[/imath] is analytic, it satisfies: [imath]u_x = v_y[/imath] and [imath]u_y = -v_x[/imath] We also see that: [imath]|f(z)|^2 = u(x,y)^2 + v(x,y)^2 = (\frac{A}{\cosh x})^2[/imath] I tried reaching a contradiction, but seem to be getting lost in a mess of reformulations. My start point is: [imath]u_x = v_y \text{ and } u_y = -v_x[/imath] along with [imath]u(x,y)u_x + v(x,y)v_x = -A^2 \frac{sinh x}{(cosh x )^3}[/imath] and [imath]u(x,y)u_y + v(x,y)v_y = 0[/imath] I would appreciate some hints!
|
2921154
|
If [imath]\gcd(m,n) = 1[/imath] then [imath]\mathbb Z_{mn} \cong \mathbb Z_m \oplus \mathbb Z_m[/imath].
I am familiar with the Chinese Remainder Theorem and I know that it must be used here in some way (Hint given by my lecturer). All I know so far is [imath]\mathbb Z_{mm} \not\cong \mathbb Z_m \oplus \mathbb Z_m[/imath]. This explains my reaction of 'no way this is true', I tried to play around with [imath]n[/imath] and [imath]m[/imath] using the fact that they're co-prime, but I really do feel like I've hit the wall on this. (EDIT) Just to elaborate on more work I've done; I played around with small examples (2,3) and now I'm actually really questioning this because how can an isomorphism exist between say [imath]\mathbb Z_{6} \cong \mathbb Z_3 \oplus \mathbb Z_3[/imath]? One clearly has 6 elements and the other 9! So how can there be a bijective mapping between these two groups? Perhaps this question has a typo? Any hints would be appreciated, so please no solutions.
|
2730144
|
Isomorphism between finite groups
What can be the general statements made, when two finite groups can be isomorphic to each other: [imath]\mathbb{Z}_{pq}[/imath] and [imath]\mathbb{Z}_{p} \times \mathbb{Z}_{q}[/imath] Say [imath]\mathbb{Z}_{pq} \simeq\mathbb{Z}_{p} \times \mathbb{Z}_{q},[/imath] what are the conditions for [imath]p[/imath] and [imath]q[/imath] and their factorization relations? How do we show this in the most elegant way? And the intuitive way?
|
2920823
|
O level Expansion problem.
yep, title sucks.. Q: The expansion of (1+px+qx2)8 = 1+8x+52x2+kx3. Find the values of p, q and k. I found the values of [imath]p[/imath] and [imath]q[/imath] and they are: p = 1 q = 3 But I am unable to find the value of [imath]k[/imath]. ANS: k = 224.
|
859439
|
Expansion of [imath](1+px+qx^2)^8\equiv 1+8x+52x^2+kx^3[/imath]...
[imath](1+px+qx^2)^8 \equiv 1+8x+52x^2+kx^3...[/imath] Given that expression, calculate the values of p, q, and k. I'm guessing this is a trinomial that needs to be turned into a binomial, but how would you do it? I have no idea.
|
2921166
|
Intuition and name for [imath]a^T R a[/imath]
I have seen this operation before but cant seem to recall the name or corresponding theory. This operation changes inner product [imath]<a,b> = a^T b[/imath] to [imath]a^T R b[/imath], where [imath]R[/imath] is some matrix. What does this do intuitively and what is the name?
|
2660467
|
What is the name of the operation [imath]v^T H v[/imath]?
To find out how something is called is the one thing I don't know how to use Google for. If I have a vector [imath]v[/imath] and a symmetric matrix [imath]H[/imath], then the operation [imath]v^T H v[/imath] gives me some sort of projection of the matrix [imath]H[/imath] on the vector [imath]v[/imath]. For example if [imath]H[/imath] is the Hessian, the operation gives me the second derivative in the direction of [imath]v[/imath]. What is the name of this operation?
|
1618071
|
To evaluate the sum [imath]\frac{1}{5}-\frac{1 \cdot 4}{5 \cdot 10}+\frac{1 \cdot 4 \cdot 7}{5 \cdot 10 \cdot 15}-\ldots[/imath]
Right now I am working through archived papers of a math aptitude quiz. For some reason I seem to be haveing a hard time with these series problems. I have managed to write the above series in a compact form but thats as far as I got. [imath]\frac{1}{5}-\frac{1 \cdot 4}{5 \cdot 10}+\frac{1 \cdot 4 \cdot 7}{5 \cdot 10 \cdot 15}-\ldots = \sum\limits_{i=0}^\infty (-1)^i\frac{\prod\limits_{j=0}^i (3j+1)}{5^{i+1}(i+1)!}[/imath] Help please!
|
1036288
|
[imath]\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot\cdot\cdot\cdot\cdot[/imath]
How to find the sum of the following series: [imath]\frac {1}{5}-\frac{1.4}{5.10}+\frac{1.4.7}{5.10.15}-\cdot\cdot\cdot\cdot\cdot\cdot\cdot[/imath] Any hints.
|
2921784
|
Are [imath]\langle \mathbb R,+\rangle [/imath] and [imath]\langle \mathbb R\setminus \{0\},*\rangle [/imath] isomorphic?
I know that [imath]\langle R,+\rangle [/imath] and [imath]\langle\mathbb R^+,*\rangle[/imath] are isomorphic. And Explicit map given by Exponential . I am not able to convince myself for map [imath]\langle \mathbb R\setminus \{0\},*\rangle [/imath] and [imath]\langle \mathbb R,+\rangle[/imath] . Any Help will be appreciated
|
1221625
|
About the additive group and the multiplicative group of a field
Let [imath]F[/imath] be a field. When happens that the additive group of [imath]F[/imath] is isomorphic to the multiplicative group? It is easily to work out that [imath]F[/imath] must have characteristic [imath]0[/imath], but then what?
|
2922511
|
Why does the curl of a function provide this particular amount of information?
In a classical electrodynamics textbook (Griffiths), it is mentioned that even though the electric field function, [imath]E:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}[/imath], is a (3D) vector valued function, the amount of information needed to fully describe it is equivalent to the amount needed to describe a scalar valued function, and this is because it is the gradient of an electric potential function, [imath]U:\mathbb{R}^{3}\rightarrow \mathbb{R}[/imath] and so is completely determined by it. The author goes on to explains that this is so because [imath]E[/imath] has the property that its curl is zero evevrywhere, and this is what restricts the freedom in determining [imath]E[/imath] (and what enables it to be the gradient of a scalar function in the first place). This is all fine (and fun), but I find myself unable to answer the question: why does the imposition of zero curl provide an amount of information exactly equivalent to two scalar-valued functions (no more, no less)? An example of the sort of reasoning that leads me to other conclusions: since zero curl is equivalent to equating three pairs of partial derivatives ([imath]\partial E_{x}/\partial y = \partial E_{y}/\partial x[/imath] and so on), leaving 6 out of 9 partial derivatives "free", it seems that "one third of the amount of information" required to describe [imath]E[/imath] has been taken up, as opposed "two thirds"... I am of course assuming here that the first order partial derivatives determine the behavior of the function (perhaps up to a constant as in the 1D case?), and that these partial derivatives are independent of each other and thus provide equal amounts of information. Is either of these assumptions wrong? Is my whole reasoning off? Any insight into this question and a possible answer would be appreciated. Edit: I am not asking why (or when) a vector field having zero curl is equivalent to it being a gradient of some scalar field. I am asking about the amount of information we get about a vector field when we determine its curl.
|
2919354
|
Why does the curl of a function provide this particular amount of information?
In a classical electrodynamics textbook (Griffiths), it is mentioned that even though the electric field function, [imath]E:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}[/imath], is a (3D) vector valued function, the amount of information needed to fully describe it is equivalent to the amount needed to describe a scalar valued function, and this is because it is the gradient of an electric potential function, [imath]U:\mathbb{R}^{3}\rightarrow \mathbb{R}[/imath] and so is completely determined by it. The author goes on to explains that this is so because [imath]E[/imath] has the property that its curl is zero evevrywhere, and this is what restricts the freedom in determining [imath]E[/imath] (and what enables it to be the gradient of a scalar function in the first place). This is all fine (and fun), but I find myself unable to answer this question: why does the imposition of zero curl provide an amount of information exactly equivalent to two scalar-valued functions (no more, no less)? An example of the sort of reasoning that leads me to other conclusions: since zero curl is equivalent to equating three pairs of partial derivatives ([imath]\partial E_{x}/\partial y = \partial E_{y}/\partial x[/imath] and so on), leaving 6 out of 9 partial derivatives "free", it seems that "one third of the amount of information" required to describe [imath]E[/imath] has been taken up, as opposed "two thirds"... I am of course assuming here that the first order partial derivatives determine the behavior of the function (perhaps up to a constant as in the 1D case?), and that these partial derivatives are independent of each other and thus provide equal amounts of information. Is either of these assumptions wrong? Is my whole reasoning off? Any insight into this question and a possible answer would be appreciated. *The question is, of course, a general one about vector-valued functions, with [imath]E[/imath] just being a particular case. **I have not mentioned all the obvious assumption of smoothness necessary for everything to be defined.. ***I realise that I'm playing fast-and-loose with the word "information", and that my whole question is very un-formal. Any reference to an area of mathematics which may perhaps put such intuitions on rigorous footing are more then welcome. Edit: I am not asking why (or when) a vector field having zero curl is equivalent to it being a gradient of some scalar field. I am asking about the amount of information we get about a vector field when we determine its curl.
|
665866
|
Uniform convergence of [imath]\lim_{n \to \infty} \frac{nx}{1 + n^2x^2} \text{ where } 0 \le x \le 1[/imath]
Prepering to second exam in Calculus 2, and I remember this question from the first exam that I fell on: Let [imath]f_n(x)[/imath] be the function sequence [imath]\{\frac{nx}{1 + n^2x^2}\}_{n=1}^\infty[/imath] Let [imath]f(x)[/imath] be the limit function such that [imath]\lim_{n \to \infty} \frac{nx}{1 + n^2x^2}[/imath] convergences to. Does [imath]f_n(x)[/imath] convergents uniformly to [imath]f(x)[/imath]? As for the answers, the answer is no. Can you please explain me why? Thanks in advance!
|
1875356
|
Showing that [imath]f_n(x) = \frac{nx}{1+n^2x^2}[/imath] does not converge uniformly
I am trying to show that [imath]f_n(x) = \frac{nx}{1+n^2x^2}[/imath] where [imath]x\in \mathbb{R}[/imath] does not converge uniformly. I have made an attempt and would like to make sure that I am going about it in the right way (i.e. if my thought process is correct or I am missing something). First note that [imath]f_n(x) \rightarrow 0[/imath] pointwise and so if it converges uniformly it should converge to [imath]f(x) = 0[/imath] as well. To show this is not the case, note that [imath]f_n(\frac{1}{n}) = 1/2[/imath]. Hence, given [imath]\epsilon = \frac{1}{3}[/imath] for all [imath]N\in \mathbb{N}[/imath] we can find [imath]n\geq N[/imath] and [imath]x\in \mathbb{R}[/imath] where [imath]|f_n(x)-f(x)|\geq \epsilon[/imath], indeed we can just take [imath]n = N[/imath] and [imath]x = \frac{1}{n} \in \mathbb{R}[/imath] and [imath]|f_n(x)-f(x)| = |f_n(\frac{1}{n})| = \frac{1}{2}>\frac{1}{3} = \epsilon[/imath]. This shows that [imath](f_n)[/imath] does not converge uniformly.
|
2923006
|
"Any open subset of the reals is a countable union of disjoint open intervals". Do we need the axiom of choice to prove this?
I was reading Wilcox and Myers' "Introduction to Lebesgue Integration and Fourier Series" (Dover). On p.18, the authors try to prove that every non-empty open subset [imath]G[/imath] of the reals can be expressed uniquely as a countable union of pairwise disjoint open intervals. Their proof is quite similar to the one in this question or the one in Steven Krantz's "A Guide to Topology" (MAA). So, I guess this is a rather usual proof. At some point, the authors have shown that when [imath]G[/imath] is bounded, [imath]G[/imath] can be written as a disjoint union of non-empty open intervals [imath]\cup_{x\in G}I_x[/imath], where either [imath]I_x\cap I_y=\phi[/imath] or [imath]I_x=I_y[/imath] for any [imath]x,y\in G[/imath]. Then they write: That this collection is countable follows from the fact that one can choose a rational number in each [imath]I_x[/imath] (using the axiom of choice), and the disjointness of the intervals guarantees that no duplication will occur in the choice of rationals. Why do we need the axiom of choice? If we denote [imath]I_x=(a_x,b_x)[/imath] and define \begin{align} n_x&=\left\lceil\frac1{b_x-a_x}\right\rceil+1,\\ q_x&=\min\left\{\frac{j}{n_x}\in I_x:\ j\in\mathbb Z\right\}, \end{align} doesn't it suffice to pick [imath]q_x[/imath] as a representative of [imath]I_x[/imath]? What am I missing?
|
1089890
|
Prove that Open Sets in [imath]\mathbb{R}[/imath] are The Disjoint Union of Open Intervals Without the Axioms of Choice
There are several proofs I have seen of this, but they all seem to use choice subtely at some point. Is there any way to prove this without choice, or is it possibly unproveable?
|
2923096
|
[imath]\mathbb{R},\emptyset[/imath], and every interval [imath](-n,n)[/imath] are a topology
I am taking a course in general topology. And I am struggling with the following exercise: Let [imath]\mathbb{R}[/imath] be the set of all real numbers. Prove that each of the following subsets of [imath]\mathbb{R}[/imath] is a topology. i) [imath]\mathscr{T}[/imath] consists of [imath]\mathbb{R},\emptyset[/imath], and every interval [imath](-n,n)[/imath], for [imath]n[/imath] any positive integer. Definition of topology: [imath]X,\emptyset\in\mathscr{T}[/imath] If [imath]\{x_\lambda:\lambda\in\Lambda\}\subseteq\mathscr{T}[/imath], so that [imath]\bigcup_{\lambda\in\Lambda}^{}x_\lambda\in\mathscr{T} [/imath] If [imath]A,A'\in\mathscr{T}[/imath] then [imath]A\cap A\in\mathscr{T}[/imath] 1) The first definition is checked once [imath]\mathbb{R},\emptyset\in\mathscr{T}[/imath] 2) The second definition is the one I do not know how to tackle. It seems straight-forward to mw that the union of any [imath](-n,n)[/imath] belongs in the topology. However I do not know how to write a proof about it. Question: How should I write a proof of the second axiom?
|
1080096
|
Verifying that a certain collection of intervals of [imath]\mathbb R[/imath] forms a topology
I'm doing exercise from "Topology without tears book", page 27, exercises a and b. First one (a) reads as follows. "Let [imath]\mathbb{R}[/imath] be the set of all real numbers. Prove that each of the following collections of subsets of [imath]\mathbb{R}[/imath] is a topology. a. [imath]\tau_1[/imath] consists of [imath]\mathbb{R}[/imath] , [imath]\emptyset[/imath] and every interval [imath](-n;n)[/imath] for [imath]n[/imath] any positive integer b. [imath]\tau_1[/imath] consists of [imath]\mathbb{R}[/imath] , [imath]\emptyset[/imath] and every interval [imath][-n;n][/imath] for [imath]n[/imath] any positive integer" To prove this I noticied that [imath]S_n=(-n;n) \in \tau_1 \; \forall n \in \mathbb{N}[/imath] by definition. [imath]S_{n_1} \subset S_{n_2} \; \forall \; n_1 < n_2, n_1,\;n_2 \in \mathbb{N}[/imath] From 2 it follows that [imath]S_{n_1} \cap S_{n_2} = S_{n_1} \; \forall n_1 < n_2 [/imath] [imath]\bigcup_{i=1}^{i=n} S_i = S_n \in S_{n+1} [/imath] Sience intersection of any two sets is in [imath]\tau_1[/imath] and union of any number of sets is also in [imath]\tau_1[/imath] and noticing that [imath]S_n \subset \mathbb{R} \; \forall n \in \mathbb{N}[/imath] I can state that [imath]\tau_1[/imath] is a topology of [imath]\mathbb{R}[/imath] Is this proven right? I have the following question. Exercise b is very similar to exercise a - the only difference is that now [imath][-n; n][/imath] is a closed interval. I think that I still can use the same prove for this case too, but not sure - does closed interval [imath][-n;n][/imath] changes anything?
|
2922932
|
Why do Hilbert logic systems use the axioms that they do?
For example in a Hilbert system for propositional logic, one sample system uses modus ponens along with three axioms: I. [imath]A \to (B \to A)[/imath] II. [imath](A \to (B \to C)) \to ((A \to B) \to (A \to C))[/imath] III. [imath](\lnot B \to \lnot A) \to (A \to B)[/imath] How did Hilbert come up with these axioms as enough to represent an entire logic system? Is it mostly trial and error to come up with a set of axioms or is there a method to the madness?
|
320437
|
How to demystify the axioms of propositional logic?
How might I go about getting some intuition on the typical axiom schemes given for propositional logic? They seem rather mysterious at first glance. For example, these are taken from: http://en.wikipedia.org/wiki/Hilbert_system#Logical_axioms [imath]\phi \to \phi[/imath] [imath]\phi \to \left( \psi \to \phi \right)[/imath] [imath]\left ( \phi \to ( \psi \rightarrow \xi \right)) \to \left( \left( \phi \to \psi \right) \to \left( \phi \to \xi \right) \right)[/imath] [imath]\left ( \lnot \phi \to \lnot \psi \right) \to \left( \psi \to \phi \right)[/imath]
|
2923076
|
Strictly increasing sequence [imath]\{ a_k \}[/imath] of positive integers such that [imath]\sum 1/{a_k}[/imath] is finite and minimal with respect to this property
Is there a strictly increasing sequence [imath]\{ a_k\}[/imath] of positive integers such that [imath]\sum 1/{a_k}[/imath] is finite and minimal with respect to this property, in a sense that if [imath]\{ b_k \}[/imath] is another sequence of such, then [imath]a_k \leq b_k[/imath] for all but finitely many values of [imath]k[/imath]?? Motivation came from comparing the series [imath]\sum 1/n[/imath] and [imath]\sum 1/{n^2}[/imath], the first of which is divergent but the second of which is not; the sequence for the first is "smaller" than that of the second. If this question need be motified to make my question clearer and/or make it mathematically answerable, please comment/suggest adit as well.
|
2923073
|
Strictly increasing sequence [imath]\{ a_k \}[/imath] of positive integers such that [imath]\sum 1/{a_k}[/imath] is finite
Is there a strictky increasing sequence [imath]\{ a_k\}[/imath] of positive integers such that [imath]\sum 1/{a_k}[/imath] is finite and minimal with respect to this property, in a sense that if [imath]\{ b_k \}[/imath] is another sequence of such, then [imath]a_k \leq b_k[/imath] for all but finitely many values of [imath]k[/imath]? Motivation came from comparing the series [imath]\sum 1/n[/imath] and [imath]\sum 1/{n^2}[/imath], the first of which is divergent but the second of which is not; the sequence for the first is "smaller" than that of the second. If this question need be motified to make my question clearer, please comment/suggest adit as well.
|
2922820
|
Find the coefficient of [imath]x^9[/imath] in the expansion of [imath](1+x)(1+x^2)(1+x^3)...(1+x^{100})[/imath]
The coefficient of [imath]x^9[/imath] in the expansion of [imath](1+x)(1+x^2)(1+x^3)...(1+x^{100})[/imath]. I tried the following concept, how to sum 9 using 1-9 only without repetition. 1:9, 2:1+8,3:7+2, 4: 6+3, 5: 5+4, 6:1+2+6 ,7:1+3+5, 8:2+3+4 The answer is 8. How will it be solved using Binomial theorem.
|
1348452
|
Find the coefficient of [imath]x^9[/imath] in [imath](1+x)(1+x^2)(1+x^3)\cdots(1+x^{100})[/imath]
This question had come in jee advanced 2015. Give a hint to solve it.
|
2923716
|
Characterization of Polynomial In terms of Derivative
In many problem or theorem in which I have to show that given function is polynomial, there is the argument that for some k show that [imath]f^{(n)}(z)=0, \forall n>k [/imath] . I know that by definition of polynomial above characteristics is ok. But I am not convinced with fact that converse part. that there is no function other than polynomial which satisfies this property. Where am I missing? Any help will be appreciated.
|
762153
|
Does [imath]f^{(n)} = 0[/imath] imply that complex [imath]f[/imath] is a polynomial?
Let [imath]f[/imath] be a complex function with the property that [imath]f^{(n)} = 0[/imath]. Does this imply that [imath]f[/imath] is a polynomial? If so, why? Upon thinking about this problem myself, I can easily observe that every [imath]n[/imath] degree polynomial satisfies [imath]f^{(n+1)} = 0[/imath]. Still, this doesn't show that some non-polynomial function couldn't also have this property.
|
2923874
|
Are absolutely convergent sums exactly those which can be reordered?
If an infinite sum is absolutely convergent, its limit remains the same however the terms are permuted. Does a sequence of real numbers [imath](a_n)_{n\in\mathbb{N}}[/imath] exist such that [imath]\sum_n a_{\small P(n)}[/imath] is the same for all permutations [imath]P:\mathbb{N}\overset{\text{bijectively}}{\to}\mathbb{N}[/imath], and such that [imath]\sum_n |a_n|[/imath] diverge?
|
143160
|
Proving that unconditional convergence is equivalent to absolute convergence
Regarding this discussion here: Absolute convergence a criterion for unconditional convergence. (thank you for the great answers, by the way) I'm still trying to do Exercise 3.2.2 (b) from these notes by Dr. Pete Clark. The question states that for the set [imath]S = \mathbb{Z}^{+}[/imath], the series [imath]\sum_{i\in S}x_{i}[/imath] converges unconditionally if and only if it converges absolutely, i.e, [imath]\sum_{i\in S}|x_{i}| < \infty[/imath]. Thanks to the prior discussion, I get why unconditional convergence isn't simply the same thing as regular convergence in the case of [imath]S = \mathbb{Z}^{+}[/imath]. Unconditional convergence (to [imath]x[/imath]) even allows for "cherry picking" of terms in the tail, rather than only being allowed to take them in order, and still reguires the resulting sums to stay within the [imath]\epsilon[/imath]-band (of [imath]x[/imath]). (Again, thanks for the detailed answers to my previous question on this.) My "best" attempt hits a brick wall: Let [imath]S = \mathbb{Z}^{+}[/imath]. Write [imath]S = S^{+}\cup S^{-}[/imath], where [imath]S^{+} := \{n\in S : x_{n} \geq 0\}[/imath] and [imath]S^{-} := \{n\in S : x_{n} < 0\}[/imath] [imath](\Rightarrow)[/imath] Now if I assume that [imath]\sum_{n=1}^{\infty}|x_{n}|[/imath] is not finite ( that is, the series is not absolutely convergent ), I want to show that the unconditional convergence definition fails. I want to try to construct arbitrarily large finite subsets [imath]K[/imath] of [imath]S^{+}[/imath] or [imath]S^{-}[/imath] such that [imath]\sum_{i\in K}x_{i}[/imath] is arbitrarily large, which would prove the contrapositive. I think I can do it if one of [imath]S^{+}[/imath] or [imath]S^{-}[/imath] is finite, but this need not be the case. Any advice or suggestions? I feel like I hit a brick wall. Edit: I added the solution as an answer below for anyone who might have this same question. Thanks all for the help.
|
2923693
|
If [imath]a_n[/imath] and [imath]b_n[/imath] are converging sequences, and for all [imath]n\in\mathbb Z+[/imath], [imath]a_n \leq b_n[/imath], then [imath]\lim_{n\to\infty} a_n \leq \lim_{n\to\infty} b_n[/imath].
I am having a bit of trouble answering this question. If [imath]a_n[/imath] and [imath]b_n[/imath] are converging sequences, and for all [imath]n \in\mathbb Z+[/imath], [imath]a_n[/imath] ≤ [imath]b_n[/imath], then [imath]\lim_{n\to\infty} a_n \leq \lim_{n\to\infty} b_n[/imath]. Letting [imath]\epsilon>0[/imath], I began with showing that there exists [imath]N_1\in \mathbb N[/imath] such that for all [imath]n \in \mathbb N[/imath], if [imath]n>N_1[/imath], then [imath]|a_n - L|< \epsilon[/imath]. Similarly exists [imath]N_2\in \mathbb N[/imath] such that for all [imath]n \in \mathbb N[/imath], if [imath]n>N_2[/imath], then [imath]|b_n - M|< \epsilon[/imath]. However from here I do not know how to proceed. Do I use the triangle inequality somehow?
|
2659256
|
Given [imath]a_n, b_n[/imath] are both convergent sequences such that [imath]a_n \leq b_n[/imath] for all [imath]n[/imath], show [imath]\lim_{n\to\infty} a_n \leq \lim_{n\to\infty} b_n[/imath].
Not sure if my proof as is is correct: Since [imath]a_n \leq b_n[/imath] for all [imath]n[/imath], we have [imath]a_n - b_n \leq 0[/imath]. As such, by the limit location theorem it holds that [imath]\lim_{n\to\infty} (a_n - b_n) \leq 0[/imath]. By linearity, we have [imath]\lim_{n\to\infty} a_n - \lim_{n\to\infty} b_n \leq 0[/imath], which implies [imath]lim_{n\to\infty} a_n \leq \lim_{n\to\infty} b_n[/imath]. My one concern is using the LL Theorem on the sequence [imath]a_n - b_n[/imath]: do I need to prove this sequence is convergent first?
|
1488049
|
How to show that a multivariable function is not differentiable?
How would I show that [imath]f(x,y) = \frac{2xy}{x^2 + y^2}[/imath] is not differentiable at the origin? Is it enough to show that as the function tends to the origin along the paths [imath]y = x[/imath] and [imath]y=2x[/imath] that we get different limits and hence the function is not continuous?
|
1398279
|
Differentiability of this picewise function
[imath]f(x,y) = \left\{\begin{array}{cc} \frac{xy}{x^2+y^2} & (x,y)\neq(0,0) \\ f(x,y) = 0 & (x,y)=(0,0) \end{array}\right.[/imath] In order to verify if this function is differentiable, I tried to prove it by the theorem that says that if [imath]\frac{∂f}{∂x}[/imath] and [imath]\frac{∂f}{∂y}[/imath] exist and are continuous at the point [imath](x_0,y_0)[/imath], then the function is differentiable at this point. So I did: [imath]\frac{\partial f}{\partial x}(0,0) = \lim_{h\to 0}\frac{f(0+h,0)-f(0,0)}{h} = 0[/imath] [imath]\frac{\partial f}{\partial y}(0,0) = \lim_{h\to 0}\frac{f(0,0+h)-f(0,0)}{h} = 0[/imath] so we have that the partial derivatives at point [imath](0,0)[/imath] is [imath]0[/imath]. Now, if we take the derivative at [imath](x,y)\neq (0,0)[/imath] and then take the limit of it as [imath](x,y)\to(0,0)[/imath], we can see if the derivatives are continuous or not. So here it is: [imath]\frac{\partial f}{\partial x}(x,y) = \frac{y(y^2-x^2)}{(x^2+y^2)}[/imath] but [imath]\lim_{(x,y)\to(0,0)} \frac{y(y^2-x^2)}{(x^2+y^2)} [/imath] does not exist (by wolfram alpha... but can anybody tell me an easy way to prove this limit does not exist? easier than taking the limit in different directions?), therefore the derivative is not continuous at [imath](0,0)[/imath], so we can't say [imath]f[/imath] is differentiable at [imath](0,0)[/imath], but for [imath](x,y)\neq (0,0)[/imath] the function is continuous, as it is a quotient of continuous functions. So [imath]f[/imath] is at least differentiable at [imath](x,y)\neq (0,0)[/imath]. Now, to verify differentiability at [imath](0,0)[/imath] I think we must use the limit definition of differentiablity: A function is differentiable at [imath](0,0)[/imath] iff: [imath]\lim_{(h,k)\to (0,0)} \frac{f(0+h,0+k)-f(0,0)-\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)}{\|(h,k)\|} = 0[/imath] Let's calculate this limit: [imath]\lim_{(h,k)\to (0,0)} \frac{f(0+h,0+k)-f(0,0)-\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)}{\|(h,k)\|} = \\ \lim_{(h,k)\to (0,0)} \frac{\frac{hk}{h^2+k^2}}{\sqrt{h^2+k^2}} = \\ \lim_{(h,k)\to (0,0)} \frac{hk}{(h^2+k^2)\sqrt{h^2+k^2}}[/imath] which I think, it's a limit that does not exist, therefore the function isn't differentiable at [imath](0,0)[/imath]
|
670966
|
How can I prove continuity of this function: [imath]f(x, y)=\frac{xy}{x^2+y^2}[/imath] for [imath](x, y)\neq (0,0)[/imath]?
I have to prove that the function [imath]f:\Bbb{R}\times\Bbb{R}\to\Bbb{R}[/imath] defined as follows: [imath]f(x, y)=\frac{xy}{x^2+y^2}\text{for } (x, y)\neq (0,0)[/imath] [imath]f(x, y)=0\text{ for }(x, y)=(0,0)[/imath]when taken as a function of [imath]x[/imath], is continuous. My method: Take any [imath]\epsilon>0[/imath]. I have to prove that there exists a [imath]\delta[/imath] such that [imath]\left|\frac{(x+\delta)y}{(x+\delta)^2+y^2}-\frac{xy}{x^2+y^2}\right|<\epsilon[/imath]. I'm getting a quadratic inequality in [imath]\delta[/imath]. My question is, what if the situation was more complicated? I got easily get a very messy expression for [imath]\delta[/imath]. What should we do then? How should we prove continuity? Thanks in advance!
|
581326
|
Show that the function [imath]f = \frac{xy}{x^2 + y^2}[/imath] is continuous along every horizontal and every vertical line
Consider the function [imath] f:\mathbb{R^2} \rightarrow \mathbb{R}[/imath] given by [imath] f(x,y) = \left\{ \begin{array}{l l} \frac{xy}{x^2 + y^2} & \quad \text{if (x,y) [/imath]\neq[imath] (0,0)}\\ 0 & \quad \text{if (x,y) = (0,0)} \end{array} \right.[/imath] Show that f is continuous along every horizontal and every vertical line (i.e. for every [imath]x_0, y_0 \epsilon \mathbb{R}[/imath], the functions [imath]g,h:\mathbb{R}\rightarrow \mathbb{R}[/imath] given by [imath]g(t) = f(x_0,t)[/imath] and [imath]h(t)=f(t,y_0)[/imath] are continuous). I know that this function is not continuous at [imath](0,0)[/imath], but the horizontal/vertical line issue is proving difficult to work with. Any help/hints appreciated!
|
2924380
|
Proof [imath]\sum_{k=0}^{n}{k\binom{n}{k}}=n2^{n-1}[/imath]
[imath]\sum_{k=0}^{n}{k\binom{n}{k}}=n2^{n-1}[/imath] [imath]n2^{n-1} = \frac{n}{2}2^{n} = \frac{n}{2}(1+1)^n = \frac{n}{2}\sum_{k=0}^{n}{\binom{n}{k}}[/imath] That's all I got so far, I don't know how to proceed
|
1847183
|
Prove [imath]\sum_{k= 0}^{n} k \binom{n}{k} = n \cdot 2^{n - 1}[/imath] using the binomial theorem
I'm trying to prove that \begin{equation} \sum_{k= 0}^{n} k \binom{n}{k} = n \cdot 2^{n - 1} \end{equation} with the Binomial Theorem. I know that the B.T. states that \begin{equation} (x + y)^n = \sum_{k= 0}^{n} \binom{n}{k} x^{n-k}y^{k} \end{equation} The proof that I have tried starts with: \begin{equation} n \cdot 2^{n-1} = \end{equation} Actually I am stucked with this step.
|
2784518
|
Show using Cauchy-Schwarz inequality
Show for any real numbers [imath]a_1,a_2,...,a_n[/imath] [imath](a_1+a_2+···+a_n)^2 \leq n(a_1^2+a_2^2+...+a_n^2)[/imath] I know the definition of Cauchy-Schwarz is [imath](\sum_{i=1}^n a_ib_i)^2 \leq \sum_{i=1}^n a_i^2 \sum_{i=1}^n b_i^2[/imath] I can write the same problem with following form, but I would not know how to continue to prove it [imath](\sum_{i=1}^n a_i)^2 \leq n(\sum_{i=1}^n a_i^2)[/imath] Note: an other definition of Cauchy Schwarz [imath] \langle\ v,u\rangle^2 \leq \langle\ u,u\rangle \langle\ v,v\rangle[/imath] Could you give me a steps what to use?
|
2623423
|
Employing Cauchy-Schwartz inequality question
Let [imath]d_i\ge 0[/imath] and [imath]\sum_{i=1}^n d_i=c[/imath] and [imath]\sum_{i=1}^nd_i^2\le C[/imath] Show that [imath]\frac{c^2}{n}\le C[/imath] The text says Using Cauchy-Schwartz inequality and the first equation, show the last equaliton. I do not know how to apply C-S to this. The inequality roughly states [imath](u,v)^2\le(u,u)(v,v)[/imath] which is a trivial equality when [imath]u=v[/imath].... We have [imath](\sum_{i=1}^n d_i)^2\ge\sum_{i=1}^n d_i^2[/imath] but I dont know what to do with it...
|
2924111
|
Topology of [imath]\mathbb{R}^{\infty}[/imath]
I want to know which is the topology of [imath]\mathbb{R}^{\infty}[/imath], but I don't know even how to start to give it one. How can I give to [imath]\mathbb{R}^{\infty}[/imath] a topology? Is there a book that explains this?
|
315909
|
Definition of [imath]\mathbb{R}^\infty[/imath]
Question: Why is the topological space [imath]\mathbb{R}^\infty[/imath] defined to be the subset of [imath]\prod_{i=1}^\infty \mathbb{R}_i[/imath] consisting of sequences [imath](a_i)_{i=1} ^{\infty}[/imath] such at most finitely many [imath]a_i\neq 0[/imath]? Why does one insist on the condition that [imath]a_i\neq0[/imath] for at most finitely many [imath]i[/imath]?
|
2924125
|
if [imath]a[/imath] is algebraic, then [imath]k[a] = k(a)[/imath]
let [imath]K/k[/imath] be a field extension and [imath]a\in K[/imath] algebraic over k. Prove that the subalgebra [imath]k[a][/imath] is a field and that [imath]k[a] = k(a)[/imath] where [imath]k(a)[/imath] is the monogenic extension, i.e. the smallest subfield of [imath]K[/imath] containing [imath]k[/imath] and [imath]a[/imath]. I know that [imath]a[/imath] is algebraic on [imath]k[a][/imath] since [imath]k\subset k[a][/imath]. I don't see how to go from there. Thank you.
|
1631401
|
If [imath]\alpha[/imath] is an algebraic element and [imath]L[/imath] a field, does the polynomial ring [imath]L[\alpha][/imath] is also a field?
If [imath]\alpha[/imath] is an algebraic element and [imath]L \subset K[/imath] are both field, does the polynomial ring [imath]L[\alpha][/imath] is also a field? I am trying to prove that the ring of fraction [imath]L(\alpha)[/imath] is equal to [imath]L[\alpha][/imath]. To do this, I use an exercise done in class : Let [imath]K[/imath] a field, [imath]\alpha \in K[/imath] and [imath]L[/imath] a subfield of [imath]K[/imath]. Then [imath]L(\alpha)[/imath] is the smallest subfield of [imath]K[/imath] containing [imath]L[/imath] and [imath]\alpha[/imath]. Is anyone could help me?
|
2924421
|
Does there exist a Lebesgue measurable subset [imath]A[/imath] of [imath]R[/imath] such that for every [imath]a we have m(A\cap(a,b))=(b-a)/2?[/imath]
Does there exist a Lebesgue measurable subset [imath]A[/imath] of [imath]R[/imath] such that for every [imath]a<b[/imath] we have [imath]m(A\cap(a,b))=(b-a)/2[/imath]? I searched before posting and found a similar question here, but it isn't exactly the same as my question. I looked over the given answer there but I couldn't figure out if that answer was relevant to my question and how to adapt the answer to my problem if it is. I'm unsure at this point how to actually go about showing if such an [imath]A[/imath] exists or not.
|
933233
|
Prove Borel Measurable Set A with the following property has measure 0.
This question is exercise 4.10 of Richard F. Bass's Real Analysis for Graduate Students, 2nd edition. Let [imath]\epsilon \in (0,1)[/imath], let [imath]m[/imath] be Lebesgue measure, and suppose [imath]A[/imath] is Borel Measurable subset of [imath]\mathbb R[/imath]. Prove that if [imath] m(A\cap I)\leq (1-\epsilon)m(I)[/imath] for every interval [imath]I[/imath], then [imath]m(A)=0[/imath]. Collection of my thoughts 1. Try to prove by contradiction, suppose [imath]m(A)=a>0[/imath] Borel Measurable Set satisfies the condition given. A property of any Borel Measurable Set on [imath]\mathbb R[/imath] is [imath]A[/imath] contains a closed set [imath]F[/imath] and is contained in open set [imath]O[/imath], and the measure of [imath]F[/imath] and [imath]O[/imath] can be arbitrarily close to [imath]a[/imath]. 2. Obviously [imath]A[/imath] contains no interval, but it does not lead to contradiction, because we can find an example of [imath]m(A)>0[/imath] that contains no interval(fat Cantor Set(remove middle 1/4 each time) has measure [imath]1/2[/imath] and contains no interval). 2'. For fat Cantor Set [imath]C[/imath] and given [imath]\epsilon>0[/imath] how could we construct an interval I satisfies [imath]m(C \cap I)>(1-\epsilon)m(I) ?[/imath] Maybe try to write Borel measurable set as countable union/intersection of intervals. I'm stucked here and will appreciate any help, some examples/special case will be great too.
|
99884
|
Is there a formula for [imath](1+i)^n+(1-i)^n[/imath]?
I'm wondering if there is a formula for the value of [imath](1+i)^n+(1-i)^n[/imath]? I calculated the first terms starting with [imath]n=1[/imath] to be, in order, [imath]2[/imath], [imath]0[/imath], [imath]-4[/imath], [imath]-8[/imath], [imath]-8[/imath], [imath]0[/imath], [imath]16[/imath], [imath]\dots[/imath] So it seems to be some sequence of positive and negative powers of [imath]2[/imath] with [imath]0[/imath]s thrown in. Is there a more explicit formulation of what [imath](1+i)^n+(1-i)^n[/imath] is, based on [imath]n[/imath]? With the binomial theorem, I get it equal to [imath] \sum_{k=0}^n\binom{n}{k}i^{n-k}(1+(-1)^{n-k}). [/imath] Can this be made nicer? Thanks.
|
1108585
|
Considering [imath] (1+i)^n - (1 - i)^n [/imath], Complex Analysis
I have been working on problems from Complex Analysis by Ahlfors, and I got stuck in the following problem: Evaluate: [imath] (1 + i)^n - (1-i)^n [/imath] I have just "reduced" to: [imath] (1 + i)^n - (1-i)^n = \sum_{k=0} ^n i^k(1 - (-1)^k) [/imath] by using expansion of each term. Thanks.
|
2924142
|
Difference Between Double Implications and Equal Signs
I always thought equal signs (=), indicated double implications, however after thinking about this example, I am very confused. For example, the Theorem (in Linear Algebra): L([imath]\vec{0}[/imath]) = [imath]\vec{0}[/imath], where L is a linear mapping can be described as: In the equation L([imath]\vec{x}[/imath]) = [imath]\vec{y}[/imath], if [imath]\vec{x}[/imath] = [imath]\vec{0}[/imath], then [imath]\vec{y}[/imath] = [imath]\vec{0}[/imath]. However, the other implication is not true, which is if [imath]\vec{y}[/imath] = [imath]\vec{0}[/imath], then [imath]\vec{x}[/imath] = [imath]\vec{0}[/imath]. Do equals not always indicate double implications? Thanks
|
2924058
|
Question About equal signs in Theorems
Lets say for example, [imath]L[/imath] is a linear mapping with [imath]V[/imath] as its domain and [imath]W[/imath] as its co-domain. Theorem: [imath]L(0) = 0[/imath]. (Zero vectors) I know its wrong to assume that every zero vector in [imath]W[/imath] is mapped from a zero vector in [imath]V[/imath], but I can't logically explain why! Since [imath]L(0) = 0[/imath], and equal signs go both ways (double implication), isn't the statement [imath]L(0) = 0[/imath] implying the [imath]0[/imath] vector in [imath]V[/imath] only maps to the zero vector in [imath]W[/imath], and the zero vector in [imath]W[/imath] can be only mapped from the zero vector in [imath]V[/imath]? I know this is wrong. Can someone explain why? Thanks
|
2925507
|
Understanding a part of the proof of Cauchy's theorem in the abelian case
(Cauchy's theorem in the abelian case): If [imath]G[/imath] is a finite abelian group such that [imath]p \ | \ |G|[/imath], where [imath]p[/imath] is a prime, then [imath]\exists g \in G[/imath] such that [imath]o(g) = p[/imath]. Basically the part of the proof that I'm stuck on is the following. After choosing an [imath]h \in G[/imath] for which [imath]h \neq 1_G[/imath] and assuming [imath]o(h) = l[/imath], we form the subgroup [imath] H = \langle h \rangle[/imath]. We can then form the quotient group [imath]G / H[/imath] which has order [imath]|G/H| = \frac{|G|}{|H|} = \frac{|G|}{l}[/imath] by Lagrange's theorem. The next part of the proof says that [imath]p[/imath] divides [imath]\frac{|G|}{l}[/imath] because [imath]p[/imath] divided [imath]|G|[/imath], and I can't immediately see why this is the case. What I think is that this part of the proof must be using some argument from elementary number theory, and I think that the theorem that's being referenced must be (similar to) the following conjecture of mine: Conjecure: If [imath]p[/imath] is prime and [imath]p \ | \ x[/imath] and [imath]\frac{x}{l} = m[/imath] where [imath]m, l \in \mathbb{Z}[/imath] then [imath]p \ | \ m[/imath]. In trying to prove the above this is as far as I got. Attempted Proof: Since [imath]p | x[/imath] there exists an [imath]n \in \mathbb{Z}[/imath] such that [imath]pn =x[/imath], Then [imath]\frac{x}{l } = m \implies \frac{pn}{l} = m \implies pn = ml[/imath] and since [imath]p[/imath] divides the left hand side of the above equation it follows that the right hand side of the equation must also be divisible by [imath]p[/imath]. So [imath]p|ml[/imath] and since [imath]p[/imath] is prime [imath]p|l[/imath] or [imath]p|m[/imath]. If [imath]p|m[/imath] we are done, if [imath]p|l[/imath] then... And that's as far as I got. Is the above conjecture actually a theorem? If so is it the one being used in Cauchy's theorem and if so how could I finish proving the above conjecture?
|
2887080
|
In this case, why must [imath]p[/imath] divide [imath]|G/N|[/imath]?
I am reading the proposition on page 102 of Dummit and Foote's Algebra, 3rd edition. Could someone explain why [imath]p[/imath] divides [imath]|G/N|[/imath] on the highlighted line? Proposition 21. If [imath]G[/imath] is a finite abelian group and [imath]p[/imath] is a prime dividing [imath]|G|[/imath], then [imath]G[/imath] contains an element of order [imath]p[/imath]. Proof: The proof proceeds by induction on [imath]|G|[/imath], namely, we assume the result is valid for every group whose order is strictly smaller than the order of [imath]G[/imath] and then prove the result valid for [imath]G[/imath] (this is sometimes referred to as complete induction). Since [imath]|G| > 1[/imath], there is an element [imath]x \in G[/imath] with [imath]x \neq 1[/imath]. If [imath]|G| = p[/imath] then [imath]x[/imath] has order [imath]p[/imath] by Lagrange’s Theorem and we are done. We may therefore assume [imath]|G| > p[/imath]. Suppose [imath]p[/imath] divides [imath]|x|[/imath] and write [imath]|x| = pn[/imath]. By Proposition 2.5(3), [imath]|x^n| = p[/imath], and again we have an element of order [imath]p[/imath]. We may therefore assume [imath]p[/imath] does not divides [imath]|x|[/imath]. Let [imath]N = \langle x \rangle[/imath]. Since [imath]G[/imath] is abelian, [imath]N \trianglelefteq G[/imath]. By Lagrange’s Theorem, [imath]|G/N| = |G|/|N|[/imath] and since [imath]N \neq 1[/imath], [imath]|G/N| < |G|[/imath]. Since [imath]p[/imath] does not divide [imath]|N|[/imath], we must have [imath]p \mid |G/N|[/imath]. We can now apply the induction assumption to the smaller group [imath]G/N[/imath] to conclude it contains an element [imath]\bar{y} = yN[/imath], of order [imath]p[/imath]. Since [imath]y \notin N[/imath] ([imath]\bar{y} \neq \bar{1}[/imath]) but [imath]y^p \in N[/imath] ([imath]\bar{y}^p = \bar{1}[/imath]) we must have [imath]\langle y^p \rangle \neq \langle y \rangle[/imath], that is, [imath]|y^p| < |y|[/imath]. Proposition 2.5(2) implies [imath]p \mid |y|[/imath]. We are now in the situation described in the preceding paragraph, so that argument again produces an element of order [imath]p[/imath]. The induction is complete. (Original image here.)
|
2925754
|
prove that the Dimension for set of all linear mappings [imath]L:V \to W[/imath] is equal to [imath]\dim V \times \dim W[/imath]
I wish to prove that the dimension for the set of all linear mappings [imath]L:V \to W[/imath] is equal to [imath]\dim V \times \dim W[/imath]. I know that any general linear mapping can be represented as a matrix, so intuitevely it makes sense that the dimension should be [imath]\dim V \times \dim W[/imath] but I cannot find a way to prove this mathematically
|
764427
|
Why [imath]\dim(\hom(V,W))=\dim(V) * \dim(W)[/imath]?
I have found that the question I want to ask someone had asked, here is the website: [imath]\hom(V,W)[/imath] is canonic isomorph to [imath]\hom(W^*, V^*)[/imath] Here is my question: Why [imath]\dim(\hom(V,W))=\dim V * \dim W [/imath]? Thanks for explanation~!
|
2924878
|
Prove a mapping [imath]C(X,Y)\to \operatorname{Hom}(C(Y), C(X))[/imath] is surjective
Let [imath]X[/imath] and [imath]Y[/imath] be compact Hausdorff spaces, and let [imath]F[/imath] be a continuous function from [imath]X[/imath] to [imath]Y[/imath]. Define a function [imath]\Phi_F[/imath] from [imath]C(Y)[/imath] to [imath]C(X)[/imath] by [imath]\Phi_F(f)=f\circ F.[/imath] I have shown [imath]\Phi_F[/imath] is a unital [imath]*[/imath]-homomorphism from [imath]C(Y)[/imath] to [imath]C(X)[/imath], where [imath]*[/imath] is the usual complex conjugation of complex-valued functions and it is naturally an involution. (i.e. I proved that [imath]\Phi_F[/imath] is an algebra homomorphism that carries identity element to identity element, and respects complex conjugation) Now the problem is to prove that the mapping [imath]F\mapsto \Phi_F[/imath] is a bijection between the set of continuous functions from [imath]X[/imath] to [imath]Y[/imath] and the set of unital [imath]*[/imath]-homomorphism from [imath]C(Y)[/imath] to [imath]C(X)[/imath]. The fact that this mapping is injective is trivial. But I can not figure out the surjective part. Can someone help me? Any help will be appreciated. Thanks!
|
2917237
|
Homomorphism between compact Hausdorff spaces.
Suppose that X and Y are compact Hausdorff spaces and [imath]p:C(X)\to C(Y)[/imath] is a unital * homomorphism. Prove that there exists a continuous function [imath]h: Y \to X [/imath] such that [imath]p(f)=f\circ h [/imath] for all f in [imath]C(X) [/imath] I have managed to prove the other 3 parts ( show the statement false if [imath]p[/imath] is not unital as well as finding conditions on [imath]h[/imath] for [imath]p[/imath] to injective and surjective. I have found a function that does what I want but I can't prove it's continuous. Any help much appreciated.
|
580778
|
Proof with the definition of limit and derivative
I want to prove with the definition of limits and derivative that if [imath]\lim_{x \to \infty} f'(x) = 0[/imath] then [imath]\lim_{x \to \infty} f(x+1)-f(x) = 0[/imath] I can solve it with the Mean Value Theorem at [imath][x, x+1][/imath] but i don't understand how i must use the definitions. It's necessary to use only definitions.
|
83782
|
If [imath]f:\mathbb{R}\rightarrow \mathbb{R}[/imath] is differentiable and [imath]\lim_{x\to \infty } f^\prime(x)=0[/imath] show that [imath]\lim _{x\to \infty } (f(x+1)-f(x))=0[/imath].
Having some trouble proving this: If [imath]f:\mathbb{R}\rightarrow \mathbb{R}[/imath] is differentiable and [imath]\lim_{x\to \infty } f^\prime(x)=0[/imath] show: [imath]\lim _{x\to \infty } (f(x+1)-f(x))=0[/imath] Attempt: from [imath]\int \lim _{x\rightarrow \infty } f^\prime(x) \mathrm{d}x=0[/imath] we can say that [imath]f^\prime(x)=const[/imath] would we then have to show that [imath]\lim _{x\to \infty } (f(x+1))[/imath] is also a constant? From there if we show that both are equal then the difference is zero. I am on the right track? Thank you for the help.
|
2927863
|
Is k=[imath]\sqrt{6+\sqrt{6+...}}[/imath] monotonically increasing?
Is k=[imath]\sqrt{6+\sqrt{6+...}}[/imath] monotonically increasing or is not ? i know it converges to [imath]3[/imath].
|
458103
|
Nested roots sequence, how to prove it's monotone and bounded?
Let [imath]a\ge1[/imath] and define the sequence [imath](x_n)[/imath] recursively by: [imath]x_1 = \sqrt{a}[/imath] [imath]x_{n+1}= \sqrt{a+x_n}[/imath] Here's what I did: Plugging in some values makes it seem as if the sequence is increasing. I first solve the equation [imath]t=\sqrt{a+t}[/imath], equivalently [imath]t^2 -t -a =0[/imath], to get an idea what's going on. The roots are [imath]t_{1,2}= \frac{1 \pm \sqrt{1+4a} }{2}.[/imath] Therefore, between the roots, that is for [imath]t\in \left( \frac{1 - \sqrt{1+4a} }{2}, \frac{1 + \sqrt{1+4a} }{2}\right)[/imath] we have [imath]t^2-t-a \le 0 \iff t \le \sqrt{a+t}[/imath]. Here's my attempt at showing inductively that [imath]\forall n\in \mathbb{N}, \ x_n \le \frac{1 + \sqrt{1+4a} }{2}[/imath] For [imath]n=1[/imath], the statement is true (simple arithmetic). Suppose it's true for [imath]n[/imath]. Consider [imath]x_{n+1}= \sqrt{a+x_n}\le \sqrt{a+\frac{1 + \sqrt{1+4a} }{2}}=\frac{1 + \sqrt{1+4a} }{2},[/imath] because [imath]\frac{1 + \sqrt{1+4a} }{2}[/imath] satisfies [imath]x=\sqrt{x+a}[/imath]. So this establishes the sequence [imath](x_n)[/imath] is bounded. Now, by induction again, I try to show it's monotonically increasing. The base case [imath]x_1\le x_2[/imath] is obvious. Assume (strong induction) [imath]x_{n-1}\le x_n[/imath] for all [imath]n>1[/imath]. But since we have [imath]x_1\le x_n \le \frac{1 + \sqrt{1+4a} }{2}, [/imath] then [imath]x_n[/imath] lies in the appropriate interval, so [imath]x_n \le \sqrt{a+x_n}=x_{n+1}.[/imath] Is this proof correct? Can the condition on [imath]a[/imath] be relaxed to [imath]a>0[/imath]? Maybe even small negative [imath]a[/imath]? And could someone provide me with another proof of convergence of [imath]x_n[/imath]? I absolutely hate this one!
|
2928532
|
Find the limit of the sequence [imath]\sqrt{2} ,\sqrt{2\sqrt{2}} , \sqrt{2\sqrt{2\sqrt{2}}} ,.....[/imath] = ?
find the lim of the sequance when n [imath]\to \infty [/imath] [imath]\sqrt{2} ,\sqrt{2\sqrt{2}} , \sqrt{2\sqrt{2\sqrt{2}}} ,.....[/imath]
|
815418
|
Why does [imath]\sqrt{n\sqrt{n\sqrt{n \ldots}}} = n[/imath]?
Ok, so I've been playing around with radical graphs and such lately, and I discovered that if the nth x = √(1st x √ 2nd x ... √nth x); Then [imath]\text{the "infinith" } x = x[/imath] Example: [imath]\sqrt{4\sqrt{4\sqrt{4\sqrt{4\ldots}}}}=4[/imath] Try it yourself, type calc in Google search, hit then a number, such as [imath]4[/imath], and repeat, ending with [imath]4[/imath], (or press the buttons instead). I'm a math-head, not big enough though, I think this sequence is divergent or convergent or whatever, too lazy to search up the difference. However, can this be explained to me? Like how the Pythagorean Theorem can be explained visually.
|
409551
|
Limit of a Sequence involving cubic root
I succeed in finding the following limit applying binomials and squeeze theorem: [imath]\lim(\sqrt{n+1} - \sqrt{n}) = \lim\frac{1}{\sqrt{n+1} + \sqrt{n}} = 0[/imath] because [imath]0 \leq \frac{1}{\sqrt{n+1} + \sqrt{n}} \leq \frac{1}{\sqrt{n}}[/imath] But I need help because I'm not finding any way to simplify and solve the following limit: [imath]\lim(\sqrt[3]{1-n^3} + n)[/imath]
|
252673
|
Find $\lim_{n\to\infty}{\sqrt{n+1}-\sqrt{n}}$ with basic methods
I need to find simple limit: [imath]\lim_{n\to\infty}{\sqrt{n+1}-\sqrt{n}}[/imath] I can't use L'Hôpital's rule, which I think is natural for such problems. How to calculate such limit using only basic methods (definition and simple transformations etc.)? It's one problem (from 50 problems) I don't know how to solve from my homework. Thanks for help.
|
2928386
|
How can we prove soundness property if it's possible for our assumption set to contain false assumptions?
Soundness property, to my knowledge is the property that: [imath]\Gamma \vdash \varphi \implies \Gamma \vDash \varphi[/imath] If [imath]\varphi[/imath] is provable (a syntactic consequence) from [imath]\Gamma[/imath] then [imath]\varphi[/imath] is also a semantic consequence of [imath]\varphi[/imath], which I believe is saying [imath]\varphi[/imath] is "true". But what if, for example, [imath]\varphi \in \Gamma[/imath] and [imath]\varphi = \bot[/imath]? It appears conceivable that some of the assumptions in [imath]\Gamma[/imath] are false, and then we might be able to prove things from it, but semantically they would be false. It's possible I just have the definition of soundness wrong but how is this accounted for? We would normally say that the Hilbert system is both complete and sound but is this still the case even if we begin with a [imath]\Gamma[/imath] that contains some false premises? Or is it "sound only in certain cases"? How does this work?
|
2910909
|
Soundness vs completeness, am I understanding? And proving soundness?
Soundness: [imath]a \vdash b \implies a \vDash b[/imath], i.e. if we can prove something, it will also be true. We don't want a system where we start out with something true and dedice something false. However it is conceivable that even if our system is sound, maybe it's quite incomplete/limited regarding what we can express, which is why we also would like... Completeness: [imath]a \vDash b \implies a \vdash b[/imath], i.e. if we can show something is true, it's also provable. We want to be able to prove all true statements. However it is conceivable that even though we can prove all true statements, maybe it also proves false ones as well, which is why we'd also like the soundness property from before. Do I have the right idea? If so, how would I begin to prove soundness? If we are already given [imath]a \vdash b[/imath] I'm not sure what all we must iterate over in e.g. propositional logic to show that we always get true statements. Especially since it seems possible I could pick a false [imath]b[/imath] that contradicts.
|
2928721
|
Principal Ideal Avoidance
Let [imath]R[/imath] be a commutative ring with unit. Is it true that given an ideal [imath]I[/imath] and [imath]r \geq 2[/imath] principal ideals [imath]P_1, ..., P_r[/imath], then if [imath]I \subseteq \bigcup_{i = 1}^r P_i[/imath], [imath]I \subseteq P_i[/imath] for some [imath]i[/imath]? This would be the analogous of the Prime Avoidance Lemma for principal ideals.
|
555547
|
Counterexamples to the avoidance lemma for arbitrary ideals
Let [imath]A[/imath] be a commutative ring with [imath]1[/imath]. Let [imath]I[/imath] and [imath]J_k[/imath], [imath]k=1,\dots,n[/imath] be ideals of [imath]A[/imath] with [imath]I\subseteq \cup _{k=1}^n J_k[/imath]. Then I have obtained the following: (1) If [imath]J_k[/imath], [imath]k=1,\dots,n[/imath], are prime ideals, then there exists some [imath]j[/imath] such that [imath]I\subseteq J_j[/imath]. (2) If [imath]A[/imath] is a principal ideal ring, then there exists some [imath]j[/imath] such that [imath]I\subseteq J_j[/imath]. I want some counterexamples that for all [imath]k=1,\dots,n[/imath], [imath]I[/imath] is not contained in [imath]J_k[/imath]. How to get the counterexamples?
|
2048557
|
cauchy sequence prove
I have a question about cauchy sequences. If we think [imath]a_n[/imath] is a cauchy sequence, we ca say [imath]a_n^2[/imath] is cauchy sequence too. But how can I prove it? And if we would say [imath]a_n^2[/imath] is cauchy sequence, can we say [imath]a_n[/imath] is a cauchy sequence? Thanks.
|
1008472
|
If [imath](a_n)_{n\in\mathbb{N}}[/imath] is a Cauchy sequence, then [imath](a_n ^2)_{n\in\mathbb{N}}[/imath] is also a Cauchy sequence
If the sequence [imath]\{a_n\}[/imath] with [imath]n∈\mathbb{N}[/imath] is a Cauchy sequence, then [imath]\{a_n^2\}_{n∈\mathbb{N}}[/imath] is also a Cauchy sequence. How do we prove it?
|
1906565
|
A beautiful triangle inequality
Prove that in any triangle with side lengths [imath]a, b, c[/imath] the inequality: [imath](-\sqrt{a}+\sqrt{b}+\sqrt{c})(\sqrt{a}-\sqrt{b}+\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})\geq \sqrt{(-a+b+c)(a-b+c)(a+b-c)}.[/imath] For demonstration, I tried to use substitutions Ravi or algebraic calculation but without success. Maybe someone has an idea of saving. Thank you!
|
1876723
|
Inequality with triangle sides
Let [imath]a,b,c[/imath] be the sides of a triangle. Show that: [imath](\sqrt a + \sqrt b - \sqrt c)(\sqrt a - \sqrt b + \sqrt c)(-\sqrt a + \sqrt b + \sqrt c) \ge \sqrt {(a+b-c)(a-b+c)(-a+b+c)}.[/imath]
|
267489
|
prove [imath]\sqrt{a_n b_n}[/imath] and [imath]\frac{1}{2}(a_n+b_n)[/imath] have same limit
I am given this problem: let [imath]a\ge0[/imath],[imath]b\ge0[/imath], and the sequences [imath]a_n[/imath] and [imath]b_n[/imath] are defined in this way: [imath]a_0:=a[/imath], [imath]b_0:=b[/imath] and [imath]a_{n+1}:= \sqrt{a_nb_n}[/imath] and [imath]b_{n+1}:=\frac{1}{2}(a_n+b_n)[/imath] for all [imath]n\in\Bbb{N}[/imath] To prove is that both sequences converge and that they have the same limit. I don't know how to show this. I have spent 2 hours on this, no sign of success
|
2995129
|
Prove [imath]a_n[/imath] and [imath]b_n[/imath] have the same limit
Let x,y be positive, constant nambers so: [imath]a_1=x, b_1=y[/imath] and [imath]a_{n+1}=(a_n+b_n)/2,\; b_{n+1}=\sqrt{a_nb_n}[/imath] Prove that [imath]a_n[/imath] and [imath]b_n[/imath] have the same limit. From inequality of arithmetic and geometric means, I know that [imath]b_{n+1} \le a_{n+1}[/imath] . I assume from it that [imath]b_n \le a_n[/imath] I can prove that [imath]a_n [/imath] monotonically dicreasing and [imath]b_n[/imath] is monotonically increasing. But how can I prove that [imath]\lim_{n\to+\infty} (b_n-a_n)=0[/imath]? If I prove it I can use Cantor's intersection theorem. Sorry for inncorrect sintax and language.
|
2928660
|
Solution to [imath]u_x + y u_y = -u[/imath] with [imath]u(0, y) = \cos y[/imath]
I am studying for my PDEs test and I want to make sure I can solve this type of equations. I used the method of characteristics. [imath]\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{b}{a} = y[/imath] Integrating I get [imath]C = ye^{-x}[/imath] where [imath]C[/imath] is a constant. I can make the substitutions [imath]\xi = x[/imath] [imath]\eta = ye^{-x}[/imath] [imath]u(x,y) = w(\xi,\eta)[/imath] Substituting in the PDE, in terms of [imath]w, \eta, \xi[/imath] it becomes [imath]\frac{\mathrm{d}w}{\mathrm{d}\xi} + w = 0[/imath] I used the Integrating factor method to solve this ODE, which gives [imath]w = f(\eta) e ^{-\xi}[/imath] If we substitute back to [imath]u, x, y[/imath] this becomes [imath]w(\xi,\eta)=u(x,y)=f(ye^{-x})e^{-x}[/imath] The initial conditions give [imath]u(0,y)=\cos y =f(y)[/imath] Thus, the final solution is [imath]u(x,y) = \cos (ye^{-x})e^{-x}[/imath] Is this correct? Thanks in advance.
|
2773993
|
Transport equation [imath]u_t + xu_x + u = 0[/imath] with [imath]u(x_0, 0) = \cos(x_0)[/imath]
I have been studying PDEs using Peter Olver's textbook. I have learnt how to solve equations such as [imath]u_t + 2u_x = \sin(x)[/imath] subject to an initial condition such as [imath]u(0,x) = \sin x[/imath]. Letting [imath]\epsilon = x - 2t[/imath] and [imath]u(t,x) = v(t,\epsilon)[/imath], I then plug this into the transport equation. However, I am not sure how to define a characteristic to solve the following equation [imath]u_t + xu_x + u = 0, \qquad u(x_0, 0) = \cos(x_0)[/imath] because it has a variable 'speed' term [imath]x[/imath] and it is also not homogenous because of the term [imath]u[/imath]. A solution would be very helpful so I can see how to approach these problems.
|
2929449
|
Show that [imath]\sum_{k=-m}^n {m+k\choose r}{n-k\choose s}={m+n+1\choose r+s+1}[/imath]
Show that [imath]\sum_{k=-m}^n {m+k\choose r}{n-k\choose s}={m+n+1\choose r+s+1}[/imath] I first tried this algebraically, which was unsuccessful. Then, I tried to apply the Vandermonde Identity, but the index starts at -m so I am unsure how to continue.
|
1938753
|
[imath]\sum_{k=-m}^{n} \binom{m+k}{r} \binom{n-k}{s} =\binom{m+n+1}{r+s+1}[/imath] using Counting argument
I saw this question here:- Combinatorial sum identity for a choose function This looks so much like a vandermonde identity, I know we can give a counting argument for Vandermonde. However much I try I am not able to come up with a counting argument for this.
|
2929960
|
Prove [imath]\frac{d}{dx} x^n=nx^{n-1} : \forall n\in \mathbb{Z}_{+}[/imath] by induction
Problem Prove [imath]\frac{d}{dx} x^n=nx^{n-1} : \forall n\in \mathbb{Z}_{+}[/imath] by mathematical induction. Attempt to solve Base case when [imath]n=1[/imath] [imath] \frac{d}{dx} x^1 = 1 \cdot x ^{0}=1 [/imath] which is true Induction step [imath]\frac{d}{dx}x^{n+1}=(n+1)x^{n+1-1}= (n+1)x^{n}[/imath] At this point not quite sure how to prove this with induction without proving operator [imath]\frac{d}{dx}[/imath] with [imath] \frac{d}{dx}f(x)=\lim_{h \rightarrow 0}\frac{f(a+h)-f(a)}{h} [/imath] and then proving existence of such limit with: [imath] 0<|x-a|< \delta \implies|\frac{f(a+h)-f(a)}{h}-\frac{d}{dx}f(a)| < \epsilon [/imath] and then we can arrive at implication that [imath]\frac{d}{dx}x^n=nx^{n-1} \implies \frac{d}{dx}x^{n+1}=(n+1)x^n[/imath] Most likely there is easier by induction which is capable of showing that [imath] \frac{d}{dx}x^n=nx^{n-1} [/imath] is applicable [imath]\forall n \in \mathbb{Z}_+[/imath]?
|
966193
|
Proof by induction (power rule of the derivative)
Using the differentiation formulas [imath]\displaystyle\frac{d}{dx}x=1[/imath] and [imath]\displaystyle\frac{d}{dx}(fg)=f\frac{dg}{dx}+g \frac{df}{dx}[/imath], prove that [imath]\frac{d}{dx} x^n=nx^{n-1}[/imath] for all natural number [imath]n[/imath]. Thanks!
|
2930348
|
Prove that [imath](-a)b = -ab[/imath]
can you please tell me if my proof is right or if I'm taking some steps for granted? Thanks [imath]\forall a, b\in \Bbb R \Rightarrow (-a)b=-ab[/imath] [imath](-a)b\cdot (-a)^{-1} = -ab\cdot (-a)^{-1} \\ 1\cdot b = \frac{-ab}{-a} \\ b=b [/imath]
|
306772
|
Rings and some of their properties
So I am studying for an exam and was wondering if anyone could explain some of the reasoning being the proofs of the following properties: Let R be a ring with [imath]a,b\in R[/imath]. Then [imath]a0=0a=0[/imath] [imath]a(-b)=(-a)b=-ab[/imath] [imath](-a)(-b)=ab[/imath] Proof 1: [imath]a0=a(0+0)=a0+a0[/imath] hence [imath]a0=0[/imath] I don't really understand why [imath]a0+a0[/imath] proves that [imath]a0=0[/imath]
|
2930586
|
If [imath]R[/imath] is a finite boolean then [imath]R \cong \mathbb{Z_2} \oplus \dots \oplus \mathbb{Z_2}[/imath]
This question wants me to use a previous result, that if [imath]e[/imath] is a central idempotent in [imath]R[/imath] then [imath]R \cong Re \oplus R(1-e)[/imath]. I was thinking about maybe some sort of induction would help here but I can't see exactly how I am going to get to the result I want. I would like to know how to do the problem this way, because this is the way the book recommends. However, I also have heard that this problem can be done another way, by realizing that [imath]R[/imath] is a vector space over [imath]Z_2[/imath] and there has a finite basis, and i am under the impression that the result is not far away knowing this, but I can't quite connect the dots. If somebody could guide me along these two different routes of solving this problem I would really appreciate it! Thanks!
|
463280
|
The structure of a Noetherian ring in which every element is an idempotent.
Let [imath]A[/imath] be a ring which may not have a unity. Suppose every element [imath]a[/imath] of [imath]A[/imath] is an idempotent. i.e. [imath]a^2 = a[/imath]. It is easily proved that [imath]A[/imath] is commutative. Suppose every ideal of [imath]A[/imath] is finitely generated. Can we determine the structure of [imath]A[/imath]?
|
2930591
|
Understanding the field of fractions of [imath]F[[x]][/imath] (the ring of formal power series in the indeterminate x with coefficents in F)
Let [imath]F[[x]][/imath] be the ring of formal power series in the indeterminate x with coefficients in F. Show that the field of fractions of [imath]F[[x]][/imath] is the ring [imath]F((x))[/imath] of formal Laurent series. I've been going in circles on this one for what seems like ever now! If anyone could lend me some insight I would be exceedingly grateful! Thanks!
|
1232173
|
Prove the field of fractions of [imath]F[[x]][/imath] is the ring [imath]F((x))[/imath] of formal Laurent series.
Prove the field of fractions of [imath]F[[x]][/imath] is the ring [imath]F((x))[/imath] of formal Laurent series. [imath]F[[x]][/imath] is contained in [imath]F((x))[/imath]. So there's at least a ring homomorphism that is injective. Can also see it's injective because the kernel of such a mapping would be trivial because [imath]0[/imath] is the same in either. Not sure if showing they are isomorphic is the best way to do this. [imath]\displaystyle \sum_{n \ge N} a_nx^n \in F((x))[/imath] Im not sure how I would define the mapping. maybe theres a better way
|
2930751
|
A trivial question on the sequences [imath]\{a_n\}[/imath] and [imath]\{a_{n+1}\}[/imath] converging to the same limit using the definition of convergence only
Let's say I didn't know that if a sequence [imath]\{a_n\}[/imath] converges to some number [imath]a[/imath], then all subsequences of [imath]\{a_n\}[/imath] also converge to [imath]a[/imath]. I want to show using the definition of convergence that if [imath]\{a_n\}[/imath] converges to [imath]a[/imath], then [imath]\{a_{n+1}\}[/imath] converges to [imath]a[/imath]. Since [imath]\{a_n\}[/imath] we have for all [imath]\epsilon > 0[/imath], there exists a natural number [imath]N[/imath] such that for all [imath]n \in \mathbb{N}[/imath] [imath]n \geq N \implies |a_n - a | < \epsilon[/imath] I'm aware that some books use [imath]n > N[/imath], but this one uses [imath]n\geq N[/imath]. Since it doesn't matter, let's stick with [imath]n\geq N[/imath] to be consistent =) Clearly [imath]n+1 > n \geq N[/imath], so I should be able to deduce from the above implication that [imath]|a_{n+1} - a| < \epsilon[/imath] But I have some doubts. My main question is if I pick any index [imath]m > n[/imath], will I be able to immediately conclude that [imath]|a_m - a| < \epsilon[/imath] In other words, picking an index m > n allows me to "update" the index in [imath]|a_n - a| < \epsilon[/imath] to [imath]|a_m - a| < \epsilon[/imath]?
|
2710753
|
How to prove that [imath]\lim\limits_{n\to \infty}a_{n+1}=\lim\limits_{n\to \infty}a_{n}[/imath]
I want to prove the following theorem Let [imath]\{a_{n}\}[/imath] be a convergent sequence. Prove that [imath]\lim\limits_{n\to \infty}a_{n+1}=\lim\limits_{n\to \infty}a_{n}[/imath] Do I need to show that it is monotone increasing and then use the relation [imath]0\leq \lim_{n\to \infty}a_{n+1}- \lim_{n\to \infty}a_{n}\leq 0 \;?[/imath] Please, I need help on this! Various proofs are welcome! Thank you very much for your time!
|
2931063
|
Shortest distance between two lines and find points on each line
I'm given the following two lines: [imath]L_1[/imath]: [imath]P_1=(−13, 3, 14)[/imath] with direction vector [imath]d_1=(2, −1, −2)[/imath] [imath]L_2[/imath]: [imath]P_2=(5, 4, 4)[/imath] with direction vector [imath]d_2=(−2, 1, 0)[/imath] I'm then asked to find the shortest distance [imath]d[/imath] between these two lines, and then find a point, [imath]Q_1[/imath], on [imath]L_1[/imath], and a point, [imath]Q_2[/imath], on [imath]L_2[/imath] so that [imath]d(Q_1,Q_2) = d[/imath]. So far, I've determined that the shortest distance between these two points can be solved with a projection of the vector [imath]\vec{P_1P_2}[/imath] onto the direction vector found by the cross product of [imath]d_1[/imath] and [imath]d_2[/imath]. In this case, it is [imath](4,8,0)[/imath], or a magnitude of [imath]4\sqrt{5}[/imath]. I'm not really sure how to determine the two points now that I've gotten the distance, any tips or explanations would be appreciated.
|
2505442
|
Shortest distance between [imath]2[/imath] lines with direction vectors
Let L1 be the line passing through the point [imath]P_1=(−11, 10, −1)[/imath] with direction vector [imath]\vec d_1=\begin{bmatrix}−2\\ 3\\ −1\end{bmatrix}t[/imath], and let [imath]L_2[/imath] be the line passing through the point [imath]P_2=(−8, 9, 10)[/imath] with direction vector [imath]\vec d_2=\begin{bmatrix}−1\\ 3\\1\end{bmatrix}t[/imath]. Find the shortest distance, [imath]d[/imath], between these two lines, and find a point [imath]Q_1[/imath] on [imath]L_1[/imath] and a point [imath]Q_2[/imath] on [imath]L_2[/imath] so that [imath]d(Q_1,Q_2) = d[/imath]. I dont really understand how to properly approach this question, how do i start?
|
2930998
|
Every finite group has a a sequence of nested subgroups where adjacent elements are normal
A group [imath]Q[/imath] is called simple if [imath]|Q|>1[/imath] and the only normal subgroups of [imath]Q[/imath] are the trivial subgroups [imath]\{e\}[/imath] and [imath]Q[/imath]. Prove that for any finite group [imath]G[/imath] there exists a sequence of nested subgroups of [imath]G[/imath], [imath]\{e\}=N_0\leq N_1\leq \cdots \leq N_n=G[/imath] such that for each integer [imath]i[/imath] with [imath]1\leq i\leq n[/imath] we have [imath]N_{i-1}\trianglelefteq N_i[/imath] and the quotient group [imath]N_i/N_{i-1}[/imath] is simple. My idea is to try induction on [imath]|G|[/imath]. But I am having a hard time putting this into practice...
|
834723
|
Every finite group has a composition series
Every finite group has a composition series. The proof of this statement is as follows Proof. If [imath]|G| = 1[/imath] or [imath]G[/imath] is simple, then the result is trivial. Suppose [imath]G[/imath] is not simple, and the result holds for all groups of order [imath]< |G|[/imath]. Let [imath]H[/imath] be a maximal normal subgroup of [imath]G[/imath]. By the induction hypothesis [imath]H[/imath] has a composition series [imath]{e} \subset H_1 \subset \cdots \subset H[/imath]. Therefore, [imath]G[/imath] has a composition series [imath]{e} \subset H_1 \subset \cdots \subset H \subset G[/imath]. My question is, how can it be guaranteed that there exists a maximal normal subgroup of [imath]G[/imath] of order [imath]< G[/imath]?
|
421725
|
Square root limit problem
Please help with this problem. Somewhere my algebra gets confused. I know the answer is zero but I need the steps for the problem.I will start the problem and you can tell me where my mistake is. [imath]\lim_{x\to0}\left(\frac1{\sqrt{x}}-\frac1{\sqrt{x^2+x}}\right)=\lim_{x\to0}\frac{\sqrt{x^2+x}-\sqrt{x}}{\sqrt{x}\cdot\sqrt{x^2+x}}[/imath] Take conjugate of numerator and multiply by numerator and denominator : [imath]\frac{\sqrt{x^2+x}-\sqrt{x}}{\sqrt{x}\cdot\sqrt{x^2+x}}\cdot\frac{\sqrt{x^2+x}+\sqrt{x}}{\sqrt{x^2+x}+\sqrt{x}}=\frac{x^2+x-(\sqrt{x})^2}{\sqrt{x}\cdot\sqrt{x^2+x}\left(\sqrt{x^2+x}+\sqrt{x}\right)}[/imath] help from here
|
285885
|
Limit of square root problem
Can you please help me out with this limit problem. Actually, I tried to solve it by the conjugate method but it didn't work with me. Thank you. [imath]\lim_{x \to 0}\; \bigg( \frac{1}{\sqrt{x}} - \frac{1}{\sqrt{x²+x}} \bigg)[/imath]
|
2931703
|
Determinant of diagonal plus constant matrix
Is there a way to simplify [imath]\det(D + C)[/imath], where [imath]D,C[/imath] are square matrices of matching dimensions, [imath]D[/imath] is diagonal (with different diagonal elements, [imath]D_{ij} = \delta_{ij}d_i[/imath]), and [imath]C[/imath] is a constant matrix, that is, all entries [imath]C_{ij}=c[/imath] are equal to the same number? To be more explicit, assuming [imath]D,C\in\mathbb{R}^{n\times n}[/imath], the matrix [imath]D+C[/imath] has the form: [imath]D + C = \left(\begin{array}{ccccc} d_1 + c & c & c & \cdots & c\\ c & d_2 + c & c & \cdots & c\\ c & c & d_3 + c & \cdots & c\\ \vdots & \vdots & \vdots & & \vdots\\ c & c & c & \cdots & d_n + c \end{array}\right)[/imath]
|
2110766
|
Calculating determinant with different numbers on diagonal and x everywhere else
I'm having troubles solving the following determinant: [imath]\left| \begin{array}{cccc} a_1 & x & \ldots & x \\ x & a_2 & \ldots & \vdots \\ \vdots & \ldots & \ddots & x \\ x & \ldots & x & a_n \\ \end{array} \right|[/imath] I have tried coming up with recurrent formula, but there's always one lingering row and column. I also tried finding values of x for which determinant is 0, hoping I could get a nice polynomial solution, with no luck. By subtracting first row from every other I get somewhat nicer result, but there's still that one row/column which I can't solve. If all other columns are added to the first one and then I subtract the first row, then the column doesn't depend on x anymore, but I haven't found 1..n much more useful.
|
2931899
|
Complex Numbers: Solve for [imath]z^8-z^4+1 = 0[/imath].
This is what I currently have. I'm not sure how to continue. Can someone show or teach me how it's done? A repost because there were flaws with formatting. Work: [imath]\begin{array}{*{20}c} {x = \frac{{ 1 \pm \sqrt {1-4} }}{{2}}} \\ \end{array}[/imath] [imath]\begin{array}{*{20}c} {x = \frac{{ 1 \pm{i}\sqrt {-3} }}{{2}}} \\ \end{array}[/imath] [imath]z^4 ={r^4(\cos(4θ))(\sin(4θ)}[/imath]
|
2931838
|
Solve for z^8-z^4+1 = 0
Current Work This is what I currently have. I'm not sure how to continue. Can someone show or teach me how it's done? Edit: Work: [imath]\begin{array}{*{20}c} {z = \frac{{ - 1 \pm \sqrt {1-4} }}{{2}}} \\ \end{array}[/imath] [imath]\begin{array}{*{20}c} {z = \frac{{ - 1 \pm{i}\sqrt {-3} }}{{2}}} \\ \end{array}[/imath] [imath]z^4 ={r^4(\cos(4θ))(\sin(4θ)}[/imath]
|
2932278
|
Find an equation of the plane that passes through the point [imath](1, 3, 1)[/imath] and contains the line: [imath]x = t, y = t[/imath] and [imath]z = -2 + t[/imath].
Find an equation of the plane that passes through the point [imath](1, 3, 1)[/imath] and contains the line: [imath]x = t, y = t[/imath] and [imath]z = -2 + t[/imath]. My attempt: The points are [imath](1,3,1), (0,0,-2)[/imath] and if [imath]t = 1[/imath]: [imath](1,,1,-1)[/imath] [imath]a = (1,1,-1) - (0,0,-2) = [1,1,1][/imath] [imath]b = (1,1,-1) - (1,3,1) = [0,-2,-2][/imath] normal vector is [imath]n = a \times b = 2j - 2k = [0,2,-2][/imath] Therefore the equation is found by [imath][0,2,-2] \cdot [x-1, y-3, z-1] = 0[/imath] [imath]\leftrightarrow 2y -2z = 4[/imath] [imath]\leftrightarrow y - z = 2[/imath] right?
|
988509
|
The normal equation of the plane that contains the line [imath](1,1,1) + t(-2,0,3)[/imath]
Determine the equation of the plane that contains the point [imath](4,2,-1)[/imath] and also the line [imath]L: (1,1,1) + t(-2,0,3)[/imath] for [imath]t\in\mathbb{R}[/imath]. The direction vector [imath](-2,0,3)[/imath] of the line is also a direction vector for the plane. We can get a vector orthogonal to the plane by solving [imath](x,y,z) \cdot (-2,0,3) = 0[/imath] [imath]-2x+3z = 0[/imath] One valid solution would be [imath](3,1,2)[/imath]. So we have that [imath](3,1,2)[/imath] is orthogonal to the plane. And since [imath](4,2,-1)[/imath] belongs to it, a normal equation for the plane can be calculated this way: [imath](x,y,z)\cdot (3,1,2) = (4,2,-1) \cdot (3,1,2)[/imath] [imath]3x+y+2z = 12[/imath] But that's wrong. If the plane contains the line [imath]L[/imath], it should also contain the point [imath](1,1,1)+5\cdot(-2,0,3) = (-9,1,16)[/imath] But it doesn't: [imath]3(-9)+(1)+2(16) = -27 + 1 + 32 = 8 \not = 12[/imath] What did I do wrong?
|
2924497
|
Showing [imath]\cos \left( \frac{1}{n} \right) > 1 - \frac{1}{2n^2}[/imath]
I'm interested in showing the inequality [imath] \cos\left(\frac{1}{n}\right) > 1-\frac{1}{2n^2}[/imath] For all [imath]n[/imath] positive integer. I've tried to proceed by induction and it's easily seen (assuming properties of trigonometric functions) that it holds for [imath]n=1[/imath] but I get stuck on showing it for [imath]n+1[/imath]. Once [imath]\cos(x)[/imath] is decreasing on [imath][0,1][/imath] we have [imath]\cos\left(\frac{1}{n+1}\right) > \cos\left(\frac{1}{n}\right) > 1- \frac{1}{2 \, n^2}.[/imath] But I don't know how to get it from here.
|
1089621
|
Showing that [imath]1 - \frac{x^2}2\leq\cos x[/imath], [imath]\forall x \in \mathbb{R}[/imath]
Show that [imath]\displaystyle1 - \frac{x^2}2\leq\cos x\quad\forall x \in \mathbb{R}[/imath] Let [imath]f(x) = \cos x - 1 + \frac{x^2}2[/imath]; then we need to show that [imath]f(x) \geq 0\quad\forall x \in \mathbb{R}[/imath]. Since [imath]\lim_{x \to \pm\infty} f(x) = +\infty[/imath] and [imath]\operatorname{im}f'' = [0, 2][/imath] (as [imath]f''(x) = 1 -\cos x[/imath]), it remains to show that [imath]x = 0[/imath] is a minimum for [imath]f(x)[/imath], but I cannot do that because I would need to solve [imath]f'(x) = x -\sin x = 0[/imath], which I cannot. I think it would suffice to show that plugging in [imath]x = 0[/imath] we have that [imath]f'(x) = 0[/imath], and combining it with the fact that the function [imath]f[/imath] is convex, we would be done. Is it enough? What's a better proof? EDIT: It just occurred to me that there is a way to solve [imath]f'(x) = 0[/imath]. Just observe that [imath]f'(x)[/imath] is an odd function and that [imath]f''(x) \geq 0[/imath] implies that [imath]f'(x)[/imath] is monotonically increasing. Since an odd function, if defined at [imath]x = 0[/imath], must be [imath]0[/imath] there, [imath]x = 0[/imath] is a solution to the equation and it's the only one.
|
2932785
|
Convergence of the power series [imath]\sum_{n=1}^\infty\frac{(1+n)^n}{n!}x^n [/imath]
Consider the power series [imath]\sum_{n=1}^\infty\frac{(1+n)^n}{n!}x^n [/imath] Now the ratio test shows that the given power series converges absolutely [imath]|x|< 1/e[/imath] and diverges for [imath]|x|>1/e[/imath]. But all the tests I know(Raabe's test) are failing at [imath]|x|=1/e[/imath]. Any help will be highly apprecieated.
|
2919271
|
Discuss convergence of the series [imath]\sum \frac{(1+n)^n}{n!}x^n[/imath].
Let [imath]a_n = \dfrac{(1+n)^n}{n!}[/imath] Discuss convergence of the series [imath]\displaystyle \sum a_nx^n[/imath]. I've shown that the series converges absolutely when |x| < 1/e and diverged when |x| > 1/e. I can't seem to determine what happens at |x| = 1/e
|
2933712
|
Proving that [imath]\limsup {x_{n}} = \lim_{n \rightarrow \infty} ( \sup _ {m \geq n} x_{m} )[/imath].
Proving that [imath]\limsup {x_{n}} = \lim_{n \rightarrow \infty} ( \sup _ {m \geq n} x_{m} )[/imath]. I want to prove it using only the definition of the [imath]\limsup {x_{n}}[/imath]: that [imath]\limsup {x_{n}}[/imath] is the largest accumulation point of [imath]\{x_{n}\}[/imath]. I have seen many links for the proof here but as far as I understand not directly from this definition and after reading them all I got confused, could any one help me in proving this?
|
1515652
|
Showing that two definitions of [imath]\limsup[/imath] are equivalent
In Rudin, [imath]\limsup[/imath] is defined as follows: Let [imath]S[/imath] be the set of subsequential limits of [imath]\{s_n\}[/imath]. Then [imath]\limsup s_n = \sup S. \tag{1}[/imath] However, our real analysis instructor defined [imath]\limsup[/imath] in a different manner: [imath]\limsup s_n = \lim_{n \to \infty} \sup_{m \ge n} s_m. \tag{2}[/imath] I am having trouble understanding how these two definitions are equivalent. It would be very helpful to me if somebody could provide a proof with some explanation. My thoughts on the problem: I have noticed that the usual trend with these sort of proofs is to prove the upper bound [imath](1) \le (2)[/imath] and then the lower bound [imath](1) \ge (2)[/imath] to get the desired conclusion. However, I am unsure how to even begin.
|
2933534
|
Solve the Recurrence Relation T(n)=sum of T(i)T(n-i)
[imath]T(1)=1 \\ T(n) = \sum_{i=1}^{n-1}T(i)T(n-i) \\ [/imath] I have been trying to get rid of the Summation using the T(n+1) - T(n) technique, but I can't figure out how to apply it to this. I used recursion trees but got too messy. Now I'm wondering if there is an elegant solution to this recurrence.
|
2324176
|
Multiple choice question about generating function of a sequence
Suppose [imath]\left\{a_n\right\}_{n=0}^{\infty}[/imath] is sequence . and we have below recursion relation[imath]\begin{cases}a_n=\sum_{k=0}^{n-1}a_ka_{n-k-1}\\\\a_0=1\end{cases}[/imath] If [imath]f(x)=\sum_{n=0}^{\infty}a_nx^n[/imath] is generating function of [imath]a_n[/imath] which one is correct . [imath]1)\space f^2(x)+xf(x)-1=0\\ 2)\space f^2(x)-xf(x)+1=0 \\ 3)\space xf^2(x)+f(x)-1=0 \\ 4)\space xf^2(x)-f(x)+1=0 [/imath] I am sorry to ask this question , But as honestly as possible it was my student's question. I forgot this type of question . Can you help me or give me a clue . Thanks in advanced.
|
2933420
|
Finding a Suitable Telescoping Sum to Use in Problem
I am currently lost on the following problem: Use a telescoping sum to give a proof without induction that for each [imath]n \in \mathbb{N},[/imath] [imath]1^3+2^3+3^3+\dots +n^3=\frac{n^2(n+1)^2}{4}[/imath] I have followed other examples where they show equalities like this, but I don't understand how they seem to come up with the telescoping series they use to solve them. Thank you!
|
538425
|
Evaluate the sum of cubes
Evaluate [imath]1^3 + 2^3 + 3^3 + . . . + n^3.[/imath] Can I get a hint? I'm really stuck and don't know how to break this problem down.
|
2934395
|
All primes of the form [imath]n^3 +1[/imath]
How many primes are there of the form [imath]s^3+1[/imath], where [imath]s\in \mathbb{Z}[/imath]? I was thinking that is would be clever to look at this problem [imath]\mod n[/imath], but I am not sure what [imath]n[/imath] should be.
|
2920416
|
Take a cube, add one, when is this a prime?
Take [imath] p = n^3 + 1 [/imath] I just realised, checking some numbers, that this only occurs for [imath]n=1[/imath]. I was wondering if there is some general proof for this?
|
1521966
|
Computing Infinite Telescopic Sums
I need to compute the sum of the following infinite series: [imath]\frac{1}{1 \cdot 3} + \frac{1}{2\cdot 4} + \frac{1}{3\cdot 5} + \frac{1}{4\cdot 6} +\cdots [/imath] How would I compute this sum? My teacher didn't really go over telescopic sums, and in the book it says I need to find the closed form of this sum, which I can do and I did. [imath]1/(n)(n+2)[/imath] From here on though I get really confused, the book separates the fraction somehow and then tries to list the terms in closed form and factor something out? Not sure.
|
227711
|
Show convergence of a given series and find the limit.
Given the Series [imath]\sum_{k=1}^\infty \frac{1}{k(k+2)}[/imath] How exactly would I find out the limit is [imath]\frac34[/imath] as suggested by Wolframalpha? I already found out I can prove it actually converges by performing the comparison test and seeing that the underlying sequence isn't a null-sequence. But unfortunately I am absolutely clueless on how to prove that it converges to [imath]\frac34[/imath]. Regards, Dennis
|
2935188
|
[imath]0 and \sum_{n=1}^\infty a_n converges \Rightarrow a_n=\mathcal{o}\left(\frac{1}{n}\right)[/imath]
I'm trying to show the claim. Using the Cauchy theorem I can say that: [imath]\sum_{n=1}^\infty a_n[/imath] converges iff [imath]\sum_{n=1}^\infty 2^na_{2^n}[/imath] converges. So, by the necessary condition of convergence: [imath]\lim_{n\to \infty}2^na_{2^n}=0 \Leftrightarrow a_{2^n}\underbrace{=}_{n\to\infty}\mathcal{o}\left(\frac{1}{2^n}\right)[/imath] but now I can't simply do the substitution [imath]n=\log_{2}m[/imath] and conclude that [imath]\lim_{m\to \infty}m\cdot a_{m}=0[/imath], because such a substitution doesn't respect the hypothesis of the composite limit theorem. Any suggestion? Thanks.
|
1893621
|
A convergence problem in real non-negative sequence, [imath]\sum_{n=1}^\infty a_n[/imath].
We now have [imath]a_n\geq 0[/imath], [imath]\forall n=1,2,...,[/imath] and [imath]\sum_{n=1}^\infty a_n <\infty[/imath]. Then I guess that [imath]\lim_{n\to\infty} a_n \cdot n = 0[/imath]. But I realized that it is wrong. Since if we let [imath]a_n = 1/n [/imath] if [imath]n = 2^i[/imath] for some [imath]i=1,2,...[/imath] and [imath]a_n = 0[/imath] for the rest of the [imath]n[/imath]. Then we have that [imath]\sum_{n=1}^\infty a_n = 1/2 + 1/4 + 1/8 + \cdots < \infty[/imath], and [imath]a_n\cdot n[/imath] does not converges to [imath]0[/imath]. Now I add another condition that [imath]a_n[/imath] is non-increasing. Does this result hold this time. i.e. the formal question is as follows: [imath]a_n\geq 0[/imath], [imath]\forall n=1,2,...,[/imath] and [imath]a_n[/imath] is non-increasing, and [imath]\sum_{n=1}^\infty a_n <\infty[/imath]. Then prove [imath]a_n \cdot n \to 0[/imath], or give a counterexample that [imath]a_n\cdot n[/imath] does not necessarily converge to [imath]0[/imath]
|
2935239
|
Sum of 3 angles
If [imath]\alpha+\beta-\gamma=\pi[/imath], show that [imath]\sin^2\alpha+\sin^2\beta-\sin^2\gamma=2\sin\alpha\sin\beta\cos\gamma[/imath] Please don't ask about ideas because I really don't have one. I was initially thinking about a triangle with the given angles... But no chance EDIT. Sorry guys, now is correct.
|
177208
|
Prove [imath]\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C)[/imath] if [imath]A+B+C=180[/imath] degrees
I most humbly beseech help for this question. If [imath]A+B+C=180[/imath] degrees, then prove [imath] \sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C) [/imath] I am not sure what trig identity I should use to begin this problem.
|
2935253
|
Approximate [imath]\sqrt{7}[/imath] using binomial theorem
How does one deduce the approximation of [imath]\sqrt{7}[/imath] to be [imath]\frac{10837}{4096}[/imath] by taking [imath]x = \frac{1}{64}[/imath] in the expansion of [imath]\sqrt{1-x}[/imath]? How should you approach such a question? I assume the first step would be to expand [imath]\sqrt{1-x}[/imath] which can only be done through binomial theorem (afaik). That gives [imath]\sqrt{1-x} = 1 - \frac{1}{2}(x) - \frac{1}{8}(x^2)[/imath] + ... and so on. How do you continue? I can't seem to figure out how taking [imath]x = \frac{1}{64}[/imath] accomplishes anything.
|
2901477
|
Expand [imath]\sqrt{1-x}[/imath] up to and including the term [imath]x^2[/imath] (How should i proceed further)
I am trying to solve a 2 part question, the first part is to expand [imath]\sqrt{1-x}[/imath] up to and including the term [imath]x^2[/imath] which I did. This gives me [imath]1-(0.5)x-(0.125)x^2 + ...[/imath] However, the 2nd part of the question says that by taking [imath]x=(1/64)[/imath] in the expansion of part (i) -> which refers to the one above, deduce that [imath]\sqrt 7 = 10837/4096[/imath]. How do i do this part? EDIT (After Hint Given By SimpliFire): [imath] \sqrt{1-\frac1{64}}=\frac{\sqrt{63}}8=\frac38\sqrt7 [/imath] [imath] =\frac83\sqrt{1-\frac1{64}}\;=\sqrt7 [/imath] [imath] =\frac83\left(1-\frac12\left(\frac1{64}\right)-\frac18\left(\frac1{64}\right)^2\right)\; [/imath] [imath] =\frac83\left(1-\frac1{128}-\frac1{32768}\right)\; [/imath] [imath] =\frac83\left(\frac{32768}{32768}-\frac{256}{32768}-\frac1{32768}\right)\; [/imath] [imath] =\frac83\left(\frac{32511}{32768}\right)\; [/imath] [imath] =\frac{260088}{98304} [/imath] [imath] \frac{10837}{4096}\;\approx\sqrt7 [/imath] I'm not sure if I can present it like this, if it is okay i will post it as my answer :X
|
2935316
|
Real-life example of high order equations?
I'm helping my son with Math in high school. He is now learning to solve higher order equations, with for example [imath]x^4[/imath], [imath]x^3[/imath] or [imath]x^5[/imath]. The point is that he sees no use in solving these equations (apart from an algebraic challenge) because for him they serve no purpose. The quadratic equation is easy to illustrate by applying it to a ball you throw upward. Would you have some examples of real-life systems where higher order equations are used? Tks!
|
2197171
|
When, where and **how often** do you find polynomials of higher degrees than two in mathematical, pure/applied, research?
A formula for solving a polynomial of degree three, see this link; [imath]ax^3+bx^2+cx+d=0[/imath], is [imath]\begin{align} x\quad&=\quad \sqrt[3]{ \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right) + \sqrt{ \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right) ^2 + \left( \frac{c}{3a} - \frac{b^2}{9a^2} \right) ^3 } }\\ &+\quad \sqrt[3]{ \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right) - \sqrt{ \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right) ^2 + \left( \frac{c}{3a} - \frac{b^2}{9a^2} \right) ^3 } } \;-\;\frac{b}{3a} \end{align}[/imath] Unlike quadratic, cubic, and quartic polynomials, the general quintic cannot be solved algebraically in terms of a finite number of additions, subtractions, multiplications, divisions, and root extractions, as rigorously demonstrated by Abel (Abel's impossibility theorem) and Galois. However, certain classes of quintic equations can be solved [...] Source: http://mathworld.wolfram.com/QuinticEquation.html At levels of [imath]5^{\text{th}}[/imath] degree polynomials, things are starting to look really serious in my eyes. My question is: If it is possible to not answer subjectively: When, where and how often do you find polynomials of higher degrees than two in mathematical, pure/applied, research?
|
2935622
|
If [imath]a[/imath] and [imath]b[/imath] are positive real numbers, then [imath]a+ b \ge 2 \sqrt{ab}[/imath]
If [imath]a[/imath] and [imath]b[/imath] are positive real numbers, then [imath]a+ b \ge 2 \sqrt{ab}[/imath]. Can you guys help me with this problem? I'm still trying to learn how to do proving and I don't really know how to start or complete this problem.
|
1007013
|
If [imath]a[/imath] and [imath]b[/imath] are positive real numbers, then [imath]a + b \geq 2 \sqrt{ab}[/imath].
If [imath]a[/imath] and [imath]b[/imath] are positive real numbers, then [imath]a + b \geq 2 \sqrt{ab}[/imath]. I know how to do the direct proof, but in this case, I want to try proving it by contradiction. I have tried manipulating the inequality [imath]a + b < 2 \sqrt{ab}[/imath] after making the assumption that [imath]a,b >0[/imath] to get a contadiction [imath]a,b \leq 0[/imath]. [imath]\begin{align} a + b &< 2 \sqrt{ab} \\a^2-2ab+b^2 &< 0 \\(a-b)^2 &< 0\end{align}[/imath] How do I show that [imath]a,b \leq 0[/imath]?
|
744474
|
[imath]\int e^{\cos(x)} \cos(nx)\ dx[/imath] using the residue theorem
I am trying to evaluate the following integral using the residue theorem: [imath]\int_0^{2\pi} e^{ \cos(\theta)} \cos(n\theta) d\theta[/imath] I have already evaluated [imath]\int_0^{2\pi} e^{e^{-i\theta}} e^{i n\theta} d\theta[/imath] using the residue theorem by making the substitution [imath]z = e^{i\theta}[/imath] and then finding the coefficient of the [imath]z^{-1}[/imath] term in the Laurent series expansion. Now, however, this no longer works, so any help would be appreciated. Thank you.
|
740514
|
Residue theorem with [imath]\frac{1}{2\pi}\int_{0}^{2\pi}e^{\cos\theta}\cos(n\theta) d\theta[/imath].
The following integral should be doable using the residue theorum: [imath]\frac1{2\pi}\int_{0}^{2\pi}e^{\cos\theta}\cos(n\theta) \,d\theta[/imath]
|
2936102
|
If [imath]\frac{\cos x}{\cos y}+\frac{\sin x}{\sin y}=-1[/imath], then what's the value of [imath]4\left(\frac{\cos^3y}{\cos x}+\frac{\sin^3y}{\sin x}\right)[/imath]?
I was trying to solve this problem: If [imath]\frac{\cos x}{\cos y}+\frac{\sin x}{\sin y}=-1[/imath] then what is the value of [imath]S=4\left(\frac{\cos^3y}{\cos x}+\frac{\sin^3y}{\sin x}\right)[/imath] I tried to solve this problem using a change of variables [imath]\cos^3y=a\cos x[/imath] and [imath]\sin^3y=b \sin x[/imath]. So: [imath]\frac{S}{4}=a+b[/imath] [imath]\frac{\cos^6y}{a^2}+\frac{\sin^6y}{b^2}=1[/imath] and [imath]\frac{\cos^2y}{a} + \frac{\sin^2y}{b}=-1[/imath] I tried to manipulate these equations, but it gives no result
|
2708677
|
If [imath]\frac{\cos \alpha}{\cos \beta}+\frac{\sin \alpha}{\sin \beta}=-1[/imath], find [imath]\frac{\cos^3\beta}{\cos \alpha}+\frac{\sin^3\beta}{\sin \alpha}[/imath].
If [imath]\displaystyle \frac{\cos \alpha}{\cos \beta}+\frac{\sin \alpha}{\sin \beta}=-1[/imath], find [imath]\displaystyle \frac{\cos^3\beta}{\cos \alpha}+\frac{\sin^3\beta}{\sin \alpha}[/imath]. I tried[imath] \sin(\alpha+\beta)=-\sin \beta \cos \beta,\\ 2\sin(\alpha+\beta)=-\sin (2\beta),[/imath] and [imath]\frac{\cos^3\beta}{\cos \alpha}+\frac{\sin^3\beta}{\sin \alpha} =\frac{\sin\alpha\cos^3\beta+\sin^3\beta \cos \alpha}{\sin \alpha \sin \alpha},[/imath] but unable to find that ratio. Any help, please.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.